Operation research unit 1: LPP Big M and Two Phase method

Linear Programming
UNIT – I
OPERATION RESEARCH
Dr. Loveleen Kumar Bhagi
Associate Professor
School of Mechanical Engineering
LPU

Dr. L K Bhagi, School of Mechanical
Engineering, LPU
3

Engineering, LPU
4

5
Operation Research Models
LPP Big M Method
Simplex method Big M method

6
Zmin= 6X+8Y
X+Y ≥ 10
2X +3Y ≥ 25
X +5Y ≥ 35
X, Y ≥ 0
Add slack Surplus
variable to convert
constraint equation into
equality
Z = 6X+8Y+0S1 − 0S2
− 0S3
X+Y−S1 =10
2X +3Y−S2 = 25
X +5Y − S3 = 35
X,Y, S1, S2, S3 ≥ 0
LPP Big M Method – Prob 7

7
Ibfs (Initial Basic Feasible Solution)
DV = 0
−S =10 ; −S2 = 25; − S3 = 35 0r
S = − 10 ; S2 = − 25; S3 = − 35 0r
Z = 6X+8Y+0S1 − 0S2 − 0S3
X+Y−S1 =10
2X +3Y−S2 = 25
X +5Y − S3 = 35
X,Y, S1, S2, S3 ≥ 0

8
DV = 0 and Surplus = 0
0 =10 ; 0 = 25; 0 = 35
Z = 6X+8Y+0S1 − 0S2 − 0S3
X+Y−S1 =10
2X +3Y−S2 = 25
X +5Y − S3 = 35
X,Y, S1, S2, S3 ≥ 0

9
X+Y ≥ 10
2X +3Y ≥ 25
X +5Y ≥ 35
X, Y ≥ 0
Along with Surplus
variable add Artificial
Variable to convert
constraint equation into
equality
X+Y−S1 +A1 =10
2X +3Y−S2+A2 = 25
X +5Y − S3 +A3 = 35
X,Y, S1, S2, S3, A1, A2, A3
≥ 0
Zmin =
6X+8Y+0S1+0S2+ 0S3+MA1+MA2+MA3
Zmin=
6X+8Y

10
Zmin =
6X+8Y+0S1+0S2+ 0S3+MA1+MA2+MA3
Put DV = 0 and Surplus variables = 0
A1=10 ;
A2= 25;
A3= 35

11
Cj
6 8 0 0 0 M M M
θ
X Y S1 S2 S3 A1 A2 A3 RHS
(b)
M
M
M
A1
A2
A3
10
25
35

12
Cj
6 8 0 0 0 M M M
θ
X Y S1 S2 S3 A1 A2 A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
−1
0
0
0
−1
0
0
0
−1
1
0
0
0
1
0
0
0
1
10
25
35

13
Cj
6 8 0 0 0 M M M
θ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
−1
0
0
0
−1
0
0
0
−1
1
0
0
0
1
0
0
0
1
10
25
35
Zj
4
M
9
M
−M −M −M M M M
M+2M+M = 4M

Engineering, LPU
14
https://www.youtube.com/watch?v=tRNwz
Sr9IXg
IITM Srinivasan sir

15
Cj
6 8 0 0 0 M M M
θ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
−1
0
0
0
−1
0
0
0
−1
1
0
0
0
1
0
0
0
1
10
25
35
10
25/3
7
Zj 4M 9M −M −M −M M M M
Zj− Cj 4M−
6
9M−
8
−M −M −M 0 0 0
Largest
positive
value Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable

16
Cj
6 8 0 0 0 M M M
θ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
−1
0
0
0
−1
0
0
0
−1
1
0
0
0
1
0
0
0
1
10
25
35
10
25/3
7
Zj− Cj 4M−
6
9M−
8
−M −M −M 0 0 0
Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Key
Element
Largest
positive
value

17
Cj
6 8 0 0 0 M M M
θ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
−1
0
0
0
−1
0
0
0
−1
1
0
0
0
1
0
0
0
1
10
25
35
10
25/3
7
Zj− Cj 4M−
6
9M−
8
−M −M −M 0 0 0

18
Cj
6 8 0 0 0 M M M
θ
X Y S1 S2 S3 A
1
A
2
A3 RHS
(b)
M
M
M
A1
A2
A3
1
2
1
1
3
5
−1
0
0
0
−1
0
0
0
−1
1
0
0
0
1
0
0
0
1
10
25
35
10
25/3
7
Zj− Cj 4M−
6
9M−
8
−M −M −M 0 0 0

19
Cj
6 8 0 0 0 M M
θ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
1
2
1/5
1
3
1
−1
0
0
0
−1
0
0
0
−1/5
1
0
0
0
1
0
10
25
7
Zj
Zj− Cj
Modifie
d key
row

20
Cj
6 8 0 0 0 M M
θ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
1
2
1/5
1
3
1
−1
0
0
0
−1
0
0
0
−1/5
1
0
0
0
1
0
10
25
7
Zj
Zj− Cj
Modifie
d key
row

21
Old
Row
− KCE × MKR = NR
1 − 1 × 1/5 = 4/5
1 − 1 × 1 = 0
−1 − 1 × 0 = −1
0 − 1 × 0 = 0
0 − 1 × −1/5 = 1/5
1 − 1 × 0 1
0 − 1 × 0 0
10 − 1 × 7 = 3

22
Cj
6 8 0 0 0 M M
θ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
2
1/5
0
3
1
−1
0
0
0
−1
0
1/5
0
−1/5
1
0
0
0
1
0
3
25
7
Zj
Zj− Cj
MKR
NR

23
Cj
6 8 0 0 0 M M
θ
X Y S1 S2 S3 A
1
A2 RHS
(b)
M
M
8
A1
A2
Y
4/5
2
1/5
0
3
1
−1
0
0
0
−1
0
1/5
0
−1/5
1
0
0
0
1
0
3
25
7
Zj
Zj− Cj
MKR
NR
OR

24
Old
Row
− KCE × MKR = NR
2 − 3 × 1/5 = 7/5
3 − 3 × 1 = 0
0 − 3 × 0 = 0
−1 − 3 × 0 = −1
0 − 3 × −1/5 = 3/5
0 − 3 × 0 0
1 − 3 × 0 1
25 − 3 × 7 = 4

25
Cj
6 8 0 0 0 M M
θ
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
−1
0
0
0
−1
0
1/5
3/5
−1/5
1
0
0
0
1
0
3
4
7
Zj
Zj− Cj
MKR
NR
NR

26
Cj
6 8 0 0 0 M M
θ
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
−1
0
0
0
−1
0
1/5
3/5
−1/5
1
0
0
0
1
0
3
4
7
Zj 11𝑀
5
+
8
5
8 −M −M 4𝑀
5
−
8
5
M M
Zj− Cj

27
Cj
6 8 0 0 0 M M
θ
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
−1
0
0
0
−1
0
1/5
3/5
−1/5
1
0
0
0
1
0
3
4
7
Zj 11𝑀
5
+
8
5
8 −M −M 4𝑀
5
−
8
5
M M
Zj− Cj 11𝑀
5
+
38
5
0 −M −M 4𝑀
5
−
8
5
0 0

28
Cj
6 8 0 0 0 M M
θ
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
−1
0
0
0
−1
0
1/5
3/5
−1/5
1
0
0
0
1
0
3
4
7
Zj 11𝑀
5
+
8
5
8 −M −M 4𝑀
5
−
8
5
M M
Zj− Cj 11𝑀
5
+
38
5
0 −M −M 4𝑀
5
−
8
5
0 0
Key Column:
decides incoming
Largest
positive
value

29
Cj
6 8 0 0 0 M M
θ
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
−1
0
0
0
−1
0
1/5
3/5
−1/5
1
0
0
0
1
0
3
4
7
15/4
20/7
35
Zj 11𝑀
5
+
8
5
8 −M −M 4𝑀
5
−
8
5
M M
Zj− Cj 11𝑀
5
+
38
5
0 −M −M 4𝑀
5
−
8
5
0 0
Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Largest
positive
value

30
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A1
X
Y
4/5
7/5
1/5
0
0
1
−1
0
0
0
−1
0
1/5
3/5
−1/5
1
0
0
3
4
7
15/4
20/7
35
Zj 11𝑀
5
+
8
5
8 −M −M 4𝑀
5
−
8
5
M
Zj− Cj 11𝑀
5
+
38
5
0 −M −M 4𝑀
5
−
8
5
0 0

31
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A1
X
Y
4/5
7/5
1/5
0
0
1
−1
0
0
0
−1
0
1/5
3/5
−1/5
1
0
0
3
4
7
Zj
Zj− Cj

32
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A1
X
Y
4/5
1
1/5
0
0
1
−1
0
0
0
−5/7
0
1/5
3/7
−1/5
1
0
0
3
20/7
7
Zj
Zj− Cj
MKR

33
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A1
X
Y
4/5
1
1/5
0
0
1
−1
0
0
0
−5/7
0
1/5
3/7
−1/5
1
0
0
3
20/7
7
Zj
Zj− Cj
MKR

34
Old
Row
− KCE × MKR = NR
4/5 − 4/5 × 1 = 0
0 − 4/5 × 0 = 0
−1 − 4/5 × 0 = −1
0 − 4/5 × −5/7 = 4/7
1/5 − 4/5 × 3/7 = −1/7
1 − 4/5 × 0 = 1
3 − 4/5 × 20/7 = 5/7

35
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A
1
RHS
(b)
M
6
8
A
1
X
Y
0
1
1/5
0
0
1
−1
0
0
4/7
−5/7
0
−1/7
3/7
−1/5
1
0
0
5/7
20/7
7
Zj
Zj
− Cj
MKR
NR

36
Old
Row
− KCE × MKR = NR
1/5 − 1/5 × 1 = 0
1 − 1/5 × 0 = 1
0 − 1/5 × 0 = 0
0 − 1/5 × −5/7 = 1/7
−1/5 − 1/5 × 3/7 = −2/7
0 − 1/5 × 0 = 0
7 − 1/5 × 20/7 = 45/7

37
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A1 RHS
(b)
M
6
8
A1
X
Y
0
1
0
0
0
1
−1
0
0
4/7
−5/7
1/7
−1/7
3/7
−2/7
1
0
0
5/7
20/7
45/7
Zj
Zj− Cj
MKR
NR
NR

38
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A1 RHS
(b)
M
6
8
A1
X
Y
0
1
0
0
0
1
−1
0
0
4/7
−5/7
1/7
−1/7
3/7
−2/7
1
0
0
5/7
20/7
45/7
Zj 6 8 −M
4𝑀
7
−
22
7
−
𝑀
7
+
2
7
M
Zj− Cj
MKR
NR
NR

39
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A1 RHS
(b)
M
6
8
A1
X
Y
0
1
0
0
0
1
−1
0
0
4/7
−5/7
1/7
−1/7
3/7
−2/7
1
0
0
5/7
20/7
45/7
Zj
6 8 −M
4𝑀
7
−
22
7
−
𝑀
7
+
2
7
M
Zj− Cj
0 0 −M 4𝑀
7
−
22
7
−
𝑀
7
+
2
7
0
Key Column: decides
incoming variable
Largest
positive
value

40
Cj
6 8 0 0 0 M
θ
X Y S1 S2 S3 A1 RHS
(b)
M
6
8
A1
X
Y
0
1
0
0
0
1
−
1
0
0
4/7
−5/7
1/7
−1/7
3/7
−2/7
1
0
0
5/7
20/7
45/7
5/4
−4
45
Zj
6 8 −M
4𝑀
7
−
22
7
−
𝑀
7
+
2
7
M
Zj− Cj
0 0 −M 4𝑀
7
−
22
7
−
𝑀
7
+
2
7
0
Key Column:
decides incoming
Largest
positive
value
Least
positive
Key
Row: decides
outgoing
variable
Key
Element

41
Cj
6 8 0 0 0
θ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
−
1
0
0
4/7
−5/7
1/7
−1/7
3/7
−2/7
5/7
20/7
45/7
Zj
Zj− Cj

42
Cj
6 8 0 0 0
θ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
−
1
0
0
4/7
−5/7
1/7
−1/7
3/7
−2/7
5/7
20/7
45/7
Zj
Zj− Cj

43
Cj
6 8 0 0 0
θ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
−7/4
0
0
1
−5/7
1/7
−1/4
3/7
−2/7
5/4
20/7
45/7
Zj
Zj− Cj
MKR

44
Cj
6 8 0 0 0
θ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
−7/4
0
0
1
−5/7
1/7
−1/4
3/7
−2/7
5/4
20/7
45/7
Zj
Zj− Cj
MKR
OR

45
Old
Row
− KCE × MKR = NR
1 − −5/7 × 0 = 1
0 − −5/7 × 0 = 0
0 − −5/7 × −7/4 = −5/4
−5/7 − −5/7 × 1 = 0
3/7 − −5/7 × −1/4 = 17/28
20/7 − −5/7 × 5/4 15/4

46
Cj
6 8 0 0 0
θ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
−7/4
−5/4
0
1
0
−1/4
17/28
−2/7
5/4
15/4
45/7
Zj
Zj− Cj
MKR
NR

47
Cj
6 8 0 0 0
θ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
−7/4
−5/4
0
1
0
1/7
−1/4
17/28
−2/7
5/4
15/4
45/7
Zj
Zj− Cj
MKR
NR
OR

48
Old
Row
− KCE × MKR = NR
0 − 1/7 × 0 = 0
1 − 1/7 × 0 = 1
0 − 1/7 × −7/4 = 1/4
1/7 − 1/7 × 1 = 0
−2/7 − 1/7 × −1/4 = −1/4
45/7 − 1/7 × 5/4 = 25/4

49
Cj
6 8 0 0 0
θ
X Y S1 S2 S3 RHS
(b)
0
6
8
S2
X
Y
0
1
0
0
0
1
−7/4
−5/4
1/4
1
0
0
−1/4
17/28
−1/4
5/4
15/4
25/4
Zj 6 8 −22/4 0 23/19
Zj− Cj
0 0 −22/4 0 23/19
MKR
NR
NR

50
𝑍𝑚𝑖𝑛=4𝑥1+𝑥2
3𝑥1+4𝑥2 ≥ 20
−𝑥1− 5𝑥2 ≤ −15
𝑥1, 𝑥2 ≥ 0

51
3𝑥1+4𝑥2 ≥ 20
−𝑥1− 5𝑥2 ≤ −15
𝑥1, 𝑥2 ≥ 0
−𝑥1− 5𝑥2 + 𝑆2 = −15
3𝑥1+4𝑥2 − 𝑆1 = 20
Add slack Surplus variable to convert
constraint equation into equality

52
3𝑥1+4𝑥2 ≥ 20
−𝑥1− 5𝑥2 ≤ −15
𝑥1, 𝑥2 ≥ 0
−𝑥1− 5𝑥2 + 𝑆2 = −15
3𝑥1+4𝑥2 − 𝑆1 = 20
DV and Surplus variables = 0
0 = 20
𝑆2 = −15

53
3𝑥1+4𝑥2 ≥ 20
𝑥1+ 5𝑥2 ≥ 15
𝑥1, 𝑥2 ≥ 0

54
3𝑥1+4𝑥2 ≥ 20
𝑥1+ 5𝑥2 ≥ 15
𝑥1, 𝑥2 ≥ 0
𝑥1+ 5𝑥2 − 𝑆2 = −15
3𝑥1+4𝑥2 − 𝑆1 = 20

55
3𝑥1+4𝑥2 ≥ 20
𝑥1+ 5𝑥2 ≥ 15
𝑥1, 𝑥2 ≥ 0
𝑥1+ 5𝑥2 − 𝑆2 = 15
3𝑥1+4𝑥2 − 𝑆1 = 20
0 = 20
0 = 15
𝑍𝑚𝑖𝑛=4𝑥1+𝑥2 + 0𝑆1 + 0𝑆2

56
𝒁𝒎𝒊𝒏=4𝒙𝟏+𝒙𝟐
3𝑥1+4𝑥2 ≥ 20
𝑥1+ 5𝑥2 ≥ 15
𝑥1, 𝑥2 ≥ 0
0 = 20 ⇒ 𝑎𝑑𝑑 𝐴𝑟𝑡𝑖𝑓𝑖𝑐𝑖𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝐴1
0 = 15 ⇒ 𝑎𝑑𝑑 𝐴𝑟𝑡𝑖𝑓𝑖𝑐𝑖𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝐴2
𝑍𝑚𝑖𝑛=4𝑥1+𝑥2 + 0𝑆1 + 0𝑆2
3𝑥1+4𝑥2 − 𝑆1 + A1 = 20
𝑥1+ 5𝑥2 − 𝑆2+A2 = 15

57
3𝑥1+4𝑥2 ≥ 20
𝑥1+ 5𝑥2 ≥ 15
𝑥1, 𝑥2 ≥ 0
𝑍𝑚𝑖𝑛=4𝑥1+𝑥2 + 0𝑆1 + 0𝑆2
3𝑥1+4𝑥2 − 𝑆1 + A1 = 20
𝑥1+ 5𝑥2 − 𝑆2+A2 = 15
𝐴1 = 20
𝐴2 = 15

58
3𝑥1+4𝑥2 ≥ 20
𝑥1+ 5𝑥2 ≥ 15
𝑥1, 𝑥2 ≥ 0
Sign of A1 & A2 𝑖𝑠 + or −
𝑣𝑒 𝑖𝑛 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑢𝑝𝑜𝑛 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚
𝑖. 𝑒. 𝑍𝑚𝑖𝑛 𝑜𝑟 𝑍𝑚𝑎𝑥
𝑍𝑚𝑖𝑛=4𝑥1+𝑥2 + 0𝑆1 + 0𝑆2
3𝑥1+4𝑥2 − 𝑆1 + A1 = 20
𝑥1+ 5𝑥2 − 𝑆2+A2 = 15
𝐴1 = 20
𝐴2 = 15

59
3𝑥1+4𝑥2 ≥ 20
𝑥1+ 5𝑥2 ≥ 15
𝑥1, 𝑥2 ≥ 0
𝑥1+ 5𝑥2 − 𝑆2+A2 = 15
3𝑥1+4𝑥2 − 𝑆1 + A1 = 20
𝑍𝑚𝑖𝑛 ⇒ MA1+MA2
𝑍𝑚𝑎𝑥 ⇒ −MA1−MA2

60
3𝑥1+4𝑥2 ≥ 20
𝑥1+ 5𝑥2 ≥ 15
𝑥1, 𝑥2 ≥ 0
𝑥1+ 5𝑥2 − 𝑆2+A2 = 15
3𝑥1+4𝑥2 − 𝑆1 + A1 = 20
𝑍𝑚𝑖𝑛=4𝑥1+𝑥2 + 0𝑆1 + 0𝑆2
+ MA1+MA2
𝐴1 = 20
𝐴2 = 15

61
Cj
4 1 0 0 M M
θ
𝑥1 𝑥2 S1 S2 A
1
A
2
RHS
(b)
M
M
A1
A2
20
15
Zj

62
Cj
4 1 0 0 M M
θ
𝑥1 𝑥2 S1 S2 A
1
A
2
RHS
(b)
M
M
A1
A2
3
1
4
5
−1
0
0
−1
1
0
0
1
20
15
Zj

63
Cj
4 1 0 0 M M
θ
𝑥1 𝑥2 S1 S2 A
1
A
2
RHS
(b)
M
M
A1
A2
3
1
4
5
−1
0
0
−1
1
0
0
1
20
15
Zj
4
M
9
M
−M −M M M

64
Cj
4 1 0 0 M M
θ
𝑥1 𝑥2 S1 S2 A
1
A
2
RHS
(b)
M
M
A1
A2
3
1
4
5
−1
0
0
−1
1
0
0
1
20
15
5
3
Zj
4M 9M −M −M M M
Zj−Cj 4M−4 9M−1 −M −M 0 0
Max
positive
value
Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Key
Element

65
Cj
4 1 0 0 M
θ
𝑥1 𝑥2 S1 S2 A
1
RHS
(b)
M
1
A1
𝑥2
3
1
4
5
−1
0
0
−1
1
0
0
15
Zj
Zj−Cj

66
Cj
4 1 0 0 M
θ
𝑥1 𝑥2 S1 S2 A
1
RHS
(b)
M
1
A1
𝑥2
3
1/5
4
1
−1
0
0
−1/5
1
0
20
15/5=3
Zj
Zj−Cj
Old Row
MKR

67
Old
Row
− KCE × MKR = NR
3 − 4 × 1/5 = 11/5
4 − 4 × 1 = 0
−1 − 4 × 0 = −1
0 − 4 × −1/5 = 4/5
1 − 4 × 0 = 1
20 − 4 × 3 = 8

68
Cj
4 1 0 0 M
θ
𝑥1 𝑥2 S1 S2 A
1
RHS
(b)
M
1
A1
𝑥2
11/5
1/5
0
1
−1
0
4/5
−1/5
1
0
8
3
Zj
Zj−Cj
New
Row
MKR

69
Cj
4 1 0 0 M
θ
𝑥1 𝑥2 S1 S2 A
1
RHS
(b)
M
1
A1
𝑥2
11/5
1/5
0
1
−1
0
4/5
−1/5
1
0
8
3
Zj
11𝑀
5
+
1
5
1 −M 4𝑀
5
−
1
5
M
Zj−Cj

70
Cj
4 1 0 0 M
θ
𝑥1 𝑥2 S1 S2 A
1
RHS
(b)
M
1
A1
𝑥2
11/5
1/5
0
1
−1
0
4/5
−1/5
1
0
8
3
Zj
11𝑀
5
+
1
5
1 −M 4𝑀
5
−
1
5
M
Zj−Cj
11𝑀 + 1 − 20
5
0 −M 4𝑀 − 1
5
0

71
Cj
4 1 0 0 M
θ
𝑥1 𝑥2 S1 S2 A
1
RHS
(b)
M
1
A1
𝑥2
11/5
1/5
0
1
−1
0
4/5
−1/5
1
0
8
3
Zj
11𝑀
5
+
1
5
1 −M 4𝑀
5
−
1
5
M
Zj−Cj
11𝑀 − 19
5
0 −M 4𝑀 − 1
5
0

72
Cj
4 1 0 0 M
θ
𝑥1 𝑥2 S1 S2 A
1
RHS
(b)
M
1
A
1
𝑥2
11/5
1/5
0
1
−1
0
4/5
−1/5
1
0
8
3
Zj
11𝑀
5
+
1
5
1 −M 4𝑀
5
−
1
5
M
Zj−Cj
11𝑀 − 19
5
0 −M 4𝑀 − 1
5
0
Max
positive
value Key Column:
decides incoming

73
Cj
4 1 0 0 M
θ
𝑥1 𝑥2 S1 S2 A
1
RHS
(b)
M
1
A
1
𝑥2
11/5
1/5
0
1
−1
0
4/5
−1/5
1
0
8
3
40/11
15
Zj
11𝑀
5
+
1
5
1 −M 4𝑀
5
−
1
5
M
Zj−Cj
11𝑀 − 19
5
0 −M 4𝑀 − 1
5
0
Max
positive
value Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Key
Element

74
Cj
4 1 0 0
θ
𝑥1 𝑥2 S1 S2 RHS
(b)
4
1
𝑥1
𝑥2
11/5
1/5
0
1
−1
0
4/5
−1/5
8
3
40/11
15
Zj
Zj−Cj

75
Cj
4 1 0 0
θ
(b)
4
1
𝑥1
𝑥2
1
1/5
0
1
−5/11
0
4/11
−1/5
40/11
3
15
Zj
Zj−Cj
MKR

76
Cj
4 1 0 0
θ
(b)
4
1
𝑥1
𝑥2
1
1/5
0
1
−5/11
0
4/11
−1/5
40/11
3
15
Zj
Zj−Cj
Old Row
MKR

77
Old Row − KCE × MKR = NR
1/5 − 1/5 × 1 = 0
1 − 1/5 × 0 = 1
0 − 1/5 × −5/11 = 1/11
−1/5 − 1/5 × 4/11 = −15/55 = −3/11
3 − 1/5 × 40/11 = 125/55 = 25/11

78
Cj
4 1 0 0
θ
(b)
4
1
𝑥1
𝑥2
1
0
0
1
−5/11
1/11
4/11
−3/11
40/11
25/11
15
Zj
Zj−Cj
New
Row
MKR

79
Cj
4 1 0 0
θ
(b)
4
1
𝑥1
𝑥2
1
0
0
1
−5/11
1/11
4/11
−3/11
40/11
25/11
15
Zj
4 1 −19/11 13/11
Zj−Cj 0 0 −19/11 13/11
Max
positive
value Key Column:
decides incoming
variable

80
Cj
4 1 0 0
θ
(b)
4
1
𝑥1
𝑥2
1
0
0
1
−5/11
1/11
4/11
−3/11
40/11
25/11
10
−
Zj
4 1 −19/11 13/11
Zj−Cj 0 0 −19/11 13/11
Max
positive
value Key Column:
decides incoming
variable
Least
positive
Key
Row: decides
outgoing
variable
Key
Element

81
Cj
4 1 0 0
θ
(b)
0
1
S2
𝑥2
1
0
0
1
−5/11
1/11
4/11
−3/11
40/11
25/11
Zj
Zj−Cj

82
Cj
4 1 0 0
θ
(b)
0
1
S2
𝑥2
11/4
0
0
1
−5/4
1/11
1
−3/11
10
25/11
Zj
Zj−Cj
MKR

83
Cj
4 1 0 0
θ
(b)
0
1
S2
𝑥2
11/4
0
0
1
−5/4
1/11
1
−3/11
10
25/11
Zj
Zj−Cj
MKR
Old
Row

84
Old Row − KCE × MKR = NR
0 − −3/11 × 11/4 = 3/4
1 − −3/11 × 0 = 1
1/11 − −3/11 × −5/4 = 1
11
−
3
11
×
5
4
=
1
11
−
15
44
= −
11
44
=−
1
4
−3/11 − −3/11 × 1 = 0
25/11 − −3/11 × 10 25
11
+
30
11
=
55
11
= 5

85
Cj
4 1 0 0
θ
(b)
0
1
S2
𝑥2
11/4
3/4
0
1
−5/4
−1/4
1
0
10
5
Zj
Zj−Cj
MKR
New
Row

86
Cj
4 1 0 0
θ
(b)
0
1
S2
𝑥2
11/4
3/4
0
1
−5/4
−1/4
1
0
10
5
Zj
3/4 1 −1/4 0
Zj−Cj

87
Cj
4 1 0 0
θ
(b)
0
1
S2
𝑥2
11/4
3/4
0
1
−5/4
−1/4
1
0
10
5
Zj
3/4 1 −1/4 0 5
Zj−Cj −13/4 0 −1/4 0
So, the optimal Solution we get without
𝑥1
The optimal Solution
𝑥1 = 0; 𝑥2 = 5 𝑎𝑛𝑑 𝑍𝑚𝑖𝑛= 4𝑥1+𝑥2= 4×0 + 5 = 5

88
𝑍𝑚𝑎𝑥 = 𝑥1+5𝑥2
3𝑥1+ 4𝑥2 ≤ 6
𝑥1+3𝑥2 ≥ 2
𝑥1, 𝑥2 ≥ 0

89
𝑥1+ 3𝑥2 − 𝑆2 + 𝐴1 = 2
𝑥1, 𝑥2, 𝑆1 , 𝑆2 , 𝐴1 ≥ 0
3𝑥1+4𝑥2 + 𝑆1 = 6
Add slack and Surplus variable to
convert constraint equation into equality
𝑍𝑚𝑎𝑥 = 𝑥1+5𝑥2
3𝑥1+ 4𝑥2 ≤ 6
𝑥1+3𝑥2 ≥ 2
𝑥1, 𝑥2 ≥ 0

90
𝑆1 = 6
𝐴1 = 2
𝑍𝑚𝑎𝑥= 𝑥1+ 5𝑥2 + 0𝑆1 + 0𝑆2 − MA1

91
Cj
1 5 0 0 −M
θ
𝑥1 𝑥2 S1 S2 A1 RHS (b)
0
−M
S1
A1
6
2
Zj
𝑍𝑚𝑎𝑥= 𝑥1+ 5𝑥2 + 0𝑆1 + 0𝑆2 − MA1

92
Cj
1 5 0 0 −M
θ
0
−M
S1
A1
3
1
4
3
1
0
0
−1
0
1
6
2
Zj
𝑍𝑚𝑎𝑥= 𝑥1+ 5𝑥2 + 0𝑆1 + 0𝑆2 − MA1

93
Cj
1 5 0 0 −M
θ
0
−M
S1
A1
3
1
4
3
1
0
0
−1
0
1
6
2
Zj
−M −3M 0 M −M
𝑍𝑚𝑎𝑥= 𝑥1+ 5𝑥2 + 0𝑆1 + 0𝑆2 − MA1

94
Cj
1 5 0 0 −M
θ
0
−M
S1
A1
3
1
4
3
1
0
0
−1
0
1
6
2
Zj
−M −3M 0 M −M
Cj − Zj
1+M 5+3M 0 −M 0
𝑍𝑚𝑎𝑥= 𝑥1+ 5𝑥2 + 0𝑆1 + 0𝑆2 − MA1

95
Cj
1 5 0 0 −M
θ
0
−M
S1
A1
3
1
4
3
1
0
0
−1
0
1
6
2
Zj
−M −3M 0 M −M
Cj − Zj
1+M 5+3M 0 −M 0
Max
positive
value
Key Column:
decides incoming
variable

96
Cj
1 5 0 0 −M
θ =
𝒃
𝒂
𝑥1 𝑥2 S1 S2 A1 RHS
(b)
0
−M
S1
A1
3
1
4
3
1
0
0
−1
0
1
6
2
6/4 = 3/2
2/3
Zj
−M −3M 0 M −M
Cj − Zj
1+M 5+3M 0 −M 0
Max
positive
value
Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Key
Element
Least
positive
Key
Row: decides
outgoing
variable
Key
Element

97
Cj
1 5 0 0
θ =
𝒃
𝒂
𝑥1 𝒙𝟐 S1 S2 RHS
(b)
0
5
S1
𝒙𝟐
3
1
4
3
1
0
0
−1
6
2
Zj
Cj − Zj

98
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S1
𝒙𝟐
3
1/3
4
1
1
0
0
−1/3
6
2/3
Zj
Cj − Zj
MKR

99
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S1
𝒙𝟐
3
1/3
4
1
1
0
0
−1/3
6
2/3
Zj
Cj − Zj
MKR
OR

100
Old
Row
− KCE × MKR = NR
3 − 4 × 1/3 = 5/3
4 − 4 × 1 = 0
1 − 4 × 0 = 1
0 − 4 × −1/3 = 4/3
6 − 4 × 2/3 = 10/3

101
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S1
𝒙𝟐
5/3
1/3
0
1
1
0
4/3
−1/3
10/3
2/3
Zj
5/3 5 0 −5/3
Cj − Zj
−2/3 0 0 5/3
MKR
NR
Max
positive
value Key Column:
decides incoming
variable

102
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S1
𝒙𝟐
5/3
1/3
0
1
1
0
4/3
−1/3
10/3
2/3
5/2
−
Zj
5/3 5 0 −5/3
Cj − Zj
−2/3 0 0 5/3
Max
positive
value Key Column:
decides incoming
variable
Least
positive
Key
Row: decides
outgoing
variable
Key
Element

103
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S2
𝒙𝟐
5/3
1/3
0
1
1
0
4/3
−1/3
10/3
2/3
5/2
−
Zj
5/3 5 0 −5/3
Cj − Zj
−2/3 0 0 5/3

104
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S2
𝒙𝟐
5/3
1/3
0
1
1
0
4/3
−1/3
10/3
2/3
5/2
−
Zj
5/3 5 0 −5/3
Cj − Zj
−2/3 0 0 5/3

105
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S2
𝒙𝟐
5/3
1/3
0
1
1
0
4/3
−1/3
10/3
2/3
Zj
Cj − Zj

106
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S2
𝒙𝟐
5/4
1/3
0
1
3/4
0
1
−1/3
5/2
2/3
Zj
Cj − Zj
MKR

107
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S2
𝒙𝟐
5/4
1/3
0
1
3/4
0
1
−1/3
5/2
2/3
Zj
Cj − Zj
MKR
OR

108
Old
Row
− KCE × MKR = NR
1/3 − −1/3 × 5/4 = 3/4
1 − −1/3 × 0 = 1
0 − −1/3 × 3/4 = 1/4
−1/3 − −1/3 × 1 = 0
2/3 − −1/3 × 5/2 = 3/2
2
3
+
1
3
×
5
2
=
2
3
+
5
6
=
4 + 5
6
=
9
6
=
3
2

109
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S2
𝒙𝟐
5/4
3/4
0
1
3/4
1/4
1
0
5/2
3/2
Zj
Cj − Zj
MKR
NR

110
Cj
1 5 0 0
θ =
𝒃
𝒂
(b)
0
5
S2
𝒙𝟐
5/4
3/4
0
1
3/4
1/4
1
0
5/2
3/2
Zj
15/4 5 5/4 0
Cj − Zj
−11/4 0 −5/4 0
MKR
NR
So, the optimal Solution we get without
𝑥1
𝑥1 = 0; 𝑥2 =
3
2
and 𝑍𝑚𝑎𝑥 = 𝑥1 + 5𝑥2 = 0 +
15
2
, 𝑍𝑚𝑎𝑥 =
15
2

Engineering, LPU
111

Engineering, LPU
112

Engineering, LPU
113

Engineering, LPU
114

Engineering, LPU
115
There is no region that satisfies both the
constraints.
This LP is infeasible
Refer: Slide no. 97

117
LPP Two Phase Method
LPP using Graphical
Method
LPP using iteration Method
≤
inequality
= and ≥
inequality
Simplex
Method
Big M
Method
Two Phase
Method

118
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 −1 to each of the artificial variables
𝑍𝑚𝑖𝑛 or 𝑍𝑚𝑎𝑥 . 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.

119
𝑍𝑚𝑖𝑛 = 𝑥1+𝑥2
2𝑥1+ 𝑥2 ≥ 4
𝑥1+7𝑥2 ≥ 7
𝑥1, 𝑥2 ≥ 0
LPP Two Phase Method – Prob 10

120
𝑥1+ 7𝑥2 − 𝑆2 + 𝐴2 = 7
𝑥1, 𝑥2, 𝑆1 , 𝑆2 , 𝐴1 , 𝐴2 ≥ 0
2𝑥1+𝑥2 − 𝑆1 + 𝐴1 = 4
Add Surplus variable and artificial
variable to convert constraint equation
into equality
𝑍𝑚𝑖𝑛 = 𝑥1+𝑥2
2𝑥1+ 𝑥2 ≥ 4
𝑥1+7𝑥2 ≥ 7
𝑥1, 𝑥2 ≥ 0
𝐴1 = 4
𝐴2 = 7
𝑍𝑚𝑖𝑛= 0 × 𝑥1+ 0 × 𝑥2 + 0 × 𝑆1 + 0 × 𝑆2 + 1 × A1 + 1 × A2
Create Auxiliary
function

121
Cj
0 0 0 0 1 1
θ
𝑥1 𝑥2 S1 S2 A1 A2 RHS (b)
1
1
A1
A2
4
7
Zj
𝑍𝑚𝑖𝑛= 0 × 𝑥1+ 0 × 𝑥2 + 0 × 𝑆1 + 0 × 𝑆2 + 1 × A1 + 1 × A2

122
Cj
0 0 0 0 1 1
θ
1
1
A1
A2
2
1
1
7
−1
0
0
−1
1
0
0
1
4
7
Zj
𝑍𝑚𝑖𝑛= 0 × 𝑥1+ 0 × 𝑥2 + 0 × 𝑆1 + 0 × 𝑆2 + 1 × A1 + 1 × A2

123
Cj
0 0 0 0 1 1
θ
1
1
A1
A2
2
1
1
7
−1
0
0
−1
1
0
0
1
4
7
Zj
3 8 −1 −1 1 1

124
Cj
0 0 0 0 1 1
θ
1
1
A1
A2
2
1
1
7
−1
0
0
−1
1
0
0
1
4
7
Zj
3 8 −1 −1 1 1
Zj−Cj
3 8 −1 −1 0 0
Max
positive
value Key Column:
decides incoming

125
Cj
0 0 0 0 1 1
θ=
𝒃
𝒂
1
1
A1
A2
2
1
1
7
−1
0
0
−1
1
0
0
1
4
7
4
1
Zj
3 8 −1 −1 1 1
Zj−Cj
3 8 −1 −1 0 0
Max
positive
value Key Column:
decides incoming
Least
positive
Key
Row: decides
outgoing
variable
Key
Element

126
Cj
0 0 0 0 1
θ=
𝒃
𝒂
1
0
A1
x2
2
1/7
1
1
−1
0
0
−1/7
1
0 1
Zj
Zj−Cj

127
Cj
0 0 0 0 1
θ=
𝒃
𝒂
1
0
A1
x2
2
1/7
1
1
−1
0
0
−1/7
1
0
4
1
Zj
Zj−Cj
MKR
OR

128
Old
Row
− KCE × MKR = NR
2 − 1 × 1/7 = 13/7
1 − 1 × 1 = 0
−1 − 1 × 0 = −1
0 − 1 × −1/7 = 1/7
1 − 1 × 0 = 1
4 − 1 × 1 = 3

129
Cj
0 0 0 0 1
θ=
𝒃
𝒂
1
0
A1
x2
13/7
1/7
0
1
−1
0
1/7
−1/7
1
0
3
1
Zj
Zj−Cj
MKR
NR

130
Cj
0 0 0 0 1
θ=
𝒃
𝒂
1
0
A1
x2
13/7
1/7
0
1
−1
0
1/7
−1/7
1
0
3
1
Zj
13/7 0 −1 1/7 1
Zj−Cj
13/7 0 −1 1/7 0
Max
positive
value Key Column:
decides incoming

131
Cj
0 0 0 0 1
θ=
𝒃
𝒂
1
0
A1
x2
13/7
1/7
0
1
−1
0
1/7
−1/7
1
0
3
1
21/13
7
Zj
13/7 0 −1 1/7 1
Zj−Cj
13/7 0 −1 1/7 0
Max
positive
value Key Column:
decides incoming

132
Cj
0 0 0 0 1
θ=
𝒃
𝒂
1
0
A1
x2
13/7
1/7
0
1
−1
0
1/7
−1/7
1
0
3
1
21/13
7
Zj
13/7 0 −1 1/7 1
Zj−Cj
13/7 0 −1 1/7 0
Max
positive
value Key Column:
decides incoming
Least
positive
Key
Element

133
Cj
0 0 0 0
θ=
𝒃
𝒂
𝒙𝟏 𝑥2 S1 S2 RHS (b)
0
0
𝒙𝟏
𝑥2
13/7
1/7
0
1
−1
0
1/7
−1/7
3
1
21/13
7
Zj
13/7 0 −1 1/7
Zj−Cj
13/7 0 −1 1/7

134
Cj
0 0 0 0
θ=
𝒃
𝒂
0
0
𝒙𝟏
𝑥2
1
1/7
0
1
−7/13
0
1/13
−1/7
21/13
1
7
Zj
Zj−Cj
MKR

135
Cj
0 0 0 0
θ=
𝒃
𝒂
0
0
𝒙𝟏
𝑥2
1
1/7
0
1
−7/13
0
1/13
−1/7
21/13
1
7
Zj
Zj−Cj
MKR
OR

136
Cj
0 0 0 0
θ=
𝒃
𝒂
1
1
𝒙𝟏
𝑥2
1
0
0
1
−7/13
1/13
1/13
−2/13
21/13
70/91
7
Zj
0 0 0 0
Zj−Cj
0 0 0 0
MKR
NR

137
Cj
1 1 0 0
θ=
𝒃
𝒂
1
1
𝒙𝟏
𝑥2
1
0
0
1
−7/13
1/13
1/13
−2/13
21/13
70/91
7
Zj
1 1 −6/13 −1/13
Zj−Cj
0 0 −6/13 −1/13
MKR
NR
𝑥1 =
21
13
; 𝑥2 =
70
90
=
10
13
and 𝑍𝑚𝑖𝑛 =
21
13
+
10
13
=
31
13

138
Cj
1 1 0 0
θ=
𝒃
𝒂
1
1
𝒙𝟏
𝑥2
1
0
0
1
−7/13
1/13
1/13
−2/13
21/13
70/91
7
Zj
1 1 −6/13 −1/13
Zj−Cj
0 0 −6/13 −1/13
MKR
NR
𝑥1 =
21
13
; 𝑥2 =
70
90
=
10
13
and 𝑍𝑚𝑖𝑛 =
21
13
+
10
13
=
31
13

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

Engineering, LPU
141
In graphical representation the bounded region is known as _________ region.
A. Solution
B. basic solution
C. feasible solution
D. optimal

Engineering, LPU
142
The optimal value of the objective function for the following L.P.P.
Max z = 4X1 + 3X2
subject to
X1 + X2 ≤ 50
X1 + 2X2 ≤ 80
2X1 + X2 ≥ 20
X1, X2 ≥ 0
is
(a) 200
(b) 330
(c) 420
(d) 500

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

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

Engineering, LPU
145

Engineering, LPU
146
https://www.aplustopper.com/section-
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Operation research unit 1: LPP Big M and Two Phase method

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Operation research unit 1: LPP Big M and Two Phase method