Sensitivity Analysis

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B (1.6, 2.4) 24.8 (Maximum Value)

C (2.3, 1.7) 23.4

D (0,3) 21

Therefore, the maximum value of Z occurs at B (1.6, 2.4) and the optimal solution is X1=1.6 and X2=2.4

Problem 3.3 a

A company produces 2 types of hats. Every hat A requires twice as much labor as the second hat B. If the company produces only hat B it can produce a total of 500 hats a day. The market limits daily sales of the hat A and hat B to 150 and 200 hats. The profits of hat A and B are Rs.8 and Rs.5 respectively. Solve graphically to get the optimal solution.

Solution

Let X1 and X2 be the number of hats of type A and type B respectively

Max Z=8X1 + 5X2

Subject to 2X1 + 2X2 ≤ 500

X1 ≥ 150

X2 ≥ 250

X1, X2 ≥ 0

First rewrite the inequality of the constraint into an equation and plot the lines in the graph

2X1 + X2 = 500 passes through (0,500) (250, 0)

X1 = 150 passes through (150, 0)

X2 = 250 passes through (0, 250)

We mark the region below the lines lying in the first quadrant as the inequality of the constraints are ≤. The feasible region is OABCD. B and C are the points of intersection of lines.

2X1 + X2 = 500 X1 =150 and

2X1 + X2 =500 X2 = 250

[pic 5]

Corner Points Value of Z = 8X1 + 5X2

O (0, 0) 0

A (150, 0) 1200

B (150, 200) 2200

C (125, 250) 2250 (Maximum)

D (0, 250) 1250

The maximum value of Z is attained at C (125, 250)

Therefore, the optimal solution is X1 = 125, X2 = 250

i.e. the company should produce 125 hats of type A and 250 hats of type B in order to get a maximum profit of Rs. 2250/-

Example 3.4

By Graphical Method solve the following LPP.

Max Z = 3X1 + 4X2

Subject to 5X1 + 4X2 ≤ 200

3X1 + 5X2 ≤ 150

5X1 + 4X2 ≥ 100

8X1 + 4X2 ≥ 80

And X1, X2 ≥ 0

[pic 6]

Corner Points Value of Z = 3X1 + 4X2

O (20, 0) 60

A (40, 0) 120

B (30.8, 11.5) 138.4 (Maximum)

C (0, 30) 120

D (0, 25) 100

The maximum value of Z is attained at B (30.8, 11.5)

Therefore, the optimal solution is X1 = 30.8, X2 = 11.5

Example 3.5

Use graphical Method to solve the LPP

Maximize Z = 6X1+4X2

Subject to -2X1+X2 ≤ 2

X1- X2 ≤ 2

3X1 + 2X2 ≤ 9

X1, X2 ≥ 0

Solution

[pic 7]

Feasible region is given by ABC

Corner Points Value of Z = 6X1 + 4X2

A (2, 0) 12

B (3, 0) 18

C (13/5, 3/5) 98/5 = 19.6 (Maximum)

The maximum value of Z is attained at C (13/5, 3/5)

Therefore, the optimal solution is X1 = 13/5, X2 = 3/5.

Example 3.6 for Unbounded

Use graphical Method to solve the LPP

Minimize Z = 3X1 + 2X2

Subject to 5X1 + X2 ≥ 10

X1 + X2 ≥ 6

X1 + 4X2 ≥ 12

X1, X2 ≥ 0

Solution

Corner Points Value of Z = 3X1 + 2X2

A (0, 10) 20

B (1, 5) 13 (Minimum)

C (4, 2) 16

D (12, 0) 36

[pic 8]

Since the minimum value is attained at B (1,5) the optimum solution is X1 = 1, X2 = 5.

Note: In the above problem if the objective function is maximization then the solution is unbounded, as maximum value of Z occurs at infinity.

3.1.1 Some more cases

There are some linear programming problems which may have:

- a unique optimal solution

- an infinite number of optimal solutions

- an unbounded solution

- no solution.

- Redundant Case

The following problems will illustrate these cases.

Example 3.7 for Multiple/Infinite Solutions case and Redundant Constrain case.

Solve