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OPERATIONAL RESEARCH
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)
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 LPP
 Objective function
 Constraints
 Non-Negativity restrictions
 Solution
 Feasible Solution
 Optimum Feasible Solution
SOLVED EXAMPLE -1
 A 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.
SOLUTION
 From 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
              x 2 > 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 , x 2 > 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
40
                        Scale : Xaxis = 1 cm = 5 units
35                              Yaxis = 1 cm = 5 units

30      D

25

20 B

15

10                  p

5
                       C   A                             X axis
    O       5 10 15 20 25 30    35 40
 OBPC is the feasible region
Points       x1     x2    z
O             0      0    z=0
B             0     20    z = 40(0) + 35 (20) = 700
P            18      8    z = 40(18) + 35(8) = 1000
C            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 LPP
Minimize Z = 6x + 5y
Subject to = x + y > 7
            x<3,y<4
            x<0,y>0
Solution : Removing Inequality in given equation
1. x + y > 7
Put y = 0 : x = 7
Put x = 0 : y = 7
The two points are : A = (7 , 0) & B = (0 , 7)
Further,
X=3,y=4
Y axis

8

7   B                               Scale : X axis = 1cm = 1 unit
                                          Y axis = 1cm = 1 unit
6

5
                    P
4

3

2

1

                                        A
        1   2   3       4   5   6   7       8                       X axis
O
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
CONCLUSION
 Linear 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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Linear Programming Feasible Region

  • 1. OPERATIONAL RESEARCH 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)
  • 2. 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”.
  • 3. TERMINOLOGY OF LINEAR PROGRAMMING A typical linear program has the following components  An objective Function.  Constraints or Restrictions.  Non-negativity Restrictions.
  • 4. TERMS USED TO DESCRIBE LINEAR PROGRAMMING PROBLEMS  Decision variables.  Objective function.  Constraints.  Linear relationship.  Equation and inequalities.  Non-negative restriction.
  • 5. FORMATION OF LPP  Objective function  Constraints  Non-Negativity restrictions  Solution  Feasible Solution  Optimum Feasible Solution
  • 6. SOLVED EXAMPLE -1  A 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.
  • 7. SOLUTION  From 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
  • 8. CONTI…..  Subject to: x₁ ≥ 0 x₂ ≥ 0 3x₁ + 2x₂ ≤ 1800 2x₁ + 7x₂ ≤ 1400
  • 9. 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.
  • 10. 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
  • 11. The LPP is formulated as follows Maximise ; Z = 1800 X1 + 3000 x Subject to ; X1 + x2 < 50 X1 > 5 x 2 > 3 X1
  • 12. 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 , x 2 > 0
  • 13. 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)
  • 14. 2) 4x1 + 3x2 < 96 Put x2 = 0 : 4x1 = 96 x1 = 24 C = (24 , 0) Put x1 = 0 : 3x2 = 96 x2 = 32 D = (0 , 32)
  • 15. Y axis 40 Scale : Xaxis = 1 cm = 5 units 35 Yaxis = 1 cm = 5 units 30 D 25 20 B 15 10 p 5 C A X axis O 5 10 15 20 25 30 35 40
  • 16.  OBPC is the feasible region Points x1 x2 z O 0 0 z=0 B 0 20 z = 40(0) + 35 (20) = 700 P 18 8 z = 40(18) + 35(8) = 1000 C 24 0 z = 40(24) + 35(0) = 960 Thus, the optimal feasible solution is x1 = 18 , x2 = 8 and z = 1000
  • 17. CONTI…..  Find the feasible solution to following LPP Minimize Z = 6x + 5y Subject to = x + y > 7 x<3,y<4 x<0,y>0
  • 18. Solution : Removing Inequality in given equation 1. x + y > 7 Put y = 0 : x = 7 Put x = 0 : y = 7 The two points are : A = (7 , 0) & B = (0 , 7) Further, X=3,y=4
  • 19. Y axis 8 7 B Scale : X axis = 1cm = 1 unit Y axis = 1cm = 1 unit 6 5 P 4 3 2 1 A 1 2 3 4 5 6 7 8 X axis O
  • 20. 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
  • 21. CONCLUSION  Linear 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.