Question

Given the following LP problem formulation and output data, perform the analysis below.

Max.    100X1 + 120X2 + 150X3 + 125X4

s.t        X1 + 2X2 + 2X3 + 2X4 < 108            (C1)

           

            3X1 + 5X2 + X4 < 120                       (C2)

            X1 + X3 < 25                                      (C3)

X2 + X3 + X4 > 50                             (C4)

OPTIMAL SOLUTION:

Objective Function Value = 7475.000

Variable		Value	Reduced Costs
X1		8.000	0.500
X2		0.000	5.000
X3		17.000	0.000
X4		*A	0.000
	Constraint	Slack / Surplus	Dual Prices
	1	0.000	*B
	2	63.000	0.000
	3	*C	25.000
	4	0.000	-25.000

            Objective Co-efficient Ranges

Variable	Lower Limit	Current Value	Upper Limit
X1	87.500	100.000	No Upper Limit
X2	No lower limit	120.000	125.000
X3	125.000	150.000	162.500
X4	120.000	125.000	150.000

            Right Hand Side Ranges

Constraint	Lower Limit	Current Value	Upper Limit
1	100.000	108.000	123.750
2	57.000	120.000	No Upper Limit
3	8.000	25.000	58.000
4	41.000	50.000	54.000

1. Find *A

2. How much of constraint C2 is used?

3. In each of the following, calculate the resulting change in the objective function, if possible and show you calculations. If not possible, explain why not.

1. The obj.func.co-eff of X4 increases from 125 to 135.

2. The level of C4 decreases from 50 to 42.

Answer 1

Question – a:

The objective function is: 100X1 + 120*X2 + 150*X3 + 125*X4

Here,

X1 = 8

X2 = 0

X3 = 17

X4 = *A

Objective function value = 7475

Putting values in equation:

100*8 + 120*0 + 150*17 + 125*(*A) = 7475

800 + 2550 + 125*(*A) = 7475

125*(*A) + 3350 = 7475

125*(*A) = 7475 – 3350 = 4125

(*A) = 4125/125 = 33

Answer is: *A = 33

Question – b:

Constraint 2 is: 3X1 + 5x2 + X4

Putting values in equation:

3*8 + 0 + 33 = 24 + 33 = 57

Hence, 57 units of Constraint 2 have been used.

Question – c:

1. As the objective coefficient upper limit is 150, increasing it from 125 to 135 wont affect the optimal solution. The current solution will remain optimal.

Objective function will be: 100X1 + 120X2 + 150X3 + 135X4

X1 = 8, X2 = 0, X3 = 17, X4 = 33

Objective function value = 100*8 + 120*0 + 150*17 + 135*33 = 7805

2. The current solution will remain optimal by decreasing the C4 from 50 to 42 as the lower limit is 41. But we wont be able to calculate the objective function value as the shadow price has not been provided.

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