L.P.P-Graphical Method

1. Operations Research
Dr.P.Geetha

2. Linear Programming:
• Linear Programming is a mathematical technique
for finding the best use of limited resources of a
concern. This is a technique to allocate scarce
available resources under conditions of certainty
in an optimum manner.
• By using linear programming technique, a
production manager can allocate the limited
amount of machine time, labour hours and raw
material available with him to the different
activities so as to maximise the output/profit.

3. For solving a problem by linear programming, following
conditions must be fulfilled:
• i. Objective i.e., reduction in cost or to
maximise the profit, be stated mathematically.
• ii. Resources can be measured as quantities
i.e., in number, weight, volume or Rupees etc.
• iii. There may be many alternative solutions.
• iv. Relationship between factors should be
linear.
• v. Restrictions of the resources must be fully
spelt out.

4. Graphical Method
To understand graphical method, let us take following example:
• Objective of a firm is to maximise profit by producing product
A and / or product B both, of which have to be processed on
machines 1 and 2. Product A requires 2 hours on both
machines 1 end 2, while product B needs 3 hours on machine
1 and only 1 hour on machine 2. There are only 12 and 8
hours available on machine 1 and 2 respectively. The profit
per unit is estimated at Rs. 600 and Rs. 700 in case of A and B
respectively. Find out number of units of product A and B, he
should produce to maximise the profit?

5. • First Step:
• Formulation of Linear programming problems.
Let Z is the profit. The maximum profit by
manufacturing A and B.
• 600 A + 700 B = Max. profit Z …(1)
• This equation is called objective function.
• ADVERTISEMENTS:
• Now set up the equations for the process times.
• Machine I – 2A + 3B ≤ 12 …(2)
• Machine II – 2A + B ≤ 8 …(3)

6. • Second Step:
• ADVERTISEMENTS:
• Plotting of the equations on graph . Suppose
product A is shown on X-axis and product B is
shown on the Y-axis of the graph. Now to plot
the equation 2A + 3B ≤ 12 ,first find the two
terminal points, and then joining these points
by a straight line.

7. • Now the question is how to find the two terminal points:
• (a) Suppose all the time available on machine 1 is used for
making product A, this means the production of product B
is zero, this means 6 units of product A would be made,
then the first terminal point is (6, 0).
• (b) Now supposing that all the time available on machine 1
is utilised for making product B then the production of
product A is zero, and only 4 units of product B would be
made. The second terminal point is (0, 4).
• ADVERTISEMENTS:
• By joining (6, 0) and (0, 4), we get a straight line BC.
• In the same way, the other equation 2A + B ≤ 8 can also be
drawn by a line EF by terminal points (4, 0) and (0, 8).

9. • Third Step:
• Identify Feasibility Region and ascertain co-ordinates of its corner
points. Under this step we have to identify the cross shaded area
BOFD (Fig. 30.1), generally known as Feasibility Region and then
finding on the co-ordinates of its corner points. These co-ordinates
can directly be read with the help of accurately drawn graph or we
can also find the co-ordinate with the help of mathematics such as;
• O (0, 0)
• B (0, 4)
• F (4, 0)
• and for point D, solve the two equations
• 2A + 3B = 12
• 2A + B = 8
• On solving, we get A = 3, B = 2,
• so the co-ordinate of point D is (3, 2).

10. • Fourth Step:
• Test which corner point is most profitable. Putting
different co-ordinates of corner points in the
objective function Eq. (1), amount of profit in
hundred Rupees.
• Corner point O (0, 0) = (6 × 0) + (7 × 0) = 0
• Corner point B (0, 5) = (6 × 0) + (4 × 4) = 16
• Corner point F (4, 0) = (6 × 4) + (7 × 0) = 24
• Corner point D (3, 2) = (6 × 3) + (7 × 2) = 32
• Thus corner point D is yielding maximum profit of
Rs. 3200 and the firm must produce 3 units of
product A, and 2 units of product B.