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# Solution of LPP by graphical method

After formulating the linear programming problem, our aim is to determine the values of decision variables to find the optimum (maximum or minimum) value of the objective function.

Solution of LPP by graphical method

After formulating the linear programming problem, our aim is to determine the values of decision variables to find the optimum (maximum or minimum) value of the objective function. Linear programming problems which involve only two variables can be solved by graphical method. If the problem has three or more variables, the graphical method is impractical.

The major steps involved in this method are as follows

(i) State the problem mathematically

(ii) Write all the constraints in the form of equations and draw the graph

(iii) Find the feasible region

(iv) Find the coordinates of each vertex (corner points) of the feasible region. The coordinates of the vertex can be obtained either by inspection or by solving the two equations of the lines intersecting at the point

(v) By substituting these corner points in the objective function we can get the values of the objective function

(vi) If the problem is maximization then the maximum of the above values is the optimum value. If the problem is minimization then the minimum of the above values is the optimum value

Example 10.5

Solve the following LPP

Maximize Z = 2 x1 +5x2

subject to the conditions x1+ 4x2 ≤ 24

3x1+x2 ≤ 21

x1+x2 ≤ 9and x1, x2 ≥ 0

Solution:

First we have to find the feasible region using the given conditions.

Since both the decision variables x1 and x2 are non-negative ,the solution lies in the first quadrant.

Write all the inequalities of the constraints in the form of equations.

Therefore we have the lines x1+ 4x2=24 ; 3x1 + x2 = 21; x1 + x2= 9 x1+ 4x2= 24 is a line passing through the points (0 , 6) and (24 , 0). [(0,6) is obtained by taking x1=0 in x1 + 4x2 = 24 , (24 , 0) is obtained by taking x2 = 0 in x1+ 4x2 = 24].

Any point lying on or below the line x1 + 4x2 = 24 satisfies the constraint x1 + 4x2≤ 24 .

3x1 +x2= 21 is a line passing through the points (0, 21) and (7, 0). Any point lying on or below the line 3 x1 + x2 = 21 satisfies the constraint 3 x1 + x2 ≤ 21.

x1+ x2 = 9 is a line passing through the points (0 , 9) and ( 9 , 0) .Any point lying on or below the line x1 + x2 = 9 satisfies the constraint x1+ x2 ≤ 9.

Now we draw the graph.

The feasible region satisfying all the conditions is OABCD.The co-ordinates of the points are O(0,0) A(7,0);B(6,3) [ the point B is the intersection of two lines x1+ x2= 9 and 3 x1+ x2= 21];C(4,5) [ the point C is the intersection of two lines

x1+ x2 = 9 and x1+ 4x2 = 24] and D(0,6).

Maximum value of Z occurs at C. Therefore the solution is x1 =4, x2 = 5, Z max = 33

Example 10.6

Solve the following LPP by graphical method Minimize z = 5x1+4x2 Subject to constraints 4x1+ x2 ≥ 40 ; 2x1+3x2 ≥ 90 and x1, x2 > 0

Solution:

Since both the decision variables x1 and x2 are non-negative, the solution lies in the first quadrant of the plane.

Consider the equations 4x1+x2  = 40 and 2 x1+3 x2 = 90

4x1+x 2 = 40 is a line passing through the points (0,40) and (10,0).Any point lying on or above the line 4x1+x2= 40 satisfies the constraint 4x1+ x2 ≥ 40.

2x1+3x2 = 90 is a line passing through the points (0,30) and (45,0). Any point lying on or above the line 2 x1+3x2= 90 satisfies the constraint 2x1+3x2 ≥ 90.

Draw the graph using the given constraints.

The feasible region is ABC (since the problem is of minimization type we are moving towards the origin.

The minimum value of Z occurs at B(3,28).

Hence the optimal solution is x1 = 3, x2 = 28 and Zmin=127

Example 10.7

Solve the following LPP.

Maximize Z= 2 x1 +3x2 subject to constraints x1 + x2 ≤ 30 ; x2 ≤ 12; x1 ≤ 20 and x1, x2≥ 0

Solution:

We find the feasible region using the given conditions.

Since both the decision variables x1 and x2 are non-negative, the solution lies in the first quadrant of the plane.

Write all the inequalities of the constraints in the form of equations.

Therefore we have the lines

x1+x2=30; x2 =12; x1= 20

x1+x2  =30 is a line passing through the points (0,30) and (30,0)

x2  = 12 is a line parallel to x1–axis

x1  = 20 is a line parallel to x2–axis.

The feasible region satisfying all the conditions x1+ x2≤ 30; x2≤ 12 ; x1≤ 20 and x1, x2 ≥ 0 is shown in the following graph.

The feasible region satisfying all the conditions is OABCD.

The co-ordinates of the points are O(0,0) ; A(20,0); B(20,10) ; C(18,12) and D(0,12).

Maximum value of Z occurs at C. Therefore the solution is x1 = 18 , x2= 12, Z max = 72

Example 10.8

Maximize Z = 3x1 + 4x2  subject to x1x2 < –1; –x1+x2 < 0 and x1, x2 ≥ 0

Solution:

Since both the decision variables x1, x2 are non-negative ,the solution lies in the first quadrant of the plane.

Consider the equations x1x2 = –1 and – x1 + x2 = 0

x1 x2 =–1 is a line passing through the points (0,1) and (–1,0)

x1 + x2 = 0 is a line passing through the point (0,0)

Now we draw the graph satisfying the conditions x1x2 < –1; –x1+x2 < 0 and x1, x2≥0

There is no common region(feasible region) satisfying all the given conditions.

Hence the given LPP has no solution.

Exercise 10.1

1. A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality . Profits on pens A and B are Rs 5 and Rs 3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day . Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day . Formulate this problem as a linear programming problem.

2. A company produces two types of products say type A and B. Profits on the two types of product are Rs.30/- and Rs.40/- per kg respectively. The data on resources required and availability of resources are given below.

Formulate this problem as a linear programming problem to maximize the profit.

3. A company manufactures two models of voltage stabilizers viz., ordinary and auto-cut. All components of the stabilizers are purchased from outside sources , assembly and testing is carried out at company’s own works. The assembly and testing time required for the two models are 0.8 hour each for ordinary and 1.20 hours each for auto-cut. Manufacturing capacity 720 hours at present is available per week. The market for the two models has been surveyed which suggests maximum weekly sale of 600 units of ordinary and 400 units of auto-cut . Profit per unit for ordinary and auto-cut models has been estimated at Rs 100 and Rs 150 respectively. Formulate the linear programming problem.

4. Solve the following linear programming problems by graphical method.

(i) Maximize Z = 6x1 + 8x2 subject to constraints 30x1+20x2 300;5x1+10x2 110; and x1, x2 > 0 .

(ii) Maximize Z = 22x1 + 18x2 subject to constraints 960x1 + 640x2 15360 ; x1 + x2 20 and x1 , x2 0 .

(iii) Minimize Z = 3x1 + 2x2 subject to the constraints 5x1+ x2≥10; x1+ x2≥6; x1+ 4 x2 ≥12 and x1, x2≥0.

(iv)  Maximize  Z = 40x1 + 50x2 subject to constraints 30x1 + x2 9 ; x1 + 2x2 8 and       x1 , x2 0

(v) Maximize Z = 20x1 + 30x2  subject to constraints 3x1 + 3x2 36 ;  5x1 + 2x2 50 ; 2x1 + 6x2 60 and x1 , x2 0

(vi)  Minimize Z = 20x1 + 40x2 subject to the  constraints  36x1 + 6x2 108, 3x1 + 12x2 36, 20x1 + 10x2100 and x1 , x2 0

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11th Business Mathematics and Statistics(EMS) : Chapter 10 : Operations Research : Solution of LPP by graphical method | Linear programming problem