Solution Manual for Spreadsheet Modeling and Decision Analysis

1. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-1
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Chapter 3
Modeling & Solving LP Problems In A Spreadsheet
1. In general, it does not matter what is placed in a variable (changing) cell. Ultimately, Solver will
determine the optimal values for these cells. If the model builder places formulas in changing cells,
Solver will replace the formulas with numeric constants representing the optimal values of the decision
variables. An exception to this general principle is found in Chapter 8 where, when solving nonlinear
programming problems, the values placed in the changing cells represent the initial starting point for the
optimizer.
2. Communication - once the user understands the first formula in a range of copied cells, he or she
should understand all the formulas in the range.
Reliability - assuming the first formula is entered correctly, all the copied formulas should be correct
also.
Auditability - once the user understands the first formula in a range of copied cells, he or she should
understand (and audit) all the formulas in the range
Maintainability - if a change needs to be made, it can be made in one formula and then copied as
necessary.
3. TV ads = 10, Magazine ads =25, Maximum profit = $775,000
See file: Prb3_3.xls
4. Ore 1 = 28, Ore 2 = 8, Minimum cost = $3,480
See file: Prb3_4.xls
5. Beef = 50%, Pork = 50%, Minimum cost per pound = $0.75
See file: Prb3_5.xls
6. Razors = 240, Zoomers = 420, Maximum profit = $33,600
See file: Prb3_6.xls
7. Executive desks = 100, Senator desks = 500, Maximum profit = $59,300
See file: Prb3_7.xls
8. Acres planted in watermelons = 60, Acres planted in cantaloupes = 40, Maximum profit = $26,740
See file: Prb3_8.xls
9. Doors = 20, Windows = 40 , Maximum profit = $26,000
See file: Prb3_9.xls
10. Desktops = 46.15, Laptops = 69.23, Maximum profit = $90,000 (alternate optimal solutions exist)
See file: Prb3_10.xls
11. TV = 20, Managize = 2 , Minimal cost = $3.5 million
See file: Prb3_11.xls
12. a. X1 = Number of country tables to produce
X2 = Number of contemporary tables to produce
MAX 350 X1 + 450 X2
ST 1.5 X1 + 2 X2 ≤ 1,000
3 X1 + 4.5 X2 ≤ 2,000
2.5 X1 + 1.5 X2 ≤ 1,500

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X1/ ( X1 +X2) ≥ 0.20 (implement as X1 ≥ 0.2* ( X1 +X2) )
X2/ ( X1 +X2) ≥ 0.30 (implement as X2 ≥ 0.3* ( X1 +X2) )
Xi ≥ 0
Many students attempt to implement the ratio constraints in their original form; resulting in a
division by zero error at the null solution and a message from Solver that the model is not linear.
The algebraic equivalence of the alternate form of these constraints (given parenthetically above)
should be noted.
b. See file: Prb3_12.xls
c. X1 = 405.80, X2 = 173.91, Maximum revenue = $220,290
13. a. X1,j = Number of dehumidifiers made in Atlanta in month j
X2,j = Number of dehumidifiers made in Phoenix in month j
Bj = Beginning inventory in month j
MIN 400 (X11 + X12 + X13 ) +360 (X21 + X22 + X23 ) + 30 ( B1 + B2 + B3 )
ST B1 + X11 + X21 – 300 ≥ 0
B2 + X12 + X22 – 400 ≥ 0
B3 + X13 + X23 – 500 ≥ 0
Xij ≤ 300
Xij ≥ 0
Where
B1 = 0
B2 = B1 + X11 + X21 – 300
B3 = B2 + X12 + X22 – 400
b. See file: Prb3_13.xls
c. X11 = 0, X12 = 100, X13 = 200, X21 = 300, X22 = 300, X23 = 300, Maximum revenue = $444,000
14. a. X1 = pounds of Whole product to produce
X2 = pounds of Cluster product to produce
X3 = pounds of Crunch product to produce
X4 = pounds of Roasted product to produce
MAX 1.85 X1 + 1.4 X2 + 1.04 X3 + 1.40 X4
ST 1 X1 + 1 X2 + 1 X3 + 1 X4 < 3600
2 X1 + 1.5 X2 + 1 X3 + 1.75 X4 < 3600
1 X1 + 0.7 X2 + 0.2 X3 + 0.00 X4 < 3600
2.5 X1 + 1.6 X2 + 1.25 X3 + 1 X4 < 3600
0.6 X1 + 0.4 X2 + 0.2 X3 + 1 X4 < 1100
0.4 X1 + 0.6 X2 + 0.8 X3 + 0 X4 < 800
1,000 < X1 < 99,999
400 < X2 < 500
0 < X3 < 150
0 < X4 < 200
b. See file: Prb3_14.xls
c. X1= 1000, X2= 500, X3= 80, X4= 200, Maximum profit = $2,913.2
15. a. N1 = number of Newspaper ads to run at $1,000 each
N2 = number of Newspaper ads to run at $900 each

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N3 = number of Newspaper ads to run at $800 each
T1 = number of Television ads to run at $12,000 each
T2 = number of Television ads to run at $10,000 each
T3 = number of Television ads to run at $8,000 each
MAX 900 N1 + 700 N2 + 400 N3 + 10,000 T1 + 7,500 T2 + 5,000 T3
ST 1,000 N1 + 900 N2 + 800 N3 + 12,000 T1 + 10,000 T2 + 8,000 T3 ≤ 145,000
Ni ≤ 10
Ti ≤ 5
Ti, Ni > 0
b. See file: Prb3_12.xls
c. N1 = 10, N2 = 10, N3 = 0, T1 = 5 , T2 = 5 , T3 = 2
New Customers = 113,500
d. N3 is used before N2.
16. a. Xij = Square feet of space rented in month i (i=1, 2, 3, 4, 5) through month j (j=i, i+1, …, 5)
MIN 55X11 + 95X12 + 130X13 + 155X14 + 185X15 + 55X22 + 95X23 + 130X24 + 155X25 + 55X33 +
95X34 + 130X35 + 55X44 + 95X45 + 55X55
ST X11 + X12 + X13 + X14 + X15 > 20000
X12 + X13 + X14 + X15 + X22 + X23 + X24 + X25> 30000
X13 + X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 > 40000
X14 + X15 + X24 + X25 + X34 + X35 + X44 + X45 > 35000
X15 + X25 + X35 + X45 + X55 > 50000
Xij > 0
b. See file: Prb3_16.xls
c. X15 = 20000, X25 = 10000, X33 = 5000, X35 = 5000, X55 = 15000
Total leasing cost = $7 million
d. $9.625 million
17. a. X1 = Amount invested in Bonds
X2 = Amount invested in Mortgages
X3 = Amount invested in Car loans
X4 = Amount invested in Personal Loans
MAX 10 X1 + 8.5 X2 + 9.5 X3 + 12.5 X4
ST X1 + X2 + X3 + X4 = 650,000
X4 ≤ .25*(650000)
X4 ≤ X2
X4 ≤ X1
X1 , X2 , X3 , X4 ≥ 0
b. See file: Prb3_17.xls
c. X1 = 325,000, X2 = 162,500, X3 = 0, X4 = 162,500, Maximum return = 10.25%
18. a. X1 = number of HyperLink cards to produce
X2 = number of FastLink cards to produce
X3 = number of SpeedLink cards to produce
X4 = number of MicroLink cards to produce
X5 = number of EtherLink cards to produce
MAX 53 X1 + 48 X2 + 33 X3 + 32 X4 + 38 X5
ST 20 X1 + 15 X2 + 10 X3 + 8 X4 + 5 X5 ≤ 80,000
28 X1 + 24 X2 + 18 X3 + 12 X4 + 16 X5 ≤ 100,000
8 X1 + 8 X2 + 4 X3 + 4 X4 + 6 X5 ≤ 30,000

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0.75 X1 + 0.6 X2 + 0.5 X3 + 0.65 X4 + 1 X5 ≤ 5,000
2 X1 - 1 X2 ≤ 0
Xi ≥ 500
b. See file: Prb3_18.xls
c. X1 = 500, X2 = 1000, X3 = 1500, X4 = 2250, X5 = 500, Total Profit = $215,000
d. No. The assembly constraint is nonbinding.
19. a. A = amount to invest in bond A
B = amount to invest in bond B
C = amount to invest in bond C
D = amount to invest in bond D
E = amount to invest in bond E
MAX 0.095A + 0.08B + 0.09C + 0.09D + 0.09E
ST A + B + C + D + E = 100,000
B + E ≥ 50,000
A + D + E ≤ 50,000
A + B + D ≥ 30,000
0.095A + 0.08B + 0.09D ≥ 0.4* (0.095A + 0.08B + 0.09C + 0.09D + 0.09E)
A, B, C, D, E ≥ 0
b. See file Prb3_19.xls
c. A=20,339, B=20,339, C=29,661, D=0 , E=29,661
Maximum return = $8,898 (or 8.898%)
20. a. M1= number of electric trimmers to make
M2= number of gas trimmers to make
B1= number of electric trimmers to buy
B2= number of gas trimmers to buy
MIN 55M1 + 85 M2 + 67 B1 + 95 B2
ST M1 + B1 = 30,000
M2 + B2 = 15,000
0.20M1 + 0.40M2 ≤ 10,000
0.30M1 + 0.50M2 ≤ 15,000
0.10M1 + 0.10M2 ≤ 5,000
Mi, Bi ≥ 0
b. See file: Prb3_20.xls
c. M1=30,000, M2=10,000, B1=0, B2=5,000
Minimum cost = $2,975,000
21. a. Xij = 1 if component i is assigned to company j; 0, otherwise
MIN 185 X1A +225 X1B +193 X1C +207 X1D
+200 X2A +190 X2B +175 X2C +225 X2D
+330 X3A +320 X3B +315 X3C +300 X3D
+375 X4A +389 X4B +425 X4C +445 X4D
ST X1A + X1B + X1C + X1D = 1
X2A + X2B + X2C + X2D = 1

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X3A + X3B + X3C + X3D = 1
X4A + X4B + X4C + X4D = 1
X1A + X2A + X3A + X4A = 1
X1B + X2B + X3B + X4B = 1
X1C + X2C + X3C + X4C = 1
X1D + X2D + X3D + X4D = 1
b. See file: Prb3_21.xls
c. X1A = X2C = X3D = X4B = 1, Minimum cost = $1,049 (in $1,000s)
22. a. Pi = proportion of compound i to include in the mix
MIN 5.00 P1 + 5.25 P2 + 5.50 P3
ST 0.20 P1 + 0.40 P2 + 0.10 P3 ≥ .20
0.60 P1 + 0.30 P2 + 0.40 P3 ≥ .30
0.20 P1 + 0.30 P2 + 0.50 P3 ≥ .30
0.20 P1 + 0.30 P2 + 0.50 P3 ≤ .45
P1 + P2 + P3 = 1.0
Pi ≥ 0
b. See file: Prb3_22.xls
c. P1=0.5714, P2=0.1429, P3=0.2857
Minimum cost per pound = $5.18
23. a. Bi = pounds of grade i fruit used in baskets
Ji = pounds of grade i fruit used in juice
MAX: $2.50 (B1 + B2 + B3 + B4 + B5 ) + $1.75 (J1 + J2 + J3 + J4 + J5 )
S.T.: B1 + J1 ≤ 90
B2 + J2 ≤ 225
B3 + J3 ≤ 300
B4 + J4 ≤ 100
B5 + J5 ≤ 75
1 B1 + 2 B2 + 3 B3 + 4 B4 + 5 B5 ≥ 3.75 (B1 + B2 + B3 + B4 + B5 )
1 J1 + 2 J2 + 3 J3 + 4 J4 + 5 J5 ≥ 2.50 (J1 + J2 + J3 + J4 + J5 )
Bi , Ji ≥ 0
b. See file: Prb3_23.xls
c. B1 = 0, B2 = 46.67, B3 =0, B4 = 100 B5 = 45.33,
J1 = 90, J2 = 178.33, J3 =300, J4 = 0 J5 = 29.67,
Profit = $1,526,500
24. a. XiR = barrels of input i used to produce regular
XiS = barrels of input i used to produce supreme
MAX: (21-17.25)X1R+(21-15.75)X2R+(21-17.75)X3R+(25-17.25)X1S+(25-15.75)X2S+(25-17.75)X3S
ST: X1R + X1S ≤ 150
X2R + X2S ≤ 350
X3R + X3S ≤ 300
X1R + X2R + X3R = 300
X1S + X2S + X3S = 450
(100X1R + 87X2R + 110X3R)/300 ≥ 90

6. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-6
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(100X1S + 87X2S + 110X3S)/450 ≥ 97
Xij ≥ 0
b. See file Prb3_24.xls
c. X1R=0, X2R=260.87, X3R=39.13, X1S=150, X2S=89.13, X3S=210.87 (alternate optimal exist)
Maximum Profit = $5,012.5 (in $1,000s)
25. a. X1 = number of workers starting at 12 am
X2 = number of workers starting at 4 am
X3 = number of workers starting at 8 am
X4 = number of workers starting at 12 pm
X5 = number of workers starting at 4 pm
X6 = number of workers starting at 8 pm
MIN X1+ X2 + X3+ X4 + X5+ X6
ST X6 + X1 ≥ 90
X1 + X2 ≥ 215
X2 + X3 ≥ 250
X3 + X4 ≥ 165
X4 + X5 ≥ 300
X5 + X6 ≥ 125
Xi ≥ 0
b. See file: Prb3_25.xls
c. X1=90, X2 =250, X3=0, X4 =175, X5=125, X6=0 (alternate optimal solutions exist)
Minimum number of employees = 640
Teaching Note: As an interesting extension to this problem, ask students to consider how to minimize
the maximum number of excess employees on any shift while holding the total number of employees
used at its optimal value of 640.
26. a. Xij = number of units of specimen i assigned to machine j
MIN 3 X1A + 4 X2A + 4 X3A + 5 X4A + 3 X5A
+ 5 X1B + 3 X2B + 5 X3B + 4 X4B + 5 X5B
+ 2 X1C + 5 X2C + 3 X3C + 3 X4C + 4 X5C
ST 3 X1A + 4 X2A + 4 X3A + 5 X4A + 3 X5A ≤ 480
5 X1B + 3 X2B + 5 X3B + 4 X4B + 5 X5B ≤ 480
2 X1C + 5 X2C + 3 X3C + 3 X4C + 4 X5C ≤ 480
X1A + X1B + X1C = 80
X2A + X2B + X2C = 75
X3A + X3B + X3C = 80
X4A + X4B + X4C = 120
X5A + X5B + X5C = 60
Xij ≥ 0
b. See file: Prb3_26.xls
c. X1C = 80, X2B = 75, X3A = 75, X3C = 5, X4B = 18.33, X4C = 101.67, X5A = 60
Minimum processing time = 1258.33 minutes. (If an integer solution is needed the LP solution can
be

7. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-7
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rounded to yield the optimal integer solution.)
d. Machine A & C are used all 480 minutes, machine B is used 298.33 minutes
e. A solution exists where all machine are used for an equal amount of time (425.5 minutes each).
This
increases the total time used to 1276.5 minutes.
27. a. Pi = proportion of coal i to include in the mix
MAX 24,000 P1 + 36,000 P2 + 28,000 P3
ST 1,100 P1 + 3,500 P2 + 1,300 P3 ≤ 2,500
1.7 P1 + 3.2 P2 + 2.4 P3 ≤ 2.8
P1 + P2 + P3 = 1.0
Pi ≥ 0
b. See file: Prb3_27.xls
c. P1=0.058, P2=0.5507, P3=0.3913
Maximum steam production = 32,174 pounds per ton
d. 32,174 × 30 = 965,217 pounds of steam
28. a. X1 = number of CD players to produce
X2 = number of tape decks to produce
X3 = number of stereo tuners to produce
MAX 75 X1 + 50 X2 + 40 X3
ST 3 X1 + 2 X2 + 1 X3 ≤ 400,000
50,000 ≤ X1 ≤ 150,000
50,000 ≤ X2 ≤ 100,000
50,000 ≤ X3 ≤ 90,000
b. See file: Prb3_28.xls
c. X1= 70,000, X2 = 50,000, X3 = 90,000
Maximum profit = $11,350,000
29. a. Xij = number of cars shipped from location i to location j
MIN 54 X13 + 17 X14 + 23 X15 + 30 X16 + 24 X23 + 18 X24 + 19 X25 + 31 X26
ST X13 + X14 + X15 + X16 = 16
X23 + X24 + X25 + X26 = 18
5 ≤ X13 + X23 ≤ 10
5 ≤ X14 + X24 ≤ 10
5 ≤ X15 + X25 ≤ 10
5 ≤ X16 + X26 ≤ 10
Xij ≥ 0
b. See file: Prb3_29.xls
c. X23 = 9, X14 = 10, X15 = 1, X25 = 9, X16 = 5
Minimum transportation cost = $730
30. a. See file: Prb3_30.xls
b. Minimum cost = $3,011,360, optimal shipping plan (using all production capacity) is:

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FromTo Tacoma San Diego Dallas Denver St. Louis Tampa Baltimore
Macon 0 0 0 0 0 12000 6000
Louisville 600 0 0 0 14400 0 0
Detroit 400 0 10800 12600 0 0 1200
Phoenix 5800 14200 0 0 0 0 0
31. a. See file: Prb3_31.xls
c. Minimum cost = $44,067.67, recycling plan is:
Newsprint Packaging Print Stock
Newspaper 588.24 11.76 0.00
Mixed Paper 0.00 71.43 428.57
White Office Paper 0.00 300.00 0.00
Cardboard 0.00 397.78 0.00
32. a. Xij = number of bottles produced at vineyard i sold to restaurant j
MAX 39X11 + 36X12 + 34X13 + 34X14 + 32X21 + 36X22 + 37X23 + 34X24
ST X11 + X12 + X13 + X14 = 3,500
X21 + X22 + X23 + X24 = 3,100
X11 + X21 ≤ 1800
X12 + X22 ≤ 2300
X13 + X23 ≤ 1250
X14 + X24 ≤ 1750
Xij ≥ 0
b. See file: Prb3_32.xls
c. X11 = 1,800, X12 = 1,700, X22 = 600, X23 = 1,250, X24 = 1,250, Maximum profit = $241,750
(Alternate optima exist.)
33. a. X1 = Cases of Extra Hot sauce to produce
X2 = Cases of Hot sauce to produce
X3 = Cases of Mild sauce to produce
A1 = Advertising dollars spent promoting Extra Hot sauce
A2 = Advertising dollars spent promoting Hot sauce
A3 = Advertising dollars spent promoting Mild sauce
MAX 4 X1 + 4.5 X2 + 4.75 X3 - A1 - A2 - A3
ST X1 = 8,000 + 10 A1
X2 = 10,000 + 8 A2
X3 = 12,000 + 5 A3
A1 + A2 + A3 ≤ 25,000
Ai ≥ 5,000
Note that the Xi can be computed directly from the Ai. Therefore, the Ai are the only decision
variables (changing cells) in the model. The Xi can be computed in the spreadsheet using the
conditions imposed by the first three constraints. Therefore, it is not necessary to indicate these as
constraints cells for Solver.

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b. See file: Prb3_33.xls
c. X1 = 158,000, X2 = 50,000, X3 = 37,000, A1 = 15,000, A2 = A3 = 5000
Maximum profit = $1,007,750
34. a. Pi = Number of units to produce in month i
Ii = Inventory held at the end of month i
MIN 49 X1 + 45 X2 + 46 X3 + 47 X4 - 1.5 (120 + 2I1 + 2I2 + 2I3 + I4 )/2
ST I1 =120 + P1 - 420
I2 = I1+ P2 - 580
I3 = I2+ P3 - 310
I4 = I3+ P4 - 540
400 ≤ P1 ≤ 500
400 ≤ P2 ≤ 520
400 ≤ P3 ≤ 450
400 ≤ P4 ≤ 550
Ii ≥ 50
Note that the Ii can be computed directly from the Pi. Therefore, the Pi are the only decision
variables (changing cells) in the model. The Ii can be computed in the spreadsheet using the
conditions imposed by the first four constraints. Therefore, it is not necessary to indicate these as
constraints cells for Solver. However, note that lower bounds of 50 must be indicated for these
cells.
b. See file: Prb3_34.xls
c. X1 = 410, X2 = 520, X3 = 400, X4 = 450, I1 = 110, I2 = 50, I3 = 140, I4 = 50
Minimum cost = $83,617
35. a. Xij = tons of commodity i stored in hold j
MAX 70(X11+X12+X13) + 50(X21+X22+X23) + 60(X31+X32+X33) + 80(X41+X42+X43)
ST X11 + X12 + X13 < 4800
X21 + X22 + X23 < 2500
X31 + X32 + X33 < 1200
X41 + X42 + X43 < 1700
X11 + X21 + X31 + X41 < 3000
X12 + X22 + X32 + X42 < 6000
X13 + X23 + X33 + X43 < 4000
40X11 + 25X21 + 60X31 + 55X41 < 145000
40X12 + 25X22 + 60X32 + 55X42 < 180000
40X13 + 25X23 + 60X33 + 55X43 < 155000
0.9(X13 + X23 + X33 + X43 ) < X11 + X21 + X31 + X41 < 1.1(X13 + X23 + X33 + X43)
0.4 × Total < X12 + X22 + X32 + X42 < 0.6 × Total
Xij > 0
b. See file: Prb3_35.xls
c. Profit = $669,000
1 2 3 4
Forward X11=1287.5 X21=0 X31=0 X41=1700

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Center X12=1580 X22=2500 X32=0 X42=0
Rear X13=1932.5 X23=0 X33=1200 X43=0
36. a. Xij = Square feet (in 1000s) leased at beginning of month i for j months
MIN 300 (X11 + X21 + X31 + X41 + X51) + 525 (X12 + X22 + X32 + X42 ) + 775 (X13 + X23 + X33 )
+ 850 (X14 + X24) + 975 X15
ST X11 + X12 + X13 + X14 + X15 > 25
X21 + X22 + X23 + X24 + X12 + X13 + X14 + X15 > 10
X31 + X32 + X33 + X22 + X23 + X24 + X13 + X14 + X15 > 20
X41 + X42 + X32 + X33 + X23 + X24 + X14 + X15 > 10
X51 + X42 + X33 + X24 + X15 > 5
Xij > 0
b. See file: Prb3_36.xls
c. X11 = 15, X14 = 5, X15 = 5, X31 = 10
Total cost = $16,625
37. a. Pi = Number of tons purchased in month i
Si = Number of tons sold in month i
Ii = Inventory held at the end of month i
MAX 135 S1 + 110 S2 + 150 S3 + 175 S4 + 130 S5 + 145 S6
-135 P1 - 110 P2 - 150 P3 - 175 P4 - 130 P5 - 145 P6
- 10 ( 70 + 2I1 + 2I2 + 2I3 + 2I4 + 2I5 + 2I6)/2
ST I1 = 70 + P1 - S1
I2 = I1 + P2 - S2
I3 = I2 + P3 - S3
I4 = I3 + P4 - S4
I5 = I4 + P5 - S5
I6 = I5 + P6 - S6
I6 = 0
0 ≤ Ij ≤ 400
Pj ≥ 0
Sj ≥ 0
Note that the Ij can be computed directly from the Pj and Sj . Therefore, the Pj and Sj are the only
decision variables (changing cells) in the model. The Ij can be computed in the spreadsheet using
the conditions imposed by the first six constraints. Therefore, it is not necessary to indicate these
as constraints cells for Solver. However, note that lower bounds of 0 and upper bounds of 400
apply to these cells. Some students will want to know how much Earl paid for the 70 tons of
soybeans in the beginning inventory. This represents a sunk cost that is irrelevant for the problem
at hand.
b. See file: Prb3_37.xls
c. P1 = 0, P2 = 400, P3 = 0, P4 = 0, P5 = 400, P6 = 0
S1 = 70, S2 = 0, S3 = 0, S4 = 400, S5 = 0, S6 = 400
Maximum profit = $29,100
38. a. A = Amount to invest in investment A

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B = Amount to invest in investment B
C = Amount to invest in investment C
D = Amount to invest in investment D
E = Amount to invest in investment E
S98 = Amount to invest in savings in 1998
S99 = Amount to invest in savings in 1999
S00 = Amount to invest in savings in 2000
MAX 1.25 B + 1.35 C + 1.13 D + 1.08 S00
ST A + C + E + S98 = 1,000,000
0.5 A + 1.08 S98 - B - S99 = 0
0.8 A + 1.27 E + 1.08 S99 - D - S00 = 0
0 ≤ A ≤ 500,000
0 ≤ B ≤ 500,000
0 ≤ C ≤ 500,000
0 ≤ D ≤ 500,000
0 ≤ E ≤ 500,000
50,000 ≤ S98 ≤ 500,000
50,000 ≤ S99 ≤ 500,000
50,000 ≤ S00 ≤ 500,000
b. See file: Prb3_38.xls
c. A=500,000, B= 275,685, C=0, D=500,000, E=429,921, S98=70,079, S99=50,000, S00=500,000
Maximum amount of money at the beginning of 2001 = $1,449,606
39. a. MIN A12 + B13 + C14 + D18
ST 1.06A12 - A23 = 0
1.06A23 + 1.14B13 - A34 - B35 = 0
1.06A34 + 1.18C14 - A45 - C47 = 0
1.06A45 + 1.14B35 - A56 - B57 = 0
1.06A56 - A67 = 12
1.06A67 + 1.14B57 + 1.18C47 - A78 - B79 = 14
1.06A78 + 1.65D18 - A89 = 16
1.06A89 + 1.14B79 = 18
Aij, Bij, Cij, Dij ≥ 0
b. See file Prb3_39.xls
c. A56 = $11,321, A89 = $16,981, B13 = $18,161, B35 = $20,703, B57 = $12,281, D18 = $19,989
Minimum investment = $38,149
40. a. Same as in problem 38 above with the following additional constraints:
-3A12 -1B13 + 2C14 + 4D18 ≤ 0
-3A23 -1B13 + 2C14 + 4D18 ≤ 0
-3A34 -1B35 + 2C14 + 4D18 ≤ 0
-3A45 -1B35 + 2C47 + 4D18 ≤ 0
-3A56 -1B57 + 2C47 + 4D18 ≤ 0
-3A67 -1B57 + 2C47 + 4D18 ≤ 0
-3A78 -1B79 + 4D18 ≤ 0
-3A89 -1B79 ≤ 0
b. See file Prb3_40.xls
c. A56 = $11,321, A78 = $4,945, A89 = $1,345, B13 = $31,306, B35 = $35,688, B57 = $29,364,
B79 = $14,531, D18 = $7,341
Minimum investment = $38,647
41. a. See file Prb3_41.xls

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b. Borrow $220,000 for 5 months in month 1.
Borrow $275,000 for 4 months in month 2.
Borrow $49,505 for 1 months in month 3.
Borrow $270,270 for 2 months in month 3.
Borrow $5,225 for 3 months in month 3.
Total Finance Charge = $22,878.
c. There is no feasible solution if the company is restricted to borrowing no more than $100,000 at
each level in the term / rate structure. Ask students to determine what the borrowing limit would
need to be increase to in order to obtain a feasible solution. The answer (of $101,257) can be
obtained by: 1) making the borrowing limit a changing cell, and 2) also making it the set cell and
minimizing its value.
42. a. Ti = Number of people in group i surveyed by telephone
Wi = Number of people in group i surveyed in “person” vai web-cam
MIN: 18 T1 + 14T2 + 25T3 + 20T4 + 40W1 + 35W2 + 60W3 + 45W4
ST
2000 < T1 + T2 + W1 + W2 < 4000
1000 < W1 + W2 + W3 + W4 < 4000
1000 < W1 + W2 + W3 + W4 < 4000
-4000 < T1 – W1 < 0
0 < W2 + W4 + T2 + T4 < 1600
0 < 0.25W2 + 0.25W4 – 0.75T2 – 0.75T4 < 4000
400 < Ti , Wi < 2000
b. See file: Prb4_42.xls
c. T1 = 1000, T2 = 1200, T3 = 400, T4 = 400, W1 = 1000, W2 =W3=W4= 0 ;
Minimum cost = $92,800
43. See file: Prb3_43.xls
Take a six month loan for $48,000.
Borrow $27,200 against receivables in February and $105,000 in March.
Defer $3,000 in payments in March.
44. a. See file: Prb3_44.xls
b. Purchase 79.5337 units on bond 1, 82.8987 units of bond 2, and 35.023 units of bond 3 and invest
$52,482 in the savings account. Total investment = $246,769.
45. a. See file: Prb3_45.xls
b. Total Profit = $1,309,900
Thousand cubic feet
Day Bought Sold
1 200.00 0.00
2 0.00 170.00
3 0.00 180.00
4 160.00 0.00
5 200.00 0.00
6 0.00 180.00
7 0.00 180.00
8 0.00 0.00
9 180.00 0.00
10 0.00 180.00
c. The prices and inventory level should be updated and re-solved everyday with the decisions for
"Day 1" implemented each day.

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Teaching note: Some students may want to know how much money the company paid for the
150,000 cf of gas in storage. Because this represents a sunk cost, it is irrelevant for decision
making purposes.
46. a. See file: Prb3_44.xls
b. Sheritown Inn, Merrylot, FairPrice Inn, and Western Hotels are efficient.
c. Target values:
Satisfact. Value Price Conv. Comfort Climate Service Food
88 86.1474 64.8421 1.38947 1.01895 0.18526 0.46316 0.46316
47. a. See file: Prb3_45.xls
b. Branches 1, 2, 6 & 8 are efficient.
c. Target values:
ROA New Loans Satisfaction Labor Hrs Op. Costs
6.4000 928.9375 98.0000 5.5692 6.8260
Case 3-1: Putting the Link in the Supply Chain
a. See file: Case3_1.xls; Maximum profit = $1,304,544
b. The solution uses all the aluminum in Daytona and Memphis and all the wood in Tempe.
c. Maximum profit = $1,349,439
d. If 80% of the demand must be met the company could earn $1,329,986. This is $25,422 more than
under the original 90% scenario. Thus, after paying the $10,000 penalty the company would still be
$15,422 ahead under the 80% scenario. However, the increased profit may or may not offset the
potential good will that may be lost if the customers are not happy having less of their desired order
quantities met.
Case 3-2: Baldwin Enterprises
a. See file: Case3_2.xls
b. In millions:
Sell Buy USD EUR GBP HKD JPY
USD 0.000 2.945 0.000 0.000 0.000
EUR 0.000 0.000 0.000 0.000 0.000
GBP 0.000 0.000 0.000 0.000 0.840
HKD 9.328 0.000 0.000 0.000 0.000
JPY 0.000 0.000 0.000 131.314 0.000
c. Transaction cost: $27,867
d. The transaction cost barely changes to $27,860.
e. This creates an unbounded solution – or an arbitrage opportunity.
Case 3-3: The Wolverine Retirement Fund
See file: Case3_3.xls
a. Buy 411 share of AC&C, 169 share of MicroHard (fractional solution rounded up), Total Cost =
$495,892
b. Cost of stipulation (1): $506,590, Cost of stipulaiton (2): $500,736
Case 3-4: Saving the Manatees
a. See file: Case3_4.xls. Maximum impact rating = 23,523
b. Minimum Full-Page ads in the daily papers, minimum full-page Sunday ads, minimum evening TV
spots, maximum highway billboards, maximum 15 & 30 second radio ads, minimum full-page magazine
ads, maximum number of total daily paper ads, maximum Sunday paper ads, maximum magazine ads.

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c. Not at all. These constraints are not binding.
d. Maximum impact rating = 29,289