Graphical Methods Of Unbounded Solution In Linear Programming Model

Introduction:

Linear programming is the method which deals lots of management problem via mathematical way easily. Mathematical programming includes many more optimization models known as Non - linear Programming, Stochastic programming, Integer Programming and Dynamic Programming - each one of them is an efficient optimization technique to solve the problem with a specific structure, which depends on the assumptions made in formulating the model.

EXAMPLE Maximise Z = 0.75 a 1b S.T.

1a 1b ≥ 0

–0.5 a 1b ≤ 1

 and both a and b are ≥ 0.

Solution: Writing the inequalities as equations,

1a 1b = 0 i.e., a = b = 1 which is a line passing through origin at 45°

0.5 a 1 b = 1 and both a and b are ≥ 0.

The polygon is not closed one i.e., the feasible area is unbound. When Iso-profit line is drawn, it passes through open side of the polygon and it does not coincide with any corner or any line. Hence the line can be moved indefinitely, still containing a part of the feasible area. Thus there is no finite maximum value of Z. That the value of Z can be increased indefinitely.

 When the value of Z can be increased indefinitely, the problem is said to have an UNBOUND solution.

Example: Formulate the l.p.p. and solve the below given problem graphically. Old hens can be bought for Rs.2.00 each but young ones costs Rs. 5.00 each. The old hens lay 3 eggs per week and the young ones lay 5 eggs per week. Each egg costs Rs. 0.30. A hen costs Rs.1.00 per week to feed. If the financial constraint is to spend Rs.80.00 per week for hens and the capacity constraint is that total number of hens cannot exceed 20 hens and the objective is to earn a profit more

than Rs.6.00 per week, find the optimal combination of hens.

Solution: Let x be the number of old hens and y be the number of young hens to be bought. Now the old hens lay 3 eggs and the young one lays 5 eggs per week. Hence total number of eggs one get is 3x 5y.

Total revenues from the sale of eggs per week is Rs. 0.30 (3x 5y) i.e., 0.90 x 1.5 y

Now the total expenses per week for feeding hens is Re.1 (1x 1y) i.e., 1x 1y.

Hence the net income = Revenue – Cost = (0.90 x 1.5 y) – (1 x 1 y) = – 0.1 x 0.5 y or 0.5y– 0.1 x. Hence the desired l.p.p. is

Maximise Z = 0.5 y – 0.1 × S.T.2 x 5 y ≤ 80

1x 1y ≤ 20 and both x and y are ≥ 0The equations are:Maximise Z = 0.5 y – 0.1 × S.T.2 x 5 y = 801x 1y = 20 and both x and y are ≥ 0 Hence Zc = Z (0, 16) = Rs.8.00. Hence, one has to buy 16 young hens and his weekly profit will be Rs.8.00