This document summarizes the two phase simplex method for solving linear programming problems. In phase I, artificial variables are introduced to convert infeasible problems into feasible problems. The objective is to minimize the artificial variables. If the minimum is zero, the original problem is feasible and phase II begins. Phase II uses the original objective function and simplex method to find an optimal solution. An example problem is provided to illustrate the two phase method.

1. Operations Research
CHAPTER 07 - THE TWO PHASE SIMPLEX METHOD

2. The Two Phase Simplex Method
Phase I: We create an artificial objective function as the sum of all
the artificial variables, and we minimize this objective function using
the tableau simplex method.
If the minimum value of this artificial objective function is zero, then
this means that all the artificial variables have been reduced to zero,
and we have a basic feasible solution to the original problem, and
we proceed to phase II. If this minimum value is positive, then this
means that the original problem is infeasible, so we terminate.
Phase II: The final tableau of phase I becomes the first tableau of
phase II using the original objective function now. Use the tableau
method again to obtain an optimal solution.

3. Example 1
Min Z= -3x1 + x2 + x3
S.T. x1 – 2x2 + x3 <= 11
-4x1 + x2 +2x3 >= 3
-2x1 + x3 = 1

4. Solution
Write the problem in standard form
Min Z= -3x1 + x2 + x3
S.T. x1 – 2x2 + x3 + x4 = 11
-4x1 + x2 +2x3 - x5 = 3
-2x1 + x3 = 1

5. Solution
Write the problem in canonical form (add artificial variable to the 2nd
and to the 3rd constraints)
Min Z= -3x1 + x2 + x3
S.T. x1 – 2x2 + x3 + x4 = 11
-4x1 + x2 +2x3 - x5 + x6 = 3
-2x1 + x3 + x7 = 1

6. Solution
Phase 1: -Change Z to W and add to it the artificial variables
-W is always minimized
-Keep the constraints
Min W= x6 + x7
S.T. x1 – 2x2 + x3 + x4 = 11
-4x1 + x2 +2x3 - x5 + x6 = 3
-2x1 + x3 + x7 = 1

7. Solution
Iteration 1: Basic: x4=11, x6=3, x7=1. W=4
0 0 0 0 0 1 1
x1 x2 x3 x4 x5 x6 x7
0 x4 1 -2 1 1 0 0 0 11
1 x6 -4 1 2 0 -1 1 0 3
1 x7 -2 0 1 0 0 0 1 1
CJ 0 0 0

8. Solution
C1= 0 – (0,1,1) = 6
C2= 0 – (0,1,1) = -1
C3= 0 – (0,1,1) = -3
C5= 0 – (0,1,1) = 1
x3 is entering, x3= min{11,3/2,1}=1, x7 is leaving.
1
-4
-2
-2
1
0
1
2
1
0
-1
0

9. Solution
Iteration 2: Basic: x4=11-1=10, x6=3-2=1, x3=1. W=1
0 0 0 0 0 1 1
x1 x2 x3 x4 x5 x6 x7
0 x4 3 -2 0 1 0 0 -1 10
1 x6 0 1 0 0 -1 1 -2 1
0 x3 -2 0 1 0 0 0 1 1
CJ 0 0 0 0

10. Solution
C1= 0 – (0,1,0) = 0
C2= 0 – (0,1,0) = -1
C5= 0 – (0,1,0) = 1
x2 is entering, x2= min{-5,1,inf.}=1, x6 is leaving.
Basic: x4=12, x2=1, x3=1. W=0 [End of phase 1 because W is zero].
3
0
-2
-2
1
0
0
-1
0

11. Solution
Phase 2: -Change W to Z and remove the artificial variables
-The same last table but remove the artificial variables
columns

12. Solution
-3 1 1 0 0
x1 x2 x3 x4 x5
0 x4 3 -2 0 1 0 10
1 x2 0 1 0 0 -1 1
1 x3 -2 0 1 0 0 1
CJ 0 0 0

13. Solution
-3 1 1 0 0
x1 x2 x3 x4 x5
0 x4 3 0 0 1 -2 12
1 x2 0 1 0 0 -1 1
1 x3 -2 0 1 0 0 1
CJ 0 0 0

14. Solution
C1= -3 – (0,1,1) = -1
C5= 0 – (0,1,1) = 1
x1 is entering, x1= min{4,inf.,-1/2}=4, x4 is leaving.
Basic: x1=4, x2=1, x3=9. Z=-12+1+9=-2.
3
0
-2
-2
-1
0

15. Solution
-3 1 1 0 0
x1 x2 x3 x4 x5
-3 x1 1 0 0 1/3 -2/3 4
1 x2 0 1 0 0 -1 1
1 x3 0 0 1 2/3 -4/3 9
CJ 0 0 0

16. Solution
C4= 0 – (-3,1,1) = 1/3 > 0
C5= 0 – (-3,1,1) = 1/3 > 0
No entering variables, and the current solution is optimal.
Basic: x1=4, x2=1, x3=9. Z=-12+1+9=-2.
1/3
0
2/3
-2/3
-1
-4/3

17. Example 2
Max Z= x1 + 2x2 + x3
S.T. x1 + x2 + x3 = 7
2x1 – 5x2 +x3 >= 10
In standard form: Max Z= x1 + 2x2 + x3
S.T. x1 + x2 + x3 = 7
2x1 – 5x2 +x3 – x4 = 10
In chanocial form: Min W= x5 + x6
S.T. x1 + x2 + x3 + x5 = 7
2x1 – 5x2 +x3 – x4 + x6 = 10

18. Solution
Iteration 1: Basic: x5=7, x6=10, W=17
0 0 0 0 1 1
x1 x2 x3 x4 x5 x6
1 x5 1 1 1 0 1 0 7
1 x6 2 -5 1 -1 0 1 10
CJ 0 0

19. Solution
C1= 0 – (1,1) = -3
C2= 0 – (1,1) = 4
C3= 0 – (1,1) = -2
C4= 0 – (1,1) = 1
x1 is entering, x1= min{7,5}=5, x6 is leaving.
1
2
1
-5
1
1
0
-1

20. Solution
Iteration 2: Basic: x5=2, x1=5, W=2
0 0 0 0 1 1
x1 x2 x3 x4 x5 x6
1 x5 0 7/2 1/2 1/2 1 -1/2 2
0 x1 1 -5/2 1/2 -1/2 0 1/2 5
CJ 0 0 0

21. Solution
C2= 0 – (1,0) = -7/2
C3= 0 – (1,0) = -1/2
C4= 0 – (1,0) = -1/2
x2 is entering, x2= min{4/7,-2}=4/7, x5 is leaving.
7/2
-
5/2
1/2
1/2
1/2
-
1/2

22. Solution
Iteration 3: Basic: x2=4/7, x1=45/7, W=0 (Move to phase 2)
1 2 1 0
x1 x2 x3 x4
2 x2 0 7/2 1/2 1/2 2
1 x1 1 -5/2 1/2 -1/2 5
CJ 0 0

23. Solution
Iteration 3: Basic: x2=4/7, x1=45/7, W=0 (Move to phase 2)
1 2 1 0
x1 x2 x3 x4
2 x2 0 1 1/7 1/7 4/7
1 x1 1 0 6/7 -1/7 45/7
CJ 0 0 53/7

24. Solution
C3= 1 – (2,1) = -1/7
C4= 0 – (2,1) = -1/7
No entering variables, the current solution is optimal.
x2=4/7, x1=45/7, z= 53/7
1/7
6/7
1/7
-
1/7

25. Note
In chapter 9 there is example of solving problem with unrestricted
variables using two phase method