Part 1: Solutions Manual Chapter 1 Introduction Learning Objectives 1.
Develop a general understanding of the management science/operations research approach to decision making.
2.
Realize that quantitative applications begin with a problem situation.
3.
Obtain a brief introduction to quantitative techniques and their frequency of use in practice.
4.
Understand that managerial problem situations have both quantitative and qualitative considerations that are important in the decision making process.
5.
Learn about models in terms of what they are and why they are useful (the emphasis is on mathematical models).
6.
Identify the step-by-step procedure that is used in most quantitative approaches to decision making.
7.
Learn about basic models of cost, revenue, and profit and be able to compute the breakeven point.
8.
Obtain an introduction to the use of computer software packages such as Microsoft Excel in applying quantitative methods to decision making.
9.
Understand the following terms: model objective function constraint deterministic model stochastic model feasible solution
infeasible solution management science operations research fixed cost variable cost breakeven point
Solutions:
1-1
Chapter 1
1.
Management science and operations research, terms used almost interchangeably, are broad disciplines that employ scientific methodology in managerial decision making or problem solving. Drawing upon a variety of disciplines (behavioral, mathematical, etc.), management science and operations research combine quantitative and qualitative considerations in order to establish policies and decisions that are in the best interest of the organization.
2.
Define the problem Identify the alternatives Determine the criteria Evaluate the alternatives Choose an alternative For further discussion see section 1.3
3.
See section 1.2.
4.
A quantitative approach should be considered because the problem is large, complex, important, new and repetitive.
5.
Models usually have time, cost, and risk advantages over experimenting with actual situations.
6.
Model (a) may be quicker to formulate, easier to solve, and/or more easily understood.
7.
Let
d = distance m = miles per gallon c = cost per gallon,
2d Total Cost = c m We must be willing to treat m and c as known and not subject to variation. 8.
a. Maximize 10x + 5y s.t. 5x + 2y 40 x 0, y 0 b.
Controllable inputs: x and y Uncontrollable inputs: profit (10,5), labor hours (5,2) and labor-hour availability (40)
c.
1-2
Introduction
Profit:
$10/unit for x $ 5/ unit for y
Labor Hours:
5/unit for x 2/ unit for y 40 labor-hour capacity
Uncontrollable Inputs
Max 10 x + 5 y s.t. 10 x + 5 y 40 x 0 y 0
Production Quantities x and y Controllable Input
Projected Profit and check on production time constraint Output
Mathematical Model d.
x = 0, y = 20 Profit = $100 (Solution by trial-and-error)
e.
Deterministic - all uncontrollable inputs are fixed and known.
9.
If a = 3, x = 13 1/3 and profit = 133 If a = 4, x = 10 and profit = 100 If a = 5, x = 8 and profit = 80 If a = 6, x = 6 2/3 and profit = 67 Since a is unknown, the actual values of x and profit are not known with certainty.
10. a.
Total Units Received = x + y
b.
Total Cost = 0.20x +0.25y
c.
x + y = 5000
d. x 4000 Kansas City Constraint y 3000 Minneapolis Constraint e. Min s.t.
0.20x + 0.25y x+ x
y y
=
5000 4000 3000
x, y 0 11. a.
at $20 d = 800 - 10(20) = 600
1-3
Chapter 1
at $70 d = 800 - 10(70) = 100 b.
at $26 d = 800 - 10(26) = 540 at $27 d = 800 - 10(27) = 530 If the firm increases the per unit price from $26 to $27, the number of units the firm can sell falls by 10. at $42 d = 800 - 10(42) = 380 at $43 d = 800 - 10(43) = 370 If the firm increases the per unit price from $426 to $43, the number of units the firm can sell falls by 10. This suggests that the relationship between the per-unit price and annual demand for the product in units is linear between $20 and $70 and that annual demand for the product decreases by 10 units when the price is increased by $1.
c.
TR = dp = (800 - 10p)p = 800p - 10p2
d.
at $30 TR = 800(30) - 10(30)2 = 15,000 at $40 TR = 800(40) - 10(40)2 = 16,000 at $50 TR = 800(50) - 10(50)2 = 15,000 Total Revenue is maximized at the $40 price.
e.
d = 800 - 10(40) = 400 units TR = $16,000
12. a.
TC = 2000 + 60x
b.
P = 80x - (2000 + 60x) = 20x - 2000
c.
Breakeven point is the value of x when P = 0 Thus 20x - 2000 = 0 20x = 2000 x = 100
13. a.
Total cost = 9600 + 60x
b.
Total profit = total revenue - total cost = 600x - (9600 + 120x) = 480x - 9600
c.
Total profit = 480(30) - 9600 = 4800
d.
480x - 9600 = 0 x = 9600/480 = 20 The breakeven point is 20 students.
14. a.
Profit = Revenue - Cost = 46x - (160,000 + 6x) = 40x - 160,000 40x - 160,000 = 0 40x = 160,000 x = 4000 Breakeven point = 4000
1-4
Introduction
b.
Profit = 40(3800) - 16,000 = -8000 Thus, a loss of $8000 is anticipated.
c.
Profit = px - (160,000 + 6x) = 3800p - (160,000 + 6(3800)) = 0 3800p = 182,800 p = 48.105 or $48.11
d.
Profit = $50.95 (3800) - (160,000 + 6 (3800)) = $10,810 Probably go ahead with the project although the $10,810 is only a 5.98% return on the total cost of $182,800.
15. a. b.
Profit
= 300,000x - (4,500,000 + 150,000x) = 0 150,000x = 4,500,000 x = 30 Build the luxury boxes. Profit = 300,000 (50) - (4,500,000 + 150,000 (50)) = $3,000,000
16. a.
Max
50x + 30y 800,000 50x 500,000 30y 450,000 x, y 0
b.
17. a.
6x + 4y
sj = sj - 1 + xj - dj or sj - sj-1 - xj + dj = 0
b.
xj cj
c. sj Ij 18. a. maximize (3.10 – 0.30)x1 + (3.10 – 0.40) x2 + (3.10 – 0.48)x3 + (3.20 – 0.30)y1 + (3.20 – 0.40)y2 +(3.20-0.48)y3 = maximize 2.80x1 + 2,70 x2 + 2.62x3 +2.90y1 + 2.80y2 +2.72y3 b. (1) x1 + y1 ≤ 12,000 (2) x2 + y2 ≤ 20,000 (3) x3 + y3 ≤ 24,000 c. x1 ≥ .35(x1 + x2 + x3) or .65x1 - .35x2 - .35x3 ≥ 0 x2 ≤ .50(x1 + x2 + x3) or -.50x1 + .50x2 - .50x3 ≤ 0 x3 ≥ .15(x1 + x2 + x3) or -.15x1 - .15x2 + .85x3 ≥ 0 y1 ≥ .20(y1 + y2 + y3) or .80y1 - .20y2 - .20y3 ≥ 0
(Blend A must be composed of at least 35% of oil from the Texas well) (Blend A must be composed of no more than 50% of oil from the Oklahoma well) (Blend A must be composed of at least 15% of oil from the California well) (Blend B must be composed of at least 20% of oil from the Texas well)
1-5
Chapter 1
y2 ≥ .30(y1 + y2 + y3) or -.30y1 + .70y2 - .30y3 ≥ 0 y3 ≤ .40(y1 + y2 + y3) or -.40y1 - .40y2 + .60y3 ≤ 0 x1 + x2 + x3 ≥ 20,000 y1 + y2 + y3 ≥ 20,000
19. a.
Profit per contemporary cabinet is $90 – [2.0($15) + 1.5($12) + 1.3($18)] = $18.60 Profit per farmhouse cabinet is $85 – [2.5($15) + 1.0($12) + 1.2($18)] = $13.90 So the objective function is maximize 18.6x + 13.9y
b.
2.0x1 + 2.5y1 ≤ 3000 1.5x2 + 1.0y2 ≤ 1500 1.3x3 + 1.2y3 ≤ 1500
c.
x ≥ 500 y ≥ 650
20. a.
hours available in carpentry hours available in carpentry hours available in carpentry
maximize 7000x + 4000y
b.
500x + 250y ≤ 100,000
c.
x ≤ 20
d.
y ≥ 50
e. f.
(Blend B must be composed of at least 30% of oil from the Oklahoma well) (Blend B must be composed of no more than 40% of oil from the California well) (Long-term contracts require at least 20,000 gallons of Blend A to be produced) (Long-term contracts require at least 20,000 gallons of Blend B to be produced)
, or 2x – y ≥ 0 If the constraints for the maximum number of television ads to be used from part(c) and the minimum number of internet ads to be used from part (d) are both satisfied, then television ads can be at most . That is, the constraint for the stipulated ratio of television ads to Internet ads cannot be satisfied. Therefore, the problem as stated is infeasible.
1-6
Chapter 2 An Introduction to Linear Programming Learning Objectives 1.
Obtain an overview of the kinds of problems linear programming has been used to solve.
2.
Learn how to develop linear programming models for simple problems.
3.
Be able to identify the special features of a model that make it a linear programming model.
4.
Learn how to solve two variable linear programming models by the graphical solution procedure.
5.
Understand the importance of extreme points in obtaining the optimal solution.
6.
Know the use and interpretation of slack and surplus variables.
7.
Be able to interpret the computer solution of a linear programming problem.
8.
Understand how alternative optimal solutions, infeasibility and unboundedness can occur in linear programming problems.
9.
Understand the following terms: problem formulation constraint function objective function solution optimal solution nonnegativity constraints mathematical model linear program linear functions feasible solution
feasible region slack variable standard form redundant constraint extreme point surplus variable alternative optimal solutions infeasibility unbounded
2-1
Chapter 2
Solutions: 1.
a, b, and e, are acceptable linear programming relationships. c is not acceptable because of −2B 2 d is not acceptable because of 3 A f is not acceptable because of 1AB c, d, and f could not be found in a linear programming model because they have the above nonlinear terms.
2.
a. B 8
4
0
4
8
4
8
A
b.
B 8
4
0
A
c.
B Points on line are only feasible points
8 4
0
4
2-2
8
A
An Introduction to Linear Programming
3.
a. B (0,9)
A
0
(6,0)
b. B (0,60)
A
0
(40,0)
c. B Points on line are only feasible solutions (0,20) A (40,0)
0
4.
a. B
(20,0)
(0,-15)
2-3
A
Chapter 2
b. B
(0,12) (-10,0)
A
c. B
(10,25)
Note: Point shown was used to locate position of the constraint line A
0
5. B
a
300
c 200
100 b A 0
100
200
2-4
300
An Introduction to Linear Programming
6.
7A + 10B = 420 is labeled (a) 6A + 4B = 420 is labeled (b) -4A + 7B = 420 is labeled (c)
B 100 80 60 (b)
(c)
40 20
(a) A
-100
-80
-60
-40
-20
0
20
40
60
80
100
7. B 100
50
A
0 50
100
150
2-5
200
250
Chapter 2
8. B 200
133 1/3
(100,200)
A -200
0
-100
100
200
9.
B (150,225) 200
100
0
(150,100)
100
-100
-200
2-6
200
300
A
An Introduction to Linear Programming
10.
B 5
4
Optimal Solution A = 12/7, B = 15/7
3 Value of Objective Function = 2(12/7) + 3(15/7) = 69/7 2
1
A
0 1
2
(1) × 5 (2) - (3)
4
5
2B 3B 10B 7B B
= 6 = 15 = 30 = -15 = 15/7
3
A 5A 5A
+ + + -
From (1), A = 6 - 2(15/7) = 6 - 30/7 = 12/7
2-7
6
(1) (2) (3)
Chapter 2
11.
B A = 100
Optimal Solution A = 100, B = 50 Value of Objective Function = 750
100 B = 80
A
0 100
200
12. a.
B 6
5
4
Optimal Solution A = 3, B = 1.5 Value of Objective Function = 13.5
3
(3,1.5)
2
1
A
(0,0) 1
2
3
2-8
4 (4,0)
5
6
An Introduction to Linear Programming
b.
B 3
Optimal Solution A = 0, B = 3 Value of Objective Function = 18
2
1
A
(0,0) 1 c.
2
3
4
5
6
7
8
There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3).
13. a.
B 8
6 Feasible Region consists of this line segment only
4
2
0
A 2
b.
4
The extreme points are (5, 1) and (2, 4).
2-9
6
8
9
10
Chapter 2
c.
B 8
6
Optimal Solution A = 2, B = 4
4
2
0
A 2
14. a.
4
6
Let F = number of tons of fuel additive S = number of tons of solvent base Max s.t.
40F
+
30S
2/5F
+
1/ S 200 Material 1 2 1/ S 5 Material 2 5
3/ F 5 F, S 0
+
3/ S 10
21
Material 3
2 - 10
8
An Introduction to Linear Programming
b. S
F
c.
Material 2: 4 tons are used, 1 ton is unused.
d.
No redundant constraints.
15. a.
2 - 11
Chapter 2
b.
Similar to part (a): the same feasible region with a different objective function. The optimal solution occurs at (708, 0) with a profit of z = 20(708) + 9(0) = 14,160.
c.
The sewing constraint is redundant. Such a change would not change the optimal solution to the original problem.
16. a.
b.
A variety of objective functions with a slope greater than -4/10 (slope of I & P line) will make extreme point (0, 540) the optimal solution. For example, one possibility is 3S + 9D. Optimal Solution is S = 0 and D = 540.
c. Department Cutting and Dyeing Sewing Finishing Inspection and Packaging
Hours Used 1(540) = 540 5 /6(540) = 450 2 /3(540) = 360 1 /4(540) = 135
Max. Available 630 600 708 135
17. Max s.t.
5A
+ 2B
+
0S1
1A 2A 6A
- 2B + 3B - 1B
+
1S1
+ 0S2 +
0S3
+ 1S2
= 420 = 610 1S3 = 125
+ A, B, S1, S2, S3 0
18. a. Max s.t.
4A + 1B
+ 0S1
10A + 2B 3A + 2B 2A + 2B
+ 1S1
+ 0S2
+ 0S3
+ 1S2 + 1S3 A, B, S1, S2, S3 0
2 - 12
= 30 = 12 = 10
Slack 90 150 348 0
An Introduction to Linear Programming
b.
B 14
12
10
8
6 Optimal Solution A = 18/7, B = 15/7, Value = 87/7 4
2
0 c.
A 2
4
6
8
3A +
4B
+ 0S1
+ 0S2
+ 0S3
-1A + 1A + 2A +
2B 2B 1B
+ 1S1
10
S1 = 0, S2 = 0, S3 = 4/7
19. a. Max s.t.
+ 1S2 + 1S3 A, B, S1, S2, S3 0
2 - 13
= 8 = 12 = 16
(1) (2) (3)
Chapter 2
b.
B 14 (3) 12
10
(1)
8
6 Optimal Solution A = 20/3, B = 8/3 Value = 30 2/3
4
2 (2) 0
A 2
c.
4
6
8
10
12
S1 = 8 + A – 2B = 8 + 20/3 - 16/3 = 28/3 S2 = 12 - A – 2B = 12 - 20/3 - 16/3 = 0 S3 = 16 – 2A - B = 16 - 40/3 - 8/3 = 0
20. a. Max s.t.
3A
+ 2B
A 3A A A
+ B + 4B
- S1 + S2 - S3
-
- S4
B
A, B, S1, S2, S3, S4 0
2 - 14
= = = =
4 24 2 0
An Introduction to Linear Programming
b.
c.
S1 = (3.43 + 3.43) - 4 = 2.86 S2 = 24 - [3(3.43) + 4(3.43)] = 0 S3 = 3.43 - 2 = 1.43 S4 = 0 - (3.43 - 3.43) = 0
2 - 15
Chapter 2
21. a. and b. B
90
80
70 Constraint 2
60
50
40
Optimal Solution
Constraint 1
Constraint 3
30
Feasible Region
20
10
2A + 3B = 60
A
0 10 c.
20
30
40
50
60
70
80
90
100
Optimal solution occurs at the intersection of constraints 1 and 2. For constraint 2, B = 10 + A Substituting for B in constraint 1 we obtain 5A + 5(10 + A) 5A + 50 + 5A 10A A
= 400 = 400 = 350 = 35
B = 10 + A = 10 + 35 = 45 Optimal solution is A = 35, B = 45 d.
Because the optimal solution occurs at the intersection of constraints 1 and 2, these are binding constraints.
2 - 16
An Introduction to Linear Programming
e.
Constraint 3 is the nonbinding constraint. At the optimal solution 1A + 3B = 1(35) + 3(45) = 170. Because 170 exceeds the right-hand side value of 90 by 80 units, there is a surplus of 80 associated with this constraint.
22. a.
C 3500
3000
2500
Inspection and Packaging
2000
Cutting and Dyeing
5 1500
Feasible Region
4 1000
Sewing 3 5A + 4C = 4000
500
0
2 1
500
1000 1500 2000 2500 Number of All-Pro Footballs
A 3000
b. Extreme Point 1 2 3 4 5
Coordinates (0, 0) (1700, 0) (1400, 600) (800, 1200) (0, 1680)
Profit 5(0) + 4(0) = 0 5(1700) + 4(0) = 8500 5(1400) + 4(600) = 9400 5(800) + 4(1200) = 8800 5(0) + 4(1680) = 6720
Extreme point 3 generates the highest profit. c.
Optimal solution is A = 1400, C = 600
d.
The optimal solution occurs at the intersection of the cutting and dyeing constraint and the inspection and packaging constraint. Therefore these two constraints are the binding constraints.
e.
New optimal solution is A = 800, C = 1200 Profit = 4(800) + 5(1200) = 9200
2 - 17
Chapter 2
23. a.
Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced Max s.t.
2400E
+
6E
+
2E
+
1800L 3L 2100 L 280 2.5L 1000 E, L 0
Engine time Lady-Sport maximum Assembly and testing
b. L 700
Number of EZ-Rider Produced
600
Engine Manufacturing Time
500
400
Frames for Lady-Sport
300
Optimal Solution E = 250, L = 200 Profit = $960,000
200
100 Assembly and Testing 0
E 100
300
200
400
500
Number of Lady-Sport Produced
c. 24. a.
The binding constraints are the manufacturing time and the assembly and testing time. Let R = number of units of regular model. C = number of units of catcher’s model. Max s.t.
5R
+
8C
1R
+
1/ R 2 1/ R 8
+
3/ C 2 1/ C 3 1/ C 4
+
900
Cutting and sewing
300
Finishing
100
Packing and Shipping
R, C 0
2 - 18
.
An Introduction to Linear Programming
b.
C 1000
F
Catcher's Model
800 600
C&
400
S
P&
Optimal Solution (500,150) S
200 R 0
200
400
600
800
1000
Regular Model c.
5(500) + 8(150) = $3,700
d.
C&S
1(500) + 3/2(150) = 725
F
1/ (500) + 1/ (150) = 300 2 3
P&S
1/ (500) + 1/ (150) = 100 8 4
e. Department C&S F P&S 25. a.
Usage 725 300 100
Slack 175 hours 0 hours 0 hours
Let B = percentage of funds invested in the bond fund S = percentage of funds invested in the stock fund Max s.t.
b.
Capacity 900 300 100
0.06 B
+
0.10 S
B 0.06 B B
+ +
0.10 S S
=
0.3 0.075 1
Optimal solution: B = 0.3, S = 0.7 Value of optimal solution is 0.088 or 8.8%
2 - 19
Bond fund minimum Minimum return Percentage requirement
Chapter 2
26. a.
a.
Let N = amount spent on newspaper advertising R = amount spent on radio advertising Max s.t.
50N
+ 80R
N N
+
N
R = 1000 Budget 250 Newspaper min. R 250 Radio min. -2R 0 News 2 Radio
N, R 0 b. R
1000
Radio Min
Optimal Solution N = 666.67, R = 333.33 Value = 60,000
Budget
N = 2R
500
Newspaper Min Feasible region is this line segment N 0
27.
5 00
1000
Let I = Internet fund investment in thousands B = Blue Chip fund investment in thousands Max s.t.
0.12I
+
0.09B
1I 1I 6I
+
1B
+ 4B I, B 0
50 35 240
Available investment funds Maximum investment in the internet fund Maximum risk for a moderate investor
2 - 20
An Introduction to Linear Programming
B
Blue Chip Fund (000s)
60
Risk Constraint Optimal Solution I = 20, B = 30 $5,100
50
40
Maximum Internet Funds
30
20
10
Objective Function 0.12I + 0.09B
Available Funds $50,000
0
I 10
30
20
40
50
60
Internet Fund (000s)
Internet fund Blue Chip fund Annual return b.
$20,000 $30,000 $ 5,100
The third constraint for the aggressive investor becomes 6I + 4B 320 This constraint is redundant; the available funds and the maximum Internet fund investment constraints define the feasible region. The optimal solution is: Internet fund Blue Chip fund Annual return
$35,000 $15,000 $ 5,550
The aggressive investor places as much funds as possible in the high return but high risk Internet fund. c.
The third constraint for the conservative investor becomes 6I + 4B 160 This constraint becomes a binding constraint. The optimal solution is Internet fund Blue Chip fund Annual return
$0 $40,000 $ 3,600
2 - 21
Chapter 2
The slack for constraint 1 is $10,000. This indicates that investing all $50,000 in the Blue Chip fund is still too risky for the conservative investor. $40,000 can be invested in the Blue Chip fund. The remaining $10,000 could be invested in low-risk bonds or certificates of deposit. 28. a.
Let W = number of jars of Western Foods Salsa produced M = number of jars of Mexico City Salsa produced Max s.t.
1W
+
1.25M
5W 3W + 2W + W, M 0
7M 1M 2M
4480 2080 1600
Whole tomatoes Tomato sauce Tomato paste
Note: units for constraints are ounces b.
Optimal solution: W = 560, M = 240 Value of optimal solution is 860
29. a.
Let B = proportion of Buffalo's time used to produce component 1 D = proportion of Dayton's time used to produce component 1
Buffalo Dayton
Maximum Daily Production Component 1 Component 2 2000 1000 600 1400
Number of units of component 1 produced: 2000B + 600D Number of units of component 2 produced: 1000(1 - B) + 600(1 - D) For assembly of the ignition systems, the number of units of component 1 produced must equal the number of units of component 2 produced. Therefore, 2000B + 600D = 1000(1 - B) + 1400(1 - D) 2000B + 600D = 1000 - 1000B + 1400 - 1400D 3000B + 2000D = 2400 Note: Because every ignition system uses 1 unit of component 1 and 1 unit of component 2, we can maximize the number of electronic ignition systems produced by maximizing the number of units of subassembly 1 produced. Max 2000B + 600D In addition, B 1 and D 1.
2 - 22
An Introduction to Linear Programming
The linear programming model is: Max s.t.
2000B
+ 600D
3000B B
+ 2000D
= 2400 1 1 0
D B, D
The graphical solution is shown below. D 1.2
1.0
30
.8
00 B+ 20
.6
00 D =2 40
.4
0
Optimal Solution
2000B + 600D = 300
.2
B 0
.2
.4
.6
.8
1.0
Optimal Solution: B = .8, D = 0 Optimal Production Plan Buffalo - Component 1 Buffalo - Component 2 Dayton - Component 1 Dayton - Component 2
.8(2000) = 1600 .2(1000) = 200 0(600) = 0 1(1400) = 1400
Total units of electronic ignition system = 1600 per day.
2 - 23
1.2
Chapter 2
30. a.
Let
E = number of shares of Eastern Cable C = number of shares of ComSwitch
Max s.t.
15E
+ 18C
40E 40E
+ 25C
25C 25C E, C 0
50,000 15,000 10,000 25,000
Maximum Investment Eastern Cable Minimum ComSwitch Minimum ComSwitch Maximum
b. C
Number of Shares of ComSw itch
2000
Minimum Eastern Cable
1500
Maximum Comswitch
1000
Maximum Investment 500 Minimum Conswitch
0
500 1000 1500 Number of Shares of Eastern Cable
E
c.
There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000)
d.
Optimal solution is E = 625, C = 1000 Total return = $27,375
2 - 24
An Introduction to Linear Programming
31.
B 6
Feasible Region
4
2
A 0
2
4
6 3A + 4B = 13
Optimal Solution A = 3, B = 1 Objective Function Value = 13 32. A B A
B A
2 - 25
Chapter 2
Extreme Points (A = 250, B = 100) (A = 125, B = 225) (A = 125, B = 350)
Objective Function Value 800 925 1300
Surplus Demand 125 — —
Surplus Total Production — — 125
33. a.
xB2 6
4
2
0
xA1 2
4
6
Optimal Solution: A = 3, B = 1, value = 5 b. (1) (2) (3) (4)
3 + 4(1) = 7 2(3) + 1 = 7 3(3) + 1.5 = 10.5 -2(3) +6(1) = 0
Slack = 21 - 7 = 14 Surplus = 7 - 7 = 0 Slack = 21 - 10.5 = 10.5 Surplus = 0 - 0 = 0
2 - 26
Slack Processing Time — 125 —
An Introduction to Linear Programming
c. B
A
Optimal Solution: A = 6, B = 2, value = 34 34. a. B x2 4
3 Feasible Region
2
(21/4, 9/4)
1
(4,1) x1A 0
1
2
3
4
5
b.
There are two extreme points: (A = 4, B = 1) and (A = 21/4, B = 9/4)
c.
The optimal solution is A = 4, B = 1
2 - 27
6
Chapter 2
35. a. Min s.t.
6A
+
4B
+
0S1
2A 1A
+ +
1B 1B 1B
-
S1
+
-
0S2
+
0S3
S2 +
S3
= = =
A, B, S1, S2, S3 0 b.
The optimal solution is A = 6, B = 4.
c.
S1 = 4, S2 = 0, S3 = 0.
36. a.
Let
Max s.t.
T = P =
number of training programs on teaming number of training programs on problem solving
10,000T
+
8,000P
+ +
P P 2P
T T 3T
8 10 25 84
T, P 0
2 - 28
Minimum Teaming Minimum Problem Solving Minimum Total Days Available
12 10 4
An Introduction to Linear Programming
b. P Minimum Teaming
Number o f Probl em-Solv ing Pro grams
40
30 Minimum Total 20
Days Available
Minimum Problem Solving
10
0
10 20 Number of Teaming Programs
c.
There are four extreme points: (15,10); (21.33,10); (8,30); (8,17)
d.
The minimum cost solution is T = 8, P = 17 Total cost = $216,000
30
37. Mild Extra Sharp
Regular 80% 20%
Zesty 60% 40%
8100 3000
Let R = number of containers of Regular Z = number of containers of Zesty Each container holds 12/16 or 0.75 pounds of cheese Pounds of mild cheese used
= =
0.80 (0.75) R + 0.60 (0.75) Z 0.60 R + 0.45 Z
Pounds of extra sharp cheese used = =
0.20 (0.75) R + 0.40 (0.75) Z 0.15 R + 0.30 Z
2 - 29
T
Chapter 2
Cost of Cheese
= = = =
Cost of mild + Cost of extra sharp 1.20 (0.60 R + 0.45 Z) + 1.40 (0.15 R + 0.30 Z) 0.72 R + 0.54 Z + 0.21 R + 0.42 Z 0.93 R + 0.96 Z
Packaging Cost = 0.20 R + 0.20 Z Total Cost
= (0.93 R + 0.96 Z) + (0.20 R + 0.20 Z) = 1.13 R + 1.16 Z
Revenue
= 1.95 R + 2.20 Z
Profit Contribution = Revenue - Total Cost = (1.95 R + 2.20 Z) - (1.13 R + 1.16 Z) = 0.82 R + 1.04 Z Max s.t.
0.82 R
+
1.04 Z
0.60 R + 0.15 R + R, Z 0
0.45 Z 0.30 Z
8100 3000
Mild Extra Sharp
Optimal Solution: R = 9600, Z = 5200, profit = 0.82(9600) + 1.04(5200) = $13,280 38. a.
Let
S = yards of the standard grade material per frame P = yards of the professional grade material per frame
Min s.t.
7.50S
+ 9.00P
0.10S + 0.30P 0.06S + 0.12P S + P S, P 0
6 3 = 30
carbon fiber (at least 20% of 30 yards) kevlar (no more than 10% of 30 yards) total (30 yards)
2 - 30
An Introduction to Linear Programming
b. P
Professional Grad e (yards)
50
40
total Extreme Point S = 10 P = 20
30
Feasible region is the line segment
20
kevlar carbon fiber
10
Extreme Point S = 15 P = 15 S 0
10
20
30
40
50
60
Standard Grade (yards)
c. Extreme Point (15, 15) (10, 20)
Cost 7.50(15) + 9.00(15) = 247.50 7.50(10) + 9.00(20) = 255.00
The optimal solution is S = 15, P = 15 d.
Optimal solution does not change: S = 15 and P = 15. However, the value of the optimal solution is reduced to 7.50(15) + 8(15) = $232.50.
e.
At $7.40 per yard, the optimal solution is S = 10, P = 20. The value of the optimal solution is reduced to 7.50(10) + 7.40(20) = $223.00. A lower price for the professional grade will not change the S = 10, P = 20 solution because of the requirement for the maximum percentage of kevlar (10%).
39. a.
Let S = number of units purchased in the stock fund M = number of units purchased in the money market fund Min s.t.
8S
+
50S 5S
+ +
3M
100M 4M M S, M, 0
1,200,000 Funds available 60,000 Annual income 3,000 Minimum units in money market
2 - 31
Chapter 2
Units of Money Market Fund
x2 M
20000 + 3M = 62,000 8x8S 1 + 3x2 = 62,000
15000 Optimal Solution .
10000
5000
0
5000
10000
15000
20000
x1S
Units of Stock Fund Optimal Solution: S = 4000, M = 10000, value = 62000
40.
b.
Annual income = 5(4000) + 4(10000) = 60,000
c.
Invest everything in the stock fund. Let P1 = gallons of product 1 P2 = gallons of product 2 Min s.t.
1P1
+
1P2
1P1
+
1P1
1P2 2P2 + P1, P2 0
2 - 32
30 20 80
Product 1 minimum Product 2 minimum Raw material
An Introduction to Linear Programming
P2
Feasible Region
P +1
1
60
1P =
2
55
Number of Gallons of Product 2
80
40
20
Use 8 (30,25)
0
0g als.
40 20 60 80 Number of Gallons of Product 1
P1
Optimal Solution: P1 = 30, P2 = 25 Cost = $55 41. a.
Let R = number of gallons of regular gasoline produced P = number of gallons of premium gasoline produced Max s.t.
0.30R
+
0.50P
0.30R 1R
+ +
0.60P 1P 1P
18,000 50,000 20,000
R, P 0
2 - 33
Grade A crude oil available Production capacity Demand for premium
Chapter 2
b. P
Gallons o f Premiu m Gasoli ne
60,000
50,000 Production Capacity 40,000
30,000 Maximum Premium
20,000
Optimal Solution R = 40,000, P = 10,000 $17,000
10,000
Grade A Crude Oil 0
R 10,000 20,000 30,000 40,000 50,000 60,000 Gallons of Regular Gasoline
Optimal Solution: 40,000 gallons of regular gasoline 10,000 gallons of premium gasoline Total profit contribution = $17,000 c. Constraint 1 2 3
d.
Value of Slack Variable 0 0 10,000
Interpretation All available grade A crude oil is used Total production capacity is used Premium gasoline production is 10,000 gallons less than the maximum demand
Grade A crude oil and production capacity are the binding constraints.
2 - 34
An Introduction to Linear Programming
42. B x2
14 Satisfies Constraint # 2
12 10 8
Infeasibility
6 4
Satisfies Constraint # 1
2 0
4
2
6
8
10
12
x1A
43. Bx 2 4 Unbounded
3 2 1
0
1
2
3
x1A
44. a.
xB2 Objective Function Optimal Solution (30/16, 30/16) Value = 60/16
4
2 0 b.
2
New optimal solution is A = 0, B = 3, value = 6.
2 - 35
4
xA1
Chapter 2
45. a. B
A A
A
46.
B A
b.
Feasible region is unbounded.
c.
Optimal Solution: A = 3, B = 0, z = 3.
d.
An unbounded feasible region does not imply the problem is unbounded. This will only be the case when it is unbounded in the direction of improvement for the objective function. Let
N = number of sq. ft. for national brands G = number of sq. ft. for generic brands
Problem Constraints: N N
+
G G
2 - 36
200 120 20
Space available National brands Generic
An Introduction to Linear Programming
Extreme Point 1 2 3
N 120 180 120
G 20 20 80
a.
Optimal solution is extreme point 2; 180 sq. ft. for the national brand and 20 sq. ft. for the generic brand.
b.
Alternative optimal solutions. Any point on the line segment joining extreme point 2 and extreme point 3 is optimal.
c.
Optimal solution is extreme point 3; 120 sq. ft. for the national brand and 80 sq. ft. for the generic brand.
2 - 37
Chapter 2
47. Bx2
e Tim ing cess Pro
600
500 400
300 Alternate optima (125,225)
200
100
(250,100)
0
100
200
300
400
xA 1
Alternative optimal solutions exist at extreme points (A = 125, B = 225) and (A = 250, B = 100). Cost
= 3(125) + 3(225) = 1050
Cost
= 3(250) + 3(100) = 1050
or
The solution (A = 250, B = 100) uses all available processing time. However, the solution (A = 125, B = 225) uses only 2(125) + 1(225) = 475 hours. .
Thus, (A = 125, B = 225) provides 600 - 475 = 125 hours of slack processing time which may be used for other products.
2 - 38
An Introduction to Linear Programming
48.
Possible Actions: i.
Reduce total production to A = 125, B = 350 on 475 gallons.
ii.
Make solution A = 125, B = 375 which would require 2(125) + 1(375) = 625 hours of processing time. This would involve 25 hours of overtime or extra processing time.
iii. Reduce minimum A production to 100, making A = 100, B = 400 the desired solution. 49. a.
Let
P = number of full-time equivalent pharmacists T = number of full-time equivalent physicians
The model and the optimal solution are shown below: MIN 40P+10T S.T. 1) 2) 3)
P+T >=250 2P-T>=0 P>=90
Optimal Objective Value 5200.00000 Variable P T
Value 90.00000 160.00000
2 - 39
Reduced Cost 0.00000 0.00000
Chapter 2
Constraint 1 2 3
Slack/Surplus 0.00000 20.00000 0.00000
Dual Value 10.00000 0.00000 30.00000
The optimal solution requires 90 full-time equivalent pharmacists and 160 full-time equivalent technicians. The total cost is $5200 per hour. b. Pharmacists Technicians
Current Levels 85 175
Attrition 10 30
Optimal Values 90 160
New Hires Required 15 15
The payroll cost using the current levels of 85 pharmacists and 175 technicians is 40(85) + 10(175) = $5150 per hour. The payroll cost using the optimal solution in part (a) is $5200 per hour. Thus, the payroll cost will go up by $50 50.
Let
M = number of Mount Everest Parkas R = number of Rocky Mountain Parkas Max s.t.
100M
+
150R
30M 45M 0.8M
+ + -
20R 15R 0.2R
7200 Cutting time 7200 Sewing time 0 % requirement
Note: Students often have difficulty formulating constraints such as the % requirement constraint. We encourage our students to proceed in a systematic step-by-step fashion when formulating these types of constraints. For example: M must be at least 20% of total production M 0.2 (total production) M 0.2 (M + R) M 0.2M + 0.2R 0.8M - 0.2R 0
2 - 40
An Introduction to Linear Programming
The optimal solution is M = 65.45 and R = 261.82; the value of this solution is z = 100(65.45) + 150(261.82) = $45,818. If we think of this situation as an on-going continuous production process, the fractional values simply represent partially completed products. If this is not the case, we can approximate the optimal solution by rounding down; this yields the solution M = 65 and R = 261 with a corresponding profit of $45,650. 51.
Let
C = number sent to current customers N = number sent to new customers
Note: Number of current customers that test drive = .25 C Number of new customers that test drive = .20 N Number sold = .12 ( .25 C ) + .20 (.20 N ) = .03 C + .04 N Max s.t.
.03C
+
.04N
.25 C .20 N .25 C - .40 N 4C + 6N C, N, 0
30,000 10,000 0 1,200,000
2 - 41
Current Min New Min Current vs. New Budget
Chapter 2
Current Min.
N 200,000
Current 2 New
Budget
.03
C
+. 04
N
=6
00 0
100,000
Optimal Solution C = 225,000, N = 50,000 Value = 8,750 New Min.
0
52.
Let
100,000
200,000
S = number of standard size rackets O = number of oversize size rackets Max s.t.
10S
+
15O
0.8S 10S 0.125S
+ + S, O, 0
0.2O 12O 0.4O
2 - 42
0 4800 80
% standard Time Alloy
300,000
C
An Introduction to Linear Programming
53. a.
Let
R = time allocated to regular customer service N = time allocated to new customer service Max s.t.
1.2R
+
N
R 25R -0.6R
+ + +
N 8N N
80 800 0
R, N, 0 b. Optimal Objective Value 90.00000 Variable R N
Value 50.00000 30.00000
Reduced Cost 0.00000 0.00000
Constraint 1 2 3
Slack/Surplus 0.00000 690.00000 0.00000
Dual Value 1.12500 0.00000 -0.12500
Optimal solution: R = 50, N = 30, value = 90 HTS should allocate 50 hours to service for regular customers and 30 hours to calling on new customers. 54. a.
Let
M1 = number of hours spent on the M-100 machine M2 = number of hours spent on the M-200 machine
Total Cost 6(40)M1 + 6(50)M2 + 50M1 + 75M2 = 290M1 + 375M2 Total Revenue 25(18)M1 + 40(18)M2 = 450M1 + 720M2 Profit Contribution (450 - 290)M1 + (720 - 375)M2 = 160M1 + 345M2
2 - 43
Chapter 2
Max s.t.
160 M1
+
345M2
M1 M2 M1 40 M1
+
M2 50 M2
15 10 5 5 1000
M-100 maximum M-200 maximum M-100 minimum M-200 minimum Raw material available
M1, M2 0 b. Optimal Objective Value 5450.00000 Variable M1 M2
Value 12.50000 10.00000
Reduced Cost 0.00000 145.00000
Constraint 1 2 3 4 5
Slack/Surplus 2.50000 0.00000 7.50000 5.00000 0.00000
Dual Value 0.00000 145.00000 0.00000 0.00000 4.00000
The optimal decision is to schedule 12.5 hours on the M-100 and 10 hours on the M-200.
55. Mr. Krtick’s solution cannot be optimal. Every department has unused hours, so there are no binding constraints. With unused hours in every department, clearly some more product can be made. 56. No, it is not possible that the problem is now infeasible. Note that the original problem was feasible (it had an optimal solution). Every solution that was feasible is still feasible when we change the constraint to lessthan-or-equal-to, since the new constraint is satisfied at equality (as well as inequality). In summary, we have relaxed the constraint so that the previous solutions are feasible (and possibly more satisfying the constraint as strict inequality). 57. Yes, it is possible that the modified problem is infeasible. To see this, consider a redundant greater-thanor-equal to constraint as shown below. Constraints 2,3, and 4 form the feasible region and constraint 1 is redundant. Change constraint 1 to less-than-or-equal-to and the modified problem is infeasible.
2 - 44
An Introduction to Linear Programming
Original Problem:
Modified Problem:
58. It makes no sense to add this constraint. The objective of the problem is to minimize the number of products needed so that everyone’s top three choices are included. There are only two possible outcomes relative to the boss’ new constraint. First, suppose the minimum number of products is <= 15, then there was no need for the new constraint. Second, suppose the minimum number is > 15. Then the new constraint makes the problem infeasible.
2 - 45
Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution Learning Objectives 1.
Understand what happens in graphical solutions when coefficients of the objective function change.
2.
Be able to interpret the range for an objective function coefficient.
3.
Understand what happens in graphical solutions when right-hand sides change.
4.
Be able to interpret the dual value.
5.
Be able to interpret the range for a right-hand side.
6.
Learn how to formulate, solve and interpret the solution for linear programs with more than two decision variables.
7.
Understand the following terms: sensitivity analysis dual value reduced cost sunk cost relevant cost
3-1
Chapter 3
Solutions: 1.
a. B 10
)+ 3(7
8
)= 2(3 27
6
4
Optimal Solution A = 7, B = 3
A = 4, B = 6
2
A 2
b.
6
4
8
10
The same extreme point, A = 7 and B = 3, remains optimal. The value of the objective function becomes 5(7) + 2(3) = 41
2.
c.
A new extreme point, A = 4 and B = 6, becomes optimal. The value of the objective function becomes 3(4) + 4(6) = 36.
d.
The objective coefficient range for variable A is 2 to 6. Since the change in part (b) is within this range, we know the optimal solution, A = 7 and B = 3, will not change. The objective coefficient range for variable B is 1 to 3. Since the change in part (c) is outside this range, we have to re-solve the problem to find the new optimal solution.
a. + .5) 3(6
B 10
= .5) 2(4
8
.5 28
6 Optimal Solution A = 6.5, B = 4.5 4
Enlarged Feasible Region
2
A 2
4
6
3-2
8
10
Linear Programming: Sensitivity Analysis and Interpretation of Solution
3.
b.
The value of the optimal solution to the revised problem is 3(6.5) + 2(4.5) = 28.5. The one-unit increase in the right-hand side of constraint 1 has improved the value of the optimal solution by 28.5 - 27 = 1.5. Thus, the dual value for constraint 1 is 1.5.
c.
The right-hand-side range for constraint 1 is 8 to 11.2. As long as the right-hand side stays within this range, the dual value of 1.5 is applicable.
d.
The improvement in the value of the optimal solution will be 0.5 for every unit increase in the righthand side of constraint 2 as long as the right-hand side is between 18 and 30.
a. Y 10
8
6
4
2
X = 2, Y = 3
8(3 )+
12 (2) =
Optimal Solution X = 3, Y = 2
48 X
2
b.
4
6
8
10
The same extreme point, X = 3 and Y = 2, remains optimal. The value of the objective function becomes 6(3) + 12(2) = 42.
c.
A new extreme point, X = 2 and Y = 3, becomes optimal. The value of the objective function becomes 8(2) + 6(3) = 34.
d.
The objective coefficient range for variable X is 4 to 12. Since the change in part (b) is within this range, we know that the optimal solution, X = 3 and Y = 2, will not change. The objective coefficient range for variable Y is 8 to 24. Since the change in part (c) is outside this range, we have to re-solve the problem to find the new optimal solution.
3-3
Chapter 3
4.
a. Y 10
8
6 Optimal Solution X = 2.5, Y = 2.5
4
2
8(2 .5) +
12 (2. 5) =
50 X
2
5.
4
6
8
10
b.
The value of the optimal solution to the revised problem is 8(2.5) + 12(2.5) = 50. Compared to the original problem, the value of the optimal solution has increased by 50 - 48 = 2. Thus, the dual value is 2.
c.
The right-hand side range for constraint 1 is 5 to 11. As long as the right-hand side stays within this range, the dual value of 2 is applicable. Since increasing the right-hand side does not improve the value of the optimal solution, decreasing the right-hand side of constraint 1 would b desirable.
d.
As long as the right-hand side of constraint 2 is between 9 and 18, a unit increase in the right-hand side will cause the value of the optimal solution to increase by 3.
a.
Regular Glove = 500 Catcher’s Mitt = 150 Value = 3700
b.
The finishing and packaging and shipping constraints are binding.
c.
Cutting and Sewing = 0 Finishing = 3 Packaging and Shipping = 28 Additional finishing time is worth $3 per unit and additional packaging and shipping time is worth $28 per unit.
d. 6.
In the packaging and shipping department. Each additional hour is worth $28.
a. Variable Regular Glove Catcher’s Mitt b.
Objective Coefficient Range 4 to 12 3.33 to 10
As long as the profit contribution for the regular glove is between $4.00 and $12.00, the current
3-4
Linear Programming: Sensitivity Analysis and Interpretation of Solution
solution is optimal. As long as the profit contribution for the catcher's mitt stays between $3.33 and $10.00, the current solution is optimal. The optimal solution is not sensitive to small changes in the profit contributions for the gloves. c.
The dual values for the resources are applicable over the following ranges: Constraint Cutting and Sewing Finishing Packaging
7.
Right-Hand-Side Range 725 to No Upper Limit 133.33 to 400 75 to 135
d.
Amount of increase = (28) (20) = $560
a.
U = 800 H = 1200 Estimated Annual Return = $8400
b.
Constraints 1 and 2. All funds available are being utilized and the maximum permissible risk is being incurred.
c. Constraint Funds Avail. Risk Max U.S. Oil Max
8.
9.
Dual Values 0.09 1.33 0
d.
No, the optimal solution does not call for investing the maximum amount in U.S. Oil.
a.
By more than $7.00 per share.
b.
By more than $3.50 per share.
c.
None. This is only a reduction of 100 shares and the allowable decrease is 200. management may want to address.
a.
Optimal solution calls for the production of 560 jars of Western Foods Salsa and 240 jars of Mexico City Salsa; profit is $860.
b. Variable Western Foods Salsa Mexico City Salsa
c. Constraint 1
Dual Value 0.125
2
0.000
Objective Coefficient Range 0.893 to 1.250 1.000 to 1.400
Interpretation One more ounce of whole tomatoes will increase profits by $0.125 Additional ounces of tomato sauce will not improve profits;
3-5
Chapter 3
3
slack of 160 ounces. One more ounce of tomato paste will increase profits by $0.187
0.187
d. Constraint 1 2 3 10. a.
Right-Hand-Side Range 4320 to 5600 1920 to No Upper Limit 1280 to 1640
S = 4000 M = 10,000 Total risk = 62,000
b. Variable S M
Objective Coefficient Range 3.75 to No Upper Limit No Upper Limit to 6.4
c.
5(4000) + 4(10,000) = $60,000
d.
60,000/1,200,000 = 0.05 or 5%
e.
-0.057 risk units
f.
-0.057(100) = -5.7%
11. a.
No change in optimal solution; there is no upper limit for the range of optimality for the objective coefficient for S.
b.
No change in the optimal solution; the objective coefficient for M can increase to 6.4.
c.
The optimal solution does not change.
12. a.
E = 80, S = 120, D = 0 Profit = $16,440
b.
Fan motors and cooling coils
c.
Labor hours; 320 hours available.
d.
Objective function coefficient range of optimality No lower limit to 159. Since $150 is in this range, the optimal solution would not change.
13. a.
Range of optimality E S D
47.5 to 75 87 to 126 No lower limit to 159.
b. The optimal solution does not change, but the change in total profit will be: E 80 unit @ + $6 =
$480
3-6
Linear Programming: Sensitivity Analysis and Interpretation of Solution S 120 unit @ - $2 =
-240 $240 Profit = $16,440 + 240 = 16,680. c.
Range of feasibility Constraint 1 Constraint 2 Constraint 3
d. 14. a.
b.
160 to 180 200 to 400 2080 to No Upper Limit
Yes, the dual value will change since 100 is greater than the allowable increase of 80. Manufacture 100 cases of model A Manufacture 60 cases of model B Purchase 90 cases of model B Total Cost = $2170 Demand for model A Demand for model B Assembly time
c. Constraint 1 2 3 4
Dual Value 12.25 9.0 0 -.375
If demand for model A increases by 1 unit, total cost will increase by $12.25 If demand for model B increases by 1 unit, total cost will increase by $9.00 If an additional minute of assembly time is available, total cost will decrease by $.375 d.
The assembly time constraint. Each additional minute of assembly time will decrease costs by $.375. Note that this will be true up to a value of 1133.33 hours. Some students may say that the demand constraint for model A should be selected because decreasing the demand by one unit will decrease cost by $12.25. But, carrying this argument to the extreme would argue for a demand of 0.
15. a. Decision Variable AM BM AP BP
Ranges of Optimality No lower limit to 11.75 3.667 to 9 12.25 to No Upper Limit 6 to 11.333
Provided a single change of an objective function coefficient is within its above range, the optimal solution AM = 100, BM = 60, AP = 0, and BP = 90 will not change. b.
This change is within the allowable increase. The optimal solution remains AM = 100, BM = 60, AP = 0, and BP = 90. The $11.20 - $10.00 = $1.20 per unit cost increase will increase the total cost to $2170 = $1.20(100) = $2290.
c.
Yes, it is necessary to solve the model again. Resolving the problem provides the new optimal solution: AM = 0, BM = 135, AP = 100, and BP =
3-7
Chapter 3 15; the total cost is $22,100. 16. a.
The optimal solution calls for the production of 100 suits and 150 sport coats. Forty hours of cutting overtime should be scheduled, and no hours of sewing overtime should be scheduled. The total profit is $40,900.
b.
The objective coefficient range for suits shows and upper limit of $225. Thus, the optimal solution will not change. But, the value of the optimal solution will increase by ($210-$190)100 = $2000. Thus, the total profit becomes $42,990.
c.
The slack for the material coefficient is 0. Because this is a binding constraint, Tucker should consider ordering additional material. The dual value of $34.50 is the maximum extra cost per yard that should be paid. Because the additional handling cost is only $8 per yard, Tucker should order additional material. Note that the dual value of $34.50 is valid up to 1333.33 -1200 = 133.33 additional yards.
d.
The dual value of $35 for the minimum suit requirement constraint tells us that lowering the minimum requirement by 25 suits will increase profit by $35(25) = $875.
17. a.
Produce 1000 units of model DRB and 800 units of model DRW Total profit contribution = $424,000
b.
The dual value for constraint 1 (steel available) is 8.80. Thus, each additional pound of steel will increase profit by $8.80. At $2 per pound Deegan should purchase the additional 500 pounds. Note: the upper limit on the right hand side range for constraint 1 is approximately 40,909. Thus, the dual value of $8.80 is applicable for an increase of as much as 909 pounds.
c.
Constraint 3 (assembly time) has a slack of 4000 hours. Increasing the number of hours of assembly time is not worthwhile.
d.
The objective coefficient range for model DRB shows a lower limit of $112. Thus, the optimal solution will not change; the value of the optimal solution will be $175(1000) + $280(800) = $399,000.
e.
An increase of 500 hours or 60(500) = 30,000 minutes will result in 150,000 minutes of manufacturing time being available. Because the allowable increase is 40,000 minutes, the dual value of $0.60 per minute will not change.
18. a.
The linear programming model is as follows: Min s.t.
30AN
+ 50AO
AN
+
+
25BN
AO
AN
+
BN BN
AO b.
+ 40BO +
BO
+
BO
Optimal solution: Model A Model B
New Line 50,000 30,000
Old Line 0 40,000
Total Cost $3,850,000
3-8
50,000 70,000 80,000 60,000
Linear Programming: Sensitivity Analysis and Interpretation of Solution c.
The first three constraints are binding because the values in the Slack/Surplus column for these constraints are zero. The fourth constraint, with a slack of 0 is nonbinding.
d.
The dual value for the new production line capacity constraint is -15. Because the dual value is negative, increasing the right-hand side of constraint 3 will cause the objective function value to decrease. Thus, every one unit increase in the right hand side of this constraint will reduce the total production cost by $15. In other words, an increase in capacity for the new production line is desirable.
e.
Because constraint 4 is not a binding constraint, any increase in the production line capacity of the old production line will have no effect on the optimal solution. Thus, there is no benefit in increasing the capacity of the old production line.
f.
The reduced cost for Model A made on the old production line is 5. Thus, the cost would have to decrease by at least $5 before any units of model A would be produced on the old production line.
g.
The right hand side range for constraint 2 shows an allowable decrease of 20,000. Thus, if the minimum production requirement is reduced 10,000 units to 60,000, the dual value of 40 is applicable. Thus, total cost would decrease by 10,000(40) = $400,000.
19. a.
Let
P1 = units of product 1 P2 = units of product 2 P3 = units of product 3
Max s.t.
30P1 + 50P2 0.5P1 + 2P2 P1 + P2 2P1 + 5P2 0.5P1 0.5P2 -0.2P1 0.2P2 P1, P2, P3 0
+ + + + +
20P3 0.75P3 0.5P3 2P3 0.5P3 0.8P3
40 40 100 0 0
The optimal solution is
Optimal Objective Value 1250.00000 Variable
Value
Reduced Cost
P1
25.00000
0.00000
P2
0.00000
-7.50000
P3
25.00000
0.00000
Constraint
Slack/Surplus
Dual Value
1
8.75000
0.00000
2
2.50000
0.00000
3
0.00000
12.50000
4
0.00000
10.00000
5
15.00000
0.00000
3-9
Machine 1 Machine 2 Labor Max P1 Min P3
Chapter 3
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
30.00000
Infinite
10.00000
50.00000
7.50000
Infinite
20.00000
10.00000
4.28571
RHS
Allowable
Allowable
Value
Increase
Decrease
40.00000
Infinite
8.75000
40.00000
Infinite
2.50000
100.00000
6.66667
100.00000
0.00000
5.00000
25.00000
0.00000
15.00000
Infinite
b.
Machine Hours Schedule: Machine 1 31.25 Hours Machine 2 37.50 Hours
c.
$12.50
d.
Increase labor hours to 120; the new optimal product mix is P1 = 24 P2 = 8 P3 = 16 Profit = $1440
20. a.
Let
Max s.t.
H = amount allocated to home loans P = amount allocated to personal loans A = amount allocated to automobile loans 0.07H
+
0.12P
+
0.09A
H 0.6H
+ -
P 0.4P P
+ -
A 0.4A 0.6A
=
1,000,000 0 0
Amount of New Funds Minimum Home Loans Personal Loan Requirement
b.
H = $400,000 P = $225,000 A = $375,000 Total annual return = $88,750 Annual percentage return = 8.875%
c.
The objective coefficient range for H is No Lower Limit to 0.101. Since 0.09 is within the range, the solution obtained in part (b) will not change.
d.
The dual value for constraint 1 is 0.089. The right-hand-side range for constraint 1 is 0 to No Upper Limit. Therefore, increasing the amount of new funds available by $10,000 will increase the total annual return by 0.089 (10,000) = $890.
3 - 10
Linear Programming: Sensitivity Analysis and Interpretation of Solution e.
The second constraint now becomes -0.61H - 0.39P - 0.39A 0 The new optimal solution is H = $390,000 P = $228,750 A = $381,250 Total annual return = $89,062.50, an increase of $312.50 Annual percentage return = 8.906%, an increase of approximately 0.031%.
21. a.
Let S1 S2 = D1 = D2 = B1 =
= SuperSaver rentals allocated to room type I SuperSaver rentals allocated to room type II Deluxe rentals allocated to room type I Deluxe rentals allocated to room type II Business rentals allocated to room type II
The linear programming formulation and solution is given. MAX 30S1+20S2+35D1+30D2+40B2 S.T. 1) 2) 3) 4) 5)
1S1+1S2<130 1D1+1D2<60 1B2<50 1S1+1D1<100 1S2+1D2+1B2<120
The optimal solution is
Optimal Objective Value 7000.00000 Variable
Value
Reduced Cost
S1
100.00000
0.00000
D1
0.00000
-5.00000
S2
10.00000
0.00000
D2
60.00000
0.00000
B2
50.00000
0.00000
Constraint
Slack/Surplus
Dual Value
1
20.00000
0.00000
2
0.00000
10.00000
3
0.00000
20.00000
4
0.00000
30.00000
5
0.00000
20.00000
3 - 11
Chapter 3
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
30.00000
Infinite
5.00000
35.00000
5.00000
Infinite
20.00000
5.00000
20.00000
30.00000
Infinite
5.00000
40.00000
Infinite
20.00000
RHS
Allowable
Allowable
Value
Increase
Decrease
130.00000
Infinite
20.00000
60.00000
10.00000
20.00000
50.00000
10.00000
20.00000
100.00000
20.00000
100.00000
120.00000
20.00000
10.00000
20 SuperSaver rentals will have to be turned away if demands materialize as forecast. b.
RoundTree should accept 110 SuperSaver reservations, 60 Deluxe reservations and 50 Business reservations.
c.
Yes, the effect of a person upgrading is an increase in demand for Deluxe accommodations from 60 to 61. From constraint 2, we see that such an increase in demand will increase profit by $10. The added cost of the breakfast is only $5.
d.
Convert to a Type I room. From the dual value to constraint 4 we see that this will increase profit by $30.
e.
Yes. We would need the forecast of demand for each rental class on the next night. Using the demand forecasts, we would modify the right-hand sides of the first three constraints and resolve.
22. a.
Let
Max s.t.
L = number of hours assigned to Lisa D = number of hours assigned to David S = amount allocated to Sarah 30L
+
25D
+
18S
L 0.6L -0.15L -0.25L L
+ -
D 0.4D 0.15D 0.25D
+
S
+ +
0.85S S
=
100 0 0 0 50
Total Time Lisa 40% requirement Minimum Sarah Maximum Sarah Maximum Lisa
b.
L = 48 hours D = 72 Hours S = 30 Hours Total Cost = $3780
c.
The dual value for constraint 5 is 0. Therefore, additional hours for Lisa will not change the solution.
3 - 12
Linear Programming: Sensitivity Analysis and Interpretation of Solution
d.
23. a.
The dual value for constraint 3 is 0. Because there is No Lower Limit on the right-hand-side range, the optimal solution will not change. Resolving the problem without this constraint will also show that the solution obtained in (b) does not change. Constraint 3, therefore, is really a redundant constraint. Let
C1 = units of component 1 manufactured C2 = units of component 2 manufactured C3 = units of component 3 manufactured
Max s.t.
8C1 + 6C1 + 4C1 +
6C2 + 4C2 + 5C2 +
C1 C2 C1
9C3 4C3 2C3 C3
7200 6600 200 1000 1000 600 C1, C2, C3 0
The optimal solution is C1 = 600 C2 = 700 C3 = 200
b. Variable C1
Objective Coefficient Range No Lower Limit to 9.0
C2
5.33 to 9.0
C3
6.00 to No Lower Limit
Individual changes in the profit coefficients within these ranges will not cause a change in the optimal number of components to produce. Constraint 1 2 3 4 5 6
Right-Hand-Side Range 4400 to 7440 6300 to No Upper Limit 100 to 900 600 to No Upper Limit 700 to No Upper Limit 514.29 to 1000
These are the ranges over which the dual values for the associated constraints are applicable.
24.
d.
Nothing, since there are 300 minutes of slack time on the grinder at the optimal solution.
e.
No, since at that price it would not be profitable to produce any of component 3.
Let
A = number of shares of stock A
3 - 13
Chapter 3 B = number of shares of stock B C = number of shares of stock C D = number of shares of stock D a.
To get data on a per share basis multiply price by rate of return or risk measure value. Min s.t.
10A
+ 3.5B
+
4C
+ 3.2D
100A 12A 100A
+ +
+ +
80C 4.8C
+ 40D + 4D
50B 4B 50B
80C 40D A, B, C, D 0
=
200,000 18,000 100,000 100,000 100,000 100,000
(9% of 200,00)
Solution: A = 333.3, B = 0, C = 833.3, D = 2500 Risk: 14,666.7 Return: 18,000 (9%) from constraint 2
b. Max s.t.
12A
+
4B
+
4.8C
+
4D
100A 100A
+
50B
+
80C
+
40D
50B 80C 40D
=
200,000 100,000 100,000 100,000 100,000
A, B, C, D 0 Solution: A = 1000, B = 0, C = 0, D = 2500 Risk: 10A + 3.5B + 4C + 3.2D = 18,000 Return: 22,000 (11%) c.
The return in part (b) is $4,000 or 2% greater, but the risk index has increased by 3,333. Obtaining a reasonable return with a lower risk is a preferred strategy in many financial firms. The more speculative, higher return investments are not always preferred because of their associated higher risk.
25. a.
Let O2 O3 C1 C2 C3
O1 = percentage of Oak cabinets assigned to cabinetmaker 1 = percentage of Oak cabinets assigned to cabinetmaker 2 = percentage of Oak cabinets assigned to cabinetmaker 3 = percentage of Cherry cabinets assigned to cabinetmaker 1 = percentage of Cherry cabinets assigned to cabinetmaker 2 = percentage of Cherry cabinets assigned to cabinetmaker 3
3 - 14
Linear Programming: Sensitivity Analysis and Interpretation of Solution Min 1800 O1 + 1764 O2 + 1650 O3 + 2160 C1 + 2016 C2 + 1925 C3 s.t. 50 O1 + 60 C1 40 42O2 + 48 C2 30 30 O3 + 35 C3 35 O1 + O2 + O3 =1 C1 + C2 + C3 = 1 O1, O2, O3, C1, C2, C3 0
Hours avail. 1 Hours avail. 2 Hours avail. 3 Oak Cherry
Note: objective function coefficients are obtained by multiplying the hours required to complete all the oak or cherry cabinets times the corresponding cost per hour. For example, 1800 for O1 is the product of 50 and 36, 1764 for O2 is the product of 42 and 42 and so on. b. Cabinetmaker 1 O1 = 0.271 C1 = 0.000
Oak Cherry
Cabinetmaker 2 O2 = 0.000 C2 = 0.625
Cabinetmaker 3 O3 = 0.729 C3 = 0.375
Total Cost = $3672.50 c.
No, since cabinetmaker 1 has a slack of 26.458 hours. Alternatively, since the dual value for constraint 1 is 0, increasing the right hand side of constraint 1 will not change the value of the optimal solution.
d.
The dual value for constraint 2 is -1.750. The upper limit on the right-hand-side range is 41.143. Therefore, each additional hour of time for cabinetmaker 2 will reduce total cost by $1.75 per hour, up to a maximum of 41.143 hours.
e.
The new objective function coefficients for O2 and C2 are 42(38) = 1596 and 48(38) = 1824, respectively. The optimal solution does not change but the total cost decreases to $3552.50.
26. a.
Let
M1 =
units of component 1 manufactured
M2 = units of component 2 manufactured M3 = units of component 3 manufactured P1 = units of component 1 purchased P2 = units of component 2 purchased P3 = units of component 3 purchased Min s.t.
4.50 M1 + 5.00M2 2M1 + 1M1 +
1.5M1 +
+ 2.75M3
3M2 + 1.5M2 + 2M2 +
+ 6.50P1
+ 8.80P2 +
7.00P3 21,600 Production
4M3
15,000 Assembly
3M3 5M3
M1
18,000 Testing/Packaging +
1M2 1M3 M1, M2, M3, P1, P2, P3 0 b.
3 - 15
1P1
+
1P2
+
=
6,000 Component 1
=
4,000 Component 2
1P3 =
3,500 Component 3
Chapter 3 Source Manufacture Purchase
Component 1 2000 4000
Component 2 4000 0
Component 3 1400 2100
Total Cost: $73,550 c.
Since the slack is 0 in the production and the testing & packaging departments, these department are limiting Benson's manufacturing quantities. Dual value information: Production $-0.906/minute x 60 minutes = $-54.36 per hour Testing/Packaging $-0.125/minute x 60 minutes = $ -7.50 per hour
d.
27.
The dual value is $7.969. This tells us that the value of the optimal solution will worsen (the cost will increase) by $7.969 for an additional unit of component 2. Note that although component 2 has a purchase cost per unit of $8.80, it would only cost Benson $7.969 to obtain an additional unit of component 2.
a. LP Formulation The objective function maximizes the total revenue generated from wet-harvested cranberries and dry-harvested cranberries. Each barrel of wet-harvested cranberries generates $17.50 in revenue; dry-harvested cranberries earn $32.50 per barrel. The first constraint ensures that no more than 5,000 total barrels of cranberries are harvested. The second constraint states that every barrel of wetharvested cranberries requires 0.04 hours in the dechaffing operation, every barrel of dry-harvested cranberries requires 0.18 hours in the dechaffing operation, and there are 1008 dechaffing hours available . The third and fourth constraints represent similar requirements for the cleaning and drying operations. Let W = barrels of cranberries harvested using wet method Let D = barrels of cranberries harvested using dry method Max
17.5W
+
32.5D
W
+
D
≤
5,000 Total Harvest
.04W
+
.18D
≤
1,008 Dechaffing
.10W
+
.32D
≤
1,008 Cleaning
≤
1,008 Drying
,
D
≥
.22W W
b.
0 Nonegativity
The optimal solution is to harvest 2690.909 barrels of wet cranberries and 2309.091 barrels of dry cranberries.
c. OPTIMAL SOLUTION Optimal Objective Value 122136.360 Variable
Value
3 - 16
Reduced Cost
Linear Programming: Sensitivity Analysis and Interpretation of Solution
W
2690.909
0.00000
D
2309.091
0.00000
Constraint
Slack/Surplus
Dual Value
1
0.00000
10.682
2
484.727
0.000
3
0.00000
68.182
4
416.000
0.000
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
17.500
15.000
7.340
32.500
23.500
15.000
RHS
Allowable
Allowable
Value
Increase
Decrease
5000.00000
1300.000
1850.000
523.273
Infinite
484.727
1008.000
592.000
416.000
582.000
Infinite
416.000
From this output, we see that dechaffing (Constraint 2) is not a binding constraint because only 523.273 of the 1008 available hours are being used. Fresh Made should not purchase additional dechaffing capacity. d.
28. a.
Assuming everything else remains unchanged, each additional hour of cleaning capacity (Constraint 3) added (subtracted) from the 1,008 available hours increases (decreases) revenues $68.18 based on the Dual Values shown here. This is true for any cleaning capacity value between 1008 – 416 = 592 and 1008 + 592 = 1600. Let
G = amount invested in growth stock fund S = amount invested in income stock fund M = amount invested in money market fund
Max s.t.
0.20G +
0.10S +
0.06M
0.10G + G
0.05S +
0.01M M M
S G
+
S
+
G, S, M 0 b.
The solution to Hartmann's portfolio mix problem is given. Optimal Objective Value
3 - 17
(0.05)(300,000) (0.10)(300,000) (0.10)(300,000) (0.20)(300,000) 300,000
Hartmann's max risk Growth fund min. Income fund min. Money market min, Funds available
Chapter 3
36000.00000
c.
Value
G
120000.00000
Reduced Cost 0.00000
S
30000.00000
0.00000
M
150000.00000
0.00000
Constraint
Slack/Surplus
Dual Value
1
0.00000
1.55556
2
90000.00000
0.00000
3
0.00000
-0.02222
4
90000.00000
0.00000
5
0.00000
0.12000
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
0.20000
Infinite
0.05000
0.10000
0.02222
1.20000
0.06000
0.14000
0.04000
RHS
Allowable
Allowable
Value
Increase
Decrease
0.00000
8100.00000
8100.00000
0.00000
90000.00000
Infinite
0.00000
162000.00000
30000.00000
0.00000
90000.00000
Infinite
300000.00000
Infinite
300000.00000
These are given by the objective coefficient ranges. The portfolio above will be optimal as long as the yields remain in the following intervals: Growth stock Income stock Money Market
d.
Variable
0.15 No Lower Limit 0.02
c1 0.60 < c2 0.122 c3 0.20
The dual value for the first constraint provides this information. A change in the risk index from 0.05 to 0.06 would increase the constraint RHS by 3000 (from 15,000 to 18,000). This is within the right-hand-side range, so the dual value of 1.556 is applicable. The value of the optimal solution would increase by (3000)(1.556) = 4668. Hartmann's yield with a risk index of 0.05 is 36,000 / 300,000 = 0.12
3 - 18
Linear Programming: Sensitivity Analysis and Interpretation of Solution His yield with a risk index of 0.06 would be 40,668 / 300,000 = 0.1356 e.
This change is outside the objective coefficient range so we must re-solve the problem. The solution is shown below.
LINEAR PROGRAMMING PROBLEM MAX .1G
+ .1S
+ .06M
S.T. 1) 2) 3) 4) 5)
.1G + .05S + .01M < 15000 G > 30000 S > 30000 M > 60000 G + S + M < 300000
OPTIMAL SOLUTION Optimal Objective Value 27600.00000 Variable
Value
Reduced Cost
G
30000.00000
0.00000
S
210000.00000
0.00000
M
60000.00000
0.00000
Constraint
Slack/Surplus
Dual Value
1
900.00000
0.00000
2
0.00000
0.00000
3
180000.00000
0.00000
4
0.00000
-0.04000
5
0.00000
0.09200
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
0.10000
0.00000
0.92000
0.10000
Infinite
0.00000
0.06000
0.04000
0.33498
RHS
Allowable
Allowable
Value
Increase
Decrease
0.00000
Infinite
900.00000
0.00000
18000.00000
30000.00000
3 - 19
Chapter 3
0.00000
180000.00000
Infinite
0.00000
180000.00000
22500.00000
300000.00000
Infinite
299999.99999
f.
The client's risk index and the amount of funds available.
g.
With the new yield estimates, Pfeiffer would solve a new linear program to find the optimal portfolio mix for each client. Then by summing across all 50 clients he would determine the total amount that should be placed in a growth fund, an income fund, and a money market fund. Pfeiffer then would make the necessary switches to have the correct total amount in each account. There would be no actual switching of funds for individual clients.
29. a.
b.
Relevant cost since LaJolla Beverage Products can purchase wine and fruit juice on an as - needed basis. Let
Max s.t.
W = gallons of white wine R = gallons of rose wine F = gallons of fruit juice 1.5 W + 0.5W -0.2W -0.3W -0.2W W
+ + -
1R
+
2F
0.5R 0.8R 0.7R 0.2R
+
0.5F 0.2F 0.3F 0.8F
R
=
0 0 0 0 10000 8000
% white % rose minimum % rose maximum % fruit juice Available white Available rose
W, R, F 0 Optimal Solution: W = 10,000, R = 6000, F = 4000 profit contribution = $29,000. c.
Since the cost of the wine is a relevant cost, the dual value of $2.90 is the maximum premium (over the normal price of $1.00) that LaJolla Beverage Products should be willing to pay to obtain one additional gallon of white wine. In other words, at a price of $3.90 = $2.90 + $1.00, the additional cost is exactly equal to the additional revenue.
d.
No; only 6000 gallons of the rose are currently being used.
e.
Requiring 50% plus one gallon of white wine would reduce profit by $2.40. Note to instructor: Although this explanation is technically correct, it does not provide an explanation that is especially useful in the context of the problem. Alternatively, we find it useful to explore the question of what would happen if the white wine requirement were changed to at least 51%. Note that in this case, the first constraint would change to 0.49W - 0.51R - 0.51F 0. This shows the student that the coefficients on the left-hand side are changing; note that this is beyond the scope of sensitivity analysis discussed in this chapter. Resolving the problem with this revised constraint will show the effect on profit of a 1% change.
f.
Allowing the amount of fruit juice to exceed 20% by one gallon will increase profit by $1.00.
30. a.
Let
L = minutes devoted to local news
3 - 20
Linear Programming: Sensitivity Analysis and Interpretation of Solution N = minutes devoted to national news W = minutes devoted to weather S = minutes devoted to sports Min s.t.
300L
+
200N +
100W +
L L L
+
N +
W +
+
N
-L
-
N
100S
S = W S + S W L, N, W, S 0
20 3 10 0 0 4
Time available 15% local 50% requirement Weather - sports Sports requirement 20% weather
Optimal Solution: L = 3, N = 7, W = 5, S = 5 Total cost = $3,300 b.
Each additional minute of broadcast time increases cost by $100; conversely, each minute reduced will decrease cost by $100. These interpretations are valid for increase up to 10 minutes and decreases up to 2 minutes from the current level of 20 minutes.
c.
If local coverage is increased by 1 minute, total cost will increase by $100.
d.
If the time devoted to local and national news is increased by 1 minute, total cost will increase by $100.
e.
Increasing the sports by one minute will have no effect for this constraint since the dual value is 0.
31. a.
Let
Min s.t.
B = number of copies done by Benson Printing J = number of copies done by Johnson Printing L = number of copies done by Lakeside Litho 2.45B +
2.5J +
B J 0.9B + B -
0.99J + 0.1J
2.75L 30,000 50,000 L 50,000 0.995L = 75,000 0 L 30,000 B, J, L 0
Benson Johnson Lakeside # useful reports Benson - Johnson % Minimum Lakeside
Optimal Solution: B = 4,181, J = 41,806, L = 30,000 b.
Suppose that Benson printing has a defective rate of 2% instead of 10%. The new optimal solution would increase the copies assigned to Benson printing to 30,000. In this case, the additional copies assigned to Benson Printing would reduce on a one-for-one basis the number assigned to Johnson Printing.
c.
If the Lakeside Litho requirement is reduced by 1 unit, total cost will decrease by $0.2210.
32. a.
Let P1 = number of PT-100 battery packs produced at the Philippines plant P2 = number of PT-200 battery packs produced at the Philippines plant P3 = number of PT-300 battery packs produced at the Philippines plant M1 = number of PT-100 battery packs produced at the Mexico plant
3 - 21
Chapter 3 M2 = number of PT-200 battery packs produced at the Mexico plant M3 = number of PT-300 battery packs produced at the Mexico plant Total production and shipping costs ($/unit) Philippines 1.13 1.16 1.52
PT-100 PT-200 PT-300 Min s.t.
1.13P1
+
P1
+
Mexico 1.08 1.16 1.25
1.16P2
+
P2
+
1.52P3
+
1.08M1
+
1.16M2
+
1.25M3
M1 M2 P3
P1
+
+
M3
P2 M1
+
M2
P3 M3
= = =
200,000 100,000 150,000 175,000 160,000 75,000 100,000
P1, P2, P3, M1, M2, M3 0 b.
The optimal solution is as follows: PT-100 PT-200 PT-300
Philippines 40,000 100,000 50,000
Mexico 160,000 0 100,000
The total production and transportation cost is $535,000. c.
The range of optimality for the objective function coefficient for P1 shows a lower limit of $1.08. Thus, the production and/or shipping cost would have to decrease by at least 5 cents per unit.
d.
The range of optimality for the objective function coefficient for M2 shows a lower limit of $1.11. Thus, the production and/or shipping cost could have to decrease by at least 5 cents per unit.
3 - 22
Chapter 4 Linear Programming Applications in Marketing, Finance and Operations Management Learning Objectives 1.
Learn about applications of linear programming that have been encountered in practice.
2.
Develop an appreciation for the diversity of problems that can be modeled as linear programs.
3.
Obtain practice and experience in formulating realistic linear programming models.
4.
Understand linear programming applications such as: media selection portfolio selection blending problems
production scheduling work force assignments
Note to Instructor The application problems of Chapter 4 have been designed to give the student an understanding and appreciation of the broad range of problems that can be approached by linear programming. While the problems are indicative of the many linear programming applications, they have been kept relatively small in order to ease the student's formulation and solution effort. Each problem will give the student an opportunity to practice formulating a linear programming model. However, the solution and the interpretation of the solution will require the use of a software package such as Microsoft Excel's Solver or LINGO.
4-1
Chapter 4
Solutions: 1.
a.
Let T = number of television spot advertisements R = number of radio advertisements N = number of newspaper advertisements
Max s.t.
100,000T
+
18,000R
+
40,000N
2,000T T
+
300R
+
600N
-
N 0.5N 0.1N
R -0.5T 0.9T
+ -
0.5R 0.1R
18,200 10 20 10 0 0
T, R, N, 0
Solution:
Budget $ $8,000 4,200 6,000 $18,200
T=4 R = 14 N = 10
Audience = 1,052,000.
OPTIMAL SOLUTION Optimal Objective Value 1052000.00000 Variable
Value
Reduced Cost
T
4.00000
0.00000
R
14.00000
0.00000
N
10.00000
11826.08695
Constraint
Slack/Surplus
Dual Value
1
0.00000
51.30430
2
6.00000
0.00000
3
6.00000
0.00000
4
0.00000
11826.08695
5
0.00000
5217.39130
6
1.20000
0.00000
4-2
Budget Max TV Max Radio Max News Max 50% Radio Min 10% TV
Linear Programming Applications in Marketing, Finance and Operations Management
2.
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
100000.00000
20000.00000
118000.00000
18000.00000
Infinite
3000.00000
40000.00000
Infinite
11826.08695
RHS
Allowable
Allowable
Value
Increase
Decrease
18200.00000
13800.00000
3450.00000
10.00000 20.00000
Infinite Infinite
6.00000 6.00000
10.00000
2.93600
10.00000
0.00000
2.93617
8.05000
0.00000
1.20000
Infinite
b.
The dual value for the budget constraint is 51.30. Thus, a $100 increase in budget should provide an increase in audience coverage of approximately 5,130. The right-hand-side range for the budget constraint will show this interpretation is correct.
a.
Let
x1 = units of product 1 produced x2 = units of product 2 produced Max s.t.
30x1
+
15x2
x1 0.30x1 0.20x1
+ + +
0.35x2 0.20x2 0.50x2
100 36 50
Dept. A Dept. B Dept. C
x1 , x2 0 Solution: x1 = 77.89, x2 = 63.16 Profit = 3284.21 b.
The dual value for Dept. A is $15.79, for Dept. B it is $47.37, and for Dept. C it is $0.00. Therefore we would attempt to schedule overtime in Departments A and B. Assuming the current labor available is a sunk cost, we should be willing to pay up to $15.79 per hour in Department A and up to $47.37 in Department B.
c.
Let
xA = hours of overtime in Dept. A xB = hours of overtime in Dept. B xC = hours of overtime in Dept. C
Max
30x1
+
15x2
-
18xA
4-3
-
22.5xB
-
12xC
Chapter 4
s.t. x1 0.30x1 0.20x1
+ + +
0.35x2 0.20x2 0.50x2
xA
-
xB
-
-
xC
xA xB xC
100 36 50 10 6 8
x1, x2, xA, xB, xC 0 x1 = 87.21 x2 = 65.12 Profit = $3341.34 Overtime Dept. A Dept. B Dept. C
10 hrs. 3.186 hrs 0 hours
Increase in Profit from overtime = $3341.34 - 3284.21 = $57.13 3.
x1 = $ automobile loans x2 = $ furniture loans x3 = $ other secured loans x4 = $ signature loans x5 = $ "risk free" securities
Max s.t.
0.08x1 +
0.10x2 +
0.11x3
+
0.12x4
+
0.09x5 x5
or
-0.10x1 x1 +
0.10x2 x2 + x2 +
or
-
x1 +
+ x2 +
or
0.10x3 x3 x3 x3 x3 x3
+
x4 0.90x4
x4 x4 x4
+ + +
x5 x5
+
600,000 0.10(x1 + x2 + x3 + x4) 0 x1 0 x5
[1]
0 = 2,000,000
[4] [5]
x1, x2, x3, x4, x5 0 Solution: Automobile Loans Furniture Loans Other Secured Loans Signature Loans Risk Free Loans
(x1) (x2) (x3) (x4) (x5)
Annual Return $188,800 (9.44%)
4-4
= = = = =
$630,000 $170,000 $460,000 $140,000 $600,000
[2] [3]
Linear Programming Applications in Marketing, Finance and Operations Management 4.
a.
x1 = pounds of bean 1 x2 = pounds of bean 2 x3 = pounds of bean 3
Min s.t.
0.50x1 + 0.70x2
+
0.45x3
75x1 + 85x2 + 60 x3 x1 + x2 + x3 10x2 - 15x3
or
75
86 x1 + 88 x2 + 75x3 x1 + x2 + x3 or
6x1 + x1
8x2
-
x2
Aroma
80 5x3
x2 x1 +
0
+ x1 , x2 , x3 0
x3 x3
0 500 600 400 = 1000
Taste Bean 1 Bean 2 Bean 3 1000 pounds
Optimal Solution: x1 = 500, x2 = 300, x3 = 200, Cost: $550 b.
Cost per pound = $550/1000 = $0.55
c.
Surplus for aroma: s1 = 0; thus aroma rating = 75 Surplus for taste: s2 = 4400; thus taste rating = 80 + 4400/1000 lbs. = 84.4
d. 5.
Dual value = $0.60. Extra coffee can be produced at a cost of $0.60 per pound.
a. Let W = # of servings of Wimpy to make D = # of servings of Dial 911 to make Max .45 W + .58 D s.t. .25 W + .25 D ≤ 20 .25 W + .4 D ≤ 15 5 W + 2 D ≤ 88 5 D ≤ 60 W, D ≥ 0
6.
(Beef) (Onions) (Special Sauce) (Hot Sauce)
b.
Solution: W = 12.8, D = 12, Profit = $12.72.
c.
Dual value for special sauce = $.09
d.
Solution: W=13, D=12, Profit = $12.81. Note: $12.81-$12.72 = $.09. Dual value confirmed. Let
x1 = units of product 1 x2 = units of product 2 b1 = labor-hours Dept. A
4-5
Chapter 4 b2 = labor-hours Dept. B Max s.t.
25x1
+ 20x2
+
0b1
6x1 12x1
+ 8x2 + 10x2
-
1b1 1b1
+
0b2
+
1b2 1b2
= =
0 0 900
x1 , x2 , b 1 , b 2 0 Solution: x1 = 50, x2 = 0, b1 = 300, b2 = 600 Profit: $1,250 7.
a.
Let
F G1 G2 Si
= total funds required to meet the six years of payments = units of government security 1 = units of government security 2 = investment in savings at the beginning of year i
Note: All decision variables are expressed in thousands of dollars MIN F S.T. 1) 2) 3) 4) 5) 6)
F - 1.055G1 - 1.000G2 - S1 = 190 .0675G1 + .05125G2 +1.04S1 - S2 = 215 .0675G1 + .05125G2 + 1.04S2 - S3 = 240 1.0675G1 + .05125G2 + 1.04S3 - S4 = 285 1.05125G2 + 1.04S4 - S5 = 315 1.04S5 = 460
OPTIMAL SOLUTION Optimal Objective Value 1484.96660 Variable
Value
Reduced Cost
G1
232.39356
0.00000
G2
720.38782
0.00000
F
1484.96655
0.00000
S1
329.40353
0.00000
S2
180.18611
0.00000
S3
0.00000
0.02077
S4
0.00000
0.01942
S5
442.30769
0.00000
Constraint 1
Slack/Surplus 0.00000
Dual Value 1.00000
4-6
Linear Programming Applications in Marketing, Finance and Operations Management
2
0.00000
0.96154
3
0.00000
0.92456
4
0.00000
0.86903
5
0.00000
0.81693
6
0.00000
0.78551
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
0.00000
0.02132
0.01973
0.00000
0.01963
Infinite
1.00000
Infinite
1.00000
0.00000
0.65430
0.01980
0.00000
1.28344
0.02026
0.00000
Infinite
0.02077
0.00000
Infinite
0.01942
0.00000
Infinite
Infinite
RHS
Allowable
Allowable
Value
Increase
Decrease
0.00000
Infinite
1484.96655
0.00000
Infinite
342.57967
0.00000
Infinite
187.39356
0.00000
2762.03307
248.08012
0.00000
3824.23689
757.30769
0.00000
3977.20636
460.00000
The current investment required is $1,484,967. This calls for investing $232,394 in government security 1 and $720,388 in government security 2. The amounts, placed in savings are $329,404, $180,186 and $442,308 for years 1,2 and 5 respectively. No funds are placed in savings for years 3, and 4. b.
The dual value for constraint 6 indicates that each $1 increase in the payment required at the beginning of year 6 will increase the amount of money Hoxworth must pay the trustee by $0.78551. A per unit decrease would therefore decrease the cost by $0.78551. The allowable decrease on the right-hand-side range is $460,000 so a reduction in the payment at the beginning of year 6 by $60,000 will save Hoxworth $60,000 (0.78551) = $47,131.
c.
The dual value for constraint 1 shows that every dollar of reduction in the initial payment leads to a drop in the objective function value of $1.00. So Hoxworth should be willing to pay anything less than $40,000.
d.
To reformulate this problem, one additional variable needs to be added, the right-hand sides for the original constraints need to be shifted ahead by one, and the right-hand side of the first constraint needs to be set equal to zero. The value of the optimal solution with this formulation is $1,417,739. Hoxworth will save $67,228 by having the payments moved to the end of each year.
4-7
Chapter 4
The revised formulation is shown below: MIN F S.T. 1) 2) 3) 4) 5) 6) 7) 8.
Let
F - 1.055G1 - 1.000G2 - S1 = 0 .0675G1 + .05125G2 + 1.04S1 - S2 = 190 .0675G1 + .05125G2 + 1.04S2 - S3 = 215 1.0675G1 + .05125G2 + 1.04S3 - S4 = 240 1.05125G2 +1.04S4 - S5 = 285 1.04S5 - S6 = 315 1.04S6 - S7 = 460 x1 = the number of officers scheduled to begin at 8:00 a.m. x2 = the number of officers scheduled to begin at noon x3 = the number of officers scheduled to begin at 4:00 p.m. x4 = the number of officers scheduled to begin at 8:00 p.m. x5 = the number of officers scheduled to begin at midnight x6 = the number of officers scheduled to begin at 4:00 a.m.
The objective function to minimize the number of officers required is as follows: Min
x1 + x2 + x3 + x4 + x5 + x6
The constraints require the total number of officers of duty each of the six four-hour periods to be at least equal to the minimum officer requirements. The constraints for the six four-hour periods are as follows: Time of Day 8:00 a.m. - noon noon to 4:00 p.m. 4:00 p.m. - 8:00 p.m. 8:00 p.m. - midnight midnight - 4:00 a.m. 4:00 a.m. - 8:00 a.m.
x1 x1
+ x6 + x2 x2
+ x3 x3
+ x4 x4
x1, x2, x3, x4, x5, x6 0
Schedule 19 officers as follows: x1 = 3 begin at 8:00 a.m. x2 = 3 begin at noon x3 = 7 begin at 4:00 p.m. x4 = 0 begin at 8:00 p.m.
4-8
+ x5 x5
+ x6
5 6 10 7 4 6
Linear Programming Applications in Marketing, Finance and Operations Management x5 = 4 begin at midnight x6 = 2 begin at 4:00 a.m. 9.
Let Xi = the number of call center employees who start work on day i (i=1 = Monday, i=2=Tuesday…) Min X1 + X 2 + X 3 + X 4 + X 5 + X 6 + X 7 s.t. X1 + X4 + X5 + X6 + X7 ≥ 75 X1 + X 2 + X5 + X6 + X7 ≥ 50 X1 + X 2 + X 3 + X6 + X7 ≥ 45 X1 + X 2 + X 3 + X4 +X7 ≥ 60 X1 + X 2 + X 3 + X4 + X 5 ≥ 90 X2 + X 3 + X4 + X 5 + X 6 ≥ 75 X3 + X4 + X5 + X6 + X7 ≥ 45 X1 , X2 , X3 , X4 , X5 , X6 , X7 ≥ 0 Solution: X1 = 20 , X2 = 20 , X3 = 0 , X4 = 45, X5 = 5, X6 = 5, X7 = 0 Total Number of Employees = 95 Excess employees: Thursday = 25, Sunday = 10, all others = 0. Note: There are alternative optima to this problem (Number of employees may differ from above, but will have objective function value = 95).
10. a.
Let S = the proportion of funds invested in stocks B = the proportion of funds invested in bonds M = the proportion of funds invested in mutual funds C = the proportion of funds invested in cash The linear program and optimal solution are as follows: MAX 0.1S+0.03B+0.04M+0.01C S.T. 1) 2) 3) 4) 5) 6)
1S+1B+1M+1C=1 0.8S+0.2B+0.3M<0.4 1S<0.75 -1B+1M>0 1C>0.1 1C<0.3
OPTIMAL SOLUTION Optimal Objective Value 0.05410 Variable
Value
Reduced Cost
S
0.40909
0.00000
B
0.14545
0.00000
M
0.14545
0.00000
4-9
Chapter 4
C
0.30000
Constraint
Slack/Surplus
0.00455 Dual Value
1
0.00000
0.00000
2
0.00000
0.11818
3
0.34091
0.00000
4
0.00000
-0.00091
5
0.20000
0.00000
6
0.00000
0.00545
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
0.10000
Infinite
0.01000
0.03000
0.00625
0.00200
0.04000
0.00167
Infinite
0.01000
Infinite
0.00455
RHS
Allowable
Allowable
Value
Increase
Decrease
1.00000
0.90000
0.20000
0.40000
0.16000
0.22500
0.75
Infinite
0.341
0.00000
0.32000
0.26667
0.10000
0.20000
Infinite
0.30000
0.20000
0.20000
The optimal allocation among the four investment alternatives is Stocks Bonds Mutual Funds Cash
40.9% 14.5% 14.5% 30.0%
The annual return associated with the optimal portfolio is 5.4% The total risk = 0.409(0.8) + 0.145(0.2) + 0.145(0.3) + 0.300(0.0) = 0.4 b.
Changing the right-hand-side value for constraint 2 to 0.18 and resolving we obtain the following optimal solution: Stocks Bonds Mutual Funds Cash
0.0% 36.0% 36.0% 28.0%
The annual return associated with the optimal portfolio is 2.52%
4 - 10
Linear Programming Applications in Marketing, Finance and Operations Management
The total risk = 0.0(0.8) + 0.36(0.2) + 0.36(0.3) + 0.28(0.0) = 0.18 c.
Changing the right-hand-side value for constraint 2 to 0.7 and resolving we obtain the following optimal solution: The optimal allocation among the four investment alternatives is Stocks Bonds Mutual Funds Cash
75.0% 0.0% 15.0% 10.0%
The annual return associated with the optimal portfolio is 8.2% The total risk = 0.75(0.8) + 0.0(0.2) + 0.15(0.3) + 0.10(0.0) = 0.65
11.
d.
Note that a maximum risk of 0.7 was specified for this aggressive investor, but that the risk index for the portfolio is only 0.65. Thus, this investor is willing to take more risk than the solution shown above provides. There are only two ways the investor can become even more aggressive: increase the proportion invested in stocks to more than 75% or reduce the cash requirement of at least 10% so that additional cash could be put into stocks. For the data given here, the investor should ask the investment advisor to relax either or both of these constraints.
e.
Defining the decision variables as proportions means the investment advisor can use the linear programming model for any investor, regardless of the amount of the investment. All the investor advisor needs to do is to establish the maximum total risk for the investor and resolve the problem using the new value for maximum total risk. Let
xij = units of component i purchased from supplier j
Min s.t.
12x11 + 13x12 + 14x13 x11 +
x12 +
+ 10x21 +
11x22 + 10x23
x13
x11
+ x12
x21 + x21 +
= 1000 = 800 600 1000 x23 800
x22 +
x23
x22
x13
+
x11, x12, x13, x21, x22, x23 0 Solution: Supplier 2
1 Component 1 Component 2
12.
600 0
3
400 0 0 800 Purchase Cost = $20,400
Let Bi = pounds of shrimp bought in week i, i = 1,2,3,4 Si = pounds of shrimp sold in week i, i = 1,2,3,4 Ii = pounds of shrimp held in storage (inventory) in week i Total purchase cost = 6.00B1 + 6.20B2 + 6.65B3 + 5.55B4
4 - 11
Chapter 4
Total sales revenue = 6.00S1 + 6.20S2 + 6.65S3 + 5.55S4 Total storage cost = 0.15I1 + 0.15I2 + 0.15I3 + 0.15I4 Total profit contribution = (total sales revenue) - (total purchase cost) - (total storage cost) Objective: maximize total profit contribution subject to balance equations for each week, storage capacity for each week, and ending inventory requirement for week 4. Max
6.00S1 + 6.20S2 + 6.65S3 + 5.55S4 - 6.00B1 - 6.20B2 - 6.65B3 - 5.55B4 - 0.15I1 - 0.15I2 0.15I3 - 0.15I4
s.t. 20,000 I1 I2 I3
+ + + +
B1 B2 B3 B4
-
S1 S2 S3 S4 I1 I2 I3 I4 I4 all variables 0
= = = =
I1 I2 I3 I4 100,000 100,000 100,000 100,000 25,000
Balance eq. - week 1 Balance eq. - week 2 Balance eq. - week 3 Balance eq. - week 4 Storage cap. - week 1 Storage cap. - week 2 Storage cap. - week 3 Storage cap. - week 4 Req'd inv. - week 4
Note that the first four constraints can be written as follows: I1 - B1 + S1 = 20,000 I1 - I2 + B2 - S 2 = 0 I2 - I3 + B3 - S 3 = 0 I3 - I4 + B4 - S 4 = 0 The optimal solution obtained follows: Week (i) 1 2 3 4
Bi 80,000 0 0 25,000
Si 0 0 100,000 0
Ii 100,000 100,000 0 25,000
Total profit contribution = $12,500 Note however, ASC started week 1 with 20,000 pounds of shrimp and ended week 4 with 25,000 pounds of shrimp. During the 4-week period, ASC has taken profits to reinvest and build inventory by 5000 pounds in anticipation of future higher prices. The amount of profit reinvested in inventory is ($5.55 + $0.15)(5000) = $28,500. Thus, total profit for the 4-week period including reinvested profit is $12,500 + $28,500 = $41,000. 13.
Let BR BD CR CD
= = = =
pounds of Brazilian beans purchased to produce Regular pounds of Brazilian beans purchased to produce DeCaf pounds of Colombian beans purchased to produce Regular pounds of Colombian beans purchased to produce DeCaf
Type of Bean
Cost per pound ($)
4 - 12
Linear Programming Applications in Marketing, Finance and Operations Management
Brazilian Colombian
1.10(0.47) = 0.517 1.10(0.62) = 0.682
Total revenue = 3.60(BR + CR) + 4.40(BD + CD) Total cost of beans = 0.517(BR + BD) + 0.682(CR + CD) Total production cost = 0.80(BR + CR) + 1.05(BD + CD) Total packaging cost = 0.25(BR + CR) + 0.25(BD + CD) Total contribution to profit = (total revenue) - (total cost of beans) - (total production cost)
Total contribution to profit = 2.033BR + 2.583BD + 1.868CR + 2.418CD Regular % constraint BR = 0.75(BR + CR) 0.25BR - 0.75CR = 0 DeCaf % constraint BD = 0.40(BD + CD) 0.60BD - 0.40CD = 0 Pounds of Regular: BR + CR = 1000 Pounds of DeCaf: BD + CD = 500 The complete linear program is Max s.t.
2.033BR
+ 2.583BD
0.25BR
+ 1.868CR -
+ 2.418CD
0.75CR
0.60BD
-
BR
0.40CD
= 0 = 0 = 1000 = 500
+ CR BD + CD BR, BD, CR, CD 0 The optimal solution is BR = 750, BD = 200, CR = 250, and CD = 300. The value of the optimal solution is $3233.75 14. a.
Let
xi = number of Classic 2l boats produced in Quarter i; i = 1,2,3,4 si = ending inventory of Classic 2l boats in Quarter i; i = 1,2,3,4
Min
10,000x1 + 11,000x2 + 12,100x3 + 13,310x4 + 250s1 + 250s2 + 300s3 + 300s4
s.t. x1 - s1 = 1900 s1 + x2 - s2 = 4000 s2 + x3 - s3 = 3000 s3 + x4 - s4 = 1500
Quarter 1 demand Quarter 2 demand Quarter 3 demand Quarter 4 demand
s4 500
Ending Inventory
4 - 13
Chapter 4 x1 4000
Quarter 1 capacity
x2 3000
Quarter 2 capacity
x3 2000
Quarter 3 capacity
x4 4000
Quarter 4 capacity
b. Quarter 1 2 3 4
15.
Production 4000 3000 2000 1900
Ending Inventory 2100 1100 100 500
Cost 40,525,000 33,275,000 24,230,000 25,439,000 $123,469,000
c.
The dual values tell us how much it would cost if demand were to increase by one additional unit. For example, in Quarter 2 the dual value is 12,760; thus, demand for one more boat in Quarter 2 will increase costs by $12,760.
d.
The dual value of 0 for Quarter 4 tells us we have excess capacity in Quarter 4. The negative dual values in Quarters 1-3 tell us how much increasing the production capacity will improve the objective function. For example, the dual price of -$2510 for Quarter 1 tells us that if capacity is increased by 1 unit for this quarter, costs will go down $2510. Let
x11 = gallons of crude 1 used to produce regular x12 = gallons of crude 1 used to produce high-octane x21 = gallons of crude 2 used to produce regular x22 = gallons of crude 2 used to produce high-octane
Min 0.10x11 + 0.10x12 + 0.15x21 + 0.15x22 s.t. Each gallon of regular must have at least 40% A. x11 + x21 = amount of regular produced 0.4(x11 + x21) = amount of A required for regular 0.2x11 + 0.50x21 = amount of A in (x11 + x21) gallons of regular gas 0.2x11 + 0.50x21 0.4x11 + 0.40x21 -0.2x11 + 0.10x21 0 Each gallon of high octane can have at most 50% B. x12 + x22 0.5(x12 + x22)
= amount high-octane = amount of B required for high octane
4 - 14
[1]
Linear Programming Applications in Marketing, Finance and Operations Management 0.60x12 + 0.30x22
= amount of B in (x12 + x22) gallons of high octane.
0.60x12 + 0.30x22 0.1x12 -
0.5x12 + 0.5x22 0
[2]
x11 + x21
800,000
[3]
x12 + x22
500,000
[4]
0.2x22
x11, x12, x21, x22 0 Optimal Solution: x11 = 266,667, x12 = 333,333, x21 = 533,333, x22 = 166,667 Cost = $165,000 16.
Let
Min s.t.
x1
xi = number of 10-inch rolls of paper processed by cutting alternative i; i = 1,2...,7 +
x2 +
6x1
x3
x4
+
+ 2x3 4x2 2x3
+ x4 + 2x4
x5 +
+
x6 +
x7
+ x5 + x6 + 4x7 + 3x5 + 2x6 + x6 + x7
1000 2000 4000
1 1/2" production 2 1/2" production 3 1/2" production
x1, x2, x3, x4, x5, x6, x7 0 x1 = 0 x2 = 125 x3 = 500 x4 = 1500 x5 = 0 x6 = 0 x7 = 0
2125 Rolls Production: 1 1/2" 1000 2 1/2" 2000 3 1/2" 4000
Waste: Cut alternative #4 (1/2" per roll) 750 inches. b.
Only the objective function needs to be changed. An objective function minimizing waste production and the new optimal solution are given. Min x1 + 0x2 + 0x3 + 0.5x4 + x5 + 0x6 + 0.5x7 x1 = 0 x2 = 500 x3 = 2000 2500 Rolls x4 = 0 x5 = 0 Production: x6 = 0 1 1/2" 4000 x7 = 0 2 2/1" 2000 3 1/2" 4000
4 - 15
Chapter 4
Waste is 0; however, we have over-produced the 1 1/2" size by 3000 units. Perhaps these can be inventoried for future use. c.
17. a.
Minimizing waste may cause you to over-produce. In this case, we used 375 more rolls to generate a 3000 surplus of the 1 1/2" product. Alternative b might be preferred on the basis that the 3000 surplus could be held in inventory for later demand. However, in some trim problems, excess production cannot be used and must be scrapped. If this were the case, the 3000 unit 1 1/2" size would result in 4500 inches of waste, and thus alternative a would be the preferred solution. Let
FM FP SM SP TM TP
= number of frames manufactured = number of frames purchased = number of supports manufactured = number of supports purchased = number of straps manufactured = number of straps purchased
Min s.t.
38FM
+ 51FP
3.5FM 2.2FM 3.1FM FM
+
+ 11.5SM + 15SP
+ 6.5TM + 7.5TP
+ + +
+ 0.8TM
1.3SM 1.7SM 2.6SM
+ 1.7TM
FP SM
+
SP
TM FM, FP, SM, SP, TM, TP 0.
+
TP
21,000 25,200 40,800 5,000 10,000 5,000
Solution: Frames Supports Straps
Manufacture 5000 2692 0
Purchase 0 7308 5000
b.
Total Cost = $368,076.91
c.
Subtract values of slack variables from minutes available to determine minutes used. Divide by 60 to determine hours of production time used. Constraint 1 2 3
Cutting: Milling: Shaping:
Slack = 0 350 hours used (25200 - 9623) / 60 = 259.62 hours (40800 - 18300) / 60 = 375 hours
d.
Nothing, there are already more hours available than are being used.
e.
Yes. The current purchase price is $51.00 and the reduced cost of 3.577 indicates that for a purchase price below $47.423 the solution may improve. Resolving with the coefficient of FP = 45 shows that 2714 frames should be purchased. The optimal solution is as follows:
OPTIMAL SOLUTION
4 - 16
Linear Programming Applications in Marketing, Finance and Operations Management
Optimal Objective Value 361500.00000 Variable
Value
FM
2285.71429
0.00000
FP
2714.28571
3.57692
SM
10000.00000
0.00000
SP
0.00000
0.00000
TM
0.00000
1.15385
TP
5000.00000
0.00000
Constraint
18. a.
Let
Reduced Cost
Slack/Surplus
Dual Value
1
0.00000
-2.69231
2
3171.42857
0.00000
3
7714.28571
0.00000
4
0.00000
47.42308
5
0.00000
15.00000
6
0.00000
7.50000
x1 = number of Super Tankers purchased x2 = number of Regular Line Tankers purchased x3 = number of Econo-Tankers purchased
Min s.t.
550x1
+
425x2 +
350x3
6700x1 15(5000)x1
+ +
55000x2 + 20(2500)x2 +
4600x3 25(1000)x3
75000x1 x1
+ +
50000x2 + x2 +
25000x3 x3 x3
600,000 Budget 550,000
or 550,000 Meet Demand 15 Max. Total Vehicles 3 Min. Econo-Tankers
x1 1/2(x1 + x2 + x3) or 1/2x1 - 1/2x2 - 1/2x3 0
No more than 50% Super Tankers
x1 , x 2 , x3 0
Solution: 5 Super Tankers, 2 Regular Tankers, 3 Econo-Tankers Total Cost: $583,000 Monthly Operating Cost: $4,650
4 - 17
Chapter 4
b.
The last two constraints in the formulation above must be deleted and the problem resolved. The optimal solution calls for 7 1/3 Super Tankers at an annual operating cost of $4033. However, since a partial Super Tanker can't be purchased we must round up to find a feasible solution of 8 Super Tankers with a monthly operating cost of $4,400. Actually this is an integer programming problem, since partial tankers can't be purchased. We were fortunate in part (a) that the optimal solution turned out integer. The true optimal integer solution to part (b) is x1 = 6 and x2 = 2 with a monthly operating cost of $4150. This is 6 Super Tankers and 2 Regular Line Tankers.
19. a.
Let
x11 = amount of men's model in month 1 x21 = amount of women's model in month 1 x12 = amount of men's model in month 2 x22 = amount of women's model in month 2 s11 = inventory of men's model at end of month 1 s21 = inventory of women's model at end of month 1 s12 = inventory of men's model at end of month 2 s22 = inventory of women's model at end of month
The model formulation for part (a) is given. Min
120x11 + 90x21 + 120x12 + 90x22 + 2.4s11 + 1.8s21 + 2.4s12 + 1.8s22
s.t. 20 + x11 - s11= 150 or x11 - s11 = 130
Satisfy Demand
[1]
x21 - s21 = 95
Satisfy Demand
[2]
s11 + x12 - s12= 200 s21 + x22 - s22= 150
Satisfy Demand Satisfy Demand
[3] [4]
25
Ending Inventory
[5]
25
Ending Inventory
[6]
Labor Smoothing for
[7]
Month 1
[8]
3.5 x11 + 2.6 x21 - 3.5 x12 - 2.6 x22 100
Labor Smoothing for
[9]
-3.5 x11 - 2.6 x21 + 3.5 x12 + 2.6 x22 100
Month 2
[10]
30 + x21 - s21= 125 or
s12 s22 Labor Hours:
Men’s = 2.0 + 1.5 = 3.5 Women’s = 1.6 + 1.0 = 2.6
3.5 x11 + 2.6 x21 900
3.5 x11 + 2.6 x21 1100
4 - 18
Linear Programming Applications in Marketing, Finance and Operations Management x11, x12, x21, x22, s11, s12, s21, s22 0 The optimal solution is to produce 193 of the men's model in month 1, 162 of the men's model in month 2, 95 units of the women's model in month 1, and 175 of the women's model in month 2. Total Cost = $67,156 Inventory Schedule Month 1 Month 2
63 Men's 25 Men's
0 Women's 25 Women's
Labor Levels Previous month Month 1 Month 2 b.
1000.00 hours 922.25 hours 1022.25 hours
To accommodate this new policy the right-hand sides of constraints [7] to [10] must be changed to 950, 1050, 50, and 50 respectively. The revised optimal solution is given. x11 = 201 x21 = 95 x12 = 154 x22 = 175
Total Cost = $67,175
We produce more men's models in the first month and carry a larger men's model inventory; the added cost however is only $19. This seems to be a small expense to have less drastic labor force fluctuations. The new labor levels are 1000, 950, and 994.5 hours each month. Since the added cost is only $19, management might want to experiment with the labor force smoothing restrictions to enforce even less fluctuations. You may want to experiment yourself to see what happens. 20.
Let
xm = number of units produced in month m Im = increase in the total production level in month m Dm = decrease in the total production level in month m sm = inventory level at the end of month m
where m = 1 refers to March m = 2 refers to April m = 3 refers to May Min
1.25 I1 + 1.25 I2 + 1.25 I3 + 1.00 D1 + 1.00 D2 + 1.00 D3
s.t. Change in production level in March x1 - 10,000 = I1 - D1 or x1 - I1 + D1 = 10,000 Change in production level in April
4 - 19
Chapter 4 x 2 - x 1 = I 2 - D2 or x 2 - x 1 - I 2 + D2 = 0 Change in production level in May x 3 - x 2 = I 3 - D3 or x 3 - x 2 - I 3 + D3 = 0 Demand in March 2500 + x1 - s1 = 12,000 or x1 - s1 = 9,500 Demand in April s1 + x2 - s2 = 8,000 Demand in May s2 + x3 = 15,000 Inventory capacity in March s1 3,000 Inventory capacity in April s2 3,000 Optimal Solution: Total cost of monthly production increases and decreases = $2,500 x1 = 10,250 x2 = 10,250 x3 = 12,000 s1 = 750 s2 = 3000 21.
I1 = 250 I2 = 0 I3 = 1750
D1 = 0 D2 = 0 D3 = 0
Decision variables : Regular Model Bookshelf Floor
Month 1 B1R F1R
Decision variables : Overtime
4 - 20
Month 2 B2R F2R
Linear Programming Applications in Marketing, Finance and Operations Management
Model Bookshelf Floor Labor costs per unit Model Bookshelf Floor
Month 1 B1O F1O
Month 2 B2O F2O
Regular .7 (22) = 15.40 1 (22) = 22
Overtime .7 (33) = 23.10 1 (33) = 33
IB = Month 1 ending inventory for bookshelf units IF = Month 1 ending inventory for floor model Objective function Min + + + +
15.40 B1R + 15.40 B2R + 22 F1R + 22 F2R 23.10 B1O + 23.10 B2O + 33 F1O + 33 F2O 10 B1R + 10 B2R + 12 F1R + 12 F2R 10 B1O + 10 B2O + 12 F1O + 12 F2O 5 IB + 5 IF
or Min
25.40 B1R + 25.40 B2R + 34 F1R + 34 F2R + 33.10 B1O + 33.10 B2O + 45 F1O + 45 F2O + 5 IB + 5 IF
s.t. .7 B1R + 1 F1R .7 B2R + 1 F2R .7B1O + 1 F1O .7B2O + 1 F2O B1R + B1O - IB = IB + B2R + B2O = F1R + F1O - IF = IF + F2R + F2O =
2400 2400 1000 1000 2100 1200 1500 2600
Regular time: month 1 Regular time: month 2 Overtime: month 1 Overtime: month 2 Bookshelf: month 1 Bookshelf: month 2 Floor: month 1 Floor: month 2
OPTIMAL SOLUTION Optimal Objective Value 241129.99940 Variable
Value
Reduced Cost
B1R
1228.57143
0.00000
B1O
871.42855
0.00000
B2R
0.00000
0.00000
B2O
1200.00000
0.00000
IB
0.00000
1.50000
F1R
1540.00000
0.00000
F1O
0.00000
0.00000
F2R
2400.00000
0.00000
4 - 21
Chapter 4
F2O
160.00000
0.00000
IF
40.00000
0.00000
Constraint
22.
Let
SM1 SM2 SM3 LM1 LM2
= = = = =
Slack/Surplus
Dual Value
1
0.00000
-11.00000
2
0.00000
-16.00000
3
390.00001
0.00000
4
0.00000
-5.00000
5
0.00002
33.10000
6
0.00000
36.60000
7
0.00000
45.00000
8
0.00000
50.00000
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
25.40000
0.00000
1.50000
33.10000
Infinite
0.00000
25.40000
0.00000
Infinite
33.10000
Infinite
0.00000
5.00000
Infinite
1.50000
34.00000
2.14286
0.00000
45.00000
0.00000
5.00000
34.00000
16.00000
0.00000
45.00000
0.00000
Infinite
5.00000
2.14286
5.00000
RHS
Allowable
Allowable
Value
Increase
Decrease
2400.00000
610.00000
390.00000
2400.00000
40.00000
390.00000
1000.00000
Infinite
390.00000
1000.00000
40.00000
390.00000
2100.00000
557.14286
871.42857
1200.00000
557.14286
57.14286
1500.00000
390.00000
610.00000
2600.00000
390.00000
40.00000
No. of small on machine M1 No. of small on machine M2 No. of small on machine M3 No. of large on machine M1 No. of large on machine M2
4 - 22
Linear Programming Applications in Marketing, Finance and Operations Management
LM3 = No. of large on machine M3 MM2 = No. of meal on machine M2 MM3 = No. of meal on machine M3 The formulation and solution follows. Note that constraints 1-3 guarantee that next week's schedule will be met and constraints 4-6 enforce machine capacities. LINEAR PROGRAMMING PROBLEM MIN 20SM1+24SM2+32SM3+15LM1+28LM2+35LM3+18MM2+36MM3 S.T. 1) 2) 3) 4) 5) 6)
1SM1+1SM2+1SM3>80000 +1LM1+1LM2+1LM3>80000 +1MM2+1MM3>65000 0.03333SM1+0.04LM1<2100 +0.02222SM2+0.025LM2+0.03333MM2<2100 +0.01667SM3+0.01923LM3+0.02273MM3<2400
OPTIMAL SOLUTION Optimal Objective Value 5516000.00000 Variable
Value
Reduced Cost
1
0.00000
4.66667
2
52500.00000
0.00000
3
0.00000
4.00000
4
0.00000
6.50000
5
80000.00000
0.00000
6
27500.00000
0.00000
7
63000.00000
0.00000
8
2000.00000
0.00000
Constraint
Slack/Surplus
Dual Value
1
0.00000
32.00000
2
0.00000
35.00000
3
0.00000
36.00000
4
0.00000
-500.00000
5
0.00000
-540.00000
6
492.36597
0.00000
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
20.00000
Infinite
4.66667
15.00000
5.60000
Infinite
4 - 23
Chapter 4
24.00000
Infinite
4.00000
28.00000
Infinite
6.50000
32.00000
4.00000
32.00000
35.00000
6.50000
5.60000
18.00000
6.00000
Infinite
36.00000
Infinite
6.00000
RHS
Allowable
Allowable
Value
Increase
Decrease
80000.00000
29541.95804
80000.00000
80000.00000
25603.03030
27500.00000
65000.00000
21664.10256
2000.00000
2100.00000
1100.00000
1024.12121
2100.00000
66.66667
722.13675
2400.00000
Infinite
492.36597
Note that 5,516,000 square inches of waste are generated. Machine 3 has 492 minutes of idle capacity.
23.
Let
F M A Im Dm sm
= = = = = =
number of windows manufactured in February number of windows manufactured in March number of windows manufactured in April increase in production level necessary during month m decrease in production level necessary during month m ending inventory in month m
Min
1I1 + 1I2 + 1I3 + 0.65D1 + 0.65D2 + 0.65D3
s.t. 9000 + F - s1 = 15,000
February Demand
or (1)
F1 - s1 = 6000
(2)
s1 + M - s2 = 16,500
March Demand
(3)
s2 + A - s3 = 20,000
April Demand
F - 15,000 = I1 - D1
Change in February Production
or (4)
F - I1 + D1 = 15,000 M - F = I 2 - D2
Change in March Production
or (5)
M - F - I 2 + D2 = 0
4 - 24
Linear Programming Applications in Marketing, Finance and Operations Management
A - M = I 3 - D3
Change in April Production
or (6)
A - M - I 3 + D3 = 0
(7)
F 14,000
February Production Capacity
(8)
M 14,000
March Production Capacity
(9)
A 18,000
April Production Capacity
(10)
s1 6,000
February Storage Capacity
(11)
s2 6,000
March Storage Capacity
(12)
s3 6,000
April Storage Capacity
Optimal Solution: Cost = $6,450
Production Level Increase in Production Decrease in Production Ending Inventory 24.
Let
February 12,000 0 3,000 6,000
March 14,000 2,000 0 3,500
April 16,500 2,500 0 0
x1 = proportion of investment A undertaken x2 = proportion of investment B undertaken s1 = funds placed in savings for period 1 s2 = funds placed in savings for period 2 s3 = funds placed in savings for period 3 s4 = funds placed in savings for period 4 L1 = funds received from loan in period 1 L2 = funds received from loan in period 2 L3 = funds received from loan in period 3 L4 = funds received from loan in period 4
Objective Function: In order to maximize the cash value at the end of the four periods, we must consider the value of investment A, the value of investment B, savings income from period 4, and loan expenses for period 4. Max
3200x1 + 2500x2 + 1.1s4 - 1.18L4
Constraints require the use of funds to equal the source of funds for each period. Period 1:
4 - 25
Chapter 4 1000x1 + 800x2 + s1 = 1500 + L1 or 1000x1 + 800x2 + s1 - L1 = 1500 Period 2: 800x1 + 500x2 + s2 + 1.18L1 = 400 + 1.1s1 + L2 or 800x1 + 500x2 - 1.1s1 + s2 + 1.18L1 - L2 = 400 Period 3 200x1 + 300x2 + s3 + 1.18L2 = 500 + 1.1s2 + L3 or 200x1 + 300x2 - 1.1s2 + s3 + 1.18L2 - L3 = 500 Period 4 s4 + 1.18L3 = 100 + 200x1 + 300x2 + 1.1s3 + L4 or -200x1 - 300x2 - 1.1s3 + s4 + 1.18L3 - L4 = 100
Limits on Loan Funds Available L1 200 L2 200 L3 200 L4 200 Proportion of Investment Undertaken x1 1 x2 1 Optimal Solution: $4340.40 Investment A Investment B
x1 = 0.458 x2 = 1.0
or or
45.8% 100.0%
Savings/Loan Schedule:
Savings Loan 25.
Let
Period 1
Period 2
Period 3
Period 4
242.11 —
— 200.00
— 127.58
341.04 —
x1 = number of part-time employees beginning at 11:00 a.m. x2 = number of part-time employees beginning at 12:00 p.m. x3 = number of part-time employees beginning at 1:00 p.m. x4 = number of part-time employees beginning at 2:00 p.m. x5 = number of part-time employees beginning at 3:00 p.m.
4 - 26
Linear Programming Applications in Marketing, Finance and Operations Management x6 = number of part-time employees beginning at 4:00 p.m. x7 = number of part-time employees beginning at 5:00 p.m. x8 = number of part-time employees beginning at 6:00 p.m. Each part-time employee assigned to a four-hour shift will be paid $7.60 (4 hours) = $30.40. Min 30.4x1 + 30.4x2 + 30.4x3 + 30.4x4 + 30.4x5 + 30.4x6 + 30.4x7 + 30.4x8
Part-Time Employees Needed
s.t. x1 x1 + x1 + x1 +
x2 x2 + x2 + x2 +
x3 x3 + x3 + x3 +
x4 x4 + x4 + x4 +
x5 x5 + x5 + x5 +
x6 x6 + x6 + x6 +
x7 x7 + x7 + x7 +
8 8 7 1 2 1 5 x8 10 x8 10 x8 6 x8 6
-
xj 0 j = 1,2,...8 Full-time employees reduce the number of part-time employees needed. A portion of the solution output to the model follows. OPTIMAL SOLUTION Optimal Objective Value 608.00000 Variable
Value
Reduced Cost
X1
8.00000
0.00000
X2
0.00000
0.00000
X3
0.00000
0.00000
X4
0.00000
0.00000
X5
2.00000
0.00000
X6
0.00000
0.00000
X7
4.00000
0.00000
X8
6.00000
0.00000
Constraint
Slack/Surplus
Dual Value
1
0.00000
30.40000
2
0.00000
0.00000
4 - 27
11:00 a.m. 12:00 p.m. 1:00 p.m. 2:00 p.m. 3:00 p.m. 4:00 p.m. 5:00 p.m. 6:00 p.m. 7:00 p.m. 8:00 p.m. 9:00 p.m.
Chapter 4
3
1.00000
0.00000
4
7.00000
0.00000
5
0.00000
30.40000
6
1.00000
0.00000
7
1.00000
0.00000
8
2.00000
0.00000
9
0.00000
30.40000
10
4.00000
0.00000
11
0.00000
0.00000
The optimal schedule calls for 8 starting at 11:00 a.m. 2 starting at 3:00 p.m. 4 starting at 5:00 p.m. 6 starting at 6:00 p.m. b.
Total daily salary cost = $608 There are 7 surplus employees scheduled from 2:00 - 3:00 p.m. and 4 from 8:00 - 9:00 p.m. suggesting the desirability of rotating employees off sooner.
c.
Considering 3-hour shifts Let x denote 4-hour shifts and y denote 3-hour shifts where y1 = number of part-time employees beginning at 11:00 a.m. y2 = number of part-time employees beginning at 12:00 p.m. y3 = number of part-time employees beginning at 1:00 p.m. y4 = number of part-time employees beginning at 2:00 p.m. y5 = number of part-time employees beginning at 3:00 p.m. y6 = number of part-time employees beginning at 4:00 p.m. y7 = number of part-time employees beginning at 5:00 p.m. y8 = number of part-time employees beginning at 6:00 p.m. y9 = number of part-time employees beginning at 7:00 p.m. Each part-time employee assigned to a three-hour shift will be paid $7.60 (3 hours) = $22.80 New objective function: 8
9
j =1
i =1
min 30.40 x j + 22.80 yi Each constraint must be modified with the addition of the yi variables. For instance, the first constraint becomes x1 + y1 8 and so on. Each yi appears in three constraints because each refers to a three hour shift. The optimal
4 - 28
Linear Programming Applications in Marketing, Finance and Operations Management
solution is shown below. x8 = 6
y1 = 8 y3 = 1 y5 = 1 y7 = 4
Optimal schedule for part-time employees: 4-Hour Shifts x8 = 6
3-Hour Shifts y1 = 8 y3 = 1 y5 = 1 y7 = 4
Total cost reduced to $501.60. Still have 20 part-time shifts, but 14 are 3-hour shifts. The surplus has been reduced by a total of 14 hours.
4 - 29
Chapter 5 Advanced Linear Programming Applications Learning Objectives 1.
Learn about applications of linear programming that are solved in practice.
2.
Develop an appreciation for the diversity of problems that can be modeled as linear programs.
3.
Obtain practice and experience in formulating realistic linear programming models.
4.
Understand linear programming applications such as: data envelopment analysis revenue management portfolio selection game theory
5.
Know what is meant by a two-person, zero-sum game.
6.
Be able to identify a pure strategy for a two-person, zero-sum game.
7.
Be able to use linear programming to identify a mixed strategy and compute optimal probabilities for the mixed strategy games.
8.
Understand the following terms: game theory two-person, zero-sum game saddle point pure strategy mixed strategy
Note to Instructor The application problems of Chapter 5 are designed to give the student an understanding and appreciation of the broad range of problems that can be approached by linear programming. While the problems are indicative of the many linear programming applications, they have been kept relatively small in order to ease the student's formulation and solution effort. Each problem will give the student an opportunity to practice formulating a linear programming model. However, the solution and the interpretation of the solution will require the use of a software package such as Microsoft Excel's Solver or LINGO.
5-1
Chapter 5
Solutions: 1.
a. Min s.t.
E wg 48.14wg 43.10wg 253wg 41wg -285.2E + 285.2wg -123.80E + 1123.80wg -106.72E + 106.72wg
+ wu + wc + 34.62wu + 36.72wc + 27.11wu + 45.98wc + 148wu + 175wc + 27wu + 23wc + 162.3wu + 275.7wc + 128.70wu + 348.50wc + 64.21wu + 104.10wc
+ + + + + + + +
ws 33.16ws 56.46ws 160ws 84ws 210.4ws 154.10ws 104.04ws
=
1 48.14 43.10 253 41 0 0 0
wg, wu, wc, ws 0
2.
b.
Since wg = 1.0, the solution does not indicate General Hospital is relatively inefficient.
c.
The composite hospital is General Hospital. For any hospital that is not relatively inefficient, the composite hospital will be that hospital because the model is unable to find a weighted average of the other hospitals that is better.
a. Min E s.t. wa + 55.31wa + 49.52wa + 281wa + 47wa + -250E+310wa + -316E+134.6wa + -94.4E+116wa +
wb + 37.64wb + 55.63wb + 156wb + 3wb + 278.5wb + 114.3wb + 106.8wb +
wc + 32.91wc + 25.77wc + 141wc + 26wc + 165.6wc + 131.3wc + 65.52wc +
wd + 33.53wd + 41.99wd + 160wd + 21wd + 250wd + 316wd + 94.4wd +
we + 32.48we + 55.30we + 157we + 82we + 206.4we + 151.2we + 102.1we +
wa, wb, wc, wd, we, wf, wg 0 b.
E = 0.924 wa = 0.074 wc = 0.436 we = 0.489 All other weights are zero.
c.
D is relatively inefficient Composite requires 92.4 of D's resources.
d.
34.37 patient days (65 or older)
5-2
wf + 48.78wf + 81.92wf + 285wf + 92wf + 384wf + 217wf + 153.7wf +
wg 58.41wg 119.70wg 111wg 89wg 530.1wg 770.8wg 215wg
= 1 33.53 41.99 160 21 0 0 0
Advanced Linear Programming Applications
41.99 patient days (under 65)
3.
e.
Hospitals A, C, and E.
a.
Make the following changes to the model in problem 2. New Right-Hand Side Values for Constraint 2 32.48 Constraint 3 55.30 Constraint 4 157 Constraint 5 82 New Coefficients for E in Constraint 6 Constraint 7 Constraint 8
4.
-206.4 -151.2 -102.1
b.
E = 1; we = 1; all other weights = 0
c.
No; E = 1 indicates that all the resources used by Hospital E are required to produce the outputs of Hospital E.
d.
Hospital E is the only hospital in the composite. If a hospital is not relatively inefficient, the hospital will make up the composite hospital with weight equal to 1.
a. Min s.t.
E wb 3800wb 25wb 8wb - 110E + 96wb - 22E + 16wb -1400E + 850wb
+ + + + + + +
wc 4600wc 32wc 8.5wc 110wc 22wc 1400wc
+ wj + + 4400wj + + 35wj + + 8wj + + 100wj + + 18wj + + 1200wj +
wb, wc, wj, wn, ws 0 b. Optimal Objective Value 0.95955
5-3
wn 6500wn 30wn 10wn 125wn 25wn 1500wn
+ + + + + + +
ws 6000ws 28ws 9ws 120ws 24ws 1600ws
=
1 4600 32 8.5 0 0 0
Chapter 5
Variable
Value
Reduced Cost
E
0.95955
0.00000
WB
0.17500
0.00000
WC
0.00000
0.04045
WJ
0.57500
0.00000
WN
0.25000
0.00000
WS
0.00000
0.08455
Constraint
Slack/Surplus
Dual Value
1
0.00000
-0.20000
2
220.00000
0.00000
3
0.00000
0.00364
4
0.00000
0.12273
5
0.00000
-0.00909
6
1.71000
0.00000
7
129.61364
0.00000
c.
Yes; E = 0.960 indicates a composite restaurant can produce Clarksville's output with 96% of Clarksville's available resources.
d.
More Output (Constraint 2 Surplus) $220 more profit per week. Less Input Hours of Operation 110E = 105.6 hours FTE Staff 22 - 1.71 (Constraint 6 Slack) = 19.41 Supply Expense 1400E - 129.614 (Constraint 7 Slack) = $1214.39 The composite restaurant uses 4.4 hours less operation time, 2.6 less employees and $185.61 less supplies expense when compared to the Clarksville restaurant.
5.
e.
wb = 0.175, wj = 0.575, and wn = 0.250. Consider the Bardstown, Jeffersonville, and New Albany restaurants.
a.
If the larger plane is based in Pittsburgh, the total revenue increases to $107,849. If the larger plane is based in Newark, the total revenue increases to $108,542. Thus, it would be better to locate the larger plane in Newark. Note: The optimal solution to the original Leisure Air problem resulted in a total revenue of $103,103. The difference between the total revenue for the original problem and the problem that has a larger plane based in Newark is $108,542 - $103,103 = $5,439. In order to make the decision to change to a larger plane based in Newark, management must determine if the $5,439 increase in revenue is sufficient to cover the cost associated with changing to the larger plane.
b.
Using a larger plane based in Newark, the optimal allocations are:
5-4
Advanced Linear Programming Applications
PCQ = 33 PCY = 16 NCQ = 26 NCY = 15 CMQ = 32 COQ = 46
PMQ = 23 PMY= 6 NMQ = 56 NMY = 7 CMY = 8 COY = 10
POQ = 43 POY = 11 NOQ = 39 NOY = 9
The differences between the new allocations above and the allocations for the original Leisure Air problem involve the five ODIFs that are boldfaced in the solution shown above. c.
Using a larger plane based in Pittsburgh and a larger plane based in Newark, the optimal allocations are: PCQ = 33 PCY = 16 NCQ = 26 NCY = 15 CMQ = 37 COQ = 44
PMQ = 44 PMY= 6 NMQ = 56 NMY = 7 CMY = 8 COY = 10
POQ = 45 POY = 11 NOQ = 39 NOY = 9
The differences between the new allocations above and the allocations for the original Leisure Air problem involve the four ODIFs that are boldfaced in the solution shown above. The total revenue associated with the new optimal solution is $115,073, which is a difference of $115,073 - $103,103 = $11,970.
6.
d.
In part (b), the ODIF that has the largest bid price is COY, with a bid price of $443. The bid price tells us that if one more Y class seat were available from Charlotte to Myrtle Beach that revenue would increase by $443. In other words, if all 10 seats allocated to this ODIF had been sold, accepting another reservation will provide additional revenue of $443.
a.
The calculation of the number of seats still available on each flight leg is shown below: ODIF Original ODIF Code Allocation 1 PCQ 33 2 PMQ 44 3 POQ 22 4 PCY 16 5 PMY 6 6 POY 11 7 NCQ 26 8 NMQ 36 9 NOQ 39 10 NCY 15 11 NMY 7 12 NOY 9 13 CMQ 31 14 CMY 8 15 COQ 41 16 COY 10 Flight Leg 1: 8 + 0 + 4 + 4 + 1 + 2 = 19 Flight Leg 2: 6 + 3 + 2 + 4 + 2 + 1 = 18 Flight Leg 3: 0 + 1 + 3 + 2 + 4 + 2 = 12 Flight Leg 4: 4 + 2 + 2 + 1 + 6 + 3 = 18
Seats Sold 25 44 18 12 5 9 20 33 37 11 5 8 27 6 35 7
5-5
Seats Available 8 0 4 4 1 2 6 3 2 4 2 1 4 2 6 3
Chapter 5
Note: See the demand constraints for the ODIFs that make up each flight leg. b.
The calculation of the remaining demand for each ODIF is shown below:
ODIF 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 c.
ODIF Code PCQ PMQ POQ PCY PMY POY NCQ NMQ NOQ NCY NMY NOY CMQ CMY COQ COY
Original Allocation 33 44 45 16 6 11 26 56 39 15 7 9 64 8 46 10
Seats Sold 25 44 18 12 5 9 20 33 37 11 5 8 27 6 35 7
Seats Available 8 0 27 4 1 2 6 23 2 4 2 1 37 2 11 3
The LP model and solution are shown below: MAX 178PCQ+268PMQ+228POQ+380PCY+456PMY+560POY+199NCQ+249NMQ+349NOQ+ 385NCY+444NMY+580NOY+179CMQ+380CMY+224COQ+582COY S.T. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20)
1PCQ+1PMQ+1POQ+1PCY+1PMY+1POY<19 1NCQ+1NMQ+1NOQ+1NCY+1NMY+1NOY<18 1PMQ+1PMY+1NMQ+1NMY+1CMQ+1CMY<12 1POQ+1POY+1NOQ+1NOY+1COQ+1COY<18 1PCQ<8 1PMQ<1 1POQ<27 1PCY<4 1PMY<1 1POY<2 1NCQ<6 1NMQ<23 1NOQ<2 1NCY<4 1NMY<2 1NOY<1 1CMQ<37 1CMY<2 1COQ<11 1COY<3 Optimal Objective Value 15730.00000 Variable PCQ
Value
Reduced Cost
8.00000
5-6
0.00000
Advanced Linear Programming Applications
PMQ
1.00000
0.00000
POQ
3.00000
0.00000
PCY
4.00000
0.00000
PMY
1.00000
0.00000
POY
2.00000
0.00000
NCQ
6.00000
0.00000
NMQ
3.00000
0.00000
NOQ
2.00000
0.00000
NCY
4.00000
0.00000
NMY
2.00000
0.00000
NOY
1.00000
0.00000
CMQ
3.00000
0.00000
CMY
2.00000
0.00000
COQ
7.00000
0.00000
COY
3.00000
0.00000
Constraint
7.
a.
Slack/Surplus
Dual Value
1
0.00000
4.00000
2
0.00000
70.00000
3
0.00000
179.00000
4
0.00000
224.00000
5
0.00000
174.00000
6
0.00000
85.00000
7
24.00000
0.00000
8
0.00000
376.00000
9
0.00000
273.00000
10
0.00000
332.00000
11
0.00000
129.00000
12
20.00000
0.00000
13
0.00000
55.00000
14
0.00000
315.00000
15
0.00000
195.00000
16
0.00000
286.00000
17
34.00000
0.00000
18
0.00000
201.00000
19
4.00000
0.00000
20
0.00000
358.00000
Let CT = number of convention two-night rooms CF = number of convention Friday only rooms CS = number of convention Saturday only rooms RT = number of regular two-night rooms
5-7
Chapter 5
RF = number of regular Friday only rooms RS = number of regular Saturday only room b./c. The formulation and output are shown below. LINEAR PROGRAMMING PROBLEM MAX 225CT+123CF+130CS+295RT+146RF+152RS S.T. 1) 1CT<40 2) 1CF<20 3) 1CS<15 4) 1RT<20 5) 1RF<30 6) 1RS<25 7) 1CT+1CF>48 8) 1CT+1CS>48 9) 1CT+1CF+1RT+1RF<96 10) 1CT+1CS+1RT+1RS<96 Optimal Objective Value 25314.00000 Variable
Value
Reduced Cost
CT
36.00000
0.00000
CF
12.00000
0.00000
CS
15.00000
0.00000
RT
20.00000
0.00000
RF
28.00000
0.00000
RS
25.00000
0.00000
Constraint
Slack/Surplus
Dual Value
1
4.00000
0.00000
2
8.00000
0.00000
3
0.00000
28.00000
4
0.00000
47.00000
5
2.00000
0.00000
6
0.00000
50.00000
7
0.00000
-23.00000
8
3.00000
0.00000
9
0.00000
146.00000
10
0.00000
102.00000
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
225.00000
28.00000
5-8
102.00000
Advanced Linear Programming Applications
d.
8.
123.00000
23.00000
28.00000
130.00000
Infinite
28.00000
295.00000
Infinite
47.00000
146.00000
47.00000
23.00000
152.00000
Infinite
50.00000
RHS
Allowable
Allowable
Value
Increase
Decrease
40.00000
Infinite
4.00000
20.00000
Infinite
8.00000
15.00000
8.00000
4.00000
20.00000
3.00000
2.00000
30.00000
Infinite
2.00000
25.00000
3.00000
4.00000
48.00000
8.00000
2.00000
48.00000
3.00000
Infinite
96.00000
2.00000
28.00000
96.00000
4.00000
3.00000
The dual value for constraint 10 shows an added profit of $50 if this additional reservation is accepted. To determine the percentage of the portfolio that will be invested in each of the mutual funds we use the following decision variables: FS = proportion of portfolio invested in a foreign stock mutual fund IB = proportion of portfolio invested in an intermediate-term bond fund LG = proportion of portfolio invested in a large-cap growth fund LV = proportion of portfolio invested in a large-cap value fund SG = proportion of portfolio invested in a small-cap growth fund SV = proportion of portfolio invested in a small-cap value fund
a.
A portfolio model for investors willing to risk a return as low as 0% involves 6 variables and 6 constraints. Max 12.03FS + 6.89IB + 20.52LG + 13.52LV + 21.27SG + 13.18SV
s.t. 10.06FS 13.12FS 13.47FS 45.42FS -21.93FS FS
+ + + + +
17.64IB 3.25IB 7.51IB 1.33IB 7.36IB IB
+ + + + +
32.41LG + 32.36LV + 18.71LG + 20.61LV + 33.28LG + 12.93LV + 41.46LG + 7.06LV + 23.26LG 5.37LV LG + LV + FS, IB, LG, LV, SG, SV ≥ 0
5-9
33.44SG 19.40SG 3.85SG 58.68SG 9.02SG SG
+ + + + +
24.56SV 25.32SV 6.70SV 5.43SV 17.31SV SV
≥ ≥ ≥ ≥ ≥ =
0 0 0 0 0 1
Chapter 5
b. Optimal Objective Value 18.49994 Variable FS IB LG LV SG SV
Value 0.00000 0.00000 0.00000 0.00000 0.65742 0.34258
Reduced Cost -13.20669 -9.35367 -5.12313 -6.63108 0.00000 0.00000
Constraint 1 2 3 4 5 6
Slack/Surplus 30.39793 21.42804 0.23583 40.43788 0.00000 0.00000
Dual Value 0.00000 0.00000 0.00000 0.00000 -0.30710 18.49994
The recommended allocation is to invest 65.7% of the portfolio in a small-cap growth fund and 34.3% of the portfolio in a small-cap value fund. The expected return for this portfolio is 18.499%. c.
One constraint must be added to the model in part a. It is FS ≥ .10 The solution is given. Optimal Objective Value 17.17927 Variable FS IB LG LV SG SV
Value 0.10000 0.00000 0.00000 0.00000 0.50839 0.39161
Reduced Cost -13.20669 -9.35367 -5.12313 -6.63108 0.00000 0.00000
Constraint 1 2 3 4 5 6 7
Slack/Surplus 27.62453 21.09031 0.68055 36.50095 0.00000 0.00000 0.00000
Dual Value 0.00000 0.00000 0.00000 0.00000 -0.30710 18.49994 0.00000
5 - 10
Advanced Linear Programming Applications
The recommended allocation is to invest 10% of the portfolio in the foreign stock fund, 50.8% of the portfolio in the small-cap growth fund, and 39.2 percent of the portfolio in the small-cap value fund. The expected return for this portfolio is 17.178%. The expected return for this portfolio is 1.321% less than for the portfolio that does not require any allocation to the foreign stock fund. 9.
To determine the percentage of the portfolio that will be invested in each of the mutual funds we use the following decision variables: LS = proportion of portfolio invested in a large-cap stock mutual fund MS = proportion of portfolio invested in a mid-cap stock fund SS = proportion of portfolio invested in a small-cap growth fund ES = proportion of portfolio invested in an energy sector fund HS = proportion of portfolio invested in a health sector fund TS = proportion of portfolio invested in a technology sector fund RS = proportion of portfolio invested in a real estate sector fund a.
A portfolio model for investors willing to risk a return as low as 2% involves 7 variables and 6 constraints. Max 9.68LS + 5.91MS + 15.20SS + 11.74ES + 7.34HS + 16.97TS + 15.44RS s.t. 35.3LS 20.0LS 28.3LS 10.4LS -9.3LS LS
+ + + +
32.3MS 23.2MS 0.9MS 49.3MS 22.8MS MS
+ 20.8SS + 22.5SS + 6.0SS + 33.3SS + 6.1SS + SS
+ 25.3ES + 33.9ES + 20.5ES + 20.9ES - 2.5ES + ES
+ 49.1HS + 5.5HS + 29.7HS + 77.7HS - 24.9HS + HS
+ + + + +
46.2TS 21.7TS 45.7TS 93.1TS 20.1TS TS
LS, MS, SS, ES, HS, TS, RS ≥ 0 b.
The solution is. Optimal Objective Value 15.56789 Variable LS MS SS ES HS SS RS
Constraint 1 2 3 4 5 6
Value Reduced Cost 0.00000 -6.68713 0.00000 -11.76067 0.18824 0.00000 0.43266 0.00000 0.00000 -10.33130 0.00000 -0.20550 0.37910 0.00000
Slack/Surplus 20.63325 33.58301 0.00000 14.29661 0.00000 0.00000
5 - 11
Dual Value 0.00000 0.00000 -0.00575 0.00000 -0.08412 15.74763
+ 20.5RS + 44.0RS - 21.1RS + 2.6RS + 5.1RS + RS
≥ ≥ ≥ ≥ ≥ =
2 2 2 2 2 1
Chapter 5
The recommended allocation is to invest approximately 18.9% of the portfolio in the small-cap stock fund, 43.3% of the portfolio in the energy sector fund, and 38% of the portfolio in the real estate sector fund. The expected portfolio return is 15.57%. c.
The portfolio model is modified by changing the right-hand side of the first 5 constraints from 2 to 0.
d.
The solution is. Optimal Objective Value 15.72794 Variable LS MS SS ES HS TS RS Constraint 1 2 3 4 5 6
Value Reduced Cost 0.00000 -6.60214 0.00000 -11.37785 0.00000 -0.09410 0.35336 0.00000 0.00000 -10.00367 0.09581 0.00000 0.55083 0.00000 Slack/Surplus 24.65850 38.29446 0.00000 17.73749 0.00000 0.00000
Dual Value 0.00000 0.00000 -0.00286 0.00000 -0.06830 15.72794
The recommended allocation is to invest 35.3% of the portfolio is the energy sector fund, 9.6% of the portfolio in the technology sector fund, and 55.1% of the portfolio in the real estate sector fund. The expected portfolio return is 15.73%. This is an increase of .16% over the portfolio that limits risk to a return of at least 2%. Most investors would conclude that the small increase in the portfolio return is not enough to justify the increased risk.
10.
Player A
b1
Player B b2
b3
Minimum
a1
8
5
7
5
a2
2
4
10
2
Maximum
8
5
7
The maximum of the row minimums is 5 and the minimum of the column maximums is 5. The game has a pure strategy. Player A should take strategy a1 and Player B should take strategy b2 . The value of the game is 5.
5 - 12
Advanced Linear Programming Applications
11.
By definition, a pure-strategy solution means each player selects a single strategy with probability 1. Thus, if a linear programming solution has a probability decision variable equal to 1, the game has a pure-strategy solution.
12. a.
The payoff table is: Blue Army Attack
Defend
Minimum
Attack
30
50
30
Defend
40
0
0
Maximum
40
50
Red Army
The maximum of the row minimums is 30 and the minimum of the column maximums is 40. Because these values are not equal, a mixed strategy is optimal. Therefore, we must determine the best probability, p, for which the Red Army should choose the Attack strategy. Assume the Red Army chooses Attack with probability p and Defend with probability 1-p. If the Blue Army chooses Attack, the expected payoff is 30p + 40(1-p). If the Blue Army chooses Defend, the expected payoff is 50p + 0*(1-p) Setting these equations equal to each other and solving for p, we get p = 2/3. Red Army should choose to Attack with probability 2/3 and Defend with probability 1/3. b.
13.
Assume the Blue Army chooses Attack with probability q and Defend with probability 1-q. If the Red Army chooses Attack, the expected payoff for the Blue Army is 30q + 50*(1-q). If the Red Army chooses Defend, the expected payoff for the Blue Army is 40q + 0*(1-q). Setting theses equations equal to each other and solving for q we get q = 0.833. Therefore the Blue Army should choose to Attack with probability 0.833 and Defend with probability 1-0.833 = 0.167. The row minimums are -5, 6, and 4. Station A prefers the maximin strategy a2 to obtain a gain of at least 6. The column maximums are 10, 8 and 7. Station B prefers the minimax strategy b3 to limit its maximum loss to no more than 7. However, because the maximum of the row minimums is not equal to the minimum of the row maximums, the game does not have a pure strategy. A mixedstrategy solution with a value of the game between 6 and 7 exists. The linear programming formulation and solution for Station A follows. Max
GAINA
s.t. 10PA1
+
8PA2
+
4PA3
-
GAINA
> 0
-5PA1
+
7PA2
+
8PA3
-
GAINA
> 0
Station B Strategy (Strategy b1 ) (Strategy b2 )
3PA1
+
6PA2
+
7PA3
-
GAINA
> 0
(Strategy b3 )
PA1
+
PA2
+
PA3
= 1
PA1, PA2, PA3 > 0 Optimal Objective Value 6.40000
5 - 13
Chapter 5
Variable PA1 PA2 PA3 GAINA
Value 0.00000 0.60000 0.40000 6.40000
Reduced Cost -2.00000 0.00000 0.00000 0.00000
Constraint 1 2 3 4
Slack/Surplus 0.00000 1.00000 0.00000 0.00000
Dual Value -0.20000 0.00000 -0.80000 6.40000
The optimal strategy is for Station A to implement strategy a 2 with probability 0.6 and strategy a3 with probability 0.4. Using the absolute values of the dual values, we see that it is optimal for Station B to implement strategy b1 with probability 0.2 and strategy b3 with probability 0.8. The expected value of the game is 6.4. This is an expected increase of 6400 viewers for Station A. 14.
The row minimums are -15, -10, -25, and 10. The Republican candidate prefers the maximin strategy a 4 to obtain a gain of at least 10. The column maximums are 30, 20, 10, and 20. Station B prefers the minimax strategy b3 to limit its maximum loss to no more than 10. The maximum of the row minimums is equal to the minimum of the row maximums. The game has a pure strategy. The Republican candidate goes to South Bend and the Democratic candidate goes to Fort Wayne. The value of the game shows a 10,000 voter increase for the Republican candidate.
15. a.
Player A
$1 $5
Player B $1 $5 -1 5 1 -5
$1 $5 Maximum
Player B $1 $5 -1 5 1 -5 1 5
b.
Player A
Maximum -1 -5
The maximum of the row minimums is -1 and the minimum of the column maximums is 1. The game does not have a pure strategy. This makes sense because if Player A adopted a pure strategy such as always selecting $1, Player B would learn and always select $1 in order to win Player A’s $1. The players must adopt a mixed strategy in order to play the game. c.
For Player A, let p = probability of $1 and (1 – p) = probability of $5 If b1 = $1, EV = -1p + 1(1 – p) If b2 = $5, EV = 5p – 5(1 – p) 5 - 14
Advanced Linear Programming Applications
-1p + 1(1 – p) -1p + 1 – 1p 12p p (1 – p) = 1 - 0.50
= = = =
5p – 5(1 – p) 5p – 5 + 5p 6 0.50
= 0.50
For Player B, let q = probability of $1 and (1 – q) = probability of $5 If a1 = $1, EV = -1q + 5(1 – q) If a2 = $5, EV = 1q – 5(1 – q) -1q + 5(1 – q) -1q + 5 – 5q 12q q (1 – q) = 1 – 5/6 or q = 0.8333
= = = =
1q – 5(1 – q) 1q – 5 + 5q 10 5/6
= 1/6
(1-q) = 0.1667
Value of game using Player A EV = -1p + 1(1 – p)
= -1(0.50) + 1(0.50) = 0
This is a fair game. Neither player is favored. d.
If Player A realizes Player B is using a 50/50 strategy, we can use an expected value with these probabilities to show: EV(a1 = $1) EV(a2 = $5)
= -1(0.50) + 5(0.50) = 2.00 = -1(0.50) - 5(0.50) = -2.00
Player A should see that the expected value of a1 is now larger than the expected value of a2. If Player A believes Player B will continue with a 50/50 strategy, then Player A should always play strategy a1: reveal $1. But, if Player A does this, Player B will catch on and begin revealing a $1 bill all the time. The only way for a player to protect against the opponent taking advantage is to play the optimal strategy all the time. 16.
The row minimums are 0, -2, 2 and -2. Company A prefers the maximin strategy a3 to obtain a payoff of at least 2. The column maximums are 4, 6, 5 and 6. Player B prefers the minimax strategy b1 to limit its maximum loss to no more than 4. However, because the maximum of the row minimums is not equal to the minimum of the row maximums, the game does not have a pure strategy. A mixed-strategy solution with a value of the game between 2 and 4 exists. The linear programming formulation and solution for Player A is as follows. Min
LOSSB
s.t.
5 - 15
Chapter 5
3PB1
+
2PB3
2PB1
-
2PB2
+
1PB3
4PB1
+
2PB2
+
5PB3
-2PB1
+
6PB2
-
1PB3
PB1
+
PB2
+
PB3
+ +
4PB4 6PB4
+
-
LOSSB
≤ 0
-
LOSSB
≤ 0
-
LOSSB
≤ 0
-
LOSSB
≤ 0
PB4
Company A Strategy (Strategy a1 ) (Strategy a 2 ) (Strategy a3 ) (Strategy a 4 )
= 1
PB1, PB2, PB3, PB4 > 0 Optimal Objective Value 2.80000 Variable PB1 PB2 PB3 PB4 LOSSB
Value 0.40000 0.60000 0.00000 0.00000 2.80000
Reduced Cost 0.00000 0.00000 1.40000 2.00000 0.00000
Constraint 1 2 3 4 5
Slack/Surplus 1.60000 3.20000 0.00000 0.00000 0.00000
Dual Value 0.00000 0.00000 -0.80000 -0.20000 2.80000
The optimal mixed strategy solution for Company B is to select strategy b1 with probability 0.4 and strategy b2 with probability 0.6. Using the dual values, the optimal mixed strategy for Company A is to select strategy a3 with a probability 0.8 and a strategy a 4 with a probability 0.2. The expected gain for Company A is 2.8%. 17.
The payoff table is as follows: GB Packers Run Defense
Pass Defense
Minimum
Chicago
Run
2
6
2
Bears
Pass
11
-1
-1
Maximum
11
6
The Bears prefer the maximin strategy run to obtain a payoff of at least 2 yards. The Packers prefer the minimax strategy pass defense to limit its maximum loss to no more than 6 yards. However, because the maximum of the row minimums is not equal to the minimum of the row maximums, the game does not have a pure strategy. A mixed-strategy solution with a value of the game between 2 and 6 exists. The linear programming formulation and solution for the Bears is as follows. Max
GAINBEARS
s.t.
5 - 16
Advanced Linear Programming Applications
2PA1 6PA1 PA1
+ 11PA2 1PA2 + PA2
-
GAINA GAINA
≥ 0 ≥ 0 = 1
Green Bay Strategy (Run Defense) (Pass Defense)
Optimal Objective Value 4.25000 Variable PA1 PA2 GAINA
Value 0.75000 0.25000 4.25000
Reduced Cost 0.00000 0.00000 0.00000
Constraint 1 2 3
Slack/Surplus 0.00000 0.00000 0.00000
Dual Value -0.43750 -0.56250 4.25000
The optimal mixed strategy is for the Bears to run with a 0.75 probability and pass with a 0.25 probability. Using the absolute value of the dual values, the optimal mixed strategy for the Packers is to use a run defense with a 0.437 probability and a pass defense with a 0.563 probability. The expected value of the game shows that with the mixed-strategy solution, the Bears average 4.25 yards per play.
5 - 17
Chapter 6 Distribution and Network Models Learning Objectives 1.
Understand the usefulness of using optimization for supply chain problems.
2.
Be able to identify the special features of the transportation problem.
3.
Become familiar with the types of problems that can be solved by applying a transportation model.
4.
Be able to develop network and linear programming models of the transportation problem.
5.
Know how to handle the cases of (1) unequal supply and demand, (2) unacceptable routes, and (3) maximization objective for a transportation problem.
6.
Be able to identify the special features of the assignment problem.
7.
Become familiar with the types of problems that can be solved by applying an assignment model.
8.
Be able to develop network and linear programming models of the assignment problem.
9.
Be familiar with the special features of the transshipment problem.
10.
Become familiar with the types of problems that can be solved by applying a transshipment model.
11
Be able to develop network and linear programming models of the transshipment problem.
12.
Know the basic characteristics of the shortest route problem.
13.
Be able to develop a linear programming model and solve the shortest route problem.
14.
Know the basic characteristics of the maximal flow problem.
15.
Be able to develop a linear programming model and solve the maximal flow problem.
16.
Know how to structure and solve a production and inventory problem as a transshipment problem.
17.
Understand the following terms:
network flow problem transportation problem origin destination capacitated transportation problem assignment problem transshipment problem Solutions:
capacitated transshipment problem shortest route maximal flow source node sink node arc flow capacities supply chain
6-1
Chapter 6
1.
The network model is shown.
Atlanta
1400
Dallas
3200
Columbus
2000
Boston
1400
2 5000
Phila.
6 6 2
1 2 3000
New Orleans
5 7
2.
a. Let
x11 : Amount shipped from Jefferson City to Des Moines x12 : Amount shipped from Jefferson City to Kansas City • • • 9x12 Min 14x11 + + 7x13 + 8x21 + 10x22 + s.t. x11 x12 x13 + + x21 + x22 + x11 x21 + x12 x22 + x13 + x11, x12, x13, x21, x22, x23, 0
b.
Optimal Solution: Amount 5 15 10 20
Jefferson City - Des Moines Jefferson City - Kansas City Jefferson City - St. Louis Omaha - Des Moines Total 3.
a.
6-2
Cost 70 135 70 160 435
5x23 x23 = = x23 =
30 20 25 15 10
Distribution and Network Models
b.
Let xij = amount shipped from supply node i to demand node j. Min s.t.
10x11
+
20x12
+
15x13 +
x11
+
x12
+
x13
x11
+ x12
12x21 + 15x22 + 18x23
x21 + x21 +
x13
x22 + x22 +
x23 = = x23 =
500 400 400 200 300
xij 0 for all i, j c.
Optimal Solution Amount 200 300 200 200
Southern - Hamilton Southern - Clermont Northwest - Hamilton Northwest - Butler Total Cost d.
Cost $ 2000 4500 2400 3000 $11,900
To answer this question the simplest approach is to increase the Butler County demand to 300 and to increase the supply by 100 at both Southern Gas and Northwest Gas. The new optimal solution is: Amount 300 300 100 300
Southern - Hamilton Southern - Clermont Northwest - Hamilton Northwest - Butler Total Cost
Cost $ 3000 4500 1200 4500 $13,200
From the new solution we see that Tri-County should contract with Southern Gas for the additional 100 units.
4.
6-3
Chapter 6
a.
The optimization model can be written as
x ij = Red GloFish shipped from i to j i = M for Michigan, T for Texas; j = 1, 2, 3. y ij = Blue GloFish shipped from i to j , i = M for Michigan, T for Texas; j = 1, 2, 3. Minimize x M1 + 2.50x M2 + 0.50x M3 + y M1 + 2.50y M2 + 0.50y M3 + 2.00y T1 + 1.50y T2 + 2.80y T3 subject to x M1 + x M2 + x M3 ≤ 1,000,000 y M1 + y M2 + y M3 ≤ 1,000,000 y T1 +y T2 + y T3 ≤ 600,000 x M1 ≥ 320,000 x M2 ≥ 300,000 x M3 ≥ 160,000 y M1 + y T1 ≥ 380,000 y M2 + y T2 ≥ 450,000 y M3 + y T3 ≥ 290,000 x ij ≥ 0 Solving this linear program we find that we should produce 780,000 red GloFish in Michigan, 670,000 blue GloFish in Michigan, and 450,000 blue GloFish in Texas. Using the notation in the model, the number of GloFish shipped from each farm to each retailer can be expressed as: xM1 = 320,000 xM2 = 300,000 xM3 = 160,000 yM1 = 380,000 yM2 = 0 yM3 = 290,000 yT1 = 0 yT2 = 450,000 yT3 = 0 b.
The minimum transportation cost is $2.35 million.
c.
We have to add variables xT1, xT2, and xT3 for Red GloFish shipped between Texas and Retailers 1, 2 and 3. The revised objective function is
Minimize x M1 + 2.50x M2 + 0.50x M3 + y M1 + 2.50y M2 + 0.50y M3 + 2.00y T1 + 1.50y T2 + 2.80y T3 + x T1 + 2.50x T2 + 0.50x T3 We replace the third constraint above with
6-4
Distribution and Network Models
x T1 +x T2 + x T3 + y T1 +y T2 + y T3 ≤ 600,000
And we change the constraints xM1 ≥ 320,000 xM2 ≥ 300,000 xM3 ≥ 160,000 to xM1 + xT1 ≥ 320,000 xM2 + xT2 ≥ 300,000 xM3 + xT3 ≥ 160,000 Using this new objective function and constraint the optimal solution is $2.2 million, so the savings are $150,000. 5.
a.
b.
Let xij = number of hours from consultant i assigned to client j
6-5
Chapter 6
Max 100x11 + 125x12 + 115x13 + 100x14 + 120x21 + 135x22 + 115x23 + 120x24 + 155x31 + 150x32 + 140x33 + 130x34 s.t. x11 + x12 + x13 + x14 x21 + x22 + x23 + x24 x31 + x32 + x33 + x34 x11 x21 x31 + + = x12 x22 x32 + + = x13 x23 x33 + + = x14 x24 + + x34 =
160 160 140 180 75 100 85
xij 0 for all i, j Optimal Solution
Avery - Client B Avery - Client C Baker - Client A Baker - Client B Baker - Client D Campbell - Client A Total Billing c.
Billing $ 5,000 11,500 4,800 4,725 10,200 21,700 $57,925
New Optimal Solution
Avery - Client A Avery - Client C Baker - Client B Baker - Client D Campbell - Client A Total Billing 6.
Hours Assigned 40 100 40 35 85 140
Hours Assigned 40 100 75 85 140
Billing $ 4,000 11,500 10,125 10,200 21,700 $57,525
The network model, the linear programming formulation, and the optimal solution are shown. Note that the third constraint corresponds to the dummy origin. The variables x31, x32, x33, and x34 are the amounts shipped out of the dummy origin; they do not appear in the objective function since they are given a coefficient of zero.
6-6
Distribution and Network Models
Demand
Supply
D1
2000
D2
5000
D3
3000
D4
2000
32
5000
34
C.S.
32 40 34 30 3000
D.
28 38 0
4000
0 0
Dum
0
Note: Dummy origin has supply of 4000. Max
32x11 + 34x12 + 32x13 + 40x14 + 34x21 + 30x22 + 28x23 + 38x24
s.t. x11 +
x12 +
x13 +
x31 + x11
+
x14
x32 +
x21
x12
+ +
x21 +
x22 +
x33 +
x34
+ +
x24
x31
x22
x13
x23 +
x32
x23
x14
+ +
x24
xij 0 for all i, j
6-7
x33 +
x34
5000
3000
4000
=
2000
=
5000
=
3000
= 2000
Dummy
Chapter 6
Optimal Solution
Units
Cost
Clifton Springs - D2 Clifton Springs - D4 Danville - D1 Danville - D4
4000 $136,000 1000 40,000 2000 68,000 1000 38,000 Total Cost: $282,000
Customer 2 demand has a shortfall of 1000 Customer 3 demand of 3000 is not satisfied.
7. a. Let xij = MW produced at plant i for city j; i = L for Los Angeles, T for Tulsa, S for Seattle, j = 1, …, 10. Min 356.26xL1 + 356.25xL2 + 178.13xL3 + 356.25xL4 + 237.50xL5 + 415.63xL6 + 356.25xL7 + 356.25xL8 + 178.13xL9 + 356.25xL10 + 593.75xT1 + 593.75xT2 + 475.00xT3 + 475.00xT4 + 475.00xT5 + 415.63xT6 + 415.63xT7 + 356.25xT8 + 475.00xT9 + 296.88xT10 + 59.38xS1 + 178.13xS2 + 298.88xS3 + 298.88xS4 + 356.25xS5 + 296.88xS6 + 356.25xS7 + 475.00xS8 + 593.75xS9 + 593.75xS10 subject to xL1 + xT1 + xS1 ≥ 950.00 xL2 + xT2 + xS2 ≥ 831.25 xL3 + xT3 + xS3 ≥ 2375.00 xL4 + xT4 + xS4 ≥ 593.75 xL5 + xT5 + xS5 ≥ 950.00 xL6 + xT6 + xS6 ≥ 593.75 xL7 + xT7 + xS7 ≥ 1187.50 xL8 + xT8 + xS8 ≥ 712.50 xL9 + xT9 + xS9 ≥ 1187.50 xL10 + xT10 + xS10 ≥ 1543.75 xij ≥ 0, i = L, T, S j = 1,…, 10 By solving this linear program, we find the optimal solution is to produce 6412.50 MWs in Los Angeles, 1543.75 MWs in Tulsa, and 2968.75 MWs in Seattle. The total distribution cost of this solution is $2,552,382.81. b.
We must add the following constraints to the linear program shown above: xL1 + xL2 + xL3 + xL4 + xL5 + xL6 + xL7 + xL8 + xL9 + xL10 ≤ 4000 xT1 + xT2 + xT3 + xT4 + xT5 + xT6 + xT7 + xT8 + xT9 + xT10 ≤ 4000 xS1 + xS2 + xS3 + xS4 + xS5 + xS6 + xS7 + xS8 + xS9 + xS10 ≤ 4000
Solving the LP with these added constraints yields an optimal production of 4000 MWs in Los Angeles, 2925 MWs in Tulsa, and 4000 MWs in Seattle at a total cost of $2,652,949.22. Therefore, the increase in cost associated with the additional constraints is $2,652,949.22 - $2,552,382.81 = $100,566.41.
8.
a.
6-8
Distribution and Network Models
1 Boston
7 1 100
50
11
Denver 8 13
2 70
Dallas
20 17 2 100
12
Atlanta
10 8 3
150
3 Los Angeles
60
18 13
Chicago 16 4 St. Paul
b.
80
There are alternative optimal solutions. Solution #1 Denver to St. Paul: Atlanta to Boston: Atlanta to Dallas: Chicago to Dallas: Chicago to Los Angeles: Chicago to St. Paul:
Solution # 2 10 50 50 20 60 70
Denver to St. Paul: Atlanta to Boston: Atlanta to Los Angeles: Chicago to Dallas: Chicago to Los Angeles: Chicago to St. Paul:
10 50 50 70 10 70
Total Profit: $4240 If solution #1 is used, Forbelt should produce 10 motors at Denver, 100 motors at Atlanta, and 150 motors at Chicago. There will be idle capacity for 90 motors at Denver. If solution #2 is used, Forbelt should adopt the same production schedule but a modified shipping schedule.
6-9
Chapter 6
9.
The linear programming formulation and optimal solution are shown. x1A x1B
Let
= =
Units of product A on machine 1 Units of product B on machine 1
=
Units of product C on machine 3
• • •
x3C Min
x1A + 1.2x1B + 0.9x1C + 1.3x2A + 1.4x2B + 1.2x2C + 1.1x3A + x3B + 1.2x3C
s.t. x1A +
x1B +
x1C x2A +
x2B +
x2C x3A + x3B +
x1A
+
x2A
x1B
+
x3A
x2B
+ x1C
+ x3B x2C
+
+
1500
1500
x3C
1000
=
2000
=
500
x3C =
1200
xij 0 for all i, j
Optimal Solution
Units
Cost
1-A 1-C 2-A 3-A 3-B
300 1200 1200 500 500
$ 300 1080 1560 550 500 Total: $3990
Note: There is an unused capacity of 300 units on machine 2. 10. a.
The total cost is the sum of the purchase cost and the transportation cost. We show the calculation for Division 1 - Supplier 1 and present the result for the other Division-Supplier combinations. Division 1 - Supplier 1 Purchase cost (40,000 x $12.60) Transportation Cost (40,000 x $2.75) Total Cost:
6 - 10
$504,000 110,000 $614,000
Distribution and Network Models
Cost Matrix ($1,000s)
Supplier 1
2
3
4
5
6
1
614
660
534
680
590
630
2
603
639
702
693
693
630
3
865
830
775
850
900
930
4
532
553
511
581
595
553
5
720
648
684
693
657
747
Division
b.
Optimal Solution: Supplier 1 - Division 2 Supplier 2 - Division 5 Supplier 3 - Division 3 Supplier 5 - Division 1 Supplier 6 - Division 4
$ 603 648 775 590 553 Total $3,169
11. a. Network Model
Demand
Supply 450
1 P1
4
600
380
4
4 W1
8
7 C2
300
8 C3
300
9 C4
400
4
8 5
3 5 W2
5 3 P3
300
6 7
2 P2
6 C1
6
7 7
6
6 - 11
Chapter 6
b. & c. The linear programming formulation and solution is shown below. LINEAR PROGRAMMING PROBLEM MIN 4X14 + 7X15 + 8X24 + 5X25 + 5X34 + 6X35 + 6X46 + 4X47 + 8X48 + 4X49 + 3X56 + 6X57 + 7X58 + 7X59 S.T. 1) X14 + X15 < 450 2) X24 + X25 < 600 3) X34 + X35 < 380 4) X46 + X47 + X48 + X49 - X14 - X24 - X34 = 0 5) X56 + X57 + X58 + X59 - X15 - X25 - X35 = 0 6) X46 + X56 = 300 7) X47 + X57 = 300 8) X48 + X58 = 300 9) X49 + X59 = 400 OPTIMAL SOLUTION Objective Function Value = Variable -------------X14 X15 X24 X25 X34 X35 X46 X47 X48 X49 X56 X57 X58 X59
Value --------------450.000 0.000 0.000 600.000 250.000 0.000 0.000 300.000 0.000 400.000 300.000 0.000 300.000 0.000
11850.000 Reduced Costs -----------------0.000 3.000 3.000 0.000 0.000 1.000 3.000 0.000 1.000 0.000 0.000 2.000 0.000 3.000
There is an excess capacity of 130 units at plant 3.
6 - 12
Distribution and Network Models
12. a.
Three arcs must be added to the network model in problem 23a. The new network is shown. Demand
Supply 450
1 P1
4
600
380
4
4 W1
8
7 C2
300
8 C3
300
9 C4
400
4
8 5
2
2 5 W2
5 3 P3
300
6 7
2 P2
6 C1
3 6 7 7
6 7
b.&c. The linear programming formulation and optimal solution is shown below. LINEAR PROGRAMMING PROBLEM MIN 4X14 + 7X15 + 8X24 + 5X25 + 5X34 + 6X35 + 6X46 + 4X47 + 8X48 + 4X49 + 3X56 + 6X57 + 7X58 + 7X59 + 7X39 + 2X45 + 2X54 S.T. 1) X14 + X15 < 450 2) X24 + X25 < 600 3) X34 + X35 + X39 < 380 4) X45 + X46 + X47 + X48 + X49 - X14 - X24 - X34 - X54 = 0 5) X54 + X56 + X57 + X58 + X59 - X15 - X25 - X35 - X45 = 0 6) X46 + X56 = 300 7) X47 + X57 = 300 8) X48 + X58 = 300 9) X39 + X49 + X59 = 400
OPTIMAL SOLUTION
6 - 13
Chapter 6
Objective Function Value = Variable -------------X14 X15 X24 X25 X34 X35 X46 X47 X48 X49 X56 X57 X58 X59 X39 X45 X54
Value --------------320.000 0.000 0.000 600.000 0.000 0.000 0.000 300.000 0.000 20.000 300.000 0.000 300.000 0.000 380.000 0.000 0.000
11220.000 Reduced Costs -----------------0.000 2.000 4.000 0.000 2.000 2.000 2.000 0.000 0.000 0.000 0.000 3.000 0.000 4.000 0.000 1.000 3.000
The value of the solution here is $630 less than the value of the solution for problem 23. The new shipping route from plant 3 to customer 4 has helped (x39 = 380). There is now excess capacity of 130 units at plant 1.
6 - 14
Distribution and Network Models
13. a.
Network Model DC Capacity
Supply
Demand
500 25
1 Detroit
350
4 Iowa
30
500
30
35
2 LA
200
9 Sports Stuff
500
10 Sports Dude
650
20
40
350
27.5
25
8 Just Sports
35 6
45
5 Maryland
32.5 40
35
500
42.5 40
6 Idaho
42.5
500
35 40 32.5
40 3 Austin
700
32.5
b.
Let
7 Ark
27.5
25
42.5
= units shipped from node i to node j
Min 25x1,4 + 25x1,5 + 35x1,6 + 40x1,7 + 35x2,4 + 45x2,5 + 35x2,6 + 42.5x2,7 + 40x3,4 + 40x3,5 + 42.5x3,6 + 32.5x3,7 + 30x4,8 + 27.5x4,9 + 30x4,10 + 20x5,8 + 32.5x5,9 + 40x5,10 + 35x6,8 + 40x6,9 + 32.5x6,10 + 27.5x7,8 + 25x7,9 + 42.5x 7,10 subject to x1,4 + x1,5 + x1,6 + x1,7 ≤ 350 x2,4 + x2,5 + x2,6 + x2,7 ≤ 350 x3,4 + x3,5 + x3,6 + x3,7 ≤ 700 x1,4 + x2,4 + x3,4 = x4,8 + x4,9 + x4,10 x1,5 + x2,5 + x3,5 = x5,8 + x5,9 + x5,10 x1,6 + x2,6 + x3,6 = x6,8 + x6,9 + x6,10 x1,7 + x2,7 + x3,7 = x7,8 + x7,9 + x7,10 x1,4 + x2,4 + x3,4 ≤ 500 x1,5 + x2,5 + x3,5 ≤ 500 x1,6 + x2,6 + x3,6 ≤ 500 x1,7 + x2,7 + x3,7 ≤ 500 x4,8 + x5,8 + x6,8 + x7,8 ≥ 200 x4,9 + x5,9 + x6,9 + x7,9 ≥ 500 x4,10 + x5,10 + x6,10 + x7,10 ≥ 650 xij ≥ 0 for all i and j Solving this formulation we get an optimal cost of $79,625 per week.
6 - 15
Chapter 6
c. Replace the seventh constraint above with x1,4 + x2,4 + x3,4 ≤ 800. The optimal cost associated with this increased capacity at the Iowa distribution center is $79,250 per week. Therefore, the savings over 50 weeks is ($79,625 - $79,250) * 50 = $18,750 per year. The $40,000 cost of expansion exceeds the potential savings, so Sports of All Sorts should not expand. 14.
6 - 16
Distribution and Network Models
A linear programming model is Min 8x14 + 6x15 + 3x24 + 8x25 + 9x34 + 3x35 + 44x46 + 34x47 + 34x48 + 32x49 + 57x56 + 35x57 + 28x58 + 24x59 s.t. 3
x14 + x15
6
x24 + x25
5
x34 + x35 - x24
-x 14 - x15
- x34 - x25
+
x46 +
x47 +
x48 +
= 0
x49
- x35 x46 x47
+
x56 +
+
x56
x57 + x58 +
x59 = 0 = 2
+
x57
x48
= 4 + x58
x49
= 3 +
x59 = 3
xij 0 for all i, j
Optimal Solution Units Shipped Muncie to Cincinnati 1 Cincinnati to Concord 3 Brazil to Louisville 6 Louisville to Macon 2 Louisville to Greenwood 4 Xenia to Cincinnati 5 Cincinnati to Chatham 3
Cost 6 84 18 88 136 15 72 419
Two rail cars must be held at Muncie until a buyer is found. 15.
The positive numbers by nodes indicate the amount of supply at that node. The negative numbers by
6 - 17
Chapter 6
nodes indicate the amount of demand at the node. 16. a.
Min 20x12 + 25x15 + 30x25 + 45x27 + 20x31 + 30x42 + 25x53 + 15x54 + 28x56
+ 35x36 + 12x67 + 27x74
s.t. x31
-
x12
-
x31 + x53 +
x54 +
x15
= 8
x25 +
x27
x36
x53
-
x56
-
-
x12
-
x27
-
x54 +
x74
x15
x25
x67
xij 0 for all i, j b.
x12 = 0 x15 = 0 x25 = 8 x27 = 0 x31 = 8 x36 = 0 x42 = 3
x53 = 5 x54 = 0 x56 = 5 x67 = 0 x74 = 6 x56 = 5
Total cost of redistributing cars = $917
6 - 18
x42
= 5 = 3
-
x36 x74
-
-
x42
= 3 = 2
+
x56 -
x67
= 5 = 6
Distribution and Network Models
17. a.
b. Min 10x11 + 16x12 + 32x13 + 14x21 + 22x22 + 40x23 + 22x31 + 24x32 + 34x33 s.t. x11 + x12 + x13 x21 + x22 + x23 x31 + x32 + x33 x11 + x21 + x31 x12 + x22 + x32 x13 + x23 + x33 xij 0 for all i, j Solution x12 = 1, x21 = 1, x33 = 1
Total completion time = 64
6 - 19
1 1 1 = 1 = 1 = 1
Chapter 6 18. a.
Crews 1
Jobs 30 44
1 Red 31
38 47
25
1
1
1
2 White
2
1
1
3 Blue
3
1
1
4 Green
4
1
5
1
26 1
5 Brown
34 44 43 28
6 - 20
Distribution and Network Models
b. Min s.t.
30x11 x11
x11
+ 44x12 + x21
+ 38x13
x12 + + x22 x31 + x41
+ x12
x13 + x32 + x51 x31 + x23 + x15
x21 + + x22 x13 + x14
+ 47x14
+ 31x15
+ x23 + x42 + + x32 + x24 +
+ x24 + x43 + + x42 + x34 +
x14 + x33 + x52 x41 + x33 + x25
x15 + x34 + x53 x51 + x43 + x35
+ 25x21
+
x25 + x44 +
x35 + x45 x54 + x55
x52 + x44 +
x53 + x54 x45 + x55
+ 28x55
= = = = =
1 1 1 1 1 1 1 1 1 1
xij 0, i = 1, 2,.., 5; j = 1, 2,.., 5 Optimal Solution: Green to Job 1 Brown to Job 2 Red to Job 3 Blue to Job 4 White to Job 5
$26 34 38 39 25 $162
Since the data is in hundreds of dollars, the total installation cost for the 5 contracts is $16,200. 19.
This can be formulated as a linear program with a maximization objective function. There are 24 variables, one for each program/time slot combination. There are 10 constraints, 6 for the potential programs and 4 for the time slots. Optimal Solution: NASCAR Live Hollywood Briefings World News Ramundo & Son
5:00 – 5:30 p.m. 5:30 – 6:00 p.m. 7:00 – 7:30 p.m. 8:00 – 8:30 p.m.
Total expected advertising revenue = $30,500 20. a.
This is the variation of the assignment problem in which multiple assignments are possible. Each distribution center may be assigned up to 3 customer zones. The linear programming model of this problem has 40 variables (one for each combination of distribution center and customer zone). It has 13 constraints. There are 5 supply ( 3) constraints and 8 demand (= 1) constraints. The optimal solution follows.
Plano:
Assignments Kansas City, Dallas
Cost ($1000s) 34
6 - 21
Chapter 6 Flagstaff: Springfield: Boulder:
21.
Los Angeles Chicago, Columbus, Atlanta Newark, Denver Total Cost -
15 70 97 $216
b.
The Nashville distribution center is not used.
c.
All the distribution centers are used. Columbus is switched from Springfield to Nashville. Total cost increases by $11,000 to $227,000. A linear programming formulation and the optimal solution are given. For the decision variables, xij, we let the first subscript correspond to the supplier and the second subscript correspond to the distribution hub. Thus, xij = 1 if supplier i is awarded the bid to supply hub j and xij = 0 if supplier i is not awarded the bid to supply hub j. Min 190x11 +175x12 + 125x13 + 230x14 + 150x21 + 235x22 + 155x23 + 220x24 + 210x31 + 225x32 + 135x33 +260x34 + 170x41 + 185x42 + 190x43 + 280x44 + 220x51 + 190x52 + 140x53 + 240x54 + 270x61 + 200x62 + 130x63 + 260x64 s.t. x11
+ x12 + x21
+ x13 + x22
x31
+ x14 + x23
+ x32 x41
+ x33 + x42
x51 x11
+ x21
+ x31 x12
+ x41 + x22
+ x24
+ x51 + x32 x13
+ x34 + x43
+ x52 x61 + x61 + x42 + x23
+ x62
+ x52 + x62 + x33 + x43 x14 + x24 xij 0 for all i, j
Optimal Solution
Bid
Martin – Hub 2 Schmidt Materials – Hub 4 D&J Burns – Hub 1 Lawler Depot – Hub 3
175 220 170 130 695
6 - 22
+ x44 + x53
+ x54 + x63
+ x53 + x34
+ x63 + x44
+ x64
+ x45
+ x46
≤ ≤ ≤ ≤ ≤ ≤ = = = =
1 1 1 1 1 1 1 1 1 1
Distribution and Network Models
22.
A linear programming formulation of this problem can be developed as follows. Let the first letter of each variable name represent the professor and the second two the course. Note that a DPH variable is not created because the assignment is unacceptable.
Max 2.8AUG
+ 2.2AMB
+ 3.3AMS
+ 3.0APH
+
+ +
AMS BMB CUG
+ + +
CUG BMB AMS
+ + +
+ 3.2BUG
+
···
+ 2.5DMS
s.t. AUG
AUG
+
AMB BUG
BUG AMB
+ +
APH BMS CMB DUG DUG CMB BMS APH
+ + +
BPH CMS DMB
+ +
CPH DMS
+ + +
DMB CMS BPH
+ +
DMS CPH
All Variables 0 Optimal Solution: A to MS course B to Ph.D. course C to MBA course D to Undergraduate course Max Total Rating 23.
Rating 3.3 3.6 3.2 3.2 13.3
Origin – Node 1 Transshipment Nodes 2 to 5 Destination – Node 7 The linear program will have 14 variables for the arcs and 7 constraints for the nodes. 1 if the arc from node i to node j is on the shortest route Let xij = 0 otherwise
Min 7 x12 + 9 x13 + 18 x14 + 3x23 + 5 x25 + 3x32 + 4 x35 + 3x46 + 5 x52 + 4 x53 + 2 x56 +6 x57 + 2 x65 + 3 x67 s.t. Flow Out
Flow In
Node 1
x12 + x13 + x14
Node 2
x23 + x25
− x12 − x32 − x52 = 0
Node 3
x32 + x35
− x13 − x23 − x53 = 0
Node 4
x46
− x14
Node 5
x52 + x53 + x56 + x57
− x25 − x35 − x65 = 0
Node 6
x65 + x67
− x46 − x56
=0
+ x57 + x67
=1
Node 7
=1
6 - 23
=0
= = = =
1 1 1 1 1 1 1 1
Chapter 6
xij > 0 for all i and j
Optimal Solution: x12 = 1 , x25 = 1 , x56 = 1 , and x67 = 1 Shortest Route 1-2-5-6-7 Length = 17 24.
The linear program has 13 variables for the arcs and 6 constraints for the nodes. Use same six constraints for the Gorman shortest route problem as shown in the text. The objective function changes to travel time as follows. Min 40 x12 + 36 x13 + 6 x23 + 6 x32 + 12 x24 + 12 x42 + 25 x26 + 15 x35 + 15 x53 + 8 x45 + 8 x54 + 11 x46 + 23 x56 Optimal Solution: x12 = 1 , x24 = 1 , and x46 = 1 Shortest Route 1-2-4-6 Total Time = 63 minutes
25. a.
Origin – Node 1 Transshipment Nodes 2 to 5 Destination – Node 6 The linear program will have 13 variables for the arcs and 6 constraints for the nodes. 1 if the arc from node i to node j is on the shortest route Let xij = 0 otherwise
Min 35 x12 + 30 x13 + 12 x23 + 18 x24 + 39 x26 + 12 x32 + 15 x35 + 18 x42 + 12 x45 +16x46 +15x53 +12x54 +30x56
s.t. Flow Out
Flow In
Node 1
x12 + x13
Node 2
x23 + x24 + x26
− x12 − x32 − x42
=0
Node 3
x32 + x35
− x13 − x23 − x53
=0
Node 4
x42 + x45 + x46
− x24 − x54
=0
Node 5
+ x53 + x54 + x56
− x35 − x45
=0
+ x26 + x46 + x56
=1
=1
Node 6
xij > 0 for all I and j
b.
Optimal Solution: x12 = 1 , x24 = 1 , and x46 = 1 Shortest Route 1-2-4-6
6 - 24
Distribution and Network Models
Total time = 69 minutes c.
26.
Allowing 8 minutes to get to node 1 and 69 minutes to go from node 1 to node 6, we expect to make the delivery in 77 minutes. With a 20% safety margin, we can guarantee a delivery in 1.2(77) = 92 minutes. It is 1:00 p.m. now. Guarantee delivery by 2:32 p.m. Origin – Node 1 Transshipment Nodes 2 to 5 and node 7 Destination – Node 6 The linear program will have 18 variables for the arcs and 7 constraints for the nodes. 1 if the arc from node i to node j is on the shortest route Let xij = 0 otherwise
Min
35 x12 + 30 x13 + 20 x14 + 8 x23 + 12 x25 + 8 x32 + 9 x34 + 10 x35 + 20 x36 +9 x43 + 15 x47 + 12 x52 + 10 x53 + 5 x56 + 20 x57 + 15 x74 + 20 x75 + 5 x76
s.t. Flow Out
Flow In
Node 1
x12 + x13 + x14
Node 2
x23 + x25
− x12 − x32 − x52
Node 3
x32 + x34 + x35 + x36
− x13 − x23 − x43 − x53 = 0
Node 4
x43 + x47
− x14 − x34 − x74
=0
Node 5
x52 + x53 + x56 + x57
− x25 − x35 − x75
=0
+ x36 + x56 + x76
=1
− x47 − x57
=0
Node 6 Node 7
x74 + x75 + x76
=1 =0
xij > 0 for all i and j
Optimal Solution: x14 = 1 , x47 = 1 , and x76 = 1 Shortest Route 1-4-7-6 Total Distance = 40 miles 27.
Origin – Node 1 Transshipment Nodes 2 to 9 Destination – Node 10 (Identified by the subscript 0) The linear program will have 29 variables for the arcs and 10 constraints for the nodes.
6 - 25
Chapter 6
1 if the arc from node i to node j is on the shortest route Let xij = 0 otherwise
Min
8 x12 + 13 x13 + 15 x14 + 10 x15 + 5 x23 + 15 x27 + 5 x32 + 5 x36 + 2 x43 + 4 x45 +3 x46 + 4 x54 + 12 x59 + 5 x63 + 3 x64 + 4 x67 + 2 x68 + 5 x69 + 15 x72 + 4 x76 +2 x78 + 4 x70 + 2 x86 + 5 x89 + 7 x80 + 12 x95 + 5 x96 + 5 x98 + 5 x90
s.t. Flow Out
Flow In
Node 1
x12 + x13 + x14 + x15
Node 2
x23 + x27
− x12 − x32 − x72
=0
Node 3
x32 + x36
− x13 − x23 − x43 − x63
=0
Node 4
x43 + x45 + x46
− x14 − x54 − x64
=0
Node 5
x54 + x59
− x15 − x45 − x95
=0
Node 6
x63 + x64 + x67 + x68 + x69
− x36 − x46 − x76 − x86 − x96 = 0
Node 7
x72 + x76 + x78 + x70
− x27 − x67
=0
Node 8
x86 + x89 + x80
− x68 − x78 − x98
=0
Node 9
x95 + x96 + x98 + x90
− x59 − x69 − x89
=0
+ x70 + x80 + x90
=1
=1
Node 10
xij > 0 for all i and j
Optimal Solution: x15 = 1 , x54 = 1 , x46 = 1 , x67 = 1 , and x70 = 1 Shortest Route 1-5-4-6-7-10 Total Time = 25 minutes 28.
Origin – Node 0 Transshipment Nodes 1 to 3 Destination – Node 4 The linear program will have 10 variables for the arcs and 5 constraints for the nodes. 1 if the arc from node i to node j is on the minimum cost route Let xij = 0 otherwise
Min
600 x01 + 1000 x02 + 2000 x03 + 2800 x04 + 500 x12 + 1400 x13 + 2100 x14 +800 x23 + 1600 x24 + 700 x34
s.t. Flow Out Node 0
Flow In
x01 + x02 + x03 + x04
=1
6 - 26
Distribution and Network Models − x01
=0
x23 + x24
− x02 − x12
=0
x34
− x03 − x13 − x23
=0
− x04 − x14 − x24 − x34
=1
Node 1
x12 + x13 + x14
Node 2 Node 3 Node 4
xij > 0 for all i and j
Optimal Solution: x02 = 1 , x23 = 1 , and x34 = 1 Shortest Route 0-2-3-4 Total Cost = $2500 29.
The capacitated transshipment problem to solve is given: Max x61 s.t. x12 + x13 + x14 - x61 = 0 x24 + x25 - x12 - x42 = 0 x34 + x36 - x13 - x43 = 0 x42 + x43 + x45 + x46 - x14 - x24 - x34 - x54 = 0 x54 + x56 - x25 - x45 =0 x61 - x36 + x46 - x56 =0 x12 2 x13 6 x14 3 x24 1 x25 4 x34 3 x36 2 x42 1 x43 3 x45 1 x46 3 x54 1 x56 6 xij 0 for all i, j 3 2
5
2
1
3
3
1
4
1
4 2
4
6 2
3
The system cannot accommodate a flow of 10,000 vehicles per hour.
30.
6 - 27
Maximum Flow 9,000 Vehicles Per Hour
Chapter 6
4
2 3
1 3
1
5
3
4 3
5
6
2
11,000
6
2
3
31.
The maximum number of messages that may be sent is 10,000.
32. a.
10,000 gallons per hour or 10 hours
b. 33.
Flow reduced to 9,000 gallons per hour; 11.1 hours. Current Max Flow = 6,000 vehicles/hour. With arc 3-4 at a 3,000 unit/hour flow capacity, total system flow is increased to 8,000 vehicles/hour. Increasing arc 3-4 to 2,000 units/hour will also increase system to 8,000 vehicles/hour. Thus a 2,000 unit/hour capacity is recommended for this arc.
34.
Maximal Flow = 23 gallons / minute. Five gallons will flow from node 3 to node 5.
35. a.
Modify the problem by adding two nodes and two arcs. Let node 0 be a beginning inventory node with a supply of 50 and an arc connecting it to node 5 (period 1 demand). Let node 9 be an ending inventory node with a demand of 100 and an arc connecting node 8 (period 4 demand to it).
b. 2x15
Min s.t.
+ 5x26
+ 3x37
+ 3x48
+ 0.25x56
+ 0.25x67
+ 0.25x78
+ 0.25x89
x05 x15 x26 x37 x48 x05
+ x15
+
x26 x37
x56 x56
+
x48 xij 0 for all i and j
Optimal Solution:
6 - 28
x67 x67
+
x78 x78
-
x89 x89
= = = = = =
50 600 300 500 400 400 500 400 400 100
Distribution and Network Models x05 = 50 x15 = 600 x26 = 250 x37 = 500 x48 = 400
x56 = 250 x67 = 0 x78 = 100 x89 = 100
Total Cost = $5262.50 36. a.
Let R1, R2, R3 O1, O2, O3 D1, D2, D3
represent regular time production in months 1, 2, 3 represent overtime production in months 1, 2, 3 represent demand in months 1, 2, 3
Using these 9 nodes, a network model is shown.
b.
Use the following notation to define the variables: first two letters designates the "from node" and the second two letters designates the "to node" of the arc. For instance, R1D1 is amount of regular time production available to satisfy demand in month 1, O1D1 is amount of overtime production in
6 - 29
Chapter 6
month 1 available to satisfy demand in month 1, D1D2 is the amount of inventory carried over from month 1 to month 2, and so on. MIN 50R1D1 + 80O1D1 + 20D1D2 + 50R2D2 + 80O2D2 + 20D2D3 + 60R3D3 + 100O3D3 S.T. 1) R1D1 275 2) O1D1 100 3) R2D2 200 4) O2D2 50 5) R3D3 100 6) O3D3 50 7) R1D1 + O1D1 - D1D2 = 150 8) R2D2 + O2D2 + D1D2 - D2D3 = 250 9) R3D3 + O3D3 + D2D3 = 300 c.
Optimal Solution:
Variable -------------R1D1 O1D1 D1D2 R2D2 O2D2 D2D3 R3D3 O3D3
Value --------------275.000 25.000 150.000 200.000 50.000 150.000 100.000 50.000
Value = $46,750 Note: Slack variable for constraint 2 = 75. d.
The values of the slack variables for constraints 1 through 6 represent unused capacity. The only nonzero slack variable is for constraint 2; its value is 75. Thus, there are 75 units of unused overtime capacity in month 1.
6 - 30
Chapter 7 Integer Linear Programming Learning Objectives 1.
Be able to recognize the types of situations where integer linear programming problem formulations are desirable.
2.
Know the difference between all-integer and mixed integer linear programming problems.
3.
Be able to solve small integer linear programs with a graphical solution procedure.
4.
Be able to formulate and solve fixed charge, capital budgeting, distribution system, and product design problems as integer linear programs.
5.
See how zero-one integer linear variables can be used to handle special situations such as multiple choice, k out of n alternatives, and conditional constraints.
6.
Be familiar with the computer solution of MILPs.
7.
Understand the following terms: all-integer mixed integer zero-one variables LP relaxation multiple choice constraint
mutually exclusive constraint k out of n alternatives constraint conditional constraint co-requisite constraint
Solutions:
7-1
Chapter 7
1.
a.
This is a mixed integer linear program. Its LP Relaxation is Max s.t.
30x1 +
25x2
3x1 + 1.5x1 + x1 +
1.5x2 2x2 x2
400 250 150
x1 , x 2 0 b.
2.
This is an all-integer linear program. Its LP Relaxation just requires dropping the words "and integer" from the last line.
a.
b.
The optimal solution to the LP Relaxation is given by x1 = 1.43, x2 = 4.29 with an objective function value of 41.47. Rounding down gives the feasible integer solution x1 = 1, x2 = 4. Its value is 37.
c.
7-2
Integer Linear Programming x2 6
Optimal Integer Solution (0,5)
5 4 5x
3
1 +
8x
2 =
40
2
1 x1
0
0 1 2 3 4 5 6 7 8 The optimal solution is given by x1 = 0, x2 = 5. Its value is 40. This is not the same solution as that
found by rounding down. It provides a 3 unit increase in the value of the objective function. 3.
a.
b.
The optimal solution to the LP Relaxation is shown on the above graph to be x1 = 4, x2 = 1. Its value is 5.
7-3
Chapter 7
c.
4.
The optimal integer solution is the same as the optimal solution to the LP Relaxation. This is always the case whenever all the variables take on integer values in the optimal solution to the LP Relaxation.
a.
The value of the optimal solution to the LP Relaxation is 36.7 and it is given by x1 = 3.67, x2 = 0.0. Since we have all less-than-or-equal-to constraints with positive coefficients, the solution obtained by "rounding down" the values of the variables in the optimal solution to the LP Relaxation is feasible. The solution obtained by rounding down is x1 = 3, x2 = 0 with value 30. Thus a lower bound on the value of the optimal solution is given by this feasible integer solution with value 30. An upper bound is given by the value of the LP Relaxation, 36.7. (Actually an upper bound of 36 could be established since no integer solution could have a value between 36 and 37.)
b.
7-4
Integer Linear Programming
x2 7 6 5
10 x1 + 3 x 2 = 36
4 3 2 1 0
x1 0
1
2
3
4
5
The optimal solution to the ILP is given by x1 = 3, x2 = 2. Its value is 36. The solution found by "rounding down" the solution to the LP relaxation had a value of 30. A 20% increase in this value was obtained by finding the optimal integer solution - a substantial difference if the objective function is being measured in thousands of dollars.
7-5
Chapter 7
c.
x2 7 ..
6 5
Optimal solution to LP relaxation (0,5.71) Optimal integer solutions (2.47,3.60)
4
3x1 + 6x2 = 34.26
3
3x1 + 6x2 = 30
2
(3.67,0)
1 0
x1
0 1 2 3 4 5 The optimal solution to the LP Relaxation is x1= 0, x2 = 5.71 with value = 34.26. The solution obtained by "rounding down" is x1 = 0, x2 = 5 with value 30. These two values provide an upper bound of 34.26 and a lower bound of 30 on the value of the optimal integer solution. There are alternative optimal integer solutions given by x1 = 0, x2 = 5 and x1 = 2, x2 = 4; value is 30. In this case rounding the LP solution down does provide the optimal integer solution.
7-6
Integer Linear Programming
5.
a.
The feasible mixed integer solutions are indicated by the boldface vertical lines in the graph above. b.
The optimal solution to the LP relaxation is given by x1 = 3.14, x2 = 2.60. Its value is 14.08. Rounding the value of x1 down to find a feasible mixed integer solution yields x1 = 3, x2 = 2.60 with a value of 13.8. This solution is clearly not optimal. With x1 = 3 we can see from the graph that x2 can be made larger without violating the constraints.
c.
Optimal mixed integer solution (3, 2.67)
The optimal solution to the MILP is given by x1 = 3, x2 = 2.67. Its value is 14.
7-7
Chapter 7
6.
a.
x2 8 7 Optimal solution to LP relaxation (1.96, 5.48)
6 5
x1 + x 2 = 7.44
4 3 2 1
x1
0 0 b.
1
2
3
4
5
6
7
8
The optimal solution to the LP Relaxation is given by x1 = 1.96, x2 = 5.48. Its value is 7.44. Thus an upper bound on the value of the optimal is given by 7.44. Rounding the value of x2 down yields a feasible solution of x1 = 1.96, x2 = 5 with value 6.96. Thus a lower bound on the value of the optimal solution is given by 6.96.
7-8
Integer Linear Programming
c.
x2 8 7
Optimal mixed integer solution (1.29, 6)
6 5 x1 + x 2 = 7.29
4 3 2 1 0
x1 0
1
2
3
4
5
6
7
The optimal solution to the MILP is x1 = 1.29, x2 = 6. Its value is 7.29. The solution x1 = 2.22, x2 = 5 is almost as good. Its value is 7.22. 7.
a.
x1 + x3 + x5 + x6 = 2
b.
x3 - x5 = 0
c.
x1 + x4 = 1
d.
x4 x1 x4 x3
e.
x4 x1 x4 x3 x4 x1 + x3 - 1
7-9
8
Chapter 7
8.
a.
1 if investment alternative i is selected Let xi = 0 otherwise max s.t.
4000x1 + 6000x2 +
10500x3 + 4000x4 + 8000x5
+ 3000x6
3000x1 + 2500x2 + 1000x1 + 3500x2 + 4000x1 + 3500x2 +
6000x3 + 2000x4 + 5000x5 4000x3 + 1500x4 + 1000x5 5000x3 + 1800x4 + 4000x5
+ 1000x6 + 500x6 + 900x6
x1, x2, x3, x4, x5, x6 = 0, 1 The optimal solution is x3 = 1 x4 = 1 x6 = 1 Value = 17,500 b.
The following mutually exclusive constraint must be added to the model. x1 + x2 1 No change in optimal solution.
c.
The following co-requisite constraint must be added to the model in b. x3 - x4 = 0. No change in optimal solution.
9.
a.
x4 8000 s4
b.
x6 6000 s6
c.
x4 8000 s4 x6 6000 s6 s4 + s6 = 1
d.
10. a.
Min 15 x4 + 18 x6 + 2000 s4 + 3500 s6
Let xi = 1 if a substation is located at site i, 0 otherwise
7 - 10
10,500 7,000 8,750
Integer Linear Programming
min s.t.
xA +
xB +
xC +
xA +
xB + xB
xC
xD +
b. 11. a.
xD
+
xA +
xB
+ xD + xD
xC +
xE xE + + xE +
xF xF + xF + +
xG xG 1 1 1 1 xG 1 xG 1 xG 1
(area 1 covered) (area 2 covered) (area 3 covered) (area 4 covered) (area 5 covered) (area 6 covered) (area 7 covered)
Choose locations B and E. Let Pi = units of product i produced Max s.t.
b.
xB +
xF + +
xC xA +
xE +
25P1
+
28P2
1.5P1 2P1 .25P1
+ + +
3P2 1P2 .25P2
+
30P3
2P3 + + 2.5P3 + .25P3 P1, P2, P3 0
450 350 50
The optimal solution is P1 = 60 P2 = 80 P3 = 60
Value = 5540
This solution provides a profit of $5540. c.
Since the solution in part (b) calls for producing all three products, the total setup cost is $1550 = $400 + $550 + $600. Subtracting the total setup cost from the profit in part (b), we see that Profit = $5540 - 1550 = $3990
d.
We introduce a 0-1 variable yi that is one if any quantity of product i is produced and zero otherwise. With the maximum production quantities provided by management, we obtain 3 new constraints: P1 175y1 P2 150y2 P3 140y3 Bringing the variables to the left-hand side of the constraints, we obtain the following fixed charge formulation of the Hart problem.
7 - 11
Chapter 7
Max s.t.
25P1
+
28P2
+
30P3
1.5P1 2P1 .25P1 P1
+ + +
3P2 1P2 .25P2
2P3 + + 2.5P3 + .25P3
-
-
400y1
-
550y2
-
600y3
175y1
P2
-
150y2
P3
-
140y3
450 350 50 0 0 0
P1, P2, P3 0; y1, y2, y3 = 0, 1 e.
The optimal solution is P1 = 100 P2 = 100 P3 = 0
y1 = 1 y2 = 1 y3 = 0
Value = 4350
The profit associated with this solution is $4350. This is an improvement of $360 over the solution in part (c). 12. a. We use the following data: i=1,2,…20 i=1,2,…20 i=1,2,…20
b[i,j] = the bid for city i from carrier j, dem[i] = the demand in truckloads for city i c[i,j] = the cost of assigning city i to carrier j
j=1,2,…7 j=1,2,…7
Note: cij = (demi)(bij) used in the objective function. For example, c[8,2] = dem[8]*b[8,2] = 25*1800 = 45000. Let
y[j] = 1 if carrier j is selected, 0 if not x[i,j] = 1 if city i is assigned to carrier j, 0 if not
j=1,2,…7 i=1,2,…20
j=1,2,…7
Minimize the cost of city-carrier assignments (note: for brevity, zeros are not shown). Minimize 65640x[1,5] + 49980x[1,6] + 53700x[1,7] + 14530x[2,2] + 26020x[2,5] + 17670x[2,6] + 30680x[3,2] + 45660x[3,5] + 37140x[3,6] + 37400x[3,7] + 67480x[4,2] + 104680x[4,5] + 69520x[4,6] + 15230x[5,2] + 22390x[5,5] + 17710x[5,6] + 18550x[5,7] + 15210x[6,2] + 15710x[6,5] + 15450x[6,7] + 25200x[7,2] + 23064x[7,4] + 23256x[7,5] + 24600x[7,7] + 45000x[8,2] + 35800x[8,4] + 35400x[8,5] + 43475x[8,7] + 28350x[9,2] + 30825x[9,4] + 29525x[9,5] + 28750x[9,7] + 22176x[10,2] + 20130x[10,4] + 22077x[10,5] + 22374x[10,7] + 7964x[11,1] + 7953x[11,3] + 6897x[11,4] + 7227x[11,5] + 7766x[11,7] + 22214x[12,1] + 22214x[12,3] + 20909x[12,4] + 19778x[12,5] + 21257x[12,7] + 8892x[13,1] + 8940x[13,3] + 8184x[13,5] + 8796x[13,7] + 19560x[14,1] + 19200x[14,2] + 19872x[14,3] + 17880x[14,5] + 19968x[14,7] + 9040x[15,1] + 8800x[15,3] + 8910x[15,5] + 9140x[15,7] + 9580x[16,1] + 9330x[16,3] + 8910x[16,5] + 9140x[16,7] + 21275x[17,1] + 21367x[17,3] + 21551x[17,5] + 22632x[17,7] + 22300x[18,1] + 21725x[18,3] + 20550x[18,4] + 20725x[18,5] + 21600x[18,7] + 11124x[19,1] + 11628x[19,3] + 11604x[19,5] + 12096x[19,7] + 9630x[20,1] + 9380x[20,3] + 9550x[20,5] + 9950x[20,7]
7 - 12
Integer Linear Programming
subject to Every City is assigned to exactly one carrier: x[1,1] + x[1,2] + x[1,3] + x[1,4] + x[1,5] + x[1,6] + x[1,7] = 1 x[2,1] + x[2,2] + x[2,3] + x[2,4] + x[2,5] + x[2,6] + x[2,7] = 1 x[3,1] + x[3,2] + x[3,3] + x[3,4] + x[3,5] + x[3,6] + x[3,7] = 1 x[4,1] + x[4,2] + x[4,3] + x[4,4] + x[4,5] + x[4,6] + x[4,7] = 1 x[5,1] + x[5,2] + x[5,3] + x[5,4] + x[5,5] + x[5,6] + x[5,7] = 1 x[6,1] + x[6,2] + x[6,3] + x[6,4] + x[6,5] + x[6,6] + x[6,7] = 1 x[7,1] + x[7,2] + x[7,3] + x[7,4] + x[7,5] + x[7,6] + x[7,7] = 1 x[8,1] + x[8,2] + x[8,3] + x[8,4] + x[8,5] + x[8,6] + x[8,7] = 1 x[9,1] + x[9,2] + x[9,3] + x[9,4] + x[9,5] + x[9,6] + x[9,7] = 1 x[10,1] + x[10,2] + x[10,3] + x[10,4] + x[10,5] + x[10,6] + x[10,7] = 1 x[11,1] + x[11,2] + x[11,3] + x[11,4] + x[11,5] + x[11,6] + x[11,7] = 1 x[12,1] + x[12,2] + x[12,3] + x[12,4] + x[12,5] + x[12,6] + x[12,7] = 1 x[13,1] + x[13,2] + x[13,3] + x[13,4] + x[13,5] + x[13,6] + x[13,7] = 1 x[14,1] + x[14,2] + x[14,3] + x[14,4] + x[14,5] + x[14,6] + x[14,7] = 1 x[15,1] + x[15,2] + x[15,3] + x[15,4] + x[15,5] + x[15,6] + x[15,7] = 1 x[16,1] + x[16,2] + x[16,3] + x[16,4] + x[16,5] + x[16,6] + x[16,7] = 1 x[17,1] + x[17,2] + x[17,3] + x[17,4] + x[17,5] + x[17,6] + x[17,7] = 1 x[18,1] + x[18,2] + x[18,3] + x[18,4] + x[18,5] + x[18,6] + x[18,7] = 1 x[19,1] + x[19,2] + x[19,3] + x[19,4] + x[19,5] + x[19,6] + x[19,7] = 1 x[20,1] + x[20,2] + x[20,3] + x[20,4] + x[20,5] + x[20,6] + x[20,7] = 1 If a Carrier is selected, it can be assigned only the number of bids made: Note: The idea here is that if carrier j is not chosen, then no cities can be assigned to that carrier. Hence if y[j] = 0, the sum must be less than or equal to zero and hence all the associated x’s must be zero. If y[j] = 1 then the constraint becomes redundant. It could also be modeled as x[i,j] ≤ y[j] but this would generate more constraints x[1,1] + x[2,1] + x[3,1] + x[4,1] + x[5,1] + x[6,1] + x[7,1]+ x[8,1] + x[9,1] + x[10,1] + x[11,1] + x[12,1] + x[13,1] + x[14,1] + x[15,1] + x[16,1] + x[17,1] + x[18,1] + x[19,1] + x[20,1] ≤ 10y[1] x[1,2] + x[2,2] + x[3,2] + x[4,2] + x[5,2] + x[6,2] + x[7,2] + x[8,2] + x[9,2] + x[10,2] + x[11,2] + x[12,2] + x[13,2] + x[14,2] + x[15,2] + x[16,2] + x[17,2] + x[18,2] + x[19,2] + x[20,2] ≤ 10y[2] x[1,3] + x[2,3] + x[3,3] + x[4,3] + x[5,3] + x[6,3] + x[7,3] + x[8,3] + x[9,3] + x[10,3] + x[11,3] + x[12,3] + x[13,3] + x[14,3] + x[15,3] + x[16,3] + x[17,3] + x[18,3] + x[19,3] + x[20,3] ≤ 10y[3] x[1,4] + x[2,4] + x[3,4] + x[4,4] + x[5,4] + x[6,4] + x[7,4]+ x[8,4] + x[9,4] + x[10,4] + x[11,4] + x[12,4] + x[13,4] + x[14,4] + x[15,4] + x[16,4] + x[17,4] + x[18,4] + x[19,4] + x[20,4] ≤ 7y[4] x[1,5] + x[2,5] + x[3,5] + x[4,5] + x[5,5] + x[6,5] + x[7,5] + x[8,5] + x[9,5] + x[10,5] + x[11,5] + x[12,5] + x[13,5] + x[14,5] + x[15,5] + x[16,5] + x[17,5] + x[18,5] + x[19,5] + x[20,5] ≤ 20y[5] x[1,6] + x[2,6] + x[3,6] + x[4,6] + x[5,6] + x[6,6] + x[7,6]+ x[8,6] + x[9,6] + x[10,6] + x[11,6] + x[12,6] + x[13,6] + x[14,6] + x[15,6] + x[16,6] + x[17,6] + x[18,6] + x[19,6] + x[20,6] ≤ 5y[6]
7 - 13
Chapter 7
x[1,7] + x[2,7] + x[3,7] + x[4,7] + x[5,7] + x[6,7] + x[7,7] + x[8,7] + x[9,7] + x[10,7] + x[11,7] + x[12,7] + x[13,7] + x[14,7] + x[15,7] + x[16,7] + x[17,7] + x[18,7] + x[19,7] + x[20,7] ≤ 18y[7] Non-bids must be set to 0: x[1,1] + x[2,1] + x[3,1] + x[4,1] + x[5,1] + x[6,1] + x[7,1] + x[8,1] + x[9,1] + x[10,1] = 0 x[1,2] + x[11,2] + x[12,2] + x[13,2] + x[15,2] + x[16,2] + x[17,2] + x[18,2] + x[19,2] + x[20,2] = 0 x[1,3] + x[2,3] + x[3,3] + x[4,3] + x[5,3] + x[6,3] + x[7,3] + x[8,3] + x[9,3] + x[10,3] = 0 x[1,4] + x[2,4] + x[3,4] + x[4,4] + x[5,4] + x[6,4] + x[13,4] + x[14,4] + x[15,4] + x[16,4] + x[17,4] + x[19,4] + x[20,4] = 0 x[6,6] + x[7,6] + x[8,6] + x[9,6] + x[10,6] + x[11,6] + x[12,6] + x[13,6] + x[14,6] + x[15,6] + x[16,6] + x[17,6] + x[18,6] + x[19,6] + x[20,6] = 0 x[2,7] + x[4,7] = 0 No more than 3 carriers y[1] + y[2] + y[3] + y[4] + y[5] + y[6] + y[7] ≤ 3 Solution:
Total Cost = $436,512 Carrier 2: assigned cities 2, 3, 4, 5, 6, and 9 Carrier 5: assigned cities 7, 8 and 10-20 Carrier 6: assigned city 1
7 - 14
Integer Linear Programming
b.
Given the incremental drop in cost, three seems like the correct number of carriers (the curve flattens considerably after three carriers).
13. a.
One just needs to add the following multiple choice constraint to the problem. y1 + y2 = 1 New Optimal Solution: y1 = 1, y3 = 1, x12 = 10, x31 = 30, x52 = 10, x53 = 20 Value = 940
b.
Since one plant is already located in St. Louis, it is only necessary to add the following constraint to the model y3 + y4 1 New Optimal Solution: y4 = 1, x42 = 20, x43 = 20, x51 = 30 Value = 860
14. a.
Let 1 denote the Michigan plant 2 denote the first New York plant 3 denote the second New York plant 4 denote the Ohio plant 5 denote the California plant It is not possible to meet needs by modernizing only one plant.
7 - 15
Chapter 7
The following table shows the options which involve modernizing two plants.
1
2
Plant 3
4
5
Transmission Capacity
Engine Block Capacity
Feasible ?
700 1100 900 600 1200 1000 700 1400 1100 900
1300 900 1400 700 1200 1700 1000 1300 600 1100
No Yes Yes No Yes Yes No Yes No Yes
b.
Modernize plants 1 and 3 or plants 4 and 5.
c.
1 if plant i is modernized Let xi = 0 if plant i is not modernized 25x1 +
Min s.t.
35x2 +
35x3 +
40x4 +
Cost
60 65 70 75 75 60
25x5
300x1 + 400x2 + 800x3 + 600x4 + 300x5 900 Transmissions 500x1 + 800x2 + 400x3 + 900x4 + 200x5 900 Engine Blocks d.
Optimal Solution: x1 = x3 = 1.
15. a. Let xj = 1 if place a PBB in county j , 0 if not
j=1,2,…88
Min x1 + x2 + x3 +….+ x88 s.t. x1 + x2 + x18 + x19 x1 + x2 + x3 + x17 + x18
≥ 1 ≥ 1
● ● ●
● ● ●
● ● ●
● ● ●
≥ 1
x71 + x72 + x87 + x88
County 1 County 2
County 88
Solution: Minimum number of counties is 15.
7 - 16
Integer Linear Programming
16. a. min 105x9 x9 x9 x9 x9
+ 105x10 + 105x11 + 32y9 + 32y10 + 32y11 + 32y12 + 32y1 y9 + x10 y9 + y10 + + x10 + x11 + y9 + y10 + y11 + x10 + x11 + y9 + y10 + y11 + y12 + x10 + x11 y10 + y11 + y12 + y1 + x9 x11 y y y1 + 11 + 12 + x9 + x10 y12 + y1 + x9 + x10 + x11 y1 + x10 + x11 x11
+ 32y2 + 32y3
+ + + +
y2 y2 + y2 + y2 + +
y3 y3 y3 y3
xi, yj 0 and integer for i = 9, 10, 11 and j = 9, 10, 11, 12, 1, 2, 3 b.
Solution to LP Relaxation obtained using LINDO/PC: y9 = 6 y11 = 2
c.
y12 = 6 y1 = 1
y3 = 6
All other variables = 0. Cost: $672.
The solution to the LP Relaxation is integral therefore it is the optimal solution to the integer program. A difficulty with this solution is that only part-time employees are used; this may cause problems with supervision, etc. The large surpluses from 5, 12-1 (4 employees), and 3-4 (9 employees) indicate times when the tellers are not needed for customer service and may be reassigned to other tasks.
d.
Add the following constraints to the formulation in part (a).
7 - 17
6 4 8 10 9 6 4 7 6 6
Chapter 7
x9 1 x11 1
x9 +x10 + x11 5 The new optimal solution, which has a daily cost of $909 is x9 = 1 x11 = 4
y9 = 5 y12 = 5 y3 = 2
There is now much less reliance on part-time employees. The new solution uses 5 full-time employees and 12 part-time employees; the previous solution used no full-time employees and 21 part-time employees. 17.
a.
Let xj = 1 if place a PBB in county j , 0 if not j=1,2,…88 Let yi = 1 if county i is covered (that is, shares a border with a county with a PPB or has a PBB located in it), 0 if not, j=1,2,…88 Max 36396y1 + 39987y2 + ….+ 17382y88 s.t. x1 + x2 + x18 + x19 x1 + x2 + x3 + x17 + x18
≥ y1 ≥ y2
● ● ●
● ● ●
● ● ●
● ● ●
x71 + x72 + x87 + x88 x1 + x2 + ……+ x88
18. a.
≥ y88 ≤ k
County 1 County 2 ● ● ●
County 88 No more than k PPBs
b.
Solution for k=2 is place PPBs in counties 11 and 77 with a population reach of 5,239,214.
c.
Based on the graph below, 7 or 8 PPBs seems reasonable, as the incremental reach decreases significantly at that point.
Add the part-worths for Antonio's Pizza for each consumer in the Salem Foods' consumer panel.
7 - 18
Integer Linear Programming
Consumer 1 2 3 4 5 6 7 8 b.
Max s.t.
Overall Preference for Antonio's 2 + 6 + 17 + 27 = 52 7 + 15 + 26 + 1 = 49 5 + 8 + 7 + 16 = 36 20 + 20 + 14 + 29 = 83 8 + 6 + 20 + 5 = 39 17 + 11 + 30 + 12 = 70 19 + 12 + 25 + 23 = 79 9 + 4 + 16 + 30 = 59
Let lij = 1 if level i is chosen for attribute j, 0 otherwise yk = 1 if consumer k chooses the Salem brand, 0 otherwise y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 11l11 + 11l11 + 7l11 + 13l11 + 2l11 + 12l11 + 9l11 + 5l11 + l11 +
2l21 + 7l21 + 5l21 + 20l21 + 8l21 + 17l21 + 19l21 + 9l21 + l21
6l12 + 15l12 + 8l12 + 20l12 + 6l12 + 11l12 + 12l12 + 4l12 +
7l22 + 17l22 + 14l22 + 17l22 + 11l22 + 9l22 + 16l22 + 14l22 +
l12 +
l22
3l13 + 16l13 + 16l13 + 17l13 + 30l13 + 2l13 + 16l13 + 23l13 +
17l23 + 26l23 + 7l23 + 14l23 + 20l23 + 30l23 + 25l23 + 16l23 +
l13 +
l23
26l14 + 14l14 + 29l14 + 25l14 + 15l14 + 22l14 + 30l14 + 16l14 +
27l24 + 1l24 + 16l24 + 29l24 + 5l24 + 12l24 + 23l24 + 30l24 +
8l34 10l34 19l34 10l34 12l34 20l34 19l34 3l34 -
l14 +
l24 +
l34
52y1 49y2 36y3 83y4 39y5 70y6 79y7 59y8
1 1 1 1 1 1 1 1 =1 =1 =1 =1
The optimal solution shows l21 = l22 = l23 = l24 = 1. This calls for a pizza with a thick crust, a cheese blend, a chunky sauce, and medium sausage. With y1 = y2 = y3 = y5 = y7 = y8 = 1, we see that 6 of the 8 people in the consumer panel will prefer this pizza to Antonio's. 19. a.
Let lij = 1 if level i is chosen for attribute j, 0 otherwise yk = 1 if child k prefers the new cereal design, 0 otherwise The share of choices problem to solve is given below:
Max s.t.
y1 + y2 + y3 + y4 + y5 + y6 15l11 + 30l11 + 40l11 + 35l11 + 25l11 + 20l11 + 30l11 + l11 +
35l21 + 20l21 + 25l21 + 30l21 + 40l21 + 25l21 + 15l21 + l21
30l12 + 40l12 + 20l12 + 25l12 + 40l12 + 20l12 + 25l12 +
40l22 + 35l22 + 40l22 + 20l22 + 20l22 + 35l22 + 40l22 +
25l32 + 25l32 + 10l32 + 30l32 + 35l32 + 30l32 + 40l32 +
l12 +
l22 +
l32
15l13 + 8l13 + 7l13 + 15l13 + 18l13 + 9l13 + 20l13 +
9l23 11l23 14l23 18l23 14l23 16l23 11l23 -
l13 +
l23
75y1 75y2 75y3 75y4 75y5 75y6 75y7
1 1 1 1 1 1 1 =1 =1 =1
The optimal solution obtained using LINDO on Excel shows l11 = l32 = l13 = 1. This indicates that a cereal with a low wheat/corn ratio, artificial sweetener, and no flavor bits will maximize the share of choices.
7 - 19
Chapter 7
The optimal solution also has y4 = y5 = y7 = 1 which indicates that children 4, 5, and 7 will prefer this cereal. b.
The coefficients for the yi variable must be changed to -70 in constraints 1-4 and to -80 in constraints 5-7. The new optimal solution has l21 = l12 = l23 = 1. This is a cereal with a high wheat/corn ratio, a sugar sweetener, and no flavor bits. Four children will prefer this design: 1, 2, 4, and 5.
20. a.
Objective function changes to Min 25x1 + 40x2 + 40x3 + 40x4 + 25x5
b.
x4 = x5 = 1; modernize the Ohio and California plants.
c.
Add the constraint x2 + x3 = 1
d.
x1 = x3 = 1; modernize the Michigan plant and the first New York plant.
21. a.
Let
1 if a camera is located at opening i xi = 0 if not
min x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 s.t. x1 + x4 + x6
1
Room 1
x6 + x8 + x12
1
Room 2
x1 + x2 + x3
1
Room 3
x3 + x4 + x5 + x7
1
Room 4
x7 + x8 + x9 + x10 1
Room 5
1
Room 6
x2 + x5 + x9 + x11 1
Room 7
1
Room 8
x10 + x12 + x13 x11 + x13 b.
x1 = x5 = x8 = x13 = 1. Thus, cameras should be located at 4 openings: 1, 5, 8, and 13. An alternative optimal solution is x1 = x7 = x11 = x12 = 1.
c.
Change the constraint for room 7 to x2 + x5 + x9 + x11 2
d.
x3 = x6 = x9 = x11 = x12 = 1. Thus, cameras should be located at openings 3, 6, 9, 11, and 12. An alternate optimal solution is x2 = x4 = x6 = x10 = x11 = 1. Optimal Value = 5
22.
Note that Team Size = x1 + x2 + x3 The following two constraints will guarantee that the team size will be 3, 5, or 7.
7 - 20
Integer Linear Programming x1 + x2 + x3 = 3y1 + 5y2 + 7y3 y1 + y2 + y3 = 1 Of course, the variables in the first constraint will need to be brought to the left hand side if a computer solution is desired. 23. a.
A mixed integer linear program can be set up to solve this problem. Binary variables are used to indicate whether or not we setup to produce the subassemblies. Let SB STVC SVCRC STVP SVCRP BM BP TVCM
= 1 if bases are produced; 0 if not = 1 if TV cartridges are produced; 0 if not = 1 if VCR cartridges are produced; 0 if not = 1 if TV keypads are produced; 0 if not = 1 if VCR keypads are produced; 0 if not = No. of bases manufactured = No. of bases purchased = No. of TV cartridges made
• • •
VCRPP
= No. of VCR keypads purchased
A mixed integer linear programming model for solving this problem follows. There are 11 constraints. Constraints (1) to (5) are to satisfy demand. Constraint (6) reflects the limitation on manufacturing time. Finally, constraints (7) - (11) are constraints not allowing production unless the setup variable equals 1. Variables SB, STVC, SVCRC, STVP, and SVCRP must be specified as 0/1. LINEAR PROGRAMMING PROBLEM MIN 0.4BM+2.9TVCM+3.15VCRCM+0.3TVPM+0.55VCRPM+0.65BP+3.45TVCP+3.7VCRCP+ 0.5TVPP+0 .7VCRPP+1000SB+1200STVC+1900SVCRC+1500STVP+1500SVCRP S.T. 1) 1BM+1BP=12000 2) +1TVCM+1TVCP=7000 3) +1VCRCM+1VCRCP=5000 4) +1TVPM+1TVPP=7000 5) +1VCRPM+1VCRPP=5000 6) 0.9BM+2.2TVCM+3VCRCM+0.8TVPM+1VCRPM <= 30000 7) 1BM-12000SB <= 0 8) +1TVCM-7000STVC <=0 9) +1VCRCM-5000SVCRC <=0 10) +1TVPM-7000STVP <=0 11) +1VCRPM-5000SVCRP <=0 OPTIMAL SOLUTION Objective Function Value = 52800.00 Variable Value ---------------------------BM 12000.000 TVCM 7000.000 VCRCM 0.000
7 - 21
Chapter 7 TVPM VCRPM BP TVCP VCRCP TVPP VCRPP SB STVC SVCRC STVP SVCRP Constraint -------------1 2 3 4 5 6 7 8 9 10 11 b.
0.000 0.000 0.000 0.000 5000.000 7000.000 5000.000 1.000 1.000 0.000 0.000 0.000 Slack/Surplus --------------0.000 0.000 0.000 0.000 0.000 3800.000 0.000 0.000 0.000 0.000 0.000
This can be solved by changing appropriate coefficients in the formulation for part (a). The SVCRC coefficient becomes 3000 and the VCRCM coefficient becomes 2.6 in the objective function. Also, the coefficient of VCRCM becomes 2.5 in constraint (6). The new optimal solution follows. OPTIMAL SOLUTION Objective Function Value = Variable -------------BM TVCM VCRCM TVPM VCRPM BP TVCP VCRCP TVPP VCRPP SB STVC SVCRC STVP SVCRP Constraint -------------1 2
52300.00
Value --------------0.000 7000.000 5000.000 0.000 0.000 12000.000 0.000 0.000 7000.000 5000.000 0.000 1.000 1.000 0.000 0.000 Slack/Surplus --------------0.000 0.000
7 - 22
Integer Linear Programming 3 4 5 6 7 8 9 10 11 24. Let
Max
0.000 0.000 0.000 2100.000 0.000 0.000 0.000 0.000 0.000
Xi = the amount (dollars) to invest in alternative i Yi = 1 if Dave invests in alternative i, 0 if not
i=1,2,…10 i = 1,2…10
.067X1 + .0765X2 + .0755X3 + .0745X4 + .075X5 + .0645X6 + .0705X7 + .069X8 + .052X9 + .059X10
Subject to X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 = 100,000
Invest $100,000
Xi ≤ 25,000Yi i=1,2,..10
Invest no more than $25,000 in any one fund
Xi ≥ 10,000Yi
If invest in a fund, invest at least $10,000 in a fund
i=1,2,..10
Y1 + Y2 + Y3 + Y4 ≤ 2
No more than 2 pure growth funds
Y9 + Y10 ≥ 1
At least 1 must be a pure bond fund
X9 + X10 ≥ X1 + X2 + X3 + X4
Amount in pure bonds must be at least that invested in pure growth funds
Xi ≥ 0
i=1,2,..10
The optimal solution follows:
Fund
Type
Amount Invested
Fund 1
growth
$0
Fund 2
growth
$12,500
Fund 3
growth
$0
Fund 4
growth
$0
Fund 5
growth & income
$25,000
Fund 6
growth & income
$0
Fund 7
growth & income
$25,000
Fund 8
stock & bond
$25,000
Fund 9
bond
$0
Fund 10
bond
$12,500
Total Invested
$100,000
The optimal return on investment is $7,056.25 Assumptions: (1) the expected annual returns are valid for the future. (2) All $100,000 will be invested. (3) These are the only alternatives for this $100,000.
7 - 23
Chapter 7
Since these are annual returns, we would expect to run this no more often than once per year. 25. a.
Let
1 if a service facility is located in city i xi = 0 otherwise
x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 Min s.t. 1 x1 + x2 + x3 (Boston) 1 (New York) x1 + x2 + x3 + x4 + x5 + x6 1 (Philadelphia) x1 + x2 + x3 + x4 + x5 + x6 + x7 1 x2 + x3 + x4 + x5 + x6 + x7 (Baltimore) 1 x + x + x + x + x + x (Washington) 1 2 3 4 5 6 7 x2 + x3 + x4 + x5 + x6 + x7 + x8 (Richmond) 1 x3 + x4 + x5 + x6 + x7 + x8 + x9 (Raleigh) 1 x6 + x7 + x8 + x9 + x10 (Florence) 1 x7 + x8 + x9 + x10 + x11 (Savannah) 1 x8 + x9 + x10 + x11 (Jacksonville) 1 x9 + x10 + x11 + x12 1 (Tampa) x11 + x12 1 (Miami) xi = 0, 1 b.
3 service facilities: Philadelphia, Savannah and Tampa. Note: alternate optimal solution is New York, Richmond and Tampa.
c.
4 service facilities: New York, Baltimore, Savannah and Tampa. Note: alternate optimal solution: Boston, Philadelphia, Florence and Tampa.
7 - 24
Nonlinear Optimization Models
Chapter 8 Nonlinear Optimization Models Learning Objectives 1.
Learn about applications of nonlinear programming that have been encountered in practice.
2.
Develop an appreciation for the diversity of problems that can be modeled as nonlinear programs.
3.
Obtain practice and experience in formulating realistic nonlinear programming models.
4.
Learn to use software such as LINGO and Excel Solver to solve nonlinear programming models.
5.
Understand nonlinear programming applications such as: Allocating a media budget Markowitz portfolio selection Minimizing value at risk The Black-Scholes option pricing model Selecting the smoothing parameter used in the exponential smoothing model Selecting the parameters in the Bass forecasting model The quadratic assignment problem The Cobb-Douglass production function Production scheduling between multiple plants or facilities
Note to Instructor The application problems of Chapter 8 have been designed to give the student an understanding and appreciation of the broad range of problems that can be modeled as a nonlinear program. While the problems are indicative of actual nonlinear programming applications, they have been kept relatively small in order to ease the student’s formulation and solution effort. Each problem will give the student an opportunity to practice formulating a nonlinear programming model. However, the solution and the interpretation of the solution will require the use of a software package such as LINGO or Microsoft Excel’s Solver. Note that computer output for many of these solutions are from LINGO. For maximization problems, LINGO gives the opposite sign for reduced costs than the standard output discussed in the chapter.
Chapter 8
Solutions: 1.
2.
Using LINGO or Excel Solver, the optimal solution is X = 2, Y = -4, for an optimal solution value of 0. a.
b.
c.
3.
4.
Using LINGO or Excel Solver, the optimal solution is X = 4.32, Y = 0.92, for an optimal solution value of 4.84 The dual value on the constraint X + 4Y 8 is -0.88 so we expect the optimal objective function value to decrease by 0.88 if we increase the right-hand-side from 8 to 9. If we resolve the problem with a new right-hand-side of 9 the new optimal objective function value is 4.0 so the actual decrease is only .84 rather than 0.88. The revenue function is: PsDs + PHDH = Ps( 222 - 0.6Ps + 0.35PH ) + PH(270 + 0.1Ps - 0.64PH). This is an example of an unconstrained optimization problem because no constraints are required here. The optimal solution is: Ps = $304.21 and PH = $317.89 with an optimal revenue of $76,681.48.
a.
q1 = 950 – 1.5(2000) + .7(6000) = 2150 q2 = 2500 + .3(2000) - .5(6000) = 100 Cost of residential mower = c1 = 10,000 + 1500q1 = 10,000 + 1500(2150) = 3,235,000 Cost of industrial mower = c2 = 30,000 + 4000q2 = 30,000 + 4000(100) = 430,000 Gross Profit = G = p1q1 + p2 q2 − c1 − c2 = 2000(2150) + 6000(100) – 3,235,000 – 430,000 = 1,235,000
b.
Gross Profit = G = p1q1 + p2 q2 − c1 − c2 Substituting for q1 and q2 will allow us to write G as a function of p1 and p 2 .
G = p1 (950 − 1.5 p1 + .7 p2 ) + p2 (2500 + .3 p1 − .5 p2 ) − (10, 000 + 1500(950 − 1.5 p1 + .7 p2 )) − (30, 000 + 4000(2500 + .3 p1 − .5 p2 )) = −1.5 p12 − .5 p22 + p1 p2 + 2000 p1 + 3450 p2 − 11, 465, 000 c.
Using LINGO or Excel Solver we can maximize the gross profit function developed in part (b). The optimal prices are p1 =$2725 and p 2 =$6175. The quantities sold at these prices will be q1 = 1185 and q 2 =230. The gross profit is G =$1,911,875.
Nonlinear Optimization Models
d.
This formulation involves 6 variables and 4 constraints. M ax p1q1 + p2 q2 − c1 − c2 s.t. c1 = 10000 + 1500q1 c2 = 30000 + 4000q2 q1 = 950 − 1.5 p1 + .7 p2 q2 = 2500 + .3 p1 − .5 p2
The advantage of this formulation is that the optimal solution provides the values for the prices, quantities and costs. With the previous formulation the solution only provided the values of the prices. The other information was computed from that. This also shows that there can be more than one correct formulation of a problem. The formulation in part (c) is an unconstrained problem. The formulation in part (d) is a constrained problem. 5.
a.
With $1000 being spent on radio and $1000 being spent on direct mail we can simply substitute those values into the sales function. S = −2 R 2 − 10 M 2 − 8 RM + 18 R + 34 M = −2(22 ) − 10(12 ) − 8(2)(1) + 18(2) + 34(1) = 18
Sales of $18,000 will be realized with this allocation of the media budget. b.
We simply add a budget constraint to the sales function that is to be maximized. max − 2 R 2 − 10M 2 − 8RM + 18R + 34M s.t. R+M 3
c.
6.
Using LINGO or Excel Solver, we find that the optimal solution is to invest $2,500 in radio advertising and $500 in direct mail advertising. The total sales generated will be $37,000. Substituting the given data into the model formulation gives us 150∗2000 𝑄 135∗2000 𝑄 Min[100 ∗ 2000 + + 0.20 ∗ 100 ∗ 1 + 50 ∗ 2000 + + 0.20 ∗ 50 ∗ 2 𝑄1 2 𝑄2 2 125 ∗ 1000 𝑄3 +80 ∗ 1000 + + 0.20 ∗ 80 ∗ ] 𝑄3 2 s.t. [100 ∗ 𝑄1 + 50 ∗ 𝑄2 + 80 ∗ 𝑄3 ] ≤ 20,000 𝑄1 , 𝑄2 , 𝑄3 ≥ 0 Using LINGO or Excel Solver, we find that the optimal solution is Q1 = 52.223, Q2 = 70.065, Q3 = 37.689 with a total cost of $25,830.
Chapter 8
7.
a.
The optimization model is
Max 5 L.25C .75 s.t. 25L + 75C 75000 L, C 0
8.
b.
Using LINGO or Excel Solver, the optimal solution to this is L = 750 and C = 750 for an optimal objective function value of 3750. If Excel Solver is used for this problem we recommend starting with an initial solution that has L > 0 and C > 0.
a.
The optimization model is
min 50L + 100C s.t. 20 L0.30 C 0.70 = 50000 L, C 0
9.
b.
Using LINGO or Excel Solver, the optimal solution to this problem is L = 2244.281 and C = 2618.328 for an optimal solution of $374,046.9. If Excel Solver is used for this problem we recommend starting with an initial solution that has L > 0 and C > 0.
a.
Let OT be the number of overtime hours scheduled. Then the optimization model is
max − 3 x12 + 42 x1 − 3 x2 2 + 48 x2 + 700 − 5OT s.t. 4 x1 + 6 x2 24 + OT x1 , x2 , OT 0 b.
10. a.
Using LINGO or Excel Solver, the optimal solution is to schedule OT = 8.66667 overtime hours and produce x1 = 3.66667 units of product 1 and x2 = 3.00000 units of product 2 for a profit of 887.3333. If X is the weekly production volume in thousand of units at the Dayton plant and Y is the weekly production volume in thousands of units at the Hamilton plant, then the optimization model is
min X 2 − X + 5 + Y 2 + 2Y + 3 s.t. X +Y = 8 X ,Y 0 b.
Using LINGO or Excel Solver, the optimal solution is X = 4.75 and Y = 3.25 for an optimal objective value of 42.875.
Nonlinear Optimization Models
11.
Define the variables to be the dollars invested the in mutual fund. For example, IB = 500 means that $500 is invested in the Intermediate-Term bond fund. The LINGO formulation is
MIN = (1/5)*((R1 - RBAR)^2 + (R2 - RBAR)^2 + (R3 - RBAR)^2 + (R4 RBAR)^2 + (R5 - RBAR)^2); .1006*FS + .1764*IB + .3241*LG + .3236*LV + .3344*SG + .2456*SV = R1; .1312*FS + .0325*IB + .1871*LG + .2061*LV + .1940*SG + .2532*SV = R2; .1347*FS + .0751*IB + .3328*LG + .1293*LV + .0385*SG - .0670*SV = R3; .4542*FS - .0133*IB + .4146*LG + .0706*LV + .5868*SG + .0543*SV = R4; -.2193*FS + .0736*IB - .2326*LG - .0537*LV - .0902*SG + .1731*SV = R5; FS + IB + LG + LV + SG + SV = 50000; (1/5)*(R1 + R2 + R3+ R4 + R5) = RBAR; RBAR > RMIN; RMIN = 5000; @FREE(R1); @FREE(R2); @FREE(R3); @FREE(R4); @FREE(R5); Recall that the individual scenario returns can take on negative values. By default LINGO assumes all variables are nonnegative so it is necessary to “free” the variables using the @FREE function. The optimal solution to this model using LINGO is: Local optimal solution found. Objective value: Total solver iterations: Model Title: MARKOWITZ Variable R1 RBAR R2 R3 R4 R5 FS IB LG LV SG SV RMIN
Value 9478.492 5000.000 5756.023 2821.951 4864.037 2079.496 7920.372 26273.98 2103.251 0.000000 0.000000 13702.40 5000.000
Excel Solver will also produce the same optimal solution.
6784038. 19
Reduced Cost 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 208.2068 78.04764 0.000000 0.000000
Chapter 8
12.
Below is a screen capture of an Excel Spreadsheet Solver model for this problem.
In cell B1 is the smoothing parameter α. The forecasts in Column C are a function of the smoothing parameter, that is the forecast in time t + 1 is Ft +1 = Yt + (1 − ) Ft
where Yt is the actual value of sales in period t and Ft is the forecast of sales for period t. In Cell E16 is the sum of squared errors. This is what we minimize by using Solver. In the Solver parameters dialog box, we use cell E16 as the Target Cell and cell B1 as the By Changing Cells. After clicking Solve, the result is that the optimal value of α is 0.1743882 and the resulting sum of squared errors is 98.56. This problem is also easily solved using LINGO. Here is a LINGO formulation. In the LINGO formulation, the Y variables correspond to observed sales. For example, in Week 1, Y1 = 17. F1 = Y1; F2 = F1 + ALPHA*( Y1 - F1); F3 = F2 + ALPHA*( Y2 - F2); F4 = F3 + ALPHA*( Y3 - F3); F5 = F4 + ALPHA*( Y4 - F4); F6 = F5 + ALPHA*( Y5 - F5); F7 = F6 + ALPHA*( Y6 - F6); F8 = F7 + ALPHA*( Y7 - F7); F9 = F8 + ALPHA*( Y8 - F8); F10 = F9 + ALPHA*( Y9 - F9); F11 = F10 + ALPHA*( Y10 - F10); F12 = F11 + ALPHA*( Y11 - F11); MIN = (Y2 - F2)^2 + (Y3 - F3)^2 + (Y4 - F4)^2 + (Y5 - F5)^2 + (Y6 - F6)^2 + (Y7 - F7)^2 + (Y8 - F8)^2 + (Y9 - F9)^2 + (Y10 - F10)^2 + (Y11 - F11)^2 + (Y12 - F12)^2 ;
Nonlinear Optimization Models
Y1 = 17; Y2 = 21; Y3 = 19; Y4 = 23; Y5 = 18; Y6 = 16; Y7 = 20; Y8 = 18; Y9 = 22; Y10 = 20; Y11 = 15; Y12 = 22; Solving this in LINGO also produces ALPHA = 0.1743882. 13.
Here are the returns calculated from the Yahoo stock data. AAPL
AMD
ORCL
AAPL
AMD
ORCL
Year
Adj. Close*
Adj. Close*
Adj. Close*
Return
Return
Return
1
4.16
17.57
4.32
0.0962
-0.5537
-0.1074
2
4.58
10.1
3.88
0.8104
0.1272
0.8666
3
10.30
11.47
9.23
0.9236
0.4506
0.9956
4
25.94
18
24.98
-0.8753
0.3124
0.1533
5
10.81
24.6
29.12
0.1340
-0.4270
-0.5230
6
12.36
16.05
17.26
-0.5432
-1.1194
-0.3610
7
7.18
5.24
12.03
0.4517
1.0424
0.1416
8
11.28
14.86
13.86
1.2263
0.0613
-0.0065
9
38.35
15.8
13.77
0.6749
0.9729
-0.0912
10
75.51
41.8
12.57
Data Source: CSI Web site: http://www.csidata.com 14.
MODEL: TITLE MARKOWITZ; ! MINIMIZE VARIANCE OF THE PORTFOLIO; MIN = (1/9)*((R1 - RBAR)^2 + (R2 - RBAR)^2 + (R3 - RBAR)^2 + (R4 - RBAR)^2 + (R5 - RBAR)^2 + (R6 - RBAR)^2 + (R7 - RBAR)^2 + (R8 - RBAR)^2 + (R9 - RBAR)^2); ! SCENARIO 1 RETURN; 0.0962*AAPL - 0.5537*AMD - 0.1074*ORCL = R1; ! SCENARIO 2 RETURN; 0.8104*AAPL + 0.1272*AMD + 0.8666*ORCL = R2; ! SCENARIO 3 RETURN; 0.9236*AAPL + 0.4506*AMD + 0.9956*ORCL = R3; ! SCENARIO 4 RETURN; -0.8753*AAPL + 0.3124*AMD + 0.1533*ORCL = R4; ! SCENARIO 5 RETURN; 0.1340*AAPL - 0.4270*AMD - 0.5230*ORCL = R5;
Chapter 8
! SCENARIO 6 RETURN; -0.5432*AAPL - 1.1194*AMD - 0.3610*ORCL = R6; !SCENARIO 7 RETURN; 0.4517*AAPL + 1.0424*AMD + 0.1416*ORCL = R7; !SCENARIO 8 RETURN; 1.2263*AAPL + 0.0613*AMD - 0.0065*ORCL = R8; !SCENARIO 9 RETURN; 0.6749*AAPL + 0.9729*AMD - 0.0912*ORCL = R9; ! MUST BE FULLY INVESTED IN THE MUTUAL FUNDS; AAPL + AMD + ORCL = 1; ! DEFINE THE MEAN RETURN; (1/9)*(R1 + R2 + R3+ R4 + R5 + R6 + R7 + R8 + R9) = RBAR; ! THE MEAN RETURN MUST BE AT LEAST 10 PERCENT; RBAR > .12; ! SCENARIO RETURNS MAY BE NEGATIVE; @FREE(R1); @FREE(R2); @FREE(R3); @FREE(R4); @FREE(R5); @FREE(R6); @FREE(R7); @FREE(R8); @FREE(R9); END Local optimal solution found. Objective value: Total solver iterations:
0.1990478 12
Model Title: MARKOWITZ Variable R1 RBAR R2 R3 R4 R5 R6 R7 R8 R9 AAPL AMD ORCL
Value -0.1457056 0.1518649 0.7316081 0.8905417 -0.6823468E-02 -0.3873745 -0.5221017 0.3499810 0.2290317 0.2276271 0.1817734 0.1687534 0.6494732
Reduced Cost 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
Nonlinear Optimization Models
15.
MODEL: TITLE MATCHING S&P INFO TECH RETURNS; ! MINIMIZE SUM OF SQUARED DEVIATIONS FROM S&P INFO TECH RETURNS; MIN = ((R1 - .2854)^2 + (R2 - .7814)^2 + (R3 - .7874)^2 + (R4 + .4090)^2 + (R5 + .2587)^2 + (R6 + .3741)^2 + (R7 - .4840)^2 + (R8 - .0256)^2 + (R9 - 0.0099)^2); ! SCENARIO 1 RETURN; 0.0962*AAPL - 0.5537*AMD - 0.1074*ORCL = R1; ! SCENARIO 2 RETURN; 0.8104*AAPL + 0.1272*AMD + 0.8666*ORCL = R2; ! SCENARIO 3 RETURN; 0.9236*AAPL + 0.4506*AMD + 0.9956*ORCL = R3; ! SCENARIO 4 RETURN; -0.8753*AAPL + 0.3124*AMD + 0.1533*ORCL = R4; ! SCENARIO 5 RETURN; 0.1340*AAPL - 0.4270*AMD - 0.5230*ORCL = R5; ! SCENARIO 6 RETURN; -0.5432*AAPL - 1.1194*AMD - 0.3610*ORCL = R6; !SCENARIO 7 RETURN; 0.4517*AAPL + 1.0424*AMD + 0.1416*ORCL = R7; !SCENARIO 8 RETURN; 1.2263*AAPL + 0.0613*AMD - 0.0065*ORCL = R8; !SCENARIO 9 RETURN; 0.6749*AAPL + 0.9729*AMD - 0.0912*ORCL = R9; ! MUST BE FULLY INVESTED IN THE MUTUAL FUNDS; AAPL + AMD + ORCL = 1; ! SCENARIO RETURNS MAY BE NEGATIVE; @FREE(R1); @FREE(R2); @FREE(R3); @FREE(R4); @FREE(R5); @FREE(R6); @FREE(R7); @FREE(R8); @FREE(R9); END Local optimal solution found. Objective value: Total solver iterations:
0.4120213 8
Model Title: MATCHING S&P INFO TECH RETURNS Variable R1 R2 R3 R4 R5 R6 R7 R8
Value -0.5266475E-01 0.8458175 0.9716207 -0.1370104 -0.3362695 -0.4175977 0.2353628 0.3431437
Reduced Cost 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
Chapter 8 R9 AAPL AMD ORCL 16.
0.1328016 0.2832558 0.6577707E-02 0.7101665
0.000000 0.000000 0.000000 0.000000
In order to measure the semi-variance it is necessary to measure only the downside (deviation below the mean). Do this by introducing two new variables for each scenario. For example, for scenario 1 define D1P as the deviation of return 1 above the mean, and D1N as the deviation of return 1 below the mean. That is
D1P − DIN = R1 − RBAR and D1P and D1N are required to be nonnegative. Then in the objective function, we minimize the average of the negative deviations squared, i.e. the semi-variance. The complete LINGO model is TITLE MARKOWITZ WITH SEMIVARIANCE; ! MINIMIZE THE SEMI-VARIANCE MIN = (1/5)*((D1N)^2 + (D2N)^2 + (D3N)^2 + (D4N)^2 + (D5N)^2); ! DEFINE THE EXPECTED RETURNS; 10.06*FS + 17.64*IB + 32.41*LG + 32.36*LV + 33.44*SG + 24.56*SV = R1; 13.12*FS + 3.25*IB + 18.71*LG + 20.61*LV + 19.40*SG + 25.32*SV = R2; 13.47*FS + 7.51*IB + 33.28*LG + 12.93*LV + 3.85*SG - 6.70*SV = R3; 45.42*FS - 1.33*IB + 41.46*LG + 7.06*LV + 58.68*SG + 5.43*SV = R4; -21.93*FS + 7.36*IB - 23.26*LG - 5.37*LV - 9.02*SG + 17.31*SV = R5; ! INVESTMENT LEVELS SUM TO 1; FS + IB + LG + LV + SG + SV = 1; ! DEFINE EXPECTED RETURN; (1/5)*(R1 + R2 + R3+ R4 + R5) = RBAR; ! DEFINE THE MINIMUM ; RBAR > RMIN; RMIN = 10; ! MEASURE POSITIVE AND NEGATIVE DEVIATION FROM MEAN; D1P - D1N = R1 - RBAR; D2P - D2N = R2 - RBAR; D3P - D3N = R3 - RBAR; D4P - D4N = R4 - RBAR; D5P - D5N = R5 - RBAR; ! MAKE THE RETURN VARIABLES UNRESTRICTED; @FREE(R1); @FREE(R2); @FREE(R3); @FREE(R4); @FREE(R5); Local optimal solution found. Objective value: Total solver iterations: Model Title: MARKOWITZ WITH SEMIVARIANCE Variable Value D1N 0.000000 D2N 0.8595142 D3N 3.412762 D4N 2.343876 D5N 4.431505 FS 0.000000 IB 0.6908001
7.503540 18 Reduced Cost 0.000000 0.000000 0.000000 0.000000 0.000000 6.491646 0.000000
Nonlinear Optimization Models LG LV SG SV R1 R2 R3 R4 R5 RBAR RMIN D1P D2P D3P D4P D5P
0.6408726E-01 0.000000 0.8613837E-01 0.1589743 21.04766 9.140486 6.587238 7.656124 5.568495 10.00000 10.00000 11.04766 0.000000 0.000000 0.000000 0.000000
0.000000 14.14185 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.3438057 1.365105 0.9375505 1.772602
The solution calls for investing 69.1% of the portfolio in the intermediate-term bond fund, 6.4% of the portfolio in the large-cap growth fund, 8.6% of the portfolio in the small-cap growth fund, and 15.9% of the portfolio in the small-cap value fund. Excel Solver may have trouble with this problem, depending upon the starting solution that is used. A starting solution of each fund at 0.167 will produce the optimal value. 17. a.
Let:
FS = proportion of portfolio invested in the foreign stock mutual fund IB = proportion of portfolio invested in the intermediate-term bond fund LG = proportion of portfolio invested in the large-cap growth fund LV = proportion of portfolio invested in the large-cap value fund SG = proportion of portfolio invested in the small-cap growth fund SV = proportion of portfolio invested in the small-cap value fund
R = the expected return of the portfolio Rs = the return of the portfolio in year s
Max R s.t. 10.06 FS + 17.64 IB + 32.41LG + 32.36 LV + 33.44 SG + 24.56 SV = R1 13.12 FS + 3.25 IB + 18.71LG + 20.61LV + 19.40 SG + 25.32 SV = R2 13.47 FS + 7.51IB + 33.28 LG + 12.93LV + 3.85SG − 6.70 SV = R3 45.42 FS − 1.33IB + 41.46 LG + 7.06 LV + 58.68SG + 5.43SV = R4 −21.93FS + 7.36 IB − 23.26 LG − 5.37 LV − 9.02 SG + 17.31SV = R5 FS +
IB +
LG +
LV +
SG +
SV = 1 5
1
5
R = R s =1
5
1
5
(R − R) s =1
s
2
s
30
FS , I B, LG, LV , SG , SV 0
Chapter 8
b.
Using LINGO or Excel Solver, we find that the optimal solution is FS = 0.133 IB = 0.496 LG = 0.074 LV = 0.000 SG = 0.000 SV = 0.298 which results in an expected return for the portfolio of 𝑅̅ = 10.449.
18.
The frontier is shown below. As the maximum variance increases the expected return increases but at a decreasing rate. In Figure 8.10, as the required expected return increases, the variance increases at an increasing rate. These curves are known as the efficient frontier and are really the same curve from two different perspectives. For example, a variance of 40 with an expected return of 11.835 is on both curves.
19.
The objective is to minimize the Value at Risk at 1%. The selection of stocks that go into the portfolio determine the mean return of the portfolio and the standard deviation of portfolio returns. Let μ denote the mean return of the portfolio and σ the standard deviation. If returns are normally distributed this means that 99% of the returns will be above the value
− 2.33 Another way to state this is to say that there is a 99% probability that a portfolio return will exceed − 2.33 and a 1% probability that the portfolio return will be less than − 2.33 . The number − 2.33 is called the Value at Risk at 1%. The number 2.33 can be found in a Normal probability table, or by using LINGO and setting @PSN(-Z) = 0.01, or by using Excel Solver and setting NORMSDIST(-Z) = 0.01 and observing that Z = 2.33 (or more accurately, 2.32638013). Although people commonly talk about minimizing the Value at Risk of a portfolio, this terminology is misleading. Minimizing the Value at Risk of a portfolio at 1 percent really means making the number − 2.33 as large as possible – in other words we want 99% of the returns to be as large as possible.
Nonlinear Optimization Models
a.
The LINGO model and solution is given below.
MAX = VaR; 10.06*FS + 17.64*IB + 32.41*LG + 32.36*LV + 33.44*SG + 24.56*SV = R1; 13.12*FS + 3.25*IB + 18.71*LG + 20.61*LV + 19.40*SG + 25.32*SV = R2; 13.47*FS + 7.51*IB + 33.28*LG + 12.93*LV + 3.85*SG - 6.70*SV = R3; 45.42*FS - 1.33*IB + 41.46*LG + 7.06*LV + 58.68*SG + 5.43*SV = R4; -21.93*FS + 7.36*IB - 23.26*LG - 5.37*LV - 9.02*SG + 17.31*SV = R5; FS + IB + LG + LV + SG + SV = 1; (1/5)*(R1 + R2 + R3+ R4 + R5) = RBAR; STD = ((1/5)*((R1 - RBAR)^2 + (R2 - RBAR)^2 + (R3 - RBAR)^2 + (R4 RBAR)^2 + (R5 - RBAR)^2))^0.5; MU = RBAR; PROB = 0.01; PROB = @PSN( -Z); VaR = MU - Z*STD; @FREE(R1); @FREE(R2); @FREE(R3); @FREE(R4); @FREE(R5); @FREE(VaR); END The solution is Local optimal solution found. Objective value: Total solver iterations:
2.118493 35
Model Title: MARKOWITZ Variable VAR FS IB LG LV SG SV R1 R2 R3 R4 R5 RBAR RMIN STD MU PROB Z
Value -2.118493 0.1584074 0.5254795 0.4206502E-01 0.000000 0.000000 0.2740480 18.95698 11.51205 5.643902 9.728075 4.158993 10.00000 10.00000 5.209237 10.00000 0.1000000E-01 2.326347
Reduced Cost 0.000000 0.000000 0.000000 0.000000 9.298123 3.485460 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
Chapter 8
b.
No, minimizing risk is not the same thing as minimizing VaR. Minimizing σ is not the same thing as maximizing Value at Risk. If we maximize Value at Risk, the objective function is Max − 2.33 = Min 2.33 −
and the objective has two variables, σ and μ. c.
If we fix mean return then it is a constant and Max − 2.33 = Min 2.33 − = − + Min 2.33 = − + 2.33Min
Finally observe that minimizing σ is the same as minimizing σ2 since the standard deviation is always nonnegative. This problem may also be solved with Excel Solver. 20.
Black Scholes Model !PRICE OF CALL OPTION; C = S*@PSN(Z) - X*@EXP(-R*T)*@PSN(Z - YSD *T^.5); !BLACK-SCHOLES OPTION FORMULA; ! THE PARAMETERS; X = 60; !STRIKE OR EXERCISE PRICE; S = 60.87; !CURRENT STOCK PRICE; !YVAR = .3^2; !YEARLY VARIANCE; !YSD = .3; !YEARLY STANDARD DEVIATION; R = 0.0494; !YEARLY RISKFR-FREE RATE (THREE MONTH TBILL); !TIME TO MATURITY IN YEARS ; !WE ARE LOOKIN AT THE CLOSE ON AUGUST 25 UNTIL CLOSE ON SEPT 15; T = 21/365; ! P&G WEEKLY VOLITILITY ! NORMALIZED STANDARD DEVIATION; Z = (( R + YVAR/2)*T + @LOG( S/X))/( YSD * T^.5); WVAR = 0.000479376; YVAR = 52 * WVAR; YSD = YVAR^.5; @FREE(Z); !BID - 1.35; !ASK - 1.45; Variable C S Z X R T YSD YVAR WVAR This problem may also be solved with Excel Solver.
Value 1.524709 60.87000 0.4741179 60.00000 0.4940000E-01 0.5753425E-01 0.1578846 0.2492755E-01 0.4793760E-03
Nonlinear Optimization Models
21.
This is a nonlinear 0/1 integer programming problem. Let Xij = 1 if tanker i is assigned loading dock j and 0 if not. First consider the constraints. Every tanker must be assigned to a loading dock. These constraints are as follows. ! EACH TANKER MUST BE ASSIGNED A DOCK; X11 + X12 + X13 = 1; !TANKER 1; X21 + X22 + X23 = 1; !TANKER 2; X31 + X32 + X33 = 1; !TANKER 3; Since there are three tankers and three loading docks each loading dock must be assigned to a tanker. ! EACH LOADING DOCK MUST BE ASSIGNED A TANKER; X11 + X21 + X31 = 1; !DOCK 1; X12 + X22 + X32 = 1; !DOCK 2; X13 + X23 + X33 = 1; !DOCK 3; The constraints that require each tanker to be assigned a loading dock, and each loading dock assigned a tanker form the constraint set for the assignment problem. The assignment problem was introduced in Chapter 6. However, unlike the assignment problem the objective function in this problem is nonlinear. Consider, for example, the result of assigning tanker 1 to dock 2 and tanker 3 to dock 1. The distance between loading docks 1 and 2 is 100 meters. Also, tanker 1 must transfer 80 tons of goods to tanker 3. This means that 80 tons must be moved 100 meters. To capture this in the objective function, we write 100*80*X12*X31. Since both X12 = 1 and X31 = 1 their product is 1 the product of distance and tonnage moved (100*80) is captured in the objective function. The complete objective function is MIN = 100*60*X11*X22 + 150*60*X11*X23 + 100*80*X11*X32 + 150*80*X11*X33 + 100*60*X12*X21 + 50*60*X12*X23 + 100*80*X12*X31 + 50*80*X12*X33 + 150*60*X13*X21 + 50*60*X13*X22 + 150*80*X13*X31 + 50*80*X13*X32; The solution to this model is Global optimal solution found. Objective value: Extended solver steps: Total solver iterations: Variable X11 X22 X23 X32 X33 X12 X21 X31 X13
Value 0.000000 0.000000 0.000000 0.000000 1.000000 1.000000 1.000000 0.000000 0.000000
10000.00 0 38 Reduced Cost 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
Thus tanker 1 should be assigned to dock 2, tanker 2 to dock 1 and tanker 3 to dock 3. Depending on the starting point, Excel Solver will likely get stuck at a local optimum and not the find the optimal solution that LINGO finds.
Chapter 8
22.
The objective is to minimize total production cost. To minimize total product cost minimize the production cost at Aynor plus the production cost at Spartanburg. Minimize the production cost at the two plants subject to the constraint that total production of kitchen chairs is equal to 40. The model is: Min (75Q1 + 5Q12 + 100) + (25Q2 + 2.5Q2 2 + 150) s.t. Q1 + Q 2 = 40 Q1, Q 2 0
The optimal solution to this model using either LINGO or Excel Solver is to produce 10 chairs at Aynor for a cost of $1350 and 30 chairs at Spartanburg for a cost of $3150. The total cost is $4500. 23.
Duplicate the objective function and constraints given in Section 8.5. The objective function which is to minimize the sum of forecast errors is: MIN = (E1)^2 + (E2)^2 + (E3)^2 + (E4)^2 + (E5)^2 + (E6)^2 + (E7)^2 + (E8)^2 + (E9)^2 + (E10)^2 + (E11)^2 + (E12)^2 ; ! CALCULATE THE FORECAST IN EACH PERIOD; F1 = (P + Q*C0/M)*(M - C0); F2 = (P + Q*C1/M)*(M - C1); F3 = (P + Q*C2/M)*(M - C2); F4 = (P + Q*C3/M)*(M - C3); F5 = (P + Q*C4/M)*(M - C4); F6 = (P + Q*C5/M)*(M - C5); F7 = (P + Q*C6/M)*(M - C6); F8 = (P + Q*C7/M)*(M - C7); F9 = (P + Q*C8/M)*(M - C8); F10 = (P + Q*C9/M)*(M - C9); F11 = (P + Q*C10/M)*(M - C10); F12 = (P + Q*C11/M)*(M - C11); ! CALCULATE THE FORECAST ERROR IN EACH PERIOD; E1 = F1 - S1; E2 = F2 - S2; E3 = F3 - S3; E4 = F4 - S4; E5 = F5 - S5; E6 = F6 - S6; E7 = F7 - S7; E8 = F8 - S8; E9 = F9 - S9; E10 = F10 - S10; E11 = F11 - S11; E12 = F12 - S12; ! SALES DATA; S1 = 72.39; S2 = 37.93; S3 = 17.58; S4 = 9.57; S5 = 5.39; S6 = 3.13; S7 = 1.62; S8 = .87; S9 = .61; S10 = .26; S11 = 0.19; S12 = 0.35; ! CUMULATIVE SALES DATA; C0 = 0; C1 = 72.39; C2 = 110.32; C3 = 127.9; C4 = 137.47; C5 = 142.86; C6 = 145.99; C7 = 147.61; C8 = 148.48; C9 = 149.09; C10 = 149.35;
Nonlinear Optimization Models C11 = 149.54; ! START WITH A P OF AT LEAST .1; P >= .1. ! ALLOW THE FORECAST ERROR TO BE NEGATIVE; @FREE(E1); @FREE(E2); @FREE(E3); @FREE(E4); @FREE(E5); @FREE(E6); @FREE(E7); @FREE(E8); @FREE(E9); @FREE(E10); @FREE(E11); @FREE(E12); ! ALLOW WORD OF MOUTH PARAMETER TO BE NEGATIVE. “IT WAS A BAD MOVIE, DON’T! GO AND SEE IT”; @FREE(Q); The optimal solution is Local optimal solution found. Objective value: Total solver iterations: Variable Value E1 0.2194821 E2 -1.126291 E3 0.9546390 E4 0.6119302 E5 0.2755269 E6 0.1286099E-02 E7 0.4277524E-01 E8 0.3362006E-01 E9 -0.1138197 E10 -0.4938969E-01 E11 -0.1010815 E12 -0.3500000 F1 72.60948 P 0.4855522 Q -0.1758224E-01 C0 0.000000 M 149.5400 F2 36.80371 C1 72.39000 F3 18.53464 C2 110.3200 F4 10.18193 C3 127.9000 F5 5.665527 C4 137.4700 F6 3.131286 C5 142.8600 F7 1.662775 C6 145.9900 F8 0.9036201 C7 147.6100 F9 0.4961803 C8 148.4800 F10 0.2106103 C9 149.0900 F11 0.8891854E-01 C10 149.3500
2.829486 47 Reduced Cost 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
Chapter 8 F12 C11 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 S12
0.000000 149.5400 72.39000 37.93000 17.58000 9.570000 5.390000 3.130000 1.620000 0.8700000 0.6100000 0.2600000 0.1900000 0.3500000
0.7973561 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
For this problem, Excel Solver is not as robust as LINGO. We recommend a start value for p of at least 0.2 and a value of 100 for m. If Excel Solver is used, one should utilize the Multistart option in Excel Solver’s GRG Nonlinear set of Options.
Chapter 9 Project Scheduling: PERT/CPM Learning Objectives 1.
Understand the role and application of PERT/CPM for project scheduling.
2.
Learn how to define a project in terms of activities such that a network can be used to describe the project.
3.
Know how to compute the critical path and the project completion time.
4.
Know how to convert optimistic, most probable, and pessimistic time estimates into expected activity time estimates.
5.
With uncertain activity times, be able to compute the probability of the project being completed by a specific time.
6.
Understand the concept and need for crashing.
7.
Be able to formulate the crashing problem as a linear programming model.
8.
Learn how to schedule and control project costs with PERT/Cost.
9.
Understand the following terms: network PERT/CPM activities event optimistic time most probable time pessimistic time
beta distribution path critical path critical activities slack crashing
9-1
Chapter 9
Solutions: 1. C
A
F
D
G
Start
H B
Finish
E
2.
Start
A
E
G
B
D
F
C
J
Finish
H
I 3.
D
G
A
Start
E
C
B
F
9-2
Finish
Project Scheduling: PERT/CPM
4.
a.
Critical Path: A-D-G b.
The critical path activities require 15 months to complete. Thus the project should be completed in 1-1/2 years.
5. UPDATED ACTIVITY SCHEDULE FOR THE WESTERN HILLS SHOPPING CENTER PROJECT Earliest
Latest
Earliest
Latest
Start
Start
Finish
Finish
Slack
Critical
(ES)
(LS)
(EF)
(LF)
(LS – ES)
Path?
A
0
0
5
5
0
Yes
B
7
11
8
12
4
C
7
8
11
12
1
D
7
7
10
10
0
Yes
E
5
5
6
6
0
Yes
F
7
7
10
10
0
Yes
G
10
10
24
24
0
Yes
H
11
12
23
24
1
Activity
9-3
Chapter 9
6.
a.
Critical path: A-D-F-H
b.
22 weeks
c.
No, it is a critical activity
d.
Yes, 2 weeks
e.
Schedule for activity E: Earliest Start Latest Start Earliest Finish Latest Finish
7.
3 4 10 11
a. A
F G
Start
B
Finis h
D H
C b.
A-D-E-H
c.
Activity F: 4 weeks
E
9-4
Project Scheduling: PERT/CPM
d.
Yes, project completion time is 16 weeks
Activity A B C D E F G H 8.
Earliest Start 0 0 0 3 7 3 7 12
Latest Start 0 2 1 3 7 7 10 12
Earliest Finish 3 1 2 7 12 6 13 16
Latest Finish 3 3 3 7 12 10 16 16
Slack 0 2 1 0 0 4 3 0
Earliest Start 0 0 8 20 20 26 26 41
Latest Start 2 0 8 22 20 26 29 41
Earliest Finish 6 8 20 24 26 41 38 49
Latest Finish 8 8 20 26 26 41 41 49
Slack 2 0 0 2 0 0 3 0
Critical Activity Yes
Yes Yes
Yes
a.
b.
B-C-E-F-H
c.
Activity A B C D E F G H d. 9.
Critical Activity Yes Yes Yes Yes Yes
Yes. Project Completion Time 49 weeks.
a. A
C
E
Star t
G Finish
B
D
F
9-5
Chapter 9
b. Activity A B C D E F G
Expected Time 10.5 8 6.5 3 5 7 7.5
Variance 1.36 0.11 0.69 0 1.78 1 1.36
Activity
ES
LS
EF
LF
Slack
Critical Activity
A B C D E F G
0 0 10.5 8 17 11 22
0 9 10.5 19.5 17 22.5 22
10.5 8 17 11 22 18 29.5
10.5 17 17 22.5 22 29.5 29.5
0 9 0 11.5 0 11.5 0
Yes
c.
The critical path A – C – E – G:
d.
Expected duration = 10.5 + 6.5 + 5 + 7.5 = 29.5
Yes Yes Yes
Variance = 1.36 + 0.69 + 1.78 + 1.36 = 5.19 z = (30 – 29.5) / √5.19 = 0.219. P(Z ≤ 0.219) ≈ .5871
e.
f. No effect on expected project duration since slack of Activity B (9 days) exceeds the delay beyond the early start time. 10. a. Activity A B C D E F b.
Optimistic 4 8 7 6 6 5
Most Probable 5 9 7.5 9 7 6
Pessimistic 6 10 11 10 9 7
Expected Times 5.00 9.00 8.00 8.83 7.17 6.00
Critical activities: B-D-F Expected project completion time: 9.00 + 8.83 + 6.00 = 23.83. Variance of projection completion time: 0.11 + 0.25 + 0.11 = 0.47
9-6
Variance 0.11 0.11 0.44 0.25 0.25 0.11
Project Scheduling: PERT/CPM
11.
12. a.
Activity A B C D E F G H I
Activity
Expected Time
Variance
A B C D E F G H I
4.83 4.00 6.00 8.83 4.00 2.00 7.83 8.00 4.00
0.25 0.44 0.11 0.25 0.44 0.11 0.69 0.44 0.11
Earliest Start 0.00 0.00 4.83 4.83 4.00 10.83 13.67 13.67 21.67
Latest Start 0.00 0.83 5.67 4.83 17.67 11.67 13.83 13.67 21.67
Earliest Finish 4.83 4.00 10.83 13.67 8.00 12.83 21.50 21.67 25.67
Critical Path: A-D-H-I b.
c.
E(T) = tA + tD + tH + tI = 4.83 + 8.83 + 8 + 4 = 25.66 days
2 = 2A + 2D + 2H + 2I = 0.25 + 0.25 + 0.44 + 0.11 = 1.05
9-7
Latest Finish 4.83 4.83 11.67 13.67 21.67 13.67 21.67 21.67 25.67
Slack 0.00 0.83 0.83 0.00 13.67 0.83 0.17 0.00 0.00
Critical Activity Yes
Yes
Yes Yes
Chapter 9
Using the normal distribution, z=
25 − E (T )
=
25 − 25.66
= −0.65
105 . From the table for standard normal distribution, the cumulative probability for z = -0.65 is 0.2578. Probability of 25 days or less is 0.2578. 13. Activity A B C D E F G H
Expected Time 5 3 7 6 7 3 10 8
Variance 0.11 0.03 0.11 0.44 0.44 0.11 0.44 1.78
From problem 6, A-D-F-H is the critical path. E (T) = 5 + 6 + 3 + 8 = 22 2 = 0.11 + 0.44 + 0.11 + 1.78 = 2.44 z =
Time - E (T) = Time - 22 2.44
a.
Time = 21
z = -0.64 Cumulative Probability = 0.2611 P(21 weeks) = 0.2611
b.
Time = 22
z = 0.00 Cumulative Probability = 0.5000 P(22 weeks) = 0.5000
c.
Time = 25
z = +1.92 Cumulative Probability = 0.9726 P(25 weeks) = 0.9726
14. a. Activity A B C D E F G H
Expected Time 8 7 12 5 10 9 15 7
9-8
Variance 1.78 0.11 4.00 0.44 4.00 4.00 2.78 1.78
Project Scheduling: PERT/CPM
Activity A B C D E F G H
Earliest Start 0 8 8 8 20 20 30 45
Latest Start 0 14 8 15 20 21 30 45
Earliest Finish 8 15 20 13 30 29 45 52
Latest Finish 8 21 20 20 30 30 45 52
Slack 0 6 0 7 0 1 0 0
Critical Activity Yes Yes Yes Yes Yes
Critical Path: A-C-E-G-H b.
E(T) = tA + tC + tE + tG + tH = 8 + 12 + 10 + 15 + 7 = 52 weeks (1 year)
c.
2 = A2 + C2 + E2 + G2 + H2 = 1.78 + 4.00 + 4.00 + 2.78 + 1.78 = 14.34 Using the normal distribution,
z=
44 − E (T )
=
44 − 52 14.34
= −2.11
From the table for standard normal distribution, the cumulative probability for z = -2.11 is 0.0174. Probability of 44 weeks or less is 0.0174 d.
Using the normal distribution,
z=
57 − E (T )
=
57 − 52 14.34
= 1.32
From the table for standard normal distribution, the cumulative probability for z = 1.32 is 0.9066. Probability of more than 57 weeks = 1.0000 - 0.9066 = 0.0934 e.
It is very unlikely that the project can be completed in 10 months. But it is also unlikely the project will take longer than 13 months. Therefore, Davison Construction might prefer to tell the owner that the best estimate is a completion time of one year.
9-9
Chapter 9
15. a.
A
B
E
F
Start
I
C
D
G
Finish
H
b. Activity A B C D E F G H I
Activity A B C D E F G H I c.
Earliest Start 0 2 0 2 5 6 5 9 13
Expected Time 2 3 2 2 1 2 4 4 2
Variance 0.03 0.44 0.11 0.03 0.03 0.11 0.44 0.11 0.03
Latest Start 0 2 1 3 10 11 5 9 13
Latest Finish 2 5 3 5 11 13 9 13 15
Earliest Finish 2 5 2 4 6 8 9 13 15
Slack 0 0 1 1 5 5 0 0 0
Critical Activity Yes Yes
Yes Yes Yes
Critical Path: A-B-G-H-I E(T) = 2 + 3 + 4 + 4 + 2 = 15 weeks
d.
Variance on critical path 2 = 0.03 + 0.44 + 0.44 + 0.11 + 0.03 = 1.05 From the table for standard normal distribution, the cumulative probability of 0.99 occurs at z = 2.33. Thus z = T - E (T) = T - 15 = 2.33 1.05 or T = 15 + 2.33 1.05 = 17.4 weeks
9 - 10
Project Scheduling: PERT/CPM
16. a.
b. Activity Expected Time A 6.67 B-C B 7.5 C 1 Expected Time = 7.5 + 1 = 8.5 weeks
Variance 2.78 1.36 0
𝜎 2 = 1.36 + 0 = 1.36 𝑧=
10−8.5 √1.36
= 1.29
Cumulative probability = 0.90 c.
A-C Expected Time = 6.67 + 1 = 7.67 weeks 𝜎 2 = 2.78 + 0 = 2.78 𝑧=
10−7.67 √2.78
= 1.40
Cumulative probability = 0.92 P(entire project completed) = 0.90 × 0.92 = 0.828 d.
The probability estimate from (c) based on both paths is more accurate. Both paths must be completed for the entire project to be completed. These two paths only share one activity (C), which has no variability, and thus are effectively independent.
9 - 11
Chapter 9
17. a.
B
Start
A
D
Finish
C
b.
Critical Path A-B-D Expected Time = 4.5 + 8.0 + 6.0 = 18.5 weeks
c.
Material Cost = $3000 + $5000 = $8000 Best Cost (Optimistic Times) 3 + 5 + 2 + 4 = 14 days Total Cost = $8000 + 14($400) = $12,800 Worst Case (Pessimistic Times) 8 + 11 + 6 + 12 = 37 days Total Cost = $8000 + 37($400) = $22,800
d.
Bid Cost = $8000 + 18.5($400) = $15,400 .50 probability time and cost will exceed the expected time and cost.
e.
= 3.47 = 1.86 Bid
= $16,800 = $8,000 + Days ($400) 400 Days = 16,800 - 8000 = 8,800 Days = 22
The project must be completed in 22 days or less. The probability of a loss = P (T > 22)
z=
22 − 18.5 = 188 . 186 .
From the table for standard normal distribution, the cumulative probability for z = 1.88 is 0.9699. Probability of 22 weeks or less is 0.9699 The probability of a loss is 1.0000 - 0.9699 = 0.0301
9 - 12
Project Scheduling: PERT/CPM
18. a.
b. Activity A B C D E F G H I
Activity A B C D E F G H I c.
Earliest Start 0.00 1.17 1.17 7.17 7.17 1.17 9.17 11.17 13.17
Expected Time 1.17 6.00 4.00 2.00 3.00 2.00 2.00 2.00 1.00
Latest Start 0.00 1.17 3.17 7.17 10.17 11.17 9.17 11.17 13.17
Earliest Finish 1.17 7.17 5.17 9.17 10.17 3.17 11.17 13.17 14.17
Variance 0.03 0.44 0.44 0.11 0.11 0.11 0.11 0.11 0.00 Latest Finish 1.17 7.17 7.17 9.17 13.17 13.17 11.17 13.17 14.17
Slack 0.00 0.00 2.00 0.00 3.00 10.00 0.00 0.00 0.00
Critical Activity Yes Yes Yes
Yes Yes Yes
Critical Path: A-B-D-G-H-I Expected Project Completion Time = 1.17 + 6 + 2 + 2 + 2 + 1 = 14.17 weeks
d.
Compute the probability of project completion in 13 weeks or less.
2 = 2A + 2B + 2D + 2G+ 2H + 2I = 0.03 + 0.44 + 0.11 + 0.11 + 0.11 + 0.00 = 0.80 z =
13 - E (T) = 13 - 14.17 = -1.31 0.80
From the table for standard normal distribution, the cumulative probability for z = -1.31 is 0.0951. Probability of 13 weeks or less is 0.0951 With this low probability, the manager should start prior to February 1.
9 - 13
Chapter 9
19. a. Activity A B C D E F G H I J
Expected Time 4 4 5 3 10 9 6 7 3 5
A 0 4 4 16 20
Variance 0.11 0.44 0.11 0.11 1.78 0.69 0.25 1.78 0.44 0.11
D 9 12 3 20 23 I 3 C 5
Start
4 9 15 20
Finish Completion Time = 26
G 14 20 6 17 23 B 4
0 0
4 4
E 10
4 14 4 14 H 7
14 21 14 21
F 4 13 9 12 21
Activity A B C D E F G H I J
Earliest Start 0 0 4 9 4 4 14 14 20 21
20 23 23 26
Latest Start 16 0 15 20 4 12 17 14 23 21
J 5
Earliest Finish 4 4 9 12 14 13 20 21 23 26
Critical Path: B-E-H-J
9 - 14
Latest Finish 20 4 20 23 14 21 23 21 26 26
Slack 16 0 11 11 0 8 3 0 3 0
21 26 21 26
Critical Activity Yes
Yes
Yes Yes
Project Scheduling: PERT/CPM b.
E(T) = tB + tE + tH + tJ = 4 + 10 + 7 + 5 = 26
2 = B2 + E2 + H2 + J2 = 0.44 + 1.78 + 1.78 + 0.11 = 4.11
z=
z=
T − E (T )
=
T − E (T )
=
25 − 26
= −0.49
Cumulative Probability P(25 weeks) = 0.3121
= 1.97
Cumulative Probability P(30 weeks) = 0.9756
4.11
30 − 26 4.11
20. a. Activity A B C D E F G H
Maximum Crash 2 3 1 2 1 2 5 1
Crash Cost/Week 400 667 500 300 350 450 360 1000
Min 400YA + 667YB + 500YC + 300YD + 350YE + 450YF + 360YG + 1000YH s.t. xA + yA 3 xB + yB 6
xE + yE - xD 4 xF + yF - xE 3
xC + yC - xA 2
xG + y G - xC 9
xD + yD - xB 5
xH + yH - xF 3
xD + yD - xC 5
xG + y G - xB 9
Maximum Crashing: yA 2 yB 3 yC 1
yD 2 yE 1 yF 2
yG 5
yH 1
9 - 15
xH + yH - xG
3
xH
16
Chapter 9
b.
Linear Programming Solution Activity A B C D E F G H
Crash Time 0 1 0 2 1 1 1 0
New Time 3 5 2 3 3 2 8 3 Total Crashing Cost
Crash Cost — 667 — 600 350 450 360 — $2,427
c. Activity A B C D E F G H
Earliest Start 0 0 3 5 8 11 5 13
Latest Start 0 0 3 5 8 11 5 13
Earliest Finish 3 5 5 8 11 13 13 16
Latest Finish 3 5 5 8 11 13 13 16
Slack 0 0 0 0 0 0 0 0
Latest Start 0 1 3 3 8 10 12
Earliest Finish 3 2 8 7 14 10 12
Latest Finish 3 3 8 8 14 12 14
Slack 0 1 0 1 0 2 2
Critical Activity Yes Yes Yes Yes Yes Yes Yes Yes
All activities are critical. 21. a. Activity A B C D E F G
Earliest Start 0 0 3 2 8 8 10
Critical Path: A-C-E Project Completion Time = tA + tC + tE = 3 + 5 + 6 = 14 days b.
Total Cost = $8,400
9 - 16
Critical Activity Yes Yes Yes
Project Scheduling: PERT/CPM
22. a. Activity A B C D E F G Min s.t.
Max Crash Days 1 1 2 2 2 1 1
Crash Cost/Day 600 700 400 400 500 400 500
600yA + 700yB + 400yC + 400yD + 500yE + 400yF + 400yG xA + yA 3 xB + yB 2
xC + yC - xA 5
xD + yD - xB 5 xE + yE - xC 6
xE + yE - xD 6 xF + yF - xC 2
xF + yF - xD 2
xG + y G - xF 2 xFIN - xE 0
xFIN - xG 0 xFIN 12 yA 1 yB 1 yC 2
yD 2 yE 2 yF 1
yG 1
All x, y 0 b.
Activity C E
c.
Total Cost
Crash 1 day 1 day Total
= Normal Cost + Crashing Cost = $8,400 + $900 = $9,300
9 - 17
Crashing Cost $400 500 $900
Chapter 9
23.
This problem involves the formulation of a linear programming model that will determine the length of the critical path in the network. Since xI, the completion time of activity I, is the project completion time, the objective function is: Min xI Constraints are needed for the completion times for all activities in the project. The optimal solution will determine xI which is the length of the critical path. Activity A
xA A
B
xB B
C
xC - xA C
D
xD - xA D
E
xE - xA E
F
xF - xE F xG - xD G
G
xG - xF G H
xH - xB H xH - xC H xI - xG I
I
xI - xH I All x 0 24. a.
b.
c.
Earliest Latest Earliest Latest Activity Start Start Finish Finish Slack A 0 0 10 10 0 B 10 10 18 18 0 C 18 18 28 28 0 D 10 11 17 18 1 E 17 18 27 28 1 F 28 28 31 31 0 Activities A, B, C, and F are critical. The expected project completion time is 31 weeks.
9 - 18
Project Scheduling: PERT/CPM
d. Crash Activities A B C D E
Number of Weeks 2 2 1 1 1
Cost $ 40 30 20 10 12.5 $ 112.5
e. Earliest Start 0 8 14 8 14 23
Activity A B C D E F
Latest Start 0 8 14 8 14 23
Earliest Finish 8 14 23 14 23 26
Latest Finish 8 14 23 14 23 26
Slack 0 0 0 0 0 0
All activities are critical. f.
Total added cost due to crashing $112,500 (see part d.)
25. a. Let
Min s.t.
Ki Mi Yi Xi Ti
= = = = =
Cost to crash activity i one week Maximum crash time in weeks for activity i Number of weeks activity i is crashed Completion time for activity i Normal completion time for activity i
KAyA + KByB + KCyC + KDyD + KEyE + KFyF + KGyG + KHyH + KIyI + KJyJ xA + yA A xB + yB B
xC + yC - xB C
xD + yD - xA D xD + yD - xC D xE + yE - xB E xF + yF - xB F
xG + yG - xE G xH + yH - xE H xI + yI - xD I xI + yI - xG I
xJ + yJ - xF J
xJ + yJ - xH J xFIN - xI 0 xFIN - xJ 0 xFIN T
9 - 19
Chapter 9 J values can be computed from data given in problem 19. (See solution to problem 19.) yA MA
yF MF
yB MB
yG M G
yD MD
yI MI
yC MC
yH MH
yE ME
b.
yJ MJ
Information needed: 1. Maximum crash time for each activity (M i) 2. Crashing cost for each activity (Ki) 3. Desired project completion time (T)
9 - 20
Chapter 10 Inventory Models Learning Objectives 1.
Learn where inventory costs occur and why it is important for managers to make good inventory policy decisions.
2.
Learn the economic order quantity (EOQ) model.
3.
Know how to develop total cost models for specific inventory systems.
4.
Be able to use the total cost model to make how-much-to-order and when-to-order decisions.
5.
Extend the basic approach of the EOQ model to inventory systems involving production lot size, planned shortages, and quantity discounts.
6.
Be able to make inventory decisions for single-period inventory models.
7.
Know how to make order quantity and reorder point decisions when demand must be described by a probability distribution.
8.
Learn about lead time demand distributions and how they can be used to meet acceptable service levels.
9.
Be able to develop order quantity decisions for periodic review inventory systems.
10.
Understand the following terms: inventory holding costs cost of capital ordering costs economic order quantity (EOQ) constant demand rate reorder point lead time lead time demand cycle time safety stock
backorder quantity discounts goodwill costs probabilistic demand lead time demand distribution service level single-period inventory model periodic review
10 - 1
Chapter 10
Solutions: 1.
2 DCo = Ch
a.
Q* =
b.
r = dm =
c.
T=
d.
TC =
2.
2(3600)(20) = 438.18 0.25(3)
3600 (5) = 72 250
250Q* 250(438.18) = = 30.43 days D 3600
1 D 1 3600 QCh + Co = (438.18)(0.25)(3) + (20) = $328.63 2 Q 2 438.18
Annual Holding Cost
1 1 QCh = (438.18)(0.25)(3) = $164.32 2 2 Annual Ordering Cost D 3600 Co = (20) = 164.32 Q 438.18
Total Cost = $328.64. Q* =
3.
d=
a.
2 DCO = Ch
2(5000)(32) = 400 2
D 5000 = = 20 units per day 250 250
r = dm = 20(5) = 100 Since r Q*, both inventory position and inventory on hand equal 100.
b.
r = dm = 20(15) = 300 Since r Q*, both inventory position and inventory on hand equal 300.
c.
r = dm = 20(25) = 500 Inventory position reorder point = 500. One order of Q* = 400 is outstanding. The on-hand inventory reorder point is 500 - 400 = 100.
d.
r = dm = 20(45) = 900 Inventory position reorder point = 900. Two orders of Q* = 400 are outstanding. The on-hand inventory reorder point is 900 - 2(400) = 100.
10 - 2
Inventory Models
4.
2 DCo = Ch
a.
Q* =
b.
r = dm =
c.
T=
d.
Holding
2(12, 000)(25) = 1095.45 (0.20)(2.50)
1200 (5) = 240 250
250Q* 250(1095.45) = = 22.82 D 12, 000
Ordering
1 2
1
𝑄𝐶ℎ = (1095.45)(0.20)(2.50) = $273.86 2
D 12, 000 Co = (25) = 273.86 Q 1095.45
Total Cost = $547.72 5.
a.
Q* =
b.
T=
2 DCo = Ch
2(150, 000)(250) = 12,500 0.48
300Q* 300(12,500) = = 25 days D 150, 000
300 days/25 days = 12 orders per year One order per month c.
Maximum inventory = Q* = 12,500 Metropolitan Bus Company does not need to expand its storage tanks. r = dm =
d. 6.
150, 000 (10) = 5, 000 gallons 300
a.
Q *pens =
2 DCo = Ch
2 (1500 )( 20 ) 240Q * 240 ( 632 ) = 632 pens, Tpens = = = 101 days D 1500 (1.50 )(.10 )
Q *pencils =
2 DCo = Ch
2 ( 400 )( 20 ) 240Q * 240 ( 224 ) = 200 pencils, Tpencils = = = 120 days D 400 ( 4 )(.10 )
1 D 1 1500 TC pens = Q pensCh + Co = ( 632 )(.1)(1.5) + 20 = $94.87 2 Q pens 2 632 1 D 1 400 TC pencils = Q pencilsCh + Co = ( 200)(.1)( 4 ) + 20 = $80 2 Q pencils 2 200 Thus, the total cost is $94.87 + $80 = $174.87.
10 - 3
Chapter 10
b.
Setting the cycle times of pens and pencils equal:
240Q pens D pens
=
240Q pencils D pencils
which implies Q pens = 3.75Q pencils . The total cost (for both pens and pencils) is:
D 1 1 D TC = Q pensCh , pens + pens Co, pens + Q pencilsCh , pencils + Co, pencils 2 Q pens 2 Q pencils Substituting Q pens = 3.75Q pencils , we obtain
TC =
D pens D 1 1 3.75Q pencilsCh , pens + Co, pens + Q pencilsCh , pencils + pencils Co, pencils 2 3.75Q pencils 2 Q pencils
Combining like terms:
D pensCo , pens + D pencilsCo , pencils 1 3.75 TC = Q pencils ( 3.75Ch , pens + Ch , pencils ) + 2 Q pencils Solving for Qpencils by observing that this total cost equation is the same as
( DCo ) 1 TC = Q pencilsCh' + 2 Q pencils
'
where Ch' = 3.75Ch , pens + Ch , pencils and ( DCo ) = '
D pensCo , pens 3,75
+ D pencilsCo , pencils
Thus,
2 ( DCo ) = Ch' '
Q pencils =
(1500 )(15) 2 + ( 400 )(15) 3.75 = 158 pencils 3.75 (.1)(1.5) + (.1)( 4 )
Q pens = 3.75 (158) = 593 pens These quantities are ordered every T = (240)(593)/1500 = (240)(158)/400 = 95 days. The total cost (for both pens and pencils) is:
TC =
1 1500 1 400 ( 593)(.1)(1.5) + (15) + (158)(.1)( 4 ) + (15) = $151.99 2 593 2 158
10 - 4
Inventory Models
Thus, the consolidate shipments result in annual savings of $174.87 - $151.99 = $22.88. 7.
a.
We cannot directly use the EOQ equation because we are not given the annual demand, fixed ordering cost, and ordering price. However, we can use the original value of Q and knowledge of how we compute it to determine the revised economic order quantity corresponding to the updated holding cost rate. That is, we know: Q* =
2 DCo = Ch
2 DCo IC
Let Q ' be the revised order quantity for the new carrying charge I ' . Thus 𝑄′ = √
Q '/ Q* =
2 DCo / I ' C
Q ' =
I * Q I'
Q' =
2 DCo / IC
=
2𝐷𝐶0 𝐼′ 𝐶
I I'
0.22 (80) = 72 0.27
b. 𝐼 𝑄′ = 𝑄∗ √ ′ 𝐼 8.
Annual Demand D = (5/month)(12 months) = 60 Ordering Cost = Fixed Cost per class = $22,000 Holding Cost = ($1,600/month)(12 months) = $19,200 per year for one driver Q* =
2 DCo = Ch
2(60)(22, 000) = 11.73 (19, 200)
Use 12 as the class size. D/Q* = 60/12 = 5 classes per year Driver holding cost =
1 1 QCh = (12)(19, 200) = $115, 200 2 2
Class ordering cost = ( D / Q)CO = (60 /12)(22, 000) = 110, 000 Total cost = $225,200
10 - 5
Chapter 10
9.
2 DCo = Ch
a.
Q* =
b.
r = dm =
2(5000)(80) = 400 (0.25)(20)
5000 (12) = 240 250
5000 (35) = 700 . The reorder point of 700 exceeds the order quantity of 400. This means 250 one order will be outstanding when the reorder point is reached.
c. r = dm =
d.
10.
Since r = 700 and Q* = 400, one order will be outstanding when the reorder point is reached. Thus the inventory on hand at the time of reorder will be 700 - 400 = 300. This is a production lot size model. However, the operation is only six months rather than a full year. The basis for analysis may be for periods of one month, 6 months, or a full year. The inventory policy will be the same. In the following analysis we use a monthly basis.
Q* =
T=
2 DCo = (1 − D / P )Ch
20Q 20(1414.21) = = 28.28 days D 1000
Production run length =
11.
2(1000)(150) = 1414.21 1000 1 − (0.02)(10) 4000
Q* =
2 DCo = (1 − D / P )Ch
Q 1414.21 = = 7.07 days P / 20 4000 / 20
2(6400)(100) 6400 1 − P 2
a. b. c. d.
P = 8,000 P = 10,000 P = 32,000 P = 100,000
Q* = 1789 Q* = 1333 Q* = 894 Q* = 827
EOQ Model: Q* =
2 DCo = Ch
2(6400)(100) = 800 2
Production Lot Size Q* is always greater than the EOQ Q* with the same D, C0, and Ch values. As the production rate P increases, the recommended Q* decreases, but always remains greater than the EOQ Q*.
10 - 6
Inventory Models
12. a.
b.
Q* =
2 DCo = (1 − D / P)Ch
2(6000)(2345) = 1500 6000 1 − 16000 20
D/Q* = 6000/1500 = 4 production runs 12 months/4 = 3 month cycle time
c.
Current total cost using Q = 500 is as follows: TC =
D 1 D 1 6000 6000 1 − QCh + Co = 1 − 500(20) + 500 (2345) = 3125 + 28140 = $31, 265 2 P Q 2 16000
Proposed Total Cost using Q* = 1500 is as follows:
TC =
1 6000 6000 1− 1500(20) + (2345) = 9375 + 9380 = $18, 755 2 16000 1500
Change to Q* = 1500 Savings = $31,265 - $18,755 = $12,510 12,510/31,265 = 40% savings over current policy 13. a.
Q* =
2 DCo 2(7200)(150) = = 1078.12 7200 (1 − D / P )Ch 1 − (0.18)(14.50) 25000
b.
Number of production runs = D / Q* = 7200 / 1078.12 = 6.68
c.
T=
d.
Production run length =
e.
Maximum Inventory
250Q 250(1078.12) = = 37.43 days D 7200
Q 1078.12 = = 10.78 days P / 250 25000 / 250
7200 D 1 − P Q = 1 − 25000 (1078.12) = 767.62 f.
Holding Cost
1 D 1 7200 1 − QCh = 1 − (1078.12)(0.18)(14.50) = $1001.74 2 P 2 25000
10 - 7
Chapter 10
Ordering cost =
D 7200 Co = (150) = $1001.74 Q 1078.12
Total Cost = $2,003.48 g. 14.
7200 D r = dm = m= (15) = 432 250 250 C = current cost per unit C ' = 1.23 C new cost per unit
2 DCo 2 DC0 = = 5000 (1 − D / P)Ch (1 − D / P) IC
Q* =
Let Q' = new optimal production lot size Q' =
Q' = Q*
2 DCo (1 − D / P) IC '
2 DCo (1 − D / P) IC ' 2 DCo (1 − D / P) IC
=
1 C C 1 C' = = = = 0.9017 C' 1.23C 1.23 1 C
Q' = 0.9017(Q*) = 0.9017(5000) = 4509
15. a.
Q* =
2 DCo Ch + Cb = Ch Cb
2(1200)(25) 0.50 + 5 0.50 = 1148.91 0.50
b.
Ch 0.50 S* = Q * = 1148.91 = 104.45 C + C 0.50 + 5 b h
c.
Max inventory = Q* - S* = 1044.46
d.
T=
e.
Holding:
250Q * 250(1148.91) = = 23.94 D 12000
(Q − S )2 Ch = $237.38 2Q
Ordering:
D Co = 261.12 Q
Backorder:
S2 Cb = 23.74 2Q
10 - 8
Inventory Models
Total Cost: $522.24 The total cost for the EOQ model in problem 4 was $547.72. Allowing backorders reduces the total cost. 16.
12000 r = dm = 5 = 240 250 With backorder allowed the reorder point should be revised to r = dm - S = 240 - 104.45 = 135.55 The reorder point will be smaller when backorders are allowed.
17.
EOQ Model Q* =
2 DCo = Ch
Total Cost =
2(800)(150) = 282.84 3
1 D 800 282.84 QCh + C0 = 3+ (150) = $848.53 2 Q 282.84 2
Planned Shortage Model
Q* =
2 DCo Ch + Cb = Ch Cb
2(800)(150) 3 + 20 20 = 303.32 3
Ch 3 S* = Q * = 303.32 = 39.56 C + C 3 + 20 b h
Total Cost =
(Q − S )2 D S2 Ch + Co + Cb = 344.02 + 395.63 + 51.60 = $791.25 2Q Q 2Q
Cost Reduction with Backorders allowed $848.53 - 791.25 = $57.28 (6.75%) Both constraints are satisfied: 1. S / Q = 39.56 / 303.32 = 0.13 Only 13% of units will be backordered. 2. Length of backorder period = S / d = 39.56 / (800/250) = 12.4 days 18.
Reorder points:
10 - 9
Chapter 10
800 EOQ Model: r = dm = 20 = 64 250 Backorder Model: r = dm - S = 24.44 19. a.
Q* =
2 DCo = Ch
Total Cost: =
b.
Q* =
2(480)(15) = 34.64 (0.20)(60)
1 D QCh + C0 = 207.85 + 207.85 = $415.70 2 Q
2 DCo Ch + Cb = Ch Cb
2(480)(15) 0.20(60) + 45 = 39 (0.20)(60) 45
Ch S* = Q * = 8.21 Ch + Cb
Total Cost =
(Q − S ) 2 D S2 Ch + Co + Cb = 145.80 + 184.68 + 38.88 = $369.36 2 Q 2Q S 8.21 = = 5.13 days d 480 / 300
c.
Length of backorder period =
d.
Backorder case since the maximum wait is only 5.13 days and the cost savings is $415.70 - 369.36 = $46.34 (11.1%)
e.
480 EOQ: r = dm = 6 = 9.6 300 Backorder: r = dm - S = 1.39
20.
Q=
2 DCo Ch
Q1 =
2(120)(20) = 25.30 0.25(30)
Q1 = 25
Q2 =
2(120)(20) = 25.96 0.25(28.5)
Q2 = 50 to obtain 5% discount
Q3 =
2(120)(20) = 26.67 0.25(27)
Q3 = 100 to obtain 10% discount
Order
Holding
10 - 10
Purchase
Total
Inventory Models
Category
Unit Cost
Quantity
Cost
1 30.00 25 93.75 2 28.50 50 178.13 3 27.00 100 337.50 Q = 100 to obtain the lowest total cost.
Order Cost
Cost
Cost
96 48 24
3600 3420 3240
$3,789.75 $3,646.13 $3,601.50
The 10% discount is worthwhile.
21.
Q=
2 DCo Ch
Q1 =
2(500)(40) = 141.42 0.20(10)
Q2 =
2(500)(40) = 143.59 0.20(9.7)
Since Q1 is over its limit of 99 units, Q1 cannot be optimal (see problem 23). Use Q2 = 143.59 as the optimal order quantity. Total Cost: = 22.
1 D QCh + Co + DC = 139.28 + 139.28 + 4,850.00 = $5,128.56 2 Q
D = 4(500) = 2,000 per year Co = $30 I = 0.20 C = $28 Annual cost of current policy: (Q = 500 and C = $28) TC = 1/2(Q)(Ch) + (D/Q)Co + DC = 1/2(500)(0.2)(28) + (2000/500)(30) + 2000(28) = 1400 + 120 + 56,000 = 57,520 Evaluation of Quantity Discounts Q* =
2 DCo Ch
Order Quantity 0-99 100-199 200-299
Ch (0.20)(36) = 7.20 (0.20)(32) = 6.40 (0.20)(30) = 6.00
Q*
Q to obtain Discount
TC
129 137 141
* 137 200
— 64,876 60,900
10 - 11
Chapter 10
300 or more
(0.20)(28) = 5.60
146
300
57,040
*Cannot be optimal since Q* > 99. Reduce Q to 300 pairs/order. Annual savings is $480; note that shoes will still be purchased at the lowest possible cost ($28/pair). 23.
TC =
1 D QIC + Co + DC 2 Q
At a specific Q (and given I, D, and C0), since C of category 2 is less than C of category 1, the TC for 2 is less than TC for 1.
category 1 category 2 TC
minimum for group 2 Q
Thus, if the minimum cost solution for category 2 is feasible, there is no need to search category 1. From the graph we can see that all TC values of category 1 exceed the minimum cost solution of category 2. 24.
a. Cost of overestimation, co = $9 Cost of underestimation, cu = $10 - $9 -$0.50 = $0.50 P( demand ≤ Q*) =
𝑐𝑢 𝑐𝑢 +𝑐𝑜
0.5
= 9.5 = 0.0526
From the normal table, a cumulative probability of 0.0526 corresponds to z = -1.62 Thus, Q* = µ - 1.62σ = 9000 – (1.62)(400) = 8352 magazines. b. Cost of overestimation, co = $9 - $8 = $1 Cost of underestimation, cu = $10 - $9 -$0.50 = $0.50 P(demand ≤ Q*) =
Cu .5 = = 0.3333 Cu + Co 1.5
From the normal table, a cumulative probability of 0.3333 corresponds to z = -0.43
10 - 12
Inventory Models Thus, Q* = µ - 0.43σ = 9000 – (0.43)(400) = 8828 magazines. 25. a.
co = 80 - 50 = 30 cu = 125 - 80 = 45 P( D Q*) =
cu 45 = = 0.60 cu + co 45 + 30
=8
P(D Q*) = 0.60
20 Q* For the cumulative standard normal probability 0.60, z = 0.25
Q* = 20 + 0.25(8) = 22 b. 26. a. b.
P(Sell All) = P(D Q*) = 1 - 0.60 = 0.40 co = $150 The city would have liked to have planned for more additional officers at the $150 per officer rate. However, overtime at $240 per officer will have to be used. cu = $240 - $150 = $90
c.
P(Demand Q*) =
cu 90 = = 0.375 cu + co 90 + 150
For the cumulative standard normal probability 0.375, z = -0.32
Q* = + z = 50 - 0.32(10) = 46.8 Recommend 47 additional officers d.
P(Overtime) = 1 - 0.375 = 0.625
10 - 13
Chapter 10
27. a.
co = 1.19 - 1.00 = 0.19 cu = 1.65 - 1.19 = 0.46
P( D Q*) =
cu 0.46 = = 0.7077 cu + co 0.46 + 0.19
= 30
P(D Q*) = 0.7077
150 Q* For the cumulative standard normal probability 0.7077, z = 0.55
Q* = 150 + 0.55(30) = 166.5 b.
P(Stockout) = P(D Q*) = 1 - 0.7077 = 0.2923
c.
co = 1.19 - 0.25 = 0.94 P( D Q*) =
cu 0.46 = = 0.3286 cu + co 0.46 + 0.94
For the cumulative standard normal probability 0.3286, z = -0.45 Q* = 150 - 0.45(30) = 136.50 The higher rebate increases the quantity that the supermarket should order. 28. a.
co = 8 - 5 = 3 cu = 10 - 8 = 2
P( D Q*) =
cu 2 = = 0.40 cu + co 2 + 3
Q* is 40% of way between 700 and 800 Q* = 200 + 0.40(600) = 440 b.
P(stockout) = P(D Q*) = 1 - 0.40 = 0.60
10 - 14
Inventory Models
c.
P(D Q*) = 0.85 P(Stockout) = 0.15 Q* = 200 + 0.85(600) = 710
d.
Let g = goodwill cost cu = lost profit + goodwill cost = (10 - 8) + g = 2 + g P( D Q*) =
cu = 0.85 cu + co
Solve for cu = 17 cu = 2 + g = 17 g = 15 29. a. b.
r = dm = (200/250)15 = 12 D / Q = 200 / 25 = 8 orders / year The limit of 1 stockout per year means that P(Stockout/cycle) = 1/8 = 0.125
= 2.5 P(Stockout) = 0.125
12
r
P(No Stockout/cycle) = 1 – 0.125 = 0 .875 For the cumulative standard normal probability 0.875, z = 1.15 Thus, z =
r − 12 = 0.15 2.5
or r = 12 + 1.15(2.5) = 14.875 c.
Use 15
Safety Stock = 3 units Added Cost = 3($5) = $15/year
30.
a. Q* = 2 DC0 = Ch
800 2 ( 52 ) 15 100 = 55.86 ≈ 56 boxes. (10)(.15)
10 - 15
Chapter 10 b. r = µ + zσ = 300 + (2.33)(75) = 474.75 ≈ 475 cups. 31. a.
Q* =
2 DCo = Ch
2(1000)(25.5) = 79.84 8
b.
=5 P(Stockout) = 0.02
25
r
P(No Stockout/cycle) = 1 – 0.02 = 0 .98 For the cumulative standard normal probability 0.98, z = 2.05 Thus, z =
r − 25 = 2.05 5
or r = 25 + 2.05(5) = 35.3
Use 35
Safety Stock = 35 - 25 = 10 Safety Stock Cost = (10)($8) = $80/year c.
=5 P(Stockout) = ?
25
30
z = r - 25 = 30 - 25 = 1 5 5 The cumulative standard normal probability for z = 1 is 0.8413. Thus, P(Stockout/cycle) =1.0.8413 = 0.1587 Number of Stockouts/year = 0.1587 (Number of Orders) = 0.1587 D/Q = 2
10 - 16
Inventory Models
32. a.
b.
Q* =
2 DCo = Ch
2(300)(5) = 31.62 (0.15)(20)
D / Q* = 9.49 orders per year
P(Stockout ) =
2 = 0.2108 D / Q*
=6 P(Stockout) = 0.2108
25
r
P(No Stockout) = 1 – 0.2108 = 0.7892 For the cumulative standard normal probability 0.7892, z = 0.80 Thus, z =
r − 25 = 0.80 6
or r = 15 + 0.80(6) = 19.8 c.
Use 20
Safety Stock = 20 - 15 = 5 Safety Stock Cost =5(0.15)(20) = $15
33. a. b.
1/52 = 0.0192 P(No Stockout) = 1 – 0.0192 = 0.9808 For the cumulative standard normal probability 0.9808, z = 2.07 Thus, z =
M − 60 = 2.07 12
or M = µ + z = 60 + 2.07(12) = 85
c. 34. a.
M = 35 + (0.9808)(85-35) = 84 P(Stockout) = 0.01 z = 2.33 P(No Stockout) = 1 – 0.01 = 0.99
10 - 17
Chapter 10
For the cumulative standard normal probability 0.99, z = 2.33 Thus, z =
r − 150 = 2.33 40
or r = µ + z = 150 + 2.33(40) = 243 b.
Safety Stock = 243 - 150 = 93 units Annual Cost = 93(0.20)(2.95) = $54.87
c.
z = 2.33 M = µ + z = 450 + 2.33(70) = 613 units
d.
Safety Stock = 613 - 450 = 163 units Annual Cost = 163(0.20)(2.95) = $96.17
e.
The periodic review model is more expensive ($96.17 - $54.87) = $41.30 per year. However, this added cost may be worth the advantage of coordinating orders for multi products. Go with the periodic review system.
f.
Unit Cost = $295 Annual Difference = $4,130 Use continuous review for the more expensive items.
35. a.
24 − 18 = 1.0 6 The cumulative standard normal probability for z = 1.0 is 0.8413. z=
P(Stockout) = 1.0000 - 0.8413 = 0.1587 b.
P(Stockout) = 0.025 P(No Stockout) = 1 – 0.025 = 0.975 For the cumulative standard normal probability 0.975, z = 1.96 Thus, z =
M − 18 = 1.96 6
or M = µ + z = 18 + 1.96(6) = 29.76 Use M = 30. The manager should have ordered Q = 30 - 8 = 22 units. 36. a.
b.
µ = Week 1 demand + Week 2 demand + Lead Time demand = 16 + 16 + 8 = 40 2 = Var (Week 1) + Var (Week 2) + Var (Lead Time) = 25 + 25 + 12.25 = 62.25
10 - 18
Inventory Models
= 62.25 = 7.9 c.
26 orders per year P(Stockout) = 1/26 = 0.0385 per replenishment P(No Stockout) = 1 – 0.0385 = 0.9615 For the cumulative standard normal probability 0.9615, z = 1.77 Thus, z =
M − 40 = 1.77 7.9
or M = µ + z = 40 + 1.77(7.9) = 54 d.
54 - 18 = 36
10 - 19
Chapter 11 Waiting Line Models Learning Objectives 1.
Be able to identify where waiting line problems occur and realize why it is important to study these problems.
2.
Know the difference between single-channel and multiple-channel waiting lines.
3.
Understand how the Poisson distribution is used to describe arrivals and how the exponential distribution is used to describe services times.
4.
Learn how to use formulas to identify operating characteristics of the following waiting line models: a. Single-channel model with Poisson arrivals and exponential service times b. Multiple-channel model with Poisson arrivals and exponential service times c. Single-channel model with Poisson arrivals and arbitrary service times d. Multiple-channel model with Poisson arrivals, arbitrary service times, and no waiting e. Single-channel model with Poisson arrivals, exponential service times, and a finite calling population
5.
Know how to incorporate economic considerations to arrive at decisions concerning the operation of a waiting line.
6.
Understand the following terms: queuing theory queue single-channel multiple-channel arrival rate service rate queue discipline
steady state utilization factor operating characteristics blocking infinite calling population finite calling population
11 - 1
Chapter 11
Solutions: 1.
a.
= 5(0.4) = 2 per five minute period
b.
x - x -2 P (x) = e = 2 e x! x!
x 0 1 2 3
2.
P(x) 0.1353 0.2707 0.2707 0.1804
c.
P(Delay Problems) = P(x > 3) = 1 - P(x 3) = 1 - 0.8571 = 0.1429
a.
µ = 0.6 customers per minute P(service time 1) = 1 - e-(0.6)1 = 0.4512
3.
b.
P(service time 2) = 1 - e-(0.6)2 = 0.6988
c.
P(service time > 2) = 1 - 0.6988 = 0.3012
a.
P0 = 1 - = 1 - 0.4 = 0.3333 0.6 2 (0.4)2 Lq = = = 1.3333 0.6 (0.6 - 0.4) ( - )
b. c.
L = L q + = 1.3333 + 0.4 = 2 0.6
d.
= 1.3333 = 3.3333 min. 0.4 W = W q + 1 = 3.3333 + 1 = 5 min. 0.6
e. f.
Wq =
Lq
Pw = = 0.4 = 0.6667 0.6 n n Pn = P0 = 0.4 (0.3333) 0.6 n
4.
Pn
0 1 2 3
0.3333 0.2222 0.1481 0.0988
P(n > 3) = 1 - P(n 3) = 1 - 0.8024 = 0.1976 5.
a.
P0 = 1 - = 1 - 10 = 0.1667 12
b.
Lq =
102 2 = = 4.1667 12 (12 - 10) ( - )
11 - 2
Waiting Line Models
c.
6.
7.
Wq =
Lq
= 0.4167 hours (25 minutes)
d.
W = W q + 1 = .5 hours (30 minutes)
e.
Pw =
a.
P0 = 1 −
b.
Lq =
c.
Wq =
d.
Pw =
e.
Average one customer in line with a 50 second average wait appears reasonable.
a.
Lq =
10 = = 0.8333 12 1.25 = 1− = 0.375 2
2 1.252 = = 1.0417 ( − ) 2(2 − 1.25) Lq
=
1.0417 = 0.8333 minutes (50 seconds) 1.25
1.25 = = 0.625 2
2 (2.5) 2 = = 0.5000 ( − ) 5(5 − 2.5)
L = Lq +
Lq
b.
Wq =
c.
W = Wq +
d.
Pw =
2.5 = 0.5000 + =1 5
=
0.5000 = 0.20 hours (12 minutes) 2.5
1
= 0.20 +
1 = 0.40 hours (24 minutes) 5
2.5 = = 0.50 5
11 - 3
Chapter 11
= 1 and = 1.25
8.
P0 = 1 −
Lq =
1 = 1− = 0.20 1.25
2 1 = = 3.2 ( − ) 1.25(0.25)
L = Lq +
Wq =
Lq
1 = 3.2 + =4 1.25 =
1
W = Wq +
Pw =
3.2 = 3.2 minutes 1
= 3.2 +
1 = 4 minutes 1.25
1 = = 0.80 1.25
Even though the services rate is increased to µ = 1.25, this system provides slightly poorer service due to the fact that arrivals are occurring at a higher rate. The average waiting times are identical, but there is a higher probability of waiting and the number waiting increases with the new system. 9.
2.2 = 1− = 0.56 5
a.
P0 = 1 −
b.
2.2 P1 = P0 = (0.56) = 0.2464 5
c.
2.2 P2 = P0 = (0.56) = 0.1084 5
d.
2.2 P3 = P0 = (0.56) = 0.0477 5
2
2
3
3
e.
P(More than 2 waiting) = P(More than 3 are in system) = 1 - (P0 + P1 + P2 + P3) = 1 - 0.9625 = 0.0375
f.
Lq =
Wq =
2 2.22 = = 0.3457 ( − ) 5(5 − 2.2) Lq
= 0.157 hours
(9.43 minutes)
11 - 4
Waiting Line Models
10. a.
b.
=2 Average number waiting (Lq) Average number in system (L) Average time waiting (Wq) Average time in system (W) Probability of waiting (Pw) New mechanic
µ=3 1.3333 2.0000 0.6667 1.0000 0.6667
µ=4 0.5000 1.0000 0.2500 0.5000 0.5000
= $30(L) + $14 = 30(2) + 14 = $74 per hour
Experienced mechanic = $30(L) + $20 = 30(1) + 20 = $50 per hour Hire the experienced mechanic 11. a.
= 2.5 = 60/10 = 6 customers per hour 2
2
Lq =
2.5 = = 0.2976 6 (6 - 2.5) ( - )
L = L q + = 0.7143
Wq =
Lq
= 0.1190 hours (7.14 minutes)
W = W q + 1 = 0.2857 hours
Pw = = 2.5 = 0.4167 6
b.
No; Wq = 7.14 minutes. Firm should increase the mean service rate (µ) for the consultant or hire a second consultant.
c.
= 60/8 = 7.5 customers per hour Lq = Wq =
2
2
2.5 = = 0.1667 7.5 (7.5 - 2.5) ( - ) Lq
= 0.0667 hours (4 minutes)
The service goal is being met.
11 - 5
Chapter 11
12. a.
P0 = 1 - = 1 - 15 = 0.25 20 Lq =
152 2 = = 2.25 20 (20 - 15) ( - )
L = L + = 3
Wq =
Lq
= 0.15 hours (9 minutes)
W = W q + 1 = 0.20 hours (12 minutes)
Pw = = 15 = 0.75 20 b. With Wq = 9 minutes, the checkout service needs improvements. 13.
Average waiting time goal: 5 minutes or less. a.
One checkout counter with 2 employees
= 15 = 30 per hour Lq = Wq = b.
152 2 = = 0.50 30 (30 - 15) ( - ) Lq
= 0.0333 hours (2 minutes)
Two channel-two counter system
= 15 = 20 per hour for each From Table, P0 = 0.4545
Lq = Wq =
( / )2 P = (15 / 20)2 (15) (20) (0.4545) = 0.1227 0 1! (2 (20) - 15)2 (40 - 15)2 Lq
= 0.0082 hours (0.492 minutes)
Recommend one checkout counter with two people. This meets the service goal with Wq = 2 minutes. The two counter system has better service, but has the added cost of installing a new counter. 14. a.
=
60 = 8 customers per hour 7.5
11 - 6
Waiting Line Models
5 = 1 − = 0.3750 8
b.
P0 = 1 −
c.
Lq =
d.
Wq =
e.
Pw =
f.
62.5% of customers have to wait and the average waiting time is 12.5 minutes. Ocala needs to add more consultants to meet its service guidelines.
2 52 = = 10417 . ( − ) 8(8 − 5) Lq
=
10417 . = 0.2083 hours (12.5 minutes) 5
5 = = 0.6250 8
k = 2, = 5, = 8
15.
Using the equation for P0, P0 = 0.5238 𝐿𝑞 =
Wq = 𝑃1 =
(𝜆⁄𝜇)2 𝜆𝜇 1!(𝑘𝜇−𝜆)2
Lq
=
𝜆⁄𝜇1 1!
𝑃0 = 0.0676
0.0676 = 0.0135 hours (0.81 minutes) 5 5
𝑃0 = (. 5238) = 0.3274 8
Pw = P(n 2) = 1 - P(n 1) = 1 - 0.5238 - 0.3274 = 0.1488 Two consultants meet service goals with only 14.88% of customers waiting with an average waiting time of 0.81 minutes (49 seconds). 16. a.
P0 = 1 - = 1 - 5 = 0.50 10
2 52 = = 0.50 ( − ) 10(10 − 5)
b.
Lq =
c.
Wq =
d.
W = Wq +
e.
Yes, unless Wq = 6 minutes is considered too long.
17. a. b.
Lq
= 0.1 hours (6 minutes)
1
= 0.2 hours (12 minutes)
From Table, P0 = 0.60 2 L q = ( / ) P0 = 0.0333 1! (k - )2
11 - 7
Chapter 11
c.
Wq =
Lq
= 0.0067 hours (24.12 seconds)
d.
W = W q + 1 = 0.1067 (6.4 minutes)
e.
This service is probably much better than necessary with average waiting time only 24 seconds. Both channels will be idle 60% of the time.
Arrival rate: = 5.4 per minute
18.
Service rate: = 3 per minute for each station a.
Using the table of values of P0, / = 1.8, k = 2, and P0 = 0.0526
( / ) k
Lq =
(k − 1)!(k − )
L = Lq +
Wq =
Lq
2
P0 =
= 7.67 + 1.8 = 9.47 =
W = Wq +
7.67 = 1.42 minutes 5.4
1
= 1.42 + 0.33 = 1.75 minutes k
Pw =
(1.8) 2 (5.4)(3) (0.0526) = 7.67 (2 − 1)!(6 − 5.4) 2
1 k 1 5.4 2(3) 0.0526 = 0.8526 P0 = k ! k − 2! 3 2(3) − 5.4 2
b.
The average number of passengers in the waiting line is 7.67. Two screening stations will be able to meet the manager’s goal.
c.
The average time for a passenger to move through security screening is 1.75 minutes.
19. a.
For the system to be able to handle the arrivals, we must have k > , where k is the number of channels. With = 2 and = 5.4, we must have at least k = 3 channels. To see if 3 screening stations are adequate, we must compute Lq. Using the table of values of P0, / = 2.7, k = 3, and P0 = 0.02525 (halfway between / = 2.6 and / = 2.8)
( / ) k
Lq =
(k − 1)!(k − )
P = 2 0
(2.7)3 (5.4)(2) (0.02525) = 7.45 2!(6 − 5.4) 2
Having 3 stations open satisfies the manager’s goal to limit the average number of passengers in the waiting line to at most 10.
11 - 8
Waiting Line Models
b.
The average time required for a passenger to pass through security screening is Wq =
Lq
=
W = Wq +
7.45 = 1.38 5.4 1
= 1.38 + 0.5 = 1.88 minutes
Note: The above results are based on using the tables of P0 and an approximate value for P0. If a computer program is used, we obtain exact results as follows: P0 = 0.0249 Lq = 7.35 W = 1.86 minutes 20. a.
Note
1.2 = = 1.60 1. Thus, one postal clerk cannot handle the arrival rate. 0.75
Try k = 2 postal clerks From Table with
Lq =
( / ) 2 P0 = 2.8444 1!(2 − ) 2
L = Lq +
Wq =
= 1.60 and k = 2, P0 = 0.1111
Lq
= 4.4444 = 2.3704 minutes
W = Wq +
1
= 3.7037 minutes
Pw = 0.7111 Use 2 postal clerks with average time in system 3.7037 minutes. No need to consider k = 3. b.
Try k = 3 postal clerks. From Table with
2.1 = = 2.80 and k = 3, P0 = 0.0160 .75
11 - 9
Chapter 11
Lq =
( / )3 P0 = 12.2735 2(3 − ) 2
L = Lq +
Wq =
Lq
= 15.0735 = 5.8445 minutes
W = Wq +
1
= 7.1778 minutes
Pw = 0.8767 Three postal clerks will not be enough in two years. Average time in system of 7.1778 minutes and an average of 15.0735 customers in the system are unacceptable levels of service. Post office expansion to allow at least four postal clerks should be considered. 21.
From question 11, a service time of 8 minutes has µ = 60/8 = 7.5 2
2
(2.5) = = 0.1667 7.5 (7.5 - 2.5) ( - )
Lq =
L = L q + = 0.50
Total Cost
= $25L + $16 = 25(0.50) + 16 = $28.50
Two channels: = 2.5 µ = 60/10 = 6 Using equation, P0 = 0.6552 L q = ( / ) P0 = 0.0189 1! (2 - )2 2
L = L q + = 0.4356
Total Cost = 25(0.4356) + 2(16) = $42.89 Use the one consultant with an 8 minute service time. 22.
= 24 Characteristic a. b. c. d. e.
P0 Lq Wq W L
System A (k = 1, µ = 30)
System B (k = 1, µ = 48)
System C (k = 2, µ = 30)
0.2000 3.2000 0.1333 0.1667 4.0000
0.5000 0.5000 0.0200 0.0417 1.0000
0.4286 0.1524 0.0063 0.0397 0.9524
11 - 10
Waiting Line Models f.
Pw
0.8000
0.5000
0.2286
System C provides the best service. Service Cost per Channel
23.
System A: System B: System C:
6.50 2(6.50) 6.50
+ + +
20.00 20.00 20.00
= = =
$26.50/hour $33.00/hour $26.50/hour
Total Cost = cwL + csk System A: System B: System C:
25(4) 25(1) 25(0.9524)
+ + +
26.50(1) 33.00(1) 26.50(2)
= = =
$126.50 $ 58.00 $ 76.81
System B is the most economical. 24. = 4, W = 10 minutes a.
µ = 1/2 = 0.5
b.
Wq = W - 1/µ = 10 - 1/0.5 = 8 minutes
c.
L = W = 4(10) = 40
25. A two-server system with a queue dedicated to each server effectively operates as two one-server waiting line systems each with λ = 0.75 / 2 = 0.375 customers arriving per hour and µ = 1 customer processed per hour. Thus for each server: 𝜆2
0.3752
𝐿𝑞 = 𝜇(𝜇−𝜆) = 1(1−0.375) = 0.225 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠 𝜆
𝐿 = 𝐿𝑞 + 𝜇 = 0.225 + 𝐿𝑞
0.375 1
= 0.6 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠
0.225
𝑊𝑞 = 𝜆 = 0.375 = 0.6 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 1
1
𝑊 = 𝑊𝑞 + 𝜇 = 0.6 + 1 = 1.6 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 For the entire system (both servers and their respective dedicated queue) on average there are 2 × 0.225 = 0.45 customers waiting and 2 × 0.6 = 1.2 customers in the system. On average, each customer experiences 0.6 minutes of waiting time and is in the system 1.6 minutes. As shown in Section 15.3, in the two-server system with a single shared queue, there are 0.1227 customers waiting and 0.8737 customers in the system on average. Furthermore, each customer experiences 0.1636 minutes of waiting time and is in the system 1.1636 minutes. Comparing these numbers, it is clear that the “shared single queue” results in better process performance than the two “dedicated queues.”
11 - 11
Chapter 11 It is important to note that we have made a number of assumptions, namely, that the customers are equally split and that customers never switch lines.
26. a.
Express and µ in mechanics per minute
= 4/60 = 0.0667 mechanics per minute µ = 1/6 = 0.1667 mechanics per minute Lq = Wq = 0.0667(4) = 0.2668 W = Wq + 1/µ = 4 + 1/0.1667 = 10 minutes L = W = (0.0667)(10) = 0.6667 b.
Lq = 0.0667(1) = 0.0667 W = 1 + 1/0.1667 = 7 minutes
c.
L = W = (0.0667)(7) = 0.4669 One-Channel Total Cost = 20(0.6667) + 12(1) = $25.33 Two-Channel Total Cost = 20(0.4669) + 12(2) = $33.34 One-Channel is more economical.
27. a.
2/8 hours = 0.25 per hour
b.
1/3.2 hours = 0.3125 per hour
c.
2 2 2 2 2 2 L q = + ( / ) = (0.25) (2) + (0.25 / 0.3125) = 2.225 2 (1 - / ) 2 (1 - 0.25 / 0.3125)
d.
= 2.225 = 8.9 hours 0.25 1 1 W = Wq + = 8.9 + = 12.1 hours 1.3125
e.
f.
Wq =
Lq
Same at Pw = =
0.25
= 0.80
0.3125 80% of the time the welder is busy.
=5
28. a.
Design A B
µ 60/6 = 10 60/6.25 = 9.6
11 - 12
Waiting Line Models
b.
Design A with µ = 10 jobs per hour.
c.
3/60 = 0.05 for A 0.6/60 = 0.01 for B
d. Characteristic P0 Lq L Wq W Pw
e.
Design B 0.4792 0.2857 0.8065 0.0571 0.1613 0.5208
Design B is slightly better due to the lower variability of service times. System A: System B:
29. a.
Design A 0.5000 0.3125 0.8125 0.0625 0.1625 0.5000
W = 0.1625 hrs W = 0.1613 hrs
(9.75 minutes) (9.68 minutes)
= 3/8 = .375 µ = 1/2 = .5
b.
Lq =
2 2 + ( / ) 2 = (.375)2 (1.5) 2 + (.375/ .5)2 = 1.7578 2 (1 - / ) 2 (1 - .375 / .5)
L = Lq + / µ = 1.7578 + .375 / .5 = 2.5078 TC = cwL + csk = 35 (2.5078) + 28 (1) = $115.71 c. Current System ( = 1.5)
New System ( = 0) Lq = 1.7578 L = 2.5078 Wq = 4.6875 W = 6.6875 TC = $115.77
Lq = 1.125 L = 1.875 Wq = 3.00 W = 5.00
TC = cwL + csk = 35 (1.875) + 32 (1) = $97.63 d.
Yes; Savings = 40 ($115.77 - $97.63) = $725.60 Note: Even with the advantages of the new system, Wq = 3 shows an average waiting time of 3 hours. The company should consider a second channel or other ways of improving the emergency repair service.
30. a.
= 42 µ = 20 i 0
(/µ)i / i ! 1.0000
11 - 13
Chapter 11
1 2 3
2.1000 2.2050 1.5435 6.8485
j
Pj
0 1 2 3
1/6.8485 2.1/6.8485 2.2050/6.8485 1.5435/6.8485
= = = =
0.1460 0.3066 0.3220 0.2254 1.0000
b.
0.2254
c.
L = /µ(1 - Pk) = 42/20 (1 - 0.2254) = 1.6267
d.
Four lines will be necessary. The probability of denied access is 0.1499.
31. a.
= 20 µ = 12 i 0 1 2
j 0 1 2
b.
c. 32. a.
P2 = 0.3425
34.25%
k=3
P3 = 0.1598
k=4
P4 = 0.0624
(/µ)i / i ! 1.0000 1.6667 1.3889 4.0556 Pj 1/4.0556 1.6667/4.0556 1.3889/4.0556
Must go to k = 4.
L = /µ(1 - P4) = 20/12(1 - 0.0624) = 1.5626
= 40 µ = 30 i 0 1 2
b.
= = =
(/µ)i / i ! 1.0000 1.3333 0.8888 3.2221
P0 = 1.0000/3.2221 = 0.3104
31.04%
P2 = 0.8888/3.2221 = 0.2758
27.58%
11 - 14
0.2466 0.4110 0.3425
Waiting Line Models
c. (/µ)i / i ! 0.3951 0.1317
i 3 4 P2 = 0.2758 P3 = 0.3951/(3.2221 + 0.3951) = 0.1092
P4 = 0.1317/(3.2221 + 0.3951 + 0.1317) = 0.0351 d. 33. a.
k = 3 with 10.92% of calls receiving a busy signal.
= 0.05 µ = 0.50 /µ = 0.10 N = 8 n 0 1 2 3 4 5 6 7 8
N! n (N - n) ! 1.0000 0.8000 0.5600 0.3360 0.1680 0.0672 0.0202 0.0040 0.0004 2.9558
P0 = 1/2.9558 = 0.3383 𝜆+𝜇 0.55 𝐿𝑞 = 𝑁 − ( ) (1 − 𝑃0 ) = 8 − ( ) (1 − 0.3383) = 0.7215 𝜆 0.05 L = Lq + (1 - P0) = 0.7213 + (1 - 0.3383) = 1.3832
Wq =
Lq ( N − L)
W = Wq + b.
1
=
0.7215 = 2.1808 hours (8 − 13832 . )(0.05)
= 2.1808 +
1 = 4.1808 hours 0.50
P0 = 0.4566 Lq = 0.0646 L = 0.7860 Wq = 0.1791 hours W = 2.1791 hours
11 - 15
Chapter 11
c.
One Employee Cost = 80L + 20 = 80(1.3832) + 20 = $130.65 Two Employees Cost = 80L + 20(2) = 80(0.7860) + 40 = $102.88 Use two employees. N = 5 = 0.025 µ = 0.20 /µ = 0.125
34. a.
n 0 1 2 3 4 5
N! n (N - n) ! 1.0000 0.6250 0.3125 0.1172 0.0293 0.0037 2.0877
P0 = 1/2.0877 = 0.4790 b.
+ 0.225 Lq = N − (1 − P0 ) = 5 − 0.025 (1 − 0.4790) = 0.3110
c.
L = Lq + (1 - P0) = 0.3110 + (1 - 0.4790) = 0.8321
d.
Wq =
e.
W = Wq +
f.
Trips/Days = (8 hours)(60 min/hour) () = (8)(60)(0.025) = 12 trips
Lq ( N − L)
1
=
0.3110 = 2.9854 min (5 − 0.8321)(0.025)
= 2.9854 +
Time at Copier: Wait Time at Copier: g.
1 = 7.9854 min 0.20
12 x 7.9854 = 95.8 minutes/day 12 x 2.9854 = 35.8 minutes/day
Yes. Five administrative assistants x 35.8 = 179 min. (3 hours/day) 3 hours per day are lost to waiting. (35.8/480)(100) = 7.5% of each administrative assistant's day is spent waiting for the copier.
11 - 16
Waiting Line Models N = 10 = 0. 25 µ = 4 /µ = 0.0625
35. a.
n 0 1 2 3 4 5 6 7 8 9 10
N! n (N - n) ! 1.0000 0.6250 0.3516 0.1758 0.0769 0.0288 0.0090 0.0023 0.0004 0.0001 0.0000 2.2698
P0 = 1/2.2698 = 0.4406 𝜆+𝜇
4.25
b.
𝐿𝑞 = 𝑁 − (
c.
L = Lq + (1 - P0) = 0.4895 + (1 - 0.4406) = 1.0490
d.
0.4895 = = 0.2188 ( N − L) (10 − 10490 . )(0.25) 1 1 W = Wq + = 0.2188 + = 0.4688 4
e.
TC
g.
k=2
= =
TC
= = 50L = L = h.
) (1 − 𝑃0 ) = 10 − (
0.25
) (1 − 0.4406) = 0.4895
Lq
Wq =
f.
𝜆
cw L + cs k 50 (1.0490) + 30 (1) = $82.45
cw L + cs k 50L + 30(2) = $82.45 22.45 0.4490 or less.
Using the Excel worksheet template Finite with k = 2, L
=
0.6237
TC
= =
cw L + cs k 50 (0.6237) + 30 (2) = $84.95
The company should not expand to the two-channel truck dock.
11 - 17
Chapter 12 Simulation Learning Objectives 1.
Understand what simulation is and how it aids in the analysis of a problem.
2.
Learn why simulation is a significant problem-solving tool.
3.
Understand the difference between static and dynamic simulation.
4.
Identify the important role probability distributions, random numbers, and the computer play in implementing simulation models.
5.
Realize the relative advantages and disadvantages of simulation models.
6.
Understand the following terms: simulation simulation model
Monte Carlo simulation discrete-event simulation
12 - 1
Chapter 12
Solutions: 1.
a.
Profit = (249 - c1 - c2 ) x - 1,000,000 = (249 - 45 - 90) (20,000) - 1,000,000 = $1,280,000 (Engineer's)
2.
b.
Profit = (249 - 45 - 100) (10,000) - 1,000,000 = $40,000 (Financial Analyst)
c.
Simulation will provide probability information about the various profit levels possible. What if scenarios show possible profit outcomes but do not provide probability information.
a.
Let
c x
= =
variable cost per unit demand
Profit = 50x - cx - 30,000 = (50 - c) x - 30,000 b.
Base case: Profit Worst case: Profit Best case: Profit
c.
The possibility of a $41,400 profit is interesting, but the worst case loss of $22,200 is risky. Risk analysis would be helpful in evaluating the probability of a loss.
3. a.
= (50 - 20) 1200 - 30,000 = 6,000 = (50 - 24) 300 - 30,000 = -22,200 = (50 - 16) 2100 - 30,000 = 41,400
Using the random number intervals Direct Labor Cost $43 $44 $45 $46 $47
Interval of Random Numbers 0.0 but less than 0.1 0.1 but less than 0.3 0.3 but less than 0.7 0.7 but less than 0.9 0.9 but less than 1.0
Random Number 0.3753 0.9218 0.0336 0.5145 0.7000 b.
Direct Labor Cost $45 $47 $43 $45 $46
Parts cost = a + r(b – a) = 80 + r(100 – 80) = 80 + r(20) r = 0.6221 Parts cost = 80 + (0.6221)(20) = $92.44 r = 0.3418 Parts cost = 80 + (0.3418)(20) = $86.84 r = 0.1402 Parts cost = 80 + (0.1402)(20) = $82.80 r = 0.5198 Parts cost = 80 + (0.5198)(20) = $90.40
c.
r = 0.9375 Parts cost = 80 + (0.9375)(20) = $98.75 First-year demand = µ + zσ = 15,000 + z(4500)
12 - 2
Simulation
r = 0.8531 z = 1.05
Demand = 15,000 + 1.05(4500) = 19,725
r = 0.1762 z = -0.93
Demand = 15,000 - 0.93(4500) = 10,815
r = 0.5000 z = 0.00
Demand = 15,000 + 0.00(4500) = 15,000
r = 0.6810 z = 0.47
Demand = 15,000 + 0.47(4500) = 17,115
r = 0.2879 z = -0.56
Demand = 15,000 - 0.56(4500) = 12,480 Random Number 0.3753 0.9218 0.0336 0.5145 0.7000
4.
Direct Labor Cost $45 $47 $43 $45 $46
a. Number of New Accounts Opened 0 1 2 3 4 5 6
Probability .01 .04 .10 .25 .40 .15 .05
Interval of Random Numbers .00 but less than .01 .01 but less than .05 .05 but less than .15 .15 but less than .40 .40 but less than .80 .80 but less than .95 .95 but less than 1.00
Random Number 0.7169 0.2186 0.2871 0.9155 0.1167 0.9800 0.5029 0.4154 0.7872 0.0702
Number of New Accounts Opened 4 3 3 5 2 6 4 4 4 2
b. Trial 1 2 3 4 5 6 7 8 9 10 c.
5.
For the 10 trials Gustin opened 37 new accounts. With an average first year commission of $5000 per account, the total first year commission is $185,000. The cost to run the 10 seminars is $35,000, so the net contribution to profit for Gustin is $150,000 or $15,000 per seminar. Because the seminars are a very profitable way of generating new business, Gustin should continue running the seminars.
a. Stock Price Change -2 -1 0 +1 +2
Probability .05 .10 .25 .20 .20
12 - 3
Interval .00 but less than .05 .05 but less than .15 .15 but less than .40 .40 but less than .60 .60 but less than .80
Chapter 12
+3 +4
.10 .10
.80 but less than .90 .90 but less than 1.00
Price Change -1 +4 0 +3
Ending Price Per Share $38 $42 $42 $45
b. Random Number 0.1091 0.9407 0.1941 0.8083 Ending price per share = $45 6.
a. Payment 0 500 1,000 2,000 5,000 8,000 10,000
Probability .83 .06 .05 .02 .02 .01 .01
Interval of Random Numbers .00 but less than .83 .83 but less than .89 .89 but less than .94 .94 but less than .96 .96 but less than .98 .98 but less than .99 .99 but less than 1.00
b. Trial 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Random Number 0.9850 0.2122 0.5590 0.0004 0.5177 0.0064 0.2044 0.2390 0.7305 0.6351 0.4971 0.9569 0.1091 0.9907 0.1900 0.9403 0.3545 0.8083 0.4704 0.5274
Payment 8,000 0 0 0 0 0 0 0 0 0 0 2,000 0 10,000 0 2,000 0 0 0 0
4 claims paid out of 20 for a total of $22,000 in payments to policyholders. 7.
Time = µ + zσ = 15 + z(3) r = 0.1562
z = -1.01
Time = 15 - 1.01(3) = 11.97 minutes
12 - 4
Simulation
8.
a.
b.
9.
r = 0.9821
z = 2.10
Time = 15 + 2.10(3) = 21.30 minutes
r = 0.3409
z = -0.41
Time = 15 - 0.41(3) = 13.77 minutes
r = 0.5594
z = 0.15
Time = 15 + 0.15(3) = 15.45 minutes
r = 0.7758
z = 0.76
Time = 15 + 0.76(3) = 17.28 minutes
The following table can be used to simulate a win for Atlanta Game Interval for Atlanta Win 1 .00 but less than .60 2 .00 but less than .55 3 .00 but less than .48 4 .00 but less than .45 5 .00 but less than .48 6 .00 but less than .55 7 .00 but less than .50 Using the random numbers in column 6 beginning with 0.3813, 0.2159 and so on, Atlanta wins games 1 and 2, loses game 3, wins game 4, loses game 5 and wins game 6. Thus, Atlanta wins the 6-game World Series 4 games to 2 games.
c.
Repeat the simulation many times. In each case, record who wins the series and the number of games played, 4, 5, 6 or 7. Count the number of times Atlanta wins. Divide this number by the total number of simulation runs to estimate the probability that Atlanta will win the World Series. Count the number of times the series ends in 4 games and divide this number by the total number of simulation runs to estimate the probability of the World Series ending in 4 games. This can be repeated for 5-game, 6-game and 7-game series.
a.
Base case using most likely completion times. A B C D
Worst case: Best case:
6 5 14 8 33
weeks
8 + 7 + 18 + 10 = 43 weeks 5 + 3 + 10 + 8 = 26 weeks
b. Activity A B C D
c.
Random Number 0.1778 0.9617 0.6849 0.4503 Total:
Completion Time 5 7 14 8 34
Weeks
Simulation will provide a distribution of project completion time values. Calculating the percentage of simulation trials with completion times of 35 weeks or less can be used to estimate the probability
12 - 5
Chapter 12
of meeting the completion time target of 35 weeks. 10. a. Hand Value 17 18 19 20 21 Broke
Probability .1654 .1063 .1063 .1017 .0972 .4231
Interval .0000 but less than .1654 .1654 but less than .2717 .2717 but less than .3780 .3780 but less than .4797 .4797 but less than .5769 .5769 but less than 1.000
b/c. Hand 1 2 3 4 5 6 7 8 9 10 d.
Dealer Value Broke 18 21 17 21 17 18 18 Broke Broke
Player Value Broke Broke 17 Broke 21 17 17 Broke 17 Broke
Hand 11 12 13 14 15 16 17 18 19 20
Dealer Value 21 Broke 17 Broke 18 Broke 19 Broke 20 21
Player Value 17 Broke Broke 20 20 18 Broke 20 Broke Broke
Dealer wins 13: 1-4, 7, 8, 10-13, 17, 19, 20 Pushes = 2: 5, 6 Player wins 5: 9, 14, 15, 16, 18 At a bet of $10 per hand, the player loses $80.
e.
Player wins 7: 1, 9, 10, 12, 14, 16, 18 At a bet of $10 per hand, the player loses $60. On the basis of these results, we would not recommend the player take a hit on 16 when the dealer is showing a 6.
11. a.
Let r = random number a = smallest value = -8 b = largest value = 12 Return % = a + r(b - a) = -8 + r(12-(-8)) = -8 + r20
12 - 6
Simulation
1st Quarter
r = .52 Return % = -8 + .52(20) = 2.4%
For all quarters: Quarter 1 2 3 4 5 6 7 8 b.
r 0.52 0.99 0.12 0.15 0.50 0.77 0.40 0.52
Return % 2.4% 11.8% -5.6% -5.0% 2.0% 7.4% 0.0% 2.4%
For each quarter, Ending price = Beginning price + Change For Quarter 1: Ending price = $80.00 + .024($80.00) = $80.00 + $1.92 = $81.92 For Quarter 2: Ending price = $81.92 + .118($81.92) = $81.92 + $9.67 = $91.59
Quarter 1 2 3 4 5 6 7 8
Starting Price/Share $80.00 $81.92 $91.59 $86.46 $82.13 $83.78 $89.98 $89.98
Return % 2.4% 11.8% -5.6% -5.0% 2.0% 7.4% 0.0% 2.4%
Change $ $1.92 $9.67 -$5.13 -$4.32 $1.64 $6.20 $0.00 $2.16
Ending Price/Share $81.92 $91.59 $86.46 $82.13 $83.78 $89.98 $89.98 $92.14
Price per share at the end of two years = $92.14 c.
12. a.
Conducting a risk analysis would require multiple simulations of the eight-quarter, two-year period. For each simulation, the price per share at the end of two years would be recorded. The distribution of the ending price per share values would provide an indication of the maximum possible gain, the maximum possible loss and other possibilities in between. Profit = Selling Price - Purchase Cost - Labor Cost - Transportation Cost Base Case using most likely costs Profit = 45 - 11 - 24 - 3 = $7/unit Worst Case Profit = 45 - 12 - 25 - 5 = $3/unit Best Case Profit = 45 - 10 - 20 - 3 = $12/unit
12 - 7
Chapter 12
b. Purchase Cost $10 11 12
Labor Cost $20 22 24 25
Interval .00 but less than .10 .10 but less than .35 .35 but less than .70 .70 but less than 1.00
Transportation Cost $3 5
Interval .00 but less than .75 .75 but less than 1.00
c.
Profit = 45 - 11 - 24 - 5 = $5/unit
d.
Profit = 45 - 10 - 25 - 3 = $7/unit
e.
Simulation will provide a distribution of the profit per unit values. Calculating the percentage of simulation trials providing a profit less than $5 per unit would provide an estimate of the probability the profit per unit will be unacceptably low. Use the PortaCom spreadsheet. Simulation results will vary. Mean profit of approximately equal to $710,000. Probability of a loss in the 0.07 to 0.10 range can be anticipated.
13. a. b. 14.
Interval .00 but less than .25 .25 but less than .70 .70 but less than 1.00
The Excel worksheet for this problem is as follows:
12 - 8
Simulation
Selected cell formulas are as follows: Cell
Formula
B12 C12 D12 C514 C515 C516 C517 C518 C519
=$B$7 + RAND() * ($B$8 - $B$7) =NORMINV(RAND(), $E$7, $E$8) =($B$3 - B12) * C12 - $B$4 =AVERAGE(D12:D511) =STDEV(D12:D511) =MIN(D12:D511) =MAX(D12:D511) =COUNTIF(D12:D511, “<0”) =C519 / COUNT(D12:D511)
a.
The mean profit should be approximately $6,000. Simulation results will vary with most simulations having a mean profit between $5,500 and $6,500.
b.
120 to 150 of the 500 simulation trails should show a loss. Thus, the probability of a loss should typically be between 0.24 and 0.30.
12 - 9
Chapter 12
c.
15.
This project appears too risky. The relatively high probability of a loss and only roughly $6,000 as a mean profit indicate that the potential gain is not worth the risk of a loss. More precise estimates of the variable cost per unit and the demand could help determine a more precise profit estimate. The Excel worksheet for this problem is as follows:
Selected cell formulas are as follows: Cell
Formula
B15 C15 D15
=VLOOKUP(RAND(), $A$5:$C$10, 3, TRUE) =VLOOKUP(RAND(), $A$5:$C$10, 3, TRUE) =B15 + C15
12 - 10
Simulation
B1017 =COUNTIF(D15:D1014, 7) B1018 =H1017 / COUNT(D15:D1014) Simulation results will vary with most simulations showing between 155 and 180 7’s. The probability of a 7 should be approximately 0.1667. 16.
Target Answers: a.
Simulation runs will vary. Generally, 340 to 380, or roughly 36% of the simulation runs will show $130,000 to be the highest and winning bid.
b.
$150,000. Profit = $160,000 = $150,000 = $10,000
c.
Again, simulation results will vary. Simulation results should be consistent with the following: Amount Bid $130,000 $140,000 $150,000
Win the Bid 340 to 380 times 620 to 660 times 1000 times
Profit per Win $30,000 $20,000 $10,000
Average Profit Approx. $10,800 Approx. $12,800 $10,000
Using an average profit criterion, both the $130,000 and $140,000 bids are preferred to the $150,000 bid. Of the three alternatives, $140,000 is the recommended bid.
12 - 11
Chapter 12
17.
The Excel worksheet for this problem is as follows:
Selected cell formulas are as follows:
a.
Cell
Formula
B9 C512 D512
=NORMINV(RAND(), $B$4, $B$5) =COUNTIF(B9:B508, “>40000”) =C512 / COUNT(B9:B508)
Most simulations will provide between 105 and 130 tires exceeding 40,000 miles. should be roughly 24%.
b. Mileage 32,000 30,000 28,000
In Most Simulations Number of Tires 80 to 100 42 to 55 18 to 30
12 - 12
Approximate Percentage 18% 10% 4%
The percentage
Simulation
c.
18.
Of mileages considered, 30,000 miles should come closest to meeting the tire guarantee mileage guideline. The Excel worksheet with data in thousands of dollars is as follows:
Selected cell formulas are as follows: Cell
Formula
B10 C10 D10 C1013 D1013
=$B$4 + RAND() * ($B$5-$B$4) =NORMINV(RAND(), $E$4, $E$5) =MAX(B10:C10) =COUNTIF(D10:D1009, “<750000”) =C1013 / COUNT(D10:D1009)
a.
Cell C1013 provides the number of times the contractor’s bid of $750,000 will beat the highest competitive bid shown in column D. Simulation results will vary but the bid of $750,000 should win roughly 600 to 650 of the 1000 times. The probability of winning the bid should be between 0.60 and 0.65.
b.
Cells C1014 and C1015 provide the number of times the bids of $775,000 and $785,000 win. Again, simulation results vary but the probability of $750,000 winning should be roughly 0.82 and the probability of $785,000 winning should be roughly 0.88. Given these results, a contractor’s bid of $775,000 is recommended.
12 - 13
Chapter 12
19.
Butler Inventory simulation spreadsheet. The shortage cost has been eliminated so $0 can be entered in cell C5. Trial replenishment levels of 110, 115, 120 and 125 can be entered in cell C7. Since the shortage cost has been eliminated, Butler can be expected to reduce the replenishment level. This will allow more shortages. However, since the cost of a stockout is only the lost profit and not the lost profit plus a goodwill shortage cost, Butler can permit more shortages and still show an improvement in profit. A replenishment level of 115 should provide a mean profit of approximately $4600. The replenishment levels of 110, 115 and 120 all provide near optimal results.
20.
The Excel worksheet for this problem is as follows:
Selected cell formulas are as follows:
a.
Cell
Formula
B12 C12 D12 E12 F12 G12 H12 C518
=NORMINV(RAND(), $E$4, $E$5) =MIN($B$8, B12) =$B$5*C12 =MAX(0, $B$8 – B12) =E12 * $B$6 =$B$3 + $B$4 * $B$8 =D12 + F12 - G12 =COUNTIF(E12:E511, “=0”)
The simulated mean profit with a production quantity of 60,000 units should be in the $170,000 to $210,000 range. The probability of a stockout is about 0.50.
12 - 14
Simulation
21.
b.
The conservative 50,000 unit production quantity is recommended with a simulated mean profit of approximately $230,000. The more aggressive 70,000 unit production quantity should show a simulated mean profit less than $100,000.
c.
When a 50,000 unit production quantity is used, the probability of a stockout should be approximately 0.75. This is a relative high probability indicating that Mandrell has a good chance of being able to sell all the dolls it produces for the holiday season. As a result, a shortage of dolls is likely to occur. However, this production strategy will enable the company to avoid the high cost associated with selling excess dolls for a loss after the first of the year. The Excel worksheet for this problem is as follows:
Selected cell formulas are as follows: Cell
Formula
B14 C14 D14 E14 F14 G14 B520
=VLOOKUP(RAND(),$A$5:$C$5,3,TRUE) =MIN($F$9, B14) =$F$6*C14 =B14-C14 =$F$7*E14 =D14-F14 =1 – (SUM(E14:E513) / SUM(B14:B513))
12 - 15
Chapter 12
a.
Without overbooking, the problem states that South Central has a mean profit of $2,800 per flight. The overbooking simulation model with a total of 32 reservations (2 overbookings) projects a mean profit of approximately $2925. This is an increase in profit of $125 per flight (4.5%). The overbooking strategy appears worthwhile. The simulation spreadsheet indicates a service level of approximately 99.2% for all passenger demand. This indicates that only 0.8% of the passengers would encounter an overbooking problem. The overbooking strategy up to a total of 32 reservations is recommended.
b.
The same spreadsheet design can be used to simulate other overbooking strategies including accepting 31, 33 and 34 passenger reservations. In each case, South Central would need to obtain data on the passenger demand probabilities. Changing the passenger demand table and rerunning the simulation model would enable South Central to evaluate the other overbooking alternatives and arrive at the most beneficial overbooking policy.
22.
Use the Black Sheep Scarves spreadsheet. Changing the interarrival times to a uniform distribution between 0 and 4 is the only change needed for each spreadsheet. The mean time between arrivals is 2 minutes and the mean service time is 2 minutes. On the surface it appears that there is an even balance between the arrivals and the services. However, since both arrivals and services have variability, simulated system performance with 1 ATM will probably be surprisingly poor. Simulation results can be expected to show some waiting times of 30 minutes or more near the end of the simulation period. One ATM is clearly not acceptable.
23.
Use the Black Sheep Scarves spreadsheet. a.
The interarrival times and service times section of the spreadsheet will need to be modified. Assume that the mean interarrival time of 0.75 is placed in cell B4 and that the mean service time of 1 is placed in cell B8. The following cell formulas would be required. Cell
Formula
B16
=(1/$B$4)*LN(RAND())
F16
=(1/$B$8)*LN(RAND())
The simulation results will vary but most should show an average waiting time in a 2 to 4 minute range. b.
The service time mean and standard deviation would be entered in cells B8 and B9 as in the original Black Sheep 1 spreadsheet. Cell F16 would have its original cell formula =NORMINV(RAND(),$B$8,$B$9). Again simulation results will vary. The lower variability of the normal probability distribution should improve the performance of the waiting line by reducing the average waiting time. An average waiting time in the range 1.4 to 2 minutes should be observed for most simulation runs.
24.
Use the Black Sheep 2 spreadsheet included on the companion website for this book. The interarrival times section of the spreadsheet will need to be modified. Assume that the mean interarrival time of 4 is placed in cell B4. The following cell formula would be placed in cell B16: =(1/$B$4)*LN(RAND()) a. b.
Both the mean interarrival time and the mean service time should be approximately 4 minutes. Simulation results should provide a mean waiting time of approximately .8 minutes (48 seconds).
12 - 16
Simulation
c.
25.
Simulation results should predict approximately 150 to 170 customers had to wait. Generally, the percentage should be 30 to 35%. The Excel worksheet for this problem is as follows:
Cell C27 contains the formula: “=IF(B27=0,0,IF(B27=1,VLOOKUP(RAND(),$A$10:$C$12,3,TRUE),IF(B27=2,VLOOKUP(RAND(),$A$ 16:$C$18,3,TRUE),IF(B27="Unknown",VLOOKUP(RAND(),$A$22:$C$24,3,TRUE)))))” This formula is a nested IF command that checks the number of RSVP’ed Attendees (cell B27) and depending on this value in cell B27, uses a VLOOKUP command in concert with a randomly-generated number between 0 and 1 to generate a random number of Actual Attendees. Similarly, cell C28 through cell 251 contains the same formula as C27 except referencing C27, D27, … respectively instead of B27. Cells F1:H1001 contains the data table used to generate 1000 replications of the wedding reception attendance data. Cell G1 contains “=D27” to link the data table to the total attendance of a single replication.
Cells G2: G1001 contain “{=TABLE(,D3)}” created by the data table function. Cell H2 contains “=IF(G2<=$J$3,1,0)” and cells H3 through H1001 contain the same formula except referencing G3, G4, … respectively instead of G2.
12 - 17
Chapter 12
Cell J2 contains “=AVERAGE(G2:G1001)”, and cell J4 contains “=SUM(H2:H1001)/COUNT(H2:H1001)”. a.
Most simulations will provide an average of about 233 guests.
b.
Most simulations will require approximately X=242 guests to ensure more than a 90% probability that the actual attendance is less than or equal to X. That is, the caterer should prepare for 242 guests in order to have a 90% chance of having enough food.
12 - 18
Chapter 13 Decision Analysis Learning Objectives 1.
Learn how to describe a problem situation in terms of decisions to be made, chance events and consequences.
2.
Be able to analyze a simple decision analysis problem from both a payoff table and decision tree point of view.
3.
Be able to develop a risk profile and interpret its meaning.
4.
Be able to use sensitivity analysis to study how changes in problem inputs affect or alter the recommended decision.
5.
Be able to determine the potential value of additional information.
6.
Learn how new information and revised probability values can be used in the decision analysis approach to problem solving.
7.
Understand what a decision strategy is.
8.
Learn how to evaluate the contribution and efficiency of additional decision making information.
9.
Be able to use a Bayesian approach to computing revised probabilities.
10.
Be able to use TreePlan software for decision analysis problems.
11.
Understand the following terms: decision alternatives chance events states of nature influence diagram payoff table decision tree optimistic approach conservative approach minimax regret approach opportunity loss or regret expected value approach expected value of perfect information (EVPI)
Solutions: 1.
a.
13 - 1
decision strategy risk profile sensitivity analysis prior probabilities posterior probabilities expected value of sample information (EVSI) efficiency of sample information Bayesian revision
Chapter 13
s1 d1
s2 s3 s1
d2
s2 s3
250 100 25 100 100 75
b. Decision d1 d2
Maximum Profit 250 100
Minimum Profit 25 75
Optimistic approach: select d1 Conservative approach: select d2 Regret or opportunity loss table: s1 0 150
d1 d2
s2 0 0
s3 50 0
Maximum Regret: 50 for d1 and 150 for d2; select d1 2.
a. Decision d1 d2 d3 d4
Maximum Profit 14 11 11 13
Minimum Profit 5 7 9 8
Optimistic approach: select d1 Conservative approach: select d3
Regret or Opportunity Loss Table with the Maximum Regret
d1
s1 0
s2 1
s3 1
s4 8
Maximum Regret 8
13 - 2
Decision Analysis d2 d3 d4
3 5 6
0 0 0
3 1 0
6 2 0
6 5 6
Minimax regret approach: select d3 b.
The choice of which approach to use is up to the decision maker. Since different approaches can result in different recommendations, the most appropriate approach should be selected before analyzing the problem.
c. Decision d1 d2 d3 d4
Minimum Cost 5 7 9 8
Maximum Cost 14 11 11 13
Optimistic approach: select d1 Conservative approach: select d2 or d3 Regret or Opportunity Loss Table
d1 d2 d3 d4
s1 6 3 1 0
s2 0 1 1 1
s3 2 0 2 3
s4 0 2 6 8
Maximum Regret 6 3 6 8
Minimax regret approach: select d2 3.
a.
The decision to be made is to choose the best plant size. There are 2 alternatives to choose from: a small plant or a large plant. The chance event is the market demand for the new product line. It is viewed as having 3 possible outcomes (states of nature): low, medium and high.
b.
Influence Diagram: Plant Size
c.
Market Demand
Profit
13 - 3
Chapter 13
Low
Small
Medium
High
Low
Large
Medium
High
150
200
200
50
200
500
d. Decision Small Large
Maximum Profit 200 500
Minimum Profit 150 50
Maximum Regret 300 100
Optimistic approach: select Large plant Conservative approach: select Small plant Minimax regret approach: select Large plant 4.
a.
The decision faced by Amy is to select the best lease option from three alternatives (Hepburn Honda, Midtown Motors, and Hopkins Automotive). The chance event is the number of miles Amy will drive.
b.
The payoff for any combination of alternative and chance event is the sum of the total monthly charges and total additional mileage cost, i.e., for the Hepburn Honda lease option: 36000 miles (12000 miles for 3 years): 36($299) + $0.15(36000 - 36000) = $10,764 45000 miles (15000 miles for 3 years): 36($299) + $0.15(45000 - 36000) = $12,114 54000 miles (18000 miles for 3 years): 36($299) + $0.15(54000 - 36000) = $13,464 for the Midtown Motors lease option: 36000 miles (12000 miles for 3 years): 36($310) + $0.20*max(36000 – 45000,0) = $11,160.00 45000 miles (15000 miles for 3 years): 36($310) + $0.20*max(45000 – 45000,0) = $11,160.00 54000 miles (18000 miles for 3 years): 36($310) + $0.20*max(54000 – 45000,0) = $12,960.00 for the Hopkins Automotive lease option: 36000 miles (12000 miles for 3 years): 36($325) + $0.15*max(36000 – 54000, 0) = $11,700 45000 miles (15000 miles for 3 years): 36($325) + $0.15*max(45000 – 54000, 0) = $11,700 54000 miles (18000 miles for 3 years): 36($325) + $0.15*max(54000 – 54000, 0) = $11,700
So the payoff table for Amy’s problem is:
13 - 4
Decision Analysis
Actual Miles Driven Annually
c.
Dealer
12000
15000
18000
Hepburn Honda
$10,764
$12,114
$13,464
Midtown Motors
$11,160
$11,160
$12,960
Hopkins Automotive
$11,700
$11,700
$11,700
The minimum and maximum payoffs for each of Amy’s three alternatives are: Dealer
Minimum Cost
Maximum Cost
Hepburn Honda
$10,764
$13,464
Midtown Motors
$11,160
$12,960
Hopkins Automotive
$11,700
$11,700
Thus: The optimistic approach results in selection of the Hepburn Automotive lease option (which has the smallest minimum cost of the three alternatives - $10,764). The conservative approach results in selection of the Hopkins Automotive lease option (which has the smallest maximum cost of the three alternatives - $11,700). To find the lease option to select under the minimax regret approach, we must first construct an opportunity loss (or regret) table. For each of the three chance events (driving 12000 miles, driving 15000 miles, driving 18000 miles) subtract the minimum payoff from the payoff for each decision alternative. The regret table for this problem is State of Nature (Actual Miles Driven Annually) Decision Alternative
12000
15000
18000
Maximum Regret
Hepburn Honda
$0
$954
$1,764
$1,764
Midtown Motors
$396
$0
$1,260
$1,260
Hopkins Automotive
$936
$540
$0
$936
The minimax regret approach results in selection of the Hopkins Automotive lease option (which has the smallest regret of the three alternatives: $936).
13 - 5
Chapter 13
d.
We first find the expected value for the payoffs associated with each of Amy’s three alternatives: EV(Hepburn Honda) = 0.5($10,764) + 0.4($12,114) + 0.1($13,464) = $11,574 EV(Midtown Motors) = 0.5($11,160) + 0.4($11,160) + 0.1($12,960) = $11,340 EV(Hopkins Automotive) = 0.5($11,700) + 0.4($11,700) + 0.1($11,700) = $11,700
e.
The expected value approach results in selection of the Midtown Motors lease option (which has the minimum expected value of the three alternatives - $11,340). The risk profile for the decision to lease from Midtown Motors is:
Note that although we have three chance outcomes (drive 12000 miles annually, drive 15000 miles annually, and drive 18000 miles annually), we only have two unique costs on this graph. This is because for this decision alternative (lease from Midtown Motors) there are only two unique payoffs associated with the three chance outcomes – the payoff (cost) associated with the Midtown Motors lease is the same for two of the chance outcomes (whether Amy drives 12000 miles or 15000 miles annually, her payoff is $11,160). f.
We first find the expected value for the payoffs associated with each of Amy’s three alternatives: EV(Hepburn Honda) = 0.3($10,764) + 0.4($12,114) + 0.3($13,464) = $12,114 EV(Midtown Motors) = 0.3($11,160) + 0.4($11,160) + 0.3($12,960) = $11,700 EV(Hopkins Automotive) = 0.3($11,700) + 0.4($11,700) + 0.3($11,700) = $11,700 The expected value approach results in selection of either the Midtown Motors lease option or the Hopkins Automotive lease option (both of which have the minimum expected value of the three alternatives - $11,700).
5.
Using the expected value approach yields the following results: EV(d1) = 0.65(250) + 0.15(100) + 0.20(25) = 182.5
13 - 6
Decision Analysis EV(d2) = 0.65(100) + 0.15(100) + 0.20(75) = 95.0 The optimal decision using the expected value approach is d1. 6.
a.
EV(C) = 0.2(10) + 0.5(2) + 0.3(-4) = 1.8 EV(F) = 0.2(8) + 0.5(5) + 0.3(-3) = 3.2 EV(M) = 0.2(6) + 0.5(4) + 0.3(-2) = 2.6 EV(P) = 0.2(6) + 0.5(5) + 0.3(-1) = 3.4 Pharmaceuticals recommended 3.4%
b.
Using probabilities 0.4, 0.4, 0.2. EV(C) = 4.0 EV(F) = 4.6 EV(M) = 3.6 EV(P) = 4.2 Financial recommended 4.6%
7.
a.
EV(own staff) = 0.2(650) + 0.5(650) + 0.3(600) = 635 EV(outside vendor) = 0.2(900) + 0.5(600) + 0.3(300) = 570 EV(combination) = 0.2(800) + 0.5(650) + 0.3(500) = 635 The optimal decision is to hire an outside vendor with an expected annual cost of $570,000.
b.
The risk profile in tabular form is shown. Cost 300 600 900
Probability 0.3 0.5 0.2 1.0
A graphical representation of the risk profile is also shown:
Probability
0.5 0.4 0.3 0.2 0.1 300
600 Cost
8.
a.
EV(d1) = p(10) + (1 - p) (1) = 9p + 1 EV(d2) = p(4) + (1 - p) (3) = 1p + 3
13 - 7
900
Chapter 13
10
p
1
0 Value of p for which EVs are equal 9p + 1 = 1p + 3 and hence p = .25 d2 is optimal for p 0.25; d1 is optimal for p 0.25. b.
The best decision is d2 since p = 0.20 < 0.25. EV(d1) = 0.2(10) + 0.8(1) = 2.8 EV(d2) = 0.2(4) + 0.8(3) = 3.2
c.
The best decision in part (b) is d2 with EV(d2) = 3.2. Decision d2 will remain optimal as long as its expected value is higher than that for d1 (EV(d1) = 2.8). Let s = payoff for d2 under state of nature s1. Decision d2 will remain optimal provided that EV(d2) = 0.2(s) + 0.8(3) 2.8 0.2s 2.8 - 2.4 0.2s 0.4 s2 As long as the payoff for s1 is 2, then d2 will be optimal.
9.
a.
The decision to be made is to choose the type of service to provide. The chance event is the level of demand for the Myrtle Air service. The consequence is the amount of quarterly profit. There are two decision alternatives (full price and discount service). There are two outcomes for the chance event (strong demand and weak demand).
b. Type of Service Full Price Discount
Maximum Profit $960 $670
Minimum Profit -$490 $320
Optimistic Approach: Full price service Conservative Approach: Discount service
13 - 8
Decision Analysis
Opportunity Loss or Regret Table
Full Service Discount Service
High Demand 0 290
Low Demand 810 0
Maximum Regret 810 290
Minimax Regret Approach: Discount service c.
EV(Full) = 0.7(960) + 0.3(-490) = 525 EV (Discount) = 0.7(670) + 0.3(320) = 565 Optimal Decision: Discount service
d.
EV(Full) = 0.8(960) + 0.2(-490) = 670 EV (Discount) = 0.8(670) + 0.2(320) = 600 Optimal Decision: Full price service
e.
Let p = probability of strong demand EV(Full) = p(960) + (1- p)(-490) = 1450p - 490 EV (Discount) = p(670) + (1- p)(320) = 350p + 320 EV (Full) = EV(Discount) 1450p - 490 = 350p + 320 1100p = 810 p = 810/1100 = 0.7364 If p = 0.7364, the two decision alternatives provide the same expected value. For values of p below 0.7364, the discount service is the best choice. For values of p greater than 0.7364, the full price service is the best choice.
13 - 9
Chapter 13
10. a.
Battle Pacific
2
High 0.2
1000
Medium 0.5
700
Low 0.3
300
1 With Competition 0.6
Space Pirates
800
Medium 0.4
400
Low 0.3
200
High 0.5
1600
Medium 0.3
800
Low 0.2
400
3
Without Competition 0.4
b.
4
High 0.3
5
EV(node 2) = 0.2(1000) + 0.5(700) + 0.3(300) = 640 EV(node 4) = 0.3(800) + 0.4(400) + 0.3(200) = 460 EV(node 5) = 0.5(1600) + 0.3(800) + 0.2(400) = 1120 EV(node 3) = 0.6EV(node 4) + 0.4EV(node 5) = 0.6(460) + 0.4(1120) = 724 Space Pirates is recommended. Expected value of $724,000 is $84,000 better than Battle Pacific.
c.
Risk Profile for Space Pirates Outcome: 1600 800 400 200
(0.4)(0.5) = 0.20 (0.6)(0.3) + (0.4)(0.3) = 0.30 (0.6)(0.4) + (0.4)(0.2) = 0.32 (0.6)(0.3) = 0.18
13 - 10
Decision Analysis
Probability
0.30
0.20
0.10
200
400
800
1600
Profit ($ thousands)
d.
Let p = probability of competition p=0
EV(node 5) = 1120
p=1
EV(node 4) = 460
1120
Expected Value
Space Pirates
640 Battle Pacific
460
0
1 p
1120 - p(1120 - 460) 660p p
= 640 = 480 = 480/660 = 0.7273
The probability of competition would have to be greater than 0.7273 before we would change to the Battle Pacific video game. 11. a.
Currently, the large complex decision is optimal with EV(d3) = 0.8(20) + 0.2(-9) = 14.2. In order for d3 to remain optimal, the expected value of d2 must be less than or equal to 14.2. Let s = payoff under strong demand EV(d2) = 0.8(s) + 0.2(5) 14.2 0.8 s + 1 14.2
13 - 11
Chapter 13 0.8 s 13.2 s 16.5 Thus, if the payoff for the medium complex under strong demand remains less than or equal to $16.5 million, the large complex remains the best decision. b.
A similar analysis is applicable for d1 EV(d1) = 0.8(s) + 0.2(7) 14.2 0.8 s + 1.4 14.2 0.8 s 12.8 s 16 If the payoff for the small complex under strong demand remains less than or equal to $16 million, the large complex remains the best decision.
12. a.
There is only one decision to be made: whether or not to lengthen the runway. There are only two decision alternatives. The chance event represents the choices made by Air Express and DRI concerning whether they locate in Potsdam. Even though these are decisions for Air Express and DRI, they are chance events for Potsdam. The payoffs and probabilities for the chance event depend on the decision alternative chosen. If Potsdam lengthens the runway, there are four outcomes (both, Air Express only, DRI only, neither). The probabilities and payoffs corresponding to these outcomes are given in the tables of the problem statement. If Potsdam does not lengthen the runway, Air Express will not locate in Potsdam so we only need to consider two outcomes: DRI and no DRI. The approximate probabilities and payoffs for this case are given in the last paragraph of the problem statements. The consequence is the estimated annual revenue.
b.
Runway is Lengthened New Air Express Center Yes Yes No No
New DRI Plant Yes No Yes No
Probability 0.3 0.1 0.4 0.2
Annual Revenue $600,000 $150,000 $250,000 -$200,000
EV (Runway is Lengthened) = 0.3($600,000) + 0.1($150,000) + 0.4($250,000) - 0.2($200,000) = $255,000 c.
EV (Runway is Not Lengthened) = 0.6($450,000) + 0.4($0) = $270,000
d.
The town should not lengthen the runway.
e.
EV (Runway is Lengthened) = 0.4(600,000) + 0.1($150,000) + 0.3($250,000) - 0.2(200,000) = $290,000 The revised probabilities would lead to the decision to lengthen the runway.
13. a.
The decision is to choose what type of grapes to plant, the chance event is demand for the wine and the consequence is the expected annual profit contribution. There are three decision alternatives
13 - 12
Decision Analysis
(Chardonnay, Riesling and both). There are four chance outcomes: (W,W); (W,S); (S,W); and (S,S). For instance, (W,S) denotes the outcomes corresponding to weak demand for Chardonnay and strong demand for Riesling. b.
In constructing a decision tree, it is only necessary to show two branches when only a single grape is planted. But, the branch probabilities in these cases are the sum of two probabilities. For example, the probability that demand for Chardonnay is strong is given by: P (Strong demand for Chardonnay) = P(S,W) + P(S,S) = 0.25 + 0.20 = 0.45 Weak for Chardonnay 0.55
20
Plant Chardonnay 2
EV = 42.5 Strong for Chardonnay 0.45
70
Weak for Chardonnay, Weak for Riesling 0.05 Weak for Chardonnay, Strong for Riesling 0.50 1
Plant both grapes
3
22
40
EV = 39.6 Strong for Chardonnay, Weak for Riesling 0.25 Strong for Chardonnay, Strong for Riesling
26
60
0.20 Weak for Riesling 0.30
25
Plant Riesling 4
EV = 39 Strong for Riesling
45
0.70
c.
EV (Plant Chardonnay) = 0.55(20) +0.45(70) = 42.5 EV (Plant both grapes) = 0.05(22) + 0.50(40) + 0.25(26) + 0.20(60) = 39.6 EV (Plant Riesling) = 0.30(25) + 0.70(45) = 39.0 Optimal decision: Plant Chardonnay grapes only.
d.
This changes the expected value in the case where both grapes are planted and when Riesling only is
13 - 13
Chapter 13
planted. EV (Plant both grapes) = 0.05(22) + 0.50(40) +0.05(26) + 0.40(60) = 46.4 EV (Plant Riedling) = 0.10(25) + 0.90(45) = 43.0 We see that the optimal decision is now to plant both grapes. The optimal decision is sensitive to this change in probabilities. e.
Only the expected value for node 2 in the decision tree needs to be recomputed. EV (Plant Chardonnay) = 0.55(20) + 0.45(50) = 33.5 This change in the payoffs makes planting Chardonnay only less attractive. It is now best to plant both types of grapes. The optimal decision is sensitive to a change in the payoff of this magnitude.
14. a.
If s1 then d1 ; if s2 then d1 or d2; if s3 then d2
b.
EVwPI = .65(250) + .15(100) + .20(75) = 192.5
c.
From the solution to Problem 5 we know that EV(d1) = 182.5 and EV(d2) = 95; thus, the recommended decision is d1. Hence, EVwoPI = 182.5.
d.
EVPI = EVwPI - EVwoPI = 192.5 - 182.5 = 10
15. a.
EV (Small)
= 0.1(400) + 0.6(500) + 0.3(660) = 538
EV (Medium) = 0.1(-250) + 0.6(650) + 0.3(800) = 605 EV (Large)
= 0.1(-400) + 0.6(580) + 0.3(990) = 605
Best decision: Build a medium or large-size community center. Note that using the expected value approach, the Town Council would be indifferent between building a medium-size community center and a large-size center. Risk profile for medium-size community center: 0.6
Probability
b.
0.4
0.2
-400
0 400 Net Cash Flow
Risk profile for large-size community center:
13 - 14
800
Decision Analysis
Probability
0.6
0.4
0.2
-400
0 400 Net Cash Flow
800
Given the mayor's concern about the large loss that would be incurred if demand is not large enough to support a large-size center, we would recommend the medium-size center. The large-size center has a probability of 0.1 of losing $400,000. With the medium-size center, the most the town can loose is $250,000. c.
The Town's optimal decision strategy based on perfect information is as follows: If the worst-case scenario, build a small-size center If the base-case scenario, build a medium-size center If the best-case scenario, build a large-size center Using the consultant's original probability assessments for each scenario, 0.10, 0.60 and 0.30, the expected value of a decision strategy that uses perfect information is: EVwPI = 0.1(400) + 0.6(650) + 0.3(990) = 727 In part (a), the expected value approach showed that EV(Medium) = EV(Large) = 605. Therefore, EVwoPI = 605 and EVPI = 727 - 605 = 122 The town should seriously consider additional information about the likelihood of the three scenarios. Since perfect information would be worth $122,000, a good market research study could possibly make a significant contribution.
d.
EV (Small)
= 0.2(400) + 0.5(500) + 0.3(660) = 528
EV (Medium) = 0.2(-250) + 0.5(650) + 0.3(800) = 515 EV (Small)
= 0.2(-400) + 0.5(580) + 0.3(990) = 507
Best decision: Build a small-size community center. e.
If the promotional campaign is conducted, the probabilities will change to 0.0, 0.6 and 0.4 for the worst case, base case and best case scenarios respectively. EV (Small)
= 0.0(400) + 0.6(500) + 0.4(660) = 564
EV (Medium) = 0.0(-250) + 0.6(650) + 0.4(800) = 710 EV (Small)
= 0.0(-400) + 0.6(580) + 0.4(990) = 744
13 - 15
Chapter 13 In this case, the recommended decision is to build a large-size community center. Compared to the analysis in Part (a), the promotional campaign has increased the best expected value by $744,000 605,000 = $139,000. Compared to the analysis in part (d), the promotional campaign has increased the best expected value by $744,000 - 528,000 = $216,000. Even though the promotional campaign does not increase the expected value by more than its cost ($150,000) when compared to the analysis in part (a), it appears to be a good investment. That is, it eliminates the risk of a loss, which appears to be a significant factor in the mayor's decision-making process. 16. a.
s1 d1 F
3
Research
7
2
s2 s1
d1 U
8
4
s2 s1
d2
9
1
s2 s1
d1 No Market
10
5
Research
EV (node 6)
= 0.57(100) + 0.43(300)
=
186
EV (node 7)
= 0.57(400) + 0.43(200)
=
314
EV (node 8)
= 0.18(100) + 0.82(300)
=
264
13 - 16
s2 s1
d2
b.
s2 s1
d2 Market
6
11
s2
Profit Payoff 100
300
400 200
100
300 400
200 100
300 400 200
Decision Analysis EV (node 9)
= 0.18(400) + 0.82(200)
=
236
EV (node 10) = 0.40(100) + 0.60(300)
=
220
EV (node 11) = 0.40(400) + 0.60(200)
=
280
EV (node 3)
= Max(186,314) = 314
d2
EV (node 4)
= Max(264,236) = 264
d1
EV (node 5)
= Max(220,280) = 280
d2
EV (node 2)
= 0.56(314) + 0.44(264)
=
EV (node 1)
= Max(292,280) = 292
292
Market Research If Favorable, decision d2 If Unfavorable, decision d1 17. a.
EV(node 4) = 0.5(34) + 0.3(20) + 0.2(10) = 25 EV(node 3) = Max(25,20) = 25
Decision: Build
EV(node 2) = 0.5(25) + 0.5(-5) = 10 EV(node 1) = Max(10,0) = 10
Decision: Start R&D
Optimal Strategy: Start the R&D project If it is successful, build the facility Expected value = $10M b.
At node 3, payoff for sell rights would have to be $25M or more. In order to recover the $5M R&D cost, the selling price would have to be $30M or more.
c. Possible Profit $34M $20M $10M -$5M 18. a.
(0.5)(0.5) = (0.5)(0.3) = (0.5)(0.2) =
0.25 0.15 0.10 0.50 1.00
Outcome 1 ($ in 000s) Bid Contract Market Research High Demand
-$200 -2000 -150 +5000 $2650
Outcome 2 ($ in 000s) Bid Contract
-$200 -2000
13 - 17
Chapter 13 Market Research Moderate Demand b.
-150 +3000 $650
EV (node 8)
= 0.85(2650) + 0.15(650) = 2350
EV (node 5)
= Max(2350, 1150) = 2350
EV (node 9)
= 0.225(2650) + 0.775(650) = 1100
EV (node 6)
= Max(1100, 1150) = 1150
Decision: Build Decision: Sell
EV (node 10) = 0.6(2800) + 0.4(800)= 2000 EV (node 7)
= Max(2000, 1300) = 2000
Decision: Build
EV (node 4)
= 0.6 EV(node 5) + 0.4 EV(node 6) = 0.6(2350) + 0.4(1150) = 1870
EV (node 3)
= MAX (EV(node 4), EV (node 7)) = Max (1870, 2000) = 2000 Decision: No Market Research
EV (node 2)
= 0.8 EV(node 3) + 0.2 (-200) = 0.8(2000) + 0.2(-200) = 1560
EV (node 1)
= MAX (EV(node 2), 0) = Max (1560, 0) = 1560 Decision: Bid on Contract
Decision Strategy: Bid on the Contract Do not do the Market Research Build the Complex Expected Value is $1,560,000 c.
Compare Expected Values at nodes 4 and 7. EV(node 4) = 1870 Includes $150 cost for research EV (node 7) = 2000 Difference is 2000 - 1870 = $130 Market research cost would have to be lowered $130,000 to $20,000 or less to make undertaking the research desirable.
d.
Shown below is the reduced decision tree showing only the sequence of decisions and chance events for Dante's optimal decision strategy. If Dante follows this strategy, only 3 outcomes are possible with payoffs of -200, 800, and 2800. The probabilities for these payoffs are found by multiplying the probabilities on the branches leading to the payoffs. A tabular presentation of the risk profile is: Payoff ($million) -200 800 2800
Probability .20 (.8)(.4) = .32 (.8)(.6) = .48
Reduced Decision Tree Showing Only Branches for Optimal Strategy Win Contract 0.8 Bid
3 Build Complex
2 No Market Research
13 - 18
1 Lose Contract 0.2
High Demand 0.6
2800
Moderate Demand 0.4
800
10
7
-200
Decision Analysis
19. a. s1 d1
6
s2 s3
-100 50 150
Favorable 3 s1 d2 7
s2 s3
Agency
100 100 100
2 s1 d1
8
s2 s3
-100 50 150
Unfavorable 4 s1 d2 9
s2 s3
1
s1 d1
10
s2 s3
No Agency
s1 11
s2 s3
Using node 5, EV (node 10) = 0.20(-100) + 0.30(50) + 0.50(150)
= 70
EV (node 11) = 100 Decision Sell Expected Value = $100 c.
EVwPI = 0.20(100) + 0.30(100) + 0.50(150) = $125 EVPI
d.
100 100
-100 50 150
5 d2
b.
100
= $125 - $100 = $25
EV (node 6) = 0.09(-100) + 0.26(50) + 0.65(150) = 101.5 EV (node 7) = 100
13 - 19
100 100 100
Chapter 13 EV (node 8) = 0.45(-100) + 0.39(50) + 0.16(150) = -1.5 EV (node 9) = 100 EV (node 3) = Max(101.5,100)
=
101.5
Produce
EV (node 4) = Max(-1.5,100)
=
100
Sell
EV (node 2) = 0.69(101.5) + 0.31(100) = 101.04 If Favorable, Produce If Unfavorable, Sell EV = $101.04 e.
EVSI = $101.04 - 100 = $1.04 or $1,040.
f.
No, maximum Hale should pay is $1,040.
g.
No agency; sell the pilot.
20. a. Accept
Favorable 0.7
Success 0.75
750
Failure 0.25
-250
7
4 Reject
Review
0
2 Accept
Unfavorable 0.3
Success 0.417
750
Failure 0.583
-250
8
5 Reject
0
1
Accept
Do Not Review
750
Failure 0.35
-250
6
3 Reject
b.
Success 0.65
0
EV (node 7) = 0.75(750) + 0.25(-250) = 500 EV (node 8) = 0.417(750) + 0.583(-250) = 167
13 - 20
Decision Analysis
Decision (node 4) → Accept EV = 500 Decision (node 5) → Accept EV = 167 EV(node 2) = 0.7(500) + 0.3(167) = $400 Note: Regardless of the review outcome F or U, the recommended decision alternative is to accept the manuscript. EV(node 3) = .65(750) + .35(-250) = $400 The expected value is $400,000 regardless of review process. The company should accept the manuscript. c.
The manuscript review cannot alter the decision to accept the manuscript. Do not do the manuscript review.
d.
Perfect Information. If s1, accept manuscript $750 If s2, reject manuscript -$250 EVwPI = 0.65(750) + 0.35(0) = 487.5 EVwoPI = 400 EVPI = 487.5 - 400 = 87.5 or $87,500. A better procedure for assessing the market potential for the textbook may be worthwhile.
21.
The decision tree is as shown in the answer to problem 16a. The calculations using the decision tree in problem 16a with the probabilities and payoffs here are as follows: a,b. EV (node 6)
= 0.18(600) + 0.82(-200) =
EV (node 7)
= 0
EV (node 8)
= 0.89(600) + 0.11(-200) =
EV (node 9)
= 0
EV (node 10) = 0.50(600) + 0.50(-200) =
-56 512 200
EV (node 11) = 0 EV (node 3) EV (node 4) EV (node 5)
= Max(-56,0) = = Max(512,0) = = Max(200,0) =
0 512 200
d2 d1 d1
EV (node 2)
= 0.55(0) + 0.45(512) = 230.4
Without the option, the recommended decision is d1 purchase with an expected value of $200,000. With the option, the best decision strategy is If high resistance H, d2 do not purchase If low resistance L, d1 purchase
13 - 21
Chapter 13 Expected Value = $230,400 c. 22. a.
EVSI = $230,400 - $200,000 = $30,400. Since the cost is only $10,000, the investor should purchase the option. EV (1 lot)
= 0.3(60) + 0.3(60) + 0.4(50)
= 56
EV (2 lots) = 0.3(80) + 0.3(80) + 0.4(30)
= 60
EV (3 lots) = 0.3 (100) + 0.3(70) + 0.4(10) = 55 Decision: Order 2 lots Expected Value $60,000 b.
The following decision tree applies.
s1 d1
6
s2 s3 s1
d2
Excellent
7
3
s3 s1
d3 8 V.P. Prediction
s2
s2 s3
2
s1 d1
9
s2 s3
Very Good
s1
d2 10
4
s2 s3 s1
d3 11
1
s2 s3 s1
d1
12
s2 s3
No V.P. Prediction
s1
d2 13
5
s2 s3 s1
d3 14
s2 s3
13 - 22
60 60 50 80 80 30 100 70 10 60 60 50 80 80 30 100 70 10 60 60 50 80 80 30 100 70 10
Decision Analysis
Calculations EV (node 6)
= 0.34(60) + 0.32(60) + 0.34(50)
=
56.6
EV (node 7)
= 0.34(80) + 0.32(80) + 0.34(30)
=
63.0
EV (node 8)
= 0.34(100) + 0.32(70) + 0.34(10) =
59.8
EV (node 9)
= 0.20(60) + 0.26(60) + 0.54(50)
=
54.6
EV (node 10) = 0.20(80) + 0.26(80) + 0.54(30)
=
53.0
EV (node 11) = 0.20(100) + 0.26(70) + 0.54(10) =
43.6
EV (node 12) = 0.30(60) + 0.30(60) + 0.40(50)
=
56.0
EV (node 13) = 0.30(80) + 0.30(80) + 0.40(30)
=
60.0
EV (node 14) = 0.30(100) + 0.30(70) + 0.40(10) =
55.0
EV (node 3)
= Max(56.6,63.0,59.8) = 63.0
2 lots
EV (node 4)
= Max(54.6,53.0,43.6) = 54.6
1 lot
EV (node 5)
= Max(56.0,60.0,55.0) = 60.0
2 lots
EV (node 2)
= 0.70(63.0) + 0.30(54.6) =
60.5
EV (node 1)
= Max(60.5,60.0) = 60.5 Prediction
Optimal Strategy: If prediction is excellent, 2 lots If prediction is very good, 1 lot c.
EVwPI = 0.3(100) + 0.3(80) + 0.4(50) = 74 EVPI
= 74 - 60 = 14
EVSI
= 60.5 - 60 = 0.5
Efficiency =
EVSI 0.5 (100) = (100) = 3.6% EVPI 14
The V.P.’s recommendation is only valued at EVSI = $500. The low efficiency of 3.6% indicates other information is probably worthwhile. The ability of the consultant to forecast market conditions should be considered. 23. State of Nature s1 s2 s3
24.
P(sj) 0.2 0.5 0.3 1.0
P(I sj) 0.10 0.05 0.20 P(I) =
a. Using Bayes’ Theorem:
13 - 23
P(I sj) 0.020 0.025 0.060 0.105
P(sj I) 0.1905 0.2381 0.5714 1.0000
Chapter 13
b. d1 0.695C
d1 Weather Check
2
0.215O
d1 0.09R
No Weather Check
EV (node 7) = 30 EV (node 8) = 0.98(25) + 0.02(45) = 25.4
13 - 24
30
s1
0.98
25
s2 s1
0.02
s2
0.21
s1
0.79
s2
0.21
s1
0.00
s2
1.00
s1
0.00
s2
1.00
s1
0.85
s2
0.15
s1
0.85
s2
0.15
0.79
45 30
30 25
45 30
11 30 25
12 45 30
13
6 d2
0.02
10
1 d1
s2
9
5 d2
30
8
4 d2
0.98
7
3 d2
s1
30 25
14 45
Decision Analysis EV (node 9) = EV (node 10) = EV (node 11) = EV (node 12) = EV (node 13) = EV (node 14) =
30 0.79(25) + 0.21(45) = 29.2 30 0.00(25) + 1.00(45) = 45.0 30 0.85(25) + 0.15(45) = 28.0
EV (node 3) EV (node 4) EV (node 5) EV (node 6)
Min(30,25.4) = Min(30,29.2) = Min(30,45) = Min(30,28) =
= = = =
25.4 29.2 30.0 28.0
Expressway Expressway Queen City Expressway
EV (node 2) = 0.695(25.4) + 0.215(29.2) + 0.09(30.0) = 26.6 EV (node 1) = Min(26.6,28) = 26.6 Weather c. Optimal strategy: Check the weather, take the expressway unless there is rain. If rain, take Queen City Avenue. Expected time: 26.6 minutes. 25. a.
d1 = Manufacture component d2 = Purchase component
s1 = Low demand s2 = Medium demand s3 = High demand
s1 d1
.35 2
s2 .35
s3 .30
1
s1 .35
d2
3
s2 .35
s3 .30
EV(node 2) = (0.35)(-20) + (0.35)(40) + (0.30)(100) = 37 EV(node 3) = (0.35)(10) + (0.35)(45) + (0.30)(70) = 40.25 Recommended decision: d2 (purchase component) b.
Optimal decision strategy with perfect information: If s1 then d2 If s2 then d2
13 - 25
-20 40 100 10 45 70
Chapter 13
If s3 then d1 Expected value of this strategy is 0.35(10) + 0.35(45) + 0.30(100) = 49.25 EVPI = 49.25 - 40.25 = 9 or $9,000 c.
If F - Favorable
State of Nature s1 s2 s3
P(sj) 0.35 0.35 0.30
P(F sj)
P(F sj)
0.10 0.40 0.60 P(F) =
0.035 0.140 0.180 0.355
P(sj) 0.35 0.35 0.30
P(U sj)
P(U sj)
P(sj F) 0.0986 0.3944 0.5070
If U - Unfavorable State of Nature s1 s2 s3
0.90 0.60 0.40 P(U) =
0.315 0.210 0.120 0.645
P(sj U) 0.4884 0.3256 0.1860
The probability the report will be favorable is P(F ) = 0.355 d.
Assuming the test market study is used, a portion of the decision tree is shown below.
13 - 26
Decision Analysis
s1 d1
4
s2 s3
F
2
s1 d2
5
s2 s3
1
s1 d1
6
s2 s3
U
3
s1 d2
7
s2 s3
Summary of Calculations Node 4 5 6 7
Expected Value 64.51 54.23 21.86 32.56
Decision strategy: If F then d1 since EV(node 4) > EV(node 5) If U then d2 since EV(node 7) > EV(node 6) EV(node 1) = 0.355(64.51) + 0.645(32.56) = 43.90 e.
With no information: EV(d1) = 0.35(-20) + 0.35(40) + 0.30(100) = 37 EV(d2) = 0.35(10) + 0.35(45) + 0.30(70) = 40.25
13 - 27
-20 40 100 10 45 70 -20 40 100 10 45 70
Chapter 13 Recommended decision: d2 f.
Optimal decision strategy with perfect information: If s1 then d2 If s2 then d2 If s3 then d1 Expected value of this strategy is 0.35(10) + 0.35(45) + 0.30(100) = 49.25 EVPI = 49.25 - 40.25 = 9 or $9,000 Efficiency = (3650 / 9000)100 = 40.6%
26. a.
EV(d1) = 10,000 EV(d2) = 0.96(0) + 0.03(100,000) + 0.01(200,000) = 5,000 Using EV approach, we should choose No Insurance (d2).
b.
Lottery: p = probability of a $0 Cost 1 - p = probability of a $200,000 Cost
c.
Insurance No Insurance
s1 None 9.9 10.0
d1 d2
s2 Minor 9.9 6.0
s3 Major 9.9 0.0
EU(d1) = 9.9 EU(d2) = 0.96(10.0) + 0.03(6.0) + 0.01(0.0) = 9.78 Using EU approach → Insurance (d1)
d.
27. a.
Use expected utility approach. The EV approach results in a decision that can be very risky since it means that the decision maker could lose up to $200,000. Most decision makers (particularly those considering insurance) are risk averse. P(Win) = 1/250,000
P(Lose) = 249,999/250,000
EV(d1) = 1/250,000(300,000) + 249,999/250,000(-2) = -0.80 EV(d2) = 0 Therefore, the best decision under the EV approach is d2 - Do not purchase lottery ticket. b.
Purchase
s1 Win 10
d1
13 - 28
s2 Lose 0
Decision Analysis d2
Do Not Purchase
0.00001
0.00001
EU(d1) = 1/250,000(10) + 249,999/250,000(0) = 0.00004 EU(d2 ) = 0.00001 Therefore, the best decision under the EV approach is d1 - purchase lottery ticket. 28. a.
1.0 .9 C
Probability
.8 .7
A
.6
B
.5 .4 .3 .2 .1 -100
-50
0 Payoff
b.
A - Risk avoider B - Risk taker C - Risk neutral
c.
Risk avoider A, at $20 payoff p = 0.70 Thus, EV(Lottery) = 0.70(100) + 0.30(-100) = $40 Therefore, will pay 40 - 20 = $20 Risk taker B, at $20 payoff p = 0.45 Thus, EV(Lottery) = 0.45(100) + 0.55(-100) = -$10 Therefore, will pay 20 - (-10) = $30
29.
Decision Maker A
EU(d1 ) = 0.25(7.0) + 0.50(9.0) + 0.25(5.0) = 7.5 d1 EU(d 2 ) = 0.25(9.5) + 0.50(10.0) + 0.25(0.0) = 7.375 Decision Maker B
13 - 29
50
100
Chapter 13
EU(d1 ) = 0.25(4.5) + 0.50(6.0) + 0.25(2.5) = 4.75 d2 EU(d 2 ) = 0.25(7.0) + 0.50(10.0) + 0.25(0.0) = 6.75 Decision Maker C
EU(d1 ) = 0.25(6.0) + 0.50(7.5) + 0.25(4.0) = 6.175 d2 EU(d 2 ) = 0.25(9.0) + 0.50(10.0) + 0.25(0.0) = 7.25 30.
31.
Monetary Payoff, x
Utility, U(x)
-200
-1.226
-100
-0.492
0
0.000
100
0.330
200
0.551
300
0.699
400
0.798
500
0.865
a. The exponential utility function for this decision maker is b. The utility function values and plot of x
U(x)
-20000
-1.226
-15000
-0.822
-10000
-0.49
-5000
-0.221
0
0
5000
0.181
10000
0.330
15000
0.451
20000
0.551
25000
0.632
30000
0.699
35000
0.753
13 - 30
. is shown below.
Decision Analysis
Utility, U(x)
1 0.5 0 10000
0
-30000 -20000 -10000
20000
30000
-0.5
40000 x
-1 -1.5
c. The decision maker is risk averse. d. The plots of the two utility functions appear below. We see that the new utility function (represented by the dashed curve) is “flatter” than the utility function from part b. This means that, while the decision maker is still, in general, risk averse, she is willing to accept more risk than the decision maker in part b. Therefore, the decision maker here is being more risk seeking (less risk averse) than in part b.
Utility, U(x)
1 0.5
0
-30000 -20000 -10000
0
-0.5 -1 -1.5
13 - 31
10000
20000
30000
40000 x
Chapter 14 Multicriteria Decision Problems Learning Objectives 1.
Understand the concept of multicriteria decision making and how it differs from situations and procedures involving a single criterion.
2.
Be able to develop a goal programming model of a multiple criteria problem.
3.
Know how to use the goal programming graphical solution procedure to solve goal programming problems involving two decision variables.
4.
Understand how the relative importance of the goals can be reflected by altering the weights or coefficients for the decision variables in the objective function.
5.
Know how to develop a solution to a goal programming model by solving a sequence of linear programming models using a general purpose linear programming package.
6.
Know what a scoring model is and how to use it to solve a multicriteria decision problem.
7.
Understand how a scoring model uses weights to identify the relative importance of each criterion.
8.
Know how to apply the analytic hierarchy process (AHP) to solve a problem involving multiple criteria.
9.
Understand how AHP utilizes pairwise comparisons to establish priority measures for both the criteria and decision alternatives.
10.
Understand the following terms: multicriteria decision problem goal programming deviation variables priority levels goal equation preemptive priorities scoring model
analytic hierarchy process (AHP) hierarchy pairwise comparison matrix synthesization consistency consistency ratio
14 - 1
Chapter 14
Solutions: 1.
a. Raw Material
Amount Needed to Achieve Both P1 Goals
1
2/ (30) + 1/ (15) = 12 + 7.5 = 19.5 5 2 1/ (15) = 3 5
2
3/ (30) + 3/ (15) = 18 + 4.5 = 22.5 5 10
3
Since there are only 21 tons of Material 3 available, it is not possible to achieve both goals. b.
Let x1 = the number of tons of fuel additive produced x2 = the number of tons of solvent base produced d 1+ = the amount by which the number of tons of fuel additive produced exceeds the target value of 30 tons − d 1 = the amount by which the number of tons of fuel additive produced is less than the target of 30 tons + d 2 = the amount by which the number of tons of solvent base produced exceeds the target value of 15 tons − d 2 = the amount by which the number of tons of solvent base is less than the target value of 15 tons Min
d 1−
+
d 2−
x1
+
1 x 2 2 1 x 5 2 3 x 10 2
s.t. 2 5 3
5
x1
+
x1 x1 , x2 ,
14 - 2
Material 1
5
Material 2
21
Material 3
d 1−
=
30
Goal 1
-
d
+ 2
+
d
− 2
=
15
Goal 2
d ,
− 1
+ 2
d
− 2
0
+ 1
d 1+
20
+
x2
-
d ,
d ,
Multicriteria Decision Problems
c.
In the graphical solution, point A minimizes the sum of the deviations from the goals and thus provides the optimal product mix.
x2 70
Tons of Solvent Base
Goal 1
l3 teria Ma
60 50
Ma teri al 1
40 30
Material 2 20
Goal 2
10
A (30, 10)
B (27.5, 15) 0
10
20
30
40
50
x1
Tons of Fuel Additive
2.
d.
In the graphical solution shown above, point B minimizes 2d1− + d 2− and thus provides the optimal product mix.
a.
Let x1 = number of shares of AGA Products purchased x2 = number of shares of Key Oil purchased To obtain an annual return of exactly 9% 0.06(50)x1 + 0.10(100)x2 = 0.09(50,000) 3x1 + 10x2 = 4500 To have exactly 60% of the total investment in Key Oil 100x2 = 0.60(50,000) x2 = 300 Therefore, we can write the goal programming model as follows: Min
P1( d 1− )
+
50x1 3x1
+ +
P2( d 2+ )
s.t. 100x2 10x2 -
d
x2 -
d 2+
+ 1
b.
− 1
+ 2
+ 1
+ d 1− + d 2− − 2
x1 , x2 , d , d , d , d 0 In the graphical solution shown below, x1 = 250 and x2 = 375.
14 - 3
=
50,000 4,500
=
300
Funds Available P1 Goal P2 Goal
Chapter 14
3.
a.
Let x1 = number of units of product 1 produced x2 = number of units of product 2 produced Min
P1( d 1+ )
+ P1( d 1− )
+
P1( d 2+ )
+
P1( d 2− )
+ P2( d 3− )
+
1x2
-
d 1+
+
d 1−
=
350 Goal 1
+
5x2
-
d
+ 2
+
d
− 2
=
1000 Goal 2
+
2x2
d
+ 3
+
d
− 3
=
1300 Goal 3
+ 2
− 2
+ 3
0
s.t. 1x1 2x1 4x1
+ 1
− 1
− 3
x1 , x2 , d , d , d , d , d , d
14 - 4
Multicriteria Decision Problems b.
In the graphical solution, point A provides the optimal solution. Note that with x1 = 250 and x2 = 100, this solution achieves goals 1 and 2, but underachieves goal 3 (profit) by $100 since 4(250) + 2(100) = $1200.
x2 700 600 500 al3 Go
400 300
Go al1
200
Goal 2
B (281.25, 87.5)
100 A (250, 100) 0
100
200
14 - 5
300
400
500
x1
Chapter 14
c. Max s.t.
4x1
+
2x2
1x1 2x1
+ + x1 ,
1x2 5x2 x2
350 1000 0
Dept. A Dept. B
The graphical solution indicates that there are four extreme points. The profit corresponding to each extreme point is as follows: Extreme Point 1 2 3 4
Profit 4(0) + 2(0) = 0 4(350) + 2(0) = 1400 4(250) + 2(100) = 1200 4(0) + 2(200) = 400
Thus, the optimal product mix is x1 = 350 and x2 = 0 with a profit of $1400.
x2
.
400
tA en rtm pa De
300
4 (0,250) Dep artm ent B
200
3 (250,100)
100 Feasible Region (0,0) 1
0
100
200
300
2
x1 400
500
(350,0) d.
The solution to part (a) achieves both labor goals, whereas the solution to part (b) results in using only 2(350) + 5(0) = 700 hours of labor in department B. Although (c) results in a $100 increase in profit, the problems associated with underachieving the original department labor goal by 300 hours may be more significant in terms of long-term considerations.
e.
Refer to the graphical solution in part (b). The solution to the revised problem is point B, with x1 = 281.25 and x2 = 87.5. Although this solution achieves the original department B labor goal and the profit goal, this solution uses 1(281.25) + 1(87.5) = 368.75 hours of labor in department A, which is 18.75 hours more than the original goal.
4.
a.
Let
14 - 6
Multicriteria Decision Problems x1 = number of gallons of IC-100 produced x2 = number of gallons of IC-200 produced P1( d 1− )
Min
+ P1( d 2+ )
+ P2( d 3− )
+ P2( d 4− )
+
P3( d 5− )
+
d 1−
=
4800
Goal 1
+
d
− 2
=
6000
Goal 2
d
− 3
=
100
Goal 3
d
− 4
=
120
Goal 4
d
− 5
=
300
Goal 5
s.t. 20x1 20x1
+
30x2
+
30x2
x1
-
d 1+
-
d
+ 2
d
+ 3
d
+ 4
d
+ 5
x2
x1
+
x2
-
+ + +
x1, x2, all deviation variables 0 b.
In the graphical solution, the point x1 = 120 and x2 = 120 is optimal.
x2 Goal 3 300
Go al 5
200
Go al 2
Solution points that satisfy goals 1-4
Optimal solution (120, 120)
Go al 1
Goal 4 100
x1 0 5.
a.
100
May:
x1 − s1
= 200
June:
s1 + x2 − s2
= 600
July:
s2 + x3 − s3
= 600
August:
s3 + x4
= 600
200
300
(no need for ending inventory)
b.
14 - 7
Chapter 14
c.
May to June:
x2 − x1 − d1+ + d1− = 0
June to July:
x3 − x2 − d 2+ + d 2− = 0
July to August:
x4 − x3 − d3+ + d3− = 0
No. For instance, there must be at least 200 pumps in inventory at the end of May to meet the June requirement of shipping 600 pumps. The inventory variables are constrained to be nonnegative so we only need to be concerned with positive deviations.
d.
e.
June:
s1 − d 4+ = 0
July:
s2 − d5+ = 0
August:
s3 − d 6+ = 0
Production capacity constraints are needed for each month. May:
x1 500
June:
x2 400
July:
x3 800
August:
x4 500
Min d1+ + d1− + d 2+ + d 2− + d3+ + d3− + d 4+ + d 5+ + d 6+ s.t. 3 Goal equations in (b) 3 Goal equations in (c) 4 Demand constraints in (a) 4 Capacity constraints in (d) x1 , x2 , x3 , x4 , s1 , s 2 , s 3 , d1+ , d1− , d 2+ , d 2− , d3+ , d3− , d 4+ , d5+ , d 6+ 0
Optimal Solution: x1 = 400, x2 = 400, x3 = 700, x4 = 500, s1 = 200, s 2 = 0, s 3 = 100, d 2+ = 300, d3− = 200, d 4+ = 200, d 6+ = 100
f.
Yes. Note in part (c) that the inventory deviation variables are equal to the ending inventory variables. So, we could eliminate those goal equations and substitute s1 , s 2 , and s 3 for d 4+ , d 5+ and d 6+ in the objective function. In this case the inventory variables themselves represent the deviations
from the goal of zero. 6.
a.
Note that getting at least 10,000 customers from group 1 is equivalent to x1 = 40,000 (25% of 40,000 = 10,000) and getting 5,000 customers is equivalent to x2 = 50,000 (10% of 50,000 = 5,000). Thus, to satisfy both goals, 40,000 + 50,000 = 90,000 letters would have to be mailed at a cost of 90,000($1) = $90,000.
14 - 8
Multicriteria Decision Problems
Let x1 = number of letters mailed to group 1 customers x2 = number of letters mailed to group 2 customers d 1+ = number of letters mailed to group 1 customers over the desired 40,000
d 1− = number of letters mailed to group 1 customers under the desired 40,000 d 2+ = number of letters mailed to group 2 customers over the desired 50,000 d 2− = number of letters mailed to group 2 customers under the desired 50,000
d 3+ = the amount by which the expenses exceeds the target value of $70,000 d 3− = the amount by which the expenses falls short of the target value of $70,000
Min
P1( d 1− )
+ P1( d 2− )
+
P2( d 3+ )
-
d 1+
s.t. x1
+
d 1−
x2 1x1 +
1x2
-
d
+ 3
+
d
1d
+ 2
+
1d
− 2
− 3
=
40,000
Goal 1
=
50,000
Goal 2
=
70,000
Goal 3
x1, x2, d 1+ , d 1− , d 2+ , d 2− , d 3+ , d 3− 0 b.
Optimal Solution: x1 = 40,000, x2 = 50,000
c.
Objective function becomes min P1( d 1− ) + P1(2 d 2− ) + P2( d 3+ ) Optimal solution does not change since it is possible to achieve both goals 1 and 2 in the original problem.
7.
a.
Let x1 = number of TV advertisements x2 = number of radio advertisements x3 = number of newspaper advertisements Min
P1( d 1− )
+ P2( d 2− )
+ P3( d 3+ )
+ P4( d 4+ )
s.t.
d +1
+
d 1−
=
+
d 2−
=
0 Goal 2
+
d
− 3
=
0 Goal 3
d
− 4
=
200 Goal 4
x1 x2 20x1
+
5x2 +
x3 10x3 -
0.7x1
-
0.3x2 -
0.3x3 -
d 2+
+
0.8x2 -
0.2x3 -
d
+ 3
+
4x2 +
5x3 -
d
+ 4
-0.2x1 25x1
+
x1, x2, x3, d 1+ , d 1− , d 2+ , d 2− , d 3+ , d 3− , d 4+ , d 4− 0
14 - 9
10 15 20 400
TV Radio Newspaper Goal 1
Chapter 14 b. The optimal solution to the Priority Level 4 problem shown above is: x1 = 9.474, x2 = 2.105, x3 = 20. Note that this requires solving the problem for Priority Levels 1-3, adding the following constraints to the above formulation: d 1− = 0, d 2− = 0, and d 3+ = 0 and changing the objective to simply Min d 4+ . Rounding down leads to a recommendation of 9 TV advertisements, 2 radio advertisements, and 20 newspaper advertisements. Note, however, that rounding down results in not achieving goals 1 and 2. 8.
Let x1 x2
= =
d i+ =
first coordinate of the new machine location second coordinate of the new machine location amount by which x1 coordinate of new machine exceeds x1 coordinate of machine i
− i
=
amount by which x1 coordinate of machine i exceeds x1 coordinate of new machine
+ i
e
=
amount by which x2 coordinate of new machine exceeds x2 coordinate of machine i
− i
=
amount by which x2 coordinate of machine i exceeds x2 coordinate of new machine
d e
The goal programming model is given below. d –1 + d +1 + e –1 + e +1 + d –2 + d +2 + e –2 + e +2 + d –3 + d +3 + e –3 + e +3
Min s.t.
+ d–
x1
1
x2
d +1
= 1 + e –1
= 7
e +1 + d–
x1
2
= 5
d +2 + e–
x2
2
= 9
e +2 + d–
x1
3
d +3
+ e– - e+ = 2 3 3
x2 x1 , x2 , d –1 , d +1 , e –1 , e +1 , d –2 , d +2 , e –2 , e +2 , d –3 , d +3 , e –3 , e +3 0
b.
= 6
The optimal solution is given by x1 = 5 x2 = 7 d i+ = 4 e2− = 2 d 3− = 1
e3+ = 5
The value of the solution is 12.
9.
14 - 10
Multicriteria Decision Problems A
1 2 3 4 5 6 7 8 9 10 11 12
Scoring Calculations
B
Criteria Career Advancement Location Management Salary Prestige Job Security Enjoy the Work
C
D
E
Analyst Chicago 35 10 30 28 32 8 28
Accountant Denver 20 12 25 32 20 10 20
Auditor Houston 20 8 35 16 24 16 20
171
139
139
Score
The analyst position in Chicago is recommended. The overall scores for the accountant position in Denver and the auditor position in Houston are the same. There is no clear second choice between the two positions. 10.
A
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
B
C
D
E
Weight 4 3 5 4 3 5 4
Georgetown Kentucky 7 6 7 6 5 6 7
Ratings Marysville Ohio 4 5 8 7 7 8 6
Clarksville Tennessee 5 8 6 5 4 5 5
Georgetown Kentucky 28 18 35 24 15 30 28
Marysville Ohio 16 15 40 28 21 40 24
Clarksville Tennessee 20 24 30 20 12 25 20
178
184
151
Kenyon Manufacturing P lant Location
Criteria Land Cost Labor Cost Labor Availability Construction Cost Transportation Access to Customers Long Range Goals Scoring Calculations Criteria Land Cost Labor Cost Labor Availability Construction Cost Transportation Access to Customers Long Range Goals Score
Marysville, Ohio (184) is the leading candidate. However, Georgetown, Kentucky is a close second choice (178). Kenyon Management may want to review the relative advantages and disadvantages of these two locations one more time before making a final decision. 11. A 1 2 3 4 5 6 7 8 9 10
B
C
Criteria Travel Distance Vacation Cost Entertainment Available Outdoor Activities Unique Experience Family Fun
D
E
Myrtle Beach South Carolina 10 25 21 18 24 40
Smokey Mountains 14 30 12 12 28 35
Branson Missouri 6 20 24 10 32 35
138
131
127
Score
Myrtle Beach is the recommended choice.
14 - 11
Chapter 14
12.
A 1 2 3 4 5 6 7 8 9 10 11 12
Criteria School Prestige Number of Students Average Class Size Cost Distance From Home Sports Program Housing Desirability Beauty of Campus
B
C
D
E
F
Midwestern University 24 12 20 25 14 36 24 15
State College at Newport 18 20 25 40 16 20 20 9
Handover College 21 32 40 15 14 16 28 24
Techmseh State 15 28 35 30 12 24 24 15
170
168
190
183
Score
Handover College is recommended. However Tecumseh State is the second choice and is less expensive than Handover. If cost becomes a constraint, Tecumseh State may be the most viable alternative. 13.
A
1 2 3 4 5 6 7 8 9 10 11 12
Criteria Cost Location Appearance Parking Floor P lan Swimming Pool View Kitchen Closet Space
B
C
D
E
Park Shore 25 28 35 10 32 7 15 32 18
The Terrace 30 16 20 16 28 2 12 28 24
Gulf View 25 36 35 10 20 3 27 24 12
202
176
192
Score
Park Shore is the preferred condominium. 14. a. 1 2 3 4 5 6 7 8 9 10 11
A
Criteria Cost Overnight Capability Kitchen/Bath Facilities Appearance Engine/Speed Towing/Handling Maintenance Resale Value
B
C
D
E
220 Bowrider 40 6 2 35 30 32 28 21
230 Overnighter 25 18 8 35 40 20 20 15
240 Sundancer 15 27 14 30 20 8 12 18
194
181
144
Score
Clark Anderson prefers the 220 Bowrider.
14 - 12
Multicriteria Decision Problems
b. 1 2 3 4 5 6 7 8 9 10 11
A
B
Criteria Cost Overnight Capability Kitchen/Bath Facilities Appearance Engine/Speed Towing/Handling Maintenance Resale Value
C
D
230 Overnighter 18 30 15 28 10 12 5 12
240 Sundancer 15 40 35 28 6 4 4 12
91
130
144
Score
Julie Anderson prefers the 240 Sundancer. 15.
Synthesization Step 1: Column totals are 8, 10/3, and 7/4 Step 2: Price Accord Saturn Cavalier
Accord 1/8 3/8 4/8
Saturn 1/10 3/10 6/10
Cavalier 1/7 2/7 4/7
Accord 0.125 0.375 0.500
Saturn 0.100 0.300 0.600
Cavalier 0.143 0.286 0.571
Step 3: Price Accord Saturn Cavalier
E
220 Bowrider 21 5 5 20 8 16 6 10
Consistency Ratio Step 1: 1 1/ 3 1/ 4 0.123 3 + 0.320 1 + 0.557 1/ 2 4 2 1 0.123 0.107 0.139 0.369 0.369 + 0.320 + 0.279 = 0.967 0.492 0.640 0.557 1.688
Step 2:
0.369/0.123 = 3.006 0.967/0.320 = 3.019 1.688/0.557 = 3.030
Step 3:
max = (3.006 + 3.019 + 3.030)/3 = 3.02
Step 4:
CI = (3.02 - 3)/2 = 0.010
Step 5:
CR = 0.010/0.58 = 0.016
14 - 13
Row Average 0.123 0.320 0.557
Chapter 14
Since CR = 0.016 is less than 0.10, the degree of consistency exhibited in the pairwise comparison matrix for price is acceptable. 16.
Synthesization Step 1: Column totals are 17/4, 31/21, and 12 Step 2: Style Accord Saturn Cavalier
Accord 4/17 12/17 1/17
Saturn 7/31 21/31 3/31
Cavalier 4/12 7/12 1/12
Step 3: Style Accord Saturn Cavalier
Accord 0.235 0.706 0.059
Saturn 0.226 0.677 0.097
Cavalier 0.333 0.583 0.083
1
1/3
4
1
+ 0.080 7
1/7
1
0.320
0.802
Row Average 0.265 0.656 0.080
Consistency Ratio Step 1:
0.265
3
+ 0.656
1/4
0.265
0.219
0.795 + 0.656 + 0.560 0.066
0.094
0.080
=
2.007 0.239
Step 2:
0.802/0.265 = 3.028 2.007/0.656 = 3.062 0.239/0.080 = 3.007
Step 3:
max = (3.028 + 3.062 + 3.007)/3 = 3.032
Step 4:
CI = (3.032 - 3)/2 = 0.016
Step 5:
CR = 0.016/0.58 = 0.028
Since CR = 0.028 is less than 0.10, the degree of consistency exhibited in the pairwise comparison matrix for style is acceptable.
14 - 14
Multicriteria Decision Problems
17. a. Reputation School A School B b.
School A 1 1/6
School B 6 1
School A 6/7 1/7
School B 6/7 1/7
Step 1: Column totals are 7/6 and 7 Step 2: Reputation School A School B Step 3: Reputation School A School B
18. a.
School A 0.857 0.143
School B 0.857 0.143
Row Average 0.857 0.143
City 2 15/19 3/19 1/19
City 3 7/11 3/11 1/11
Step 1: Column totals are 47/35, 19/3, 11 Step 2: Desirability City 1 City 2 City 3
City 1 35/47 7/47 5/47
Step 3: Desirability City 1 City 2 City 3 b.
City 1 0.745 0.149 0.106
City 2 0.789 0.158 0.053
City 3 0.636 0.273 0.091
1
5
7
0.724 1/5 + 0.193
1
+ 0.083 3
1/7
1/3
1
0.581
2.273
Step 1:
0.723
0.965
0.145 + 0.193 + 0.249 0.103
0.064
0.083
=
0.588 0.251
Step 2:
2.273/0.724 = 3.141 0.588/0.193 = 3.043 0.251/0.083 = 3.014
Step 3:
max = (3.141 + 3.043 + 3.014)/3 = 3.066
Step 4:
CI = (3.066 - 3)/2 = 0.033
Step 5:
CR = 0.033/0.58 = 0.057
14 - 15
Row Average 0.724 0.193 0.083
Chapter 14
Since CR = 0.057 is less than 0.10, the degree of consistency exhibited in the pairwise comparison matrix is acceptable. 19. a.
Step 1: Column totals are 4/3 and 4 Step 2: A 3/4 1/4
A B
B 3/4 1/4
Step 3: A 0.75 0.25
A B b.
B 0.75 0.25
Row Average 0.75 0.25
The individual's judgements could not be inconsistent since there are only two programs being compared.
20. a. Flavor A B C b.
A 1 1/3 1/2
B 3 1 1/5
C 2 5 1
A 6/11 2/11 3/11
B 15/21 5/21 1/21
C 2/8 5/8 1/8
B 0.714 0.238 0.048
C 0.250 0.625 0.125
Row Average 0.503 0.348 0.148
Step 1: Column totals are 11/6, 21/5, and 8 Step 2: Flavor A B C Step 3: Flavor A B C
c.
A 0.545 0.182 0.273
Step 1: 1
3
2
0.503 1/3 + 0.348
1
+ 0.148 5
1/2
1/5
1
0.296
1.845
Weighted Sum: 0.503
1.044
0.168 + 0.348 + 0.740 0.252
0.070
0.148
14 - 16
=
1.258 0.470
Multicriteria Decision Problems
Step 2:
1.845/0.503 = 3.668 1.258/0.348 = 3.615 0.470/0.148 = 3.123
Step 3:
max = (3.668 + 3.615 + 3.123)/3 = 3.469
Step 4:
CI = (3.469 - 3)/2 = 0.235
Step 5:
CR = 0.235/0.58 = 0.415
Since CR = 0.415 is greater than 0.10, the individual's judgements are not consistent. 21. a. Flavor A B C b.
A 1 2 1/5
B 1/2 1 1/5
C 5 5 1
B 5/17 10/17 2/17
C 5/11 5/11 1/11
Step 1: Column totals are 16/5, 17/10, and 11 Step 2: Flavor A B C
A 5/16 10/16 1/16
Step 3: Flavor A B C c.
A 0.313 0.625 0.063
B 0.294 0.588 0.118
C 0.455 0.455 0.091
Row Average 0.354 0.556 0.090
1
1/2
5
1
+ 0.090 5
1/5
1
0.450
1.083
Step 1:
0.354
2
+ 0.556
1/5
0.354
0.278
0.708 + 0.556 + 0.450 0.071
0.111
0.090
=
1.715 0.272
Step 2:
1.083/0.354 = 3.063 1.715/0.556 = 3.085 0.272/0.090 = 3.014
Step 3:
max = (3.063 + 3.085 + 3.014)/3 = 3.054
Step 4:
CI = (3.054 - 3)/2 = 0.027
14 - 17
Chapter 14
Step 5:
CR = 0.027/0.58 = 0.046
Since CR = 0.046 is less than 0.10, the individual's judgements are consistent. 22. a.
Let
D = Dallas S = San Francisco N = New York Location D S N
b.
D 1 4 7
S 1/4 1 3
N 1/7 1/3 1
S 1/17 4/17 12/17
N 3/31 7/31 21/31
Row Average 0.080 0.265 0.656
Step 1: Column totals are 12, 17/4, and 31/21 Step 2: Location D S N
D 1/12 4/12 7/12
Step 3: Location D S N c.
D 0.083 0.333 0.583
S 0.059 0.235 0.706
N 0.097 0.226 0.677
1
1/4
1/7
0.080 4 + 0.265
1
+ 0.656 1/3
7
3
1
Step 1:
0.080
0.066
0.094
0.320 + 0.265 + 0.219 0.560
0.795
0.656
0.239 =
0.802 2.007
Step 2:
0.239/0.080 = 3.007 0.802/0.265 = 3.028 2.007/0.656 = 3.062
Step 3:
max = (3.007 + 3.028 + 3.062)/3 = 3.035
Step 4:
CI = (3.035 - 3)/2 = 0.017
Step 5:
CR = 0.017/0.58 = 0.028
Since CR = 0.028 is less than 0.10, the manager's judgements are consistent.
14 - 18
Multicriteria Decision Problems
23. a.
Step 1: Column totals are 94/21, 33/4, 18, and 21/12 Step 2: Performance 1 2 3 4
1 21/94 7/94 3/94 63/94
2 12/33 4/33 1/33 16/33
3 7/18 4/18 1/18 6/18
4 4/21 3/21 2/21 12/21
Performance 1 2 3 4
1 0.223 0.074 0.032 0.670
2 0.364 0.121 0.030 0.485
3 0.389 0.222 0.056 0.333
4 0.190 0.143 0.095 0.571
3
7
1/3
Step 3:
b.
Step 1: 1 0.292
1/3 1/7
1
+ 0.140
1/4
3
0.292 0.097 0.042 0.876
4
0.420 +
+ 0.053
0.140 0.035 0.560
0.371 +
0.212 0.053 0.318
+
4 1
+ 0.515
1/6
6
1
0.172
1.257
0.129 0.086
=
0.515
0.579 0.216 2.270
Step 2:
1.257/0.292 = 4.305 0.579/0.140 = 4.136 0.216/0.053 = 4.075 2.270/0.515 = 4.408
Step 3:
max = (4.305 + 4.136 + 4.075 + 4.408)/4 = 4.231
Step 4:
CI = (4.231 - 4)/3 = 0.077
Step 5:
CR = 0.077/0.90 = 0.083
Since CR = 0.083 is less than 0.10, the judgements are consistent. 24. a.
1/4
Criteria:
Yield and Risk
Step 1:
Column totals are 1.5 and 3
Step 2:
14 - 19
Row Average 0.292 0.140 0.053 0.515
Chapter 14
Criterion Yield Risk
Yield 0.667 0.333
Risk 0.667 0.333
Priority 0.667 0.333
With only two criteria, CR = 0 and no computation of CR is made. The same calculations for the Yield and the Risk pairwise comparison matrices provide the following: Stocks CCC SRI b.
Yield Priority 0.750 0.250
Risk Priority 0.333 0.667
Overall Priorities: CCC 0.667(0.750) + 0.333(0.333) = 0.611 SRI 0.667(0.250) + 0.333(0.667) = 0.389 CCC is preferred.
25. a.
Criteria: Leadership, Personal, Administrative Step 1: Column Totals are 8, 11/6 and 13/4 Step 2: Criterion Leadership Personal Administrative
Leader 0.125 0.375 0.500
Personal 0.182 0.545 0.273
Administrative 0.077 0.615 0.308
Priority 0.128 0.512 0.360
CR = 0.094 if computed. The same calculations for the leadership, personal and administrative pairwise comparison matrices provide the following. Candidate
Leadership Priority 0.800 0.200
Jacobs Martin b.
Personal Priority 0.250 0.750
Overall Priorities: Jacobs 0.128(0.800) + 0.512(0.250) + 0.360(0.667) = 0.470 Martin 0.128(0.200) + 0.512(0.250) + 0.360(0.333) = 0.530 Martin is preferred.
26. a.
Criteria:
Price, Sound and Reception
Step 1:
Column totals are 19/12, 13/3 and 8
Step 2:
14 - 20
Administrative Priority 0.667 0.333
Multicriteria Decision Problems
Criterion Price Sound Reception
Price 0.632 0.211 0.158
Sound 0.692 0.231 0.077
Reception 0.500 0.375 0.125
Priority 0.608 0.272 0.120
CR = 0.064 The same calculations for the price, sound and reception pairwise comparison matrices provide the following: System System A System B System C CR b.
Price Priority 0.557 0.123 0.320 0.016
Sound Priority 0.137 0.239 0.623 0.016
Reception Priority 0.579 0.187 0.234 0.046
Overall Priorities: System A 0.608(0.557) + 0.272(0.137) + 0.120(0.579) = 0.446 System B 0.608(0.123) + 0.272(0.239) + 0.120(0.187) = 0.162 System C 0.608(0.320) + 0.272(0.623) + 0.120(0.046) = 0.392 System A is preferred.
14 - 21
Chapter 15 Time Series Analysis and Forecasting Learning Objectives 1.
Understand that the long-run success of an organization is often closely related to how well management is able to predict future aspects of the operation.
2.
Know the various components of a time series.
3.
Be able to use smoothing techniques such as moving averages and exponential smoothing.
4.
Be able to use the least squares method to identify the trend component of a time series.
5.
Understand how the classical time series model can be used to explain the pattern or behavior of the data in a time series and to develop a forecast for the time series.
6.
Be able to determine and model seasonal patterns for forecasting a time series.
7.
Know how curve-fitting can be used in forecasting.
8.
Know the definition of the following terms: time series forecast trend component cyclical component seasonal component irregular component
mean squared error moving averages weighted moving averages smoothing constant seasonal index
15 - 1
Chapter 15
Solutions: 1.
The following table shows the calculations for parts (a), (b), and (c).
Time Series Value Forecast
Week
Absolute Value of Forecast Error
Forecast Error
Squared Forecast Error
Percentage Error
Absolute Value of Percentage Error
1
18
2
13
18
-5
5
25
-38.46
38.46
3
16
13
3
3
9
18.75
18.75
4
11
16
-5
5
25
-45.45
45.45
5
17
11
6
6
36
35.29
35.29
6
14
17
-3
3
9
-21.43
21.43
Totals
22
104
-51.30
159.38
2.
a.
MAE = 22/5 = 4.4
b.
MSE = 104/5 = 20.8
c.
MAPE = 159.38/5 = 31.88
d.
Forecast for week 7 is 14
The following table shows the calculations for parts (a), (b), and (c).
Time Series Value Forecast
Week
Forecast Error
Absolute Value of Forecast Error
Squared Forecast Error
Percentage Error
Absolute Value of Percentage Error
1
18
2
13
18.00
-5.00
5.00
25.00
-38.46
38.46
3
16
15.50
0.50
0.50
0.25
3.13
3.13
4
11
15.67
-4.67
4.67
21.81
-42.45
42.45
5
17
14.50
2.50
2.50
6.25
14.71
14.71
6
14
15.00
-1.00
1.00
1.00
-7.14
7.14
Totals
13.67
54.31
-70.21
105.86
3.
a.
MAE = 13.67/5 = 2.73
b.
MSE = 54.31/5 = 10.86
c.
MAPE = 105.89/5 = 21.18
d.
Forecast for week 7 is (18 + 13 + 16 + 11 + 17 + 14) / 6 = 14.83
The following table shows the measures of forecast error for both methods.
15 - 2
Forecasting
MAE MSE MAPE
Exercise 1 4.40 20.80 31.88
Exercise 2 2.73 10.86 21.18
For each measure of forecast accuracy the average of all the historical data provided more accurate forecasts than simply using the most recent value. 4.
a.
Month
Time Series Value Forecast
Squared Forecast Error
Forecast Error
1
24
2
13
24
-11
121
3
20
13
7
49
4
12
20
-8
64
5
19
12
7
49
6
23
19
4
16
7
15
23
-8
64
Total
363
MSE = 363/6 = 60.5 Forecast for month 8 = 15 b.
Week
Time Series Value Forecast
Forecast Error
Squared Forecast Error
1
24
2
13
24.00
-11.00
121.00
3
20
18.50
1.50
2.25
4
12
19.00
-7.00
49.00
5
19
17.25
1.75
3.06
6
23
17.60
5.40
29.16
7
15
18.50
-3.50
12.25
Total
216.72
MSE = 216.72/6 = 36.12 Forecast for month 8 = (24 + 13 + 20 + 12 + 19 + 23 + 15) / 7 = 18 The average of all the previous values is better because MSE is smaller. c.
The average of all the previous values is better because MSE is smaller.
15 - 3
Chapter 15
5.
a.
The data appear to follow a horizontal pattern. b.
Three-week moving average.
Time Series Value Forecast
Week 1
18
2
13
3
16
4
11
15.67
5
17
6
14
Forecast Error
Squared Forecast Error
-4.67
21.78
13.33
3.67
13.44
14.67
-0.67
0.44
Total
35.67
MSE = 35.67/3 = 11.89 The forecast for week 7 = (11 + 17 + 14) / 3 = 14
c.
Smoothing constant = .2
15 - 4
Forecasting
Time Series Value Forecast
Week
Squared Forecast Error
Forecast Error
1
18
2
13
18.00
-5.00
25.00
3
16
17.00
-1.00
1.00
4
11
16.80
-5.80
33.64
5
17
15.64
1.36
1.85
6
14
15.91
-1.91
3.66
Total
65.15
MSE = 65.15/5 = 13.03 The forecast for week 7 is .2(14) + (1 - .2)15.91 = 15.53 d.
The three-week moving average provides a better forecast since it has a smaller MSE.
e. Alpha
0.367694922 Squared
Time Series
Forecast
Forecast
Forecast
Error
Error
13
18
-5.00
25.00
3
16
16.16
-0.16
0.03
4
11
16.10
-5.10
26.03
5
17
14.23
2.77
7.69
6
14
15.25
-1.25
1.55
Total
60.30
Week
Value
1
18
2
MSE =
12.060999
15 - 5
Chapter 15
6.
a.
The data appear to follow a horizontal pattern. b.
Week
Three-week moving average.
Time Series Value Forecast
Forecast Error
Squared Forecast Error
1
24
2
13
3
20
4
12
19.00
-7.00
49.00
5
19
15.00
4.00
16.00
6
23
17.00
6.00
36.00
7
15
18.00
-3.00
9.00
Total
110.00
MSE = 110/4 = 27.5. The forecast for week 8 = (19 + 23 + 15) / 3 = 19
15 - 6
Forecasting
c.
Smoothing constant = .2
Time Series Value Forecast
Week
Forecast Error
Squared Forecast Error
1
24
2
13
24.00
-11.00
121.00
3
20
21.80
-1.80
3.24
4
12
21.44
-9.44
89.11
5
19
19.55
-0.55
0.30
6
23
19.44
3.56
12.66
7
15
20.15
-5.15
26.56
Total
252.87
MSE = 252.87/6 = 42.15 The forecast for week 8 is .2(15) + (1 - .2)20.15 = 19.12 d.
The three-week moving average provides a better forecast since it has a smaller MSE.
e. Alpha
0.351404848 Squared
Time Series
Forecast
Forecast
Forecast
Error
Error
Month
Value
1
24
2
13
24
-11.00
121.00
3
20
20.13
-0.13
0.02
4
12
20.09
-8.09
65.40
5
19
17.25
1.75
3.08
6
23
17.86
5.14
26.40
7
15
19.67
-4.67 Total
21.79 237.69
MSE = 237.69/6 = 39.61428577
15 - 7
Chapter 15
7.
a.
Week 1 2 3 4 5 6 7 8 9 10 11 12
b.
Time-Series Value 17 21 19 23 18 16 20 18 22 20 15 22
4-Week Moving Average Forecast
(Error)2
20.00 20.25 19.00 19.25 18.00 19.00 20.00 18.75 Totals
4.00 18.06 1.00 1.56 16.00 1.00 25.00 10.56 77.18
5-Week Moving Average Forecast
19.60 19.40 19.20 19.00 18.80 19.20 19.00
(Error)2
12.96 0.36 1.44 9.00 1.44 17.64 9.00 51.84
MSE(4-Week) = 77.18 / 8 = 9.65 MSE(5-Week) = 51.84 / 7 = 7.41
c. 8.
For the limited data provided, the 5-week moving average provides the smallest MSE.
a. Week 1 2 3 4 5 6 7 8 9 10 11 12
b.
Time-Series Value 17 21 19 23 18 16 20 18 22 20 15 22
Weighted Moving Average Forecast
19.33 21.33 19.83 17.83 18.33 18.33 20.33 20.33 17.83
Forecast Error
3.67 -3.33 -3.83 2.17 -0.33 3.67 -0.33 -5.33 4.17 Total
(Error)2
13.47 11.09 14.67 4.71 0.11 13.47 0.11 28.41 17.39 103.43
MSE = 103.43 / 9 = 11.49 Prefer the unweighted moving average here; it has a smaller MSE.
c.
9.
You could always find a weighted moving average at least as good as the unweighted one. Actually the unweighted moving average is a special case of the weighted ones where the weights are equal.
The following tables show the calculations for = .1.
15 - 8
Forecasting
Time Series Value Forecast 17 21 17.00 19 17.40 23 17.56 18 18.10 16 18.09 20 17.88 18 18.10 22 18.09 20 18.48 15 18.63 22 18.27
Week 1 2 3 4 5 6 7 8 9 10 11 12
Forecast Error 4.00 1.60 5.44 -0.10 -2.09 2.12 -0.10 3.91 1.52 -3.63 3.73 Totals
Absolute Value of Forecast Error
Squared Forecast Error
4.00 1.60 5.44 0.10 2.09 2.12 0.10 3.91 1.52 3.63 3.73 28.24
16.00 2.56 29.59 0.01 4.37 4.49 0.01 15.29 2.31 13.18 13.91 101.72
Percentage Error 19.05 8.42 23.65 -0.56 -13.06 10.60 -0.56 17.77 7.60 -24.20 16.95 65.67
Absolute Value of Percentage Error 19.05 8.42 23.65 0.56 13.06 10.60 0.56 17.77 7.60 24.20 16.95 142.42
The following tables show the calculations for = .2
Time Series Value Forecast 17 21 17.00 19 17.80 23 18.04 18 19.03 16 18.83 20 18.26 18 18.61 22 18.49 20 19.19 15 19.35 22 18.48
Week 1 2 3 4 5 6 7 8 9 10 11 12
a.
Forecast Error 4.00 1.20 4.96 -1.03 -2.83 1.74 -0.61 3.51 0.81 -4.35 3.52 Totals
Absolute Value of Forecast Error
Squared Forecast Error
4.00 1.20 4.96 1.03 2.83 1.74 0.61 3.51 0.81 4.35 3.52 28.56
MSE for = .1 = 101.72/11 = 9.25 MSE for = .2 = 98.80/11 = 8.98
= .2 provides more accurate forecasts based upon MSE
b.
MAE for = .1 = 28.24/11 = 2.57
15 - 9
16.00 1.44 24.60 1.06 8.01 3.03 0.37 12.32 0.66 18.92 12.39 98.80
Percentage Error 19.05 6.32 21.57 -5.72 -17.69 8.70 -3.39 15.95 4.05 -29.00 16.00 35.84
Absolute Value of Percentage Error 19.05 6.32 21.57 5.72 17.69 8.70 3.39 15.95 4.05 29.00 16.00 147.44
Chapter 15 MAE for = .2 = 28.56/11 = 2.60
= .1 provides more accurate forecasts based upon MAE; but, they are very close. c.
MAPE for = .1 = 142.42/11 = 12.95% MAPE for = .2 = 147.44/11 = 13.40%
= .1 provides more accurate forecasts based upon MAPE. 10. a.
Yˆ13 = .2Y12 + .16Y11 + .64(.2Y10 + .8 Yˆ10 ) = .2Y12 + .16Y11 + .128Y10 + .512 Yˆ10 Yˆ13 = .2Y12 + .16Y11 + .128Y10 + .512(.2Y9 + .8 Yˆ9 ) = .2Y12 + .16Y11 + .128Y10 + .1024Y9 + .4096 Yˆ9 Yˆ13 = .2Y12 + .16Y11 + .128Y10 + .1024Y9 + .4096(.2Y8 + .8 Yˆ8 ) = .2Y12 + .16Y11 + .128Y10 + .1024Y9 +
.08192Y8 + .32768 Yˆ8 b.
The more recent data receives the greater weight or importance in determining the forecast. The moving averages method weights the last n data values equally in determining the forecast.
11. a.
The first two time series values may be an indication that the time series has shifted to a new higher level, as shown by the remainig 10 values. But, overall, the time series plot exhibits a horizontal pattern.
15 - 10
Forecasting
b.
Month 1 2 3 4 5 6 7 8 9 10 11 12
Yt 80 82 84 83 83 84 85 84 82 83 84 83
3-Month Moving Averages Forecast
82.00 83.00 83.33 83.33 84.00 84.33 83.67 83.00 83.00
(Error)2
α=2 Forecast
1.00 0.00 0.45 2.79 0.00 5.43 0.45 1.00 0.00 11.12
80.00 80.40 81.12 81.50 81.80 82.24 82.79 83.03 82.83 82.86 83.09
(Error)2 4.00 12.96 3.53 2.25 4.84 7.62 1.46 1.06 0.03 1.30 0.01 39.06
MSE(3-Month) = 11.12 / 9 = 1.24 MSE(α = .2) = 39.06 / 11 = 3.55 A 3-month moving average provides the most accurate forecast using MSE. c.
We will use the 3-month moving average because we have found that it yields a superior MSE over our time series. The 3-month moving average3-month moving average forecast = (83 + 84 + 83) / 3 = 83.3.
12. a.
The data appear to follow a horizontal pattern.
b.
15 - 11
Chapter 15
Month 1 2 3 4 5 6 7 8 9 10 11 12
Time-Series Value 9.5 9.3 9.4 9.6 9.8 9.7 9.8 10.5 9.9 9.7 9.6 9.6
3-Month Moving Average Forecast
9.40 9.43 9.60 9.70 9.77 10.00 10.07 10.03 9.73
(Error)2
0.04 0.14 0.01 0.01 0.53 0.01 0.14 0.18 0.02 1.08
4-Month Moving Average Forecast
9.45 9.53 9.63 9.73 9.95 9.98 9.97 9.92
(Error)2
0.12 0.03 0.03 0.59 0.00 0.08 0.14 0.10 1.09
MSE(3-Month) = 1.08 / 9 = .12 MSE(4-Month) = 1.09 / 8 = .14 Use 3-Month moving averages. c.
We will use the 3-month moving average because we have found that it yields a superior MSE over our time series. The 3-month moving average3-month moving average forecast = (9.7 + 9.6 + 9.6) / 3 = 9.63.
13. a.
The data appear to follow a horizontal pattern.
15 - 12
Forecasting
b. Month 1 2 3 4 5 6 7 8 9 10 11 12
Time-Series Value 240 350 230 260 280 320 220 310 240 310 240 230
3-Month Moving Average Forecast
273.33 280.00 256.67 286.67 273.33 283.33 256.67 286.67 263.33
(Error)2
177.69 0.00 4010.69 4444.89 1344.69 1877.49 2844.09 2178.09 1110.89 17,988.52
α = .2 Forecast 240.00 262.00 255.60 256.48 261.18 272.95 262.36 271.89 265.51 274.41 267.53
(Error)2 12100.00 1024.00 19.36 553.19 3459.79 2803.70 2269.57 1016.97 1979.36 1184.05 1408.50 27,818.49
MSE(3-Month) = 17,988.52 / 9 = 1998.72 MSE(α = .2) = 27,818.49 / 11 = 2528.95 Based on the above MSE values, the 3-month moving averages appears to be superior. However, exponential smoothing was penalized by including month 2 which was difficult for any method to forecast. Using only the errors for months 4 to 12, the MSE for exponential smoothing is: MSE(α = .2) = 14,694.49 / 9 = 1632.72 Thus, exponential smoothing was better considering months 4 to 12. c.
Using exponential smoothing Yˆ13 = α Y12 + (1 - α) Yˆ12 = .20(230) + .80(267.53) = 260
14. a.
15 - 13
Chapter 15
The data appear to follow a horizontal pattern. b.
Smoothing constant = .3.
Month t 1 2 3 4 5 6 7 8 9 10 11 12
Time-Series Value Yt 105 135 120 105 90 120 145 140 100 80 100 110
Forecast Yˆt 105.00 114.00 115.80 112.56 105.79 110.05 120.54 126.38 118.46 106.92 104.85
Forecast Error Squared Error Yt - Yˆt (Yt - Yˆt )2 30.00 6.00 -10.80 -22.56 14.21 34.95 19.46 -26.38 -38.46 -6.92 5.15 Total
900.00 36.00 116.64 508.95 201.92 1221.50 378.69 695.90 1479.17 47.89 26.52 5613.18
MSE = 5613.18 / 11 = 510.29 Forecast for month 13: Yˆ13 = .3(110) + .7(104.85) = 106.4 c. Alpha
0.032564518 Squared
Time Series
Forecast
Forecast
Forecast
Error
Error
Month
Value
1
105
2
135
105
30.00
900.00
3
120
105.98
14.02
196.65
4
105
106.43
-1.43
2.06
5
90
106.39
-16.39
268.53
6
120
105.85
14.15
200.13
7
145
106.31
38.69
1496.61
8
140
107.57
32.43
1051.46
9
100
108.63
-8.63
74.47
10
80
108.35
-28.35
803.65
11
100
107.43
-7.43
55.14
12
110
107.18
2.82
7.93
Total
5056.62
MSE = 5056.62 / 11 = 459.6929489
15 - 14
Forecasting
15. a.
You might think the time series plot shown above exhibits some trend. But, this is simply due to the fact that the smallest value on the vertical axis is 7.1, as shown by the following version of the plot.
In other words, the time series plot shows an underlying horizontal pattern.
15 - 15
Chapter 15
b. Alpha
0.910230734 Squared
Futures
Forecast
Forecast
Forecast
Error
Error
Week
Index
1
7.35
2
7.4
7.35
0.05
0.00
3
7.55
7.40
0.15
0.02
4
7.56
7.54
0.02
0.00
5
7.6
7.56
0.04
0.00
6
7.52
7.60
-0.08
0.01
7
7.52
7.53
-0.01
0.00
8
7.7
7.52
0.18
0.03
9
7.62
7.68
-0.06
0.00
10
7.55
7.63
-0.08
0.01
Total
0.08
MSE = .08/9 = .008507355 16. a.
The number of homes is generally increasing over time, so we see a positive trend over the years the Super Bowl has been played. The trend appears to be linear with some random variation around the positive trend from year to year.
b.
Because this time series plot indicates a possible linear trend in the data, so forecasting methods
15 - 16
Forecasting
discussed in this chapter are appropriate to develop forecasts for this time series. c.
The following values are needed to compute the slope and intercept:
t = 1128
t = 35720 2
Y = 1808715
tY = 48566536
t
t
Computation of slope: b1 =
tY − ( t Y ) / n = 48566536 − (1128)(1808715) / 47 = 596.3663 35720 − (1128 ) / 47 t − ( t ) / n t
t
2
2
2
Computation of intercept: b0 = Y − b1 t = (38483.30/47) – (596.366)(1128/47) = 24170.506
Equation for linear trend: yˆ t = 24170.506 + 596.366t The annual increase in households viewing the Super Bowl is approximately 596,366.
17. a.
The time series plot shows a linear trend. b.
Minimizing SSE is the same as minimizing MSE: b0
4.70
b1
2.10 Squared
Year
Sales
Forecast
15 - 17
Forecast
Forecast
Error
Error
Chapter 15
1
6.00
6.80
-0.80
0.64
2
11.00
8.90
2.10
4.41
3
9.00
11.00
-2.00
4.00
4
14.00
13.10
0.90
0.81
5
15.00
15.20
-0.20
0.04
6
17.30
Total
9.9
MSE = 9.9/5 = 1.982.475 c.
Yˆ6 = 4.7 + 2.1(6) = 17.3
18. a.
b. Alpha
0.467307293 Squared Forecast
Forecast
Forecast
Error
Error
Period
Stock %
1st-2012
29.8
2nd-2012
31
29.8
1.20
1.44
3rd-2012
29.9
30.36
-0.46
0.21
4th-2012
30.1
30.15
-0.05
0.00
1st-2013
32.2
30.12
2.08
4.31
2nd-2013
31.5
31.09
0.41
0.16
3rd-2013
32
31.28
0.72
0.51
4th-2013
31.9
31.62
0.28
0.08
1st-2014
30
31.75
-1.75
3.06
30.93
Total
9.78
2nd-2014 MSE =
1.222838367
15 - 18
Forecasting
c.
Forecast for second quarter 2014 = 30.93
19. a.
The time series plot shows a linear trend. b. b0
119.71
b1
-4.9286 Squared
Observed Period
Value
Forecast
Error
Error
1
120.00
114.79
5.21
27.19
2
110.00
109.86
0.14
0.02
3
100.00
104.93
-4.93
24.29
4
96.00
100.00
-4.00
16.00
5
94.00
95.07
-1.07
1.15
6
92.00
90.14
1.86
3.45
7
88.00
85.21
2.79
7.76
8 c.
Forecast
Forecast
80.29
Yˆ8 = 119.714 − 4.9286(8) = 80.29
20. a.
15 - 19
Total
79.857143
Chapter 15
The time series plot exhibits a curvilinear trend. b. b0 b1 c.
4.7167 1.4567
Yˆ10 =4.7167 + 1.4567(10) = 19.28
21. a.
This time series plot indicates a possible negative linear trend in the data.
b.
The following values are needed to compute the slope and intercept:
15 - 20
Forecasting
t = 66
t = 506 2
Y = 228.7
tY = 1335.2
t
t
Computation of slope: b1 =
tY − ( t Y ) / n = 1335.2 − ( 66 )( 228.7 ) / 11 = 0.3364 506 − ( 66 ) / 11 t − ( t ) / n t
t
2
2
2
Computation of intercept: b0 = Y − b1 t = (228.7/11) – (-0.3364)(66/11) = 22.8909
Equation for linear trend: Yˆt = 22.89096 − 0.3364t The annual decrease in the percent of adults who smoke is approximately 0.3364%. c.
The forecast of the percent of adults who smoke for 2020 is Yˆ20 = 22.89096 − 0.3364 ( 20 ) = 16.0818
The regression model from part (b) does suggest that the OSH is not on target to meet this goal. The forecast of the percent of adults who smoke falls below 12% in 2033: Yˆ33 = 22.89096 − 0.3364 ( 33 ) = 11.702
15 - 21
Chapter 15
22. a.
The time series plot shows an upward linear trend b.
b0
19.9928
b1
1.7738 Squared
Year
Cost/Unit($)
Forecast
Forecast
Forecast
Error
Error
1
20.00
21.77
-1.77
3.12
2
24.50
23.54
0.96
0.92
3
28.20
25.31
2.89
8.33
4
27.50
27.09
0.41
0.17
5
26.60
28.86
-2.26
5.12
6
30.00
30.64
-0.64
0.40
7
31.00
32.41
-1.41
1.99
8
36.00
34.18
1.82
3.30
9
35.96 MSE =
Total
2.9183
c. The average cost/unit has been increasing by approximately $1.77 per year. d. T9 = 19.9928 + 1.7738(9) = 35.96
15 - 22
23.34619
Forecasting
23.
a.
This time series plot indicates a possible positive linear trend in the data. b. The following values are needed to compute the slope and intercept:
t = 120
t = 1240 2
Y = 723.8 t
tY = 5990.7 t
Computation of slope: b1 =
tY − ( t Y ) / n = 5990.7 − (120 )( 723.8) / 15 = 0.7154 1240 − (120 ) / 15 t − ( t ) / n t
t
2
2
2
Computation of intercept: b0 = Y − b1 t = (723.8/15) – (0.7154)(120/15) = 42.5305
Equation for linear trend: Yˆt = 42.5305 + 0.7154t The annual increase in the percent of adults who report that they exercise for 30 or more minutes at least three times per week is approximately 0.7154%. c. The forecast of the percent of adults who will report that they exercise for 30 or more minutes at least three times per week smoke next year (year 16 of the study) is Yˆ16 = 42.5305 + 0.7154 (16 ) = 53.9762 or 53.98%. d. The linear trend we observed in the time series plot from part (a) appears to be stable, so the
15 - 23
Chapter 15
trend equation from part (b) can be used to forecast the percentage of adults three years from now (year 18 of the study) who will report that they exercise for 30 or more minutes at least three times per week. The forecast of the percent of adults who will report that they exercise for 30 or more minutes at least three times per week smoke three years from now (year 18 of the study) is Yˆ18 = 42.5305 + 0.7154 (18 ) = 55.4069 or 55.41% 24. a.
The time series plot shows a horizontal pattern. But, there is a seasonal pattern in the data. For instance, in each year the lowest value occurs in quarter 2 and the highest value occurs in quarter 4. b.
The fitted equation is: Value = 77.0 - 10.0 Qtr1 - 30.0 Qtr2 - 20.0 Qtr3
c.
The quarterly forecasts for next year are as follows: Quarter 1 forecast = 77.0 - 10.0(1) - 30.0(0) - 20.0(0) = 67 Quarter 2 forecast = 77.0 - 10.0(0) - 30.0(1) - 20.0(0) = 47 Quarter 3 forecast = 77.0 - 10.0(0) - 30.0(0) - 20.0(1) = 57 Quarter 4 forecast = 77.0 - 10.0(0) - 30.0(0) - 20.0(0) = 77
15 - 24
Forecasting
25. a.
The time series plot shows a linear trend and a seasonal pattern in the data. b.
The fitted regression model is: Value = 3.42 + 0.219 Qtr1 - 2.19 Qtr2 - 1.59 Qtr3 + 0.406 t
c.
The quarterly forecasts for next year (t = 13, 14, 15, and 16) are as follows: Quarter 1 forecast = 3.42 + 0.219(1) - 2.19(0) - 1.59(0) + 0.406(13) = 8.92 Quarter 2 forecast = 3.42 + 0.219(0) - 2.19(1) - 1.59(0) + 0.406(14) = 6.92 Quarter 3 forecast = 3.42 + 0.219(0) - 2.19(0) - 1.59(1) + 0.406(15) = 7.92 Quarter 4 forecast = 3.42 + 0.219(0) - 2.19(0) - 1.59(0) + 0.406(16) = 9.92
26. a.
There appears to be a seasonal pattern in the data and perhaps a moderate upward linear trend.
15 - 25
Chapter 15
b.
The fitted regression model is: Value = 2492 - 712 Qtr1 - 1512 Qtr2 + 327 Qtr3
c.
The quarterly forecasts for next year are as follows: Quarter 1 forecast = 2492 – 712(1) – 1512(0) + 327(0) = 1780 Quarter 2 forecast = 2492 – 712(0) – 1512(1) + 327(0) = 980 Quarter 3 forecast = 2492 – 712(0) – 1512(0) + 327(1) = 2819 Quarter 4 forecast = 2492 – 712(0) – 1512(0) + 327(0) = 2492
d.
The fitted regression model is: Value = 2307 - 642 Qtr1 - 1465 Qtr2 + 350 Qtr3 + 23.1 t The quarterly forecasts for next year are as follows: Quarter 1 forecast = 2307 – 642(1) – 1465(0) + 350(0) + 23.1(17) = 2058 Quarter 2 forecast = 2307 – 642(0) – 1465(1) + 350(0) + 23.1(18) = 1258 Quarter 3 forecast = 2307 – 642(0) – 1465(0) + 350(1) + 23.1(19) = 3096 Quarter 4 forecast = 2307 – 642(0) – 1465(0) + 350(0) + 23.1(20) = 2769
27. a.
The time series plot indicates a seasonal pattern in the data and perhaps a slight upward linear trend. b.
The fitted regression model is: Level = 21.7 + 7.67 Hour1 + 11.7 Hour2 + 16.7 Hour3 + 34.3 Hour4 + 42.3 Hour5 + 45.0 Hour6 + 28.3 Hour7 + 18.3 Hour8 + 13.3 Hour9 + 3.33 Hour10 + 1.67 Hour11 c. The hourly forecasts for the next day can be obtained very easily using the estimated regression equation. For instance, setting Hour1 = 1 and the rest of the dummy variables equal to 0 provides the forecast for the first hour; setting Hour2 = 1 and the rest of the dummy variables equal to 0 provides the forecast for the second hour; and so on.
15 - 26
Forecasting
Forecast for hour 1 = 21.667 + 7.667(1) + 11.667(0) + 16.667(0) + 34.333 (0) + 42.333(0) + 45.000(0) + 28.333(0) + 18.333(0) + 13.333(0) + 3.333(0) + 1.667(0) = 29.33 Forecast for hour 2 = 21.667 + 7.667(0) + 11.667(1) + 16.667(0) + 34.333 (0) + 42.333(0) + 45.000(0) + 28.333(0) + 18.333(0) + 13.333(0) + 3.333(0) + 1.667(0) = 33.33 The forecasts for the remaining hours can be obtained similarly. But, since there is no trend the data the hourly forecasts can also be computed by simply taking the average of the three time series values for each hour. Hour
July 15
July 16
July 17
Average
1
25
28
35
29.33
2
28
30
42
33.33
3
35
35
45
38.33
4
50
48
70
56.00
5
60
60
72
64.00
6
60
65
75
66.67
7
40
50
60
50.00
8
35
40
45
40.00
9
30
35
40
35.00
10
25
25
25
25.00
11
25
20
25
23.33
12
20
20
25
21.67
In other words, the forecast for hour 1 is the average of the three observations for hour 1 on July 15, 16, and 17, or 29.33; the forecast for hour 2 is the average of the three observations for hour 1 on July 15, 16, and 17, or 33.33; and so on. Note that the forecast for the last hour is 21.67, the value of b0 in the estimated regression equation. d.
The fitted regression model is: Level = 11.2 + 12.5 Hour1 + 16.0 Hour2 + 20.6 Hour3 + 37.8 Hour4 + 45.4 Hour5 + 47.6 Hour6 + 30.5 Hour7 + 20.1 Hour8 + 14.6 Hour9 + 4.21 Hour10 + 2.10 Hour11 + 0.437 t Hour 1 on July 18 corresponds to Hour1 = 1 and t = 37. Forecast for hour 1 on July 18 = 11.167 + 12.479(1) + .4375(37) = 39.834 Hour 2 on July 18 corresponds to Hour2 = 1 and t = 38. Forecast for hour 2 on July 18 = 11.167 + 16.042(1) + .4375(38) = 43.834 The forecasts for the other hours are computed in a similar manner. The following table shows the forecasts for the 12 hours on July 18.
Hour 1
39.834
15 - 27
Chapter 15
Hour 2
43.834
Hour 3
48.834
Hour 4
66.500
Hour 5
74.501
Hour 6
77.167
Hour 7
60.501
Hour 8
50.500
Hour 9
45.501
Hour 10
35.500
Hour 11
33.834
Hour 12
32.167
28. a.
The time series plot shows both a linear trend and seasonal effects. b.
The fitted regression model is: Revenue = 70.0 + 10.0 Qtr1 + 105 Qtr2 + 245 Qtr3 Quarter 1 forecast = 70.0 + 10.0(1) + 105(0) + 245(0) = 80 Quarter 2 forecast = 70.0 + 10.0(0) + 105(1) + 245(0) = 175 Quarter 3 forecast = 70.0 + 10.0(0) + 105(0) + 245(1) = 315 Quarter 4 forecast = 70.0 + 10.0(0) + 105(0) + 245(0) = 70
c.
The fitted regression model is: Revenue = - 70.1 + 45.0 Qtr1 + 128 Qtr2 + 257 Qtr3 + 11.7 Period Quarter 1 forecast = -70.1 + 45.0(1) + 128(0) + 257(0) + 11.7(21) = 221 Quarter 2 forecast = -70.1 + 45.0(0) + 128(1) + 257(0) + 11.7(22) = 315
15 - 28
Forecasting
Quarter 3 forecast = -70.1 + 45.0(0) + 128(0) + 257(1) + 11.7(23) = 456 Quarter 4 forecast = -70.1 + 45.0(0) + 128(0) + 257(0) + 11.7(24) = 211
15 - 29
Chapter 16 Markov Processes Learning Objectives 1.
Learn about the types of problems that can be modeled as Markov processes.
2.
Understand the Markov process approach to the market share or brand loyalty problem.
3.
Be able to set up and use the transition probabilities for specific problems.
4.
Know what is meant by the steady-state probabilities.
5.
Know how to solve Markov processes models having absorbing states.
6.
Understand the following terms: state of the system transition probability state probability steady-state probability absorbing state fundamental matrix
16 - 1
Solutions: 1. State Probability 1 (n)
0 0.5
1 0.55
2 0.585
3 0.610
4 0.627
5 0.639
6 0.647
7 0.653
8 0.657
9 0.660
10 0.662
Large →n → 2/3
2 (n)
0.5
0.45
0.415
0.390
0.373
0.361
0.353
0.347
0.343
0.340
0.338
→ 1/3
Probabilities are approaching 1 = 2/3 and 2 = 1/3. 2.
3.
4.
a.
b.
1 = 0.5, 2 = 0.5
c.
𝑃=[
a.
0.10
b.
1 = 0.75, 2 = 0.25
a.
1 = 0.92, 2 = 0.08
b.
Without component:
0.90 0.10
0.10 ] 𝜋1 = .6 0.85
With component:
𝜋2 = .4
Expected Cost = 0.25 ($500) = $125 Expected Cost = 0.08 ($500) = $ 40
Breakeven Cost = $125 - $40 = $85 per hour. 5.
No Traffic Delay
Traffic Delay
No Traffic Delay
0.85
0.15
Traffic Delay
0.25
0.75
16 - 2
a. .75
.75
Delay
Delay
.25 No Delay
Delay .25 .15
No Delay
Delay
.85 No Delay
P (Delay 60 minutes) = (0.75)(0.75) = 0.5625 b.
1 = 0.85 1 + 0.25 2 2 = 0.15 1 + 0.75 2 1 + 2 = 1 Solve for 1 = 0.625 and 2 = 0.375
c. 6.
This assumption may not be valid. The transition probabilities of moving to delay and no delay states may actually change with the time of day. a. Given the player last chose Rock, the transition matrix shows that she is most likely to choose
Paper next (with probability 0.42). Therefore, you should choose Scissors (because Scissors beats Paper).
16 - 3
Rock .27 Rock
b.
Paper
.42 .31
Scissors
.27 Rock .36 .42
Rock
Paper
Paper
.15 .49
Scissors
Rock
.31
.18 .55 Scissors
c.
P =
.27
Paper Scissors
The one step probability matrix is
0.27
0.42
0.31
0.36
0.15
0.49
0.18
0.55
0.27
The probability your opponent will choose Paper in the second round is given by 𝜋2 (2). This can be found from Π(2) by first finding Π(1) as shown below.
Π(1) = [1
0
0.27 0] [0.36 0.18
0.42 0.15 0.55
0.31 0.49] = [0.27 0.27
16 - 4
0.42
0.31]
Π(2) = [0.27
0.42
0.27 0.31] [0.36 0.18
0.42 0.15 0.55
0.31 0.49] = [0.28 0.27
City
Suburbs
City
0.98
0.02
Suburbs
0.01
0.99
0.35
0.37]
So 𝜋2 (2) = 0.35 7.
a.
b.
1 = 0.98 1 + 0.01 2
(1)
2 = 0.02 1 + 0.99 2
(2)
1 + 2 = 1
(3)
Solving equations (1) and (3) provides 0.02 1 - 0.01 2 = 0 2 = 1 - 1 Thus,
0.02 - 0.01 (1 - 1) = 0 0.03 1 - 0.01 = 0 1 = 0.333
and
8.
2 = 1 - 0.333 = 0.667
c.
The area will show increases in the suburb population and decreases in the city population. The current 40% in the city is expected to drop to around 33%.
a.
Let
1 = Murphy’s steady-state probability 2 = Ashley’s steady-state probability 3 = Quick Stop’s steady-state probability
1 = 0.85 1 + 0.20 2 + 0.15 3
(1)
2 = 0.10 1 + 0.75 2 + 0.10 3
(2)
3 = 0.05 1 + 0.05 2 + 0.75 3
(3)
1 + 2 + 3 = 1 Using 1, 2, and 4 we have 0.15 1 - 0.20 2 - 0.15 3
= 0
16 - 5
-0.10 1 - 0.25 2 - 0.10 3
= 0
1 + 2 + 3
= 1
Solving three equations and three unknowns gives 1 = 0.548, 2 = 0.286, and 3 = 0.166
9.
b.
16.6%
c.
Murphy’s 548, Ashley’s 286, and Quick Stop’s 166. Quick Stop should take 667 - 548 = 119 Murphy’s customers and 333 - 286 = 47 Ashley’s customers.
a.
Only 5% of MDA customers switch to Special B while 10% of Special B customers switch to MDA
b.
1 = 0.333, 2 = 0.667
10. 1
=
2 3
+
=
0.801 0.101
+
+
0.052 0.752
=
0.101
+
0.202
also 1 + 2 + 3 = 1
[1]
+
0.403 0.303
+
0.303
[3]
[2]
[4]
Using equations 1,2 and 4, we have 1 = 0.442, 2 = 0.385, and 3 = 0.173. The Markov analysis shows that Special B now has the largest market share. In fact, its market share has increased by almost 11%. The MDA brand will be hurt most by the introduction of the new brand, T-White. People who switch from MDA to T-White are more likely to make a second switch back to MDA.
16 - 6
11.
WIN
1 a.
Tree Diagram
-6 0.5
WIN
0.5
LOSE
0.5
WIN
0.5
LOSE
0 p
0
-14
0 p
1-p
-8
1-p
b.
Transition Probability Matrix -14 -8 -6 0 WIN LOSE
-14 0 0 0 0 0 0
-8 1-p 0 0 0 0 0
-6 p 0 0 0 0 0
0 0 p 0 0 0 0
WIN 0 0 1 0.5 1 0
LOSE 0 1-p 0 0.5 0 1
16 - 7
LOSE
c. 1(1)
=
1
2(1)
0
3(1)
0
4(1)
0
5(1)
0
6(1)
0
= 1(2)
=
0
2(2)
1-p
3(2)
p
4(2)
0
5(2)
0
=
0
2(3)
0
3(3)
0
4(3)
(1-p )p
5(3)
p
1-p 0 0 0 0 0
p 0 0 0 0 0
0 p 0 0 0 0
0 0 1 0.5 1 0
0 1-p 0 0.5 0 1
0
1-p
p
0
0
0
0 0 0 0 0 0
1-p 0 0 0 0 0
p 0 0 0 0 0
0 p 0 0 0 0
0 0 1 0.5 1 0
0 1-p 0 0.5 0 1
0
0
0
(1-p )p
p
(1-p )2
0 0 0 0 0 0
1-p 0 0 0 0 0
p 0 0 0 0 0
0 p 0 0 0 0
0 0 1 0.5 1 0
0 1-p 0 0.5 0 1
0
0
0
0
0.5(1-p )p+p
(1-p)2+0.5(1-p)p
6(2)
0
= 1(3)
0 0 0 0 0 0
6(3)
(1-p )2
=
The probability of winning is 0.5(1-p )p + p = 1.5p - 0.5p 2 . Temple's coach should go for 2 points if this probability is greater than the probability of winning by going for a one-point conversion after each touchdown, which is 0.5. Therefore, he should go for a two-point conversion when p is such that 1.5p - 0.5p 2 > 0.5. Solving this as a quadratic equation, and recalling that p must be between 0 and 1, we find that this inequality is satisfied when p > 0.382.
16 - 8
12. (I-Q) = 1 0
0 - 0.4 1 0.1
0.3 0.5
N = (1-Q)-1 = 1.85 0.37 NR = 1.85 0.37
1.11 2.22
=
0.6 -0.3 -0.1 0.5
1.11 2.22
0.2 0.2
0.1 0.2
= 0.59 0.52
0.41 0.48
0.59 probability state 3 units end up in state 1; 0.52 probability state 4 units end up in state 1. 13. (I-Q) =
0.75 -0.25 -0.05 0.75
N = 1.36 0.09
0.45 1.36
NR = 0.909 0.727
0.091 0.273
BNR = [4000, 5000] NR = [7271, 1729]
$1729 in bad debts. 14.
(I-Q) = 0.5 0.0
-0.2 0.5
-1 N = (I-Q) = 2 0
15. a.
b.
NR = 0.52 0.80
0.48 0.20
BNR = 1500
3500
0.8 2.0
0.52 0.80
0.48 0.20
= 3580
1420
Retirement and leaves for personal reasons are the two absorbing states since both result in the manager leaving the company. Middle Managers: Probability of retirement = 0.03 Probability of leaving (personal) = 0.07 Probability of staying middle manager = 0.80 Probability of promotion to senior manager = 0.10
c.
Senior Managers: Probability of retirement = 0.08 Probability of leaving (personal) = 0.01 Probability of staying middle manager = 0.03 Probability of promotion to senior manager = 0.88
16 - 9
d.
(I-Q) = 1 0
0 - 0.80 1 0.03
0.10 0.88
=
0.20 -0.10 -0.03 0.12
-1 N = (I-Q) = 5.714 4.762 1.429 9.524
NR = 0.552 0.448 0.805 0.195 55.2% will retire and 44.8% will leave for personal reasons.
e. 𝐵𝑁𝑅 = [640 280] [0.552 0.448] = [579 341] 0.805
0.195
579 will retire (63%) and 341 will leave (37%). 16. a.
The Injured and Retired states are absorbing states.
b.
Rearrange the transition probability matrix to: Injured
Retired
Backup
Starter
Injured
1
0
0
0
Retired
0
1
0
0
Backup
0.1
0.1
0.4
0.4
Starter
0.15
0.25
0.1
0.5
I-Q =
N=
NR =
0.6 -0.1
-0.4 0.5
(I - Q)-1 =
1.923 1.538 0.385 2.308
0.423 0.577 0.385 0.615
38.5% of Starters will eventually be Injured and 61.5% will be Retired.
16 - 10
c.
BNR = [8
5] [0.423 0.385
0.577 ] = [5.308 0.615
16 - 11
7.691]
Chapter 17 Linear Programming: The Simplex Method Learning Objectives 1.
Learn how to find basic and basic feasible solutions to systems of linear equations when the number of variables is greater than the number of equations.
2.
Learn how to use the simplex method for solving linear programming problems.
3.
Obtain an understanding of why and how the simplex calculations are made.
4.
Understand how to use slack, surplus, and artificial variables to set up tableau form to get started with the simplex method for all types of constraints.
5.
Understand the following terms: simplex method basic solution basic feasible solution tableau form simplex tableau
6.
net evaluation row basis iteration pivot element artificial variable
Know how to recognize the following special situations when using the simplex method to solve linear programs. infeasibility unboundedness alternative optimal solutions degeneracy
17 - 1
Chapter 17
Solutions: 1.
a.
With x1 = 0, we have x2 4x2
+ x3
= 6 = 12
(1) (2)
From (1), we have x2 = 6. Substituting for x2 in (2) yields 4(6)
+ x3 x3
= 12 = 12 - 24 = -12
Basic Solution: x1 = 0, x2 = 6, x3 = -12 b.
With x2 = 0, we have 3x1 2x1
+ x3
= 6 = 12
(3) (4)
From (3), we find x1 = 2. Substituting for x1 in (4) yields 2(2)
+ x3 x3
= 12 = 12 - 4 = 8
Basic Solution: x1 = 2, x2 = 0, x3 = 8 c.
With x3 = 0, we have 3x1 2x1
+ x2 + 4x2
= 6 = 12
(5) (6)
Multiplying (6) by 3/2 and Subtracting form (5) yields 3x1 -(3x1
+ +
x2 6x2 ) -5x 2 x2
= 6 = -18 = -12 = 12/5
Substituting x2 = 12/5 into (5) yields 3x1 3x1 x1
+ 12/5 = 6 = 18/5 = 6/5
Basic Solution: x1 = 6/5, x2 = 12/5, x3 = 0
2.
d.
The basic solutions found in (b) and (c) are basic feasible solutions. The one in (a) is not because x3 = -12.
a.
Standard Form: Max
x1
+
2x2
17 - 2
Linear Programming: The Simplex Method s.t. x1 2x1
+
5x2 + s1 6x2 + s2 x1, x2, s1, s2 0
= 10 = 16
b.
We have n = 4 and m = 2 in standard form. So n - m = 4 - 2 = 2 variables must be set equal to zero in each basic solution.
c.
There are 6 combinations of the two variables that may be set equal to zero and hence 6 possible basic solutions. x1 = 0, x2 = 0 s1 = 10 s2 = 16 This is a basic feasible solution. x1 = 0, s1 = 0 5x2 6x2
+
s2
= 10 = 16
(1) (2)
From (1) we have x2 = 2. And substituting for x2 in (2) yields 6(2)
+
s2 s2
= 16 = 16 - 12 = 4
s1
= 10 = 16
This is a basic feasible solution. x1 = 0, s2 = 0 5x2 6x2
+
(3) (4)
From (4), we have x2 = 8/3. Substituting for x2 in (3) yields 5(8/3)
+
s1 s1
= 10 = 10 - 40/3 = -10/3
This is not a basic feasible solution. x2 = 0, s1 = 0 x1 2x1
+
s2
= 10 = 16
(5) (6)
From (5) we have x1 = 10. And substituting for x1 in (6) yields 2(10)
+
s2 s2
= 16 = 16 - 20 = -4
This is not a basic feasible solution. x2 = 0, s2 = 0
17 - 3
Chapter 17
x1 2x1
+
s1
= 10 = 16
(7) (8)
From (8) we find x1 = 8. And substituting for x1 in (7) yields 8
+
s1 s1
= 10 = 2
5x2 6x2
= 10 = 16
This is a basic feasible solution s1 = 0, s2 = 0 x1 2x1
+ +
(9) (10)
From (9) we have x1 = 10 - 5x2. Substituting for x1 in (10) yields 2(10 - 5x2) 20 -10x2
+ + -
6x2 6x2 4x2 4x2 x2
= 16 = 16 = 16 - 20 = -4 = 1
Then, x1 = 10 - 5(1) = 5 This is a basic feasible solution. d.
The optimal solution is the basic feasible solution with the largest value of the objective function. There are 4 basic feasible solutions from part (c) to evaluate in the objective function. x1 = 0, x2 = 0, s1 = 10, s2 = 16 Value = 1(0) + 2(0) = 0 x1 = 0, x2 = 2, s1 = 0, s2 = 4 Value = 1(0) + 2(2) = 4 x1 = 8, x2 = 0, s1 = 2, s2 = 0 Value = 1(8) + 2(0) = 8
x1 = 5, x2 = 1, s1 = 0, s2 = 0 Value = 1(5) + 2(1) = 7 The optimal solution is x1 = 8, x2 = 0 with value = 8. 3.
a. Max s.t.
5x1
+
9x2
+ 0s1
/2x1 1x1
+ +
1x2 1x2
+ 1s1 - 1s2
1
+ 0s2
+ 0s3 = 8 = 10
17 - 4
Linear Programming: The Simplex Method 1
/4x1 + 3/2 x2 x1, x2, s1, s2, s3, 0
4.
- 1s3
= 6
b.
2
c.
x1 = 4, x2 = 6, and s3 = 4.
d.
x2 = 4, s1 = 4, and s2 = -6.
e.
The answer to part c is a basic feasible solution and an extreme point solution. The answer to part d is not a basic feasible solution because s2 is negative.
f.
The graph below shows that the basic solution for part c is an extreme point and the one for part d is
a.
Standard Form: Max s.t.
b.
60x1
+
15x1 5x1
+
90x2
45x2 + s1 5x2 + s2 x1, x2, s1, s2 0
= 90 = 20
Partial initial simplex tableau: x1
x2
s1
s2
60
90
0
0
15
45
1
0
90
5
5
0
1
20
17 - 5
not
Chapter 17
5.
a.
b.
Initial Tableau x1
x2
s1
s2
Basis cB
5
9
0
0
s1
0
10
9
1
0
90
s2
0
-5
3
0
1
15
zj
0
0
0
0
0
cj - zj
5
9
0
0
We would introduce x2 at the first iteration.
c.
Max s.t.
5x1 + 9x2 10x1 + 9x2 -5x1 + 3x2 x1, x2
6.
90 15 0
a. x1
x2
x3
s1
s2
s3
Basis
cB
5
20
25
0
0
0
s1
0
2
1
0
1
0
0
40
s2
0
0
2
1
0
1
0
30
s3
0
3
0
-1/2
0
0
1
15
zj
0
0
0
0
0
0
0
cj - zj
5
20
25
0
0
0
b. Max s.t.
c.
5x1
+ 20x2
2x1
+ 1x2 2x2
+ 25x3
+ 0s1
+ 0s2
+ 0s3
+ 1s1 = 40 1x3 + 1s2 = 30 3x1 1/2x3 + 1s3 = 15 x1, x2, x3, s1, s2, s3, 0. The original basis consists of s1, s2,, and s3. It is the origin since the nonbasic variables are x1, x2, and x3 and are all zero. + -
d.
0.
e.
x3 enters because it has the largest cj - zj and s2 will leave because row 2 has the only positive coefficient.
f.
30; objective function value is 30 times 25 or 750.
g.
Optimal Solution: x1 = 10
s1 = 20
x2 = 0
s2 = 0
17 - 6
Linear Programming: The Simplex Method
x3 = 30
s3 = 0
z = 800. 7.
Sequence of extreme points generated by the simplex method: (x1 = 0, x2 = 0) (x1 = 0, x2 = 6) (x1 = 7, x2 = 3) 8.
a.
Initial simplex tableau x1
x2
s1
s2
s3
s4
Basis
cB
10
9
0
0
0
0
s1
0
7/10
1
1
0
0
0
630
s2
0
1/2
5/6
0
1
0
0
600
s3
0
1
2/3
0
0
1
0
708
s4
0
1/10
1/4
0
0
0
1
135
zj
0
0
0
0
0
0
0
cj - zj
10
9
0
0
0
0
Final simplex tableau
17 - 7
Chapter 17
x1
x2
s1
s2
s3
s4
0
0
0
Basis
cB
10
9
0
x2
9
0
1
30/16
s2
0
0
x1
10
s4
0 -21/16 0
252
0 -15/16
1
5/32
0
120
1
0 -20/16
0
30/16
0
540
0
0
0 -11/32
0
9/64
1
18
zj
10
9
70/16
0 111/16 0
7668
cj - zj
0
0 -70/16
0 -111/16 0
x1 = 540 standard bags x2 = 252 deluxe bags b.
$7668
c. & d. Slack s1 = 0 s2 = 120 s3 = 0 s4 = 18
Production Time Cutting and dyeing time = 630 hours Sewing time = 600 - 120 = 480 hours Finishing time = 708 hours Inspection and Packaging time = 135 - 18 = 117 hours
17 - 8
Linear Programming: The Simplex Method 9.
Note: Refer to Chapter 2, problem 21 for a graph showing the location of the extreme points. Initial simplex tableau (corresponds to the origin) x1
x2
s1
s2
s3
Basis
cB
40
30
0
0
0
s1
0
2/5
1/2
1
0
0
20
s2
0
0
1/5
0
1
0
5
s3
0
3/5
3/10 0
0
1
21
zj
0
0
0
0
0
0
cj - zj
40
30
0
0
0
bi
ai1
20/(2/5) = 50
21/(3/5) = 35
First iteration: x1 enters the basis and s3 leaves (new basic feasible solution) x1
x2
s1
s2
s3
Basis
cB
40
30
0
0
0
s1
0
0
3/10
1
0
-2/3
6
6/(3/10) = 20
s2
0
0
1/5
0
1
0
5
5/(1/5) = 25
x1
40
1
1/2
0
0
5/3
35
35/(1/2) = 70
zj
40
20
0
0 200/3
1400
cj - zj
0
10
0
0 -200/3
bi
ai 2
Next iteration: x2 enters the basis and s1 leaves (new basic feasible solution) x1
x2
s1
s2
s3 0
Basis
cB
40 30
0
0
x2
30
0
1
10/3
0 -20/9
20
s2
0
0
0
-2/3
1
4/9
1
x1
40
1
0
-5/3
0
25/9
25
zj
40 30 100/3 0 400/9
1600
cj - zj
0
0 -100/3 0 -400/9
Optimal Solution: x1 = 25 s1 = 0
10.
x2 = 20 s2 = 1 s3 = 0.
Initial simplex tableau:
17 - 9
Chapter 17
x1
x2
x3
s1
s2
s3
Basis
cB
5
5
24
0
0
0
bi
s1
0
15
4
12
1
0
0
2800 2800/12 = 233.33
s2
0
15
8
0
0
1
0
6000
s3
0
1
0
8
0
0
1
1200 1200/8 = 150
zj
0
0
0
0
0
0
cj - zj
5
5
24
0
0
0
ai 3
0
First iteration: x3 enters, s3 leaves x1
x2
x3
s1
s2
s3
Basis
cB
5
5
24
0
0
0
s1
0
27/2
4
0
1
0
-3/2 1000 1000/4 = 250
s2
0
15
8
0
0
1
0
x3
24
1/8
0
1
0
0
1/8
6000 6000/8 = 750 150
zj
3
0
24
0
0
3
3600
cj - zj
2
5
0
0
0
-3
x1
x2
x3
s1
s2
s3 0
bi
Second iteration: x2 enters, s1 leaves
Basis
cB
5
5
0
0
0
x2
5
27/8
1
0
1/4
0 -3/8
250
s2
0
-12
0
0
-2
1
3
4000
x3
24
1/8
0
1
0
0
1/8
150
zj
159/8 5
24
5/4
0
9/8
4850
cj - zj
-119/8 0
0
-5/4
0 -9/8
Optimal Solution: x2 = 250, x3 = 150, s2 = 4000, Value = 4850
11.
17 - 10
ai 2
Linear Programming: The Simplex Method
Extreme Points: (x1 = 6, x2 = 0), (x1 = 4.2, x2 = 3.6), (x1 = 15, x2 = 0) Simplex Solution Sequence: (x1 = 0, x2 = 0) (x1 = 6, x2 = 0) (x1 = 4.2, x2 = 3.6) 12. Let
x1 x2 x3
= units of product A. = units of product B. = units of product C.
Max s.t.
20x1
+ 20x2
7x1 5x1
+ 6x2 + 3x3 + 4x2 + 2x3 x1, x2, x3 0
+ 15x3 100 200
Optimal Solution: x1 = 0, x2 = 0, x3 = 33 1/3 Profit = 500. 13. Let
x1 x2
= number of units of Grade A Plywood produced = number of units of Grade B Plywood produced
17 - 11
Chapter 17
x3 Max s.t.
= number of units of Grade X Plywood produced
40x1
+ 30x2
+ 20x3
2x1 + 5x2 2x1 + 5x2 4x1 + 2x2 x1, x2, x3 0
+ 10x3 + 3x3 - 2x3
900 400 600
Optimal Solution: x1 = 137.5, x2 = 25, x3 = 0 Profit = 6250. 14. Let
x1 x2 x3
Max s.t.
= = = 1.00x1
gallons of Heidelberg Sweet produced gallons of Heidelberg Regular produced gallons of Deutschland Extra Dry produced + 1.20x2
+ 2.00x3
1x1 + 2x2 1x1 + + 2x1 + 1x2 2x1 + 3x2 + x1, x2, x3, x4, 0
2x3 1x3
150 150 80 225
Grapes Grade A Grapes Grade B Sugar Labor-hours
a. x1 = 0 x2 = 50 x3 = 75
s1 = 50 s2 = 0 s3 = 30 s4 = 0
Profit = $210 b.
s1 = unused bushels of grapes (Grade A) s2 = unused bushels of grapes (Grade B) s3 = unused pounds of sugar s4 = unused labor-hours
c.
s2 = 0 and s4 = 0. Therefore the Grade B grapes and the labor-hours are the binding resources. Increasing the amounts of these resources will improve profit.
15. Max s.t.
4x1
+ 2x2
- 3x3
+ 5x4
+ 0s1
- Ma1
2x1 3x1
- 1x2
+ 1x3 - 1x3
+ 2x4 + 2x4
- 1s1
+
+ 0s2
1a1 + 1s2
17 - 12
- Ma3 = 50 = 80
Linear Programming: The Simplex Method
1x1
+ 1x2
-4x1
- 5x2
+ 1x4 x1, x2, x3, x4, s1, s2, a1, a3 0
+ 1a3
= 60
16. Max s.t.
4x1 -1x1 2x1 17.
- 1x2 + 2x2 + 1x2
- 3x3
+ 0s1
+ 2x3 + 1x3
- 1s1
+ 0s2
+ 0s4
- Ma1
- Ma2
- Ma3
+ 1a1 - 1s2
+ 1a2 + 1a3
+ 1x3 + 1s4 x1, x2, x3, s1, s2, s4, a1, a2, a3 0
= 20 = 8 = 5 = 12
x1 = 1, x2 = 4, z = 19 Converting to a max problem and solving using the simplex method, the final simplex tableau is:
18.
x1
x2
x3
s1
s2
Basis
cB
-3
-4
-8
0
0
x1
-3
1
0
-1
-1/4
1/8
1
x2
-4
0
1
2
0
-1/4
4
zj
-3
-4
-5
3/4
5/8
-19
cj - zj
0
0
-3
-3/4 -5/8
Initial tableau (Note: Min objective converted to Max.) x1
x2
x3
s1
s2
s3
a2
a3
Basis
cB
-84
-4
-30
0
0
0
-M
-M
s1
0
8
1
3
1
0
0
0
0
240
240/8 = 30
a2
-M
16
1
7
0
-1
0
1
0
480
480/16 = 30
a3
-M
8
-1
4
0
0
-1
0
1
160
160/8 = 20
zj
-24M
0
-11M
0
M
M
-M
-M
-640M
cj - zj
-84+24M
-4
-30+11M
0
-M
-M
0
0
Iteration 1: x1 enters, a3 leaves (Drop a3 column)
17 - 13
bi
ai1
Chapter 17
x1
x2
x3
s1
s2
s3
a2
Basis
cB
-84
-4
-30
0
0
0
-M
s1
0
0
2
-1
1
0
1
0
80
a2
-M
0
3
-1
0
-1
2
1
160
x1
-84
1
-1/8
1/2
0
0
-1/8
0
20
zj
-84
cj - zj
0
21/ - 3M 42+M 2 -29/ + 3M -72-M 2
0
M
0
-M
21/ - 2M -M 2 -21/ +2M 0 2
-1680-160M
Iteration 2: x2 enters, s1 leaves x1
x2
x3
s1
s2
s3
a2
Basis
cB
-84 -4
-30
0
0
0
-M
x2
-4
0
1
-1/2
1/2
0
1/2
0
40
a2
-M
0
0
1/2
-3/2
-1
1/2
1
40
x1
-84
1
0
7/16
1/16
0
-1/16
0
25
zj
-84 -4
-139 4 19 + 4
-29 + 3M 4 2 29 3M 4 2
M
13 - M 4 2 -13 M + 4 2
-M
-2260-100M
cj - zj
0
0
M 2 M 2
-M
0
Iteration 3: x3 enters, x1 leaves x1
x2
x3
s1
s2
s3
a2
Basis
cB
-84
-4
-30
0
0
0
-M
x2
-4
8/7
1
0
4/7
0
3/7
0
480/7
a2
-M
-8/7
0
0
-11/7
-1
4/7
1
80/7
x3
-30
16/7
0
1
1/7
0
-1/7
0
400/7
zj
-512+8M 7 -76-8M 7
-4
-30
M
-13920 - 80M 7
0
-42-4M 7 42+4M 7
-M
0
-46+11M 7 46-11M 7
cj - zj
Iteration 4: s3 enters, a2 leaves (Drop a2 column)
17 - 14
-M
0
Linear Programming: The Simplex Method
x1
x2
x3
s1
s2
s3
Basis
cB
-84
-4
-30
0
0
0
x2
-4
2
1
0
7/4
3/4
0
60
s3
0
-2
0
0
-11/4 -7/4
1
20
x3
-30
2
0
1
-1/4 -1/4
0
60
zj
-68
-4
-30
1/2
9/2
0
-2040
cj - zj
-16
0
0
-1/2 -9/2
0
Optimal Solution: x2 = 60, x3 = 60, s3 = 20 Value = 2040 19.
Let x1 x2 x3
= no. of sailboats rented = no. of cabin cruisers rented = no. of luxury yachts rented
The mathematical formulation of this problem is: Max s.t.
50x1
+ 70x2
+ 100x3
x1 x2 x1 x1
x3 + x2 + x3 + 2x2 + 3x3 x1, x2, x3, 0
4 8 3 10 18
Optimal Solution: x1 = 4, x2 = 4, x3 = 2 Profit = $680. 20. Let
x1 x2 x3
= = =
Max s.t.
0.10x1 2x1 2x1 3x1
number of 20-gallon boxes produced number of 30-gallon boxes produced number of 33-gallon boxes produced + 0.15x2 + + +
+ 0.20x3
3x2 + 3x3 2x2 + 3x3 4x2 + 5x3 x1, x2, x3, 0
7200 Cutting 10800 Sealing 14400 Packaging
Optimal Solution x1 = 0, x2 = 0, x3 = 2400
17 - 15
Chapter 17
Profit = $480. 21. Let
Max s.t.
x1 x2 x3
= = =
no. of gallons of Chocolate produced no. of gallons of Vanilla produced no. of gallons of Banana produced
1.00x1
+ .90x2
+
.95x3
.45x1 .50x1 .10x1
+ .50x2 + .40x3 + .40x2 + .40x3 + .15x2 + .20x3 x1, x2, x3, 0
200 Milk 150 Sugar 60 Cream
Optimal Solution x1 = 0, x2 = 300, x3 = 75 Profit = $341.25. Additional resources: Sugar and Cream. 22. Let
x1 x2 x3 x4 x5 x6
= = = = = =
number of cases of Incentive sold by John number of cases of Temptation sold by John number of cases of Incentive sold by Brenda number of cases of Temptation sold by Brenda number of cases of Incentive sold by Red number of cases of Temptation sold by Red
Max s.t.
30x1
+ 25x2
10x1
+ 15x2
+ 30x3
+ 25x4
15x3
+ 10x4
+ 30x5
12x5 x1, x2, x3, x4, x5, x6, 0 Optimal Solution: x1 = 480
x4 = 480
x2 = 0
x5 = 0
x3 = 0
x6 = 800
Objective Function maximized at 46,400.
Time Allocation: John Brenda Red
Incentive 4800 min. no time no time
Temptation no time 4800 min. 4800 min.
17 - 16
+ 25x6
+ 6x6
4800 4800 4800
Linear Programming: The Simplex Method 23.
Final simplex tableau x1
x2
s1
s2
a2
Basis
cB
4
8
0
0
-M
x2
8
1
1
1/2
0
0
5
a2
-M
-2
0
-1/2
-1
1
3
zj
8+2M
8 4+M/2 +M -M 40-3M
cj - zj
-4-2M
0 -4-M/2 -M
0
Infeasible; optimal solution condition is reached with the artificial variable a2 still in the solution. 24.
Alternative Optimal Solutions x1
x2
s1
s2
s3
Basis
cB
-3
-3
0
0
0
s2
0
0
0
-4/3
1
1/6
4
x1
-3
1
0
-2/3
0
1/12
4
x2
-3
0
1
2/3
0
-1/3
4
zj
-3
-3
0
0
3/4
-24
cj - zj
0
0
0
0
-3/4
indicates alternative optimal solutions exist x1 = 4, x2 = 4, z = 24 x1 = 8, x2 = 0, z = 24
17 - 17
Chapter 17
25.
Unbounded Solution x1
x2
s1
s2
s3
Basis
cB
1
1
0
0
0
s3
0
8/3
0
-1/3
0
1
4
s2
0
4
0
-1
1
0
36
x2
1
4/3
1
-1/6
0
0
4
zj
4/3
1
-1/6
0
0
4
cj - zj
-1/3
0
1/6
0
0
Incoming Column
26.
Alternative Optimal Solutions x1
x2
x3
s1
s2
s3
Basis
cB
2
1
1
0
0
0
x1
2
1
2
1/2
0
0
1/4
4
s2
0
0
0
-1
0
1
-1/2
12
s1
0
0
6
0
1
0
1
12
zj
2
4
1
0
0
1/4
8
cj - zj
0
-3
0
0
0
-1/4
Two possible solutions: x1 = 4, x2 = 0, x3 = 0 or x1 = 0, x2 = 0, x3 = 8 27.
The final simplex tableau is given by: x1
x2
s1
s2
Basis
cB
2
4
0
0
s1
0
1/2
0
1
0
4
x2
4
1
1
0
0
12
s3
0
-1/2
0
0
1
0
zj
4
4
0
0
48
cj - zj
-2
0
0
0
This solution is degenerate since the basic variable s3 is in solution at a zero value.
28.
The final simplex tableau is:
17 - 18
Linear Programming: The Simplex Method
x1
x2
x3
s1
s2
s3
a1
a3
Basis
cB
+4
-5
-5
0
0
0
-M
-M
a1
-M
1
-2
0
-1
1
0
1
0
1
x3
-5
-1
1
1
0
-1
0
0
0
1
a3
-M
-1
1
0
0
-1
-1
0
1
2
zj
+5
-5+M
-5
+M
+5
+M
-M
-M
-5-3M
cj - zj
-1
-M
0
-M
-5
-M
0
0
Since both artificial variables a1 and a3 are contained in this solution, we can conclude that we have an infeasible problem. 29.
We must add an artificial variable to the equality constraint to obtain tableau form. Tableau form: Max s.t.
120x1 + 80x2 + 14x3 + 0s1 4x1
+
32x1 +
8x2 + 2x2 + 4x2 +
+ 0s2
- Ma3
x3 + 1s1 x3 + s2 2x3 + x1, x2, x3, s1, s2, a3 0
a3
= 200 = 300 = 400
Initial Tableau: x1
x2
x3
s1
s2
a3
Basis
cB
120
80
14
0
0
-M
s1
0
4
8
1
1
0
0
200
s2
0
0
2
1
0
1
0
300
a3
-M
32
4
2
0
0
1
400
zj
-32M
-4M
-2M
0
0
-M -400M
120+32M 80+4M 14+2M
0
0
0
cj - zj
17 - 19
bi
ai1
200/4 = 50
400/32 = 12.5
Chapter 17
Iteration 1: x1 enters, a3 leaves (drop a3 column) x1
x2
x3
s1
s2
Basis
cB
120
80
14
0
0
s1
0
0
15/2
3/4
1
0
150
150/ 15/2 = 20
s2
0
0
2
1
0
1
300
300/2 = 150
x1
120
1
1/8
1/16
0
0
12.5
12.5 / 1 /8 = 100
zj
120
15
15/2
0
0
1500
cj - zj
0
65
13/2
0
0
x2
x3
s1
s2
bi
ai 2
Iteration 2: x2 enters, s1 leaves x1 Basis
cB
120 80
14
0
0
x2
80
0
1
1/10
2/15
0
20
s2
0
0
0
8/10 -4/15 1
260
x1
120
1
0
1/20 -1/60 0
10
zj
120 80
14
26/3
2800
0
0
-26/3 0
cj - zj
0
0
Optimal solution: x1 = 10, x2 = 20, and s2 = 260, Value = 2800 30. a.
Note: This problem has alternative optimal solutions; x3 may be brought in at a value of 200. The mathematical formulation of this problem is: Max s.t.
3x1
+ 5x2
+
4x3
12x1 15x1 3x1 x1
+ 10x2 + 15x2 + 4x2
+ 8x3 + 12x3 + 2x3
18,000 12,000 6,000 1,000
C&D S I and P
x1, x2, x3, 0 There is no feasible solution. Not enough sewing time is available to make 1000 All-Pro footballs. b.
The mathematical formulation of this problem is now Max s.t.
3x1 +
5x2 +
4x3
12x +
10x2 +
8x3
18,000
C&D
15x2 +
12x3
18,000
S
4x2 +
2x3
9,000 1,000
I&P
1
15x + 1
3x1 + x1
x1, x2, x3, 0
17 - 20
Linear Programming: The Simplex Method
Optimal Solution x1 = 1000, x2 = 0, x3 = 250 Profit = $4000 There is an alternative optimal solution with x1 = 1000, x2 = 200, and x3 = 0. Note that the additional Inspection and Packaging time is not needed.
17 - 21
Chapter 18 Simplex-Based Sensitivity Analysis and Duality Learning Objectives 1.
Be able to use the final simplex tableau to compute ranges for the coefficients of the objective function.
2.
Understand how to use the optimal simplex tableau to identify dual prices.
3.
Be able to use the final simplex tableau to compute ranges on the constraint right-hand sides.
4.
Understand the concepts of duality and the relationship between the primal and dual linear programming problems.
5.
Know the economic interpretation of the dual variables.
6.
Be able to convert any maximization or minimization problem into its associated canonical form.
7.
Be able to obtain the primal solution from the final simplex tableau of the dual problem.
Solutions:
18 - 1
Chapter 18
1.
a.
Recomputing the cj - zj values for the nonbasic variables with c1 as the coefficient of x1 leads to the following inequalities that must be satisfied. For x2, we get no inequality since there is a zero in the x2 column for the row x1 is a basic variable in. For s1, we get c1 c1
0 4
12 + 2c1 2c1 c1 Range 4 c1
0 12 6 6
0
+
4
-
For s2, we get 0
b.
-
Since x2 is nonbasic we have c2 8
c.
Since s1 is nonbasic we have cs1 1
2.
a. For s1 we get 0 - c2 (8/25) - 50 (-5/25) 0 c2 (8/25) 10 c2 31.25 For s3 we get 0 - c2 (-3/25) - 50 (5/25) 0 c2 (3/25) 10 c2 83.33 Range: 31.25 c2 83.33 b.
For s1 we get 0 - 40 (8/25) - cS2 (-8/25) - 50 (-5/25) 0 -64/5 + cS2 (8/25) + 10 0 cS2 25/8 (14/5) = 70/8 = 8.75 For s3 we get 0 - 40 (-3/25) - cS2 (3/25) - 50 (5/25) 0
18 - 2
Simplex-Based Sensitivity Analysis and Duality 24/5 - cS2 (3/25) - 10 0 cS2 (25/3) (-26/5) = -130/3 = -43.33 Range: -43.33 cS2 8.75 c.
cS3 - 26 / 5 0 cS3 26/5
3.
d.
No change in optimal solution since c2 = 35 is within range of optimality. Value of solution decreases to $35 (12) + $50 (30) = $1920.
a.
It is the zj value for s1. Dual Price = 1.
b.
It is the zj value for s2. Dual Price = 2.
c.
It is the zj value for s3. Dual Price = 0.
d.
s3
= 80 + 5(-2) = 70
x3
= 30 + 5(-1) = 25
x1
= 20 + 5(1) = 25
Value = 220 + e.
5(1)
=
225
s3 = 80 - 10(-2) = 100 x3 = 30 - 10(-1) = 40 x1 = 20 - 10(1) = 10 Value = 220 -
4.
10(1)
=
210
a. 80 30 20
+ + +
b1 (-2) b1 (-1) b1 (1)
0 → 0 → 0 →
b1 b1 b1
40 30 -20
-20 b1 30 100 b1 150
18 - 3
Chapter 18
b. 80 30 20
+ b2 (7) + b2 (3) + b2 (-2)
→ b2 → b2 → b2
0 0 0
-80/7 -10 10
b3
80
→ → →
100 b3 b3 -66 2/3 b3 -150
-10 b2 10 40 b2 60 c. 80 30 20 -
b3 (1) 0 → b3 (0) 0 b3 (0) 0
b3 80 b3 110 5
a. 12 8 30
+ + +
b2 (0) b2 (1) b2 (0)
0 0 0
therefore b2 -8 Range: b2 12 b. 12 8 30
+ b3 (-3/25) + b3 (3/25) + b3 (5/25)
0 0 0
therefore -66 2/3 b3 100 Range: 233 1/3 b3 400 c.
The dual price for the warehouse constraint is 26/5 and the 20 unit increase is within the range of feasibility, so the dual price is applicable for the entire increase. Profit increase = 20 (26/5) = 104
18 - 4
Simplex-Based Sensitivity Analysis and Duality
6.
a.
The final simplex tableau with c1 shown as the coefficient of x1 is x1 x2
s1
s2
s3
s4
Basis
cB
c1
9
0
0
0
0
x2
0
0
1
30/16
0
-21/16
0
252
s2
0
0
0
-15/16
1
5/32
0
120
x1
c1
1
0
-20/16
0
30/16
0
540
s4
0
0
0
-11/32
0
9/64
1
18
zj
c1
9 (270-20c1 )/16
0 (30c 1 -189)/16 0
cj - zj
0
0 (20c 1 -270)/16
0 (189-30c1 )/16 0
(20c1 - 270) / 16 0
→
c1 13.5
(189 - 30c1) / 16 0
→
c1 6.3
2268+540c1
Range: 6.3 c1 13.5 b.
Following a similar procedure for c2 leads to (200 - 30c2) / 16 0
→
c2 6 2/3
(21c2 - 300) / 16 0
→
c2 14 2/7
Range : 6 2/3 c2 14 2/7 c.
There would be no change in product mix, but profit will drop to 540 (10) + 252 (7) = 7164.
d.
It would have to drop below $6 2/3 or increase above $14 2/7.
e.
We should expect more production of deluxe bags since its profit contribution has increased. The new optimal solution is given by x1 = 300, x2 = 420 Optimal Value: $9300
7.
a. 252 120 540 18
+ + + +
b1 (30/16) b1 (-15/16) b1 (-20/16) b1 (-11/32)
0 0 0 0
→ → → →
therefore -134.4 b1 52.36 Range: 495.6 b1 682.36 b.
480 b2
c.
580 b3 900
18 - 5
b1 b1 b1 b1
-134.4 128 432 52.36
Chapter 18
8.
9.
d.
117 b4
e.
The cutting and dyeing and finishing since the dual prices and the allowable increases are positive for both.
a. x1
x2
s1
s2
s3
s4
Basis
cB
10
9
0
0
0
0
x2
9
0
1
30/16
0
-21/16
0
3852/11
s2
0
0
0
-15/16
1
5/32
0
780/11
x1
10
1
0
-20/16
0
30/16
0
5220/11
s4
0
0
0
-11/32
0
9/64
1
0
zj
10
9
70/16
0
111/16
0
86,868/11 = 7897 1 /11
cj - zj
0
0
-70/16
0
-111/16
0
b.
No, s4 would become nonbasic and s1 would become a basic variable.
a.
Since this is within the range of feasibility for b1, the increase in profit is given by
2100 70 16 30 = 16 b.
It would not decrease since there is already idle time in this department and 600 - 40 = 560 is still within the range of feasibility for b2.
c.
Since 570 is within the range of feasibility for b1, the lost profit would be equal to
4200 70 16 60 = 16 10. a. b.
The value of the objective function would go up since the first constraint is binding. When there is no idle time, increased efficiency results in increased profits. No. This would just increase the number of idle hours in the sewing department.
18 - 6
Simplex-Based Sensitivity Analysis and Duality
11. a. x1
x2
s1
s2
s3
Basis
cB
c1
30
0
0
0
x2
30
0
1
10/3
0
-20/9
20
s2
0
0
0
-2/3
1
4/9
1
x1
c1
1
0
-5/3
0
25/9
25
zj
c1
30
100-(5/3c 1 )
0
cj - zj
0
0
5 / c -100 3 1
0
-200 25 + c 3 9 1 200 - 25 c 3 9 1
600 + 25c1
Hence /3c1 - 100 0
5
and 200/3 - 25/9c1 0 Using the first inequality we obtain /3c1 100 or c1 60
5
Using the second inequality we obtain /9c1 200/3
25
c1 (9/25) (200/3) c1 24 Thus the range of optimality for c1 is given by 24 c1 60 A similar approach for c2 leads to (200 - 10c2) / 3 0 →
c2 20
(20c2 - 1000) / 9 0 →
c2 50
Range: 20 c2 50 b.
Current solution is still optimal. However, the total profit has been reduced to $30 (25) + $30 (20) = $1350.
c.
From the zj entry in the s1 column we see that the dual price for the material 1 constraint is $33.33. It is the increase in profit that would result from having one additional ton of material one.
d. 12. a.
Material 3 is the most valuable and RMC should be willing to pay up to $44.44 per ton for it. 20
+
b1 (10/3)
0
→
b1
18 - 7
-6
Chapter 18
1 25
+ +
b1 (-2/3) b1 (-5/3)
0 0
→ →
b1 b1
0 0 0
→ → →
no restriction b2 -1 no restriction
0 0 0
→ → →
b3 b3 b3
3/2 15
therefore -6 b1 1 1/2 Range: 14 b1 21 1/2 b. 20 1 25
+ + +
b2 (0) b2 (1) b2 (0)
Range: b2 4 c. 20 1 25
+ + +
b3 (-20/9) b3 (4/9) b3 (25/9)
9 -9/4 -9
therefore -2 1/4 b3 9 Range: 18 3/4 b3 30 d.
Dual price: 400/9 Valid for 18 3/4 b3 30
13. a.
b.
The final simplex tableau is given by x1
x2
x3
x4
s2
s3
Basis
cB
3
1
5
3
0
0
s2
0
5/2
7/6
0
0
1
1/3
115/3
x3
5
3/2
1/2
1
0
0
0
15
x4
3
0
2/3
0
1
0
1/3
25/3
zj
15/2 9/2
5
3
0
1
100
cj - zj
-9/2 -7/2
0
0
0
-1
Range: 2 c3
18 - 8
Simplex-Based Sensitivity Analysis and Duality
c.
Since 1 is not contained in the range of optimality, a new basis will become optimal. The new optimal solution and its value is x1 = 10 x4 = 25/3 s2 = 40/3 (Surplus associated with constraint 2)
d.
Since x2 is a nonbasic variable we simply require c2 - 9/2 0 Range: c2 4 1/2
e. 14. a.
Since 4 is contained in the range, a three unit increase in c2 would have no effect on the optimal solution or on the value of that solution. 400/3 b1 800
b.
275 b2
c.
275/2 b3 625
15. The final simplex tableau is given: x1
x2
x3
s1
s2
s3
Basis
cB
15
30
20
0
0
0
x1
15
1
0
1
1
0
0
4
x2
30
0
1
1/4
-1/4 1/2
0
1/2
s3
0
0
0
3/4
-3/4 -1/2 1
3/2
zj
15
30 45/2
15/2 15
0
75
cj - zj
0
0
-5/2 -15/2 -15
0
a.
x1 = 4, x2 = 1/2
Optimal value: 75
b.
75
c.
Constraints one and two.
d.
There are 1 1/2 units of slack in constraint three.
e.
Dual prices: 15/2, 15, 0 Increasing the right-hand side of constraint two would have the greatest positive effect on the objective function.
f. 12.5 20
c1 c2
60
18 - 9
Chapter 18
c3
22.5
The optimal values for the decision variables will not change as long as the objective function coefficients stay in these intervals. g.
For b1 4 + 1/2 + 3/2 +
b1 (1) b1 (-1/4) b1 (-3/4)
0 0 0
→ → →
b1 b1 b1
-4 2 2
0 0 0
→ → →
no restriction b2 − b2
→ → →
no restriction no restriction b3 -3/2
therefore -4 b1 2 Range: 0 b1 6 For b2 4 + 1/2 + 3/2 +
b2 (0) b2 (1/2) b2 (-1/2)
therefore -1 b2 3 Range: 2 b2 6 For b3 4 + b3 (0) 1/2 + b3 (0) 3/2 + b3 (1)
0 0 0
therefore -3/2 b3 Range: 4 1/2 b3 The dual prices accurately predict the rate of change of the objective function with respect to an increase in the right-hand side as long as the right-hand side remains within its range of feasibility. 16. a.
After converting to a maximization problem by multiplying the objective function by (-1) and solving we obtain the optimal simplex tableau shown. x1
x2
s1
s2
s3
Basis
cB
-8
-3
0
0
0
s3
0
0
0
1/60
1/6
1
7,000
x1
-8
1
0
-1/75
-1/3
0
4,000
x2
-3
0
1
1/60
1/6
0
10,000
zj
-8
-3 17/300
13/6
0
-62,000
cj - zj
0
0 -17/300 -13/6
0
Total Risk = 62,000 b.
The dual price for the second constraint is -13/6 = -2.167. So, every $1 increase in the annual income requirement increases the total risk of the portfolio by 2.167.
18 - 10
Simplex-Based Sensitivity Analysis and Duality
c. 7000 4000 10,000
d.
-
b2 (1/6) b2 (-1/3) b2 (1/6)
0 0 0
So,
-12,000 b2 42,000
and
48,000 b2
→ → →
b2 b2 b2
42,000 -12,000 60,000
37000/6 17,000/3 55,000/6
= = =
6,166.667 5,666.667 9,166.67
102,000
The new optimal solution and its value are s3 x1 x2
= = =
7000 4000 10,000 -
5000(1/6) 5000(-1/3) 5000(1/6)
= = =
Value = -62,000 - 5000(13/6) = -437,000/6 = -72,833.33 Since, this is a min problem being solved as a max, the new optimal value is 72,833.33 e.
There is no upper limit in the range of optimality for the objective function coefficient of the stock fund. Therefore, the solution will not change. But, its value will increase to: 9(4,000) + 3(10,000) = 66,000
17. a.
The dual is given by: Min s.t.
b.
550u1
+
700u2
+
1.5u1 2u1 4u1 3u1
+ + + +
4u2 + 1u2 + 2u2 + 1u2 + u1, u2, u3, 0
200u3 2u3 3u3 1u3 2u3
4 6 3 1
Optimal solution: u1 = 3/10, u2 = 0, u3 = 54/30 The zj values for the four surplus variables of the dual show x1 = 0, x2 = 25, x3 = 125, and x4 = 0.
c.
Since u1 = 3/10, u2 = 0, and u3 = 54/30, machines A and C (uj > 0) are operating at capacity. Machine C is the priority machine since each hour is worth 54/30.
18 - 11
Chapter 18
18.
The dual is given by: Max s.t.
19.
5u1
+
5u2
+
24u3
15u1 + 4u2 + 15u1 + 8u2 u1 + u1, u2, u3 0
12u3
2800 6000 1200
8u3
The canonical form is Max s.t.
3x1
+
3x1 -3x1 -2x1
+ -
x2
+
5x3
+
1x2 + 2x3 1x2 - 2x3 1x2 - 3x3 2x2 x1, x2, x3, x4, 0.
3x4
+
x4 3x4
30 -30 -15 25
3 1 5 3
The dual is Max s.t.
30u1'
-
30u1''
-
15u2
3u1' u 1' 2u1'
-
3u1'' u1'' 20u1''
-
2u2 u2 3u2 u2
+
25u3
+
2u3
+
3u3
u1', u1'', u2, u3 0 20. a. Max s.t.
30u1
+
20u2
u1 u2 u1, u2, u3 0 b.
+
80u3
+ +
u3 2u3
1 1
The final simplex tableau for the dual problem is given by u1
u2
u3
s1
s2
Basis
cB
30
20
80
0
0
u1
30
1
-1/2
0
1
-1/2
1/2
u3
80
0
1/2
1
0
1/2
1/2
zj
30
25
80
30
25
55
cj - zj
0
-5
0
-30
-25
The zj values for the two slack variables indicate x1 = 30 and x2 = 25. c. 21. a.
With u3 = 1/2, the relaxation of that constraint by one unit would reduce costs by $.50. Max
15u1
+
30u2
+
20u3
18 - 12
Simplex-Based Sensitivity Analysis and Duality
s.t. u1 0.5u1 + 2u2 u1 + u2 u1, u2, u3 0 b.
+ + +
4 3 6
u3 u3 2u3
The optimal simplex tableau for the dual is u1
u2
u3
s1
s2
s3
Basis
cB
15
30
20
0
0
0
u1
15
1
0
1
1
0
0
4
u2
30
0
1
1/4
-1/4 1/2
0
1/2
s3
0
0
0
3/4
-3/4 -1/2
1
3/2
zj
15
30 45/2
15/2 15
0
75
cj - zj
0
0
-5/2 -15/2 -15
0
c.
From the zj values for the surplus variables we see that the optimal primal solution is x1 = 15/2, x2 = 15, and x3 = 0.
d.
The optimal value for the dual is shown in part b to equal 75. Substituting x1 = 15/2 and x2 = 15 into the primal objective function, we find that it gives the same value. 4(15/2) + 3(15) = 75
22. a. Max s.t.
10x1
+
5x2
x1 x1 3x1 + x1, x2 0 b.
x2 x2 x2
20 20 100 100 175
The dual problem is Min s.t.
-20u1 - 20u2 + 100u3+ 100u4 -u1
+ u3 u2 + u1, u2, u3, u4, u5 0
u4
+
175u5
+ +
3u5 u5
10 5
The optimal solution to this problem is given by: u1 = 0, u2 = 0, u3 = 0, u4 = 5/3, and u5 = 10/3. c.
The optimal number of calls is given by the negative of the dual prices for the dual: x1 = 25 and x2 = 100. Commission = $750.
18 - 13
Chapter 18
d.
u4 = 5/3: $1.67 commission increase for an additional call for product 2. u5 = 10/3: $3.33 commission increase for an additional hour of selling time per month.
23. a.
b.
Extreme point 1:
x1 = 0, x2 = 0
value = 0
Extreme point 2:
x1 = 5, x2 = 0
value = 15
Extreme point 3:
x1 = 4, x2 = 2
value = 16
Dual problem: Min s.t.
c.
8u1 + 10u2 u1 + 2u2 2u1 + u2 u1, u2, 0
Extreme Point 1: u1 = 3, u2 = 0
3 2
value = 24
Extreme Point 2: u1 = 1/3, u2 = 4/3 value = 16 Extreme Point 3: u1 = 0, u2 = 2
value = 20
d.
Each dual extreme point solution yields a value greater-than-or-equal-to each primal extreme point solution.
e.
No. The value of any feasible solution to the dual problem provides an upper bound on the value of any feasible primal solution.
24. a.
If the current optimal solution satisfies the new constraints, it is still optimal. Checking, we find 6(10) +
4(30) - 15 = 165
170
ok
1
30
= 32.5
25
ok
/4(10) +
Both of the omitted constraints are satisfied.
18 - 14
Simplex-Based Sensitivity Analysis and Duality
Therefore, the same solution is optimal.
18 - 15
Chapter 19 Solution Procedures for Transportation and Assignment Problems Learning Objectives 1.
Be able to use the transportation simplex method to solve transportation problems.
2.
Know how to set up a transportation tableau.
3.
Be able to use the minimum cost method to find an initial feasible solution to a transportation problem.
4.
Know what the MODI method is and how it is use it to determine the incoming transportation route that will cause the largest reduction per unit in the total cost of the current solution.
5.
Understand the stepping-stone method and be able to use it to improve the transportation solution.
6.
Be able to handle degenerate solutions by keeping m + n – 1 cells occupied.
7.
Know how to adapt the transportation simplex method to solve transportation problems involving a maximization objective and unacceptable routes.
8.
Be able to use the Hungarian method to solve assignment problems.
9.
Know how to adapt the Hungarian method to solve assignment problems involving a maximization objective, unacceptable assignments, and the number of agents not equal to the number of tasks.
10.
Understand the following terms: transportation simplex method transportation tableau minimum cost method MODI method stepping-stone method dummy destination
Hungarian method matrix reduction opportunity loss net-evaluation index degenerate solution dummy origin
19 - 1
Chapter 19
Solutions: 1.
a.
This is the minimum cost solution since e ij 0 for all i, j. Solution: Shipping Route (Arc) O1 - D1 O1 - D3 O2 - D3 O2 - D4 O3 - D1 O4 - D2 O4 - D4 b.
Units Unit Cost Arc Shipping Cost 25 5 $ 125 50 10 500 100 8 800 75 2 150 100 6 600 100 5 500 50 4 200 Total Transportation Cost: $2875
Yes, e32 = 0. This indicates that we can ship over route O3 - D2 without increasing the cost. To find the alternative optimal solution identify cell (3, 2) as the incoming cell and make appropriate adjustments on the stepping stone path. The increasing cells on the path are O4 - D4, O2 - D3, and O1 - D1. The decreasing cells on the path are O4 - D2, O2 - D4, O1 - D3, and O3 - D1. The decreasing cell with the smallest number of units is O1 - D3 with 50 units. Therefore, 50 units is assigned to O3 - D2. After making the appropriate increases and decreases on the stepping stone path the following alternative optimal solution is identified.
19 - 2
Solution Procedures for Transportation and Assignment Problems
Note that all eij 0 indicating that this solution is also optimal. Also note that e13 = 0 indicating there is an alternative optimal solution with cell (1, 3) in solution. This is the solution we found in part (a). An initial solution is given below.
San Jose
Las Vegas
4
10
6
8
16
6
100
100
200 14
Tucson
San Diego
San Francis co
a.
Los Angeles
2.
18
10
300
Total Cost: $7800 b.
Note that the initial solution is degenerate. A zero is assigned to the cell in row 3 and column 1 so that the row and column indices can be computed.
19 - 3
Chapter 19
vj ui
8
4 4
0
10
100
4 16
4
100 14
10
6
2 8
4
2
0
6 200 10
18 300
0
-2
Cell in row 3 and column 3 is identified as an incoming cell.
Stepping-stone path shows cycle of adjustments. Outgoing cell is in row 3 and column 1.
vj ui
10
4 4
0
10 4 16 2
100 14
8
6
0
100 8
4
2
2
6 200 10
18 300
0
Solution is recognized as optimal. It is degenerate. Thus, the initial solution turns out to be the optimal solution; total cost = $7800.
19 - 4
Solution Procedures for Transportation and Assignment Problems
c.
To begin with we reduce the supply at Tucson by 100 and the demand at San Diego by 100; the new solution is shown below:
vj ui
4 4
0
6 4
-2 -
8
100
16
100
4
6 100
14 6
2 10
100 +
4
12
M
18 200
M-8
vj ui
10
2 4 2
0
10 6
200
16 0
14 8
4
6
100 8
6
0
100 18
200
19 - 5
6
M M-8
Chapter 19
Optimal Solution: recall, however, that the 100 units shipped from Tucson to San Diego must be added to obtain the total cost. San Jose to San Francisco: Las Vegas to Los Angeles: Las Vegas to San Diego: Tucson to San Francisco: Tucson to San Diego: Total Cost: $7800
100 200 100 200 100
Note that this total cost is the same as for part (a); thus, we have alternative optima. d.
3.
The final transportation tableau is shown below. The total transportation cost is $8,000, an increase of $200 over the solution to part (a).
a.
b.
19 - 6
Solution Procedures for Transportation and Assignment Problems
14
9
7 30
8
10
10 25 15
5 10
15
10
This is an initial feasible solution with a total cost of $475.
19 - 7
20 10 0
Chapter 19
4.
An initial feasible solution found by the minimum cost method is given below.
Computing row and column indexes and evaluating the unoccupied cells one identifies the cell in row 2 and column 1 as the incoming cell.
The +'s and -'s above show the cycle of adjustments necessary on the stepping-stone path as flow is allocated to the cell in row 2 and column 1. The cell in row 1 and column 1 is identified as corresponding to the outgoing arc. The new solution follows. vj
ui
16
16
20 0
4
16 10
100
10 100
100
8 300
12 -4
24
300 10
-6
14
10
18 6
0
Since all per-unit costs are 0, this solution is optimal. However, an alternative optimal solution can be found by shipping 100 units over the P3 - W3 route.
19 - 8
Solution Procedures for Transportation and Assignment Problems
5.
a.
Initial Solution:
Total Cost: $4600 b.
Incoming arc: O1 - D3
Outgoing arc: O2 - D3
19 - 9
Chapter 19
vj ui
6
8
8
6 0
150
8 50
50
18 8
4
12
14 2
150 8
12
0
2
8
2
10 100
Since all cell evaluations are non-negative, the solution is optimal; Total Cost: $4500. c.
At the optimal solution found in part (b), the cell evaluation for O3 - D1 = 0. Thus, additional units can be shipped over the O3 - D1 route with no increase in total cost.
D1 _ O1
D2 6
150
D3 8
O2
8
50
50 18
+
12
14
150 8
12
O3
_ 100
19 - 10
10
Solution Procedures for Transportation and Assignment Problems
Thus, an alternative optimal solution is
6.
Subtract 10 from row 1, 14 from row 2, and 22 from row 3 to obtain: 1
2
3
Jackson
0
6
22
Ellis
0
8
26
Smith
0
2
12
Subtract 0 from column 1, 2 from column 2, and 12 from column 3 to obtain:
Two lines cover the zeros. The minimum unlined element is 4. Step 3 yields: 1
2
3
Jackson
0
0
6
Ellis
0
2
10
Smith
0
0
0
19 - 11
Chapter 19
Optimal Solution: Jackson - 2 Ellis -1 Smith - 3 Time requirement is 64 days. 7.
Subtract 30 from row 1, 25 from row 2, 23 from row 3, 26 from row 4, and 26 from row 5 to obtain: 1
2
3
4
5
Red
0
14
8
17
1
White
0
7
20
19
0
Blue
0
17
14
16
6
Green
0
12
11
19
2
Brown
0
8
18
17
2
Subtract 0 from column 1, 7 from column 2, 8 from column 3, 16 from column 4, and 0 from column 5 to obtain: 1
2
3
4
5
Red
0
7
0
1
1
White
0
0
12
3
0
Blue
0
10
6
0
6
Green
0
5
3
3
2
Brown
0
1
10
1
2
Four lines cover the zeroes. The minimum unlined element is 1. Step 3 of the Hungarian algorithm yields: 1
2
3
4
5
Red
1
7
0
1
1
White
1
0
12
3
0
Blue
1
10
6
0
6
Green
0
4
2
2
1
Brown
0
0
9
0
1
19 - 12
Solution Procedures for Transportation and Assignment Problems
Optimal Solution: Green to Job 1 Brown to Job 2 Red to Job 3 Blue to Job 4 White to Job 5
$26 34 38 39 25 $162
Total cost is $16,200. 8.
After adding a dummy column, we get an initial assignment matrix.
Applying Steps 1 and 2 we obtain:
Applying Step 3 followed by Step 2 results in:
19 - 13
Chapter 19
Finally, application of Step's 3 and 2 lead to the optimal solution shown below.
Terry: Carle: McClymonds: Higley:
Client 2 (15 days) Client 3 ( 5 days) Client 1 ( 6 days) Not accepted Total time = 26 days
Note: An alternative optimal solution is Terry: Client 2, Carle: unassigned, McClymonds: Client 3, and Higley: Client 1. 9.
We start with the opportunity loss matrix.
Optimal Solution
1
2
3
4 Dummy
Shoe
4
11
0
5
0
Toy :
2
18
Toy
0
0
8
3
1
Auto :
4
16
Auto
0
10
2
0
3
Houseware :
3
13
Houseware
1
6
0
4
1
Video :
1
14
Video
0
1
6
1
0
19 - 14
Profit
61
Solution Procedures for Transportation and Assignment Problems
10.
Subtracting each element from the largest element in its column leads to the opportunity loss matrix.
7
10
1
4
0
2
M
8
1
0
0
6
0
M
0
3
4
0
2
0
3
0
7
0
0
Optimal Solution
1
2
3
4 Dummy
Shoe
6
9
1
3
0
Toy :
4
11
Toy
1
M
8
0
0
Auto :
1
17
Auto
0
6
1
M
1
Houseware :
3
13
Houseware
2
3
0
1
0
Video :
2
16
Video
3
0
8
0
1
19 - 15
Profit
57
Chapter 20 Minimal Spanning Tree Learning Objectives 1.
Know the concept of a spanning tree for a network.
2.
Understand the minimal spanning tree problem.
3.
Be able to identify the minimal spanning tree solution for a network.
Solutions: 1.
20 - 1
Chapter 20
1-3 3-4 4-5 5-2 5-7 7-8 7-6
2 2 3 2 3 2 3 17 miles
2. Start Node 1 6 7 7 10 9 9 3 4 7 8 14 15 14
End Node 6 7 8 10 9 4 3 2 5 11 13 15 12 13
Distance 2 3 1 2 3 2 3 1 3 4 4 2 3 4
Total length = 37
3.
20 - 2
Minimal Spanning Tree
Minimum length of connections = 2 + 0.5 + 1 + 1 + 2 + 0.5 + 1 = 8 8000 feet 4.
Minimum length of cable lines = 2 + 2 + 2 + 3 + 3 + 2 +3 + 3 + 4 + 4 = 28 miles
20 - 3
Chapter 21 Dynamic Programming Learning Objectives 1.
Understand the basics of dynamic programming and its approach to problem solving.
2.
Learn the general dynamic programming notation.
3.
Be able to use the dynamic programming approach to solve problems such as the shortest route problem, the knapsack problem and production and inventory control problems.
4.
Understand the following terms: stages state variables principle of optimality stage transformation function return function knapsack problem
21 - 1
Chapter 21
Solutions: 1. Route (1-2-5-8-10) (1-2-5-9-10) (1-2-6-8-10) (1-2-6-9-10) (1-2-7-8-10) (1-2-7-9-10) (1-3-5-8-10) (1-3-5-9-10)
Value 22 25 24 20 25 24 19 22
Route (1-3-6-8-10) (1-3-6-9-10) (1-3-7-8-10) (1-3-7-9-10) (1-4-5-8-10) (1-4-5-9-10) (1-4-6-8-10) (1-4-6-9-10)
Value 26 22 22 21 22 25 27 23
The route (1-3-5-8-10) has the smallest value and is thus the solution to the problem. The dynamic programming approach results in fewer computations because all 16 paths from node 1 to node 10 need not be computed. For example, at node 1 we considered only 3 paths: the one from 1-2 plus the shortest path from node 2 to node 10, the one from 1-3 plus the shortest path from node 3 to node 10, and the one from 1-4 plus the shortest path from node 4 to node 10. 2.
a.
The numbers in the squares above each node represent the shortest route from that node to node 10. 18
8
2
7
19
8
10 26 1
8
8
5
3 6
5
21
7
7
11
4
17
9
10
10
8 5
6
6
11
4
10 10 6
9
The shortest route is given by the sequence of nodes (1-4-6-9-10). b.
The shortest route from node 4 to node 10 is given by (4-6-9-10).
c. Route (1-2-5-7-10) (1-2-5-8-10) (1-2-5-9-10) (1-3-5-7-10) (1-3-5-8-10) (1-3-5-9-10)
Value 32 36 28 31 35 27
Route (1-3-6-8-10) (1-3-6-9-10) (1-4-6-8-10) (1-4-6-9-10)
Value 34 31 29 26
See 1 above for an explanation of how the computations are reduced.
21 - 2
Dynamic Programming
3.
Use 4 stages; one for each type of cargo. Let the state variable represent the amount of cargo space remaining. a.
In hundreds of pounds we have up to 20 units of capacity available. Stage 1 (Cargo Type 1) x1
0
1
2
d1*
f1(x1)
x0
0-7
0
0
0
0-7
8-15
0
22
1
22
0-7
16-20
0
22
44
2
44
0-4
Stage 2 (Cargo Type 2) x2
0
1
2
d 2*
f2(x2)
x1
0-4
0
0
0
0-4
5-7
0
12
1
12
0-2
8-9
22
12
0
22
8-9
10-12
22
12
24
2
24
0-2
13-15
22
34
24
1
34
8-10
16-17
44
34
24
0
44
16-17
18-20
44
34
46
2
46
8-10
Stage 3 (Cargo Type 3)
21 - 3
Chapter 21
x3
0
1
2
3
4
d 3*
f3(x3)
x2
0-2
0
0
0
0-2
3-4
0
7
1
7
0-1
5
12
7
0
12
5
6-7
12
7
14
2
14
0-1
8
22
19
14
0
22
8
9
22
19
14
21
0
22
9
10
24
19
14
21
0
24
10
11
22
29
26
21
1
29
8
12
24
29
26
21
28
1
29
9
13
34
31
26
21
28
0
34
13
14-15
34
31
36
33
28
2
36
8-9
16
44
41
38
33
28
0
44
16
17
44
41
38
43
40
0
44
17
18
46
41
38
43
40
0
46
18
19
46
51
48
45
40
1
51
16
20
46
51
48
45
50
1
51
17
Stage 4 (Cargo Type 4) x4
0
1
2
3
d 4*
f4(x4)
x3
20
51
49
50
45
0
51
20
Tracing back through the tables we find
Stage 4
State Variable Entering 20
Optimal Decision 0
State Variable Leaving 20
3
20
1
17
2
17
0
17
1
17
2
1
Load 1 unit of cargo type 3 and 2 units of cargo type 1 for a total return of $5100. b.
Only the calculations for stage 4 need to be repeated; the entering value for the state variable is 18. x4
0
1
2
3
d 4*
f4(x4)
x3
18
46
47
42
38
1
47
16
Optimal solution: d4 = 1, d3 = 0, d2 = 0, d1 = 2
21 - 4
Dynamic Programming
Value = 47 4.
a.
There are two optimal solutions each yielding a total profit of 186. # of Employees
Return
3 2 0 3
44 48 46 48 186
# of Employees
Return
2 3 0 3
37 55 46 48 186
# of Employees
Return
1 2 0 3
30 48 46 48 172
Activity 1 Activity 2 Activity 3 Activity 4
Activity 1 Activity 2 Activity 3 Activity 4
b.
Activity 1 Activity 2 Activity 3 Activity 4
5.
a.
Set up a stage for each possible length the log can be cut into: Stage 1 - 16 foot lengths, Stage 2 - 11 foot lengths, Stage 3 - 7 foot lengths, and Stage 4 - 3 foot lengths. Stage 1 d1 x1
0
1
d1*
f1(x1)
x0
0-15
0
0
0
x1
16-20
0
8
1
8
x1-16
21 - 5
Chapter 21
Stage 2 d2 x2
0
1
d 2*
f2(x2)
x1
0-10
0
0
0
x2
11-15
0
5
1
5
x2-11
16-20
8
5
0
8
x2
x3
0
d3 1
2
d 3*
f3(x3)
x2
0-6
0
0
0
x3
7-10
0
3
1
3
x3-7
11-13
5
3
0
5
x3
14-15
5
3
6
2
6
x3-14
16-17
8
3
6
0
8
x3
18-20
8
8
8
1
8
x3-7
x4
0
1
2
d4 3
4
5
6
d 4*
f4(x4)
x3
20
8
6
1
9
17
Stage 3
Stage 4
Tracing back through the tableau, we see that our optimal decisions are d 4* = 1, d 3* = 0, d 2* = 0, d1* = 1.
The total return per log when this pattern is used is $9. b.
We simply create a stage for every length.
21 - 6
Dynamic Programming
6.
a.
In the network representation below each node represents the completion of an assignment. The numbers above the arcs coming into the node represent the time it takes to carry out the assignment.
Once the network is set up we can solve as we did in section 18.1 for the shortest route problem. The optimal sequence of assignments is A-G-J-M. The total training period for this solution is 30 months. b. 7.
He should choose H. The training program can then be completed in 17 months. Let each stage correspond to a store. Let the state variable xn represent the number of golf balls remaining to allocate at store n and dn represent the number allocated to store n. Stage 1 (Store 1)
x1
0
100
200
d1 300
400
500
d1*
f1(x1)
x0
0
0
0
0
0
100
0
600
100
600
0
200
0
600
1100
200
1100
0
300
0
600
1100
1550
300
1550
0
400
0
600
1100
1550
1700
400
1700
0
500
0
600
1100
1550
1700
1800
500
1800
0
21 - 7
Chapter 21
Stage 2 (Store 2) d2 x2
0
100
200
300
400
500
d 2*
f2(x2)
x1
0
0
0
0
0
100
600
500
100
600
100
200
1100
1100
1200
200
1200
0
300
1550
1600
1800
1700
200
1800
100
400
1700
2050
2300
2300
2000
200 or 300
2300
200 or 100
500
1800
2200
2750
2800
2600
2100
300
2800
200
Stage 3 (Store 3) d3 x3
0
100
200
300
400
500
d 3*
f3(x3)
x2
500
2800
2850
2900
2700
2450
1950
200
2900
300
Robin should ship 200 dozen to store 3, 200 dozen to store 2 and 100 dozen to store 1. He can expect a profit of $2900. 8.
a.
Let each of the media represent a stage in the dynamic programming formulation. Further, let the state variable, xn, represent the amount of budget remaining with n stages to go and the decision variable, dn, the amount of the budget allocated to media n. Stage 1 (Internet)
x1
0
1
2
3
d1 4
5
6
7
8
d1*
f1(x1)
x0
0
0
0
0
0
1
0
24
1
24
0
2
0
24
37
2
37
0
3
0
24
37
46
3
46
0
4
0
24
37
46
59
4
59
0
5
0
24
37
46
59
72
5
72
0
6
0
24
37
46
59
72
80
6
80
0
7
0
24
37
46
59
72
80
82
7
82
0
8
0
24
37
46
59
72
80
82
82
7 or 8
82
0 or 1
Stage 2 (Print)
21 - 8
Dynamic Programming
x2
0
1
2
3
d2 4
5
6
7
8
d 2*
f2(x2)
x1
0
0
0
0
0
1
24
15
0
24
1
2
37
39
55
2
55
0
3
46
52
79
70
2
79
1
4
59
61
92
94
75
3
94
1
5
72
74
101
107
99
90
3
107
2
6
80
87
114
116
112
114
95
3
116
3
7
82
95
127
129
121
127
119
95
3
129
4
8
82
97
135
142
134
136
132
119
95
3
142
5
Stage 3 (Radio)
x3
0
1
2
3
d3 4
5
6
7
8
d 3*
f3(x3)
x2
0
0
0
0
0
1
24
20
0
24
1
2
55
44
30
0
55
2
3
79
75
54
45
0
79
3
4
94
99
85
69
55
1
99
3
5
107
114
109
100
79
60
1
114
4
6
116
127
124
124
110
84
62
1
127
5
7
129
136
137
139
134
115
86
63
3
139
4
8
142
149
146
152
149
139
117
87
63
3
152
5
Stage 4 (Television)
x4
0
1
2
3
d4 4
5
6
7
8
d 4*
f4(x4)
x3
8
152
159
167
169
164
149
125
94
70
3
169
5
Tracing back through the tables, we find the optimal decision is d4 = 3 d3 = 1 d2 = 3 d1 = 1 b.
x3 = x4 - d 4 = 8 - 3 = 5 x2 = x3 - d 3 = 5 - 1 = 4 x1 = x2 - d 2 = 4 - 3 = 1 x0 = x1 - d 1 = 1 - 1 = 0
This gives the agency a maximum exposure of 169. We simply redo the calculations for stage 4 with x4 = 6.
21 - 9
Chapter 21
x4
0
1
2
d4 3
4
5
6
d 4*
f4(x4)
x3
6
127
134
139
134
120
94
70
2
139
4
Tracing back through the tables, we find a new optimal solution of d4 = 2 d3 = 1 d2 = 2 d1 = 1
x3 = x4 - d 4 = 6 - 2 = 4 x2 = x3 - d 3 = 4 - 1 = 3 x1 = x2 - d 2 = 3 - 2 = 1 x0 = x1 - d 1 = 1 - 1 = 0
This gives the agency a maximum value of 139. c.
We can simply return to the stage 3 table and read off the answers. If $8,000 was available, the optimal decisions are d3 = 3 d2 = 3 d1 = 2. The maximum exposure value is 152. If $6,000 was available, the optimal decisions are d3 = 1 d2 = 3 d1 = 2. The maximum exposure is 127.
9.
a.
21 - 10
Dynamic Programming d3
d2
d1
100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 500 500 500 500 500 500 500 500 500
100 100 100 100 100 300 300 300 300 300 500 500 500 500 500 600 600 600 600 800 800 100 100 100 100 100 300 300 300 500
0 100 200 300 400 0 100 200 300 400 0 100 200 300 400 0 100 200 300 0 100 0 100 200 300 400 0 100 200 0
r1(d1) + r2(d2) + r3(d3) 295 405 595 695 720 575 685 875 975 1000 825 935 1125 1225 1250 875 985 1175 1275 1150 1260 820 930 1120 1220 1245 1100 1210 1400 1350
The optimal solution is d1 = 200, d2 = 300, d3 = 500. b.
Stage 1
x1
0
100
d1 200
300
400
d1*
f1(x1)
x0
0
0
0
0
0
100
0
110
100
110
0
200
0
110
300
200
300
0
300
0
110
300
400
300
400
0
Stage 2 d2
21 - 11
Chapter 21
x2
0
300
500
600
800
d 2*
f2(x2)
x1
500
545
700
650
300
700
200
900
545
825
1075
1100
1085
600
1100
300
Stage 3 d3 x3
100
500
d 3*
f3(x3)
x2
1000
1275
1400
500
1400
500
Tracing back through the tableaus, we see we get the same solution as in (a) with much less effort. 10.
The optimal production schedule is given below.
Month 1 2 3
Beginning Inventory 10 10 0
Production 20 20 30
21 - 12
Ending Inventory 10 0 0
Appendix A Building Spreadsheet Models Learning Objectives 1. Learn the basic fundamentals of spreadsheet modeling. 2. Learn how to create and save a spreadsheet model in Microsoft Excel. 3. Learn to use functions that are useful in quantitative modeling. These functions include, SUMPRODUCT, IF, COUNTIF and VLOOKUP. 4. Learn the principles of building good spreadsheet models. 5. Learn to how to audit a spreadsheet model in Excel to ensure that the model is error free and accurate.
1.
2. The formula in cell E6 is as follows:
3.
=F6*$F$3
4.
5. a.
b.
c.
6.
7.
8.
9.
a. A portion of the spreadsheet is shown below. b. See column I below.
10.
11. Error #1:
The formula in cell C17 is: =SUMPRODUCT(C8:G11,B22:F25) but should be =SUMPRODUCT(C8:F11,B22:F25)
Error #2:
The formula in cell C17 is: The formula in cell G22 is: =SUM(B22:E22) But should be =SUM(B22:F22)
Part 2: Case Solutions Chapter 1 Introduction Case Problem: Scheduling a Golf League Note to Instructor: This case problem illustrates the value of the rational management science approach. The problem is easy to understand and, at first glance, appears simple. But, most students will have trouble finding a solution. The solution procedure suggested involves decomposing a larger problem into a series of smaller problems that are easier to solve. The case provides students with a good first look at the kinds of problems where management science is applied in practice. The problem is a real one that one of the authors was asked by the Head Professional at Royal Oak Country Club for help with. Solution: Scheduling problems such as this occur frequently, and are often difficult to solve. The typical approach is to use trial and error. An alternative approach involves breaking the larger problem into a series of smaller problems. We show how this can be done here using what we call the Red, White, and Blue algorithm. Suppose we break the 18 couples up into 3 divisions, referred to as the Red, White, and Blue divisions. The six couples in the Red division can then be identified as R1, R2, R3, R4, R5, R6; the six couples in the White division can be identified as W1, W2,…, W6; and the six couples in the Blue division can be identified as B1, B2,…, B6. We begin by developing a schedule for the first 5 weeks of the season so that each couple plays every other couple in its own division. This can be done fairly easily by trial and error. Shown below is the first 5-week schedule for the Red division. Week 1 R1 vs. R2 R3 vs. R4 R5 vs. R6
Week 2 R1 vs. R3 R2 vs. R5 R4 vs. R6
Week 3 R1 vs. R4 R2 vs. R6 R3 vs. R5
Week 4 R1 vs. R5 R2 vs. R4 R3 vs. R6
Week 5 R1 vs. R6 R2 vs. R3 R4 vs. R5
Similar 5-week schedules can be developed for the White and Blue divisions by replacing the R in the above table with a W or a B. To develop the schedule for the next 3 weeks, we create 3 new six-couple divisions by pairing 3 of the teams in each division with 3 of the teams in another division; for example, (R1, R2, R3, W1, W2, W3), (B1, B2, B3, R4, R5, R6), and (W4, W5, W6, B4, B5, B6). Within each of these new divisions, matches can be scheduled for 3 weeks without any couples playing a couple they have played before. For instance, a 3-week schedule for the first of these divisions is shown below: Week 6 R1 vs. W1 R2 vs. W2 R3 vs. W3
Week 7 R1 vs. W2 R2 vs. W3 R3 vs. W1
Week 8 R1 vs. W3 R2 vs. W1 R3 vs. W2
A similar 3-week schedule can be easily set up for the other two new divisions. This will provide us with a schedule for the first 8 weeks of the season. For the final 9 weeks, we continue to create new divisions by pairing 3 teams from the original Red, White and Blue divisions with 3 teams from the other divisions that they have not yet been paired with. Then a 3week schedule is developed as above. Shown below is a set of divisions for the next 9 weeks.
CP - 1
Chapter 1
Weeks 9-11 (R1, R2, R3, W4, W5, W6)
(W1, W2, W3, B1, B2, B3)
(R4, R5, R6, B4, B5, B6)
(W1, W2, W3, B4, B5, B6)
(W4, W5, W6, R4, R5, R6)
(W1, W2, W3, R4, R5, R6)
(W4, W5, W6, B1, B2, B3)
Weeks 12-14 (R1, R2, R3, B1, B2, B3) Weeks 15-17 (R1, R2, R3, B4, B5, B6)
This Red, White and Blue scheduling procedure provides a schedule with every couple playing every other couple over the 17-week season. If one of the couples should cancel, the schedule can be modified easily. Designate the couple that cancels, say R4, as the Bye couple. Then whichever couple is scheduled to play couple R4 will receive a Bye in that week. With only 17 couples a Bye must be scheduled for one team each week. This same scheduling procedure can obviously be used for scheduling sports teams and or any other kinds of matches involving 17 or 18 teams. Modifications of the Red, White and Blue algorithm can be employed for 15 or 16 team leagues and other numbers of teams.
CP - 2
Chapter 2
An Introduction to Linear Programming Case Problem 1: Workload Balancing 1. Model DI-910 DI-950
Production Rate (minutes per printer) Line 1 Line 2 3 4 6 2
Profit Contribution ($) 42 87
Capacity: 8 hours 60 minutes/hour = 480 minutes per day Let
D1 = number of units of the DI-910 produced D2 = number of units of the DI-950 produced
Max s.t.
42D1
+ 87D2
3D1 + 6D2 4D1 + 2D2 D 1, D 2 0
480 480
Line 1 Capacity Line 2 Capacity
The optimal solution is D1 = 0, D2 = 80. The value of the optimal solution is $6960. Management would not implement this solution because no units of the DI-910 would be produced. 2.
Adding the constraint D1 D2 and resolving the linear program results in the optimal solution D1 = 53.333, D2 = 53.333. The value of the optimal solution is $6880.
3.
Time spent on Line 1: 3(53.333) + 6(53.333) = 480 minutes Time spent on Line 2: 4(53.333) + 2(53.333) = 320 minutes Thus, the solution does not balance the total time spent on Line 1 and the total time spent on Line 2. This might be a concern to management if no other work assignments were available for the employees on Line 2.
4.
Let
T1 = total time spent on Line 1 T2 = total time spent on Line 2
Whatever the value of T2 is, T1 T2 + 30 T1 T2 - 30 Thus, with T1 = 3D1 + 6D2 and T2 = 4D1 + 2D2 3D1 + 6D2 4D1 + 2D2 + 30 3D1 + 6D2 4D1 + 2D2 − 30
CP - 3
Chapter 2
Hence, −1D1 + 4D2 30 −1D1 + 4D2 −30 Rewriting the second constraint by multiplying both sides by -1, we obtain −1D1 + 4D2 30 1D1 − 4D2 30 Adding these two constraints to the linear program formulated in part (2) and resolving we obtain the optimal solution D1 = 96.667, D2 = 31.667. The value of the optimal solution is $6815. Line 1 is scheduled for 480 minutes and Line 2 for 450 minutes. The effect of workload balancing is to reduce the total contribution to profit by $6880 - $6815 = $65 per shift. 5.
The optimal solution is D1 = 106.667, D2 = 26.667. The total profit contribution is 42(106.667) + 87(26.667) = $6800 Comparing the solutions to part (4) and part (5), maximizing the number of printers produced (106.667 + 26.667 = 133.33) has increased the production by 133.33 - (96.667 + 31.667) = 5 printers but has reduced profit contribution by $6815 - $6800 = $15. But, this solution results in perfect workload balancing because the total time spent on each line is 480 minutes.
Case Problem 2: Production Strategy 1.
Let Max s.t.
BP100 = the number of BodyPlus 100 machines produced BP200 = the number of BodyPlus 200 machines produced 371BP100 +
461BP200
8BP100 5BP100 2BP100 -0.25BP100
12BP200 10BP200 2BP200 0.75BP200
+ + + +
600 450 140 0
BP100, BP200 0
CP - 4
Machining and Welding Painting and Finishing Assembly, Test, and Packaging BodyPlus 200 Requirement
Solutions to Case Problems
BP200 80
Number of BodyPlus 200
.
70
Assembly, Test, and Packaging
60 50
Machining and Welding
40 BodyPlus 200 Requirement
30 20 10
Painting and Finishing BP100 30 40 50 60 70 80 90 100 Number of BodyPlus 100 Optimal Solution
0
10
20
Optimal solution: BP100 = 50, BP200 = 50/3, profit = $26,233.33. Note: If the optimal solution is rounded to BP100 = 50, BP200 = 16.67, the value of the optimal solution will differ from the value shown. The value we show for the optimal solution is the same as the value that will be obtained if the problem is solved using a linear programming software package. 2.
In the short run the requirement reduces profits. For instance, if the requirement were reduced to at least 24% of total production, the new optimal solution is BP100 = 1425/28, BP200 = 225/14, with a total profit of $26,290.18; thus, total profits would increase by $56.85. Note: If the optimal solution is rounded to BP100 = 50.89, BP200 = 16.07, the value of the optimal solution will differ from the value shown. The value we show for the optimal solution is the same as the value that will be obtained if the problem is solved using a linear programming software package such as Excel Solver.
3.
If management really believes that the BodyPlus 200 can help position BFI as one of the leader's in high-end exercise equipment, the constraint requiring that the number of units of the BodyPlus 200 produced be at least 25% of total production should not be changed. Since the optimal solution uses all of the available machining and welding time, management should try to obtain additional hours of this resource.
CP - 5
Chapter 2
Case Problem 3: Hart Venture Capital 1.
Let S = fraction of the Security Systems project funded by HVC M = fraction of the Market Analysis project funded by HVC Max s.t.
1,800,000S
+
1,600,000M
600,000S 600,000S 250,000S S
+ + +
500,000M 350,000M 400,000M
S,M
M 0
800,000 700,000 500,000 1 1
Year 1 Year 2 Year 3 Maximum for S Maximum for M
The solution obtained is shown below: OPTIMAL SOLUTION Optimal Objective Value 2486956.52174 Variable
Value
Reduced Cost
S
0.60870
0.00000
M
0.86957
0.00000
Constraint
Slack/Surplus
Dual Value
1
0.00000
2.78261
2
30434.78261
0.00000
3
0.00000
0.52174
4
0.39130
0.00000
5
0.13043
0.00000
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
1800000.00000
120000.00000
800000.00000
1600000.00000 1280000.00000
100000.00000
RHS
Allowable
Allowable
Value
Increase
Decrease
800000.00000
22950.81967
60000.00000
700000.00000
Infinite
30434.78261
500000.00000
25000.00000
38888.88889
1.00000
Infinite
0.39130
1.00000
Infinite
0.13043
CP - 6
Solutions to Case Problems
Thus, the optimal solution is S = 0.609 and M = 0.870. In other words, approximately 61% of the Security Systems project should be funded by HVC and 87% of the Market Analysis project should be funded by HVC. The net present value of the investment is approximately $2,486,957. 2. Year 1 $365,400 $435,000 $800,400
Security Systems Market Analysis Total
Year 2 $365,400 $304,500 $669,900
Year 3 $152,250 $348,000 $500,250
Note: The totals for Year 1 and Year 3 are greater than the amounts available. The reason for this is that rounded values for the decision variables were used to compute the amount required in each year. 3.
If up to $900,000 is available in year 1 we obtain a new optimal solution with S = 0.689 and M = 0.820. In other words, approximately 69% of the Security Systems project should be funded by HVC and 82% of the Market Analysis project should be funded by HVC. The net present value of the investment is approximately $2,550,820. The solution follows:
OPTIMAL SOLUTION Optimal Objective Value 2550819.67213 Variable
Value
Reduced Cost
S
0.68852
0.00000
M
0.81967
0.00000
Constraint
Slack/Surplus
Dual Value
1
77049.18033
0.00000
2
0.00000
2.09836
3
0.00000
2.16393
4
0.31148
0.00000
5
0.18033
0.00000
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
1800000.00000
942857.14286
800000.00000
1600000.00000 1280000.00000
550000.00000
CP - 7
Chapter 2
4.
Allowable
Allowable
Value
Increase
Decrease
900000.00000
Infinite
77049.18033
700000.00000
102173.91304
110000.00000
500000.00000
45833.33333
135714.28571
1.00000
Infinite
0.31148
1.00000
Infinite
0.18033
If an additional $100,000 is made available, the allocation plan would change as follows:
Security Systems Market Analysis Total 5.
RHS
Year 1 $413,400 $410,000 $823,400
Year 2 $413,400 $287,000 $700,400
Year 3 $172,250 $328,000 $500,250
Having additional funds available in year 1 will increase the total net present value. The value of the objective function increases from $2,486,957 to $2,550,820, a difference of $63,863. But, since the allocation plan shows that $823,400 is required in year 1, only $23,400 of the additional $100,00 is required. We can also determine this by looking at the slack variable for constraint 1 in the new solution. This value, 77049.180, shows that at the optimal solution approximately $77,049 of the $900,000 available is not used. Thus, the amount of funds required in year 1 is $900,000 - $77,049 = $822,951. In other words, only $22,951 of the additional $100,000 is required. The differences between the two values, $23,400 and $22,951, is simply due to the fact that the value of $23,400 was computed using rounded values for the decision variables.
CP - 8
Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution Case Problem 1: Product Mix Note to Instructor: The difference between relevant and sunk costs is critical. The cost of the shipment of nuts is a sunk cost. Practice in applying sensitivity analysis to a business decision is obtained. You may want to suggest that sensitivity analyses other than the ones we have suggested be undertaken. 1.
Cost per pound of ingredients Almonds $7500/6000 = $1.25 Brazil $7125/7500 = $.95 Filberts $6750/7500 = $.90 Pecans $7200/6000 = $1.20 Walnuts $7875/7500 = $1.05 Cost of nuts in three mixes: Regular mix: .15($1.25) + .25($.95) + .25($90) + .10($1.20) + .25($1.05) = $1.0325 Deluxe mix
.20($1.25) + .20($.95) + .20($.90) + .20($1.20) + .20($1.05) = $1.07
Holiday mix: .25($1.25) + .15($.95) + .15($.90) + .25($1.20) + .20($1.05) = $1.10 2.
Let R = pounds of Regular Mix produced D = pounds of Deluxe Mix produced H = pounds of Holiday Mix produced Note that the cost of the five shipments of nuts is a sunk (not a relevant) cost and should not affect the decision. However, this information may be useful to management in future pricing and purchasing decisions. A linear programming model for the optimal product mix is given. The following linear programming model can be solved to maximize profit contribution for the nuts already purchased. Max s.t.
1.65R
+
2.00D
+
2.25H
0.15R 0.25R 0.25R 0.10R 0.25R R
+ + + + +
0.20D 0.20D 0.20D 0.20D 0.20D
+ + + + +
0.25H 0.15H 0.15H 0.25H 0.20H
D H R, D, H 0
CP - 9
6000 7500 7500 6000 7500 10000 3000 5000
Almonds Brazil Filberts Pecans Walnuts Regular Deluxe Holiday
Chapter 3
Optimal Objective Value 61375.00000 Variable
Value
R
17500.00000
0.00000
D
10625.00000
0.00000
H
5000.00000
0.00000
Constraint
Slack/Surplus
Reduced Cost
Dual Value
1
0.00000
8.50000
2
250.00000
0.00000
3
250.00000
0.00000
4
875.00000
0.00000
5
0.00000
1.50000
6
7500.00000
0.00000
7
7625.00000
0.00000
8
0.00000
-0.17500
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
1.65000
0.35000
0.15000
2.00000
0.20000
0.10769
2.25000
0.17500
Infinite
RHS
Allowable
Allowable
Value
Increase
Decrease
6000.00000
583.33333
610.00000
7500.00000
Infinite
250.00000
7500.00000
Infinite
250.00000
6000.00000
Infinite
875.00000
7500.00000
250.00000
750.00000
10000.00000
7500.00000
Infinite
3000.00000
7625.00000
Infinite
5000.00000
4692.30769
5000.00000
3.
From the dual values it can be seen that additional almonds are worth $8.50 per pound to TJ. Additional walnuts are worth $1.50 per pound. From the slack variables, we see that additional Brazil nut, Filberts, and Pecans are of no value since they are already in excess supply.
4.
Yes, purchase the almonds. The dual value shows that each pound is worth $8.50; the dual value is applicable for increases up to 583.33 pounds.
CP - 10
Solutions to Case Problems Resolving the problem by changing the right-hand side of constraint 1 from 6000 to 7000 yields the following optimal solution. The optimal solution has increased in value by $4958.34. Note that only 583.33 pounds of the additional almonds were used, but that the increase in profit contribution more than justifies the $1000 cost of the shipment. Optimal Objective Value 66333.33333 Variable
Value
R
11666.66667
0.00000
D
17916.66667
0.00000
H
5000.00000
0.00000
Constraint
5.
Slack/Surplus
Reduced Cost
Dual Value
1
416.66667
0.00000
2
250.00000
0.00000
3
250.00000
0.00000
4
0.00000
5.66667
5
0.00000
4.33333
6
1666.66667
0.00000
7
14916.66667
0.00000
8
0.00000
-0.03333
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
1.65000
0.10000
0.65000
2.00000
1.30000
0.02353
2.25000
0.03333
Infinite
RHS
Allowable
Allowable
Value
Increase
Decrease
7000.00000
Infinite
416.66667
7500.00000
Infinite
250.00000
7500.00000
Infinite
250.00000
6000.00000
250.00000
1790.00000
7500.00000
250.00000
250.00000
10000.00000
1666.66667
Infinite
3000.00000
14916.66667
Infinite
5000.00000
10529.41176
5000.00000
From the dual values it is clear that there is no advantage to not satisfying the orders for the Regular and Deluxe mixes. However, it would be advantageous to negotiate a decrease in the Holiday mix requirement.
CP - 11
Chapter 3
Case Problem 2: Investment Strategy 1.
The first step is to develop a linear programming model for maximizing return subject to constraints for funds available, diversity, and risk tolerance. Let G = Amount invested in growth fund I = Amount invested in income fund M = Amount invested in money market fund The LP formulation and optimal solution found using The Management Scientist are shown.
MAX .18G +.125I +.075M S.T. 1) 2) 3) 4) 5) 6) 7)
G + I + M < 800000 .8G -.2I -.2M > 0 .6G -.4I -.4M < 0 -.2G +.8I -.2M > 0 -.5G +.5I -.5M < 0 -.3G -.3I +.7M > 0 .05G + .02I -.04M < 0
Funds Available Min growth fund Max growth fund Min income fund Max income fund Min money market fund Max risk
Optimal Objective Value 94133.33333 Variable
Value
Reduced Cost
G
248888.88889
0.00000
I
160000.00000
0.00000
M
391111.11111
0.00000
Constraint
Slack/Surplus
Dual Value
1
0.00000
0.11767
2
88888.88889
0.00000
3
71111.11111
0.00000
4
0.00000
-0.02000
5
240000.00000
0.00000
6
151111.11111
0.00000
7
0.00000
1.16667
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
0.18000
Infinite
0.03000
0.12500
0.02000
0.58833
0.07500
0.10500
0.06000
RHS
Allowable
Allowable
CP - 12
Solutions to Case Problems
Value
Increase
Decrease
800000.00000
Infinite
800000.00000
0.00000
88888.88889
Infinite
0.00000
Infinite
71111.11111
0.00000
133333.33333
106666.66667
0.00000
Infinite
240000.00000
0.00000
151111.11111
Infinite
0.00000
6400.00000
8000.00000
Rounding to the nearest dollar, the portfolio recommendation for Langford is as follows. Fund: Growth Income Money Market Total
Amount Invested $248,889 160,000 391,111 $800,000
Yield = 94,133 / 800,000 = .118 The portfolio yield is .118 or 11.8%. Note that the portfolio yield equals the dual value for the funds available constraint. 2.
If Langford’s risk index is increased by .005 that is the same as increasing the right-hand side of constraint 7 by .005 (800,000) = 4000. Since this amount of increase is within the right-hand-side range, we would expect an increase in return of 1.167 (4000) = 4668. The revised formulation and new optimal solution are shown below. Except for rounding, the value has increased as predicted; the new optimal allocation is Fund: Growth Income Money Market Total
Amount Invested $293,333 160,000 346,667 $800,000
The portfolio yield becomes 98,800/800,000 = .124 or 12.4% MAX .18G +.125I +.075M S.T. 1) 2) 3) 4) 5) 6) 7)
G + I + M < 800000 .8G -.2I -.2M > 0 .6G -.4I -.4M < 0 -.2G +.8I -.2M > 0 -.5G +.5I -.5M < 0 -.3G -.3I +.7M > 0 .045G + .015I-.045M < 0
CP - 13
Chapter 3
Optimal Objective Value 98800.00000
3.
Variable
Value
Reduced Cost
G
293333.33333
0.00000
I
160000.00000
0.00000
M
346666.66667
0.00000
Constraint
Slack/Surplus
Dual Value
1
0.00000
0.12350
2
133333.33333
0.00000
3
26666.66667
0.00000
4
0.00000
-0.02000
5
240000.00000
0.00000
6
106666.66667
0.00000
7
0.00000
1.16667
Since .16 is in the objective coefficient range for the growth fund return, there would be no change in allocation. However, the return would decrease by (.02) ($248,889) = $4978. A decrease to .14 is outside the objective function coefficient range forcing us to resolve the problem. The new formulation and optimal solution is as follows.
MAX .14G +.125I +.075M S.T. 1) 2) 3) 4) 5) 6) 7)
G + I + M < 800000 .8G -.2I -.2M > 0 .6G -.4I -.4M < 0 -.2G +.8I -.2M > 0 -.5G +.5I -.5M < 0 -.3G -.3I +.7M > 0 .05G + .02I-.04M < 0
Optimal Objective Value 85066.66667 Variable
Value
Reduced Cost
G
160000.00000
0.00000
I
293333.33333
0.00000
M
346666.66667
0.00000
Constraint
Slack/Surplus
Dual Value
CP - 14
Solutions to Case Problems
4.
1
0.00000
0.10633
2
0.00000
-0.01000
3
160000.00000
0.00000
4
133333.33333
0.00000
5
106666.66667
0.00000
6
106666.66667
0.00000
7
0.00000
0.83333
Since the current optimal solution has more invested in the growth fund than the income fund, adding this requirement will force us to resolve the problem with a new constraint. We should expect a decrease in return as is shown in the following optimal solution.
MAX .18G +.125I +.075M S.T. 1) 2) 3) 4) 5) 6) 7) 8)
G + I + M < 800000 .8G -.2I -.2M > 0 .6G -.4I -.4M < 0 -.2G +.8I -.2M > 0 -.5G +.5I -.5M < 0 -.3G -.3I +.7M > 0 .05G + .02I-.04M < 0 G - I < 0
Optimal Objective Value 93066.66667 Variable
Value
1
213333.33333
0.00000
2
213333.33333
0.00000
3
373333.33333
0.00000
Constraint
Slack/Surplus
1
Reduced Cost
Dual Value
0.00000
0.11633
2
53333.33333
0.00000
3
106666.66667
0.00000
4
53333.33333
0.00000
5
186666.66667
0.00000
6
133333.33333
0.00000
7
0.00000
1.03333
8
0.00000
0.01200
Note that the value of the solution has decreased from $94,133 to $93,067. This is only a decrease of 0.2% inyield. Since the yield decrease is so small, Williams may prefer this portfolio for Langford.
CP - 15
Chapter 3 5.
It is possible a model such as this could be developed for each client. The changed yield estimates would require a change in the objective function coefficients and resolving the problem if the change was outside the objective coefficient range.
Case Problem 3: Truck Leasing Strategy 1.
Let xij = number of trucks obtained from a short term lease signed in month i for a period of j months and yi = number of trucks obtained from the long-term lease that are used in month i Monthly fuel costs are 20 ($100) = $2000.
Monthly Costs for Short-Term Leased Trucks Note: the costs shown here include monthly fuel costs of $2000. Decision Variables x11 , x21 , x31 , x41
Cost $4000 + $2000 = $6000
x12 , x22 , x32
2 ($3700 + $2000) = $11,400
x13 , x23
3 ($3225 + $2000) = $15,675
x14
4 ($3040 + $2000) = $20,160
Monthly Costs for Long-Term Leased Trucks Since Reep Construction is committed to the long-term lease and since employees cannot be laid off, the only relevant cost for the long-term leased trucks is the monthly fuel cost of $2000. LINEAR PROGRAMMING PROBLEM MIN 6000X11+11400X12+15675X13+20160X14+6000X21+11400X22+15675X23+6000X31+ 11400X32+6000X41+2000Y1+2000Y2+2000Y3+2000Y4 S.T. 1) 2) 3) 4) 5) 6) 7) 8)
1X11+1X12+1X13+1X14+1Y1=10 1X12+1X13+1X14+1X21+1X22+1X23+1Y2=12 1X13+1X14+1X22+1X23+1X31+1X32+1Y3=14 1X14+1X23+1X32+1X41+1Y4=8 1Y1<1 1Y2<2 1Y3<3 1Y4<1
CP - 16
Solutions to Case Problems
Optimal Objective Value 203660.00000 Variable
Value
Reduced Cost
X11
0.00000
1515.00000
X12
0.00000
1725.00000
X13
3.00000
0.00000
X14
6.00000
0.00000
X21
0.00000
810.00000
X22
0.00000
210.00000
X23
1.00000
0.00000
X31
1.00000
0.00000
X32
0.00000
915.00000
X41
0.00000
1515.00000
Y1
1.00000
0.00000
Y2
2.00000
0.00000
Y3
3.00000
0.00000
Y3
1.00000
0.00000
Constraint
Slack/Surplus
Dual Value
1
0.00000
4485.00000
2
0.00000
5190.00000
3
0.00000
6000.00000
4
0.00000
4485.00000
5
0.00000
-2485.00000
6
0.00000
-3190.00000
7
0.00000
-4000.00000
8
0.00000
-2485.00000
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
6000.00000
Infinite
1515.00000
11400.00000
Infinite
1725.00000
15675.00000
210.00000
915.00000
20160.00000
915.00000
210.00000
6000.00000
Infinite
810.00000
11400.00000
Infinite
210.00000
15675.00000
210.00000
1515.00000
6000.00000
915.00000
810.00000
11400.00000
Infinite
915.00000
CP - 17
Chapter 3
6000.00000
Infinite
1515.00000
2000.00000
2485.00000
Infinite
2000.00000
3190.00000
Infinite
2000.00000
4000.00000
Infinite
2000.00000
2485.00000
Infinite
RHS
Allowable
Allowable
Value
Increase
Decrease
10.00000
1.00000
6.00000
12.00000
1.00000
1.00000
14.00000
Infinite
1.00000
8.00000
3.00000
6.00000
1.00000
6.00000
1.00000
2.00000
1.00000
1.00000
3.00000
1.00000
3.00000
1.00000
6.00000
1.00000
2.
The total cost associated with the leasing plan is $203,660.
3.
If Reep Construction is willing to consider the possibility of layoffs, we need to include driver costs of $3200 per month. Replacing the coefficients for y1, y2, y3, and y4 in our previous linear program with $5200 and resolving resulted in the following leasing plan: Month Leased 1 2 3 4
1 0 0 0 0
Length of Lease (Months) 2 3 0 3 0 1 0
4 7
In addition, in month 2, one of the trucks from the long-term leases was used and in month 3 three of the trucks from the long-term leases were used. The total cost of this leasing plan is $224,620. To see what effect a no layoff policy has, we can set y1 = 1, y 2 = 2, y3 = 3, y 4 = 1 and resolve the linear program using objective coefficients of $5200 for y1 , y 2 , y3 , and y 4 . The new optimal solution forces us to use all the available trucks from the long-term lease; the optimal leasing plan is shown below. Month Leased 1 2 3 4
1 0 0 1 0
Length of Lease (Months) 2 3 0 3 0 1 0
4 6
The total cost associated with this solution is $226,060. Thus, if Reep maintains their current policy of no layoffs they will incur an additional cost of $226,060 - $224,620 = $1,440.
CP - 18
Chapter 4 Linear Programming Applications in Marketing, Finance and Operations Management Case Problem 1: Planning an Advertising Campaign The decision variables are as follows: T1 = number of television advertisements with rating of 90 and 4000 new customers T2 = number of television advertisements with rating of 40 and 1500 new customers R1 = number of radio advertisements with rating of 25 and 2000 new customers R2 = number of radio advertisements with rating of 15 and 1200 new customers N1 = number of newspaper advertisements with rating of 10 and 1000 new customers N2 = number of newspaper advertisements with rating of 5 and 800 new customers The Linear Programming Model and solution are as follows: MAX 90T1+55T2+25R1+20R2+10N1+5N2 S.T. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10)
1T1<=10 1R1<=15 1N1<=20 10000T1+10000T2+3000R1+3000R2+1000N1+1000N2<=279000 4000T1+1500T2+2000R1+1200R2+1000N1+800N2>=100000 -2T1-2T2+1R1+1R2>=0 1T1+1T2<=20 10000T1+10000T2>=140000 3000R1+3000R2<=99000 1000N1+1000N2>=30000
OPTIMAL SOLUTION Optimal Objective Value 2160.00000 Variable
Value
Reduced Cost
T1
10.00000
0.00000
T2
5.00000
0.00000
R1
15.00000
0.00000
R2
18.00000
0.00000
N1
20.00000
0.00000
N2
10.00000
0.00000
Slack/Surplus
CP - 19
Dual Value
Chapter 4
Constraint
1.
1
0.00000
35.00000
2
0.00000
5.00000
3
0.00000
5.00000
4
0.00000
0.00550
5
27100.00000
0.00000
6
3.00000
0.00000
7
5.00000
0.00000
8
10000.00000
0.00000
9
0.00000
0.00117
10
0.00000
-0.00050
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
90.00000
Infinite
35.00000
55.00000
11.66667
5.00000
25.00000
Infinite
5.00000
20.00000
5.00000
3.50000
10.00000
Infinite
5.00000
5.00000
0.50000
Infinite
RHS
Allowable
Allowable
Value
Increase
Decrease
10.00000
5.00000
10.00000
15.00000
18.00000
15.00000
20.00000
10.00000
20.00000
279000.00000
15000.00000
10000.00000
100000.00000
27100.00000
Infinite
0.00000
3.00000
Infinite
20.00000
Infinite
5.00000
140000.00000
10000.00000
Infinite
99000.00000
10000.00000
5625.00000
30000.00000
10000.00000
10000.00000
Summary of the Optimal Solution T1 + T2 = 10 + 5 = 15 Television advertisements R1 + R2 = 15 + 18 = 33 Radio advertisements N1 + N2 = 20 + 10 = 30 Newspaper advertisements
CP - 20
Solutions to Case Problems
Advertising Schedule: Media Television Radio Newspaper Totals
Number of Ads 15 33 30 78
Total Exposure Rating: Total New Customers Reached: 2.
Budget $150,000 99,000 30,000 $279,000 2,160 127,100 (Surplus constraint 5)
The dual value shows that total exposure increases 0.0055 points for each one dollar increase in the advertising budget. Right Hand Side Ranges show this dual value applies for a budget increase of up to $15,000. Thus the dual value applies for the $10,000 increase. Total Exposure Rating would increase by 10,000(0.0055) = 55 points A $10,000 increase in the advertising budget is a 3.6% increase. But, it only provides a 2.54% increase in total exposure. Management may decide that the additional exposure is not worth the cost. This is a discussion point.
3.
The ranges for the exposure rating of 90 for the first 10 television ads show that the solution remains optimal as long as the exposure rating is 55 or higher. This indicates that the solution is not very sensitive to the exposure rating HJ has provided. Indeed, we would draw the same conclusion after reviewing the next four ranges. We could conclude that Flamingo does not have to be concerned about the exact exposure rating. The only concern might be the newspaper exposure rating of 5. A rating of 5.5 or better can be expected to alter the current optimal solution.
4.
Remove constraint #5 for the linear programming model and use it to develop the objective function: MAX 4000T1+1500T2+2000R1+1200R2+1000N1+800N2 Solving provides the following Optimal Solution T1 + T2 = 10 + 4 = 14 Television advertisements R1 + R2 = 15 + 13 = 28 Radio advertisements N1 + N2 = 20 + 35 = 55 Newspaper advertisements Advertising Schedule: Media Television Radio Newspaper Totals
Number of Ads 14 28 55 97
Total New Customers Reached:
Budget $140,000 83,000 55,000 $279,000 139,600
Total Exposure Rating 90(10) + 55(4) + 25(15) + 20(13) + 10(20) + 5(35) = 2130
CP - 21
Chapter 4
5.
The solution with the objective to maximize the number of potential new customers reached looks attractive. The total number of ads is increased from 78 to 97 (24%) and the number of potential new customers reached is increased by 139,600 – 127,100 = 12,500 (9.8%). Maximizing total exposure may seem to be the preferred objective because it is a more general measure of advertising effectiveness. Exposure includes issues of image, message recall and appeal to repeat customers. However, in this case, many more potential new customers will be reached with the objective of maximizing reach, and the total exposure is only reduced by 2160 – 2130 = 30 points (1.4%). At this point, we would expect some discussion concerning which solution is preferred: the one obtained by maximizing total exposure or the one obtained by maximizing potential new customers reached. Expect students to have differing opinions on the final recommendation. Basically, there are two good media allocation solutions for this problem.
Case Problem 2: Schneider’s Sweet Shop Papa Jack’s recipe can just be calculated, whereas the optimized and flexible optimized recipes require linear programming. The following linear programs may be used to calculate the latter two: Let Xi = amount (in pounds) of ingredient i to use in making 50 pounds of ice cream, i = 1,2,3…14 Min 1.19X1 + .70X2 + 2.32X3 + …… + 1.68X13 + 0.0X14 s.t. .4X1 + .2X2 + .8X3 + .8X4 + .9X5 + .1X6 + .5X10 + .6X11 = .16(50) .1X1 +.1X4 + .1X6 + .3X7 + 1X8 = .08(50) .7X9 + .1X10 = .16(50) .4X10 + .4X11 = .0035(50) X12 = .0025(50) X13 = .0015(50) .5X1 + .8X2 + .2X3 + .1X4 + .1X5 +.8X6 + .7X7 + .3X9 + X14 = .5925(50) X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 + X11 + X12 + X13 + X14 = 50 X1 , X2 , X3 , X4 , X5 , X6 , X7 , X8 , X9 , X10 , X11 , X12 , X13 , X14 ≥ 0 And for the flexible recipe: Min 1.19X1 + .70X2 + 2.32X3 + …… + 1.68X13 + 0.0X14 s.t. .15(50) ≤ .4X1 + .2X2 + .8X3 + .8X4 + .9X5 + .1X6 + .5X10 + .6X11 ≤ .17(50) .07(50) ≤ .1X1 +.1X4 + .1X6 + .3X7 + 1X8 ≤ .09(50) .155(50) ≤ .7X9 + .1X10 ≤ .165(50) .003(50) ≤ .4X10 + .4X11 ≤ .004(50) .002(50) ≤ X12 ≤ .003(50) .001(50) ≤ X13 ≤ .002(50) .58(50) ≤ .5X1 + .8X2 + .2X3 + .1X4 + .1X5 +.8X6 + .7X7 + .3X9 + X14 ≤ .595(50) X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 + X11 + X12 + X13 + X14 = 50 X1 , X2 , X3 , X4 , X5 , X6 , X7 , X8 , X9 , X10 , X11 , X12 , X13 , X14 ≥ 0
CP - 22
Solutions to Case Problems
The recipes and costs are as follows:
Ingredient Name 40% Cream 23% Cream Butter Plastic Cream Butter Oil 4% Milk Skim Condensed Milk Skim Milk Powder Liquid Sugar Sugared Frozen Fresh Egg Yolk Powdered Egg Yolk Stabilizer Emulsifier Water
Fat Serum Solids Sugar Solids Egg Solids Stabilizer Emulsifier Water
Ingredient 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Total Cost
Papa Jack Optimized 16.0% 16.0% 8.0% 8.0% 16.0% 16.0% 0.4% 0.4% 0.2% 0.3% 0.1% 0.2% 59.2% 59.3%
Papa Jack 0 0 0 9.73 0 0 0 3.03 11.37 0.44 0 0.12 0.07 25.24 50 $28.37
Optimized 0 0 0 5.72 0 32.05 0 0.22 11.37 0.44 0 0.13 0.08 0 50 $25.35
Optimized Flexible 0 0 0 5.12 0 32.15 0 0.47 11.73 0.38 0 0.10 0.05 0 50 $24.04
Optimized Flexible 15.0% 8.4% 16.5% 0.3% 0.2% 0.1% 59.5%
The shaded areas in the recipe table indicate larger differences. The recipes generated by the cost-minimizing linear program use no water and instead get the required water from other ingredients that provide water and other requirements. In fact, the table shows that even though water is (essentially) free, it is not used because it does not contribute to other requirements. The optimized recipe saves $28.37 - $25.35 = $3.02 per $50 batch and the optimized flexible recipe saves an additional $25.35-$24.04 = $1.31 per 50 pound batch. Note that the flexible recipe increases the amount of sugar solids and that may or may not be accepted well by customers. Since the optimized recipe follows the original requirements and saves $3.02 per 50 pound batch, we recommend going with that and maybe test marketing the optimized flexible recipe to better understand its acceptance by the market.
CP - 23
Chapter 4
Case Problem 3: Textile Mill Scheduling Let X3R = Yards of fabric 3 on regular looms X4R = Yards of fabric 4 on regular looms X5R = Yards of fabric 5 on regular looms X1D = Yards of fabric 1 on dobbie looms X2D = Yards of fabric 2 on dobbie looms X3D = Yards of fabric 3 on dobbie looms X4D = Yards of fabric 4 on dobbie looms X5D = Yards of fabric 5 on dobbie looms Y1 = Yards of fabric 1 purchased Y2 = Yards of fabric 2 purchased Y3 = Yards of fabric 3 purchased Y4 = Yards of fabric 4 purchased Y5 = Yards of fabric 5 purchased Profit Contribution per Yard
Fabric
1 2 3 4 5
Manufactured 0.33 0.31 0.61 0.73 0.20
Purchased 0.19 0.16 0.50 0.54 0.00
1 2 3 4 5
Regular — — 0.1912 0.1912 0.2398
Dobbie 0.21598 0.21598 0.1912 0.1912 0.2398
Production Times in Hours per Yard
Fabric
Model may use a Max Profit or Min Cost objective function. Max
0.61X3R + 0.73X4R + 0.20X5R + 0.33X1D + 0.31X2D + 0.61X3D + 0.73X4D + 0.20X5D + 0.19Y1 + 0.16Y2 + 0.50Y3 + 0.54Y4
or Min
0.49X3R + 0.51X4R + 0.50X5R + 0.66X1D + 0.55X2D + 0.49X3D + 0.51X4D + 0.50X5D + 0.80Y1 + 0.70Y2 + 0.60Y3 + 0.70Y4 + 0.70Y5
Regular Hours Available 30 Looms x 30 days x 24 hours/day = 21600 Dobbie Hours Available 8 Looms x 30 days x 24 hours/day = 5760
CP - 24
Solutions to Case Problems
Constraints: Regular Looms: 0.1912X3R + 0.1912X4R + 0.2398X5R 21600 Dobbie Looms: 0.21598X1D + 0.21598X2D + 0.1912X3D + 0.1912X4D + 0.2398X5D 5760 Demand Constraints X1D + Y1 X2D + Y2 X3R + X3D + Y3 X4R + X4D + Y4 X5R + X5D + Y5
= 16500 = 22000 = 62000 = 7500 = 62000
OPTIMAL SOLUTION Optimal Objective Value 62531.49090 Variable X3R
Value
Reduced Cost
27707.80815
0.00000
X4R
7500.00000
0.00000
X5R
62000.00000
0.00000
X1D
4668.80000
0.00000
X2D
22000.00000
0.00000
X3D
0.00000
-0.01394
X4D
0.00000
-0.01394
X5D
0.00000
-0.01748
Y1
11831.20000
0.00000
Y2
0.00000
-0.01000
Y3
34292.19185
0.00000
Y4
0.00000
-0.08000
Y5
0.00000
-0.06204
Constraint
Slack/Surplus
Dual Value
1
0.00000
0.57530
2
0.00000
0.64820
3
0.00000
0.19000
4
0.00000
0.17000
5
0.00000
0.50000
6
0.00000
0.62000
7
0.00000
0.06204
CP - 25
Chapter 4
Objective
Allowable
Allowable
Coefficient
Increase
Decrease
0.61000
0.01394
0.11000
0.73000
Infinite
0.01394
0.20000
Infinite
0.01748
0.33000
0.01000
0.01575
0.31000
Infinite
0.01000
0.61000
0.01394
Infinite
0.73000
0.01394
Infinite
0.20000
0.01748
Infinite
0.19000
0.01575
0.01000
0.16000
0.01000
Infinite
0.50000
0.11000
0.01394
0.54000
0.08000
Infinite
0.00000
0.06204
Infinite
RHS
Allowable
Allowable
Value
Increase
Decrease
21600.00000
6556.82444
5297.86007
5760.00000
2555.33477
1008.38013
16500.00000
Infinite
11831.20000
22000.00000
4668.80000
11831.20000
62000.00000
Infinite
34292.19185
7500.00000
27707.80815
7500.00000
62000.00000
22092.07648
27341.95794
Production/Purchase Schedule (Yards) Regular Looms
Fabric
1 2 3 4 5
27711 7500 62000
Projected Profit: $62,531.49 Value of 9th Dobbie Loom Dual Value (Constraint 2) = 0.6482 per hour dobbie
CP - 26
Dobbie Looms 4669 22000
Purchased 11831 34289
Solutions to Case Problems
Monthly Value of 1 Dobbie Loom (30 days)(24 hours/day)($0.6482) = $466.70 Note: This change is within the Right-Hand Side Ranges for Constraint 2. Discussion of Objective Coefficient Ranges For example, fabric one on the dobbie loom shares ranges of 0.31426 to 0.34 for the profit maximization model or 0.64426 to 0.67 for the cost minimization model. Note here that since demand for the fabrics is fixed, both the profit maximization and cost minimization models will provide the same optimal solution. However, the interpretation of the ranges for the objective function coefficients differ for the two models. In the profit maximization case, the coefficients are profit contributions. Thus, the range information indicates how price per unit and cost per unit may vary simultaneously. That is, as long as the net changes in price per unit and cost per unit keep the profit contributions within the ranges, the solution will remain optimal. In the cost minimization model, the coefficients are costs per unit. Thus, the range information indicates that assuming price per unit remains fixed how much the cost per unit may vary and still maintain the same optimal solution.
Case Problem 4: Workforce Scheduling 1.
Let tij = number of temporary employees hired under option i (i = 1, 2, 3) in month j (j = 1 for January, j = 2 for February and so on) The following table depicts the decision variables used in this case problem.
Option 1 Option 2 Option 3
Jan. t11 t21 t31
Feb. t12 t22 t32
Mar. t13 t23 t33
Apr. t14 t24 t34
May t15 t25
June t16
Costs: Contract cost plus training cost Option 1 2 3
Contract Cost $2000 $4800 $7500
Training Cost $875 $875 $875
Total Cost $2875 $5675 $8375
Min. 2875(t11 + t12 + t13 + t14 + t15 + t16) + 5675(t21 + t22 + t23 + t24 + t25) + 8375(t31 + t32 + t33 + t34) One constraint is required for each of the six months. Constraint 1: Need 10 additional employees in January t11 = number of temporary employees hired under Option 1 (one-month contract) in January t21 = number of temporary employees hired under Option 2 (two-month contract) in January t31 = number of temporary employees hired under Option 3 (three-month contract) in January t11 + t21 + t31 = 10
CP - 27
Chapter 4
Constraint 2: Need 23 additional employees in February t12 , t22 and t32 are the number of temporary employees hired under Options 1, 2 and 3 in February. But, temporary employees hired under Option 2 or Option 3 in January will also be available to satisfy February needs. t21 + t31 + t12 + t22 + t32 = 23 Note: The following table shows the decision variables used in this constraint Jan. Option 1 Option 2 Option 3
t21 t31
Feb. t12 t22 t32
Mar.
Apr.
May
June
Constraint 3: Need 19 additional employees in March
Option 1 Option 2 Option 3
Jan.
Feb.
t31
t22 t32
Mar. t13 t23 t33
Apr.
May
June
t31 + t22 + t32 + t13 + t23 + t33 = 19 Constraint 4: Need 26 additional employees in May Jan. Option 1 Option 2 Option 3
Feb.
Mar.
t32
t23 t33
Apr. t14 t24 t34
May
June
t32 + t23 + t33 + t14 + t24 + t34 = 26 Constraint 5: Need 20 additional employees in May Jan.
Feb.
Option 1 Option 2 Option 3
Mar.
Apr.
t33
t24 t34
May t15 t25
June
t33 + t24 + t34 + t15 + t25 = 20 Constraint 6: Need 14 additional employees in June Jan. Option 1 Option 2 Option 3
Feb.
Mar.
Apr.
May t25
t34
t34 + t25 + t16 = 14
CP - 28
June t16
Solutions to Case Problems
Optimal Solution: Total Cost = $313,525
Option 1 Option 2 Option 3
Jan. 0 3 7
Feb. 1 0 12
Mar. 0 0 0
Apr. 0 0 14
May 6 0
June 0
2. Option 1 2 3
3.
Number Hired 7 3 33 Total:
Contract Cost $14,000 $14,400 $247,500 $275,900
Training Cost $6,125 $2,625 $28,875 $37,625
Total Cost $20,125 $17,025 $276,375 $313,525
Hiring 10 full-time employees at the beginning of January will reduce the number of temporary employees needed each month by 10. Using the same linear programming model with the righthand sides of 0, 13, 9, 16, 10 and 4, provides the following schedule for temporary employees:
Option 1 Option 2 Option 3 Option 1 2 3 Total:
Jan. 0 0 0
Feb. 4 0 9
Number Hired 7 3 13 23
Mar. 0 0 0
Apr. 0 3 4
May 3 0
Contract Cost $14,000 $14,400 $97,500
June 0
Training Cost $6,125 $2,625 $11,375
Total Cost $20,125 $17,025 $108,875 $146,025
Full-time employees cost: Training cost: 10($875) = $8,750 Salary: 10(6)(168)($16.50) = $166,320 Total Cost = $146,025 + $8750 + $166,320 = $321,095 Hiring 10 full-time employees is $321,095 - $313,525 = $7,570 more expensive than using temporary employees. Do not hire the 10 full-time employees. Davis should continue to contract with WorkForce to obtain temporary employees. 4.
With the lower training costs, the costs per employee for each option are as follows: Option 1 2 3
Cost $2000 $4800 $7500
Training Cost $700 $700 $700
Total Cost $2700 $5500 $8200
Resolving the original linear programming model with the above costs indicates that Davis should hire all temporary employees on a one-month contract specifically to meet each month's employee needs. Thus, the monthly temporary hire schedule would be as follows: January - 10; February - 23; March - 19; April - 26; May - 20; and June - 14. The total cost of this strategy is $302,400. Note that if training costs were any lower, this would still be the optimal hiring strategy for Davis.
CP - 29
Chapter 4
Case Problem 5: Duke Energy Coal Allocation A linear programming model can be used to determine how much coal to buy from each of the mining companies and where to ship it. Let xij = tons of coal purchased from supplier i and used by generating unit j
The objective function minimizes the total cost to buy and burn coal. The objective function coefficients, cij , are the cost to buy coal at mine i, ship it to generating unit j, and burn it at generating unit j. Thus, the objective function is cij xij . In computing the objective function coefficients three inputs must be added: the cost of the coal, the transportation cost to the generating unit, and the cost of processing the coal at the generating unit. There are two types of constraints: supply constraints and demand constraints. The supply constraints limit the amount of coal that can be bought under the various contracts. For the fixedtonnage contracts, the constraints are equalities. For the variable-tonnage contracts, any amount of coal up to a specified maximum may be purchased. Let Li represent the amount that must be purchased under fixed-tonnage contract i and Si represent the maximum amount that can be purchased under variable-tonnage contract i. Then the supply constraints can be written as follows:
x = L ij
i
for all fixed-tonnage contracts
x S
i
for all variable-tonnage contracts
j ij
j
The demand constraints specify the number of mWh of electricity that must be generated by each generating unit. Let aij = mWh hours of electricity generated by a ton of coal purchased from supplier i and used by generating unit j, and Dj = mWh of electricity demand at generating unit j. The demand constraints can then be written as follows:
a x = D ij ij
j
for all generating units
i
Note: Because of the large number of calculations that must be made to compute the objective function and constraint coefficients, we developed an Excel spreadsheet model for this problem. Copies of the data and model worksheets are included after the discussion of the solution to parts (a) through (f). 1.
The number of tons of coal that should be purchased from each of the mining companies and where it should be shipped is shown below: Miami Fort # 5 Miami Fort # 7 RAG
Beckjord
East Bend
Zimmer
0
0
61,538
288,462
0
Peabody
217,105
11,278
71,617
0
0
American
0
0
0
0
275,000
Consol
0
0
33,878
0
166,122
Cyprus
0
0
0
0
0
Addington
0
200,000
0
0
0
Waterloo
0
0
98,673
0
0
CP - 30
Solutions to Case Problems
The total cost to purchase, deliver, and process the coal is $53,407,243. 2.
The cost of the coal in cents per million BTUs for each generating unit is as follows: Miami Fort #5 111.84
3.
Miami Fort #7 136.97
Beckjord 127.24
East Bend 103.85
Zimmer 114.51
The average number of BTUs per pound of coal received at each generating unit is shown below: Miami Fort #5 13,300
Miami Fort #7 12,069
Beckjord 12,354
East Bend 13,000
Zimmer 12,468
4.
The sensitivity report shows that the dual value per ton of coal purchased from American Coal Sales is -$13 per ton and the allowable increase is 88,492 tons. This means that every additional ton of coal that Duke Energy can purchase at the current price of $22 per ton will decrease cost by $13. So even paying $30 per ton, Duke Energy will decrease cost by $5 per ton. Thus, they should buy the additional 80,000 tons; doing so will save them $5(80,000) = $400,000.
5.
If the energy content of the Cyprus coal turns out to be 13,000 BTUs per ton the procurement plan changes as shown below: Miami Fort # 5 Miami Fort # 7 Beckjord RAG
6.
East Bend
Zimmer
0
0
61,538
288,462
0
Peabody
36,654
191,729
71,617
0
0
American
0
0
0
0
275,000
Consol
0
0
33,878
0
166,122
Cyprus
0
0
85,769
0
0
Addington
200,000
0
0
0
0
Waterloo
0
0
0
0
0
The dual values for the demand constraints are as follows: Miami Fort #5 21
Miami Fort #7 20
Beckjord 20
East Bend 18
Zimmer 19
The East Bend unit is the least cost producer at the margin ($18 per mWh), and the allowable increase is 160,000 mWh. Thus, Duke Energy should sell the 50,000 mWh over the grid. The additional electricity should be produced at the East Bend generating unit. Duke Energy’s profit will be $12 per mWh.
CP - 31
Chapter 4
The Excel data and model worksheets used to solve the Duke Energy coal allocation problem are as follows: Duke Energy Coal Allocation Model (Data)
Duke Energy Coal Allocation Model (Solution)
CP - 32
Chapter 6 Distribution and Network Models Case Problem 1: Solutions Plus 1.
This case can be formulated as a transportation problem with the Cincinnati and Oakland production facilities as the origins. The locations of the railway stations are the destinations. Each objective function coefficient is the sum of the production cost at an origin and the freight cost to ship from the origin to a destination. The minimum cost solution has a value of $1,318,985 and 773,522 gallons of cleaning fluid are produced and shipped. So, the average cost per gallon of producing the cleaning fluid and shipping it to the railway stations is $1.705168. Shown below is the optimal shipping plan.
Santa Ana El Paso Pendleton Houston Kansas City Los Angeles Glendale Jacksonville Little Rock Bridgeport Sacramento Total
Cincinnati 0 6800 39636 100447 24570 0 0 68486 148586 111475 0 500000
Oakland 22418 0 40654 0 0 64761 33689 0 0 0 112000 273522
From the shipping plan it can be seen that the Cincinnati plant is at capacity. The shadow price for the Cincinnati capacity constraint indicates that if more capacity can be made available the total cost can be decreased by $0.11 for every additional gallon up to 40,654. 2.
Since it costs Solutions Plus $1.705168 per gallon to produce and deliver the cleaning fluid to the railroad, this is the breakeven point. If Mr. Miller bids any less, Solutions Plus will lose money on the contract.
3.
A 15% profit margin corresponds to a price of 1.15($1.705168) = $1.9609432 per gallon. Rounding up, we would recommend the president bid $1.97 per gallon if a 15% profit margin is desired.
4.
If management of Solutions Plus expects oil prices to go up, they can also expect freight rates to go up. The breakeven point found by solving the minimum cost transportation problem is using freight rates for the first year. So, management should perhaps bid a bit higher to allow for an increase in freight rates in the second year of the contract. You might offer to resolve the problem assuming, say, a 10% increase in freight rates to see what the breakeven point would be in this situation. On the other hand, if freight rates are expected to go down in the second year, management might want to bid a bit lower to increase the chances of winning the contract.
CP - 33
CP - 34
Santa Ana Origin 0 Cincinnati 22418 Oakland 22418 Total = 22418
Pendleton El Paso 39636 6800 40654 0 80290 6800 = = 80290 6800
Per Gallon Total Min Cost 1318984.93 1.705167959 Destination Kansas City Los Angeles Glendale Jacksonville Little Rock Bridgeport Sacramento Total Houston 0 500000 <= 500000 111475 148586 68486 0 0 24570 100447 112000 273522 <= 500000 0 0 0 33689 64761 0 0 112000 773522 111475 148586 68486 33689 64761 24570 100447 = = = = = = = = 112000 111475 148586 68486 33689 64761 24570 100447
Pendleton El Paso 2.03 2.04 2.14 2.39 80290 6800
Santa Ana Origin 11.2 Cincinnati 1.87 Oakland 22418 Demand
Model
FREIGHT + PRODUCTION COST Destination Kansas City Los Angeles Glendale Jacksonville Little Rock Bridgeport Sacramento Supply Houston 11.2 500000 1.54 1.54 1.54 11.2 11.2 1.56 1.65 1.8 500000 11.65 11.65 11.65 1.87 1.87 11.65 11.65 112000 111475 148586 68486 33689 64761 24570 100447
Pendleton El Paso 0.83 0.84 0.49 0.74 80290 6800
Cincinnati Oakland 1.65 1.2
Santa Ana Origin 10 Cincinnati 0.22 Oakland 22418 Demand
Production Cost
FREIGHT COST Destination Kansas City Los Angeles Glendale Jacksonville Little Rock Bridgeport Sacramento Supply Houston 10 500000 0.34 0.34 0.34 10 10 0.36 0.45 0.15 500000 10 10 10 0.22 0.22 10 10 112000 111475 148586 68486 33689 64761 24570 100447
Solutions Plus
Chapter 10
Solutions to Case Problems
Case Problem 2: Supply Chain Design Three related linear programming models were developed and solved to answer the questions in this case. First, we developed a linear programming formulation of the network model shown on the following page; we refer to this network model and the corresponding linear program as the Part 2 Network Model and the Part 2 Linear Program, respectively. Variations of the Part 2 Linear Program were then used to answer the questions in parts 1 and 3. For instance, to answer the question concerning the existing system in part 1 we added constraints to the Part 2 Linear Program which set the flow equal to 0 over the distribution centercustomer arcs that Darby does not currently use. For Part 3, we added three plant-to-customer arcs to the Part 2 Network Model and corresponding linear program: El Paso - San Antonio, San Bernardino - Los Angeles, and San Bernardino - San Diego. The decision variables used in each of the linear programs use 4 letters to describe the amount of flow over each arc. The first two letters in each variable name identify the “from” node and the second two letters identify the “to” nodes. For instance, EPFW represents the amount shipped from El Paso to Ft. Worth and LVDE represents the amount shipped from Las Vegas to Denver. A description of the LP models that provide the basis for answering the questions in the managerial report follows the network model for the questions in part 2.
CP - 35
Chapter 10
Part 2 Network Model: Dallas
6300
San Antonio
4880
Wichita
2130
Kansas City
1210
Denver
6120
Salt Lake City
4830
Phoenix
2750
0.3 2.1 Ft. Worth
3.1 4.4 6.0
3.20 30,000
5.4
El Paso 2.20
4.5
5.2
6.0 4.20
2.7
Santa Fe
4.7 3.90 20,000
3.4 3.3
San 2.7
Bernardino 1.20
5.4
3.3
Las Vegas
2.4 2.1 2.5
Los Angeles
San Diego
CP - 36
8580
4460
Solutions to Case Problems
LP Model and Solution for Part 2 MIN 13.7EPFW + 12.7EPSF + 14.7EPLV + 13.9SBSF + 11.2SBLV + .3FWDA + 2.1FWSA + 3.1FWWI + 4.4FWKC + 6.0FWDE + 5.2SFDA + 5.4SFSA + 4.5SFWI + 6.0SFKC + 2.7SFDE + 4.7SFSL + 3.4SFPH + 3.3SFLA + 2.7SFSD + 5.4LVDE + 3.3LVSL + 2.4LVPH + 2.1LVLA + 2.5LVSD S.T.
SFSD
1) EPFW + EPSF + EPLV < 30000 2) SBSF + SBLV < 20000 3) FWDA + FWSA + FWWI + FWKC + FWDE - EPFW = 0 4) SFDA + SFSA + SFWI + SFKC + SFDE + SFSL + SFPH + SFLA + - EPSF - SBSF = 0 5) LVDE + LVSL + LVPH + LVLA + LVSD - EPLV - SBLV = 0 6) FWDA + SFDA = 6300 7) FWSA + SFSA = 4880 8) FWWI + SFWI = 2130 9) FWKC + SFKC = 1210 10) FWDE + SFDE + LVDE = 6120 11) SFSL + LVSL = 4830 12) SFPH + LVPH = 2750 13) SFLA + LVLA = 8580 14) SFSD + LVSD = 4460
Objective Function Value = Variable -------------EPFW EPSF EPLV SBSF SBLV FWDA FWSA FWWI FWKC FWDE SFDA SFSA SFWI SFKC SFDE SFSL SFPH SFLA SFSD LVDE LVSL
600942.000
Value --------------14520.000 6740.000 0.000 0.000 20000.000 6300.000 4880.000 2130.000 1210.000 0.000 0.000 0.000 0.000 0.000 6120.000 0.000 0.000 0.000 620.000 0.000 4830.000
CP - 37
Reduced Costs -----------------0.000 0.000 1.800 2.900 0.000 0.000 0.000 0.000 0.000 4.300 3.900 2.300 0.400 0.600 0.000 1.200 0.800 1.000 0.000 2.900 0.000
Chapter 10
Variable -------------LVPH LVLA LVSD
Value --------------2750.000 8580.000 3840.000
Reduced Costs -----------------0.000 0.000 0.000
LP Model and Solution for Part 1 Questions (Existing System) The following constraints were added to the above LP Model so that each customer can only be served by the distribution center it is currently served by. 15) 16) 17) 18) 19) 20) 21) 22) 23) 24)
SFDA SFSA SFWI SFKC SFLA SFSD LVDE FWDE LVSL LVPH
= = = = = = = = = =
0 0 0 0 0 0 0 0 0 0
The optimal solution shows that the total cost (manufacturing plus distribution) of the existing system is $620,770. Details of this solution and the modifications suggested in questions 2 and 3 are contained in the “Summary of Optimal Solutions.” LP Model and Solution for Part 3 (Plant-Customer Shipments) The network model for the part 2 questions must be modified by adding 3 arcs: San Bernardino - Los Angeles, San Bernardino - San Diego, and El Paso - San Antonio. The corresponding linear program must also be modified to incorporate the new variables.
CP - 38
Solutions to Case Problems
A summary of the optimal solutions suggested by questions (1), (2), and (3) follows.
Summary o f Optimal So lutions Part 1
Part 2
Costs
Existing System
Any Distribution Center
Part 3 Plant Customer Shipments
Manufacturing Cost
$426,710
$423,230
$423,230
Shipping Cost
194,060
177,712
130,304
Total Cost
$620,770
$600,942
$553,534
EPFW
14520
14520
9640
EPSF
13700
6740
6740
EPLV
0
0
0
SBSF
0
0
0
SBLV
13040
20000
6960
FWDA
6300
6300
6300
FWSA
4880
4880
0
FWWI
2130
2130
2130
FWKC
1210
1210
1210
FWDE
0
0
0
SFDA
0
0
0
SFSA
0
0
0
SFWI
0
0
0
SFKC
0
0
0
SFDE
6120
6120
6120
SFSL
4830
0
0
SFPH
2750
0
620
SFLA
0
0
0
SFSD
0
620
0
LVDE
0
0
0
LVSL
0
4830
4830
LVPH
0
2750
2130
Decision Variables
CP - 39
Chapter 10
Discussion 1.
With the supply chain, the minimum total cost is $620,770; the total cost is the sum of $426,710 for manufacturing and $194,060 for shipping. One might suspect that restricting customers to a single distribution center is not optimal.
2.
Allowing customers to be serviced by any distribution center reduces total cost to $600,942; the total cost consists of $423,230 of manufacturing cost and $177,712 of shipping cost. This is a 3.19% decrease in total cost and an 8.42% decrease in shipping cost. Only one customer, San Diego, is served by more than one distribution center.
3.
Allowing some direct shipments from plants to customers causes a substantial reduction in total cost. The total cost is $553,534, a 10.83% reduction over the current plan. Note, however, there is a 32.85% reduction in shipping cost. Again, one customer, Phoenix, receives shipments from 2 distribution centers.
4.
Total capacity at the two plants (50,000 units) is adequate to handle the growth in demand. However, the company might want to consider expanding the more efficient San Bernardino plant. The shadow price from Part 3 output (not shown here) indicates that each unit of added capacity will reduce distribution costs by $2.50. However, the range of feasibility shows this shadow price is only applicable for the net 620 units of capacity. Additional computer runs with higher capacity levels for San Bernardino are recommended.
CP - 40
Chapter 7 Integer Linear Programming Case Problem 1: Textbook Publishing An integer programming model can be used advantageously to assist in developing recommendations.
The subscripts correspond to the books as follows: i 1 2 3 4 5 6 7 8 9 10
Book Business Calculus Finite Math General Statistics Mathematical Statistics Business Statistics Finance Financial Accounting Managerial Accounting English Literature German
An integer programming model for maximizing projected sales (thousands of units) subject to the restrictions mentioned is given. Max 20x1 + 30x2 + 15x3 + 10x4 + s.t. 30x1 + 16x2 + 24x3 + 20x4 + 40x1 + 24x2 30x3 + 24x4 + x3 + x4 +
25x5 +
18x6 + 25x7 + 50x8 + 20x9 +
30x10
10x5
+ 40x9 + 24x7 + 28x8 + 34x9 + 14x6 + 26x7 + 30x8 + 30x9 +
60 John 50x10 40 Susan 36x10 40 Monica 2 No. of Stat Books 1 Account Book = 1 Math Book
16x5 + x5
x7 + x1
+
x2
x8
xi = 0, 1 for all i The optimal solution (x2 = x5 = x6 = 1) calls for publishing the finite math, the business statistics and the finance books. Projected sales are 73,000 copies. (1)
If Susan can be made available for another 12 days, the optimal solution is x2 = x8 = 1. This calls for publishing the finite math and managerial accounting texts for projected sales of 80,000 copies.
(2)
If Monica is also available for 10 more days, a big improvement can be made. The new optimal solution calls for producing the finite math book, the business statistics book, and the managerial accounting book. Projected sales are 105,000 copies.
CP - 41
Chapter 7
(3)
The solution in (2) above does not include any new books. In the long run this would appear to be a bad strategy for the company. A variety of modifications can be made to the model to examine the short run impact of postponing a revision. For instance, a constraint could be added to require publication of at least one new book.
Case Problem 2: Yeager National Bank A mixed integer linear programming (MILP) model can be used advantageously to assist in preparing a report for Mr. Wolff. Since the annual fixed costs of operating the lockboxes are not known exactly we formulate a model that can be used without that information. Then determine if information more precise than the $20,000 - $30,000 estimate is needed to make a final decision. We formulate a model that can be solved to find the best set of locations with a given number of lockboxes. It is then solved 4 times to find the best set of locations with 1 lockbox, up to 2 lockboxes, up to 3 lockboxes, and up to 4 lockboxes. We then can address the question of whether annual operating cost information is needed from some or all of the potential lockbox locations. We use the following variable definitions: P = 1 if a lockbox is in Phoenix; 0 otherwise S = 1 if a lockbox is in Salt Lake City; 0 otherwise A = 1 if a lockbox is in Atlanta; 0 otherwise B = 1 if a lockbox is in Boston; 0 otherwise PNW = 1 if the Northwest region is assigned to Phoenix; 0 otherwise PSW = 1 if the Southwest region is assigned to Phoenix; 0 otherwise . . . BSE = 1 if the Southeast region is assigned to Boston; 0 otherwise The MILP model has 24 variables and 26 constraints. The objective function calls for minimizing lost interest income. To see how the objective function coefficients are computed, consider assigning the Northwest region to Phoenix. Daily collections are $80,000 and it takes 4 days to receive and process payments. At a 15% rate Yeager could save $48,000 = (.15)(4)($80,000) annually over this assignment if the Northwest collections could be credited to their account instantaneously. A similar calculation is made for all the other potential assignments. The objective function coefficients for P, S, A, and B are zero since we are not including the cost of operating the lockboxes at this stage. Constraints (1) to (5) ensure that each region is assigned a lock box. Constraints (6) to (25) ensure that a region is only assigned to a lockbox location that is open. And, constraint (26) is a limitation on the number of lockbox locations that may be chosen. The right-hand-side of that constraint will be varied from 1 to 4 as the problem is solved 4 times. Shown below is the MILP model that is solved to find the optimal solution when up to 2 lockboxes may be used.
CP - 42
Solutions to Case Problems INTEGER LINEAR PROGRAMMING PROBLEM MIN 48PNW+27PSW+112.5PCE+135PNE+60PSE+24SNW+40.5SSW+67.5SCE+108SNE+90SSE+48 ANW+5 4ASW+67.5ACE+81ANE+30ASE+48BNW+81BSW+90BCE+54BNE+45BSE S.T. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26)
1PNW+1SNW+1ANW+1BNW=1 +1PSW+1SSW+1ASW+1BSW=1 +1PCE+1SCE+1ACE+1BCE=1 +1PNE+1SNE+1ANE+1BNE=1 +1PSE+1SSE+1ASE+1BSE=1 1PNW-1P<0 +1PSW-1P<0 +1PCE-1P<0 +1PNE-1P<0 +1PSE-1P<0 +1SNW-1S<0 +1SSW-1S<0 +1SCE-1S<0 +1SNE-1S<0 +1SSE-1S<0 +1ANW-1A<0 +1ASW-1A<0 +1ACE-1A<0 +1ANE-1A<0 +1ASE-1A<0 +1BNW-1B<0 +1BSW-1B<0 +1BCE-1B<0 +1BNE-1B<0 +1BSE-1B<0 +1P+1S+1A+1B<2
OPTIMAL SOLUTION Objective Function Value = 231.000 Variable -------------PNW PSW PCE PNE PSE SNW SSW SCE SNE SSE ANW ASW ACE ANE ASE
Value --------------0.000 0.000 0.000 0.000 0.000 1.000 1.000 1.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
CP - 43
Chapter 7 BNW BSW BCE BNE BSE P S A B
0.000 0.000 0.000 1.000 1.000 0.000 1.000 0.000 1.000
Constraint -------------1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
Slack/Surplus --------------0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 0.000 0.000 0.000 1.000 1.000 1.000 0.000 0.000 0.000
With 2 lockbox locations, the lost interest income is $231,000. Resolving with values of 1, 3, and 4 on the right-hand-side of constraint 26 provides the following: No. of Lockboxes 1 2 3 4
Locations Atlanta Salt Lake City & Boston Salt Lake City, Atlanta & Boston Salt Lake City, Phoenix, Atlanta & Boston
Lost Interest Income ($) 280,500 231,000 216,000 202,500
From this information, it is clear that if the fixed cost per lockbox location is anywhere between $20,000 and $30,000 two locations (Salt Lake City & Boston) should be chosen. Mr. Wolff can be told that he only needs to collect information and negotiate on the cost to choose a site in those 2 locations.
CP - 44
Solutions to Case Problems
Case Problem 3: Production Scheduling with Changeover Costs A mixed integer programming model can be used advantageously to assist in developing recommendations. We describe such a model here; it has 48 decision variables and 64 constraints. We show here how to use Microsoft Excel to formulate and solve the problem. The spreadsheet at the end shows how we set up the problem and the optimal solution. We describe the model now. Variables There are variables for production, inventory, setup, and changeover in each week. Pi = number of P-Heads produced in week i Hi = number of H-Heads produced in week i IPi = number of P-Heads in inventory at end of week i IHi = number of H-Heads in inventory at end of week i SPi = 1 if line is setup for P in week i, 0 if setup for H Changei = 1 if a changeover occurs at the beginning of week i, 0 otherwise Constraints There are constraints for production capacity, inventory balance, maintenance of safety stock, and enforcement of changeovers. Also, Excel requires that you identify the 0-1 (binary) variables in the Solver dialog box. The constraints as specified in the Excel Solver dialog box are as follows (references are to cells of the spreadsheet): B20:C27 B34:C41
production capacity, or nonnegativity of slack
G20:G27 H34:H41 G20:G27 I34:I41
forces Changei to 1 when a changeover occurs
D20:E27 D6:E13
ending inventory safety stock
D34:E41 = B6:C13
beg. inv. + production - end inv. = demand
F18:F25 = Bin
setup variables must be binary
Note that even though the Changei variable must also be integer it is not necessary to require it because minimization will never let it be any bigger than it has to be. And, the second set of constraints force it to be 1 whenever the setup variable changes from 1 to 0 or from 0 to 1. Objective We want to minimize total cost which is represented by cell J23 in the spreadsheet. It is the sum of production cost, inventory cost, and changeover cost. The Spreadsheet The first 14 rows of the spreadsheet contain the data for the problem; information on demand, safety stock, various costs and beginning inventories are given. Cells B20:G27, as shown contain the optimal solution to the problem. Before solving, those cells were empty.
CP - 45
Chapter 7
The spreadsheet formulation and solution are shown. A 1
B
C
D
Product Demand P H
P
E
F
G
H
I
J
Production Scheduling
2 3 Safety Stock
4 5
Week
6
1
55
38
44
7
2
55
38
35.2
8
3
44
30
9
4
0
0
10
5
45
48
11
6
45
12
7
13 14
8 9
P H
H
Production Cost
225
30.4
Max Weekly Rate
100
80
24
Changeover Cost
500
500
0
0
Weekly Inv. Rate
0.00375
0.00375
36
38.4
Weekly Inv. Cost
0.84375
1.1625
36
38.4
Beginning Inv.
125
143
48
28.8
46.4
36
58
28
45.6
35 35
57 58
28
46.4
310
15 16
Model
17 18 19
Week
20
1
18.00
0.00
88.00
105.00
1.00
0.00
Prod. Cost
117374
21
2
100.00
0.00
133.00
67.00
1.00
0.00
Inv. Cost
1280.35125
22
3
100.00
0.00
189.00
37.00
1.00
0.00
Changeover Cost
500
23
4
0.00
1.40
189.00
38.40
0.00
1.00
Min Total Cost
119154.3513
24
5
0.00
48.00
144.00
38.40
0.00
0.00
25
6
0.00
56.00
99.00
46.40
0.00
0.00
26 27
7 8
0.00 0.00
57.20 57.80
63.00 28.00
45.60 46.40
0.00 0.00
0.00 0.00
P
H
Inv. P
Inv. H
Setup P
Changeover
28 29 Inventory Balance
30
Beginning Inv. + Prod.
31 Production Capacity
32
- Ending Inv.
Changeover Def.
33
Week
34
1
100
0
55
38
1
0.00
0.00
35
2
100
0
55
38
2
0.00
0.00
36
3
100
0
44
30
3
0.00
0.00
37
4
0
80
0
0
4
-1.00
1.00
38
5
0
80
45
48
5
0.00
0.00
39
6
0
80
45
48
6
0.00
0.00
40
7
0
80
36
58
7
0.00
0.00
41
8
0
80
35
57
8
0.00
0.00
P
H
P
H
Week
To P if 1
To H if 1
Solution Comments The spreadsheet contains the optimal solution. The minimum total cost is $119,154.35. The components of that cost are production: $117,374, inventory: $1280.35 and changeover: $500. From cells F20:F27 we see that the line will be setup to produce P-Heads in weeks 1-3 and H-Heads in weeks 4-8. Cell G23 shows that there will be a changeover from producing P-Heads to H-Heads at the beginning of week 4. By adjusting the data for this problem (e.g. beginning inventories and the various costs) a number of variations of this problem can be created with the same basic model. Also, one might want to vary the safety stock requirements and the number of weeks in the planning horizon to create other variations of the problem.
CP - 46
Solutions to Case Problems
Case Problem 4: Applecore Children’s Clothing The Model: Let
i=1,2,…53 j=1,2,..10
yi = 1 if customer i is reached, 0 if not, xj = 1 if we place an ad on website j, Max
y1 + Y2 + Y3 + … + Y53
s.t ≥ y1 ≥ y2 ≥ y3
x10 x2 + x4 x1 ● ●
● ●
● ●
x2 + x5 + x8 + x9 + x10 ≥ y53 5x1 + 8x2 + 3.5x3 + 5.5x4 + 7.0x5 + 4.5x6 + 6.0x7 + 5.0x8 + 3.0x9 + 2.2x10 ≤ 10 The optimal solution for a Budget of $10,000 is to place ads on websites 6, 9 and 10. This will reach 23 of the 53 customers or 43% of the customer base. A spreadsheet model follows. Because of size, we only show portions.
CP - 47
Chapter 7
CP - 48
Solutions to Case Problems
CP - 49
Chapter 7
We solve the model for various budget levels:
Budget ($000) 5
Reach
% Reach
Incremental Reach
12
22.6%
10
23
43.4%
20.8%
15
31
58.5%
15.1%
20
35
66.0%
7.5%
25
41
77.4%
11.3%
30
45
84.9%
7.5%
35
47
88.7%
3.8%
Note from the incremental reach column in the preceding table, each additional $5000 in ad investment does not result in consistently diminishing returns in percentage reach. This is because the decision to place ads on a web site is a “yes-or-no” decision represented with a binary variable. The discrete nature of the decisions results in a percentage reach versus budget curve that is relatively flat for some budget ranges and makes “jumps” across other ranges. Significant incremental reach is achieved for a budget up to $25,000.
CP - 50
Chapter 8 Nonlinear Optimization Models Case Problem: Portfolio Optimization with Transaction Costs 1.
If $41,268.51are spent purchasing the Intermediate-Term Bond fund, and the transaction cost is 1 percent (i.e. one cent on the dollar), then the transaction fee is .01 ($41, 268.51) = $412.6851
2.
The total dollar amount spent on transaction costs is $1090.311.
3.
After rebalancing, Ms. Delgado has $100,000 - $1090.311 = $98,909.689
4.
To calculate the expected amount in the Intermediate-Term bond fund at year end, figure out the year-end amount in each scenario and then weight each scenario by a factor of 1/5 = .2. Since Ms. Delgado started with $51,268.51 after rebalancing, this gives $51,268.51 + (.2)(.1764)( $51,268.51) + (.2)(.0325)( $51,268.51) + (.2)(.0751)( $51,268.51) - (.2)(.0133)( $51,268.51) +(.2)(.0736)( $51,268.51) = $51,268.51 + $3530.35 = $54,798.86
5.
From the LINGO solution we see that Ms. Delgado can expect an average return of $10,000 on her investment. After rebalancing (see question 3) Ms. Delgado has $98,909.689 in her portfolio to start the year, so she is getting a return of greater than 10 percent on the amount of money she starts with after rebalancing! However, this is not what she wants or expects. Since she starts the year with $98,909.689 after rebalancing and earns (in expectation) $10,000 she can expect $98,909.689 + $10,000 = $108,909.689 in her portfolio. However, this does not give a return of 10 percent on her original portfolio of $100,000 which is what she expected. In order to end up with a return of at least 10 percent on her original portfolio she must have at least $110,000.00 in her portfolio at the end of the year.
6.
Here is the formulation that provides an expected value at least $110,000 at year end.
MIN = (1/5)*((R1 - RBAR)^2 + (R2 - RBAR)^2 + (R3 - RBAR)^2 + (R4 RBAR)^2 + (R5 - RBAR)^2); ! THE SCENARIO RETURNS; 1.1006*FS + 1.1764*IB + 1.3241*LG + 1.3236*LV + 1.3344*SG + 1.2456*SV = R1; 1.1312*FS + 1.0325*IB + 1.1871*LG + 1.2061*LV + 1.1940*SG + 1.2532*SV = R2; 1.1347*FS + 1.0751*IB + 1.3328*LG + 1.1293*LV + 1.0385*SG + (1.0670)*SV = R3; 1.4542*FS + (1 - .0133)*IB + 1.4146*LG + 1.0706*LV + 1.5868*SG + 1.0543*SV = R4; (1-.2193)*FS + 1.0736*IB + (1- .2326)*LG + (1- .0537)*LV + (1.0902)*SG + 1.1731*SV = R5; ! PORTFOLIO AVERAGE RETURN; (1/5)*(R1 + R2 + R3+ R4 + R5) = RBAR; ! UNITY CONSTRAINT; FS + IB + LG + LV + SG + SV + TRANS_COST = 100000;
CP - 51
Chapter 6 RBAR > RMIN; RMIN = 10000 + 100000; ! DEFINE BUY AND SELL QUANTITIES; FS = FS_START + FS_BUY - FS_SELL; IB = IB_START + IB_BUY - IB_SELL; LG = LG_START + LG_BUY - LG_SELL; LV = LV_START + LV_BUY - LV_SELL; SG = SG_START + SG_BUY - SG_SELL; SV = SV_START + SV_BUY - SV_SELL; ! DEFINE TOTAL TRANSACTION COSTS; TRANS_COST = TRANS_FEE*(FS_BUY + FS_SELL + IB_BUY + IB_SELL + LG_BUY + LG_SELL + LV_BUY + LV_SELL + SG_BUY + SG_SELL + SV_BUY + SV_SELL); FS_START = 10000; IB_START = 10000; LG_START = 10000; LV_START = 40000; SG_START = 10000; SV_START = 20000; TRANS_FEE = 0.01; @FREE(R1); @FREE(R2); @FREE(R3); @FREE(R4); @FREE(R5);
CP - 52
Solutions to Case Problems
7.
Here is the solution for the formulation in Question 7 using LINGO.
Local optimal solution found. Objective value: Total solver iterations: Variable R1 RBAR R2 R3 R4 R5 FS IB LG LV SG SV TRANS_COST RMIN FS_START FS_BUY FS_SELL IB_START IB_BUY IB_SELL LG_START LG_BUY LG_SELL LV_START LV_BUY LV_SELL SG_START SG_BUY SG_SELL SV_START SV_BUY SV_SELL TRANS_FEE
0.3380209E+08 46 Value 119713.7 110000.0 112235.3 105190.3 109673.8 103186.9 10000.00 44769.45 10870.66 0.000000 719.8605 32664.18 975.8443 110000.0 10000.00 0.000000 0.000000 10000.00 34769.45 0.000000 10000.00 870.6613 0.000000 40000.00 0.000000 40000.00 10000.00 0.000000 9280.140 20000.00 12664.18 0.000000 0.1000000E-01
CP - 53
Reduced Cost 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 327.7706 0.000000 0.000000 0.000000 0.000000 0.000000 41.03809 106.0720 0.000000 0.000000 147.1100 0.000000 0.000000 147.1100 0.000000 147.1100 0.000000 0.000000 147.1100 0.000000 0.000000 0.000000 147.1100 0.000000
Chapter 6
Case Problem 2
CAFÉ COMPLIANCE IN THE AUTO INDUSTRY
Note that the solution to this case uses LINGO output. 1.
The model is
! PASSENGR CAR MILES PER GALLON; PASSMPG = 30; ! LIGHT TRUCK MILES PER GALLON; LTRUCKMPG = 20; ! PP - PASSENGER CAR PRICE; ! TP - LIGHT TRUCK PRICE; ! PD - PASSENGER CAR DEMAND ! TD - LIGHT TRUCK DEMAND; PD = 750 - 20*PP; TD = 830 - 17*TP; PCOST = 15; TCOST = 17; MAX = PD*(PP - PCOST) + TD*(TP - TCOST); The objective function PD*(PP - PCOST) + TD*(TP - TCOST) is the desired contribution margin. 2.
The LINGO solution to the above model is Local optimal solution found. Objective value: Extended solver steps: Total solver iterations:
6835.382 5 37
Variable PASSMPG LTRUCKMPG PD PP TD TP PCOST TCOST Row 1 2 3 4 5 6
Value 30.00000 20.00000 225.0000 26.25000 270.5000 32.91176 15.00000 17.00000
Reduced Cost 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
Slack or Surplus 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
Dual Price 0.000000 0.000000 11.25000 15.91176 -225.0000 -270.5000
CP - 54
Solutions to Case Problems 7
6835.382
1.000000
3.
As seen from the LINGO solution in question 2, 225 passenger cars are sold and 270.5 light trucks are sold.
4.
See Figure 8.17.
5.
The Café average is
CAFEAVG = (PD + TD)/((PD/PASSMPG)+(TD/LTRUCKMPG)); which for the solution in 2 is 23.56718. 6.
Adding
CAFEAVG = (PD + TD)/((PD/PASSMPG)+(TD/LTRUCKMPG)); and CAFEAVG > 25; to the LINGO model gives Local optimal solution found. Objective value: Extended solver steps: Total solver iterations:
6274.514 5 39
CP - 55
Chapter 9 Project Scheduling: PERT/CPM Case Problem: R.C. Coleman 1.
R.C. Coleman's Project Network
Activity A B C D E F G H I J K
Activity A B C D E F G H I J K
Earliest Start 0 0 9 13 13 23 13 29 29 35 39
Expected Time 6 9 4 12 10 6 8 6 7 4 4 Latest Start 3 0 9 17 13 23 21 29 32 35 39
Earliest Finish 6 9 13 25 23 29 21 35 36 39 43
CP - 59
Variance 0.44 2.78 0.44 7.11 1.00 0.44 7.11 0.44 2.78 0.11 0.44 Latest Finish 9 9 13 29 23 29 29 35 39 39 43
Slack 3 0 0 4 0 0 8 0 3 0 0
Critical Activity Yes Yes Yes Yes Yes Yes Yes
Chapter 9
The expected project completion time is 43 weeks. The critical path activities are B-C-E-F-H-J-K. The variance of the critical path is 5.67.
z=
40 − 43 5.67
= −1.26
Area = 0.3962 P(T 40) = 0.5000 - 0.3962 = 0.1038 Given the above calculations, we can conclude that there is about a 10% chance that the project can be completed in 40 weeks or less. Coleman should consider crashing project activities. 2.
20% 80%
Planned project completion time
Desires 40-week completion time
For 80% chance, z = +0.84 Thus
40 − E (T ) 5.67
= 0.84
Solve for E(T) = 38 weeks. R.C. Coleman should crash activities to reduce the expected project completion time to 38 weeks. 3.
In this section, we will use expected activity times as normal times and use a linear programming model based on expected times to make the crashing decisions. Let
xi = the completion time for activity i yi = the amount of crash time for activity i
CP - 60
Solutions to Case Problems
Min 450yA + 400yB + 600yC + 300yD + 1000yE + 550yF + 750yG + 700yH + 800yI + 400yJ + 500yK s.t. xA + yA 6
xK 38
xB + yB 9
yA 2
xC + yC – xA 4
yB 2
xC + yC – xB 4
yC 2
xD + yD – xC 12
yD 4
xE + yE – xC 10
yE 3
xF + yF – xE 6
yF 2
xG + y G – xC 8
yG 3
xH + yH – xF 6
yH 2
xH + yH – xG 6
yI 3
xI + yI – xD 7
yJ 1
xI + yI – xF 7
yK 1
xJ + yJ – xH 4
All xi,yi 0
xK + yK – xI 4 xK + yK – xJ 4 The optimal crashing decisions are as follows: Crash Activity B F J K
Weeks 2 1 1 1 Total
CP - 61
Cost 800 550 400 500 2250
Chapter 9
A revised activity schedule based on these crashing decisions is as follows: Activity A B C D E F G H I J K
Earliest Start 0 0 7 11 11 21 11 26 26 32 35
Latest Start 1 0 7 14 11 21 18 26 28 32 35
Earliest Finish 6 7 11 23 21 26 19 32 33 35 38
Latest Finish 7 7 11 26 21 26 26 32 35 35 38
Slack 1 0 0 3 0 0 7 0 2 0 0
Critical Activity Yes Yes Yes Yes Yes Yes Yes
The student should comment on the fact that the crashing decisions may alter the variance in the project completion time. By defining revised optimistic, most probable, and pessimistic times for crashed activities B, F, J, and K, a revised variance in the project completion time can be found. Using this result, a revised probability of a 40-week completion time can be computed.
CP - 62
Chapter 10 Inventory Models Case Problem 1: Wagner Fabricating Company 1.
2.
Holding Cost Cost of capital Taxes/Insurance (24,000/600,000) Shrinkage (9,000/600,000) Warehouse overhead (15,000/600,000) Annual rate Ordering Cost 2 hours at $28.00 Other expenses (2,375/125) Cost per order
3.
14.0% 4.0% 1.5% 2.5% 22.0%
$56.00 19.00 $75.00
Set-up Cost 8 Hours at $50.00 $400 per set-up
4. & 5. a. Order from Supplier - EOQ model Ch = IC = 0.22 ($18.00) = $3.96
Q* =
2 DCo = Ch
2(3200)75 = 348.16 units 3.96
Number of orders = D/Q = 9.19/year Cycle time = 250(Q) / D = 250(348.16) / 3200 = 27.2 days Reorder Point: P(Stockout) = 1 / 9.19 = 0.1088
Stockout
64
CP - 63
z = 1.24
Chapter 10
r = 64 + 1.24(10) = 76.4 Safety stock = 76.4 - 64 = 12.4 Maximum inventory = Q + 12.4 = 360.56 Average inventory = Q/2 + 12.4 = 186.48 Annual holding cost = 186.48(3.96) = $738.46 Annual ordering cost = 9.19(75) = $689.35 Purchase cost = 3200($18) = $57,600 Total annual cost = $59,027.81 b. Manufacture - Production lot size model Ch = IC = 0.22($17.00) = $3.74 P = 1000(12) = 12,000/year Note: The five-month capacity of 5,000 units is sufficient to handle annual demand of 3,200 units.
Q* =
2 DCo = (1 − D / P)Ch
2(3200)(400) = 966.13 (1 − 3200 /1200)3.74
Number of production runs = D/Q = 3.31/year Cycle Time = 250(Q) / D = 250(966.13) / 3200 = 75.48 days Reorder point: P(Stockout) = 1 / 3.31 = 0.3021 Stockout
128
r = 128 + 0.52(20) = 138.4 Safety stock = 138.4 - 128 = 10.4 Maximum inventory = (1 - 3200/12000)966.13 + 10.4 = 718.89 Annual holding cost = (354.25 + 10.4)(3.74) = $1363.79 Annual set up cost = 3.31(400) = $1363.79 Manufacturing cost = 3200($17) = $54,400 Total Annual Cost = $57,088.67
CP - 64
z = 0.52
Solutions to Case Problems
6.
Recommend manufacturing the part Savings: $59,027.81 - 57,088.67 = $1,939.14 (3.3%)
Case Problem 2: River City Fire Department 1.
During a three-week scheduling period, a unit is scheduled seven days and must have at least 186 firefighters on duty each day. Thus, each unit must provide staffing for 7 186 = 1302 firefighter days During a three-week scheduling period, each firefighter provides six days of service. Thus, the base number of firefighters per unit must be
1302 = 217 firefighters 6 2.
The single-period inventory model can be used to determine the number of additional firefighters to be added to the unit to cover daily absences. Let Q* = minimum cost number of additional firefighters If Demand < Q*, Q* has overestimated demand. The cost of overestimating demand is the daily wage rate (d) for a firefighter. co = d If Demand > Q*, Q* has underestimated demand and overtime is needed. The cost of underestimating demand is the overtime wage (1.55d) minus the wage rate (d) that would have been required if we had hired enough additional firefighters. cu = 1.55d - d = 0.55d Using the single period inventory model,
P(Demand Q*) =
cu 0.55d 0.55 = = = 0.3548 cu + co 0.55 + d 1.55
Using the normal distribution, area = 0.5000 - 0.3548 = 0.1452. z = -0.37 Q* = + z = 20 − 0.37(5) = 18.14
Use 18 additional firefighters.
CP - 65
Chapter 10
3.
From part 2, this is the probability demand will exceed 18.14 is 1 - 0.3548 = 0.6452. This is an acceptable approximation of the probability overtime will be needed. A more complete analysis would include the fact that we have rounded the number of firefighters needed to 18. Then allowing for continuity correction, overtime would only be needed if demand exceeded 18.5 firefighters. Thus a revised estimate of the probability of overtime is:
z=
18.5 − 20 = −.30 Area = .1179 5
P(Overtime) = .5000 + .1179 = .6179 4.
Firefighters for each unit 217 + 18 additional = 235 Firefighters for the department (3 units) 3 235 = 705 firefighters
CP - 66
Chapter 11 Waiting Line Models Case Problem 1: Regional Airlines 1.
Single-Server Waiting Line Analysis The analysis that follows is based upon the assumptions of Poisson arrivals and exponential service times. With one call every 3.75 minutes, we have an arrival rate of
= 60/3.75 = 16 calls per hour Similarly, with an average service time of 3 minutes, we have a service rate of µ = 60/3 = 20 calls per hour The operating characteristics of a single server system with = 16 and µ = 20 are as follows:
16 P0 = 1 − = 1 − = 0.20 20 Lq =
2 162 = = 3.2 ( − ) 20(20 − 16)
L = Lq +
Wq +
Lq
16 = 3.2 + =4 20 =
W = Wq +
Pw =
3.2 = 0.20 hours (12 minutes) 16
1
= 0.20 +
1 = 0.25 hours (15 minutes) 20
16 = = 0.80 20
Operating the telephone reservation service with only one ticket agent appears unacceptable. With 80% of incoming calls waiting (Pw = 0.80) and an average waiting time of 12 minutes (Wq = 12), the company clearly needs to consider using two or more agents. 2.
Multiple-Server Waiting Line Analysis Since Regional's management team agreed that an acceptable service goal was to immediately answer and process at least 85% of the incoming calls, the probability of waiting must be 15% or less. Computing Pw for k = 2 agents and k = 3 agents provides the following. k k P Pw = 1 0 k! k -
CP - 67
Chapter 15
For k = 2 2 2(20) (0.4286) = 0.2286 Pw = 1 16 2! 20 2(20) - 16
For k = 3 3 3(20) (0.4472) = 0.0520 Pw = 1 16 3! 20 3(20) - 16
Based on the value of Pw, 3 ticket agents will be required to meet the service goal. Other operating characteristics of the 3-ticket-agent system are as follows: P0 = 0.4472 Lq = 0.0189 L = 0.8189 Wq = 0.0012 hours = 0.07 minutes W = 0.0512 hours = 3.07 minutes 3.
The primary advantage of the expanded system that allows a caller to wait is that, with three ticket agents, the average wait time will be short (less than 5 seconds), but in the current system these customers would have been rejected (and the business may have been lost). The disadvantage of the expanded system would be that any single wait time could be considerable, especially if there is not sufficient staffing, and customers may hang up. When a customer hangs up after waiting too long (called “reneging”) these is effectively the same as the original system blocking the customer.
4.
We would need to know the arrival rate for each hourly period throughout the day. An analysis similar to the one above would determine the recommended number of reservation agents each hour. This information could then be used to develop full-time and part-time shift schedules which would meet the service goals.
Case Problem 2: Office Equipment, Inc. 1.
= 1 call/50 hours = 0.02 calls per hour
2.
Mean service time = travel time + repair time = 1 + 1.5 = 2.5 hours = 1 / 2.5 hours = 0.4 customers per hour
3.
The travel time is 1 hour. While this is considered part of the service time it actually means that the customer will be waiting during the first hour of the service time (while the service technician travels to the customer location). That is, the predicted time spent in line provided by the waiting line model only includes the time until the service technician is available. Thus, to determine the amount of time the customer waits until the service technician arrives, the travel time must be added to the predicted time spent in line.
4.
Using the Excel worksheet template Finite, we have the following: Probability that no customers are in the system Average number of customers waiting
CP - 68
0.5380 0.2972
Solutions to Case Problems
Average number of customers in the system Average time a customer spends in the waiting line Average time until the machine is back in operation Probability of a wait more than one hour Total cost per hour for the service operation
0.7593 1.6082 hours* 4.1082 hours 0.4620 $155.93
*The average time a customer spends in the waiting line is 1.6082 hours. This is the average time for the service technician to complete all previous service call commitments and be ready to travel to the new customer. Since the average travel time is 1 hour for the service technician to reach the new customer's office, the total customer waiting time is 1.6082 + 1 = 2.6082 hours. Thus, the one technician is able to meet the company's 3-hour service guideline. The total cost is $155.93 per hour. Note that the waiting line model indicates the probability that a customer has to wait is 0.4620. Since all customers wait an average of 1-hour of travel time whenever the service technician is free, this probability is actually the probability that a customer will have to wait more than 1-hour for a service technician to arrive. 5.
If the company continues to use one technician when the customer base expands to 20 customers, the average time in the waiting line will increase to 6.9454 hours. With an average travel time of 1 hour, the average time a customer waits until the service technician arrives will be 6.9454 + 1 = 7.9454 hours. The total cost will be $397.78 per hour. This average total waiting time is too long and a second technician is definitely necessary. Using the Excel worksheet template Finite, two service technicians provide the following: Probability that no customers are in the system Average number of customers in the waiting line Average number of customers in the system Average time a customer spends in the waiting line Average time until the machine is back in operation Probability of a wait more than one hour Total cost per hour of service operation
0.3525 0.2104 1.1527 0.5581 hours* 3.0581 hours 0.2949 $275.27
*The average time a customer spends in the waiting line is 0.5581 hours. This is the average time for the service technician to complete all previous service call commitments and be ready to travel to the new customer. Since the average travel time is 1 hour for the service technician to reach the new customer's office, the total customer waiting time is 0.5581 + 1 = 1.5581 hours. Thus, two technicians are needed to meet the company's 3-hour service guideline when the company reaches 20 customers. The total cost is $275.27 per hour. 6.
Using the Excel worksheet template Finite, two service technicians provide the following: Probability that no customers are in the system Average number of customers in the waiting line Average number of customers in the system Average time a customer spends in the waiting line Average time until the machine is back in operation Probability of a wait more than one hour Total cost per hour of service operation
0.1760 0.9353 2.3194 1.6895 hours* 4.1895 hours 0.5600 $391.94
*The average time a customer spends in the waiting line is 1.6895 hours. While the average travel time is 1-hour for the service technician to reach the new customer's office, the average total customer waiting time is 1.6895 + 1 = 2.6895 hours. Using the Excel worksheet template Finite, three service technicians provide the following:
CP - 69
Chapter 15
Probability that no customers are in the system Average number of customers in the waiting line Average number of customers in the system Average time a customer spends in the waiting line Average time until the machine is back in operation Probability of a wait more than one hour Total cost per hour of service operation
0.2227 0.1493 1.5708 0.2626 hours* 2.7626 hours 0.2010 $397.08
For 30 customers and two technicians, the average customer waiting time until a service technician arrives is 1.6895 + 1 = 2.6895 hours. For 30 customers and three technicians, the average customer waiting time until a service technician arrives is 0.2626 + 1 = 1.2626 hours. The hourly cost with two technicians is $391.94 and the hourly cost with three technicians is $397.08. While three technicians provide a smaller waiting time, two technicians are able to meet the 3-hour service guideline for a total lower cost. Thus, the company should continue to use two technicians when the customer base expands to 30 customers. 7.
The OEI planning committee’s proposal anticipated that three technicians would be needed at a total cost of $397.08 per hour. Thus, the recommendation to stay with two technicians has as annual savings of (397.08 – 391.94) x 8 hours/day x 250 days/year = $10,280.
CP - 70
Chapter 12 Simulation Case Problem 1: Tri-State Corporation With the specific financial analysis data input into cells D3:D8, the formulas used to develop the portfolio projection worksheet are shown below. The rows are copied to extend the worksheet to the desired 30-year projection.
1.
Increasing the annual investment rate in cell D7 will generate the following 30-year portfolio projections: Rate 5% 6% 7% 8%
Projected Portfolio $ 721,667 815,397 909,127 1,002,857
A 1% increase in the annual investment rate increases the projected 30-year portfolio by $93,730. The annual investment rate would have to be increased to 8% in order to achieve the $1,000,000 goal. 2.
The simulation worksheet that we developed placed the simulated annual salary growth rate in column D and the simulated annual portfolio growth rate in column G. The revised data input section and the cell formulas used to develop the simulation worksheet are as follows.
CP - 71
Chapter 16
Data Input The simulated 30-year portfolio amounts will vary considerably from trial to trial. The uncertainty associated with the annual salary growth rate and the annual portfolio growth rate will show that there is uncertainty in reaching the $1,000,000 even if the annual investment rate is increased to 8%. A few simulation trials will point out that the 30-year portfolio variability and the uncertainty of reaching the $1,000,000 goal. However, repeating the simulation numerous times will be necessary to provide an objective basis for estimating the probability of reaching $1,000,000. We preformed the simulation of 1000 trials. The probability of reaching $1,000,000 was estimated to be 0.48. Thus, the simulation conclusion is that there is less than a 50% chance of reaching $1,000,000 even if an 8% investment rate is used. During the 1000 trials, the 30-year portfolio varied from $550,000 to $1,900,000. There was a 30% chance that the portfolio would not reach $900,000. 3.
The simulation model suggests additional strategies should be considered to obtain a reasonably high probability of reaching the $1,000,000 portfolio goal. Increasing the annual investment rate to 9% or 10% may be worth considering. If this rate is getting too high, then extending the 30-year period by adding years may be the best strategy for reaching the $1,000,000 goal.
4.
The longer 35-year period is definitely a good idea. Expanding the simulation spreadsheet by five years will show that almost every simulated 35-year portfolio exceeded $1,000,000. 1000 simulated trials with the 35-year period showed better than a 99% chance that the portfolio would exceed $1,000,000. In fact, the extra five years increased the expected portfolio from $1,002,857 to $1,697,622, or almost $700,000. This result demonstrates the long-term advantages of investing.
CP - 72
Solutions to Case Problems
5.
The simulation worksheet can be used for any employee. The employee may enter his/her age, current salary, current portfolio, and any assumptions he/she cares to make about salary growth rate, investment rate, and portfolio growth rate. These assumptions in cells D3:D8 would be different for each employee and rows would be added to the worksheet to reflect the number of years appropriate for the employee. By varying the inputs and projecting the ultimate portfolio value, the employees should be able to make investment observations such as the following: • •
• •
Begin early with an investment program. The more years the better the ending portfolio value. Make new contributions to an investment program at the highest possible rate. Increasing the rate 1% or 2% will have a significant impact on the long-term portfolio. Resist the temptation to use assets in investment programs for short-term personal expenditures. Early withdraws can be shown to have a major impact the long-term portfolio value. Be patient. Benefits of investment programs may not be significant for the first 5 to 10 years. It is really in the later years where a consistent investment program pays its biggest dividends.
CP - 73
Chapter 16
Case Problem 2: Harbor Dunes Golf Course The Excel worksheet with Analytic Solver Platform simulation model is as follows:
CP - 74
Solutions to Case Problems
1.
Selected statistical results: Simulation results for various. Below is one set of possible results: Statistics Mean Median Standard Deviation Minimum Maximum
Option 1 $11,062 $11,160 $ 1,802 $ 5,940 $14,400
Option 2 $11,173 $11,360 $ 1,838 $ 5,760 $14,400
Using the mean, the options are similar, but Option 2 is preferred with a daily revenue advantage of $11,173 - $11,062 = $111.
CP - 75
Chapter 16
2.
Go with Option 2: The $50 per replay option.
3.
Without the replay option, Harbor Dunes reported $10,240 daily revenue. Thus the Option 2 replay policy is estimated to generate an additional revenue of $11,173 - $10,240 = $933 per day. Over a 90day spring season, the estimated revenue increase is 90 * $933 = $83,970. The replay policy is definitely worthwhile.
4.
One suggestion is that Harbor Dunes wait until mid-morning to determine the afternoon replay option. If by mid-morning, the afternoon tee time reservations are relatively low, Harbor Dunes may want to offer the Option 1 replay policy which generates more demand for the afternoon tee times. However, if by mid-morning, the afternoon tee time reservations were relatively high, Harbor Dunes would want to continue the Option 2 replay policy. Simulation runs could help determine a level of advance afternoon tee time reservations that would indicate whether to implement the Option 1 or Option 2 replay policy.
Case Problem 3: County Beverage Drive-Thru 1. Excel worksheets patterned after those used for the Black Sheep Scarves one- and two-server simulations in Figure 15.15 and 15.17 may be used to solve this case problem. We recommend using one workbook with three worksheets, one for each of the following Drive-Thru designs: single-server with one clerk, single-server with two clerks, and two-server with 2 clerks. One difference between the Black Sheep Scarves simulation and the County Beverage Drive-Thru simulation is the treatment of the transient case. In the statistical summary, Black Sheep Scarves excluded the first 100 customers since the simulation was intended to analyze the long-run effect of the quality inspection systems. For County Beverage Drive-Thru, management wants to analyze the system only during daily peak hours (4 PM to 10 PM). The analysis of the simulation results should include all customer arrivals in the first 360 minutes of the simulation. The logic of the spreadsheet should be modified accordingly. The simulation models track the number of customers arriving between 4 PM and 10 PM, the probability that a customer waits, the average waiting time, the clerk utilization, the probability of waiting more than six minutes, and the probability of waiting more than 10 minutes. 2.
CP - 76
Solutions to Case Problems
Single-Server System Operated by 1 Clerk Use the format of the BlackSheep1 simulation model and set up the spreadsheet as follows:
Cell formulas modified from BlackSheep1 are as follows: Cell
Formula
B20 F20 C1022 C1023 C1024 C1025 C1026 C1027 C1028 C1029 C1030
=$B$4 * -1 * LN(RAND()) =VLOOKUP(RAND(),$A$8:$C$15,3,TRUE) =COUNTIF(C20:C1019,"<360") =COUNTIFS(E20:E1019,">0",C20:C1019,"<360") =C1023/C1022 =AVERAGEIF(C20:C1019,"<360",E20:E1019) =SUMIF(C20:C1019,"<360",F20:F1019)/360 =COUNTIFS(E20:E1019,">6",C20:C1019,"<360") =C1027/C1022 =COUNTIFS(E20:E1019,">10",C20:C1019,"<360") =C1029/C1022
Cells B20 and F20 can be copied to fill columns B and F. Note that in the computation of the single-run summary statistics, only the customers that arrive within the first 360 minutes of the simulation are included (as these are the ones arriving between 4 PM and 10 PM).
CP - 77
Chapter 16
Single-Server System Operated by 2 Clerks Similar to the single-server system with one clerk, we use the format of the BlackSheep1 simulation model (only the service time distribution differs from the single-server with one clerk case). For instance, in cell F17, the service time is now computed by “=VLOOKUP(RAND(),$A$8:$C$12,3,TRUE)”
CP - 78
Solutions to Case Problems
Two-Server System Operated by 2 Clerks We use the format of the BlackSheep2 simulation model. Customer interarrival times and service times are the same as shown for the single-server operated by one clerk design.
This part of the case is optional in that significant spreadsheet modeling skills are required to duplicate the 2-server simulation model. Selected cell formulas area as follows: Cell I20 J20 I21 J21
Formula =G16 =0 =IF(I20=MIN(I20,J20),G21,I20) =IF(J20=MIN(I20,J20),G21,J20)
Cells I21 and J21 can be copied to fill columns I and J. A quick way to generate multiple runs of each of these waiting line simulations is to use a data table and track the corresponding summary statistics across each replication. Simulation results will vary but one set of results over 1000 runs are as follows:
CP - 79
Chapter 16
Characteristic Probability of Waiting Average Waiting Time Utilization of Drive Thru Probability Waiting > 6 Minutes Probability Waiting > 10 Minutes
Single-Server System, 1 Clerk 0.68 5.32 0.71 0.33 0.19
Single-Server System, 2 Clerks
Two-Server System, Two Clerks
0.38 0.84 0.43 0.00 0.00
0.17 0.34 0.36 0.00 0.00
3. The single-server system with 1 clerk system appears unacceptable. The mean waited time is over 5 minutes which exceeds the company guideline of 1.5 minutes. In addition, 33% of customers waited over 6 minutes and 19% waited over 10 minutes. The expected number of unsatisfied customers is then: ((0.33 − 0.19) × 60 × 0.3) + (0.19 × 60 × 0.9) = 12.78. The company must do something to improve the service characteristics of its drive-thru operation or face the loss of substantial business. The single-server system with 2 clerks appears to be the most appropriate design. The mean waiting time of less than 1 minute is with the company’s guideline of 1.5 minutes. Very few customers experienced wait times exceeding 6 or 10 minutes. The expected number of unsatisfied customers is then: (0 × 60 × 0.3) + (0 × 60 × 0.9) = 0. The performance of the two-server system with 2 clerks is the best overall, but the added cost may not justify the expansion to the two server operation. The mean waiting time of less than 1 minute is with the company’s guideline of 1.5 minutes. Very few customers experienced wait times exceeding 6 or 10 minutes. . The expected number of unsatisfied customers is then: (0 × 60 × 0.3) + (0 × 60 × 0.9) = 0.
CP - 80
Chapter 13 Decision Analysis Case Problem 1: Property Purchase Strategy The decision tree for the Oceanview decision problem is shown below. Note that the final outcome of whether the zoning change is approved or rejected by the voters occurs of Oceanview decides to bid and then wins the bid. Otherwise, the outcome of the vote on the zoning change does not enter into the problem. Payoff*
Highest Bid
Bid
Predicts Approval
Zoning Approved
1,985,000
Zoning Rejected
-515,000
9
6 Not Highest Bid
3
-15,000
Do Not Bid Zoning Approved Market Research
Highest Bid
2
Predicts Rejection
10 Zoning Rejected
Bid
Not Highest Bid
-15,000
Do Not Bid Zoning Approved Highest Bid
-15,000 2,000,000
11 Zoning Rejected
Bid
No Market Research
-515,000
7
4
1
-15,000 1,985,000
-500,000
8 Not Highest Bid
5 Do Not Bid
0
0
* Payoff values are the profit computed from the given revenue, property cost, construction expenses, property cost deposit and marketing research cost given in the problem.
CP - 81
Chapter 13
The decision tree with branch probabilities and expected values is as follows: Approved .6585 Highest Bid .2 Bid
Predicts Approval .41
6
9
EV = 1,131,250 Rejected -515,000 .3415
EV = 214,250 Not Highest Bid .8
3 Do Not Bid
Approved .0508 Market Research
2
Highest Bid .2
EV = 78,993 Bid
Predicts Rejection .59
7
10
EV = -89,600
-15,000
Do Not Bid Approved .3 Highest Bid .2 Bid
No Market Research
8
11
-15,000 2,000,000
EV = 250,000 Rejected -500,000 .7
EV = 50,000 Not Highest Bid .8
5
-15,000 1,985,000
EV = -388,000 Rejected -515,000 .9492
Not Highest Bid .8
4
1
1,985,000
Do Not Bid
0
0
If the market research is not conducted, Oceanview should go ahead and bid. The expected value of this action is $50,000 as shown at node 8 of the decision tree. However, the analysis shows that the recommended strategy is to conduct the market research first. the value of the market research exceeds its cost. The optimal strategy is as follows: If the market research predicts approval of the zoning change, bid. If the market research predicts rejection of the zoning change, do not bid. The expected value of this strategy, including the cost of conducting the market research, is $78,993. The market research has a value of $78,993 - $50,000 = $28,993 over and above the cost of the research only.
CP - 82
Solutions to Case Problems
Case Problem 2: Lawsuit Defense Strategy 1. Decision Tree Allied Accepts John's Offer of $750,000
$750 John Accepts Allied's Counteroffer 0.10
$400 Allied Loses Damages of $1,500,000 0.3
$1500
1 Allied Counteroffers with $400,000
John Rejects Allied's Allied Loses Damages Counteroffer of $750,000 2 3 0.4 0.5 Allied Wins No Damages 0.2 Allied Accepts John's Counteroffer
$750
$0
$600 Allied Loses Damages of $1,500,000 0.3
John Counteroffers with $600,000 4 0.5 Allied Rejects John's Counteroffer
Allied Loses Damages of $750,000 5 0.5 Allied Wins No Damages 0.20
To find the optimal decision strategy, we must fold the tree back computing expected values at the chance nodes and choosing the least cost alternative at the decision nodes. Shown below are the values computed at each node. Node 1 2 3 4 5
Value 670 670 825 600 825
CP - 83
$1500
$750
$0
Chapter 13
Shown below is the decision tree with all nonoptimal branches eliminated. John Accepts Allied's Counteroffer 0.10
$400 Allied Loses Damages of $1,500,000 0.3
$1500
1 Allied Counteroffers with $400,000
John Rejects Allied's Allied Loses Damages Counteroffer of $750,000 2 3 0.4 0.5 Allied Wins No Damages 0.2 Allied Accepts John's Counteroffer
$750
$0
$600
John Counteroffers with $600,000 4 0.5
2.
Allied should not accept John's offer to settle for $750,000. The strategy associated with a counteroffer of $400,000 has an expected value of $670,000.
3.
If John accepts Allied's counteroffer of $400,000, no further action is required. If John rejects Allied's counteroffer and elects to have a jury decide the settlement amount, Allied must prepare for a trial. If John counteroffers with $600,000, Allied should accept John's counteroffer.
4.
To develop the risk profile for the optimal strategy, we simply multiple the branch probabilities on all paths to the end points of the decision tree. The risk profile is given below in tabular form. Settlement Amount ($1000s) 0 400 600 750 1500
Probability 0.08 0.10 0.50 0.20 0.12 1.00
CP - 84
Chapter 14 Multicriteria Decision Problems Case Problem: EZ Trailers, Inc. Let x11 = number of EZ-190 trailers produced in March x12 = number of EZ-190 trailers produced in April x21 = number of EZ-250 trailers produced in March x22 = numb r of EZ-250 trailers produced in April s11 = EZ-190 ending inventory in March s12 = EZ-190 ending inventory in April s21 = EZ-250 ending inventory in March s22 = EZ-250 ending inventory in April P1 Goal: Meet demand for the EZ-250: March:
300 + x21 - s21 - d1+ + d1− = 1000
(1)
April:
s21 + x22 - s22 - d 2+ + d 2− = 1200
(2)
P2 Goal: Meet demand for the EZ-190: March: April:
200 + x11 - s11 - d 3+ + d 3− = 800
(3)
s11 + x12 - s12 - d 4+ + d 4− = 600
(4)
P3 Goal: Limit labor fluctuations from month to month to at most 1000: March: 5300 4 x11 + 6 x21 7300 4 x11 + 6 x21 - d 5+ + d 5− = 5300 + 6
(5)
− 6
4 x11 + 6 x21 - d + d = 7300
(6)
April: (4 x11 + 6 x21) - 1000 4 x12 + 6 x22 (4 x11 + 6 x21) + 1000 4 x12 + 6 x22 = [(4 x11+ 6 x21) - 1000] + d 7+ + d 7− + 7
− 7
4 x12 + 6 x22 - 4 x11 - 6 x21 - d + d = -1000
or (7)
4 x12 + 6 x22 = [(4 x11 + 6 x21) + 1000] + d8+ + d8− or 4 x12 + 6 x22 - 4 x11 - 6 x21 - d8+ + d8− = 1000
CP - 85
(8)
Chapter 14
The complete goal programming model is
( )
( )
( )
( )
( )
( )
( )
( )
Min P1 d1− + P1 d 2− + P2 d 3− + P2 d 4− + P3 d 5− + P3 d 6+ + P3 d 7− + P3 d 8+ s.t.
+
–
x 21 – s21 – d1 + d 1 = 800 +
–
x22 + s 21 – s 22 – d 2 + d 2 = 1200 +
–
x11 – s 11 – d3 + d 3 = 600 x12 + s 11 – s 12 – d4+ + d –4 = 600 +
–
+
–
4x11 + 6x21 – d 5 + d 5 = 5300 4x11 + 6x21 – d 6 + d 6 = 7300 –4x11 + 4x12 – 6 x21 +6x22 – d +7 + d –7 = – 1000 –4x11 + 4 x12 – 6x21 +6x 22 – d +8 + d –8 = 1000 all variables 0 Information Needed for the Managerial Report 1.
EZ-190 EZ-250
2.
March 775 800
No changes since the ending inventories for the optimal production schedule are as follows:
EZ-190 EZ-250 3.
April 425 1200
March 175 0
April 0 0
The following constraints must be added to the model: April ending inventory: s12 100, s22 100 Maximum storage of 300 units in each month: s11 300, s12 300, s21 300, s22 300 The new optimal production schedule is as follows:
EZ-190 EZ-250
March 900 800
April 400 1300
March 300 0
April 100 100
The corresponding ending inventories are:
EZ-190 EZ-250
CP - 86
Solutions to Case Problems
4.
The new optimal production schedule is:
EZ-190 EZ-250
March 625 800
April 275 1200
March 25 0
April 0 0
The corresponding ending inventories are:
EZ-190 EZ-250
CP - 87
Chapter 15 Time Series Analysis and Forecasting Case Problem 1: Forecasting Sales 1.
Month 1 corresponds to January for year 1, month 2 corresponds to February for year 1, and so on. A graph of the time series is shown below. It appears that there is seasonality on a monthly basis with a slight upward trend.
300
250
Sales
200
150
100
50 0
5
10
15
20
25
30
35
40
Month
2. Building an Excel model with dummy data using parameters Month1, Month2…Month11 along with the period (t) for trend, we fit a model as shown below (for brevity, we round to 2 digits): Sales = 199.25 + 49.85819 Month1 + 29.17 Month2 + 33.49 Month3 – 23.53 Month4 - 20.88Month5 67.90 Month6 – 62.58 Month7 - 57.26 Month8 -101.28 Month9 – 85.63 Month10 - 58.65 Month10 + 1.02t The MSE= 12.6. The following graph shows Sales (observed) and fitted values with this model.
CP - 88
Chapter 15 The data set below with Excel Solver were used to fit this model:
Using this fitted model to forecast for year 4 yields in the following forecasts: January
$286,750
February
$267,084
March
$272,417
April
$216,417
May
$220,084
June
$174,084
July
$180,417
August
$186,750
September
$143,750
October
$160,417
November
$188,417
December
$248,083
Forecast error = $295,000 - $286,750 = $8,250. The forecast we developed under predicted by $8,250; this represents a very small error (about 3%). To ease any existing concern, the analysis can be easily updated each month.
CP - 89
Solutions to Case Problems Case Problem 2: Forecasting Lost Sales 1.
The data used for the forecast is the Carlson sales data for the 48 months preceding the storm. Using dummy variables with a model for trend and seasonal components assuming the hurricane had not occurred the forecasts are as are as follows: Month September October November December
Forecast ($ million) 2.23 2.55 3.11 4.52
The fitted model is: Sales = 2.009 - .323 Sept - .016 Oct + .533 Nov + 1.934 Dec - .107 Jan - .286 Feb + .003 Mar .011 April +.193 May - .037 Jun - .054 July - .011t where the month variables are dummy variables and t is the period. 2.
The data used for this forecast is the total sales for the 48 months preceding the storm for all department sores in the county. Using dummy variables with a model for trend and seasonal components assuming the hurricane had not occurred the forecasts are as are as follows: Month September October November December
Forecast ($ million) 49.89 51.99 66.24 105.692
The fitted model is: Sales = 63.812 - 9.413 Sept - 7.221 Oct + 7.121 Nov + 46.663 Dec - 15.344 Jan - 13.152 Feb 2.860 Mar - 4.568 April - .1626 May - 5.434 Jun - 6.392 July - .092 t where the month variables are dummy variables and t is the period. 3.
By comparing the forecast of county-wide department store sales with actual sales, one can determine whether or not there are excess storm-related sales. We have computed a "lift factor" as the ratio of actual sales to forecast sales as a measure of the magnitude of excess sales. Forecast Sales ($ million) 49.89 51.99 66.24 105.69 273.81
Actual Sales ($ million) 69.0 75.0 85.2 121.8 351.0
Lift Factor 1.383 1.443 1.286 1.152 1.282
From the analysis a strong case can be made for excess storm related sales. For each month, actual sales exceed the forecast of what sales would have been without the hurricane. For the 4-month total, actual sales exceeded the forecast by 28.2%. The explanation for the increase is that people had to replace real and personal property damaged by the storm. In addition, the additional construction workers, the disaster relief teams, and so on, created additional commercial activity in the area.
CP - 90
Chapter 15 4.
One approach would be to use the forecast of what sales would have been without the hurricane and then multiply by the lift factor to account for the excess storm-related sales. Such an estimate of lost sales is developed below: Forecast ($ million)
Lift Factor
2.23 2.55 3.11 4.52
1.383 1.443 1.286 1.152
Lost Sales ($ million)
Total
3.084 3.679 4.000 5.209 15.972
Based on this analysis, Carlson Department Stores can make a case to the insurance company for a business interruption claim of $15,972,000. Another approach would be to use the 48 months of historical data to compute a market share for Carlson. That is, compute Carlson’s sales as a fraction of county-wide department store sales. Then you could develop a forecast of Carlson’s market share for September through December. Finally, an estimate of lost sales for each of the four months can be obtained by multiplying the forecasts of market share by the actual department store sales.
CP - 91
Chapter 16 Markov Processes Case Problem: Dealer’s Absorbing State Probabilities in Black Jack 1.
The first step is to create the transition probability matrix for the dealer’s hand. The states are defined by the value of the dealer’s hand after the up card is dealt, and the values the dealer’s hand may assume after the down card is revealed, and after taking hits. The absorbing states are 17, 18, 19, 20, 21, and bust. According to the house rules, once the dealer’s hand takes on one of these values she quits taking hits and either pays or collects the bets that are on the table. The states S12, S13, S14, S15, and S16 represent what are called soft hands; they include an ace that may be played as 1 or 11. There are no S17, S18, S19, S20, and S21 states because the dealer plays these the same as hard hands of the same value. There are 27 states so, due to space restrictions, the transition matrix that is shown below is printed in 3 parts. All of the rows and a subset of the columns are shown in each part.
A S12 S13 S14 S15 S16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Bust
A 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
S12 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
S13 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
S14 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
S15 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
S16 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
`CP - 92
2 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
3 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
4 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
5 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
6 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
Solutions to Case Problems
A S12 S13 S14 S15 S16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Bust
7 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
8 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
9 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
10 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
11 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
12
13
14
15
16
0.3077 0.0769 0.0769 0.0769 0.0769 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
0.3077 0.0769 0.0769 0.0769 0.0000 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0 0 0 0 0 0
0.3077 0.0769 0.0769 0.0000 0.0000 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0 0 0 0 0 0
0.3077 0.0769 0.0000 0.0000 0.0000 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0 0 0 0 0 0
0.3077 0.0000 0.0000 0.0000 0.0000 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0 0 0 0 0 0
CP - 93
17 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0769 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 1 0 0 0 0 0
Chapter 16
A S12 S13 S14 S15 S16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Bust
18 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0 1 0 0 0 0
19 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0 0 1 0 0 0
20 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0769 0 0 0 1 0 0
21 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0769 0.3077 0.0769 0.0769 0.0769 0.0769 0.0769 0 0 0 0 1 0
Bust 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.3077 0.3846 0.4615 0.5385 0.6154 0 0 0 0 0 1
CP - 94
Solutions to Case Problems
The Q matrix is defined as the first 21 rows (through state 16) and the first 21 columns (through state 16) of the transition matrix. We must now compute I – Q and (I – Q)-1. We did this using Excel and Excel’s matrix inversion procedure described in the chapter appendix. To find the absorption probabilities we multiply (I – Q)-1 times the R matrix. The R matrix is a 21 x 6 matrix. It is identified as the first 21 rows and the last 6 columns of the transition matrix. We show (I – Q)-1R below. Probability of Dealer’s Finishing State Given Starting State
A S12 S13 S14 S15 S16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
17 0.1308 0.1510 0.1455 0.1400 0.1346 0.1292 0.1398 0.1350 0.1305 0.1223 0.1654 0.3686 0.1286 0.1200 0.1114 0.1114 0.1035 0.0961 0.0892 0.0828 0.0769
18 0.1308 0.1510 0.1455 0.1400 0.1346 0.1292 0.1349 0.1305 0.1259 0.1223 0.1063 0.1378 0.3593 0.1200 0.1114 0.1114 0.1035 0.0961 0.0892 0.0828 0.0769
19 0.1308 0.1510 0.1455 0.1400 0.1346 0.1292 0.1297 0.1256 0.1214 0.1177 0.1063 0.0786 0.1286 0.3508 0.1114 0.1114 0.1035 0.0961 0.0892 0.0828 0.0769
20 0.1308 0.1510 0.1455 0.1400 0.1346 0.1292 0.1240 0.1203 0.1165 0.1131 0.1017 0.0786 0.0694 0.1200 0.3422 0.1114 0.1035 0.0961 0.0892 0.0828 0.0769
21 0.3616 0.1510 0.1455 0.1400 0.1346 0.1292 0.1180 0.1147 0.1112 0.1082 0.0972 0.0741 0.0694 0.0608 0.1114 0.3422 0.1035 0.0961 0.0892 0.0828 0.0769
Bust 0.1153 0.2450 0.2725 0.3000 0.3272 0.3541 0.3536 0.3739 0.3945 0.4164 0.4232 0.2623 0.2447 0.2284 0.2121 0.2121 0.4827 0.5196 0.5539 0.5858 0.6154
This matrix shows the probabilities for the ending values of the dealer’s hand given any of the starting states identified by each row. For instance, if the dealer has a 6 as an up card, the probability of finishing with 17 is .1654, the probability of finishing with 18 is .1063 and so on. The probability of the dealer busting is .4232. For this reason, black jack players recommend not taking a hit with a hand of 12 or better. Why should the player take a chance of busting when the dealer has a high probability of busting?
CP - 95
Chapter 16
2.
For this situation another row and column must be added to the transition matrix to account for the soft 17 state, S17. When the dealer has to stay on S17, there was no reason to differentiate soft 17 and hard 17. So only the state, 17 was used. For space reasons, we do not show the new transition matrix here. But, after computing (I – Q)-1R, the absorbing state probabilities are given below. Probability of Dealer's Finishing State Given Starting State
A S12 S13 S14 S15 S16 S17 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
17 0.0575 0.0829 0.0823 0.0813 0.0801 0.0786 0.3422 0.1301 0.1263 0.1224 0.1184 0.1148 0.3686 0.1286 0.1200 0.1114 0.1114 0.1035 0.0961 0.0892 0.0828 0.0769
18 0.1432 0.1625 0.1562 0.1500 0.1438 0.1377 0.1114 0.1365 0.1320 0.1273 0.1229 0.1148 0.1378 0.3593 0.1200 0.1114 0.1114 0.1035 0.0961 0.0892 0.0828 0.0769
19 0.1432 0.1625 0.1562 0.1500 0.1438 0.1377 0.1114 0.1313 0.1271 0.1228 0.1184 0.1148 0.0786 0.1286 0.3508 0.1114 0.1114 0.1035 0.0961 0.0892 0.0828 0.0769
20 0.1432 0.1625 0.1562 0.1500 0.1438 0.1377 0.1114 0.1257 0.1218 0.1179 0.1138 0.1103 0.0786 0.0694 0.1200 0.3422 0.1114 0.1035 0.0961 0.0892 0.0828 0.0769
21 Bust 0.3740 0.1389 0.1625 0.2669 0.1562 0.2929 0.1500 0.3189 0.1438 0.3448 0.1377 0.3704 0.1114 0.2121 0.1196 0.3567 0.1162 0.3767 0.1126 0.3971 0.1089 0.4177 0.1057 0.4395 0.0741 0.2623 0.0694 0.2447 0.0608 0.2284 0.1114 0.2121 0.3422 0.2121 0.1035 0.4827 0.0961 0.5196 0.0892 0.5539 0.0828 0.5858 0.0769 0.6154
This matrix shows the probabilities for the ending values of the dealer’s hand given any of the starting states identified by each row for the case when the dealer hits soft 17. For instance, if the dealer has a 6 as an up card, the probability of finishing with 17 is .1148, the probability of finishing with 18 is .1148, and so on. The probability of the dealer busting is .4395. 3.
Mathematicians have shown that when the dealer stays on soft 17 it is better for the player. We can provide the following argument using the finishing values for the dealer’s hand shown in part 2 above. Note that when the dealer gets S17 and hits it, the finishing values and probabilities for the dealer’s hand are Ending Value 17 18 19 20 21 Bust
Probability .3422 .1114 .1114 .1114 .1114 .2121
CP - 96
Solutions to Case Problems
If the player has 17, the probability of a tie is .3422, the probability of losing is 4(.1114) = .4456, and the probability of winning is .2121. So, when the player has 17, she will do better (a tie) if the dealer stays on S17. Similarly, if the player has 18, 19, 20, or 21, she will do much better if the dealer stays on S17. She will always win. But, if the player has 12, 13, 14, 15, or 16, she will be better off if the dealer hits soft 17. This is because when the dealer busts she will win (if the dealer does not hit S17 she loses). And, this will happen with probability .2121. But, the player will lose with probability 1 - .2121 = .7879 when the dealer hits S17. This small probability of the player winning with 12, 13, 14, 15, and 16 when the dealer hits S17 is not enough to offset the large increase in the probability of the player winning with 17, 18,19, 20, and 21 when the dealer stays on S17.
CP - 97
Chapter 21 Dynamic Programming Case Problem: Process Design The optimal solution is to operate the heater at a temperature of 800°, use catalyst C2 for the reactor, and purchase separator S1. This results in a weekly profit of $9,580 calculated as follows. Weekly revenue (2700 lbs. x $6.00) Weekly costs
$16,200
Payback on heater (12,000/100) Payback on reactor ( 50,000/100) Payback on separator (20,000/100) Operate heater Operate reactor Operate separator (2700 x 0.10) Raw material (4500 x 1.00) Net Weekly Profit
120 500 200 380 650 270 4,500
6,620 $ 9,580
Calculations Let stage 1 be separation, stage 2 reactor, and stage 3 heating. Stage 1 d1 x1
s1
s2
d1*
f1(d1)
900 lb.
$5,110
$5,170
s2
$5,170
1800 lb.
10,420
10,390
s1
10,420
2700 lb.
15,730
15,610
s1
15,730
Stage 2 d2 x2
c1
c2
d 2*
f2(d2)
x1
4500 lb., 700
$4,220
$9,270
c2
$9,270
1800
4500 lb., 800
9,470
14,580
c1
14,580
2700
Stage 3 d3 x3
700
800
d 3*
f3(d3)
x1
4500 lb.
$4,370
$9,580
800
$9,580
(4500, 800)
CP - 98