TaylMod-Aff.qxd
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8:50 PM
Page 277
16. cj
Basic Variables
300
400
0
0
0
Quantity
x1
x2
s1
s2
s3
0
s1
18
3
2
1
0
0
0
s2
20
2
4
0
1
0
0
s3
4
0
1
0
0
1
zj
0
0
0
0
0
0
cj – zj
300
400
0
0
0
Basic Variables
300
400
0
0
0
Quantity
x1
x2
s1
s2
s3
cj
0
s1
10
3
0
1
0
–2
0
s2
4
2
0
0
1
–4
400
x2
4
0
1
0
0
1
zj
1,600
0
400
0
0
400
cj – zj
300
0
0
0
–400
Basic Variables
300
400
0
0
0
Quantity
x1
x2
s1
s2
s3
cj
0
s1
4
0
0
1
–3/2
4
300
x1
2
1
0
0
1/2
–2
400
x2
4
0
1
0
0
1
zj
2,200
300
400
0
150
–200
0
0
0
–150
200
300
400
0
0
0
Quantity
x1
x2
s1
s2
s3
cj – zj cj
Basic Variables 0
s3
1
0
0
1/4
–3/8
1
300
x1
4
1
0
1/2
–1/4
0
400
x2
3
0
1
–1/4
3/8
0
zj
2,400
300
400
50
75
0
0
0
–50
–75
0
cj – zj Optimal
277
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Page 278
17. cj
Basic Variables
Quantity
5
4
0
0
x1
x2
s1
s2
0
s1
150
3/10
1/2
1
0
0
s2
2,000
10
4
0
1
zj
0
0
0
0
0
cj – zj
5
4
0
0
Basic Variables
5
4
0
0
x1
x2
s1
s2
cj
Quantity
0
s1
90
0 19/50
1 –3/100
5
x1
200
1
2/5
0
1/10
zj
1,000
5
2
0
1/2
0
2
0
–1/2
5
4
0
0
Quantity
x1
x2
s1
s2
cj – zj cj
Basic Variables
4
x2
4,500/19
0
1
5
x1
2,000/19
1
0 –20/19
zj
28,000/19
5
4 100/19 13/28
0
0 –100/19 –13/38
cj – zj
50/19 –3/38 5/38
Optimal 18. cj
Basic Variables
100
150
0
0
0
Quantity
x1
x2
s1
s2
s3
0
s1
160
10
4
1
0
0
0
s2
20
1
1
0
1
0
0
s3
300
10
20
0
0
1
zj
0
0
0
0
0
0
100
150
0
0
0
cj – zj
(continued)
278
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cj
8:50 PM
Basic Variables
Page 279
100
150
0
0
0
Quantity
x1
x2
s1
s2
s3
0
s1
100
8
0
1
0
–1/5
0
s2
5
1/2
0
0
1
–1/20
150
x2
15
1/2
1
0
0
1/20
zj
2,250
75
150
0
0
15/2
cj – zj
25
0
0
0
–15/2
Basic Variables
100
150
0
0
0
Quantity
x1
x2
s1
s2
s3
cj
0
s1
20
0
0
1
–16
3/5
100
x1
10
1
0
0
2
–1/10
150
x2
10
0
1
0
–1
1/10
zj
2,500
100
150
0
50
5
0
0
0
–50
–5
100
20
60
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
cj – zj Optimal 19. cj
Basic Variables
0
s1
60
3
5
0
1
0
0
0
s2
100
2
2
2
0
1
0
0
s3
40
0
0
1
0
0
1
zj
0
0
0
0
0
0
0
cj – zj
100
20
60
0
0
0
Basic Variables
100
20
60
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
cj
100
x1
20
1
5/3
0
1/3
0
0
0
s2
60
0
–4/3
2
–2/3
1
0
0
s3
40
0
0
1
0
0
1
zj
2,000
100
500/3
0
100/3
0
0
60 –100/3
0
0
cj – zj
0 –440/3
(continued)
279
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cj
8:50 PM
Basic Variables
Page 280
100
20
60
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
100
x1
20
1
5/3
0
1/3
0
0
60
x3
30
0
–2/3
1
–1/3
1/2
0
0
s3
10
0
2/3
0
1/3
–1/2
1
zj
3,800
100
380/3
60
40/3
30
0
0 –320/3
0
–40/3
–30
0
cj – zj Optimal 20.
a. b. c. d. e.
maximization, because cj – zj x2 = 10, s2 = 20, x1 = 10, Z = 30 maximize Z = x1 + 2x2 – x3 3 No, because there are three constraints and three “slack” variables f. s1 = 0 g. yes, because cj – zj = 0 for s1 h. x2 = 3 1/3, s2 = 26 2/3, x1 = 23 1/3, Z = 30
21. cj
Basic Variables
120
40
240
0
0
M
M
Quantity
x1
x2
x3
s1
s2
A1
A2
M
A1
27
4
1
3
–1
0
1
0
M
A2
30
2
6
3
0
–1
0
1
zj
54M
6M
7M
6M
–M
–M
M
M
7M – 40 6M – 240 –M
–M
0
0
zj – cj cj
Basic Variables
6M – 120
Quantity
120
40
240
0
0
M
x1
x2
x3
s1
s2
A1
M
A1
22
11/3
0
5/2
–1
1/6
1
40
x2
5
1/3
1
1/2
0
–1/6
0
zj
19M + 200
11M/3 + 40/3
40
5M/2 + 20
–M
M/6 – 20/3
M
11M/3 – 320/3
0
5M/2 – 220
–M
M/6 – 20/3
0
zj – cj cj
Basic Variables
120
40
240
0
0
Quantity
x1
x2
x3
s1
s2
120
x1
6
1
0
15/22
–3/11
1/22
40
x2
3
0
1
6/22
1/11
–2/11
zj
840
120
40
1,020/11 –320/11
–20/11
0 –1,620/11 –320/11
–20/11
cj – zj
0
Optimal 280
TaylMod-Aff.qxd
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Page 281
22. cj
Basic Variables
.05
.10
0
0
M
M
Quantity
x1
x2
s1
s2
A1
A2
M
A1
36
6
2
–1
0
1
0
M
A2
50
5
5
0
–1
0
1
86M
11M
7M
–M
–M
M
M
11M – .05
7M – .10
–M
–M
0
0
zj zj – cj
cj
Basic Variables
.05
.10
0
0
M
Quantity
x1
x2
s1
s2
A2
.05
x1
6
1
1/3
–1/6
0
0
M
A2
20
0
10/3
5/6
–1
1
zj
20M + .3
zj – cj cj
Basic Variables
.05
10M/3 + .02
5M/6 – .01
–M
M
0
10M/3 – .08
5M/6 – .01
–M
0
.05
.10
0
0
Quantity
x1
x2
s1
s2
.05
x1
4
1
0
–1/4 1/10
.10
x2
6
0
1
1/4 –3/10
zj
.80
.05
.10
.0125 –.025
0
0
.0125 –.025
.05
.10
0
0
Quantity
x1
x2
s1
s2
zj – cj cj
Basic Variables
.05
x1
10
1
1
0 –1/5
0
s1
24
0
4
1 –6/5
zj
.50
.05
.05
0 –.01
0
–.05
0 –.01
zj – cj Optimal
281
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Page 282
23. x2 24 20 16 12 Third tableau
8 4
0
4 First tableau
8
12 16 20 Fourth tableau Second tableau
24
28
32
x1
24. cj
Basic Variables
10
12
7
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
0
s1
300
20
15
10
1
0
0
0
s2
120
10
5
0
0
1
0
0
s3
40
1
0
2
0
0
1
zj
0
0
0
0
0
0
0
cj – zj
10
12
7
0
0
0
Basic Variables
10
12
7
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
cj
12
x2
20
4/3
1
2/3
1/15
0
0
0
s2
20
10/3
0
–10/3
–1/3
1
0
0
s3
40
1
0
2
0
0
1
zj
240
16
12
8
4/5
0
0
–6
0
–1
–4/5
0
0
cj – zj Optimal
282
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Page 283
25. cj
Basic Variables
6
2
12
0
0
Quantity
x1
x2
x3
s1
s2
0
s1
24
4
1
3
1
0
0
s2
30
2
6
3
0
1
zj
0
0
0
0
0
0
cj – zj
6
2
12
0
0
Basic Variables
6
2
12
0
0
Quantity
x1
x2
x3
s1
s2
cj
12
x3
8
4/3
1/3
1
1/3
0
0
s2
6
–2
5
0
–1
1
zj
96
16
4
12
4
0
–10
–2
0
–4
0
cj – zj Optimal
26. cj
Basic Variables
100
75
90
95
0
0
0
0
Quantity
x1
x2
x3
x4
s1
s2
s3
s4
0
s1
40
3
2
0
0
1
0
0
0
0
s2
25
0
0
4
1
0
1
0
0
0
s3
2,000
200
0
250
0
0
0
1
0
0
s4
2,200
100
0
0
200
0
0
0
1
zj
0
0
0
0
0
0
0
0
0
cj – zj
100
75
90
95
0
0
0
0
Basic Variables
100
75
90
95
0
0
0
0
Quantity
x1
x2
x3
x4
s1
s2
s3
s4
cj
0
s1
10
0
2 –3.75
0
1
0 –.015
0
0
s2
25
0
0
4
1
0
1
0
0
100
x1
10
1
0
1.25
0
0
0
.005
0
0
s4
1,200
0
0 –1.25
200
0
0
–.50
1
zj
1,000
100
0
1.25
0
0
0
.50
0
0
75
–35
95
0
0
–.50
0
cj – zj
(continued)
283
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8:50 PM
Basic Variables
Page 284
100
75
90
95
0
0
0
0
Quantity
x1
x2
x3
x4
s1
s2
s3
s4 0
0
s1
10
0
2 –3.75
0
1
0 –.015
0
s2
19
0
0
4.6
0
0
1
.002 –.005
100
x1
10
1
0
1.25
0
0
0
.005
0
95
x4
6
0
0 –.625
1
0
0 –.002
.005
zj
1,570
100
0
65.6
95
0
0
.475
0
75
24.4
0
0
0
100
75
90
95
0
0
0
0
Quantity
x1
x2
x3
x4
s1
s2
s3
s4
0 –.007
0
cj – zj cj
Basic Variables
.26
–.26 –.475
75
x2
5
0
1 –1.875
0
.50
0
s2
19
0
0
4.6
0
0
1
.002 –.005
100
x1
10
1
0
1.25
0
0
0
.005
0
95
x4
6
0
0 –.625
1
0
0 –.002
.005
zj
1,945
100
75
–75
95
37.5
0
.475
0
0
165
0 –37.5
0
100
75
90
95
0
0
0
0
x1
x2
x3
x4
s1
s2
s3
s4
cj – zj cj
Basic Variables
Quantity
–.30
.30 –.475
75
x2
12.7
0
1
0
0
.5
.405 –.006 –.002
90
x3
4.1
0
0
1
0
0
.216
100
x1
4.9
1
0
0
0
0 –.270
95
x4
8.6
0
0
0
1
zj
2,623
100
75
90
95
0
0
0
100
75
90
95
0
0
0
0
Quantity
x1
x2
x3
x4
s1
s2
s3
s4
0
0
cj – zj cj
Basic Variables
.001 –.001 .004
.001
0
.135 –.002
.004
37.5
35.7 –.211
.297
0 –37.5 –35.7
.211 –.297
75
x2
20
1.5
1
0
0
.50
0
90
x3
3.5
–.125
0
1
0
0
.25
0
s3
1,125
231.25
0
0
0
0 –62.5
1
.313
95
x4
11
.50
0
0
1
0
0
0
.005
zj
2,860
148.75
75
90
95
37.5
22.5
0
.36
–48.75
0
0
0 –37.5 –22.5
0
–.36
cj – zj Optimal
284
0 –.001
TaylMod-Aff.qxd
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Page 285
27. cj
Basic Variables
20
16
0
0
0
M
M
M
Quantity
x1
x2
s1
s2
s3
A1
A2
A3
M
A1
6
3
1
–1
0
0
1
0
0
M
A2
4
1
1
0
–1
0
0
1
0
M
A3
12
2
6
0
0
–1
0
0
1
22M
6M
8M
–M
–M
–M
M
M
M
8M – 16
–M
–M
–M
0
0
0
zj zj – cj
cj
Basic Variables
6M – 20
20
16
0
0
0
M
M
Quantity
x1
x2
s1
s2
s3
A1
A2
M
A1
4
8/3
0
–1
0
1/6
1
0
M
A2
2
2/3
0
0
–1
1/6
0
1
16
x2
2
1/3
1
0
0
–1/6
0
0
10M/3 + 16/3
16
–M
–M
M/3 – 8/3
M
M
10M/3 – 44/3
0
–M
–M
M/3 – 8/3
0
0
zj
32 + 6M
zj – cj
cj
Basic Variables
20
16
0
0
0
M
Quantity x1
x2
s1
s2
s3
A2
20
x1
3/2
1
0
–3/8
0
1/16
0
M
A2
1
0
0
1/4
–1
1/8
1
16
x2
3/2
0
1
1/8
0
–3/16
0
zj
M + 27
20
16
M/4 – 53/6
–M + 16/3
M/8 – 25/12
M
0
0
M/4 – 53/6
–M + 16/3
M/8 – 25/12
0
zj – cj
(continued)
285
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8:50 PM
Basic Variables
Page 286
20
16
0
0
0
Quantity
x1
x2
s1
s2
s3
20
x1
3
1
0
0
–3/2
1/4
0
s1
4
0
0
1
–4
1/2
16
x2
1
0
1
0
1/2
–1/4
zj
76
20
16
0
–22
1
zj – cj
0
0
0
–22
1
Basic Variables
20
16
0
0
0
Quantity
x1
x2
s1
s2
s3
cj
20
x1
1
1
0
–1/2
1/2
0
0
s3
8
0
0
2
–8
1
16
x2
3
0
1
1/2
–3/2
0
zj
68
20
16
–2
–14
0
0
0
–2
–14
0
zj – cj Optimal
28. x2 12 10 8 6 4
Fifth tableau Third tableau
2
Fourth tableau
Second tableau 0
2 4 First tableau
6
8
10
12
14
16
x1
29. Minimize Z = 8x1 + 2x2 + 7x3 + 0s1 + 0s2 + 0s3 – MA1 – MA2 – MA3 subject to 2x1 + 6x2 + x3 + A1 = 30 3x2 + 4x3 – s1 + A2 = 60 4x1 + x2 + 2x3 + s2 = 50 x1 + 2x2 – s3 + A3 = 20 x1, x2, x3 ≥ 0
286
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Page 287
30. Minimize Z = 40x1 + 55x2 + 30x3 + 0s1 + 0s2 + 0s3 + MA1 + MA2 + MA3 subject to x1 + 2x2 + 3x3 + s1 = 60 2x1 + x2 + x3 + A1 = 40 x1 + 3x2 + x3 – s2 + A2 = 50 5x2 –3x3 – s3 + A3 = 100 x1, x2, x3 ≥ 0
31. cj
Basic Variables
40
60
0
0
0
0
–M
–M
Quantity
x1
x2
s1
s2
s3
s4
A1
A2
0
s1
30
1
2
1
0
0
0
0
0
0
s2
72
4
4
0
1
0
0
0
0
–M
A1
5
1
0
0
0
–1
0
1
0
–M
A2
12
0
1
0
0
0
–1
0
1
–17M
–M
–M
0
0
M
M
–M
–M
0
0
–M
–M
0
0
zj cj – zj cj
Basic Variables
M + 40
M + 60
40
60
0
0
0
0
–M
Quantity
x1
x2
s1
s2
s3
s4
A1
0
s1
6
1
0
1
0
0
2
0
0
s2
24
4
0
0
1
0
4
0
–M
A1
5
1
0
0
0
–1
0
1
60
x2
12
0
1
0
0
0
–1
0
zj
–5M + 720
–M
60
0
0
M
–60
–M
0
0
0
–M
60
0
cj – zj cj
Basic Variables
M + 40 40
60
0
0
0
0
Quantity
x1
x2
s1
s2
s3
s4
0
s1
1
0
0
1
0
1
2
0
s2
4
0
0
0
1
4
4
40
x1
5
1
0
0
0
–1
0
60
x2
12
0
1
0
0
0
–1
zj
920
40
60
0
0
–40
–60
0
0
0
0
40
60
cj – zj
(continued)
287
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cj
8:50 PM
Basic Variables
Page 288
40
60
0
0
0
0
Quantity
x1
x2
s1
s2
s3
s4
0
s4
1/2
0
0
1/2
0
1/2
1
0
s2
2
0
0
–2
1
2
0
40
x1
5
1
0
0
0
–1
0
60
x2
25/2
0
1
1/2
0
1/2
0
zj
950
40
60
30
0
–10
0
cj – zj
0
0
–30
0
10
0
Basic Variables
40
60
0
0
0
0
Quantity
x1
x2
s1
s2
s3
s4
cj
0
s4
0
0
0
1
–1/4
0
1
0
s3
1
0
0
–4
1/2
1
0
40
x1
6
1
0
–1
1/2
0
0
60
x2
12
0
1
1
–1/4
0
0
zj
960
40
60
20
5
0
0
0
0
–20
–5
0
0
cj – zj
Tie
Optimal 32. cj
Basic Variables
1
5
0
0
0
–M
Quantity
x1
x2
s1
s2
s3
A1
–M
A1
25
5
5
–1
0
0
1
0
s2
16
2
4
0
1
0
0
0
s3
5
1
0
0
0
1
0
zj
–25M
–5M
–5M
M
0
0
–M
5M + 1
5M + 5
–M
0
0
0
cj – zj cj
Basic Variables
1
5
0
0
0
–M
Quantity
x1
x2
s1
s2
s3
A1
–M
A1
5
5/2
0
–1
–5/4
0
1
5
x2
4
1/2
1
0
1/4
0
0
0
s3
5
1
0
0
0
1
0
–5M + 20 –5M/2 + 5/2
5
M
–5M/4 + 5/4
0
–M
5M/2 – 3/2
0
–M
5M/4 – 5/4
0
0
zj cj – zj
(continued) 288
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cj
8:50 PM
Basic Variables
Page 289
1
5
0
0
0
Quantity
x1
x2
s1
s2
s3
1
x1
2
1
0
–2/5
–1/2
0
5
x2
3
0
1
1/5
1/2
0
0
s3
3
0
0
2/5
1/2
1
zj
17
1
5
3/5
2
0
0
0
–3/5
–2
0
cj – zj Optimal
33. cj
Basic Variables
3
6
0
0
0
0
M
Quantity
x1
x2
s1
s2
s3
s4
A1
0
s1
18
3
2
1
0
0
0
0
M
A1
5
1
1
0
–1
0
0
1
0
s3
4
1
0
0
0
1
0
0
0
s4
7
0
1
0
0
0
1
0
zj
5M
M
M
0
–M
0
0
M
zj – cj
M–3
M–6
0
–M
0
0
0
Basic Variables
3
6
0
0
0
0
M
Quantity
x1
x2
s1
s2
s3
s4
A1
cj
0
s1
6
0
2
1
0
–3
0
0
M
A1
1
0
1
0
–1
–1
0
1
3
x1
4
1
0
0
0
1
0
0
0
s4
7
0
1
0
0
0
1
0
zj
M + 12
3
M
0
–M
–M + 3
0
M
zj – cj
0
M–6
0
–M
–M + 3
0
0
Basic Variables
3
6
0
0
0
0
Quantity
x1
x2
s1
s2
s3
s4
cj
0
s1
4
0
0
1
2
–1
0
6
x2
1
0
1
0
–1
–1
0
3
x1
4
1
0
0
0
1
0
0
s4
6
0
0
0
1
1
1
zj
18
3
6
0
–6
–3
0
0
0
0
–6
–3
0
zj – cj Optimal
289
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8:50 PM
Page 290
34. cj
Basic Variables
10
5
0
0
–M
–M
Quantity
x1
x2
s1
s2
A1
A2
–M
A1
10
2
1
–1
0
1
0
–M
A2
4
0
1
0
0
0
1
0
s2
20
1
4
0
1
0
0
zj
–14M
–2M
–2M
M
0
–M
–M
2M + 10
2M + 5
–M
0
0
0
cj – zj cj
Basic Variables
10
5
0
0
–M
Quantity
x1
x2
s1
s2
A2
10
x1
5
1
1/2
–1/2
0
0
–M
A2
4
0
1
0
0
1
0
s2
15
0
7/2
1/2
1
0
zj
–4M + 50
10
–M + 5
–5
0
–M
5
0
0
cj
cj – zj
0
M
Basic Variables
10
5
0
0
Quantity
x1
x2
s1
s2
10
x1
3
1
0
–1/2
0
5
x2
4
0
1
0
0
0
s2
1
0
0
1/2
1
zj
50
10
5
–5
0
cj – zj
0
0
5
0
Basic Variables
10
5
0
0
Quantity
x1
x2
s1
s2
cj
10
x1
4
1
0
0
1
5
x2
4
0
1
0
0
0
s1
2
0
0
1
2
zj
60
10
5
0
10
0
0
0 –10
cj – zj Optimal
290
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Page 291
35. cj
Basic Variables
1
2
–1
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
0
s1
40
0
4
1
1
0
0
0
s2
20
1
–1
0
0
1
0
0
s3
60
2
4
3
0
0
1
zj
0
0
0
0
0
0
0
cj – zj
1
2
–1
0
0
0
Basic Variables
1
2
–1
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
cj
2
x2
10
0
1
1/4
1/4
0
0
0
s2
30
1
0
1/4
1/4
1
0
0
s3
20
2
0
2
–1
0
1
zj
20
0
2
1/2
1/2
0
0
cj – zj
1
0
–3/2
–1/2
0
0
Basic Variables
1
2
–1
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
cj
2
x2
10
0
1
1/4
1/4
0
0
0
s2
30
0
0
–3/4
3/4
1
–1/2
1
x1
10
1
0
1
–1/2
0
1/2
zj
30
1
2
3/2
0
0
1/2
0
0
–5/2
0
0
–1/2
1
2
–1
0
0
0
x1
x2
x3
s1
s2
s3
cj – zj Multiple optimum solution Alternate solution: cj
Basic Variables
Quantity
2
x2
10/3
0
1
1/2
0
–1/3
1/6
0
s1
80/3
0
0
–1
1
4/3
–2/3
1
x1
70/3
1
0
1/2
0
2/3
1/6
zj
30
1
2
3/2
0
0
1/2
0
0
–5/2
0
0
–1/2
cj – zj
291
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Page 292
36. cj
Basic Variables
1
2
2
0
0
–M
–M
Quantity
x1
x2
x3
s1
s2
A1
A2
0
s1
12
1
1
2
1
0
0
0
–M
A1
20
2
1
5
0
0
1
0
–M
A2
8
1
1
–1
0
–1
0
1
–28M
–3M
–2M
–4M
0
M
–M
–M
0
–M
0
0
zj
cj
cj – zj
1 + 3M
2 + 2M 2 + 4M
Basic Variables
1
2
2
0
0
–M
Quantity
x1
x2
x3
s1
s2
A2
0
s1
4
1/5
3/5
0
1
0
0
2
x3
4
2/5
1/5
1
0
0
0
–M
A2
12
7/5
6/5
0
0
–1
1
zj
cj
–12M + 8 –4/5 – 7M/5
2/5 – 6M/5
2
0
M
–M
cj – zj
1/5 + 7M/5
8/5 + 6M/5
0
0
–M
0
Basic Variables
1
2
2
0
0
x1
x2
x3
s1
s2
Quantity
0
s1
16/7
0
3/7
0
1
1/7
2
x3
4/7
0
–1/7
1
0
2/7
1
x1
60/7
1
6/7
0
0
–5/7
zj
68/7
1
4/7
2
0
–1/7
cj – zj
0
10/7
0
0
1/7
Basic Variables
1
2
2
0
0
x1
x2
x3
s1
s2
cj
Quantity
2
x2
16/3
0
1
0
7/3
1/3
2
x3
4/3
0
0
1
1/3
1/3
1
x1
4
1
0
0
–2
–1
zj
52/3
1
2
2
10/3
1/3
0
0
0
–10/3
–1/3
cj – zj Optimal
292
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Page 293
37. cj
Basic Variables
400
350
450
0
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
s4
0
s1
120
2
3
2
1
0
0
0
0
s2
160
4
3
1
0
1
0
0
0
s3
100
3
2
4
0
0
1
0
0
s4
40
1
1
1
0
0
0
1
zj
0
0
0
0
0
0
0
0
cj – zj
400
350
450
0
0
0
0
Basic Variables
400
350
450
0
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
s4
cj
0
s1
70
1/2
2
0
1
0
–1/2
0
0
s2
135
13/4
5/2
0
0
1
–1/4
0
450
x3
25
3/4
1/2
1
0
0
1/4
0
0
s4
15
–1/4
1/2
0
0
0
–1/4
1
zj
11,250
1,350/4 450/2
450
0
0 450/4
0
250/4 250/2
0
0
0 –450/4
0
cj – zj cj
Basic Variables
Quantity
400
350
450
0
0
0
0
x1
x2
x3
s1
s2
s3
s4
0
s1
10
–1/2
0
0
1
0
1/2
–4
0
s2
60
2
0
0
0
1
1
–5
450
x3
10
1/2
0
1
0
0
1/2
–1
350
x2
30
1/2
1
0
0
0
–1/2
2
zj
15,000
400
350
450
0
0
50
250
0
0
0
0
0
–50
–250
400
350
450
0
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
s4
cj – zj Multiple optimum solution at x1 Alternate solution: cj
Basic Variables 0
s1
20
0
0
1
1
0
1
–5
0
s2
20
0
0
–4
0
1
–1
–1
400
x1
20
1
0
2
0
0
1
–2
350
x2
20
0
1
–1
0
0
–1
3
zj
15,000
400
350
450
0
0
50
250
0
0
0
0
0
–50
–250
cj – zj
293
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Page 294
38. (a) x2 5
Infeasible-no common solution space 4 3 2 1
0
1
2
3
4
5
6
7
x1
(b) cj
Basic Variables
3
2
0
0
–M
Quantity
x1
x2
s1
s2
A1
0
s1
1
1
1
1
0
0
–M
A1
2
1
1
0
–1
1
–2M
–M
–M
0
M
–M
cj – zj
M+3
M+2
0
–M
0
Basic Variables
3
2
0
0
–M
Quantity
x1
x2
s1
s2
A1
zj
cj
3
x1
1
1
1
1
0
0
–M
A1
1
0
0
–1
–1
1
zj
3–M
3
3
M
M
–M
0
–1
–M
–M
0
cj – zj Infeasible solution
39. (a) x2 5 4 3
Unbounded solution
2 1
-4
-3
-2
-1
Z 0
1
2
3
4
5
6
x1
294
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Page 295
(b) cj
Basic Variables
1
1
0
0
Quantity
x1
x2
s1
s2
0
s1
1
–1
1
1
0
0
s2
4
–1
2
0
1
zj
0
0
0
0
0
1
1
0
0
cj – zj
Tie for entering variable; if x1 is chosen, the solution is unbounded. Select x2 arbitrarily. cj
Basic Variables
1
1
0
0
Quantity
x1
x2
s1
s2
1
x2
1
–1
1
1
0
0
s2
3
1
0
–1
1
zj
1
–1
1
1
0
cj – zj
2
0
–1
0
Basic Variables
1
1
0
0
Quantity
x1
x2
s1
s2
cj
1
x2
4
0
1
0
1
1
x1
3
1
0
–1
1
zj
7
1
1
–1
2
0
0
1
–2
cj – zj Unbounded; no pivot row available
40. cj
Basic Variables
7
5
5
0
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
s4
0
s1
25
1
1
1
1
0
0
0
0
s2
40
2
1
1
0
1
0
0
0
s3
25
1
1
0
0
0
1
0
0
s4
6
0
0
1
0
0
0
1
zj
0
0
0
0
0
0
0
0
7
5
5
0
0
0
0
cj – zj
(continued)
295
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cj
8:50 PM
Basic Variables
Page 296
7
5
5
0
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
s4
0
s1
5
0
1/2
1/2
1
–1/2
0
0
7
x1
20
1
1/2
1/2
0
1/2
0
0
0
s3
5
0
1/2
–1/2
0
–1/2
1
0
0
s4
6
0
0
0
0
0
0
1
zj
140
7
7/2
7/2
0
7/2
0
0
0
3/2
3/2
0
–7/2
0
0
cj – zj
Tie cj
Basic Variables
7
5
5
0
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
s4
5
x2
10
0
1
1
2
–1
0
0
7
x1
15
1
0
0
–1
1
0
0
0
s3
0
0
0
–1
–1
0
1
0
0
s4
6
0
0
1
0
0
0
1
zj
155
7
5
5
3
2
0
0
0
0
0
–3
–2
0
0
cj – zj Multiple optimum
(continued)
Alternate Solution: cj
Basic Variables
7
5
5
0
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
s4
5
x2
4
0
1
0
2
–1
0
–1
7
x1
15
1
0
0
–1
1
0
0
0
s3
6
0
0
0
–1
0
1
1
5
x3
6
0
0
1
0
0
0
1
zj
155
7
5
5
3
2
0
0
0
0
0
–3
–2
0
0
cj – zj
296
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Page 297
41. cj
Basic Variables
15
25
0
0
0
M
Quantity
x1
x2
s1
s2
s3
A1
M A2
M
A1
12
3
4
–1
0
0
1
0
M
A2
6
2
1
0
–1
0
0
1
0
s3
9
3
2
0
0
1
0
0
zj
18M
5M
5M
–M
–M
0
M
M
5M – 25
–M
–M
0
0
zj – cj cj
Basic Variables
5M – 15
0
15
25
0
0
0
M
Quantity
x1
x2
s1
s2
s3
A1
M
A1
3
0
5/2
–1
1
0
1
15
x1
3
1
1/2
0
0
0
0
0
s3
0
0
1/2
0
0
1
0
15
5M/2 + 15/2
–M
3M/2 – 15/2
0
M
0
5M/2 – 35/2
–M
3M/2 – 15/2
0
0
zj
3M + 45
zj – cj cj
Basic Variables
15
25
0
0
0
M
Quantity
x1
x2
s1
s2
s3
A1
M
A1
3
0
0
–1
–6
–5
1
15
x1
3
1
0
0
–2
–1
0
25
x2
0
0
1
0
3
2
0
15
15
–M
–6M +45
–5M +45
M
0
0
–M
–6M +45
–5M +45
0
zj
3M + 45
zj – cj Infeasible solution
42.
a). minimize Zd = 90y1 + 60y2 subject to y1 + 2y2 ≥ 6 4y1 + 2y2 ≥ 10 y1, y2 ≥ 0 b) y1 = the marginal value of one additional lb of brass = $1.33 y2 = the marginal value of one additional hr of labor = $2.33
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c) c1, basic: cj
6+∆
10
0
0
Quantity
x1
x2
s1
s2
x2
20
0
1
1/3
–1/6
x1
10
1
0
–1/3
2/3
6+∆
10
4/3 – ∆/3
7/3 + 2∆/3
0
0
–4/3 + ∆/3
–7/3 – 2∆/3
Basic Variables
10 6+∆
zj
260 + 10∆
cj – zj
Since c2 = 10 + ∆; ∆ = c2 – 10. Thus
Solving for the cj – zj inequalities:
c2 – 10 ≤ 14 c2 ≤ 24
–4/3 + ∆/3 ≤ 0 ∆/3 ≤ 4/3 ∆≤4
Summarizing, 6 ≤ c2 ≤ 24.
Since c1 = 6 + ∆; ∆ = c1 – 6. Thus c1 – 6 ≤ 4 c1 ≤ 10 –7/3 – 2∆/3 ≤ 0 –2∆/3 ≤ 7/3 –2∆ ≤ 7 ∆ ≥ –7/2
d) q1: 20 + ∆/3 ≥ 0 x1: ∆/3 ≥ –20 ∆ ≥ –60
x2:
10 – ∆/3 ≥ 0 –∆/3 ≥ –10 ∆ ≤ 30
Therefore, –60 ≤ ∆ ≤ 30. Since q1 = 90 + ∆ ∆ = q1 – 90 –60 ≤ q1 – 90 ≤ 30 30 ≤ q1 ≤ 120
Since c1 = 6 + ∆; ∆ = c1 – 6. Thus c1 – 6 ≥ –7/2 c1 ≥ 5/2 Summarizing, 5/2 ≤ c1 ≤ 10. c2, basic: 6
10 + ∆
0
0
Quantity
x1
x2
s1
s2
10 + ∆ x2
20
0
1
1/3
–1/6
6
10
1
0
–1/3
2/3
6
10 + ∆
4/3 + ∆/3
7/3 – ∆/6
0
0
–4/3 – ∆/3
–7/3 + ∆/6
cj
Basic Variables
x1 zj
280 + 20∆
cj – zj
Solving for the cj – zj inequalities:
q2:
–4/3 – ∆/3 ≤ 0 –∆/3 ≤ 4/3 ∆ ≥ –4
x2:
Since c2 = 10 + ∆; ∆ = c2 – 10. Thus
20 – ∆/6 ≥ 0 – ∆/6 ≥ –20 ∆ ≤ 120
x1: 10 + 2∆/3 ≥ 0 2∆/3 ≥ –10 ∆ ≥ –15
Therefore, –15 ≤ ∆ ≤ 120. Since
c2 – 10 ≥ –4 c2 ≥ 6 –7/3 + ∆/6 ≤ 0 ∆/6 ≤ 7/3 ∆ ≤ 14
q2 = 60 + ∆ ∆ = q2 – 60 –15 ≤ q2 – 60 ≤ 120 45 ≤ q2 ≤ 180
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e) The marginal value of 1 hr of labor is $2.33. From part d, the sensitivity range for q2, labor, is 45 ≤ q2 ≤ 180. Thus, the company would purchase up to 180 hr at the marginal value price.
c)
x2 60 50 40
43.
a) Minimize Zd = 500y1 + 800y2 subject to
30
10y1 + 34y2 ≥ 200 50y1 + 20y2 ≥ 300 y1, y2 ≥ 0
20
y1 = $4.13 = the marginal value of one additional lb of chili beans; y2 = $4.67 = the marginal value of one additional lb of ground beef. b)
50 40 30 20
A B 10
20 C 30
40
50
60
10
20 C 30
40
50
60
x1
The constraint line for chili beans rotates, creating a new, smaller solution space, and the optimal solution shifts from point B to point B´ where x1 = 21.43 and x2 = 3.57.
60
0
B B´ 0
x2
10
A
10
x1
Point C must become the optimal solution for x2 = 0; therefore the slope of the objective function must be greater than the slope of the constraint for ground beef, –34/20. Solving the following for the profit, p, of Razorback chili yields –p/300 = –34/20 p = $510 Thus, if the profit of Razorback chili is greater than $510, no Longhorn chili will be produced. The new optimal solution will be x1 = 23.5 and x2 = 0.
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d) c1, basic: cj
Basic Variables
200 + ∆
300
0
0
Quantity
x1
x2
s1
s2
300
x2
6
0
1
17/750
–1/150
200 + ∆
x1
20
1
0
–1/75
1/30
zj
5,800 + 20∆
200 + ∆ 300 310/75 – ∆/75
cj – zj
70/15 + ∆/30
0 –310/75 + ∆/75 –70/15 – ∆/30
0
Since c2 = 300 + ∆; ∆ = c2 – 300. Thus
Solving for the cj – zj inequalities:
c2 – 300 ≤ 700 c2 ≤ 1,000
–310/75 + ∆/75 ≤ 0 ∆/75 ≤ 3310/75 ∆ ≤ 310
Summarizing, 117.65 ≤ c2 ≤ 1,000.
Since c1 = 200 + ∆; ∆ = c1 – 200. Thus c1 – 200 ≤ 310 c1 ≤ 510 –70/15 – ∆/30 ≤ 0 –∆/30 ≤ 70/15 ∆ ≥ –140
e) q1: x2: 6 + 17∆/750 ≥ 0 x1: 20 – ∆/75 ≥ 0 17∆/750 ≥ –6 –∆/75 ≥ –20 ∆ ≥ –264.7 ∆ ≤ 1,500 Therefore, –264.7 ≤ ∆ ≤1,500. Since
Since c1 = 200 + ∆; ∆ = c1 – 200. Thus
q1 = 500 + ∆ ∆ = q1 – 500 –264.7 ≤ q1 – 500 ≤ 1,500 235.3 ≤ q1 ≤ 2,000
c1 – 200 ≥ –140 c1 ≥ 60 Summarizing, 60 ≤ c1 ≤ 510. c2, basic:
cj
Basic Variables
200
300 + ∆
0
0
Quantity
x1
x2
s1
s2
300 + ∆
x2
6
0
1
17/750
–1/150
200
x1
20
1
0
–1/75
1/30
zj
5,800 + 6∆
200
cj – zj
300 + ∆
0
0
310/75 + 17∆/750
70/15 – ∆/150
–310/75 – 17∆/750 –70/15 + ∆/150
Solving for the cj – zj inequalities:
q2:
–310/75 – 17∆/750 ≤ 0 –17∆/750 ≤ 310/75 ∆ ≥ –182.35
x2:
Since c2 = 300 + ∆; ∆ = c2 – 300. Thus
Therefore, –600 ≤ ∆ ≤ 900. Since
c2 – 300 ≥ –182.35 c2 ≥ 117.65 –70/15 + ∆/150 ≤ 0 ∆/150 ≤ 70/15 ∆ ≤ 700
6 – ∆/150 ≥ 0 x1: 20 + ∆/30 ≥ 0 – ∆/150 ≥ –6 ∆/30 ≥ –20 ∆ ≤ 900 ∆ ≥ –600 q2 = 800 + ∆ ∆ = q2 – 800 –600 ≤ q2 – 800 ≤ 900 200 ≤ q2 ≤ 1,700
300
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f) The marginal value of 1 lb of chili beans is $4.13. The sensitivity range for q1, chili beans, is 235.3 ≤ q1q ≤ 2,000. Thus, the company would purchase up to 2,000 lb at the marginal value price. g) Groundbeef h) No effect 44. a) Minimize Zd = 60y1 + 40y2 subject to 12y1 + 4y2 ≥ 9 4y1 + 8y2 ≥ 7 y1, y2 ≥ 0 b) y1 = the marginal value of one additional hr of process 1; y2 = the marginal value of one additional hr of process 2 For the s1 column, the cj – zj value of $.55 is the marginal value of 1 hr of process 1 production time. For the s2 column, the cj – zj value of $0.60 is the marginal value of 1 hr of process 2 production time.
c) cj, basic: cj
Basic Variables
9+∆
7
0
0
Quantity
x1
x2
s1
s2
9+∆
x1
4
1
0
1/10
–1/20
7
x2
3
0
1
–1/20
3/20
zj
57 + 4∆
9+∆
7
cj – zj
0
11/20 – ∆/10
12/20 + ∆/20
0 –11/20 – ∆/10 –12/20 + ∆/20
301
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Since c2 = 7 + ∆; ∆ = c2 – 7. Thus
Solving for the cj – zj inequalities:
c2 – 7 ≥ –4 c2 ≥ 3
–11/20 – ∆/10 ≤ 0 –∆/10 ≤ 11/20 –∆ ≤ 11/2 ∆ ≥ –11/2
Summarizing, 3 ≤ c2 ≤ 18.
Since c1 = 9 + ∆ , ∆ = c1 – 9. Thus
d) q1:
c1 – 9 ≥ 11/2 c1 ≥ 7/2 –12/20 + ∆/20 ≤ 0 ∆/20 ≤ 12/20 ∆ ≤ 12
x1: 4 + ∆/10 ≥ 0 ∆/10 ≥ –4 –∆ ≥ –80 ∆ ≥ –40
x2: 3 – ∆/20 ≥ 0 –∆/20 ≥ –3 –∆ ≥ –60 ∆ ≤ 60
Therefore, –40 ≤ ∆ ≤60. Since
Since c1 = 9 + ∆ , ∆ = c1 – 9. Thus
q1 = 60 + ∆ ∆ = q1 – 60 –40 ≤ q1 – 60 ≤ 60 20 ≤ q1 ≤ 120
c1 – 9 ≤ 12 c1 ≤ 12 Summarizing, 7/2 ≤ c1 ≤ 12. c2, basic: cj
Basic Variables
9
7+∆
0
0
Quantity
x1
x2
s1
s2
9
x1
4
1
0
1/10
–1/20
7+∆
x2
3
0
1
–1/20
3/20
zj
57 + 3∆
9
7+∆
11/20 – ∆/20
12/20 + 3∆/20
0
0
–11/20 + ∆/20
–12/20 – 3∆/20
cj – zj
Solving for the cj – zj inequalities:
q2:
–11/20 + ∆/20 ≤ 0 ∆/20 ≤ 11/20 ∆ ≤ 11
x1: 4 – ∆/20 ≥ 0 –∆/20 ≥ –4 –∆ ≥ –80 ∆ ≤ 80
Since c2 = 7 + ∆; ∆ = c2 – 7. Thus
x2: 3 +3 ∆/20 ≥ 0 3∆/20 ≥ –3 ∆ ≥ –20
Therefore, –20 ≤ ∆ ≤ 80. Since
c2 – 7 ≤ 11 c2 ≤ 18 –12/20 – 3∆/20 ≤ 0 –3∆/20 ≤ 12/20 ∆ ≥ –4
q2 = 40 + ∆ ∆ = q2 – 40 –20 ≤ q2 – 40 ≤ 80 20 ≤ q2 ≤ 120
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e) The marginal value of 1 hr of process 1 production time is $.55. The sensitivity range for q1, production hours, is 20 ≤ q1 ≤ 120. Thus, the company would purchase up to 120 hr at the marginal value price.
Point A must become the optimal solution for x1 = 0; therefore the slope of the objective function must be less than the slope of the constraint for labor, –2/5. Solving the following for the profit, p, of coffee tables gives –200/p = –2/5 p = $500
45. a) Minimize Zd = 180y1 + 135y2 subject to
Thus, if the profit for coffee tables is greater than $500, no end tables will be produced. The new optimal solution will be x1 = 0 and x2 = 36.
2y1 + 3y2 ≥ 200 5y1 + 3y2 ≥ 300 y1, y2 ≥ 0 d)
b) y1 = $33.33 = the marginal value of an additional hr of labor; y2 = $44.44 = the marginal value of an additional bd. ft. of wood
x2 60
c)
50
x2
40
A
60
30
B
B´
50 20 40 10
A
B
30
C 0
20 10
C 0
10
20
30
40
50
60
70
80
90
100
x1
303
10
20
30
40
50
C´ 60
70
80
90
100
The constraint line for wood moves outward, creating a new solution space, and the optimal solution point shifts from point B to point B´ where x1 = 31.67 and x2 = 23.33.
x1
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e) c1, basic: cj
Basic Variables
200 + ∆
300
0
0
Quantity
x1
x2
s1
s2
300
x2
30
0
1
1/3
–2/9
200 + ∆
x1
15
1
0
–1/3
5/9
zj
12,000 + 15∆
200 + ∆ 300
cj – zj
0
0
100/3 – ∆/3
400/9 + 5∆/9
–100/3 + ∆/3 –400/9 – 5∆/9 Since c2 = 300 + ∆; ∆ = c2 – 300. Thus
Solving for the cj – zj inequalities: –100/3 + ∆/3 ≤ 0 ∆/3 ≤ 100/3 ∆ ≤ 100
c2 – 300 ≤ 200 c2 ≤ 500 Summarizing, 200 ≤ c2 ≤ 500.
Since c1 = 200 + ∆; ∆ = c1 – 200. Thus c1 – 200 ≤ 100 c1 ≤ 300 –400/9 – 5∆/9 ≤ 0 –5∆/9 ≤ 400/9 –∆ ≤ 80 ∆ ≥ –80
f) q1: x2: 30 + ∆/3 ≥ 0 ∆/3 ≥ –30 ∆ ≥ –90
x1: 15 – ∆/3 ≥ 0 –∆/3 ≥ –15 –∆ ≤ –45 ∆ ≤ 45
Therefore, –90 ≤ ∆ ≤ 45. Since
Since c1 = 200 + ∆; ∆ = c1 – 200. Thus
q1 = 180 + ∆ ∆ = q1 – 180 –90 ≤ q1 – 180 ≤ 45 90 ≤ q1 ≤ 225
c1 – 200 ≥ –80 c1 ≥ 120 Summarizing, 120 ≤ c1 ≤ 300. c2, basic: cj
Basic Variables
200
300 + ∆
0
0
Quantity
x1
x2
s1
s2
300 + ∆
x2
30
0
1
1/3
–2/9
200
x1
15
1
0
–1/3
5/9
zj
12,000 + 300∆
cj – zj
200
300 + ∆ 100/3 + ∆/3
400/9 – 2∆ /9
–100/3 – ∆/3
–400/9 + 2∆ /9
0
0
Solving for the cj – zj inequalities:
q2:
–100/3 – ∆/3 ≤ 0 –∆/3 ≤ 100/3 –∆ ≤ 100 ∆ ≥ –100
x2: 30 – 2∆/9 ≥ 0 x1: 15 + 5∆/9 ≥ 0 –2∆/9 ≥ –30 5∆/9 ≥ –15 –∆ ≥ –135 ∆ ≥ –27 ∆ ≤ 135
Since c2 = 300 + ∆; ∆ = c2 – 300. Thus
Therefore, –27 ≤ ∆ ≤ 135. Since
c2 – 300 ≥ –100 c2 ≥ 200 –400/9 + 2∆/9 ≤ 0 2∆/9 ≤ 400/9 ∆ ≤ 200
q2 = 135 + ∆ ∆ = q2 – 135 –27 ≤ q2 – 135 ≤ 135 108 ≤ q2 ≤ 270
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g) The marginal value of 1 lb of wood is $44.44. From part f, the sensitivity range for q2, wood, is 108 ≤ q2 ≤ 270. Thus, the company would purchase up to 270 bd. ft. of wood at the marginal value price. g) The marginal value of labor is $33.33 and the marginal value of wood is $44.44; thus, wood should be purchased. 46. a) Minimize Zd = 19y1 + 14y2 + 20y3 subject to 2y1 + y2 + y3 ≥ 70 y1 + y2 + 2y3 ≥ 80 y1, y2, y3 ≥ 0 b) y1 =$20 = the marginal value of an additional hr of production time; y2 =$0 = the marginal value of an additional lb of steel; y3 = $30 = the marginal value of an additional ft of wire
c) c1, basic: 70 + ∆
80
0
0
0
Quantity
x1
x2
s1
s2
s3
x1
6
1
0
2/3
0
–1/3
0
s2
1
0
0
–1/3
1
–1/3
80
x2
7
0
1
–1/3
0
2/3
zj
980 + 6∆
70 + ∆
80
20 + 2∆/3
0
30 – ∆/3
0
0
–20 – 2∆/3
0
–30 + ∆/3
cj
Basic Variables
70 + ∆
cj – zj
Solving for the cj – zj inequalities:
Solving for the cj – zj inequalities:
–20 – 2∆/3 ≤ 0 –2∆/3 ≤ 20 –∆ ≤ 30 ∆ ≥ –30
–20 + ∆/3 ≤ 0 ∆/3 ≤ 20 ∆ ≥ 60 Since c2 = 80 + ∆; ∆ = c2 – 80. Thus
Since c1 = 70 + ∆; ∆ = c1 – 70. Thus
c2 – 80 ≤ 60 c2 ≤ 140 –30 – 2∆/3 ≤ 0 –2∆/3 ≤ 30 –∆ ≤ 45 ∆ ≥ –45
c1 – 70 ≥ –30 c1 ≥ 40 –30 + ∆/3 ≤ 0 ∆/3 ≤ 30 ∆ ≤ 90 Since c1 = 70 + ∆; ∆ = c1 – 70. Thus
Since c2 = 80 + ∆; ∆ = c2 – 80. Thus
c1 – 70 ≤ 90 c1 ≤ 160
c2 – 80 ≥ –45 c2 ≥ 35
Summarizing, 40 ≤ c1 ≤ 160.
Summarizing, 35 ≤ c2 ≤ 140.
c2, basic: 305
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Basic Variables
Page 306
70
80 + ∆
0
0
0
Quantity
x1
x2
s1
s2
s3
70
x1
6
1
0
2/3
0
–1/3
0
s2
1
0
0
–1/3
1
–1/3
x2
7
0
1
–1/3
0
2/3
zj
980 + 7∆
70
80 + ∆
20 – ∆/3
0
30 + 2∆/3
0
0
–20 + ∆/3
0
–30 – 2∆/3
80 + ∆
cj – zj d) q1: x1:
6 + 2∆/3 ≥ 0 2∆/3 ≥ –6 ∆ ≥ –9
7 – ∆/3 ≥ 0 –∆/3 ≥ –7 –∆ ≥ –21 ∆ ≤ 21
x2:
s2:
1 – ∆/3 ≥ 0 –∆/3 ≥ –1 –∆ ≥ –3 ∆≤3
Therefore, –9 ≤ ∆ ≤ 3. Since q1 = 19 + ∆ ∆ = q1 – 19 –9 ≤ q1 – 19 ≤ 3 10 ≤ q1 ≤ 22 q2: x1:
6 + 0∆ ≥ 0
7 + 0∆ ≥ 0
x2:
s2:
1+∆≥0 ∆ ≥ –1
s2:
1 – ∆/3 ≥ 0 –∆/3 ≥ –1 –∆ ≥ –3 ∆≤3
Therefore, ∆ ≥ –1. Since q2 = 14 + ∆ ∆ = q2 – 14 q2 – 14 ≥ –1 q2 ≥ 13 q3: x1:
6 – ∆/3 ≥ 0 –∆/3 ≥ –6 –∆ ≥ –18 ∆ ≤ 18
x2:
7 + 2∆/3 ≥ 0 2∆/3 ≥ –7 ∆ ≥ –21/2
Therefore, –21/2 ≤ ∆ ≤ 3. Since q3 = 20 + ∆ ∆ = q3 – 20
–21/2 ≤ q3 – 20 ≤ 3 19/2 ≤ q3 ≤ 23
e) The sensitivity range for production hours is 10 ≤ q1 ≤ 22. Since 25 hr exceeds the upper limit of the range, it would change the optimal solution.
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47. a) Minimize Zd = 64y1 + 50y2 + 120y3 + 7y4 + 7y5 subject to 4y1 + 5y2 + 15y3 + y4 ≥ 9 8y1 + 5y2 + 8y3 + y5 ≥ 12 y1, y2, y3, y4, y5 ≥ 0 b) y1 = $.75 = the marginal value of one additional hr of labor for process 1; y2 = $1.20 = the marginal value of one additional hr of labor for process 2; y3, y4, y5 = $0; these resources have no value since there were units available which were not used.
c) c1, basic: Solving for the cj – zj inequalities: –3/4 + ∆/4 ≤ 0 ∆/4 ≤ 3/4 ∆≤3
–6/5 – 2∆/5 ≤ 0 –2∆/5 ≤ 6/5 ∆ ≥ –3
Since c1 = 9 + ∆; ∆ = c1 – 9. Thus –3 ≤ ∆ ≤ 3 –3 ≤ c1 – 9 ≤ 3 6 ≤ c1 ≤ 12 c2, basic: Solving for the cj – zj inequalities: –3/4 + ∆/4 ≤ 0 –∆/4 ≤ 3/4 ∆ ≥ –3
–6/5 + ∆/5 ≤ 0 ∆/5 ≤ 6/5 ∆≤6
Since c2 = 12 + ∆; ∆ = c1 – 12. Thus –3 ≤ c2 – 12 ≤ 6 9 ≤ c2 ≤ 18
d) q1: x1:
4 – ∆/4 ≥ 0 –∆/4 ≥ –4 ∆ ≤ 16
s5:
1 – ∆/4 ≥ 0 –∆/4 ≥ –1 ∆≤4
s4:
3 + ∆/4 ≥ 0 ∆/4 ≥ –3 ∆ ≥ –12
x2:
6 + ∆/4 ≥ 0 ∆/4 ≥ –6 ∆ ≥ –24
s3:
307
12 + 7∆/4 ≥ 0 7∆/4 ≥ –12 ∆ ≥ –6.86
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Summarizing, –24 < –12 < –6.86 ≤ ∆ ≤ 4 ≤ 16 and, therefore, –6.86 ≤ ∆ ≤ 4 Since q1 = 64 + ∆; ∆ = q1 – 64. Therefore, –6.86 ≤ ∆ q1 – 64 ≤ 4 57.14 ≤ q1 ≤ 68 e) q3: x1:
4 + 0∆ ≥ 0 0∆ ≥ –4 ∆≤∞
s4:
3 + 0∆ ≥ 0 ∆≤∞
s5:
1 + 0∆ ≥ 0 ∆≤∞
s3:
6 + 0∆ ≥ 0 ∆≤∞ –12 ≤ ∆ ≤ ∞ x2:
Since q3 = 120 + ∆; ∆ = q3 – 120. Therefore, –12 ≤ ∆ ≤ ∞ –12 ≤ q3 – 120 ≤ ∞ 108 ≤ q3 ≤ ∞ Since 100 pounds is less than the lower limit of the range, the optimal solution mix will change. s2 enters the solution and s3 leaves. The new solution is, x1 = 3.27 s2 = 1.82 s5 = 0.64 s4 = 3.73 x2 = 6.36 Z = 105.82 48. a) Minimize Zd = 120y1 + 160y2 + 100y3 + 40y4 subject to 2y1 + 4y2 + 3y3 + y4 ≥ 40 3y1 + 3y2 + 2y3 + y4 ≥ 35 2y1 + y2 + 4y3 + y4 ≥ 45 y1, y2, y3, y4 ≥ 0 b) y1, y2 = 0; y3 = $5 = the marginal value of 1 hr of operation 3 time; y4 = $25 = the marginal value of 1 ft2 of storage space c) It does not have an effect. In the alternate solution the dual values remain the same, i.e., y3 = $5 and y4 = $25.
308
12 + ∆ ≥ 0 ∆ ≥ –12
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d) c2, basic: cj
Basic Variables
Quantity
40
35 + ∆
45
0
0
0
0
x1
x2
x3
s1
s2
s3
s4
0
s1
10
–1/2
0
0
1
0
1/2
–4
0
s2
60
2
0
0
0
1
1
–5
45
x3
10
1/2
0
1
0
0
1/2
–1
x2
30
1/2
1
0
0
0
–1/2
2
zj
1,500 + 35∆
45
0
0
5 – ∆/2
0
0
0
35 + ∆
cj – zj
40 + ∆/2 35 + ∆ –∆/2
0
25 + 2∆
–5 + ∆/2 –25 – 2∆
Since c2 = 35 + ∆; ∆ = c2 – 35. Thus
Solving for the cj – zj inequalities: –∆/2 ≤ 0 ∆≥0
c2 – 35 ≥ –12.5 c2 ≥ 22.5
Since c2 = 35 + ∆; ∆ = c2 – 35. Thus
Summarizing, 35 ≤ c2 ≤ 45.
c2 – 35 ≥ 0 c2 ≥ 35 –5 + ∆/2 ≤ 0 ∆/2 ≤ 5 ∆ ≤ 10
e) q4:
Since c2 = 35 + ∆; ∆ = c2 – 35. Thus c2 – 35 ≤ 10 c2 ≤ 45 –25 – 2∆ ≤ 0 –2∆ ≤ 25 ∆ ≥ –12.5
s1:
10 – 4∆ ≥ 0 –4∆ ≥ –10 ∆ ≤ 5/2
s2:
60 – 5∆ ≥ 0 –5∆ ≥ –60 ∆ ≤ 12
x3:
10 – ∆ ≥ 10 –∆ ≥ –10 ∆ ≤ 10
x2:
30 + 2∆ ≥ 0 2∆ ≥ –30 ∆ ≥ –15
Therefore, –15 ≤ ∆ ≤ 5/2. Since q4 = 40 + ∆ ∆ = q4 – 40 –15 ≤ q4 – 40 ≤ 5/2 25 ≤ q4 ≤ 42.5
f) The marginal value of 1 ft2 of storage is $25. From part e, the sensitivity range for q4 is 25 ≤ q4 ≤ 42.5. Thus, the company would purchase up to 42.5 ft2 of storage space at the marginal value price. 49. a) Maximize Zd = 20y1 + 30y2 + 12y3 subject to 4y1 + 12y2 + 3y3 ≤ .03 5y1 + 3y2 + 2y3 ≤ .02 y1, y2, y3 ≥ 0 b) y1 = $0 = marginal value of 1 mg of protein; y2 = $0 = marginal value of 1 mg of iron; y3 = $.01 = marginal value of 1 mg of carbohydrate (i.e., if one less mg of carbohydrate was required, it would be worth $.01 to the dietitian)
309
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c) c1, basic: cj
Basic Variables
.03 + ∆
.02
0
0
0
Quantity
x1
x2
s1
s2
s3
.02
x2
3.6
0
1
0
.20
–.80
.03 + ∆
x1
1.6
1
0
0
–.13
.20
s1
4.4
0
0
1
.47
–3.2
zj
.12 + 1.6∆
.02
0
0 – .133∆ –.01 + .2∆
0
0
0 – .133∆ –.01 + .2∆
0
.03 + ∆
zj – cj
0
Since c2 = .02 + ∆; ∆ = c1 – .02. Thus
Solving for the zj – cj inequalities: 0 – .133∆ ≤ 0 –.133∆ ≤ 0 –∆ ≤ 0 ∆≥0
c2 – .02 ≤ 0 c2 ≤ .02 –.01 + .8∆ ≤ 0 .8∆ ≤ .01 ∆ ≤ .0125
Since c1 = .03 + ∆; ∆ = c1 – .03. Thus
Since c2 = .02 + ∆; ∆ = c1 – .02. Thus
c1 – .03 ≥ 0 c1 ≥ .03 –.01 + .2∆ ≤ 0 .2∆ ≤ .01 ∆ ≤ .05
c2 – .02 ≤ .0125 c2 ≥ .0075 Summarizing, c2 ≤ .0375. d) When determining sensitivity ranges for qi values in a minimization problem, since artificial variables are eliminated, the surplus variable column coefficients must be used. This corresponds to a qi – ∆ change.
Since c1 = .03 + ∆; ∆ = c1 – .03. Thus c1 – .03 ≤ .05 c1 ≤ .08 Summarizing, .03 ≤ c1 ≤ .08. c2, basic:
cj
Basic Variables
.03
.02 + ∆
0
0
0
Quantity
x1
x2
s1
s2
s3
.02 + ∆
x2
3.6
0
1
0
.20
–.80
.03
x1
1.6
1
0
0
–.13
.20
0
s1
4.4
0
0
1
.47
–3.2
zj
.12 + 3.6∆
zj – cj
.03 0
.02 + ∆ 0
0
0 + .2∆
–.01 + .8∆
0
0 + .2∆
–.01 + .8∆
Solving for the zj – cj inequalities: 0 + .2∆ ≤ 0 .2∆ ≤ 0 ∆≤0
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q1: x2:
3.6 + 0∆ ≥ 0
x1:
1.6 + 0∆ ≥ 0
s1:
Therefore, ∆ ≥ –4.4. Since q1 = 20 – ∆ ∆ = 20 – q1 20 – q1 ≥ –4.4 q1 ≤ 24.4 q2: x2:
s1:
3.6 + .2∆ ≥ 0 x1: 1.6 – .133∆ ≥ 0 .2∆ ≥ –3.6 –.133∆ ≥ –1.6 ∆ ≥ –18 –∆ ≥ –12 ∆ ≤ 12 4.4 + .47∆ ≥ 0 .47∆ ≥ –4.4 ∆ ≥ –9.36
Therefore, –9.36 ≤ ∆ ≤ 12. Since q2 = 30 – ∆ ∆ = 30 – q2 –9.36 ≤ 30 – q2 ≤ 12 18 ≤ q2 ≤ 39.36 q3: x2:
s1:
3.6 – .8∆ ≥ 0 x1: –.8∆ ≥ –3.6 –∆ ≥ –4.5 ∆ ≤ 4.5
1.6 + .2∆ ≥ 0 .2∆ ≥ –1.6 ∆ ≥ –8
4.4 – 3.2∆ ≥ 0 –3.2∆ ≥ –4.4 –∆ ≥ –1.375 ∆ ≤ –1.375
Therefore, –8 ≤ ∆ ≤ 1.375. Since q3 = 12 – ∆ ∆ = 12 – q3 –8 ≤ 12 – q3 ≤ 1.375 10.625 ≤ q3 ≤ 20 e) The marginal value of 1 mg of carbohydrates is $.01. From part d, the sensitivity range for q3, carbohydrates, is 10.625 ≤ q3 ≤ 20. Thus, the dietitian could lower the requirements for carbohydrates to10.625 at the marginal value without the solution becoming infeasible.
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4.4 + ∆ ≥ 0 ∆ ≥ –4.4
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50. a) Minimize Zd = 1,200y1 + 500y3 subject to .50y1 + y2 ≥ 1.25 1.2y1 – y2 + y3 ≥ 2.00 .80y1 + y2 ≥ 1.75 y1, y2, y3 ≥ 0 y1 = the marginal value of an additional hour of production time = $0 y2 = the marginal value of increasing the combined demand for cheese sandwiches by one sandwich = $1.75 y3 = the marginal value of producing an additional ham salad sandwich = $3.75 b) c1, non-basic: –5 + ∆ ≤ 0 ∆ ≤ .50 Since c1 = 1.25 + ∆; ∆ = c1 – 1.25. Therefore, c1 – 1.25 ≤ .50 c1 ≤ 1.75 c2, basic: –3.75 – ∆ ≤ 0 –∆ ≤ 3.75 ∆ ≥ –3.75 Since c2 = 2 + ∆; ∆ = c2 – 2. Therefore, c2 –2 ≥ –3.75 c2 ≥ –1.75 c3, basic: –.5 – ∆ ≤ 0 –∆ ≤ .5 ∆ ≥ –.5
–3.75 – ∆ ≤ 0 –∆ ≤ 3.75 ∆ ≥ –3.75
Summarizing, –3.75 ≤ –.5 ≤ ∆ and, therefore, –.5 ≤ ∆ Since c3 = 1.75 + ∆; ∆ = c3 – 1.75. Therefore, –.5 ≤ ∆ –.5 ≤ c3 – 1.75 1.25 ≤ c3
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c) q3: s1:
200 – 2∆ ≥ 0 x3: –2∆ ≥ –200 ∆ ≤ 100
500 + ∆ ≥ 0 x2: ∆ ≥ –500
500 + ∆ ≥ 0 ∆ ≥ –500
Summarizing, –500 ≤ ∆ ≤ 100 Since q3 = 500 + ∆, ∆ = q3 – 500. Therefore, –500 ≤ q3 – 500 ≤ 100 0 ≤ q3 ≤ 600 d) The marginal value for demand for cheese sandwiches is $1.75. The range for q2 is computed as follows. s1:
200 – .8∆ ≥ 0 x3: –.8∆ ≥ –200 ∆ ≤ 250
500 + ∆ ≥ 0 x2: ∆ ≥ –500
Summarizing, –500 ≤ ∆ ≤ 250 Since q2 = 0 + ∆, ∆ = q2. Therefore, –500 ≤ q2 ≤ 250 Thus, the demand for cheese sandwiches can be increased up to a maximum of 250 sandwiches.
The additional profit for 200 more cheese sandwiches would be, ($1.75) (200) = $350 Since the cost of advertising is $100, a $250 profit would result, therefore the company should advertise.
51. a) c3, nonbasic: –13 – ∆ ≤ 0 –∆ ≤ 13 ∆ ≥ –13 Since c3 = 2 + ∆, ∆ = c3 – 2. And c3 – 2 ≥ –13 c3 ≥ –11
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500 + 0∆ ≥ 0 0∆ ≥ –500 ∆≤∞
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c1, basic: cj
Basic Variables
3+∆
5
2
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
0
s2
15
0
0
–4
–3/2
1
–1/2
3+∆
x1
5
1
0
–2
–1/2
0
1/2
5
x2
30
0
1
–1
–1/2
0
–1/2
zj
165 + 5∆
3+∆
5
–11 – 2∆
–4 – ∆/2
0
–1 + ∆/2
0
0
–13 – 2∆
–4 – ∆/2
0
–1 + ∆/2
zj – cj
Solving for the zj – cj inequalities:
Solving for the zj – cj inequalities:
–13 – 2∆ ≤ 0 –2∆ ≤ 13 ∆ ≥ –13/2
–13 – 2∆ ≤ 0 –2∆ ≤ 13 ∆ ≥ –13/2
Since c1 = 3 + ∆, ∆ = c1 – 3. Thus
Since c2 = 5 + ∆, ∆ = c2 – 5. Thus
c1 – 3 ≥ –13/2 c1 ≥ –7/2 –4 – 2∆ ≤ 0 –2∆ ≤ 4 ∆ ≥ –2
c2 – 5 ≥ –13/2 c2 ≥ –3/2 –4 – ∆/2 ≤ 0 –∆/2 ≤ 4 ∆ ≥ –8
Since c1 = 3 + ∆, ∆ = c1 – 3. Thus
Since c2 = 5 + ∆, ∆ = c2 – 5. Thus
c1 – 3 ≥ –2 c1 ≥ 1 –1 + ∆/2 ≤ 0 ∆/2 ≤ 1 ∆≤2
c2 – 3 ≥ –8 c2 ≥ –3 –1 – ∆/2 ≤ 0 –∆/2 ≤ 1 ∆ ≥ –2
Since c1 = 3 + ∆, ∆ = c1 – 3. Thus
Since c2 = 5 + ∆, ∆ = c2 – 5. Thus
c1 – 3 ≤ 2 c1 ≤ 5
c2 – 5 ≥ –2 c2 ≥ 3
Summarizing, 1 ≤ c1 ≤ 5.
Summarizing, c2 > 3.
c2: basic:
cj
Basic Variables
3
5+∆
2
0
0
0
Quantity
x1
x2
x3
s1
s2
s3
0
s2
15
0
0
–4
–3/2
1
–1/2
3
x1
5
1
0
–2
–1/2
0
1/2
5+∆
x2
30
0
1
–1
–1/2
0
–1/2
zj
165 + 30∆
3
5+∆
–11 – 2∆
–4 – ∆/2
0
–1 – ∆/2
0
0
–13 – 2∆
–4 – ∆/2
0
–1 – ∆/2
zj – cj
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b) When determining sensitivity ranges for qi values, since artificial values are eliminated, the surplus variable column coefficients must be used. This corresponds to a qi – ∆ change. q1: s2:
15 – 3∆/2 ≥ 0 –3∆/2 ≥ –15 ∆ ≤ 10
x1:
5 – ∆/2 ≥ 0 –∆/2 ≥ –5 ∆ ≤ 10
x2:
30 – ∆/2 ≥ 0 –∆/2 ≥ –30 ∆ ≤ 60
x1:
5 – 0∆ ≥ 0
x2:
30 – 0∆ ≥ 0
x1:
5 + ∆/2 ≥ 0 ∆/2 ≥ –5 ∆ ≥ –10
x2:
30 – ∆/2 ≥ 0 –∆/2 ≥ –30 ∆ ≤ 60
Therefore, ∆ ≤ 10. Since q1 = 35 – ∆ ∆ = 35 – q1 35 – q1 ≤ 10 q1 ≥ 25 q2: s2:
15 + ∆ ≥ 0 ∆ ≥ –15
Therefore, ∆ ≤ –15. Since q2 = 50 – ∆ ∆ = 50 – q2 50 – q2 ≤ –15 q2 ≤ 65 q3: s2:
15 – ∆/2 ≥ 0 –∆/2 ≥ –15 ∆ ≤ 30
Therefore, –10 ≤ ∆ ≤ 30. Since q3 = 25 – ∆ and ∆ = 25 – q3 –10 ≤ 25 – q3 ≤ 30 –5 ≤ q3 ≤ 35
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52. a) y1 = 7/15 = $.467
d) Range for q5:
Range for q1:
s5:
x1: 8 + ∆/15 ≥ 0 x2: 16 – ∆/30 ≥ 0 ∆/15 ≥ –8 –∆/30 ≥ –16 ∆ ≥ –120 ∆ ≤ 480
Since q5 = 40 + ∆, ∆ = q5 – 40 q5 – 40 ≥ –8 q5 ≥ 32
x3: 3 – ∆/40 ≥ 0 ∆ ≥ 120 s4: 36 – ∆/30 ≥ 0 –∆/30 ≥ – 36 ∆ ≤ 1,080
s5: 8 – ∆/10 ≥ 0 –∆/10 ≥ –8 ∆ ≤ 80 –120 ≤ ∆ ≤ 80
No, increasing q5 from 40 to 50 will have no affect on the optimal solution. e) Since y1 = 7/15 = $.467, pears should be secured.
Since q1 = 320 + ∆,
53. a) y2 = $0, spruce has no marginal value
∆ = q1 – 320 –120 ≥ q1 – 320 ≤ 80 200 ≤ q1 ≤ 400
q2: s2:
As many as 400 pears can be purchased.
∆ = q2 – 160 q2 – 160 ≥ –70 q2 ≥ 90
Range for q2: 8 + ∆/30 ≥ 0 x2: 16 + ∆/15 ≥ 0 ∆/30 ≥ –8 ∆/15 ≥ –16 ∆ ≥ –240 ∆ ≥ –240
3 – ∆/10 ≥ 0 –∆/10 ≥ –3 ∆ ≤ 30 –240 ≤ ∆ ≤ 30 x3:
b) y3 = $2, marginal value of cutting hours q3:
s4: 36 – ∆/30 ≥ 0 –∆/30 ≥ –36 ∆ ≤ 1,080
Since q2 = 400 + ∆, ∆ = q2 – 400 –240 ≤ q2 – 400 ≤ 30 160 ≤ q2 ≤ 430
s1: 80 – 4∆/3 ≥ 0 s2: –4∆/3 ≥ –80 ∆ ≤ 60
70 + ∆/3 ≥ 0 ∆/3 ≥ –70 ∆ ≥ –210
x3: 20 + 2∆/3 ≥ 0 x2: 2∆/3 ≥ –20 ∆ ≥ –30
10 – ∆/3 ≥ 0 –∆/3 ≥ –10 ∆ ≤ 30
Therefore, –30 ≤ ∆ ≤ 30. Since q3 = 50 + ∆, ∆ = q3 – 50 20 ≤ q3 ≤ 80
Range over which the value of peaches is valid c) Range for q3: s3:
70 + ∆ ≥ 0 ∆ ≥ –70
Since q2 = 160 + ∆,
b) y2 = 1/15 = $.067
x1:
8+∆≥0 ∆ ≥ –8
c) y3 = $2, cutting hours; y4 = $2, pressing hours. Since they both have the same marginal value, management could choose either.
3+∆≥0 ∆ ≥ –3
Since q3 = 43 + ∆, ∆ = q3 – 43 q3 – 43 ≥ –3 q3 ≥ 40 No, increasing q3 from 43 to 60 will not affect the optimal solution.
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d) From part a, q2 ≥ 90; thus, a decrease from 160 to 100 lb of spruce will not affect the solution. e) Compute the range for c1, a nonbasic cj value. c1 = 4 + ∆ –2 + ∆ ≤ 0 ∆≤2 c1 – 4 ≤ 2 c1 ≤ 6 The unit profit from Western paneling would have to be $6 or more before it would be produced. f) Compute the range for c3. c3 = 8 + ∆ –2 – ∆/3 ≤ 0 –2 – 2∆/3 ≤ 0 –∆/3 ≤ 2 –2∆/3 ≤ 2 ∆ ≥ –6 ∆ ≥ –3
–2 +∆/6 ≤ 0 ∆/6 ≤ 2 ∆ ≤ 12
Since ∆ = c3 – 8, c3 – 8 ≥ –6 c3 ≥ 2
c3 – 8 ≥ –3 c3 ≥ 5
c3 – 8 ≤ 12 c3 ≤ 20
Summarizing, 5 ≤ c3 ≤ 20. If the unit profit of Colonial paneling is increased to $13, the percent solution would not be affected. 54.
y1 = $1.33 q1: x4: 80 + 2∆/3 ≥ 0 x2: 2∆/3 ≥ –80 ∆ ≥ –120
40 – ∆/3 ≥ 0 –∆/3 ≥ –40 ∆ ≤ 120
–120 ≤ ∆ ≤ 120 Since q1 = 200 + ∆, ∆ = q1 – 200 –120 ≤ q1 – 200 ≤ 120 80 ≤ q1 ≤ 320
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