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8. Graphical Methods to Solve Linear Programming

Problems.

9. Convex sets, Extreme point.

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In an LPP, if the objective function be a function of two variables only we

can solve it graphically.

Let us consider the following LPP in two variables only:

Maximize z = c1x +c2y

subject to a11 x + a12y (or ) b1

a21 x + a22y (or ) b2

x, y 0

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Step 1: Consider the LP (Linear programming) problem in two

variables only.

Step 2: Construct a graph and plot the constraint lines.

Step 3: Determine the valid side of each constraint line.

Step 4: Identify the feasible solution region.

Step 5: Plot the objective function on the graph.

Step 6: Find the optimum point.

Graphical method of linear programming is used to solve problems by finding the

highest or lowest point of intersection between the objective function line and the feasible

region on a graph.

This process can be broken down into 6 simple steps explained below.

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Methods of Finding Optimal Solutions

There are two techniques to find the optimal solution of an LPP.

1.Corner point method

The optimal solution to a LPP, if it exists, occurs at the corners of the

feasible region.

2. ISO- PROFIT (OR ISO-COST) Line Method

Let the objective function be Z = ax + by. Draw a dotted line for the equation ax + by = k,

where k is any constant. Sometimes it is convenient to take k as the LCM of a and b.

To maximize Z draw a line parallel to ax + by = k and farthest from the origin.

This line should contain at least one point of the feasible region. Find the

coordinates of this point by solving the equations of the lines on which it lies.

To minimize Z draw a line parallel to ax + by = k and nearest to the origin. This

line should contain at least one point of the feasible region. Find the co-

ordinates of this point by solving the equation of the line on which it lies

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Graphical Solution by Corner Point Method

Let us consider the following LPP:

Maximize z = 50x +18y

subject to 2x + y 100

x + y 80

x, y 0

Problem 1.

The corner points of feasible region are O(0,0), A(0,80), B(20,,60), C(50,0)

At O z = 50.0 + 18.0 = 0

At A z = 50.0 + 18.80 = 1440

At B z = 50.20 + 18.60 = 2080

At C z = 50.50 + 18.0 = 2500 maximum

Hence z is maximum at C & maximum value of z is 2500.

The optimal solution is x = 50, y = 0 6Bulbul Sen SNC

Graphical Solution by ISO Profit (or, ISO Cost) Line Method


Maximize z = 120x +100y

subject to 2x + y 16

x + y 11

x + 2y 6

5x +6y 90

x, y 0

Problem 2.

Give a constant value 600 to Z in the objective function, then we have an equation of

the line 120x + 100y = 600

or 6x + 5y = 30 (Dividing both sides by 20)

P1Q1 is the line corresponding to the equation 6x + 5y = 30. (1)7Bulbul Sen SNC


We give a constant 1200 to Z then the P2Q2

represents the line.

120x + 100y = 1200

or, 6x + 5y = 60

P2Q2 is a line parallel to P1Q1 and has one

point 'M' which belongs to feasible region

and farthest from the origin. If we take any

line P3Q3 parallel to P2Q2 away from the

origin, it does not touch any point of the

feasible region.

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The co-ordinates of the point M can be

obtained by solving the equation 2x + y =

16

x + y =11 which give

x = 5 and y = 6

The optimal solution for the objective

function is x = 5 and y = 6

The optimal value of z=

120 (5) + 100 (6) = 600 + 600

= 1200


Solution of LPP Graphically Using

TORA

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Click here 11Bulbul Sen SNC

Main Menu

Click here

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Select

Input

Click here

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Maximize z = 150x + 100y

subject to 8x + 5y 60

4x + 5y 40

x, y 0

Input

INPUT SCREEN

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INPUT GRID

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Editing Input Grid

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Maximize z = 150x + 100y


4x + 5y 40

x, y 0

Input data

Click here 17Bulbul Sen SNC

If you wish to save data click ‘Yes’, otherwise ‘No’

Click here

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To solve the problem

graphically click here

Solve Menu

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Input data

Select

Click here

Formatting Output

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Click here

Output Screen

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Solution

Optimal Solution

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Assignment

Solve the following Linear Programming Problems graphically:

1. Maximize z = 2x + y

subject to x + 3y 15

3x 4y 12

x, y 0

2. Minimize z = 2x y

subject to x + y 5

x + 2y 8

4x + 3y 12

x, y 0

3. Maximize z = 6x +10 y


5x + 3y 15

x, y 0

4. Maximize z = 3x +2y

subject to x y 1

x + y 3

x, y 0

5. Maximize z = 3x +2y

subject to 2x + y 2

3x + 4y 12

x, y 023Bulbul Sen SNC

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