Linear Programming Applications in Marketing, Finance and Operations Management
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 Brazilian Colombian
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Cost per pound ($) 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 2.033BR + 2.583BD + 1.868CR + 2.418CD s.t.
The optimal solution is BR = 750, BD = 200, CR = 250, and CD = 300. The value of the optimal solution is $3233.75
= 0 = 0 = 1000
BR, BD, CR, CD 30
17. a. Let FM = number of frames manufactured
FP SM SP TM TP
Min 38FM s.t.
3.5FM 2.2FM 3.1FM
= number of frames purchased
= number of supports manufactured = number of supports purchased
= number of straps manufactured
= number of straps purchased
Linear Programming Applications in Marketing, Finance and Operations Management
11.5SM + 15SP + 6.5TM + 7.5TP
1.3SM 1.7SM 2.6SM
SM + SP FM,FP,SM,SP,TM,TP 3 0.
¡ê 21,000 ¡ê 25,200 ¡ê 40,800 3 5,000 3 10,000 3 5,000
Manufacture
+ 0.8TM + 1.7TM
b. Total Cost = $368,076.91
Constraint
Milling: Shaping:
Slack = 0 350 hours used
(25200 – 9623) / 60 = 259.62 hours (40800 – 18300) / 60 = 375 hours
Frames Supports Straps
5000 0 2692 7308 0 5000
c. Subtract values of slack variables from minutes available to determine minutes used. Divide by 60 to determine hours of production time used.
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
Optimal Objective Value 361500.00000
Variable FM FP SM SP TM TP
Constraint 1
Value 2285.71429 2714.28571
10000.00000 0.00000 0.00000 5000.00000
Slack/Surplus 0.00000 3171.42857 7714.28571 0.00000 0.00000 0.00000
Reduced Cost 0.00000 3.57692 0.00000 0.00000 1.15385 0.00000
Dual Value -2.69231
0.00000 47.42308 15.00000
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