Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-11
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19. a.
Xij = number of units of specimen i assigned to machine j
MIN
ST
b.
c.
d.
e.
23. a.
See file: PRB3_19.XLS
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
rounded to yield the optimal integer solution.)
Machine A & C are used all 480 minutes, machine B is used 298.33 minutes
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.
XiR = barrels of input i used to produce regular
XiS = barrels of input i used to produce supreme
MAX:
ST:
b.
c.
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
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
(21-17.25)X1R+(21-15.75)X2R+(21-17.75)X3R+(25-17.25)X1S+(25-15.75)X2S+(25-17.75)X3S
X1R + X1S  150
X2R + X2S  350
X3R + X3S  300
X1R + X2R + X3R = 300
X1S + X2S + X3S = 450
(100X1R + 87X2R + 110X3R)/300  90
(100X1S + 87X2S + 110X3S)/450  97
Xij  0
See file PRB3_23.XLS
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)
S-12 : Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet
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35. 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
ST
b.
c.
d.
29. a.
See file: PRB3_35.XLS
X1 = 500, X2 = 1000, X3 = 1500, X4 = 2250, X5 = 500, Total Profit = $215,000
No. The assembly constraint is nonbinding.
Xij = 1 if component i is assigned to company j; 0, otherwise
MIN
ST
b.
c.
33. a.
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
X1A + X1B + X1C + X1D = 1
X2A + X2B + X2C + X2D = 1
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
See file: PRB3_29.XLS
X1A = X2C = X3D = X4B = 1, Minimum cost = $1,049 (in $1,000s)
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
ST
b.
c.
53 X1 + 48 X2 + 33 X3 + 32 X4 + 38 X5
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
0.75 X1 + 0.6 X2 + 0.5 X3 + 0.65 X4 + 1 X5  5,000
2 X1 - 1 X2  0
Xi  0
0.095A + 0.08B + 0.09C + 0.09D + 0.09E
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
See file PRB3_33.XLS
A=20,339, B=20,339, C=29,661, D=0 , E=29,661
Maximum return = $8,898 (or 8.898%)
Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-13
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47. B. 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
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.
c.
See file: PRB3_47.XLS
X1 = 405.80, X2 = 173.91, Maximum revenue = $220,290
12. a.
A = Amount to invest in investment A
B = Amount to invest in investment B
C = Amount to invest in investment C
D = Amount to invest in investment D
MAX
ST
b.
c.
E = Amount to invest in investment E
M1 = Amount to invest in savings year 1
M2 = Amount to invest in savings year 2
M3 = Amount to invest in savings year 3
1.25 B + 1.35 C + 1.13 D + 1.08 M3
A + C + E + M1 = 1,000,000
0.5 A + 1.08 M1 - B – M2 = 0
0.8 A + 1.27 E + 1.08 M2 - D – M3 = 0
0  A  500,000
0  B  500,000
0  C  500,000
0  D  500,000
0  E  500,000
50,000  M1  500,000
50,000  M2  500,000
50,000  M3  500,000
See file: PRB3_12.XLS
A=500,000, B= 275,685, C=0, D=500,000, E=429,921, M1=70,079, M2=50,000, M3=500,000
Maximum amount of money at the beginning of 2001 = $1,449,606