Mina KhairzadaEGR 549 Homework # 2

Page 233 Example 2 (Tucker)

Parisay’s Comments are in red.

Solve using WinQSB. Print input and summary report. Answer the following questions without using the software. Refer to the transparencies on summary of sensitivity analysis.

a) Suppose Tucker must produce exactly 1050 cars. What is the new optimal solution?

Ans: C1= 1000 is binding Bb=1050, 900 <=Bb<=1,066.6670 it is inside the range and the basis (var) will remain the same and the bfs (value)will change. Then the Z-value changes so that the new optimal solution is: Z= $11,600 +(1050-1000) *30 = $ 13,100.

b)Suppose that 3900 units of raw material are available. What is the new optimal solution?

Ans: C4= 4000 is binding Bb=3900, 3800 <=Bb<=4,300 it is inside the range and the basis (var) will remain the same and the bfs (value) will change. Then the Z-value changes so that the new optimal solution is: Z= $11,600 +(3900-4000) *(-5) = $ 12100.

c) Suppose Tucker must produce exactly 1300 cars. What is the new optimal solution?

C1= 1000 is binding Bb=1300, 900 <=Bb<=1,066.6670 it is outside the range and the basis (var) changes. Also the bfs (value) changes and you must solve it again.

d) Suppose that 3500 hours of labor is available. What is the new optimal solution to the LP?

C3=3300 is non binding constraint Bb=3500, 3000 <=Bn<=M it is inside the range and the basis (var) will remain the same and the bfs (value) and Z will remain the same too.

e) What should we do so that we can produce at Plant 4?

X4 is NBV,Cnbv=7 and RC=7. That means the cost for each X4 should be less than 7-7=0 to be able to produce at Plant 4. Therefore, there is no way that we can produce at Plant 4 because we can not have negative cost (profit).

f) Suppose the unit cost in Plant 2 is $11 (in thousands). What is the new optimal solution to the LP?

X2 is BV, Cbv=9, .Cbv’=11 -M<=Cbv<=12, and is within the range. Basis is the same and bfs is the same and the new Z-value is $11,600 + (11-10)*200= $ 11,800.

g) Suppose the unit cost in Plant 3 is $3.5 (in thousands). What is the new optimal solution to the LP?

X3 is BV Cbv=9 now Cbv’=3.5 and is out of the range. 5<=Cbv<=M, Basis must be changed and BFS is going to change and we must resolve this problem.

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Spring 2006Dr.Parisay