GRAPHICAL METHOD - Quantitative Techniques for management

Graphical method to solve Linear Programming problem (LPP) helps to visualize the procedure explicitly. It also helps to understand the different terminologies associated with the solution of LPP. Linear programming problems with two variables can be represented and solved graphically with ease. Though in real-life, the two variable problems are practiced very little, the interpretation of this method will help to understand the simplex method. The solution method of solving the problem through graphical method is discussed with an example given below.

Example:

A company manufactures two types of boxes, corrugated and ordinary cartons. The boxes undergo two major processes: cutting and pinning operations. The profits per unit are Rs. 6 and Rs. 4 respectively. Each corrugated box requires 2 minutes for cutting and 3 minutes for pinning operation, whereas each carton box requires 2 minutes for cutting and 1 minute for pinning. The available operating time is 120 minutes and 60 minutes for cutting and pinning machines. Determine the optimum quantities of the two boxes to maximize the profits.

Solution:

Key Decision:
To determine how many (number of) corrugated and carton boxes are to be manufactured.

Decision variables:
Let x1 be the number of corrugated boxes to be manufactured.
x2 be the number of carton boxes to be manufactured

Objective Function:
The objective is to maximize the profits. Given profits on corrugated box and carton box are Rs. 6 and Rs. 4 respectively.

The objective function is,

Zmax = 6x1 + 4x2

Constraints:
The available machine-hours for each machine and the time consumed by each product are given.
Therefore, the constraints are,

2x1 + 3x2≤ 120 ..........................(i)
2x1+ x2≤ 60 ..........................(ii)
where x1, x2≥0

Graphical Solution:
As a first step, the inequality constraints are removed by replacing ‘equal to’ sign to give the following equations:

2x1 + 3x2 = 120 ......................(1)
2x1 + x2 = 60 .......................(2)

Find the co-ordinates of the lines by substituting x1 = 0 and x2 = 0 in each equation. In equation (1), put x1 = 0 to get x2 and vice versa

2x1 + 3x2 = 120
2(0) + 3x2 = 120, x2 = 40

Similarly, put x2 = 0,
2x1 + 3x2 = 120
2x1 + 3(0) = 120, x1 = 60

The line 2x1 + 3x2 = 120 passes through co-ordinates (0, 40) (60, 0).
The line 2x1 + x2 = 60 passes through co-ordinates (0, 60) (30, 0).

The lines are drawn on a graph with horizontal and vertical axis representing boxes x1 and x2 respectively. Figure shows the first line plotted.

Graph Considering First Constraint The inequality constraint of the first line is (less than or equal to) ≤ type which means the feasible solution zone lies towards the origin. The no shaded portion can be seen is the feasible area shown in Figure (Note: If the constraint type is ≥ then the solution zone area lies away from the origin in the opposite direction). Now the second constraints line is drawn.

Graph Showing Feasible Area When the second constraint is drawn, you may notice that a portion of feasible area is cut. This indicates that while considering both the constraints, the feasible region gets reduced further. Now any point in the shaded portion will satisfy the constraint equations. For example, let the solution point be (15,20) which lies in the feasible region. If the points are substituted in all the equations, it should satisfy the conditions.

2x1 + 3x2≤ 120 = 30 + 60 ≤ 120 = 90 ≤ 120
2x1 + x2≤ 60 = 30 + 20 ≤ 60 = 50 ≤ 60

Now, the objective is to maximize the profit. The point that lies at the furthermost point of the feasible area will give the maximum profit. To locate the point, we need to plot the objective function (profit) line.

Equate the objective function for any specific profit value Z,
Consider a Z-value of 60, i.e.,

6x1+ 4x2= 60

Substituting x1 = 0, we get x2 = 15 and
if x2 = 0, then x1 = 10

Therefore, the co-ordinates for the objective function line are (0, 15), (10, 0) as indicated by dotted line L1 in Figure. The objective function line contains all possible combinations of values of xl and x2.

The line L1 does not give the maximum profit because the furthermost point of the feasible area lies above the line L1. Move the line (parallel to line L1) away from the origin to locate the furthermost point. The point P, is the furthermost point, since no area is seen further. Take the corresponding values of x1 and x2 from point P, which is 15 and 30 respectively, and are the optimum feasible values of x1 and x2.

Therefore, we conclude that to maximize profit, 15 numbers of corrugated boxes and 30 numbers of carton boxes should be produced to get a maximum profit. Substituting x1 = 15 and x2= 30 in objective function, we get

Zmax = 6x1 + 4x2
= 6(15) + 4(30)

Maximum profit: Rs. 210.00

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