Graph can illustrate every problem in lpp maths easily and properly. Here discussed the solution of LPP maths using Graphical Method.
Steps involved in this method are as follows:
- Analyse the problem(decision variables, objective function & restriction, etc.)
- Set up mathematical formulation related to the problem.
- Plot a graph representing all the constraints of the problem and identify feasible region(where solution lies) it is the intersection of all the regions of the constraints. The problem is restricted to first quadrant only.
- Compute the coordinates of the limit points of the feasible region obtained in previous step.
- Select the point that gives the required optimized(maximized or minimized) value of objective function. This is the optimum feasible solution.
One example of graphical method in lpp maths is illustrated below for easy understanding:
Example: A company makes two kinds of leather belts. Belt A is high quality belt & B is low quality belt. The respective profits are, Rs. 4 & Rs. 3 respectively. Each type of belt A requires twice the time as for belt B & if all belts of type B are made then company could make 1000 belts per day. The supply of leather is sufficient for only 800 belts per day. Belt A requires fancy buckle & only 400 buckles per day are available. There are only 700 buckles per day are available for belt B. Determine the optimal product mix.
Solution: Initially, we consider the variables. Assume,
x1 be No. of belts of type A
x2 be No. of belts of type B
Then, the objective function is,
subject to the constraints;
2x1+x2 = 1000 —-(no. of belts)—(1)
x1+x2 = 800 —-(leather supply)—(2)
also x1 = 400 x1 = 700—-(3)
and x2 = x1 = 0——(4)
Graphical representation of given problem is as follows:
From constraint (1);
if x1=0 ? x2=1000
For x2=0 ? x1=500
From constraint (2);
if x1=0 ? x2= 800
For x2=0 ? x1= 800
The representation of the data from above graph is given in the table below (please see the image below):
The maximum value from the above graph is, Z= 2600.