1. Simplex method : Tableau Form
Maximise Z= 10X1+9X2+0S1+0S2+0S3+0S4
subject to:
7/10X1+1X2+1S1 =630
1/2X1+5/6X2+ +1S2 =600
1X1+2/3X2 +1S3 =708
1/10X1+1/4X S4 =135
X1,X2,X3,X4,S1,S2,S3,S4 > 0
Constraints equations make more variables (six) than equations (four), the simplex method find solutions for these equations by assigning zero values to variables
A basic solution can be either feasible or infeasible. A basic feasible solution is a solution which is both basic and satisfy non negativity condition.

2. cj = objective function coefficient for variables j
Simplex Form
cj = objective function coefficient for variables j
bj = right hand side coefficient for constraints i
aij = Coefficient associated with variable j in constraint i
c1 c2 …………………………………………….cn
a a12 ……………………………………………… a1n b1
a a22 ………………………………………………..a2n b2
………………………………………………………………………….
am1 am2 amn bm
C(row) = row of objective function coefficient
b(column)= column of right hand side value of the constraint equation
A(matrix) = m-row and n-column of coefficient of the variables in the constraint equations

3. Simplex Tableau X1 X2 S1 S2 S3 S4 bj/aij Basis Cj 10 9 7/10 1 630 ½
7/10 1
630
5/6
600
2/3
708
1/10
135
Profit Zj
Cj-Zj

4. Improving the Solution
Zj represents the decrease in profit that will result if one unit of X1 is brought into the solution Cj-Zj - net evaluation row Values in Zj row is calculated by multiplying the element in the cj column by the corresponding elements in the column of the A matrix and summing them. The process of moving from one basic feasible solution to another is called iteration. We select the entering variable that has the highest coefficient in the net evaluation row. Criteria of moving a variable from the current basic: For each row i compute the ratio bj/aij> 0. This ratio tells us the maximum amount of the variable Xj that can be brought into the solution and still satisfy the constraint equation represented by that row. Select the basic variable corresponding to the minimum of these ratios as the variable to leave.

5. Optimality
Optimality Condition: The optimal solution to a LP problem has been reached when there are no positive values in the net evaluation row of the simplex tableau.
Pivot Row :
Replace the leaving variable In the basic column with the entering variable.
New pivot row = current row divided by pivot element
All other rows including Z :
New row : Current row – (pivot column element) X ( New pivot row)

6. X1 X2 S1 S2 S3 S4
bj/aij
Basis Cj 10 9
16/30 1
-7/10
134.4
-1/2
246
2/3
708
22/120
-1/10
64.2
Profit Zj
20/3
7080
Cj-Zj
7/3
-10

7. X1 X2 S1 S2 S3 S4
bj/aij
Basis Cj 10 9 1
30/16
-21/16
252
-15/16
5/32
120
-20/16
50/16
540
-11/32
9/64 18
Profit Zj
70/16
111/16
7668
Cj-Zj
-70/16
-111/16

8. Condition for Optimal Solution
Cj-Zj< 0 for both of the non-basic variables S1 & S2, an attempt to bring non- basic variables into basic will lower the current value of the objective function. The optimal solution of a LP problem has been reached when there are no positive values at the net evaluation row.
X1 540
X2 252
S1 0
S2 120
S3 0
S4 18

9. Sample Problem Max. 4X1+6X2+3X3+1X4 s.t 3/2X1+2X2+4X3+3X4<550
Problem in standard form:
Max> 4X1+6X2+3X3+1X4+0S1+0S2+0S3
s t 3/2X1+2X2+4X3+3X4+1S1 = 550
4X1+1X2+2X3+1X S2 = 700
2X1+3X2+1X3+2X S3 = 200
X1, X2, X3, X4, S1, S2, S3 > 0

10. Initial Simplex Tableau
Basis Cj 4 6 3 1
3/2 2
550
700
200 Zj
Cj-Zj

11. X1 X2 X3 X4 S1 S2 S3 Cj 4 6 3 1 1/6 5/3 -2/3 1/3 -1/3 x2 2/3 Zj 6/3
Iteration :1… X1 X2 X3 X4 S1 S2 S3
Basis Cj 4 6 3 1
1/6
10/3
5/3
-2/3
416(2/3)
1/3
-1/3
633(1/3) x2
2/3
66(2/3) Zj
12/3
6/3
400
Cj-Zj
3/3
-9/3
-6/3

12. x1 x2 x3 x4 S1 S2 S3 Cj 4 6 3 1 Zj 525 3/60 3/10 Basis -5/10 425 -1/10
Iteration 2…. x1 x2 x3 x4 S1 S2 S3
Basis Cj 4 6 3 1
3/60
5/10
3/10
-2/10
125
39/12
-15/30
-5/10
425
39/60
15/30
-1/10
12/30 25 Zj
81/20
9/2
54/30
525
Cj-Zj
-1/20
-7/2
-3/10
-54/30

13. Eliminating negative right hand side:
Treating Constraints
Eliminating negative right hand side:
The number of std bag had to be less or
X1 < X equal to the number of deluxe bag after 25
1X1 - 1X2 < bags had been set aside for displaying
-X1 + X2 > purpose
Greater than equal to constraint :
6X1 + 3 X2 -4X3 > -20
-6X1 – 3X2 + 4 X3 < 20

14. Simplex Problem Max. 10X1 + 9 X2 s t 7/10X1+1X2 < 630
1X1 > a5, a6 (artificial variables)
1X2 > very large cost -M
Max. 10X1+9X2+ 0S1+S2+0S3+0S4+0S5+0S6-Ma5-Ma6
s t. 7/10X1+1X2+1S1 = 630
1/2X1+5/6X S =600
1X1+2/3X S3 =708
1/10X1+1/4X S4 =135
1X S5+1a5 =100
1X S6+1a =100

15. X1 X2 S1 S2 S3 S4 S5 S6 a5 a6 630 600 708 135 Basis Cj 10 9 -M 7/10 1-M
7/10 1
630
1/2
5/6
600
2/3
708
1/10
135 -1
100 Zj
Cj-Zj
10+M
9+M M
-200M

16. X1 X2 S1 S2 S3 S4 S5 S6 a5 a6 Cj 10 9 -M 1 -1 ¼ Zj M 560 5/6 550 2/3
Basis Cj 10 9 -M 1
7/10
-7/10
560
5/6
1/2
-1/2
550
2/3 -1
608
1/10
-1/10
125
100 Zj
Cj-Zj
9+M
-10 M
-M-10
1000-
100M

17. X1 X2 S1 S2 S3 S4 S5 S6 a5 a6
Basis Cj 10 9 -M
30/16
-210/160 1 -1
152
-15/16
25/160
120
-20/16
300/160
440
-11/32
45/320 18
540
252 Zj
Cj-Zj
9+M
70/16
-70/16
111/16
-111/16
7668

18. Max. 6X1+3X2+4X3+1X4 s t -2X1-1/2X2+1X3+-6X4 = -60
Equality Constraints
Max. 6X1+3X2+4X3+1X4
s t -2X1-1/2X2+1X3+-6X4 = -60
1X X3+2/3X4<20
-1X2-5X <-50
X1, X2, X3, X4> 0
Max. 6X1+3X2+4X3+1X4+0S2+0S3-Ma1-Ma2
st 2X1+1/2X2-1X3+6X4 +1a = 60
1X X3-2/3X4 +1S1 = 20
1X2+5X S a5= 50
X1, X2, X3, X4, S2, S3, a1, a3 > 0

19. Sensitivity Analysis
Study of how optimal solution and value of optimal solution to a LP given a changes in the various coefficients of the problem:
1. What effect on the optimal solution will a change in a coefficient of the objective function (Cj) have?
2. What effect on the optimal solution will a change in constraint in right hand side value (bj) have?
3. What effect on the optimal solution will a change in a coefficient of a constraint equation (aij) have?