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+1/4X2 +1S4 =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.

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 c1c2 …………………………………………….cn

--------------------------------------------------------------------------------------

a11 a12 ……………………………………………… a1n b1

a21 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

Simplex Tableau

X1 X2 S1 S2 S3 S4 bj/aij

Basis Cj 10 9 0 0 0 0

S1 0 7/10 1 1 0 0 0 630

S2 0 ½ 5/6 0 1 0 0 600

S3 0 1 2/3 0 0 1 0 708

S4 0 1/10 ¼ 0 0 0 1 135

Profit Zj

0 0 0 0 0 0 0

Cj-Zj 10 9 0 0 0 0

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.

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)

X1 X2 S1 S2 S3 S4 bj/aij

Basis Cj 10 9 0 0 0 0

S1 0 0 16/30 1 0 -7/10 0 134.4

S2 0 0 ½ 0 1 -1/2 0 246

X1 10 1 2/3 0 0 1 0 708

S4 0 0 22/120

0 0 -1/10 1 64.2

Profit Zj

10 20/3 0 0 10 0 7080

Cj-Zj 0 7/3 0 0 -10 0

X1 X2 S1 S2 S3 S4 bj/aij

Basis Cj 10 9 0 0 0 0

X2 9 0 1 30/16 0 -21/16 0 252

S2 0 0 0 -15/16 1 5/32 0 120

X1 10 1 0 -20/16 0 50/16 0 540

S4 0 0 0 -11/32 0 9/64 1 18

Profit Zj

10 9 70/16 0 111/16 0 7668

Cj-Zj 0 0 -70/16 0 -111/16

0

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 540X2 252S1 0S2 120S3 0S4 18

Sample Problem

Max. 4X1+6X2+3X3+1X4

s.t 3/2X1+2X2+4X3+3X4<550

4X1+1X2+2X3+1X4<700

2X1+3X2+1X3+2X4<200

X1, X2, X3, X4 >0

Problem in standard form:

Max> 4X1+6X2+3X3+1X4+0S1+0S2+0S3

s t 3/2X1+2X2+4X3+3X4+1S1 = 550

4X1+1X2+2X3+1X4 +1S2 = 700

2X1+3X2+1X3+2X4 +1S3 = 200

X1, X2, X3, X4, S1, S2, S3 > 0

Initial Simplex Tableau

X1 X2 X3 X4 S1 S2 S3

Basis

Cj 4 6 3 1 0 0 0

S1 0 3/2 2 4 3 1 0 0 550

S2 0 4 1 2 1 0 1 0 700

S3 0 2 3 1 2 0 0 1 200

Zj 0 0 0 0 0 0 0 0Cj-Zj 4 6 3 1 0 0 0

Iteration :1…

X1 X2 X3 X4 S1 S2 S3

Basis Cj 4 6 3 1 0 0 0

S1 0 1/6 0 10/3 5/3 1 0 -2/3 416(2/3)

S2 0 10/3 0 5/3 1/3 0 1 -1/3 633(1/3)

x2 6 2/3 1 1/3 2/3 0 0 1/3 66(2/3)

Zj 12/3 6 6/3 12/3 0 0 6/3 400

Cj-Zj 0 0 3/3 -9/3 0 0 -6/3

Iteration 2….

x1 x2 x3 x4 S1 S2 S3

Basis

Cj 4 6 3 1 0 0 0

x3 3 3/60 0 1 5/10 3/10 0 -2/10 125

S2 0 39/12 0 0 -15/30

-5/10 1 0 425

x2 6 39/60 1 0 15/30 -1/10 0 12/30 25

Zj 81/20 6 3 9/2 3/10 0 54/30 525Cj-Zj -1/20 0 0 -7/2 -3/10 0 -54/30

Treating Constraints Eliminating negative right hand side:

The number of std bag had to be less or

X1 < X2-25 equal to the number of deluxe bag after 25

1X1 - 1X2 < -25 bags had been set aside for displaying

-X1 + X2 > 25 purpose

Greater than equal to constraint :

6X1 + 3 X2 -4X3 > -20

-6X1 – 3X2 + 4 X3 < 20

Simplex ProblemMax. 10X1 + 9 X2

s t 7/10X1+1X2 < 6301/2X1+5/6X2 < 6001X1+2/3X2 < 7081/10X1+1/4X2 <1351X1 >100 a5, a6 (artificial variables)

1X2 >100 very large cost -M Max. 10X1+9X2+ 0S1+S2+0S3+0S4+0S5+0S6-Ma5-Ma6

s t. 7/10X1+1X2+1S1 = 6301/2X1+5/6X2 +1S2 =6001X1+2/3X2 +1S3 =7081/10X1+1/4X2 +1S4 =1351X1 -1S5+1a5 =100 1X2 -S6+1a6 =100

X1 X2 S1 S2 S3 S4 S5 S6 a5 a6Basis

Cj 10 9 0 0 0 0 0 0 -M -M

S1 0 7/10 1 1 0 0 0 0 0 0 0 630

S2 0 1/2 5/6 0 1 0 0 0 0 0 0 600

S3 0 1 2/3 0 0 1 0 0 0 0 0 708

S4 0 1/10 ¼ 0 0 0 1 0 0 0 0 135

a5 -M 1 0 0 0 0 0 -1 0 1 0 100

a6 -M 0 1 0 0 0 0 0 -1 0 1 100

Zj

Cj-Zj

-M

10+M

-M

9+M

0

0

0

0

0

0

0

0

M

-M

M

-M

-M

0

-M

0

-200M

X1 X2 S1 S2 S3 S4 S5 S6 a5 a6Basis Cj 10 9 0 0 0 0 0 0 -M -M

S1 0 0 1 1 0 0 0 7/10 0 -7/10 0 560

S2 0 0 5/6 0 1 0 0 1/2 0 -1/2 0 550

S3 0 1 2/3 0 0 1 0 1 0 -1 0 608

S4 0 0 ¼ 0 0 0 1 1/10 0 -1/10 0 125

X1 10 1 0 0 0 0 0 -1 0 1 0 100

a6 -M 0 1 0 0 0 0 0 -1 0 1 100

ZjCj-Zj

10

0

-M

9+M

0

0

0

0

0

0

0

0

-10

-10

M

-M

10-M-10

-M

0

1000-

100M

X1 X2 S1 S2 S3 S4

S5 S6 a5 a6

Basis Cj 10 9 0 0 0 0 0 0 -M -M

S6 0 0 0 30/16 0-210/160 0 0 1 0 -1 152

S2 0 0 0 -15/16

125/160

0 0 0 0 0 120

S3 0 0 0 -20/16 0 300/1

60 0 1 0 -1 0 440

S4 0 0 0 -11/32 0

45/320

1 0 0 0 0 18

X1 10 1 0 -20/16 0 300/1

60 0 0 0 0 0 540

X2 9 0 1 30/16 0

-210/160

0 0 0 0 0 252

ZjCj-Zj

10

0

9

9+M

70/16

-70/16

0

0

111/16

-111/16

0

0

0

0

0

0

0

-M

0

-M

7668

Equality Constraints

Max. 6X1+3X2+4X3+1X4s t -2X1-1/2X2+1X3+-6X4 = -60 1X1 +1X3+2/3X4<20

-1X2-5X3 <-50X1, X2, X3, X4> 0

Max. 6X1+3X2+4X3+1X4+0S2+0S3-Ma1-Ma2st 2X1+1/2X2-1X3+6X4 +1a1 = 60

1X1 +1X3-2/3X4 +1S1 = 20 1X2+5X3 -1S3 +1a5= 50X1, X2, X3, X4, S2, S3, a1, a3 > 0

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?