Solution manual for spreadsheet modeling and decision analysis a practical introduction to managemen

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Solution Manual for Spreadsheet Modeling and Decision Analysis A Practical Introduction to Management Science 6th Edition by Ragsdale

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Chapter 7

Goal Programming & Multiple Objective Optimization

1. Both are mathematical functions of the decision variables. However, there is not predetermined target level for an objective function we want to either maximize it or minimize it. A goal has some predetermined target level and we want to find a solution that comes as close as possible to achieving this target.

2. A GP or MOLP has no optimal solution in the sense of an LP problem where we can mechanically apply an algorithm to locate the optimal solution. Instead, both of these procedures require that we search for an optimal or "most satisfying" solution. This solution may differ from one decision maker to the next.

3. The developers stated goals were making a profit and preserving scenic views. These goals may conflict in the sense that preserving scenic views may restrict the number of lots developed (where fewer lots means fewers houses and less revenue). However, some consumers may be willing to pay a premium for a larger lot with a more scenic view.

4. a. Objectives include: Maximizing lives saved, minimizing property damage, minimizing costs, minimizing political damage, etc.

b. Resources: money, food, shelter, water, manpower, transportation services, medical assistance

c. Most of the resources impact several objectives.

d. Minimzing costs is at odds with the other objectives

e. MOLP could be used to help decision makers analyze the trade-offs between the various objectives under different resource allocation scenarios.

5 a. W1=1, W2=0, W3=0

b. W1=0, W2=1, W3=0

c. W1=0, W2=0, W3=1

d. None. The points along this edge are inefficient (i.e., not Pareto Optimal).

6 a. 2 X1 + 5 X2 + 1 d-1 +d = 25

b. 1 d- = 2, 1 +d =0

c. No. Both deviational variables could be made smaller.

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-1
7 a. 2 4 6 8 10 12 2 4 6 8 10 12 14 16 18 0 0 b. X1 X2 Obj 1 Obj 2

c. If both objectives are to be maximized, the only Pareto optimal solution is X1 =8.57 and X2 = 0.857. This point simultaneously maximizes both objectives. If the second objective is minimized, all points from X1 =4 , X2 =0 to X1 =9, X2 = 0 are Pareto optimal.

8 There are two potential problems. First, the solution to this problem will only generate corner point solutions to the feasible region of the LP. Second, this approach mixes units in the objective function, making the numerical value of the objective difficult to interpret.

9 a. MIN Q

 12W + 4G - 48  Q

0  4W + 4G - 28  Q

0  10W + 20G - 100  Q

W,G  0

b. See file: Prb7 9.xlsm

c. Wythe = 3.33, Giles = 3.67, Maximum excess = 6.667

d. Wythe = 3.06, Giles = 3.94, Maximum percentage excess = 9.37%

10. a. Ai = amount of money (in $1000s) allocated to department i

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-2 4 0 16 4 9 0 36 9 8.57 0.857 36 11.1 1.33 2.66 10.66 9.33
ST 0
MIN 4d1/450 + 2d2/310 + 2d3/275 + 1d4/187.5 + 1d5/135 ST A1 + d1- d1 + = 450 A2 + d2- d2 + = 310 A3 + d3- d3 + = 275 A4 + d4- d4 + = 187.5 A5 + d5- d5 + = 135 A1 + A2 + A3 + A4 + A5 = 1200 A1 > 0.7(450) A2 > 0.7(310) A3
0.7(275)
4
5
b. See file: Prb7_10.xlsm. A1 = 450, A2 = 217, A3 = 266.75, A4 = 131.25, A5 = 135. Obj =0.96 c. See file: Prb7_10.xlsm. A1 = 450, A2 = 217, A3 = 275, A4 = 131.25, A5 = 135. Obj =0.955 d. See file: Prb7_10.xlsm. A1 = 422.29, A2 = 290.91, A3 = 258.06, A4 = 175.95, A5 = 126.69. Obj =0.0616 11. a. MIN d1 + + d1+ d2 + + d2+ d3 + + d3ST 3  Xij  6, j = 1, 2, 3 Xij = 1, i = 1, 2, …, 13 Xij + dj-dj + =325, j = 1, 2, 3 Xij, dj, dj + binary
See file Prb7 11.xlsm There are multiple alternate solutions with minimum objective value = 2
>
A
> 0.7(187.5) A
> 0.7(135) Ai, di, di + > 0
b.

b. See file: Prb7 12.xlsm

c. Aqua-Spas = 61, Hydro-Luxes = 134

d. The Pareto optimal solutions fall on the lines from (0,0) to (174,0) and from (174,0) to (122,78)

13 a. Let Pi = percentage of meat i to include in the mix

b. See file: Prb7_13.xlsm. Min cost = $0.865 per pound, Min Fat Content = 5%

c. ($0.98-$0.865)*500 = $57.5 d.

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-3 12 a.
ST
2 ) 
MIN Q
1(66100 - 350X1 - 300X
Q 2(15X1 + 10X2 - 0)  Q X1 + X2  200 9X1 + 6X2  1566 12X1 + 16X2  2880 X1, X2  0
MIN 0.75P1 + 0.87P2 + 0.98P3 MIN 0.15P1 + 0.10P2 + 0.05P3 ST 0.70P1 + 0.75P2 + 0.80P3  .75 0.12P1 + 0.10P2 + 0.08P3  .1 0.03P1 + 0.05P2 + 0.07P3  .1 P1 + P2 + P3 = 1 P1, P2 , P3  0
MIN Q ST (0.75P1 + 0.87P2 + 0.98P3 - .865 )/0.865  Q (0.15P1 + 0.10P2 + 0.05P3 - .05 )/0.05  Q 0.70P1 + 0.75P2 + 0.80P3  .75 0.12P1 + 0.10P2 + 0.08P3  .1 0.03P1 + 0.05P2 + 0.07P3  .1 P1 + P2 + P3 = 1 P1, P2 , P3  0 The solution is: P1 = 5.9%, P3=94.1% e. MIN Q ST 2(0.75P1 + 0.87P2 + 0.98P3 - .865 )/0.865  Q (0.15P1 + 0.10P2 + 0.05P3 - .05 )/0.05  Q 0.70P1 + 0.75P2 + 0.80P3  .75 0.12P1 + 0.10P2 + 0.08P3  .1 0.03P1 + 0.05P2 + 0.07P3  .1 P1 + P2 + P3 = 1 P1, P2 , P3  0 The solution is: P1 = 3.1%, P3=96.9%

14 a. Let X1 = number of magazine ads, X2 = number of TV ads

MIN Q

ST 250 X1 + 1,200 X2 + 1d - 1d + = 17,250

2,000 X1 + 12,000 X2 + 2d2d + = 12,000

X1  10

X2  5

Xi integer

0

 i i d ,d+  Q

b. See file: Prb7_14.xlsm

c. The solution is: X1=37, X2=5, Q = 11.67%.

d. Place a constraint on 2d + representing the largest cost overrun the manager can tolerate.

15. a. See file: Prb7_15.xlsm

b. Sales = 2%, Property = 1.2%, Food = 1%, Utility = 3%; Maximum Deviation = 13.53%

16 a. MIN

c+/900 + m-/45 + w-/60 + r-/50

ST 120 P + 85 S + 100 N + c- c+ = 900

6 P + 3 S + 6 N + m- m+ = 45

3 P + 4 S + 4 N + w- w+ = 60

4 P + 7 S + 3 N + r- r+ = 50

P, S, N  0 and integer

m, m+, w, w+, r, r+, c, c+  0

b. See file: Prb7_16.xlsm

c. Primetime=0, Soaps=5, News=5

d. Primetime=1, Soaps=8, News=1

e. Primetime=0, Soaps=10, News=0

f. The correct answer to this question will vary from person to person.

17 a. See file Prb7_17.xlsm

b.

b. See file: Prb7_18.xlsm

c. Robo-I = 2, Robo-II = 5, Robo-III = 8

d. Robo-I = 3, Robo-II = 4, Robo-III = 8

e. Robo-I = 1, Robo-II = 4, Robo-III = 8

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-4
Sulfur 1100, Coal dust 1.7, Steam 32,174
2 = .0593, Coal 3 = .3351, Max
= 19.03%
c. Coal 1 = .6056, Coal
deviation
1 = .0161, Coal 2 = .1479, Coal 3 = .836, Max deviation = 47.48% 18. a. MIN g1+/400 + g2-/3200 + g3-/2500 + g4-/3500 + g5-/3000 + g6-/2500 ST 18.5R1 + 25R2 + 35R3 + g1- g1+ = 400 100R1 + 265R2 + 200R3 + g2- g2+ = 3200 130R1 + 235R2 + 160R3 + g3- g3+ = 2500 140R1 + 170R2 + 260R3 + g4- g4+ = 3500 210R1 + 220R2 + 180R3 + g5- g5+ = 3000 80R1 + 120R2 + 220R3 + g6- g6+ = 2500 R1, R2, R3  0
gi-, gi+ 
d. Coal
and integer
0 for all i

19. a. See file: Prb7_19.xlsm

b. Total Cost = $3470

c. See file: Prb7_19.xlsm Maximum deviation = 75%

20. a. Let X1=number of lunch counter units built

X2=number of mall units built

X3=number of standalone units built

MAX 85 X1 + 125 X2 + 175 X3 (returns) MAX 9 X1 + 17 X2 + 35 X3 (jobs created)

150 X1 + 275 X2 + 450 X3  2,000

b. See file: Prb7 20.xlsm

Max return = $965,000

Max jobs = 141

21 a. Let A=funds received by city A (in $1,000s)

B=funds received by city B (in $1,000s)

C=funds received by city C (in $1,000s)

MAX 485/750 A + 850/1,200 B + 1,500/2,500 C

deviation =

b. Best possible value for objective 1 = 1965

Best possible value for objective 2 = 67.4%

c. A=493, B=863, C=1,644. Maximum deviation from optimal on both objectives is 2.5%.

See file: Prb7 21.xlsm

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-5
0  X1  5 0  X2  7 0  X3  3
ST
Xi integer
1
3, X2
4, X3
1,
1
3, X2= 4, X3
1,
c. The solution is: X
=
=
=
Minimax percentage
7.8%. d. The solution is: X
=
=
Minimax weighted percentage deviation = 10.88%.
ST A + B
C
0  A  750 0  B  1,200 0  C  2,500 A/750  Q B/1,200  Q C/2,500  Q
MAX Q
+
= 3,000

22 a. Let E = the number of weekly contacts with existing customers N = the number of weekly contacts with new customers

b. E=10, N=5 See file: Prb7 22.xlsm

23 a. See file Prb7 23.xlsm

b. Cost=44,067.74, Sludge=163,839.29

c. Max Deviation = 1.97%

24 a. See file Prb7 24.xlsm

b. Yield = 11%, Maturity = 1, Risk = 1

c. Maximum deviation = 0.8571

d. Maximum deviation = 1.0392

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-6
MIN d1/6000 + d2/10 + d3/5 + d4/45 ST 425 E + 350 N + d1  6000 E + d2  10 N + d3  5 3 E + 4 N - d4  45 E, N  0 and integer di  0
Newsprint Packaging Print Stock Newspaper 21.09 578.91 0.00 Mixed Paper 414.29 0.00 85.71 White Office Paper 0.00 0.00 300.00 Cardboard 136.52 195.53 0.00
Deviation
3.75% Newsprint Packaging Print Stock Newspaper 53.34 546.66 0.00 Mixed Paper 414.29 0.00 85.71 White Office Paper 0.00 0.00 300.00 Cardboard 102.26 232.38 0.00
d. Max
=
Bond Amount A 0.00% B 85.71% C 14.29% D 0.00% E 0.00%
Bond Amount A 0.00% B 82.35% C 6.86% D 0.00% E 10.78% 25 a.

26.

b. MIN i i (d d -  + +)

ST S + 1d - 1 +d = 7

S + 2d - 2 +d = 3

S + 3d - 3 +d = 1

S +

d -

+d = 6 A +

A, B  0 and integer

i + d d ,  0

Optimal solution: S = 6 , A= 6

See file: Prb7_25.xlsm

a. See file: Prb7_26.xlsm

b. Build towers in areas 8, 11, 19 & 22; Profit = $377 thousand.

c. Build towers in areas 3, 5, 6, 14, 17, 22 & 25; Profit = $219 thousand.

d. Build towers in areas 5, 8, 11, 19 & 22 ; Profit = $368 thousand, Coverage = 21 blocks.

e. Build towers in areas 5, 8, 11, 19 & 22 ; Profit = $368 thousand, Coverage = 21 blocks.

27 a. MIN i i (d d -  + +)

ST A + 43,890 B + 1d - 1 +d = 12,500

A + 35,750 B + 2d - 2 +d = 13,350

A + 27,300 B + 3d - 3 +d = 14,600

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-7 street avenue 1 2 3 4
5
5
6
6
7
7
4
4
d -
+d = 3 A +
d -
+d = 1 A +
d -
+d = 6 A + 8d8 +d = 9
i

b. A = 18,055.70, B = -0.1266, See file Prb7_27.xlsm

c.

28 A = 13,368.36, B = -0.1429, See file Prb7_28.xlsm

29 Let Xij = 1 if machine i performs operation j, 0 otherwise

Case 7-1: Removing Snow In Montreal

See file: Case7_1.xlsm

1 Minimum distance = 4784.9

2. $167,472

3 Maximum contaminant removal = 612.8

4. Distance = 5043.6, Contaminants removed = 579.6

5 Distance = 5161.1, Contaminants removed = 603.16

6. Transporting snow greater distances for contaminant removal costs money. Perhaps the city can improve its contaminant removal technology at key sites at less cost than they will incur to transport snow.

Case 7-2: Planning Diets for the Food Stamp Program

See file: Case7_2.xlsm

1 Cost = $12.952, Preference Rating = 232.698

2. Cost = $109.695, Preference Rating = 2851.55

3 Cost = $19.616, Preference Rating = 1384.31

4. Cost = $16.839, Preference Rating = 1140.03

Case 7-3: Sales Territory Planning At Caro-Life

1. Average population = 690,200, Average area = 4660.34 sq miles. See file: Case7_3.xlsm

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-8
+
B
4d -
A + 8,900 B + 5d - 5 +d =
i i + d d ,  0
A
15,500
+
4 +d = 15,750
17,500
Mileage Estimated Price 43,890 12,500 35,750 13,530 27,300 14,600 15,500 16,094 8,900 16,929
MIN d ST 7 X1A + 8 X1B + 4 X1C + 4 X1D + 9 X1E  d 5 X2A + 3 X2B + 9 X2C + 6 X2D + 8 X2E  d X1A + X2A = 1 X1B + X2B = 1 X1C + X2C = 1 X1D + X2D = 1 X1E + X2E = 1 Xij binary
2A= X2B = X1C = X2D = X1E = 1, d=14 See
The optimal solution is: X
file: Prb7_29.xlsm

2. Average population = 452,107, Average area = 3296.29 sq miles. See file: Case7_3.xlsm

3. Average population = 584,850, Average area = 3803.12 sq miles. See file: Case7_3.xlsm

4. Average population = 619,340, Average area = 3977.63 sq miles. See file: Case7_3.xlsm

5 Some counties are coverd by more than one office – effectively reducing the sale potential of those counties for the agents assigned to them. This could be adjusted for using nonlinear programming techniques.

Chapter 7 - Goal Programming & Multiple Objective Optimization : S-9
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