Some Exceptional Cases

Ask a Question, Get an Answer!

Type your question here

Hundreds of tutors are online and ready to help you right now!

We may come across LPP which may have no feasible (infeasible) solution or may have unbounded solution. We conclude the chapter by giving one example for each of these exceptional cases.

Example:

Show graphically that the model

Maximise Z = -5y

Subject to

y0 has no feasible solution.

Suggested answer:

Draw the graphs x + y = 1

- 0.5 -5y = - 10

Shade the half planes of the constraints x + y 1 …(1)

-0.5x - 5y -10 …(2)

Note that the origin (0, 0) does not satisfy the inequation (2) hence the required region is the upper half plane.

From the graph, it is clear that the intersection of the constraints is empty. Hence the given problem has no feasible solution. Therefore the given L.P.P has no solution.

Example:

Show graphically that the L.P.P

Maximize

Z = 6x + y

Subject to the constraints

x 0, y 0 has an unbounded solution.

Suggested answer:

The intersection of the half planes 2x + y 3 and x - y 0 is shown as shaded region in the figure.

Feasible region is an unbounded convex region

at A (0, 3), Z = 6 (0) + 3 = 3

At B (1, 1), Z = 6(1) + 1 = 7

Consider any point say (4, 5), the value of Z = 24 + 5 = 29.

Hence the maximum flow is from (1, 1) to (4, 5)

Hence the given problem has an unbounded solution.