6. 6
Zmin= 6X+8Y
X+Y β₯ 10
2X +3Y β₯ 25
X +5Y β₯ 35
X, Y β₯ 0
Add slack Surplus
variable to convert
constraint equation into
equality
Z = 6X+8Y+0S1 β 0S2
β 0S3
X+YβS1 =10
2X +3YβS2 = 25
X +5Y β S3 = 35
X,Y, S1, S2, S3 β₯ 0
Operation Research Models
LPP Big M Method β Prob 7
7. 7
Ibfs (Initial Basic Feasible Solution)
DV = 0
βS =10 ; βS2 = 25; β S3 = 35 0r
S = β 10 ; S2 = β 25; S3 = β 35 0r
Z = 6X+8Y+0S1 β 0S2 β 0S3
X+YβS1 =10
2X +3YβS2 = 25
X +5Y β S3 = 35
X,Y, S1, S2, S3 β₯ 0
Operation Research Models
LPP Big M Method β Prob 7
8. 8
Ibfs (Initial Basic Feasible Solution)
DV = 0 and Surplus = 0
0 =10 ; 0 = 25; 0 = 35
Z = 6X+8Y+0S1 β 0S2 β 0S3
X+YβS1 =10
2X +3YβS2 = 25
X +5Y β S3 = 35
X,Y, S1, S2, S3 β₯ 0
Operation Research Models
LPP Big M Method β Prob 7
9. 9
X+Y β₯ 10
2X +3Y β₯ 25
X +5Y β₯ 35
X, Y β₯ 0
Along with Surplus
variable add Artificial
Variable to convert
constraint equation into
equality
X+YβS1 +A1 =10
2X +3YβS2+A2 = 25
X +5Y β S3 +A3 = 35
X,Y, S1, S2, S3, A1, A2, A3
β₯ 0
Operation Research Models
LPP Big M Method β Prob 7
Zmin =
6X+8Y+0S1+0S2+ 0S3+MA1+MA2+MA3
Zmin=
6X+8Y
10. 10
Operation Research Models
LPP Big M Method β Prob 7
Zmin =
6X+8Y+0S1+0S2+ 0S3+MA1+MA2+MA3
Ibfs (Initial Basic Feasible Solution)
Put DV = 0 and Surplus variables = 0
A1=10 ;
A2= 25;
A3= 35
11. 11
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M M
ΞΈ
X Y S1 S2 S3 A1 A2 A3 RHS
(b)
M
M
M
A1
A2
A3
10
25
35
12. 12
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M M
ΞΈ
X Y S1 S2 S3 A1 A2 A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
β1
0
0
0
β1
0
0
0
β1
1
0
0
0
1
0
0
0
1
10
25
35
13. 13
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M M
ΞΈ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
β1
0
0
0
β1
0
0
0
β1
1
0
0
0
1
0
0
0
1
10
25
35
Zj
4
M
9
M
βM βM βM M M M
M+2M+M = 4M
14. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
14
https://www.youtube.com/watch?v=tRNwz
Sr9IXg
IITM Srinivasan sir
15. 15
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M M
ΞΈ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
β1
0
0
0
β1
0
0
0
β1
1
0
0
0
1
0
0
0
1
10
25
35
10
25/3
7
Zj 4M 9M βM βM βM M M M
Zjβ Cj 4Mβ
6
9Mβ
8
βM βM βM 0 0 0
Largest
positive
value Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
16. 16
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M M
ΞΈ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
β1
0
0
0
β1
0
0
0
β1
1
0
0
0
1
0
0
0
1
10
25
35
10
25/3
7
Zj 4M 9M βM βM βM M M M
Zjβ Cj 4Mβ
6
9Mβ
8
βM βM βM 0 0 0
Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Key
Element
Largest
positive
value
17. 17
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M M
ΞΈ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
β1
0
0
0
β1
0
0
0
β1
1
0
0
0
1
0
0
0
1
10
25
35
10
25/3
7
Zj 4M 9M βM βM βM M M M
Zjβ Cj 4Mβ
6
9Mβ
8
βM βM βM 0 0 0
18. 18
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M M
ΞΈ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
β1
0
0
0
β1
0
0
0
β1
1
0
0
0
1
0
0
0
1
10
25
35
10
25/3
7
Zj 4M 9M βM βM βM M M M
Zjβ Cj 4Mβ
6
9Mβ
8
βM βM βM 0 0 0
19. 19
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
1
2
1/5
1
3
1
β1
0
0
0
β1
0
0
0
β1/5
1
0
0
0
1
0
10
25
7
Zj
Zjβ Cj
Modifie
d key
row
20. 20
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
1
2
1/5
1
3
1
β1
0
0
0
β1
0
0
0
β1/5
1
0
0
0
1
0
10
25
7
Zj
Zjβ Cj
Modifie
d key
row
22. 22
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
2
1/5
0
3
1
β1
0
0
0
β1
0
1/5
0
β1/5
1
0
0
0
1
0
3
25
7
Zj
Zjβ Cj
MKR
NR
23. 23
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
2
1/5
0
3
1
β1
0
0
0
β1
0
1/5
0
β1/5
1
0
0
0
1
0
3
25
7
Zj
Zjβ Cj
MKR
NR
OR
25. 25
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
7/5
1/5
0
0
1
β1
0
0
0
β1
0
1/5
3/5
β1/5
1
0
0
0
1
0
3
4
7
Zj
Zjβ Cj
MKR
NR
NR
26. 26
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
7/5
1/5
0
0
1
β1
0
0
0
β1
0
1/5
3/5
β1/5
1
0
0
0
1
0
3
4
7
Zj 11π
5
+
8
5
8 βM βM 4π
5
β
8
5
M M
Zjβ Cj
27. 27
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
7/5
1/5
0
0
1
β1
0
0
0
β1
0
1/5
3/5
β1/5
1
0
0
0
1
0
3
4
7
Zj 11π
5
+
8
5
8 βM βM 4π
5
β
8
5
M M
Zjβ Cj 11π
5
+
38
5
0 βM βM 4π
5
β
8
5
0 0
28. 28
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
7/5
1/5
0
0
1
β1
0
0
0
β1
0
1/5
3/5
β1/5
1
0
0
0
1
0
3
4
7
Zj 11π
5
+
8
5
8 βM βM 4π
5
β
8
5
M M
Zjβ Cj 11π
5
+
38
5
0 βM βM 4π
5
β
8
5
0 0
Key Column:
decides incoming
Largest
positive
value
29. 29
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M M
ΞΈ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
7/5
1/5
0
0
1
β1
0
0
0
β1
0
1/5
3/5
β1/5
1
0
0
0
1
0
3
4
7
15/4
20/7
35
Zj 11π
5
+
8
5
8 βM βM 4π
5
β
8
5
M M
Zjβ Cj 11π
5
+
38
5
0 βM βM 4π
5
β
8
5
0 0
Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Largest
positive
value
30. 30
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A1
X
Y
4/5
7/5
1/5
0
0
1
β1
0
0
0
β1
0
1/5
3/5
β1/5
1
0
0
3
4
7
15/4
20/7
35
Zj 11π
5
+
8
5
8 βM βM 4π
5
β
8
5
M
Zjβ Cj 11π
5
+
38
5
0 βM βM 4π
5
β
8
5
0 0
31. 31
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A1
X
Y
4/5
7/5
1/5
0
0
1
β1
0
0
0
β1
0
1/5
3/5
β1/5
1
0
0
3
4
7
Zj
Zjβ Cj
32. 32
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A1
X
Y
4/5
1
1/5
0
0
1
β1
0
0
0
β5/7
0
1/5
3/7
β1/5
1
0
0
3
20/7
7
Zj
Zjβ Cj
MKR
33. 33
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A1
X
Y
4/5
1
1/5
0
0
1
β1
0
0
0
β5/7
0
1/5
3/7
β1/5
1
0
0
3
20/7
7
Zj
Zjβ Cj
MKR
35. 35
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A
1
X
Y
0
1
1/5
0
0
1
β1
0
0
4/7
β5/7
0
β1/7
3/7
β1/5
1
0
0
5/7
20/7
7
Zj
Zj
β Cj
MKR
NR
37. 37
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A1 RHS
(b)
M
6
8
A1
X
Y
0
1
0
0
0
1
β1
0
0
4/7
β5/7
1/7
β1/7
3/7
β2/7
1
0
0
5/7
20/7
45/7
Zj
Zjβ Cj
MKR
NR
NR
38. 38
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A1 RHS
(b)
M
6
8
A1
X
Y
0
1
0
0
0
1
β1
0
0
4/7
β5/7
1/7
β1/7
3/7
β2/7
1
0
0
5/7
20/7
45/7
Zj 6 8 βM
4π
7
β
22
7
β
π
7
+
2
7
M
Zjβ Cj
MKR
NR
NR
39. 39
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A1 RHS
(b)
M
6
8
A1
X
Y
0
1
0
0
0
1
β1
0
0
4/7
β5/7
1/7
β1/7
3/7
β2/7
1
0
0
5/7
20/7
45/7
Zj
6 8 βM
4π
7
β
22
7
β
π
7
+
2
7
M
Zjβ Cj
0 0 βM 4π
7
β
22
7
β
π
7
+
2
7
0
Key Column: decides
incoming variable
Largest
positive
value
40. 40
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0 M
ΞΈ
X Y S1 S2 S3 A1 RHS
(b)
M
6
8
A1
X
Y
0
1
0
0
0
1
β
1
0
0
4/7
β5/7
1/7
β1/7
3/7
β2/7
1
0
0
5/7
20/7
45/7
5/4
β4
45
Zj
6 8 βM
4π
7
β
22
7
β
π
7
+
2
7
M
Zjβ Cj
0 0 βM 4π
7
β
22
7
β
π
7
+
2
7
0
Key Column:
decides incoming
Largest
positive
value
Least
positive
Key
Row: decides
outgoing
variable
Key
Element
41. 41
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0
ΞΈ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
β
1
0
0
4/7
β5/7
1/7
β1/7
3/7
β2/7
5/7
20/7
45/7
Zj
Zjβ Cj
42. 42
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0
ΞΈ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
β
1
0
0
4/7
β5/7
1/7
β1/7
3/7
β2/7
5/7
20/7
45/7
Zj
Zjβ Cj
43. 43
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0
ΞΈ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
β7/4
0
0
1
β5/7
1/7
β1/4
3/7
β2/7
5/4
20/7
45/7
Zj
Zjβ Cj
MKR
44. 44
Operation Research Models
LPP Big M Method β Prob 7
Cj
6 8 0 0 0
ΞΈ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
β7/4
0
0
1
β5/7
1/7
β1/4
3/7
β2/7
5/4
20/7
45/7
Zj
Zjβ Cj
MKR
OR
62. 62
Operation Research Models
LPP Big M Method β Prob 8
Cj
4 1 0 0 M M
ΞΈ
π₯1 π₯2 S1 S2 A
1
A
2
RHS
(b)
M
M
A1
A2
3
1
4
5
β1
0
0
β1
1
0
0
1
20
15
Zj
63. 63
Operation Research Models
LPP Big M Method β Prob 8
Cj
4 1 0 0 M M
ΞΈ
π₯1 π₯2 S1 S2 A
1
A
2
RHS
(b)
M
M
A1
A2
3
1
4
5
β1
0
0
β1
1
0
0
1
20
15
Zj
4
M
9
M
βM βM M M
64. 64
Operation Research Models
LPP Big M Method β Prob 8
Cj
4 1 0 0 M M
ΞΈ
π₯1 π₯2 S1 S2 A
1
A
2
RHS
(b)
M
M
A1
A2
3
1
4
5
β1
0
0
β1
1
0
0
1
20
15
5
3
Zj
4M 9M βM βM M M
ZjβCj 4Mβ4 9Mβ1 βM βM 0 0
Max
positive
value
Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Key
Element
65. 65
Operation Research Models
LPP Big M Method β Prob 8
Cj
4 1 0 0 M
ΞΈ
π₯1 π₯2 S1 S2 A
1
RHS
(b)
M
1
A1
π₯2
3
1
4
5
β1
0
0
β1
1
0
0
15
Zj
ZjβCj
66. 66
Operation Research Models
LPP Big M Method β Prob 8
Cj
4 1 0 0 M
ΞΈ
π₯1 π₯2 S1 S2 A
1
RHS
(b)
M
1
A1
π₯2
3
1/5
4
1
β1
0
0
β1/5
1
0
20
15/5=3
Zj
ZjβCj
Old Row
MKR
109. 109
Operation Research Models
LPP Big M Method β Prob 9
Cj
1 5 0 0
ΞΈ =
π
π
π₯1 ππ S1 S2 RHS
(b)
0
5
S2
ππ
5/4
3/4
0
1
3/4
1/4
1
0
5/2
3/2
Zj
Cj β Zj
MKR
NR
110. 110
Operation Research Models
LPP Big M Method β Prob 9
Cj
1 5 0 0
ΞΈ =
π
π
π₯1 ππ S1 S2 RHS
(b)
0
5
S2
ππ
5/4
3/4
0
1
3/4
1/4
1
0
5/2
3/2
Zj
15/4 5 5/4 0
Cj β Zj
β11/4 0 β5/4 0
MKR
NR
So, the optimal Solution we get without
π₯1
The optimal Solution
π₯1 = 0; π₯2 =
3
2
and ππππ₯ = π₯1 + 5π₯2 = 0 +
15
2
, ππππ₯ =
15
2
111. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
111
112. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
112
113. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
113
114. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
114
115. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
115
There is no region that satisfies both the
constraints.
This LP is infeasible
Refer: Slide no. 97
117. 117
Operation Research Models
LPP Two Phase Method
LPP using Graphical
Method
LPP using iteration Method
β€
inequality
= and β₯
inequality
Simplex
Method
Big M
Method
Two Phase
Method
118. 118
Operation Research Models
LPP Two Phase Method
In Two Phase Method, the whole procedure of solving a linear
programming problem (LPP) involving artificial variables is
divided into two phases.
In phase I, we form a new objective function (Auxiliary
function) by assigning zero to all variable (slack and surplus
variables) and +1 or β1 to each of the artificial variables
ππππ or ππππ₯ . Then we try to eliminate the artificial variables
from the basis. The solution at the end of phase I serves as a
basic feasible solution for phase II.
In phase II, Auxiliary function is replaced by the original
objective function and the usual simplex algorithm is used to
find an optimal solution.
140. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
140
The objective function and constraints are functions of two types of variables,
_______________ variables and ____________ variables.
A. Positive and negative
B. Controllable and uncontrollable.
C. Strong and weak
D. None of the above
141. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
141
In graphical representation the bounded region is known as _________ region.
A. Solution
B. basic solution
C. feasible solution
D. optimal
142. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
142
The optimal value of the objective function for the following L.P.P.
Max z = 4X1 + 3X2
subject to
X1 + X2 β€ 50
X1 + 2X2 β€ 80
2X1 + X2 β₯ 20
X1, X2 β₯ 0
is
(a) 200
(b) 330
(c) 420
(d) 500
143. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
143
To formulate a problem for solution by the Simplex method, we must add artificial
variable to
(a) only equality constraints
(b) only LHS is greater than equal to RHS constraints
(c) both (a) and (b)
(d) None of the above
144. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
144
In graphical method of LPP, If at all there is a feasible solution (feasible area of polygon)
exists then, the feasible area region has an important property known as ____________in
geometry
a) convexity Property
b) Convex polygon
c) Both of the above
d) None of the above
145. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
145
146. Dr. L K Bhagi, School of Mechanical
Engineering, LPU
146
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