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 *x*_{1} +5*x*_{2}

subject to the conditions *x*_{1}+ 4*x*_{2} ≤ 24

3*x*_{1}+*x*_{2}
≤ 21

*x*_{1}+*x*_{2}* *≤ 9and* x*_{1},* x*_{2}* *≥ 0

*Solution:*

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

Since
both the decision variables *x*_{1}
and *x*_{2} 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 *x*_{1}+ 4*x*_{2}=24 ; 3*x*_{1}
+ *x*_{2} = 21; *x*_{1} + *x*_{2}= 9 *x*_{1}+
4*x*_{2}= 24 is a line passing
through the points (0 , 6) and (24 , 0). [(0,6) is obtained by taking *x*_{1}=0 in *x*_{1} + 4*x*_{2}
= 24 , (24 , 0) is obtained by taking *x*_{2}
= 0 in *x*_{1}+ 4*x*_{2} = 24].

Any point lying on or below the
line *x*_{1} + 4*x*_{2} = 24 satisfies the
constraint *x*_{1}* *+ 4*x*_{2}≤
24 .

* 3x*_{1}* *+*x*_{2}= 21 is a line passing through the points (0, 21) and
(7, 0). Any point lying* *on or below
the line 3 *x*_{1} + *x*_{2} = 21 satisfies the
constraint 3 *x*_{1} + *x*_{2} ≤ 21.

*x*_{1}+* x*_{2}* *= 9 is a line passing through the points (0 , 9) and ( 9 , 0) .Any
point lying on* *or below the line *x*_{1} + *x*_{2} = 9 satisfies the constraint *x*_{1}+ *x*_{2}
≤ 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 *x*_{1}+ *x*_{2}= 9 and 3 *x*_{1}+
*x*_{2}= 21];C(4,5) [ the point
C is the intersection of two lines

*x*_{1}+* x*_{2}* *= 9 and* x*_{1}+ 4*x*_{2}* *= 24] and D(0,6).

Maximum
value of Z occurs at C. Therefore the solution is *x*_{1} =4, *x*_{2}
= 5, Z _{max} = 33

**Example 10.6**

Solve the
following LPP by graphical method Minimize *z*
**= 5***x*_{1}+4*x _{2}* Subject to constraints 4

*Solution:*

Since both the decision variables
*x*_{1} and *x*_{2} are non-negative, the
solution lies in the first quadrant of the plane.

Consider the equations 4*x*_{1}+*x*_{2} = 40 and 2 *x*_{1}+3 *x*_{2} = 90

4*x*_{1}+*x* _{2}
= 40 is a line passing through the points (0,40) and (10,0).Any point lying on
or above the line 4*x*_{1}+*x*_{2}= 40 satisfies the
constraint 4*x*_{1}+ *x*_{2} ≥ 40.

2*x*_{1}+3*x*_{2}
= 90 is a line passing through the points (0,30) and (45,0). Any point lying on
or above the line 2 *x*_{1}+3*x*_{2}= 90 satisfies the constraint
2*x*_{1}+3*x*_{2} ≥ 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 *x*_{1} =
3, *x*_{2} = 28 and *Z*_{min}=127

**Example 10.7**

Solve the following LPP.

Maximize *Z*= 2 *x*_{1}
+3*x*_{2} subject to
constraints *x*_{1} + *x*_{2} ≤ 30 ; *x*_{2} ≤ 12; *x*_{1} ≤ 20 and *x*_{1},* x*_{2}≥ 0

*Solution:*

We find the feasible region using
the given conditions.

Since both the decision variables
*x*_{1} and *x*_{2} 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

*x*_{1}+*x*_{2}=30;* x*_{2}* *=12;* x*_{1}= 20

*x*_{1}+*x*_{2}* *=30 is a line passing
through the points (0,30) and (30,0)

*x*_{2}* *= 12 is a
line parallel to* x*_{1}–axis

*x*_{1}* *= 20 is a
line parallel to* x*_{2}–axis.

The
feasible region satisfying all the conditions *x*_{1}+ *x*_{2}≤
30; *x*_{2}≤ 12 ; *x*_{1}≤ 20 and *x*_{1},* x*_{2}* *≥ 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 *x*_{1} = 18 , *x*_{2}= 12, *Z* _{max} = 72

**Example 10.8**

Maximize
Z = 3*x*_{1} + 4*x*_{2} subject to *x*_{1} – *x*_{2}
__<__ –1; –*x*_{1}+*x*_{2} __<__ 0 and *x*_{1}, *x*_{2} ≥ 0

*Solution:*

Since both the decision variables
*x*_{1}, *x*_{2} are non-negative ,the solution lies in the first
quadrant of the plane.

Consider the equations *x*_{1}– *x*_{2} = –1 and – *x*_{1}
+ *x*_{2} = 0

*x*_{1}–* x*_{2}* *=–1 is a line passing through the points (0,1) and (–1,0)

–*x*_{1} + *x*_{2}
= 0 is a line passing through the point (0,0)

Now we
draw the graph satisfying the conditions *x*_{1}
– *x*_{2} __<__ –1; –*x*_{1}+*x*_{2} __<__ 0 and *x*_{1},
*x*_{2}≥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 = 6*x*_{1} + 8*x*_{2} subject to constraints 30*x*_{1}+20*x*_{2} ≤300;5*x*_{1}+10*x*_{2} ≤110; and *x*_{1}, *x*_{2} __>__ 0 .

(ii)
Maximize *Z *=* *22*x*_{1}* *+*
*18*x*_{2}* *subject to constraints* *960*x*_{1}* *+*
*640*x*_{2}* *≤* *15360* *;* x*_{1}* *+*
x*_{2}* *≤* *20* *and*
x*_{1}* *,* x*_{2}* *≥* *0* *.

(iii) Minimize
*Z* = 3*x*_{1} + 2*x*_{2}
subject to the constraints 5*x*_{1}+
*x*_{2}≥10; *x*_{1}+ *x*_{2}≥6; *x*_{1}+
4 *x*_{2}* *≥12 and* x*_{1},* x*_{2}≥0.

(iv) Maximize
*Z* = 40*x*1 + 50*x*2 subject to
constraints 30*x*1 + *x*2 ≤ 9 ; *x*1*
*+* *2*x*2* *≤* *8 and *x*1* *,*
x*2* *≥* *0

(v) Maximize
*Z* = 20*x*_{1} + 30*x*_{2} subject to constraints 3*x*_{1} + 3*x*_{2}
≤ 36 ; 5*x*_{1}
+ 2*x*_{2} ≤ 50 ; 2*x*_{1} + 6*x*_{2}
≤ 60 and *x*_{1} , *x*_{2}
≥ 0

(vi) Minimize *Z
*=* *20*x*_{1}* *+* *40*x*_{2}* *subject to the constraints
36*x*_{1} + 6*x*_{2 }≥ 108, 3*x*_{1} + 12*x*_{2
}≥ 36, 20*x*_{1} + 10*x*_{2} ≥ 100 and *x*_{1}* *,* x*_{2 }≥ 0

Tags : Linear programming problem , 11th Business Mathematics and Statistics(EMS) : Chapter 10 : Operations Research

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