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Production and Operations Management (29:623:311) Spring 2019 3. A linear programming model is given as follows: maximize Z =
0 0
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Answer #1

a)

The solver setup is shown below

X3 64 Obj func 0 X2 2 Coeff 3 Dec var 4 5 Const 1 6 Const 2 7 Const 3 8 Const 4 9 Const 5 10 50 80 80 Value 0 0 Limit 2.5 3.9

The spreadsheet formulas are shown below

A1 x4 Obj func 2 80 3 0 4 5 3.99 6 1.9 7 25 8 10.55 9 10 -SUMPRODUCT(B2:E2,SB$3:ŞE$3) Value Limit -SUMPRODUCT(B5:E5,SBS3:$E$3

The solver parameters are shown below

Solver Parameters Obj func Set Objective: SG$2 2 Coeff 3 Dec var 50 80 Max Yalue Of: 0 0 Value 0 0 0 By Changing Variable Cel

The result is shown below. Here, x2 = 18.18 and x1, x3, x4 = 0

Obj func 1454.55 2 Coeff 3 Dec var 50 80 0 18.1818 0 0 Value 45.4545 47.2727 400 229.091 18.1818 Limit 2.5 2.6 5 Const 1 6 Co

The sensitivity analysis is shown below

Microsoft Excel 15.0 Sensitivity Report A B 6 Variable Cells Final Value Reduced objective Allowable Allowable Decrease Cell

b)

Sensitivity ranges for obj function coefficients (high to low)

X1 = 54.54 to infinity

X2 = Infinity to 78.22

X3 = 65.45 to infinity

X4 = 90.90 to infinity

c)

Sensitivity ranges for constraint quantity values is (high to low)

Const 1 = Infinity to 45.45

Const 2 = Infinity to 47.27

Const 3 = 2968.25 to 1700 (current value)

Const 4 = Infinity to 229.09

d)

There will be no effect. The second constraint is not a binding constraint

e)

There will be no effect. The allowable decease of x4 coefficient is infinity.

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