chap 3 simplex

CHAPTER 3: SIMPLEX METHOD

Introduction:

In the simplex method, the model is put into the form of table,

& then a number of mathematical steps are performed on the

table (simplex tableau).

The simplex method moves from one better solution to another

until the best one is found, & then it stops.

Example of simplex method:

Max Z = 40 X1 + 50X2

s.t X1 + 2X2 ≤ 40 (labor)

4X1 + 3X2 ≤ 120 (clay)

X1, X2 ≥ 0

1

Initial tableau:

Basic Quantity 40 50 0 0Cj Variables (Zj) X1 X2 S1 S2

0 S1 40 1 2 1 00 S2 120 4 3 0 1

Zj 0 0 0 0 0Cj - Zj 40 50 0 0


50 X2 20 0.5 1 0.5 00 S2 60 2.5 0 -1.5 1

Zj 1000 25 50 25 0Cj - Zj 15 0 -25 0

Optimal tableau:


50 X2 8 0 1 0.8 -0.240 X1 24 1 0 -0.6 0.4

Zj 1360 40 50 16 6Cj - Zj 0 0 -16 -6

Solution:

X1 = 24, X2 = 8

S1 = 0, S2 = 0

Z = 1360

2

Steps:

1. Transform the LP model into standard form.

2. Set up the initial tableau for the basic feasible solution at

the origin & compute the zj & cj – zj (max) @ zj – cj (min)

value.

3. Determine the pivot column by selecting the column with

the highest positive value in the cj – zj @ zj – cj row.

4. Determine the pivot row by dividing the quantity column

values by the pivot column values & select the row with

the minimum nonnegative quotient.

5. Compute the new pivot row values using the formula:

new tableau pivot row values =

old tableau pivot row valuespivot number

6. Compute all other row values using the formula:

7. Compute the new zj & cj – zj @ zj – cj row.

New tableau row values

Old tableau row values

Corresponding coefficient in pivot column

Corresponding new tableau in pivot row value

= X

3

8. Determine whether the new solution is optimal by

checking the cj – zj @ zj – cj row. If all the values are 0 or

negative, the solution is optimal. If not, return to step 3 &

repeat the simplex steps.

Converting the Model into Standard Form

Standard form requires that all constraints be in the form of

equations rather than inequalities.

A set of rules for transforming all 3 types of model constraints:

Constraint Adjustment Objective function coefficientMaximization Minimization

≤ (+) a slack variable

0 0

= (+) an artificial variable

- M M

≥ (-) a surplus variable

0 0

(+) an artificial variable

- M M

4

Artificial variable (Ai) does not have a meaning as slack variable

or surplus variable does.

It is inserted into the equation simply to give a positive solution

at the origin.

M is a very large positive number.

Example 1: Maximize Z = 400x1 + 200x2

s.t x1 + x2 = 30

2x1 + 8x2 ≥ 80

x1 ≤ 20

x1, x2 ≥ 0

Standard form:

5

Example 2: Minimize Z = 400x1 + 200x2

s.t x1 + x2 = 30

2x1 + 8x2 ≥ 80

x1 ≤ - 20

x1, x2 ≥ 0

Standard form:

6

Example 3:

Z = 3X1 + 5X2 + 10X3 + 0S1 + 0S2 – MA1 – MA2

s.t X1 – X3 + A1 = 50

6X2 + S1 = 380

- 40X2 + 60X3 – S2 + A2 = 300

X1, X2, X3, S1, S2, A1, A2 ≥ 0

LP model:

7

Exercises 1:

Transform the following LP model into standard form model.

1. Max Z = 3X1 + 6X2 + 5X3

s.t 3X1 + 2X2 ≤ 18

X1 + X2 ≥ 5

X1 + X3 ≤ 10

X1, X2, X3 ≥ 0

2. Max Z = 8X1 + 7X2

s.t 10X1 + 8X2 = 40

6X1 + 14X2 ≤ 48

X2/X1 ≥ 7/8

X1, X2 ≥ 0

3. Min Z = 4X1 + 8X2

s.t -6X1 + 5X2 ≥ -16

7X1 + 15X2 ≥ 18

X1, X2 ≥ 0

8

Exercises 2:

Transform the following standard form model into LP model.

1. Z = 70X1 + 50X2 + 0S1 + MA1 + MA2 + MA3

s.t 5X1 + 3X2 + A1 = 300

4X1 – S1 + A2 = 40

7X2 + A3 = 500

X1, X2, S1, A1, A2, A3 ≥ 0

2. Z = 150X1 + 200X2 + 180X3 + 0S1 + 0S2 + MA1

s.t X1 + S1 = 40

-7X1 – 3X2 + 5X3 + A1 = 350

X2 + X3 + S2 = 100

X1, X2, X3, S1, S2, A1 ≥ 0

3. Z = 4X1 + 8X2 + 0S1 + 0S2 – MA1 – MA2

s.t -6X1 + 5X2 – S1 + A1 = 16

7X1 + 15X2 – S2 + A2 = 18

X1, X2, S1, S2, A1, A1 ≥ 0

9

Simplex Method For ≤ Constraints

Example:

Max Z = 5X1 + 10X2 + 15X3

s.t X1 + X2 ≤ 7

X2 + X3 ≤ 10

2X1 + 2X2 + 2X3 ≤ 30

X1, X2, X3 ≥ 0

Standard form:

10

Initial tableau:

Cj BasicVar.

Quantity

Zj

Cj BasicVar.

Quantity

Zj

11

Cj BasicVar.

Quantity

Zj

12

Corresponding New tableau

Old row - Coefficient in X Pivot = New tableau

Column Value Pivot column Row value Row value




13







14

Simplex Method For ≥ Constraints

Example:

Min Z = 6X1 + 3X2

s.t 2X1 + 4X2 ≥ 16

4X1 + 3X2 ≥ 24

X1, X2 ≥ 0

Standard form:

15

Initial tableau:

Cj BasicVar.

Quantity

Zj

Cj BasicVar.

Quantity

Zj

16

Cj BasicVar.

Quantity

Zj

Cj BasicVar.

Quantity

Zj

17







18




19

Simplex Method For Mixed Constraints ( ≥, =, ≤ )

Example:

Max Z = 400X1 + 200X2

s.t X1 + X2 = 30

2X1 + 8X2 ≥ 80

X1 ≤ 20

X1, X2 ≥ 0

Standard form:

20

Initial tableau:

Cj BasicVar.

Quantity

Zj

Cj BasicVar.

Quantity

Zj

21

Cj BasicVar.

Quantity

Zj

Cj BasicVar.

Quantity

Zj

22







23







24







25

The basic feasible solution in the initial simplex tableau is the

origin where all the decision variables equal zero.

The quantity column values are the solution values for the

variables in the basic feasible solution.

The cj values are the contribution to profit (or cost) for each

variable.

The zj row values are computed by multiplying the cj column

values by the variable column values and summing.

The variable with the largest positive cj – zj @ zj – cj value is the

entering variable (pivot column).

The leaving variable is determined by dividing the quantity

values by the pivot column values & selecting the minimum

possible value or zero (pivot row).

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The pivot number is the number at the intersection of the pivot

column & row.

The solution is optimal when all cj – zj @ zj – cj ≤ 0.

Artificial variables are assigned a large cost in the objective

function to eliminate them from the final solution.

Once an artificial variable is selected as the leaving variable, it

will never reenter the tableau, so it can be eliminated.

27

Irregular types of LP Problems

1. Multiple Optimal Solutions

Alternate optimal solutions have the same Z value but

different variables values.

Example:

Max Z=40 X1+30 X2

s .t :

X1+2 X2≤404 X1+3 X2≤120X1 , X2≥0

The optimal tableau for this problem is as follows:


0 S1 10 0 5/4 1 -1/440 X1 30 1 ¾ 0 ¼

Zj 1200 40 30 0 10Cj - Zj 0 0 0 -10

To get the alternate solution, proceed the tableau i.e. X2 enter

the tableau and X1 leave.

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2. An Infeasible Problem

An infeasible problem has an artificial variable in the final

simplex tableau.

Example:

Basic Quantity 5 3 0 0 0 -M -MCj Var. X1 X2 S1 S2 S3 A1 A2

3 X2 4 2 1 ½ 0 0 0 0-M A1 4 1 0 0 -1 0 1 0-M A2 2 -2 0 -1/2 0 -1 0 1

Zj 12-6M 6 + M 3 3/2 +

M/2

M M -M -M

Cj – Zj -1-M 0 -3/2 –

M/2

-M -M 0 0

3. An Unbounded Problem

A pivot row cannot be selected for an unbounded problem.

Basic 4 2 0 0 MCj Variables Quantit

yX1 X2 S1 S2 A1

M A1 4 1 0 -1 0 10 S2 2 0 1 0 1 0

Zj 4M M 0 -M 0 MCj – Zj 4-M 2 M 0 0

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4. Tie For the Pivot Column

Sometimes when selecting the pivot column, the greatest

positive cj – zj (or zj – cj) row values are the same.

5. Tie For the Pivot Row

Example:

Basic Quantity 4 6 0 0 0Cj Var. X1 X2 S1 S2 S3

0 S1 12 6 0 1 -4 06 X2 3 0 1 0 1 00 S3 10 5 0 0 -10 1

Zj 18 0 6 0 6 0Cj – Zj 4 0 0 -6 0

6. Negative Quantity Values

Standard form for simplex solution requires positive (right-

hand-side) RHS values.

Example:

-6x1 + 2x2 ≥ -30

(-1) (-6x1 + 2x2 ≥ -30)

6x1 - 2x2 ≤ 30

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chap 3 simplex

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