simplex method for solving lp problems with two variables

Simplex MethodSimplex Method for solving LP problems for solving LP problems with two variableswith two variables

1- Introduction Graphical method presented in last chapter is fine for 2

variables. But most LP problems are too complex for simple graphical procedures.

The Simplex Method:o is appropriate for problems with more than 2 variables;o uses algebra rules, to find optimal solutions; o is an algorithm and is a series of steps that will

accomplish a certain task.

In this chapter will introduce the simplex method for problems with 2 variables.

2- Simplex algorithm through an example:

Assume one company producing flair furniture: Tables (T) and Chairs (C). The following table provides the information available:

Department

Production Available

Working

Hours

Tables Chairs

Carpentry 4 3 240

Painting & varnishing

2 1 100

Profit 7SR 5SR

Formulation:

• Decision variables:T = Number of tablesC = Number of chairs

• Objective function:Maximize Z = 7T + 5C

• Constraints:2T + 1C 100 (Painting & varnishing) 4T + 3C 240 (Carpentry)T, C 0 (non-negativity constraints)

Graphical method solutionN

um

ber

of

Ch

airs 100

80

60

40

20

0 20 40 60 80 100 T

C

Number of Tables

FeasibleRegion

4T + 3C

2T + 1C

Optimal Solution:

T=30, C=40 Z= 410 SR

11stst Step: Step: Built initial Simplex tableau Less-than-or-equal-to constraints (≤) are converted

to equations by adding a variable called “slack variable”.

o Slack variables represent unused resources.

For the flair furniture problem, define the slacks as:o S1 = unused hours in the painting department

o S2 = unused hours in the carpentry department

The constraints are now written as:o 2T + 1C + S1 = 100

o 4T + 3C + S2 = 240

Slack variables not appearing in an equation are added with a coefficient of 0.This allows all the variables to be monitored at all times.

The final Simplex equations appear as:o 2T + 1C + 1S1 + 0S2 = 100

o 4T + 3C + 0S1 + 1S2 = 240

o T, C, S1, S2 0

The slacks are added to the objective coefficient with 0 profit coefficients. The objective function, then, is: Max. Z= 7T + 5C + 0S1 + 0S2

11stst Step, continued Step, continued

Initial Simplex tableau

T C S1 S2 Quantity (Q)

S1 2 1 1 0 100

S2 4 3 0 1 240

Z 7 5 0 0 0

We start by a basic solution Z=0.

Non-Basic variables

Bas

ic

vari

able

s

22ndnd Step: Step: Entering variable

Choose one entering variable from non-basic variables (T or C) for which we have the largest positive coefficient in the objective function. Here the entering variable will be T and the corresponding column is called pivot column.

T C S1 S2 Quantity (Q)

S1 2 1 1 0 100S2 4 3 0 1 240Z 7 5 0 0 0

Pivot column

33rd rd Step: Step: Leaving variable

Choose one leaving variable from the basic variables (S1 or S2) for which we have the smallest value of quantities (Q) divided by items of pivot column:

for S1 we have 100/2=50for S2 we have 240/4=60

Then the leaving variable will be S1 and the corresponding row is called pivot row.


T C S1 S2 (Q)

S1 2 1 1 0 100S2 4 3 0 1 240Z 7 5 0 0 0

Pivot column

Pivot row

Pivot item

44th th Step: Step: Pivoting

The pivoting is the changing of simplex tableau values as follow:

The entering variable (T) takes the place of leaving variable (S1).

All items of pivot column are =0 except pivot item =1.All items of pivot row are divided by pivot item.Other items of the tableau are calculated as follow:

A B

C D

Pivot column

Pivot rowNew A = A – B*C/D


The New value of the objective function is calculated as follow:

Capacity of pivot rowZ’ = Z + Largest coefficient * Pivot item

If all coefficients of objective function are negatives or equal to zero the optimal solution is found. Otherwise go to step2.


T C S1 S2 Q

S1 2 1 1 0 100S2 4 3 0 1 240

Z 7 5 0 0 0

Pivot column

Pivot row

Pivot item

Pivoting


All coefficients of objective function aren’t negatives or equal to zero, the optimal solution is not found. Then go to step2.

T C S1 S2 Q

T 1 1/2 1/2 0 50S2 0 1 -2 1 40Z 0 3/2 -7/2 0 350


The new largest positive coefficient in the objective function is 3/2 then the entering variable will be C.


The new smallest value of quantities (Q) divided by items of pivot column is 40/1=40 then the leaving variable will be S2 .


T C S1 S2 QT 1 1/2 1/2 0 50S2 0 1 -2 1 40Z 0 3/2 -7/2 0 350

Pivot column

Pivot row

Pivot item


T C S1 S2 QT 1 0 3/2 -1/2 30C 0 1 -2 1 40Z 0 0 -1/2 -3/2 410

All coefficients of objective function are now equal to zero. then the optimal solution is found: T=30, C=40; Z=410

Using Simplex method, solve the LP problems modeled in chapter1.

3- Applications

The mathematical model of problem is:

Max Z= 3x1 + 5x2

S.t. x1 4 (Plant 1 constraint)

2x2 12 (Plant 2 constraint)3x1+ 2x2 18 (Plant 3 constraint)x1, x2 0

3.1- Solving of problem1

11stst Step: Step: Initial Simplex tableau


Non-Basic variables

Bas

ic

vari

able

s

X1 X2 S1 S2 S3 Quantity (Q)

S1 1 0 1 0 0 4

S2 0 2 0 1 0 12

S3 3 2 0 0 1 18

Z 3 5 0 0 0 0


The largest positive coefficient in the objective function is 5 then the entering variable will be X2.


The smallest value of quantities (Q) divided by items of pivot column is 12/2=6 then the leaving variable will be S2 .


Pivot column

Pivot row

Pivot item X1 X2 S1 S2 S3 (Q)

S1 1 0 1 0 0 4

S2 0 2 0 1 0 12

S3 3 2 0 0 1 18

Z 3 5 0 0 0 0


X1 X2 S1 S2 S3 (Q)

S1 1 0 1 0 0 4

X2 0 1 0 1/2 0 6

S3 3 0 0 -1 1 6

Z 3 0 0 -5/2 0 30



The new largest positive coefficient in the objective function is 3 then the entering variable will be X1.




Pivot column

Pivot row

Pivot item X1 X2 S1 S2 S3 (Q)

S1 1 0 1 0 0 4

X2 0 1 0 1/2 0 6

S3 3 0 0 -1 1 6

Z 3 0 0 -5/2 0 30


All coefficients of objective function are now equal to zero. then the optimal solution is found. X1=2, X2=6, Z=36.

X1 X2 S1 S2 S3 (Q)

S1 0 0 1 1/3 -1/3 2

X2 0 1 0 1/2 0 6

X1 1 0 0 -1/3 1/3 2

Z 0 0 0 7/2 -1 36

The mathematical model of problem is:

Max Z= 10x1 + 15x2

S.t. 2x1 + 4x2 100 ( aluminum constraint) 3x1 + 2x2 80 ( steel constraint)

x1 , x2 0

3.2- Solving of problem2

11stst Step: Step: Initial Simplex tableau

X1 X2 S1 S2 Quantity (Q)

S1 2 4 1 0 100

S2 3 2 0 1 80

Z 10 15 0 0 0


Non-Basic variables

Bas

ic

vari

able

s


The largest positive coefficient in the objective function is 15 then the entering variable will be X2.




Pivot column

Pivot row

Pivot item X1 X2 S1 S2 Q

S1 2 4 1 0 100

S2 3 2 0 1 80

Z 10 15 0 0 0


X1 X2 S1 S2 Q

X2 1/2 1 1/4 0 25

S2 2 0 -1/2 1 30

Z 5/2 0 15/4 0 375



The new largest positive coefficient in the objective function is 5/2 then the entering variable will be X1.


The smallest value of quantities (Q) divided by items of pivot column is 30/2=15 then the leaving variable will be S2.


Pivot column

Pivot row

Pivot item

X1 X2 S1 S2 Q

X2 1/2 1 1/4 0 25

S2 2 0 -1/2 1 30

Z 5/2 0 15/4 0 375


All coefficients of objective function are now equal to zero, the optimal solution is found: X1=15, X2=17.5, Z= 412.5.

X1 X2 S1 S2 Q

X2 0 1 3/8 -1/4 17.5

X1 1 0 -1/4 1/2 15

Z 0 0 30/8 -5/4 412.5

simplex method for solving lp problems with two variables

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c s1

tables t

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