The document describes the two phase simplex method for solving linear programming problems (LPP) with greater than or equal to constraints. In phase I, an auxiliary LPP is constructed by introducing artificial variables and slack/surplus variables. The objective is to minimize the artificial variables using the simplex method. If an artificial variable is positive or zero in the optimal basis, phase II is executed to solve the original LPP. Phase I provides a starting feasible solution for phase II to obtain the actual optimal solution.

2. Simplex Method
When decision variables are more than 2, we always use
Simplex Method
Slack Variable: Variable added to a £ constraint to
convert it to an equation (=).
A slack variable represents unused resources
A slack variable contributes nothing to the objective
function value.
Surplus Variable: Variable subtracted from a ³
constraint to convert it to an equation (=).
A surplus variable represents an excess above a
constraint requirement level.
Surplus variables contribute nothing to the calculated
value of the objective function.

3. Cont….
Basic Solution(BS) : This solution is obtained by setting
any n variables (among m+n variables) equal to zero and
solving for remaining m variables, provided the
determinant of the coefficients of these variables is non-zero.
Such m variables are called basic variables and
remaining n zero valued variables are called non basic
variables.
Basic Feasible Solution(BFS) : It is a basic solution
which also satisfies the non negativity restrictions.

4. Cont…..
BFS are of two types:
Degenerate BFS: If one or more basic
variables are zero.
Non-Degenerate BFS: All basic variables are
non-zero.
Optimal BFS: BFS which optimizes the
objective function.

5. Example
Max. Z = 13x1+11x2
Subject to constraints:
4x1+5x2 << 1550000
55xx1++33xx22 << 1557755
xx1++22xx22 << 442200
xx1,, xx22 > 0

6. Solution :
Step 1: Convert all the inequality constraints into equalities
by the use of slack variables.
Let S1, S2 , S3 be three slack variables.
Introducing these slack variables into the inequality constraints
and rewriting the objective function such that all variables are
on the left-hand side of the equation. Model can rewritten as:
Z - 13x1 -11x2 = 0
Subject to constraints:
4x1+5x2 + S1 = 1550000
55xx1++33xx22 ++SS22== 1557755
xx1++22xx22 ++SS33 == 442200
xx1,, xx22,, SS1,, SS22,, SS33 > 0

7. Cont…
Step II: Find the Initial BFS.
One Feasible solution that satisfies all the
constraints is: x1= 0, x2= 0, S1= 1500,
S2= 1575, S3= 420 and Z=0.
Now, S1, S2, S3 are Basic variables.
Step III: Set up an initial table as:

8. Cont…
Row
NO.
Basic
Variable
Coefficients of: Sol. Rati
Z x o 1 x2 S1 S2 S3
A1 Z 1 -13 -11 0 0 0 0
B1 S1 0 4 5 1 0 0 1500 375
C1 S2 0 5 3 0 1 0 1575 315
D1 S3 0 1 2 0 0 1 420 420
Step IV: a) Choose the most negative number from row A1(i.e Z row). Therefore,
x1 is a entering variable.
b) Calculate Ratio = Sol col. / x1 col. (x1 > 0)
c) Choose minimum Ratio. That variable(i.e S2) is a departing
variable.

9. Cont….
Step V: x1 becomes basic variable and S2 becomes non basic
variable. New table is:
Row
NO.
Basic
Varia
ble
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
A1 Z 1 0 -16/5 0 13/5 0 4095
B1 S1 0 0 13/5 1 -4/5 0 240 92.3
C1 x1 0 1 3/5 0 1/5 0 315 525
D1 S3 0 0 7/5 0 -1/5 1 105 75

10. Cont…
Next Table is :
Row
NO.
Basic
Varia
ble
Coefficients of: Sol.
Z x1 x2 S1 S2 S3
A1 Z 1 0 0 0 15/7 16/7 4335
B1 S1 0 0 0 1 -3/7 -13/7 45
C1 x1 0 1 0 0 2/7 -3/7 270
D1 x2 0 0 1 0 -1/7 5/7 75
Optimal Solution is : x1= 270, x2= 75, Z= 4335

11. Example
Max. Z = 3x1+5x2+4x3
Subject to constraints:
2x1+3x2 << 88
22xx22++55xx33 << 100
33xx1++22xx22++44xx33 << 155
xx1,, xx22,, xx33 > 0

12. Cont…
Let S1, S2, S3 be the three slack variables.
Modified form is:
Z - 3x1-5x2-4x3 =0
2x1+3x2 +S1= 88
22xx22++55xx33 ++SS22== 100
33xx1++22xx22++44xx33++SS33== 155
xx1,, xx22,, xx33,, S1, SS22,, SS33 > 0
Initial BFS is : x1= 0, x2= 0, x3=0, S1= 8,
S2= 10, S3= 15 and Z=0.

13. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 x3 S1 S2 S3
Z 1 -3 -5 -4 0 0 0 0
S1 0 2 3 0 1 0 0 8 8/3
S2 0 0 2 5 0 1 0 10 5
S3 0 3 2 4 0 0 1 15 15/2
Therefore, x2 is the entering variable and S1 is the
departing variable.

14. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 x3 S1 S2 S3
Z 1 1/3 0 -4 5/3 0 0 40/3
x2 0 2/3 1 0 1/3 0 0 8/3 -
S2 0 -4/3 0 5 -2/3 1 0 14/3 14/15
S3 0 5/3 0 4 -2/3 0 1 29/3 29/12
Therefore, x3 is the entering variable and S2 is the
departing variable.

15. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 x3 S1 S2 S3
Z 1 -11/15 0 0 17/15 4/5 0 256/15
x2 0 2/3 1 0 1/3 0 0 8/3 4
x3 0 -4/15 0 1 -2/15 1/5 0 14/15 -
S3 0 41/15 0 0 2/15 -4/5 1 89/15 89/41
Therefore, x1 is the entering variable and S3 is the
departing variable.

16. Cont…
Basic
Variable
Coefficients of: Sol.
Z x1 x2 x3 S1 S2 S3
Z 1 0 0 0 45/41 24/41 11/41 765/41
x2 0 0 1 0 15/41 8/41 -10/41 50/41
x3 0 0 0 1 -6/41 5/41 4/41 62/41
x1 0 1 0 0 -2/41 -12/41 15/41 89/41
Optimal Solution is : x1= 89/41, x2= 50/41, x3=62/41, Z= 765/41

17. Example
Min.. Z = x1 - 3x2 + 2x3
Subject to constraints:
3x1 - x2 + 3x3 << 7
--22xx11 ++ 44xx22 << 1122
--44xx11 ++ 33xx22 ++ 88xx33 << 1100
xx11,, xx22,, xx33 > 0

18. Cont…
Convert the problem into maximization problem
Max.. Z’ = -x1 + 3x2 - 2x3 where Z’= -Z
Subject to constraints:
3x1 - x2 + 3x3 << 7
--22xx11 ++ 44xx22 << 1122
--44xx11 ++ 33xx22 ++ 88xx33 << 1100
xx11,, xx22,, xx33 > 0

19. Cont…
Let S1, S2 and S3 be three slack variables.
Modified form is:
Z’ + x1 - 3x2 + 2x3 = 0
3x1 - x2 + 3x3 +S1 = 7
--22xx11 ++ 44xx22 ++ SS22 == 1122
--44xx11 ++ 33xx22 ++ 88xx33 ++SS33 == 1100
xx11,, xx22,, xx33 > 0
Initial BFS is : x1= 0, x2= 0, x3=0, S1= 7, S2= 12, S3 = 10
and Z=0.

20. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 1 -3 2 0 0 0 0
S1 0 3 -1 3 1 0 0 7 -
S2 0 -2 4 0 0 1 0 12 3
S3 0 -4 3 8 0 0 1 10 10/3
Therefore, x2 is the entering variable and S2 is the
departing variable.

21. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 -1/2 0 2 0 3/4 0 9
S1 0 5/2 0 3 1 1/4 0 10 4
x2 0 -1/2 1 0 0 1/4 0 3 -
S3 0 -5/2 0 8 0 -3/4 1 1 -
Therefore, x1 is the entering variable and S1 is the
departing variable.

22. Cont…
Basic
Variable
Coefficients of: Sol.
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 0 0 13/5 1/5 8/10 0 11
x1 0 1 0 6/5 2/5 1/10 0 4
x2 0 0 1 3/5 1/5 3/10 0 5
S3 0 0 0 11 1 -1/2 1 11
Optimal Solution is : x1= 4, x2= 5, x3= 0,
Z’ = 11 Z = -11

23. Example
Max.. Z = 3x1 + 4x2
Subject to constraints:
x1 - x2 << 1
--xx11 ++ xx22 << 22
xx11,, xx22 > 0

24. Cont…
Let S1 and S2 be two slack variables
.
Modified form is:
Z -3x1 - 4x2 = 0
x1 - x2 +S1 = 1
--xx11 ++ xx22 ++SS22 == 22
xx11,, xx22,, SS11,, SS22 > 0
Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2
and Z=0.

25. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -3 -4 0 0 0
S1 0 1 -1 1 0 1 -
S2 0 -1 1 0 1 2 2
Therefore, x2 is the entering variable and S2 is the
departing variable.

26. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -7 0 0 4 8
S1 0 0 0 1 1 3 -
x2 0 -1 1 0 1 2 -
x1 is the entering variable, but as in x1 column every no. is less
than equal to zero, ratio cannot be calculated. Therefore given
problem is having a unbounded solution.

28. Introduction
LPP, in which constraints may also have > and = signs, we
introduce a new type of variable , called the artificial
variable. These variables are fictitious and cannot
have any physical meaning. Two Phase Simplex
Method is used to solve a problem in which some
artificial variables are involved. The solution is
obtained in two phases.

29. Example
Min.. Z = 15/2 x1 - 3x2
Subject to constraints:
3x1 - x2 - x3 > 33
xx11 -- xx22 ++ xx33 > 2
xx11,, xx22,, xx33 > 0

30. Cont…
Convert the objective function into the maximization form
Max. Z’ = -15/2 x1 + 3x2 where Z’= -Z
Subject to constraints:
3x1 - x2 - x3 > 33
xx11 -- xx22 ++ xx33 > 2
xx11,, xx22,, xx33 > 0

31. Cont…
Introduce surplus variables S1 and S2, and artificial variables
a1 and a2
Modified form is :
Z’ + 15/2 x1 - 3x2 = 0
Subject to constraints:
3x1 - x2 - x3 –S1 + a1 = 33
xx11 -- xx22 ++ xx33 ––SS22 ++ aa22 == 2
xx11,, xx22,, xx33 ,, S1, SS22,, a1, aa22 > 0

32. Cont…
Phase I : Simplex method is applied to a specially constructed
Auxiliary LPP leading to a final simplex table containing a BFS
to the original problem.
•Step 1: Assign a cost –1 to each artificial variable and a cost
0 to all other variables in the objective function.
•Step 2: Construct the auxiliary LPP in which the new
objective function Z* is to be maximized subject to the given
set of constraints.

33. Cont…
Auxiliary LPP is:
Max. Z* = -a1 –a2
Z* + a1 + a2 = 0
Subject to constraints:
3x1 - x2 - x3 –S1 + a1 = 33
xx11 -- xx22 ++ xx33 ––SS22 ++ aa22 == 2
xx11,, xx22,, xx33 ,, S1, SS22,, a1, aa22 > 0
Initial solution is a1 = 3, a2 = 2 and Z* = 0

34. Cont…
•Step 3: Solve the auxiliary problem by simplex method until either
of the following three possibilities arise:
•Max Z* < 0 and at least one artificial variable appear in the
optimum basis at a positive level. In this case given problem
does not have any feasible solution.
•Max Z* = 0 and at least one artificial variable appear in the
optimum basis at a zero level. In this case proceed to Phase II.
•Max Z* = 0 and no artificial variable appear in the optimum
basis. In this case also proceed to Phase II.

35. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 x3 S1 S2 a1 a2
Z* 1 0 0 0 0 0 1 1 0
a1 0 3 -1 -1 -1 0 1 0 3
a2 0 1 -1 1 0 -1 0 1 2
This table is not feasible as a1 and a2 has non zero coefficients in
Z* row. Therefore next step is to make the table feasible.

36. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z* x1 x2 x3 S1 S2 a1 a2
Z* 1 -4 2 0 1 1 0 0 -5
a1 0 3 -1 -1 -1 0 1 0 3 1
a2 0 1 -1 1 0 -1 0 1 2 2
Therefore, x1 is the entering variable and a1 is the
departing variable.

37. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z* x1 x2 x3 S1 S2 a2
Z* 1 0 2/3 -4/3 -1/3 1 0 -1
x1 0 1 -1/3 -1/3 -1/3 0 0 1 -
a2 0 0 -2/3 4/3 1/3 -1 1 1 3/4
Therefore, x3 is the entering variable and a2 is the
departing variable.

38. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 x3 S1 S2
Z* 1 0 0 0 0 0 0
x1 0 1 -1/2 0 -1/4 -1/4 5/4
x3 0 0 -1/2 1 1/4 -3/4 3/4
As there is no variable to be entered in the basis, this table is
optimum for Phase I. In this table Max. Z* = 0 and no artificial
variable appears in the optimum basis, therefore we can proceed
to Phase II.

39. Cont…
Phase II: The artificial variables which are non basic at the end of
Phase I are removed from the table and as well as from the
objective function and constraints. Now assign the actual costs to
the variables in the Objective function. That is, Simplex method is
applied to the modified simplex table obtained at the Phase I.
Basic
Variable
Coefficients of: Sol.
Z’ x1 x2 x3 S1 S2
Z’ 1 15/2 -3 0 0 0 0
x1 0 1 -1/2 0 -1/4 -1/4 5/4
x3 0 0 -1/2 1 1/4 -3/4 3/4
Again this table is not feasible as basic variable x1 has a non zero
coefficient in Z’ row. So make the table feasible.

40. Cont…
Basic
Variable
Coefficients of: Sol.
Z’ x1 x2 x3 S1 S2
Z’ 1 0 3/4 0 15/8 15/8 -75/8
x1 0 1 -1/2 0 -1/4 -1/4 5/4
x3 0 0 -1/2 1 1/4 -3/4 3/4
Optimal Solution is : x1= 5/4, x2= 0, x3= 3/4,
Z’ = -75/8 Z = 75/8

41. Example
Min.. Z = x1 - 2x2 –3x3
Subject to constraints:
-2x1 + x2 + 3x3 = 22
22xx11 ++ 33xx22 ++ 44xx33 == 1
xx11,, xx22,, xx33 > 0

42. Cont…
Phase I:
Introducing artificial variables a1 and a2
Auxiliary LPP is: Max. Z* = -a1 - a2
Z* + a1 + a2 = 0
Subject to constraints:
-2x1 + x2 + 3x3 + a1 = 22
22xx11 ++ 33xx22 ++ 44xx33 ++ aa22 == 1
xx11,, xx22,, xx33 > 0

43. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 x3 a1 a2
Z* 1 0 0 0 1 1 0
a1 0 -2 1 3 1 0 2
a2 0 2 3 4 0 1 1
This table is not feasible as a1 and a2 has non zero coefficients in
Z* row. Therefore next step is to make the table feasible.

44. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z* x1 x2 x3 a1 a2
Z* 1 0 -4 -7 0 0 -3
a1 0 -2 1 3 1 0 2 2/3
a2 0 2 3 4 0 1 1 1/4
Therefore, x3 is the entering variable and a2 is the
departing variable.

45. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 x3 a1
Z* 1 7/4 5/4 0 0 -5/4
a1 0 -7/2 -5/4 0 1 5/4
x3 0 1/2 3/4 1 0 1/4
As there is no variable to be entered in the basis, therefore
Phase I ends here. But one artificial variable is present in the
basis and Z* < 0. Therefore we cannot proceed to Phase II.
Given problem is having a non-feasible solution.

46. Example
Min.. Z = 4x1 + x2
Subject to constraints:
3x1 + x2 = 33
44xx11 ++ 33xx22 > 6
x1 +2x2 << 44
xx11,, xx22 > 0

47. Cont…
Phase I:
Introducing artificial variable a1 and a2, surplus variable S1 and
slack variable S2
Auxiliary LPP is:
Max. Z* = -a1 - a2
Z* + a1 + a2 = 0
Subject to constraints:
3x1 + x2 +a1 = 33
44xx11 ++ 33xx22 ––SS11 ++ aa22 = 6
x1 +2x2 +S2 = 44

48. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 S1 S2 a1 a2
Z* 1 0 0 0 0 1 1 0
a1 0 3 1 0 0 1 0 3
a2 0 4 3 -1 0 0 1 6
S2 0 1 2 0 1 0 0 4
This table is not feasible as a1 and a2 has non zero coefficients in
Z* row. Therefore next step is to make the table feasible.

49. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z* x1 x2 S1 S2 a1 a2
Z* 1 -7 -4 1 0 0 0 -9
a1 0 3 1 0 0 1 0 3 1
a2 0 4 3 -1 0 0 1 6 3/2
S2 0 1 2 0 1 0 0 4 4
Therefore, x1 is the entering variable and a1 is the
departing variable.

50. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z* x1 x2 S1 S2 a2
Z* 1 0 -5/3 1 0 0 -2
x1 0 1 1/3 0 0 0 1 3
a2 0 0 5/3 -1 0 1 2 6/5
S2 0 0 5/3 0 1 0 3 9/5
Therefore, x2 is the entering variable and a2 is the
departing variable.

51. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 S1 S2
Z* 1 0 0 0 0 0
x1 0 1 0 1/5 0 3/5
x2 0 0 1 -3/5 0 6/5
S2 0 0 0 1 1 1
As there is no variable to be entered in the basis, this table is
optimum for Phase I. In this table Max. Z* = 0 and no artificial
variable appears in the optimum basis, therefore we can proceed
to Phase II.

52. Cont…
Phase II:
Original Objective function is:
Min.. Z = 4x1 + x2 Convert the objective function into the maximization form
Max. Z’ = -4 x1 - x2 where Z’= -Z
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 S1 S2
Z* 1 4 1 0 0 0
x1 0 1 0 1/5 0 3/5
x2 0 0 1 -3/5 0 6/5
S2 0 0 0 1 1 1
Again this table is not feasible as basic variable x1 and x2 has a non zero
coefficient in Z’ row. So make the table feasible.

53. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z’ x1 x2 S1 S2
Z’ 1 0 0 -1/5 0 -18/5
x1 0 1 0 1/5 0 3/5 3
x2 0 0 1 -3/5 0 6/5 -
S2 0 0 0 1 1 1 1
Therefore, S1 is the entering variable and S2 is the
departing variable.

54. Cont…
Basic
Variable
Coefficients of: Sol.
Z’ x1 x2 S1 S2
Z’ 1 0 0 0 1/5 -17/5
x1 0 1 0 0 -1/5 2/5
x2 0 0 1 0 3/5 9/5
S2 0 0 0 1 1 1
Optimal Solution is : x1= 2/5, x2= 9/5,
Z’ = -17/5 Z = 17/5

56. Introduction
At the stage of improving the solution during Simplex
procedure, if a tie for the minimum ratio occurs at least
one basic variable becomes equal to zero in the next
iteration and the new solution is said to be Degenerate.

57. Example
Max.. Z = 3x1 + 9x2
Subject to constraints:
x1 + 4x2 < 88
xx11 ++ 22xx22 < 4
xx11,, xx22 > 0

58. Cont…
Let S1 and S2 be two slack variables.
Modified form is:
Z - 3x1 - 9x2 = 0
x1 + 4x2 + S1 = 88
xx11 ++ 22xx22 ++SS22 == 4
xx11,, xx22,, SS11,, SS22 > 0
Initial BFS is : x1= 0, x2= 0, S1= 8, S2= 4 and Z=0.

59. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -3 -9 0 0 0
S1 0 1 4 1 0 8 2
S2 0 1 2 0 1 4 2
In this table S1 and S2 tie for the leaving variable. So
any one can be considered as leaving variable.
Therefore, x2 is the entering variable and S1 is the
departing variable.

60. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -3/4 0 9/4 0 18
x2 0 1/4 1 1/4 0 2 8
S2 0 1/2 0 -1/2 1 0 0
Therefore, x1 is the entering variable and S2 is the
departing variable.

61. Cont…
Basic
Variable
Coefficients of: Sol.
Z x1 x2 S1 S2
Z 1 0 0 3/2 3/2 18
x2 0 0 1 1/2 -1/2 2
x1 0 1 0 -1 2 0
Optimal Solution is : x1= 0, x2= 2
Z = 18
It results in a Degenerate Basic Solution.