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1. Big-M Method
Mr. P.Praveen Babu
Assistant Professor
MREC(A)

2. Big-M Method
Example: I
 Use Penalty (Big-M) method to solve the following LP problem.
 Objective function:
 Minimize Z = 5 x1 + 3 x2
 Subject to the constraints:
 (i) 2x1 + 4x2 ≤ 12, (ii) 2x1 + 2x2 = 10,
 (iii) 5x1 + 2x2 ≥ 10.
 and x1 , x2 ≥ 0

3. Standard form
 Objective Function:
 Minimize Z = 5 x1 + 3 x2 + 0.S1 + 0.S2 +M A1 +M A2
 Subject to the constraints:
2x1 + 4x2 ≤ 12
2x1 + 4x2 + S1 = 12
2x1 + 2x2 = 10
2x1 + 2x2 + A1= 10
5x1 + 2x2 ≥ 10
5x1 + 2x2 – S2 + A2 = 10

4. Simplex Table
CB
Cj
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1 A2
Zj
Cj - Zj
Z = 5 x1 + 3 x2 + 0.S1 + 0.S2 +M A1 +M A2
5 3 0 0 M M
2x1 + 4x2 + S1 = 12
2 4 1 0 0 0 12
2x1 + 2x2 + A1= 10
2 2 0 0 1 0 10
5x1 + 2x2 – S2 + A2 = 10
5 2 0 1 0 1 10
1S
1A
2A
0
M
M

5. Find Zj values
CB
Cj 5 3 0 0 M M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1 A2
0 S1 2 4 1 0 0 0 12
M A1 2 2 0 0 1 0 10
M A2 5 2 0 -1 0 1 10
Zj
Cj - Zj
M7 M4 0 M M M M20

6. Find Cj - Zj values
CB
Cj 5 3 0 0 M M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1 A2
0 S1 2 4 1 0 0 0 12
M A1 2 2 0 0 1 0 10
M A2 5 2 0 -1 0 1 10
Zj 7M 4M 0 -M M M 20M
Cj - Zj
M75 M43 0 M 0 0

7. Test for Optimality
CB
Cj 5 3 0 0 M M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1 A2
0 S1 2 4 1 0 0 0 12
M A1 2 2 0 0 1 0 10
M A2 5 2 0 -1 0 1 10
Zj 7M 4M 0 -M M M 20M
Cj - Zj 5-7M 3-4M 0 M 0 0
Smallest Negative number

8. Key Column, Key row & key element
CB
Cj 5 3 0 0 M M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1 A2
0 S1 2 4 1 0 0 0 12
M A1 2 2 0 0 1 0 10
M A2 5 2 0 -1 0 1 10
Zj 7M 4M 0 -M M M 20M
Cj - Zj 5-7M 3-4M 0 M 0 0
6
2
12

5
2
10

2
5
10


9. Finding New Values
CB
Cj 5 3 0 0 M M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1 A2
0 S1 2 4 1 0 0 0 12 6
M A1 2 2 0 0 1 0 10 5
5 2 0 -1 0 1 10 2
Zj 7M 4M 0 -M M M 20M
Cj - Zj 5-7M 3-4M 0 M 0 0
1X5

10. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
0 S1 2 4 1 0 0 12
M A1 2 2 0 0 1 10
5 X1
Zj 7M 4M 0 -M M 20M
Cj - Zj 5-7M 3-4M 0 M 0
1 5
2
0 5
1
0 2

11. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
0 S1 2 4 1 0 0 12
M A1 2 2 0 0 1 10
M A2 5 2 0 -1 0 10
Zj 7M 4M 0 -M M 20M
Cj - Zj 5-7M 3-4M 0 M 0
elementKey
valuerowKeyCorrvaluecolumnKeyCorr
ValueOldValueNew
.
...*...
.. 

12. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
0 S1
M A1
5 X1 1 2/5 0 -1/5 0 2
Zj
Cj - Zj
0 5
16 1 5
2 0 8
0 5
6
0 5
2
1 6
5 2
5
6

M
0 1
5
2

M
M M610
0 1
5
6

M
0 1
5
2

M
0
Smallest Negative number

13. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
0 S1 0 16/5 1 2/5 0 8
M A1 0 6/5 0 2/5 1 6
5 X1 1 2/5 0 -1/5 0 2
Zj 5 0 M 10+6M
Cj - Zj 0 0 0
2
5
6

M
1
5
2

M
1
5
6

M
1
5
2

M
2
5
5
16
8 
5
5
6
6 
5
5
2
2 

14. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
0 16/5 1 2/5 0 8
M A1 0 6/5 0 2/5 1 6
5 X1 1 2/5 0 -1/5 0 2
Zj 5 0 M 10+6M
Cj - Zj 0 0 0
2
5
6

M
1
5
2

M
1
5
6

M
1
5
2

M
2X3

15. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
3 X2
M A1 0 6/5 0 2/5 1 6
5 X1 1 2/5 0 -1/5 0 2
Zj 5 0 M 10+6M
Cj - Zj 0 0 0
2
5
6

M
1
5
2

M
1
5
6

M
1
5
2

M
0 1
16
5
8
1
0 2
5

16. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
0 S1 0 16/5 1 2/5 0 8
M A1 0 6/5 0 2/5 1 6
5 X1 1 2/5 0 -1/5 0 2
Zj 5 0 M 10+6M
Cj - Zj 0 0 0
2
5
6

M
1
5
2

M
1
5
6

M
1
5
2

M
elementKey
valuerowKeyCorrvaluecolumnKeyCorr
ValueOldValueNew
.
...*...
.. 

17. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
3 X2
M A1
5 X1
Zj
Cj - Zj
0 1
16
5
8
1
0 2
5
0 0
8
3
4
1
1 3
1 0
8
1
 4
1
 0 1
5 3 16
5
8
3

M
8
7
4

M
M M3
2
25

0 0 16
5
8
3

M
8
7
4

M
0
20
12
4

18. CB
Cj 5 3 0 0 M
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2 A1
3 X2
5 X1
Zj
Cj - Zj
0 1
16
5
8
1
0 2
5
0 0
8
3
4
1
1 3
1 0
8
1
 4
1
 0 1
5 3 16
5
8
3

M
8
7
4

M
M M3
2
25

0 0 16
5
8
3

M
8
7
4

M
0
20
12
4
2S0

19. CB
Cj 5 3 0 0
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2
3 X2
5 X1
Zj
Cj - Zj
0 1
16
5
8
1
2
5
0 0
8
3
4
1
3
1 0
8
1
 4
1

1
5 3 16
5
8
3

M
8
7
4

M M3
2
25

0 0 16
5
8
3

M
8
7
4

M
20
12
4
2S0

20. CB
Cj 5 3 0 0
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2
3 X2
5 X1
Zj
Cj - Zj
0 1
16
5
8
1
2
5
0 0
2
3
 1 12
1 0
8
1
 4
1

1
5 3 16
5
8
3

M
8
7
4

M M3
2
25

0 0 16
5
8
3

M
8
7
4

M
2S0
elementKey
valuerowKeyCorrvaluecolumnKeyCorr
ValueOldValueNew
.
...*...
.. 

21. CB
Cj 5 3 0 0
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2
3 X2
5 X1
Zj
Cj - Zj
0 1
16
5
8
1
2
5
0 0
8
3
4
1
3
1 0
8
1
 4
1

1
5 3 16
5
8
3

M
8
7
4

M M3
2
25

0 0 16
5
8
3

M
8
7
4

M
20
12
4
2S0
elementKey
valuerowKeyCorrvaluecolumnKeyCorr
ValueOldValueNew
.
...*...
.. 

22. CB
Cj 5 3 0 0
Solution
Minimum
RatioBasic
Variables
X1 X2 S1 S2
3 X2
5 X1
Zj
Cj - Zj
0 1
2
1
0 1
0 0
2
3
 1 12
1 0
2
1
 0 4
5 3 1 0 23
0 0 1 0
2S0

23. Result
 Optimal Solution:
 X1 = 4, X2 = 1, S1 = 0, & S2 = 12.
 Min Z = 23.

24. Transportation Problem
 It is a special kind of LPP in which goods are
transformed from a set of sources to a set of
destinations subject to the supply and demand of the
sources and destination respectively, Such that the
total cost of transportation is minimized.

25. Types
 Balanced Transportation problem
Supply = Demand
 Unbalanced Transportation problem
Supply ≠ Demand

26. Procedure
 Phase - I : Finding the initial basic feasible solution.
 Phase – II : Finding Optimization.

27. Phase - I
 Northwest corner cell method.
 Least cost cell method.
 Vogule’s Approximation method.