simplex method

Simplex MethodBy

Dr. Eng. / Tamer Atteya

Tanta Univ. – EGYPT

Facebook link

http://www.facebook.com/

Example

Max. Z = 13x1+11x2

Subject to constraints:

4x1+5x2 << 1500 1500

5x5x11+3x+3x2 2 << 1575 1575

xx11+2x+2x2 2 << 420 420

xx11, x, x2 2 > 0

دالة الهدف

القيود أو

الشروط

Rewrite objective function so it is equal to zero

We then need to rewrite Z = 13x1+11x2 as:

Z -13x1-11x2 =0

Convert all the inequality constraints into equalities by the use of slack variables S1, S2 , S3.

4x1+5x2 + S1 = 1500 1500

5x5x11+3x+3x2 2 +S+S22= 1575= 1575

xx11+2x+2x2 2 +S+S33 = 420 = 420

xx11, x, x22, S, S11, S, S22, S, S3 3 > 0

Z x1 x2 S1 S2 S3 Value Ratio

Z

S1

S2

S3

Z - 13x1 -11x2 = 0


4x1+5x2 + S1 = 1500 1500

5x5x11+3x+3x2 2 +S+S2 2 = 1575= 1575

xx11+2x+2x2 2 +S+S33 = 420 = 420

Next, these equations are placed in a tableau


Z 1 -13 -11 0 0 0 0

S1 0 4 5 1 0 0 1500

S2 0 5 3 0 1 0 1575

S3 0 1 2 0 0 1 420


Z 1 -13 -11 0 0 0 0

S1 0 4 5 1 0 0 1500

S2 0 5 3 0 1 0 1575

S3 0 1 2 0 0 1 420

a) Choose the most negative number from Z-row. That variable ( x1 ) is the entering variable.

b) Calculate Ratio = (value-col.) / (entering -col.)

c) Choose minimum +ve Ratio. That variable (S2) is the departing variable.


Z 1 -13 -11 0 0 0 0

S1 0 4 5 1 0 0 1500 1500/4=375

S2 0 5 3 0 1 0 1575 1575/5=315

S3 0 1 2 0 0 1 420 420/1=420

a) Choose the most negative number from Z-row. That variable ( x1 ) is the entering variable.

b) Calculate Ratio = (value-col.) / (entering -col.)

c) Choose minimum +ve Ratio. That variable (S2) is the departing variable.


Z 1 -13 -11 0 0 0 0

S1 0 4 5 1 0 0 1500

x1 0 5 3 0 1 0 1575

S3 0 1 2 0 0 1 420

Make the pivot element equal to 1


Z 1 -13 -11 0 0 0 0

S1 0 4 5 1 0 0 1500

x1 0 5/5 3/5 0 1/5 0 1575/5S3 0 1 2 0 0 1 420



Z 1 -13 -11 0 0 0 0

S1 0 4 5 1 0 0 1500

x1 0 1 3/5 0 1/5 0 315

S3 0 1 2 0 0 1 420


Make the pivot column values into zeros


Z 1 -13 -11 0 0 0 0

S1 0 4 5 1 0 0 1500

x1 0 1 3/5 0 1/5 0 315

S3 0 1 2 0 0 1 420


Z 1 -13 -11 0 0 0 0

S1 0 4 5 1 0 0 1500

x1 0 1 3/5 0 1/5 0 315

S3 0

-0

1

-1

2

-3/5

0

-0

0

-1/5

1

-0

420

-315



Z 1 -13 -11 0 0 0 0

S1 0

-4(0)

4

-4(1)

5

-4(3/5)

1

-4(0)

0

-4(1/5)

0

-4(0)

1500

-4(315)

x1 0 1 3/5 0 1/5 0 315

S3 0 0 7/5 0 -1/5 1 105



Z 1 -13 -11 0 0 0 0

S1 0 0 13/5 1 -4/5 0 240

x1 0 1 3/5 0 1/5 0 315

S3 0 0 7/5 0 -1/5 1 105



Z 1 0 -16/5 0 13/5 0 4095

S1 0 0 13/5 1 -4/5 0 240

x1 0 1 3/5 0 1/5 0 315

S3 0 0 7/5 0 -1/5 1 105



Z 1 0 -16/5 0 13/5 0 4095

S1 0 0 13/5 1 -4/5 0 240 92.3

x1 0 1 3/5 0 1/5 0 315 525

S3 0 0 7/5 0 -1/5 1 105 75


Z 1 0 0 0 15/7 16/7 4335

S1 0 0 0 1 -3/7 -13/7 45

x1 0 1 0 0 2/7 -3/7 270

x2 0 0 1 0 -1/7 5/7 75

Optimal Solution is : x1= 270, x2= 75, Z= 4335

Example

Max. Z = 3x1+5x2+4x3


2x1+3x2 << 8 8

2x2x22+5x+5x3 3 << 10 10

3x3x11+2x+2x22+4x+4x3 3 << 15 15

xx11, x, x22, x, x3 3 > 0

Cont…

Let S1, S2, S3 be the three slack variables.

Modified form is:

Z - 3x1-5x2-4x3 =0

2x1+3x2 +S1= 8 8

2x2x22+5x+5x3 3 +S+S22= 10= 10

3x3x11+2x+2x22+4x+4x33+S+S33= 15= 15

xx11, x, x22, x, x33,, S1, SS22,, SS3 3 > 0

Initial BFS is : x1= 0, x2= 0, x3=0, S1= 8,

S2= 10, S3= 15 and Z=0.

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.

Cont…

Basic Variable


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


Cont…

Basic Variable


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


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

Example

Min.. Z = x1 - 3x2 + 2x3


3x1 - x2 + 3x3 << 7

-2x-2x1 1 + 4x+ 4x2 2 << 12 12

-4x-4x11 + 3x + 3x22 + 8x + 8x33 < < 1010

xx11, x, x22, x, x3 3 > 0

Cont…

Convert the problem into maximization problem

Max.. Z’ = -x1 + 3x2 - 2x3 where Z’= -Z


3x1 - x2 + 3x3 << 7

-2x-2x1 1 + 4x+ 4x2 2 << 12 12

-4x-4x11 + 3x + 3x22 + 8x + 8x33 < < 1010

xx11, x, x22, x, x3 3 > 0

Cont…

Let S1, S2 and S3 be three slack variables.

Modified form is:

Z’ + x1 - 3x2 + 2x3 = 0

3x1 - x2 + 3x3 +S1 = 7

-2x-2x1 1 + 4x+ 4x2 2 + S + S22 = 12 = 12

-4x-4x11 + 3x + 3x22 + 8x + 8x33 +S +S33 = = 1010

xx11, x, x22, x, x3 3 > 0Initial BFS is : x1= 0, x2= 0, x3=0, S1= 7, S2= 12, S3 = 10

and Z=0.

Cont…

Basic Variable


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


Cont…

Basic Variable


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 -


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

Example

Max.. Z = 3x1 + 4x2


x1 - x2 << 1

-x-x1 1 + x+ x2 2 << 2 2

xx11, x, x2 2 > 0

Cont…Let S1 and S2 be two slack variables

.

Modified form is:

Z -3x1 - 4x2 = 0

x1 - x2 +S1 = 1

-x-x1 1 + x+ x2 2 +S+S2 2 = 2= 2

xx11, x, x22, S, S11, S, S2 2 > 0Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2

and Z=0.

Cont…

Basic Variable


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


Cont…

Basic Variable


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.

simplex method

Education