This document provides an example of using the simplex method to solve a linear programming minimization problem. The problem involves determining the optimal amounts of two tonics (X and Y) a patient should purchase to minimize cost while meeting daily vitamin requirements. The solution involves setting up the problem as a system of inequalities, introducing surplus and artificial variables, and performing two iterations of the simplex method to arrive at the optimal solution.

1. Faculty of Economics and Business Administration
Lebanese University
Chapter 2: Simplex Method
(Minimization Case M-Method)
Dr. Kamel ATTAR
attar.kamel@gmail.com
Lecture #3 F Monday 8/Mar/2021 F

2. 2Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
H Minimization Case M-Method H
Example (¶)
A patient visits the doctor to get treatment for ill health. The doctor examines
the patient and advises him to consume at least 40 units of vitamin A and 50
units of vitamin B daily for a specified time period. He also advises the patient
that to get vitamin A and vitamin B he has to drink tonic X and tonic Y that
have both vitamin A and vitamin B in a proportion. One unit of tonic X consists
2 units of vitamin A and 3 units of vitamin B and one unit of tonic Y consists of
4 units of vitamin A and 2 units of vitamin B. These tonics are available in
medical shops at a cost of 3$ and 2.5$ per unit of X and Y respectively. Now
the problem of patient is how much of X and how much of Y is to be purchased
from the shop to minimise the total cost and at the same time he can get
required amounts of vitamins A and B.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

3. 3Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Solution
First we shall enter all the data in the form of a table.
Vitamin Tonic Requirement
X Y
A 2 4 40
B 3 2 50
Cost in $ 3 2.5
Let the patient purchase x1 units of X and x2 units of Y then the inequalities
are:
Min : Z = 3x1 + 2.5x2
Subject to 


2x1 + 4x2 ≥ 40
3x1 + 2x2 ≥ 50
x1, x2 ≥ 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

4. 4Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
As we do not know the values of x1 and x2 to transform the inequalities to
equalities we have to subtract a SURPLUS VARIABLE, generally represented
by S1, S2, · · · etc. If we do this then we obtain



2x1 + 4x2 − 1S1 = 40
3x1 + 2x2 − 1S2 = 50
x1, x2, S1, S2 ≥ 0
Now if we allocate value zero to x1 and x2 then S1 = −40. Which is against to
the rules of l.p.p. as every l.p.problem the values of variables must be positive.
Hence in minimization problem, we introduce one more Surplus variable,
known as ARTIFICIAL SURPLUS VARIABLE generally represented by A1, A2,
A3 · · · etc. Now by introducing artificial surplus variable, we can write



2x1 + 4x2 − 1S1 + 1A1 = 40
3x1 + 2x2 − 1S2 + 1A2 = 50
x1, x2, S1, S2, A1, A2 ≥ 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

5. 5Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
However, because the artificial variables are not part of the original LP
model, they are assigned a very high penalty in the objective function, thus
forcing them (eventually) to equal zero in the optimum solution. This will always
be the case if the problem has a feasible solution. Thus the cost coefficient of
A1 is represented by a very high value represented by M (M → +∞). As we
are introducing CAPITAL ‘M’, THIS METHOD IS KNOWN AS BIG ‘M’ METHOD.
Min : Z = 3x1 + 2.5x2 + 0S1 + 0S2 + MA1 + MA2



À 2x1 + 4x2 − 1S1 + 0S2 + 1A1 + 0A2 = 40
Á 3x1 + 2x2 + 0S1 − 1S2 + 0A1 + 1A2 = 50
x1, x2, S1, S2, A1, A2 ≥ 0
Now let’s write the objective function:
Z0
= Z − M
À + Á
= (3 − 5M)x1 + (2.5 − 6M)x2 + MS1 + MS2 + 0A1 + 0A2
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

6. 6Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
A1 40 2 4 − 1 0 1 0
A2 50 3 2 0 −1 0 1
Z 3 − 5M 5
2
− 6M M M 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

7. 7Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
A1 40 2 4 − 1 0 1 0
A2 50 3 2 0 −1 0 1
Z 3 − 5M 5
2
− 6M M M 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

8. 8Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
A1 40 2 4 − 1 0 1 0 40
4
= 10
A2 50 3 2 0 −1 0 1 50
2
= 25
Z 3 − 5M 5
2
− 6M M M 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

9. 9Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
A1 40 2 4 − 1 0 1 0 40
4
= 10
A2 50 3 2 0 −1 0 1 50
2
= 25
Z 3 − 5M 5
2
− 6M M M 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

10. 10Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
A1 40 2 4 − 1 0 1 0 40
4
= 10 R0
1 ↔ 1
4
R1
A2 50 3 2 0 −1 0 1 50
2
= 25
Z 3 − 5M 5
2
− 6M M M 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

11. 11Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
A1 40 2 4 − 1 0 1 0 40
4
= 10 R0
1 ↔ 1
4
R1
A2 50 3 2 0 −1 0 1 50
2
= 25
Z 3 − 5M 5
2
− 6M M M 0 0
x2 10 1
2
1 − 1
4
0 1
4
0
A2 50 3 2 0 −1 0 1
Z 3 − 5M 2.5 − 6M M M 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

12. 12Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
A1 40 2 4 − 1 0 1 0 40
4
= 10 R0
1 ↔ 1
4
R1
A2 50 3 2 0 −1 0 1 50
2
= 25
Z 3 − 5M 5
2
− 6M M M 0 0
x2 10 1
2
1 − 1
4
0 1
4
0
A2 50 3 2 0 −1 0 1 R2 → R2 − 2R0
1
Z 3 − 5M 2.5 − 6M M M 0 0 R3 → R3 −
5
2
− 6M
R0
1
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

13. 13Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
A1 40 2 4 − 1 0 1 0 40
4
= 10 R0
1 ↔ 1
4
R1
A2 50 3 2 0 −1 0 1 50
2
= 25
Z 3 − 5M 5
2
− 6M M M 0 0
x2 10 1
2
1 − 1
4
0 1
4
0
A2 50 3 2 0 −1 0 1 R2 → R2 − 2R0
1
Z 3 − 5M 2.5 − 6M M M 0 0 R3 → R3 −
5
2
− 6M
R0
1
x2 10 1
2
1 − 1
4
0 0
A2 30 2 0 1
2
−1 1
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

14. 14Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0
A2 30 2 0 1
2
− 1 1
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

15. 15Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0 10
0.5
= 20
A2 30 2 0 1
2
− 1 1 30
2
= 15
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

16. 16Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0 10
0.5
= 20
A2 30 2 0 1
2
− 1 1 30
2
= 15
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

17. 17Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0 10
0.5
= 20
A2 30 2 0 1
2
− 1 1 30
2
= 15 R0
2 ↔ 1
2
R2
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

18. 18Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0 10
0.5
= 20
A2 30 2 0 1
2
− 1 1 30
2
= 15 R0
2 ↔ 1
2
R2
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
x2 10 1
2
1 − 1
4
0
x1 15 1 0 1
4
− 1
2
Z 7
4
− 2M 0 5
8
− 3
2
M M
.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

19. 19Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0 10
0.5
= 20
A2 30 2 0 1
2
− 1 1 30
2
= 15 R0
2 ↔ 1
2
R2
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
x2 10 1
2
1 − 1
4
0 R1 → R1 − 1
2
R0
2
x1 15 1 0 1
4
− 1
2
Z 7
4
− 2M 0 5
8
− 3
2
M M R3 → R3 −
7
4
− 2M
R0
2
.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

20. 20Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0 10
0.5
= 20
A2 30 2 0 1
2
− 1 1 30
2
= 15 R0
2 ↔ 1
2
R2
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
x2 10 1
2
1 − 1
4
0 R1 → R1 − 1
2
R0
2
x1 15 1 0 1
4
− 1
2
Z 7
4
− 2M 0 5
8
− 3
2
M M R3 → R3 −
7
4
− 2M
R0
2
x2 2.5 0 1 − 3
8
1
4
x1 15 1 0 1
4
− 1
2
Z 0 0 3
16
7
8
.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

21. 21Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0 10
0.5
= 20
A2 30 2 0 1
2
− 1 1 30
2
= 15 R0
2 ↔ 1
2
R2
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
x2 10 1
2
1 − 1
4
0 R1 → R1 − 1
2
R0
2
x1 15 1 0 1
4
− 1
2
Z 7
4
− 2M 0 5
8
− 3
2
M M R3 → R3 −
7
4
− 2M
R0
2
x2 2.5 0 1 − 3
8
1
4
x1 15 1 0 1
4
− 1
2
Z 0 0 3
16
7
8
.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

22. 22Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
B. V. Cp. x1 x2 S1 S2 A1 A2 Repl. Ratio Row Operations
x2 10 1
2
1 − 1
4
0 0 10
0.5
= 20
A2 30 2 0 1
2
− 1 1 30
2
= 15 R0
2 ↔ 1
2
R2
Z 7
4
− 2M 0 5
8
− 3
2
M M 0
x2 10 1
2
1 − 1
4
0 R1 → R1 − 1
2
R0
2
x1 15 1 0 1
4
− 1
2
Z 7
4
− 2M 0 5
8
− 3
2
M M R3 → R3 −
7
4
− 2M
R0
2
x2 2.5 0 1 − 3
8
1
4
x1 15 1 0 1
4
− 1
2
Z 0 0 3
16
7
8
Hence, optimal solution is arrived with value of variables as x1 = 15 and x2 = 2.5.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

23. 23Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Example (·)
Find the optimal solution
Min : Z = 2x1 + 10x2 + 8x3







2x1 + 4x2 + 6x3 ≥ 24
4x2 + 2x3 ≥ 8
2x1 + 4x2 + 8x3 ≥ 4
x1, x2, x3 ≥ 0
Solution
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

24. 24Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Example (·)
Find the optimal solution
Min : Z = 2x1 + 10x2 + 8x3







2x1 + 4x2 + 6x3 ≥ 24
4x2 + 2x3 ≥ 8
2x1 + 4x2 + 8x3 ≥ 4
x1, x2, x3 ≥ 0
Solution
Min : Z = 2x1 + 10x2 + 8x3 + 0S1 + 0S2 + 0S3 + MA1 + MA2 + MA3
Subject to







2x1 + 4x2 + 6x3 − 1S1 + 0S2 + 0S3 + 1A1 + 0A2 + 0A3 = 24
0x1 + 4x2 + 2x3 + 0S1 − 1S2 + 0S3 + 0A1 + 1A2 + 0A3 = 8
2x1 + 4x2 + 8x3 + 0S1 + 0S2 − 1S3 + 0A1 + 0A2 + 1A3 = 4
x1, x2, x3, S1, S2, S3, A1, A2, A3 ≥ 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

25. 25Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
BV CP x1 x2 x3 S1 S2 S3 A1 A2 A3 M.R. Row Operations
A1 24 2 4 6 −1 0 0 1 0 0 24
6
= 4
A2 8 0 4 2 0 −1 0 0 1 0 8
2
= 4
A3 4 2 4 8 0 0 −1 0 0 1 4
8
= 1/2 R0
3 ↔ 1
8
R3
Z 2 − 4M 10 − 12M 8 − 16M M M M 0 0 0 R4 − (8 − 16M)R0
3
A1 24 2 4 6 −1 0 0 1 0 24
6
= 4 R1 − 6R0
3
A2 8 0 4 2 0 −1 0 0 1 8
2
= 4 R2 − 2R0
3
x3 1/2 1/4 1/2 1 0 0 −1/8 0 0
Z 2 − 4M 10 − 12M 8 − 16M M M M 0 0 R4 − (8 − 16M)R0
3
A1 21 1/2 1 0 −1 0 3/4 1 0
A2 7 −1/2 3 0 0 −1 1/4 0 1
x3 1/2 1/4 1/2 1 0 0 −1/8 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

26. 26Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Second - iteration
BV CP x1 x2 x3 S1 S2 S3 A1 A2 A3 M.R. Row Operations
A1 21 1/2 1 0 −1 0 3/4 1 0 21
1
= 21
A2 7 −1/2 3 0 0 −1 1/4 0 1 7
3
x3 1/2 1/4 1/2 1 0 0 −1/8 0 0
1/2
1/2
= 1 R0
3 ↔ 2R3
Z 0 6 − 4M 0 M M 1 − M 0 0
A1 21 1/2 1 0 −1 0 3/4 1 0 21
1
= 21 R1 − R0
3
A2 7 −1/2 3 0 0 −1 1/4 0 1 7
3
R2 − 3R0
3
x2 1 1/2 1 2 0 0 −1/4 0 0
Z 0 6 − 4M 0 M M 1 − M 0 0 R4 − (6 − 4M)R0
3
Z 0 6 − 4M 0 M M 1 − M 0 0
A1 20 0 0 −2 −1 0 1 1 0
A2 4 −2 0 −6 0 −1 1 0 1
x2 1 1/2 1 2 0 0 −1/4 0 0
Z −3 + 2M 0 −12 + 8M M M 5
2
− 2M 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

27. 27Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Third - iteration
BV CP x1 x2 x3 S1 S2 S3 A1 A2 A3 M.R. Row Operations
A1 20 0 0 −2 −1 0 1 1 0 20
1
= 20 R1 − R3
A2 4 −2 0 −6 0 −1 1 0 1 4
1
= 4
x2 1 1/2 1 2 0 0 −1/4 0 0 · · · R3 + 1
4
R2
Z −3 + 2M 0 −12 + 8M M M 5
2
− 2M 0 0 R4 −
5
2
− 2M
R3
A1 16 2 0 4 −1 1 0 1
S3 4 −2 0 −6 0 −1 1 0
x2 2 0 1 1/2 0 −1/4 0 0
Z 2 − 2M 0 3 − 4M M 5
2
− M 0 0
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

28. 28Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Forth - iteration
BV CP x1 x2 x3 S1 S2 S3 A1 A2 A3 M.R. Row Operations
A1 16 2 0 4 −1 1 0 1 16
4
= 4 R0
1 ↔ 1
4
R1
S3 4 −2 0 −6 0 −1 1 0 · · ·
x2 2 0 1 1/2 0 −1/4 0 0 2
1/2
= 4
Z 2 − 2M 0 3 − 4M M 5
2
− M 0 0
x3 4 1/2 0 1 −1/4 1/4 0 16
4
= 4
S3 4 −2 0 −6 0 −1 1 0 · · · R2 → R2 + 6R0
1
x2 2 0 1 1/2 0 −1/4 0 0 2
1/2
= 4 R3 → R3 − 1
2
R0
1
Z 2 − 2M 0 3 − 4M M 5
2
− M 0 0 R4 → R4 − (3 − 4M) R0
1
x3 4 1/2 0 1 −1/4 1/4 0
S3 28 1 0 0 −3/2 1/2 1
x2 0 −1/4 1 0 1/8 −3/8 0
Z 1/2 0 0 3/4 7/4 0
Hence, optimal solution is arrived with value of variables as :
x1 = 0, x2 = 0, x3 = 4 and Z = 32
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

29. 29Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Example (¸)
Find the optimal solution for
Min : Z = 20T + 8C
Subject to 






4T + 2C ≤ 60 Modeling
2T + 4C ≤ 48 Finishing
T ≥ 2
C ≥ 4
Here T indicate the unit number of Table and C number unit of Chair.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

30. 30Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Solution
The standard form
Min : Z = 20T + 8C + 0S1 + 0S2 + 0S3 + 0S4 + MA1 + MA2
Subject to









À 4T + 2C + 1S1 + 0S2 + 0S3 + 0S4 + 0A1 + 0A2 = 60
Á 2T + 4C + 0S1 + 1S2 + 0S3 + 0S4 + 0A1 + 0A2 = 48
 T + 0C + 0S1 + 0S2 − 1S3 + 0S4 + 1A1 + 0A2 = 2
à 0T + C + 0S1 + 0S2 + 0S3 − 1S4 + 0A1 + 1A2 = 4
Let us enter the data in the Initial table of Simplex method. We have
Min : Z0
= Z − M(Â + Ã)
= T(20 − M) + C(8 − M) + MS3 + MS4
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

31. 31Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. T C S1 S2 S3 S4 A1 A2 Repl. Ratio Row Operations
S1 60 4 2 1 0 0 0 0 0 60
2
= 30 R1 → R1 − 2R4
S2 48 2 4 0 1 0 0 0 0 48
4
= 12 R2 → R2 − 4R4
A1 2 1 0 0 0 −1 0 1 0 2
0
= ∞
A2 4 0 1 0 0 0 −1 0 1 4
1
= 4
Z 20 − M 8 − M 0 0 M M 0 0 R5 → R5 − (8 − M)R4
Second - iteration
B. V. Cp. T C S1 S2 S3 S4 A1 A2 Repl. Ratio Row Operations
S1 52 4 0 1 0 0 2 0 −2 52
4
= 13 R1 → R1 − 4R3
S2 32 2 0 0 1 0 4 0 −4 32
3
= 16 R2 → R2 − 2R3
A1 2 1 0 0 0 −1 0 1 0 2
1
= 2
C 4 0 1 0 0 0 −1 0 1 4
0
= ∞
Z 20 − M 0 0 0 M 8 0 M − 8 R5 → R5 − (20 − M)R3
S1 44 0 0 1 0 4 2 −4 −2
S2 28 0 0 0 1 2 4 −2 −4
T 2 1 0 0 0 −1 0 1 0
C 4 0 1 0 0 0 −1 0 1
Z 0 0 0 0 20 8 M − 20 M − 8
We reach the optimal solution. Then the answer is T = 2 and C = 4.
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

32. 32Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Example (¹)
Find the optimal solution for
Min : Z = 2x1 − 3x2 − 4x3 Subject to







x1 + x2 + x3 ≤ 30
2x1 + x2 + 3x3 ≥ 60
−x1 + x2 − 2x3 = −20
x1, x2, x3 ≥ 0
Solution
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

33. 33Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
Example (¹)
Find the optimal solution for
Min : Z = 2x1 − 3x2 − 4x3 Subject to







x1 + x2 + x3 ≤ 30
2x1 + x2 + 3x3 ≥ 60
−x1 + x2 − 2x3 = −20
x1, x2, x3 ≥ 0
Solution
Min : Z = 2x1 − 3x2 − 4x3 + 0S1 + 0S2 + MA1 + MA2









À x1 + x2 + x3 + 1S1 + 0S2 + 0A1 + 0A2 = 30
Á 2x1 + x2 + 3x3 + 0S1 − 1S2 + 1A1 + 0A2 = 60
 x1 − x2 + 2x3 + 0S1 + 0S2 + 0A1 + 1A2 = 20
x1, x2, x3, S1, S2, A1, A2 ≥ 0
Min : Z0
= Z − M(Á + Â)
= (2 − M)x1 − 3x2 − (4 + 5M)x3 + 0S1 + MS2 + 0A1 + 0A2
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

34. 34Ú35
Minimization Case M-Method Example ¶
Example ·
Example ¸
Example ¹
First - iteration
B. V. Cp. x1 x2 x3 S1 S2 A1 A2 Repl. Ratio Row Operations
S1 30 1 1 1 1 0 0 0 30
1
= 30 R1 − R0
3
A1 60 2 1 3 0 −1 1 0 60
3
= 20 R2 − 3R0
3
A2 20
↓
1
↓
−1
↓
2
↓
0
↓
0
↓
0
↓
1
↓
20
2
= 10 R0
3 ↔ 1
2
R3
10 1/2 −1/2 1 0 0 0 1/2 New third row
Z 2 − 3M −3 −4 − 5M 0 M 0 0 R4 + (4 + 5M)R0
3
Second - iteration
B. V. Cp. x1 x2 x3 S1 S2 A1 A2 Repl. Ratio Row Operations
S1 20 1
2
3
2
0 1 0 0 − 1
2
20
3/2
= 13.3 R1 − 3
2
R0
2
A1 30
↓
1/2
↓
5/2
↓
0
↓
0
↓
−1
↓
1
↓
−3/2
↓
30
5/2
= 12 R0
2 ↔ 5
2
R2
12 1/5 1 0 0 −2/5 2/5 −3/5 New second row
x3 10 1
2
− 1
2
1 0 0 0 1
2
10
1/2
= 20 R3 + 1
2
R0
2
Z 4 − 1/2M −5 − 5/2M 0 0 M 0 2 + 5/2M R5 + (5 + 5/2M)R0
2
Third - iteration
B. V. Cp. x1 x2 x3 S1 S2 A1 A2 Repl. Ratio Row Operations
S1 2
↓
1/5
↓
0
↓
0
↓
1
↓
3/5
↓
−3/5
↓
2/5
↓
2
3/5
= 10
3
R0
1 ↔ 5
3
R1
10/3 1/3 0 0 5/3 1 −1 2/3 New first row
x2 12 1/5 1 0 0 −2/5 2/5 −3/5 12
2/5
= 30 R2 + 2
5
R0
1
x3 16 3/5 0 1 0 −1/5 1/5 1/5 16
1/5
= 80 R3 + 1
5
R0
1
Z 5 0 0 0 −2 M + 2 −1 + M R4 + 2R0
1
S2 10/3 1/3 0 0 5/3 1 −1 2/3
x2 40/3 1/3 1 0 2/3 0 0 −1/3
x3 50/3 10/15 0 0 1/5 0 0 7/25
Z 14/3 0 0 4/3 0 M 1/3 + M
Dr. Kamel ATTAR | Chapter 2: Simplex Method | Minimization

35. Thank you! Questions?