Sensitivity Analysis

Graphical Method in LP

Problem 3.1: Solve the following LPP by graphical method

                Minimize Z= 20X1 + 10X2

                Subject to: X1 + 2X2 ≤ 40

                              3X1 + X2 ≥ 30

                              4X1 + 3X2 ≥ 60

                              X1,X2 ≥ 0

Solution

         Replace all the inequalities of the constraints by equation

                So, X1 + 2X2 = 40

                If X1 = 0 => X2 = 20

                If X2 = 0 => X1 = 40

Therefore,       X1 + 2X2 = 40    passes through (0,20) (40,0)

                3X1 + X2 = 30    passes through (0,30) (10,0)

                4X1 + 3X2 = 60  passes through (0,20) (15,0)

Plotting each equation on the graph:

[pic 1][pic 2]

The feasible region is ABCD.

C and D are points of intersection of lines.

From,

X1 + 2X2 = 40, 3X1 + X2 = 30 and 4X1 + 3X2 = 60

X1 + X2 = 30

On solving we get         C = (4,18)

                        D = (6,12)

Corner points                 Value of Z = 20X1 + 10X2

A(15,0)                300

B(40,0)                800

C(4,18)                260

D(6,12)                240 (Minimum Value)

Therefore, the minimum value of Z occurs at D (6, 12). Hence, the optimal solution is X1=6, X2=12.

Problem 3.2 :

Solve graphically the following LPP

Maximize         Z = 3X1 + 2X2

Subject to         X1 + 2X2 ≤ 6

                2X1 + X2 ≤ 8

                X2 – X1 ≤ 1

                X2 ≤ 2

                X1 ≥ 0

                X2 ≥ 0

Solution

The following graph shows all the six constraints plotted as straight lines. The region in which each constraint holds when the inequality is actuated by the direction of the arrow on the associated straight line is the solution space. The solution space is thus determined.

[pic 3]

The optimal solution can always be identified with one of the feasible corner points A,B,C,D,E and F of the solution space .Thus

At A         X1=0                X2=0                Z=0

At B        X1=4                X2=0                Z=12

At C        X1=3.33        X2=1.33        Z=12.66

At D        X1=2                X2=2                Z=10

At E        X1=1                X2=2                Z=7

At F        X1=0                X2=1                Z=2

Thus maximum value of Z occurs at C. The optimal solution is

X1=3.33, X2=1.33, Zmax = 12.66

The optimal solution may be determined by another way. Plot the objective line passing through the origin. Move this line as far away from the origin as possible and yet within or touching the boundary of the solution space as in the graph. The optimum solution occurs at the point C. The coordinates of the point C give the optimum values of X1 and X2.

The coordinates of C may be noted from the graph. The coordinates may also be determined analytically. Since C is the intersection of lines (1) and (2), the values of X1 and X2 may be determined by solving simultaneously the equations

        X1+2X2 = 6        and         2X1+X2 = 8

The values are 3.33 and 1.33. Zmax = 3*3.33 + 2*1.33 = 12.66.

Problem 3.3

Find the maximum value of Z = 5X1 + 7X2

        Subject to                X1 + X2 ≤ 4

                                3X1 + 8X2 ≤ 24

                                10X1 + 7X2 ≤ 35

                                X1, X2 > 0

Solution

Replace all the inequalities of the constraints by forming the equations

X1 + X2 = 4                        passes through (0, 4) (4, 0)

3X1 + 8X2 = 24                passes through (0, 2) (8, 0)

10X1 + 7X2 = 35                passes through (0, 5) (3.5, 0)

Plot these lines in the graph and mark the region below the line as the inequality of the constraint is ≤ and is also lying in the first quadrant.

[pic 4]

The feasible region is OABCD. B and C are points of intersection of lines.

X1+X2 = 4, 10X1+7X2 = 35 and 3X1+8X2 = 24, X1+X2 = 4

On solving, we get

B = (1.6, 2.4)

C = (2.3, 1.7)

Corner points                        Value of Z=5X1 + 7X2

O (0, 0)                        0

A (3.5, 0)                        17.5

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