SOLVING LINEAR PROGRAMMING GRAPHICALLY USING COMPUTER - Quantitative Techniques for management

The above problem is solved using computer with the help of TORA. Open the TORA package and select LINEAR PROGRAMMING option. Then press Go to Input and enter the input data as given in the input screen shown below, in Figure.

Linear Programming, TORA Package (Input Screen)

Linear Programming, TORA Package (Input Screen)

Now, go to Solve Menu and click Graphical in the 'solve problem' options. Then click Graphical, and then press Go to Output. The output screen is displayed with the graph grid on the right hand side and equations in the left hand side. To plot the graphs one by one, click the first constraint equation. Now the line for the first constraint is drawn connecting the points (40, 60). Now, click the second equation to draw the second line on the graph.

You can notice that a portion of the graph is cut while the second constraint is also taken into consideration. This means the feasible area is reduced further. Click on the objective function equation.

Linear Programming graphical method Maximization problem

The linear programming graphical method of maximization problem are explained below

The objective function line locates the furthermost point (maximization) in the feasible area which is (15,30) shown in Figure below.

Graph Showing Feasible Area

Graph Showing Feasible Area

Example:

A soft drink manufacturing company has 300 ml and 150 ml canned cola as its products with profit margin of Rs. 4 and Rs. 2 per unit respectively. Both the products have to undergo process in three types of machine. The following table indicates the time required on each machine and the available machine-hours per week.

Available Data

Available Data

Formulate the linear programming problem specifying the product mix which will maximize the profits within the limited resources. Also solve the problem using computer.

Solution:
Let x1 be the number of units of 300 ml cola and x2 be the number of units of 150 ml cola to be produced respectively. Formulating the given problem, we get

Objective function:

Zmax = 4x1 + 2x2

Subject to constraints,

3x1 + 2x2≤300 ............................(i)
2x1 +4x2≤ 480 ............................(ii)
5x1 +7x2≤ 560 ............................(iii)
where x1 , x2≥ 0

The inequalities are removed to give the following equations:

3x1 + 2x2 = 300 ............................(iv)
2x1 + 4x2 = 480 ............................(v)
5x1 + 7x2 = 560 ............................(vi)

Find the co-ordinates of lines by substituting x1 = 0 to find x2 and x2 = 0 to find x1.

Therefore,

Line 3x2 + 2x2 = 300 passes through (0,150), (100, 0)
Line 2x1 + 4x2 = 480 passes through (0,120), (240, 0)
Line 5x1 + 7x2 = 650 passes through (0, 80), (112, 0)

Graphical Presentation of lines (TORA, Output Screen)

Graphical Presentation of lines (TORA, Output Screen)

For objective function,

The Line 4x1 + 2x2 = 0 passes through (–10, 20), (10,–20)

Plot the lines on the graph as shown in the computer output Figure.

The objective is to maximize the profit. Move the objective function line away from the origin by drawing parallel lines. The line that touches the furthermost point of the feasible area is (100, 0). Therefore, the values of x1 and x2 are 100 and 0 respectively.

Maximum Profit, Zmax = 4x1 + 2x2
= 4(100) + 2(0)
= Rs. 400.00

Example 10:
Solve the following LPP by graphical method.
Minimize Z = 18x1+ 12x2

Subject to constraints,

2x1+ 4x2≤ 60 ........................(i)
3x1 + x2≥ 30 ........................(ii)
8x1 + 4x2≥ 120 ........................(iii)

where x1 , x2≥ 0

Solution:
The inequality constraints are removed to give the equations,

2x1 + 4x2 = 60 ........................(iv)
3x1 + x2 = 30 ........................(v)
8x1 + 4x2 = 120 ........................(vi)

The equation lines pass through the co-ordinates as follows:

For constraints,

2x1 + 4x2 = 60 passes through (0,15), (30,0).
3x1 + x2 = 30 passes through (0,30), (10,0).
8x1 + 4x2 = 120 passes through (0,30), (15,0).

The objective function,
18x1 + 12x2 = 0 passes through (–10,15), (10,–15).

Plot the lines on the graph as shown in Figure

Here the objective is minimization. Move the objective function line and locate a point in the feasible region which is nearest to the origin, i.e., the shortest distance from the origin. Locate the point P, which lies on the x – axis. The co-ordinates of the point P are (15,0) or x1= 15 and x2= 0.

The minimum value of Z

Zmin = 18 x1 + 12x2
= 18 (15) + 12 (0)
= Rs. 270.00

Graphical Presentation (Output Screen, TORA)

Graphical Presentation (Output Screen, TORA)


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