Big-M method or Penalty Method
Definition 1.3.1
= Add artificial variable only
≥ Subtract surplus variable + Add artificial variable
≤ Add slack variable only
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
66 / 349
Problem 1.3.2
Minimize z = 4x1 + x2 using penalty method.
Subject to the constrains
3x1 + x2 = 3
4x1 + 3x2 ≥ 6
x1 + 2x2 ≤ 4
x1 , x2 ≥ 0
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
67 / 349
Therefore the objective function is
min z = 4x1 + x2 + 0s1 + 0s2 + MA1 + MA2
Subject to the constrains
3x1 + x2 + A1 = 3
4x1 + 3x2 − s1 + A2 = 6
x1 + 2x2 + s2 = 4
x1 , x2 , s1 , s2 , A1 , A2 ≥ 0
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
68 / 349
Step 1: Initial simplex table
cBi
M
M
0
cj
BV
A1
A2
s2
zj
zj − cj
4
1
0
0 M M
Solution Min Ratio
x1
x2
s1 s2 A1 A2
3
1
0
0 1 0
3
4
3
−1 0 0 1
6
1
2
0
1 0 0
4
7M
4M
−M 0 M M
7M − 4 4M − 1 −M 0 0 0
To fill zj :
M(3) + M(4) + 0(1) = 7M M(0) + M(−1) + 0(0) = −M M(1)M(0) + 0(0)
M(1) + M(3) + 0(2) = 4M
M(0) + M(0) + 0(1) = 0
M(0)M(1) + 0(0)
Positive maximum in zj − cj , let us substitute M some least positive element 1,
then amoung {7M − 4, 4M − 1, −M, 0, 0, 0}, −7M + 4 is the most positive.
So, x1 is the pivot column.
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
69 / 349
cBi
M
M
0
cj
BV
A1
A2
s2
zj
zj − cj
4
1
0
0 M
Solution
x1
x2
s1 s2 A2
3
1
0
0 0
3
4
3
−1 0 1
6
1
2
0
1 0
4
7M
4M
M
0 M
7M − 4 4M − 1 −M 0 0
Ratio
3
3
6
4
4
1
=1
= 23
=4
To choose pivot row, the minimum positive value in ration column is 1. So, A1
is the pivot row.
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
70 / 349
cBi
M
M
0
cj
BV
A1
A2
s2
zj
zj − cj
4
1
0
0 M
Solution
x1
x2
s1 s2 A2
3
1
0
0 0
3
4
3
−1 0 1
6
1
2
0
1 0
4
7M
4M
−M 0 M
7M − 4 4M − 1 −M 0 0
Entering variable
Leaving variable
Pivot element
Leaving basis variable
Dr.A.Benevatho Jaison
Ratio
3
3
6
4
4
1
=1
= 23
=4
x1
A1
3
A1
MAT3002 - Applied Linear Algebra
January 25, 2022
71 / 349
To make pivot element into 1, we divide pivot row A1 by 3, then we have,
cBi
4
cj
BV
x1
A2
s2
4 1 0 0 M
Solution Ratio
x1 x2 s1 s2 A2
1 13 0 0 0
1
zj
zj − cj
Calculation:
Old value of A2
4 3 −1 0 1 6
(−)4 × New value of x1 4 34
0 0 0 4
5
0 3 −1 0 1 2
Old value of s2
1 2
0 1 0 4
1
(−)New value of x1
1 3
0 0 0 1
0 1 0 3
0 53
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
72 / 349
cBi
4
M
0
cj
BV
x1
A2
s2
4 1
0 0
x1 x2 s1 s2
1 13
0 0
5
0 3 −1 0
0 53
0 1
M
A1
1
3
−4
3
−1
3
M
Solution Ratio
A2
0
1
1
2
0
3
zj
zj − cj
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
73 / 349
cBi
4
M
0
cj
BV
x1
A2
s2
zj
zj − cj
4
x1
1
0
0
4
0
1
x2
4
3
1
3
0
0 M
Sol Ratio
s1 s2 A2
1
0
0 0
1
3
5
−1
0
1
2
3
5
0
1 0
3
3
+ 53 M −M 0 M
+ 53 M −M 0 0
1
5
5
4 5
4(1) + M(0) + 0(0) = 4;
4
+M
+0
= + M;
3
3
3
3 3
4(0) + M(−1) + 0(0) = −M
4(0) + M(0) + 0(1) = 0
4(0) + M(1) + 0(0) = M
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
74 / 349
cBi
4
M
0
cj
BV
x1
A2
s2
zj
zj − cj
4
x1
1
0
0
4
0
1
x2
4
3
1
3
0
0
Sol
s1 s2
1
0
0
1
3
5
−1
0
2
3
5
0
1
3
3
+ 53 M −M 0
+ 53 M −M 0
Ratio
6
5
9
5
3
= 1.2
= 1.8
Entering variable
x2
Leaving variable
A2
5
Pivot element
3
Leaving basis variable A2
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
75 / 349
cBi
4
1
0
cj
BV
x1
x2
s2
4 1
x1 x2
0
s1
0
Sol Ratio
s2
0
−3
5
0
1
6
5
zj
zj − cj
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
76 / 349
cBi
4
1
0
cj
BV
x1
x2
s2
4 1
x1 x2
0
s1
0
Sol Ratio
s2
0
−3
5
0
1
6
5
zj
zj − cj
Old value of x1
1 31 0 0
1
(−) 3 × New value of x2 0 13 −1
0
5
1
1 0 5 0
Old value of s2
0 35 0 1
(−) 53 × New value of x2 0 53 −1 0
0 0 1 1
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
1
2
5
3
5
3
2
1
January 25, 2022
77 / 349
cBi
4
1
0
cj
BV
x1
x2
s2
zj
zj − cj
4 1
x1 x2
1 0
0 1
0 0
4 1
0 0
0
s1
1
5
−3
5
1
1
5
1
5
0
Sol Ratio
s2
3
0
3
5
6
0
−2
5
1
1
1
0
0
In zj − cj , 15 is the most positive. So, s1 is the pivot column. In ratio column 1
is the least positive element. So, s2 is the pivot row.
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
78 / 349
cBi
4
1
0
cj
BV
x1
x2
s2
4 1 0 0
Sol Ratio
x1 x2 s1 s2
0
0
1
1
1
1
zj
zj − cj
Entering variable s1
Leaving variable s2
Pivot element
1
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
January 25, 2022
79 / 349
cBi
4
1
0
cj
BV
x1
x2
s1
zj
zj − cj
4 1 0
x1 x2 s1
1 0 0
0 1 0
0 0 1
4 1 0
0 0 0
0
s2
Sol Ratio
−1
5
3
5
2
5
9
5
1
1
−1
5
−1
5
Old value of x1
1 0
(−) 15 × New value of s1 0 0
1 0
Old value of x2
0 1
(+) 35 New value of s1
0 0
0 1
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
1
5
1
5
0
−3
5
3
5
0
0
1
5
−1
5
0
3
5
3
5
3
5
1
5
2
5
6
5
3
5
9
5
January 25, 2022
80 / 349
Since all Zj − Cj ≤ 0
Hence, optimal solution is arrived with value of variables as : x1 =
x2 = 95
Min Z = 17
5
Dr.A.Benevatho Jaison
MAT3002 - Applied Linear Algebra
2
5
January 25, 2022
81 / 349