Operation and management
leosdesire
assignmentquestion1.docx1)
Linear Program is formulated as below:
Let
B = number of copies to be printed by BP
J = number of copies to be printed by JP
L = number of copies to be printed by LL
Min 2.5B+2.6J+2.8L
s.t.
0.9B+0.98J+0.99L >= 80000
B >= 0.15J or, B-0.15J >= 0
B <= 25000
J <= 45000
L <= 55000
L >= 25000
B, J, L >= 0
b)
Create Excel model and enter Solver Parameters as follows:
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B |
J |
L |
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Cost |
2.5 |
2.6 |
2.8 |
=SUMPRODUCT(B3:D3,$B$13:$D$13) |
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0.9 |
0.98 |
0.99 |
=SUMPRODUCT(B5:D5,$B$13:$D$13) |
>= |
80000 |
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1 |
-0.15 |
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=SUMPRODUCT(B6:D6,$B$13:$D$13) |
>= |
0 |
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1 |
=SUMPRODUCT(B7:D7,$B$13:$D$13) |
>= |
25000 |
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1 |
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=SUMPRODUCT(B8:D8,$B$13:$D$13) |
<= |
25000 |
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1 |
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=SUMPRODUCT(B9:D9,$B$13:$D$13) |
<= |
45000 |
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1 |
=SUMPRODUCT(B10:D10,$B$13:$D$13) |
<= |
55000 |
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Result: |
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EXCEL FORMULAS:
Click Solve on the Solver Parameters window to generate the optimal solution
c)
Refer sensitivity report
As the quality level for BP improves, value of B in the optimal decreases, but the value of J and L remains same.
Allowable decrease for cost of B is 0.112244898
Therefore lower limit of cost is 2.5-0.112244898 = 2.387755102
lower limit of cost per unit non-defective = 2.387755102/0.9 = 2.653061224
Allowable upper limit of quality level of B = 2.5/2.653061224 = 0.9423
Therefore, as long as the quality level of B is less than 0.9423, the values of J and L remain same, but value of B changes in order to make 11150 non-defective reports. (80000-45000*0.98-25000*0.99 = 11150)
d)
Shadow price of this constraint is 0.05
If this requirement is relaxed, then the optimal solution will change. and total cost will reduce by 0.05 for every unit decrease in minimum requirement.
