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Topic: Linear Programming Problem
Submitted To : Prof. Nilesh
Coordinators : Zeel Mathkiya (19)
Dharmik Mehta (20)
Sejal Mehta (21)
Hirni Mewada (22)
Varun Modi (23)
Siddhi Nalawade (24)
OPERATIONAL RESEARCH
DEFINITION OF LINEAR PROGRAMMING
The Mathematical Definition of LP:
“It is the analysis of problem in which a linear function of a number of variables is to maximised (minimised), when those variables are subject to a number of restraints in the form of linear inequalities”.
TERMINOLOGY OF LINEAR PROGRAMMING
A typical linear program has the following components
An objective Function.Constraints or Restrictions.Non-negativity Restrictions.
TERMS USED TO DESCRIBE LINEAR PROGRAMMING PROBLEMS
Decision variables.Objective function.Constraints.Linear relationship.Equation and inequalities.Non-negative restriction.
FORMATION OF LPPObjective functionConstraintsNon-Negativity restrictionsSolutionFeasible SolutionOptimum Feasible Solution
SOLVED EXAMPLE -1A Company manufactures 2 types of product H₁ & H₂. Both the
product pass through 2 machines M₁,M₂.The time requires for processing each unit of product H₁,H₂.On each machine & the available capacity of each machine is given below:
Product Machine
M₁ M₂
H₁ 3 2
H₂ 2 7
Available Capacity(hrs) 1800 1400
The availability of materials is sufficient to product 350 unit of H₁ & 150 of H₂.Each unit of H₁ gives a profit of Rs.25,each unit of H₂ gives profit of Rs.20.Formulate the above problem as LPP.
SOLUTIONFrom manufactures point of view we need to maximise the profit.The
profit depend upon the number of unit of product H₁ &H₂ produced.
Let x₁= no of unit of H₁ produce
x₂=no of unit of H₂ produce
x₁ ≥ 0 1
x₂ ≥ 0 2
3x₁ + 2x₂ ≤ 1800 3
2x₁ + 7x₂ ≤ 1400 4
Z= 25x₁ + 20x₂
LPP is formed as follows:
Maximise Z= 25x₁ + 20x₂
CONTI…..Subject to:
x₁ ≥ 0 x₂ ≥ 0 3x₁ + 2x₂ ≤ 1800 2x₁ + 7x₂ ≤ 1400
CONTI…...A Manager of hotel dreamland plans and extancison
not more than 50 groups attleast 5 must be executive single rooms the number of executive double rooms should be atleast 3 times the number of executive single rooms. He charges Rs.3000 for executive double rooms and Rs.1800 executive single rooms per day.
CONTI…..
Formulate the above problume for LPP
SOLUTION →
The LPP is formulated as follows ;
Let X1 = Total No. of single executive rooms
Let X2 = Total No. of Double executive rooms
... X1 + x2 < 50
X1 > 5
x2 > 3 X1
Maximise ; Z = 1800 X1 + 3000 x2
The LPP is formulated as follows
Maximise ; Z = 1800 X1 + 3000 x
Subject to ; X1 + x2 < 50
X1 > 5
x2 > 3 X1
GRAPHICAL METHOD
1. Arrive at a graphical solution for the following LPP.Maximize Z = 40x1 + 35x2
Subject to : 2x1 + 3x2 < 60
4x1 + 3x2 < 96
x1 , x2 > 0
Solution : Let us consider the equation
1) 2x1 + 3x2 = 60
Put x2 = 0: 2x1 = 60
x1 = 30
A = (30 , 0)Put x1 = 0 : 3x2 = 60
x2 = 20
B = (0 , 20)
2) 4x1 + 3x2 < 96
Put x2 = 0 : 4x1 = 96
x1 = 24
C = (24 , 0)Put x1 = 0 : 3x2 = 96
x2 = 32
D = (0 , 32)
Y axis
Scale : Xaxis = 1 cm = 5 units Yaxis = 1 cm = 5 units D
B
p
C A X axis O
5 10 15 20 25 30 35 40
40
35
30
25
20
15
10
5
OBPC is the feasible regionPoints x1 x2 z
O 0 0 z = 0B 0 20 z = 40(0) + 35 (20) = 700P 18 8 z = 40(18) + 35(8) = 1000C 24 0 z = 40(24) + 35(0) = 960
Thus, the optimal feasible solution is x1 = 18 , x2 = 8 and z = 1000
CONTI…..Find the feasible solution to following LPPMinimize Z = 6x + 5ySubject to = x + y > 7
x < 3 , y < 4x < 0 , y > 0
Solution : Removing Inequality in given equation1. x + y > 7Put y = 0 : x = 7Put x = 0 : y = 7The two points are : A = (7 , 0) & B = (0 , 7)Further,X = 3 , y = 4
Y axis
B Scale : X axis = 1cm = 1 unit Y axis = 1cm = 1 unit
P
A X axis O
1 2 3 4 5 6 7 8
8
7 6
5 4
3
2
1
CONTI…..As all the 3 lines intersect each other at a
common point P( 3 , 4) it is the feasible solution to LPP
Z = 6(3) + 5(4) = 18 + 20 = 38
CONCLUSIONLinear programming is very important
mathematical technique which enables managers to arrive at proper decisions regarding his area of work. Thus it is very important part of operations research.
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