Question

Figure 1 provides the Excel Sensitivity output for the following LP model. 10x1 + 8x2 Max Z= subject to: 31 +2x2 < 24 2x1 + 4x2 = 12 -2x1 + 2 x2 56 X1, X2 > 0 Variable Cells Cell Name $B$13 Solution x1 $C$13 Solution x2 Final Reduced Objective Allowable Allowable Value Cost Coefficient Increase Decrease 6 0 10 1E+30 0 -12 8 12 1E+30 6 Constraints Cell $D$6 $D$7 $D$8 Name C1 Totals C2 Totals C3 Totals Final Shadow Value Price 18 0 12 5 -12 0 Constraint Allowable Allowable R.H. Side Increase Decrease 24 1E+30 6 12 4 12 6 1E+30 18 Figure. 1 a) (10 pts) Determine the objective function value when we increased the unit of resource of 1 from 24 to 30. Can you find the new optimal solution X1 and X2? Explain. b) (10 pts) Determine the objective function value if the profit of x2 is increased from 8 to 16? Does the solution change? Explain. c) (10 pts) Determine the objective function value if the profit of xl is decreased from 10 to 3? What can you say about the solution? Explain. d) (10 pts) Determine the objective function value when we increased the unit of resource of 2 from 6 to 12. What can you say about the solution? Explain.

Answer 1

Question – a:

As unit of resource 1 has an allowable increase of infinite, the current solution will remain optimal solution no matter how much the rhs increase.

Answer is:

X1 = 6

X2 = 0

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Question – b:

X2 has an allowable increase of 12, till the profit is (8 + 12 =) 20, this solution will remain optimal.

Objective function value will be: Z = 10*6 + 16*0 = 60

Answer is: 60

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Question – c:

x1 profit has an allowable decrease of 6. If profit from x1 reduces from 10 to 3 (a decrease of 7), the current solution wont be optimal anymore.

Answer is: The optimal solution will change. But, it is not possible to determine the value of objective function value only from the sensitivity report.

(***Extra answer – If we solve the program then the optimal profit will be 24)

***

Question – d:

Resource 2 has an allowable increase of infinity. So if we increase it from 6 to 12, this solution will remain optimal.

The solution wont change. X1 = 6, X2 = 0

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