SOLUTIONS MANUAL For EXCURSIONS IN MODERN MATHEMATICS 10TH EDITION Peter Tannenbaum by StudyGuide

Table of Contents Solutions Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

104

Chapter 7

120

Chapter 8

130

Chapter 9

148

Chapter 10

160

Chapter 11

173

Chapter 12

191

Chapter 13

211

Chapter 14

224

Chapter 15

233

Chapter 16

247

Chapter 17

261

iii

Chapter 1 WALKING 1.1. Ballots and Preference Schedules 1.

Number of voters 5 3 5 3 2 3 1st choice A A C D D B 2nd choice B D E C C E 3rd choice C B D B B A 4th choice D C A E A C 5th choice E E B A E D This schedule was constructed by noting, for example, that there were five ballots listing candidate C as the first preference, candidate E as the second preference, candidate D as the third preference, candidate A as the fourth preference, and candidate B as the last preference.

Number of voters 4 1st choice A 2nd choice D 3rd choice B 4th choice C 3. (a) 5 + 5 + 3 + 3 + 3 + 2 = 21

5 B C D A

6 C A D B

2 A C D B

(b) 11. There are 21 votes all together. A majority is more than half of the votes, or at least 11. (c) Chavez. Argand has 3 last-place votes, Brandt has 5 last-place votes, Chavez has no last-place votes, Dietz has 3 last-place votes, and Epstein has 5 + 3 + 2 = 10 last-place votes. 4. (a) 202 + 160 + 153 + 145 + 125 + 110 + 108 + 102 + 55 = 1160 (b) 581; There are 1160 votes all together. A majority is more than half of the votes, or at least 581. (c) Alicia. She has no last-place votes. Note that Brandy has 125 + 110 + 55 = 290 last-place votes, Cleo has 202 + 145 + 102 = 449 last-place votes, and Dionne has 160 + 153 + 108 = 421 last-place votes. 5.

Number of voters 37 36 24 13 5 1st choice B A B E C 2nd choice E B A B E 3rd choice A D D C A 4th choice C C E A D 5th choice D E C D B Here Brownstein was listed first by 37 voters. Those same 37 voters listed Easton as their second choice, Alvarez as their third choice, Clarkson as their fourth choice, and Dax as their last choice. Number of voters 1st choice 2nd choice 3rd choice 4th choice 5th choice

14 B A E C D

10 B D A E C

8 A B E D C

7 D C B E A

4 E B A C D

Chapter 1: The Mathematics of Elections 7.

Number of voters 14 10 8 7 4 A 2 3 1 5 3 B 1 1 2 3 2 C 5 5 5 2 4 D 4 2 4 1 5 E 3 4 3 4 1 Here 14 voters had the same preference ballot listing B as their first choice, A as their second choice, E as their third choice, D as their fourth choice, and C as their fifth and last choice.

Number of voters A B C D E

37 1 3 2 5 4

36 2 1 4 3 5

24 5 2 3 1 4

13 2 4 1 5 3

5 4 1 5 2 3

Number of voters 255 480 765 1st choice L C M 2nd choice M M L 3rd choice C L C (0.17)(1500) = 255; (0.32)(500) = 480; The remaining voters (51% of 1500 or 1500-255-480=765) prefer M the most, C the least, so that L is their second choice.

10.

Number of voters 450 900 225 675 1st choice A B C C 2nd choice C C B A 3rd choice B A A B 100% - 20% - 40% = 40% of the voters number 225 + 675 = 900. So, if N represents the total number of voters, then (0.40) N = 900 . This means there are N = 2250 total voters. 20% of 2250 is 450 (these voters have preference ballots A, C, B). 40% of 2250 is 900 (these voters have preference ballots B, C, A).

1.2. Plurality Method 11. (a) C. A has 15 first-place votes. B has 11 + 8 + 1 = 20 first-place votes. C has 27 first-place votes. D has 9 first-place votes. C has the most first-place votes with 27 and wins the election. (b) C, B, A, D. Candidates are ranked according to the number of first-place votes they received (27, 20, 15, and 9 for C, B, A, and D respectively). 12. (a) D. A has 21 first-place votes. B has 18 first-place votes. C has 10 + 1 = 11 first-place votes. D has 29 first-place votes. D has the most first-place votes with 29 and wins the election. (b) D, A, B, C. 13. (a) C. A has 5 first-place votes. B has 4 + 2 = 6 first-place votes. C has 6 + 2 + 2 + 2 = 12 first-place votes. D has no first-place votes. C has the most first-place votes with 12 and wins the election. (b) C, B, A, D. Candidates are ranked according to the number of first-place votes they received (12, 6, 5, and 0 for C, B, A, and D respectively). 14. (a) B. A has 6 + 3 = 9 first-place votes. B has 6 + 5 + 3 = 14 first-place votes. C has no first-place votes. D has 4 first-place votes. B has the most first-place votes with 14 and wins the election. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

(b) B, A, D, C. Candidates are ranked according to the number of first-place votes they received (14, 9, 4 and 0 for B, A, D, and C respectively). 15. (a) D. A has 11% of the first-place votes. B has 14% of the first-place votes. C has 24% of the first-place votes. D has 23% + 19% + 9% = 51% of the first-place votes. E has no first-place votes. D has the largest percentage of first-place votes with 51% and wins the election. (b) D, C, B, A, E. Candidates are ranked according to the percentage of first-place votes they received (51%, 24%, 14%, 11% and 0% for D, C, B, A, and E respectively). 16. (a) C. A has 12% of the first-place votes. B has 15% of the first-place votes. C has 25% + 10% + 9% + 8% = 52% of the first-place votes. D has no first-place votes. E has 21% of the first-place votes. C has the largest percentage of first-place votes with 52% and wins the election. (b) C, E, B, A, D. Candidates are ranked according to the percentage of first-place votes they received (52%, 21%, 15%, 12%, and 0% for C, E, B, A and D respectively). 17. (a) A. A has 5 + 3 = 8 first-place votes. B has 3 first-place votes. C has 5 first-place votes. D has 3 + 2 = 5 first-place votes. E has no first-place votes. A has the most first-place votes with 8 and wins the election. (b) A, C, D, B, E. Candidates are ranked according to the number of first-place votes they received (8, 5, 5, 3, and 0 for A, C, D, B, and E respectively). Since both candidates C and D have 5 first-place votes, the tie in ranking is broken by looking at last-place votes. Since C has no last-place votes and D has 3 lastplace votes, candidate C is ranked above candidate D. 18. (a) A. A has 153 + 102 + 55 = 310 first-place votes. B has 202 + 108 = 310 first-place votes. C has 160 + 110 = 270 first-place votes. D has 145 + 125 = 270 first-place votes. Both A and B have the most firstplace votes with 310 so the tie is broken using last-place votes. A has no last-place votes. B has 125 + 110 + 55 = 290 last-place votes. So A wins the election. (b) A, B, D, C. Candidates are ranked according to the number of first-place votes they received (310, 310, 270 and 270 for A, B, D, and C respectively). In part (a), we saw that the tie between A and B was broken in favor of A. Since both candidates C and D have 270 first-place votes, the tie in ranking is broken by looking at last-place votes. Since C has 202 + 145 + 102 = 449 last-place votes and D has 160 + 153 + 108 = 421 last-place votes, candidate D is ranked above candidate C. 19. (a) A. A has 5 + 3 = 8 first-place votes. B has 3 first-place votes. C has 5 first-place votes. D has 3 + 2 = 5 first-place votes. E has no first-place votes. A has the most first-place votes with 8 and wins the election. (Note: This is exactly the same as in Exercise 17(a).) (b) A, C, D, B, E. Candidates are ranked according to the number of first-place votes they received (8, 5, 5, 3, and 0 for A, C, D, B, and E respectively). Since both candidates C and D have 5 first-place votes, the tie in ranking is broken by a head-to-head comparison between the two. But candidate C is ranked higher than D on 5 + 5 + 3 = 13 of the 21 ballots (a majority). Therefore, candidate C is ranked above candidate D. 20. (a) B. A has 153 + 102 + 55 = 310 first-place votes. B has 202 + 108 = 310 first-place votes. C has 160 + 110 = 270 first-place votes. D has 145 + 125 = 270 first-place votes. Both A and B have the most firstplace votes with 310 so the tie is broken by head-to-head comparison. But candidate A is ranked higher than B on 153 + 125 + 110 + 102 + 55 = 545 of the 1160 ballots (less than a majority). So B wins the tiebreaker and the election.

Chapter 1: The Mathematics of Elections

(b) B, A, D, C. Candidates are ranked according to the number of first-place votes they received (310, 310, 270 and 270 for B, A, D, and C respectively). In part (a), we saw that the tie between A and B was broken in favor of B. Since both candidates C and D have 270 first-place votes, the tie in ranking is broken by head-to-head comparison. Candidate C is ranked higher than D on 160 + 153 + 110 + 108 = 531 of the 1160 ballots (less than a majority). So in the final ranking, candidate D is ranked above candidate C.

1.3. Borda Count 21. (a) A has 4 × 15 + 3 × (9 + 8 + 1) + 2 × 11 + 1× 27 = 163 points. B has 4 × (11 + 8 + 1) + 3 × 15 + 2 × (27 + 9) + 1× 0 = 197 points. C has 4 × 27 + 3 × 0 + 2 × 8 + 1× (15 + 11 + 9 + 1) = 160 points. D has 4 × 9 + 3 × (27 + 11) + 2 × (15 + 1) + 1× 8 = 190 points. The winner is B. (b) B, D, A, C. Candidates are ranked according to the number of Borda points they received. 22. (a) A has 4 × 21 + 3 × 18 + 2 × (29 + 10) + 1× 1 = 217 points. B has 4 × 18 + 3 × (10 + 1) + 2 × 21 + 1× 29 = 176 points. C has 4 × (10 + 1) + 3 × (29 + 21) + 2 × 18 + 1× 0 = 230 points. D has 4 × 29 + 3 × 0 + 2 × 1 + 1× (21 + 18 + 10) = 167 points. The winner is C. (b) C, A, B, D. Candidates are ranked according to the number of Borda points they received. 23. (a) A has 4 × 5 + 3 × 2 + 2 × (6 + 2) + 1× (4 + 2 + 2) = 50 points. B has 4 × (4 + 2) + 3 × (2 + 2) + 2 × 2 + 1× (6 + 5) = 51 points. C has 4 × (6 + 2 + 2 + 2) + 3 × 0 + 2 × (5 + 4 + 2) + 1× 0 = 70 points. D has 4 × 0 + 3 × (6 + 5 + 4 + 2) + 2 × 2 + 1× (2 + 2) = 59 points. The winner is C. (b) C, D, B, A. Candidates are ranked according to the number of Borda points they received. 24. (a) A has 4 × (6 + 3) + 3 × (4 + 3) + 2 × 6 + 1× 5 = 74 points. B has 4 × (6 + 5 + 3) + 3 × 3 + 2 × 0 + 1× (6 + 4) = 75 points. C has 4 × 0 + 3 × (6 + 6 + 5) + 2 × (4 + 3 + 3) + 1× 0 = 71 points. D has 4 × 4 + 3 × 0 + 2 × (6 + 5) + 1× (6 + 3 + 3) = 50 points. The winner is B. (b) B, A, C, D. Candidates are ranked according to the number of Borda points they received. 25. Here we can use a total of 100 voters for simplicity. A has 5 × 11 + 4 × (24 + 23 + 19) + 3 × (14 + 9) + 2 × 0 + 1× 0 = 388 points. B has 5 × 14 + 4 × 0 + 3 × (24 + 11) + 2 × 23 + 1× (19 + 9) = 249 points. C has 5 × 24 + 4 × (14 + 11 + 9) + 3 × 23 + 2 × 19 + 1× 0 = 363 points. D has 5 × (23 + 19 + 9) + 4 × 0 + 3 × 0 + 2 × 14 + 1× (24 + 11) = 318 points. E has 5 × 0 + 4 × 0 + 3 × 19 + 2 × (24 + 11 + 9) + 1× (23 + 14) = 182 points. The ranking (according to Borda points) is A, C, D, B, E.

ISM: Excursions in Modern Mathematics, 10E

26. We use a total of 100 voters for simplicity. A has 5 × 12 + 4 × 0 + 3 × 9 + 2 × (25 + 21 + 10) + 1× (15 + 8) = 222 points. B has 5 × 15 + 4 × 9 + 3 × (21 + 12) + 2 × 8 + 1× (25 + 10) = 261 points. C has 5 × (25 + 10 + 9 + 8) + 4 × 0 + 3 × 0 + 2 × 15 + 1× (21 + 12) = 323 points. D has 5 × 0 + 4 × (21 + 15 + 12 + 10) + 3 × (25 + 8) + 2 × 0 + 1× 9 = 340 points. E has 5 × 21 + 4 × (25 + 8) + 3 × (15 + 10) + 2 × (12 + 9) + 1× 0 = 354 points. The ranking (according to Borda points) is E, D, C, B, A. 27. Haskins had 3 × 46 + 2 × 111 + 1× 423 = 138 + 222 + 423 = 783 points. Murray had 3 × 517 + 2 × 278 + 1× 60 = 1551 + 556 + 60 = 2167 points. Tagovailoa had 3 × 299 + 2 × 431 + 1× 112 = 897 + 862 + 112 = 1871 points. The ranking (according to Borda points) is Murray (2167), Tagovailoa (1871), and Haskins (783). 28. Bieber had 7 × 0 + 4 × 0 + 3 × 11 + 2 × 13 + 1× 5 = 64 points. Cole had 7 × 13 + 4 × 17 + 3 × 0 + 2 × 0 + 1× 0 = 159 points. Lynn had 7 × 0 + 4 × 0 + 3 × 0 + 2 × 3 + 1× 12 = 18 points. Morton had 7 × 0 + 4 × 0 + 3 × 18 + 2 × 10 + 1× 1 = 75 points. Verlander had 7 × 17 + 4 × 13 + 3 × 0 + 2 × 0 + 1× 0 = 171 points. As in Example 1.12, this uses a modified Borda count. In this case, first-place votes count 7 points rather than the usual 5. The ranking (according to modified Borda points) is Verlander (171), Cole (159), Morton (75), Bieber (64), and Lynn (18). 29. Each ballot has 4 × 1 + 3 × 1 + 2 × 1 + 1× 1 = 10 points that are awarded to candidates according to the Borda count. With 110 voters, there are a total of 110 × 10 = 1100 Borda points. So D has 1100 – 320 – 290 – 180 = 310 Borda points. The ranking is thus A (320), D (310), B (290), and C (180). 30. Each ballot has 7 × 1 + 4 × 1 + 3 × 1 + 2 × 1 + 1× 1 = 17 points that are awarded to candidates according to the Borda count. With 50 voters, there are a total of 50 × 17 = 850 Borda points. So E has 850 – 152 – 133 – 191 – 175 = 199 Borda points. The ranking is thus E (199), C (191), D (175), A (152), and B (133).

1.4. Plurality-with-Elimination 31. (a) A is the winner. Round 1: Candidate A B C D Number of first-place votes 15 20 27 9 D is eliminated. Round 2: The 9 first-place votes originally going to D now go to A. Candidate A B C D Number of first-place votes 24 20 27 B is eliminated. Round 3: There are 8 + 1 = 9 first-place votes originally going to B that now go to A. There are also 11 first-place votes going to B that would now go to D. But, since D is already eliminated, these 11 firstplace votes go to A. Candidate A B C D Number of first-place votes 44 27 Candidate A now has a majority of the first-place votes and is declared the winner. (b) A complete ranking of the candidates can be found by noting in part (a) when each candidate was eliminated. Since D was eliminated first, it is ranked last. Since B was eliminated next, it is ranked next to last. The final ranking is hence A, C, B, D.

Chapter 1: The Mathematics of Elections 32. (a) B is the winner. Round 1: Candidate A B C D Number of first-place votes 21 18 11 29 C is eliminated. Round 2: The 11 first-place votes originally going to C would next go to B. Candidate A B C D Number of first-place votes 21 29 29 Round 3: The 21 first-place votes going to A would next go to C. But C has been eliminated. So these 21 first-place votes go to B. Candidate A B C D Number of first-place votes 50 29 Candidate B now has a majority of the first-place votes and is declared the winner. (b) A complete ranking of the candidates can be found by noting in part (a) when each candidate was eliminated. Since C was eliminated first, it is ranked last. Since A was eliminated next, it is ranked next to last. The final ranking is hence B, D, A, C. 33. (a) C is the winner. Round 1: Candidate A B C D Number of first-place votes 5 6 12 0 Candidate C has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that C wins and at the end of round 1, D is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D Number of first-place votes 5 6 12 A is eliminated. Round 3: There are 5 first-place votes originally going to A that now go to C. Candidate Number of first-place votes The final ranking is C, B, A, D.

B 6

C 17

34. (a) B is the winner. Round 1: Candidate A B C D Number of first-place votes 9 14 0 4 Candidate B has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that B wins and at the end of round 1, C is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D Number of first-place votes 9 14 4 D is eliminated. Round 3: There are 4 first-place votes originally going to D that now go to A. Candidate Number of first-place votes The final ranking is B, A, D, C.

A 13

B 14

ISM: Excursions in Modern Mathematics, 10E

35. Round 1: Candidate A Number of first-place votes 8 E is eliminated. Round 2: No first-place votes change.

B 8

C 7

D 6

E 0

Candidate A B C D E Number of first-place votes 8 8 7 6 D is eliminated. Round 3: There are 4 first-place votes for D that move to candidate C and there are 2 first-place votes for D that move to candidate B. Candidate A B C D E Number of first-place votes 8 10 11 A is eliminated. Round 4: There are 5 first-place votes for A that move to candidate B and there are 3 first-place votes for A that move to candidate C (since D and E have both been eliminated). Candidate A B C D E Number of first-place votes 15 14 B now has a majority of the first-place votes and is declared the winner. The final ranking is B, C, A, D, E. 36. Round 1: Candidate A B C D E Number of first-place votes 8 4 11 12 5 B is eliminated. Round 2: There are 4 first-place votes for B that move to candidate E. Candidate A B C D E Number of first-place votes 8 11 12 9 A is eliminated. Round 3: There are 5 first-place votes for A that move to candidate E (since B is eliminated), 2 first-place votes for A that move to candidate D (since B is eliminated), and 1 first-place vote for A that moves to C. Candidate A B C D E Number of first-place votes 12 14 14 C is eliminated. Round 4: There are 6 first-place votes for C that move to candidate E (since A is eliminated), 5 first-place votes for C that move to candidate D (since both A and B are eliminated), and there is 1 first-place vote for C that moves to candidate E (C earned this vote earlier when candidate A was eliminated). Candidate A B C D E Number of first-place votes 19 21 E now has a majority of the first-place votes and is declared the winner. The final ranking is E, D, C, A, B. 37. (a) D is the winner. Round 1: Candidate A B C D E Percentage of first-place votes 11 14 24 51 0 Candidate D has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that D wins and at the end of round 1, E is the first candidate eliminated. Round 2: No first-place votes are changed.

Chapter 1: The Mathematics of Elections Candidate A B C D E Percentage of first-place votes 11 14 24 51 A is eliminated. Round 3: The 11% of the first-place votes that went to A now go to C. Candidate A B C D E Percentage of first-place votes 14 35 51 B is eliminated. Round 4: The 14% of the first-place votes that went to B now go to C. Candidate A B C D E Percentage of first-place votes 49 51 A complete ranking of the candidates can be found by noting when each candidate was eliminated. The final ranking is hence D, C, B, A, E. 38. (a) C is the winner. Round 1: Candidate A B C D E Percentage of first-place votes 12 15 52 0 21 Candidate C has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that C wins and at the end of round 1, D is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D E Percentage of first-place votes 12 15 52 21 A is eliminated. Round 3: The 12% of the first-place votes that went to A now go to B (since D has already been eliminated). Candidate A B C D E Percentage of first-place votes 27 52 21 E is eliminated. Round 4: The 14% of the first-place votes that went to E now go to B (since D has already been eliminated). Candidate A B C D E Percentage of first-place votes 48 52 A complete ranking of the candidates can be found by noting when each candidate was eliminated. The final ranking is hence C, B, E, A, D. 39. Round 1: Candidate A B C D E Number of first-place votes 8 8 7 6 0 Candidates E, D, and C are all eliminated. Round 2: There are 4 first-place votes for D that go to B (since C has been eliminated). There are 2 firstplace votes for D that go to B. There are 7 first-place votes for C that go to A (since both D and E are eliminated). Candidate A B C D E Number of first-place votes 15 14 A now has a majority of the first-place votes and is declared the winner.

ISM: Excursions in Modern Mathematics, 10E

40. Round 1: Candidate A B C D E Number of first-place votes 8 4 11 12 5 Candidates B, E, and A are all eliminated. Round 2: There are 4 first-place votes for B that go to C (since E has been eliminated). There are 5 first-place votes for E that go to D (since A has been eliminated). There are 5 first-place votes for A that go to C (since both B and E are eliminated), there are 2 first-place votes for A that go to D (since B is eliminated), and 1 first-place vote for A that goes to C. Candidate A B C D E Number of first-place votes 21 19 C now has a majority of the first-place votes and is declared the winner.

1.5. Pairwise Comparisons 41. (a) Candidate D is the winner. A versus B: 15 + 9 = 24 votes to 27 + 11 + 8 + 1 = 47 votes (B wins). B gets 1 point. A versus C: 15 + 11 + 9 + 8 + 1 = 44 votes to 27 votes (A wins). A gets 1 point. A versus D: 15 + 8 + 1 = 24 votes to 27 + 11 + 9 = 47 votes (D wins). D gets 1 point. B versus C: 15 + 11 + 9 + 8 + 1 = 44 votes to 27 votes (B wins). B gets 1 point. B versus D: 15 + 11 + 8 + 1 = 35 votes to 27 + 9 = 36 votes (D wins). D gets 1 point. C versus D: 27 + 8 = 35 votes to 15 + 11 + 9 + 1 = 36 votes. (D wins). D gets 1 point. The final tally is 1 point for A, 2 points for B, 0 points for C, and 3 points for D. (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is D (3 points), B (2 points), A (1 point), and C (0 points). 42. (a) Candidate C is the winner. A versus B: 29 + 21 = 50 votes to 18 + 10 + 1 = 29 votes (A wins). A gets 1 point. A versus C: 21 + 18 = 39 votes to 29 + 10 + 1 = 40 votes (C wins). C gets 1 point. A versus D: 21 + 18 + 10 = 49 votes to 29 + 1 = 30 votes (A wins). A gets 1 point. B versus C: 18 votes to 29 + 21 + 10 + 1 = 61 votes (C wins). C gets 1 point. B versus D: 21 + 18 + 10 + 1 = 50 votes to 29 votes (B wins). B gets 1 point. C versus D: 21 + 18 + 10 + 1 = 50 votes to 29 votes. (C wins). C gets 1 point. The final tally is 2 points for A, 1 point for B, 3 points for C, and 0 points for D. (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is C (3 points), A (2 points), B (1 point), and D (0 points). 43. (a) Candidate C is the winner. A versus B: 6 + 5 = 11 votes to 4 + 2 + 2 + 2 + 2 = 12 votes (B wins). B gets 1 point. A versus C: 5 + 2 = 7 votes to 6 + 4 + 2 + 2 + 2 = 16 votes (C wins). C gets 1 point. A versus D: 5 + 2 + 2 = 9 votes to 6 + 4 + 2 + 2 = 14 votes (D wins). D gets 1 point. B versus C: 4 + 2 = 6 votes to 6 + 5 + 2 + 2 + 2 = 17 votes (C wins). C gets 1 point. B versus D: 4 + 2 + 2 + 2 = 10 votes to 6 + 5 + 2 = 13 votes (D wins). D gets 1 point. C versus D: 6 + 2 + 2 + 2 + 2 = 10 votes to 5 + 4 = 9 votes. (C wins). C gets 1 point. The final tally is 0 points for A, 1 point for B, 3 points for C, and 2 points for D. (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is C (3 points), D (2 points), B (1 point), and A (0 points).

Chapter 1: The Mathematics of Elections

44. (a) Candidate B is the winner. A versus B: 6 + 4 + 3 = 13 votes to 6 + 5 + 3 = 14 votes (B wins). B gets 1 point. A versus C: 6 + 4 + 3 + 3 = 16 votes to 6 + 5 = 11 votes (A wins). A gets 1 point. A versus D: 6 + 6 + 3 + 3 = 18 votes to 5 + 4 = 9 votes (A wins). A gets 1 point. B versus C: 6 + 5 + 3 = 14 votes to 6 + 4 + 3 = 13 votes (B wins). B gets 1 point. B versus D: 6 + 5 + 3 = 14 votes to 6 + 4 + 3 = 13 votes (B wins). B gets 1 point. C versus D: 6 + 6 + 5 + 3 + 3 = 23 votes to 4 votes. (C wins). C gets 1 point. The final tally is 2 points for A, 3 points for B, 1 point for C, and 0 points for D. (b) The final ranking is B (3 points), A (2 points), C (1 point), and D (0 points). 45. Candidate D is the winner. A versus B: 24% + 23% + 19% + 11% + 9% = 86% of the votes to 14% of the votes (A wins). A versus C: 23% + 19% + 11% = 53% of the votes to 24% + 14% + 9% = 47% of the votes (A wins). A versus D: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (D wins). A versus E: 24% + 23% + 19% + 14% + 11% + 9% = 100% of the votes to 0% of the votes (A wins). B versus C: 14% of the votes to 24% + 23% + 19% + 11% + 9% = 86% of the votes (C wins). B versus D: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (D wins). B versus E: 24% + 23% + 14% + 11% = 72% of the votes to 19% + 9% = 28% of the votes (B wins). C versus D: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (D wins). C versus E: 24% + 23% + 14% + 11% + 9% = 81% of the votes to 19% of the votes (C wins). D versus E: 23% + 19% + 14% + 9% = 65% of the votes to 24% + 11% = 35% of the votes (D wins). The final tally is 3 points for A, 1 point for B, 2 points for C, 4 points for D, and 0 points for E. 46. Candidate C is the winner. A versus B: 25% + 12% + 10% = 47% of the votes to 21% + 15% + 9% + 8% = 53% of the votes (B wins). A versus C: 21% + 12% = 33% of the votes to 25% + 15% + 10% + 9% + 8% = 67% of the votes (C wins). A versus D: 12% + 9% = 21% of the votes to 25% + 21% + 15% + 10% + 8% = 79% of the votes (D wins). A versus E: 12% + 9% = 21% of the votes to 25% + 21% + 15% + 10% + 8% = 79% of the votes (E wins). B versus C: 21% + 15% + 12% = 48% of the votes to 25% + 10% + 9% + 8% = 52% of the votes (C wins). B versus D: 15% + 9% = 24% of the votes to 25% + 21% + 12% + 10% + 8% = 76% of the votes (D wins). B versus E: 15% + 12% + 9% = 36% of the votes to 25% + 21% + 10% + 8% = 64% of the votes (E wins). C versus D: 25% + 10% + 9% + 8% = 52% of the votes to 21% + 15% + 12% = 48% of the votes (C wins). C versus E: 25% + 10% + 9% + 8% = 52% of the votes to 21% + 15% + 12% = 48% of the votes (C wins). D versus E: 15% + 12% + 10% = 37% of the votes to 25% + 21% + 9% + 8% = 63% of the votes (E wins). The final tally is 0 points for A, 1 point for B, 4 points for C, 2 points for D, and 3 points for E. 47. A versus B: 7 + 5 + 3 = 15 votes to 8 + 4 + 2 = 14 votes (A wins). A gets 1 point. A versus C: 8 + 5 + 3 = 16 votes to 7 + 4 + 2 = 13 votes (A wins). A gets 1 point. A versus D: 8 + 5 + 3 = 16 votes to 7 + 4 + 2 = 13 votes (A wins). A gets 1 point. A versus E: 5 + 3 + 2 = 10 votes to 8 + 7 + 4 = 19 votes (E wins). E gets 1 point. B versus C: 8 + 5 + 2 = 15 votes to 7 + 4 + 3 = 14 votes (B wins). B gets 1 point. B versus D: 8 + 5 = 13 votes to 7 + 4 + 3 + 2 = 16 votes (D wins). D gets 1 point. B versus E: 8 + 5 + 4 + 2 = 19 votes to 7 + 3 = 10 votes (B wins). B gets 1 point. C versus D: 8 + 7 + 5 = 20 votes to 4 + 3 + 2 = 9 votes (C wins). C gets 1 point. C versus E: 7 + 5 + 4 + 2 = 18 votes to 8 + 3 = 11 votes (C wins). C gets 1 point. D versus E: 5 + 4 + 3 + 2 = 14 votes to 8 + 7 = 15 votes (E wins). E gets 1 point. The final tally is 3 points for A, 2 points for B, 2 points for C, 1 point for D, and 2 points for E. Now B, C, and E each have 2 points. In head-to-head comparisons, B beats C and B beats E so that B is ranked higher than C and E. Also, C beats E so C is ranked higher than E as well. The final ranking is thus A, B, C, E, D.

ISM: Excursions in Modern Mathematics, 10E

48. A versus B: 6 + 5 + 5 + 5 + 5 + 2 + 1 = 29 votes to 11 votes (A wins). A gets 1 point. A versus C: 7 + 5 + 5 + 2 + 1 = 20 votes to 20 votes (tie). A and C each get ½ point. A versus D: 6 + 5 + 5 + 5 + 2 + 1 = 24 votes to 16 votes (A wins). A gets 1 point. A versus E: 7 + 6 + 5 + 5 + 5 + 2 + 1 = 31 votes to 9 votes (A wins). A gets 1 point. B versus C: 7 + 5 + 5 + 4 + 2 = 23 votes to 17 votes (B wins). B gets 1 point. B versus D: 6 + 5 + 5 + 4 + 2 + 1 = 23 votes to 17 votes (B wins). B gets 1 point. B versus E: 7 + 5 + 5 + 4 + 2 = 23 votes to 17 votes (B wins). B gets 1 point. C versus D: 6 + 5 + 5 + 4 + 1 = 21 votes to 19 votes (C wins). C gets 1 point. C versus E: 7 + 6 + 5 + 5 + 1 = 24 votes to 16 votes (C wins). C gets 1 point. D versus E: 7 + 5 + 5 + 2 = 19 votes to 21 votes (E wins). E gets 1 point. The final tally is 3.5 points for A, 3 points for B, 2.5 points for C, 0 points for D, and 1 point for E. The final ranking is thus A, B, C, E, D. 49. (a) With five candidates, there are a total of 4 + 3 + 2 + 1 = 10 pairwise comparisons. Each candidate is part of 4 of these (one against each other candidate). So, to find the number of points each candidate earns, 1 we simply subtract the losses from 4. The 10 points are distributed as follows: E wins 1 points, D 2 1 1 1 wins 2 points, C gets 3 points, B gets 2 points, and A gets the remaining 10 − 1 − 2 − 3 − 2 = 1 point. 2 2 2 So A loses 3 pairwise comparisons. (b) Candidate C, with 3 points, is the winner. [The complete ranking is C, D, B, E, A.] 1 50. (a) Since there are a total of (6 × 5) / 2 = 15 pairwise comparisons, F must have won 15 – 1 – 2 – 2 - 3 2 1 1 1 2 = 4 of them (A earned 1 point, B and C each earned 2 points, D earned 3 , and E earned 2 2 2 2 points). This means F lost 1 pairwise comparison.

(b) Candidate F, with 4 points, is the winner.

1.6. Fairness Criteria 51. First, we determine the winner using the Borda count. A has 4 × 6 + 3 × 0 + 2 × 0 + 1 × (2 + 3) = 29 points. B has 4 × 2 + 3 × 6 + 2 × 3 + 1 × 0 = 32 points. C has 4 × 3 + 3 × 2 + 2 × 6 + 1 × 0 = 30 points. D has 4 × 0 + 3 × 3 + 2 × 2 + 1 × 6 = 19 points. So candidate B is the winner using Borda count. However, candidate A has 6 of the 11 votes (a majority) and beats all three other candidates (B, C, and D) in head-to-head comparisons. That is, candidate A is a Condorcet candidate. Since this candidate did not win using the Borda count, this is a violation of the Condorcet criterion. 52. First, we determine the winner is candidate B using the plurality-with-elimination method (see Exercise 32(a)). In Exercise 42, however, we saw that candidate C was a Condorcet candidate (beating each other candidate in head-to-head comparisons and earning 3 points in the process). Since candidate C did not win using plurality-with-elimination, this is a violation of the Condorcet criterion.

Chapter 1: The Mathematics of Elections

53. The winner of this election is candidate R using the plurality method. Now F is clearly a nonwinning candidate. Removing F as a candidate leaves the following preference table. Number of voters 49 48 3 1st choice R H H 2nd choice H S S 3rd choice O O O 4th choice S R R In a recount, candidate H would be the winner using the plurality method. This is a violation of the IIA criterion. 54. First, we determine the winner using the Borda count. A has 4 × 14 + 3 × 0 + 2 × 0 + 1 × (10 + 8 + 4 + 1) = 79 points. B has 4 × 4 + 3 × (14 + 10) + 2 × (8 + 1) + 1 × 0 = 106 points. C has 4 × (10 + 1) + 3 × 8 + 2 × (14 + 4) + 1 × 0 = 104 points. D has 4 × 8 + 3 × (4 + 1) + 2 × 10 + 1 × 14 = 81 points. So candidate B (Boris) is the winner using Borda count. Now D (Dave) is clearly a nonwinning (irrelevant) candidate. Removing D as a candidate leaves the following preference table. Number of voters 14 10 8 4 1 1st choice A C C B C 2nd choice B B B C B 3rd choice C A A A A We now recount using the Borda count. A has 3 × 14 + 2 × 0 + 1 × (10 + 8 + 4 + 1) = 65 points. B has 3 × 4 + 2 × (14 + 10 + 8 + 1) + 1 × 0 = 78 points. C has 3 × (10 + 8 + 1) + 2 × 4 + 1 × 14 = 79 points. In a recount, candidate C (Carmen!) would be the winner using Borda count, a violation of the IIA criterion. 55. First, we determine the winner using plurality-with-elimination. Round 1: Candidate A B C D Number of first-place votes 8 3 5 5 Candidate E is eliminated. Round 2: No votes are shifted.

E 0

Candidate A B C D E Number of first-place votes 8 3 5 5 Candidate B is eliminated. Round 3: The 3 first-place votes for B now go to A (since E has been eliminated). Candidate A B C D E Number of first-place votes 11 5 5 Candidate A now has a majority (11 of the 21 votes) and is declared the winner. Now C is clearly a nonwinning (irrelevant) candidate. Removing C as a candidate leaves the following preference table. Number of voters 5 5 1st choice A E 2nd choice B D 3rd choice D B 4th choice E A We now recount using the plurality-with-elimination.

3 A D B E

3 D B E A

3 B E A D

2 D B A E

ISM: Excursions in Modern Mathematics, 10E

Round 1: Candidate A B D E Number of first-place votes 8 3 5 5 Candidate B is eliminated. Round 2: The 3 first-place votes that went to B now shift to E. Candidate A B D E Number of first-place votes 8 5 8 Candidate D is eliminated. Round 3: D had 5 first-place votes. Of these, 3 go to E and 2 go to A (since B was eliminated). Candidate A B D E Number of first-place votes 10 11 Candidate E now has a majority (11 of the 21 votes) and is declared the winner. Remember that candidate A was the winner before candidate C was removed. This is a violation of the IIA criterion. 56. If X has a majority of the first-place votes, then X will win every pairwise comparison (it is ranked above all other candidates on more than half the ballots) and is, therefore, the winner under the method of pairwise comparisons. 57. If X is the Condorcet candidate, then by definition X wins every pairwise comparison and is, therefore, the winner under the method of pairwise comparisons. 58. When a voter moves a candidate up in his or her ballot the number of first place votes for that candidate either increases or stays the same. It follows that if X had a plurality of the first place votes and a voter changes his or her ballot to rank X higher, then X still has a plurality. 59. When a voter moves a candidate up in his or her ballot, that candidate’s Borda points increase. It follows that if X had the most Borda points and a voter changes his or her ballot to rank X higher, then X still has the most Borda points. 60. When a voter moves a candidate up in his or her ballot it can’t hurt the candidate in a pairwise comparison— the candidate wins the same pairwise comparisons as before and possibly a few more. It follows that if X won the most pairwise comparisons and a voter changes his or her ballot to rank X higher, then X still wins the most pairwise comparisons.

JOGGING 61. Suppose the two candidates are A and B and that A gets a first-place votes and B gets b first-place votes and suppose that a > b. Then A has a majority of the votes and the preference schedule is Number of voters a b 1st choice A B 2nd choice B A It is clear that candidate A wins the election under the plurality method, the plurality-with-elimination method, and the method of pairwise comparisons. Under the Borda count method, A gets 2a + b points while B gets 2b + a points. Since a > b, 2a + b > 2b + a and so again A wins the election. 62. In this variation of the Borda count each candidate gets 1 less point per ballot. It follows that if N is the number of voters, each candidate gets N fewer points than he or she would under the standard Borda count method. Since each candidate’s total points gets decreased by the same number N, the ranking of the candidates remains the same. 63. The number of points under this variation is complementary to the number of points under the standard Borda count method: a first place is worth 1 point instead of N, a second place is worth 2 points instead of N – 1,…, a last place is worth N points instead of 1. It follows that having the fewest points here is equivalent to having the most points under the standard Borda count method, having the second fewest is equivalent to having the second most, and so on. Copyright © 2022 Pearson Education, Inc.

Chapter 1: The Mathematics of Elections

As another way to see this, suppose candidates C1 and C2 receive p1 and p2 points respectively using the Borda count as originally described in the chapter and r1 and r2 points under the variation described in this exercise. Then, for an election with k voters, we have p1 + r1 = p2 + r2 = k ( N + 1) . So if p1 < p2, we have –p1 > –p2 and so k(N + 1) – p1 > k(N + 1) – p2 which implies r1 > r2. Consequently the relative ranking of the candidates is not changed. 64. Use the Reverse Borda count method described in Exercise 63. In this variation the ranking of a candidate on a ballot equals the number of Borda points the candidate gets from that ballot. It follows that the average ranking of a candidate equals the total Borda points for that candidate divided by N (the number of voters). The candidate with the lowest average ranking is the candidate with the least total points and thus the winner of the election. 65. (a) Each of the voters gave Ohio State 25 points, so we compute 1625/25 to find 65 voters in the poll. (b) Let x = the number of Florida’s second-place votes. Then 1529 = 24 ⋅ x + 23 ⋅ ( 65 − x ) and so x = 34. This leaves 65 – 34 = 31 third-place votes for Florida. (c) Let y = the number of Michigan’s second-place votes. Then 1526 = 24 ⋅ y + 23 ⋅ ( 65 − y ) and so y = 31. This leaves 65 – 31 = 34 third-place votes for Michigan. 66. (a) Under the Borda count, each voter gives X more points than Y. Hence, the Borda count for X will be greater than the Borda count for Y and so X will rank above Y. (b) Since every voter prefers candidate X to candidate Y, a pairwise comparison between X and Y results in a win for X. If Y wins a pairwise comparison against Z, then X must also win against Z (by transitivity – since X is above Y on each ballot, when Y is above Z it must be that X is also above Z). Therefore X will have at least one more point than Y (from the head-to-head between X and Y). Thus, X will rank above Y under pairwise comparisons. 67. By looking at Deandre Ayton’s vote totals, it is clear that 1 point is awarded for each third-place vote (0 points is too few and 2 points is too many for each third-place vote). Let x = points awarded for each secondplace vote. Then, 1x + 63 = 66 gives x = 3 points for each second-place vote. Next, let y = points awarded for each first-place vote. Looking at Luka Doncic’s votes, we see that 98 y + 2 × 3 = 496 . Solving this equation gives y = 5 points for each first-place vote. 68. (a) Recall that using plurality-with-elimination in the Math Club election resulted in B being eliminated in the first round, C being eliminated second, A being eliminated third, and D winning (see Example 1.13). If top-two IRV is used, candidates B and D are both eliminated in round one. The 4 first-place votes for B would transfer to candidate C. The 8 first-place votes for D would also transfer to candidate C. The new preference table is shown below. Number of voters

1st choice

2nd choice C A We then have C winning the election -- clearly different than the plurality-with-elimination outcome. (b) Many examples are possible. (Example 1.21 shows that plurality-with-elimination violates the monotonicity criterion and in that example, since there are only 3 candidates from the start, top-two IRV and plurality-withelimination are identical methods and give the same result.)

ISM: Excursions in Modern Mathematics, 10E

1st choice

2nd choice

3rd choice C A A In this election B is a Condorcet candidate (since B beats A by a ‘score’ of14 to 10 and B beats C by a score of 16 to 8) and yet B is eliminated first using top-two IRV making C the winner. 69. (a) Round 1: Number of voters

1st choice

2nd choice

3rd choice

4th choice D A A A Candidate A with 23 last-place votes is eliminated. Round 2:

Number of voters

1st choice

2nd choice

3rd choice D D B C B Candidate B has 8 + 1 = 9 last-place votes, C has 4 last-place votes, and D has 14 + 10 = 24 last-place votes. So candidate D is eliminated. Round 3: Number of voters

1st choice

2nd choice C B B C B B has more last-place votes (meaning C has more first-place votes) and C is declared the winner. (b) Consider the election given by the preference schedule below. Here A is a Condorcet candidate but, having the most last-place votes, is eliminated in the first round. Number of voters

1st choice

2nd choice

3rd choice

4 choice A C D B D (c) Consider the election given by the preference schedule below. Here B wins under the Coombs method (C is eliminated first and B is preferred to A by 19 voters). However, if 8 voters move B from their 3rd choice to their 2nd choice, then C wins (since A would be eliminated first and C is preferred head-to-head over B). Number of voters

1st choice

2nd choice

3rd choice

Chapter 1: The Mathematics of Elections

70. (a) In round 1, no candidate has a majority of the 37 votes. Number of voters

1st choice

In round 2, counting 1st and 2nd place votes, A totals 14 votes, B totals 14 + 10 + 4 = 28 votes, C totals 10 + 8 + 1 = 19 votes, and D total 8 + 4 + 1 = 13 votes. Number of voters

1st choice

2nd choice

Both B and C have a majority of the 37 votes. However, candidate B has more and is declared the winner. (b) C is a Condorcet candidate in the Math Club election. However, as shown in part (a), B wins the election using the Bucklin method. (c) Say a ballot has candidate X ranked in the kth position. If and when the election gets to round k that vote contributes to X’s total under the Bucklin method. It follows that if X wins in round k and a voter moves X up in his or her ballot then X would still win (in round k or possibly in an earlier round).

RUNNING 71. One reasonable approach would be to use variables xk , 1 ≤ k ≤ 5 , to represent the number of points handed out for kth place. With so many variables, there is some flexibility in determining the values of these variables. To start, focus attention on Antetonkoumpo and Harden and variables x1 and x2 . The following two equations must hold: 78 x1 + 23 x2 = 941 23 x1 + 78 x2 = 776 Multiplying the first equation by 78 and the second equation by -23 produces the two equations: 6084 x1 + 1794 x2 = 73398 −529 x1 − 1794 x2 = −17848 Adding these equations eliminates the variable x2 and gives 5555 x1 = 55550 which has solution x1 = 10 . The first equation then gives 780 + 23 x2 = 941 so that x2 = 7 . We then turn our attention to Stephen Curry. His votes yield equation 16 x3 + 25 x4 + 20 x5 = 175 in which three of the four terms are multiples of 5. This observation means that variable x3 must also be a multiple of 5. Since the value of x3 is a positive integer strictly less than 7, we deduce that x3 = 5 . From that, we have 16(5) + 25 x4 + 20 x5 = 175 or 25 x4 + 20 x5 = 95 . Since x4 and x5 are both positive integers strictly less than 5, we can easily guess and check that x4 = 3 and x5 = 1 . The point values on each ballot are thus determined to be 10 points for 1st, 7 points for 2nd, 5 points for 3rd, 3 points for 4th, and 1 point for 5th. 72. (a) In this election, C is the winner under the plurality method yet a majority of the voters (4 out of 7) prefer both A and B over C. Number of voters

1st choice

2nd choice

3rd choice

ISM: Excursions in Modern Mathematics, 10E

(b) We may use the same example as in part (a). Both A and B are eliminated in the first round and so C is the winner under the plurality-with-elimination method. (c) Suppose there are k voters and N candidates. Under the Borda count method, the total number of points for all the candidates is kN ( N + 1) / 2 (each voter contributes 1 + 2 +  + N = N ( N + 1) / 2 points to the total), so the average number of points per candidate is k ( N + 1) / 2 . Now suppose that X is a candidate that loses to every other candidate in a one-to-one comparison. The claim is that under the Borda count method, X will receive less than the average k ( N + 1) points and therefore cannot be the winner. (One way to see this is as follows: Suppose that the N candidates are presented to the voters one at a time, with each voter keeping an updated partial ranking of the candidates as they are being presented. Suppose, moreover, that X is the first candidate to be presented. When this happens, X starts with kN points (X is the only candidate in every voter’s ballot and therefore has all the first-place votes.) As soon as the next candidate (say Y) is presented, X’s points will drop by more than k/2 (X loses to Y in a one-to-one comparison means that more than k/2 of the voters have ranked Y above X.) By the same argument each time one of the subsequent candidates is presented X’s points drop by more than k/2. By the time we are done, X must have less than kN − k2 ( N − 1) = k2 ( N + 1) points.) 73. (a) The Math Club election serves as an example. Candidate A has a majority of last-place votes (23 of the 37) and yet wins the election when the plurality method is used. Number of voters

1st choice

2nd choice

3rd choice

4 choice D A A A A (b) In this election, C is the winner under the plurality method (in the case of the tie in round 1, eliminate all candidates with the fewest number of votes). However, a majority of the voters (4 out of 7) prefer both A and B over C. Number of voters

1st choice

2nd choice

3rd choice C C B (c) Under the method of pairwise comparisons, if a majority of voters have candidate X ranked last on their ballot, then candidate X will never win a head-to-head comparison (since any other candidate Y is preferred to X by a majority of voters). Thus, X will end up with no points under the method and cannot win the election. (d) Suppose there are N candidates, x is the number of voters placing X last, and y is the remaining number of voters. Since a majority of the voters place X last, x > y. The maximum number of points that X can receive using a Borda count is x + Ny. (One point for each of the x last place votes and, assuming all other voters place X in first-place, N points for each of the other y voters.) The total number of points given out by each voter is 1 + 2 + 3 + ... + N = N(N + 1)/2 and so the total number of points given out by all x + y voters is N(N + 1)(x + y)/2. Since some candidate must receive at least 1/N of the total points, some candidate must receive at least (N + 1)(x + y)/2 points. Since x > y, we have (N + 1)(x + y)/2 = (Nx + x + Ny + y)/2 > (Ny + x + Ny + y)/2 > x + Ny, and consequently, X cannot be the winner of the election using the Borda count method.

Chapter 1: The Mathematics of Elections

74. In this election, D is preferred over A by a majority of the voters (12 to 10) and yet all four complete rankings rank A above D. Number of voters

1st choice

2nd choice

3rd choice

4th choice

Method

Winner

2nd place

3rd place

Last place

Plurality

Plurality with elimination

D D

Borda count

C,B (tied)

Pairwise comparison

A,B (tied)

C,D (tied)

75. Suppose there are k voters and N candidates. Case 1: k is odd, say k = 2t + 1. Suppose the candidate with a majority of the first-place votes is X. The fewest possible Borda points X can have is F(t + 1) + t [when there are (t + 1) votes that place X first and the remaining votes place X last]. The most Borda points that any other candidate can have is (N – 1)(t + 1) + Ft [when there are (t + 1) voters that place the candidate second and the remaining voters place that candidate first]. Thus, the majority criterion will be satisfied when F(t+ 1) + t > (N – 1)(t + 1) + Ft, which after simplification implies F > N(t + 1) – (2t + 1), or F > N ( k 2+1 ) − k . Case 2: k is even, say k = 2t. An argument similar to the one given in case 1 gives the inequality F(t + 1) + (t – 1) > (N – 1)(t + 1) + F(t – 1) which after simplification implies F > [N(t + 1) – 2t]/2, or F >  N ( k2 + 1) − k  / 2 .

APPLET BYTES 76. (a) Plurality method: A has 89 first-place votes, B has 91 first-place votes, and C has 73 first-place votes. So, using the plurality method, B wins.

Borda count method: A has 3 × 49 + 3 × 40 + 2 × 40 + 1× 51 + 2 × 43 + 1× 30 = 514 points. B has 2 × 49 + 1× 40 + 3 × 40 + 3 × 51 + 1× 43 + 2 × 30 = 514 points. C has 1× 49 + 2 × 40 + 1× 40 + 2 × 51 + 3 × 43 + 3 × 30 = 490 points. So, using the Borda count method, there is a tie between candidates A and B (each with 514 points). Plurality-with-elimination method: In round 1, B has 91 votes, A has 89 votes, and C has 73 votes. So candidate C is eliminated. In an election between A and B, A has 49 + 40 + 43 = 132 votes and B has 40 + 51 + 30 = 121 votes so B is eliminated and A is declared the winner. Pairwise comparisons method: A vs. B – A wins 132-121 and gets 1 point. A vs. C – A wins 129-124 and gets 1 point. B vs. C – B wins 140-113 and gets 1 point.

ISM: Excursions in Modern Mathematics, 10E

The final tally is A – 2 points, B – 1 point, C – 0 points. So, using the pairwise comparisons method, candidate A wins. (b) One possible manipulation is to change two votes from B>C>A to A>B>C. This makes A the winner under all four voting methods studied. Number of voters

1st choice

2nd choice

3rd choice

77. Many preference schedules are possible. Giving candidate C more strength relative to A and B is one approach to take. One such schedule moves 9 votes away from A>B>C (4 go to A>C>B and 5 go to C>B>A) and 5 votes away from B>C>A to C>B>A. In this new schedule, candidate B wins using the plurality method, candidate C wins using the Borda count method, candidate A wins using the plurality-with-elimination method, and candidate C wins using pairwise comparisons. Number of voters

1st choice

2nd choice

3rd choice

78. (a) Plurality method: A has 93 first-place votes, E has 81 first-place votes, B has 44 first-place votes, D has 42 first-place votes, and C has 40 first-place votes. So, using the plurality method, A wins.

Borda count method: A has 5 × 93 + 1× 44 + 4 × 10 + 2 × 30 + 2 × 42 + 1× 81 = 774 points. B has 4 × 93 + 5 × 44 + 2 × 10 + 1× 30 + 1× 42 + 2 × 81 = 846 points. C has 3 × 93 + 2 × 44 + 5 × 10 + 5 × 30 + 4 × 42 + 3 × 81 = 978 points. D has 2 × 93 + 4 × 44 + 1× 10 + 3 × 30 + 5 × 42 + 4 × 81 = 996 points. E has 1× 93 + 3 × 44 + 3 × 10 + 4 × 30 + 3 × 42 + 5 × 81 = 906 points. So, using the Borda count method, candidate D wins with 996 points. Plurality-with-elimination method: Candidate E wins (C is eliminated first, followed by D, B, and then A). Pairwise comparisons: A vs. B – A wins 175-125 and gets 1 point. A vs. C – C wins 93-207 and gets 1 point. A vs. D – D wins 103-197 and gets 1 point. A vs. E – E wins 103-197 and gets 1 point. B vs. C – C wins 137-163 and gets 1 point. B vs. D – D wins 147-153 and gets 1 point. B vs. E – E wins 137-163 and gets 1 point. C vs. D – D wins 133-167 and gets 1 point. C vs. E – C wins 175-125 and gets 1 point. D vs. E – D wins 179-121 and gets 1 point. So, using the pairwise comparisons method, candidate D wins (4 points for D, 3 points for C, 2 points for E, 1 point for A, 0 points for B). Since candidate D is a Condorcet candidate, we conclude that plurality method and the plurality-withelimination method both violate the Condorcet criterion in this election. Copyright © 2022 Pearson Education, Inc.

Chapter 1: The Mathematics of Elections (b) We aim to make D a winning candidate when using the plurality method. One way to do this is to count all first place votes for candidate A, D, and E and divide that number by 3. We then make sure candidate D has at least that many first-place votes and candidates A and E have less than that many. Since 93 + 42 + 81 = 216, we will take 21 votes away from the 93 voters ranking A>B>C>D>E and take 10 votes away from the voters ranking E>D>C>B>A giving these votes to the 42 voters ranking D>C>E>A>B. This makes D the winner under all four voting methods studied in this chapter. Number of voters

1 choice 2 choice 3 choice 4 choice 5 choice

79. D is preferred over A by a majority of the voters (11 to 10) but all four complete rankings rank A above D. Number of voters

1st choice

2nd choice

3rd choice

4th choice

Method

Winner

2nd place

3rd place

Last place

Plurality

Borda count

Plurality with elimination

Pairwise comparison

A,B (tied)

C,D (tied)

Chapter 2 WALKING 2.1. Weighted Voting 1. (a) A generic weighted voting system with N = 5 players is described using notation [q : w1 , w2 , w3 , w4 , w5 ] where q represents the value of the quota and wi represents the weight of player Pi . In this case, the players are the partners, w1 = 15 , w2 = 12 , w3 = w4 = 10 , and w5 = 3 . Since the total number of votes is 15 + 12 + 10 + 10 + 3 = 50 and the quota is determined by a simple majority (more than 50% of the total number of votes), q = 26. That is, the partnership can be described by [26 :15,12,10,10,3] . 1 2 (b) Since   50 = 33 , we choose the quota q as the smallest integer larger than this value which is 34. 3 3 The partnership can thus be described by [34 :15,12,10,10,3] .

2. (a) There are 100 votes (each one representing 1% ownership). A simple majority requires more than half of that or q = 51. That is, the partnership can be described by [51: 30, 25, 20,16,9] . 2 2 (b) Since  100 = 66 , we choose the quota q as the smallest integer larger than this value which is 67. 3 3 The partnership can thus be described by [67 : 30, 25, 20,16,9] .

3. (a) The quota must be more than half of the total number of votes. This system has 6 + 4 + 3 + 3 + 2 + 2 = 20 total votes. Since 50% of 20 is 10, the smallest possible quota would be 11. Note: q = 10 is not sufficient. If 10 votes are cast in favor and 10 cast against a motion, that motion should not pass. (b) The largest value q can take is 20, the total number of votes. (c)

3 × 20 = 15, so the value of the quota q would be 15. 4

(d) The value of the quota q would be strictly larger than 15. That is, 16. 4. (a) The quota must be more than half of the total votes. This system has 10 + 6 + 5 + 4 + 2 = 27 total votes. 1 × 27 = 13.5, so the smallest value q can take is 14. 2 (b) The largest value q can take is 27, the total number of votes. (c)

2 × 27 = 18 3

(d) 19 5. (a) P1 , the player with the most votes, does not have veto power since the other players combined have 3 + 3 + 2 = 8 votes and can successfully pass a motion without him. The other players can’t have veto power either then as they have fewer votes and hence less (or equal) power. So no players have veto power in this system.

Chapter 2: The Mathematics of Power (b) P1 does have veto power here since the other players combined have only 3 + 3 + 2 = 8 votes and cannot successfully pass a motion without him. P2 , on the other hand, does not have veto power since the other players combined have the 4 + 3 + 2 = 9 votes necessary to meet the quota. Since P3 and P4 have the same or fewer votes than P2 , it follows that only P1 has veto power. (c) P4 , the player with the fewest votes, does not have veto power here since the other players combined have 4 + 3 + 3 = 10 votes and can pass a motion without him. P3 , on the other hand, does have veto power since the other players combined have only 4 + 3 + 2 = 9 votes. Since P1 and P2 have the same or more votes than P3 , it follows that P1 , P2 , and P3 all have veto power. (d) P4 , the player with the fewest votes, has veto power since the other players combined have only 4 + 3 + 3 = 10 votes and cannot pass a motion without him. Since the other players all have the same or more votes than P4 , it follows that all players have veto power in this system. 6. (a) P1 does have veto power here since the other players combined have only 4 + 2 + 1 = 7 votes and cannot pass a motion without him. P2 , on the other hand, does not have veto power since the other players combined have the 8 + 2 + 1 = 11 votes necessary to meet the quota. Since P3 and P4 have the same or fewer votes than P2 , it follows that only P1 has veto power. (b) Based on (a), we know that P1 still has veto power since only the quota has changed (increased). It would now be even more difficult for the other players to pass a motion without P1 . P2 also has veto power here since the other players combined have 8 + 2 + 1 = 11 votes and cannot pass a motion without him. P3 , on the other hand, does not have veto power since the other players combined have 8 + 4 + 1 = 13 votes. Since P4 has fewer votes than P3 , it follows that only P1 and P2 have veto power. (c) Based on (a) and (b), we know that P1 and P2 still have veto power since only the quota has changed (increased). P3 now also has veto power here since the other players combined have only 8 + 4 + 1 = 13 votes and cannot pass a motion without him. P4 , on the other hand, does not have veto power since the other players combined have 8 + 4 + 2 = 14 votes. (d) P4 , the player with the fewest votes, has veto power since the other players combined have only 8 + 4 + 2 = 14 votes and cannot pass a motion without him. Since the other players all have the same or more votes than P4 , it follows that all four players have veto power in this system. 7. (a) In order for all three players to have veto power, the player having the fewest votes (the weakest) must have veto power. In order for that to happen, the quota q must be strictly larger than 7 + 5 = 12. The smallest value of q for which this is true is q = 13. (b) In order for P2 to have veto power, the quota q must be strictly larger than 7 + 3 = 10. The smallest value of q for which this is true is q = 11. [Note: When q = 11, we note that P3 does not have veto power since the other two players have 7 + 5 = 12 votes.] 8. (a) In order for all five players to have veto power, the player having the fewest votes (the weakest) must have veto power. In order for that to happen, the quota q must be strictly larger than 10 + 8 + 6 + 4 = 28. The smallest value of q for which this is true is q = 29. (b) In order for P3 to have veto power, the quota q must be strictly larger than 10 + 8 + 4 + 2 = 24. The smallest value of q for which this is true is q = 25. [Note: When q = 25, we note that P4 does not have veto power since the other four players have 10 + 8 + 6 + 2 = 26 votes.] Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

9. To determine the number of votes each player has, write the weighted voting system as [49: 4x, 2x, x, x]. (a) If the quota is defined as a simple majority of the votes, then x is the largest integer satisfying 4x + 2x + x + x 49 > which means that 8 x < 98 or x < 12.25 . So, x = 12 and the system can be 2 described as [49: 48, 24, 12, 12]. (b) If the quota is defined as more than two-thirds of the votes, then x is the largest integer satisfying 2 ( 4x + 2x + x + x) 49 > which means that 16 x < 147 or x < 9.1875 . So, x = 9 and the system can be 3 described as [49: 36, 18, 9, 9]. (c) If the quota is defined as more than three-fourths of the votes, then x is the largest integer satisfying 3(4x + 2x + x + x) which means that 24 x < 196 or x < 8.167 . So, x = 8 and the system can be 49 > 4 described as [49: 32, 16, 8, 8]. 10. (a) [121: 96, 48, 48, 24, 12, 12]; if the quota is defined as a simple majority of the votes, then x is the largest 8x + 4 x + 4 x + 2 x + x + x integer satisfying 121 > . 2 (b) [121: 72, 36, 36, 18, 9, 9]; if the quota is defined as more than two-thirds of the votes, then x is the largest 2 (8 x + 4 x + 4 x + 2 x + x + x) . integer satisfying 121 > 3 (c) [121: 64, 32, 32, 16, 8, 8]; if the quota is defined as more than three-fourths of the votes, then x is the 3(8x + 4 x + 4 x + 2 x + x + x) . largest integer satisfying 121 > 4

2.2. Banzhaf Power 11. (a) w1 + w3 = 7 + 3 = 10 (b) Since (7 + 5 + 3) / 2 = 7.5, the smallest allowed value of the quota q in this system is 8. For {P1 , P3 } to be a winning coalition, the quota q could be at most 10. So, the values of the quota for which {P1 , P3 } is winning are 8, 9, and 10. (c) Since 7 + 5 + 3 = 15, the largest allowed value of the quota q in this system is 15. For {P1 , P3 } to be a losing coalition, the quota q must be strictly greater than its weight, 10. So, the values of the quota for which {P1 , P3 } is losing are 11, 12, 13, 14, and 15. 12. (a) 8 + 6 + 4 = 18 (b) Since (10 + 8 + 6 + 4 + 2) / 2 = 15, the smallest allowed value of the quota q in this system is 16. For {P2 , P3 , P4 } to be a winning coalition, the quota q could be at most 18. So, the values of the quota for which {P2 , P3 , P4 } is winning are 16, 17, and 18. (c) Since 10 + 8 + 6 + 4 + 2 = 30, the largest allowed value of the quota q in this system is 30. For {P2 , P3 , P4 } to be a losing coalition, the quota q must be strictly greater than its weight, 18. So, the values of the quota for which {P2 , P3 , P4 } is losing are integer values from 19 to 30. Copyright © 2022 Pearson Education, Inc.

Chapter 2: The Mathematics of Power

13. P1 is critical (underlined) three times; P2 is critical three times; P3 is critical once ; P4 is critical once. The total number of times the players are critical is 8 (number of underlines). The Banzhaf power distribution is β1 = 3 / 8; β 2 = 3 / 8; β 3 = 1/ 8; β 4 = 1/ 8 . 14. P1 is critical (underlined) seven times; P2 is critical five times; P3 is critical three times; P4 is critical three times; P5 is critical one time. The total number of times the players are critical (all underlines) is 19. The Banzhaf power distribution is β1 = 7 /19; β 2 = 5 /19; β3 = 3 /19; β4 = 3 /19; β5 = 1/19. 15. (a) P1 is critical since the other players only have 5 + 2 = 7 votes. P2 is also critical since the other players only have 6 + 2 = 8 votes. However, P4 is not critical since the other two players have 6 + 5 = 11 (more than q = 10). (b) The winning coalitions are those whose weights are 10 or more. These are: {P1 , P2 }, {P1 , P3 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P2 , P3 , P4 }, {P1 , P2 , P3 , P4 }. (c) We underline the critical players in each winning coalition: {P1 , P2 }, {P1 , P3 }, {P1 , P2 , P3 }, {P1 , P2 , P4 },

{P1 , P3 , P4 }, {P2 , P3 , P4 }, {P1 , P2 , P3 , P4 }. Then, it follows that β1 =

5 3 1 ; β 2 = β3 = ; β 4 = . 12 12 12

16. (a) All the players are critical in this coalition since the total weight of the coalition (3 + 1 + 1 = 5) is exactly the same as the quota. If any one player were to leave the coalition, the remaining players would not have enough votes to meet the quota. (b) The winning coalitions are those whose weights are 5 or more. These are: {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P1 , P2 , P3 , P4 }. (c) We underline the critical players in each winning coalition: {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 },

{P1 , P3 , P4 }, {P1 , P2 , P3 , P4 }. Then, it follows that β1 =

5 3 1 1 ; β2 = ; β3 = ; β 4 = . 10 10 10 10

17. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , and {P1 , P2 , P3 } . P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times the players are critical is 5. The Banzhaf power distribution is β1 = 3 / 5; β2 = 1/ 5; β3 = 1/ 5. (b) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , and {P1 , P2 , P3 } . P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times the players are critical is 5. The Banzhaf power distribution is β1 = 3/ 5; β2 = 1/ 5; β3 = 1/ 5. The distributions in (a) and (b) are the same. 18. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 } and {P1 , P2 , P3 } . P1 is critical twice; P2 is critical twice; P3 is never critical. The total number of times the players are critical is 4. The Banzhaf power distribution is β1 = 2 / 4; β 2 = 2 / 4; β 3 = 0 / 4.

ISM: Excursions in Modern Mathematics, 10E

(b) The winning coalitions (with critical players underlined) are: {P1 , P2 } and {P1 , P2 , P3 } . P1 is critical twice; P2 is critical twice; P3 is never critical. The total number of times the players are critical is 4. The Banzhaf power distribution is β1 = 2 / 4; β 2 = 2 / 4; β 3 = 0 / 4. The distributions in (a) and (b) are the same. 19. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P2 , P5 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P3 , P4 , P5 } , {P2 , P3 , P4 , P5 } ,

{P1 , P2 , P3 , P4 , P5 } . The Banzhaf power distribution is β1 = 8 / 24; β 2 = 6 / 24; β 3 = 4 / 24; β 4 = 4 / 24;

β 5 = 2 / 24. (b) The situation is like (a) except that {P1 , P2 , P5 } , {P1 , P3 , P4 } and {P2 , P3 , P4 , P5 } are now losing coalitions. In addition, P2 is now critical in {P1 , P2 , P3 , P4 } , P3 is now critical in {P1 , P2 , P3 , P5 } , P4 is now critical in {P1 , P2 , P4 , P5 } , P5 is now critical in {P1 , P3 , P4 , P5 } , and P1 is now critical in the grand coalition {P1 , P2 , P3 , P4 , P5 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P3 , P4 , P5 } , {P1 , P2 , P3 , P4 , P5 } .

The Banzhaf power distribution is β1 = 7 /19; β 2 = 5 /19; β 3 = 3 /19; β 4 = 3 /19; β 5 = 1/19. (c) This situation is like (b) with the following exceptions: {P1 , P2 , P4 } and {P1 , P3 , P4 , P5 } are now losing coalitions; P3 is critical in {P1 , P2 , P3 , P4 } ; P5 is critical in {P1 , P2 , P4 , P5 } ; and P2 is critical in {P1 , P2 , P3 , P4 , P5 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P2 , P3 , P4 , P5 } . The Banzhaf power distribution

is β1 = 5 /15; β 2 = 5 /15; β 3 = 3 /15; β 4 =1/15; β 5 = 1/15. (d) Since the quota equals the total number of votes in the system, the only winning coalition is the grand coalition and every player is critical in that coalition. The Banzhaf power distribution is easy to calculate in the case since all players share power equally. It is β1 = 1/ 5; β 2 =1/ 5; β 3 =1/ 5; β 4 =1/ 5; β 5 =1/ 5. 20. (a) P1 is a dictator and the other players are dummies. Thus P1 is the only critical player in each winning coalition. The Banzhaf power distribution is β1 = 1; β 2 = β 3 = β 4 = 0. (b) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , {P1 , P4 } , {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } . The Banzhaf power distribution is β1 =

β2 =

7 = 70%; 10

1 1 1 = 10%; β 3 = = 10%; β 4 = = 10%. 10 10 10

(c) This situation is like (b) with the following exceptions: {P1 , P4 } is now a losing coalition; P1 and P2 are both critical in {P1 , P2 , P4 }; P1 and P3 are both critical in {P1 , P3 , P4 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 } , {P1 , P3 } , {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } . The Banzhaf power distribution is β1 =

6 2 2 = 60%; β 2 = = 20%; β 3 = = 20%; 10 10 10

β 4 = 0.

Chapter 2: The Mathematics of Power (d) The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P3 , P4 } . The

Banzhaf power distribution is β1 =

2 2 2 ; β 2 = ; β 3 = ; β 4 = 0. 6 6 6

21. (a) A player is critical in a coalition if that coalition without the player is not on the list of winning coalitions. So, in this case, the critical players are underlined below. {P1 , P2 } , {P1 , P3 } , {P1 , P2 , P3 } (b) P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times all players are critical is 3 + 1 + 1 = 5. The Banzhaf power distribution is β1 = 3 / 5; β 2 = 1/ 5; β3 = 1/ 5. 22. (a) A player is critical in a coalition if that coalition without the player is not on the list of winning coalitions. {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P2 , P3 , P4 } (b) β1 = 4 / 8 , β 2 = 4 / 8 , β 3 = β 4 = 0 23. (a) The winning coalitions (with critical players underlined) are {P1 , P2 } , {P1 , P3 } , {P2 , P3 } , {P1 , P2 , P3 } ,

{P1 , P2 , P4 } , {P1 , P2 , P5 } , {P1 , P2 , P6 } , {P1 , P3 , P4 } , {P1 , P3 , P5 } , {P1 , P3 , P6 } , {P2 , P3 , P4 } , {P2 , P3 , P5 } , {P2 , P3 , P6 } . (b) These winning coalitions (with critical players underlined) are {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P2 , P3 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P4 , P5 } , {P1 , P2 , P4 , P6 } , {P1 , P3 , P4 , P5 } , {P1 , P3 , P4 , P6 } , {P2 , P3 , P4 , P5 } ,

{P2 , P3 , P4 , P6 } , {P1 , P2 , P3 , P4 , P5 } , {P1 , P2 , P3 , P4 , P6 } , {P1 , P2 , P4 , P5 , P6 } , {P1 , P3 , P4 , P5 , P6 } , {P2 , P3 , P4 , P5 , P6 } , {P1 , P2 , P3 , P4 , P5 , P6 } . (c) P4 is never a critical player since every time it is part of a winning coalition, that coalition is a winning coalition without P4 as well. So, β 4 = 0. (d) A similar argument to that used in part (c) shows that P5 and P6 are also dummies. One could also argue that any player with fewer votes than P4 , a dummy, will also be a dummy. So, P4 , P5 , and P6 will never be critical -- they all have zero power.

The only winning coalitions with only two players are {P1 , P2 }, {P1 , P3 }, and {P2 , P3 }; and both players are critical in each of those coalitions. All other winning coalitions consist of one of these coalitions plus additional players, and the only critical players will be the ones from the two-player coalition. So P1 , P2 , and P3 will be critical in every winning coalition they are in, and they will all be in the same number of winning coalitions, so they all have the same power. Thus, the Banzhaf power distribution is β1 = 1/ 3; β 2 = 1/ 3; β3 = 1/ 3; β 4 = 0; β 5 = 0; β 6 = 0. 24. (a) Three-player winning coalitions (with critical players underlined): {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P2 , P5 },

{P1 , P3 , P4 }, {P2 , P3 , P4 }. (b) Four-player winning coalitions (with critical players underlined): {P1 , P2 , P3 , P4 }, {P1 , P2 , P3 , P5 },

{P1 , P2 , P3 , P6 }, {P1 , P2 , P4 , P5 }, {P1 , P2 , P4 , P6 }, {P1 , P2 , P5 , P6 }, {P1 , P3 , P4 , P5 }, {P1 , P3 , P4 , P6 }, {P1 , P3 , P5 , P6 }, {P2 , P3 , P4 , P5 }, {P2 , P3 , P4 , P6 }. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

(c) Five-player winning coalitions (with critical players underlined): {P1 , P2 , P3 , P4 , P5 }, {P1 , P2 , P3 , P4 , P6 }, {P1 , P2 , P3 , P5 , P6 }, {P1 , P2 , P4 , P5 , P6 }, {P1 , P3 , P4 , P5 , P6 }, {P2 , P3 , P4 , P5 , P6 }. 15 13 11 9 3 1 (d) β1 = 52 ; β2 = 52 ; β3 = 52 ; β4 = 52 ; β5 = 52 ; β6 = 52 . Note: The grand coalition is also a

winning coalition. However, it has no critical players. 25. (a) { A, B} , { A, C} , {B, C} , { A, B, C} , { A, B, D} , { A, C , D} , {B, C , D} , { A, B, C , D} (b) A, B, and C have Banzhaf power index of 4/12 each; D is a dummy. (D is never a critical player, and the other three clearly have equal power.) 26. (a) { A, B} , { A, C} , { A, D} , { A, B, C} , { A, B, D} , { A, C , D} , {B, C , D} , { A, B, C , D} (b) A has Banzhaf power index of 6/12; B, C, and D have Banzhaf power index of 2/12 each.

2.3. Shapley-Shubik Power 27. P1 is pivotal (underlined) ten times; P2 is pivotal (again, underlined) ten times; P3 is pivotal twice ; P4 is pivotal twice. The total number of times the players are pivotal is 4! = 24 (number of underlines). The Shapley-Shubik power distribution is σ 1 = 10 / 24; σ 2 = 10 / 24; σ 3 = 2 / 24; σ 4 = 2 / 24 . 28. P1 is pivotal (underlined) ten times; P2 is pivotal six times; P3 is pivotal six times ; P4 is pivotal twice. The total number of times the players are pivotal is 4! = 24. The Shapley-Shubik power distribution is σ 1 = 10 / 24; σ 2 = 6 / 24; σ 3 = 6 / 24; σ 4 = 2 / 24 . 29. (a) There are 3! = 6 sequential coalitions of the three players. Each pivotal player is underlined. < P1 , P2 , P3 >, < P1 , P3 , P2 >, < P2 , P1 , P3 >, < P2 , P3 , P1 >, < P3 , P1 , P2 >, < P3 , P2 , P1 > (b) P1 is pivotal four times; P2 is pivotal one time; P3 is pivotal one time. The Shapley-Shubik power distribution is σ 1 = 4 / 6; σ 2 = 1/ 6; σ 3 = 1/ 6. 30. (a) There are 3! = 6 sequential coalitions of the three players. Each pivotal player is underlined. < P1 , P2 , P3 >, < P1 , P3 , P2 >, < P2 , P1 , P3 >, < P2 , P3 , P1 >, < P3 , P1 , P2 >, < P3 , P2 , P1 > (b) σ 1 = 2 / 6; σ 2 = 2 / 6; σ 3 = 2 / 6. 31. (a) Since P1 is a dictator, σ 1 = 1, σ 2 = 0, σ 3 = 0, and σ 4 = 0 . (b) There are 4! = 24 sequential coalitions of the four players. Each pivotal player is underlined. < P1 , P2 , P3 , P4 >, < P1 , P2 , P4 , P3 >, < P1 , P3 , P2 , P4 >, < P1 , P3 , P4 , P2 >,

< P1 , P4 , P2 , P3 >, < P1 , P4 , P3 , P2 >, < P2 , P1 , P3 , P4 >, < P2 , P1 , P4 , P3 >, < P2 , P3 , P1 , P4 >, < P2 , P3 , P4 , P1 >, < P2 , P4 , P1 , P3 >, < P2 , P4 , P3 , P1 >, < P3 , P1 , P2 , P4 >, < P3 , P1 , P4 , P2 >, < P3 , P2 , P1 , P4 >, < P3 , P2 , P4 , P1 >, < P3 , P4 , P1 , P2 >, < P3 , P4 , P2 , P1 >, < P4 , P1 , P2 , P3 >, < P4 , P1 , P3 , P2 >, < P4 , P2 , P1 , P3 >, < P4 , P2 , P3 , P1 >, < P4 , P3 , P1 , P2 >, < P4 , P3 , P2 , P1 > P1 is pivotal 16 times; P2 is pivotal 4 times; P3 is pivotal 4 times; P4 is pivotal 0 times. The ShapleyShubik power distribution is σ 1 = 16 / 24; σ 2 = 4 / 24; σ 3 = 4 / 24; σ 4 = 0. Copyright © 2022 Pearson Education, Inc.

Chapter 2: The Mathematics of Power (c) The only way a motion will pass is if P1 and P2 both support it. In fact, the second of these players that appears in a sequential coalition will be the pivotal player in that coalition. It follows that σ 1 = 12 / 24 = 1/ 2, σ 2 = 12 / 24 = 1/ 2, σ 3 = 0, and σ 4 = 0 . (d) Because the quota is so high, the only way a motion will pass is if P1 , P2 , and P3 all support it. In fact, the third of these players that appears in a sequential coalition will always be the pivotal player in that coalition. It follows that σ 1 = 8 / 24 = 1/ 3, σ 2 = 8 / 24 = 1/ 3, σ 3 = 8 / 24 = 1/ 3, and σ 4 = 0 .

32. (a) Since P1 is a dictator, σ 1 = 1 , σ 2 = 0 , σ 3 = 0 , σ 4 = 0 . (b) < P1 , P2 , P3 , P4 >, < P1 , P2 , P4 , P3 >, < P1 , P3 , P2 , P4 >, < P1 , P3 , P4 , P2 >,

The Shapley-Shubik power distribution is σ 1 = 9 /12 , σ 2 = 1/12 , σ 3 = 1/12 , σ 4 = 1/12 . (c) σ 1 = 1/ 2 , σ 2 = 1/ 2 , σ 3 = 0 , σ 4 = 0 (The second of P1 or P2 to vote is always the pivotal player in any of the 24 sequential coalitions. This happens 12 times in each case.) (d) σ 1 = 1 / 3 , σ 2 = 1/ 3 , σ 3 = 1/ 3 , σ 4 = 0 (The third of P1 , P2 , or P3 to vote is always the pivotal player in any of the 24 sequential coalitions. This happens 8 times in each case.) 33. (a) There are 4! = 24 sequential coalitions of the four players. Each pivotal player is underlined. < P1 , P2 , P3 , P4 >, < P1 , P2 , P4 , P3 >, < P1 , P3 , P2 , P4 >, < P1 , P3 , P4 , P2 >,

< P1 , P4 , P2 , P3 >, < P1 , P4 , P3 , P2 >, < P2 , P1 , P3 , P4 >, < P2 , P1 , P4 , P3 >, < P2 , P3 , P1 , P4 >, < P2 , P3 , P4 , P1 >, < P2 , P4 , P1 , P3 >, < P2 , P4 , P3 , P1 >, < P3 , P1 , P2 , P4 >, < P3 , P1 , P4 , P2 >, < P3 , P2 , P1 , P4 >, < P3 , P2 , P4 , P1 >, < P3 , P4 , P1 , P2 >, < P3 , P4 , P2 , P1 >, < P4 , P1 , P2 , P3 >, < P4 , P1 , P3 , P2 >, < P4 , P2 , P1 , P3 >, < P4 , P2 , P3 , P1 >, < P4 , P3 , P1 , P2 >, < P4 , P3 , P2 , P1 > P1 is pivotal in 10 coalitions; P2 is pivotal in six coalitions; P3 is pivotal in six coalitions; P4 is pivotal in two coalitions. The Shapley-Shubik power distribution is σ1 = 10 / 24; σ 2 = 6 / 24; σ 3 = 6 / 24; σ 4 = 2 / 24.

(b) This is the same situation as in (a) – there is essentially no difference between 51 and 59 because the players’ votes are all multiples of 10. The Shapley-Shubik power distribution is thus still σ1 = 10 / 24; σ 2 = 6 / 24; σ 3 = 6 / 24; σ 4 = 2 / 24. (c) This is also the same situation as in (a) – any time a group of players has 51 votes, they must have 60 votes. The Shapley-Shubik power distribution is thus still σ1 = 10 / 24; σ 2 = 6 / 24; σ 3 = 6 / 24; σ 4 = 2 / 24.

ISM: Excursions in Modern Mathematics, 10E

34. (a) σ 1 = 18 / 24; σ 2 = 2 / 24; σ 3 = 2 / 24; σ 4 = 2 / 24. The sequential coalitions (with pivotal player in each coalition underlined) are: < P1 , P2 , P3 , P4 >, < P1 , P2 , P4 , P3 >, < P1 , P3 , P2 , P4 >, < P1 , P3 , P4 , P2 >,

< P1 , P4 , P2 , P3 >, < P1 , P4 , P3 , P2 >, < P2 , P1 , P3 , P4 >, < P2 , P1 , P4 , P3 >, < P2 , P3 , P1 , P4 >, < P2 , P3 , P4 , P1 >, < P2 , P4 , P1 , P3 >, < P2 , P4 , P3 , P1 >, < P3 , P1 , P2 , P4 >, < P3 , P1 , P4 , P2 >, < P3 , P2 , P1 , P4 >, < P3 , P2 , P4 , P1 >, < P3 , P4 , P1 , P2 >, < P3 , P4 , P2 , P1 >, < P4 , P1 , P2 , P3 >, < P4 , P1 , P3 , P2 >, < P4 , P2 , P1 , P3 >, < P4 , P2 , P3 , P1 >, < P4 , P3 , P1 , P2 >, < P4 , P3 , P2 , P1 > (b) σ 1 = 18 / 24; σ 2 = 2 / 24; σ 3 = 2 / 24; σ 4 = 2 / 24. This is the same situation as in (a). (c) This is also the same situation as in (a) – any time a group of players has 41 votes, they must have 50 votes. The Shapley-Shubik power distribution is thus still σ 1 = 12 / 24; σ 2 = 4 / 24; σ 3 = 4 / 24; σ 4 = 4 / 24. 35. (a) There are 3! = 6 sequential coalitions of the three players. < P1 , P2 , P3 >, < P1 , P3 , P2 >, < P2 , P1 , P3 >, < P2 , P3 , P1 >, < P3 , P1 , P2 >, < P3 , P2 , P1 >

(The second player listed will always be pivotal unless the first two players listed are P2 and P3 since that is not a winning coalition. In that case, P1 is pivotal.) (b) P1 is pivotal four times; P2 is pivotal one time; P3 is pivotal one time. The Shapley-Shubik power distribution is σ1 = 4 / 6; σ 2 = 1/ 6; σ 3 = 1/ 6. 36. (a) < P1 , P2 , P3 > , < P1 , P3 , P2 > , < P2 , P1 , P3 > , < P2 , P3 , P1 > , < P3 , P1 , P2 > , < P3 , P2 , P1 >

(The third player listed will always be pivotal unless the first two players listed are P1 and P2 since that is a winning coalition. In that case, the second of these players is pivotal.) (b) σ 1 = 1/ 2 , σ 2 = 1/ 2 , σ 3 = 0 37. We proceed to identify pivotal players by moving down each column. We start with the first column. Since P2 is pivotal in the sequential coalition < P1 , P2 , P3 , P4 > , we know that { P1 , P2 } form a winning

coalition. This tells us that P2 is also pivotal in the second sequential coalition < P1 , P2 , P4 , P3 > listed in the first column. A similar argument tells us that P3 is pivotal in the fourth sequential coalition

< P1 , P3 , P4 , P2 > listed in the first column. In the second column, P1 is clearly pivotal in the sequential coalition < P2 , P1 , P3 , P4 > since we know that

{P1 , P2 } form a winning coalition. Similarly, P1 is pivotal in the sequential coalition < P2 , P1 , P4 , P3 > . In the third column, P1 is pivotal in the sequential coalition < P3 , P1 , P2 , P4 > since the third sequential coalition in the first column ( < P1 , P3 , P2 , P4 > ) identified { P1 , P3 } as a winning coalition. Similarly, P1 is pivotal in the sequential coalition < P3 , P1 , P4 , P2 > . Since P1 was pivotal in the sequential coalition

< P2 , P3 , P1 , P4 > back in the second column, it must also be the case that P1 is pivotal in the sequential coalition < P3 , P2 , P1 , P4 > . Similarly, P4 is pivotal in the sequential coalition < P3 , P2 , P4 , P1 > since P4 was Copyright © 2022 Pearson Education, Inc.

Chapter 2: The Mathematics of Power

also pivotal in the sequential coalition < P2 , P3 , P4 , P1 > back in the second column. In the fourth column, the third player listed will always be pivotal since the first two players when listed first in the opposite order (in an earlier column) never contain a pivotal player and the fourth player was never pivotal in the first three columns. The final listing of pivotal players is found below. < P1 , P2 , P3 , P4 >, < P2 , P1 , P3 , P4 >, < P3 , P1 , P2 , P4 >, < P4 , P1 , P2 , P3 >,

< P1 , P2 , P4 , P3 >, < P2 , P1 , P4 , P3 >, < P3 , P1 , P4 , P2 >, < P4 , P1 , P3 , P2 >, < P1 , P3 , P2 , P4 >, < P2 , P3 , P1 , P4 >, < P3 , P2 , P1 , P4 >, < P4 , P2 , P1 , P3 >, < P1 , P3 , P4 , P2 >, < P2 , P3 , P4 , P1 >, < P3 , P2 , P4 , P1 >, < P4 , P2 , P3 , P1 >, < P1 , P4 , P2 , P3 >, < P2 , P4 , P1 , P3 >, < P3 , P4 , P1 , P2 >, < P4 , P3 , P1 , P2 >, < P1 , P4 , P3 , P2 >, < P2 , P4 , P3 , P1 >, < P3 , P4 , P2 , P1 >, < P4 , P3 , P2 , P1 > The Shapley-Shubik distribution is then σ 1 = 10 / 24; σ 2 = 6 / 24; σ 3 = 6 / 24; σ 4 = 2 / 24. 38. We identify pivotal players by moving down each column. Since P2 is pivotal in the sequential

coalition < P1 , P2 , P3 , P4 > , we know that { P1 , P2 } form a winning coalition. This tells us that P2 is also pivotal in the second sequential coalition < P1 , P2 , P4 , P3 > . In the second column, P1 is clearly pivotal in the sequential coalition < P2 , P1 , P3 , P4 > since we know that

{P1 , P2 } form a winning coalition. Similarly, P1 is pivotal in the sequential coalition < P2 , P1 , P4 , P3 > . Since P1 is pivotal in < P2 , P3 , P4 , P1 > , we know that P1 must be pivotal in < P2 , P4 , P3 , P1 > too. In the third column, P2 is pivotal in the sequential coalition < P3 , P1 , P2 , P4 > according to the third sequential coalition in the first column ( < P1 , P3 , P2 , P4 > ). Similarly, P4 is pivotal in the sequential coalition < P3 , P1 , P4 , P2 > according to the fourth sequential coalition in the first column ( < P1 , P3 , P4 , P2 > ). P1 is pivotal in the sequential coalition < P3 , P2 , P1 , P4 > according to the third sequential coalition in the second column ( < P2 , P3 , P1 , P4 > ). P1 is pivotal in the sequential coalitions

< P3 , P2 , P4 , P1 > and < P3 , P4 , P2 , P1 > according to the fourth sequential coalition in the second column ( < P2 , P3 , P4 , P1 > ). In the fourth column, P2 is pivotal in the sequential coalition < P4 , P1 , P2 , P3 > according to the fifth sequential coalition in the first column ( < P1 , P4 , P2 , P3 > ). Also, P3 is pivotal in the sequential coalition

< P4 , P1 , P3 , P2 > according to the sixth sequential coalition in the first column ( < P1 , P4 , P3 , P2 > ). We have P1 pivotal in the sequential coalition < P4 , P2 , P1 , P3 > according to the fifth sequential coalition in the

second column ( < P2 , P4 , P1 , P3 > ). We also have P1 pivotal in the sequential coalitions

< P4 , P2 , P3 , P1 > and < P4 , P3 , P2 , P1 > according to the fourth sequential coalition in the second column ( < P2 , P3 , P4 , P1 > ).

ISM: Excursions in Modern Mathematics, 10E

Lastly, we see that P1 pivotal in the sequential coalition < P4 , P3 , P1 , P2 > according to the fifth sequential coalition in the third column ( < P3 , P4 , P1 , P2 > ). The final listing of pivotal players is found below.

< P1 , P2 , P3 , P4 >, < P2 , P1 , P3 , P4 >, < P3 , P1 , P2 , P4 >, < P4 , P1 , P2 , P3 >, < P1 , P2 , P4 , P3 >, < P2 , P1 , P4 , P3 >, < P3 , P1 , P4 , P2 >, < P4 , P1 , P3 , P2 >, < P1 , P3 , P2 , P4 >, < P2 , P3 , P1 , P4 >, < P3 , P2 , P1 , P4 >, < P4 , P2 , P1 , P3 >, < P1 , P3 , P4 , P2 >, < P2 , P3 , P4 , P1 >, < P3 , P2 , P4 , P1 >, < P4 , P2 , P3 , P1 >, < P1 , P4 , P2 , P3 >, < P2 , P4 , P1 , P3 >, < P3 , P4 , P1 , P2 >, < P4 , P3 , P1 , P2 >, < P1 , P4 , P3 , P2 >, < P2 , P4 , P3 , P1 >, < P3 , P4 , P2 , P1 >, < P4 , P3 , P2 , P1 > The Shapley-Shubik distribution is then σ 1 = 14 / 24; σ 2 = 6 / 24; σ 3 = 2 / 24; σ 4 = 2 / 24.

2.4. Subsets and Permutations 39. (a) A set with N elements has 2 N subsets. So, A has 210 = 1024 subsets. (b) There is only one subset, namely the empty set { }, having less than one element. So, using part (a), the number of subsets of A having one or more elements is 1024 – 1 = 1023. (c) Each element of A can be used to form a subset containing exactly one element. So, there are exactly 10 such subsets. (d) The number of subsets of A having two or more elements is the number of subsets of A not having either 0 or 1 element. From parts (a), (b) and (c), we calculate this number as 1024 – 1 – 10 = 1013. 40. (a) A set with N elements has 2 N subsets. So, A has 212 = 4096 subsets. (b) There is only one subset, namely the empty set { }, having less than one element. So, using part (a), the number of subsets of A having one or more elements is 4096 – 1 = 4095. (c) Each element of A can be used to form a subset containing exactly one element. So, there are exactly 12 such subsets. (d) The number of subsets of A having two or more elements is the number of subsets of A not having either 0 or 1 element. From parts (a), (b) and (c), we calculate this number as 4096 – 1 – 12= 4083. 41. (a) A weighted voting system with N players has 2 N − 1 coalitions. So, a system with 10 players has 210 − 1 = 1023 coalitions. (b) The number of coalitions with two or more players is all of the coalitions minus the number of those having at most one player. But we know that there are exactly 10 coalitions consisting of exactly one player. So, the number of coalitions with two or more players is 1023 – 10 = 1013. See also Exercise 39(d). 42. (a) A weighted voting system with N players has 2 N − 1 coalitions. So, a system with 12 players has 212 − 1 = 4095 coalitions. (b) 4095 – 12 = 4083 43. (a) 26 − 1 = 63 coalitions (b) There are 25 − 1 = 31 coalitions of the remaining five players P2 , P3 , P4 , P5 , and P6 . These are exactly those coalitions that do not include P1 . Copyright © 2022 Pearson Education, Inc.

Chapter 2: The Mathematics of Power (c) As in (b), there are 25 − 1 = 31 coalitions of the remaining five players P1 , P2 , P4 , P5 , and P6 . These are exactly those coalitions that do not include P3 . (d) There are 2 4 − 1 = 15 coalitions of the remaining four players P2 , P4 , P5 , P6 . These are exactly those coalitions that do not include either P1 or P3 . (e) 16 coalitions include P1 and P3 (all 2 4 − 1 = 15 coalitions of the remaining four players P2 , P4 , P5 , P6 together with the empty coalition could be combined with these two players to form such a coalition).

44. (a) 25 − 1 = 31 coalitions (b) There are 24 − 1 = 15 coalitions of the remaining four players P2 , P3 , P4 , and P5 . These are exactly those coalitions that do not include P1 . (c) As in (b), there are 24 − 1 = 15 coalitions of the remaining four players P1 , P2 , P3 , and P4 . These are exactly those coalitions that do not include P5 . (d) There are 23 − 1 = 7 coalitions of the remaining three players P2 , P3 , P4 . These are exactly those coalitions that do not include P1 or P5 . (e) 8 coalitions include P1 and P5 (all 23 − 1 = 7 coalitions of the remaining three players P2 , P3 , P4 together with the empty coalition could be combined with these two players to form such a coalition). 45. (a) 13! = 6,227,020,800 (b) 18! = 6,402,373,705,728,000 ≈ 6.402374 ×1015 (c) 25! = 15,511,210,043,330,985,984,000,000 ≈ 1.551121× 1025 (d) There are 25! sequential coalitions of 25 players. 1 second 1 hour 1 day 1 year × × × 25! sequential coaltions × 1, 000, 000, 000, 000 sequential coalitions 3600 seconds 24 hours 365 days ≈ 491,857 years 46. (a) 12! = 479,001,600 (b) 15! = 1,307,674,368,000 ≈ 1.307674 × 1012 (c) 20! = 2,432,902,008,176,640,000 ≈ 2.432902 ×1018 (d) There are 20! sequential coalitions of 20 players. 1 second 1 hour 1 day 1 year × × × 20! sequential coalitions × 1, 000, 000, 000 sequential coalitions 3600 seconds 24 hours 365 days ≈ 77 years 47. (a)

(b)

ISM: Excursions in Modern Mathematics, 10E (c)

13! 6, 227, 020,800 = = 715 4!9! 24 × 362,880

(d)

13! 6, 227, 020,800 = = 1287 5!8! 120 × 40,320

48. (a)

12! 479, 001, 600 = = 239,500,800 2! 2

(b)

12! 479, 001, 600 = = 66 2!10! 2 × 3, 628,800

(c)

12! 479, 001, 600 = = 220 3!9! 6 × 362,880

(d)

12! 479, 001, 600 = = 495 4!8! 24 × 40,320

49. (a) 10! = 10 × 9 × 8 ×  × 3 × 2 × 1 = 10 × 9! 10! 3, 628,800 So, 9! = = = 362,880 . 10 10 (b) 11! = 11× 10 × 9 ×  × 3 × 2 × 1 = 11× 10! 11! 11 × 10! So, = = 11 . 10! 10! (c) 11! = 11×10 × ( 9 ×× 3 × 2 ×1) = 11×10 × 9!

So,

11! 11 × 10 × 9! = = 11 × 10 = 110 . 9! 9!

(d)

9! 9 × 8 × 7 × 6! = = 9 × 8 × 7 = 504 6! 6!

(e)

101! 101 × 100 × 99! = = 101 × 100 = 10,100 99! 99!

50. (a) 20! = 20 × 19! so 19! =

20! 2, 432, 902, 008,176, 640, 000 = = 121,645,100,408,832,000 20 20

(b) 20 (c)

201! 201 × 200 × 199! = = 201 × 200 = 40, 200 199! 199!

(d) 990 11!/ 8! = (11× 10 × 9 × 8!) / 8! = 11× 10 × 9 = 990 51. (a) 7! = 5040 sequential coalitions

Chapter 2: The Mathematics of Power (b) There are 6! = 720 sequential coalitions of the remaining six players P1 , P2 , P3 , P4 , P5 and P6 . These correspond exactly to those seven-player coalitions that have P7 as the first player. (c) As in (b), there are 6! = 720 sequential coalitions of the remaining six players P1 , P2 , P3 , P4 , P5 and P6 . These correspond exactly to those seven-player coalitions that have P7 as the last player. (d) 5040 – 720 = 4320 sequential coalitions do not have P1 listed as the first player.

52. (a) 6! = 720 (b) There are 5! = 120 sequential coalitions of the remaining five players P1 , P2 , P3 , P5 and P6 . These correspond exactly to those six-player coalitions that have P4 as the last player. (c) 5! = 120; see (b) (d) 720 – 120 = 600 53. (a) First, note that P1 is pivotal in all sequential coalitions except when it is the first player. By 51(d), P1 is pivotal in 4320 of the 5040 sequential coalitions. (b) Based on the results of (a), σ 1 = 4320 / 5040 = 6 / 7 . (c) Since players P2 , P3 , P4 , P5 , P6 and P7 share the remaining power equally, σ 1 = 4320 / 5040 = 6 / 7 , σ 2 = σ 3 = σ 4 = σ 5 = σ 6 = σ 7 = 120 / 5040 = 1/ 42 . [Note also that 1/42 is 1/7 divided into 6 equal parts.] 54. (a) 600 (see Exercise 52(d)) (b) 600/720 = 5/6 (c) σ 1 = 5 / 6 , σ 2 = σ 3 = σ 4 = σ 5 = σ 6 = 1/ 30 (the other players share power equally)

JOGGING 55. (a) Suppose that a winning coalition that contains P is not a winning coalition without P. Then P would be a critical player in that coalition, contradicting the fact that P is a dummy. (b) P is a dummy ⇔ P is never critical ⇔ the numerator of its Banzhaf power index is 0 ⇔ it’s Banzhaf power index is 0. (c) Suppose P is not a dummy. Then, P is critical in some winning coalition. Let S denote the other players in that winning coalition. The sequential coalition with the players in S first (in any order), followed by P and then followed by the remaining players has P as its pivotal player. Thus, P’s Shapley-Shubik power index is not 0. Conversely, if P’s Shapley-Shubik power index is not 0, then P is pivotal in some sequential coalition. A coalition consisting of P together with the players preceding P in that sequential coalition is a winning coalition and P is a critical player in it. Thus, P is not a dummy. 56. (a) P5 is a dummy. It takes (at least) three of the first four players to pass a motion. P5 ' s vote doesn’t make any difference. (b) β1 = σ 1 = 1/ 4; β2 = σ 2 = 1/ 4; β3 = σ 3 = 1/ 4; β4 = σ 4 = 1/ 4; β5 = σ 5 = 0 (c) q = 21, q = 31, q = 41. (d) Since we assume that the weights are listed in non-increasing order, P5 is a dummy if w = 1, 2, 3. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

57. (a) The quota must be at least half of the total number of votes and not more than the total number of votes. 7 ≤ q ≤ 13 . (b) For q = 7 or q = 8, P1 is a dictator because {P1} is a winning coalition. (c) For q = 9, only P1 has veto power since P2 and P3 together have just 5 votes. (d) For 10 ≤ q ≤ 12, both P1 and P2 have veto power since no motion can pass without both of their votes. For q = 13, all three players have veto power. (e) For q = 7 or q = 8, both P2 and P3 are dummies because P1 is a dictator. For 10 ≤ q ≤ 12, P3 is a dummy since all winning coalitions contain {P1 , P2 } which is itself a winning coalition. 58. (a) 5 ≤ w ≤ 9 ; Since we assume that the weights are listed in non-increasing order, we have 5 ≤ w . In w + 5 + 2 +1 < 9 ≤ w + 5 + 2 + 1 , i.e., w + 8 < 18 and 9 ≤ w + 8 . Since w is a whole addition, we have 2 number, this gives 5 ≤ w ≤ 9 . (b) P1 is a dictator if w = 9. (c) For w = 5, P1 and P2 have veto power. If 6 ≤ w ≤ 8 , then P1 alone has veto power. (d) If w = 5, all winning coalitions contain {P1 , P2 } which is itself a winning coalition and so both P3 and P4 are dummies. If w = 7, then P4 is a dummy. If w = 9, then P2 , P3 , and P4 are all dummies since P1 is a dictator. 59. (a) The winning coalitions for both weighted voting systems [8: 5, 3, 2] and [2: 1, 1, 0] are {P1 , P2 } and {P1 , P2 , P3 } . (b) The winning coalitions for both weighted voting systems [7: 4, 3, 2, 1] and [5: 3, 2, 1, 1] are {P1 , P2 } , {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, and {P1 , P2 , P3 , P4 } . (c) The winning coalitions for both weighted voting systems are those consisting of any three of the five players, any four of the five players, and the grand coalition. (d) If a player is critical in a winning coalition, then that coalition is no longer winning if the player is removed. An equivalent system (with the same winning coalitions) will find that player critical in the same coalitions. When calculating the Banzhaf power indexes in two equivalent systems, the same players will be critical in the same coalitions. Thus, the numerators and denominators in each player’s Banzhaf power index will be the same. (e) If a player, P, is pivotal in a sequential coalition, then the players to P’s left in the sequential coalition do not form a winning coalition, but including P does make it a winning coalition. An equivalent system (with the same winning coalitions) will find the player P pivotal in that same sequential coalition. When calculating the Shapley-Shubik indexes in two equivalent systems, the same players will be pivotal in the same sequential coalitions. Thus, the numerators and denominators in each Shapley-Shubik index will be the same. 60. (a) P has veto power ⇔ the coalition formed by all players other than P is a losing coalition ⇔ P must be a member of any winning coalition. (b) P has veto power ⇔ the coalition formed by all players other than P is a losing coalition ⇔ P is a critical player in the grand coalition. Copyright © 2022 Pearson Education, Inc.

Chapter 2: The Mathematics of Power

61. (a) V / 2 < q ≤ V − w1 (where V denotes the sum of all the weights). The idea here is that if P1 does not have veto power, neither will any of the other players (since weights decrease in value). (b) V − wN < q ≤ V ; The idea here is that if PN has veto power, so will all the other players. (c) V − wi < q ≤ V − wi +1 62. (a) [7: 3, 3, 2, 2, 2] or any answer of the form [7k: 3k, 3k, 2k, 2k, 2k] for positive integer k is correct. One process to solve this problem is start with small values of c and experiment. It is important that the sum of the children’s votes (3c) not meet the quota and that sum of the parent’s votes (2p) also not meet the quota. (b) The winning coalitions in this voting system (with critical players underlined) are {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P2 , P5 } , {P1 , P3 , P4 } , {P1 , P3 , P5 } , {P1 , P4 , P5 } , {P2 , P3 , P4 } , {P2 , P3 , P5 } , {P2 , P4 , P5 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P3 , P4 , P5 } , {P2 , P3 , P4 , P5 } , {P1 , P2 , P3 , P4 , P5 } . So, the Banzhaf power distribution is β1 = β 2 = 7 / 29 , β 3 = β 4 = β5 = 5 / 29 .

63. You should buy your vote from P3 . The following table explains why.

Buying a vote from

Resulting weighted voting system

Resulting Banzhaf power distribution

Your power

[8: 5, 4, 2, 2]

β1 = 124 ; β2 = 124 ; β3 = 122 ; β 4 = 122

2 12

[8: 6, 3, 2, 2]

β1 = 107 ; β2 = 101 ; β3 = 101 ; β 4 = 101

1 10

[8: 6, 4, 1, 2]

β1 = 106 ; β2 = 102 ; β3 = 0; β 4 = 102

2 10

64. You should buy your vote from P4 . The following table explains why.

Buying a vote from

Resulting weighted voting system

Resulting Banzhaf power distribution

Your power

[27: 9, 8, 6, 4, 3]

β1 = 14 ; β 2 = 14 ; β3 = 14 ; β 4 = 14 ; β5 = 0

[27: 10, 7, 6, 4, 3]

β1 = 14 ; β 2 = 14 ; β3 = 14 ; β 4 = 14 ; β5 = 0

[27: 10, 8, 5, 4, 3]

β1 = 14 ; β 2 = 14 ; β3 = 14 ; β 4 = 14 ; β5 = 0

[27: 10, 8, 6, 3, 3]

β1 = 113 ; β 2 = 113 ; β3 = 113 ; β 4 = 111 ; β5 = 111

1 11

ISM: Excursions in Modern Mathematics, 10E

65. (a) You should buy your vote from P2 . The following table explains why.

Buying a vote from

Resulting weighted voting system

Resulting Banzhaf power distribution

Your power

[18: 9, 8, 6, 4, 3]

β1 = 134 ; β 2 = 133 ; β3 = 133 ; β 4 = 132 ; β5 = 131

1 13

[18: 10, 7, 6, 4, 3]

β1 = 259 ; β 2 = 15 ; β3 = 15 ; β 4 = 253 ; β5 = 253

3 25

[18: 10, 8, 5, 4, 3]

β1 = 125 ; β 2 = 14 ; β3 = 16 ; β 4 = 121 ; β5 = 121

1 12

[18: 10, 8, 6, 3, 3]

β1 = 125 ; β 2 = 14 ; β3 = 16 ; β 4 = 121 ; β5 = 121

1 12

(b) You should buy 2 votes from P2 . The following table explains why.

Buying a vote from

Resulting weighted voting system

Resulting Banzhaf power distribution

Your power

[18: 8, 8, 6, 4, 4]

β1 = 277 ; β 2 = 277 ; β3 = 277 ; β 4 = 19 ; β5 = 19

1 9

[18: 10, 6, 6, 4, 4]

β1 = 135 ; β 2 = 132 ; β3 = 132 ; β 4 = 132 ; β5 = 132

2 13

[18: 10, 8, 4, 4, 4]

β1 = 11 ; β 2 = 15 ; β3 = 253 ; β 4 = 253 ; β5 = 253 25

3 25

[18: 10, 8, 6, 2, 4]

β1 = 259 ; β 2 = 257 ; β3 = 15 ; β 4 = 251 ; β5 = 253

3 25

1 3 = 4% to = 12%. Buying a second vote 25 25

2 ≈ 15.4%. The increase in power is less with the second vote, but if you 13 value power over money, it might still be worth it to you to buy that second vote.

from P2 raises your power to

66. (a) Before the merger: β1 = 53 ; β 2 = 15 ; β3 = 15 ; after the merger β1 : 12 ; β* : 12 . (b) Before the merger: β1 = 12 ; β 2 = 12 ; β3 = 0; after the merger β1 = 12 ; β* = 12 . (c) Before the merger: β1 = 13 ; β 2 = 13 ; β3 = 13 ; after the merger β1 = 12 ; β* = 12 . (d) The Banzhaf power index of the merger of two players can be greater than, equal to, or less than the sum of Banzhaf power indices of the individual players. 67. (a) [24: 14, 8, 6, 4] is just [12: 7, 4, 3, 2] with each value multiplied by 2. Both have Banzhaf power distribution β1 = 2 / 5; β2 = 1/ 5; β3 = 1/ 5; β4 = 1/ 5. (b) In the weighted voting system [q : w1 , w2 , , wN ], if Pk is critical in a coalition then the sum of the weights of all the players in that coalition (including Pk ) is at least q, but the sum of the weights of all the players in the coalition except Pk is less than q. Consequently, if the weights of all the players in that coalition are multiplied by c > 0 (c = 0 would make no sense), then the sum of the weights of all the players in the coalition (including Pk ) is at least cq but the sum of the weights of all the players in the coalition except Pk is less than cq. Therefore Pk is critical in the same coalition in the weighted voting system [cq : cw1 , cw2 ,  , cwN ]. Since the critical players are the same in both weighted voting systems, the Banzhaf power distributions will be the same.

Chapter 2: The Mathematics of Power

68. (a) Both have Shapley-Shubik power distribution σ 1 = 1/ 2; σ 2 = 1/ 6; σ 3 = 1/ 6; σ 4 = 1/ 6. (b) In the weighted voting system [q : w1 , w2 , , wN ], Pk is pivotal in the sequential coalition

P1 , P2 ,, Pk ,, PN means w1 + w2 +  + wk ≥ q but w1 + w2 +  + wk −1 < q . In the weighted voting system [cq : cw1 , cw2 ,  , cwN ] , Pk is pivotal in the sequential coalition P1 , P2 ,, Pk ,, PN means cw1 + cw2 +  + cwk ≥ cq but cw1 + cw2 +  + cwk −1 < cq . These two statements are equivalent since

cw1 + cw2 +  + cwk = c ( w1 + w2 +  + wk ) ≥ cq if and only if w1 + w2 +  + wk ≥ q , and

cw1 + cw2 +  + cwk −1 = c ( w1 + w2 +  + wk −1 ) < cq if and only if w1 + w2 +  + wk −1 < q . This same reasoning applies to any sequential coalition and so the pivotal players are exactly the same. 69. (a) A player is critical in a coalition if that coalition without the player is not on the list of winning coalitions. The critical players in the winning coalitions are underlined: {P1 , P2 , P3 }, {P1 , P2 , P4 },

{P1 , P3 , P4 }, {P1 , P2 , P3 , P4 } . So the Banzhaf power distribution is β1 = 4 /10 , β2 = β3 = β4 = 2 /10 . (b) A player P is pivotal in a sequential coalition if the players that appear before P in that sequential coalition do not, as a group, appear on the list of winning coalitions, but would appear on the list of winning coalitions if P were included. For example, P3 is pivotal in < P1 , P2 , P3 , P4 > since {P1 , P2 } is

not on the list of winning coalitions but {P1 , P2 , P3 } is on that list. The pivotal players in each sequential coalition is underlined: < P1 , P2 , P3 , P4 >, < P1 , P2 , P4 , P3 >, < P1 , P3 , P2 , P4 >, < P1 , P3 , P4 , P2 >,

< P1 , P4 , P2 , P3 >, < P1 , P4 , P3 , P2 >, < P2 , P1 , P3 , P4 >, < P2 , P1 , P4 , P3 >, < P2 , P3 , P1 , P4 >, < P2 , P3 , P4 , P1 >, < P2 , P4 , P1 , P3 >, < P2 , P4 , P3 , P1 >, < P3 , P1 , P2 , P4 >, < P3 , P1 , P4 , P2 >, < P3 , P2 , P1 , P4 >, < P3 , P2 , P4 , P1 >, < P3 , P4 , P1 , P2 >, < P3 , P4 , P2 , P1 >, < P4 , P1 , P2 , P3 >, < P4 , P1 , P3 , P2 >, < P4 , P2 , P1 , P3 >, < P4 , P2 , P3 , P1 >, < P4 , P3 , P1 , P2 >, < P4 , P3 , P2 , P1 > So, the Shapley-Shubik power indices are σ 1 = 12 / 24 = 1/ 2 , σ 2 = σ 3 = σ 4 = 4 / 24 = 1/ 6 . 70. (a) [5: 2,1,1,1,1,1,1] (b) The mayor has a Shapley-Shubik power index of 2/7. (Of the 7! sequential coalitions, the mayor is pivotal anytime he/she is in the 4th slot or in the 5th slot, and there are 6! of each kind.) Each of the 1  2 5 council members has a Shapley-Shubik power index of 1 −  = . 6  7  42

RUNNING 71. Write the weighted voting system as [q: 8x, 4x, 2x, x]. The total number of votes is 15x. If x is even, then so 15 x 15 x is 15x. Since the quota is a simple majority, q = + 1 . But, + 1 = 7.5 x + 1 ≤ 8 x since x ≥ 2 . So, P1 2 2 15 x + 1 (having 8x votes) is a dictator. If x is odd, then so is 15x. Since the quota is a simple majority, q = . 2 15 x + 1 1 = 7.5 x + ≤ 8 x since x ≥ 1 . So, again, P1 (having 8x votes) is a dictator. But, 2 2

ISM: Excursions in Modern Mathematics, 10E

72. Since P2 is a pivotal player in the sequential coalition < P1 , P2 , P3 > , it is clear that {P1 , P2 } and {P1 , P2 , P3 } are winning coalitions, but {P1} is a losing coalition. Similarly, we construct the following chart. Pivotal Player and Sequential Coalition

Winning Coalitions

Losing Coalitions

< P1 , P2 , P3 >

{P1 , P2 } , {P1 , P2 , P3 }

{P1}

< P1 , P3 , P2 >

{P1 , P3 } , {P1 , P2 , P3 }

{P1}

< P2 , P1 , P3 >

{P1 , P2 } , {P1 , P2 , P3 }

{P2 }

< P2 , P3 , P1 >

{P1 , P2 , P3 }

{P2 } , {P3 } , {P2 , P3 }

< P3 , P1 , P2 >

{P1 , P3 } , {P1 , P2 , P3 }

{P3 }

< P3 , P2 , P1 >

{P1 , P2 , P3 }

{P2 } , {P3 } , {P2 , P3 }

Based on this, the winning coalitions with critical players underlined are {P1 , P2 } , {P1 , P3 } , and {P1 , P2 , P3 } . It follows that the Banzhaf power distribution is β1 = 3 / 5; β 2 = 1/ 5; β3 = 1/ 5 . 73. (a) [4: 2, 1, 1, 1] or [9: 5, 2, 2, 2] are among the possible answers. (b) The sequential coalitions (with pivotal players underlined) are: < H , A1 , A2 , A3 >, < H , A1 , A3 , A2 >, < H , A2 , A1 , A3 >, < H , A2 , A3 , A1 >,

< H , A3 , A1 , A2 >, < H , A3 , A2 , A1 >, < A1 , H , A2 , A3 >, < A1 , H , A3 , A2 >, < A1 , A2 , H , A3 >, < A1 , A2 , A3 , H >, < A1 , A3 , H , A2 >, < A1 , A3 , A2 , H >,

< A2 , H , A1 , A3 >, < A2 , H , A3 , A1 >, < A2 , A1 , H , A3 >, < A2 , A1 , A3 , H >, < A2 , A3 , H , A1 >, < A2 , A3 , A1 , H >, < A3 , H , A1 , A2 >, < A3 , H , A2 , A1 >, < A3 , A1 , H , A2 >, < A3 , A1 , A2 , H >, < A3 , A2 , H , A1 >, < A3 , A2 , A1 , H > .

H is pivotal in 12 coalitions; A1 is pivotal in four coalitions; A2 is pivotal in four coalitions; A3 is pivotal in four coalitions. The Shapley-Shubik power distribution is σ H = 12 / 24 = 1/ 2;

σ A = σ A = σ A = 4 / 24 = 1/ 6 . 1

74. There are N! ways that P1 (the senior partner) can be the first player in a sequential coalition (this is the number of sequential coalitions consisting of the other N players). When this happens, the senior partner is not pivotal (the second player listed in the sequential coalition is instead). There are ( N + 1)!− N ! = N !( N + 1 − 1) = N ⋅ N ! ways that P1 is not the first player in a sequential coalition. In each of these cases, P1

is the pivotal player. It follows that the Shapley-Shubik power index of P1 (the senior partner) is

N ⋅ N! = ( N + 1)!

N ⋅ N! N = . Since the other N players divide the remaining power equally, the junior partners each ( N + 1) ⋅ N ! N +1 N N +1 N 1− − 1 N N N + + +1 = 1 1 have a Shapley-Shubik power index of . = N N N ( N + 1) 75. (a) The losing coalitions are {P1}, {P2 }, and {P3 }. The complements of these coalitions are {P2 , P3 }, {P1 , P3 }, and {P1 , P2 } respectively, all of which are winning coalitions.

Chapter 2: The Mathematics of Power (b) The losing coalitions are {P1}, {P2 }, {P3 }, {P4 }, {P2 , P3 }, {P2 , P4 }, and {P3 , P4 }. The complements of these coalitions are {P2 , P3 , P4 }, {P1 , P3 , P4 }, {P1 , P2 , P4 }, {P1 , P2 , P3 }, {P1 , P4 }, {P1 , P3 }, and {P1 , P2 } respectively, all of which are winning coalitions. (c) If P is a dictator, the losing coalitions are all the coalitions without P; the winning coalitions are all the coalitions that include P. The complement of any coalition without P (losing) is a coalition with P (winning). (d) Take the grand coalition out of the picture for a moment. Of the remaining 2 N − 2 coalitions, half are losing coalitions and half are winning coalitions, since each losing coalition pairs up with a winning coalition (its complement). Half of 2 N − 2 is 2 N −1 − 1. In addition, we have the grand coalition (always a winning coalition). Thus, the total number of winning coalitions is 2 N −1.

76. The mayor has power index 5/13 and each of the four other council members has a power index of 2/13. The

{

} { M , P , P } , {M , P , P } , {M , P , P } , {M , P , P } , { M , P , P } , { M , P , P , P } , { M , P , P , P } , { M , P , P , P } , { M , P , P , P } , { P , P , P , P } ,

winning coalitions (with critical players underlined) are: M , P2 , P3 , 3

{M , P2 , P3 , P4 , P5 } .

77. (a) According to Table 2-11, the Banzhaf index of California in the Electoral College is β California = 0.114 .

The relative voting weight of California is wCalifornia = for California is then π California =

55 ≈ 0.1022 . The relative Banzhaf voting power 538

0.114 ≈ 1.115 . 0.1022

1/ 3 115 1/ 3 115 1/ 3 115 0 , πH2 = , π OB = , π NH = = = = =0, 31/115 93 31/115 93 28 /115 84 21/115 0 0 π LB = = 0 , π GC = =0. 2 /115 2 /115

(b) π H 1 =

78. (a) The possible coalitions are all coalitions with A but not P in them or with P but not A in them. If we call B and C the players with 4 and 3 votes respectively, the possible coalitions are: { A} ,{ A, B} , { A, C} ,{ A, B, C} and {P} ,{ P, B} , {P, C} ,{P, B, C} . The winning coalitions (with critical

players underlined) are: { A, B} , { A, C} , { A, B, C} , { P, B, C} . The Banzhaf power distribution in this case is β A = 3 / 8; β B = 2 / 8; βC = 2 / 8; β D = 1/ 8 .

(b) The possible coalitions under these circumstances are all subsets of the players that contain either A or P but not both. There are 2 N − 2 subsets that contain neither A nor P (all possible subsets of the N – 2 other players). If we throw A into each of these subsets, we get all the coalitions that have A but not P in them—a total of 2 N − 2 coalitions. If we throw P into each of the 2 N − 2 subsets we get all the coalitions that have P but not A in them. Adding these two lists gives 2 ⋅ 2 N − 2 = 2 N −1 coalitions. (c) Consider the same weighted voting system discussed in part (a) but without restrictions. The Banzhaf 5 3 3 1 power distribution in this case is β A = ; β B = ; βC = ; β D = . When A becomes the antagonist 12 12 12 12 of P, A’s Banzhaf power index goes down (to 3/8) and P’s Banzhaf power index goes up (to 1/8). If we reverse the roles of A and P (A becomes the “player” and P his “antagonist”, the Banzhaf power index calculations remain the same, and in this case it is the antagonist’s (P) power index that goes up and the player’s (A) power that goes down.

ISM: Excursions in Modern Mathematics, 10E

(d) Once P realizes that A will always vote against him, he can vote exactly the opposite way of his true opinion. This puts A’s votes behind P’s true opinion. If A has more votes than P, this strategy essentially increases P’s power at the expense of A. 79. (a) [4: 3,1,1,1]; P1 is pivotal in every sequential coalition except the six sequential coalitions in which P1 is the first member. So, the Shapley-Shubik power index of P1 is (24 – 6)/24 = 18/24 = 3/4. P2 , P3 , and P4 share power equally and have Shapley-Shubik power index of 1/12. (b) In a weighted voting system with 4 players there are 24 sequential coalitions — each player is the first member in exactly 6 sequential coalitions. The only way the first member of a coalition can be pivotal is if she is a dictator. Consequently, if a player is not a dictator, she can be pivotal in at most 24 – 6 = 18 sequential coalitions and so that player’s Shapley-Shubik power index can be at most 18/24 = 3/4. (c) In a weighted voting system with N players there are N! sequential coalitions — each player is the first member in exactly (N–1)! sequential coalitions. The only way the first member of a coalition can be pivotal is if she is a dictator. Consequently, if a player is not a dictator, she can be pivotal in at most N! – (N–1)! = (N–1)!(N–1) sequential coalitions and so that player’s Shapley-Shubik power index can be at most (N–1)!(N–1)/N! = (N–1)/N. (d) [N: N–1, 1, 1, 1,…, 1]; Here N is the number of players as well as the quota. P1 is pivotal in every sequential coalition except for the (N–1)! sequential coalitions in which P1 is the first member. So the Shapley-Shubik power index of P1 is [N! – (N–1)!]/N! = (N–1)/N. 80. (a) A player with veto power is critical in every winning coalition. Therefore that player must have Banzhaf power index of at least as much as any other player. Since the total Banzhaf power indexes of all N players is 1, they cannot all be less than 1/N. (b) A player with veto power is pivotal in every sequential coalition in which that player is the last player. There are (N – 1)! such sequential coalitions. Consequently, the player must have Shapley-Shubik power index of at least (N – 1)!/N! = 1/N.

Chapter 2: The Mathematics of Power

APPLET BYTES 81. Power in the first Electoral College is summarized in the table below. A quota of q = 67 is used. State

Virginia

Number of Electors 21

Banzhaf Power 17.2034%

ShapleyShubik Power 17.3049%

Massachusetts

12.2842%

12.4892%

Pennsylvania

11.4728%

11.6567%

New York

8.9970%

9.0568%

North Carolina

8.9970%

9.0568%

Connecticut

6.6458%

6.5945%

Maryland

5.9173%

5.8680%

South Carolina

5.9173%

5.8680%

New Jersey

5.1474%

5.0971%

New Hampshire

4.2761%

4.2236%

Georgia

2.9160%

2.8355%

Kentucky

2.9160%

2.8355%

Rhode Island

2.9160%

2.8355%

Vermont

2.1968%

2.1390%

Delaware

2.1968%

2.1390%

82. Slightly increasing the quota from 58 produces more desirable results (in terms of reducing the gap). One quota to be considered is 65 (Note: This is slightly larger than 31 + 31 + 2 = 64). q = 65 produces the following gaps which are within the 4% threshold. Another interesting quota, though not one meeting the 4% threshold for gaps, is q = 67. Player

Weight

Weight as a % 26.96%

Banzhaf Power 26%

Hempstead 1

Gap

0.96%

Hempstead 2

26.96%

26%

0.96%

Oyster Bay

24.35%

22%

2.35%

N. Hempstead

18.26%

22%

3.74%

Long Beach

1.74%

0.26%

Glen Cove

1.74%

0.26%

Quota

ISM: Excursions in Modern Mathematics, 10E

83. (a) The following table describes Banzhaf power vs. the weight of each player (as a percentage) in the weighted voting system [65: 30, 28, 22, 15, 7, 6]. Player

Weight

Weight as a % 27.7778%

Banzhaf Power 28.8462%

Hempstead 1

Gap

1.0684%

Hempstead 2

25.9259%

25.0000%

0.9259%

Oyster Bay

20.3704%

21.1538%

0.7834%

N. Hempstead

13.8889%

17.3077%

3.4188%

Long Beach

6.4815%

5.7692%

0.7123%

Glen Cove

5.5556%

1.9231%

3.6325%

Quota

(b) The largest gap for q = 65 is 3.6325%. The average value of the gaps for q = 65 is 1.7569%. (c) Experimentation with various quotas produce the results given in the table below. The best largest and average gaps occur for any value of the quota between 60 and 63. Quotas below q = 59 or above q = 65 produce even larger gaps.

Weight

Weight as a %

Banzhaf Power (q=65)

Gap

Banzhaf Power (q=64)

Gap

Banzhaf Power (q=60-63)

Gap

Banzhaf Power (q=59)

Gap

27.78%

28.85%

1.07%

30.77%

2.99%

27.78%

0.00%

29.63%

1.85%

25.93%

25.00%

0.93%

26.92%

1.00%

27.78%

1.85%

25.93%

0.00%

20.37%

21.15%

0.78%

19.23%

1.14%

20.37%

0.00%

22.22%

1.85%

13.89%

17.31%

3.42%

15.38%

1.49%

12.96%

0.93%

11.11%

2.78%

6.48%

5.77%

0.71%

3.85%

2.64%

5.56%

0.92%

7.41%

0.93%

5.56%

1.92%

3.64%

3.85%

1.71%

5.56%

0.00%

3.70%

1.86%

MAX

3.64%

2.99%

1.85%

2.78%

AVG

1.76%

1.83%

0.62%

1.54%

Chapter 3 WALKING 3.1

Fair-Division Games

1. (a) When three players are dividing assets, a fair share to any player is one worth at least 1/3 of the total. So 1 any share worth more than 33 % would be considered a fair share. In this case, s2 and s3 (since, in 3 Henry’s eyes, they are worth 40% and 35% respectively) are fair.

1 (b) s2 and s3 (35% and 37% are each valued by Tom as at least 33 % ) 3 (c) s1 , s2 and s3 (d) Since Fred is the only player to consider s1 to be fair, any fair division of the assets must allocate s1 to Fred. Since Henry and Tom each consider both s2 and s3 to be fair shares, there are exactly two possible divisions: (1) Henry gets s2 ; Tom gets s3 ; Fred gets s1 . (2) Henry gets s3 ; Tom gets s2 ; Fred gets s1 . (e) Since Henry values s2 more than s3 and Tom values s3 more than s2 , the best division is (1) Henry gets s2 ; Tom gets s3 ; Fred gets s1 . 2. (a) When three players are dividing assets, a fair share to any player is one worth at least 1/3 of the total. So 1 any share worth more than 33 % would be considered a fair share. In this case, s1 and s3 (since, in 3 Alice’s eyes, they are worth 38% and 34% respectively) are fair. (b) s1 , s2 and s3 (c) s1 and s2 (d) Suppose we allocate share s1 to Alice. Since Carlos only considers s1 and s2 to be fair shares, Carlos must be allocated s2 . This leaves Bob to be allocated s3 . This is the first of our fair divisions. (1) Alice gets s1 ; Bob gets s3 ; Carlos gets s2 . Suppose instead that we allocate s3 to Alice. Since Bob is happy with any share, we have two options for how to allocate a share to Carlos. Carlos could fairly receive either share s1 or s2 (with Bob getting whatever share Carlos does not receive). This gives two other fair divisions. (2) Alice gets s3 ; Bob gets s2 ; Carlos gets s1 . (3) Alice gets s3 ; Bob gets s1 ; Carlos gets s2 . (e) Since Alice values s1 more than any other share and Carlos values s2 more than any other share, the best division is (1) Alice gets s1 ; Bob gets s3 ; Carlos gets s2 . 3. (a) When four players are dividing assets, a fair share to any player is one worth at least 1/4 or 25% of the total. In this case, s2 and s3 (since they are worth 26% and 28% respectively in Angie’s eyes) are fair. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

(b) s1 , s2 and s4 (c) s2 and s3 (d) s1 , s2 , s3 and s4 (e) The following table can help in determining possible fair divisions. A check denotes a fair share. s1

Angie Bev



Ceci Dina



Suppose we allocate share s2 to Angie. Since Ceci only considers s2 and s3 to be fair shares, Ceci must be allocated s3 . Since Dina finds all shares fair, this leaves two choices for the share we may allocate to Bev - s1 or s4 . These are the first two of our fair divisions. (1) Angie gets s2 ; Bev gets s1 ; Ceci gets s3 ; Dina gets s4 . (2) Angie gets s2 ; Bev gets s4 ; Ceci gets s3 ; Dina gets s1 . Suppose instead that we allocate s3 to Angie. Since Ceci only considers s2 and s3 to be fair shares, Ceci must be allocated s2 . Since Dina finds all shares fair, this leaves two choices for the share we may allocate to Bev - s1 or s4 . This gives our other two fair divisions. (3) Angie gets s3 ; Bev gets s1 ; Ceci gets s2 ; Dina gets s4 . (4) Angie gets s3 ; Bev gets s4 ; Ceci gets s2 ; Dina gets s1 . 4. (a) When four players are dividing assets, a fair share to any player is one worth at least 1/4 or 25% of the total. In this case, s2 and s3 (since they are worth 32% and 28% respectively in Mark’s eyes) are fair. (b) s1 , s2 , s3 and s4 (c) s3 and s4 (d) s2 and s4 (e) The following table can help in determining possible fair divisions. A check denotes a fair share. s1

Mark Tim



Maia Kelly



Chapter 3: The Mathematics of Sharing

Suppose we allocate share s2 to Mark. Since Kelly only considers s2 and s4 to be fair shares, Kelly must be allocated s4 . But then since Maia only considers s3 and s4 to be fair shares, Maia must be allocated s3 . This leaves Tim with share s1 . This is our first possible fair division. (1) Mark gets s2 ; Tim gets s1 ; Maia gets s3 ; Kelly gets s4 . Suppose instead that we allocate s3 to Mark. Since Maia considers only s3 and s4 to be fair shares, Maia must be allocated s4 . But then since Kelly only considers s2 and s4 to be fair shares, Kelly must be allocated s2 . This leaves Tim with share s1 . This is our second and last possible fair division. (2) Mark gets s3 ; Tim gets s1 ; Maia gets s4 ; Kelly gets s2 . 5. (a) When four players are dividing assets, a fair share to any player is one worth at least 1/4 of the total values of the assets in their eyes (the total value for each player, in dollars, may differ). In this case, Allen values this cake at $4 + $5 + $6 + $5 = $20. So fair shares to Allen are those worth at least $5.00. These are s2 , s3 , and s4 . (b) Brady sees the cake as being worth a total of $16. So fair shares to Brady are those worth at least $4.00. He sees s3 and s4 as fair. (c) Cody sees the cake as being worth a total of $18. So fair shares to Cody are those worth at least $4.50. He sees s1 and s2 as fair. (d) Diane sees the cake as being worth a total of $20. So fair shares to Diane are those worth at least $5.00. She sees s1 and s4 as fair. (e) The following table can help in determining possible fair divisions. A check denotes a fair share. s1

Allen



Brady Cody



Diane



 

Suppose we allocate share s2 to Allen. Since Cody only considers s1 and s2 to be fair shares, Cody must be allocated s1 . But then since Diane only considers s1 or s4 to be fair, Diane must be allocated s4 . This leaves Brady with s3 . This gives the first of our fair divisions. (1) Allen gets s2 ; Brady gets s3 ; Cody gets s1 ; Diane gets s4 . Suppose now that we allocate s3 to Allen. Since Brady only considers s3 and s4 to be fair shares, Brady must be allocated s4 . But then since Diane only considers s1 or s4 to be fair, Diane must be allocated s1 . This leaves Cody with s2 . This gives the second of our fair divisions. (2) Allen gets s3 ; Brady gets s4 ; Cody gets s2 ; Diane gets s1 . Lastly, suppose that we allocate s4 to Allen. Since Brady only considers s3 and s4 to be fair shares, Brady must be allocated s3 . Also, since Diane only considers s1 or s4 to be fair, Diane must be allocated s1 . This leaves Cody with s2 . This gives the third and last of our fair divisions. (3) Allen gets s4 ; Brady gets s3 ; Cody gets s2 ; Diane gets s1 .

ISM: Excursions in Modern Mathematics, 10E

6. (a) When four players are dividing assets, a fair share to any player is one worth at least 1/4 of the total values of the assets in their eyes (the total value for each player, in dollars, may differ). In this case, Carlos values this cake at $3 + $5 + $5 + $3 = $16. So fair shares to Carlos are those worth at least $4.00. These are s2 and s3 . (b) Sonya sees the cake as being worth a total of $18. So fair shares to Sonya are those worth at least $4.50. She sees s1 , s3 and s4 as fair. (c) Tanner sees the cake as being worth a total of $16. So fair shares to Cody are those worth at least $4.00. He sees s1 and s2 as fair. (d) Wen sees the cake as being worth a total of $20. So fair shares to Wen are those worth at least $5.00. She sees s1 and s4 as fair. (e) The following table can help in determining possible fair divisions. A check denotes a fair share. s1

Carlos Sonya



Tanner



Wen



 



 

Suppose we allocate share s2 to Carlos. Since Tanner only considers s1 and s2 to be fair shares, Tanner must be allocated s1 . But then since Wen only considers s1 or s4 to be fair, Wen must be allocated s4 . This leaves Sonya with s3 . This gives the first of our fair divisions. (1) Carlos gets s2 ; Sonya gets s3 ; Tanner gets s1 ; Wen gets s4 . Suppose now that we allocate s3 to Carlos. This only limits what Sonya may be allocated. If Sonya is allocated s1 , then Tanner must be allocated s2 , and Wen would be allocated s4 . However, if Sonya is allocated s4 , then Wen must be allocated s1 , and Tanner must be allocated s2 . These give the last of our three possible fair divisions. (2) Carlos gets s3 ; Sonya gets s1 ; Tanner gets s2 ; Wen gets s4 . (3) Carlos gets s3 ; Sonya gets s4 ; Tanner gets s2 ; Wen gets s1 . 7. (a) Let x denote the value of slices s2 and s3 to Adams. Then,

( x + $40, 000 ) + x + x + ( x + $60, 000 ) = $400, 000 and so x = $75,000. Since a fair share is worth $100,000, Adams considers slices s1 ($115,000) and s4 ($135,000) to be fair shares. (b) Let x denote the value of slice s3 to Benson. Then, since s4 is worth $8000 more than that and the sum of

these two is worth (0.40)$400, 000 = $160, 000 , it follows that x + ( x + $8, 000 ) = $160, 000 and so x = $76,000. So, s3 is worth $76,000 and s4 is worth $84,000. Now let y denote the value of slice s2 to Benson. The value of s1 is then y + $40, 000 . So,

( y + $40, 000 ) + y + $76, 000 + $84, 000 = $400, 000 and y = $100, 000 . Since a fair share is $100,000, Benson considers slices s1 ($140,000) and s2 ($100,000) to be fair shares.

Chapter 3: The Mathematics of Sharing (c) Let x denote the value of slice s4 to Cagle. Then, s3 is worth 2x and s1 is worth x + $20,000. Since s2 is worth $40,000 less than s1 , the value of s2 is x - $20,000. Hence,

( x + $20, 000 ) + ( x − $20, 000 ) + 2 x + x = $400, 000 and x = $80,000. Since a fair share is $100,000, Cagle considers slices s1 ($100,000) and s3 ($160,000) to be fair shares. (d) Let x denote the value of slices s2 and s3 to Duncan. Then, ( x + $4, 000) + x + x = ( 0.70)( $400, 000) =

$280,000 so that x = $92,000. Since a fair share is $100,000, Duncan only considers slice s4 ($120,000) to be a fair share. (e) Duncan must receive s4 . Then, Adams must receive s1 . After that, Benson must receive s2 leaving Cagle with s3 . 8. (a) Let x denote the value of slices s2 and s3 to Abe. Then, $3.60 + x + x + $3.50 = $15.00 and so x = $3.95. Since a fair share is worth $3.75, Abe considers slices s2 ($3.95) and s3 ($3.95) to be fair shares. (b) Let x denote the value of slice s1 to Betty. Then, x + 2 x + 3x + 4 x = 1 and so x = 1/10. So, s2 is worth 2/10, s3 is worth 3/10 and s4 is worth 4/10. Betty considers slices s3 and s4 to be fair shares. (c) Let x denote the value of slice s4 to Cory. Then, s3 is worth 3x and x + x + 3x + x = 1 and x = 1/6. Cory only considers s3 to be a fair share. (d) Let x denote the value of slice s2 to Dana. Then, ( x + $1.00) + x + ( x + $2.00) + $3.00 = $18.00 so that x

= $4.00. Since a fair share is $4.50, Dana only considers slices s1 ($5.00) and s3 ($6.00) to be a fair shares. (e) Cory must receive s3 . Then Betty must receive s4 , Dana must receive s1 , and Abe must receive s2 . 9. (a) Let C = the value of the chocolate half (in Angelina's eyes) H = the value of the hibiscus half (in Angelina’s eyes) It is known that H = 2C (Angelina likes the hibiscus half twice as much as the chocolate half) and C + H = $72 (the entire cake is worth $72). Substituting, C + 2C = $72 3C = $72 C = $24 So, the chocolate part is valued by Angelina at $24 and the hibiscus part is valued by Angelina at $7260° 1 × $48 = × $48 = $16 by Angelina. $24 = $48. The hibiscus wedge is then valued as ° 3 180 (b) By (a), the chocolate piece shown is worth

45° °

180

× $24 =

12° 180°

1 × $24 = $6 in Angelina’s eyes. 4

× $48 = $3.20 in Angelina’s eyes.

ISM: Excursions in Modern Mathematics, 10E

10. (a) Let H = the value of the hibiscus half (in Brad's eyes) C = the value of the chocolate half (in Brad’s eyes) It is known that C = 3H (Brad likes the chocolate half three times as much as the hibiscus half) and C + H = $72 (the entire cake is worth $72). Substituting, 3H + H = $72 4H = $72 H = $18 So, the hibiscus half is valued by Brad at $18 and the chocolate part is valued by Brad at $72-$18 = $54. 60° 1 × $18 = × $18 = $6 by Brad. The hibiscus wedge is then valued as 3 180° (b) By (a), the chocolate piece shown is worth

45° 180°

× $54 =

12° 180°

1 × $54 = $13.50 in Brad’s eyes. 4

× $18 = $1.20 in Brad’s eyes.

11. (a) Let H = the value of the hibiscus half (in Brad's eyes) C = the value of the chocolate half (in Brad’s eyes) It is known that C = 4H (Brad likes the chocolate half four times as much as the hibiscus half) and C + H = $72 (the entire cake is worth $72). Substituting, 4 H + H = $72 5 H = $72 H = $14.40 So, the hibiscus half is valued by Brad at $14.40 and the chocolate half is valued by Brad at $72-$14.40 60° 1 × $14.40 = × $14.40 = $4.80 by Brad. = $57.60. The hibiscus wedge is then valued as ° 3 180 (b) By (a), the chocolate piece shown is worth

45° °

180

× $57.60 =

12° 180°

1 × $57.60 = $14.40 in Brad’s eyes. 4

× $14.40 = $0.96 in Brad’s eyes.

12. (a) Let C = the value of the chocolate half (in Angelina's eyes) H = the value of the hibiscus half (in Angelina’s eyes) It is known that H = 5C (Angelina likes the hibiscus half five times as much as the chocolate half) and C + H = $72 (the entire cake is worth $72). Substituting, C + 5C = $72 6C = $72 C = $12 So, the chocolate part is valued by Angelina at $12 and the hibiscus part is valued by Angelina at $7260° 1 × $60 = × $60 = $20 by Angelina. $12 = $60. The hibiscus wedge is then valued as ° 3 180 (b) By (a), the chocolate piece shown is worth

45° °

180

× $12 =

1 × $12 = $3 in Angelina’s eyes. 4

Chapter 3: The Mathematics of Sharing

12° 180°

× $60 = $4 in Angelina’s eyes.

13. (a) Let C = the value of the chocolate part (in Karla's eyes) S = the value of the strawberry part (in Karla's eyes) V = the value of the vanilla part (in Karla's eyes) It is known that S = 2V (Karla likes strawberry twice as much as vanilla), C = 3V (Karla likes chocolate three times as much as vanilla), and C + S + V = $30 (the entire cake is worth $30). Substituting, 3V + 2V + V = $30 6V = $30 V = $5 S = 2 × ($5) = $10 C = 3 × ($5) = $15 That is, the vanilla part is worth $5, the strawberry part $10, and the chocolate part $15 in Karla’s eyes. The value of each slice is then found as follows. 60° 30° 30° s1 : × $5 = $2.50 ; s2 : × $5 + × $10 = $1.25 + $2.50 = $3.75 ; 120° 120° 120° 60° 30° 30° s3 : × $10 = $5.00 ; s4 : × $10 + × $15 = $2.50 + $3.75 = $6.25 ; 120° 120° 120° 60° 30° 30° s5 : × $15 = $7.50 ; s6 : × $15 + × $5 = $3.75 + $1.25 = $5.00 120° 120° 120° (b) A fair share would need to be worth at least $30/6 = $5.00. So, slices s3 , s4 , s5 , and s6 are each fair shares. 14. (a) Let C = the value of the chocolate part (in Marla's eyes) S = the value of the strawberry part (in Marla's eyes) V = the value of the vanilla part (in Marla's eyes) It is known that V = 2C (Marla likes vanilla twice as much as chocolate), C = 3S (Marla likes chocolate three times as much as strawberry), and C + S + V = $30 (the entire cake is worth $30). Substituting, S + 3S + 6S = $30 10S = $30 S = $3 C = 3 × ($3) = $9 V = 6 × ($3) = $18 That is, the strawberry part is worth $3, the chocolate part $9, and the vanilla part $18 in Marla’s eyes. The value of each slice is then found as follows. 60° 30° 30° s1 : × $18 = $9.00 ; s2 : × $18 + × $3 = $4.50 + $0.75 = $5.25 ; 120° 120° 120° 60° 30° 30° s3 : × $3 = $1.50 ; s4 : × $3 + × $9 = $0.75 + $2.25 = $3.00 ; 120° 120° 120° 60° 30° 30° s5 : × $9 = $4.50 ; s6 : × $9 + × $18 = $2.25 + $4.50 = $6.75 120° 120° 120° (b) A fair share would need to be worth at least $30/6 = $5.00. So, slices s1 , s2 , and s6 are each fair shares.

ISM: Excursions in Modern Mathematics, 10E

3.2

The Divider–Chooser Method

15. (a) Since Jared likes meatball subs three times as much as vegetarian subs, he values each of the pieces [0,6], [6,8], [8,10], and [10,12] the same. So, with one cut Jared would divide the sandwich into parts s1 : [0,8] and s2 :[8,12]. That is, one piece containing the entire vegetarian half and one-third of the meatball half ( s1 ) and one piece consisting of two-thirds of the meatball half ( s2 ). (b) Karla would clearly choose piece s1 since she only values vegetarian. This piece contains the entire value of the sandwich to Karla ($8.00). 16. (a) Since Karla only likes vegetarian, she would cut that part of the sub in half. That is, s1 : [0,3]; s2 : [3,12] (b) Jared would choose the piece that is larger in size ( s2 ); To Jared, if each inch of vegetarian sub is worth $x, then each inch of meatball sub is worth $3x. So, we have $6x + $6(3x) = $8.00. This means that each inch of vegetarian sub is worth x = $0.33 to Jared and each inch of meatball sub is worth $1.00. So, s2 : [3,12] is worth $1.00 + $6.00 = $7.00 to Jared. 17. (a) Let x denote the value that Martha places on one inch of turkey (in dollars). Then, Martha places that same value on one inch of roast beef. However, she places a value of 2x on each inch of ham. Since the total value of the sandwich is $9, we have 8 ( 2 x ) + 12 ( x ) + 8 ( x ) = $9 . That is, 36x = $9 or x = $0.25. Each inch of ham is worth $0.50 to Martha and each inch of turkey and roast beef is worth $0.25. To have two pieces each worth $4.50, she would cut the sandwich into pieces s1 : [0,10] and s2 :[10,28]. (b) Let y denote the value that Nick places on one inch of turkey (in dollars). Then, Nick places that same value on one inch of ham. However, he places a value of 2y on each inch of roast beef. Since the total value of the sandwich is $9, we have 8 ( y ) + 12 ( y ) + 8 ( 2 y ) = $9 . That is, 36y = $9 or y = $0.25. Each inch of roast beef is worth $0.50 to Nick and each inch of turkey and ham is worth $0.25. Nick would clearly take piece s2 :[10,28] which, in his eyes, has a value of 10 ( $0.25 ) + 8 ( $0.50 ) = $6.50 . 18. (a) Let x denote the value that Nick places on one inch of turkey (in dollars). Then, Nick places that same value on one inch of ham. However, he places a value of 2x on each inch of roast beef. Since the total value of the sandwich is $9, we have 8 ( x ) + 12 ( x ) + 8 ( 2 x ) = $9 . That is, 36x = $9 or x = $0.25. Each inch of roast beef is worth $0.50 to Nick and each inch of turkey and ham is worth $0.25. To have two pieces each worth $4.50, he would cut the sandwich into pieces s1 : [0,18]; s2 : [18,28]. (b) Let y denote the value that Martha places on one inch of turkey (in dollars). Then, Martha places that same value on one inch of roast beef. However, she places a value of 2y on each inch of ham. Since the total value of the sandwich is $9, we have 8 ( 2 y ) + 12 ( y ) + 8 ( y ) = $9 . That is, 36y = $9 or y = $0.25. Each inch of ham is worth $0.50 to Martha and each inch of turkey and roast beef is worth $0.25. Martha would clearly take piece s1 : [0,18] which, in her eyes, has a value of 8 ( $0.50) + 10 ( $0.25) = $6.50 .

19. (a) Yes; David values chocolate and vanilla equally and values each twice as much as strawberry. Paula would choose s2 since she is allergic to chocolate ( s2 contains ¾ of the value of the entire cake in her eyes). (b) No; David values s2 more than he values s1 . In fact, he values it 1.5 times more. That is, he values at 60% and s2 and s1 at 40%. Such a division is not rational since there is a risk that Paula, whom he knows nothing about, would choose that piece.

Chapter 3: The Mathematics of Sharing (c) Yes; Based on David’s value system, the chocolate and vanilla parts are each worth twice as much as the strawberry part. So, if the entire cake is worth $5 (hypothetically), then, the chocolate and vanilla parts would each be worth $2 and the strawberry part would be valued at $1. In that case, s1 is worth 90 60 $2 ) + ( $2 ) = $2.50 , or half the full value of the cake. Paula would choose the piece with the  ( 120 120 least chocolate (piece s2 ).

20. (a) No; Paula values s2 more than s1 . In fact, since Paula is allergic to chocolate, s2 is worth 75% and s1 is worth 25% to Paula. (b) Yes; David would choose s2 which is worth 60% to him. [Think of the cake as worth $5. Then, the chocolate and vanilla parts are each worth $2 to David while the strawberry part is worth $1. From this, 60 s2 is worth ( $2 ) + $2 = $3 which is 60% of the full value of the cake.] 120 (c) No; Paula values s2 more than s1 . In fact, since she is allergic to chocolate, s2 is worth 75% and s1 is worth 25% to Paula.

3.3

The Lone-Divider Method

21. (a) D must be given s1 . But C1 and C2 are happy with either of the other shares. So, the following are two fair divisions of the land. C1 : s2 , C2 : s3 , D : s1 ; C1 : s3 , C2 : s2 , D : s1 (b) If C1 is assigned s2 , then C2 can be assigned either of the other pieces. However, if C1 is assigned s3 , then C2 must be assigned s1 . So, the following are three fair divisions. C1 : s2 , C2 : s1 , D : s3 ; C1 : s2 , C2 : s3 , D : s1 ; C1 : s3 , C2 : s1 , D : s2 22. (a) C1 must be assigned s2 . Then, C2 can be assigned to either of the other shares. C1 : s2 , C2 : s1 , D : s3 ; C1 : s2 , C2 : s3 , D : s1 (b) If C1 is assigned s1 , then C2 can be assigned either of the other pieces. However, if C1 is assigned s2 , then C2 must be assigned s3 . So, the following are three fair divisions. C1 : s1 , C2 : s2 , D : s3 ; C1 : s1 , C2 : s3 , D : s2 ; C1 : s2 , C2 : s3 , D : s1 23. (a) First, C1 must receive s2 . Then, C3 must receive s3 (the only other share C3 considers fair). So, C2 receives s1 and D receives s4 . (b) If C1 receives s2 , then C3 must receive s1 . But then C2 must receives s3 leaving D with s4 . If, on the other hand, C1 receives s3 , then C2 must receive s1 . But then C3 must receives s2 leaving D with s4 . Hence, two fair divisions are as follows.

ISM: Excursions in Modern Mathematics, 10E

C1 : s2 , C2 : s3 , C3 : s1 , D : s4 ; C1 : s3 , C2 : s1 , C3 : s2 , D : s4

(c) C1 must receive s2 . If C2 receives s1 , then C3 must receive s4 leaving D with s3 . On the other hand, if C2 receives s3 , then C3 could receive either s1 or s4 leaving D with the remaining piece. Three fair divisions follow below. C1 : s2 , C2 : s1 , C3 : s4 , D : s3 ; C1 : s2 , C2 : s3 , C3 : s1 , D : s4 ; C1 : s2 , C2 : s3 , C3 : s4 , D : s1 24. (a) First, C1 must receive s2 . Then, C3 must receive s3 . So, C2 receives s1 and D receives s4 . C1 : s2 , C2 : s1 , C3 : s3 , D : s4 (b) C1 : s2 , C2 : s1 , C3 : s4 , D : s3 ; C1 : s2 , C2 : s3 , C3 : s1 , D : s4 ; C1 : s2 , C2 : s3 , C3 : s4 , D : s1 (c) C1 : s2 , C2 : s1 , C3 : s3 , D : s4 ; C1 : s2 , C2 : s1 , C3 : s4 , D : s3 ; C1 : s2 , C2 : s3 , C3 : s4 , D : s1 25. (a) Since Tim valued each piece at 25% (and was the only player to do so), he must have been the divider. (b) Based on the percentages listed, the following would be the bids made. Mark :{s2 , s3 } , Maia :{s3 , s4 } , Kelly :{s4 } . So, a fair division of the cake must have Kelly with slice s4 . But then Maia must get slice s3 . That would require Mark to receive slice s2 . This leaves the divider, Tim, with slice s1 . 26. (a) Since Cody valued each piece at $5.00 (and was the only player to value each piece equally), he must have been the divider. (b) Based on the dollar values listed, the following would be the bids made: Allen :{s2 , s4 } ,

Brady :{s1 , s2 } , Diane :{s1} . Note that players bid on any slice worth $5.00 or more. A fair division of the cake must have Diane with slice s1 . But then Brady must get slice s2 . That would require Allen to receive slice s4 . This leaves the divider, Cody, with slice s3 . 27. (a) Fair; none of the bidders considered s3 as being worth ¼ of the land. So, these three shares must be worth at least ¾ of the land in each of their eyes. If they each wind up with at least 1/3 of the newly combined piece, that would constitute at least 1/3 of 3/4 (i.e. ¼) of the value of the original plot of land. (b) Not fair; {s2 , s3 , s4 } may be worth less than ¾ of the value of the original plot of land to C2 and C3 . Even if they wind up with 1/3 of the newly combined piece, it may not be a fair share. (c) Fair; see (a). (d) Not fair; {s1 , s4 } may be worth less than ½ of the value of the original plot of land to C2 and C3 . That is, s2 may be of a lot of value to C2 and C3 .

Chapter 3: The Mathematics of Sharing

28. (a) Fair; none of the bidders considered s1 as being worth ¼ of the land. So, these three shares must be worth at least ¾ of the land in each of their eyes. If they each wind up with at least 1/3 of the newly combined piece, that would constitute at least 1/3 of 3/4 (i.e. ¼) of the value of the original plot of land. (b) Not fair; {s1 , s2 , s4 } is worth less than 75% to C3 . (c) Fair; same reason as in (a). (d) Not fair; C1 values {s1 , s2 } as less than 50%. 29. (a) C1 : s2 , C2 : s4 , C3 : s5 , C4 : s3 , D : s1 ; C1 : s4 , C2 : s2 , C3 : s5 , C4 : s3 , D : s1 ; If C1 is to receive s2 , then C2 must receive s4 (and conversely). So, C4 must receive s3 . It follows that C3 must receive s5 . This leaves D with s1 . (b) C1 : s2 , C2 : s4 , C3 : s5 , C4 : s3 , D : s1 ; If C1 is to receive s2 , then C2 must receive s4 . So, C4 must receive s3 . It follows that C3 must receive s5 . This leaves D with s1 . 30. (a) C1 : s2 , C2 : s4 , C3 : s1 , C4 : s3 , D: s5 ; C1 : s3 , C2 : s2 , C3 : s1 , C4 : s4 , D: s5 ; C1 : s3 , C2 : s4 , C3 : s2 , C4 : s1 , D: s5 ; These three divisions are completely determined by the share given to C4 . (b) C1 : s1 , C2 : s4 , C3 : s5 , C4 : s2 , D: s3 ; If C4 is to receive s2 , then C2 must receive s4 . So, C1 must receive s1 . It follows that C3 must receive s5 . This leaves D with s3 . 31. (a) The Divider was Gong since that is the only player that could possibly value each piece equally. (b) To determine the bids placed, each player’s bids should add up to $480,000. So, for example, Egan would bid $480,000 - $80,000 - $85,000 - $195,000 = $120,000 on parcel s3 . Further, one player must bid the same on each parcel (the divider). The following table shows the value of the four parcels in the eyes of each partner.

Egan Fine Gong Hart

s1 $80,000 $125,000 $120,000 $95,000

s2 $85,000 $100,000 $120,000 $100,000

s3 $120,000 $135,000 $120,000 $175,000

s4 $195,000 $120,000 $120,000 $110,000

For a player to bid on a parcel, it would need to be worth at least $480,000/4 = $120,000. Based on this table, the choosers would make the following declarations: Egan: {s3 , s4 } ; Fine: { s1 , s3 , s4 } ; Hart: {s3 } . (c) One possible fair division of the land is Egan: s4 ; Fine: s1 , Gong: s2 ; Hart: s3 . In fact, this is the only possible division of the land since Hart must receive s3 , so that Egan in turn must receive s4 . It follows that Fine must receive s1 so that Gong (the divider) winds up with s2 . Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

32. (a) The Divider was Betty since that is the only player that could possibly value each piece equally. (b) To determine the bids placed, each player’s bids should add up to $18.

Abe Betty Cory Dana

s1 $5.00 $4.50 $4.80 $4.00

s2 $5.00 $4.50 $4.20 $3.75

s3 $3.50 $4.50 $4.00 $4.25

s4 $4.50 $4.50 $5.00 $6.00

For a player to bid on a share, it would need to be worth at least $18/4 = $4.50. Based on this table, the choosers would make the following declarations: Abe: {s1 , s2 , s4 } , Cory: {s1 , s4 } , Dana: {s4 } . (c) One possible fair division of the pizza is Abe: s2 , Betty: s3 , Cory: s1 , Dana: s4 . This is the only possible division since Dana must receive s4 , so that Cory in turn must receive s1 . It follows that Abe must receive s2 so that Betty (the divider) winds up with s3 . 33. (a) s1 : [0,4]; s2 : [4, 8]; s3 : [8, 12] (b) Karla values [0,2], [2,4], and [4,6] equally. She places no value on the segment [6,12]. So, Karla would consider any piece that contains [0,2] or [2,4] or [4,6] to be a fair share. It follows that Karla views s1 and s2 as fair shares. (c) Suppose that Lori places a value of $1 (or 1 cent, or one peso, or whatever) on one inch of vegetarian sub. To her then the entire sub is worth $6 + $12 = $18. Also, she values s1 at $4, s2 at $6, and s3 at $8. Since a fair share must be worth $6 to Lori, it follows that s2 and s3 are fair shares. (d) Three possible fair divisions are Jared: s2 , Karla: s1 , Lori: s3 ; Jared: s1 , Karla: s2 , Lori: s3 ; Jared: s3 , Karla: s1 , Lori: s2 34. (a) s1 : [0,6]; s2 : [6, 9]; s3 : [9, 12] (b) Jared values each inch of sandwich equally. So, to be a fair share, a piece would need to be at least 4 inches long. It follows that Jared views only s1 as a fair share. (c) Karla values [0,2], [2,4], and [4,6] equally. She places no value on the segment [6,12]. So, Karla would consider any piece that contains [0,2] or [2,4] or [4,6] to be a fair share. It follows that Karla views only s1 as a fair share. (d) Since neither Jared nor Karla consider s3 to be a fair share, s1 and s2 are recombined into a single piece and divided fairly by Jared and Karla using the divider-chooser method.

Chapter 3: The Mathematics of Sharing

3.4

The Lone-Chooser Method

35. (a) Angela sees s1 as being worth $18. In her second division, she will create three $6 pieces.

Since the strawberry quarter at the lower left is worth $13.50 to Angela, the size of the central angle in one pie slice she will cut will be determined by x ( $13.50 ) = $6 so that x = 40° . 90 In fact, Angela cuts two 40° pieces from the strawberry, and the remaining strawberry plus the vanilla makes up the third piece. The pieces can be described as 40° strawberry, 40° strawberry, and 90° vanilla-10° strawberry. (b) Boris sees s2 as being worth $15. In his second division, he will create three $5 pieces.

Since the strawberry quarter at the lower right is worth $9.00 to Boris, the size of the central angle in one pie slice he will cut will be determined by x ( $9.00 ) = $5.00 so that x = 50° . So Boris cuts one 50° piece from the 90 strawberry part of the cake. Also, since the vanilla quarter at the upper right x is worth $6.00 to Boris, we see that  ( $6.00 ) = $5.00 tells us that 90 x = 75° is the size of the central angle in a second pie slice (vanilla this time) Boris makes. The remaining 15° piece of vanilla plus the 40° piece of strawberry makes up the third piece. The pieces can be described as 75° vanilla, 15° vanilla-40° strawberry, 50° strawberry. (c) Since Carlos values vanilla twice as much as strawberry, Carlos would clearly take the vanilla part of Angela’s division (the 90° vanilla-10° strawberry wedge). He would also want to take the most vanilla that he could from Boris (the 75° vanilla wedge). One possible fair division is Angela: 80° strawberry, Boris: 15° vanilla-90° strawberry, and Carlos: 165° vanilla10° strawberry.

(d) Since she receives two of her $6.00 pieces, the value of Angela’s final share (in Angela’s eyes) is $12.00. Similarly, the value of Boris’ final share (in Boris’ eyes) is $10.00. The 90 10 75 value of Carlos’ final share (in Carlos’ eyes) is  ( $12) +  ( $6) +  ( $12) ≈ $22.67 . 90 90 90 36. (a) Since Carlos likes vanilla twice as much as strawberry, one possible second division by Carlos is 45° vanilla, 45° vanilla, 90° strawberry.

ISM: Excursions in Modern Mathematics, 10E (b) One possible second division by Angela is 90° vanilla-10° strawberry, 40° strawberry, 40° strawberry

8 × $13.50 = $12.00 . 9 1 The value of Boris’ final share (in Boris’ eyes) is $6 + × $9 + $9 = $16.00 . 9 The value of Carlos’ final share (in Carlos’ eyes) is $12.00.

(d) The value of Angela’s final share (in Angela’s eyes) is

37. First, note that the value of s1 (in Angela’s eyes) is

120° × $27 = $18. She values s2 at $18 too. 180°

(a) Boris chooses s2 because he views it as being worth $12 +

60° × $18 = $12 + $6 = $18 ( s1 is only worth $12.00 to him). 180°

Boris views s2 as being worth $18 and divides it into three pieces each worth $6.00. Thus, one possible second division of s2 by Boris is 90° vanilla, 90° vanilla, 60° strawberry. (b) Angela divides s1 evenly (40° strawberry, 40° strawberry, 40° strawberry).

(c) Carlos will select any one of Angela’s pieces since they are identical in value to him. He will also select one of Boris’s vanilla wedges since he values vanilla twice as much as he values strawberry.

One possible fair division is Angela: 80° strawberry, Boris: 90° vanilla-60° strawberry, and Carlos: 90° vanilla-40° strawberry.

Chapter 3: The Mathematics of Sharing 80° × $27 = $12.00 . 180° 90° 60° × $12 + × $18 = $12.00 . Boris thinks his share is worth 180° 180° 40° 90° × $12 + × $24 = $14.67 . Carlos thinks his share is worth 180° 180°

(d) Angela thinks her share is worth

38. (a) Angela’s choice in the first division is s2 . One possible second division by Angela is 65° strawberry, 65° strawberry, 50° strawberry-45° vanilla.

(b) One possible second division by Carlos is 45° vanilla, 45° vanilla, 45° vanilla.

(c) Boris will select any one of Carlos’s pieces since they are identical in value to him. He will also select Angela’s largest wedge since he calculates it to have the most value to him (see (d)).

One possible fair division is Angela: 130° strawberry, Boris: 90° vanilla-50° strawberry, Carlos: 90° vanilla.

130° × $27.00 = $19.50. 180° 45° 45° 50° × $12.00 + × $12.00 + × $18.00 = $11.00. Boris thinks his share is worth 180° 180° 180° 90° Carlos thinks his share is worth × $24.00 = $12.00. 180°

(d) Angela thinks her share is worth

39. (a) After Arthur makes the first cut, Brian chooses s1 . To him, it is worth 100% of the cake. One possible second division by Brian is to divide the piece evenly (60° chocolate, 30° chocolate-30° strawberry, 60° strawberry).

ISM: Excursions in Modern Mathematics, 10E

(b) Arthur places all of the value on the orange half of s2 , so he divides the orange evenly. The second division by Arthur can be described as 30° orange, 30° orange, 30° orange-90° vanilla.

(c) Since Carl likes chocolate and vanilla, one possible fair division is Arthur: 60° orange, Brian: 90° strawberry-30° chocolate, and Carl: 90° vanilla-30° orange-60° chocolate.

1 2 %. Brian thinks his share is worth 66 % (remember, he is getting 3 3 2/3 of s1 which he values at 100%). Carl thinks his share is worth

(d) Arthur thinks his share is worth 33

60° 90° 30° 1 × 50% + × 50% + × 0% = 83 % . 90° 90° 90° 3

40. (a) Arthurs’s choice of the first division is s2 . One possible second division by Arthur is 45° chocolate, 45° orange, 45° orange-45° vanilla.

(b) One possible second division by Carl is 30° chocolate, 15° chocolate90° strawberry-15° vanilla, 30° vanilla.

Chapter 3: The Mathematics of Sharing (c) One possible fair division is Arthur: 90° orange-45° vanilla; Brian: 60° chocolate-90° strawberry-15° vanilla; Carl: 30° chocolate-30° vanilla.

45° 45° × 100% + × 100% = 50% of the cake. 180° 180° 45° 15° 90° 1 × 100% + × 100% + × 100% = 83 % of the cake. Brian thinks his share is worth 180° 180° 180° 3 30° 30° 1 × 100% + × 100% = 33 % of the cake. Carl thinks his share is worth 180° 180° 3

(d) Arthur thinks his share is worth

41. (a) Ignoring the meatball part (having no value to her), Karla would cut the sandwich right down the middle of the vegetarian part. That is, s1 : [0,3], s2 :[3,12]. (b) Since Jared likes meatball and vegetarian the same, he will pick the larger half (by size), namely s2 . He will then divide his choice ( s2 :[3,12]) equally as J1 :[3,6], J 2 :[6,9], J 3 :[9,12]. (c) Karla’s subdivision is just as simple as Jared’s – she likes all parts of s1 equally. So her division can be described as K1 :[0,1], K 2 :[1,2], K3 :[2,3]. (d) Since Lori likes meatball twice as much as vegetarian, she will select any of Karla’s identical pieces (say, K 3 :[2,3] is selected). She will also select either J 2 or J 3 from Jared (say, J 3 :[9,12]). So, one fair division would be described by Jared: [3,9], Karla: [0,2], Lori: [2,3] and [9,12]. In this case, Jared ends up with half of the sandwich (in size) worth 50% of the value of the whole sandwich to him. Karla ends 1 up with 1/3 of the vegetarian half of the sandwich worth 33 % of the value of the whole sandwich to 3 her. Lori values the vegetarian half at 1/3 and the meatball part at 2/3 so that her piece is valued at 1 1 1 2 7 8 × + × = = 38 % . 6 3 2 3 18 9 42. (a) If Lori values each inch of vegetarian at $1, then she values each inch of meatball at $2. In this case, Lori would value the entire sandwich at $6 + $12 = $18. Lori would cut the sandwich at the $9 mark. That is, s1 : [0,7.5], s2 :[7.5,12]. (b) Since she only likes vegetarian, Karla picks s1 . Her second subdivision would divide the vegetarian part of s1 into equal sized parts: K1 :[0,2], K 2 :[2,4], K3 :[4,7.5]. (c) Since Lori values the meatball parts equally, her second subdivision would reflect that: L1 :[7.5,9], L2 :[9,10.5], L3 :[10.5,12].

ISM: Excursions in Modern Mathematics, 10E

(d) Since Jared likes meatball and vegetarian the same, he will select any of the largest of Karla’s pieces [4,7.5]. He will also select the largest piece (in size) from Lori, say [10.5,12] since they are all the same. So, one fair division would be described by Jared: [4,7.5] and [10.5,12], Karla: [0,4], Lori: [7.5,10.5]. In 2 this case, Jared ends up with 5/12 of the sandwich (in size and value to him) worth 5/12 = 41 % of the 3 value of the whole sandwich to him. Karla ends up with 2/3 of the vegetarian half of the sandwich worth 2 66 % of the value of the whole sandwich to her. Lori values the vegetarian half at 1/3 and the 3 1 2 1 1 meatball part at 2/3 so that her piece is valued at × = = 33 % . 2 3 3 3

3.5

The Method of Sealed Bids

43. (a) Ana and Belle’s fair share is $300. Chloe’s fair share is $400. Item

Ana

Belle

Chloe

Dresser

150

300

275

Desk

180

150

165

Vanity

170

200

260

Tapestry

400

250

500

Total Bids

900

1200

Fair share

300

400

(b) In the first settlement, Ana receives the desk and $120 in cash; Belle receives the dresser; Chloe gets the vanity and the tapestry and pays $360. Ana

Belle

Chloe

Total Bids

900

1200

Fair share

300

400

Value of items received

180

300

760

Prelim cash settlement

120

–360

(c) $240; $360 was paid in and $120 was paid out in the first settlement. (d) Ana, Belle, and Chloe split the $240 surplus cash three ways. This adds $80 cash to the first settlement. In the final settlement, Ana gets the desk and receives $200 in cash; Belle gets the dresser and receives $80; Chloe gets the vanity and the tapestry and pays $280. Item

Ana

Belle

Chloe

Prelim cash settlement

120

–360

Share of surplus

Final cash settlement

200

–280

Chapter 3: The Mathematics of Sharing

44. (a) In the first settlement, Andre gets the farm and pays $220,000; Bea gets the painting and $151,000; Chad gets the house and $45,000. Item

Andre

Bea

Chad

House

150000

146000

175000

Farm

430000

425000

428000

Painting

50000

59000

57000

Total Bids

630000

660000

Fair Share

210000

220000 Total

Value of items received

430000

59000

175000 Surplus

Prelim. cash settlement

–220000

151000

45000 24000

(b) In the final settlement, Andre, Bea, and Chad split the $24,000 surplus evenly. Andre gets the farm and pays $212,000; Bea gets the painting and $159,000; Chad gets the house and $53,000. 45. (a) Each player’s fair share is the sum of their bids divided by 5. Item

Item 1

$352

$295

$395

$368

$324

Item 2

$98

$102

$98

$95

$105

Item 3

$460

$449

$510

$501

$476

Item 4

$852

$825

$832

$817

$843

Item 5

$513

$501

$505

$491

Item 6

$725

$738

$750

$744

$761

Total Bids

$3000

$2910

$3090

$3030

$3000

Fair Share

$600

$582

$618

$606

$600

(b) After the first settlement, A ends up with items 4 and 5 and pays $765, B gets $582, C receives items 1 and 3 and pays $287, D gets $606, and E receives items 2 and 6 and pays $266. A

Total Bids

3000

2910

3090

3030

3000

Fair Share

600

582

618

606

600

Value of items rec’d

1365

905

866

Prelim. cash

–765

582

–287

606

–266

(c) There is a surplus of $765 - $582 + $287 - $606 + $266 = $130. (d) A, B, C, D, and E split the $130 surplus 5 ways ($26 each). In the final settlement, A ends up with items 4 and 5 and pays $739; B ends up with $608; C ends up with items 1 and 3 and pays $261; D ends up with $632; E ends up with items 2 and 6 and pays $240.

ISM: Excursions in Modern Mathematics, 10E 46. (a) Each player’s fair share is the sum of their bids divided by 3. Oscar: $6,400; Bert: $5,600; Ernie: $4,800. Item

Oscar

Bert

Ernie

Item 1

$8,600

$5,500

$3,700

Item 2

$3,500

$4,200

$5,000

Item 3

$2,300

$4,400

$3,400

Item 4

$4,800

$2,700

$2,300

Total Bids

$19,200

$16,800

$14,400

Fair Share

$6,400

$5,600

$4,800

Total

Value of items rec’d

$13,400

$4,400

$5,000

Surplus

Prelim. cash

-$7000

$1,200

-$200

$6,000

(b) Oscar receives items 1 and 4 and pays $7000, Bert receives item 3 and $1200, and Ernie ends up with item 2 and pays the estate $200. (c) There is a surplus of $6000 after the first settlement. (d) Oscar, Bert, and Ernie split the $6000 surplus 3 ways ($2000 each). Oscar ends up with items 1 and 4 and only pays the estate $5000; Bert ends up with item 3 and $3200; Ernie ends up with item 2 and $1800. [Essentially, Oscar pays Bert $3200 and Ernie $1800.] 47. Anne gets $75,000 and Chia gets $80,000. Item

Anne

Bette

Chia

Partnership

$210,000

$240,000

$225,000

Total Bids

$210,000

$240,000

$225,000

Fair Share

$70,000

$80,000

$75,000 Total

$240,000

0 Surplus

Prelim. cash settlement

$70,000 –$160,000

$75,000 $15,000

Share of surplus

$5,000

Final cash settlement

$75,000 –$155,000

$80,000

Value of items received

Chapter 3: The Mathematics of Sharing

48. Al gets $56,000 and Ben gets $54,000. Item

Ben

Cal

Partnership

$156,000

$150,000

$171,000

Total Bids

$156,000

$150,000

$171,000

Fair Share

$52,000

$50,000

$57,000 Total

Value of items received

$171,000 Surplus

Prelim. cash settlement

$52,000

$50,000

-$114,000 $12,000

Share of surplus

$4,000

Final cash settlement

$56,000

$54,000

-$110,000

49. In this case, the low bid is awarded for each chore. The total bids here represent the amount of money they are willing to be paid to do all four chores. Ali does chore 4 and pays Caren $15, Briana does chore 2 and receives no cash, and Caren does chores 1 and 3 and receives $15 from Ali. Item

Ali

Briana

Caren

Chore 1

$65

$70

$55

Chore 2

$100

$85

$95

Chore 3

$60

$50

$45

Chore 4

$75

$80

$90

Total Bids

$300

$285

Fair Share

$100

$95

$95 Total

Value of chores done

$75

$85

$100 Surplus

Prelim. cash settlement

-$25

-$10

$5 $30

Share of surplus

$10

Final cash settlement

$-15

$15

50. Anne does chore 1 and pays Cindy $12; Bess does chore 2 and pays Cindy $12; Cindy does chores 3, 4, and 5 and receives $24. Item

Anne

Bess

Cindy

Chore 1

$20

$30

$40

Chore 2

$50

$10

$22

Chore 3

$30

$20

$15

Chore 4

$30

$20

$10

Chore 5

$20

$40

$15

ISM: Excursions in Modern Mathematics, 10E

3.6

Item

Anne

Bess

Cindy

Total Bids

$150

$120

$102

Fair Share

$50

$40

$34 Total

Value of chores done

$20

$10

$40 Surplus

Prelim. cash settlement

-$30

$6 $54

Share of surplus

$18

Final cash settlement

-$12

$24

The Method of Markers

51. (a) Using the first first marker, B gets items 1, 2, 3; Then, using the first second marker, C gets items 5, 6, 7; Finally, A gets items 10, 11, 12, 13. (b) B gets items 1, 2, 3. (c) C gets items 5, 6, 7. (d) Items 4, 8, and 9 are left over. 52. (a) A gets items 4, 5, 6, 7, 8 (b) B gets items 10, 11, 12, 13 (c) C gets items 1, 2 (d) Items 3 and 9 are left over. 53. (a) Using the first first marker, A gets items 1, 2; Then, using the first second marker, C gets items 4, 5, 6, 7; Finally, B gets items 10, 11, 12. (b) B gets items 10, 11, 12 (c) C gets items 4, 5, 6, 7 (d) Items 3, 8, and 9 are left over. 54. (a) A gets items 3, 4, 5, 6, 7 (b) B gets items 10, 11, and 12 (c) C gets item 1 (d) Items 2, 8 and 9 are left over. 55. (a) First C gets items 1, 2, 3; Then, E gets items 5, 6, 7, 8; Then, D gets items 11, 12, 13; Then, B gets items 15, 16, 17; Finally, A gets items 19, 20. (b) Items 4, 9, 10, 14, and 18 are left over. 56. (a) A gets item 7; B gets item 15; C gets items 1, 2, and 3; D gets item 11. (b) Items 4, 5, 6, 8, 9, 10, 12, 13, and 14 are left over.

Chapter 3: The Mathematics of Sharing

57. (a) Quintin thinks the total value is 3 × $12 + 6 × $7 + 6 × $4 + 3 × $6 = $120 , so to him a fair share is worth $30. Ramon thinks the total value is 3 × $9 + 6 × $5 + 6 × $5 + 3 × $11 = $120 , so to him a fair share is worth $30. Stephone thinks the total value is 3 × $8 + 6 × $7 + 6 × $6 + 3 × $14 = $144 , so to him a fair share is worth $36. Tim thinks the total value is 3 × $5 + 6 × $4 + 6 × $4 + 3 × $7 = $84 , so to him a fair share is worth $21. They would place their markers as shown below.

(b)

58. (a) Each player places markers as shown below.

(b) Two Grateful Dead and two Beach Boys CDs are leftover.

(c) Queenie would pick a Beach Boys CD, Sophie would pick a Grateful Dead CD, and Roxy would pick either of the remaining Beach Boys or Grateful Dead CDs. 59. (a) Ana places her markers between the different types of candy. Belle places her markers between each of the Minto bars. Chloe places her markers thinking that each Choko and Minto bar is worth $1 and each Frooto is worth $2. Since Chloe would then value the candy to be worth $12 in total, she would place markers after each $4 worth of candy.

(b) Ana gets three Choko bars, Belle gets one Minto bar, and Chloe gets two Frooto bars. Two Minto bars and one Frooto bar are left over. (c) Belle would select one of the Minto bars, Chloe would then select the Frooto bar, and finally Ana would be left with the last Minto bar. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

60. (a) C3 can be placed anywhere between the last Choko bar and the first Minto bar.

(b) Arne gets three Minto bars, Bruno gets three Frooto bars, Chloe gets one Choko bar, and Daphne gets two Rollo bars. Two Choko and four Rollo bars are left over. (c) Arne is happy with a Rollo bar, Bruno is happy with either a Choko or a Rollo bar, Chloe is happy with a Choko bar, and Daphne is happy with a Rollo bar.

JOGGING 61. (a) C gets to choose one of the three pieces. One of the three must be worth more than 1/3 of S. (b) If C chooses either s21 or s22 , D1 gets to choose s1 (a fair share to him). If C chooses s1 , D1 gets to choose between s21 and s22 . Together they are worth 2/3 of S, so one of the two must be worth at least 1/3 of S. (c) Suppose that in D2 ’s opinion, the first cut by D1 splits S into a 40%-60% split. In this case, the value of the three pieces to D2 is s1 = 40% , s21 = 30% , and s22 = 30% . If C or D1 choose s1 , D2 will not get a fair share. 62. D1 ’s share is either s1 (worth exactly 1/3) or half of a “remainder” worth at least 2/3; D2 ’s share is either s11 (worth exactly 1/3) or half of a ‘remainder” worth at least 2/3; C’s share is either s1 or s11 (both worth at least 1/3), or half of a remainder worth at least 2/3. 63. In the first settlement A receives $x/2; B receives the partnership and pays $y/2. The surplus is y x y−x − = dollars. B must pay A’s original fair share plus half of the surplus, or (in dollars) 2 2 2 x 1  y − x x  y − x x y x x y x + y . +   = + = + − = + = 2 2 2  2  4  2 4 4 4 4 4 64. In the first settlement A receives $x/3; B receives $y/3; C receives the business and pays $2z/3. The surplus is 2z x y 2z − x − y x 2z − x − y y 2z − x − y − − = dollars. A will receive + dollars. B will receive + dollars. 3 3 3 3 3 9 3 9 x y 2z − x − y C receives the business and pays + + 2 × dollars. 3 3 9 65. B receives the strawberry part. C will receive 5/6 of the chocolate part. A receives the vanilla part and 1/6 of the chocolate part.

Chapter 3: The Mathematics of Sharing

66. (a) We solve

3 ⋅ 60 180 + y  180 = 1 for y to find y = 180°. 2

(b) We solve

3 ⋅ 72 180 + y  180 = 1 for y to find y = 144° 2

3 ⋅108 180 + y  180 = 1 for y to find y = 36° 2

(d) y = 0° 67. (a) Two, three, or four choosers bidding on the same item; three or four choosers bidding on the same two items; four choosers bidding on the same three items. (b) 1 + 2 + 3 + 4 = 10 (c) 1 + 2 + 3 +  + ( N − 2) =

( N − 1)( N − 2) 2

68. (a) A: { s1 , s2 } ; B: { s3 , s4 } ; C: { s5 , s6 } (b) The only envy-free fair division is A: { s1 , s3 } ; B: { s2 , s4 } ; C: { s5 , s6 } . Clearly, A prefers her allocation to that of B and that of C. Likewise, B and C prefer their allocations to those of the other two players. (c) Use the division from part (a). Player C will envy A's allocation since she prefers these two items to her present allocation. (d) Use the division from part (b). Note that this division is not efficient since the division in (a) gives each player a better share.

RUNNING 69. (a) Based on the M winning bids, the amount paid into the “kitty” is w1 + w2 + w3 +  + wM . Player j’s fair share (based on their bids) is c j / N . So, the total of the fair shares (the amount to be paid out of the  c + c + c +  + cN  “kitty”) is c1 / N + c2 / N + c3 / N +  + cN / N =  1 2 3  . It follows that the surplus S is  N  c + c + c +  + cN  given by S = ( w1 + w2 + w3 +  + wM ) −  1 2 3  .  N

(b) The total value of all bids is c1 + c2 + c3 +  + cN . But this is the same as r1 + r2 + r3 +  + rM (here we are just summing the rows rather than the columns in table T). So,  c + c + c +  + cN  S = ( w1 + w2 + w3 +  + wM ) −  1 2 3   N  r + r + r +  + rM  = ( w1 + w2 + w3 +  + wM ) −  1 2 3   N r   r  r    =  w1 − 1  +  w2 − 2  +  +  wM − M       N N N

ISM: Excursions in Modern Mathematics, 10E

(c) Since the winning bid on item k is always greater than or equal to the average bid on that item, r wk − k ≥ 0 . Since this is true for every item, S ≥ 0 . N (d) The surplus S = 0 exactly when the winning bid on each item k is the same as the average bid on that item. That is, when all players bid the exact same amount on each item. 70. If c1 , c2 , c3 , , cN denote the sum of each player’s bids, then define the fair shares to be

x1 x ⋅ c1 , 2 ⋅ c2 , …, 100 100

xN ⋅ cN respectively (rather than c1 / N , c2 / N ,…, cN / N ). 100

71. After the division, suppose that the divisors end up with shares s1 , s2 , , s N −1 respectively. The sum of the values of these shares in the chooser C’s eyes will be 1. Call these values x1 , x2 , , xN −1 so that x1 + x2 +  + xN −1 = 1 . After the subdivision by each divider into N subshares, the chooser will select shares 1 1 1 ⋅ x1 , ⋅ x2 , , ⋅ xN −1 . So, the total value of her selection will be worth at least N N N 1 1 1 1 1 ⋅ x1 + ⋅ x2 +  + ⋅ xN −1 = ( x1 + x2 +  + xN −1 ) = . The dividers, on the other hand, will end up N N N N N N − 1 ( ) of at least 1 of the value of the entire cake. That is, with a collection of subshares worth exactly N −1 N worth at least 1/N of the value of the cake.

worth at least

Chapter 4 WALKING 4.1. Apportionment Problems and Apportionment Methods 1. (a) To determine the standard divisor, we find the total population of the states and divide by the number of available seats. 3,310, 000 + 2, 670, 000 + 1,330, 000 + 690, 000 = 50, 000 Standard divisor = 160 (b) To determine each state’s standard quota, we divide that state’s population by the standard divisor found in part (a). 3,310, 000 2, 670, 000 = 66.2 ; Barinas: = 53.4 ; Apure: 50, 000 50, 000 1,330, 000 690, 000 Carabobo: = 26.6 ; Dolores: = 13.8 50, 000 50, 000 2. (a) Standard divisor =

(b) A:

4,360,000 + 2,280,000 + 729,000 + 2,631,000 = 50, 000 200

4,360, 000 2, 280, 000 729, 000 2, 631, 000 = 87.2 ; B: = 45.6 ; C: = 14.58 ; D: = 52.62 50, 000 50, 000 50, 000 50, 000

3. (a) Standard divisor =

45,300 + 31, 070 + 20, 490 + 14,160 + 10, 260 + 8, 720 = 1040 125

(b) The “states” in any apportionment problem are the entities that will have “seats” assigned to them according to a share rule. In this case, the states are the six bus routes and the seats are the 125 buses. The standard divisor of 1,040 represents the average number of passengers per bus per day.

45,300 31, 070 20, 490 = 43.558 ; B: = 29.875 ; C: = 19.702 ; 1040 1040 1040 14,160 10, 260 8, 720 D: = 13.615 ; E: = 9.865 ; F: = 8.385 1040 1040 1040

4. (a) Standard divisor =

21 + 19 + 35 + 30 + 25 = 0.52 250

(b) The “states” in any apportionment problem are the entities that will have “seats” assigned to them according to a share rule. In this case, the states are the five units at the Hospital (ECU, ICU, MU, PU, and SU) and the seats are the 250 nurses. The standard divisor of 0.52 represents the average number of beds per nurse.

21 19 35 30 = 40.385 ; ICU: = 36.538 ; MU: = 67.308 ; PU: = 57.692 ; 0.52 0.52 0.52 0.52 25 SU: = 48.077 0.52

ISM: Excursions in Modern Mathematics, 10E

(b) The standard divisor is the total population divided by the number of available seats (found in part (a)). 27, 400, 000 Standard divisor = = 200, 000 137 (c) Population = standard quota × standard divisor, so A: 41.2 × 200,000 = 8,240,000; B: 31.9 × 200,000 = 6,380,000; C: 24.8 × 200,000 = 4,960,000; D: 22.6 × 200,000 = 4,520,000; E: 16.5 × 200,000 = 3,300,000 6. (a) Number of faculty positions = 32.92 + 15.24 + 41.62 + 21.32 + 138.90 = 250 (b) 50. [Note: The standard divisor (SD = 12,500/250 = 50) represents the average number of students per faculty position.] (c) Agriculture: 32.92 × 50 = 1646 ; Business: 15.24 × 50 = 762 ; Education: 41.62 × 50 = 2081 ; Humanities: 21.32 × 50 = 1066 ; Science: 138.90 × 50 = 6945 7. With 8.14% of the U.S. population, Texas should receive 8.14% of the number of seats available in the House of Representatives. So, the standard quota for Texas is 0.0814 × 435 = 35.409 ≈ 35.41 . 8. With 12.07% of the U.S. population, California should receive 12.07% of the number of seats available in the House of Representatives. So, the standard quota for California is 0.1207 × 435 = 52.5045 ≈ 52.50 . 9. (a) Standard divisor =

100% = 0.5% 200

11.37% 8.07% 38.62% = 22.74 ; Beta Theta: = 16.14 ; Chi Omega: = 77.24 ; 0.5% 0.5% 0.5% 14.98% 10.42% 16.54% = 29.96 ; Epsilon Tau: = 20.84 ; Phi Sigma: = 33.08 Delta Gamma: 0.5% 0.5% 0.5%

(b) Alpha Kappa:

10. (a) Standard divisor =

(b) Aleta:

6.24 + 26.16 + 28.48 + 39.12 100 = = 0.8% 125 125

6.24% 26.16% 28.48% 39.12% = 7.8 ; Bonita: = 32.7 ; Corona: = 35.6 ; Doritos: = 48.9 .8% .8% .8% .8%

4.2. Hamilton’s Method 11. The standard quotas were calculated in Exercise 1(b) to be A: 66.2; B: 53.4; C: 26.6; D: 13.8. Lower quotas (the standard quotas rounded down) are thus A: 66; B: 53; C: 26; D: 13, and their sum is 158. So there are 160 - 158 = 2 seats remaining to allocate. These are given to D and C, since they have the largest fractional parts of the standard quota. The final Hamilton apportionment is A: 66; B: 53; C: 27; D: 14. 12. The standard quotas were calculated in Exercise 2(b) to be A: 87.2; B: 45.6; C: 14.58; D: 52.62. Lower quotas (the standard quotas rounded down) are thus A: 87; B: 45; C: 14; D: 52, and their sum is 198. So there are 200 - 198 = 2 seats remaining to allocate. These are given to D and B, since they have the largest fractional parts of the standard quota. The final Hamilton apportionment is A: 87; B: 46; C: 14; D: 53. 13. The standard quotas were calculated in Exercise 3(c) to be A: 43.558; B: 29.875; C: 19.702; D: 13.615; E: 9.865; F: 8.385. Lower quotas are thus A: 43; B: 29; C: 19; D: 13; E: 9; F: 8, and the sum is 121. So we have 125–121 = 4 buses remaining to allocate. These are given, in order, to B, E, C and D, since they have the largest fractional parts of the standard quota. The final Hamilton apportionment is A: 43; B: 30; C: 20; D: 14; E: 10; F: 8.

Chapter 4: The Mathematics of Apportionment

14. The standard quotas were calculated in Exercise 4(c) to be ECU: 40.385; ICU: 36.538; MU: 67.308; PU: 57.692; SU: 48.077. Lower quotas (the standard quotas rounded down) are thus ECU: 40; ICU: 36; MU: 67; PU: 57; SU: 48, and their sum is 248. So there are 250 - 248 = 2 nurses remaining to allocate. These are given (in order) to PU and ICU, since they have the largest fractional parts of the standard quota. The final Hamilton apportionment is ECU: 40; ICU: 37; MU: 67; PU: 58; SU: 48. 15. The standard quotas were given in Exercise 5 to be A: 41.2; B: 31.9; C: 24.8; D: 22.6; E: 16.5. Lower quotas (the standard quotas rounded down) are thus A: 41; B: 31; C: 24; D: 22; E: 16, and their sum is 134. So there are 137 - 134 = 3 seats remaining to allocate. These are given to B, C and D, since they have the largest fractional parts of the standard quota. The final Hamilton apportionment is A: 41; B: 32; C: 25; D: 23; E: 16. 16. The standard quotas were given in Exercise 6 to be A: 32.92; B: 15.24; E: 41.62; H: 21.32; S: 138.90. Lower quotas are thus A: 32; B: 15; E: 41; H: 21; S: 138, and their sum is 247. So there are 250 - 247 = 3 seats remaining to allocate. These are given to A, S and E, since they have the largest fractional parts of the standard quota. The final Hamilton apportionment is Agriculture: 33; Business: 15; Education: 42; Humanities: 21; Science: 139. 17. The standard quotas were calculated in Exercise 9(b) to be A: 22.74; B: 16.14; C: 77.24; D: 29.96; E: 20.84; F: 33.08. Lower quotas are thus A: 22; B: 16; C: 77; D: 29; E: 20, F: 33, and the sum is 197. So we have 200–197 = 3 seats remaining to allocate. These are given to D, E and A, since they have the largest fractional parts of the standard quota. The final Hamilton apportionment is A: 23; B: 16; C: 77; D: 30; E: 21; F: 33. 18. The standard quotas were calculated in Exercise 10(b) to be A: 7.8; B: 32.7; C: 35.6; D: 48.9. Lower quotas are thus A: 7; B: 32; C: 35; and D: 48, and the sum is 122. So we have 125–122 = 3 seats remaining to allocate. These are given to D, A, and B, since they have the largest fractional parts of the standard quota. The final Hamilton apportionment is A: 8; B: 33; C: 35; D: 49. 41 + 106 + 253 = 16.667 . The standard quotas are then Dunes: 2.46; Smithville: 24 6.36; Johnstown: 15.18. Lower quotas are thus Dunes: 2; Smithville: 6; Johnstown: 15, and the sum is 23. So we have 24–23 = 1 social worker remaining to allocate. This is given to Dunes, since they have the largest fractional part of the standard quota. The final Hamilton apportionment is Dunes: 3; Smithville: 6; Johnstown: 15.

19. (a) The standard divisor is

41 + 106 + 253 = 16 . The standard quotas are now Dunes: 2.56; Smithville: 25 6.63; Johnstown: 15.81. Lower quotas are now Dunes: 2; Smithville: 6; Johnstown: 15, and the sum is 23. So we have 25–23 = 2 social workers remaining to allocate. These are given to Johnstown and Smithville, since they have the largest fractional part of the standard quota. The final Hamilton apportionment is Dunes: 2; Smithville: 7; Johnstown: 16.

(b) The standard divisor is now

(c) Dunes’ apportionment went from 3 social workers to 2 even as the total number of social workers to be assigned increased from 24 to 25. This is an example of the Alabama paradox (with Dunes playing the role of Alabama). 154 + 66 + 30 = 12.5 . The standard quotas are A: 12.32; B: 5.28; C: 2.4. Lower 20 quotas are A: 12; B: 5; C: 2, and the sum is 19. So we have 20–19 = 1 nurse remaining to allocate. This is given to wing C, since it has the largest fractional part of the standard quota. The final Hamilton apportionment is A: 12; B: 5; C: 3.

20. (a) The standard divisor is

154 + 66 + 30 = 11.9 . The standard quotas are A: 12.94; B: 5.54; C: 2.52. Lower 21 quotas are A: 12; B: 5; C: 2, and the sum is 19. So we have 21–19 = 2 nurses remaining to allocate. These are given to wings A and B, since they have the largest fractional part of the standard quota. The final Hamilton apportionment is A: 13; B: 6; C: 2.

(b) The standard divisor is

ISM: Excursions in Modern Mathematics, 10E

(c) When the staff consisted of 20 nurses, wing C was assigned 3 of them; when an additional nurse was hired, wing C ended up with one fewer nurse. This is an example of the Alabama paradox (with wing C playing the role of Alabama).

4.3. Jefferson’s Method 4,500,000 + 4,900,000 + 3,900,000 +6,700,000 = 200, 000 . To determine each 100 county’s standard quota, we divide that county’s population by the standard divisor. 4,500, 000 4,900, 000 = 22.5 ; Belarmine: = 24.5 ; Arcadia: 200, 000 200, 000 3,900, 000 6, 700, 000 = 19.5 ; Dandia: = 33.5 Crowley: 200, 000 200, 000

21. (a) The standard divisor is

(b) The lower quotas are Arcadia: 22; Belarmine: 24; Crowley: 19; Dandia: 33. The sum of these is 98. (c) To determine each county’s modified quota, we divide that county’s population by the modified divisor. 4,500, 000 4,900, 000 = 22.84 ; Belarmine: = 24.87 ; Arcadia: 197, 000 197, 000 3,900, 000 6, 700, 000 = 19.80 ; Dandia: = 34.01 Crowley: 197, 000 197, 000 The modified quotas rounded down are now Arcadia: 22; Belarmine: 24; Crowley: 19; Dandia: 34. The sum of these is 99. 4,500, 000 4,900, 000 = 23.08 ; Belarmine: = 25.13 ; 195, 000 195, 000 3,900, 000 6, 700, 000 = 20 ; Dandia: = 34.36 Crowley: 195, 000 195, 000 The modified quotas rounded down are now Arcadia: 23; Belarmine: 25; Crowley: 20; Dandia: 34. The sum of these is 102.

(d) Arcadia:

4,500, 000 4,900, 000 = 22.98 ; Belarmine: = 25.03 ; 195,800 195,800 3,900, 000 6, 700, 000 = 19.92 ; Dandia: = 34.22 Crowley: 195,800 195,800 The modified quotas rounded down are now Arcadia: 22; Belarmine: 25; Crowley: 19; Dandia: 34. The sum of these is 100.

(e) Arcadia:

4,500, 000 4,900, 000 = 22.96 ; Belarmine: = 25 ; 196, 000 196, 000 3,900, 000 6, 700, 000 = 19.90 ; Dandia: = 34.18 Crowley: 196, 000 196, 000 The modified quotas rounded down are now Arcadia: 22; Belarmine: 25; Crowley: 19; Dandia: 34. The sum of these is 100.

(f) Arcadia:

(g) We know that modified divisors d = 195,800 and d = 196,000 work from parts (e) and (f). Any other divisor in between (for example, d = 195,900) will also work.

Chapter 4: The Mathematics of Apportionment 5,900,000 + 7,800,000 + 6,100,000 +6,900,000 = 534, 000 . To determine each 50 province’s standard quota, we divide that province’s population by the standard divisor. 5,900, 000 7,800, 000 = 11.05 ; Brock: = 14.61 ; Anline: 534, 000 534, 000 6,100, 000 6,900, 000 Clanwin: = 11.42 ; Drundell: = 12.92 534, 000 534, 000

22. (a) The standard divisor is

(b) The lower quotas are Anline: 11; Brock: 14; Clanwin: 11; Drundell: 12. The sum of these is 48. (c) To determine each province’s modified quota, we divide that province’s population by the modified divisor. 5,900, 000 7,800, 000 Anline: = 11.8 ; Brock: = 15.6 ; 500, 000 500, 000 6,100, 000 6,900, 000 Clanwin: = 12.2 ; Drundell: = 13.8 500, 000 500, 000 The modified quotas rounded down are now Anline: 11; Brock: 15; Clanwin: 12; Drundell: 13. The sum of these is 51. 5,900, 000 7,800, 000 = 11.13 ; Brock: = 14.72 ; 530, 000 530, 000 6,100, 000 6,900, 000 = 11.51 ; Drundell: = 13.02 Clanwin: 530, 000 530, 000 The modified quotas rounded down are now Anline: 11; Brock: 14; Clanwin: 11; Drundell: 13. The sum of these is 49.

(d) Anline:

5,900, 000 7,800, 000 = 11.35 ; Brock: = 15 ; 520, 000 520, 000 6,100, 000 6,900, 000 = 11.73 ; Drundell: = 13.27 Clanwin: 520, 000 520, 000 The modified quotas rounded down are now Anline: 11; Brock: 15; Clanwin: 11; Drundell: 13. The sum of these is 50.

(e) Anline:

5,900, 000 7,800, 000 = 11.57 ; Brock: = 15.29 ; 510, 000 510, 000 6,100, 000 6,900, 000 = 11.96 ; Drundell: = 13.53 Clanwin: 510, 000 510, 000 The modified quotas rounded down are now Anline: 11; Brock: 15; Clanwin: 11; Drundell: 13. The sum of these is 50.

(f) Anline:

(g) We know that modified divisors d = 520,000 and d = 510,000 work from parts (e) and (f). Any other divisor in between (for example, d = 511,000) will also work. 23. Any modified divisor between approximately 49,285.72 and 49,402.98 can be used for this problem. Using D = 49,300 we obtain:

State Modified Quota

67.14

54.16

26.98

13.996

ISM: Excursions in Modern Mathematics, 10E

24. Any modified divisor between approximately 49,550 and 49,565 can be used for this problem. Using D = 49,550 we obtain:

Province Modified Quota

87.99

46.01

14.71

53.10

Modified Lower Quota 87 46 14 53 The final apportionment using Jefferson’s method is the modified lower quota value in the table. 25. Any modified divisor between approximately 1012 and 1024 can be used for this problem. Using D = 1020 we obtain:

Route Modified Quota

44.412

30.461

20.088

13.882

10.059

8.549

Modified Lower Quota

44 30 20 13 10 8 The final apportionment using Jefferson’s method is the modified lower quota value in the table. 26. Any modified divisor between approximately 9258 and 9386 can be used for this problem. Using D = 9300 we obtain:

State Modified Quota

3.742

11.269

6.968

15.140

5.892

Modified Lower Quota 3 11 6 15 5 The final apportionment using Jefferson’s method is the modified lower quota value in the table. 27. Any modified divisor between approximately 196,200 and 196,500 can be used for this problem. Using D = 196,500 we obtain:

State Modified Quota

41.93

32.47

25.24

23.00

16.79

Modified Lower Quota

41 32 25 23 16 The final apportionment using Jefferson’s method is the modified lower quota value in the table. 28. Any modified divisor between approximately 49.55 and 49.60 can be used for this problem. Using D = 49.6 we obtain:

School Modified Quota

Agriculture 33.18

Business Education 15.36

41.96

Humanities

Science

21.49

140.02

Modified Lower Quota 33 15 41 21 140 The final apportionment using Jefferson’s method is the modified lower quota value in the table. 29. Any modified divisor between approximately 0.4944% and 0.4951% can be used for this problem. Using D = 0.495% we obtain:

Planet Modified Quota

22.9697

16.3030

78.0202

30.2626

21.0505

33.4141

Modified Lower Quota

Chapter 4: The Mathematics of Apportionment

30. Any modified divisor between approximately 0.785% and 0.791% can be used for this problem. Using D = 0.79% we obtain:

Island

Aleta

Bonita

Corona

Doritos

Modified Quota

7.90

33.11

36.05

49.52

Modified Lower Quota

7 33 36 49 The final apportionment using Jefferson’s method is the modified lower quota value in the table.

4.4. Adams’s and Webster’s Method 31. Any modified divisor between approximately 205,500 and 205,800 can be used for this problem. Using D = 205,600 we obtain:

State Modified Quota

40.08

31.03

24.12

21.98

16.05

Modified Upper Quota

41 32 25 22 17 The final apportionment using Adams’s method is the modified upper quota value in the table. 32. Any modified divisor between approximately 50.694 and 50.756 can be used for this problem. Using D = 50.7 we obtain:

School Modified Quota

Agriculture 32.47

Business Education 15.03

Humanities

Science

21.03

136.98

41.05

Modified Upper Quota 33 16 42 22 137 The final apportionment using Adams’s method is the modified upper quota value in the table. 33. Any modified divisor between approximately 0.5044% and 0.5081% can be used for this problem. Using D = 0.505% we obtain:

Planet Modified Quota

22.5149

15.9802

76.4752

29.6634

20.6337

32.7525

Modified Upper Quota

23 16 77 30 21 33 The final apportionment using Adams’s method is the modified upper quota value in the table. 34. Any modified divisor between approximately 0.8138% and 0.8149% can be used for this problem. Using D = 0.814% we obtain:

Island

Aleta

Bonita

Corona

Doritos

Modified Quota

7.67

32.14

34.99

48.06

Modified Upper Quota

8 33 35 49 The final apportionment using Adams’s method is the modified upper quota value in the table.

ISM: Excursions in Modern Mathematics, 10E

35. Any modified divisor between 49,907 and 50,188 can be used for this problem. Using D = 50,000 we obtain:

State

Modified Quota

66.2

53.4

26.6

13.8

Rounded Modified Quota 66 53 27 14 The final apportionment using Webster’s method is the rounded quota value in the table. 36. Any modified divisor between 50,110 and 50,114 can be used for this problem. Using D = 50,112 we obtain:

Province Modified Quota

87.01

45.498

14.55

52.50

Rounded Modified Quota 87 45 15 53 The final apportionment using Webster’s method is the rounded modified quota value in the table. 37. Any modified divisor between 200,100 and 200,800 can be used for this problem. Using D = 200,500 we obtain:

State Modified Quota

41.10

31.82

24.74

22.54

16.46

Rounded Modified Quota

41 32 25 23 16 The final apportionment using Webster’s method is the rounded modified quota value in the table. 38. Any modified divisor between 49.79 and 50.14 can be used for this problem. Using D = 50 we obtain:

School Modified Quota

Agriculture

Business

Education

Humanities

Science

32.92

15.24

41.62

21.32

138.9

Rounded Modified Quota

33 15 42 21 139 The final apportionment using Webster’s method is the rounded modified quota value in the table. 39. Any modified divisor between approximately 0.499% and 0.504% can be used for this problem. Using D = 0.5% we obtain:

State Modified Quota

22.74

16.14

77.24

29.96

20.84

33.08

Rounded Modified Quota

23 16 77 30 21 33 The final apportionment using Webster’s method is the rounded modified quota value in the table. 40. Any modified divisor between 9964 and 9969 can be used for this problem. Using D = 9965 we obtain:

State

Modified Quota

3.49

10.52

6.50

14.13

5.499

Rounded Modified Quota

3 11 7 14 5 The final apportionment using Webster’s method is the rounded modified quota value in the table.

Chapter 4: The Mathematics of Apportionment

4.5. The Huntington-Hill Method 41. (a) 2. Since 1 ≤ 1.514 ≤ 2 , we need only compare 1.514 to the geometric mean 1× 2 . Since 1.514 > 2 , we round up to 2. Note that 2 ≈ 1.414213562 . (b) 1. Since 1 ≤ 1.4 ≤ 2 , we need only compare 1.4 to the geometric mean 1× 2 . Since 1.4 < 2 , we round down to 1. (c) 1. As in (b), 1.41 < 2 so we round down to 1. (d) 2. 1.415 > 2 so we round up to 2. (e) 2. 1.449 > 2 so we round up to 2. 42. (a) 9. Since 8 ≤ 8.585 ≤ 9 , we need only compare 8.585 to the geometric mean 8 × 9 . Since 8.585 > 72 , we round up to 9. Note that 72 ≈ 8.485281374 . (b) 8. Since 8 ≤ 8.485 ≤ 9 , we need only compare 8.485 to the geometric mean 8 × 9 . Since 8.485 < 72 , we round down to 8. (c) 8. As in (b), 8.484 < 72 so we round down to 8. (d) 9. 8.486 > 72 so we round up to 9. 43. 2. In 2010, the Huntington-Hill method was used in apportionment of the U.S. House of Representatives. Since 1 ≤ 1.485313 ≤ 2 , we need only compare 1.485313 to the geometric mean 1× 2 = 2 ≈ 1.414213562 . Since 1.485313 > 2 , we round up to 2. [See also Exercise 41.] 44. 8. Since 8 ≤ 8.483 ≤ 9 , we need only compare 8.483 to the geometric mean 8 × 9 = 72 ≈ 8.485281374 . Since 8.483 < 72 , we round down to 8. [See also Exercise 42.] 45. (a) The number of seats is the sum of the standard quotas. Number of seats = 3.52 + 10.48 + 1.41 + 12.51 + 12.08 = 40. (b) We start by attempting to round the standard quotas using the Huntington-Hill method. We calculate cutoffs and determine the rounded quotas sum to 4 + 10 + 1 + 13 + 12 = 40. Therefore this is the final Huntington-Hill apportionment A: 4; B: 10; C: 1; D: 13; E: 12. [Note: We got somewhat lucky in not needing to modify the quotas.] State

Standard Quota

Cutoff

Rounded Quota

3.52

3.464101615

10.48

10.48808848

1.41

1.414213562

12.51

12.489996

12.08

12.489996

ISM: Excursions in Modern Mathematics, 10E

46. (a) The number of seats is the sum of the standard quotas 25.49 + 14.52 + 8.48 + 30.71 + 20.8 = 100. (b) We start by attempting to round the standard quotas using the Huntington-Hill method. We calculate cutoffs and determine the rounded quotas sum to 25 + 15 + 8 + 31 + 21 = 100. Therefore this is the final Huntington-Hill apportionment A: 25; B: 15; C: 8; D: 31; E: 21. [Note: We got somewhat lucky in not needing to modify the quotas.] State

Standard Quota

Cutoff

Rounded Quota

25.49

25.4951

14.52

14.49138

8.48

8.485281

30.71

30.4959

20.8

20.4939

47. (a) The number of seats is the sum of the standard quotas = 3.46 + 10.49 + 1.42 + 12.45 + 12.18 = 40. (b) We start by attempting to round the standard quotas using the Huntington-Hill method. We calculate cutoffs and determine the rounded quotas sum to 3 + 11 + 2 + 12 + 12 = 40. Therefore this is the final Huntington-Hill apportionment A: 3; B: 11; C: 2; D: 12; E: 12. [Compare this to Exercise 45.] State

Standard Quota

Cutoff

Rounded Quota

3.46

3.464101615

10.49

10.48808848

1.42

1.414213562

12.45

12.489996

12.18

12.489996

48. (a) The number of seats is the sum of the standard quotas 25.496 + 14.491 + 8.486 + 30.449 + 21.078 = 100. (b) We start by attempting to round the standard quotas using the Huntington-Hill method. We calculate cutoffs and determine the rounded quotas sum to 26 + 14 + 9 + 30 + 21 = 100. Therefore this is the final Huntington-Hill apportionment A: 26; B: 14; C: 9; D: 30; E: 21. [Compare to Exercise 46.] State

Standard Quota

Cutoff

Rounded Quota

25.496

25.4951

14.491

14.49138

8.486

8.485281

30.449

30.4959

21.078

21.49419

Chapter 4: The Mathematics of Apportionment

49. (a) A modified divisor of d = 10000 (the same as the standard divisor) works for this problem. apportionment under Webster’s method is A: 34; B: 41; C: 22; D: 59; E: 15; F: 29. Stat Population Modified Quota Rounded Quota e (d = 10000) A

344,970

34.497

408,700

40.87

219,200

21.92

587,210

58.721

154,920

15.492

285,000

28.5

The final

(b) We start by attempting to round the standard quotas using the Huntington-Hill method. We calculate cutoffs and determine the rounded quotas sum to 35 + 41 + 22 + 59 + 16 + 29 = 202 (see table below). We need to modify the standard divisor upward. Stat e

Population

344,970

34.497

34.4964

408,700

40.87

40.4969

219,200

21.92

21.4942

587,210

58.721

58.4979

154,920

15.492

15.4919

285,000

28.5

28.4956

Standard Quota

Cutoff

Rounded Quota

We modify the divisor by choosing d = 10001. We calculate cutoffs and determine the rounded quotas sum to 35 + 41 + 22 + 59 + 15 + 29 = 200. Therefore this is the final Huntington-Hill apportionment A: 34; B: 41; C: 22; D: 59; E: 15; F: 29. Population

Modified Quota (d = 10001)

Cutoff

344,970

34.4936

34.4964

408,700

40.8659

40.4969

219,200

21.9178

21.4942

587,210

58.7151

58.4979

154,920

15.4905

15.4919

285,000

28.4972

28.4956

State

Rounded Quota

ISM: Excursions in Modern Mathematics, 10E

50. (a) A modified divisor of d = 9999 (slightly less than the standard divisor of 10,000) works for this problem. The final apportionment under Webster’s method is A: 35; B: 20; C: 52; D: 8; E: 15; F: 70. Stat Population Modified Quota Rounded Quota e (d = 10000) A

344,970

34.500

204,950

20.497

515,100

51.515

84,860

8.487

154,960

15.498

695,160

69.523

(b) We start by attempting to round the standard quotas using the Huntington-Hill method. We calculate cutoffs and determine the rounded quotas sum to 35 + 21 + 52 + 9 + 16 + 70 = 203 (see table below). We need to modify the standard divisor upward. Stat e

Population

344,970

34.497

34.4964

204,950

20.495

20.4939

515,100

51.510

51.4976

84,860

8.486

8.4853

154,960

15.496

15.4919

695,160

69.516

69.4982

Standard Quota

Cutoff

Rounded Quota

We modify the divisor by choosing d = 10001. We calculate cutoffs and determine the rounded quotas sum to 34 + 20 + 52 + 8 + 16 + 70 = 200. Therefore this is the final Huntington-Hill apportionment A: 34; B: 20; C: 52; D: 8; E: 16; F: 70. Population

Modified Quota (d = 10001)

Cutoff

344,970

34.4936

34.4964

204,950

20.4930

20.4939

515,100

51.5048

51.4976

84,860

8.4852

8.4853

154,960

15.4945

15.4919

695,160

69.5090

69.4982

State

Rounded Quota

Chapter 4: The Mathematics of Apportionment

4.6

The Quota Rule and Apportionment Paradoxes

51. The correct answer is (c). Under Hamilton’s method there can be no violations of the quota rule. The only possible apportionment to state X is 35 or 36 (the value of the standard quota rounded down or up respectively). That is, the final apportionment using Hamilton’s method must give X a (whole) number of seats less than one unit from its standard quota of 35.41. 52. The correct answer is (b). Under Hamilton’s method there can be no violations of the quota rule. The only possible apportionment to Y is 78 or 79 (the value of the standard quota rounded down or up respectively). 53. The correct answer is (e). The other choices represent lower-quota violations and Jefferson’s method can produce only upper-quota violations. 54. The correct answer is (a). Choices (b) and (c) represent upper-quota violations and Adams’s method can produce only lower-quota violations. Since 78 or 79 could be possible apportionments, choices (d) and (e) are also eliminated. 55. The correct answer is (e). Under Webster’s method both lower-quota and upper-quota violations are possible. 56. The correct answer is (e). Under Webster’s method both lower-quota and upper-quota violations are possible. 57. The difference between 55 (the number of seats that California would receive under Jefferson’s method) and 52.45 (California’s standard quota) is greater than 1. This fact illustrates that Jefferson’s method violates the quota rule (upper-quota violations are possible). 58. The difference between 50 (the number of seats that California would receive under Adams’s method) and 52.45 (California’s standard quota) is greater than 1. This fact illustrates that Adams’s method violates the quota rule (lower-quota violations are possible). 59. The Alabama paradox. The apportionment to Dunes went from 3 social workers to 2 for no other reason than the fact that the county hired an additional social worker (see answer to Exercise 19). 60. The Alabama paradox. The number of nurses assign to wing C went from 3 to 2 for no other reason than the fact that the hospital hired an additional nurse (see answer to Exercise 20). 61. (a) Child

Bob

Peter

Ron

Standard quota

0.594

2.673

7.733

Lower quota

54 + 243 + 703 = 90.91 . 11 The sum of lower quotas is 9, so there are 2 remaining pieces of candy to allocate. These are given to Ron and Peter, since they have the largest fractional parts of the standard quota. The final apportionment is: Bob: 0; Peter: 3; Ron: 8.

Note: The standard divisor is

(b) Child

Bob

Peter

Ron

Study time

255

789

Standard quota

.56

2.55

7.89

Lower quota

Note: The standard divisor is

56 + 255 + 789 = 100 . 11

ISM: Excursions in Modern Mathematics, 10E

The sum of the lower quotas is 9, so there are 2 remaining pieces of candy to allocate. These are given to Ron and Bob, since they have the largest fractional parts of the standard quota. The final apportionment is: Bob: 1; Peter: 2; Ron: 8. (c) Yes. For studying an extra 2 minutes (an increase of 3.70%), Bob gets a piece of candy. However, Peter, who studies an extra 12 minutes (an increase of 4.94%), has to give up a piece. This is an example of the population paradox. 62. (a) Child

Bob

Peter

Ron

Standard quota

0.54

2.43

7.03

Lower quota

54 + 243 + 703 = 100 . 10 The sum of lower quotas is 9, so there is 1 remaining piece of candy to allocate. This is given to Bob since he has the largest fractional part of the standard quota. The final apportionment is: Bob: 1; Peter: 2; Ron: 7.

Note: The standard divisor is

(b) Child

Bob

Peter

Ron

Study time

243

703

Standard quota

.594

2.673

7.733

Lower quota

54 + 243 + 703 = 90.91 . 11 The sum of the lower quotas is 9, so there are 2 remaining pieces of candy to allocate. These are given to Peter and Ron, since they have the largest fractional parts of the standard quota. The final apportionment is: Bob: 0; Peter: 3; Ron: 8.

Note: The standard divisor is

(c) Yes. When 10 pieces of candy are divided, Bob is to get one piece, but when 11 pieces are divided, Bob ends up with none. This is an example of the Alabama paradox. 63. (a) Standard divisor =

5, 200, 000 + 15,100, 000 + 10, 600, 000 = 618, 000 . 50

(b) The standard quotas are Aila: 8.41; Balin: 24.43; Cona: 17.15. Lower quotas are Aila: 8; Balin: 24; Cona: 17 and the sum is 49. So we have 50–49 = 1 seat remaining to allocate. This is given to Balin since it has the largest fractional part of the standard quota. The final Hamilton apportionment is Aila: 8; Balin: 25; Cona: 17.

5, 200, 000 + 15,100, 000 + 10, 600, 000 + 9,500, 000 = 621,538.5 . The standard 65 quotas are Aila: 8.37; Balin: 24.29; Cona: 17.05; Dent:15.28 . Lower quotas are Aila: 8; Balin: 24; Cona: 17; Dent: 15 and the sum is 64. So we have 65–64 = 1 seat remaining to allocate. This is given to Aila since it has the largest fractional part of the standard quota. The final Hamilton apportionment is Aila: 9; Balin: 24; Cona: 17; Dent: 15.

(d) The new-states paradox. The addition of Dent (and the addition of the exact number of seats that Dent is entitled to) still impacts the other apportionments. Balin gives up one seat to Aila with the addition of a new state. 64. (a) Standard divisor =

Chapter 4: The Mathematics of Apportionment (b) The standard quotas are Alexdale: 6.40; Bromville: 13.41; Canley: 10.18. Lower quotas are Alexdale: 6; Bromville: 13; Canley: 10 and the sum is 29. So we have 30–29 = 1 food-service worker remaining to allocate. This is given to Bromville since it has the largest fractional part of the standard quota. The final Hamilton apportionment is Alexdale: 6; Bromville: 14; Canley: 10.

617 + 1292 + 981 + 885 = 96.8 . The standard quotas are Alexdale: 6.37; 39 Bromville: 13.35; Canley: 10.13; Dillwood: 9.14. Lower quotas are Alexdale: 6; Bromville: 13; Canley: 10; Dillwood: 9 and the sum is 38. So we have 39–38 = 1 seat remaining to allocate. This is given to Alexdale since it has the largest fractional part of the standard quota. The final Hamilton apportionment is Alexdale: 7; Bromville: 13; Canley: 10; Dillwood: 9.

(d) The new-states paradox. The addition of Dillwood (and the addition of the exact number of seats that Dillwood needs) forces Bromville to give up one food-service worker, who is reassigned to Alexdale. 65. (a) Choose, for example, a modified divisor of d = 990 (the standard divisor is 1000. The rounded quotas sum to 88 + 4 + 5 + 3 = 100. Therefore Webster’s apportionment is A: 88; B: 4; C: 5; D: 3. State

Population

Modified Quota (d = 990)

Cutoff

Rounded Quota

86,915

87.7929

87.4986

4,325

4.3687

4.4721

5,400

5.4545

5.4772

3,360

3.3939

3.4641

(b) We start by attempting to round the standard quotas using the Huntington-Hill method. We calculate cutoffs and determine the rounded quotas sum to 87 + 4 + 5 + 3 = 99 (see table below). We need to modify the standard divisor downward. Stat e

Population

Standard Quota

Cutoff

Rounded Quota

86,915

86.915

86.4986

4,325

4.325

4.4721

5,400

5.4

5.4772

3,360

3.36

3.4641

We modify the divisor by choosing d = 990. We calculate cutoffs and determine the rounded quotas sum to 88 + 4 + 5 + 3 = 100. Therefore this is the final Huntington-Hill apportionment A: 88; B: 4; C: 5; D: 3. State

Population

Modified Quota (d = 990)

Cutoff

Rounded Quota

86,915

87.7929

87.4986

4,325

4.3687

4.4721

5,400

5.4545

5.4772

3,360

3.3939

3.4641

ISM: Excursions in Modern Mathematics, 10E

(c) Both Webster’s method and the Huntington-Hill method violate the quota rule. (An upper-quota violation occurs with A’s apportionment – A’s standard quota is only 86.915)

JOGGING 66. (a) q1 + q2 +  + qN represents the total number of seats available. (b)

p1 + p2 +  + pN represents the total population divided by the total number of seats available which q1 + q2 +  + qN also happens to be the standard divisor.

  pN (c)   × 100 represents the percentage of the total population in state N. p p p + + +  2 N   1

67. (a) State X has received at least one more seat than the standard quota suggested; an upper quota violation. (b) State X has received at least one fewer seat than the standard quota suggested; a lower quota violation. (c) The number of seats that state X has received is within 0.5 of its standard quota. Except for the case in which the standard quota has 0.5 as its decimal part, the number of seats that state X receives is the same as that found by using conventional rounding of the standard quota. (d) The number of seats that state X has received satisfies the quota rule, but is not the result of conventional rounding of the standard quota. 68. If the modified divisor used in Jefferson’s method is d, then state A has modified quota B has modified quota

pA ≥ 3.0 . Also, state d

pB 3 < 1.0 . So, p A ≥ 3d > 3 pB . That is, more than of the total population lives in 4 d

state A. 69. If the modified divisor is d, then state A has modified quota

pA ≥ 2.5 . Also, state B has modified quota d

pB 5 < 0.5 . So, p A ≥ 2.5d > 5 pB . That is, more than of the population lives in state A. 6 d

70. (a) Since the two fractional parts add up to 1, the surplus to be allocated using Hamilton’s method is 1, and it will go to the state with larger fractional part, that is, the state with fractional part more than 0.5. This is the same result that is obtained by just rounding off in the conventional way, which in this case happens to be the result given by Webster’s method. (Anytime that rounding off the quotas the conventional way produces integers that add up to M, Webster’s method reduces to rounding off the standard quotas in the conventional way.) (b) When there are only two states, Hamilton’s and Webster’s methods agree and Webster’s method can never suffer from the Alabama or population paradox, hence neither will Hamilton’s method. (c) When there are only two states, Webster’s and Hamilton’s methods agree and Hamilton’s method can never violate the quota rule, hence neither will Webster’s method. 71. (a) In Jefferson’s method the modified quotas are larger than the standard quotas and so rounding downward will give each state at least the integer part of the standard quota for that state. (b) In Adam’s method the modified quotas are smaller than the standard quota and so rounding upward will give each state at most one more than the integer part of the standard quota for that state. Copyright © 2022 Pearson Education, Inc.

Chapter 4: The Mathematics of Apportionment (c) If there are only two states, an upper quota violation for one state results in a lower quota violation for the other state. Neither Jefferson’s nor Adams’ method can have both upper and lower quota violations. So, when there are only two states neither method can violate the quota rule.

72. Of those states for which the standard quota is not an integer, there are K that receive their lower quota. The remaining states (those with the largest fractional parts) receive their upper quota of seats. The states receiving their upper and lower quotas are the same under Hamilton’s method or this alternate version of Hamilton’s method. 73. (a) A: 33; B: 138; C: 4; D: 41; E: 14; F: 20 (b) State D is apportioned 41 seats under Lowndes’ method while all of the methods discussed in the chapter apportion 42 seats to state D. States C and E do better under Lowndes’ method than under Hamilton’s method. 74. (a) Take for example q1 = 3.9 and q2 = 10.1 (with m = 14). Under both Hamilton’s method and Lowndes’ method, A gets 4 seats and B gets 10 seats. (b) Take for example q1 = 3.4 and q2 = 10.6 (with m = 14). Under Hamilton’s method, A gets 3 seats and B gets 11 seats. Under Lowndes’ method, A gets 4 seats and B gets 10 seats. (c) Assume that f1 > f 2 , so under Hamilton’s method the surplus seat goes to A. Under Lowndes’ method,

the surplus seat would go to B if

f2 f1 > which can be simplified to f 2 (q1 − f1 ) > f1 (q2 − f 2 ) q2 − f 2 q1 − f1

or f 2 q1 − f 2 f1 > f1q2 − f1 f 2 which gives f 2 q1 > f1q2 . This is rewritten as q1 f > 1 since all values are > 0. q2 f2

RUNNING 75. (a) We use a standard divisor of SD =

State Population

200, 000 ≈ 9090.91 . The standard quotas are given below. 22

18,000

18,179

40,950

122,871

Standard Quota 1.98 1.9997 4.5045 13.5158 Each state would then receive its lower quota: A: 1; B: 1; C: 4; D: 13. This leads to 3 surplus seats to be divided among the four states. To apportion these surplus seats, we use Jefferson’s method. A modified divisor of d = 40,955 works to apportion these three surplus seats. State

Population

18,000

18,179

40,950

122,871

Modified Quota (d = 40,955)

0.4395

0.4439

0.9999

3.0001

Lower Quota 0 0 0 3 The final apportionment would have state D receiving the 3 surplus seats: A: 1; B: 1; C: 4; D: 16. (b) The Hamilton-Jefferson hybrid method can produce apportionments different from Hamilton’s method because, as is clear from (a), the quota rule can be violated in the hybrid method. For the example in (a), the apportionment due to Jefferson’s method is: A: 2; B: 2; C: 4; D: 14. This Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

example shows that the Hamilton-Jefferson hybrid method can produce apportionments different from Jefferson’s method. The bias towards large states present in Hamilton’s method and in Jefferson’s method are both being exploited to make this method even more biased towards large states. (c) The hybrid method can violate the quota rule since more than one surplus seat can be handed out in the second step of the method (the Jefferson step). 76. (a) Under Jefferson’s method, an increase in the total number of seats to be apportioned results in a decrease in the modified divisor which will result in some of the states gaining seats, but none will lose seats (since all modified quotas will increase). (b) Under Jefferson’s method, the addition of a new state with its fair share of seats may require a new modified divisor. If the modified divisor is increased, some original states may lose seats, but none will gain seats; if the modified divisor is decreased, some original states may gain seats, but none will lose seats. That is, no state will gain a seat at the expense of another. 77. (a) Under Adams’ method, an increase in the total number of seats to be apportioned results in a decrease in the modified divisor which will result in some of the states gaining seats, but none will lose seats (since all modified quotas will increase). (b) Under Adams’ method, the addition of a new state with its fair share of seats may require a new modified divisor. If the modified divisor is increased, some original states may lose seats, but none will gain seats; if the modified divisor is decreased, some original states may gain seats, but none will lose seats. That is, no state will gain a seat at the expense of another. 78. (a) Under Webster’s method, an increase in the total number of seats to be apportioned results in a decrease in the modified divisor which will result in some of the states gaining seats, but none will lose seats (since all modified quotas will increase). (b) Under Webster’s method, the addition of a new state with its fair share of seats may require a new modified divisor. If the modified divisor is increased, some original states may lose seats, but none will gain seats; if the modified divisor is decreased, some original states may gain seats, but none will lose seats. That is, no state will gain a seat at the expense of another.

APPLET BYTES 79. (a) A modified divisor of 49,400 allocates too many (251) seats. A modified divisor of 49,453 allocated too few (249) seats. (b) A modified divisor of 50,627 allocates too many (251) seats. A modified divisor of 51,000 allocated too few (248) seats. (c) A good starting point is a modified divisor of 50,000. However, a modified divisor of d = 50,000 allocates too many (251) seats. Increasing this divisor to 50,080 will allocate the required 250 seats. All modified divisors between 50,080 and 50,385 will also allocate the required 250 seats. [Hint: Rather than increasing the modified divisor by 1, increase it by 40 or 50 in the applet to save time in your search for this range.] (d) A good starting point is a modified divisor of 50,000. However, a modified divisor of d = 50,000 allocates too many (251) seats. Increasing this divisor to 50,080 will allocate the required 250 seats. All modified divisors between 50,080 and 50,389 will also allocate the required 250 seats. Note how similar this range is to that found for Webster’s method. (e) There are four divisors of particular interest – 50,386 through 50,389. Note what happens to state D’s allocation for these divisors. The modified quota is greater than the Huntington-Hill cutoff of 41.497 and yet not larger than the Webster’s method cutoff of 41.5. Once the divisor is reduced to 50,385, however, the modified quota is larger than both cutoffs. Copyright © 2022 Pearson Education, Inc.

Chapter 4: The Mathematics of Apportionment

80. Using Hamilton’s method, the applet produces the following table when allocating the 8 surplus seats. State

Population

Standard Quota 6.8774 1.6128 2.0569 1.9951 8.0876 13.8027 4.1183 5.2144 9.6288 10.2657 12.57 1.9876 5.9887 2.4837 18.3104 105

Lower Quota 6 1 2 1 8 13 4 5 9 10 12 1 5 2 18 97

Standard Quota 6.8774 1.6128 2.0569 1.9951 8.0876 13.8027 4.1183 5.2144 9.6288 10.2657 12.57 1.9876 5.9887 2.4837 18.3104 105

Modified Quota 7.1624 1.6796 2.1421 2.0777 8.4226 14.3745 4.2889 5.4304 10.0277 10.691 13.0908 2.0699 6.2369 2.5866 19.069

Residue

Surplus

Apportionment

Connecticut 236,841 0.8774 1 Delaware 55,540 0.6128 1 Georgia 70,835 0.0569 Kentucky 68,705 0.9951 1 Maryland 278,514 0.0876 Massachusetts 475,327 0.8027 1 New Hampshire 141,822 0.1183 New Jersey 179,570 0.2144 New York 331,589 0.6288 1 North Carolina 353,523 0.2657 Pennsylvania 432,879 0.57 1 Rhode Island 68,446 0.9876 1 South Carolina 206,236 0.9887 1 Vermont 85,533 0.4837 Virginia 630,560 0.3104 Total 3,615,920 8 8 Standard 34,437.33 Divisor Using Jefferson’s method, the applet produces the following apportionment (step increases of size 100 expedited the process of find a modified divisor, namely 33,067, that works). State

Connecticut Delaware Georgia Kentucky Maryland Massachusetts New Hampshire New Jersey New York North Carolina Pennsylvania Rhode Island South Carolina Vermont Virginia Total

Population

236,841 55,540 70,835 68,705 278,514 475,327 141,822 179,570 331,589 353,523 432,879 68,446 206,236 85,533 630,560 3,615,920

7 2 2 2 8 14 4 5 10 10 13 2 6 2 18 105

Modified Lower Quota

7 1 2 2 8 14 4 5 10 10 13 2 6 2 19 105

Standard Divisor 34,437.33 Using Jefferson’s method (as opposed to Hamilton’s), the state of Virginia (a large state at the time) obtains an extra seat while that state of Delaware (a small state at the time) loses a seat.

ISM: Excursions in Modern Mathematics, 10E 81. (a) Many examples will exist. It is easiest to find a two-state example. Here is one of many possibilities. [Note that an applet really wasn’t needed to construct this particular simple example.]

Hamilton’s Method: Population

Standard Quota

Lower Quota

Residue

Total

100

Standard Divisor

State

Surplus

Apportionment

Jefferson’s Method: Population

Standard Quota

Modified Quota

Modified Lower Quota

Total

100

Standard Divisor

State

(b) A small example consisting of three states should work. If one state is large and two are small, the differences between Hamilton’s method and Jefferson’s method are probably more likely to appear. Below is one possible example.

Hamilton’s Method: Population

Standard Quota

Lower Quota

Residue

7.2

0.2

1.3

0.3

1.5

0.5

Total

100

Standard Divisor

State

Surplus

Apportionment

Chapter 4: The Mathematics of Apportionment

Jefferson’s Method (using a modified divisor of 9): Population

Standard Quota

Modified Quota

7.2

1.3

1.4444

1.5

1.6667

Total

100

Standard Divisor

State

Modified Lower Quota

82. An infinite number of examples exist, but it does take some time to find one. One suggestion is to have one larger state mixed in with two or three smaller states. One possible example with 4 states and 100 seats is given below.

Jefferson’s method (with a modified divisor of 1950) produces an upper quota violation. Population

Standard Quota

Modified Quota

182,033

91.0165

93.3503

5,534

2.767

2.8379

3,433

1.7165

1.7605

9,000

4.5

4.6154

200,000

100

State

Total Standard Divisor

Modified Lower Quota

100

2,000

Adams’ method (with a modified divisor of 2030) produces a lower quota violation.

Population

Standard Quota

Modified Quota

182,033

91.0165

89.6714

5,534

2.767

2.7261

3,433

1.7165

1.6911

9,000

4.5

4.4335

200,000

100

State

Total Standard Divisor

Modified Upper Quota

2,000

100

Chapter 5 component contains vertices D and Z and the edge that connects them).

WALKING 5.2

An Introduction to Graphs

1. (a) {A, B, C, X, Y, Z} The basic elements of a graph are dots and lines (vertices and edges in mathematical language). The vertex set is a listing of the labels given to each vertex (dot). (b) {AX, AY, AZ, BB, BC, BX, CX, XY} Remember, each edge can be described by listing the pair of vertices that are connected by that edge. (c) deg(A) = 3, deg(B) = 4, deg(C) = 2, deg(X) = 4, deg(Y) = 2, deg(Z) = 1 The degree of a vertex is the number of edges meeting at that vertex.

4. (a) {A, B, C, D, E, X, Y, Z} (b) {AB, AD, AE, BE, CE, DE, XY, XY, YZ} (c) deg(A) = 3, deg(B) = 2, deg(C) = 1, deg(D) = 2, deg(E) = 4, deg(X) = 2, deg(Y) = 3, deg(Z) = 1 (d) There are 2 components in this graph (one component contains the vertices A, B, C, D, and E and the edges connected them and the other component contains vertices X, Y, and Z and the edges that connect them). 5. One possible picture of the graph:

(d) One possible answer: A second possible picture:

2. (a) {A, B, C, D, E} (b) {AB, AC, AD, AE, BC, BE, CC, CE, DE}

6. One possible picture of the graph:

3. (a) {A, B, C, D, X, Y, Z} (b) {AX, AX, AY, BX, BY, DZ, XY} Note: AX is listed twice. (c) deg(A) = 3, deg(B) = 2, deg(C) = 0, deg(D) = 1, deg(X) = 4, deg(Y) = 3, deg(Z) = 1 (d) There are 3 components in this graph (one component contains the vertices A, B, X, and Y and each edge connecting them; one component contains the vertex C; one

7. (a) A, B, D, E Remember that two vertices are adjacent in a graph if they are joined by an edge. So we may simply consider the edges that contain a D (listed either first or second). (b) AD, BC, DD, DE Two edges are adjacent if they share a common vertex (in this case, either vertex B or vertex D).

Chapter 5: The Mathematics of Getting Around (c) 5; D appears 5 times in the edge list. (d) 12; one way to easily find this is to multiply the number of edges by two (since each edge contributes two degrees to the total). 8. (a) A, C, Y

(b) C, B, D, A, H, F Again, more than one answer is possible.

(d) 16

(b)

Four embedded rectangles inside the graph in Exercise 9(a) brings the degree of each vertex up to 4.

11. (a) C, B, A, H, F Other answers such as C, B, A, H, G, F are also possible.

(b) AX, AZ, CY, YY

9. (a)

(c)

Arranging the vertices in a circular pattern and connecting them with edges that run “around the circle” will accomplish this task.

One method is to construct two copies of the graph in part (a) – each on 4 vertices.

(c) C, B, A, H, F (d) C, D, B, A, H, G, G, F (e) 4 (It can be helpful to list them from shortest to longest as C, B, A; C, D, A; C, B, D, A; C, D, B, A) (f) 3 (H, F; H, G, F; H, G, G, F) (g) 12 (Any one of the 4 paths in (e) followed by edge AH followed by any one of the 3 paths in (f).) 12. (a) D, A, H, G, F, E

(c)

One method is to arrange the vertices in two equally long rows and connect each vertex in the first row with exactly one vertex in the second.

(b) D, B, A, H, G, G, F, E (c) D, A, H, F, E (d) D, C, B, A, H, G, G, F, E

10. (a)

One method is to start by constructing a graph just as in Exercise 9(a). Then, connect any two nonadjacent vertices with an edge bringing those two vertices up to degree 3.

(e) 5 (D, A; D, B, A; D, B, C, D, A; D, C, B, A; D, C, B, D, A) (f) 3 (H, F, E; H, G, F, E; H, G, G, F, E) (g) 15 (Any one of the paths in (e) followed by AH followed by any one of the paths in (f).) 13. (a) G,G. A loop is a circuit of length 1.

(b)

One method is to start by constructing a graph just as in Exercise 9(b). Then, connect two opposite vertices in one of the components to bring their degree up to 3.

(b) There are none. Such a circuit would consist of two vertices and two different edges connecting the vertices. [This explains why G,G,G is not a circuit of length 2.] (c) A,B,D,A; B,C,D,B; F,G,H,F For starters, look for triangles consisting of three vertices and three different edges in the graph.

ISM: Excursions in Modern Mathematics, 10E (d) A,B,C,D,A; F,G,G,H,F For starters, look for rectangles consisting of four vertices and four different edges such as A,B,C,D,A in the graph. Note that other possibilities (such as the triangle with a loop F,G,G,H,F) also exist. 14. (a) E,E (b) F,H,F

would need to traverse edge CI twice which isn’t allowed). 18. (a) CI and HJ. Start by noting the clique formed by F, G, H, and I. A second clique is formed by C, D, and E. Since cliques do not have bridges, only edges AE, AB, BC, CI and HJ are possible bridges. But AE, AB, and BC are not bridges.

(b) There would be 3 components in such a graph (one component containing the vertices A, B, C, D, and E and the edges that connect them; one component containing the clique F, G, H, and I; one component containing vertex J).

(d) A,B,E,D,A (e) A,C,B,E,D,A; A,B,E,E,D,A 15. (a) AH, EF; If either of these edges were removed from the graph, the graph would be disconnected.

(b) There are none. This graph is just one circuit connecting five vertices. Circuits do not have bridges.

(d) 6; One path that does this is I, F, H, G, I, H, J. Another path that does this is I, H, G, I, F, H, J. A longer path is not possible since it cannot traverse through any of the vertices A, B, C, D, and E (if it did, it would need to traverse edge CI twice which isn’t allowed).

(c) AB, BC, BE, CD; That is, every edge is a bridge. 16. (a) BF, FG (b) There are none. The graph consists of two cycles (A,B,C,E,A and A,D,E,A). (c) AB, BC, CD, DE (i.e. every edge is a bridge) 17. (a) CI and HJ. Start by noting the large clique formed by A, B, C, D, and E. A second clique is formed by F, G, and I and a third clique is formed by G, H, and I. Since cliques do not have bridges, only edges CI and HJ are possible bridges.

19. Such a graph will have five vertices (representing North Kingsburg, South Kingsburg, and islands A, B, and C) and seven edges (representing the seven bridges).

20.

(b) There would be 3 components in such a graph. One component would contain the clique formed by A, B, C, D, and E; one component would contain vertices F, G, H, and I and the edges that connect them; one component would contain vertex J. (c) The shortest path is C, I, H, J of length 3. (d) A longest path is I, G, F, I, H, J. It has length 5. Another longest path is I, F, G, I, H, J. A longer path is not possible since it cannot traverse through any of the vertices A, B, C, D, and E (if it did, it

Chapter 5: The Mathematics of Getting Around (b) For this graph, the vertices represent the five committees. An edge appears on the graph if the two committees have a member in common. For example, vertices representing Rules and Guest Speaker have an edge connecting them since A is a member of both.

21.

22.

23. Let the vertices represent the teams A, B, C, D, E, and F. The edges will correspond to the tournament pairings. Putting the vertices at the corners of a regular hexagon will make drawing and interpreting the graph a bit easier.

24. (a) The vertices of the graph represent the members of the Lodge. The edges found in the graph correspond to whether two members are on the same committee or not. For example, since A and C are both on the Rules Committee, AC is an edge. Since A and B are not on any committee together, edge AB does not appear in the graph.

25. First, draw the vertices (these represent the people). Then, draw the edges (when present) which represent Facebook friendships.

26. First, draw the vertices (these represent the courses). Then, draw the edges (when present) which represent “significant linkages.”

27. First, draw the vertices (these represent the intersections of two streets). Then, draw the edges (representing streets).

ISM: Excursions in Modern Mathematics, 10E 28. First, draw the vertices (these represent the intersections of two streets). Then, draw the edges (representing streets).

30. (a) (B); The graph has an Euler path since it is connected and exactly two vertices are odd (Euler’s Path Theorem). (b) (C); The graph has neither an Euler circuit nor an Euler path since it is connected and exactly 10 vertices are odd (Euler’s Path Theorem).

5.3

Euler’s Theorems and Fleury’s Algorithm

29. (a) (A); The graph has an Euler circuit because it is connected and all vertices have even degree (Euler’s Circuit Theorem). (b) (C); The graph has neither an Euler circuit nor an Euler path because there are four vertices of odd degree (Euler’s Path Theorem).

(c) (C); The graph has neither an Euler circuit nor an Euler path since it is connected and exactly 4 vertices are odd (Euler’s Path Theorem). (d) (C); The graph has neither an Euler circuit nor an Euler path since it is connected and exactly 4 vertices are odd (Euler’s Path Theorem). (e) (F); The sum of the degrees of all the vertices of this graph is 9 × 3 = 27 which is odd. Euler’s Sum of Degrees Theorem tells us that this cannot happen in any graph. 31. (a) (C); The graph has neither an Euler circuit nor an Euler path because there are more than two (exactly 4 in fact) vertices of odd degree (Euler’s Path Theorem). (b) (A); The graph has an Euler circuit because all vertices have even degree (Euler’s Circuit Theorem).

(d) (D); Exercises 9(a) and 9(b) gave two possible representations of such a graph. In Exercise 9(a), the graph is connected so that Euler’s Circuit Theorem guarantees that it has an Euler Circuit.

In Exercise 9(b), the graph is not connected and does not have an Euler Circuit.

(e) (F); The sum of the degrees of all the vertices of this graph is 3 × 2 + 5 × 3 = 21 which is odd. Euler’s Sum of Degrees Theorem tells us that this cannot happen in any graph.

(c) (F); The sum of the degrees of all the vertices of this graph is 11× 3 = 33 which is odd. Euler’s Sum of Degrees Theorem tells us that this cannot happen in any graph. (d) (C); The graph has neither an Euler circuit nor an Euler path since it is connected and exactly 12 vertices are odd (Euler’s Path Theorem). 32. (a) (B); The graph has an Euler path since it is connected and exactly two vertices are odd (Euler’s Path Theorem). (b) (A); The graph has an Euler circuit because all vertices have even degree (Euler’s Circuit Theorem). (c) (F); The sum of the degrees of all the vertices of this graph is 1× 1 + 1× 2 + 1× 3 + 1× 4 + 1× 5 + 1× 6 = 21 which is odd. Euler’s Sum of Degrees

Chapter 5: The Mathematics of Getting Around Theorem tells us that this cannot happen in any graph.

33. (a) (B); The graph has an Euler path since it is connected and exactly two vertices are odd (Euler’s Path Theorem). (b) (A); The graph has an Euler circuit because all vertices have even degree (Euler’s Circuit Theorem). (c) (C); While all the vertices have even degree, the vertex of degree zero tells us that the graph is disconnected. So Euler’s Path Theorem tells us that the graph has neither an Euler circuit nor an Euler path.

38.

34. (a) (B); The graph has an Euler path since it is connected and exactly two vertices are odd (Euler’s Path Theorem). (b) (C); Since the graph is disconnected, Euler’s Path Theorem tells us that the graph has neither an Euler circuit nor an Euler path.

Note that the starting and ending vertices of the Euler path are shown in black. 39. There is more than one answer.

(c) (C); While all the vertices have even degree, the vertices of degree zero tell us that the graph is disconnected. So Euler’s Path Theorem tells us that the graph has neither an Euler circuit nor an Euler path. 35.

40. 36.

37. The starting and ending vertices of the Euler path (both having odd degree of course!) are shown in black. Naturally, other answers are possible.

ISM: Excursions in Modern Mathematics, 10E 41. (a) PD or PB; Edges PA and PE have already been traveled. Neither PD nor PB is a bridge of the yet-to-be-traveled part.

45. Since this graph has vertical symmetry, a mirror image of this answer is also an optimal eulerization.

(b) BF; BF is a bridge of the yet-to-betraveled part. If BF were removed, you would be left with the following components.

46. 42. (a) CB or CD; Edges CA and CG have already been traveled. Neither CB nor CD is a bridge of the yet-to-be-traveled part. (b) AL; AL is a bridge of the yet-to-betraveled part.

5.4

Eulerizing and Semi-Eulerizing Graphs

43. Adding as few duplicate edges as possible in eliminating the odd vertices gives an optimal eulerization.

44.

Other answers are possible. 47. In an optimal semi-eulerization two vertices will remain odd. Again, since this graph has vertical symmetry, a mirror image of this answer is also an optimal semi-eulerization. Other answers are possible.

48. In an optimal semi-eulerization two vertices will remain odd. Other answers are possible.

Chapter 5: The Mathematics of Getting Around

49. In an optimal semi-eulerization, the two vertices A and D will remain odd. The vertex between A and D then requires some work to become even (i.e. it is not next to an odd vertex). Other answers are possible.

52. In an optimal semi-eulerization, the two vertices A and D will remain odd.

Other answers are possible. One of these is shown below. 50. In an optimal semi-eulerization, the two vertices A and B will remain odd. Other answers are possible.

53. Adding as few duplicate edges as possible in eliminating the odd vertices gives an optimal eulerization. Other answers are possible. 51. In an optimal semi-eulerization, the two vertices B and C will remain odd. The vertex between B and C then requires at least two additional edges to become even (i.e. it is not right next to an odd vertex).

Other answers are possible. One of these is shown below.

54. Adding as few duplicate edges as possible in eliminating the odd vertices gives an optimal eulerization.

ISM: Excursions in Modern Mathematics, 10E 55. In an optimal semi-eulerization, the two vertices A and B will remain odd. Other answers are possible.

(b) The following graph is regular (all vertices are of degree 2) and it has an Euler circuit.

The graph below is also regular (all vertices have degree 3) but it does not have an Euler circuit.

56. In an optimal semi-eulerization, the two vertices C and D will remain odd. Other answers are possible.

61. Each component of the graph is a graph in its own right and so, according to Euler’s Sum of Degrees Theorem, the number of vertices of odd degree (in each component) must be even. Therefore the 2 vertices of odd degree must be in the same component. 62. (a) Keep adding edges as long as you can without creating any circuits. (b) N – 1 dollars; the graph can be connected with one fewer edge than vertex 63. (a) NK, C, B, NK, B, A, SK, C (b) NK, C, B, NK, B, A, SK, C, SK

JOGGING

57. 5; There are 12 odd vertices in the figure. Two vertices can be used as the starting and ending vertices. The remaining 10 vertices can be paired so that each pair forces one lifting of the pencil. 58. (a) An edge XY contributes 2 to the sum of the degrees of all vertices (1 to the degree of X, and 1 to the degree of Y). (b) If there were an odd number of odd vertices, the sum of the degrees of all the vertices would be odd. That would contradict the result found in (a). 59. None. If a vertex had degree 1, then the single edge incident to that vertex would be a bridge.

(d) NK, C, B, NK, B, A, SK, C, B, A 64. (a) The circuit kissing A, D, C, A is A, B, D, F, C, E, A. Two circuits that kiss A, B, D, A are C, D, F, C and A, C, E, A. (b) Suppose C is a circuit in G. Consider the complement of C in G. If C is an Euler circuit then the complement is empty. So when C is not an Euler circuit, the complement must contain an edge leading to a vertex in C for otherwise G is not connected. Also, the complement of C has all even vertices. So, the complement has an Euler circuit and that circuit will be one that kisses C.

60. (a) If each vertex were of odd degree, then the graph has an odd number of odd vertices. This is not possible! So, it must be that each vertex is in fact of even degree. By Euler’s Circuit Theorem, the graph would have an Euler circuit.

100

Chapter 5: The Mathematics of Getting Around

65. (a) There are many ways to implement this algorithm. The following figures illustrate one possibility.

67. Since vertices R, B, D, and L are all odd, two bridges must be crossed twice. For example, crossing the Adams Bridge twice and the Lincoln Bridge twice produces an optimal route: D, L, R, L, C, A, R, B, R, B, D, C, L, D. Listing the bridges in the order they are crossed: Adams, Washington, Jefferson, Grant, Wilson, Truman, Lincoln, Lincoln, Hoover, Kennedy, Monroe, Roosevelt, Adams. 68. In this case, the route must start and end at L. One possible optimal route would cross the Lincoln Bridge twice and the Adams Bridge twice: L, D, L, C, D, B, R, B, R, L, R, A, C, L. Listing the bridges in the order they are crossed: Adams, Adams, Roosevelt, Monroe, Kennedy, Hoover, Lincoln, Lincoln, Washington, Jefferson, Truman, Wilson, Grant. 69. The answer to Euler’s question is yes. One of the many possible journeys is given by crossing the bridges in the following order: a, b, c, d, e, f, g, h, i, l, m, n, o, p, k. Note that this journey starts at E and ends at D which is acceptable since Euler did not ask that the journey start and end at the same place.

RUNNING . (b) Step 1: Find a path C0 between the two vertices v and w of odd degree. Step 2: Find a circuit C0* that kisses C0 at any vertex that is on C0 . The path and the circuit can be combined into a larger path C1 between the two odd vertices. If there are no kissing circuits to the larger path, we are done. Step 3: Repeat Step 2 until an Euler path is found. 66. If the route must start at B and end at L, then one possible optimal route would cross the Lincoln Bridge twice and the Kennedy Bridge twice: B, R, B, R, L, R, A, C, L, C, D, B, D, L. Listing the bridges in the order they are crossed: Hoover, Lincoln, Lincoln, Washington, Jefferson, Truman, Wilson, Grant, Roosevelt, Monroe, Kennedy, Kennedy, Adams.

70. The only possible value of k is 0. The graph G must have an Euler circuit since it is connected and every vertex is even. In a circuit it is possible to get from any vertex to any other vertex two different ways — traveling the circuit forward and backward. Consequently, the removal of a single edge will not disconnect the graph. 71. The possible values of k are k = 0, 1, 2, 3, …, N-3, and N-1. To have k = 0 bridges in G, place the N vertices at the corners of a polygon, connect adjacent corners with an edge and connect one pair of nonadjacent corners.

To have k = 1 bridge, attach a single vertex to a graph formed by putting vertices at the corners of a regular N-1 polygon.

ISM: Excursions in Modern Mathematics, 10E To have k = 2 bridges, consider the following graph formed by attaching two vertices to a graph formed by putting vertices at the corners of a regular N-2 polygon. Adding more “single” vertices to the left also explains how to get k = 3, 4, …, N-3 bridges.

101

APPLET BYTES 74. (a) Note that the model need not be perfect. A reasonable approximation is shown here.

To get k = N-1 bridges, consider the following example: 72. (a) Both m and n must be even. Why? If the graph has an Euler circuit, then the degree of each vertex must be even. Suppose that m is odd (i.e. A has an odd number of vertices). Then the degree of each vertex in B would be an odd number (namely n). But this contradicts the existence of an Euler circuit. Similarly, if n is odd, the degree of each vertex in A would need to be odd (a contradiction). So, both m and n must be even. (b) Either m = 1 and n = 1 or 2 or m = 2 and n is odd. Remember that a graph will have an Euler path if it has exactly two odd vertices. If m = 1 and n = 1 or 2, then it is easy to see that the graph has an Euler path. If m = 1 and n > 2, then the vertices in B are all odd (which can’t happen since an Euler path has exactly 2 odd vertices). If m = 2, then n must be odd in order to have exactly two odd vertices (both will be in A). If m > 2, then so is n. If either is odd, then the graph has more than 2 odd vertices. If neither is odd, then the graph has no odd vertices (and hence no Euler path). So m > 2 is not possible.

(b) An optimal semi-eulerization of the graph is shown here.

73. Suppose the graph has N vertices. Since there are no multiple edges or loops, the maximum degree a vertex can have is N – 1. If the degrees of the N vertices are all different, they must be 0, 1, 2, ..., N – 1, but this is impossible because the vertex of degree N – 1 would have to be adjacent to all the other vertices and then we couldn’t have a vertex of degree 0.

102

Chapter 5: The Mathematics of Getting Around then traverse edges 1 through 4. (c) One optimal route is illustrated here. In the applet, the path will emerge in red on the right as you click on vertices in the left window.

(d) Many reasonable routes are possible. Below is one.

(b) For the route described in (a), the delays are 16-15=1, 58-18=40, 55-51=4, 4844=4, 38-37=1, 26-25=1, 22-11=11, 674=63, and 64-63=1. (c) From part (b), we find 7 short delays and 2 long delays. So the score for this route would be 4. (d) The following route is pretty good. It has delays of 40-2=38, 37-4=33, 52-7=45, 5510=45, 58-13=45, 63-20=43, 61-23=38, 66-29=37, and 68-32=36. This gives the route a score of 18.

75. (a) One possible route would be very similar to that found in Figure 5-25(d) except it would start with edge number 5, continue as in the figure until edge number 64, and

ISM: Excursions in Modern Mathematics, 10E

103

76. (a) One optimal route:

(b) For the route described in (a), there are at least 23 left turns! .

Chapter 6 WALKING 6.2

Hamilton Paths and Circuits

1. (a) Remember that a Hamilton circuit is a circuit that includes each vertex of the graph once and only once. Remember too that the circuit must return to the starting vertex. 1. A, B, D, C, E, F, G, A; 2. A, D, C, E, B, G, F, A; 3. A, D, B, E, C, F, G, A (b) Remember that a Hamilton path in a graph is a path that includes each vertex of the graph once and only once. Unlike a circuit, it need not end where it started. A, G, F, E, C, D, B

The reasons that the above are the only Hamilton circuits are as follows. First note that edges AB and BC must be a part of every Hamilton circuit and that AC cannot be a part of any Hamilton circuit.

There are two possibilities at vertex A: edge AE or edge AD. Each of these alternatives can be completed in only one way.

(c) D, A, G, B, C, E, F 2. (a) Remember that a Hamilton circuit is a circuit that includes each vertex of the graph once and only once. Remember too that the circuit must return to the starting vertex. 1. A, F, E, D, C, H, I, J, G, B, A; 2. A, F, E, I, J, G, H, D, C, B, A; 3. A, F, E, I, H, D, C, B, G, J, A (b) Remember that a Hamilton path in a graph is a path that includes each vertex of the graph once and only once. Unlike a circuit, it need not end where it started. A, F, E, D, C, H, I, J, G, B (c) F, J, A, B, G, H, C, D, E, I 3. Starting at A, suppose that you travel to B. Then, there are two choices for which vertex to travel to next: C and E. If one starts out A,B,C, then the next vertex visited must be D (do you see why? Otherwise you would eventually pass through C twice which is not allowed). If one starts out A,B,E, then the next vertex visited must also be D. After D, there is only one way to return to A to form a Hamilton circuit. So the Hamilton circuits are A,B,C,D,E,F,G,A and A,B,E,D,C,F,G,A and their mirror images. [Note: The mirror images are A,G,F,E,D,C,B,A and A,G,F,C,D,E,B,A.] 4. A,B,C,D,E,A; A,B,C,E,D,A and their mirror images

or 5. (a) A, F, B, C, G, D, E; Note that edges AF, FB, CG, and GD (or their mirror images) must be part of a Hamilton path. (b) A, F, B, C, G, D, E, A (c) One approach is to “spiral” (in a counterclockwise fashion) around from E to A. One such path is E, D, G, C, B, F, A. (d) The Hamilton path found in part (c) can be easily modified to do the job. One Hamilton circuit is E, D, G, C, B, F, A, E. 6. (a) A, D, G, C, F, B, E; Note that edges BF, FC, CG, and GD (or their mirror images) must be part of a Hamilton path. (b) A, D, G, C, F, B, E, A (c) One approach is to “spiral” (in a clockwise fashion) around from E to A. One such path is E, B, F, C, G, D, A. (d) The Hamilton path found in part (c) can be easily modified to do the job. One Hamilton circuit is E, B, F, C, G, D, A, E.

ISM: Excursions in Modern Mathematics, 10E 7. (a) 8; Knowing that in a Hamilton circuit, each vertex is listed once and only once, we simply count the number of unique vertices listed in the given Hamilton circuit. (b) A, H, C, B, F, D, G, E, A (c) We can write a Hamilton path starting at A using the circuit found in (b) that does not return back to A. That is, A, H, C, B, F, D, G, E. Of course, the path A, E, G, D, F, B, C, H found using the mirror image from (b) will also work. 8. (a) 7; Knowing that in a Hamilton circuit, each vertex is listed once and only once, we simply count the number of unique vertices listed in the given Hamilton circuit. (b) F, E, G, B, D, C, A, F (c) We can write a Hamilton path starting at F using the circuit found in (b) that does not return back to F. That is, F, E, G, B, D, C, A. Of course, the path F, A, C, D, B, G, E found using the mirror image from (b) will also work. 9. (a) Breaking between C and B gives B, A, D, E, C. Breaking between E and C gives C, B, A, D, E. Breaking between D and E gives E, C, B, A, D. Breaking between A and D gives D, E, C, B, A. Breaking between B and A gives A, D, E, C, B. (b) As described in Example 6.6, A, B, E, D, C is one such path. The mirror image C, D, E, B, A is another. Another path that does not come from a “broken” Hamilton circuit is D, A, E, C, B since we cannot get back to D directly from B. Naturally the mirror image B, C, E, A, D is another path. Rather than these “Z” patterns as in the previous paths, we can construct “backwards Z” patterns to find four additional paths: B, A, E, C, D and its mirror image D, C, E, A, B and A, D, E, B, C and its mirror image C, B, E, D, A.

105

10. (a) Be sure to write these down using a pattern to be assured all are listed. Notice how A “shifts” in the circuits below. B, A, C, D, E, F, B; B, C, A, D, E, F, B; B, C, D, A, E, F, B; B, C, D, E, A, F, B; B, C, D, E, F, A, B. The reversals of these five are given below: B, A, F, E, D, C, B; B, F, A, E, D, C, B; B, F, E, A, D, C, B; B, F, E, D, A, C, B; B, F, E, D, C, A, B. (b) The idea here is to start at B and end at either D or E so that there is no single edge that would connect back to B. B, C, A, F, E, D; B, C, D, A, F, E; and in the other direction B, F, A, C, D, E; B, F, E, A, C, D. 11. (a) Suppose that starting at A we move toward F. The next vertex visited must then be E. After that, we have two choices (D or B). But once D or B is chosen, the rest of the circuit is forced. So we get A, F, E, D, C, B, A or A, F, E, B, C, D, A (or the reversals). (b) The idea here is to start at C and end at F so that there is no single edge that would connect back to C. From C, we can travel to B and then choose to go to A or E. To end at a point other than D, the path is unique. C, B, A, D, E, F; C, B, E, D, A, F; From C we could also travel in the other direction (to D rather than B first) and get paths C, D, E, B, A, F; C, D, A, B, E, F. 12. (a) When starting at A we have two choices as to which vertex to move towards (B or D). Once this vertex is chosen, the rest of the circuit is forced. So we get A, B, C, D, E, A or A, D, C, B, E, A (or the reversals).

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Chapter 6: The Mathematics of Touring (b) The idea here is to start at C and end at either A or E so that there is no single edge that would connect back to C. From C, we can travel to B and then choose to go to A or E. To end at a point other than D, the path is unique. C, B, A, D, E; C, B, E, D, A; From C, we could also travel to D rather than B first) to get paths C, D, E, B, A; C, D, A, B, E.

13. (a) When constructing such a path, be sure to travel to all vertices A, B, C, D, and E before crossing edge CI. Also note that from A only vertices B or D (and not E) may be next. One such path is A, B, E, D, C, I, H, G, K, J, F. (b) In this case, be sure to travel to all of the vertices K, J, F, G, H, and I before crossing edge IC. Starting at K, there is only one way to travel to I. This gives one path K, J, F, G, H, I C, B, A, D, E. Another possible path is K, J, F, G, H, I C, D, A, B, E. (c) CI is a bridge of the graph connecting a “left half” and a “right half.” If you start at C and go left, there is no way to get to the right half of the graph without going through C again. On the other hand, if you start at C and cross over to the right half first (i.e. you move to vertex I), there is no way to get back to the left half without going through C again. (d) No matter where you start, you would have to cross the bridge CI twice to visit every vertex and get back to where you started. This is a problem since you would then hit either vertex C or vertex I twice within the circuit. 14. (a) Here are several possible paths: B, C, F, G, H, A, E, D; B, C, D, E, A, H, G, F; B, A, E, D, C, F, G, H; B, A, H, G, F, C, D, E.

(c) Any path passing through vertices B, H, and E must contain edges AB, AH, and AE, respectively. Consequently, any path passing through vertices B, H, and E must contain at least three edges meeting at A and, therefore, would pass through A more than once. A similar argument applies for a Hamilton path starting at C. (d) The only way to go through B is to come from A and leave for C or vice-versa. The only way to go through H is to come from A and leave for G or vice-versa. The only way to go through E is to come from A and leave for D or vice-versa. It follows that to go through B, H, and E requires going through A more than once. 15. There is no Hamilton circuit since two vertices have degree 1. There is no Hamilton path since any such path must contain edges AB, BE, and BC, which would force vertex B to be visited more than once. 16. There is no Hamilton circuit since two vertices (namely C and G) have degree 1. There is no Hamilton path because the path would have to start at G and end at C or vice versa. Either way, the path would have to visit D more than once. 17. (a) 6 (b) B, D, A, E, C, B Weight = 6 + 1 + 9 + 4 + 7 = 27

(c) The mirror image B, C, E, A, D, B Weight = 7 + 4 + 9 + 1 + 6 = 27 18. (a) 8 (b) One possible answer: A, D, C, B, E, A Weight = 8 + 1 + 6 + 4 + 3 = 22

(b) Here are two possible paths: E, D, C, B, A, H, G, F; E, D, C, F, G, H, A, B. Note that starting at E and travelling to A next is not possible.

ISM: Excursions in Modern Mathematics, 10E (c) The mirror image A, E, B, C, D, A Weight = 3 + 4 + 6 + 1 + 8 = 22

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1 sec 1, 000, 000, 000, 000 circuits 1 min 1 hr 1 day 1 year × × × × 60 sec 60 min 24 hr 365 days ≈ 492, 000 years

22. (a) 25! circuits ×

19. (a) A, D, F, E, B, C Weight = 2 + 7 + 5 + 4 + 11 = 29

1 sec 1, 000, 000, 000, 000 circuits 1 min 1 hr 1 day 1 year × × × × 60 sec 60 min 24 hr 365 days ≈ 12,800,000 years

(b) 26! circuits ×

(b) A, B, E, D, F, C Weight = 10 + 4 + 3 + 7 + 6 = 30 23. (a)

20 × 19 = 190 2

(b) K 21 has 20 more edges than K 20 . 21× 20 = 210 2

(c) There are only two Hamilton paths that start at A and end at C. The path A, D, F, E, B, C found in part (a) is the optimal such path. It has weight 29. 20. (a) B, A, E, C, D Weight = 3 + 5 + 7 + 4 = 19 (b) B, A, C, E, D Weight = 3 + 9 + 7 + 6 = 25 (c) There are only two Hamilton paths that start at B and end at D. The path B, A, E, C, D found in part (a) is the optimal such path. It has weight 19. 1 sec 1, 000, 000, 000 circuits 1 min 1 hr 1 day 1 year × × × × 60 sec 60 min 24 hr 365 days ≈ 77 years

21. (a) 20! circuits ×

1 sec 1, 000, 000, 000 circuits 1 min 1 hr 1 day 1 year × × × × 60 sec 60 min 24 hr 365 days ≈ 1622 years

(b) 21! circuits ×

(c) If one vertex is added to K 50 (making for 51 vertices), the new complete graph K 51 has 50 additional edges (one to each old vertex). In short, K 51 has 50 more edges than K50 . That is, y − x = 50 . 24. (a)

200 × 199 = 19,900 2

(b) K 201 has 200 more edges than K 200 . 201× 200 = 20,100 2

( N − 1) N 2

edges and 45 =

9 × 10 , so N = 10. 2

200 × 201 , so N = 201. 2

26. (a) 720 = 6!, so N = 7. (b) 66 =

11× 12 , so N = 12. 2

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6.3 27.

400 × 401 , so N = 401. 2

The Brute Force Algorithm Tour

Weight

Mirror-image Circuit

A, B, C, D, A

48 + 32 + 18 + 22 = 120

A, D, C, B, A

A, B, D, C, A

48 + 20 + 18 + 28 = 114

A, C, D, B, A

A, C, B, D, A

28 + 32 + 20 + 22 = 102

A, D, B, C, A

An optimal tour would be A, C, B, D, A (or its reversal) with a cost of 102. 28.

Tour

Weight

Mirror-image Circuit

A, B, C, D, A

30 + 25 + 60 + 40 = 155

A, D, C, B, A

A, B, D, C, A

30 + 20 + 60 + 80 = 190

A, C, D, B, A

A, C, B, D, A

80 + 25 + 20 + 40 = 165

A, D, B, C, A

An optimal tour would be A, B, C, D, A (or its reversal) with cost of 155. 29.

Tour

Weight

Mirror-image Circuit

A, B, C, D, E, A

35 + 9 + 25 + 32 + 18 = 119

A, E, D, C, B, A

A, B, C, E, D, A

35 + 9 + 23 + 32 + 22 = 121

A, D, E, C, B, A

A, B, D, C, E, A

35 + 21 + 25 + 23 + 18 = 122

A, E, C, D, B, A

A, B, D, E, C, A

35 + 21 + 32 + 23 + 40 = 151

A, C, E, D, B, A

A, B, E, C, D, A

35 + 18 + 23 + 25 + 22 = 123

A, D, C, E, B, A

A, B, E, D, C, A

35 + 18 + 32 + 25 + 40 = 150

A, C, D, E, B, A

A, C, B, D, E, A

40 + 9 + 21 + 32 + 18 = 120

A, E, D, B, C, A

A, C, B, E, D, A

40 + 9 + 18 + 32 + 22 = 121

A, D, E, B, C, A

A, C, D, B, E, A

40 + 25 + 21 + 18 + 18 = 122

A, E, B, D, C, A

A, C, E, B, D, A

40 + 23 + 18 + 21 + 22 = 124

A, D, B, E, C, A

A, D, B, C, E, A

22 + 21 + 9 + 23 + 18 = 93

A, E, C, B, D, A

A, E, B, C, D, A

18 + 18 + 9 + 25 + 22 = 92

A, D, C, B, E, A

An optimal tour would be A, E, B, C, D, A (or its reversal) with time of 92 hours.

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30.

Tour

Weight

Mirror-image Circuit

A, B, C, D, E, A

3.6 + 3.3 + 5.3 + 5.0 + 4.5 = 21.7

A, E, D, C, B, A

A, B, C, E, D, A

3.6 + 3.3 + 3.5 + 5.0 + 5.6 = 21.0

A, D, E, C, B, A

A, B, D, C, E, A

3.6 + 5.9 + 5.3 + 3.5 + 4.5 = 22.8

A, E, C, D, B, A

A, B, D, E, C, A

3.6 + 5.9 + 5.0 + 3.5 + 1.1 = 19.1

A, C, E, D, B, A

A, B, E, C, D, A

3.6 + 6.8 + 3.5 + 5.3 + 5.6 = 24.8

A, D, C, E, B, A

A, B, E, D, C, A

3.6 + 6.8 + 5.0 + 5.3 + 1.1 = 21.8

A, C, D, E, B, A

A, C, B, D, E, A

1.1 + 3.3 + 5.9 + 5.0 + 4.5 = 19.8

A, E, D, B, C, A

A, C, B, E, D, A

1.1 + 3.3 + 6.8 + 5.0 + 5.6 = 21.8

A, D, E, B, C, A

A, C, D, B, E, A

1.1 + 5.3 + 5.9 + 6.8 + 4.5 = 23.6

A, E, B, D, C, A

A, C, E, B, D, A

1.1 + 3.5 + 6.8 + 5.9 + 5.6 = 22.9

A, D, B, E, C, A

A, D, B, C, E, A

5.6 + 5.9 + 3.3 + 3.5 + 4.5 = 22.8

A, E, C, B, D, A

A, E, B, C, D, A

4.5 + 6.8 + 3.3 + 5.3 + 5.6 = 25.5

A, D, C, B, E, A

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An optimal tour would be A, B, D, E, C, A (or its reversal) with distance 19.1miles. 31.

Tour

Weight

Mirror-image Circuit

B, C, D, A, B

17 + 11 + 14 + 12 = 54

B, A, D, C, B

B, D, C, A, B

15 + 11 + 6 + 12 = 44

B, A, C, D, B

B, D, A, C, B

15 + 14 + 6 + 17 = 52

B, C, A, D, B

An optimal tour would be B, D, C, A, B (or its reversal) with a time of 44 hours. 32.

Tour

Weight

Mirror-image Circuit

C, D, A, B, C

7 + 16 + 4 + 17 = 44

C, B, A, D, C

C, A, B, D, C

18 + 4 + 13 + 7 = 42

C, D, B, A, C

C, B, D, A, C 17 + 13 + 16 + 18 = 64 C, A, D, B, C An optimal tour would be C, A, B, D, C (or its reversal) with a distance of 44 kilometers.

6.4

The Nearest-Neighbor and Repetitive Nearest-Neighbor Algorithms

33. (a) Starting at B, the nearest neighbor is vertex C with cost of $121 (compared to costs of $185 to vertex A, $150 to vertex D, and $200 to vertex E). Once at C, we check the cost of flying to vertices A, D, and E. Of these, flying from C to A is the cheapest at $119. From A, we check the cost of flying to D and E. Since E is cheapest, we fly from A to E ($133) and then from E to D ($199). Finally, we complete the circuit and fly back to vertex B at a cost of $150. Putting this all together gives the following circuit and cost: B, C, A, E, D, B; Cost = $121 + $119 + $133 + $199 + $150 = $722 (b) C, A, E, D, B, C; Cost = $119 + $133 + $199 + $150 + $121 = $722 (c) D, B, C, A, E, D; Cost = $150 + $121 + $119 + $133 + $199 = $722 (d) E, C, A, D, B, E; Cost = $120 + $119 + $152 + $150 + $200 = $741 Copyright © 2022 Pearson Education, Inc.

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34. (a) Starting at A, the nearest neighbor is vertex E with time of 21 minutes (compared to times of 25 minutes to vertex B, 40 minutes to vertex C, and 22 minutes to vertex D). Once at E, we check the travel time of driving to vertices B, C, and D. Of these, driving from E to B is the shortest time at 23 minutes. From B, we check the travel time to C and D. Since C is the shortest time, we drive from B to C (24 minutes) and then from C to D (23 minutes). Finally, we complete the circuit and drive back to vertex A at a travel time of 22 minutes. Putting this all together gives the following circuit and cost: A, E, B, C, D, A; Total travel time = 21 + 23 + 24 + 23 + 22 = 113 minutes (b) A, E, B, C, D, A; Total travel time = 113 minutes 35. (a) Starting at A, the nearest neighbor is vertex D with distance of 185 miles (compared to distances of 500 miles to vertex B, 200 miles to vertex C, and 205 miles to vertex E). Once at D, we check the distances of driving to vertices B, C, and E. Of these, driving from D to E is the shortest distance at 302 miles. From E, we check the distance to B and C. Since C is the shortest time, we drive from E to C (165 miles) and then from C to B (305 miles). Finally, we complete the circuit and drive back to vertex A at a distance of 500 miles. Putting this all together gives the following circuit and cost: A, D, E, C, B, A; Cost of bus tour = $8 × (185 + 302 + 165 + 305 + 500) = $11,656 (b) A, D, B, C, E, A Cost of bus tour = $8 × (185 + 360 + 305 + 165 + 205) = $9,760 36. (a) Starting at E, the nearest neighbor is vertex C with travel time of 3.1 years (compared to travel times of 3.2 years to vertex G, 3.6 years to vertex I, 8.2 years to vertex M, and 8.1 years to vertex T). Once at C, we check the travel times of visits to vertices G, I, M and T. Of these, traveling from C to I is the shortest time at 0.8 years. Continuing in this manner gives the following circuit and cost: E, C, I, G, M, T, E; Total Travel Time = 3.1 + 0.8 + 1.1 + 5.7 + 0.6 + 8.1 = 19.4 years (b) T, M, I, C, G, E, T; Total Travel Time = 8.1 + 0.6 + 4.7 + 0.8 + 1.5 + 3.2 = 18.9 years Written with starting vertex E, this is E, T, M, I, C, G, E. 37. (a) Starting at A, the nearest neighbor is vertex E with time of 18 hours (compared to times of 35 hours to vertex B, 40 hours to vertex C, and 22 hours to vertex D). Once at E, we check the travel time of driving to vertices B, C, and D. Of these, driving from E to B is the shortest time at 18 hours. From B, we check the travel time to C and D. Since C is the shortest time, we drive from B to C (9 hours) and then from C to D (25 hours). Finally, we complete the circuit and drive back to vertex A at a travel time of 22 hours. Putting this all together gives the following circuit and cost: A, E, B, C, D, A; Total travel time = 18 + 18 + 9 + 25 + 22 = 92 hours (b) A, D, B, C, E, A; Total travel time = 93 hours 38. (a) Starting at A, the nearest neighbor is vertex C with distance of 1.1 miles (compared to distances of 3.6 miles to vertex B, 5.6 miles to vertex D, and 4.5 miles to vertex E). Once at C, we check the distance to vertices B, D and E. Of these, driving from C to B is the shortest distance at 3.3 miles. From B, we check the distance to D and E. Since D is the closest, we move from B to D (5.9 miles) and then from D to E (5.0 miles). Finally, we complete the circuit and move back to vertex A at a distance of 4.5 miles. Putting this all together gives the following circuit and cost: A, C, B, D, E, A; Total distance = 1.1 + 3.3 + 5.9 + 5.0 + 4.5 = 19.8 miles (b) A, B, D, E, C, A; Total distance = 19.1 miles 39. (a) The smallest number in the Atlanta column is for Columbus; the smallest number other than Atlanta in the Columbus column is for Kansas City; the smallest number in the Kansas City column other than Atlanta and Columbus is Tulsa, etc.. So the nearest-neighbor tour for Darren is Atlanta, Columbus, Kansas City, Tulsa, Minneapolis, Pierre, Atlanta. The cost of this trip = $0.75 × (533 + 656 + 248 + 695 + 394 + 1361) = $2915.25.

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(b) The nearest-neighbor circuit with Kansas City as the starting vertex is Kansas City, Tulsa, Minneapolis, Pierre, Columbus, Atlanta, Kansas City. Written with starting city Atlanta, the circuit is Atlanta, Kansas City, Tulsa, Minneapolis, Pierre, Columbus, Atlanta. The cost of this trip = $0.75 × (798 + 248 + 695 + 394 + 1071 + 533) = $2804.25 40. (a) Nashville, Louisville, St. Louis, Pittsburgh, Boston, Dallas, Houston, Nashville Total length = 168 + 263 + 588 + 561 + 1748 + 243 + 769 = 4340 miles (b) St. Louis, Louisville, Nashville, Pittsburgh, Boston, Dallas, Houston, St. Louis Written with starting city Nashville, this is Nashville, Louisville, St. Louis, Houston, Dallas, Boston, Pittsburgh, Nashville (or its mirror-image) Total length = 168 + 263 + 779 + 243 + 1748 + 561 + 553 = 4315 miles 41.

Starting Vertex

Tour

Length of Tour

A, E, B, C, D, A

18 + 18 + 9 + 25 + 22 = 92

B, C, E, A, D, B

9 + 23 + 18 + 22 + 21 = 93

C, B, E, A, D, C

9 + 18 + 18 + 22 + 25 = 92

D, A, E, B, C, D

22 + 18 + 18 + 9 + 25 = 92

E, A, D, B, C, E

18 + 22 + 21 + 9 + 23 = 93

E, B, C, D, A, E 18 + 9 + 25 + 22 + 18 = 92 Starting with vertex A, the shortest circuit is A, E, B, C, D, A (or its reversal) with cost of 92 hours. 42.

Starting Vertex

Hamilton Circuit

Length of Tour

A, C, B, D, E, A

1.1 + 3.3 + 5.9 + 5.0 + 4.5 = 19.8

B, C, A, E, D, B

3.3 + 1.1 + 4.5 + 5.0 + 5.9 = 19.8

C, A, B, D, E, C

1.1 + 3.6 + 5.9 + 5.0 + 3.5 = 19.1

D, E, C, A, B, D

5.0 + 3.5 + 1.1 + 3.6 + 5.9 = 19.1

E E, C, A, B, D, E 3.5 + 1.1 + 3.6 + 5.9 + 5.0 = 19.1 Starting with vertex A, the shortest circuit is A, B, D, E, C, A (or its reversal) with cost of 19.1 miles.

43.

Starting Vertex

Tour

Length of Tour (Miles)

A, C, K, T, M, P, A

533 + 656 + 248 + 695 + 394 + 1361 = 3887

C, A, T, K, M, P, C

533 + 772 + 248 + 447 + 394 + 1071 = 3465

K, T, M, P, C, A, K

248 + 695 + 394 + 1071 + 533 + 798 = 3739

M, P, K, T, A, C, M

394 + 592 + 248 + 772 + 533 + 713 = 3252

P, M, K, T, A, C, P

394 + 447 + 248 + 772 + 533 + 1071 = 3465

T, K, M, P, C, A, T

248 + 447 + 394 + 1071 + 533 + 772 = 3465

The shortest circuit is M, P, K, T, A, C, M. Written starting from Atlanta, this is Atlanta, Columbus, Minneapolis, Pierre, Kansas City, Tulsa, Atlanta and has a length of 3252 miles. The total cost of this tour is $0.75 × 3252 = $2439.

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44.

Chapter 6: The Mathematics of Touring

Starting Vertex

Hamilton Circuit

Length of Circuit (Miles)

B, P, L, N, S, D, H, B

4093

P, L, N, S, D, H, B, P

4093

L, N, S, P, B, D, H, L

4535

N, L, S, P, B, D, H, N

4340

S, L, N, P, B, D, H, S

4315

D, H, N, L, S, P, B, D

4340

H, D, S, L, N, P, B, H

4222

The shortest circuit is B, P, L, N, S, D, H, B. Written starting from Nashville, this is Nashville, St. Louis, Dallas, Houston, Boston, Pittsburgh, Louisville, Nashville and has a length of 4093 miles. 45. The relative error of the nearest-neighbor tour is ε = (W − Opt ) Opt where W is the weight of the nearestneighbor tour and Opt is the weight of the optimal tour. In this case, we see that this error is ε = (W − Opt ) Opt = ( $13,500 − $12, 000 ) $12, 000 = 12.5%. 46. ε = (W − Opt ) Opt = ( 21, 400 − 20,100 ) 20,100 = 6.5%

6.5

Cheapest-Link Algorithm

47. The cheapest-link tour is created by adding edges BC, AE, BE, AD and CD in that order.

Edge BC is added to the tour first since it is the cheapest edge in the graph (weight of 9). Edges AE and BE are the next cheapest edges (weights of 18) and are added next. Edge BD is the next cheapest edge but it is not added to the tour since then three edges would meet at vertex B. Edge AD is added fourth with weight 22. Edge CE is next cheapest but it is not added to the tour since then three edges would meet at vertex E. Edge CD is added at this point and completes the tour. The shortest tour using this algorithm is A, D, C, B, E, A (or its reversal). The cost of this tour is 22 + 25 + 9 + 18 + 18 = 92 hours. 48. The cheapest-link tour is created by adding edges AC, BC, BE, AD and CD in that order (see below).

Edge AC is added to the tour first since it is the cheapest edge in the graph (weight of 1.1). Edge BC is the next cheapest edge (weight of 3.3) and is added next. Edge CE is the next cheapest edge but it is not added to the tour since then three edges would meet at vertex C. Edge AB is the next cheapest edge, but it too is not added to the tour since then a circuit would be formed. Edge AE is added third with weight 4.5. Edge DE is added next with weight 5.0. At this point, we add the last remaining edge, BD, needed to complete the tour. The shortest tour using this algorithm is A, C, B, D, E, A (or its reversal). The cost of this tour is 1.1 + 3.3 + 5.9 + 5.0 + 4.5 = 19.8 miles. Copyright © 2022 Pearson Education, Inc.

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49. The first three steps in the cheapest-link algorithm are illustrated in the following figures.

After the third step, the next cheapest edge is AE (weight 205). But that edge would make three edges come together at vertex A so we skip it and move to the next cheapest edge which is DE (weight 302). Since this edge makes a circuit, we skip this edge and try the next cheapest edge. The next cheapest edge is BC (weight 305) but that makes three edges come together at vertex C so we skip this edge and choose the next cheapest edge which is CD (weight 320). That edge makes a circuit and also makes three edges come together at vertex C. So, we skip it and choose the next cheapest edge which is BE (weight 340). After BE is added, only one edge, BD, will complete the tour.

The shortest tour found using the cheapest-link algorithm is A, D, B, E, C, A. The cost of this bus tour is $8 × (165 + 185 + 200 + 340 + 360) = $10,000. 50. The cheapest-link tour is created by adding edges DE, CD, AB, AC and BE in that order. Edges DE and CD are added to the tour first since these are the cheapest two edges in the graph. Edge AD is the next cheapest edge. However, it is not added to the tour since then three edges would meet at vertex D. So, AB is the third edge added to the tour.

Edge AC is next in line (i.e. next smallest weight) to be added and doing so violates no rules in the cheapestlink algorithm. To complete the tour at this point we have no choice but to add edge BE (having weight 3.8).

The shortest tour found using this algorithm is B, E, D, C, A, B. The weight of this tour is 2.2 + 2.4 + 3.2 + 3.3 + 3.8 = 14.9.

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51.

Link Added to Tour

Cost of Link (Miles)

Kansas City - Tulsa

248

Pierre - Minneapolis

394

Minneapolis - Kansas City

447

Atlanta - Columbus

533

Atlanta - Tulsa

772

Columbus - Pierre

1071

Darren’s cheapest-link tour is Atlanta, Columbus, Pierre, Minneapolis, Kansas City, Tulsa, Atlanta. His total mileage is 248 + 394 + 447 + 533 + 772 + 1071 = 3465 miles. His total cost is $0.75 × 3465 = $2598.75. 52.

Link Added to Circuit

Cost of Link (Miles)

Nashville - Louisville

168

Dallas - Houston

243

Louisville – St. Louis

263

Pittsburgh - Nashville

553

Boston - Pittsburgh

561

St. Louis - Dallas

630

Houston - Boston

1804

The circuit is Nashville, Louisville, St. Louis, Dallas, Houston, Boston, Pittsburgh, Nashville. Total mileage = 168 + 243 + 263 + 553 + 561 + 630 + 1804 = 4222 miles. 53. (a)

Link Added to Tour

Cost of Link (Days)

A-E

A-F

C-F

D-E

B-D

B-C 20 The rover’s cheapest-link tour is A, E, D, B, C, F, A. The length is 9 + 12 + 13 + 20 + 11 + 10 = 75 days. 54. (a)

Link Added to Tour

Cost of Link (Seconds)

A-C

0.7

B-D

0.8

C-E

0.8

D-E

0.9

ISM: Excursions in Modern Mathematics, 10E (b) The cost of the optimal tour is Opt = 1.0 + 0.8 + 1.1 + 0.8 + 0.7 = 4.4. So, the relative error is ε = (W − Opt ) / Opt = 0 / 4.4 = 0% . (c) The cost of the optimal tour is Opt = 11 + 13 + 12 + 13 + 11 + 10 = 70. So, the relative error is ε = (W − Opt ) Opt = ( 75 − 70 ) 70 ≈ 7.14%

JOGGING 55. (a) 7! = 5040 (b) 7! = 5040; this is the same as in part (a). The remaining seven letters can be rearranged in any sequence.

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Office, Deli, Hospital, Karl’s , Home. The total length of the tour is 2 + 6 + 7 + 4 + 6 + 5 = 30 miles. 60. (a) Treat the guests as the vertices and let the friendships be represented by edges.

(b) Any Hamilton circuit gives a possible seating arrangement. One possibility is A, B, C, D, E, F.

(c) 6! = 720; the remaining six letters can be arranged in any sequence. 56. (a) 9! = 362,880 (b) 9! = 362,880; the remaining nine letters can be rearranged in any sequence.

(c) 8! = 40,320; the remaining eight letters can be arranged in any sequence. 57. We solve ( $1614 − Opt ) Opt = 7.6% for Opt where Opt is the cost of the optimal solution. But this gives $1614 − Opt = 0.076 ⋅ Opt or $1614 = 0.076 ⋅ Opt + Opt . So, Opt =

$1614 = $1500 . 1.076

61. The graph describing the friendships among the guests does not have a Hamilton circuit. Thus it is impossible to seat everyone around the table with friends on both sides.

58. We solve ( $2508 − Opt ) Opt = 4.5% for Opt.

But this gives $2508 − Opt = 0.045 ⋅ Opt or $2508 = 0.045 ⋅ Opt + Opt . So, Opt =

$2508 = $2400 . 1.045

62. 1001; K 501 has 500 more edges than K 500 (the difference between these two graphs is that in K 501 the 501st vertex has an edge extending to each of the other 500 vertices). Similarly K 502 has 501 more edges than K 501 . So, K 502 has 500 + 501 = 1001 more edges than K 500 .

59. (a)

(b) Just “eyeballing it” will give the optimal tour in this case. It is: Home, Bank, Post

63. Suppose the cheapest edge in a graph is the edge joining vertices X and Y. Using the nearest neighbor algorithm we will eventually

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visit one of these vertices—suppose the first one of these vertices we visit is X. Then, since edge XY is the cheapest edge in the graph and since we have not yet visited vertex Y, the nearest neighbor algorithm will take us to Y. 64. (a) Any circuit would need to cross the bridge twice (in order to return to where it started). But then each vertex at the end of the bridge would be included twice. (b) The following graph has a Hamilton path from B to C.

RUNNING 65. (a) Julie should fly to Detroit. The optimal route will then be to drive 68 + 56 + 68 + 233 + 78 + 164 + 67 + 55 = 789 miles via Detroit-Flint-Lansing-Grand RapidsCheboygan-Sault Ste. Marie-MarquetteEscanaba-Menominee for a total cost of 789 × $0.39 + $3.00 = $310.71. Since Julie can drive from Menominee back to Detroit (via Sault Ste. Marie and Cheboygan) in a matter of 227 + 78 + 280 = 585 miles at a cost of 585 × ($0.39) + $3.00 = $231.15, she should do so and drop the rental car back in Detroit (assuming she need not pay extra for gas!). The total cost of her trip would then be $310.71 + $231.15 = $541.86. (b) The optimal route would be for Julie to fly to Detroit and drive 68 + 56 + 68 + 233 + 78 + 164 + 67 + 55 = 789 miles along the Hamilton path Detroit-FlintLansing-Grand Rapids-Cheboygan-Sault Ste. Marie-Marquette-EscanabaMenominee for a total cost of 789 × $0.49 + $3.00 = $389.61.

66. (a) & (b)

(c) We can color the vertices of the grid graph with two colors (black and white) so that adjacent vertices have different colors. This implies that any circuit in this graph must consist of vertices of alternating colors and have the same number of vertices of each color. This, of course, implies that the circuits in this graph must have an even number of vertices. An m-by-n grid graph has a total of (m + 1) × (n + 1) vertices. If m and n are both even then the number of vertices in the graph is odd and it follows that the graph cannot have a Hamilton circuit. 67. (a) If a1 , a2 , a3 , an are the vertices in set A and b1 , b2 , b3 , bn are the vertices in set B, then one possible Hamilton circuit is a1 , b1 , a2 , b2 , an , bn , a1 . (b) Suppose a1 , a2 , a3 , an are the vertices in set A and b1 , b2 , b3 , bn +1 are the vertices in set B, then one possible Hamilton path is b1 , a1 , b2 , a2 , bn , an , bn +1 . (c) If a1 , a2 , a3 , am are the vertices in set A and b1 , b2 , b3 , bn are the vertices in set B, then in any Hamilton path the a’s and b’s must alternate. This implies that either (i) there is the same number of a’s and b’s (m = n); (ii) there is one more of the a’s than there is of the b’s (m = n + 1); or (iii) there is one more of the b’s than there is of the a’s (n = m + 1). There are no other possibilities. 68. (a) The degree of each vertex in a complete bipartite graph K n , n is n. So, the sum of the degrees of any two vertices X and Y is 2n. But K n , n has 2n vertices. (b) Suppose that m < n. Then, there are at least two vertices X and Y (in vertex set B

ISM: Excursions in Modern Mathematics, 10E having n vertices) having degree m. Then, deg(X) + deg(Y) = m + m = 2m < m + n. Since K m , n has m+n vertices, the result follows.

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69. If deg( X ) ≥ N / 2 for every vertex X, then deg( X ) + deg(Y ) ≥ N / 2 + N / 2 = N for every pair of vertices X and Y.

APPLET BYTES 70. (a) Many Hamilton circuits of this graph exist. Below is one possibility.

(d) Consider the edges in the cut disconnecting the inner 5-cycle from the outer one. If there is a Hamilton circuit, an even number of these edges must be chosen. If only two of them are chosen, their end-vertices must be adjacent in the two 5-cycles, which is not possible. Hence 4 of them are chosen. Assume that the top edge of the cut is not chosen (all the other cases are the same by symmetry). Of the 5 edges in the outer cycle, the two top edges must be chosen, the two side edges must not be chosen, and hence the bottom edge must be chosen. The top two edges in the inner cycle must be chosen, but this completes a non-spanning cycle, which cannot be part of a Hamilton circuit. 71. (a) Many Hamilton circuits of this graph exist. Below is one possibility.

(b) One Hamilton circuit of this graph is shown below.

(b) Suppose we color the vertices of the grid graph with two colors [say black (B) and white (W)] with adjacent vertices having different colors (see figure).

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Since there are 5 black vertices and 4 white vertices and in any Hamilton path the vertices must alternate color, the only possible Hamilton paths are of the form B,W,B,W,B,W,B,W,B. (c) Since every boundary vertex is white, part (b) shows that it is impossible to end a Hamilton path on a boundary vertex.

height at the bottom of the original middle square. Finally, travel right to the middle square and up one unit to complete the circuit. (d) One Hamilton path is shown below. Essentially, “cycling” around will do the trick.

72. (a) Many Hamilton circuits of this graph exist. Below are two possibilities.

(b) One Hamilton circuit of this graph is shown below.

(c) For an n by n grid where n is odd, a Hamilton circuit can be constructed as follows: Focus on the “middle” square. At the left top of that middle square in the grid, travel left – all the way to the edge of the grid. Then, move up (to top) and down (as far as possible without crossing the existing edge), then up (to the top again) and down (just as far as before), etc. The top of the graph will have n-2 “tentacles” on the top. When at the right edge, travel down to level with the starting vertex. Move left as far as possible (to the middle square in the grid). Then, on that middle square, move down one vertex to the lower right of the middle square. Travel right to the right edge of the grid. Then, move down (all the way to the bottom of the grid), then left one vertex, then up as far as possible, left one vertex, then down as far as possible, and continue in that way until you travel up the left edge to the

(e) A path similar to that in part (d) will work where one takes a clockwise spiral path around the grid until the center vertex is reached. (f) Suppose we color the vertices of the grid graph with two colors [say black (B) and white (W)] with adjacent vertices having different colors. Since there are 13 black vertices and 12 white vertices and in any Hamilton path the vertices must alternate color, the only possible Hamilton paths are of the form B,W,B,W,…,B,W,B. To form a Hamilton circuit, one would then need to immediately return to the original black vertex. That is not possible since adjacent vertices have different colors. (g) The same argument for an n by n grid applies here. Such a grid would contain (n + 1)(n + 1) = n 2 + 2n + 1 vertices. This is clearly an odd number if n is even. So the grid has one more black vertex than white vertices. 73. (a) Starting at Callisto (C), the nearestneighbor algorithm gives the tour E, T, M, C, I, G, E with total length of 19.0.

ISM: Excursions in Modern Mathematics, 10E Starting at Io (I), the nearest-neighbor algorithm gives the tour E, T, M, I, C, G, E with total length of 18.9.

Starting at Ganymede (G), the nearestneighbor algorithm gives the tour E, T, M, G, I, C, E with total length of 19.4.

Starting at Mimas (M), the nearestneighbor algorithm gives the tour E, M, T, I, C, G, E with total length of 19.4.

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algorithm gives the tour E, T, M, I, C, G, E with total length of 18.9.

The tour with the least total length is E, T, M, I, C, G, E with total length of 18.9. (b) Another optimal tour is E, C, T, M, I, G, E with total length of 18.3.

74. (a) The nearest-neighbor tour using G1 as the starting vertex (G1-G6-G7-G3-G2-G4G5-G1 with weight 20,100) is shown below.

Starting at Earth (E), the nearest-neighbor algorithm gives the tour E, C, I, G, M, T, E with total length of 19.4.

(b) The optimal tour for the Curiosity rover is #410 with weight of 20,100.

Starting at Titan (T), the nearest-neighbor

Chapter 7 4. (a) 2; the shortest path connecting D and A has length 2 (DC, CA).

WALKING 7.1

Networks and Trees

1. No, the graph is disconnected. Vertices A, D, and G are connected to each other but not to the other four vertices. Looking at the graph one way, this may be difficult to see.

However, looking at a representation of the graph in another fashion, this is much easier to see.

(b) 4; the shortest path connecting A and L has length 4 (AC, CD, DG, GL). (c) 4; the shortest path connecting B and H has length 4 (BE, ED, DG, GH). 5. (a) Several answers are possible. For example, vertices A and G have five degrees of separation (AB, BC, CE, EF, FG). So do vertices A and I (AB, BC, CE, EF, FI). Vertices B and H also have five degrees of separation (BC, CE, EF, FG, GH). So do K and H. (b) Only vertices A and H have six degrees of separation (AB, BC, CE, EF, FG, GH). (c) There are not any. A and H are two vertices farthest apart in the network (having six degrees of separation).

2. No, the graph is disconnected. Vertices A, D, F, and H are connected to each other but not to the other vertices. Looking at the graph one way, this may be difficult to see.

However, looking at a representation of the graph in another fashion, this is much easier to see.

3. (a) 1; the shortest path connecting C and E has length 1 (they are connected to each other). (b) 3; the shortest path connecting A and E has length 3 (AB, BC, CE). (c) 4; the shortest path connecting A and F has length 4 (AB, BC, CE, EF).

(d) 6; the largest degree of separation between pairs of vertices is that between vertices A and H. 6. (a) Several answers are possible. For example, vertices A and I have five degrees of separation (AC, CD, DG, GH, HI). So do vertices A and K (AC, CD, DG, GL, LK). Vertices C and J also have five degrees of separation (CD, DG, GH, HI, IJ). So do E and J. (b) Vertices A and J have six degrees of separation (AC, CD, DG, GH, HI, IJ). So do vertices F and J (FE, ED, DG, GH, HI, IJ). Vertices B and J also have six degrees of separation. (c) There are not any. A and J are two vertices farthest apart in the network (having six degrees of separation). This maximum degree of separation, 6, also occurs between vertices F and J and between vertices B and J. (d) 6; the largest degree of separation between pairs of vertices is that between vertices A and J (which is the same as that between vertices F and J and between B and J).

ISM: Excursions in Modern Mathematics, 10E 7. (a) N and J; O and J; O and K. Note that six is a large separation in this network so that choosing vertices that appear to be far apart is how one should approach this problem. Vertices N and O appear to be on one “end” of the network and vertices J and K appear to be on the other “end.” (b) There are not any. The largest possible separation is between N (or O) and the cluster of vertices H, I, J, K, L, and M at the bottom right. To get to any of those vertices, one must go through H. That takes four links. From H it takes only one link to get to I, L, and M and two links to get to J and K. (c) 6; the largest degree of separation between pairs of vertices is that between vertices N and K (which is the same as that between vertices N and J, between O and J, and between O and K). 8. (a) B and J; B and K. The additional link (edge GH) shortens the path between vertices N and O and the cluster H, I, J, K, L, and M. It takes three edges to get from B to H (H must be part of a path to any vertex in this aforementioned cluster) and two edges to get from H to either J or K. Every other vertex outside this cluster can reach H with at most two links. (b) There are not any. To get to the cluster H, I, J, K, L, and M at the bottom right from a vertex not in that cluster, one must pass through vertex H. From H, one can reach any vertex in the cluster in at most two links. All vertices not in the cluster except B are either one or two degrees of separation from H. B is three degrees of separation from H. (c) 5; the largest degree of separation between pairs of vertices is that between the vertex B (not in the cluster H, I, J, K, L, and M at the bottom right) and either vertex J or K located in the cluster.

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(c) 8; the shortest path connecting M and P is MJ, JE, EB, BA, AD, DG, GL, LP. (d) 8; the largest degree of separation between A and any vertex in the top of this tree is 4 (from A to M, N, or O) and the largest degree of separation between A and any vertex in the bottom of this tree is also 4 (AD, DG, GL, LP). 10. (a) 5; the shortest path connecting A and P is AD, DG, GK, KN, NP. (b) 7; the shortest path connecting E and P is EB, BA, AD, DG, GK, KN, NP. (c) 9; the shortest path connecting L and P is LI, IE, EB, BA, AD, DG, GK, KN, NP. (d) 9; the largest degree of separation between L and any vertex in the top of this tree is 4 (from A to L) and the largest degree of separation between A and any vertex in the bottom of this tree is 5 (from A to P). 11. (B) In a tree, the number of edges must be one less than the number of vertices. However, here the number of edges is one more than the number of vertices. So the network violates the N-1 edges property of trees. 12. (B) In a tree, every edge must be a bridge. However, this network has no bridges. 13. (A) Since the network has one fewer edge than vertices, the network satisfies the N – 1 edges property. 14. (A) Since each edge is a bridge, this network satisfies the all-bridges property of trees. 15. (B) If R = 1, then M = N (the number of edges is the same as the number of vertices). So, the network is not a tree since it violates the N – 1 edges property. 16. (A) If R = 0, then M = N -1(the number of edges M is one less than the number of vertices N). So, the network is a tree.

9. (a) 3; the shortest path connecting A and J is AB, BE, EJ. (b) 5; the shortest path connecting E and L is EB, BA, AD, DG, GL.

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17. (C) We do not know if other pairs of vertices are connect by single or multiple paths. For example, the following network is a tree. On the other hand, the following network also has two vertices of odd degree but is not a tree. However, the following network, also having only one path connecting A and J, is not a tree.

7.2 Spanning Trees, MSTs, and MaxSTs 25. (a) There are many spanning trees for this network. Below is one. 18. (B) The network violates the single-path property of trees. 19. (B) The network has 5 vertices and 10 edges (every edge produces two vertex degrees and there are 20 total degrees at the vertices). Since there are not 5 - 1 = 4 edges, the network violates the N – 1 edges property of trees. 20. (B) The network has 5 vertices and 5 edges (every edge produces two vertex degrees and there are 10 total degrees at the vertices). Since there are not 5 - 1 = 4 edges, the network violates the N – 1 edges property of trees. 21. (A) The network has 5 vertices and 4 edges (every edge produces two vertex degrees and there are 4 + 4 = 8 total degrees at the vertices). The network satisfies the N – 1 edges property of trees. 22. (A) The network has 5 vertices and 4 edges (every edge produces two vertex degrees and there are 2 + 3(2) = 8 total degrees at the vertices). The network satisfies the N – 1 edges property of trees. 23. (B) The network has an Euler Circuit (see Section 5.3). Since it has a circuit, it cannot be a tree.

(b) Since this network has N = 5 vertices and M = 10 edges, the redundancy is R = M – (N – 1) = 10 – (5 – 1) = 6. (c) 1; each vertex is connect to every other vertex by an edge. 26. (a) Many answers are possible. Below is one.

(b) R = M – (N-1) = 24 – (16-1) = 9 (c) 6; the shortest path connecting vertices A and G, for example, is AB, BC, CD, DE, EF, FG. 27. (a) There is only one spanning tree for this network since it is a tree. The spanning tree is the tree itself.

24. (C) This network has a Euler path (see Section 5.3). Some trees have Euler paths; some don’t. For example, the following network has two vertices of odd degree and three of even degree. It is also a tree.

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123

(b) Since this network has N = 10 vertices and M = 9 edges, the redundancy is R = M – (N – 1) = 9 – (10 – 1) = 0. (c) 6; the shortest path connecting vertices C and J, for example, is CA, AB, BD, DE, EG, GJ. [Note: C and I are also separated by this degree.] 28. (a) The graph itself forms the only spanning tree (since it is itself a tree).

(c) 6; Any one of the edges DI, IJ, JE, EK, KL, or LD could be deleted to form a spanning tree. 30. (a) Any one of the edges AB, BC, or AC could be deleted to form a spanning tree.

(b) R = M – (N-1) = 12 – (13-1) = 0. (c) 7; the shortest path connecting vertices J and M, for example, is JF, FC, CB, BD, DG, GK, KM. [Note: J and L are also separated by this degree.] 29. (a) Any one of the three edges EI, DI, or DE could be deleted to form a spanning tree. (b) Any one of the edges AB, BC, CD, or AD could be deleted to form a spanning tree.

(b) Any one of the four edges DJ, EJ, EI, or DI could be deleted to form a spanning tree. (c) 6; Any one of the edges AB, BC, CD,DE,EF or AF could be deleted to form a spanning tree. 31. (a) Each spanning tree excludes one of the edges AB, BC, CA and one of the edges DI, IE, EJ, JD so there are 3 × 4 = 12 different spanning trees. (b) Each spanning tree excludes one of the edges AB, BC, CA, one of the edges DI, IJ, JE, EK, KL, LD, and one of the edges FG, GH, HF, so there are 3 × 6 × 3 = 54 different spanning trees. Copyright © 2022 Pearson Education, Inc.

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32. (a) Each spanning tree excludes one of the edges AB, BM, ML, or AL and one of the edges OF, FG, GH, or HO so there are 4 × 4 = 16 different spanning trees. (b) 4 × 4 × 4 × 4 × 4 = 1024 33. (a) 6 + 2 × 4 = 14 ; The key is to consider two cases: either DF is part of a spanning tree or it isn’t. There are 6 spanning trees that do not contain edge DF. They can be found by removing DF along with one of the 6 edges of the circuit A, B, C, D, E, F, A.

So, the number of different spanning trees is 6 + 8 = 14. (b) Certainly the edge FE should not be included. Not including edge FD also lengthens the path from H to G.

(c) Several answers are possible. The key idea is to be sure that edge FE is included so that the degree of separation between H and G is only 3.

There are 8 spanning trees that do contain edge DF. They can be found by removing one of the 4 edges other than DF of the circuit A, B, C, D, F, A along with one of the 2 edges (again, not DF) of the circuit D, E, F, D. Since there are 4 ways to do the former and 2 ways to do the latter, there are a total of 4 × 2 = 8 spanning trees containing the edge DF.

34. (a) 8 + 3 × 5 = 23 ; The key is to consider two cases: either CI is part of a spanning tree or it isn’t. There are 8 spanning trees that do not contain edge CI. They can be found by removing CI along with one of the 8 edges of the circuit B, C, D, E, G, H, I, K, B. Here are two examples of these types of trees:

ISM: Excursions in Modern Mathematics, 10E There are 15 spanning trees that do contain edge CI. They can be found by removing one of the 3 edges other than CI of the circuit C, B, K, I, C along with one of the 5 edges (again, not CI) of the circuit CD, DE, EG, GH, HI, IC. Since there are 3 ways to do the former and 5 ways to do the latter, there are a total of 3 × 5 = 15 spanning trees containing the edge CI. Below are two of these 15 possibilities:

7.3

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Kruskal's Algorithm

35. A spanning tree for the 20 vertices will have exactly 19 edges and so the cost will be 1 mile $40, 000 19 edges × × = $380, 000 . 2 edges 1 mile 36. A spanning tree for the 30 vertices will have exactly 29 edges and so the cost will be 1 mile $40, 000 29 edges × × = $580, 000 . 2 edges 1 mile 37. Add edges to the tree in the following order: EC, AD, AC, BC.

So, the number of different spanning trees is 8 + 15 = 23. (b) Certainly the edge HI should not be included. Not including edge CI also lengthens the path from H to J. This forces the path to go ‘around’ as much as possible.

Weight = 165 + 185 + 200 + 305 = 855 38. Add edges to the tree in the following order: DE, CD, AD, AB.

Weight = 2.2 + 2.4 + 3.1 + 3.2 = 10.9. (c) Several answers are possible. The key idea is to be sure that edges HI, HG, and KI are included so that the degree of separation between K and G is only 3.

39. Add edges to the tree in the following order: DC, EF, EC, AB, AC.

Weight = 1.2 + 1.8 + 2.0 + 2.1 + 2.2 = 9.3

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40. Several MSTs are possible. One example:

Weight = 30 + 10 + 10 + 30 + 20 + 10 + 20 + 15 + 10 + 10 + 15 + 30 + 25 + 30 + 15 = 280 41. Add edges to the tree in the following order: AB, BD, BE, CD.

Weight = 35 + 35 + 30 + 30 + 30 + 30 + 30 + 30 + 30 + 30 + 20 + 20 + 20 + 15 + 15= 400

JOGGING 45. Add edges to the tree in the following order: Kansas City – Tulsa (248), Pierre – Minneapolis (394), Minneapolis – Kansas City (447), Atlanta – Columbus (533), Columbus – Kansas City (656).

Weight = 500 + 360 + 340 + 320 = 1520

46. (a) The MST is as the same as given in the figure below. If x > 59, then edge EG will not replace any edge in the existing MST since it has a larger value.

42. Add edges to the tree in the following order: BD, CE, BE, AE. [Note: After the first three edges are added, edge BC is not next since it would create a circuit.]

(b) If x < 59, then edge EG will replace edge CE in the MST. Weight = 4.5 + 4.1 + 3.8 + 3.4 = 15.8 43. Add edges to the tree in the following order: DF, BD, AE, BE, BC. Note that edges such as BF, DE, and AD are not added to the MaxST since they would form a circuit. Instead, edge BC is the best you can do to involve vertex C.

47. (a) k = 5, 2, 1, 0 are all possible. The network could have no circuits (k=5), or a circuit of length 3 (k=2), Weight = 3.1 + 3.0 + 2.9 + 2.8 + 2.4 = 14.2 44. Add edges to the tree in the following order: KP, OF, AB, CD, AL, DE, KJ, FG, JI, HG, LM, NE, PO, BC, LK. Note: Other MaxSTs are possible.

or a circuit of length 4 (k=1),

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or a circuit of length 5 (k=0).

(b) Using the same pattern as in (a), values of k = 123 and 0 ≤ k ≤ 120 are all possible. 48. The maximum redundancy occurs when the degree of each vertex is N-1. In that case (a complete graph), the number of edges is (N-1) + (N-2) + …+ 3 + 2 + 1 = N(N-1)/2. So, the redundancy is R = N(N-1)/2 – (N-1) = (N1)(N/2 – 1) = ( N 2 − 3 N + 2 ) / 2 . 49. At each step of Kruskal’s algorithm we choose (from the available edges that do not close circuits) the edge of least weight. When there is only one choice at each step, there is only one MST possible. The same argument applies to a MaxST. 50. (a) The network

51. Step 1 of Kruskal’s algorithm (which always produces an MST) will always choose edge XY. There is no other possibility when the value of edge XY is a unique minimum. 52. (a) Each component is a tree (it has no circuits and is connected). So each component with x edges has x+1 vertices. Since the total number of vertices is 3 more than the total number of edges, we see that there are 3 components. (b) 4. Since the graph has no circuits, each component must be a tree (as in (a)). Thus, in each component the number of edges is one less than the number of vertices. Since N – M = 240 – 236 = 4, there must be 4 components in the graph.

has exactly one MST:

53. Since the graph has no circuits, each component has to be a tree. Thus, in each component the number of edges is one less than the number of vertices. It follows that M = N – K.

Note that this was constructed by taking a simple cycle with all different weights and gluing on “leaves” with duplicate weights.

54. (a) For the complete graph with 4 vertices, Cayley's theorem predicts 4 4 −2 = 16 spanning trees.

(b) The network

was constructed from a 4-cycle with three larger weights and one smaller weight. Note that it has 3 MSTs. The MSTs are:

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The complete graph with N vertices has N −2 (N – 1)! Hamilton circuits and N spanning trees. Since (N – 1)! = 2 × 3 × 4 ×  × ( N − 1) and N −2

N = N × N × N × ... × N , and both expressions have the same number of factors, with each factor in (N – 1)! smaller than the corresponding factor in N −2 N , we know that for N ≥ 3 , N−2 (N −1)!< N .

RUNNING

2e ≥ t + K + 2 ( v − t − 1) = t + K + 2 ( e − t ) , or 2e ≥ t + K + 2e − 2t which means t ≥ K .

57. Pick any vertex v in a tree T. Since T is a tree, there is one and only one path joining v with any other given vertex. So, let A be the set of vertices in T that are joined to v by a path of even length and let B be the set of vertices in T that are joined to v by a path of odd length. This shows that T is bipartite. Every edge of the graph T joins a vertex from A to a vertex from B. 58. (a) The network

55. (a) 2N - 2 A tree with N vertices has N – 1 edges, and, in any graph, the sum of the degrees of all the vertices is twice the number of edges. (b) Let v be the number of vertices in the graph, e the number of edges, and k the number of vertices of degree 1. Recall that in a tree v = e + 1 and in any graph the sum of the degrees of all the vertices is 2e. Now, since we are assuming there are exactly k vertices of degree 1, the remaining v – k vertices must have degree at least 2. Therefore, the sum of the degrees of all the vertices must be at least k + 2(v – k). Putting all this together we have 2e ≥ k + 2(v – k) = k + 2(e + 1 – k), 2e ≥ k + 2e + 2 – 2k, k ≥ 2. (c) By part (b), all vertices would need to be of degree 1. But then the graph would be disconnected. 56. Let v be the number of vertices in the graph, e the number of edges, and t the number of vertices of degree 1. Recall that in a tree v = e + 1 and in any graph the sum of the degrees of all the vertices is 2e. Now, since we are assuming there is a vertex of degree K and exactly t vertices of degree 1, the remaining v – t – 1 vertices must have degree at least 2. Therefore the sum of the degrees of all the vertices must be at least t + K + 2(v – t – 1). Putting all this together we have

has exactly one edge XY with largest weight and has two MSTs:

(b) The network

has more than one MST and edge XY is in none of the MSTs. The MSTs are:

59. Start by marking the required edge in red. Proceed with Kruskal’s algorithm as usual.

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APPLET BYTES 60. (a)(b)

61. (a)(b)

(c) The total price of the MaxST would be $54 + $46 + $39 + $35 + $34 + $33 + $32 + $32 + $31 + $31 + $30 + $26 + $23 = $446 million.

Chapter 8 WALKING 8.2

Directed Graphs

1. (a) indeg(A) = 3; outdeg(A) = 2; there are three arcs, denoted by arrows, that come into vertex A (these are from vertices B, C, and E) and two arcs, once again denoted by arrows, that leave vertex A (for vertices B and C). (b) indeg(B) = 2; outdeg(B) = 2 (c) indeg(D) = 3; outdeg(D) = 0 (d) 3 + 2 + 1 + 3 + 1 = 10; indeg(A) = 3, indeg(B) = 2, indeg(C) = 1, indeg(D) = 3, indeg(E) = 1. Note that the sum of the indegrees matches the number of arcs in the digraph. (e) 2 + 2 + 4 + 0 + 2 = 10; outdeg(A) = 2, outdeg(B) = 2, outdeg(C) = 4, outdeg(D) = 0, outdeg(E) = 2. Note that the sum of the outdegrees is the same as the sum of the indegrees and also matches the number of arcs in the digraph. 2. (a) indeg(A) = 1; outdeg(A) = 3; there is one arc, denoted by an arrow, that comes into vertex A (from vertex B) and three arcs, once again denoted by arrows, that leave vertex A (for vertices C, D, and E). (b) indeg(C) = 4; outdeg(C) = 0 (c) indeg(E) = 2; outdeg(E) = 3 (d) 1 + 1 + 4 + 2 + 2 = 10; indeg(A) = 1, indeg(B) = 1, indeg(C) = 4, indeg(D) = 2, indeg(E) = 2. Note that the sum of the indegrees matches the number of arcs in the digraph. (e) 3 + 3 + 0 + 1 + 3 = 10; outdeg(A) = 3, outdeg(B) = 3, outdeg(C) = 0, outdeg(D) = 1, outdeg(E) = 3. Note that the sum of the outdegrees is the same as the sum of the indegrees and also matches the number of arcs in the digraph. 3. (a) We look for paths from A to D that travel through exactly one other vertex (so that they are of length 2). To that end, A, B, D is one path. A, C, D is a second path. Traveling through vertex E will not work as there is no arc connecting vertex A to vertex E. Thus exactly two paths of length 2 from A to D exist: A, B, D and A, C, D. (b) In this situation, we look for all paths that start at A and end at D that go through exactly two of the vertices B, C, and E. - If we start at A and travel to B, we cannot get to either C or E with an arc, so that will not work. - If we start at A and travel to C, we have two choices to create a path of length 3. We can go to B and then D (creating path A, C, B, D) or we can go to E and then D (creating path A, C, E, D). - If we start at A and travel to E, we cannot get to either B or C with an arc, so that will not work. - Since there is no arc from A to D, we cannot take advantage of the fact that A connects to B which also connects back to A. Thus exactly two paths of length 3 from A to D exist: A, C, B, D and A, C, E, D. (c) There are a two ways to look for a path of length 4 from A to D: either find a path that goes through each vertex exactly once (which doesn’t exist) or find a path that goes through one vertex (such as A) more than once. This later approach will work to produce such a path: A, B, A, C, D. Note: Another possible path constructed in this same way would be A, C, A, B, D.

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(d) As in (c), we seek a path that goes through one vertex (such as A) more than once. This time, however, we also seek a path that not only goes through A twice, but also goes through two of the vertices in the “middle” (namely B, C, and E). Two paths that do this travel from A to B and then back to A before moving on to D. They are A, B, A, C, B, D and A, B, A, C, E, D. A third path that does this travels from A to C and then to B before going back to A and then moving along to vertex D. It is A, C, B, A, B, D. (e) Since the outdegree of D is 0, there are no paths from D to A (no matter the length). 4. (a) We look for paths from B to E that travel through no other vertices (just a single arc). In this graph, since there is an arc from B to E, such a path, namely B, E, exists. (b) We look for paths from B to E that travel through exactly one other vertex (so that they are of length 2). To that end, only vertices A and C (and not D) are reachable from B. So B, A, E is one path. Since there is no arc from vertex C to vertex E, there no other paths of length 2 from B to E. (c) There are a two ways to look for a path of length 4 from B to E: either find a path that goes through each vertex exactly once (which doesn’t exist since vertex C leads nowhere) or find a path that goes through one vertex (such as B) more than once. This later approach will work to produce two such paths: B, E, B, A, E and B, A, E, B, E. (d) There is only one possible path from E to B, namely the path of length 1 consisting of the single arc EB. - There are no paths that start with the arc EA since there is no arc EA. - There are no paths that start with the arc EC, since vertex C has outdegree of 0. - There are no paths that start with the arc ED, since vertex D only leads to vertex C which, in turn, has outdegree of 0. - There are no paths that pass through vertex E more than once. (e) There are no paths from D to E since any such path could need to start with the arc DC and vertex C has outdegree of 0. 5. (a) A, B, A and A, C, A; Remember that cycle is a sequence of distinct adjacent arcs that starts and ends at the same vertex. (b) In this situation, such a cycle must begin with the arc AC (starting with arc AB fails – it leads to vertex D which has outdegree of 0). By starting with arc AC, there are two cycles having length 3 – A, C, B, A and A, C, E, A. [Again, going to vertex D from C leads nowhere since D has outdegree of 0.] (c) Much like part (b), any such cycle must begin with either arc AC or arc AB. Since vertex D cannot be part of any cycle, we only need to investigate if vertex E could be included or not. Since there is no arc from B to C, the only cycles of length 4 cannot include vertex E. These cycles are either A, B, A, C, A or A, C, A, B, A. (d) To form a cycle of length 5, we can (and must) take the two 2-cycles found in part (c) and grow one of the 2-cycles to a 3-cycle. In particular, we grow the 2-cycle A, C, A to A, C, E, A. This gives two possible cycles of length 5 – A, C, E, A, B, A and A, B, A, C, E, A. 6. (a) B, E, B; Remember that cycle is a sequence of distinct adjacent arcs that starts and ends at the same vertex. If it is to have length 2, it must consist of two vertices. (b) There is only one cycle having length 3 – A, E, B, A. There are 10 ways to select 3 of the 5 vertices to be included in such a cycle. We look for triangles formed by these combinations of vertices that have the arcs all pointed in the same direction around a the triangle (either clockwise or counterclockwise). Only these three vertices have this property. (c) There are none. There are 5 ways to select 4 of the 5 vertices to be included in such a cycle (since we can’t glue together two distinct 2-cycles in this particular graph). Since vertex C has outdegree of 0, only the vertices A, B, D, E could possibly form such a cycle. However, vertex D only leads to vertex C and so no cycle of length 4 is possible. Copyright © 2022 Pearson Education, Inc.

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7. (a) C and E; remember that given an arc XY, X is incident to Y. So CA and EA being arcs in this digraph suggest C and E are incident to A. (b) B and C; given an arc XY, Y is incident from X. So AB and AC being arcs in this digraph suggest B and C are incident from A. (c) B, C, and E (d) No vertices are incident from D. (e) CD, CE and CA (f) No arcs are adjacent to CD. 8. (a) B (b) C, D and E (c) A and E (d) No vertices are incident from D. (e) AC, AD and AE (f) No arcs are adjacent to BC. 9. (a)

(b)

(c)

10. (a)

(b)

(c)

11. (a) 2 (these correspond to arcs AB and AE) (b) 1 (this corresponds to arc EA) (c) 1 (this corresponds to arc DB) (d) 0 (there are no arcs incident to D) 12. (a) 2 (these correspond to arcs VW and VZ) (b) 0 (there are no arcs incident to V) (c) 2 (this corresponds to arcs ZY and ZW) (d) 3 (this corresponds to arcs VZ, WZ and XZ) 13. (a) A, B, D, E, F is one possible path. (b) A, B, D, E, C, F Copyright © 2022 Pearson Education, Inc.

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(c) B, D, E, B (d) The outdegree of vertex F is 0, so it cannot be part of a cycle. (e) The indegree of vertex A is 0, so it cannot be part of a cycle. (f) B, D, E, B is the only cycle 14. (a) A, E, C, D (b) The indegree of D is 1 and hence there is only one way to get to D and that is using the arc CD. Also, the indegree of C is 1 and so there is only one way to get to vertex C and that is using the arc EC. Finally, the outdegree of vertex B is 0 and so the only way to get to D from A is the given path. (c) E, C, D, E (d) The indegree of A is 0. (e) The outdegree of B is 0. (f) E, C, D, E is the only cycle 15.

16.

17. (a) B, since B is the only person that everyone respects. (b) A, since A is the only person that no one respects. (c) The individual corresponding to the vertex having the largest indegree (B in this example) would be the most reasonable choice for leader of the group since they would be the most respected person. 18. (a) E. The outdegree of E is 4, meaning E defeated every other team. (b) A. The outdegree of A is 0, meaning A didn’t win any games. (c) The indegree of a vertex represents the number of games that team lost. The outdegree represents the number of games that team won. (d) Adding the number of games a team lost and the number of games won gives N-1, the number of games played. 19. (a) The band’s website and the TicketMonster Web site are likely to have large indegree and zero or very small outdegree (the local Smallville Web sites are likely to have hyperlinks coming from either). This makes X and Y the two likely choices. It is much more likely that the rock band’s Web site will have a hyperlink to the ticket-selling Web site than the other way around. Give that there is a hyperlink from Y to X but not the other way, the most likely choice is Y for the rock band and X for the TicketMonster Website. (b) X; See above. Copyright © 2022 Pearson Education, Inc.

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Chapter 8: The Mathematics of Scheduling (c) The radio station is likely to be a major source of information on anything related to the concert (the band, where to buy tickets, where to stay if you are from out of town, etc.), so it should be a vertex with high outdegree. V is the most likely choice. (d) Z is a vertex to only the band’s Web site and to the ticket selling Web site. None of the other Web sites are paying any attention to Z. This is most likely Joe Fan’s blog. (e) The two sister hotels would clearly have links to each other’s Web sites, and since they are offering a special package for the concert, they are likely to have a link to the ticket-selling Web site. The logical choices are U and W.

20. (a) The indegree of a vertex represents the number of Web sites pointing to that particular site, so the rankings can be determined by the indegrees: H is ranked first (indegree 8), A is second (indegree 4), and B is third (indegree 3). (b) H (indegree 8 and outdegree 2) is most likely to be the Angelface Web site. 21. The vertex set is {START, A, B, C, D, E, F, G, H, END} and the arc-set is {START-A, START-H, AF, AC, BEND, CB, CE, D-END, E-END, FB, GB, GD, HG}.

22. The vertex set is {START, A, B, C, D, E, F, G, H, END} and the arc-set is {START-C, START-G, AE, BE, CA, CB, DB, DF, E-END, F-END, GD, GH, H-END}.

23. Vertex A represents the experiment taking 10 hours to complete. Vertices B and C represent the experiments requiring 7 hours each to complete. Vertices D and E represent the experiments requiring 12 hours each to complete. Vertices F, G, and H represent the experiments requiring 20 hours to complete.

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24. Vertices A, B, and C represent the checks taking 4 hours to complete. Vertices D, E, and F represent the checks requiring 7 hours each to complete. Vertices G, H, I and J represent the checks requiring 15 hours each to complete.

25.

26.

27.

28.

8.3

Priority-List Scheduling

29. (a) There are 31 × 3 = 93 processor hours available. Of these, 10 + 7 + 11 + 8 + 9 + 5 + 3 + 6 + 4 + 7 + 5 = 75 hours are used. So, there must be 93 – 75 = 18 hours of idle time. (b) There is a total of 75 hours of work to be done. Three processors working without any idle time would 75 take = 25 hours to complete the project. 3 Copyright © 2022 Pearson Education, Inc.

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30. (a) There are 19 × 5 = 95 processor hours available. Of these, 10 + 7 + 11 + 8 + 9 + 5 + 3 + 6 + 4 + 7 + 5 = 75 hours are used. So there must be 95 – 75 = 20 hours idle time. (b) There is a total of 75 hours of work to be done. Five processors working without any idle time would 75 = 15 hours to complete the project. take 5 31. There is a total of 75 hours of work to be done. Dividing the work equally between the six processors would 75 = 12.5 hours of work. Since there are no half hour jobs, the completion time require each processor to do 6 could not be less than 13 hours. 32. (a) The completion time can never be less than 11 hours, since task C by itself takes 11 hours. (b) There are 11 tasks, so clearly, having more than 11 processors makes no sense. To argue that 10 processors are enough, consider two cases: (i) if all tasks are independent then we can give two “short” tasks [for example F(5) and G(3)] to one processor and each of the remaining 9 tasks to a different processor, and still meet the optimal completion time of 11 hours; (ii) if there is at least one precedence relation (call it X → Y), then we can give tasks X and Y to one processor and each of the remaining 9 tasks to a different processor. In this case, the finishing time will be the critical time for the project, and thus the schedule will be an optimal schedule. 33. 34. 35. (a) No. B (with processing time of 7) must be completed before C can be started. (b) No. B (with processing time of 7) must be completed before D can be started. (c) No. G is a ready task and it is ahead of H in the priority list. [Note: C, D, E, and F all must wait for B to be completed before they may be assigned.] 36. (a) No. B (with processing time of 8) must be completed before D can be started. (b) No. B (with processing time of 8) must be completed before E can be started. (c) No. F is a ready task and it is ahead of G in the priority list. [Note: D and E must wait for B to be completed before they may be assigned.] 37.

38.

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39.

40.

41. Both priority lists have all tasks in the top two paths of the digraph listed before any task in the bottom path. 42. Both priority lists have all tasks in the bottom path of the digraph listed before any task in the top two paths.

8.4

The Decreasing-Time Algorithm

43. Decreasing-Time List: D(12), C(9), A(8), E(6), B(5), G(2), F(1)

44. Decreasing-Time List: D(12), C(9), A(8), E(6), B(5), G(2), F(1)

45. Decreasing-Time List: K(20), D(15), I(15), C(10), E(7), H(5), J(5), B(4), G(4), F(3), A(2)

46. Decreasing-Time List: K(20), D(15), I(15), C(10), E(7), H(5), J(5), B(4), G(4), F(3), A(2)

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47. (a) Decreasing-Time List: E(2), F(2), G(2), H(2), A(1), B(1), C(1), D(1)

(b)

Fin − Opt 8 − 6 1 = = 33 % . Opt 6 3

48. (a) Decreasing-Time List: B(6), D(5), F(4), E(3), C(2), A(1)

(b)

Fin − Opt 14 − 11 = ≈ 27.27% . Opt 11

49. (a) Decreasing-Time List: E(5), F(5), G(5), H(5), I(5), J(5), A(4), B(4), C(4), D(4)

(b)

Fin − Opt 26 − 16 = = 62.5% . Opt 16

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50. (a) Decreasing-Time List: B(10), D(9), F(8), H(7), J(6), I(5), G(4), E(3), C(2), A(1)

(b)

8.5

Fin − Opt 22 − 17 = ≈ 29.41% . Opt 17

Critical Paths and the Critical-Path Algorithm

51. (a) As shown in the diagram, the critical time of each task is found by adding the times for each task in the shortest path formed between (and including) that task and END.

(b) The critical path is the longest time of a path between START and END. In this example, Start, A, D, F, END is the critical path, with a critical time of 21. (c) From (a), the critical path list is: A[21], B[18], C[17], D[13], E[8], G[2], F[1].

(d) There are a total of 43 work units, so the shortest time the project can be completed by 2 workers is 43 = 21.5 time units. Since there are no tasks with less than 1 time unit, the shortest time the project can 2 actually be completed is 22 hours.

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52. (a)

(b) The critical path of the project has length 21 hours, so the project cannot be completed in less than 21 hours. 53. (a) The critical path list is: B[46], A[44], E[42], D[39], F[38], I[35], C[34], G[24], K[20], H[10], J[5]. So, the critical path is START, B, E, I, K, END.

(b)

54. (a)

(b)

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55. The project digraph, with critical times listed, is shown below.

The critical-time priority list is L[13.5], P[13], K[7], B[6], F[4.5], G[4], W[2], S[1.5], C[1] and the critical path is START[13.5], L[13.5], P[13], K[7], G[4], W[2], C[1], END[0].

Since processing times are given in hours, the project finishing time is 13.5 hours. 56. The project digraph, with critical times listed, is shown below.

The critical-time priority list is TC[4.5], CF[3.7], TN[3.0], FA[2.9], PT[2.5], SS[1.4], SB[0.8], TD[0.7] and the critical path is START[4.5], TC[4.5], TN[3.0], PT[2.5], TD[0.7], END[0].

Since processing times are given in hours, the project finishing time is 4.5 hours. 57. (a)

The critical path is START[20], F[20], H[17], G[16], D[9], E[8], END[0]. The critical time is 20.

(b) The critical-time priority list is F[20], B[18], H[17], G[16], C[13], A[10], D[9], E[8], J[6].

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Chapter 8: The Mathematics of Scheduling (c)

The project finishing time is 5 + 4 + 10 + 6 = 25.

(d) An optimal schedule for N = 2 processors with finishing time Opt = 23 is shown below.

(e)

58. (a)

The relative error expressed as a percent is Fin − Opt 25 − 23 = ≈ 8.7% . ε= Opt 23 The critical path is START[19], B[19], D[12], H[7], END[0]. The critical time is 19 hours.

(b) The critical-time priority list is B[19], C[13], D[12], E[10], A[9], H[7], F[6], G[4]. (c)

The project finishing time using the critical-path algorithm is 7 + 5 + 6 + 4 = 22 hours.

(d) F, G, or H cannot be started until D has been completed, and D cannot be started until B has completed. It takes at least 12 hours to complete B followed by D. The two processors will then require at least 10 hours to complete F, G, and H. Consequently, with two processors, the project requires at least 22 hours. (e)

The timeline for three processors is shown below. The finishing time is 19 hours.

(f)

The timeline for three processors is 19 hours, so any schedule with Fin = 19 must be optimal.

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JOGGING 59. Every arc of the graph contributes 1 to the indegree sum and 1 to the outdegree sum. 60. (a) totally asymmetric; all the streets are one-way. (b) symmetric; all the streets are two-way. (c) neither (d) symmetric; if Joe is Jack’s brother, then Jack is also Joe’s brother. (e) totally asymmetric; if Joe is Jack’s father, then Jack is certainly not Joe’s father. 61. Assuming that all of the tasks are independent, then A, B, C, E, G, H, D, F, I is one possible priority list. 62. (a) The finishing time of a project is always more than or equal to the number of hours of work to be done divided by the number of processors doing the work. (b) The schedule is optimal with no idle time. (c) The total idle time in the schedule.

RUNNING 63. Consider four cases: M = 4k, 4k+1, 4k+2, and 4k+3 for some positive integer k. If M = 4k, then the optimal schedule is

and the optimal completion time is

1 + 2 + 3 +  + 4k (4k + 1)(4k ) / 2 (4k + 1)(4k ) ( M + 1)( M ) = = = . 2 2 4 4

If M = 4k+1, then the optimal schedule is

and the optimal completion time is 2 + 3 +  + (4k + 1) (4k + 3)(4k ) / 2 (4k + 3)(4k ) ( M + 2)( M − 1) 1+ = 1+ = 1+ = 1+ . 2 2 4 4 If M = 4k+2, then the optimal schedule is

and the optimal completion time is

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Chapter 8: The Mathematics of Scheduling 3 +  + (4k + 2) (4k + 5)(4k ) / 2 (4k + 5)(4k ) ( M + 3)( M − 2) = 3+ = 3+ = 3+ . 2 2 4 4 If M = 4k+3, then the optimal schedule is 1+ 2 +

and the optimal completion time is

1 + 2 + 3 +  + (4k + 3) (4k + 4)(4k + 3) / 2 ( M + 1)( M ) = = . 2 2 4

64. Consider six cases: M = 6k, 6k+1, 6k+2, 6k+3, 6k+4, and 6k+5 for some positive integer k. If M = 6k, then the optimal schedule is

1 + 2 + 3 +  + 6k (6k + 1)(6k ) / 2 (6k + 1)(6k ) ( M + 1)( M ) = = = . 2 3 6 6 In a fashion, similar to Exercise 64, the optimal completion time for the other cases is given below: If M = 6k+1, then the optimal completion time is 2 + 3 +  + (6k + 1) (6k + 3)(6k ) / 2 (6k + 3)(6k ) ( M + 2)( M − 1) 1+ = = = . 3 3 6 6 If M = 6k+2, then the optimal completion time is 3 +  + (6k + 2) (6k + 5)(6k ) / 2 (6k + 5)(6k ) ( M + 3)( M − 2) 1+ 2 + = = = . 3 3 6 6 ( M + j + 1)( M − j ) In general, if M = 6k+j (j = 0,1,2,3,4, or 5), then the optimal completion time is . 6

and the optimal completion time is

65. Consider the case in which M = 6. In that case, Opt = 64. Since all tasks are independent, the critical-time priority list is identical to a decreasing-time list.

In general, Opt = 2M and the schedule is similar.

ISM: Excursions in Modern Mathematics, 10E APPLET BYTES 66. (b)

(e) Despite the fact that there is a lot of idle time in the schedule, the result is optimal since the critical path has length 34 – the same time illustrated in the schedule.

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The resulting schedule is optimal since there is no idle time for the processors.

67. (b) This schedule is not optimal since it is clear that simply switching the assignments of H and I (between processors 1 and 2) would shorten the processing time to 12.

(d) Task I will be assigned earlier in the process since tasks E, F, G, and H all need to wait until task D is completed. With only three processors, task D cannot be completed until after task A is finished. But after A is finished, task I can be started. This is desirable in finishing the project in the shortest time possible. (e) The resulting schedule is optimal since the critical path has length 12.

68. (a) This is the same result as in 67(c). The resulting schedule is optimal.

ISM: Excursions in Modern Mathematics, 10E (b) The resulting schedule if the processing times of all tasks is one unit less is shown below.

(c) Each task was shorter and yet the finishing time of the project increased from 12 to 13. (d) One possible priority list is A, B, C, I, D, E, F, G, H. The result is optimal since the length of the critical path is 10. The key in choosing the priority list is to make sure that task I gets started right after task A is completed.

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Chapter 9 WALKING

(c) If AN = 1 , then ( −1) = 1 . This is true exactly when N+1 is an even integer (Since N ≥ 1 , this means 2, 4, 6, 8,…). But N+1 is a positive even integer exactly when N is a positive odd integer. That is, when N = 1, 3, 5, 7, … N +1

9.1. Sequences and Population Sequences 1. (a) A1 = 12 + 1 = 2

1−1

(b) A100 = 1002 + 1 = 10, 001 (c) If AN = 10 , then N 2 + 1 = 10 . So N 2 = 9 and N = ±3 . Since the subscripts N are all integers satisfying N ≥ 1 in this sequence, we have N = 3.

 −1    N

(b) A15 = 315 − 2 = 14,348,907 − 2 = 14,348,905

3N = 81 and N = 4 .

4 × 9 36 = =3 9 + 3 12

4N 5 5 = . So , then N +3 2 2 8 N = 5 ( N + 3) and 8 N = 5 N + 15 . This leads to 3N = 15 and N = 5. 2(1) + 3 5 = 3(1) − 1 2

(b) A100 =

2(100) + 3 203 = 3(100) − 1 299

2N + 3 = 1 . So 3N − 1 2 N + 3 = 3N − 1 and N = 4 .

5. (a) A1 = ( −1)

1+1

(b) A100 = ( −1)

N −1

> 0 . Since N ≥ 1 , the base is

−1 < 0 . For a power N of this base to be positive, we must have an even exponent. That is, we need to have N-1 = 0, 2, 4, 6, … In other words, we require N = 1, 3, 5, 7, …

(b) We could continue as in (a). A7 = 2 A6 + A5 = 2(41) + 17 = 99 A8 = 2 A7 + A6 = 2(99) + 41 = 239 Or, we could think using the rule “double the last term in the sequence and add the term prior to that.” That is, double the (N-1)st term and add the (N-2)nd term. Using this rule, A7 would be twice 41 added to 17 (or 99). Then A8 would be twice 99 added to 41 (or 239). 8. (a) A3 = A3−1 + 2 A3− 2 = A2 + 2 A1 = 1 + 2(1) = 3 A4 = A4 −1 + 2 A4 − 2 = A3 + 2 A2 = 3 + 2(1) = 5 A5 = A5−1 + 2 A5 − 2 = A4 + 2 A3 = 5 + 2(3) = 11 A6 = A6 −1 + 2 A6 − 2 = A5 + 2 A4 = 11 + 2(5) = 21

= ( −1) = 1

100 +1

1  1 = −  = − 64  4

7. (a) A3 = 2 A3−1 + A3− 2 = 2 A2 + A1 = 2(1) + 1 = 3 A4 = 2 A4 −1 + A4 − 2 = 2 A3 + A2 = 2(3) + 1 = 7 A5 = 2 A5−1 + A5− 2 = 2 A4 + A3 = 2(7) + 3 = 17 A6 = 2 A6 −1 + A6 − 2 = 2 A5 + A4 = 2(17) + 7 = 41

4 ×1 4 = =1 1+ 3 4

4. (a) A1 =

4 −1

negative. That is,

(b) A9 =

 −1  (b) A4 =    4 

= ( −1) = 1

2. (a) A3 = 33 − 2 = 27 − 2 = 25

3. (a) A1 =

 −1  6. (a) A1 =    1 

= ( −1)

101

ISM: Excursions in Modern Mathematics, 10E (b) We could continue as in (a). A7 = A6 + 2 A5 = 21 + 2(11) = 43 A8 = A7 + 2 A6 = 43 + 2(21) = 85 Or, we could think using the rule “add the last term in the sequence to double the term prior to that.” That is, add the (N-1)st term to double the (N-2)nd term. Using this rule, A7 would be 21 added to twice 11 (or 43). Then A8 would be 43 added to the twice 21 (or 85). 9. (a) A3 = A3−1 − 2 A3− 2 = A2 − 2 A1 = −1 − 2(1) = −3 A4 = A4 −1 − 2 A4 − 2 = A3 − 2 A2 = −3 − 2(−1) = −1 A5 = A5−1 − 2 A5− 2

= A4 − 2 A3 = −1 − 2(−3) =5 A6 = A6 −1 − 2 A6 − 2 = A5 − 2 A4 = 5 − 2(−1) =7

(b) We could continue as in (a). A7 = A6 − 2 A5 = 7 − 2(5) = −3 A8 = A7 − 2 A6 = −3 − 2(7) = −17 Or, we could think using the rule “from the last term in the sequence, subtract twice the term before that.” That is, from the (N-1)st term , subtract twice the (N2)nd term. Using this rule, A7 would be 7 minus twice 5 (or -3). Then A8 would be -3 minus twice 7 (or -17). 10. (a) A3 = 2 A3−1 − 3 A3− 2 = 2 A2 − 3 A1 = 2(1) − 3(−1) =5 A4 = 2 A4 −1 − 3 A4 − 2 = 2 A3 − 3 A2 = 2(5) − 3(1) =7

149

A5 = 2 A5−1 − 3 A5− 2 = 2 A4 − 3 A3 = 2(7) − 3(5) = −1 A6 = 2 A6 −1 − 3 A6 − 2 = 2 A5 − 3 A4 = 2(−1) − 3(7) = −23

(b) We could continue as in (a). A7 = 2 A6 − 3 A5 = 2(−23) − 3(−1) = −43 A8 = 2 A7 − 3 A6 = 2(−43) − 3(−23) = −17 Or, we could think using the rule “from twice the last term in the sequence, subtract triple the term before that.” That is, from twice the (N-1)st term , subtract triple the (N-2)nd term. Using this rule, A7 would be 2(-23) minus 3(-1) which is -43. Then A8 would be 2(-43) minus 3(23) which is -17. 11. (a) There are (at least) two different ways to observe the pattern here. The first is to notice that these are perfect squares (of integers). So the pattern can be recognized as 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 7 2 = 49, … Another way to observe the pattern is to notice that consecutive odd integers are being added. So the pattern could also be recognized as 1, 1 + 3 = 4, 4 + 5 = 9, 9 + 7 = 16, 16 + 9 = 25, 25 + 11 = 36, 36 + 13 = 49, … Either way, the next two terms could reasonably be listed as 36 and 49. (b) Recognizing the pattern (see (a)) by A1 = 12 = 1, A2 = 22 = 4, A3 = 32 = 9, A4 = 42 = 16, A5 = 52 = 25, …we can explicitly describe the sequence using AN = N 2 . [Recognize the pattern as the subscript being squared.]

(c) Recognizing the pattern (see (a)) by P0 = 12 , P1 = 22 , P2 = 32 , P3 = 42 , P4 = 52 , …we can explicitly describe the

sequence using PN = ( N + 1) . [Recognize the pattern as the square of one more than the subscript being squared.] 2

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Chapter 9: Population Growth Models

12. (a) There are (at least) two different ways to observe the pattern here. The first is to notice that these are factorials (of integers). So the pattern can be recognized as 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040, … Another way to observe the pattern is to notice that we are multiplying by consecutive integers. So the pattern could also be recognized as 1, 1× 2 = 2, (1× 2) × 3 = 6, (1× 2 × 3) × 4 = 24,

(1× 2 × 3 × 4 ) × 5 = 120, (1× 2 × 3 × 4 × 5 ) × 6 = 720, (1× 2 × 3 × 4 × 5 × 6 ) × 7 = 5040, … Either way, and they are clearly equivalent, the next two terms could reasonably be listed as 720 and 5040. (b) Recognizing the pattern (see (a)) by A1 = 1!, A2 = 2!, A3 = 3!, A4 = 4!, A5 = 5!, …we can explicitly describe the sequence using AN = N !. [Recognize the pattern as the factorial of the subscript.] (c) Recognizing the pattern (see (a)) by P0 = 1!, P1 = 2!, P2 = 3!, P3 = 4!, P4 = 5!, …we can explicitly describe the

sequence using PN = ( N + 1) !. [Recognize the pattern as the factorial of one more than the subscript.] 13. (a) There are (at least) two different ways to observe the pattern here. The first is to notice that consecutive integers are being added. So the pattern could be recognized as 1, 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10, 10 + 5 = 15, 15 + 6 = 21, 21 + 7 = 28, … Another way to observe the pattern is to look at double each value in the sequence: 0, 2, 6, 12, 20, 30, 42, … The pattern here can be recognized as 0 × 1 = 0, 1× 2 = 2, 2 × 3 = 6, 3 × 4 = 12, 4 × 5 = 20, 5 × 6 = 30, 6 × 7 = 42, 7 × 8 = 56, … So cutting these each in half gives the pattern ( 0 ×1) / 2 = 0, (1× 2 ) / 2 = 1,

( 2 × 3) / 2 = 3, ( 3 × 4 ) / 2 = 6, ( 4 × 5) / 2 = 10, ( 5 × 6 ) / 2 = 15, ( 6 × 7 ) / 2 = 21, ( 7 × 8) / 2 = 28, …

Either way, the next two terms could reasonably be listed as 21 and 28. (b) Recognizing the pattern (see (a)) by A1 = ( 0 × 1) / 2 = 0, A2 = (1× 2 ) / 2 = 1, A3 = ( 2 × 3) / 2 = 3, A4 = ( 3 × 4 ) / 2 = 6,

A5 = ( 4 × 5 ) / 2 = 10, A6 = ( 5 × 6 ) / 2 = 15, …we can explicitly describe the sequence using AN = ( ( N − 1) × N ) / 2. [The pattern

is half the product of one less than the subscript multiplied by the subscript.] (c) Recognizing the pattern (see (a)) by P0 = ( 0 × 1) / 2 = 0, P1 = (1× 2 ) / 2 = 1,

P2 = ( 2 × 3) / 2 = 3, P3 = ( 3 × 4 ) / 2 = 6,

P4 = ( 4 × 5 ) / 2 = 10, P5 = ( 5 × 6 ) / 2 = 15, …

we can explicitly describe the sequence using PN = ( N ( N + 1) ) / 2. [Recognize the pattern as half the product of the subscript multiplied by one more than the subscript.] 14. (a) There are (at least) two different ways to observe the pattern here. The first is to notice that consecutive “doubles” are being added. So the pattern could be recognized as 2, 2 + 1 = 3, 3 + 2 = 5, 5 + 4 = 9, 9 + 8 = 17, 17 + 16 = 33, 33 + 32 = 65, 65 + 64 = 129, … Another way to observe the pattern is to notice that each term is almost a power of two (actually one more than that). So, the pattern here can also be recognized as 20 + 1 = 2, 21 + 1 = 3, 22 + 1 = 5, 23 + 1 = 9, 24 + 1 = 17, 25 + 1 = 33, 26 + 1 = 65, 27 + 1 = 129, … Either way, the next two terms could reasonably be listed as 65 and 129. (b) Recognizing the pattern (see (a)) by A1 = 20 + 1, A2 = 21 + 1, A3 = 22 + 1,

A4 = 23 + 1, A5 = 24 + 1, A6 = 25 + 1, …we can explicitly describe the sequence using AN = 2 N −1 + 1. [Recognize the pattern as one more than two raised to the power of one less than the subscript.] (c) Recognizing the pattern (see (a)) by P0 = 20 + 1, P1 = 21 + 1, P2 = 22 + 1,

ISM: Excursions in Modern Mathematics, 10E P3 = 23 + 1, P4 = 24 + 1, P5 = 25 + 1, …we can explicitly describe the sequence using PN = 2 N + 1. [Recognize the pattern as one more than two raised to the subscript.]

15. (a) First, it helps to list the sequence entirely in terms of improper fractions to spot a 4 8 12 16 20 pattern: , , , , , … At this 4 5 6 7 8 point, we then notice that the numerators are integer multiples of 4 and the denominators are consecutive integers. So the pattern could be seen to continue 24 28 as , ,… 9 10 (b) Recognizing the pattern (see (a)) by 2× 4 3× 4 1× 4 , A3 = , A1 = , A2 = 5 6 4 4× 4 5× 4 6× 4 A4 = , A5 = , A6 = , 7 8 9 7×4 A7 = , … we can explicitly describe 10 N (4) 4N the sequence using AN = = . N +3 N +3 [The pattern is the subscript times four divided by three more than the subscript.] 16. (a) First, it helps to list the sequence entirely in terms of improper fractions to spot a 3 4 5 6 7 pattern: , , , , , … At this 1 2 4 8 16 point, we notice that the numerators are consecutive integers and the denominators are powers of two. So the pattern could 8 9 be seen to continue as , ,… 32 64 (b) Recognizing the pattern (see (a)) by 3 4 5 A1 = 0 , A2 = 1 , A3 = 2 , 2 2 2 6 7 8 9 A4 = 3 , A5 = 4 , A6 = 5 , A7 = 6 , … 2 2 2 2 we can explicitly describe the sequence N +2 using AN = N −1 . [Recognize the pattern 2 as two more than the subscript divided by two raised to the power of one less than the subscript.]

151

17. (a) We may recognize the pattern as similar to that found in Exercise 12: 1 1 1 1 , , , , … The next term, which 2! 3! 4! 5! represents the probability that six passengers will board in order of 1 1 . decreasing seat numbers is or 6! 720

1 , 2! 1 1 1 1 A2 = , A3 = , A4 = , A5 = , … 3! 4! 5! 6! we can explicitly describe the sequence 1 using AN = . Since AN represents N ( + 1)!

(b) Recognizing the pattern by A1 =

the probability that N+1 passengers board in order of decreasing seat numbers, we seek the value of A11 . The desired probability is A11 =

1 1 = . (12 )! 479,001,600

18. (a) We notice first that the numerators in the sequence increase by consecutive integers (see Exercise 13) and the denominators are increasing powers of two: 1 1 3 1 + 2 6 3 + 3 10 6 + 4 = 4 , = 5 , = , = 3 , 2 16 2 32 2 4 22 8 … The next term, which represents the probability in six tosses we will obtain two 10 + 5 15 heads and four tails is . = 26 64 (b) As in Exercise 13, we recognize the pattern in the numerators explicitly by 1 (1× 2 ) / 2 3 ( 2 × 3) / 2 , A1 = = , A2 = = 2 8 23 4 2 6 (3 × 4) / 2 10 ( 4 × 5 ) / 2 A3 = = , A4 = = , 16 32 24 25 15 ( 5 × 6 ) / 2 A5 = = , … We can describe 64 26 ( N ( N + 1) ) / 2 . the sequence by AN = 2 N +1 Here AN represents the probability if tossing exactly two heads in N+1 tosses of a fair coin. So to find the probability of tossing exactly two heads (and 10 tails) when 12 fair coins are tossed, we determine

152

Chapter 9: Population Growth Models

the value of A11 . The desired probability is A11 =

(11× 12 ) / 2 12

66 33 = . 12 2048 2

9.2. The Linear Growth Model 19. (a) P1 = P0 + 125 = 80 + 125 = 205 ; P2 = P1 + 125 = 205 + 125 = 330 ; P3 = P2 + 125 = 330 + 125 = 455

It follows that PN = 578 − N (25) which can also be written as PN = 578 − 25 N . (c) P23 = 578 − 23(25) = 3 22. (a) P1 = P0 − 111 = 11,111 − 111 = 11, 000 ; P2 = P1 − 111 = 11, 000 − 111 = 10,889 ; P3 = P2 − 111 = 10,889 − 111 = 10, 778

(b) Notice the pattern to construct an explicit formula. P1 = 80 + 125 ; P2 = 80 + 2(125) ; P3 = 80 + 3(125) ; P4 = 80 + 4(125) … It follows that PN = 80 + N (125) which can also be written as PN = 80 + 125 N .

(b) Notice the pattern to construct an explicit formula. P1 = 11,111 − 111 ; P2 = 11,111 − 2(111) ; P3 = 11,111 − 3(111) ; P4 = 11,111 − 4(111) … It follows that PN = 11,111 − N (111) which can also be written as PN = 11,111 − 111N .

20. (a) P1 = P0 + 23 = 57 + 23 = 80 ; P2 = P1 + 23 = 80 + 23 = 103 ; P3 = P2 + 23 = 103 + 23 = 126 (b) Notice the pattern to construct an explicit formula. P1 = 57 + 23 ; P2 = 57 + 2(23) ; P3 = 57 + 3(23) ; P4 = 57 + 4(23) … It follows that PN = 57 + N (23) which can also be written as PN = 57 + 23 N . (c) P200 = 57 + 200(23) = 4657 21. (a) P1 = P0 − 25 = 578 − 25 = 553 ; P2 = P1 − 25 = 553 − 25 = 528 ; P3 = P2 − 25 = 528 − 25 = 503 (b) Notice the pattern to construct an explicit formula. P1 = 578 − 25 ; P2 = 578 − 2(25) ; P3 = 578 − 3(25) ; P4 = 578 − 4(25) …

23. (a) Here we can use the fact that P10 = P0 + 10d where d is the common difference between terms of the sequence. So 38 = 8 + 10d and 30 = 10d. It follows that the common difference is d = 3. So P1 = 8 + 3 ; P2 = 8 + 2(3) ; P3 = 8 + 3(3) ; P4 = 8 + 4(3) … It follows that PN = 8 + N (3) which can also be written as PN = 8 + 3N . (b) P50 = 8 + 50 ( 3) = 158 (using the explicit formula found in (a)). 24. (a) Here we can use the fact that P7 = P5 + 2d where d is the common difference between terms of the sequence. So 47 = 37 + 2d and 10 = 2d. It follows that the common difference is d = 5. So P4 = P5 − 5 = 37 − 5 = 32 ; P3 = P4 − 5 = 32 − 5 = 27 ; P2 = P3 − 5 = 27 − 5 = 22 ; P1 = P2 − 5 = 22 − 5 = 17 . It follows that P0 = P1 − 5 = 17 − 5 = 12 . An easier way to see this is to realize that P0 = P5 − 5(5) = 37 − 25 = 12 .

ISM: Excursions in Modern Mathematics, 10E (b) PN = P0 + N ( 5 ) = 12 + 5 N (using the common difference d found in (a)). (c) P100 = P0 + 100 ( 5 ) = 12 + 500 = 512 .

153

P25 = 6, 000, 000, 000 + 25(76,923, 077) = 7,923,076,925 ≈ 7,923, 000, 000 when rounded to the nearest million.

25. (a) Model the unemployment rates using an arithmetic sequence with d = -0.2% where P0 = 8.9% , P1 = 8.7% , P2 = 8.5% , P3 = 8.3% and, in general, PN represents the official unemployment rate N months after October of 2019. The unemployment rate in January 2021 would then be the value of P15 (since that is the 15th month after October of 2019). But P15 = 8.9% + 15(−0.2%) = 5.9% .

27. (a) The year 2003 is 8 years after 1995. So a person born in 2003 corresponds to a value of N = 8 in this model. The life expectancy of such a person is L8 = 66.17 + 0.96(8) = 73.85 years.

(b) An explicit linear model for the unemployment rate N months after October 2019 can be given by PN = 8.9% + N ( −0.2% ) . To determine when the Republic of Parador would reach a zero unemployment rate, we determine the first N value for which PN ≤ 0 . But 8.9% + N ( −0.2% ) ≤ 0 when −0.2% N ≤ −8.9% . That is, when −8.9% . The smallest integer value N≥ −0.2% of N for which this happens is N = 45. July of 2023 is the 45th month after October of 2019. The unemployment rate for June, 2023, would be 0.1% and would, in theory, hit 0% during July of that year.

28. (a) Model the number of adult smokers using an arithmetic sequence (linear growth model) with PN representing the percentage of adult smokers N years after 2005. Then P0 = 0.209 and P13 = 0.137 . The common difference d between terms in the sequence can be found from P13 = P0 + 13d . But 0.137 = 0.209 + 13d

26. (a) Model the world population using an arithmetic sequence with PN representing the population N years after 1999. Then P0 = 6, 000, 000, 000 and P13 = 7, 000, 000, 000 . With linear growth, we would expect the population to increase by another one billion people in the next 13 years as well. This would be the year 2025. (b) The common difference d between terms in the sequence found in (a) can be found from P13 = P0 + 13d . So 7, 000, 000, 000 = 6, 000, 000, 000 + 13d . This gives d ≈ 76,923, 077 . The world population in 2024 is represented by the value P25 . But

(b) We solve LN = 66.17 + 0.96 N = 90 for N. But 0.96N = 23.83 gives N ≈ 24.82 years. The year 2020 is the 25th year after 1995 and the life expectancy in that year would be L25 = 66.17 + 0.96(25) ≈ 90.17 years.

0.137 − 0.209 ≈ −0.00554 . In 13 other words, for each year since 2005 the percentage of adult smokers decreases by roughly 0.554%. The percentage of adult smokers in 2031 is given by the value P26 in this model. But, P26 = 0.209 + 26(−0.00554) = 0.065. We predict that the percentage of adult smokers in 2031 will be roughly 6.5%.

gives d =

(b) The year 2022 is N=17 years after 2005. The percentage of adult smokers in 2022 is given by the value P17 in the model. But P17 ≈ 0.209 + 17( −0.00554) = 0.115 tells us the model predicts roughly 11.5% of the 2022 adult population will be smokers. 29. The common difference in this arithmetic sequence is d = 5. We also have P0 = 2 and P99 = 497 . The sum of 100 terms in this arithmetic sequence is then (2 + 497) × 100 2 + 7 + 12 +  + 497 = 2 = 24,950.

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Chapter 9: Population Growth Models

30. The common difference in this arithmetic sequence is d = 7. We also have P0 = 21 and P56 = 413 . The sum of 57 terms in this arithmetic sequence is then (21 + 413) × 57 21 + 28 + 35 +  + 413 = 2 = 12,369. 31. (a) If 309 is the Nth term of the sequence, 309 = 12 + 3( N − 1) 309 = 12 + 3N − 3 300 = 3 N N = 100 So, 309 is the 100th term of the sequence. (12 + 309) × 100 2 = 16, 050.

(b) 12 + 15 + 18 +  + 309 =

32. (a) The last (Nth) term is 1 + 9( N − 1) = 2701 . Thus, N = 301. So 2701 is the 301st term. (1 + 2701) × 301 2 = 406, 651.

(b) 1 + 10 + 19 +  + 2701 =

33. (a) The odd numbers form an arithmetic sequence with common difference d = 2. We first find the number of terms in the sum 1 + 3 + 5 + 7 + … + 149. Writing 149 = 1 + 2( N − 1) = 1 + 2 N − 2 = 2 N − 1 and solving for N gives N = 75. We then use the arithmetic sum formula to find (1 + 149) × 75 = 5625. 1 + 3 + 5 + 7 +  + 149 = 2 (b) This is a sum of 100 terms in the same arithmetic sequence as part (a). The 100th term is given by 1 + 2(100 − 1) = 199 . We use the arithmetic sum formula to find (1 + 199) × 100 = 10, 000. 1 + 3 + 5 +  + 199 = 2 34. (a) The even numbers form an arithmetic sequence with common difference d = 2. We first find the number of terms in the sum 2 + 4 + 6 + … + 98. Writing 98 = 2 + 2( N − 1) = 2 + 2 N − 2 = 2 N and solving for N gives N = 49. We then use the arithmetic sum formula to find

2 + 4 + 6 +  + 98 =

(2 + 98) × 49 = 2450. 2

(b) This is a sum of 75 terms in the same arithmetic sequence as part (a). The 75th term is given by 2 + 2(75 − 1) = 150 . We use the arithmetic sum formula to find (2 + 150) × 75 = 5700. 2 + 4 + 6 +  + 150 = 2 35. (a) P38 = 137 + 38(2) = 213 (b) PN = 137 + 2 N (c) 137 × $1× 52 = $7124 (d) When just counting the newly installed lights, P0 = 0 and P51 = 2 × 51 = 102. (The lights installed in the 52nd week aren’t in operation during the 52-week period.) (0 + 102) × 52 0 + 2 + 4 +  + 102 = 2 = 2652 The cost is $2,652. 36. (a) 387 + 37 × 20 = 1127 (b) 387 + 37 N (c) 387 × $0.10 × 104 = $4024.80 (d) $3.70 + $7.40 + $11.10 +  + $381.10 ($3.70 + $381.10) × 103 = 2 = $19,817.20

9.3. The Exponential Growth Model 37. (a) P1 = 11× 1.25 = 13.75 (b) P9 = 11× (1.25)9 ≈ 81.956 (c) PN = 11× (1.25) N 38. (a) R =

P1 12 = = 1.5 P0 8

(b) P9 = 8 × (1.5)9 ≈ 307.547 (c) PN = 8 × (1.5) N

ISM: Excursions in Modern Mathematics, 10E 39. (a) P1 = 4 P0 = 4 × 5 = 20 ; P2 = 4 P1 = 4 × 20 = 80 ; P3 = 4 P2 = 4 × 80 = 320

155

42. (a) PN = 0.8PN −1 where P0 = 100, 000 We multiply by 0.8 each year to account for a decrease by 20%. (b) PN = 100, 000 × 0.8N

(b) Notice the pattern: P1 = 5 × 4 ;

P2 = ( 5 × 4 ) × 4 ;

P3 = ( 5 × 4 × 4 ) × 4 ;

P4 = ( 5 × 4 × 4 × 4 ) × 4 ; …

So, explicitly we have PN = 5 × 4 N . (c) We determine the smallest value of N such that PN = 5 × 4 N ≥ 1, 000, 000 . That

is, for which 4 N ≥ 200, 000 . Trial and error suggest the smallest integer value of N that does this is N = 9. After 9 generations, in fact, the population will be over 5.2 million. 40. (a) P1 = 0.75 P0 = 0.75 × 3072 = 2304 ; P2 = 0.75 P1 = 0.75 × 2304 = 1728 ; P3 = 0.75P2 = 0.75 ×1728 = 1296 ; P4 = 0.75 P3 = 0.75 × 1296 = 972 ; P5 = 0.75P4 = 0.75 × 972 = 729 (b) Given the pattern in (a) where we repeatedly multiply by 0.75, explicitly we

have PN = 3072 × ( 0.75 ) . N

such that PN = 3072 × ( 0.75 ) ≤ 200 . That N

200 ≈ 0.065 . 3072 Trial and error suggest the smallest integer value of N that does this is N = 10. After 10 generations, in fact, the population will be roughly 173. is, for which ( 0.75 ) ≤

43. If r represents the growth rate, then the population sequence is described by P0 , P1 = (1 + r ) P0 , P2 = (1 + r ) P0 , … where 2

PN represents the number of mathematics majors N years after 2019. We have P0 = 425

and P1 = 463 . Solving 463 = (1 + r ) ( 425) gives us a growth rate of r = 38/425. This is approximately 8.94%. 44. If r represents the growth rate, then the population sequence is described by P0 , P1 = (1 + r ) P0 , P2 = (1 + r ) P0 , … where 2

PN represents the United States population on January 1st, N years after 2019. We have P0 = 327,540, 066 and P1 = 329,135, 084 . Solving

329,135, 084 = (1 + r ) ( 327,540, 066 ) gives us a 1,595, 018 ≈ 0.00487 . 327,540, 066 Rounded to the nearest tenth of a percent, this is 0.5% per year.

growth rate of r =

41. (a) PN = 1.50 PN −1 where P0 = 200 We multiply by 1.50 each year to account for an increase by 50%. (b) PN = 200 × 1.5N (c) P10 = 200 × 1.510 ≈ 11,533 A good estimate would be to say that about 11,500 crimes will be committed in 2020.

45. If r represents the growth rate, then the population sequence is described by P0 , P1 = (1 + r ) P0 , P2 = (1 + r ) P0 , … where 2

PN represents the undergraduate enrollment at Bright State University N years after 2019. We have P0 = 19, 753 and P1 = 17,389 .

Solving 17,389 = (1 + r ) (19, 753) gives us a growth rate of r = −2364 /19, 753 . This is approximately -11.97%. [The growth rate being negative tells us that the population is decreasing in size.]

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Chapter 9: Population Growth Models

46. We calculate the growth rate over this time period to be 670, 031 − 713,898 r= ≈ −0.06547. 670, 031 Rounded to the nearest tenth of a percent this is a -6.5% growth rate for Detroit’s population over this time period. We note that the population is decreasing since the growth rate is negative. 47. (a) R =

P1 6 = =3 P0 2

P0 + RP0 + R 2 P0 +  + R N −1 0

R −1 We have P0 = 2 , R = 3 , and N = 21 (there are 21 terms being summed). The 321 − 1 ( 2) = 10,460,353,202 . sum is 3 −1

)

P1 6 = = 1.5 P0 4 P2 9 = too.] P1 6

( R − 1) P P = N

P0 + RP0 + R P0 +  + R

N −1

R −1 We have P0 = 4 , R = 1.5 , and N = 25 (there are 25 terms being summed). The 1.525 − 1 ( 4) sum is ≈ 202,001.35 . 1.5 − 1

)

P2 1 = too.] P1 2

(b) The geometric sum formula:

( R − 1) P P = N

[Note that R =

P2 0.4 = too.] 2 P1

( R − 1) P

P0 + RP0 + R P0 +  + R

N −1

R −1 We have P0 = 4 , R = 0.5 , and N = 12

(( 0.2) − 1) (10) ≈ 12.5 . sum is 25

0.2 − 1

(b) This is as in (a) except we no longer have N = 16. So the sum is ( R N − 1) P0 = (2N − 1) ⋅1 = 2N − 1. 2 −1 R −1 52. (a) Observing the geometric sum formula, R N − 1) P0 ( N −1 2 + + + + = , P0 RP0 R P0  R P0 R −1 we have P0 = 1 , R = 3 , and N = 11 . That is, the first term is 1, the common ratio is 3, and there are N = 11 terms. So the sum is ( R N − 1) P0 = (311 − 1) ⋅1 = 311 − 1 = 88,573. R −1 3 −1 2

P1 2 = = 0.5 P0 4

[Note that R =

P1 2 = = 0.2 P0 10

51. (a) Observing the geometric sum formula, ( R N − 1) P0 , P0 + RP0 + R 2 P0 +  + R N −1 P0 = R −1 we have P0 = 1 , R = 2 , and N = 16 . That is, the first term is 1, the common ratio is 2, and there are N = 16 terms. So the sum is ( R N − 1) P0 = (216 − 1) ⋅1 = 216 − 1 = 65,535. 2 −1 R −1

(b) The geometric sum formula:

49. (a) R =

50. (a) R =

R −1 We have P0 = 10 , R = 0.2 , and N = 25 (there are 25 terms being summed). The N

(

0.5 − 1

( R − 1) P P =

[Note that R =

P0 + RP0 + R 2 P0 +  + R N −1 P0 =

(b) The geometric sum formula:

48. (a) R =

(( 0.5) − 1) ( 4) ≈ 7.998 . sum is

(b) The geometric sum formula:

P 18 [Note that R = 2 = too.] 6 P1

(

(there are 12 terms being summed). The

(b) This is as in (a) except we no longer have N = 11. So the sum is ( R N − 1) P0 = (3N − 1) ⋅1 = 3N − 1 . R −1 3 −1 2

ISM: Excursions in Modern Mathematics, 10E

9.4. The Logistic Growth Model 53. (a) p1 = 0.8 × (1 − 0.3) × 0.3 = 0.1680 (b) p2 = 0.8 × (1 − 0.168) × 0.168 ≈ 0.1118 (c) p3 = 0.8 × (1 − 0.11182) × 0.11182

≈ 0.07945 Thus, approximately 7.945% of the habitat’s carrying capacity is taken up by the third generation. 54. (a) p1 = 0.6 × (1 − 0.7) × 0.7 = 0.1260 (b) p2 = 0.6 × (1 − 0.126) × 0.126 ≈ 0.06607 (c) 3.702% of the habitat’s carrying capacity is taken up by the third generation. p3 = 0.6 × (1 − 0.06607) × 0.06607

≈ 0.03702 55. (a) Using the formula pN +1 = r (1 − pN ) pN and a calculator with a memory register or a spreadsheet, we get p1 = 0.1680, p2 ≈ 0.1118, p3 ≈ 0.0795, p4 ≈ 0.0585, p5 ≈ 0.0441, p6 ≈ 0.0337, p7 ≈ 0.0261, p8 ≈ 0.0203, p9 ≈ 0.0159, p10 ≈ 0.0125. (b) Since pN → 0 this logistic growth model predicts extinction for this population. 56. (a) p1 ≈ 0.1260, p2 ≈ 0.0661, p3 ≈ 0.0370, p4 ≈ 0.0214, p5 ≈ 0.0126, p6 ≈ 0.0074, p7 ≈ 0.0044, p8 ≈ 0.0026, p9 ≈ 0.0016, p10 ≈ 0.0009. (b) This logistic growth model predicts extinction for this population. 57. (a) Using the formula pN +1 = r (1 − pN ) pN and a calculator with a memory register or a spreadsheet, we get p1 = 0.4320, p2 ≈ 0.4417, p3 ≈ 0.4439, p4 ≈ 0.4443, p5 ≈ 0.4444, p6 ≈ 0.4444, p7 ≈ 0.4444, p8 ≈ 0.4444, p9 ≈ 0.4444, p10 ≈ 0.4444.

157

(b) The population becomes stable at 4 ≈ 44.44% of the habitat’s carrying 9 capacity. 58. (a) p1 = 0.2400, p2 ≈ 0.2736, p3 ≈ 0.2981, p4 ≈ 0.3139, p5 ≈ 0.3230, p6 ≈ 0.3280, p7 ≈ 0.3306, p8 ≈ 0.3320, p9 ≈ 0.3327, p10 ≈ 0.3330. (b) The population becomes stable at 33.33% of the habitat’s carrying capacity. 59. (a) p1 = 0.3570, p2 ≈ 0.6427, p3 ≈ 0.6429, p4 ≈ 0.6428, p5 ≈ 0.6429, p6 ≈ 0.6428, p7 ≈ 0.6429, p8 ≈ 0.6428, p9 ≈ 0.6429, p10 ≈ 0.6428 (b) The population becomes stable at 9 ≈ 64.29% of the habitat’s carrying 14 capacity. 60. (a) p1 = 0.4, p2 ≈ 0.6, p3 ≈ 0.6, p4 ≈ 0.6, p5 ≈ 0.6, p6 ≈ 0.6, p7 ≈ 0.6, p8 ≈ 0.6, p9 ≈ 0.6, p10 ≈ 0.6 (b) The population becomes stable at 60% of the habitat’s carrying capacity. 61. (a) p1 = 0.5200, p2 = 0.8112, p3 ≈ 0.4978, p4 ≈ 0.8125, p5 ≈ 0.4952, p6 ≈ 0.8124, p7 ≈ 0.4953, p8 ≈ 0.8124, p9 ≈ 0.4953, p10 ≈ 0.8124 (b) The population settles into a two-period cycle alternating between a highpopulation period at approximately 81.24% and a low-population period at approximately 49.53% of the habitat’s carrying capacity. 62. (a) p1 = 0.8424, p2 = 0.4660, p3 ≈ 0.8734, p4 ≈ 0.3880, p5 ≈ 0.8335, p6 ≈ 0.4872, p7 ≈ 0.8769, p8 ≈ 0.3788, p9 ≈ 0.8260, p10 ≈ 0.5045 (b) The population settles into a four-period cycle with the following approximate percentages of the habitat’s carrying capacity: 38%, 83%, 51%, 88%.

158

Chapter 9: Population Growth Models

arithmetic sum formula to find (1 + 2 N − 1) × N 1+ 3 + 5 +  + 2N −1 = 2 2N ⋅ N = . 2 = N 2.

JOGGING 63. (a) Exponential (r = 2) (b) Linear (d = 2) (c) Logistic (d) Exponential ( r =

68. (a) Let a be the length of the shortest side of such a triangle. Then the other sides are of length a + 2 and a + 4. By the Pythagorean theorem, a 2 + (a + 2) 2 = (a + 4) 2

1 ) 3

(e) Logistic (f) Linear (d = -0.15)

a 2 + a 2 + 4a + 4 = a 2 + 8a + 16

(g) Linear (d = 0), Exponential (r = 1), and/or Logistic (they all apply!) 64. (a) Linear

a 2 − 4a − 12 = 0 (a − 6)(a + 2) = 0

So, a = 6 or a = -2. Since a>0, the only right triangles for which the sides are in an arithmetic sequence having d = 2 are 68-10 triangles.

(b) Logistic (c) Exponential

(b) Suppose the shortest side of such a triangle is 1. By the Pythagorean theorem, when the sides are in a geometric sequence with common ratio R, we must have 12 + R 2 = R 4 . Let x = R 2 .

(d) Logistic (e) Linear (f) Exponential 65. The first N terms of the arithmetic sequence are P0 , P0 + d , P0 + 2d , , P0 + ( N − 1)d . Their sum is ( P0 + [ P0 + ( N − 1)d ]) × N N = [ 2 P0 + ( N − 1)d ] . 2 2

66. Similar to Exercise 34(b), we compute th 2 + 4  + 6 +  . The N term is given by N terms

2 + 2( N − 1) = 2 + 2 N − 2 = 2 N . We use the arithmetic sum formula to find (2 + 2 N ) × N 2 + 4 + 6 +  + 2N = 2 2(1 + N ) N . = 2 = ( N + 1) N .

1± 5 . Since x 2 1+ 5 (the golden > 0, we have x = R 2 = 2 ratio φ ). So R = φ and R 2 = φ . So, one right triangle for which the sides are in a geometric sequence is a 1 − φ − φ triangle. Any multiple of these three numbers also works.

Then 1 + x = x 2 . So, x =

1 69. One example is P0 = 8, R = . 2 1 1 Then P1 = 4, P2 = 2, P3 = 1, P4 = , P5 =  2 4

67. Similar to Exercise 33(b), we compute 1 + 3 + 5 + . The Nth term is given by N terms

1 + 2( N − 1) = 1 + 2 N − 2 = 2 N − 1 . We use the

ISM: Excursions in Modern Mathematics, 10E

RUNNING

r 2 p02 − (r 2 + r ) p0 + (r + 1) = 0. Solving this

70. We start with S = P0 + RP0 + R 2 P0 +  + R N −1 P0 . Then,

RS = RP0 + R 2 P0 + R 3 P0 +  + R N P0 . Subtracting the first equation from the second gives RS = RP0 + R 2 P0 + R 3 P0 +  + R N P0 S = P0 + RP0 + R 2 P0 +  + R N −1 P0

equation for p0 gives the two solutions (r + 1) ± (r + 1)(r − 3) , both of which 2r work. p0 =

75. (a) If r > 4, and pN = 0.5 , pN +1 = r (1 − pN ) pN . = 4(1 − 0.5)(0.5)

−−−−−−−−−−−−−−−−−−−− + R N P0

RS − S = − P0 N

That is, RS − S = R P0 − P0 . From this, we solve for S to find S ( R − 1) = ( R N − 1) P0 or

159

( R N − 1) P0 as desired. ( R − 1)

71. No. This would require p0 = p1 = 0.8(1 − p0 ) p0 and p0 = 0 or 1 = 0.8(1 − p0 ) . So, p0 = 0 or p0 = − 0.25 , neither of which are possible. 72. If the population is constant, p0 = p1 = 0.7 . Thus r × 0.7 × (1 − 0.7) = 0.7 , i.e., 0.3r = 1. Solving for r gives r = 10/3. 73. If the population is constant, p1 = p0 . Thus, rp0 (1 − p0 ) = p0 , i.e., r (1 − p0 ) = 1 . Solving

for p0 gives p0 =

r −1 . r

r −1 are r solutions that we don’t want, we may divide r −1  both sides of the equation by p0  p0 −  r   or alternatively by p0 [ rp0 − (r − 1) ] . This

results in the equation

>1 This is impossible since the population cannot be more than 100% of the carrying capacity.

(b) If r = 4, and pN = 0.5 , pN +1 = r (1 − pN ) pN = 4(1 − 0.5)(0.5) = 1 and hence pK = 0 for all K > N + 1 (i.e. the population becomes extinct.) (c) ¼ The graph of (1 − p) p = − p 2 + p is an inverted parabola which crosses the x-axis at 0 and 1. Thus, its maximum value 1 occurs when p = . 2 (d) From part (c), if 0 < p < 1 , 1 and so if we also have 4 0 < r < 4 , then 0 < r (1 − p ) p < 1 .

0 < (1 − p) p <

74. We have p1 = rp0 (1 − p0 ) and p0 = p2 = rp1 (1 − p1 ) . Eliminating p1 and p2 from these equations gives r 3 p04 – 2r 3 p03 + (r 3 + r 2 ) p02 + (1 − r 2 ) p0 = 0.

Since both p0 = 0 and p0 =

= 0.25r

Consequently, if we start with 0 < p0 < 1 and 0 < r < 4 , then 0 < pN < 1 for every positive integer N. 76. If P0 , P1 , P2 , is arithmetic, then PN = P0 + N ⋅ d . So, the sequence 2 P0 , 2 P1 , 2 P2 , is given by

2 P0 , 2 P0 + d , 2 P0 + 2 d , 2 P0 + 3d , which is the same as 2 P0 ⋅1, 2 P0 ⋅ ( 2d ) , 2 P0 ⋅ ( 2d ) , 2 P0 ⋅ ( 2d ) , . 2

But this is a geometric sequence with common ratio 2d .

Chapter 10 7. 215; If 14% of the pieces are missing, then 86% of the pieces are present. 86% of 250 is 0.86 × 250 = 215 .

WALKING 10.1 Percentages 1. (a) 0.0625 (drop the % sign and move the decimal point two places to the left)

8. 585; If 22% are absent, then 78% are present. 78% of 750 is 0.78 × 750 = 585 .

3   (b) 0.0375  Note that 3 % = 3.75%  4  

9. Suppose that x represents the tax rate on the earrings. Then, (1 + x ) $6.95 = $7.61 . So,

7   % = 0.7% (c) 0.007  First write 10  

$7.61 − 1 = 0.095 . It follows that the tax $6.95 rate is 9.5%.

2. (a) 0.0875

1   (b) 0.04125  Note that 4 % = 4.125%  8   4   (c) 0.008  First write % = 0.8%  5   3. (a) 0.0082 (Remember, drop the % sign and move the decimal point two places to the left.) (b) 0.0005 4. (a) 0.0025 (drop the % sign and move the decimal point two places to the left) (b) 0.00003 5. The score for Lab 1 was

61 1 = 81 % . The 75 3

17 = 85% . The score for 20 118 2 = 78 % . So a ranking Lab 3 was 150 3 1 would be Lab 2 (85%), Lab 1 ( 81 % ), Lab 3 3 2 ( 78 % ). 3 score for Lab 2 was

 45   47  ≈ 86.54% , B  ≈ 83.93% , 6. D   52   56   31   34  ≈ 75.56% . A  ≈ 81.58% , C   38   45 

10. Suppose that x represents percentage increase on tuition. Then, (1 + x ) $5760 = $6048 . So,

$6048 − 1 = 0.05 . So, tuition rose 5%. $5760

11. If T was the starting tuition, then the tuition at the end of one year is 110% of T. That is, (1.10)T. The tuition at the end of two years is 115% of what it was after one year. That is, after two years, the tuition was (1.15)(1.10)T. Likewise, the tuition at the end of the three years was (1.10)(1.15)(1.10)T = 1.3915T. The total percentage increase was 39.15%. 12. If G was the starting price of gasoline, then the price at the end of one year is 108% of G. That is, (1.08)G. The price of gasoline at the end of two years is 115% of what it was after one year. That is, after two years, the price was (1.15)(1.08)G. Likewise, the price at the end of the three years was (1.20)(1.15)(1.08)G = 1.4904G. The total percentage increase was 49.04%. 13. Suppose that the cost of the shoes is originally $C. After the mark up, the price is $2.2C. When the shoes go on sale, they are priced at 80% of that price or $(0.8)(2.2)C. When the shoes find their way to the clearance rack, they are priced at 70% of the sale price which is $(0.7)(0.8)(2.2)C. After the customer applies the coupon, the final price is 90% of the clearance price. That is, $(0.9)(0.7)(0.8)(2.2)C = $1.1088C. So, the profit on these shoes is 10.88% more than the original cost $C. [If you still don’t believe it, imagine the shoes are priced at an even $100 originally and do the computations.]

ISM: Excursions in Modern Mathematics, 10E 14. Suppose that home values are originally $H. After one year, the values are $0.93H. After two years, the values are $(0.93)(0.93)H. After four years, the values would be

$ ( 0.93) H = $0.74805201H . That is, the total percentage decrease would be 1 – 0.74805201 = 0.25194799 or 25.2% rounded to the nearest tenth of a percentage point. 4

15. Suppose that when the week began the DJIA had a value of A. The following chart describes the value at the end of each day of the week. Day

DJIA value

Mon.

(1.025)A

Tues.

(1.121) (1.025)A

Wed.

(0.953)(1.121) (1.025)A

Thur.

(1.008)(0.953)(1.121) (1.025)A

Fri.

(0.946)(1.008)(0.953)(1.121)(1.025)A

At the end of the week the DJIA has a value of approximately 1.044A, a net increase of 4.4%. 16. Suppose that when the week began the DJIA had a value of A. At the end of the week the DJIA would have a value of (0.906)(0.952)(1.007)(1.112)(0.968)A (approximately 0.935A), a net decrease of 6.5%.

10.2 Simple Interest 17. $875 (1 + 4 × 0.0428) = $1,024.80 18. $1250 (1 + 3 × 0.051) = $1,441.25 19. Suppose the purchase price is $P. When you cash in the bond after four years, it is worth $ P (1 + 4 × 0.0575) = $4920. This is the face value. Solving for $P, the purchase price for $4920 = $4000 . the bond is $ P = (1 + 4 × 0.0575)

161

20. Suppose the purchase price is $P. When you cash in the bond after 15 years, it is worth $ P (1 + 15 × 0.04) = $8000. This is the face value. Solving for $P, the purchase price for $8000 = $5000 . the bond is $ P = (1 + 15 × 0.04) 21. Using the simple interest formula, we solve $5400 (1 + 8 ⋅ r ) = $8316 for r to

$8316 - $5400 = 0.0675 . So, the $5400 × 8 APR is 6.75%.

find r =

22. Using the simple interest formula, we solve $6000 (1 + 6 ⋅ r ) = $7620 for r to

find r = is 4.5%.

$7620 - $6000 = 0.045 . So, the APR $6000 × 6

23. If the principal doubles in 12 years, then a face value of 2P can be used in the simple interest formula. We solve $ P (1 + 12 ⋅ r ) = $2 P for

$2 P - $ P 1 = ≈ 0.0833 . This $ P × 12 12 gives an APR of 8.33% (to the nearest hundredth percent). r to find r =

24. If the principal doubles in 20 years, then a face value of 2P can be used in the simple interest formula. We solve $ P (1 + 20 ⋅ r ) = $2 P for

$2 P - $ P 1 = = .05 . This gives $ P × 20 20 an APR of 5%.

r to find r =

25. We use the simple interest formula I = Prt where I = $17, P = $100, and t = 14/365 years to solve $17 = $100 ⋅ r ⋅ (14 / 365) for r. We

find r ≈ 4.432 . So, the APR is 443.2%. [Yes, that is correct. Over 400%.] 26. We use the simple interest formula I = Prt where I = $25, P = $100, and t = 14/365 years to solve $25 = $100 ⋅ r ⋅ (14 / 365) for r. We

find r ≈ 6.518 . So, the APR is 651.8%. [Yes, that is correct. Over 600%.]

162

Chapter 10: Financial Mathematics

10.3 Compound Interest 27. (a) $3250(1 + 0.09) 4 = $4587.64 (b) Since interest is credited to the account at the end of each year, no growth takes place during the last half year. The balance after this time is thus $3250(1 + 0.09)5 = $5000.53 . 28. (a) $1237.50(1 + 0.0825)3 = $1569.74 (b) Since interest is credited to the account at the end of each year, no growth takes place during the last half year. The balance after this time is thus $1237.50(1 + 0.0825) 4 = $1699.25 . 29. $25, 000(1 + 0.035) 20 = $49, 744.72 30. We solve $ P(1 + 0.03)11 = $50, 000 for P. This

gives $ P =

$50, 000 = $36,121.06 . (1 + 0.03)11

31. (a) The periodic interest rate (here r = 0.03 and n = 12) is r 0.03 p= = = 0.0025 = 0.25% . n 12 (b) The future value of $1580 in 3 years (36 months of growth at 0.25% interest each

month) is $1580 (1 + 0.0025 ) = $1728.60 . 36

32. (a) The periodic interest rate (here r = 0.036 and n = 12) is r 0.036 p= = = 0.003 = 0.3% . n 12 (b) The future value of $3250 in 4 years (48 months of growth at 0.3% interest each

month) is $3250 (1 + 0.003) = $3752.56 . 48

33. (a) The periodic interest rate (here r = 0.0365 r 0.0365 and n = 365) is p = = = 0.0001 . n 365 That is, 0.01% per day! (b) The future value of $1580 in 3 years ( 3 × 365 = 1095 days of growth) is 1095

 0.0365  $1580 1 +  365  

≈ $1762.83 .

34. (a) The periodic interest rate (here r = 0.036 r 0.0438 and n = 365) is p = = = 0.00012 . n 365 That is, 0.012% per day. (b) The future value of $3250 in 4 years ( 4 × 365 = 1460 days of growth) is 1460

 0.0438  $3250  1 +  365  

≈ $3872.28 .

35. The future value F of P dollars compounded continuously for t years with APR r is F = Pe rt . In this problem, we have P = 1580, r = 0.03, and t = 3. So, the future value 0.03 ×3 is F = $1580e( ) = $1728.80 . 36. The future value F of P dollars compounded continuously for t years with APR r is F = Pe rt . In this problem, we have P = 3250, r = 0.036, and t = 4. So, the future value 0.036 × 4 is F = $3250e( ) = $3753.37 . 37. In this exercise we have P = 1580, r = 0.0365, 500 (years). So, the future value and t = 365 ( 0.0365)×

500

365 is F = $1580e ≈ $1661.008332 . Rounded to the nearest dollar, this is $1661.

38. In this exercise we have P = 3250, r = 0.0438, 1000 (years). So, the future value and t = 365 ( 0.0438)×

1000

365 is F = $3250e ≈ $3664.364768 . Rounded to the nearest dollar, this is $3664.

39. In this exercise, we know the future value F = $868.80 and seek the original principal P. We also have r = 0.0275 and t = 3 (years). To determine the original principal P, we solve 0.0275)×3 $868.80 = Pe( for P. Solving gives 868.80 P = $ ( 0.0275)×3 ≈ $800 . e 40. In this exercise, we know the future value F = $1645.37 and seek the original principal P. We also have r = 0.0185 and t = 5 (years). To determine the original principal P, we 0.0185)×5 solve $1645.37 = Pe( for P. Solving 1645.37 gives P = $ ( 0.0185)×5 ≈ $1500 . e

ISM: Excursions in Modern Mathematics, 10E 1

 0.06  41. (a) The APY is 1 +  − 1 = 0.06 = 6% . 1   2

 0.06  (b) 1 +  − 1 = 0.0609 = 6.09% . 2   12

 0.06  (c) 1 +  − 1 ≈ 0.061678 = 6.1678% 12  

(d) 1e0.06×1 − 1 ≈ 0.061837 = 6.1837% 1

 0.036  42. (a) 1 +  − 1 = 0.036 = 3.6% . 1   2

 0.036  (b) 1 +  − 1 = 0.036324 = 3.6324% 2   12

 0.036  (c) 1 +  − 1 ≈ 0.03660 = 3.660% 12  

(d) 1e0.036×1 − 1 ≈ 0.03666 = 3.666% 43. We estimate the value of t that makes $2 P = P (1 + 0.06 ) true. That is, we estimate t

163

(b) The value of the retirement savings account is found just like in part (a) only we use P = 750:  (1 + 0.066 )45 − 1  $V = $750   0.066   ≈ $190, 283.26. Note that this is amount is exactly half of what was found in part (a) since the annual contributions are half of what they were in part (a). This illustrates the linear relationship between the value V of the retirement savings account and the annual contributions P.

(c) The value of the retirement savings account is found just like in part (a) only we use P = 2250:  (1 + 0.066 )45 − 1  $V = $2250   0.066   ≈ $570,849.78. Alternatively, one could simply add the values found in parts (a) and (b): $380,566.52 + $190,283.26 = $570,849.78.

that solves 2 = (1.06 ) . Trial and error gives t = 12 years. t

44. We estimate the value of t that makes $2 P = P (1 + 0.072 ) true. That is, we estimate t

t that solves 2 = (1.072 ) . Trial and error gives t = 10 years. t

10.4 Retirement Savings 45. (a) We use the Retirement Savings Formula (with annual contributions),  (1 + r )Y − 1  $V = $ P   , where the annual r   contribution (in dollars) made at the end of each year is P = 1500, the annual interest rate is r = 0.066, and the number of years of retirement savings is Y = 45. The value of the retirement savings account is then calculated to be  (1 + 0.066 )45 − 1  $V = $1500   0.066   ≈ $380,566.52.

46. (a) We use the Retirement Savings Formula (with annual contributions),  (1 + r )Y − 1  $V = $ P   , where the annual r   contribution (in dollars) made at the end of each year is P = 1200, the annual interest rate is r = 0.054, and the number of years of retirement savings is Y = 40. The value of the retirement savings account is then calculated to be  (1 + 0.054 )40 − 1  $V = $1200   0.054   ≈ $159,920.54.

(b) The value of the retirement savings account is found just like in part (a) only we use P = 600:  (1 + 0.054 )40 − 1  $V = $600   0.054   ≈ $79,960.27. Note that this is amount is exactly half of what was found in part (a) since the annual contributions are half of what they were in part (a). This illustrates the linear relationship between the value V of the

164

Chapter 10: Financial Mathematics

retirement savings account and the annual contributions P. (c) The value of the retirement savings account is found just like in part (a) only we use P = 1800:  (1 + 0.054 )40 − 1  $V = $1800   0.054   ≈ $239,880.81. Alternatively, one could simply add the values found in parts (a) and (b): $150,920.54 + $79,960.27 = $239,880.81. 47. (a) We use the Retirement Savings Formula (with annual contributions),  (1 + r )Y − 1  $V = $ P   , where the annual r   contribution (in dollars) made at the end of each year is P = 1500, the annual interest rate is r = 0.066, and the number of years of retirement savings is Y = 40. The value of the retirement savings account is then calculated to be  (1 + 0.066 )40 − 1  $V = $1500   0.066   ≈ $270, 250.07. (b) Again, use the Retirement Savings Formula with P = 1500, r = 0.066, and Y = 50. Then,  (1 + 0.066 )50 − 1  $V = $1500   0.066   ≈ $532, 419.17. (c) Use the Retirement Savings Formula with P = 1500, r = 0.066, and Y = 52. Then,  (1 + 0.066 )52 − 1  $V = $1500   0.066   ≈ $608,116.71. This exercise illustrates that the relationship between the value V of the retirement savings account and the number of years Y of annual contributions is not linear (as we know, this relationship is exponential). 48. (a) We use the Retirement Savings Formula (with annual contributions),

 (1 + r )Y − 1  $V = $ P   , where the annual r   contribution (in dollars) made at the end of each year is P = 1200, the annual interest rate is r = 0.054, and the number of years of retirement savings is Y = 35. The value of the retirement savings account is then calculated to be  (1 + 0.054 )35 − 1  $V = $1200   0.054   ≈ $117,803.84.

(b) Use the Retirement Savings Formula with P = 1200, r = 0.054, and Y = 45. Then,  (1 + 0.054 )45 − 1  $V = $1200   0.054   ≈ $214, 705, 01. (c) Use the Retirement Savings Formula with P = 1200, r = 0.054, and Y = 50. Then,  (1 + 0.054 )50 − 1  $V = $1200   0.054   ≈ $285,967.41. This exercise illustrates that the relationship between the value V of the retirement savings account and the number of years Y of annual contributions is not linear (as we know, this relationship is exponential). 49. (a) We use the Retirement Savings Formula (with monthly contributions), T

1 + ( r / 12 )  − 1 , where the $V = $ M  ( r / 12 )

monthly contribution (in dollars) made at the end of each month is M = 125, the annual interest rate is r = 0.066, and the number of months of retirement savings is T = 12(45) = 540 . The value of the retirement savings account is then calculated to be 540

1 + ( 0.066 / 12 )  $V = $125  ( 0.066 / 12 )

−1

≈ $416, 680.55. Compare this value to that found in Exercise 45(a). The value of the account is much more even though the annual contributions were the same (just made at

ISM: Excursions in Modern Mathematics, 10E the end of each month rather than the end of each year). (b) Use the Retirement Savings Formula (with monthly contributions) with M = 62.50, r = 0.066, and T = 12(45) = 540 . Then, 540

1 + ( 0.066 / 12 )  $V = $62.50  ( 0.066 / 12 )

−1

≈ $208,340.28. Compare this value to that found in Exercise 45(b).

(c) Use the Retirement Savings Formula (with monthly contributions) with M = 187.50, r = 0.066, and T = 12(45) = 540 . Then, 540

1 + ( 0.066 /12 )  $V = $187.50  ( 0.066 / 12 )

−1

≈ $625, 020.83. Compare this value to that found in Exercise 45(c). [Alternatively, one could simply add the values found in parts (a) and (b): $416,680.55 + $208,340.28 = $625,020.83.]

50. (a) We use the Retirement Savings Formula (with monthly contributions), T

1 + ( r / 12 )  − 1 , where the $V = $ M  ( r / 12 )

monthly contribution (in dollars) made at the end of each month is M = 100, the annual interest rate is r = 0.054, and the number of months of retirement savings is T = 12(40) = 480 . The value of the retirement savings account is then calculated to be 480

1 + ( 0.054 / 12 )  $V = $100  ( 0.054 / 12 )

−1

≈ $169,538.30. Compare this value to that found in Exercise 46(a). The value of the account is much more even though the annual contributions were the same (just made at the end of each month rather than the end of each year).

480

1 + ( 0.054 / 12 )  $V = $50  ( 0.054 / 12 )

165

−1

≈ $84, 769.15. Compare this value to that found in Exercise 46(b).

(c) Use the Retirement Savings Formula (with monthly contributions) with M = 150, r = 0.054, and T = 12(40) = 480 . Then, 480

1 + ( 0.054 / 12 )  $V = $150  ( 0.054 / 12 )

−1

≈ $254,307.45. Compare this value to that found in Exercise 46(c). [Alternatively, one could simply add the values found in parts (a) and (b): $169,538.30 + $84,769.15 = $254,307.45.]

51. (a) We use the Retirement Savings Formula (with monthly contributions), T

1 + ( r / 12 )  − 1 , where the $V = $ M  ( r / 12 )

monthly contribution (in dollars) made at the end of each year is M = 125, the annual interest rate is r = 0.066, and the number of months of retirement savings is T = 12(40) = 480 . The value of the retirement savings account is then calculated to be 480

1 + ( 0.066 /12 )  $V = $125  ( 0.066 /12 )

−1

≈ $293, 459.20. Compare this value to that found in Exercise 49(a).

(b) Use the Retirement Savings Formula with M = 125, r = 0.066, and T = 12(50) = 600 . Then, 600

1 + ( 0.066 /12 )  $V = $125  ( 0.066 /12 )

≈ $587,922.62.

−1

166

Chapter 10: Financial Mathematics (c) Use the Retirement Savings Formula with M = 125, r = 0.066, and T = 12(52.5) = 630 . Then, 630

1 + ( 0.066 /12 )  $V = $125  ( 0.066 /12 )

−1

≈ $697,143.48. This exercise illustrates that the relationship between the value V of the retirement savings account and the number of months T of monthly contributions is not linear (as we know, this relationship is exponential).

52. (a) We use the Retirement Savings Formula (with monthly contributions), T

1 + ( r / 12 )  − 1 , where the $V = $ M  ( r / 12 )

monthly contribution (in dollars) made at the end of each year is M = 100, the annual interest rate is r = 0.054, and the number of months of retirement savings is T = 12(35) = 420 . The value of the retirement savings account is then calculated to be 420

1 + ( 0.054 / 12 )  $V = $100  ( 0.054 / 12 )

−1

≈ $124, 252.52. Compare this value to that found in Exercise 50(a).

(b) Use the Retirement Savings Formula with M = 100, r = 0.054, and T = 12(45) = 540 . Then, 540

1 + ( 0.054 / 12 )  $V = $100  ( 0.054 /12 )

−1

≈ $228,825.15.

53. We use the Retirement Savings Formula (with monthly contributions), T

1 + ( r / 12 )  − 1 , where the $V = $ M  ( r / 12 )

monthly contribution M (in dollars) made at the end of each year is unknown, the annual interest rate is r = 0.066, the number of months of retirement savings is T = 12(40) = 480 , and the value of the retirement account at the end of this time is V = 500,000 (in dollars). Since all the variables in this formula are known except for M, we solve 480

1 + ( 0.066 / 12 )  $500, 000 = $M  ( 0.066 / 12 )

594

obtain a monthly contribution of $500, 000 ( 0.066 / 12 ) $M = ≈ $212.98. 480 1 + ( 0.066 / 12 )  − 1 54. We use the Retirement Savings Formula T

1 + ( r / 12 )  − 1 , where the $V = $ M  ( r / 12 )

monthly contribution M (in dollars) made at the end of each year is unknown, the annual interest rate is r = 0.054, the number of months of retirement savings is T = 12(50) = 600 , and the value of the retirement account at the end of this time is V = 400,000 (in dollars). Solving 600

1 + ( 0.054 / 12 )  $400, 000 = $M  ( 0.054 / 12 )

−1

, we

obtain a monthly contribution of $400, 000 ( 0.054 / 12 ) $M = ≈ $130.53. 600 1 + ( 0.054 / 12 )  − 1

(c) Use the Retirement Savings Formula with M = 100, r = 0.054, and T = 12(49.5) = 594 . Then, 1 + ( 0.054 / 12 )  $V = $100  ( 0.054 /12 )

−1

≈ $297, 705.95. This exercise illustrates that the relationship between the value V of the retirement savings account and the number of months T of monthly contributions is not linear (as we know, this relationship is exponential). Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

167

55. A table can help better organize the calculations needed in this automatic escalation retirement plan. Think of this as three plans (Plans A, B, and C). In Plan A, $125 is contributed each month for 480 months (40 years). This leads to an ending value of 480

1 + ( 0.066 / 12 )  $V = $125  ( 0.066 / 12 )

−1

≈ $293, 459.20.

In Plan B, $100 ($225-$125) is contributed each month for 240 months (20 years). This leads to an ending 240

1 + ( 0.066 / 12 )  value of $V = $100  ( 0.066 / 12 )

−1

≈ $49, 634.67.

In Plan C, $175 ($400-$225) is contributed each month for 60 months (5 years). This leads to an ending 60

1 + ( 0.066 / 12 )  − 1 value of $V = $100  ≈ $12,399.90. ( 0.066 / 12 )

Contribution

Years

Number of Months

Value

Plan A

$125

1 to 40

480

$293,459.20

Plan B

$100

21 to 40

240

$49,634.67

Plan C

$175

35 to 40

$12,399.90

Total

$355,493.77

56. A table can help better organize the calculations needed in this automatic escalation retirement plan. Think of this as three plans (Plans A, B, and C). In Plan A, $100 is contributed each month for 540 months (45 years). This leads to an ending value of 540

1 + ( 0.054 / 12 )  $V = $100  ( 0.054 / 12 )

−1

≈ $228,825.15.

In Plan B, $125 is contributed each month for 300 months (25 years). This leads to an ending value of 300

1 + ( 0.054 / 12 )  $V = $125  ( 0.054 / 12 )

−1

≈ $79, 048.93.

In Plan C, $200 is contributed each month for 120 months (10 years). This leads to an ending value of 120

1 + ( 0.054 / 12 )  $V = $200  ( 0.054 / 12 )

−1

≈ $31, 730.20.

Contribution

Years

Number of Months

Value

Plan A

$100

1 to 45

540

$228,825.15

Plan B

$125

21 to 45

300

$79,048.93

Plan C

$200

35 to 45

120

$31,730.20

Total

$339,604.28

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Chapter 10: Financial Mathematics

10.5 Consumer Debt 57. The amortization formula is given by p(1 + p )T 0.06 where p = = 0.005 M =P T (1 + p ) − 1 12 is the monthly interest rate, M is the monthly payment, P = $14,500 is the principal, and T = 36 is the number of monthly payments. So, the monthly payments are 0.005(1 + 0.005)36 M = $14,500 = $441.12 (1 + 0.005)36 − 1 58. The amortization formula is given by p(1 + p )T 0.048 where p = = 0.004 M =P T (1 + p ) − 1 12 is the monthly interest rate, M is the monthly payment, P = $18,700 is the principal, and T = 60 is the number of monthly payments. So, the monthly payments are 0.004(1 + 0.004)60 = $351.18. M = $18, 700 (1 + 0.004)60 − 1 59. The amortization formula is given by p(1 + p )T 0.036 where p = = 0.003 M =P T (1 + p ) − 1 12 is the monthly interest rate, M = $400 is the monthly payment, P is the principal (amount to be financed), and T = 48 is the number of monthly payments. Solving for the value of P to the nearest dollar gives (1 + p )T − 1 P=M p(1 + p)T

(1 + 0.003) − 1 48 0.003 (1 + 0.003) 48

= $400

≈ $17,857.

60. The amortization formula is given by p(1 + p )T 0.06 where p = = 0.005 M =P T (1 + p ) − 1 12 is the monthly interest rate, M = $320 is the monthly payment, P is the principal (amount to be financed), and T = 60 is the number of monthly payments. Solving for the value of P

to the nearest dollar gives (1 + p )T − 1 P=M p(1 + p)T

(1 + 0.005) − 1 60 0.005 (1 + 0.005 ) 60

= $320

≈ $16,552.

61. (a) The amortization formula is given by p(1 + p )T 0.0575 where p = is M =P 12 (1 + p )T − 1 the monthly interest rate, M is the monthly payment, P = $160,000 is the principal, and T = 30 × 12 = 360 is the number of monthly payments. So, the monthly payments are 360 0.0575  0.0575  1 +  12  12  M = $160, 000 360 0.0575   1 +   −1 12   = $933.72. (b) To determine the amount of interest paid over the life of the loan, we calculate the amount paid over the course of 360 months and subtract the original principal of $160,000. This gives 360 × $933.72 − $160, 000 = $176,139.20. 62. (a) The new monthly payments are determined by the amortization formula: 180 0.0525  0.0525  1 +   12  12  M = $95, 000 180  0.0525  1 +  −1 12   = $763.68. Since the current payment is $1104 per month, their monthly savings are $1104 − 763.68 ≈ $340. (b) To determine the amount of interest paid over the life of the new loan, we calculate the amount paid over the course of 180 months and subtract the amount that was refinanced (i.e. $95,000). This gives 180 × $763.68 − $95, 000 = $42, 462.40.

ISM: Excursions in Modern Mathematics, 10E 63. The amortization formula is given by p(1 + p )T 0.055 where p = is the M =P T 12 (1 + p ) − 1 monthly interest rate, M = $950 is the monthly payment, P is the principal (amount financed), and T = 240 is the number of monthly payments. Solving for the value of P gives (1 + p)T − 1 P=M p (1 + p)T 240

 0.055  1 +  −1 12  = $950  240 0.055  0.055  1 +   12  12  ≈ $138,104. Adding the $25,000 down payment to the amount financed P will give the selling price of the house which is $138,104 + $25,000 = $163,104.

64. The amortization formula is given by p(1 + p )T 0.0575 M =P where p = is the T (1 + p ) − 1 12 monthly interest rate, M = $877 is the monthly payment, P is the principal (amount financed), and T = 360 is the number of monthly payments. Solving for the value of P gives (1 + p )T − 1 P=M p(1 + p)T 360

 0.0575  1 +  −1 12  = $877  360 0.0575  0.0575  + 1   12  12  ≈ $150, 281. Adding the $35,000 down payment to the amount financed P will give the selling price of the house which is $150,281 + $35,000 = $185,281.

169

JOGGING 65. (a) First, organize the information by periods of constant balance. Period

Daily Balance

Days

6/19 – 6/20

$1200

6/21 – 6/29

1379.58

6/30 – 7/4

1419.58

7/5 – 7/14

1517.93

7/15 – 7/18

1017.93

The average daily balance is then easy to calculate. We simply add the daily balances and then divide by 30. The sum of the daily balances is 2(1200) + 9(1379.58) + 5(1419.58) + 10(1517.93) + 4(1017.93) = 41,165.14. Dividing by 30 gives an average daily balance of $1372.17. (b) The periodic interest rate is 30 × 0.195 = 0.016027 . So, the finance 365 charge is $1372.17 × 0.016027 = $21.99 . (c) $1017.93 + $21.99 = $1039.92 66. (a) First, organize the information by periods of constant balance. Period

Daily Balance

Days

8/26 – 8/30

$5000

8/31 – 9/11

$5148.55

9/12 – 9/18

$5178.55

9/19 – 9/21

$5282.54

9/22 – 9/25

$5082.54

To compute the average daily balance, we add the daily balances and then divide by 31. The sum of the daily balances is 5(5000) + 12(5148.55) + 7(5178.55) + 3(5282.54) + 4(5082.54) =$159,210.23. Dividing by 31 gives an average daily balance of $5135.81. Copyright © 2022 Pearson Education, Inc.

170

Chapter 10: Financial Mathematics (b) The periodic interest rate is 31 × 0.2199 = 0.018676 . So, the finance 365 charge is $5135.81× 0.018676 = $95.92 . (c) $5082.54 + $95.92 = $5178.46

67. Let m be the markup and c be the retailer’s cost. 0.75(1 + m)c = 1.5c 0.75 + 0.75m = 1.5 0.75m = 0.75 m =1 The markup should be 100%. 68. We solve (1.124864)T = (1 + p ) T for p. But, this gives p = 3 1.124864 − 1 = 0.04 . So, tuition increased 4% each year. 3

69. (a) Another way to write y% is y/100 and another way to write x% is x/100. So, taking y% off an item is the same as multiplying the price of the item by 1y/100. Taking x% off an item is the same as multiplying the price by 1-x/100. If the item originally cost $P, then the result of these two discounts would be an item x   y    priced at $ 1 −   1 −  P .  100   100  

y   x    (b) $ 1 −   1 −  P  100   100   (c) Multiplication is commutative. That is, it can be done in any order. 70. Given the initial deposit of P = $500 , we

−1000 ± 10002 − 4(500)(−61.8) 2(500)

−1000 ± 1,123, 600 1000 −1000 ± 1060 = 1000 60 −2060 , = 1000 1000 = 0.06, −2.06 Taking the positive value of r gives an annual yield of 6%. =

71. The loan (amount financed) from Bank A would be P = $90,000 with a periodic (monthly) interest rate of p = 0.10 . We use 12 p(1 + p )T the amortization formula M = P (1 + p )T − 1 with T = 360 monthly payments of $M. The loan from Bank A requires a monthly payment of p(1 + p)T M =P (1 + p)T − 1 = $90, 000

0.1  0.1  1 +  12  12   0.1  1 +   12 

360

−1

≈ $789.81 The loan from Bank B would be $92,783.51 (leaving $90,000 after paying the 3% loan fee). Using the same ideas as above, the monthly payment turns out to be $780.17 for 360 months. Consequently, the loan from Bank B is $9.64 a month less than the loan from Bank A, saving $3470.40 over the 360 months. Bank B offers the better deal.

solve F = P (1 + r ) for the annual yield r. But, 561.80 = 500(1 + r ) 2 2

561.80 = 500r 2 + 1000r + 500 0 = 500r 2 + 1000r − 61.80 and the quadratic equation gives

ISM: Excursions in Modern Mathematics, 10E 72. If you took the rebate, you would need to finance P = $24,035 - $1,500 = $22,535 at 6% interest for T = 60 months. The monthly rate is p = 0.06 /12 = 0.005 . We use the amortization formula with T = 60 payments to determine the monthly payment M: p(1 + p)T M =P (1 + p )T − 1 = $22,535

0.06  0.06  1 +  12  12 

 0.06  1 +  −1 12  

≈ $435.66 If, on the other hand, you chose 0% financing for the first 36 months, you would need to finance P = $24,035. To find the monthly payments, set up the equation 24

 0.06  1 +  −1 12  and $24, 035 = 36 M + M  24 0.06  0.06  1 +  12  12  solve for M. The first term on the right represents the contribution to the principal of 36 payments with 0% APR; the second term represents the contribution to the principal of 24 payments with 6% APR. Solving gives $24, 035 M = 24  (1.005) − 1   36 + 24  0.005 (1.005 )   ≈ $410.41 So, the better deal is to take the 0% financing for 36 months.

73. (a) First, we show that the increase in monthly payment that will pay the mortgage off in 21 years is indeed close to $100 using the amortization formula. The monthly payment on a 30-year mortgage for $100,000 with an APR of 6% would be p (1 + p)T M =P (1 + p)T − 1 = $100, 000

0.06  0.06  1 +  12  12  360

M =P

p(1 + p)T (1 + p )T − 1

= $100, 000

0.06  0.06  1 +  12  12 

252

. 252  0.06  1 +  −1 12   ≈ $698.86 (call it $700). The interest paid over 30 years would be about 360 × $600 − $100, 000 = $116, 000 . The interest paid if the mortgage were paid off in 21 years would be 252 × $700 − $100, 000 = $76, 400 which equates to a savings of $40,000. (b) To pay off a P = $100,000 in 15 years (T = 180 months) with an APR of 6% would require a monthly payment of p (1 + p)T M =P (1 + p)T − 1 0.005 (1.005 )

180

= $100, 000

(1.005)

180

−1

≈ $843.86. This is an increase of approximately $844 − $600 = $244 each month (over a 30-year mortgage payment). The total interest paid on this 15-year mortgage is 180 × $843.86 - $100,000 = $51,894.80 (call it $51,895). The amount of interest saved in going from a 30- to a 15-year loan would therefore be about $116, 000 − $51,895 = $64,105.

(c) To pay off a P = $100,000 mortgage in t (t < 30) years at an APR of 6% would require a payment of p (1 + p)T M =P (1 + p)T − 1 0.005 (1.005 )

12 t

= $100, 000

(1.005) 12 t (1.005 ) = $500 12 t (1.005 ) − 1

360

 0.06  1 +  −1 12   ≈ $599.55 (call it $600). However, to pay off a $100,000 in 21 years would require a payment of

171

12 t

−1

This is an increase of 12 t (1.005) $500 − $600 each month. 12 t (1.005) − 1

172

Chapter 10: Financial Mathematics

74. After t years, investor A will have $ Pe rt . After 2t years, investor B will have $ Pe( r / 2)( 2t ) = $ Pert . That is, both investors end up with the same amount of money.

 (1 + r )Y − 1  75. (a) Let V = P   be the value of r  

the retirement savings account with annual contribution P. Then the value of a retirement savings account with annual contribution cP is   (1 + r )Y − 1    (1 + r )Y − 1  cP   = c P    = cV .   r r      

(b) For a loan with principal P, the monthly  p (1 + r )T  payments are M = P   . For a T  (1 + p ) − 1  loan with principal Q, the payments are  p (1 + r )T  N = Q  . If the principal is T  (1 + p ) − 1  changed to (P+Q), the monthly payments  p (1 + r )T  become ( P + Q )   = T  (1 + p ) − 1   p (1 + r )T   p (1 + r )T  P + Q    = M+N. T T  (1 + p ) − 1   (1 + p ) − 1 

 (1 + r )Y − 1  (b) Let V = P   be the value of the r   retirement savings account with annual  (1 + r )Y − 1  contribution P, and let W = Q   r   be the value of the retirement savings account with annual contribution Q. Then the value of a retirement savings account with annual  (1 + r )Y − 1  contribution (P+Q) is ( P + Q )   r   Y Y  (1 + r ) − 1   (1 + r ) − 1  =P +Q  = V +W. r r     76. For all loans assume that the APR is r, that p = r/12 deontes the monthly interest rate, and that the loan is amortized over T monthly installments. (a) For a loan with principal P, the monthly  p (1 + r )T  payments are M = P   . If the T  (1 + p ) − 1 

principal is changed to cP, the monthly  p (1 + r )T  payments become cP   = T  (1 + p ) − 1    p (1 + r )T   c P    = cM .   (1 + p )T − 1     

Chapter 11 WALKING 11.2 Reflections 1. (a)

(b) F. See the figure below.

C. The image point of P under the reflection with axis l1 is found by drawing a line through P perpendicular to l1 and finding the point on this line on the opposite side of l1 which is the same distance from l1 as the point P. This point is C. (c)

E. See the figure below.

(d) B. See the figure below.

2. (a) C (b) A (c) D. Line l3 has slope of ½. So, the line segment connecting P with its image under the reflection has slope of -2. Think of moving P down 2 and right 1 to the axis and then down 2 and right 1 again to the point D. (d) E

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3. (a)

Remember that the axis of reflection needs to be perpendicular to the line segment connecting A and A’. That is, l is the perpendicular bisector of AA’.

(b) The image of A’ under the reflection is A. (A reflects to A’ which reflects about l back to A) 4. (a)

Remember that the axis of reflection needs to be perpendicular to the line segment connecting P and P’. That is, l is the perpendicular bisector of PP’.

(b) The image of P’ under the reflection is P. (P reflects to P’ which reflects about l back to P) 5. (a)

(b)

Remember that the axis of reflection needs to be perpendicular to the line segment connecting S and S’. That is, l is the perpendicular bisector of SS’.

Reflect each point P, Q and R about the line l just as you did for S in (a). Then connect the dots to form the quadrilateral P’Q’R’S’.

ISM: Excursions in Modern Mathematics, 10E 6. (a)

(b)

7. (a) (b)

175

The axis of reflection needs to be perpendicular to the line segment connecting P and P’. That is, l is the perpendicular bisector of PP’.

Reflect each point Q, R, and S about the line l just as you did for P in (a). Then connect the dots to form the quadrilateral P’Q’R’S’.

The axis of reflection needs to be perpendicular to the line segment connecting P and P’. That is, l is the perpendicular bisector of PP’. Since S is on the axis of reflection l, we see that S’, the image of S under the reflection, is at the same position as S.

(c)

Reflect each point P, Q, R, and S about the line l found in (a).

(d)

Any point that is on both the axis of reflection l and the quadrilateral PQRS is a fixed point of the reflection.

8. (a) (b)

The axis of reflection needs to be perpendicular to the line segment connecting P and P’. That is, l is the perpendicular bisector of PP’.

176

Chapter 11: The Mathematics of Symmetry (c) (d)

Reflect each point P, Q, R, and S about the line l found in (a). Any point that is on both the axis of reflection l and the quadrilateral PQRS is a fixed point of the reflection.

9. (a)

The axis of reflection is perpendicular to the line segment connecting P and P’. That is, l is the perpendicular bisector of PP’.

(b)

Reflect each “corner” of the arrow about the line l found in (a). Then connect these images with line segments.

10. (a)

The axis of reflection is perpendicular to the line segment connecting R and R’. That is, l is the perpendicular bisector of RR’.

(b)

Reflect each “corner” of the arrow about the line l found in (a). Then connect these images with line segments.

11.

Since points A and B are fixed points, the axis of reflection l must pass through these points.

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Since points A and B are fixed points, the axis of reflection l must pass through these points.

12.

11.3 Rotations 13. The following figure may be of help in solving each part of this exercise.

(a) I. Think of A as the center of a clock in which B is the “9” and I is the “12.” (b) G. Think of B as the center of a clock in which A is the “3” and G is the “6.” (c) A. Think of B as the center of a clock in which D is the “1” and A is the “3.” (d) F. Think of B as the center of a clock in which D is the “1” and F is the “5.” (e) E. Think of A as the center of a clock in which I is the “12.” Rotating 3690° has the same effect as rotating 10 × 360° + 90° . That is, as rotating 10 times around the circle and then another 90° . 14. Referring to the figure in the above exercise may be helpful. (a) G. Think of B as the center of a clock in which C is the “9” and G is the “6.” (b) B. Think of A as the center of a clock in which F is the “7” and B is the “9.” (c) C. Think of B as the center of a clock in which F is the “5” and C is the “9.” (d) B. (e) C. Think of B as the center of a clock in which G is the “6.” Rotating 3870° has the same effect as rotating 10 × 360° + 270° . That is, as rotating 10 times around the circle and then another 270° . 15. (a) 2(360°) − 710° = 10° (b) 710° − 360° = 350°

19 (c) 360 7100 with a remainder of 260 . Hence, a counterclockwise rotation of 7100 is equivalent to a clockwise rotation of 360 − 260 = 100 .

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Chapter 11: The Mathematics of Symmetry 197 (d) 360 71, 000 with a remainder of 80 . Hence, a clockwise rotation of 71, 000 is equivalent to a

clockwise rotation of 80 . 16. (a) 500° − 360° = 140° (b) 2 ( 360° ) − 500° = 220°

13 (c) 360 5000 with a remainder of 320 . Hence, a clockwise rotation of 5000 is equivalent to a

(

)

clockwise rotation of 5000 − 13 360 = 320 . 138 (d) 360 50, 000 with a remainder of 320 . Hence, a clockwise rotation of 50, 000 is equivalent to a

counterclockwise rotation of 40 . 17. (a) (b)

Since BB’ and CC’ are parallel, the intersection of BC and B’C’ locates the rotocenter O. This is a 90 clockwise rotation. [Note: The vertical line is the perpendicular bisector of both BB’ and CC’.]

18. (a)

(b)

The rotocenter is located at the intersection of the perpendicular bisectors of AA’ and BB’. This is a 90 counterclockwise rotation.

Rotate each “corner” of the arrow counterclockwise 90 .

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179

Since AA’ and BB’ are parallel, the intersection of AB and A’B’ locates the rotocenter O. This is a 90 clockwise rotation. [Note: The vertical line is the perpendicular bisector of both AA’ and BB’.]

(b)

20. (a) (b)

Rotate each “corner” of the arrow clockwise 90 .

Since QQ’ and RR’ are parallel, the intersection of QR and Q’R’ locates the rotocenter O. This is a 90 clockwise rotation. [Note: The horizontal line is the perpendicular bisector of both QQ’ and RR’.]

(c)

Rotate points P, Q, R, and S clockwise 90 .

21.

The equilateral triangles that make up the grid have interior angles that each measure 60 . So the triangle ABC is being rotated 60 clockwise.

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Chapter 11: The Mathematics of Symmetry

22.

The equilateral triangles that make up the grid have interior angles that each measure 60 .

11.4 Translations 23. (a) C. Vector v1 translates a point 4 units to the right so the image of P is C. (b) C. Vector v2 translates a point 4 units to the right so the image of P is C. (c) A. Vector v3 translates a point up 2 units and right 1 unit so the image of P is A. (d) D. Vector v4 translates a point down 2 units and left 1 unit so the image of P is D. 24. (a) D. Vector v1 translates a point 2 units left and 2 units down. (b) A. Vector v2 translates a point 2 units right and 2 units up. (c) B. Vector v3 translates a point 3 units right and 3 units up. (d) C. Vector v4 translates a point to the right 3 units. 25. (a) (b) (c)

Point E’ is located left 10 units and down 1 unit from E. So, point A’ is also located left 10 units and down 1 unit from A. Doing the same for all corners of the region and connecting them will produce the image. The translation vector v is an arrow pointing left 10 units and down 1 unit. We can illustrate it from any starting position.

26. (a) (b) (c)

Point Q’ is located right 9 units and up 2 units from Q. So, point P’ is also located right 9 units and up 2 units from P. Doing the same for all corners of the region and connecting them will produce the image. The translation vector v is an arrow pointing right 9 units and up 2 units. We can illustrate it from any starting position.

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27.

28.

11.5 Glide Reflections 29.

First, reflect the triangle ABC about the axis l (to form triangle A*B*C*). Then, glide the figure three units to the right.

30.

Normally, one would first reflect ABCD about the axis l (to form A*B*C*D*). Then, one would glide the figure according to v. In this case, however, we have translated according to v first (giving A*B*C*D*) and reflected about l after that.

31.

First reflect ABCDE about the axis l (to form A*B*C*D*E*). Then, glide the figure according to v. [Note: The glide v is determined by D*D’ which is easily seen after the reflection.]

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Chapter 11: The Mathematics of Symmetry

First reflect PQRS about the axis l (to form P*Q*R*S*). Then, glide the figure according to v.

32.

[Note: The glide v is determined by P*P’ which is easily seen after the reflection.]

33. (a) (b)

The midpoints of line segments BB’ and DD’ determine the axis of reflection for this glide reflection. To determine the location of A’, reflect A across l to a point A*. Then, glide A* down 4 units to A’. [Note: B reflects to a point 4 units above B’.]

(c)

34. (a) (b)

Reflect the figure ABCDE about the axis of reflection l (to form A*B*C*D*E*). Then, glide the figure four units down.

The midpoints of line segments AA’ and CC’ determine the axis of reflection l for this glide reflection. To determine the location of B’, reflect B across l to a point B*. Then, glide B* right 8 units to B’. [Note: A reflects to a point 8 units left of A’.]

(c)

35.

Reflect the figure ABCD about the axis of reflection (to form A*B*C*D*). Then, glide the figure 8 units right.

The midpoints of line segments PP’ and QQ’ usually determine the axis of reflection for a glide reflection. However, in this case, the midpoints are the same point. So the line the PQ is drawn and the line perpendicular to PQ that passes through the midpoints is the axis of reflection l. We then reflect the figure about this axis l and glide the result 4 units to the left (so that P* lands on P’). Copyright © 2022 Pearson Education, Inc.

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The midpoints of line segments PP’ and QQ’ determine the axis of reflection l for this glide reflection. First, reflect the figure about the axis (the result is labeled with P* and Q*). Then, glide (translate) the figure according to vector v.three diagonal units down and to the left (i.e. three units down and three units left).

36.

37.

The midpoints of line segments CC’ and DD’ determine the axis of reflection l for this glide reflection. First, reflect the figure about this 60-degree axis (the result is the figure labeled with C* and D*). Then, glide down and to the right. 38.

The midpoints of line segments AA’ and DD’ determine the axis of reflection l for this glide reflection. First, reflect the figure about this 60-degree axis (the result is the figure labeled with A* and D*). Then, glide down and to the right.

11.6 Symmetries and Symmetry Types 39. The following figure gives the axes of reflection for each figure.

(a) Reflection with axis going through the midpoints of AB and DC; reflection with axis going through the midpoints of AD and BC; rotations of 180° and 360° with rotocenter the center of the rectangle. (b) No Reflections. Rotations of 180° and 360° with rotocenter the center of the parallelogram. (c) Reflection with axis going through the midpoints of AB and DC; rotation of 360° with rotocenter the center of the trapezoid.

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40. The following figure gives the axes of reflection for each figure.

(a) Reflection with axis going through C and the midpoint of AB; rotation of 360° with rotocenter the center of the triangle. (b) Reflections (three of them) with axis going through a vertex and the midpoint of the opposite side; rotations of 120° , 240° , and 360° with rotocenter the center of the triangle. (c) Rotation of 360° with rotocenter the center of the triangle. 41. (a) Reflections (three of them) with axis going through pairs of opposite vertices; reflections (three of them) with axis going through the midpoints of opposite sides of the hexagon; rotations of 60°, 120°, 180°, 240°, 300°, 360° with rotocenter the center of the hexagon.

(b) No reflections; rotations of 72°, 144°, 216°, 288°, 360° with rotocenter the center of the star. 42. (a) Reflections (five of them) with axis going through a vertex and midpoint of the opposite side; rotations of 72°, 144°, 216°, 288°, 360° with rotocenter the center of the pentagon.

(b) No reflections; Rotations of 60°, 120°, 180°, 240°, 300°,360° with rotocenter the center of the shape. 43. (a) D2 ; the figure has exactly 2 reflections and 2 rotations. (b) Z 2 ; the figure has no reflections and exactly 2 rotations. (c) D1 ; the figure has exactly 1 reflection and 1 rotation. 44. (a) D1 ; the figure has exactly 1 reflection and 1 rotation. (b) D3 ; the figure has exactly 3 reflections and 3 rotations. (c) Z1 ; the figure has no reflections and exactly 1 rotation. 45. (a) D6 ; the figure has exactly 6 reflections and 6 rotations. Copyright © 2022 Pearson Education, Inc.

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(b) Z 5 ; the figure has exactly 5 rotations (and no reflections). 46. (a) D5 ; the figure has exactly 5 reflections and 5 rotations. (b) Z 6 ; the figure has exactly 6 rotations (and no reflections). 47. (a) D1 ; the letter A has exactly 1 reflection (vertical) and 1 rotation (identity). (b) D1 ; the letter D has exactly 1 reflection (horizontal) and 1 rotation (identity). (c) Z1 ; the letter L has no reflection and exactly 1 rotation (identity). (d) Z 2 ; the letter Z has no reflection and exactly 2 rotations (identity and 180 ). (e) D2 ; the letter H has exactly 2 reflections and 2 rotations (identity and 180 ). (f) Z 2 ; the letter N has no reflection and exactly 2 rotations (identity and 180 ). 48. (a) Z 2 ; the symbol $ has exactly 2 rotations (identity and 180 ).. (b) Z1 ; the symbol @ has exactly 1 rotation (identity). (c) Z 2 ; the symbol % has exactly 2 rotations. (d) D4 ; the symbol x has exactly 4 reflections and exactly 4 rotations. (e) Z1 ; the symbol & has 1 rotation (identity). 49. Answers will vary. (a) Since symmetry type Z1 has no reflections and exactly 1 rotation, the capital letter J is an example of this symmetry type. (b) Since symmetry type D1 has exactly 1 reflection and 1 rotation, the capital letter T is an example of this symmetry type. (c) Since symmetry type Z 2 has no reflection and exactly 2 rotations, the capital letter Z is an example of this symmetry type. (d) Since symmetry type D2 has exactly 2 reflections and 2 rotations, the capital letter I is an example of this symmetry type. 50. (a) 5

(b) 3

(d) 8

11.7 Patterns 51. (a) m1; the border pattern has translation symmetry and vertical reflection but does not have horizontal reflection, half-turn rotation, or glide reflection. (b) 1m; the border pattern has translation symmetry and horizontal reflection but does not have vertical reflection, half-turn, or glide reflection.

186

Chapter 11: The Mathematics of Symmetry (c) 12; the border pattern has translation symmetry and half-turn rotation but does not have horizontal reflection, vertical reflection, or glide reflection. (d) 11; the border pattern has translation symmetry but does not have horizontal reflection, vertical reflection, half-turn rotation, or glide reflection.

52. (a) 11; the border pattern has translation symmetry but does not have horizontal reflection, vertical reflection, half-turn rotation, or glide reflection. (b) mm; the border pattern has translation symmetry, horizontal reflection, and vertical reflections. (c) m1; the border pattern has translation symmetry and vertical symmetry but does not have horizontal reflection, glide reflection, or half turns. (d) 12; the border pattern has translation symmetry but does not have vertical reflections, horizontal reflection, or glide reflections, but does have half-turns. 53. (a) m1; the border pattern has translation symmetry and vertical reflection but does not have horizontal reflection, glide reflection, or half-turn rotation. (b) 12; the border pattern has translation symmetry and half-turn rotation but does not have horizontal reflection, vertical reflection, or glide reflection. (c) 1g; the border pattern has translation symmetry and glide reflection but does not have horizontal reflection, vertical reflection, or half-turn rotation. (d) mg; the border pattern has translation symmetry, vertical reflection, half-turn rotation, and glide reflection, but does not have horizontal reflection. 54. (a) 12; the border pattern has translation symmetry and half turns but does not have vertical reflection, horizontal reflection, or glide reflection. (b) 1g; the border pattern has translation symmetry and glide reflections but does not have vertical reflection or horizontal reflections. (c) m1; the border pattern has translation symmetry and vertical reflection but does not have horizontal reflection, glide reflections, or half-turn rotation. (d) mg; the border pattern has translation symmetry, vertical reflection, half-turn rotation, and glide reflection, but does not have horizontal reflection. 55. 12; see, for example, exercise 51(c). 56. 11; see, for example, exercise 51(d). 57. mm; consider as an example …+ + + + + + …. Since the pattern has horizontal reflection symmetry, a border pattern of type __m is guaranteed. Since the motif has symmetry type D4 , we know that we must have vertical reflection symmetry (and two diagonal symmetries) too. That rules out m1. The axes of reflection for the reflection symmetries are 90-degree rotations of each other. So, the new type is mm. 58. mm; consider as an example … H H H H H H …. Since the pattern has horizontal reflection symmetry, a border pattern of type __m is guaranteed. Since the motif has symmetry type D2 , we know that we must have vertical reflection symmetry as well (the axes of reflection for the reflection symmetries are 180-degree rotations of each other). This eliminates 1m so it must be of type mm.

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JOGGING 19 R 240 59. (a) Long division gives 360 7080 . That is, 7080° is 19 full revolutions and 240° more. This means

that 19 full hours have passed. (b) Using the remainder found in (a), the minute hand points to the number on the clock that is located 240° (clockwise) from the 9. That is, it is two-thirds of the way around the face from 9 which is a full 8 of the 12 markings on the face of the clock. Rotating 8 of these markings from the 9 leaves the minute hand on the 5. 60. (a) A vertical reflection.

(b) A glide reflection about the axis shown.

(d) A rotation having center of rotation as shown.

61. (a) rotation; the two rigid motions that are proper are rotations and translations. Of these, only a rotation can have exactly one fixed point (the identity translation has infinitely many fixed points). (b) identity motion; the two rigid motions that are proper are rotations and translations. Of these, only the identity motion has infinitely many fixed points. (c) reflection; the two rigid motions that are improper are reflections and glide reflections. But glide reflections do not have fixed points while reflections can have an infinite number of them. (d) glide reflection; the two rigid motions that are improper are reflections and glide reflections. Of these, only the glide reflection has no fixed points. 62. (a) D. The reflection of P about l1 is the point E. The reflection of E about line l2 is D. So, point D is the image of point P under the product of these reflections. (b) D. The reflection of P about l2 is the point A. The reflection of A about line l1 is D. So, again, point D is the image of point P under the product of these reflections. (c) B. The reflection of P about l2 is the point A. The reflection of A about line l3 is B. (d) F. The reflection of P about l3 is the point C. The reflection of C about line l2 is F. [Note that (a), (b), (c) and (d) suggest that the order in which reflections occur sometimes makes a difference.] (e) G. The reflection of P about l1 is the point E. The reflection of E about line l4 is G. Copyright © 2022 Pearson Education, Inc.

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63. (a) C. The reflection of P about l is the point B. The image of B under the rotation is C. (b) P. The image of P under the rotation is B. The reflection of B about l is P. (c) D. The reflection of P about l is the point B. The rotation of B is D. (d) D. The rotation of P is the point C. The reflection of C about line l is D. [Note that (a), (b), (c) and (d) suggest that the order in which reflections occur sometimes makes a difference.] 64. (a) When a proper rigid motion is combined with an improper rigid motion, the result is an improper rigid motion; that is, the left-right and clockwise-counterclockwise orientations on the final figure will be the reverse of the original figure. (b) When an improper rigid motion is combined with an improper rigid motion, the result is a proper rigid motion; that is, the left-right and clockwise-counterclockwise orientations on the final figure will be the same as the original figure. This occurs because the orientations are reversed and then reversed again. The second reversal brings the orientation back to the original orientation. (c) improper; this is an improper rigid motion combined with a proper rigid motion (see (a)). (d) proper; this is an improper rigid motion combined with an improper rigid motion (see (b)). 65. The combination of two improper rigid motions is a proper rigid motion. Since C is a fixed point, the rigid motion must be a rotation with rotocenter C. 66. M is a proper rigid motion (an improper rigid motion combined with an improper rigid motion is a proper rigid motion). Moreover M has no fixed points. It follows that M must be a translation. 67. (a) The result of applying the reflection with axis l1 , followed by the reflection with axis l2 , is a clockwise rotation with center C and angle of rotation λ + λ + β + β = 2(λ + β ) = 2α . One example is shown in the figure.

(b) The result of applying the reflection with axis l2 , followed by the reflection with axis l1 , is a counterclockwise rotation with center C and angle of rotation 2α . 68. (a)

The distance from P to P '' is x + x + y + y = 2x + 2y = 2(x + y) = 2d. This is a translation by a vector perpendicular to l1 and l3 of length 2d and going from left to right.

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(b)

The distance from P to P '' is y – (x + x – y) = 2y – 2x = 2(y – x) = 2d. This is a translation by a vector perpendicular to l1 and l3 of length 2d and going from right to left. 69. (a)

P’’ is the image of P under the product of M and N. Q’’ is the image of Q under the product of M and N.

(b)

The vector of the translation representing the product of M and N is described by the arrow from P to P’’.

70. (a) By definition, a border has translation symmetries in exactly one direction (let’s assume the horizontal direction). If the pattern had a reflection symmetry along an axis forming 45° with the horizontal direction, there would have to be a second direction of translation symmetry (vertical). (b) If a pattern had a reflection symmetry along an axis forming an angle of α ° with the horizontal direction, it would have to have translation symmetry in a direction that forms an angle of 2α ° with the horizontal. This could only happen for α = 90° or α = 180° (since the only allowable direction for translation symmetries is the horizontal). 71. Rotations and translations are proper rigid motions, and hence preserve clockwise-counterclockwise orientations. The given motion is an improper rigid motion (it reverses the clockwise-counterclockwise orientation). If the rigid motion was a reflection, then PP ′, RR ′, and QQ ′ would all be perpendicular to the axis of reflection and hence would all be parallel. It must be a glide reflection (the only rigid motion left).

RUNNING 72. mg

m1 1m

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Chapter 11: The Mathematics of Symmetry 11 12

mm r1

73.

74.

75. pm; The identity is the only rotation symmetry. There is a reflection (vertical). However, there is not a glide reflection with an axis that is not a reflection axis. 76. p2; The pattern has 2-fold rotational symmetry. There are no reflections. There are also no glide reflections. 77. cm; The identity is the only rotation symmetry. There is a reflection (vertical). There are also glide reflections that do not have the same axis as the vertical reflection axes. These vertical axes are ¼ and ¾ of the distance between two horizontal hearts. 78. pm; The identity is the only rotation symmetry. There is a reflection (vertical). However, there is not a glide reflection with an axis that is not a reflection axis. 79. p4m; The pattern has 4-fold rotational symmetry. There are reflections. In fact, there are reflections that intersect at 45° angles. 80. p6m; The pattern has 6-fold rotational symmetry. Furthermore, there are reflections.

Chapter 12 WALKING 12.1 The Koch Snowflake and Self-Similarity 1. Start

M 3

l 81 cm

P 243 cm

Step 1

27 cm

324 cm

Step 2

9 cm

432 cm

Step 3

192

3 cm

576 cm

Step 4

768

1 cm

768 cm

Step 5

3072

1/3 cm

1024 cm

At each step, the value of the number of sides M is multiplied by 4 and the value of each side length l is multiplied by 1/3. The perimeter P at each step is simply the product of M and l.

Start

M 3

l 18 cm

P 54 cm

Step 1

6 cm

72 cm

Step 2

2 cm

96 cm

Step 3

192

2/3 cm

128 cm

Step 4

768

2/9 cm

512/3 cm

Step 5

3072

2/27 cm

2048/9 cm

As in exercise 1, at each step the value of the number of sides M is multiplied by 4 and the value of each side length l is multiplied by 1/3. The perimeter P at each step is simply the product of M and l. R

Start

Step 1

108

Step 2

120

Step 3

1/9

16/3

376/3

Step 4

192

1/81

64/27

3448/27

Step 5

768

1/729

256/243

31228/243

In the first step, 3 triangles are added. At each step thereafter, R (the number of triangles added) is multiplied by 4. The area S of each triangle added at a given step is 1/9 of the area added during the previous step. The total new area T added at a given step is then the product of R and S. The area of the “snowflake” obtained at a given step (which we call Q) is the sum of T and the area Q of the snowflake in the previous step. For example, at step 1 a total of R = 3 triangles are added each having area S = 9 (S is 1/9 the area of the original triangle which was 81). So, the value of Q at step 1 is 81 + 3 × 9 = 108 . Copyright © 2022 Pearson Education, Inc.

192

Chapter 12: Fractal Geometry At step 2, R = 3 × 4 = 12 triangles are added each having area S = 1 (this can be seen as either 1/9 of the area added in step 1 or 1 9 2 the area of the original triangle). The value of Q at step 2 is then Q = 108 + 12 × 1 = 120 . At step 3, R = 3 × 42 = 48 triangles are added each having area S = 1/9 (again, this can be seen as 1/9 of the previous value of S). The value of Q at step 3 is then Q = 120 + 48 × (1/ 9 ) = 376 / 3 . R

Start

729

Step 1

243

972

Step 2

108

1080

Step 3

1128

Step 4

192

1/9

64/3

3448/3

Step 5

768

1/81

256/27

31288/27

As in exercise 3, at step 3, R = 3 × 4 = 48 triangles are added each having area S = 1 (again, this can be seen as 1/9 of the previous value of S). The value of Q at step 3 is then Q = 1080 + 48 × 1 = 1128 . 5. Start

M 4

l 81 cm

P 324 cm

Step 1

27 cm

540 cm

Step 2

100

9 cm

900 cm

Step 3

500

3 cm

1500 cm

Step 4

2500

1 cm

2500 cm

At each step, the value of the number of sides M is multiplied by 5 and the value of each side length l is multiplied by 1/3 since each side of the object is replaced by 5 sides having 1/3 the length. The perimeter P at each step is simply the product of M and l. 6. Start

M 4

l 1

P 4

Step 1

1/3

20/3

Step 2

100

1/9

100/9

Step 3

500

1/27

500/27

Step 4

2500

1/81

2500/81

At each step, the value of the number of sides M is multiplied by 5 and the value of each side length l is multiplied by 1/3. The perimeter P at each step is simply the product of M and l.

ISM: Excursions in Modern Mathematics, 10E

Start

Step 1

117

Step 2

137

Step 3

100

1/9

100/9

1333/9

Step 4

500

1/81

500/81

12497/81

193

In the step 1, 4 squares are added. At each step thereafter, R (the number of squares added) is multiplied by 5. The area S of each square added at a given step is 1/9 of the area of each square added during the previous step. The total new area T added at a given step is then the product of R and S. The area of the shape obtained at a given step (which we call Q) is the sum of T and the area Q of the shape in the previous step. At step 1, for example, a total of R = 4 squares are added each having area S = 9 (S is 1/9 the area of the original square which was 81). So, the value of Q at step 1 is 81 + 4 × 9 = 117 . At step 2, R = 4 × 5 = 20 squares are added each having area S = 1 (this can be seen as 1/9 of the area added in step 1). The value of Q at step 2 is then Q = 117 + 20 × 1 = 137 . At step 3, R = 4 × 52 = 100 squares are added each having area S = 1/9 (again, this can be seen as 1/9 of the previous value of S). The value of Q at step 3 is then Q = 137 + 100 × (1/ 9 ) = 1333 / 9 . 8.

Start

243

Step 1

108

351

Step 2

411

Step 3

100

1/3

100/3

1333/3

1/27 500/27 12497/27 500 Step 4 In the step 1, 4 squares are added. At each step thereafter, R (the number of squares added) is multiplied by 5. The area S of each square added at a given step is 1/9 of the area of each square added during the previous step. The total new area T added at a given step is then the product of R and S. The area of the shape obtained at a given step (which we call Q) is the sum of T and the area Q of the shape in the previous step. 9. Start

M 3

l 81 cm

P 243 cm

Step 1

27 cm

324 cm

Step 2

9 cm

432 cm

Step 3

192

3 cm

576 cm

Step 4

768

1 cm

768 cm

Step 5

3072

1/3 cm

1024 cm

At each step, the value of the number of sides M is multiplied by 4 (each side becomes four) and the value of each side length l is multiplied by 1/3 (each side length is divided into three equal lengths). The perimeter P at each step is simply the product of M and l.

194

Chapter 12: Fractal Geometry

10.

Start

18 cm

54 cm

Step 1

6 cm

72 cm

Step 2

2 cm

96 cm

Step 3

192

2/3 cm

128 cm

Step 4

768

2/9 cm

512/3 cm

3072 2/27 cm 2048/9 cm Step 5 At each step, the value of the number of sides M is multiplied by 4 (each side becomes four) and the value of each side length l is multiplied by 1/3 (each side length is divided into three equal lengths). The perimeter P at each step is simply the product of M and l. R

Start

Step 1

Step 2

Step 3

1/9

48/9

110/3

Step 4

192

1/81

192/81

926/27

Step 5

768

1/729

768/729

8078/243

11.

In the first step, 3 triangles are subtracted. At each step thereafter, the number of triangles subtracted (which we call R) is multiplied by 4. The area S of each triangle subtracted at a given step is 1/9 of the area subtracted during the previous step. The total new area T subtracted at a given step is then the product of R and S. The area of the shape obtained at a given step (which we call Q) is the difference of the area of the shape in the previous step and T. [Note: This is similar to exercise 3 except subtraction is used rather than addition.] For example, at step 1 a total of R = 3 triangles are subtracted each having area S = 9 (S is 1/9 the area of the original triangle which was 81). So, the value of Q at step 1 is 81 − 3 × 9 = 54 . At step 2, R = 3 × 4 = 12 triangles are subtracted each having area S = 1. The value of Q at step 2 is then Q = 54 − 12 × 1 = 42 . At step 3, R = 3 × 42 = 48 triangles are subtracted each having area S = 1 9 (again, this can be seen as 1/9 of the previous value of S). The value of Q at step 3 is then Q = 42 − 48 × (1/ 9 ) = 110 / 3 . R

Start

729

Step 1

243

486

Step 2

108

378

Step 3

330

Step 4

192

1/9

192/9

926/3

Step 5

768

1/81

768/81

8078/27

12.

ISM: Excursions in Modern Mathematics, 10E

195

multiplied by 4. The area S of each triangle subtracted at a given step is 1/9 of the area subtracted during the previous step. The total new area T subtracted at a given step is then the product of R and S. The area of the shape obtained at a given step is the difference of the area of the shape in the previous step and T. 13. (a)

(b) The seed square has perimeter of 4 × 16 = 64 . By replacing each segment with the “sawtooth” version, the length is doubled. So, the perimeter of the entire figure obtained in step 1 is 2 × 64 = 128 . (c) As in (b), replacing each segment with the “sawtooth” version doubles the length. So, since the perimeter in step 1 is 128, the perimeter of the figure obtained in step 2 is 2 × 128 = 256 . (d) The perimeter doubles at each step of the construction. 14. (a) Same as exercise 13(a). (b) The seed square has perimeter of 4a . By replacing each segment with the “sawtooth” version, the length is doubled. So, the perimeter of the entire figure obtained in step 1 is 8a . (c) As in (b), replacing each segment with the “sawtooth” version doubles the length. So, since the perimeter in step 1 is 8a, the perimeter of the figure obtained in step 2 is 16a. (d) The perimeter doubles at each step of the construction. 15. (a) 256; the area at step 1 is the same as the area of the seed (what the rule “giveth,” the rule “taketh” away). (b) 256; the area at step 2 is the same as the area after step 1. (c) At each step, the area added is the same as the area subtracted. 16. (a) a 2 ; the area at step 1 is the same as the area of the seed (what the rule “giveth,” the rule “taketh” away). (b) a 2 ; the area at step 2 is the same as the area after step 1. (c) At each step, the area added is the same as the area subtracted.

12.2 The Sierpinski Gasket and the Chaos Game 17.

Start

Step 1

Step 2

Step 3

Step 4

1/4

27/4

81/4

Step 5

1/16

81/16

243/16

196

Chapter 12: Fractal Geometry In the step 1, one triangle is subtracted. At each step thereafter, the number of triangles subtracted (R) is multiplied by 3. Starting with step 2, the area S of each triangle subtracted at a given step is 1/4 of the area of each triangle subtracted during the previous step (in step 1, the area of the triangle that is subtracted is 1/4). The total new area T subtracted at a given step is the product of R and S at that step. The area of the shape obtained at a given step (Q) is the difference between the area of the shape in the previous step and T.

For example, at step 3 shown in (d) in the figure, R = 1× 32 = 9 triangles are subtracted each having area S = 1 (this can be seen as 1/4 of the previous value of S). So, T, the total area subtracted at step 3 is 9 ×1 = 9 . The value of Q at step 3 is then Q = 36 − 9 = 27 . 18.

Start

Step 1

1/4

3/4

Step 2

1/16

3/16

9/16

Step 3

1/64

9/64

27/64

Step 4

1/256

27/256

81/256

81 1/1024 81/1024 243/1024 Step 5 In the step 1, one triangle is subtracted. At each step thereafter, the number of triangles subtracted (R) is multiplied by 3. Starting with step 2, the area S of each triangle subtracted at a given step is 1/4 of the area of each triangle subtracted during the previous step (in step 1, the area of the triangle that is subtracted is 1/4). The total new area T subtracted at a given step is the product of R and S at that step. The area of the shape obtained at a given step (Q) is the difference between the area of the shape in the previous step and T. 19.

Start

8 cm

Step 1

4 cm

12 cm

Step 2

2 cm

18 cm

Step 3

1 cm

27 cm

Step 4

1/2 cm

81/2 cm

243 243/4 cm 1/4 cm Step 5 At each step, the value of the number of solid triangles U is tripled and the value of V (the perimeter of each solid triangle) is halved (since each side length is halved). The length of the boundary of the “gasket” W at each step is simply the product of U and V.

ISM: Excursions in Modern Mathematics, 10E

20.

Start

Step 1

Step 2

Step 3

5/2

135/2

Step 4

5/4

405/4

197

243 1215/8 5/8 Step 5 At each step, the value of the number of solid triangles U is tripled and the value of V (the perimeter of each solid triangle) is halved (since each side length is halved). The length of the boundary of the “gasket” W at each step is simply the product of U and V. 21.

Start

Step 1

1/9

1/3

2/3

Step 2

1/81

2/9

4/9

Step 3

108

1/729

4/27

8/27

Step 4

648

1/6561

8/81

16/81

Step N

3 ⋅ 6 N −1

1 9N

3 ⋅ 6 N −1 / 9 N

( 2 3)N

In step 1, three triangles are subtracted. In each step thereafter, the number of triangles subtracted (R) is multiplied by 6 (since six solid triangles remain for every three that are subtracted). Starting in step 2, the area S of each triangle subtracted at a given step is 1/9 of the area of each triangle subtracted during the previous step (in step 1, the area of each triangle that is subtracted is 1/9). The total new area T subtracted at a given step is the product of R and S at that step. The area of the shape obtained at a given step (Q) is the difference between the area of the shape in the previous step and T. For example, in step 2 shown in the figure, R = 3 × 6 = 18 triangles are subtracted each having area S = 1/81 (this can be seen as 1/9 of the previous value of S). So, T, the total area subtracted in step 2 is 18 × 1 81 = 2 9 . The value of Q at step 2 is then Q = 2 / 3 − 2 / 9 = 4 9 . 22.

Start

Step 1

Step 2

Step 3

108

1/9

Step 4

648

1/81

Step N

3 ⋅ 6 N −1

1 9N −2

3 ⋅ 6 N −1 9 N − 2

81( 2 3)

198

Chapter 12: Fractal Geometry

In step 1, three triangles are subtracted. In each step thereafter, the number of triangles subtracted (R) is multiplied by 6. Starting in step 2, the area S of each triangle subtracted at a given step is 1/9 of the area of each triangle subtracted during the previous step (in step 1, the area of each triangle that is subtracted is 9). The total new area T subtracted at a given step is the product of R and S at that step. The area of the shape obtained at a given step (Q) is the difference between the area of the shape in the previous step and T.

23.

Start

9 cm

Step 1

3 cm

18 cm

Step 2

1 cm

36 cm

Step 3

216

1/3 cm

72 cm

Step 4

1296

1/9 cm

144 cm

Step N

1 3N − 2 cm

9 ⋅ 2 N cm

At each step, the value of the number of solid triangles U is multiplied by 6 and the value of V (the perimeter of each solid triangle) is 1/3 of that in the previous step (since each side length is also 1/3 of that in the previous step). The length of the boundary of the “ternary gasket” W at each step is simply the product of U and V. 24.

Start

Step 1

P/3

Step 2

P/9

Step 3

216

P/27

Step 4

1296

P/81

16P

P 3N Step N 6N 2N P At each step, the value of the number of solid triangles U is multiplied by 6 and the value of V (the perimeter of each solid triangle) is 1/3 of that in the previous step (since each side length is also 1/3 of that in the previous step). The length of the boundary of the “ternary gasket” W at each step is simply the product of U and V.

25. (a) 5/9. The solid square is divided into 9 equal subsquares and 4 of these (along the sides) are removed. (b)

( 5 / 9) . The area of each of the subsquares found in step 2 is 5/9 of its area in step 1.

(c)

( 5 / 9 ) . The area of each of the subsquares found in step N is 5/9 of its area in step N-1. Since the area

in step 1 is 5/9, the area obtained in step N is found through repeated multiplication (N-1 times) by 5/9. 26. (a) 20/3. Each side in step 1 has length 1/3 of the side length in the seed (solid square). Since each of the 5 solid subsquares in step 1 has 4 sides, there are 20 sides in step 1. These 20 sides each have length 1/3 giving the perimeter of 20/3. (b) 100/9. Each side in step 2 has length 1/3 of the common side lengths in step 1 of the construction. Since the common side length in step 1 is 1/3, the common side length in step 2 is 1/3 of 1/3 which is 1/9. The number of sides in step 2 is 5 times the number of sides in step 1. Since there were 5 × 4 = 20 sides in Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

199

step 1, there are 5 × 20 = 100 sides in step 2. Since each of these 100 sides has length 1/9, the perimeter in step 2 is 100/9. (c) 4 × ( 5 / 3) . Each side in step N has length (1/ 3) . The number of sides in step N is 5 times the number N

of sides in step N-1. Since we start with 4 sides in the seed, we see that there are 5N × 4 sides in step N.

(

)

The perimeter of the figure found in step N is the product of these: 5N × 4 (1/ 3) = 4 × ( 5 / 3) . 27.

The coordinates of each point are: P1 : (32, 0); P2 : (16, 0), the midpoint of A and P1 ; P3 : (8, 16), the midpoint of C and P2 ; P4 : (20, 8), the midpoint of B and P3 ; P5 : (10, 20), the midpoint of C and P4 ; P6 : (5, 26), the midpoint of C and P5 .

28.

The coordinates of each point are: P1 : (0, 0); P2 : (0, 16), the midpoint of C and P1 ; P3 : (0, 8), the midpoint of A and P2 ; P4 : (16, 4), the midpoint of B and P3 ; P5 : (24, 2), the midpoint of B and P4 ; P6 : (12, 17), the midpoint of C and P5 .

200

29.

Chapter 12: Fractal Geometry

Roll

Point

Coordinates

(32,0)

(16,0)

(8,0)

(20,0)

(10,16)

(5,24)

P3 = (8, 0) is the midpoint of A and P2 ; P4 = (20,0) is the midpoint of B and P3 ; P5 = (10, 16) is the midpoint

of C and P4 ; P6 = (5, 24), the midpoint of C and P5 . 30.

Roll

Point

Coordinates

(0,0)

(0,16)

(0,24)

(0,12)

(16,6)

(8,19)

P3 = (0, 24) is the midpoint of C and P2 ; P4 = (0,12) is the midpoint of A and P3 ; P5 = (16, 6) is the midpoint

of B and P4 ; P6 = (8, 19), the midpoint of C and P5 . 31. General note: Find the new coordinate by picking the new x-value to be 2/3 of way from the x-coordinate of first point to x-coordinate of the second point and picking the new y-value to be 2/3 of the way from ycoordinate of the first point to y-coordinate of the second point. (a) The coordinates of each point are: P1 : (0, 27); P2 : (18, 9), 2/3 of the way from P1 to B; P3 : (6, 3), 2/3 of the way from P2 to A; P4 : (20, 1), 2/3 of the way from P3 to B.

ISM: Excursions in Modern Mathematics, 10E (b) The coordinates of each point are: P1 : (27, 27); P2 : (27, 9), 2/3 of the way from P1 to B; P3 : (9, 3), 2/3 of the way from P2 to A; P4 : (21, 1), 2/3 of the way from P3 to B.

(c) The coordinates of each point are: P1 : (27, 27); P2 : (27, 27); P3 : (9, 9), 2/3 of the way from P2 to A; P4 : (3, 3), 2/3 of the way from P3 to A.

201

202

Chapter 12: Fractal Geometry

32. (a)

(b)

ISM: Excursions in Modern Mathematics, 10E

203

(c)

33. (a) 4,2,1,2; P1 = (0,27) corresponds to a first roll of 4. Then, since P2 = (18,9) is 2/3 of the way from P1 to B, we see that the second roll is 2. Since P3 = (6,3) is 2/3 of the way from P2 to A, we see that the third roll is 1. Lastly, since P4 = (20,1) is 2/3 of the way from P3 to B, the fourth and final roll is a 2. (b) 3,1,1,3; P1 = (27,27) corresponds to a first roll of 3. Then, since P2 = (9,9) is 2/3 of the way from P1 to A, we see that the second roll is 1. Since P3 = (3,3) is 2/3 of the way from P2 to A, we see that the third roll is 1. Lastly, since P4 = (19,19) is 2/3 of the way from P3 to C, the fourth and final roll is a 3. (c) 1,3,4,2; P1 = (0,0) corresponds to a first roll of 1. Then, since P2 = (18,18) is 2/3 of the way from P1 to C, we see that the second roll is 3. Since P3 = (6,24) is 2/3 of the way from P2 to D, we see that the third roll is 4. Lastly, since P4 = (20,8) is 2/3 of the way from P3 to B, the fourth and final roll is a 2. 34. (a) 2,3,4,1; P1 = (27,0) corresponds to a first roll of 2. Then, since P2 = (27,18) is 2/3 of the way from P1 to C, we see that the second roll is 3. Since P3 = (9,24) is 2/3 of the way from P2 to D, we see that the third roll is 4. Lastly, since P4 = (3,8) is 2/3 of the way from P3 to A, the fourth and final roll is a 1. (b) 4,2,2,4; P1 = (0,27) corresponds to a first roll of 4. Then, since P2 = (18,9) is 2/3 of the way from P1 to B, we see that the second roll is 2. Since P3 = (24,3) is 2/3 of the way from P2 to B, we see that the third roll is 2. Lastly, since P4 = (8,19) is 2/3 of the way from P3 to D, the fourth and final roll is a 4. (c) 3,1,2,4; P1 = (27,27) corresponds to a first roll of 3. Then, since P2 = (9,9) is 2/3 of the way from P1 to A, we see that the second roll is 1. Since P3 = (21,3) is 2/3 of the way from P2 to B, we see that the third roll is 2. Lastly, since P4 = (7,19) is 2/3 of the way from P3 to D, the fourth and final roll is a 4.

12.4 The Mandelbrot Set 35. (a) (b)

204

Chapter 12: Fractal Geometry (c) i 2 + ( −i ) = −1 − i

36. (a)

(1 + i ) 2 + (1 + i ) = 1 + 2i + i 2 + 1 + i = 1 + 3i

(b) (1 + 3i ) + (1 + i ) = 1 + 6i + 9i 2 + 1 + i = 1 + 6i − 9 + 1 + i = −7 + 7i 2

(c)

( −7 + 7i ) 2 + (1 + i ) = ( −7) 2 − 98i + 49i 2 + 1 + i = 1 − 97i

37. (a)

( −0.25 + 0.25i ) 2 + ( −0.25 + 0.25i ) = 0.0625 − 0.125i − 0.0625 + ( −0.25 + 0.25i ) = −0.25 + 0.125i

(b)

( −0.25 − 0.25i ) 2 + ( −0.25 − 0.25i ) = 0.0625 + 0.125i − 0.0625 + ( −0.25 − 0.25i ) = −0.25 − 0.125i

38. (a) (b) 39. (a)

( −0.25 + 0.125i ) + ( −0.25 + 0.125i ) = 0.0625 − 0.0625i − 0.015625 + ( −0.25 + 0.125i ) = −0.203125 + 0.0625i 2

( −0.2 + 0.8i ) 2 + ( −0.2 + 0.8i ) = 0.04 − 0.32i − 0.64 + ( −0.2 + 0.8i ) = −0.8 + 0.48i Since the value of i (1 + i ) = −1 + i ,

i 2 (1 + i) = −1 − i , and i 3 (1 + i) = 1 − i we plot 1 + i , −1 + i , −1 − i , and 1 − i . This is just like plotting (1,1), (-1,1), (-1,-1), and (1,-1) in the standard Cartesian plane.

(b) Since the value of i (3 − 2i ) = 2 + 3i , i 2 (3 − 2i) = −3 + 2i , and i 3 (3 − 2i ) = −2 − 3i we plot 3 − 2i , 2 + 3i , −3 + 2i , and −2 − 3i .

ISM: Excursions in Modern Mathematics, 10E 40. (a)

205

Since the value of −i (1 + i ) = 1 − i ,

( −i ) (1 + i) = −1 − i , and ( −i ) (1 + i) = −1 + i we 2

plot 1 + i , 1 − i , −1 − i and −1 + i .

(b) Since the value of −i (0.8 + 1.2i ) = 1.2 − 0.8i ,

( −i ) (0.8 + 1.2i) = −0.8 − 1.2i , and 3 ( −i ) (0.8 + 1.2i) = −1.2 + 0.8i we plot 0.8 + 1.2i , 2

1.2 − 0.8i , −0.8 − 1.2i and −1.2 + 0.8i .

(c) The effect of multiplying each point in (a) and (b) by -i is a 90-degree clockwise rotation. 41. (a) s1 = (−2)2 + (−2) = 4 − 2 = 2; s2 = (2)2 + (−2) = 4 − 2 = 2;

s3 = (2)2 + (−2) = 4 − 2 = 2; s4 = (2)2 + (−2) = 4 − 2 = 2. (b) The sequence is attracted to 2 (in fact, each term equals 2), so s100 = 2. (c) Each number in the sequence is 2. In this case, the sequence could be considered periodic (with period 1) or attracted to 2 (in the sense that a sequence of 2's could be considered getting closer and closer to the attractor 2). 42. (a) s1 = 22 + 2 = 6 ; s2 = 62 + 2 = 38 ; s3 = 382 + 2 = 1446 ; s4 = 14462 + 2 = 2,090,918 . (b) Escaping 43. (a) s1 = (−0.5)2 + (−0.5) = −0.25; s2 = (−0.25)2 + (−0.5) = −0.4375; s3 = (−0.4375)2 + (−0.5) ≈ −0.3086;

s4 = (−0.3086)2 + (−0.5) ≈ −0.4048; s5 = (−0.4048)2 + (−0.5) ≈ −0.3361. (b) sN +1 = (−0.366)2 + (−0.5) ≈ −0.366 (c) From part (b), we see that sN = sN +1 , so the sequence must be attracted to –0.3660 (to 4 decimal places). 44. (a) s1 = −0.1875 , s2 = −0.214844 , s3 = −0.203842 , s4 = −0.208448 , s5 = −0.206549 , s6 = −0.207337 , s7 = −0.207011 , s8 = −0.207146 , s9 = −0.207090 , s10 = −0.207114 . (b) –0.207107 (c) Attracted to the number –0.207107 (rounded to six decimal places). Copyright © 2022 Pearson Education, Inc.

206

Chapter 12: Fractal Geometry

45. (a) For the first three values, see Exercise 35. s1 = ( −i ) + ( −i ) = i 2 − i = −1 − i; 2

s2 = ( −1 − i ) − i = 1 + 2i + i 2 − i = i; s3 = i 2 − i = −1 − i; s4 = ( −1 − i ) − i = 1 + 2i + i 2 − i = i; 2

s5 = i 2 − i = −1 − i . (b) Periodic; the odd terms are −1 − i and the even terms are i. 46. The calculations are exactly those found in Exercise 36(a),(b),(c). So, s1 = (1 + i ) + (1 + i ) = 1 + 3i , 2

s2 = (1 + 3i ) + (1 + i ) = −7 + 7i , and s3 = ( −7 + 7i ) + (1 + i ) = 1 − 97i . 2

JOGGING 47. (a) As in Exercise 18, we see that the value of R at step N is 3N −1 , the value of S at step N is A / 4 N . So, the value of T at step N of the construction is 3N −1 ⋅

4 “gasket” obtained at step N of the construction is 1

A−

A A  3 A  3 A  3 − ⋅   − ⋅   − − ⋅   4 4  4 4  4 4  4

N −1

= N

A  3 ⋅  4  4

N −1

. It follows that Q, the area of the

N −1   A A  31 A  3 2 A  3 = A −  + ⋅  + ⋅  ++ ⋅     4  4 4  4  4 4 4 

  3 N  . 1 −  N N A   4    3  3 = A− = A − A + A⋅   = A⋅   3   4  4 4 1 −   4  

(b) The area of the Sierpinski gasket is smaller than the area of the gasket formed during any step of construction. That is, if the area of the original triangle is 1, then the area of the Sierpinski gasket is less N

3 3 than   for every positive value of N. Since 0 < 3/4 < 1, the value of   can be made smaller 4   4 than any positive quantity for a large enough choice of N. It follows that the area of the Sierpinski gasket can also be made smaller than any positive quantity. N

3 48. (a)   P ; similar to the “W” column in exercise 20 (where P replaces the starting W value 20). 2

(b) Since the perimeter grows by a factor of 1.5 at each step, the perimeter of the Sierpinski gasket becomes infinitely large. 49. (a)

Start

Step 1

1/27

Step 2

20 × 7

Step 3

202 × 7

Step 4

203 × 7

Step N

20 N −1 × 7

20/27 2

(1/ 27 ) (1/ 27 )3 (1/ 27 )4 (1/ 27 ) N

( 20 / 27 )2 ( 20 / 27 )3 ( 20 / 27 )4 ( 20 / 27 ) N

ISM: Excursions in Modern Mathematics, 10E

207

At the first step, a cube is removed from each of the six faces and from the center for a total of C = 7 cubes removed. At the second step, each of the 20 remaining cubes has 7 cubes removed (i.e, C = 20 × 7 cubes are removed). In step 3, there are 202 remaining cubes each of which has 7 cubes removed (i.e. C = 20 2 × 7 cubes are removed). Etc. Since step 1 can be thought of as 3 × 3 × 3 = 27 equally sized cubes, the volume U of the middle cube removed is 1/27. For step 2 and onward, the volume U of each cube removed is 1/27 the volume of each cube removed in the previous step. The volume V of the sponge at any particular step is simply the difference of the previous value of V (the volume at the previous step) and the product of C and U (the volume removed during the current step). (b) Since 20/27 < 1, the Menger Sponge has infinitesimally small volume. (If a number between 0 and 1 is multiplied by itself over and over again, the resulting product will be close to 0.) 50. (a) To start, a cube is removed from each of the six faces and from the center for a total of 7 cubes removed. At the next step, each of the 20 remaining cubes has 7 cubes removed (i.e, 20 × 7 cubes are removed). In the next step, there are 202 remaining cubes each of which has 7 cubes removed (i.e. 20 2 × 7 cubes are removed). Continuing in this fashion, we see that 20 N −1 × 7 are removed from the 20 N −1 remaining cubes in step N. H

Number of cubes removed

Start

Step 1

Step 2

7 + 20 × 7

20 × 7

Step 3

7 + 20 × 7 + 20 2 × 7

20 2 × 7

Step 4

7 + 20 × 7 + 20 2 × 7 + 203 × 7

203 × 7

Step 5

7 + 20 × 7 + 20 2 × 7 + 203 × 7 + 20 4 × 7

20 4 × 7

(b) Using the formula for the sum of the terms in a geometric sequence in which P0 = 7 and R = 20, we see that at the Nth step there will be 7 + 7 × 20 + 7 × 202 + 7 × 203 +  + 7 × 20 N −1 = 7 × holes.

20 N − 1 7 = ( 20 N − 1) 1 − 20 19

51. If it is attracted to a number, then we have sN +1 = ( sN )2 + (−0.75) and sN +1 = sN . Substituting, we have

sN = ( sN )2 − 0.75 so sN 2 − sN − 0.75 = 0. Solve via the quadratic equation: s N =

1 ± 1 − 4 ( − 34 ) 2

1 3 or s N = . Examining the first few terms of the sequence indicates that the 2 2 1 sequence is attracted to − . 2

Solving yields sN = −

208

Chapter 12: Fractal Geometry

52. If it is attracted, there will be a solution to sN +1 = ( sN )2 + 0.25 with sN +1 = sN . Substituting, we have 1 ± 1 − 4 ( 14 )

1± 0 1 = . 2 2 2 Examining several terms of the sequence (see below) indicates that the sequence is indeed attracted to 1/2. s1 = 0.252 + 0.25 = 0.3125 ; s2 = 0.31252 + 0.25 ≈ 0.3477 ; s3 = 0.34772 + 0.25 ≈ 0.3709 ; … s30 ≈ 0.4725 .

sN = sN 2 + 0.25 or sN 2 − sN + 0.25 = 0 . Solving via the quadratic equation, s N =

53. The first twenty steps: 0.3125, -1.15234375, 0.0778961182, -1.24393219, 0.297367305,-1.16157269, 0.0992511044, -1.24014922, 0.287970084, -1.16707323, 0.112059926, -1.23744257, 0.281264121, -1.17089049, 0.120984549, -1.23536274, 0.276121097, -1.17375714, 0.127705824, -1.23369122. The numbers oscillate, alternating between positive and negative values, but always staying within the interval -1.25 ≤ x ≤ 0.3125 . The sequence is periodic. 54. Consider the first term of the sequence: s1 =

( 2 ) + 2 = 2 + 2 . Since s > 1 and 2 > 0, we are 2

guaranteed that sN +1 > sN and so the sequence is escaping. Investigating several terms will also support this reasoned conclusion.

RUNNING 55.

The critical observation is the following: Suppose that P is an arbitrary point inside (or on the boundary) of triangle AM1M2. Then the midpoint Q of the segment PB is inside (or on the boundary) of triangle BM1M3. (Why? Because segment M1M3 is parallel to AC and therefore Y is the midpoint of segment BX, which forces Q to be between Y and B.) The argument above implies that the chaos game will force us to keep hopping among the insides (or boundaries) of the triangles AM1M2, BM1M3, and CM2M3, but never into a point inside triangle M1M2M3.

56. The Sierpinski triangle. The same points are plotted as in the standard game. It just happens to be more likely to move towards A more often. 57. (a) The first twenty steps: –0.25 + 0.125i, –0.203125 + 0.1875i, –0.243896484 + 0.173828125i, –0.220730722 + 0.165207863i, –0.228571586 + 0.177067098i, –0.229107787 + 0.169054985i, –0.22608921 + 0.172536373i, –0.228652469 + 0.171982776i, –0.227296123 + 0.171351427i, –0.227697784 + 0.17210497i, –0.22777384 + 0.17162416i, –0.22757393 + 0.171817012i, –0.227731192 + 0.171797854i, –0.227653007 + 0.17175254i, –0.227673043 + 0.171800036i, –0.227680238 + 0.171771526i, –0.227667167 + 0.171782036i, –0.227676729 + 0.171781741i, –0.227672274 + 0.17177859i, –0.22767322 + 0.171781556i. The sequence is attracted to –0.22767 + 0.17178i (rounded to 5 decimal places). (b) The first twenty steps: –0.25 – 0.125i, –0.203125 – 0.1875i, –0.243896484 – 0.173828125i, –0.220730722 – 0.165207863i, –0.228571586 – 0.177067098i, –0.229107787 – 0.169054985i, –0.22608921 – 0.172536373i, –0.228652469 – 0.171982776i, –0.227296123 – 0.171351427i, –0.227697784 – 0.17210497i, –0.22777384 – 0.17162416i, –0.22757393 – 0.171817012i, –0.227731192 – 0.171797854i, –0.227653007 – 0.17175254i, –0.227673043 – 0.171800036i, –0.227680238 – 0.171771526i, –0.227667167 – 0.171782036i, –0.227676729 – 0.171781741i, –0.227672274 – 0.17177859i, –0.22767322 – 0.171781556i. The sequence is attracted to –0.22767 – 0.17178i (rounded to 5 decimal places).

ISM: Excursions in Modern Mathematics, 10E

209

58. Suppose that s = a + bi and s = a − bi . The Mandelbrot replacement process with seed s produces the sequence of complex numbers s, s 2 + s, ( s 2 + s)2 + s,  . The Mandelbrot replacement process with seed s produces the sequence of complex numbers s , s 2 + s = s 2 + s, ( s 2 + s ) 2 + s = ( s 2 + s ) 2 + s, . (In other words, each term in the second sequence is the complex conjugate of the corresponding term in the first sequence. This follows because z 2 = z 2 and u + v = u + v for all complex numbers z, u, and v.) It follows that the two sequences either both stay bounded or both are escaping and therefore if the point z is black (white) then its reflection across the x-axis, z , is also black (white). 59. The Koch curve consists of N = 4 self-similar copies of itself with each copy having been reduced by a log 4 ≈ 1.26 . scaling factor of S = 3. So, the dimension of the Koch curve is D = log 3 60. The Sierpinski gasket consists of N = 3 self-similar copies of itself with each copy having been reduced by a log 3 scaling factor of S = 2. So, the dimension of the Sierpinski gasket is D = ≈ 1.585 . log 2

APPLET BYTES 61. (a) Start. Start with a seed – the solid square. Step 1. Subdivide the see square into nine equal sub-squares and remove the central subsquare. Step 2. Subdivide each of the remaining solid squares into nine subsquares and remove the central subsquare. Call the procedure of subdividing a solid square into nine subsquares and removing the middle square procedure SC. Steps 3, 4, etc. Apply procedure SC to each solid square of the “carpet” obtained in the previous step. (b) Start. Start with a shaded seed square. Replacement rule. In each step, replace any individual solid square by a hollowed version. 62. (a) Step 1: The area is 8A/9. One of the nine equally sized subsquares has been removed. 2

8 Step 2: The area is   A . In step 1, the 8 subsquares have a total area of 8/9 of the original square. In 9 step 2, 8/9 of this total area (i.e. 8/9 of 8/9 of the area of the original square) will remain. 3

8 Step 3: The area is   A . 9 6

8 (b) At step 6, the area is   A . 9 N

8 (c) At step N, the area is   A . 9

63. (a) Step 1: A total of 1 hole is punched. Step 2: A total of 1 + 8 = 9 holes are punched. Step 3: A total of 1 + 8 + 8 × 8 = 1 + 8 + 82 = 73 holes are punched. (b) Step 4: A total of 1 + 8 + 82 + 83 = 585 holes are punched. Step 5: A total of 1 + 8 + 82 + 83 + 84 = 4681 holes are punched.

210

Chapter 12: Fractal Geometry (c) Step N: A total of 1 + 8 + 82 + 83 +  + 8 N −1 =

8N − 1 8N − 1 = holes are punched. 8 −1 7

64. (a) Step 1: Note that the boundary will include the outside perimeter and the perimeters of all removed squares. We also note that each removed square will have sides of length 1/3 that of the square it was removed from. Since each side of the original square has length 1, the starting square has perimeter 4 and the length of the boundary of the “carpet” obtained in step 1 of the construction is 1 4 12 4 16 4 + 4 ⋅ = 4 + = + = . This represents the previous boundary length plus the boundary length of 3 3 3 3 3 the new square hole with sides of length 1/3. 16 4 80 + 8⋅ = . This is the length of the boundary in step 1 plus Step 2: The length of the boundary is 3 9 9 the perimeter of 8 new square holes, each having sides of length (1 3) = 1 9 . 2

Step 3: The length of the boundary is

80 2 4 496 +8 ⋅ = . This is the length of the boundary at step 2 9 27 27

plus the perimeter of 82 new square holes, each having sides of length (1 3) = 1 27 . 3

1  4  4  4  4  (b) Step 5: The length of the boundary is 4 + 4   + 8   + 82   + 83   + 84   . This is the  3  9   27   81   243  length of the boundary at step 4 plus the perimeter of 84 new square holes, each having sides of length

(1 3) = 1 243 . 5

1  4  4  4  4  5 4  (c) Step 6: The length of the boundary is 4 + 4   + 8   + 82   + 83   + 84  +8   . This 3  9   27   81   243   729  is the length of the boundary at step 4 plus the perimeter of 85 new square holes, each having sides of length (1 3) = 1 729 . 6

Chapter 13 WALKING

13.1 Fibonacci Numbers 1. (a) F15 = 610 (the 15th Fibonacci number) (b) F15 − 2 = 610 − 2 = 608

6. (a) Subtract 3 from the Fibonacci number in position 2N.

(d) F3 N +1 represents the Fibonacci number in position (3N + 1).

(b) The Fibonacci number in position (2N - 3)

F15 610 = = 122 5 5

(e) F15 / 5 = F3 = 2 2. (a) F16 = 987

7. (a) F38 = F37 + F36 = 24,157,817 + 14,930,352 = 39, 088,169

(b) F16 + 1 = 987 + 1 = 988 (c) F16 +1 = F17 = 1597

(b) F39 = F38 + F37

F 987 = 246.75 (d) 16 = 4 4

= 39, 088,169 + 24,157,817 = 63, 245,986

(e) F16 / 4 = F4 = 3

8. (a) F34 = F33 + F32 = 3,524,578 + 2,178,309 = 5, 702,887

3. (a) F1 + F2 + F3 + F4 + F5 = 1 + 1 + 2 + 3 + 5

= 12

(b) F35 = F34 + F33

(b) F1+ 2 + 3+ 4 + 5 = F15 = 610

= 5, 702,887 + 3,524,578 = 9, 227, 465

9. (a) F35 = F37 − F36

(d) F3× 4 = F12 = 144

= 24,157,817 − 14,930,352 = 9, 227, 465

4. (a) F1 + F3 + F5 + F7 = 1 + 2 + 5 + 13 = 21 (b) F1+ 3+ 5+ 7 = F16 = 987

(b) F34 = F36 − F35 = 14,930,352 − 9, 227, 465 = 5, 702,887

10. (a) F31 = F33 − F32 = 3,524,578 − 2,178,309 = 1,346, 269

5. (a) 3FN + 1 represents one more than three times the Nth Fibonacci number. (b) 3FN +1 represents three times the Fibonacci number in position (N + 1).

(b) F30 = F32 − F31

= 2,178,309 − 1,346, 269 = 832, 040

212

Chapter 13: Fibonacci Numbers and the Golden Ratio

11. Binet’s simplified formula is N    1 + 5    where   denotes 5 FN =    2    ‘round to the nearest integer.’ So, 20    1 + 5   F20 =  5  . To evaluate this   2   1+ 5 expression, first compute , then raise 2 that value to the 20th power, then divide the result by 5 . This gives 6764.999964. Since 6764.999964 = 6765, we have that F20 = 6765 .

12. Binet’s simplified formula is N    1 + 5   5  where   denotes FN =    2   ‘round to the nearest integer.’ So, 25    1 + 5    5  . To evaluate this F25 =   2    1+ 5 expression, first compute , then raise 2 that value to the 25th power, then divide the result by 5 . This gives 75024.99912. Since 75024.99912 = 75,025, we have that

(b) 14; This is the 12th equation in the pattern. So according to the pattern, the 14th Fibonacci number is being doubled. Note that this is one less than the subscript of the Fibonacci number being subtracted. (c) N; This is the Nth equation in the pattern. 15. (a) F8 ⋅ F11 = 21 ⋅ 89 = 1869. But we also have F102 − F92 = 552 − 342 = 1869.

(b) FN ⋅ FN + 3 = FN2+ 2 − FN2+1 16. (a)

(b)

1 + 1 + 2 +  + 55 143 = = 13 11 11 ( FN + FN +1 + FN + 2 +  + FN + 9 ) = FN + 6 11

17. (a) Here we are adding the (N+1)st and the (N+2)nd Fibonacci numbers. The result is the (N+3)rd Fibonacci number. That is, FN +1 + FN + 2 = FN + 3 . (b) Since FN − 2 + FN −1 = FN , we have FN − FN − 2 = FN −1 . (c) Add two terms at a time. FN + FN +1 + FN + 3 + FN + 5

= ( FN + FN +1 ) + FN + 3 + FN + 5 = ( FN + 2 ) + FN + 3 + FN + 5

F25 = 75, 025 .

13. (a) Fifth equation in the sequence: 1 + 2 + 5 + 13 + 34 + 89 = 144 (In words, the left hand side is the sum of every other Fibonacci number. The right hand side is the next Fibonacci after the last one appearing on the left hand side.) (b) 22. In each case the Fibonacci number that appears on the right of the equality is that Fibonacci number immediately following the largest Fibonacci number appearing on the left. (c) N+1. N+1 is one bigger than N.

= ( FN + 2 + FN + 3 ) + FN + 5 = FN + 4 + FN + 5 = FN + 6

18. (a) Here we are adding the (N-2)nd and the (N-3)rd Fibonacci numbers. The result is the (N-1)st Fibonacci number (since N-1 is the next integer after N-2 and N-3). That is, FN − 2 + FN −3 = FN −1 . (b) Since FN + FN +1 = FN + 2 , we have FN + 2 − FN = FN +1 .

14. (a) Fifth equation in the sequence: 2(13) – 21 = 5 (In words, the left hand side is twice the 7th Fibonacci number less the 8th Fibonacci number. The right hand side is the fifth Fibonacci.) Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E (c) Add two terms at a time. FN − 3 + FN − 2 + FN + FN + 2

A8 = A6 + A7 = 40 + 65 = 105 , A9 = A7 + A8 = 65 + 105 = 170 ,

= ( FN − 3 + FN − 2 ) + FN + FN + 2

A10 = A8 + A9 = 105 + 170 = 275

= ( FN −1 ) + FN + FN + 2

(b) Note that every term in the sequence found in (a) is 5 times as large as the corresponding term in the Fibonacci sequence (i.e. A1 = 5 ⋅ F1 , A2 = 5 ⋅ F2 , A3 = 5 ⋅ F3 , etc.). So, A25 = 5 ⋅ F25 = 5 × 75, 025 = 375,125 .

= ( FN −1 + FN ) + FN + 2 = FN +1 + FN + 2 = FN + 3

19. (a) 1 +

FN F F = N −1 + N FN −1 FN −1 FN −1 =

FN −1 + FN FN −1

22. (a) B1 = 2 , B2 = 4 , B3 = B1 + B2 = 2 + 4 = 6 , B4 = B2 + B3 = 4 + 6 = 10 , B5 = B3 + B4 = 6 + 10 = 16 , B6 = B4 + B5 = 10 + 16 = 26 , B7 = B5 + B6 = 16 + 26 = 42 , B8 = B6 + B7 = 26 + 42 = 68 , B9 = B7 + B8 = 42 + 68 = 110 .

F = N +1 FN −1

(b)

FN −1 F F − 1 = N −1 − N FN FN FN = =

FN −1 − FN FN − ( FN − FN −1 ) FN

(b) Note that every term in the sequence found in (a) is double a term (the next term) in the Fibonacci sequence (i.e. B1 = 2 × 1 = 2 ⋅ F2 , B2 = 2 × 2 = 2 ⋅ F3 , B3 = 2 × 3 = 2 ⋅ F4 , etc.). So, B19 = 2 ⋅ F20 = 2 × 6765 = 13,530 .

F = − N −2 FN

20. (a) 1 +

(b) 1 −

FN −1 FN FN −1 = + FN FN FN =

FN + FN −1 FN

FN +1 FN

FN F F = N −2 − N FN − 2 FN − 2 FN − 2 =

FN − 2 − FN FN − 2

=−

213

FN −1 FN − 2

21. (a) A1 = 5 , A2 = 5 , A3 = A1 + A2 = 5 + 5 = 10 , A4 = A2 + A3 = 5 + 10 = 15 , A5 = A3 + A4 = 10 + 15 = 25 , A6 = A4 + A5 = 15 + 25 = 40 , A7 = A5 + A6 = 25 + 40 = 65 ,

23. (a) L1 = 1 , L2 = 3 , L3 = L1 + L2 = 1 + 3 = 4 , L4 = L2 + L3 = 3 + 4 = 7 , L5 = L3 + L4 = 4 + 7 = 11 , L6 = L4 + L5 = 7 + 11 = 18 , L7 = L5 + L6 = 11 + 18 = 29 , L8 = L6 + L7 = 18 + 29 = 47 , L9 = L7 + L8 = 29 + 47 = 76 , L10 = L8 + L9 = 47 + 76 = 123 , L11 = L9 + L10 = 76 + 123 = 199 , L12 = L10 + L11 = 123 + 199 = 322 .

214

Chapter 13: Fibonacci Numbers and the Golden Ratio (b) N = 1: L1 = 1 = 2 × 1 − 1 = 2 F2 − F1 ; N = 2: L2 = 3 = 2 × 2 − 1 = 2 F3 − F2 ; N = 3: L3 = 4 = 2 × 3 − 2 = 2 F4 − F3 ; N = 4: L4 = 7 = 2 × 5 − 3 = 2 F5 − F4 . (c) L20 = 2 F21 − F20 = 2 × 10,946 − 6765 = 15,127

24. (a) T1 = 1 , T2 = 4 , T3 = T1 + T2 = 1 + 4 = 5 , T4 = T2 + T3 = 4 + 5 = 9 , T5 = T3 + T4 = 5 + 9 = 14 , T6 = T4 + T5 = 9 + 14 = 23 , T7 = T5 + T6 = 14 + 23 = 37 , T8 = T6 + T7 = 23 + 37 = 60 , T9 = T7 + T8 = 37 + 60 = 97 , T10 = T8 + T9 = 60 + 97 = 157 , T11 = T9 + T10 = 97 + 157 = 254 , T12 = T10 + T11 = 157 + 254 = 411 . (b) N = 1: T1 = 1 = 3 × 1 − 2 × 1 = 3F2 − 2 F1 ; N = 2: T2 = 4 = 3 × 2 − 2 × 1 = 3F3 − 2 F2 ; N = 3: T3 = 5 = 3 × 3 − 2 × 2 = 3F4 − 2 F3 ; N = 4: T4 = 9 = 3 × 5 − 2 × 3 = 3F5 − 2 F4 . (c) T20 = 3F21 − 2 F20 = 3 × 10,946 − 2 × 6765

(b)

(c) One solution is positive, the second solution is negative, and their sum equals 1. This implies that when you add the two solutions their decimal parts must cancel each other out. 26. (a) First, we put x 2 = 3 x + 1 into the standard form x 2 − 3 x − 1 = 0 . The solutions to this quadratic equation are then x= =

13.2 The Golden Ratio

3 ± 13 2

So, 3 + 13 or 2 ≈ 3.30278

(b)

3 + 13 3 − 13 + =3 2 2

−(−1) ± (−1) 2 − 4(1)(−1) 2(1) 1± 5 2

−(−8) ± (−8) 2 − 4(3)(−5) 2(3)

8 ± 124 6

4 ± 31 3

So, x =

So, 1+ 5 x= or 2 ≈ 1.61803

3 − 13 2 ≈ −0.30278

(c) One solution is positive, the second solution is negative, and their sum equals 3. This implies that when you add the two solutions their decimal parts must cancel each other out.

25. (a) First, we put x 2 = x + 1 into the standard form x 2 − x − 1 = 0 . The solutions to this quadratic equation are then

−(−3) ± (−3) 2 − 4(1)(−1) 2(1)

27. (a) First, we put 3 x 2 = 8 x + 5 into the standard form 3x 2 − 8 x − 5 = 0 . The solutions to this quadratic equation are then

= 19,308

1+ 5 1− 5 + =1 2 2

1− 5 x= 2 ≈ −0.61803

(b)

4 + 31 ≈ 3.18925, or 3 4 − 31 ≈ -0.52259 3

4 + 31 4 − 31 8 + = 3 3 3

ISM: Excursions in Modern Mathematics, 10E 28. (a) First, we put 8 x 2 = 5 x + 2 into the standard form 8 x 2 − 5 x − 2 = 0 . The solutions to this quadratic equation are then x= =

−(−5) ± (−5) 2 − 4(8)(−2) 2(8)

(b)

5 + 89 ≈ 0.90212, or 16 5 − 89 ≈ -0.27712 16

Then a = 34, b = -55, and −1 + x = that x =

29. (a) Substituting x = 1 into the equation gives 55(1) 2 = 34(1) + 21 . Since the equation is satisfied, x =1 is a solution. (b) Rewrite the equation: 55 x 2 − 34 x − 21 = 0

Then a = 55, b = -34, and 1 + x =

34 so 55

34 21 − 1 = − ≈ −0.38182 . 55 55

30. (a) Substituting x = 1 into the equation gives 89(1) 2 = 55(1) + 34 . Since the equation is satisfied, x =1 is a solution. (b) Rewrite the equation: 89 x 2 − 55 x − 34 = 0 55 Then a = 89, b = -55, and 1 + x = so 89 55 34 − 1 = − ≈ −0.38202 . that x = 89 89

31. (a) Substituting x = -1 into the equation gives 21 = 34(-1) + 55 (i.e. that 21 is the difference of the two subsequent Fibonacci numbers). Since the equation is satisfied, x = -1 is a solution. (b) Rewrite the equation: 21x 2 − 34 x − 55 = 0

34 55 +1 = ≈ 2.61905 . 21 21

(b) Rewrite the equation: 34 x 2 − 55 x − 89 = 0

5 + 89 5 − 89 10 5 + = = 16 16 16 8

that x =

34 so 21

32. (a) Substituting x = -1 into the equation gives 34 = 55(-1) + 89 (i.e. 34 is the difference of the two subsequent Fibonacci numbers). Since the equation is satisfied, x = -1 is a solution.

5 ± 89 16

So, x =

Then a = 21, b = -34, and −1 + x =

215

55 so 34

55 89 +1 = ≈ 2.61765 . 34 34

33. (a) Substituting x = 1 in the equation gives FN = FN −1 + FN − 2 which is a defining equation for the Fibonacci numbers. Therefore, the equation is satisfied and x=1 is, by definition, a solution. (b) Rewrite the equation: FN x 2 − FN −1 x − FN − 2 = 0 Then, using the hint in Exercise 31(b), F 1 + x = N −1 so that FN F  x =  N −1  − 1  FN  F − FN = N −1 FN = =

− ( FN − FN −1 ) FN − FN − 2 . FN

34. (a) Putting x = -1 in the equation gives FN − 2 = − FN −1 + FN , or FN = FN − 2 + FN −1 which is a defining equation for the Fibonacci numbers. Therefore, the equation is satisfied and x=-1 is, by definition, a solution. (b) Rewrite the equation: FN − 2 x 2 − FN −1 x − FN = 0 Then, the sum of the roots of the equation −F F is − N −1 = N −1 . Given that one root is FN − 2 FN − 2

216

Chapter 13: Fibonacci Numbers and the Golden Ratio x=-1, the other root is then found to be FN −1 F + FN − 2 F + 1 = N −1 = N . FN − 2 FN − 2 FN − 2

35. (a)

39. (a) A7 = 2 A7 −1 + A7 − 2 = 2 A6 + A5 = 2 ( 70 ) + 29 = 169

≈ 0.6180339887

(b)

(b) Start with the golden property: φ 2 = φ + 1 .

Divide both sides by φ and get φ = 1 +

≈ 2.41429

φ It follows that φ and φ must have exactly the same decimal part (they are 1 unit apart). 36. (a) φ 2 ≈ 2.6180339887

(b) The golden property states that φ 2 = φ + 1 .

It follows that φ 2 and φ must have exactly the same decimal part (they are 1 unit apart). 37. (a) F500 ≈ φ × F499

(c)

A11 5, 741 = A10 2,378 ≈ 2.41421

(d) 2.41421 (Though not obvious, the ratio actually approaches 1 + 2 as the value of N increases.) 40. (a) A6 = 3 A5 + A4 = 3 ⋅109 + 33 = 360

(b)

≈ 1.6180 × 8.617 × 10103

A7 169 = 70 A6

A6 360 = A5 109

≈ 13.94 × 10103

≈ 3.30275

104

= 1.394 × 10 Remember that when using scientific notation, only one digit may occur before the decimal point.

(b) F498 ≈

F499

8.617 × 10103 1.618 ≈ 5.326 × 10103 ≈

38. (a) F1003 ≈ φ × F1002 ≈ 1.618 ×1.138 × 10 209 ≈ 1.841× 10209

(b) F1001 ≈

F1002

1.138 × 10209 1.618 ≈ 0.7033 × 10209 ≈

= 7.033 × 10208

13.3 Gnomons 41. (a) Since R and R’ are similar, each side length of R’ is 3 times as long as the corresponding side in R. So, the perimeter of R’ will be 3 times as great as the perimeter of R. This means that the perimeter of R’ is 3 × 41.5 = 124.5 inches. (b) Since each side of R’ is 3 times as long as the corresponding side in R, the area of R’ will be 32 = 9 times as larges as the area of R. That is, the area of R’ is 9 × 105 = 945 square inches. 42. (a) Since O and O’ are similar, the outer radius of O’ is 3 times as long as the outer radius of O. Since the outer radius of O is 14π = 7 feet, the outer radius of O’ is 21 2π feet. Hence, the circumference of the outer circle of O’ is 2 ⋅ π ⋅ 21 = 42π feet. (b) The area of an O-ring having inner radius r and outer radius R is A = π R 2 − π r 2 . If

ISM: Excursions in Modern Mathematics, 10E

the inner radius and outer radius are both tripled to form a new O-ring, the area will be π ( 3R ) − π ( 3r ) = 32 [π R 2 − π r 2 ] = 9 A Thus, the number of gallons to paint O’ will be 9 × 1.5 = 13.5 . 2

43. (a) The ratio of the perimeters must be the same as the ratio of corresponding side lengths. 60 m P = 13 in. 5 in. 60 m P= × 13 in. 5 in. = 156 m

(b) We use the fact that the ratio of the areas of two similar triangles (or any two similar polygons) is the same as the ratio of their side lengths squared. So, 2 A  60 m  =  20 in.2  5 in.  2

 60 m  2 A=  × (20 in. ) 5 in.   = 2880 m 2 5 = 2.5 . Thus the 2 perimeter of P’ is 2.5 × 10 = 25 .

44. (a) The scaling factor is

5 , the areas 2 2 25 5 change by a factor of   = . Thus 2 4   25 × 30 = 187.5 . the area of P’ is 4

(b) Since the scaling factor is

45. The two rectangles that need to be similar in order for the shaded rectangle to be a gnomon are the 2 by 10 rectangle and the 10 by x+2 rectangle (the one found by ‘gluing’ the white rectangle and the shaded rectangle together). The ratios of the sides of these rectangles must be the same. 10 x + 2 = 2 10 2( x + 2) = 100 2 x + 4 = 100 2 x = 96 x = 48

217

46. The two rectangles that need to be similar in order for the shaded region to be a gnomon are the 6 by 10 white rectangle and the 6+x by 12 rectangle (the one found by ‘gluing’ the white rectangle and the shaded L-shape together). The ratios of the sides of these rectangles must be the same. 6 6+ x = 10 12 60 + 10 x = 72 10 x = 12 x = 1.2 47. The two rectangles that need to be similar in order for the shaded ring to be a gnomon are the 8 by 12 white rectangle and the 12 by 14+x rectangle (the one found by ‘gluing’ the white rectangle and the shaded ring together). The ratios of the sides of these rectangles must be the same. 8 3 + 8 +1 = 12 2 + 12 + x 8 12 = 12 14 + x 8(14 + x) = 144 112 + 8 x = 144 8 x = 32 x=4

48. The two rectangles that need to be similar in order for the shaded shape to be a gnomon are the 2 by 6 white rectangle and the 3 by 6+2x rectangle (the one found by ‘gluing’ the white rectangle and the shaded region together). The ratios of the sides of these rectangles must be the same. 2 3 = 6 2x + 6 4 x + 12 = 18 4x = 6 x = 1.5

218

Chapter 13: Fibonacci Numbers and the Golden Ratio

49. The two rectangles that need to be similar in order for the shaded shape to be a gnomon are the 5 by x white rectangle and the 9 by 4+x rectangle (the one found by ‘gluing’ the white rectangle and the shaded ring together). The ratios of the sides of these rectangles must be the same. 5 5+2+2 = x x+2+2 5 9 = x x+2 5 x + 20 = 9 x 20 = 4 x x=5 50. The two rectangles that need to be similar in order for the shaded shape to be a gnomon are the 8 by x white rectangle and the 10 by 2+x rectangle (the one found by ‘gluing’ the white rectangle and the shaded ring together). The ratios of the sides of these rectangles must be the same. 8 1+ 8 +1 = x 1+ x +1 8 10 = x x+2 8 x + 16 = 10 x 16 = 2 x x =8

51. (a) In the figure, the measure of angle CAD must be 180 − 108 = 72 . Since triangle BDC must be similar to triangle BCA (in order for triangle ACD to be a gnomon), it must be that the measure of angle BDC must be 36 . Using the fact that the sum of the measures of the angles in any triangle is 180 , it follows that the measure of angle ACD is 72 .

52. The two triangles that need to be similar in order for the shaded shape to be gnomon are the 4-6-8 white triangle and the 8-x-(y+4) triangle (the one found by ‘gluing’ the white triangle and the shaded triangle together). Note: (1) Side lengths 6 and x correspond to each other as they are both opposite the same angle. (2) Side lengths 8 and y+4 correspond to each other as they are both opposite the largest angle in their corresponding triangle (the angle between the sides of length 8 and x must be the largest since it cannot be the smallest as the smallest angle in the white triangle is a part of this angle). (3) Side lengths 4 and 8 correspond to each other as they are both opposite the smallest angle in their corresponding triangle. So, we 4 6 8 . This gives x = 12 and y solve = = 8 x y+4 = 12. 53. The two triangles that need to be similar in order for the shaded shape to be gnomon are the 3-4-5 white triangle and the 9-x-(5+y) triangle (the one found by ‘gluing’ the white triangle and the shaded region together). By comparing side lengths of the similar triangles, 3 4 3 5 . That is, 3x=36 we find = and = 9 x 9 5+ y and 3(5+y) = 45. Solving these gives x = 12 and y = 10. 54. The two triangles that need to be similar in order for the shaded shape to be gnomon are the 9-12-15 white triangle and the 15-x-(y+9) triangle (the one found by ‘gluing’ the white triangle and the shaded triangle together). By comparing side lengths of the similar triangles, 9 12 15 . This gives 9x = 180 = = we find 15 x y+9 and 9y+81 = 225. Solving yields x = 20 and y = 16.

(b) Since triangle DBC is isosceles, x = φ . Note that triangle ACD is also isosceles. Hence, y = x = φ as well. Likewise, triangle ABC is isosceles so that the length of AC is 1. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

JOGGING 55. (a)

57. (a) T1 = 1 , T2 = 1 + T3 = 1 +

F3 2 F 3 = = 0.666667 , 4 = = 0.6 , F4 3 F5 5

1 1 3 = 1+ = , T2 2 2

T4 = 1 +

F5 5 F 8 = = 0.625 , 6 = = 0.615385 , F6 8 F7 13

1 1 2 5 = 1+ = 1+ = , T3 3 3 (3 / 2)

T5 = 1 +

1 1 3 8 = 1+ = 1+ = , T4 5 / 3 5 5 ( )

T6 = 1 +

1 1 5 13 = 1+ = 1+ = . 8 8 T5 (8 / 5)

F (b) We know that N +1 → φ as N gets large. FN It follows that the reciprocals of these ratios approaches the reciprocal of this F 1 value (i.e. that N → ). Since FN +1 φ 1

(b) We know that

FN +1 → φ as N gets large. FN

But TN =

FN +1 from part (a). FN

58. U1 = 1 , U 2 =

1 1 1 = = , 1 + U1 1 + 1 2

U3 =

1 1 2 = = , 1 + U 2 1 + (1/ 2) 3

U4 =

1 1 3 = = , 1 + U 3 1 + ( 2 / 3) 5

U5 =

1 1 5 = = , 1 + U 4 1 + ( 3 / 5) 8

F3 2 F 3 = = 2 , 4 = = 3, F1 1 F2 1

U6 =

1 1 8 = = . 1 + U 5 1 + ( 5 / 8 ) 13

F5 5 F 8 = = 2.5 , 6 = = 2.666667 , F3 2 F4 3

In general, U N =

F7 13 F 21 = = 2.6 , 8 = = 2.625 , F5 5 F6 8

see that U N →

= φ − 1 , we have the result. [The fact 1

= φ − 1 follows from using the φ golden rule φ 2 = φ + 1 and dividing both sides by φ .]

that

56. (a)

1 1 = 1+ = 2 , T1 1

F1 1 F 1 = = 1 , 2 = = 0.5 , F2 1 F3 2

F7 13 = = 0.619048 , 0.617647, F8 21 0.618182, 0.61798, 0.618056, 0.618026, 0.618037, 0.618033, …

F9 34 = = 2.615385 , 2.619048, F7 13 2.617647, 2.618182, 2.617978, 2.618056, 2.618026, 2.618037, 2.618033, … FN +1 → φ as N gets large. FN By adding one to each term of the F sequence, it follows that N +1 + 1 → φ + 1 . FN FN +1 F + FN FN + 2 . + 1 = N +1 = FN FN FN

FN . By exercise 55(b), we FN +1 = φ − 1 as N gets large.

59. (a) K1 = 2 F2 − F1 = 2 × 1 − 1 = 1 , K 2 = 2 F3 − F2 = 2 × 2 − 1 = 3 , and

K N −1 + K N − 2 = ( 2 FN − FN −1 ) + ( 2 FN −1 − FN − 2 )

= 2 ( FN + FN −1 ) − ( FN −1 + FN − 2 )

(b) We know that

But

219

= 2 FN +1 − FN = KN Since the seeds are the same and the recursive rule is the same, it follows that K N = LN .

220

Chapter 13: Fibonacci Numbers and the Golden Ratio (b) From part (a), we have that LN +1 2 FN + 2 − FN +1 . If you divide the = 2 FN +1 − FN LN numerator and the denominator of the right hand expression by FN +1 you get LN +1 2 FN + 2 − FN +1 = LN 2 FN +1 − FN

( 2 FN + 2 − FN +1 ) / FN +1 ( 2 FN +1 − FN ) / FN +1 2 ( FN + 2 FN +1 ) − 1 = 2 − ( FN FN +1 )

odd numbers again like the seeds, is even. It follows that every third number is even and the others in the sequence are all odd. A nice way to visualize this and convince yourself of its truth is the following: OOEOOEOOEOOEOOEOOE… 62. FN +1 − FN = ( FN +1 − FN )( FN +1 + FN ) 2

= FN −1 ⋅ FN + 2

→

63.

2φ − 1 2 − (1 φ )

10 20 = 20 10 + x 10(10 + x) = 400

φ ( 2φ − 1) 2φ − 1 =φ =

since

100 + 10 x = 400 10 x = 300 So, x = 30. Rectangle B is 20 by 30.

FN +1 →φ . FN

60. (a) They alternate between negative and positive numbers getting smaller and smaller in absolute value and closer and closer to 0. (b) Using Binet’s formula, N

 1+ 5  1− 5   2  −  2     FN =  5 1− 5  φ −   2   = 5

64. The result of attaching a gnomon to the Lshaped figure must result in a new, larger Lshaped figure — it must be higher, wider, and thicker (all by the same proportions). The possible ways of attaching a square to the Lshaped figure and still maintaining a L-shaped figure are:

None of these figures can possibly be similar to the original L-shaped figure.

It follows from (a), that FN →

φN 5

FN +1 φ N +1 / 5 → N =φ . FN φ / 5

61. First, remember that the sum of two odd numbers is always even and the sum of an odd number and an even number is always odd. Now, since the seeds in the Fibonacci sequence (1 and 1) are both odd, the third number is even. Since the second number is odd and the third number is even, the fourth number is odd. Similarly , since the third number is even and the fourth number is odd, the fifth number is odd. But then the sixth number, the sum of two

65. Yes, but only when x = y (i.e. if the picture is a square). The reason is that for a frame to be a gnomon to the picture, we must have (x+2w)/(y+2w) = x/y. Solving for x in terms of y gives x = y. x + 2w x = y + 2w y y ( x + 2w) = x ( y + 2 w ) xy + 2wy = xy + 2wx 2wy = 2wx y=x

ISM: Excursions in Modern Mathematics, 10E 66. The two triangles that need to be similar in order for the shaded shape to be gnomon are the y-60-x white triangle and the x-156(y+144) triangle (the one found by ‘gluing’ the white triangle and the shaded triangle together). By comparing side lengths of the similar triangles, we find 60 y 60 x = and . So, = 156 x 156 y + 144 60 x = 156 y and 60 y + 8640 = 156 x . Substituting yields  60  x  + 8, 640 = 156 x . 60   156  60 Thus x = 65 so that y = (65) = 25 . 156 67. From elementary geometry ∠AEF = ∠DBA . Consequently  AEF is similar to DBA (since they are both right triangles and so have all their corresponding angles congruent). So AF : FE = DA : AB which shows rectangle ADEF is similar to rectangle ABCD. 68. (a) Since we are given that AB = BC = 1, we know that ∠BAC = 72° and so ∠BAD = 180° − 72° = 108°. This makes ∠ABD = 180° − 108° − 36° = 36° and so triangle ABD is isosceles with AD = AB = 1. Therefore AC = x – 1. Using these facts and the similarity of triangle ABC and x 1 or, triangle BCD we have = 1 x −1 x 2 = x + 1 for which we know the solution is x = φ . (b) 36°, 36°, 108° (c)

221

opposite the 72° angle is the same length as the radius of the circle, namely r. In exercise 66, it was shown that the length of the side opposite the 36° angle is r φ . Hence, the perimeter of the regular 1 10 decagon is 10 × = .

(b) Using part (a), the perimeter is found to be 10r .

RUNNING 71. (a) Using the same seeds as GN (namely a and b), we define a sequence K N = bFN −1 + aFN − 2 and show that the numbers K N satisfy the same rule as GN (namely that GN = GN −1 + G N − 2 ). But,

K N −1 + K N − 2 = ( bFN − 2 + aFN − 3 ) + ( bFN − 3 + aFN − 4 )

= b ( FN − 2 + FN − 3 ) + a ( FN − 3 + FN − 4 ) = bFN −1 + aFN − 2

= KN It follows that K N = GN .

longer side x = = x =φ shorter side 1

69. Follows from exercise 66(a) and the following figure.

222

Chapter 13: Fibonacci Numbers and the Golden Ratio (b) From part (a), we have that GN +1 bFN + aFN −1 . If you divide the = GN bFN −1 + aFN − 2 numerator and the denominator of the right hand expression by FN −1 you get GN +1 bFN + aFN −1 = GN bFN −1 + aFN − 2

( bFN + aFN −1 ) / FN −1 ( bFN −1 + aFN − 2 ) / FN −1 b ( FN FN −1 ) + a = b + a ( FN − 2 FN −1 )

→

bφ + a b + a (1 φ )

φ ( bφ + a ) bφ + a =φ =

since

FN +1 →φ . FN

72. (a) 5; The possible paths are illustrated below.

74. F1 = F2 F3 = F4 − F2 F5 = F6 − F4  FN = FN +1 − FN −1 Adding the left and right hand sides respectively, yields F1 + F3 + F5 +  + FN = FN +1 75. Suppose M is a positive integer greater than 2. If M is a Fibonacci number ( M = FN ) , then it is certainly true that M can be written as the sum of distinct Fibonacci numbers ( M = FN = FN −1 + FN − 2 ) . Otherwise, M falls somewhere between two Fibonacci numbers, say between FN and FN +1 . Write M = FN + ( M − FN ) . Now M − FN < FN +1 − FN = FN −1 is itself a positive integer smaller than M and we can repeat the argument. Eventually the process must stop. (This is an argument by induction in disguise.)

Note: This argument gives an algorithm for finding a decomposition of M as a sum of Fibonacci numbers. Find the largest Fibonacci number that fits into M (call it FN ), then find the largest Fibonacci number that fits into the difference M − FN , and so on. 76. The area of triangle II is

(b) 34 (c) FN +1 ; The path can end with either a vertical stone or two horizontal stones. There are FN ways that a path of length N feet can end in a vertical stone (count all paths of length N-1). There are FN −1 ways that a path can of length N feet can end in two horizontal stones (count all paths of length N-2). Then, use the fact that FN +1 = FN −1 + FN . 73. F1 = F3 − F2 F2 = F4 − F3  FN −1 = FN +1 − FN FN = FN + 2 − FN +1 Adding the left and right hand sides respectively, yields F1 + F2 +  + FN = FN + 2 − F2 = FN + 2 − 1

1 2 x . The area of 2

triangle III is 1 y(x + y). Setting the areas 2 1 2 1 equal gives x = y ( x + y ) or 2 2 2 x = y ( x + y ) or, equivalently, x = x + y = 1 + y . Letting r = x the equation y y x x 2 1 becomes r = 1 + or r = r + 1 which has the r solution x = r = 1 + 5 = φ . 2 y

ISM: Excursions in Modern Mathematics, 10E 77. (a) From exercise 68, x = φ y and so

x =φ . y

(See figure.) Since z = x, x+ y x+ y y 1 = = 1+ = 1+ = φ . φ z x x x+ y+z z Therefore, = 1+ = x+ y x+ y 1+

=φ .

(b) From part (a), x = φ y = φ ;

x + y = zφ = xφ = φ 2 ; x + y + z = ( x + y )φ = φ 2 ⋅φ = φ3

223

Chapter 14 WALKING

218 4965 = for N. N 900 The proportion of the 900 seals recaptured in late August that had tags (k = 218) should be the same as the proportion of the 4965 seals that were tagged during the first capture.

8. Roughly 20,500; we solve

14.1 Enumeration 1. 64; the proportion of red gumballs in the 8 = 0.32. Assuming that the sample is 25 sample is representative of the population, we estimate that there are 0.32 × 200 = 64 red gumballs in the jar. 2. 70; the proportion of red gumballs in the 14 = 0.35. Assuming that the sample is 40 sample is representative of the population, we estimate that there are 0.35 × 200 = 70 red gumballs in the jar. 17 x = for x. The 253 34,522 proportion of the 253 patients sampled having A- blood type should be the same as the proportion of the 34,522 people in Madison County having A- blood type.

3. 2320; we solve

22 x = for x. The 527 34,522 proportion of the 527 patients sampled having AB+ blood type should be the same as the proportion of the 34,522 people in Madison County having AB+ blood type.

4. 1441; we solve

121 1542 = for x. The x 200 proportion of the 200 seats sampled that were female should be the same as the proportion of the x people at the concert that were female.

5. 2549; we solve

31 150 = for x. The proportion 6. 242; we solve x 50 of the 50 gumballs sampled that were green should be the same as the proportion of the x gumballs in the jar that are green. 30 500 7. 2000; we solve = for N. The N 120 proportion of the 120 fish recaptured that had tags (k = 30) should be the same as the proportion of the 500 fish in the pond that were tagged during the first capture.

76 121 = for N. The N 172 proportion of the 172 whales recaptured that had tags (k = 76) should be the same as the proportion of the 121 whales that were tagged during the first (photographic) capture.

9. 274; we solve

16 38 = for N. The proportion 27 N of the 27 dolphins recaptured in 2016 that had tags (k = 16) should be the same as the proportion of the 38 dolphins that were tagged during the first capture in 2015.

10. 64; we solve

11. We now solve

77 122 = for N. This gives 173 N + 1

122 × 173 − 1 = 273 which is one less 77 than under the standard capture-recapture formula.

12. We now solve

13 27 = for N. This gives 28 N + 1

27 × 28 − 1 = 57 which is two less than 13 under the standard capture-recapture formula.

13. Using

n k = 1 with k = 7, n1 = 1700 , and N = n2 N

160,286, we solve

7 1700 = for n2 to n2 160, 286

find n2 » 660 . 14. Using

n k = 1 with n1 = 24, 000 , N = n2 N

1,100,000 and n2 = 10,300 we solve k 24, 000 = for k to find k » 225 . 10,300 1,100, 000 That is, about 225 of the recapture carp had tags.

ISM: Excursions in Modern Mathematics, 10E

14.2 Measurement 15. (a) B - Convenience sampling. George is selecting those units in the population that are easily accessible. (b) D - Stratified sampling. The student newspaper is dividing the population into strata and then selecting a proportionately sized random sample from each stratum (in fact, 5% from each stratum). (c) A - Simple random sampling. Every subset of three players has the same chance of being selected as any other subset of three players. (d) C - Quota sampling. The coach is attempting to force the sample to fit a particular profile. 16. (a) E - Census. A complete enumeration of the population (the 15,000 student aid packages) is taken. (b) C – Quota sampling. The sample is chosen here to fit the profile of the population. (c) B - Convenience sampling. In a very real sense, this is the most convenient sampling method for the auditor to undertake. (d) D - Stratified sampling. The first stratum is the school and the second layer of strata is the major. Note too that all samples taken from the strata and substrata are random. 17. (a) The registered voters in Cleansburg. (b) The 680 registered voters that are polled by telephone. (c) Simple random sampling. 18. (a) The sampling proportion for this survey is 680 ≈ 0.082 (approximately 8.2%). 8325 (b) 306/680 = 45%; of those sampled, this represents the percentage that indicated they planned to vote for Smith.

225

19. Of the people surveyed, 45% indicated they æ 306 ö = 0.45÷÷÷ . The would vote for Smith ççç è 680 ø

actual percentage was 42%. Since 45% – 42% = 3%, the sampling error for Smith was 3%. Similarly, the sampling error for Jones was 43% - 40% = 3% and the sampling error for Brown was 15% - 15% = 0%. 20. The sampling error should be attributed primarily to chance, since the sample was chosen randomly (this eliminates selection bias) and there was a 100% response rate (this eliminates nonresponse bias). 21. (a) The sample for this survey is the 350 students attending the Eureka High football game before the election. (b)

350 = 28% 1250

22. The sampling method used in this survey is convenience sampling. 23. (a) The population consists of all 1250 students at Eureka High School while the sampling frame consists only of those students that attended the football game a week prior to the election. (b) The sampling error is mainly a result of sampling bias. The sampling frame (and hence any sample taken from it) is not representative of the population. Students that choose to attend a football game are not representative of all Eureka High students. 24. (a) Based on the sample, we expect the following results: Tomlinson – 58%, Garcia – 12%, and Marsalis – 30%. (b) The sampling error for Tomlinson was 58% - 19% = 39%. The sampling error for Garcia was 51% 12% = 39%. The sampling error for Marsalis was 30% - 30% = 0%. 25. (a) The target population of this survey is the citizens of Cleansburg.

226

Chapter 14: Censuses, Surveys, Polls, and Studies (b) The sampling frame is limited to that part of the target population that passes by a city street corner between 4:00 p.m. and 6:00 p.m.. It excludes citizens of Cleansburg having other responsibilities during that time of day.

26. (a) 475 (b) Yes, this survey is subject to nonresponse bias. The total number of respondents was 475 (add the numbers in the first two columns). The total number of nonrespondents was 1313. The response 475 ≈ 0.266 . That is, rate was 1313 + 475 only 26.6% of the people asked to participate in this survey actually did. 27. (a) The choice of street corner could make a great deal of difference in the responses collected. (b) D (We are making the assumption that people who live or work downtown are much more likely to answer yes than people in other parts of town.) (c) Yes, the survey was subject to selection bias for two main reasons. (i) People out on the street between 4 p.m. and 6 p.m. are not representative of the population at large. For example, office and whitecollar workers are much more likely to be in the sample than homemakers and school teachers. (ii) The five street corners were chosen by the interviewers and the passersby are unlikely to represent a cross section of the city. (d) No, no attempt was made to use quotas to get a representative cross section of the population. 28. Sending people out to interview passersby in a fixed location is not a good idea. Fixing a particular time of the day made it even worse. A telephone poll, based on a random sample of 500 registered voters in the city of Cleansburg would probably produce much more accurate data. Before asking the actual question (“Are you in favor, blah, blah, blah,...”) the interviewer should introduce himself as representing the City Planning Department and, indicate that the survey consists of only one question and the survey will be used by the city to make important

planning decisions. In this case, a better response rate can be expected. 29. (a) Stratified sampling. The trees are broken into three different strata (by variety) and then a random sample is taken from each stratum. (b) Quota sampling. The grower is using a systematic method to force the sample to fit a particular profile. However, because the grower is human, sampling bias could then be introduced. When selecting 300 trees of variety A, the grower does not select them at random. Selecting 300 trees in one particular part of the orchard could bias the yield. 30. (a) 80 oranges per tree 0.50 × 100+0.25 × 50+0.25 × 70 = 80 (b) A statistic. (This is information about a sample, not about a population.) 31. (a) Label the crates with numbers 1 to 250. Select 20 of these numbers at random (put the 250 numbers in a hat and draw out 20). Sample those 20 crates. The important point is that any set of 20 crates have an equal chance of being in the sample. (b) Select the top 20 crates in the shipment (those easiest to access). (c) Randomly sample 6 crates from supplier A, 6 crates from supplier B, and 8 crates from supplier C. (d) Select any 6 crates from supplier A, any 6 crates from supplier B, and any 8 crates from supplier C. 32. (a) Select 150 undergraduate students whose names have been randomly drawn from the registrar’s database. (b) Stand in front of the student union and interview the first 150 students who walk by and are willing to be interviewed. (c) Break up the undergraduates by strata (school, major class, etc.), randomly select some of the strata, and from those randomly select the students to be interviewed.

ISM: Excursions in Modern Mathematics, 10E (d) Randomly select 95 female undergraduates (30 Hispanic, 25 white, 20 black, and 20 Asian-American), and 80 male undergraduates (23 Hispanic, 21 white, 17 black, and 19 Asian-American). (This breakdown corresponds to the gender/ethnic breakdown of the entire undergraduate student body.)

14.3 Cause and Effect

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treatment group (gets vitamin X) and a control group (gets a fake pill). (iii) Have trained professionals (nurses) measure the length of each subject’s cold. (iv) Neither the subjects nor the nurses should know who is getting the real vitamin X and who is getting a placebo. 37. (a) The target population for this study is all potential knee surgery patients.

33. (a) The target population for this study is anyone who could have a cold and would consider buying vitamin X (i.e., pretty much all adults).

(b) The sample consists of 180 potential knee surgery patients at the Houston VA Medical Center who volunteered to be in the study.

(b) The sampling frame is only a small portion of the target population. It only consists of college students in the San Diego area that are suffering from colds.

38. (a) No, the subjects volunteered. 144 of the 324 declined to participate.

(c) The sample is 500 of these San Diego area students all of whom are suffering from a cold at the start of the study. Selection bias is present here since this sample would likely under represent older adults and those living in colder climates. 34. (a) No, there was no control group. (b) 1. San Diego residents are not typical of the population at large in several critical aspects (age, health, exposure to inclement weather, etc.). 2. The volunteers were paid to participate. 3. The subjects themselves determined the length of their cold. 4. There was no control group. 35. Four different problems with this study that indicate poor design include (i) using college students (College students are not a representative cross section of the population in terms of age and therefore in terms of how they would respond to the treatment.), (ii) using subjects only from the San Diego area, (iii) offering money as an incentive to participate, and (iv) allowing self-reporting (the subjects themselves determine when their colds are over). 36. (i) Choose the subjects randomly from the population at large. (ii) Divide the subjects randomly into a

(b) Yes, this was a controlled placebo experiment. There was one control group receiving the sham-surgery (placebo) and two treatment groups. (c) The first treatment group consisted of those patients receiving arthroscopic debridement. The second treatment group consisted of those patients receiving arthroscopic lavage. 39 (a) Yes, this study could be considered a randomized controlled experiment since the 180 patients in the study were assigned to a treatment group at random. (b) This was a blind experiment. The doctors certainly knew which surgery they were performing on each patient. 40. Answers will vary. One ethical issue to consider is that some patients that really do need surgery would not receive treatment if the study were continued. On the other hand, surgery is risky business and the results of this study suggest the benefits may not be all they appear to be. 41. The target population consists of all people having a history of colorectal adenomas. The point of the study is to determine the effect that Vioxx had on this population. That is, whether recurrence of colorectal polyps could be prevented in this population. 42. The sample consisted of 2,586 participants all of whom had a history of colorectal adenomas.

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43. (a) The treatment group consisted of the 1287 patients that were given 25 daily milligrams of Vioxx. The control group consisted of the 1299 patients that were given a placebo. (b) This is an experiment since members of the population received a treatment. It is a controlled placebo experiment since there was a control group that did not receive the treatment, but instead received a placebo. It is a randomized controlled experiment since the 2586 participants were randomly divided into the treatment and control groups. The study was double blind since neither the participants nor the doctors involved in the clinical trial knew who was in each group. 44. The clinical study described is a double-blind randomized controlled placebo experiment. As such, a person suffering from arthritis and having a history of colorectal adenomas should avoid taking Vioxx as this study suggests the potential for an increase in the occurrence of cardiovascular events. 45. (a) Women; particularly young women (b) The sampling frame consists of those women between 16 and 23 years of age that are not at high risk for HPV infection (that is, those women having no prior abnormal pap smears and at most five previous male sexual partners). Pregnant women are also excluded from the sampling frame due to the risks involved. Though not perfect, it appears to be a representative subset of the target population. 46. (a) 2392 women from 16 different centers across the United States (b) The sample was not chosen using random sampling. Young women were encouraged to volunteer to participate through advertising on college campuses and in surrounding communities. 47. (a) The treatment group consists of the half of the 2392 women in the sample that received the HPV vaccine. (b) This is a controlled placebo experiment since there was a control group that did not receive the treatment, but instead received a placebo injection. It is a

randomized controlled experiment since the 2392 participants were randomly divided into the treatment and control groups. The study was likely double blind – it is probable that neither the participants nor the medical personnel giving the injections knew who was in each group. 48. This vaccine appears to be very effective at preventing HPV in young women. Therefore, widespread use of this vaccine could lead to fewer incidents of cervical cancer in women. What is not clear, however, are what side effects the HPV vaccine may have. 49. (a) The stated purpose of the study was to determine the effectiveness of a new serum for treating diphtheria. So, the target population consisted of all people suffering from diphtheria. (b) The sampling frame consists of those people having diphtheria symptoms serious enough to be admitted to one particular Copenhagen hospital (between May 1896 and May 1897). It appears to be a reasonably representative subset of the target population. This is primarily because there is likely little difference between those suffering from diphtheria in this Copenhagen hospital and those suffering from diphtheria elsewhere. 50. (a) The sample consists of 484 new diphtheria patients admitted to one particular Copenhagen hospital over a one-year period. (b) It is unlikely since humans were not involved in the selection of patients. 51. (a) The treatment group consisted of those patients admitted on the “even” days that received both the new serum and the standard treatment. The control group consisted of those patients admitted on the “odd” days that received only the standard treatment for diphtheria at the time. (b) When received together with standard treatment, this new serum appears to be somewhat effective at treating diphtheria. Significantly more patients that did not receive the treatment in this study died (presumably many of them from diphtheria or related issues).

ISM: Excursions in Modern Mathematics, 10E 52. The location of the hospital could be a confounding variable (patients at each hospital could come from different socio-economic and hereditary backgrounds). Treatment procedures at each hospital could be quite different. Patients admitted to each hospital may show different degrees of illness. The staff at each hospital could have different levels of training (e.g. teaching vs. nonteaching hospitals). 53. (a) The target population consisted of men with prostate enlargement; particularly older men. (b) The sampling frame consists of men aged 45 or older having moderately impaired urinary flow. It appears to be a representative subset of the target population. 54. (a) The sample consisted of the 369 men aged 45 or older having moderately impaired urinary flow that were volunteered for the study. (b) No. Participants volunteered. 55. (a) The treatment group consisted of the group of volunteers that received saw palmetto daily. (b) The saw palmetto odor would be a dead giveaway as to who was getting the treatment and who was getting a placebo. To make the study truly blind, the odor had to be covered up. (c) Yes. There was a control group that received a placebo instead of the treatment, and the treatment group was randomized. We can infer the study was blind from the fact that an effort was made to hide the odor of the saw palmetto. From the description of the study there is no way to tell if the study was also double blind. 56. You would probably think twice about continuing with saw palmetto as it appears to produce no benefit in treating lower urinary tract symptoms. This comes from the fact that both the treatment and control groups had similar small improvements in mean symptom scores, but saw palmetto conferred no benefit over placebo on symptom scores or on any other secondary outcomes.

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JOGGING 57. (a) Assuming that the registrar has a complete list of the 15,000 undergraduates at Tasmania State University, the target population and the sampling frame both consist of all undergraduates at TSU. (b) N = 15,000 58. (a) 1% (1 of every 100 names are selected) (b) n = 135; 1 % of 15,000 is 150. 90% of 150 is 135. 59. (a) In simple random sampling, any two members of the population have as much chance of both being in the sample as any other two. But in this sample, two people with the same last name—say Len Euler and Linda Euler—have no chance of being in the sample together. (b) Sampling variability. The students sampled appear to be a reasonably representative cross section of all TSU undergraduates that might or might not be familiar with the new financial aid program. 60. (a) The proportion of students in the sample who responded by saying that they were not familiar with the new financial aid 108 108 program is = = 0.8. 0.90 × 150 135 Assuming that those who responded are representative of the population, we estimate that there are a total of 0.8 × 15, 000 = 12, 000 TSU undergraduates that are not familiar with the new financial aid program. (b) The only possible flaw with this survey is the size of the sample (too small). A larger sample, of course, would result in a more accurate (but also more expensive) survey. The decision on how big is big enough depends on how important is accuracy and how much money one is willing to spend. In any case, a sample of 150 students is probably too small.

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61.

To estimate the number of quarters, we disregard the nickels and dimes—they are 4 36 irrelevant. To that end, we solve = for 28 N N to estimate the number of quarters in the jar. This gives N = 252 quarters. Similarly, to estimate the number of nickels, we disregard the quarters and dimes. In 5 45 particular, we solve = for N to estimate 29 N the number of nickels in the jar. This gives N = 261 nickels. 8 69 = for N to estimate the 43 N number of dimes in the jar. This gives approximately N = 371 nickels.

Lastly, we solve

Since 252 quarters, 261 nickels, and 371 dimes are estimated to be in the jar, the total amount of money in the jar is estimated to be 252´($0.25) + 261´($0.05) + 371´($0.10) = $63.00 + $13.05 + $37.10 = $113.15 62. Capture-recapture would underestimate the true population. If the fraction of those tagged in the recapture appears higher than it is in reality, then the fraction of those tagged in the initial capture will also be computed as higher than the truth. This makes the total population appear smaller than it really is.

For an example to see how this works, suppose a population had N = 1000 individuals. Imagine an initial tagging of n1 = 100 individuals (10% of the population are tagged). Then, in the recapture of n2 = 20 , we should expect k = 2 individuals are tagged. If, however, some of the individuals were “trap-happy,” then the number of individuals that are captured twice (the value of k) would be higher than expected (more than 2). Say k = 4 individuals (double the original number) were captured twice. Then, the value of N would be estimated to be n 20 N = 2 ×n1 = ×100 = 500 . This is clearly an k 4 underestimate (half the original number). 63. Capture-recapture would overestimate the true population. If the fraction of those tagged in the recapture appears lower than it is in reality,

then the fraction of those tagged in the initial capture will also be computed as lower than the truth. This makes the total population appear larger than it really is. For an example to see how this works, suppose a population had N = 1000 individuals. Imagine an initial tagging of n1 = 100 individuals (10% of the population are tagged). Then, in the recapture of n2 = 20 , we should expect k = 2 individuals are tagged. If, however, some of the individuals were more likely to be prey, then the number of individuals that are captured twice (the value of k) would be lower than expected (less than 2). Say k = 1 individuals (half the original number) were captured twice. Then, the value of N would be n 20 ×100 = 2000 . estimated to be N = 2 ×n1 = k 1 This is clearly an overestimate (twice the original N-value). 64. (a) The populations are (i) the entire sky; (ii) all the coffee in the cup; (iii) the entire Math 101 class. (b) In none of the three examples is the sample random. (c) (i) In some situations one can have a good idea as to whether it will rain or not by seeing only a small section of the sky, but in many other situations rain clouds can be patchy and one might draw the wrong conclusions by just peeking out the window. (ii) If the coffee is burning hot on top, it is likely to be pretty hot throughout, so Betty’s conclusion is likely to be valid. (iii) Since Carla used convenience sampling and those students sitting next to her are not likely to be a representative sample, her conclusion is likely to be invalid. 65. Answers will vary. For example, Al might see that Apple’s stock price was down today and infer that the entire stock market had a down day. Or Betty might look at the headlines in the New York Times and determine that her son stationed overseas must be safe and sound at his post. Or Carla might go to the mall on Black Friday and deem it always that busy.

ISM: Excursions in Modern Mathematics, 10E 66. (a) The results of this survey might be invalid because the question was worded in a way that made it almost impossible to answer yes. (b) “Will you support some form of tax increase if it can be proven to you that such a tax increase is justified?” is better, but still not neutral. “Do you support or oppose some form of tax increase?” is bland but probably as neutral as one can get. (c) Many such examples exist. A very real example is a question such as “Would you describe yourself as pro-life or not?” or “Would you describe yourself as prochoice or not?” 67. Consideration of one exception made them less likely to consider a second exception. This is why it is critical that the order a series of survey questions such as this are asked is randomized. Answers will vary. 68. (a) Under method 1, people whose phone numbers are unlisted are automatically ruled out from the sample. At the same time, method 1 is cheaper and easier to implement than method 2. Method 2 will typically produce more reliable data since the sample selected would better represent the population. (b) For this particular situation, method 2 is likely to produce much more reliable data than method 1. The two main reasons are (i) people with unlisted phone numbers are very likely to be the same kind of people that would seriously consider buying a burglar alarm, and (ii) the listing bias is more likely to be significant in a place like New York City. (People with unlisted phone numbers make up a much higher percentage of the population in a large city such as New York than in a small town or rural area. Interestingly enough, the largest percentage of unlisted phone numbers for any American city is in Las Vegas, Nevada.) 69. (a) An Area Code 900 telephone poll represents an extreme case of selection bias. People that respond to these polls usually represent the extreme view points (strongly for or strongly against), leaving out much of the middle of the road point of view. Economics also plays some role

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in the selection bias. (While 50 cents is not a lot of money anymore, poor people are much more likely to think twice before spending the money to express their opinion.) (b) This survey was based on fairly standard modern-day polling techniques (random sample telephone interviews, etc.) but it had one subtle flaw. How reliable can a survey about the conduct of the newsmedia be when the survey itself is conducted by a newsmedia organization (Times-Mirror is a now defunct print media publisher)? (“The fox guarding the chicken-coop” syndrome.) (c) Both surveys seem to have produced unreliable data — survey 1 overestimating the public’s dissapproval of the role played by the newsmedia and survey 2 overestimating the public’s support for the press coverage of the administration. In survey 1, only those with strong opinions were surveyed (i.e. called the 900 number). In survey 2, people without strong opinions were called. Further, they may have been swayed by being surveyed by a media organization. (d) Any reasoned out answer should be acceptable. (Since Area Code 900 telephone polls are particularly unreliable, survey 2 gets our vote.) 70. (a) Fridays, Saturdays, and Sundays make up 3/7 or about 43% of the week. It follows that proportionally, there are fewer fatalities due to automobile accidents on Friday, Saturday, and Sunday (42%) than there are on Monday through Thursday. (b) On Saturday and Sunday there are fewer people commuting to work. The number of cars on the road and miles driven is significantly less on weekends. The number of accidents due to fatalities should be proportionally much less. The fact that it is 42% indicates that there are other factors involved and increased drinking is one possible explanation. 71. (a) Both samples should be a representative cross section of the same population. In particular, it is essential that the first sample, after being released, be allowed to disperse evenly throughout the population, and that the population should

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not change between the time of the capture and the time of the recapture. (b) It is possible (especially when dealing with elusive types of animals) that the very fact that the animals in the first sample allowed themselves to be captured makes such a sample biased (they could represent a slower, less cunning group). This type of bias is compounded with the animals that get captured the second time around. A second problem is the effect that the first capture can have on the captured animals. Sometimes the animal may be hurt (physically or emotionally) making it more (or less) likely to be captured the second time around. A third source of bias is the possibility that some of the tags will come off. 72. (a) People are very unlikely to tell IRS representatives that they cheated on their taxes even when they are promised confidentiality. (b) The critical issue in this survey is to get the respondents to give an honest answer to the question. In this regard two points are critical: (i) the survey is much more likely to get honest responses when sponsored by a “neutral” organization (a newspaper poll, for example); (ii) a mailed questionnaire is the safest way to guarantee anonymity for the respondent and in this case it is much better than a telephone or (heaven forbid) personal interview. This is a situation where a tradeoff must be made — some nonresponse bias is still preferable to dishonest answers. The nonresponse bias can be reduced by the use of an appropriate inducement or reward to those who can show proof of having mailed back the questionnaire (for example by showing a post office receipt, etc.).

74. (a) Substituting the given capture sizes in this situation yields the equation  105   60   45   30  1 −  = 1 −  1 −  1 −  N   N  N  N  3 which can be multiplied by N on both sides to yield N 2 ( N − 105 ) = ( N − 60 )( N − 45 )( N − 30 ) which implies N 3 − 105 N 2 = N 3 − 135 N 2 + 5850 N − 81000 or 30 N 2 − 5850 N + 81000 = 0 which simplifies to N 2 − 195 + 2700 = 0 . This factors to give ( N − 15 )( N − 180 ) = 0 having solutions N=15 or N=180. Since N ≥ M , N=15 cannot be a solution. Thus, N=180. (b) With just two captures, we solve Darroch’s equation  M   n1   n2   1 −  =  1 −  1 −  for N : N N N N 2 − MN = ( N − n1 )( N − n2 ) N 2 − MN = N 2 − n1 N − n2 N + n1n2

( n1 + n2 − M ) N = n1n2

RUNNING 73. 625; we solve the two equations pN = 250 and p (1 − p ) N = 150 for p and N. But

substituting N = 250 / p into the second equation gives (1 − p ) 250 = 150 which yields p = 0.4 . Thus, N = 250/0.4 = 625.

n1n2 ( n1 + n2 − M )

n1n2 k

Chapter 15 WALKING 15.1 Graphs and Charts 1. (a)

Score

100

Frequency

(b) For each score, plot a bar with height represented by the frequency of that score in the data set.

3. (a)

Grade

Frequency

(b) A relative frequency bar graph shows the percentage of data in each category.

4. (a) 3 / 25 = 12% (b) 0.12 × 360 = 43

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Chapter 15: Graphs, Charts, and Numbers (c) A: 0.28 × 360 = 100.8 , B: 0.24 × 360 = 86.4 , C: 0.28 × 360 = 100.8 , D: 0.12 × 360 = 43 , E: 0.08 × 360 = 28.8

5. (a)

Distance

0.0

0.5

1.0

1.5

2.0

2.5

3.0

5.0

6.0

8.5

Frequency

(b)

7. Class Interval

Very close

Nearby

Not too far

Far

Frequency

ISM: Excursions in Modern Mathematics, 10E 8. Class Interval

Zone A

Zone B

Zone C

Zone D

Frequency

Slice “Very close”:

8 × 360° ≈ 106.7°; 27

10 × 360° ≈ 133.3°; 27 5 Slice “Nearby”: × 360° ≈ 66.7°; 27 1 Slice “Not too far”: × 360° ≈ 13.3°; 27 3 Slice “Far”: × 360° = 40°. 27

Slice “Close”:

10.

18 × 360° = 240°; 27 5 Slice “Zone B”: × 360° ≈ 66.7°; 27 1 Slice “Zone C”: × 360° ≈ 13.3°; 27 3 Slice “Zone D”: × 360° = 40°; 27

Slice “Zone A”:

11. (a) N = 4 + 6 + 7 + 5 + 6 + 8 + 2 + 2 = 40 (b) 0% (c)

5+6+8+ 2+ 2 = 0.575 ≈ 58% (rounded to the nearest percent) 40

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12.

13. (a) qualitative; it is a variable that describes characteristics that cannot be measured numerically. (b) 0.23 × 2,839, 205 = 653, 017.15 ≈ 653, 000 died due to heart disease 14. Heart Disease: 0.23 × 2,839, 205 ≈ 653, 017 ; Cancer: 0.21× 2,839, 205 ≈ 596, 233 ; Unintentional Injuries: 0.07 × 2,839, 205 ≈ 198, 744 ; Respitory Diseases: 0.05 × 2,839, 205 ≈ 141,960 ; Stroke: 0.05 × 2,839, 205 ≈ 141,960 ; Alzheimer Disease: 0.04 × 2,839, 205 ≈ 113,568 Other: 0.35 × 2,839, 205 ≈ 993, 722 15. Class Interval

Relative Frequency

16. Class Interval

Relative Frequency Central Angle

200-290

300-390

400-490

500-590

600-690

700-800

0.4%

9.4%

29.3%

32.2%

22.1%

6.6%

400-590

600-790

800-990

1000-1190 1200-1390 1400-1600

0.2%

10.7%

29.5%

33.2%

19.4%

7.0%

1°

38°

106°

119°

70°

25°

Note that the sum of these angles is not 360 degrees due to rounding. Here, for example, the sum of the angles can be made to be 360 degrees by labeling the 38 degree angle (that which is closest to the next whole degree) as 39 degrees in the pie chart. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E 17. (a)

(b)

18. (a)

(b)

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19. (a) 60 – 48 = 12 ounces (b) The third class interval: “more than 72 ounces and less than or equal to 84 ounces.” Values that fall exactly on the boundary between two class intervals belong to the class interval to the left. (c)

Frequency

119

184

142

Percent

2.4

3.8

6.6

10.7

19.0

29.4

22.7

4.2

0.8

0.3

20. (a) The following is a frequency table for birth weights in Cleansburg using class intervals of length equal to 24 ounces. Weight in Ounces

More than

Less than or Equal

Frequencies

108

120

303

120

144

168

144

168

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Chapter 15: Graphs, Charts, and Numbers (b)

21. (a) 8 + 11 + 2 + 1 = 22 (use histogram on the left) (b) 4 + 7 = 11 (use histogram on the right) (c) By the histogram on the left, two teams had a payroll between $200 million and $250 million. By the histogram on the right, there were two teams having a payroll between $200 and $230 million. This means 2 – 2 = 0 teams had a payroll between $230 and $250 million. 22. (a) 8 + 8 = 16 (use histogram on the left) (b) 9 + 4 + 2 + 1 = 16 (use histogram on the right) (c) By the histogram on the left, 8 + 11 + 2 + 1 = 22 teams had a payroll that was more than $100 million. By the histogram on the right, there were 3 + 9 + 4 + 2 + 1 = 19 teams with a payroll more than $110 million. This means 22 – 19 = 3 teams had a payroll between $100 and $110 million.

15.2 Means, Medians, and Percentiles 23. (a) Average A =

3−5+ 7 + 4+8+ 2+8−3−6 =2 9

(b) The ordered data set is {-6, -5, -3, 2, 3, 4, 7, 8, 8}. The locator of the 50th percentile is L = (0.50)(9) = 4.5. Since L is not a whole number, the 50th percentile is located in the 5th position in the list. Hence, the median M is 3.

(9 × 2) + 2 = 2 The new ordered data 10 . set is {-6, -5, -3, 2, 2, 3, 4, 7, 8, 8}. The locator of the 50th percentile is L = (0.50)(10) = 5. Since L is a whole number, the 50th percentile is the average of the 5th and 6th numbers (2 and 3) in the ordered list. So, the median of this new data set is 2.5.

24. (a) Average A =

−4 + 6 + 8 − 5.2 + 10.4 + 10 + 12.6 − 13 = 3.1 8

(b) The ordered data set is {-13, -5.2, -4, 6, 8, 10, 10.4, 12.6}. The locator of the 50th percentile is L = (0.50)(8) = 4. Since L is a whole number, the 50th percentile is the average of the 4th and 5th numbers (6 and 8) in the ordered list. Hence, the median M is 7. (8 × 3.1) − ( −13) = 5.4 The new 7 . ordered data set is {-5.2, -4, 6, 8, 10, 10.4, 12.6}. The locator of the 50th percentile is L = (0.50)(7) = 3.5. Since L is not a whole number, the 50th percentile is located in the 4th position in the list. Hence, the median of this new data set is 8.

ISM: Excursions in Modern Mathematics, 10E 25. (a) Average A =

239

0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 45 = = 4.5 10 10

Since the data set is already ordered, the locator of the median is L = (0.50)(10) = 5. Since L is a whole number, the median is the average of the 5th and 6th numbers (4 and 5) in the data set. The median M is 4.5. (b) Average A =

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 45 = =5 9 9

The locator of the median is L = (0.50)(9) = 4.5. Since L is not a whole number, the median is the 5th number in the data set. Hence the median M is 5. (c) Average A =

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 55 = = 5.5 10 10

The locator of the median is L = (0.50)(10) = 5. Since L is a whole number, the median is the average of the 5th and 6th numbers (5 and 6) in the data set. The median M is 5.5. 26. (a) Average A = 1.5; Median M = 1.5 Note that the ordered data set {1, 1, 1, 1, 1, 2, 2, 2, 2, 2} is ‘balanced’ making for a simple computation. (b) Average A = 2.5; Median M = 2.5 Note that the ordered data set {1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4} is ‘balanced’ making for a simple computation. (c) Average A = 3; Median M = 3 Note that the ordered data set {1, 1, 2, 2, 3, 3, 4, 4, 5, 5} is ‘balanced’ making for a simple computation. 5 + 10 + 15 + 20 + 25 + 60 135 = = 22.5 . The locator for the median is L = (0.50)(6) = 3. 6 6 Hence, the median is the average of the values in the 3rd and 4th positions. That is, the median M is 17.5.

27. (a) Average A =

105 + 110 + 115 + 120 + 125 + 160 600 + 135 = = 100 + 22.5 = 122.5 . The locator for the 6 6 median is L = (0.50)(6) = 3. Once again the median is the average of the values in the 3rd and 4th positions. That is, the median M is 117.5.

(b) Average A =

5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 275 = = 27.5 . The locator for the median is L 10 10 = (0.50)(10) = 5. Hence, the median is the average of the values in the 5th and 6th positions. That is, the median M is 27.5.

28. (a) Average A =

55 + 60 + 65 + 70 + 75 + 80 + 85 + 90 + 95 + 100 500 + 275 = = 50 + 27.5 = 77.5 . The locator 10 10 for the median is L = (0.50)(10) = 5. Hence, the median is the average of the values in the 5th and 6th positions. That is, the median M is 77.5.

(b) Average A =

29. (a) Average A =

24 × 0 + 16 ×1 + 20 × 2 + 12 × 3 + 5 × 4 + 3 × 5 127 = = 1.5875 24 + 16 + 20 + 12 + 5 + 3 80

(b) The locator for the median is L = (0.50)(80) = 40. So, the median is the average of the 40th and 41st scores. The 40th score is 1, and the 41st score is 2. Thus, the median M is 1.5.

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Chapter 15: Graphs, Charts, and Numbers

30. (a) Average A =

25 × 2 + 27 × 7 + 28 × 6 + 29 × 9 + 30 ×15 + 31×12 + 32 × 9 + 33 × 9 + 37 × 6 + 39 × 4 = 31.05 2 + 7 + 6 + 9 + 15 + 12 + 9 + 9 + 6 + 4

(b) The locator for the median is L = (0.50)(79) = 39.5. So, the median M is the 40th age which is 31. 31. (a) Since the number of quiz scores N is not given, we can compute the average as (0.07 N ) × 4 + (0.11N ) × 5 + (0.19 N ) × 6 + (0.24 N ) × 7 + (0.39 N ) × 8 A= N = 0.07 × 4 + 0.11 × 5 + 0.19 × 6 + 0.24 × 7 + 0.39 × 8 = 6.77

(As the calculation shows, the number of scores is not important. If one likes, they can assume that there were 100 scores.) (b) Since 50% of the scores were at 7 or below, the median M is 7. 32. (a) Average A = 0.08 × 3 + 0.12 × 4 + 0.16 × 5 + 0.20 × 6 + 0.18 × 7 + 0.14 × 8 + 0.12 × 9 = 6.18 (b) M = 6; at least 50% of the scores are at or below 6 (and at least 50% are at or above 6). 33. The ordered data set is {-6, -5, -3, 2, 4, 7, 8, 8}. (a) The locator of the 25th percentile is L = (0.25)(8) = 2. Since this is a whole number, the 25th percentile is the average of the 2nd and 3rd values in the list (namely -5 and -3). That is, the 25th percentile is -4 ( Q1 = −4 ). (b) The locator of the 75th percentile is L = (0.75)(8) = 6. Since this is a whole number, the 75th percentile is the average of the 6th and 7th values in the list (namely 7 and 8). That is, the 75th percentile is 7.5 ( Q3 = 7.5 ). (c) The new ordered data set is {-6, -5, -3, 2, 2, 4, 7, 8, 8}. The locator of the 25th percentile is L = (0.25)(9) = 2.25. Rounding up, we find that the first quartile is the 3rd value in the ordered list. That is, Q1 = −3 . The locator of the 75th percentile is L = (0.75)(9) = 6.75. Rounding up, we find that the third quartile is the 7th value in the ordered list. That is, Q3 = 7 . 34. The ordered data set is {-13, -5.2, -4, 6, 8, 10, 10.4, 12.6}. (a) The locator of the 25th percentile is L = (0.25)(8) = 2. Since this is a whole number, the 25th percentile is the average of the 2nd and 3rd values in the list. That is, the 25th percentile is -4.6 ( Q1 = −4.6 ). (b) The locator of the 75th percentile is L = (0.75)(8) = 6. Since this is a whole number, the 75th percentile is the average of the 6th and 7th values in the list (namely 10 and 10.4). That is, the 75th percentile is 10.2 ( Q3 = 10.2 ). (c) The new ordered data set is {-5.2, -4, 6, 8, 10, 10.4, 12.6}. The locator of the 25th percentile is L = (0.25)(7) = 1.75. Rounding up, we find that the first quartile is the 2nd value in the ordered list. That is, Q1 = −4 . The locator of the 75th percentile is L = (0.75)(7) = 5.25. Rounding up, we find that the third quartile is the 6th value in the ordered list. That is, Q3 = 10.4 . 35. (a) Since the data set is already ordered, the locator of the 75th percentile is given by L = (0.75)(100) = 75. Since this is a whole number, the 75th percentile is the average of the 75th and 76th values in the list. That is, the 75th percentile is 75.5. The locator of the 90th percentile is given by L = (0.90)(100) = 90. Since this is a whole number, the 90th percentile is the average of the 90th and 91st values in the list. That is, the 90th percentile is 90.5.

ISM: Excursions in Modern Mathematics, 10E

241

(b) The locator of the 75th percentile is given by L = (0.75)(101) = 75.75. Since this not is a whole number, the 75th percentile is the 76th value in the list. That is, the 75th percentile is 75.

The locator of the 90th percentile is given by L = (0.90)(101) = 90.9. Since this is not a whole number, the 90th percentile is the 91st value in the list. That is, the 90th percentile is 90. (c) The locator of the 75th percentile is given by L = (0.75)(99) = 74.25. Since this not is a whole number, the 75th percentile is the 75th value in the list. That is, the 75th percentile is 75.

The locator of the 90th percentile is given by L = (0.90)(99) = 89.1. Since this is not a whole number, the 90th percentile is the 90th value in the list. That is, the 90th percentile is 90. (d) The locator of the 75th percentile is given by L = (0.75)(98) = 73.5. Since this not is a whole number, the 75th percentile is the 74th value in the list. That is, the 75th percentile is 74.

The locator of the 90th percentile is given by L = (0.90)(98) = 88.2. Since this is not a whole number, the 90th percentile is the 89th value in the list. That is, the 90th percentile is 89. 36. (a) The locator of the 10th percentile is given by L = (0.10)(100) = 10. Since this is a whole number, the 10th percentile is the average of the 10th and 11th numbers in the sorted list. That is, the 10th percentile is the average of 5 and 6 or 5.5.

The locator of the 25th percentile is given by L = (0.25)(100) = 25. Since this is a whole number, the 25th percentile is the average of the 25th and 26th numbers in the sorted list. That is, the 25th percentile is the average of 13 and 13 (which is, of course, 13). (b) 10th percentile : 5; 25th percentile : 13 (c) 10th percentile : 5; 25th percentile : 13 37. (a) The Cleansburg Fire Department consists of 2 + 7 + 6 + 9 + 15 + 12 + 9 + 9 + 6 + 4 = 79 firemen. The locator of the first quartile is thus given by L = (0.25)(79) = 19.75. So, the first quartile is the 20th number in the ordered data set. That is, Q1 = 29 . (b) The locator of the third quartile is given by L = (0.75)(79) = 59.25. So, the third quartile is the 60th number in the ordered data set. That is, Q3 = 32 . (c) The locator of the 90th percentile is given by L = (0.90)(79) = 71.1. So, the 90th percentile is the 72nd number in the ordered data set or 37. 38. (a) Q1 = 5 (b) Q3 = 8 (c) 7 39. (a) The 1,068,270th score, d1,068,270 . [The locator is given by L = 0.50(2,136,539) = 1,068,269.5.] (b) The 534,135th score, d534,135 . [The locator is given by L = 0.25(2,136,539) = 534,134.75.] (c) The 1,709,232nd score, d1,709,232 . [The locator is given by L = 0.80(2,136,539) = 1,709,231.2.] 40. (a) Since the locator is given by L = 0.75(2,136,539) = 1,602,404.25, the third quartile is located at the 1,602,405th score. That is, Q3 = d1,602,405 .

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Chapter 15: Graphs, Charts, and Numbers (b) Since the locator is given by L = 0.60(2,136,539) = 1,281,923.4, the 60th percentile is located at the 1,281,924th score. That is, X 60 = d1,281,924 .

41. (a) Min = –6, Q1 = −4, M = 3, Q3 = 7.5, Max = 8 (b)

42. (a) Min = –13, Q1 = −4.6, M = 7, Q3 = 10.2, Max = 12.6 (b)

43. (a) Min = 25, Q1 = 29, M = 31, Q3 = 32, Max = 39 (b)

44. (a) Min = 3, Q1 = 5, M = 6, Q3 = 8, Max = 9 (b)

45. (a) $43,000 (corresponding the vertical line in the middle of the lower box) (b) $50,000 (c) The vertical line indicating the median salary in the engineering box plot is aligned with the right end of the box in the agriculture box plot. 46. (a) 459; $45,000 is the first quartile of engineering salaries, so 0.75 × 612 = 459 engineering majors made $45,000 or more. (b) 960; $35,000 is the first quartile of the N agriculture salaries, so 0.25 × N = 240 agriculture majors made $35,000 or less. Thus, N = 240 × 4 = 960 .

15.3 Ranges and Standard Deviations 47. (a) 8 – (–6) = 14 (b) From Exercise 33, Q1 = −4, Q3 = 7.5. IQR = 7.5 – (–4) = 11.5. 48. (a) 12.6 – (-13) = 25.6 (b) From Exercise 34, IQR = 10.2 – (-4.6) = 14.8 49. (a) $156,000 - $115,000 = $41,000 (b) At least 171 homes Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E 50. (a) Approximately $63,000 - $33,000 = $30,000.

(b) Approximately $53,000 - $45,000 = $8,000. 51. (a) Note that 1.5 × IQR = 1.5 × 3 = 4.5 . Any number bigger than Q3 + 1.5 × IQR = 12 + 4.5 = 16.5 is an outlier. (b) Any number smaller than Q1 − 1.5 × IQR = 9 – 4.5 = 4.5 is also an outlier. (c) Since 1 is the only value smaller than 4.5 and 24 is the only value larger than 16.5, the values 1 and 24 are the only outliers in this data set.

(b) A =

x – 10

( x − 10) 2

–10

100

Standard deviation = (d) A =

53. The IQR for 18-year old U.S. males is 71 – 67 = 4 inches. Since 1.5 × IQR = 1.5 × 4 = 6 , an outlier is any height more than Q3 + 1.5 × IQR = 71 + 6 = 77 inches or any height less than 67 – 6 = 61 inches.

55. (a) A = 5, so x – A = 0 for every number x in the data set. The standard deviation is 0.

0 + 10 + 10 + 20 = 10 4

200

52. There are 10 outliers: 6 ages of 37 and 4 ages of 39. Note that 1.5 × IQR = 1.5 × 3 = 4.5 . Hence, outliers are ages above 32 + 4.5 = 36.5 or below 29 – 4.5 = 24.5.

54. The IQR for 18-year old U.S. females is 66 – 62.5 = 3.5 inches. Since 1.5 × IQR = 1.5 × 3.5 = 5.25 , an outlier is any height more than Q3 + 1.5 × IQR = 66 + 5.25 = 71.25 inches or any height less than 62.5 – 5.25 = 57.25 inches.

200 = 5 2 ≈ 7.1 4

1+ 2 + 3 + 4 + 5 =3 5 x

x–3

( x − 3)2

–2

–1

4 10

Standard deviation =

10 = 2 ≈ 1.4 5

56. (a) A = 3, so x – A = 0 for every number x in the data set. The standard deviation is 0. (b) A =

0 + 5 + 5 + 10 =5 4

0+6+6+8 =5 4

x–5

( x − 5) 2

–5

x–5

( x − 5) 2

–5

25 50

Standard deviation =

243

36 Standard deviation =

36 =3 4

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Chapter 15: Graphs, Charts, and Numbers

(c)

−6 + 0 + 0 + 18 =3 4

at d N +1 . Since this is one of the test 2

scores, it must end in a 0.

x–3

( x − 3) 2

-6

–9

-3

225 324

Standard deviation =

(d)

324 =9 4

6 + 7 + 8 + 9 + 10 =8 5

(b) When N is even, the median is the average of d N / 2 and d ( N / 2 +1) , two

consecutive test scores that end in 0. The only possible way that this average could end in a 5 is if the two scores are different. For this to happen, d N / 2 would have the be the last one of the scores on the list with that numerical value and d ( N / 2 +1) the first score on the list with the next numerical value. With a very large list of scores with thousands of repetitions of each score, this is very unlikely. 61. Since the median is the average of d840,567 and

x–8

( x − 8)

–2

–1

10 Standard deviation =

10 = 2 ≈ 1.4 5

JOGGING

d840,568 , the locator of the median is L = 840,567. Since (0.5)(N) = L = 840,567, we find N = 1,681,134. 62. Since the second quartile is d854,741 , the

locator L is strictly between 857,740 and 857,741. Since L=(0.5)N, we have N = 2L is strictly between 2 × 857, 740 = 1, 715, 480 and 2 × 857, 741 = 1, 715, 482 meaning N = 1,715,481. 63. (a) {1, 1, 1, 1, 6, 6, 6, 6, 6, 6} Average = 4; Median = 6. (b) {1, 1, 1, 1, 1, 1, 6, 6, 6, 6} Average = 3; Median =1.

57. (a) 30

(b) 8 58. (a) Caucasian (has largest percent or frequency) (b) There is no mode. 59. Let x = score Mike needs on the next exam. 5 ⋅ 88 + x = 90 6 5 ⋅ 88 + x = 540 x = 100

60. (a) When N is odd, the locator for the median is L = 0.5(N) and since this is not a whole number, the median M is located

(d) {1, 1, 1, 1, 1, 1, 1, 1, 6, 6} Average = 2; Q3 = 1. 64. (a) The ten numbers add up to 75. Since Min = 3, the remaining nine largest numbers add up to 72. The smallest possible value of Max is Max = 8. (It occurs when the ten numbers are 3, 8, 8, 8, 8, 8, 8, 8, 8, 8.) (b) The largest possible value of Max is Max = 48. (It occurs when the ten numbers are 3, 3, 3, 3, 3, 3, 3, 3, 3, 48.)

ISM: Excursions in Modern Mathematics, 10E

minimum value of { x1 + c, x2 + c, x3 + c, , xN + c} is m + c.

65. Since N = 30, the median is the average of the 15th and 16th sorted data points. From the histogram on the left one can deduce that the median team salary is less than $150 million. From the histogram on the right, one can deduce that the median team salary is more than $140 million.

So, the range of { x1 + c, x2 + c, x3 + c, , xN + c} is M + c – (m + c) = M – m. That is, the same as the range of { x1 , x2 , x3 , , xN } .

66. (a) The five-number summary for the original scores was Min = 1, Q1 = 9, M = 11, Q3 = 12, and Max = 24. When 2 points are added to each test score, the five-number summary will also have 2 points added to each of its numbers (i.e., Min = 3, Q1 = 11, M = 13, Q3 = 14, and Max = 26). (b) When 10% is added to each score (i.e., each score is multiplied by 1.1) then each number in the five-number summary will also be multiplied by 1.1 (i.e., Min = 1.1, Q1 = 9.9, M = 12.1, Q3 = 13.2, and Max = 26.4). 67. (a)

(b) Adding c to each value of a data set does not change how spread out the data set is. Since the standard deviation is a measure of spread, the standard deviation does not change. Specifically, the average of the data set will increase by c, but the deviations from the mean (and hence the standard deviation) will remain the same. 69. (a) 4 Column area over interval 30 – 35 5 × h 50% = = Column area over interval 20 − 30 10 × 1 25% and so 5h = 20, h = 4. (b) 0.4 Column area over interval 35 − 45 10 × h 10% = = Column area over interval 20 − 30 10 × 1 25% and so h = 0.4.

( x1 + c) + ( x2 + c) + ( x3 + c) +  + ( xN + c) N ( x1 + x2 + x3 +  + xN ) + cN = N ( x1 + x2 + x3 +  + xN ) cN = + = A+c N N

(b) The relative sizes of the numbers are not changed by adding a constant c to every number. Thus (assuming the original data set is sorted) if the median of the data set is M = xk , then the median of the new data set will be xk + c = M + c . If the median of the original set is x + xk +1 , then the median of the M = k 2 new data set will be ( xk + c) + ( xk +1 + c) x + xk +1 2c = k + 2 2 2 = M + c . In either case, we see that the median of the new data set is M + c. 68. (a) Let M be the maximum value of { x1 , x2 , x3 , , xN } . The maximum value

of { x1 + c, x2 + c, x3 + c, , xN + c} is M + c. Similarly, if the minimum value of { x1 , x2 , x3 , , xN } is m, then the

245

(c) 0.4 Column area over interval 45 − 60 15 × h 15% = = Column area over interval 20 − 30 10 × 1 25% and so h = 0.4. 70.

Column 1 (Fast): 5% of the citizens. Column 2 (Fit): 45% of the citizens. Column 3 (Average): 40% of the citizens. Column 4 (Slow): 10% of the citizens. Let the height of the first column be 1. Col. 2 area 6 × h 45% = = so the height of Col. 1 area 4 × 1 5% the second column is h = 6. Col. 3 area 8 × h 40% = = so the height of Col. 1 area 4 × 1 5% the third column is h = 4.

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Col. 4 area 16 × h 10% = = so the height of Col. 1 area 4 ×1 5% the fourth column is h = 0.5.

RUNNING 71. If the standard deviation is 0, then the average of the squared deviations is 0. The squared deviations cannot be negative, so if their average is 0, they must all be 0. It follows that all deviations from the average must be 0. This means that each data value is equal to the average a, i.e., the data set is constant. 72. Consider a data set {x1 , x2 , x3 , , xN } and assume Min = x1 ≤ x2 ≤ x3 ≤  ≤ xN = Max . Then since the average A satisfies the inequality Min ≤ A ≤ Max , we see that (xk − A) 2 ≤ (Max − Min) 2 = Range 2 for each k = 1, 2, 3,…, N. Therefore,

σ=

74. If A is the average of x1 , x2 , , xN , then the average of ax1 , ax2 , , axN is aA. Thus, (ax1 − aA) 2 +  + (axN − aA) 2 N =

a 2 ( x1 − A) 2 +  + a 2 ( xN − A) 2 N

( x1 − A) 2 +  + ( xN − A) 2 N = aσ . =a

( x1 − A) 2 + ( x2 − A) 2 +  + ( xN − A) 2 N

N × Range2 N = Range ≤

N ( N + 1) implies the 2 1 + 2 + ... + N N +1 = . If N is average A = N 2 odd, the median M is the “middle” number N +1 M = . If N is even the median M is the 2 N N and average of + 1 , which is 2 2 N +1 M = . 2

73. 1 + 2 + ... + N =

75. (a) 8/9 or approximately 89% (b) 10 (c) Since k > 0, 1 −

case.

1 < 1 will always be the k2

Chapter 16 WALKING 16.1 Sample Spaces and Events 1. (a) {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Note that there are 23 = 8 outcomes consisting of all possible sequences of H’s and T’s of length three. (b) {sss, ssf, sfs, fss, ffs, fsf, sff, fff} Note the correspondence between the outcomes in (a) and (b). (c) {0, 1, 2, 3} (d) {0, 1, 2, 3} 2. (a) {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} (b) {FFFF, FFFT, FFTF, FFTT, FTFF, FTFT, FTTF, FTTT, TFFF, TFFT, TFTF, TFTT, TTFF, TTFT, TTTF, TTTT} (c) {0%, 25%, 50%, 75%, 100%}; the coin can land heads 0, 1, 2, 3, or 4 times out of 4. (d) {0%, 25%, 50%, 75%, 100%}; the number of correct answers on the test could be 0, 1, 2, 3, or 4 out of 4. 3. (a) S = {3, 4, 5, …, 16, 17, 18} Each number rolled is an integer between 1 and 6. The minimum total that can be rolled is 1 + 1 + 1 = 3. The maximum total that can be rolled is 6 + 6 + 6 = 18. Every integer between 3 and 18 is possible. (b) S = {5, 3, 1, -1, -3, -5} If 0 heads are tossed (so that 5 tails are tossed), then 0 – 5 = -5 is the observation. If 1 head is tossed (so that 4 tails are also tossed), the observation would be 1 – 4 = -3. If 2 heads are tossed, the observation would be 2 – 3 = -1. The following table summarizes how S is generated. Heads Tossed

Tails Tossed

Observation

-5

-3

-1

4. (a) S = {ABC, ACB, BAC, BCA, CAB, CBA} This is the list of all the possible orders in which we could observes the runner cross the finish line. Notice how these have been organized. We list the two outcomes in which A finishes first, then the two outcomes in which B finishes first, and finally the two outcomes in which C finishes first.

248

Chapter 16: Probabilities, Odds, and Expectations (b) S = {A and B, A and C, A and D, B and C, B and D, C and D} In this case, we do not care about the order in which the top two runners actually finish. We are only interested in selecting the top two runners. This is the list of all possible combinations of two runners from this group {A, B, C, D} of four runners.

5. (a) S = {AB, AC, AD, AE, BA, BC, BD, BE, CA, CB, CD, CE, DA, DB, DC, DE, EA, EB, EC, ED} Here, XY denotes the outcome that X was elected chairman and Y was elected treasurer. We have listed all possible permutations (here the order matters) of two members of the five-member board. Order clearly matters here as the outcome BD where B is elected chairman and D is elected treasurer is different from the outcome DB where D is elected chairman and B is elected treasurer. Note the manner in which we list the outcomes - it is organized alphabetically to ensure that all outcomes are indeed listed. (b) S = {ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE} We have listed all possible combinations (order does not matter here) of three for this five-member board. Here, for example, ACE denotes a search committee made up of A, C, and E (all committee members have equal roles so no order is needed in listing them). Note the manner in which we list the outcomes – it is once again organized alphabetically to ensure that all outcomes are listed. 6. S = {JJ, JK, JL, JM, KK, KL, KM, LL, LM, MM} Note that there are N = 10 outcomes in this sample space. They consist of the “doubles” (JJ, KK, LL, and MM) and the number of unordered selections of two objects from four (i.e., 4 C2 = 6 ). Another way to think of this is as unordered selections with repetition allowed. Note that there are four outcomes that start with J (JJ, JK, JL, JM), three outcomes that start with K (all but KJ which is considered the same outcome as JK), two outcomes that start with LL (all but LJ and LK), and one outcome that starts with MM (MJ, MK, ML have already been listed since order is not important here). 7. Answers will vary. A typical outcome is a string of 10 letters each of which can be either an H or a T. An answer like {HHHHHHHHHH, …,TTTTTTTTTT} is not sufficiently descriptive. An answer like{… HTTHHHTHTH, …, TTHTHHTTHT, …, HHHTHTTHHT, …} is better. An answer like { X 1 X 2 X 3 X 4 X 5 X 6 X 7 X 8 X 9 X 10 : each X i is either H or T} is best. Note: This sample space consists of N = 210 = 1024 outcomes (2 possible outcomes at each of 10 stages of the experiment).

8. A typical outcome is a string of 10 letters each of which can be either an T or a F. An answer like {FFFFFFFFFF, …,TTTTTTTTTT} is not sufficiently descriptive. An answer like{… FTTFFFTFTF, …, TTFTFFTTFT, …, FFFTFTTFFT, …} is better. An answer like { X 1 X 2 X 3 X 4 X 5 X 6 X 7 X 8 X 9 X 10 : each X i is either T or F for positive integers 1 ≤ i ≤ 10 } is best. 9. Answers will vary. An outcome is an ordered sequence of four numbers, each of which is an integer between 1 and 6 inclusive. The best answer would be something like {(n1 , n2 , n3 , n4 ) : each ni is 1, 2, 3, 4, 5, or 6}. An answer such as {(1,1,1,1), …, (1,1,1,6), …,(1,2,3,4),…, (3,2,6,2), …, (6,6,6,6)} showing a few typical outcomes is possible, but not as good. An answer like {(1,1,1,1),…, (2,2,2,2), …, (6,6,6,6)} is not descriptive enough. Note: This sample space consists of N = 64 = 1296 outcomes (6 possible outcomes at each of 4 stages of the experiment). 10. An outcome consists of ten letters. The letters can be A, B, C, D or E only. Because the order in which the letters are listed is relevant, the best answer would then be of the form { X 1 X 2 X 3 X 4 X 5 X 6 X 7 X 8 X 9 X 10 : each X i is one of the letters A, B, C, D or E}. Other acceptable answers involve listing a few typical outcomes separated by …’s. 11. (a) E1 = {HHT , HTH , THH } Remember that the event is all outcomes for which we observe two of the coins coming up heads. (b) E2 = {HHH , TTT } Copyright © 2022 Pearson Education, Inc.

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(c) E3 = {} There is no way to have half of three tosses come up heads. (d) E4 = {TTH , TTT } 12. (a) E1 = {TTFF , TFTF , TFFT , FTTF , FTFT , FFTT } Note the organized manner in which the outcomes are listed – if T came before F in a two-letter alphabet, these outcomes would be in alphabetical order. (b) E2 = {TTFF , TFTF , TFFT , FTTF , FTFT , FFTT , TTTF , TTFT , TFTT , FTTT , TTTT } Again, we list “alphabetically.” We also list all (ordered) outcomes with two T’s first, then the outcomes with three T’s next, and finally the sole outcome with four T’s. (c) E3 = {FFTT , FTFT , FTTF , TFFT , TFTF , TTFF , FFFT , FFTF , FTFF , TFFF , FFFF } This time we list all (ordered) outcomes with two T’s first, the outcomes with one T next, and finally the sole outcome with no T last. (d) E4 = {TTFF , TTFT , TTTF , TTTT } 13. (a) E1 = {(1,1), (2, 2), (3,3), (4, 4), (5,5), (6, 6)} (b) E2 = {(1,1), (1, 2), (2,1), (6, 6)} Note that outcome (1,2) and outcome (2,1) are different observations here. Both of these make up the event “a total of 3 is rolled.” (c) E3 = {(1, 6), (2,5), (3, 4), (4,3), (5, 2), (6,1), (5, 6), (6,5)} Again, outcomes (2,5) and (5,2) are different observations corresponding to a “2” being rolled on one die or the other. 14. (a) E1 = {QH , QD, QS , QC} (b) E2 = { AH , 2 H ,3H , 4 H ,5H , 6 H , 7 H ,8H ,9 H ,10 H , JH , QH , KH } (c) E3 = {JH , JD, JS , JC , QH , QD, QS , QC , KH , KD, KS , KC} 15. (a) E1 = {HHHHHHHHHH } (b) E2 = {HHHHHHHHHT , HHHHHHHHTH , HHHHHHHTHH , HHHHHHTHHH , HHHHHTHHHH , HHHHTHHHHH , HHHTHHHHHH , HHTHHHHHHH , HTHHHHHHHH , THHHHHHHHH } Note that these outcomes are written in alphabetical order (an organized list).

(c) E3 = {TTTTTTTTTH , TTTTTTTTHT , TTTTTTTHTT , TTTTTTHTTT , TTTTTHTTTT , TTTTHTTTTT , TTTHTTTTTT , TTHTTTTTTT , THTTTTTTTT , HTTTTTTTTT , TTTTTTTTTT } Note that these outcomes are also written in alphabetical order. There are really two types of outcomes here – the 10 with exactly one H and the one with no H.

16. (a) E1 = { AB, AC , AD, AE , BC , BD, BE , CD, CE , DE} (b) E2 = { ABC , ABD, ABE , ACD, ACE , ADE , BCD, BCE , BDE , CDE} (c) E3 = {BCD, BCE , BDE , CDE}

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Chapter 16: Probabilities, Odds, and Expectations

16.2 The Multiplication Rule, Permutations, and Combinations 17. (a) 9 × 263 × 103 = 158,184, 000 Stage 1: There are 9 choices for the leading digit. Stages 2-4: There are 26 choices for the letter in slot 2, 26 choices for the letter in slot 3, and 26 choices for the letter in slot 3. Stages 5-7: There are 10 choices for the digit in slot 5, 10 choices for the digit in slot 6, and 10 choices for the digit in slot 7. (b) 1× 263 × 102 × 1 = 1, 757, 600 Stage 1: There is one choice for the leading digit (it must be a 5). Stages 2-4: There are 26 choices for the letter in slot 2, 26 choices for the letter in slot 3, and 26 choices for the letter in slot 3. Stages 5-6: There are 10 choices for the digit in slot 5 and 10 choices for the digit in slot 6. Stage 7: There is one choice for the last digit (it must be a 9). (c) 9 × 26 × 25 × 24 × 9 × 8 × 7 = 70, 761, 600 Stage 1: There are 9 choices for the leading digit. Stages 2-4: There are 26 choices for the letter in slot 2, 25 choices for the letter in slot 3 (it must be different from the letter in slot 2), and 24 choices for the letter in slot 3 (it must be different than the letters in slots 2 and 3). Stages 5-7: There are 9 choices for the digit in slot 5 (it must be different from the digit in slot 1), 8 choices for the digit in slot 6 (it must be different from the digits in slots 1 and 5), and 7 choices for the digit in slot 7 (it must be different from the digits in slots 1, 5, and 6). 18. (a) 264 × 10 = 4,569, 760 (b) 264 = 456,976 (c) 25 × 263 × 10 = 4,394, 000 (d) 254 × 10 = 3,906, 250 19. (a)

( 2 + 1) × ( 3 + 4 ) × 3 = 63 Stage 1 (Footwear): Jack chooses from two pairs of shoes and a pair of boots (2 + 1 = 3 choices). Stage 2 (Lower wear): Jack chooses from three pairs of jeans and four pairs of dress pants (3 + 4 = 7 choices). Stage 3 (Upper wear): Jack chooses a dress shirt (3 options).

(b)

( 2 + 1) × ( 3 + 4 ) × ( 3 + 3 × 2 ) = 189 Stage 1 (Footwear): Jack has 2 + 1 = 3 options Stage 2 (Lower wear): Jack has 3 + 4 = 7 options. Stage 3 (Upper wear): Jack chooses a dress shirt (3 options) or both a dress shirt and a jacket ( 3 × 2 = 6 options). This totals 3 + 3 × 2 = 9 options.

20. (a) 3 × 4 × 9 = 108 Stage 1 (appetizer): 3 choices. Stage 2 (salad): 4 choices. Stage 3 (main course): 9 choices.

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(b) 3 × ( 2 + 4 ) × 9 × 5 = 810 Stage 1 (appetizer): 3 choices. Stage 2 (soup or salad): 2 + 4 = 6 choices. Stage 3 (main course): 9 choices. Stage 4 (dessert): 5 choices. (c) (2+4+2 × 4) × 9 = 126 Stage 1 (soup or salad or both): 2 + 4 + 2 × 4 = 14 choices. Stage 2 (main course): 9 choices. 21. (a) 8! = 40,320 There are 8 choices for which book to place “first” (on the far left), 7 choices for which of the remaining books to place “second” (next to the first book), 6 choices for which of the remaining books to place “third”, etc. (b) 40,320 – 1 = 40,319 (there is only one way in which all of the books are in order) 22. (a) 9! = 362,880 There are 9 choices for which person to place “first” (in front), 8 choices for which of the remaining people to place “second” in line, 7 choices for which of the remaining people to place “third” in line, etc. (b) 5 × 8! = 201, 600 There are 5 choices for which woman to place in the front of the line, and 8! = 40,320 ways to order the remaining people in line. 23. (a) 5 × 4 × 3 × 2 × 1× 4 × 3 × 2 × 1 = 2880 There are 5 choices for the first person (a woman) in line. Then, there are 4 choices for the second person (one of the remaining women) in line. Next, there are 3 choices for the third person (one of the three remaining women) in line. Etc. Once the number of ways to place the women at the front of the line have been determined (as 5!), the number of ways to place the men at the end of the line can be counted. This process is similar to that used in counting the number of ways to organize the women in line. There are 4 choices for the sixth person (a man) in line. Then, there are 3 choices for the seventh person (one of the remaining men) in line. Etc. (b) 5 × 4 × 4 × 3 × 3 × 2 × 2 × 1× 1 = 2880 There are 5 choices for the first person (a woman) in line. Then, there are 4 choices for the second person (a man) in line. The third person in line is one of the 4 remaining women. The fourth person in line is one of the 3 remaining men. And so on. 24. (a) 7 × 6 × 5 × 4 × 3 × 2 × 1 = 7! = 5, 040 There are 7 choices as to which book can be placed first (on the left). It can be any book except Volume 8 which must be placed last. There are then 6 choices as to which of the remaining books can be placed second (from the left). Etc. (b) 6 × 5 × 4 × 3 × 2 × 1 = 6! = 720 There are 6 choices as to which book can be placed third (from the left). It can be any book except for either Volume 1 or 2. Then, at the next stage, there are 5 choices as to which book can be placed fourth (from the left). It can be any book except Volume 1 or 2 or the book placed in the third position. At the next stage, there are 4 choices as to which book can be placed fifth (from the left). Etc. 25. (a) A typical outcome is a string of 10 letters each of which can be either an H or a T. This sample space consists of N = 210 = 1024 outcomes (2 possible outcomes at each of 10 stages of the experiment). (b) An outcome is an ordered sequence of four numbers, each of which is an integer between 1 and 6 inclusive. This sample space consists of N = 64 = 1296 outcomes (6 possible outcomes at each of 4 stages of the experiment). Copyright © 2022 Pearson Education, Inc.

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Chapter 16: Probabilities, Odds, and Expectations (c) An outcome is sum of four numbers, each of which is an integer between 1 and 6 inclusive. This sample space consists of N = 4 × 6 − 4 × 1 + 1 = 21 outcomes (the sum must be between 4 and 24 inclusive).

26. (a) A typical outcome is a string of 10 letters each of which can be either an T or a F. This sample space consists of N = 210 = 1024 outcomes (2 possible outcomes at each of 10 stages of the experiment). (b) An outcome is an ordered sequence of five letters (A, B, C, D, or E). This sample space consists of N = 510 = 9, 765, 625 outcomes (5 possible outcomes at each of 10 stages of the experiment). 27. (a)

P4 ; the order the candidates are selected to fill these openings matters (i.e., this is an ordered selection of 4 of the 15 board members). 15

(b) 15 C4 ; the order in which the committee is selected is irrelevant (i.e. this is an unordered selection of 4 of the 15 board members). 28. (a)

10 P3 ; the order the athletes are selected to receive the gold, silver, and bronze matters (i.e. this is an ordered selection of 3 of the 10 athletes).

(b) 10 C7 ; the order in which the athletes not earning a medal are selected is irrelevant (i.e. this is an unordered selection of 7 of the 10 athletes). 29. (a)

30 C18 ; the order the songs are selected for the set list does not matter (i.e., this an unordered selection of 18 of the 30 songs).

(b)

30 P18 ; the order the songs are selected for the CD (sequencing) matters (i.e. this is an ordered selection of 18 of the 30 songs).

30. (a)

350 P25 ; the order the 25 teams that are selected for the ranking matters (i.e. this is an ordered selection of 25 of the 350 basketball teams).

(b)

31. (a)

350 C64 ; unseeded means that order does not matter in making the selection (i.e. this is an unordered selection of 64 of the 350 basketball teams). 40 C25 ; the order the players are listed on the active roster does not matter (i.e., this an unordered selection of 25 of the 40 players on the roster).

(b)

P9 ; the order the players are selected for the lineup matters (i.e. this is an ordered selection of 9 of the 25 active players).

32. (a)

C7 ; the order the 7 shirts that are selected for the trip does not matter (i.e. this is an ordered selection of 7 of the 20 shirts).

(b)

P5 ; the order he wears the shirts matters (i.e. this is an ordered selection of 5 of the 7 shirts).

16.3 Probabilities and Odds 33. (a) Pr(o1 ) + Pr(o2 ) + Pr(o3 ) + Pr(o4 ) + Pr(o5 ) = 1 0.22 + 0.24 + 3Pr(o3 ) = 1 Pr(o3 ) = 0.18

The probability assignment is Pr(o1 ) = 0.22, Pr(o2 ) = 0.24, Pr(o3 ) = 0.18, Pr(o4 ) = 0.18, Pr(o5 ) = 0.18.

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(b) Pr(o1 ) + Pr(o2 ) + Pr(o3 ) + Pr(o4 ) + Pr(o5 ) = 1

0.22 + 0.24 + [ Pr(o4 ) + 0.1] + Pr(o4 ) + 0.1 = 1

Pr(o4 ) = 0.17 The probability assignment is Pr(o1 ) = 0.22, Pr(o2 ) = 0.24, Pr(o3 ) = 0.27, Pr(o4 ) = 0.17, Pr(o5 ) = 0.1. 1 . Since Pr(o3 ) = Pr(o2 ) = 0.35 , we have 2 Pr(o1 ) = 0.15, Pr(o2 ) = 0.35, Pr(o3 ) = 0.35, Pr(o4 ) = 0.15 .

34. (a) Pr(o2 ) = 0.35 since Pr(o1 ) + Pr(o2 ) = Pr(o3 ) + Pr(o4 ) =

(b) Pr(o1 ) = 0.15, Pr(o2 ) = 0.35, Pr(o3 ) = 0.22, Pr(o4 ) = 0.28 1 . If Pr( B ) = x , then Pr(C ) = 2 x and Pr( D) = 3x . So, since these probabilities 5 must sum to 1, we have 1/5 + x + 2x + 3x = 1 which means that 6x = 4/5 or x = 2/15. It follows that the 1 2 4 6 probability assignment is Pr( A) = , Pr( B) = , Pr(C ) = , Pr( D) = . 5 15 15 15

35. We know that Pr( A) =

1 3 1 1 1 1 1 36. Pr( A) = , Pr( B) = , Pr(C ) = , Pr( D) = , Pr( E ) = , Pr( F ) = , Pr(G ) = ; the sum of the 4 10 20 10 10 10 10 1 3 1 1 + 4 x = 1 to find x = Pr( D) = Pr( E ) = Pr( F ) = Pr(G ) = . probabilities must be 1 so we solve + + 4 10 20 10 3 = 0.375 ; the sample space has 8 equally likely outcomes (see Exercise 1(a)). Of these 8 outcomes, three of them constitute the event E1 = {HHT , HTH , THH } (see Exercise 11(a)).

37. (a) Pr( E1 ) =

(b) Pr( E2 ) =

2 = 0.25 ; see Exercises 1(a) and 11(b). 8

2 = 0.25 ; see Exercises 1(a) and 11(d). 8 6 = 0.375 (see Exercise 12 and note that the answer is the same if F is replaced by T) 16

(b) Pr( E2 ) =

11 = 0.6875 16

(d) Pr( E4 ) =

4 = 0.25 16

6 1 = ; the sample space has 62 = 36 equally likely outcomes. Of these outcomes, six of 36 6 them constitute the event E1 = {(1,1), (2, 2), (3,3), (4, 4), (5,5), (6, 6)} (see Exercise 13(a)).

39. (a) Pr( E1 ) =

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Chapter 16: Probabilities, Odds, and Expectations (b) Pr( E2 ) =

4 1 = (see Exercise 13(b)) 36 9

8 2 = (see Exercise 13(c)) 36 9

40. (a) Pr( E2 ) =

4 1 = (see Exercise 14) 52 13

(b) Pr( E3 ) =

13 1 = 52 4

12 3 = 52 13

1 ≈ 0.001 ; the sample space has 210 = 1024 equally likely outcomes (see Exercise 7). Of 1024 these outcomes, only one of them constitutes the event that none of the tosses come up tails: E1 = {HHHHHHHHHH } (see Exercise 15(a)).

41. (a) Pr( E1 ) =

(b) Pr( E2 ) =

10 5 = ≈ 0.01 ; (see Exercise 15(b)) 1024 512

11 ≈ 0.01 ; (see Exercise 15(c)) 1024

C 10 5 = 42. (a) Pr( E1 ) = 5 52 = (see also Exercise 16) 32 16 2 C 10 5 = (b) Pr( E2 ) = 5 53 = 32 16 2

4 4 1 = = (see Exercise 16) 25 32 8

43. The total number of outcomes in this random experiment is 210 = 1024 . 1 . If it helps, think of this 1024 as being similar to Exercise 41(a). If the “key” to the quiz is all “True” (or “Heads”), there is only one outcome in which the student earns exactly 10 points.

(a) There is only one way to get all ten correct. So, Pr(getting 10 points) =

(b) There is only one way to get all ten incorrect (and hence a score of –5). So, 1 Pr(getting -5 points) = . Again, you can think of this as the “key” to the quiz consisting of all 1024 “True” answers. In this case, there is only one outcome in which the student earns a score of -5 (their quiz would have all “False” answers. See Exercise 41 and use True=Heads and False=Tails. (c) In order to get 8.5 points, the student must get exactly 9 correct answers and 1 incorrect answer. There are 10 C1 = 10 ways to select which answer would be answered incorrectly. So, Pr(getting 8.5 points) =

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(d) In order to get 8 or more points, the student must get at least 9 correct answers (if they only get 8 correct answers, they lose a point for guessing two incorrect answers and score 7 points). So, 10 1 11 . See also Pr(getting 8 or more points) = Pr(getting 8.5 points) + Pr(10 points) = + = 1024 1024 1024 Exercise 41(c). (e) If the student gets 6 answers correct, they score 6 − 4 × 0.5 = 4 points. If the student gets 7 answers correct, they score 7 − 3 × 0.5 = 5.5 points. So, there is no chance of getting exactly 5 points. Hence, Pr(getting 5 points) = 0 . (f) If the student gets 8 answers correct, they score 8 − 2 × 0.5 = 7 points. There are 10 C2 = 45 ways to select 45 . In order to get 7 or 1024 more points, the student needs to answer at least 8 answers correctly. Pr(getting 7 or more points) =

which 8 answers would be answered correctly. So, Pr(getting 7 points) =

Pr(getting 7 points) + Pr(getting 8.5 points) + Pr(10 points) =

44. (a) Pr(all 4 girls) =

45 10 1 56 7 + + = = . 1024 1024 1024 1024 128

1 1 = = 0.0625 4 2 16

C 6 = 0.375 (b) Pr(2 girls and 2 boys) = 4 42 = 2 16 (c) Pr(youngest child is a girl) = (d) Pr(oldest child is a boy) =

1 2

45. (a) There are 15 C4 = 1365 ways to choose four delegates. If Alice is selected, there are 14 C3 = 364 ways to

choose the other three delegates. So, Pr(Alice selected) = (b) Pr(Alice is not selected) = 1 −

364 4 = . 1365 15

364 1001 11 = = . 1365 1365 15

Dale. Pr(Alice, Bert, Cathy, and Dale selected) =

1 1 = . 1,365 15 C4

46. (a) If A is the first team chosen then there are 9 P3 = 504 ways to choose the other three teams.

Pr( A first) =

504 = 0.1 5040

(b) There are 9 P4 = 3024 ways to have A not chosen, so there are 5040 – 3024 = 2016 ways to choose A as one of the teams. 2016 Pr( A chosen) = = 0.4 5040

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Chapter 16: Probabilities, Odds, and Expectations (c) Pr( A not chosen) =

3024 = 0.6 5040

47. The complementary event F to E = “a tail comes up at least once” is F = “a tail does not come up.” The probability of this complementary event F is Pr(F) = 1/ 210 = 1/1024 . It follows that Pr(E) = 1 – Pr(F) = 1 – 1/1024 = 1023/1024. 48. The complementary event F to E = “at least one of the rolls of the dice comes up a 6” is F = “no rolls of the

dice come up 6.” The probability of this complementary event F is Pr(F) = ( 5 / 6 ) . It follows that Pr(E) = 1 3

– Pr(F) = 1 – ( 5 / 6 ) = 91/216. 3

49. (a) a = 4, b = 7, b – a = 7 – 4 = 3. The odds in favor of E are 4 to 3. That is, for every 7 times the experiment is run, we should expect the event E to occur 4 times and to not occur 3 times. (b) a = 6, b = 10, b – a = 10 – 6 = 4. The odds in favor of E are 6 to 4, or 3 to 2. 50. (a) The odds in favor of E are 3 to 8. a = 3, b = 11, b – a = 8 (b) The odds in favor of E are 3 to 5. a = 3, b = 8, b – a = 5 51. (a) Pr( E ) =

3 3 = 3+5 8

(b) Pr( E ) = 1 − 52. (a) Pr( E ) =

8 8 15 = 1− = 8 + 15 23 23

4 4 = 4+3 7

(b) Pr( E ) =

5 5 = 12 + 5 17

1 2

16.4 Expectations 53. Using the fact that 90/120 = 75% and 144/180 = 80%, Paul’s score in the course is the weighted average 0.15 × 77% + 0.15 × 83% + 0.15 × 91% + 0.1× 75% + 0.25 × 87% + 0.2 × 80% = 82.9% . 54. 0.014 × 2 + 0.361× 3 + 0.5 × 4 + 0.111× 5 + 0.014 × 6 = 3.75 55. 100% - 7% - 22% - 24% - 23% - 19% = 5% are 19 years old. So, the average age at Thomas Jefferson HS is the weighted average 0.07 ×14 + 0.22 × 15 + 0.24 × 16 + 0.23 × 17 + 0.19 × 18 + 0.05 × 19 = 16.4 . 56. 0.47 × 4, 000 + 0.27 × 3, 000 + 0.19 × 1, 500 + 0.07 × 100 = 2,982 57. E =

1 1 1 1 1 × $1 + × $5 + × $10 + × $20 + × $100 = $7.50 2 4 8 10 40

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1 3 3 1 59. E = × 0 + × 1 + × 2 + × 3 = 1.5 heads 8 8 8 8

60. E =

1 2 3 4 5 6 5 4 3 2 1 × 2 + × 3 + × 4 + × 5 + × 6 + × 7 + × 8 + × 9 + × 10 + × 11 + × 12 = 7 36 36 36 36 36 36 36 36 36 36 36

16.5 Measuring Risk 1 1 14 61. E = × $18 + × $54 + × $(−9) = $(−1) 6 18 18 18 18 2 1 × $1 + × $(−1) + × $(−1) = $ − ≈ $ − 0.05 . That is, you can expect to lose about 5 cents for 38 38 38 19 every $1 bet. So, for every $100 bet on red, you should expect to lose about $5 ($5.26 to be more precise). For every $1,000,000 bet on red, you should expect to lose about $52,631.

62. E =

1 37 1 × $36 + × $( −1) = $ − ≈ $ − 0.0263 . So, for every $100 bet on a particular number (like ‘10’), 38 38 38 you should expect to lose $2.63. For every $1,000,000 bet, you should expect to lose about $26,316.

63. E =

1 1 1 1 1 1 64. (a) E = × $1 + × $( −2) + × $3 + × $(−4) + × $5 + × $(−6) = $ − 0.50 6 6 6 6 6 6

(b) No. On the average you will lose 50 cents for every play of this game. For it to be a fair game the house should pay you 50 cents for each play of the game. 65. There is a 24% chance that Joe will gain $400 - $80 = $320. There is a 76% chance that Joe will lose his $80. So, E = 0.24 × $320 + 0.76 × $(−80) = $16 . Joe should take this risk since his expected payoff is greater than 0 ($16). That is, if he made this transaction on thousands of laptops, in the long run he could expect to save $16 for each warranty he purchases. 66. There is a 5% chance that Jackie will gain $50 - $19 = $31. There is a 95% chance that Jackie will lose her $19. So, E = 0.05 × $31 + 0.95 × $(−19) = $ − 16.50 . Jackie should not take this risk since her expected payoff is less than 0 (-$16.50). 67.

Payoff (to insurer)

$(P-500)

$(P-1,500)

$(P-4,000)

Probability

50%

35%

12%

In order to make an average profit of $50 per policy, we solve E = 0.5 × $ P + 0.35 × $( P − 500) + 0.12 × $( P − 1,500) + 0.03 × $( P − 4, 000) = $50 for P. Doing this gives P = $525. That is, the insurance company should charge $525 per policy. 68.

Payoff (to insurer)

$(P-250,000)

Probability

2, 492,500 2,500, 000

7,500 2,500, 000

In order to make an average profit of $50 per policy, we solve 2, 492,500 7,500 E= × $P + × $( P − 250, 000) = $50 for P. Doing this we find that the insurance 2,500, 000 2,500, 000 company should charge P = $800 per policy. Copyright © 2022 Pearson Education, Inc.

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JOGGING 69. (a) 35 × 34 × 33 = 35 P3 = 39, 270 (b) 15 × 14 × 13 = 15 P3 = 2730. (c) The total number of all-female committees is 15 × 14 × 13 = 15 P3 = 2730. The total number of all-male committees is 20 × 19 × 18 = 20 P3 = 6840. So, there are 2730 + 6840 = 9570 committees of the same gender. (d) The remaining 35 × 34 × 33 − (15 × 14 × 13 + 20 × 19 × 18) = 39, 270 − (2730 + 6840) = 29,700 committees are mixed. 70. (a)

C4 = 52,360 ; this is the number of ways of selecting 4 people (without regard to order) from a club having 35 members. 35

(b) From the 20 males, we can select two members 20 C2 = 190 ways. From the 15 females, we can select two members 15 C2 = 105 ways. The total number of all committees having two male members and two female members is then 190 ×105 = 19,950. 71. (a) The event that R wins the match in game 5 can be described by {AARRR, ARARR, ARRAR, RAARR, RARAR, RRAAR}. (b) The event that R wins the match can be described by {RRR, ARRR, RARR, RRAR, AARRR, ARARR, ARRAR, RAARR, RARAR, RRAAR}. (c) The event that the match goes five games can be described by { AARRR, ARARR, ARRAR, RAARR, RARAR, RRAAR, RRAAA, RARAA, RAARA, ARRAA, ARARA, AARRA}. 72. (a) The event that X wins in 5 games can be described by {YXXXX, XYXXX, XXYXX, XXXYX}. (b) The event that the series is over in 5 games can be described by {YXXXX, XYXXX, XXYXX, XXXYX, XYYYY, YXYYY, YYXYY, YYYXY}. (c) It may be useful to describe the events (1) the series goes six games with X winning and (2) the series goes six games with Y winning.

six game series in which…

X wins

Y wins

YYXXXX, YXYXXX, YXXYXX,

XXYYYY, XYXYYY, XYYXYY,

YXXXYX, XYYXXX, XYXYXX,

XYYYXY, YXXYYY, YXYXYY,

XYXXYX, XXYYXX, XXYXYX,

YXYYXY, YYXXYY, YYXYXY,

XXXYYX

YYYXXY

The event “the series is over in game 6” can be described by {YYXXXX, YXYXXX, YXXYXX, YXXXYX, XYYXXX, XYXYXX, XYXXYX, XXYYXX, XXYXYX, XXXYYX, XXYYYY, XYXYYY, XYYXYY, XYYYXY, YXXYYY, YXYXYY, YXYYXY, YYXXYY, YYXYXY, YYYXXY} (d) If X wins the series, there are 6 C3 = 20 ways that the series can end in 7 games (X is listed last in all the outcomes of this event and wins 3 of the other 6 games). Similarly, if X wins the series there are a total of 5 C3 = 10 ways that the series can end in 6 games, 4 C3 = 4 ways that the series can end in 5 games, and there is 3 C3 = 1 way the series can end in 4 games. That is, there are 20 + 10 + 4 + 1 = 35 ways the series can end with X winning. Since there are also 35 ways the series can end with Y winning, the sample space has 70 outcomes. Copyright © 2022 Pearson Education, Inc.

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73. (a) A sample space S for this random experiment might be taken to consist of all ways that 3 balls could be selected from the 10 in the urn (without regard to order). That is N = 10 C3 = 120 . The number of outcomes in the event that two balls drawn are blue and one is red is 3 C2 × 7 C1 = 21 . So, Pr(two are blue and one is red) = 21/120 = 7/40. (b) In this random experiment, each draw is an independent event. So, we can multiply the probabilities of 3 7 3 63 × × = each event. Pr(only first and third balls are blue) = . 10 10 10 1000 (c) A sample space S for this random experiment might be taken to consist of all ways that 3 balls could be selected from the 10 in the urn (with regard to order). That is N = 10 P3 = 720 . The number of outcomes in the event that only the first and third balls drawn are blue (making the second ball drawn red) is 3 P2 × 7 P1 = 42 . So, Pr(first and third balls are blue) = 42/720 = 7/120. 74. The total number of possible outcomes (strings of 10 H’s and T’s) in this experiment is 210 = 1024. (a) Number of ways to choose the positions of 5 H’s: 10 C5 = 252.

Pr(5H’s and 5 T’s) =

252 ≈ 0.246 1024

(b) Number of ways to choose the positions of 3 H’s: 10 C3 = 120.

Pr(3 H’s and 7 T’s) =

120 ≈ 0.117 1024

75. If N is odd, the probability is 0 (there can’t be the same number of H’s as T’s). If N is even, say N = 2K, then N (2 K )! . the probability is N CK 1 = N N ! = N 2 2 K !K ! 2 K !K !

()

76. In each calculation, the denominator is 52 C5 = 2,598,960, the total number of unordered draw poker hands. (a) The numerator represents the 13 ways to choose the value (Ace, two, three, …, queen, king) of the four 13 × 48 1 of a kind times the 48 ways to choose the fifth card. So, Pr(4 of a kind) = = ≈ 0.0002 . 2,598,960 4165 (b) The numerator represents the 4 ways to choose the suit times the number of (unordered) ways to choose 5 cards from the 13 cards of the chosen suit. So, the number of hands with all 5 cards the same suit is 4 × 13 C5 4 × 13 × 12 × 11 × 10 × 9 33 = ≈ 0.00198 . = 5148. Pr(all 5 same suit) = 4 × 13 C5 = 5! 2,598,960 16, 660 (c) The number of ways to get 10, J, Q, K, A of any suit (including all the same suit) is 20 × 16 × 12 × 8 × 4 = 1024. There are 4 ways for these cards to all be the same suit. So, there are 1024 – 5! 1020 1 = ≈ 0.00039 . 4 = 1020 ways to get an ace-high straight. Thus, Pr(ace-high straight) = 2,598,960 2548 (d) There are 13 ways to choose the value of the triple and then 4 C3 = 4 ways to choose the 3 suits from the chosen value. Once that has been done, there are 12 ways to choose the value of the 2 other cards and then 4 C2 = 6 ways to choose the 2 suits from the chosen value. Thus, there are 13 × 4 × 12 × 6 = 3744

ways to choose a full house. Pr(full house) =

3744 6 = ≈ 0.00144 . 2,598,960 4165

260

Chapter 16: Probabilities, Odds, and Expectations

77. The key word in this exercise is “at least.” The sample space consists of all possible outcomes. There are 500 C5 ways in which 5 tickets can be selected from the 500 to be “winning” tickets. For the event that you win at least one prize to occur, it could happen that you win one prize, two prizes, or three prizes. So, in symbolic notation, Pr(winning at least one prize) = Pr(win 1 prize) + Pr(win 2 prizes) + Pr(win 3 prizes) C ⋅ C C ⋅ C C ⋅ C = 5 1 495 2 + 5 2 495 1 + 5 3 495 0 500 C5 500 C5 500 C5 78. (a) There are three ways to select which of the three dice will not land as a 4. There are five numbers that this die can land on (1, 2, 3, 5, or 6). The multiplication rule tells us that there are 3 × 5 = 15 ways to select an outcome in which exactly two 4’s are rolled. Since there are 63 = 216 (again by the multiplication rule) possible outcomes in this random experiment, the probability of such an outcome is 15/216. (b) There are three ways to select which of the three dice will land as a 4. There are five numbers that each other die can land on (1, 2, 3, 5, or 6). The multiplication rule tells us that there are 3 × 5 × 5 = 75 ways to select an outcome in which exactly one 4 is rolled. Hence, the probability of such an outcome is 75/216. (c) There are five numbers that each die can land on (1, 2, 3, 5, or 6). So, there are 5 × 5 × 5 = 125 ways to select an outcome in which no 4 is rolled. So, the probability of not rolling any 4’s is 125/216. 6 6 = ≈ 0.00077 5 6 7776 The denominator, 7776, is the number of ordered outcomes possible. The numerator consists of the only 6 outcomes that give YAHTZEE (six 1’s, six 2’s, etc…).

79. (a) Pr(YAHTZEE) =

5 × 6 × 5 150 = ≈ 0.019 65 7776 There are 5 choices as to which die will not match the other 4. There are 6 choices for what will constitute the four of kind (four sixes, four fives, etc…) and there are 5 choices for what will constitute the other die.

(b) Pr(four of a kind) =

2 × 5! 240 = ≈ 0.031 65 7776 There are 2 choices as to what the large straight will be – it will either be 1-5 or 2-6. For each there are 5! ways to rearrange the order that the dice are rolled.

80. (a) You should fold. You could win $150 by calling and getting a spade on the river card, or you could lose $50 by calling and not getting a spade on the river cade. The probability of getting a spade on the river card is 9/46, the probability of not getting a spade is 37/46. The expected value of your call is 9 37 E= × $150 + × ( −$50 ) ≈ −$10.87 . 46 46 (b) You should call the bet. In this case the expected value of your call is 9 37 E= × $120 + × ( −$20 ) ≈ $7.40 . 46 46

Chapter 17 WALKING

6. (a) Q3 ≈ 2354 + 0.675 × 468 ≈ 2670 points

17.2 Normal Curves and Normal Distributions 1. (a) μ = 80 lb; it is located in the exact middle of the distribution. (b) M = 80 lb; it too is located in the exact middle of the distribution. (c) σ = 90 lb. – 80 lb. = 10 lb; it is the horizontal distance between the middle of the distribution and the inflection point P. 2. (a) μ = 25.1 m; it is located in the exact middle of the distribution. (b) M = 25.1 m; it too is located in the exact middle of the distribution. (c) σ = 25.1 m – 19.0 m = 6.1 m; it is the horizontal distance between the middle of the distribution and the inflection point P. 3. (a)

32 in. + 48 in. = 40 in. 2

432.5 points + 567.5 points 2 = 500 points

7. (a) μ =

(b) 567.5 ≈ 500 + 0.675 ⋅ σ σ ≈ 100 points 8. (a) μ =

229 + 391 = 310 2

(b) 391 ≈ 310 + 0.675 ⋅ σ σ ≈ 120 9. 94.7 ≈ 81.2 + 0.675 × σ σ ≈ 20 in. 10. $15,514 ≈ $18,565 − 0.675 ⋅ σ

σ ≈ $4520 11. μ ≠ M ; in any normal distribution these two values must be the same.

(b) Using the result from (a), we see that σ = 40 in. – 32 in. = 8 in. (c) Q3 = 40 in. + 0.675 × 8 in. = 45.4 in. (d) Q1 = 40 in. − 0.675 × 8 in. = 34.6 in. 4. (a)

(b) Q1 ≈ 2354 − 0.675 × 468 ≈ 2038 points

450 cm + 560 cm = 505 cm 2

(b) 560 – 505 = 55 cm (c) Since σ = 55 cm, 0.675 ⋅ σ = 37.125 cm and Q3 ≈ 505 cm + 37 cm = 542 cm . (d) Since σ = 55 cm, 0.675 ⋅ σ = 37.125 cm and Q1 ≈ 505 cm − 37 cm = 468 cm .

12. In a normal distribution the first and third quartiles are the same distance from the mean. In this distribution, μ − Q1 = 453 − 343 = 110 and Q3 − μ = 553 − 453 = 100 . 13. In a normal distribution the first and third quartiles are the same distance from the mean. In this distribution, μ − Q1 = 195 − 180 = 15 and Q3 − μ = 220 − 195 = 25 . 14. In a normal distribution the first quartile is 0.675 standard deviations below the mean. Rather than having a value of 35, the first quartile in this distribution should be μ − 0.675 × σ = 47 − 0.675 × 10 = 40.25 . 15. (a)

5. (a) Q3 ≈ 81.2 lb. + 0.675 × 12.4 lb. ≈ 89.6 lb (b) Q1 ≈ 81.2 lb. − 0.675 × 12.4 lb. ≈ 72.8 lb

(b)

45 kg − 30 kg 15 kg = =1 15 kg 15 kg 0 kg − 30 kg −30 kg = = −2 15 kg 15 kg

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Chapter 17: The Mathematics of Normality (c)

54 kg − 30 kg 24 kg = = 1.6 15 kg 15 kg

(d)

3 kg − 30 kg −27 kg = = −1.8 15 kg 15 kg

16. (a)

98 points − 110 points = −1 12 points

(b)

110 points − 110 points =0 12 points

(c)

128 points − 110 points = 1.5 12 points

(d)

71 points − 110 points −39 = = −3.25 12 points 12

17. (a) In a normal distribution, the third quartile is about 0.675 standard deviations above the mean. That is, Q3 ≈ μ + (0.675)σ . In this case, it means 391 ≈ 310 + (0.675)σ so that the standard deviation is σ ≈ 120 . Hence, the standardized value of 490 490 − 310 points is approximately = 1.5 . 120 (b)

250 − 310 = −0.5 120

(c)

220 − 310 = −0.75 120

(d)

442 − 310 = 1.1 120

61.5 lb. − 49.5 lb. = 1.5 8 lb.

(c)

35.1 lb. − 49.5 lb. = −1.8 8 lb.

(d)

67.5 lb. − 49.5 lb. = 2.25 8 lb.

x − 183.5 =0 31.2 x − 183.5 = 0 x = 183.5 ft

(b) 152.3 feet x − 183.5 = −1 31.2 x − 183.5 = −31.2 x = 152.3 ft

20. (a) 83.2 x − 83.2 =0 4.6 x = 83.2 (b) 92.4 x − 83.2 =2 4.6 x − 83.2 = 9.2 x = 92.4

21. (a) 199.1 feet x − 183.5 = 0.5 31.2 x − 183.5 = 15.6 x = 199.1 ft

(b) 111.74 feet x − 183.5 = −2.3 31.2 x − 183.5 = −71.76 x = 111.74 ft

18. (a) 44.1 lb. ≈ 49.5 lb. − (0.675)σ so that the standard deviation is σ ≈ 8 lb. Hence, the standardized value of 41.5 lb. is 41.5 lb. − 49.5 lb. approximately = −1 8 lb. (b)

19. (a) 183.5 feet

22. (a) 76.3 x − 83.2 = −1.5 4.6 x − 83.2 = −6.9 x = 76.3 (b) 81.222 x − 83.2 = −0.43 4.6 x − 83.2 = −1.978 x = 81.222

ISM: Excursions in Modern Mathematics, 10E 23.

84 − 50

30. 3.9 kg is one standard deviation above the mean, and 2.5 kg is one standard deviation below the mean. 2.5 kg + 3.9 kg = 3.2 kg μ= 2 σ = 3.9 kg − 3.2 kg = 0.7 kg

34 = 2σ σ = 17 lb.

24. 12 −6 − 30

= −3

31. (a) 98.8 cm is two standard deviations above the mean and 85.2 cm is two standard deviations below the mean. So, 85.2 cm + 98.8 cm μ= = 92 cm 2 1 σ = (98.8 cm − 92 cm) = 3.4 cm 2

−36 = −3σ σ = 12

25.

50 − μ =3 15 50 − μ = 45 μ =5

(b) Q1 = 92 cm − 0.675 × 3.4 cm ≈ 89.7 cm Q3 = 92 cm + 0.675 × 3.4 cm = 94.3 cm

26. μ = 50 10 − μ = −2 20 10 − μ = −40 μ = 50 27.

20 − μ

= −2;

263

100 − μ

σ σ 20 − μ = −2σ ; 100 − μ = 3σ μ = 20 + 2σ , so 100 − (20 + 2σ ) = 3σ 100 − 20 − 2σ = 3σ 80 = 5σ σ = 16 μ = 20 + 2 × 16 = 52

32. (a) 10.2 oz is two standard deviations above the mean and 4.2 oz is two standard deviations below the mean. 4.2 oz + 10.2 oz μ= = 7.2 oz 2 1 σ = (10.2 oz − 7.2 oz ) = 1.5 oz 2 (b) Q1 ≈ 7.2 oz − 0.675 × 1.5 oz ≈ 6.2 oz Q3 ≈ 7.2 oz + 0.675 × 1.5 oz ≈ 8.2 oz 33. (a) The 16th percentile is one standard deviation below the 50th percentile. So, σ = 71.5 in. − 65.2 in. = 6.3 in. (b) The 84th percentile is one standard deviation above the 50th percentile. So, P84 = 71.5 in.+ 6.3 in. = 77.8 in.

28. μ = −10; σ = 30 −10 − μ 50 − μ = 0; =2

σ −10 − μ = 0; 50 − μ = 2σ μ = −10, so 50 − (−10) = 2σ 60 = 2σ σ = 30

34. (a) The 84th percentile is one standard deviation above the mean. It follows that μ = 66.7 points − 12.3 points = 54.4 points .

29. 54 inches is one standard deviation above the mean, and 42 inches is one standard deviation below the mean. 42 in. + 54 in. = 48 in. μ= 2 σ = 54 in. − 48 in. = 6 in.

(b) The 16th percentile is one standard deviation below the mean. It follows that P16 = 54.4 points - 12.3 points = 42.1 points . 35. Since 97.5% of the data lies below two standard deviations above the mean, P97.5 = μ + 2σ = μ + 2 ( 6.1 cm ) = 81.5 cm .

Thus, μ = 81.5 cm - 2 ( 6.1 cm ) = 69.3 cm .

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Chapter 17: The Mathematics of Normality

within two standard deviations of the mean, 5% of the data is not within two standard deviations of the mean. So, 2.5% of the data fall above z = 2 (i.e. above μ + 2σ ). It follows that 99.85% 2.5% = 97.35% of the data falls between z = -3 and z = 2 (i.e. between μ − 3σ and μ + 2σ ).

36. Since 2.5% of the data lies below two standard deviations below the mean, P2.5 = μ − 2σ = 56.3 cm − 2σ = 33.7 cm .

Thus, μ =

33.7 cm - 56.3 cm = 11.3 cm . −2

37. 9.9 has a standardized value of 9.9 − 12.6 = −0.675 . Also, 16.6 has a 4 16.6 − 12.6 standardized value of = 1 . So, 4 25% of the data is below the first quartile of 9.9. Also, 84% of the data is below 16.6. So, 84% - 25% = 59% of the data is between 9.9 and 16.6. 38. 465 has a standardized value of 465 − 500 = −1 . Also, 605 has a standardized 35 605 − 500 = 3 . So, 16% of the data value of 35 is below 465. Also, about 100% of the data is below 605. So, 100% - 16% = 84% of the data is between 465 and 605. 39. (a) 95% of the data lies within two standard deviations of the mean. Hence, 2.5% of the data are not within two standard deviations on each side of the mean. So, 97.5% of the data fall below μ + 2σ , the point two standard deviations above the mean. (b) 97.5% of the data lies below z = 2 (i.e. below μ + 2σ ). Since 68% of the data lies within one standard deviation of the mean, 32% of the data is not within one standard deviation of the mean. So, 68% + 16% = 84% of the data fall below z = 1(i.e. below μ + σ ). It follows that 97.5% − 84% = 13.5% of the data falls between z = 1 and z = 2 (i.e. between μ + σ and μ + 2σ ). 40. (a) 99.7% of the data lies within three standard deviations of the mean. So, 0.3% of the data are not within three standard deviations on each side of the mean. This means 99.85% of the data fall above z = -3 (i.e. above μ − 3σ ), the point three standard deviations below the mean. (b) 99.85% of the data lies above z = -3 (i.e. above μ − 3σ ). Since 95% of the data lies

17.3 Modeling Approximately Normal Distributions 41. (a) 52 points (b) 50% (c) 41 − 52 = −1, 63 − 52 = 1 11 11 So, since about 68% of data fall within one standard deviation of the mean (between standardized scores of -1 and 1) in a normal distribution, we would estimate that 68% of the students would score between 41 and 63 points. (d)

1 (100% − 68%) = 16% 2 Or, since 50% + 34% = 84% of the students score 63 points or less, approximately 16% of the students score 63 points or more.

42. (a) 30 is two standard deviations below the mean and 74 is two standard deviations above the mean. Hence, 95% of the students scored between 30 and 74. (b) 30 is two standard deviations below the mean. So 2.5% of the students scored 30 points or less. (c) 19 is three standard deviations below the mean. So 0.15% of the students scored 19 points or less. 43. (a) Q1 ≈ 52 − 0.675 × 11 ≈ 44.6 points (b) Q3 ≈ 52 + 0.675 × 11 ≈ 59.4 points 44. (a) Since the mean is 52 points, a score of 52 points is at the 50th percentile. (b) A score of 63 is one standard deviation above the mean. So, 63 points is at the 84th percentile.

ISM: Excursions in Modern Mathematics, 10E (c) Any answer above the 75th percentile (but reasonably close to it) is an acceptable estimate. (The actual value is 76.6.) (d) 99.85th percentile

99 − 125 151 − 125 = −2, =2 45. (a) 13 13 95% have blood pressure between 99 and 151 mm (i.e., between standardized scores of -2 and 2). 95% of 2000 patients is 1900 patients ( 0.95 × 2000 =1900). (b) 112 is one standard deviation below the mean. 151 is two standard deviations above the mean. The percentage of patients falling between one standard deviation below the mean and two standard deviations above the mean is 68% + 13.5% = 81.5%. 81.5% of the 2000 patients is 1630 patients. 46. (a)

99 − 125 = −2 13 So by the rule of 95 approximately 97.5% of the patients will have blood pressure above 99 mm. But 0.975(2000) = 1950 patients.

(b) 99 is two standard deviations below the mean. 138 is one standard deviation above the mean. The percentage of patients falling between two standard deviations below the mean and one standard deviation above the mean is 68% + 13.5% = 81.5%. 81.5% of the 2000 patients is 1630 patients. 112 mm − 125 mm = −1 , x = 112 13 mm mm is at the 16th percentile (one standard deviation below the mean).

47. (a) Since

138 mm − 125 mm = 1 , x = 138 mm 13 mm is at the 84th percentile (one standard deviation above the mean).

(b) Since

164 mm − 125 mm = 3 , x = 164 13 mm mm is at the 99.85th percentile (three standard deviations above the mean).

265

Here we used the fact that Min is approximately three standard deviations below the mean and Max is approximately three standard deviations above the mean. (b) Min ≈ 86 mm, Q1 ≈ 125 mm – 0.675(13 mm) = 116 mm, Q2 = μ = 125 mm, Q3 ≈ 125 + 0.675(13 mm) = 134 mm, Max ≈ 164 mm. 11 oz − 12 oz = −2 and 0.5 oz 13 oz − 12 oz = 2 , the rule of 95 tells us 0.5 oz that there is approximately a 95% chance that the bag weighs somewhere between 11 and 13 ounces.

49. (a) Since

13 oz − 12 oz = 2 , the rule of 95 0.5 oz tells us that there is approximately a 47.5% chance that the bag weighs somewhere between 12 and 13 ounces (this is half of 95%).

(b) Since

(c) Because the chance of the bag weighing between 11 and 12 ounces is the same as the chance that it weighs between 12 and 13 ounces, we can use our answer to part (b) and symmetry to obtain an answer of 47.5% + 50% = 97.5%. 11.5 oz − 12 oz = −1 and 0.5 oz 12.5 oz − 12 oz = 1 , the rule of 68 tells us 0.5 oz that there is approximately a 68% chance that the bag weighs somewhere between 11.5 and 12.5 ounces.

50. (a) Since

12.5 oz − 12 oz = 1 , the rule of 68 0.5 oz tells us that there is approximately a 34% chance that the bag weighs somewhere between 12 and 12.5 ounces (this is half of 68%).

(b) Since

266

Chapter 17: The Mathematics of Normality 11 oz − 12 oz = −2 the rule of 95 0.5 oz tells us that there is approximately a 1 (100% − 95%) = 2.5% chance that any 2 one bag weighs less than 11 ounces. But, 0.025 × 500 = 12.5 ≈ 13 bags .

51. (a) Since

11.5 oz − 12 oz = −1 the rule of 68 0.5 oz tells us that there is approximately a 1 (100% − 68%) = 16% chance that any 2 one bag weighs less than 11.5 ounces. But, 0.16 × 500 = 80 bags.

(b) Since

11.5 oz − 12 oz = −1 the rule of 68 0.5 oz tells us that there is approximately a 1 (68%) = 34% chance that any one bag 2 weighs between 11.5 and 12 ounces. But, 34% of 1500 is 510 bags.

(b) Since

(c) 510 bags (see part (b) and use symmetry) (d) 202 or 203 bags (see part (a) and use symmetry) 13 oz − 12 oz = 2 and 0.5 oz 13.5 oz − 12 oz = 3 the rules of 95 and 0.5 oz 99.7 tell us that there is approximately a 1 (99.7% − 95%) = 2.35% chance that 2 any one bag weighs between 13 and 13.5 ounces. But, 2.35% of 1500 is 35 bags.

(e) Since

(c) 0.50 × 500 = 250 bags 12.5 oz − 12 oz = 1 the rule of 68 0.5 oz tells us that there is approximately a 1 (68%) + 50% = 84% chance that any 2 one bag weighs less than 12.5 ounces. But, 0.84 × 500 = 420 bags.

(d) Since

13 oz − 12 oz = 2 the rule of 95 (e) Since 0.5 oz tells us that there is approximately a 1 (95%) + 50% = 97.5% chance that any 2 one bag weighs less than 13 ounces. But, 0.975 × 500 = 487.5 ≈ 488 bags. 13.5 oz − 12 oz = 3 the rule of 99.7 0.5 oz tells us that there is approximately a 1 (99.7%) + 50% = 99.85% chance that 2 any one bag weighs less than 13.5 ounces. But, 0.9985 × 500 = 499.25 ≈ 499 bags.

(f) Since

11 oz − 12 oz = −2 and 0.5 oz 11.5 oz − 12 oz = −1 the rules of 68 and 0.5 oz 95 tell us that there is approximately a 1 (95% − 68%) = 13.5% chance that any 2 one bag weighs between 11 and 11.5

52. (a) Since

ounces. But, 13.5% of 1500 is 202 or 203 bags.

7.21 kg − 8.16 kg = −1 , x = 7.21 kg 0.95 kg is at the 16th percentile (one standard deviation below the mean).

53. (a) Since

10 kg − 8.16 kg ≈ 2 , x = 10 kg is 0.95 kg at the 97.5th percentile (two standard deviations above the mean).

(b) Since

(c) The 75th percentile corresponds to the third quartile. This is a value located 0.675 of a standard deviation above the mean. This weight is thus x = 8.16 kg + 0.675 × 0.95 kg ≈ 8.8 kg. 7.7 kg - 9.7 kg = −2 , x = 7.7 kg is 1.0 kg at the 2.5th percentile (two standard deviations below the mean).

54. (a) Since

10.4 kg − 9.7 kg = 0.7 ≈ 0.675 , x = 1.0 kg 10.4 kg is at the 75th percentile (0.675 standard deviations above the mean).

(b) Since

ISM: Excursions in Modern Mathematics, 10E (c) The 84th percentile corresponds to one standard deviation above the mean. Her weight is thus x = 9.7 kg + 1.0 kg = 10.7 kg. 5.3 kg - 4.5 kg = 2 , x = 5.3 kg is at 0.4 kg the 97.5th percentile (two standard deviations above the mean).

55. (a) Since

5.7 kg − 4.5 kg = 3 , x = 5.7 kg is at 0.4 kg the 99.85th percentile (three standard deviations above the mean).

(b) Since

(c) The 25th percentile corresponds to 0.675 standard deviations below the mean. Her weight is thus x = 4.5 kg − 0.675 × 0.4 kg ≈ 4.23 kg. 11.3 kg - 10.5 kg ≈ 0.667 ≈ 0.675 , 1.2 kg x = 11.3 kg is at the 75th percentile (0.675 standard deviations above the mean).

56. (a) Since

8.1 kg − 10.5 kg = −2 , x = 8.1 kg is 1.2 kg at the 2.5th percentile (two standard deviations below the mean).

(b) Since

(c) The 84th percentile corresponds to one standard deviation above the mean. His weight is thus x = 10.5 kg + 1.2 kg = 11.7 kg.

17.4. Normality in Random Events 57. (a) μ =

deviation above the mean. That is, between z = 0 and z = 1. (d) 1830 corresponds to a z-value of z = 1 and 1860 corresponds to a z-value of z = 2 1860 − 1800 = 2 ). So, (since 30 1 (95%) = 47.5% of the number of tails 2 tossed should fall between z = 0 and z = 2 (i.e. between 1800 and 1860). 47.5% − 34% = 13.5% of the number of tails tossed should fall between z = 1 and z = 2 (i.e. between 1830 and 1860). 58. (a) μ =

(c) Since 3080 corresponds to a z-value of -3 and 3200 corresponds to a z-value of 0, the chances of this area approximately 49.85% (or 50%). (d) Since 3240 corresponds to a z-value of 1 and 3280 corresponds to a z-value of 2, the chances of this area approximately 47.5% − 34% = 13.5% . 59. μ =

7056 7056 = 3528, σ = = 42 2 2 3486 − 3528 = −1 and 42 3570 − 3528 = 1 there is a 68% chance 42 that the number of females will fall in this range (between one standard deviation below the mean and one standard deviation above the mean).

(a) Since

(b) Since

6400 6400 = 3200, σ = = 40 2 2

(b) Since 3120 corresponds to a z-value of -2 and 3280 corresponds to a z-value of 2, the chances of this are approximately 95%.

3600 3600 = 1800, σ = = 30 2 2

1770 − 1800 = −1 and 30 1830 − 1800 = 1 there is a 68% chance 30 that Y will fall in this range (between one standard deviation below the mean and one standard deviation above the mean).

267

(b)

1 (100% − 68%) = 16% 2

(c)

1 (68%) + 50% = 84% 2

268

Chapter 17: The Mathematics of Normality

60. (a) According to the honest coin principle μ

= 2500/2 = 1250 and σ = 2500 / 2 = 25. Not losing any money means winning 1250 times or more—a 50% chance. (b) 1300 = μ + 2σ so that there is a 95% chance that the number of rolls will fall between 1200 and 1300. The probability of rolling between 1250 and 1300 is half of this of 47.5%. (c) Winning $100 or more means winning 1300 times or more—a 2.5% chance since 1300 = μ + 2σ . 61. (a) μ = 600 × 0.4 = 240,

σ = 600 × 0.4 × (1 − 0.4) = 12 (b) Q1 ≈ 240 − 0.675 × 12 ≈ 232 Q3 ≈ 240 + 0.675 × 12 ≈ 248 216 − 240 264 − 240 = −2 and =2 12 12 there is a 95% chance that the number of coins will fall in this range (between two standard deviations below the mean and two standard deviations above the mean).

62. (a) μ = 1200 × 0.75 = 900,

σ = 1200 × 0.75 × (1 − 0.75) = 15 (b) Q1 ≈ 900 − 0.675 × 15 ≈ 890 Q3 ≈ 900 + 0.675 × 15 ≈ 910

1 1 , μ = 180 × = 30, and 6 6 1  

1

64. (a) Here n = 400 and p = 0.1. On average, we would expect μ = np = 400 ( 0.1) = 40 speakers to be defective. So, there is about a 50% chance that at most 40 will be defective. (b) Here again n = 400 and p = 0.1. So, σ = 400 × 0.1 × 0.9 = 6 . This means that 52 is two standard deviations above the mean. So, there is only a 2.5% chance that more than 52 will be defective.

JOGGING 65. 0 In a normal distribution, the mean and the median are the same values. Since the z-value of the mean is 0, so too is the z-value for the median. 66. (a) approximately -0.675 In a normal distribution, Q1 ≈ μ − 0.675 × σ . So, the approximate z-value of the first quartile is given by μ − 0.675σ − μ −0.675σ = = −0.675 .

(b) approximately 0.675 In a normal distribution, Q3 ≈ μ + 0.675 × σ . So, the approximate z-value of the third quartile is given by μ + 0.675σ − μ 0.675σ = = 0.675 .

67. (a) His weight is approximately 8.16 kg + 1.65 × 0.95 kg = 9.73 kg. (b) His weight is approximately 8.16 kg − 0.84 × 0.95 kg = 7.36 kg.

σ = 180 × × 1 −  = 5 . 6 6 

40 − 30 = 2 , x = 40 is two standard 5 deviations above the mean. The rule of 95 gives a 2.5% chance of rolling a ‘6’ more than 40 times.

(b) Since

35 − 30 = 1 , x = 35 is one standard 5 deviation above the mean. The rule of 68 gives a 34% (half of 68%) chance of rolling a ‘6’ somewhere between 30 and 35 times.

68. (a) 55 + 12 × 2.33 = 82.96 ≈ 83 points (b) 55 - 12 × 0.52 = 48.76 ≈ 49 points

ISM: Excursions in Modern Mathematics, 10E 8.4 kg − 8.16 kg ≈ 0.253 , x = 8.4 0.95 kg kg is at approximately the 60th percentile (i.e. at μ + 0.25σ ).

(b) -1.28

2 7.67 kg − 8.16 kg ≈ −0.52 , x = 7 3 0.95 kg th kg is at approximately the 30 percentile (i.e. at μ − 0.52σ ).

(e) -1.04

69. (a) Since

(b) Since

73. (a) The 90th percentile of the data is located at μ + 1.28 × σ = 65.2 + 1.28 × 10 = 78 points. (b) The 70th percentile of the data is located at μ + 0.52 × σ = 65.2 + 0.52 × 10 = 70.4 points.

70. (a) 99th percentile 55 + 12 ⋅ w = 83 w ≈ 2.33 (b)

269

20th percentile

55 + 12 ⋅ w = 45

w ≈ −0.84

(c) 5000 55 + 12 ⋅ w = 35 w ≈ −1.666 So a score of 35 is at (approximately) the 5th percentile. 5% of N = 250 gives N = 5000 students. 71. (a) A score at the 75th percentile (third quartile) would be approximately 0.675 standard deviations above the mean. This score would be 531 + (0.675)(104) = 601.2 points. Since SAT scores are always multiples of 10, we estimate that a score of 600 or 610 points would be at the third quartile. (b) A score at the 84th percentile would be approximately one standard deviation above the mean using the 68-95-99.7 rule. This score would be 531 + (1)(104) = 635 points. Since SAT scores are always multiples of 10, we estimate that a score of 630 or 640 points would be at the 84th percentile. (c) A score at the 70th percentile would be approximately 0.52 standard deviations above the mean using Table 17-9. This score would be 531 + (0.52)(104) = 585.08 points. Since SAT scores are always multiples of 10, we estimate that a score of 580 or 590 points would be at the 70th percentile. 72. (a) 1.28 (the approximate location of the 90th percentile is at μ + 1.28σ )

(d) The 5th percentile of the data is located at μ − 1.65 × σ = 65.2 − 1.65 × 10 = 48.7 points. 1 n 74. For p = , μ = n × p = , and 2 2 1 1 σ = n × p × (1 − p) = n × × = 2 2

n n = . 4 2

RUNNING n . 2 There is a 16% chance Y is more than one n standard deviation above the mean, + σ . 2 n where σ = 10 , we solve Using σ = 2 n 10 = for n. But 20 = n gives 2 n = 400 .

75. The mean number of tails tossed is μ =

n 2 There is a 95% chance when X is within 2 standard deviations of the mean. 15 = 2σ

76. μ =

σ = 7.5 7.5 =

n  n  using σ =  2  2 

15 = n n = 225

270

Chapter 17: The Mathematics of Normality

77. n = 2500 X has an approximately normal distribution with mean μ = n and standard deviation 10 σ = n ⋅ 1 ⋅ 9 = 3 n . If there is a 95% 10 10 10 chance that X will be between n − 30 and 10 n + 30 , then 30 = 2σ and σ = 15 . Solving 10 3 n = 15 gives n = 2500 . 10 78. n = 720 The probability of rolling a total of 7 is p = 6 = 1 . Thus, . Thus, Y has an 36 6 approximately normal distribution with mean μ = n and standard deviation 6 σ = n ⋅ 1 ⋅ 5 = 1 5n . If there is a 95% 6 6 6 chance that Y will be between n − 20 and 6 n + 20 , then 20 = 2σ and σ = 10 . Solving 6 1 5n = 10 gives n = 720 . 6 79. (a) The probability of losing a bet on red is 20/38 ≈ 0.53 . By the dishonest coin principle, Y has an approximately normal distribution with center μ ≈ 10, 000 × 0.53 = 5300 and standard deviation σ ≈ 10000 × 0.53 × 0.47 ≈ 50 .

(b) Since the center of the distribution is at Y = 5300, the chances that we will lose 5300 times or more ( Y ≥ 5300 ) are approximately 50%. (c) The chances that we will lose somewhere between 5150 and 5450 times (i.e., Y is within plus or minus 3 standard deviations of the center) are 99.7%. (d) To break even or win we must have Y ≤ 5000 (i.e., Y must be more than 4 standard deviations left of the center). The chances of this are essentially 0. 80. (a) The standard deviation for the number of votes for Mrs. Butterworth is 800 × 0.53 × 0.47 ≈ 14.12 . Thus, the standard error is approximately 14.12/800 = 0.0176 or 1.76%. (b) A 95% confidence interval is given by 2 standard errors above and below the statistic obtained from the sample. In this example, we have a 95% confidence interval of 53% plus or minus 3.52% which means that we can say with 95% confidence that the actual percentage of votes Mrs. Butterworth will receive is between 49.48% and 56.52%. (c) A 99.7% confidence interval is given by taking 3 standard errors below and above the sample statistic. In this example, we can say with 99.7% confidence that the actual percentage of votes Mrs. Butterworth will receive is between 47.72% and 58.28%.