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2. Presented by :-
Sonal chauhan
TOPIC OF PRESENTATION
PRESENTED TO:-
Bba 2ND YR

3. INTRODUCTION
Before understanding the meaning of linear
programming, its beneficial to understand the
words linear and programming individually. So,
LINEAR means that the relationships are represented
by straight lines i.e., the relationships are in the form of y
=ax+b.
PROGRAMMING means the optimal allocation of
resources.
Hence, linear programming can be defined as
mathematical technique for determining the optimal
allocation of resources and obtaining a particular
objective by making choices from the available sources
.The objective may be cost minimization or inversely

4. APPLICATION KINDS OF L.P.P.
There are some important L.P problem discussed
below:-
 Optimal product line problems : Production and
sale of different products by the manufacturer to
make maximum profit.
 Product mix problems: Keeping in mind the different
conditions we try to determine such a product mix
which maximizes the profit and minimizes the cost.
 Diet planning problems: We have to determine the
amount of different kinds of nutrients which should be
included in a diet so as to minimize the cost .
 Transportation problems: Taking an example ,the
finished goods from the plant are also to be
transported to warehouse in such a way that the total

5.  A set of values of variables satisfying the constraints
of a L.P.P is called a solution of a L.P.P. Say for
example:-
lets take an equation (2x + 3y = 10) here
CONSTRAINT VARIABLE
Similarly 3 is constraint and Y is variable
CONSTRAINTS can be changed but VARIABLES
remains constant.
 The variable used in linear programming problem to
convert an inequality in to an equation is known as
slack variable.
FOR EXAMPLE- 2x + 3y ≤ 100 , 3x +4y ≤ 200
These are known as inequality and for converting and
for solving them we assume some slack variables which
are always equal to 0 and which does not lays any effect
on the given inequality and converts them in a equation.
As we consider some slack variables as S1, S2, =0 .

6. 2x + 3y + S1 = 100
3x + 4y + S2 = 200
Next we will put the value of S1 and S2 i.e., 0 and finally we will get
the equations as-:
2X +3y = 100
3x + 4y = 200 which can be solved further.
• General form of L.P.P is
Maximization or minimization Z= a₁x₁ + a₂x₂…..anxn
subject to :- a₁₁x₁ + a₁₂x₂+ a₁₃x₃……..≤b₁ , a₂₁x₁ + a₂₂x₂ + ……≤b₂….
For Example
MAXIMIZE Z=25x₁ + 40x₂ MINIMIZE Z=2x₁ + 3x₂
Subject to : x₁ + 2x₂ ≤ 5 Subject to : 3x₁ + 2x₂ ≥ 5
2x₁ + 5x₂ ≤ 6 2x₁ + 5x₂ ≥ 7

7. METHODS OF SOLVING LINEAR
PROGRAMMING PROBLEMS
GRAPHICAL METHOD
SIMPLEX METHOD
(Here we will discuss
only graphical method in
detail)

8. GRAPHICAL METHOD
For a L.P.P, that have only two variables ,it is possible that
entire solution can be displayed graphically by plotting
linear constraints on a graph paper to locate best solution.
This technique is called GRAPHICAL METHOD.
 It is applicable when only two variables are involved.
 There are two types of graphs i.e., maximization in
case of profit and minimization in case of cost.
Let’s take an example to understand the
graphical method practically and more
easily.Maximize Z = f(x,y) = 3x + 2y
subject to: 2x + y ≤ 18
2x + 3y ≤ 42
3x + y ≤ 24
x ≥ 0 , y ≥ 0

9. 1. Initially ,the coordinate system is drawn and each
variable is associated to an axis.
2. A numerical scale is marked in axis.
3. In the beginning ,for changing the given inequalities
into an equation we will consider some slack variables
,say( S1,S2,S3=0) and add one slack variable in each
inequality respectively.
2x + y + S1 = 18 , 2x + 3y +S2 = 42, 3x + y +S3 = 24
Now put the value of S1, S2, S3 =0
4. After putting the value of S1, S2, S3 in the above given
equations we will get the following results :-
2x + y + S1 = 18 2x + y + 0 = 18 (a)
2x + 3y + S2 = 42 2x + 3y + 0 = 42 (b)
3x + y + S3 = 24 3x + y + 0 = 24 (c)
5. Next we will find out the coordinates from the given
above equation (a) , (b) , (c) by taking each variable x
and y one by one as 0.

10. Taking equation (a) for finding out the coordinates .
If x=0, then 2x + y = 18 will be
2 X 0 + y = 18
0 + y = 18 y = 18
If y =0 , then 2x + y = 18 will be 2x+ 0 = 18
2x = 18 x = 9, therefore the coordinates we
get will be (9,18)
Taking equation (b) for finding out the coordinates.
If x = 0, then 2x + 3y = 42 will be
2 X 0 + 3y = 42
0 + 3y = 42 y = 14
If y = 0 , then 2x + 3y = 42 will be
2x + 3 X 0 = 42
2x + 0 = 42 x = 21, therefore the
coordinates we get will be ( 21 , 14 )

11. Taking equation (c) for finding out the coordinates .
If x=0, then 3x + y = 24 will be
3 X 0 + y = 24
0 + y = 24 y = 24
If y =0 , then 3x + y = 24 will be 3x+ 0 = 24
3x = 24 x = 8, therefore the coordinates we get
will be (8,24)
6. As we have got the coordinates (9,18) , (21,14) , (8,24)
from the equation (a) , (b) , (c) respectively . We will plot
these coordinates on the graph .
7. We will plot the coordinates (9,18) first
as shown in the figure by drawing straight
line joining two points 9 on the x axis and
18 on y axis.

12. 8. Next we will plot the coordinates
(21,14) On the graph shown beside.
9. Further we will plot the coordinates
(8,24) Shown here with the red shaded
region.

13. Since we have taken the example of
maximization therefore the region
towards the origin will be covered.

14. 10. The required region is the intersection of the regions
defined by the set of constraints and the coordinate axis .
This required region is represented by the O-F-H-G-C
polygon in PURPLE color.
11. Next, the extreme values are calculated. These
vertices are the optimal solutions.
12. For point O , C and F the coordinates will be (0,0) ,
(0,14) , (8,0)

15. 13. For point G, we will solve the equations (a) and (b) ,
and after solving these equations we will get the
coordinates (3,12).
Similarly for point H, we will solve the equations (a) and (c)
and we will get the coordinates as (6,6).
14.
15. Finally, the objective function (3x + 2y) is evaluated in
each of these points (results are shown in the table
above). Since G-point provides the greatest value to the Z-
function and the objective is to maximize, this point is the
optimal solution: Z = 33 with x = 3 and y= 12.
Extreme point Z = 3x + 2y Objective value (Z)
O(0,0) Z = 0 + 0 0
C(0,14) Z = 0 + 28 28
G(3,12) Z = 9 + 24 33
H(6,6) Z = 18 + 12 30
F(8,0) Z = 24 + 0 24

16. ANY
QUESTION ?