Simplex method

Simplex Method By Dr. Eng. / Tamer Atteya Tanta Univ. – EGYPT Facebook link

Example Max. Z = 13x 1 +11x 2 Subject to constraints: 4x 1 +5x 2 < 1500 5x 1 +3x 2 < 1575 x 1 +2x 2 < 420 x 1 , x 2 > 0 دالة الهدف القيود أو الشروط

Rewrite objective function so it is equal to zero We then need to rewrite Z = 13x 1 +11x 2 as: Z -13x 1 -11x 2 =0

Convert all the inequality constraints into equalities by the use of slack variables S 1 , S 2 , S 3 . 4x 1 +5x 2 + S 1 = 1500 5x 1 +3x 2 +S 2 = 1575 x 1 +2x 2 +S 3 = 420 x 1 , x 2 , S 1 , S 2 , S 3 > 0

Z - 13x 1 -11x 2 = 0 Subject to constraints: 4x 1 +5x 2 + S 1 = 1500 5x 1 +3x 2 +S 2 = 1575 x 1 +2x 2 +S 3 = 420 Next, these equations are placed in a tableau Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z S 1 S 2 S 3 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 S 2 0 5 3 0 1 0 1575 S 3 0 1 2 0 0 1 420

a) Choose the most negative number from Z-row. That variable ( x 1 ) is the entering variable . b) Calculate Ratio = (value-col.) / (entering -col.) c) Choose minimum +ve Ratio. That variable (S 2 ) is the departing variable . Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 S 2 0 5 3 0 1 0 1575 S 3 0 1 2 0 0 1 420

a) Choose the most negative number from Z-row. That variable ( x 1 ) is the entering variable . b) Calculate Ratio = (value-col.) / (entering -col.) c) Choose minimum +ve Ratio. That variable (S 2 ) is the departing variable . Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 1500/4=375 S 2 0 5 3 0 1 0 1575 1575/5=315 S 3 0 1 2 0 0 1 420 420/1=420

Make the pivot element equal to 1 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 5 3 0 1 0 1575 S 3 0 1 2 0 0 1 420

Make the pivot element equal to 1 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 5/5 3/5 0 1/5 0 1575 / 5 S 3 0 1 2 0 0 1 420

Make the pivot element equal to 1 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 1 3/5 0 1/5 0 315 S 3 0 1 2 0 0 1 420

Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 1 3/5 0 1/5 0 315 S 3 0 1 2 0 0 1 420

Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 1 3/5 0 1/5 0 315 S 3 0 -0 1 -1 2 -3/5 0 -0 0 -1/5 1 -0 420 -315

Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 -4(0) 4 -4(1) 5 -4(3/5) 1 -4(0) 0 -4(1/5) 0 -4(0) 1500 -4(315) x 1 0 1 3/5 0 1/5 0 315 S 3 0 0 7/5 0 -1/5 1 105

Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 0 13/5 1 -4/5 0 240 x 1 0 1 3/5 0 1/5 0 315 S 3 0 0 7/5 0 -1/5 1 105

Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 0 -16/5 0 13/5 0 4095 S 1 0 0 13/5 1 -4/5 0 240 x 1 0 1 3/5 0 1/5 0 315 S 3 0 0 7/5 0 -1/5 1 105

Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 0 -16/5 0 13/5 0 4095 S 1 0 0 13/5 1 -4/5 0 240 x 1 0 1 3/5 0 1/5 0 315 S 3 0 0 7/5 0 -1/5 1 105

Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 0 -16/5 0 13/5 0 4095 S 1 0 0 13/5 1 -4/5 0 240 92.3 x 1 0 1 3/5 0 1/5 0 315 525 S 3 0 0 7/5 0 -1/5 1 105 75

Optimal Solution is : x 1 = 270, x 2 = 75, Z= 4335 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 0 0 0 15/7 16/7 4335 S 1 0 0 0 1 -3/7 -13/7 45 x 1 0 1 0 0 2/7 -3/7 270 x 2 0 0 1 0 -1/7 5/7 75

Example Max. Z = 3x 1 +5x 2 +4x 3 Subject to constraints: 2x 1 +3x 2 < 8 2x 2 +5x 3 < 10 3x 1 +2x 2 +4x 3 < 15 x 1 , x 2 , x 3 > 0

Cont… Let S 1 , S 2 , S 3 be the three slack variables. Modified form is: Z - 3x 1 -5x 2 -4x 3 =0 2x 1 +3x 2 +S 1 = 8 2x 2 +5x 3 +S 2 = 10 3x 1 +2x 2 +4x 3 +S 3 = 15 x 1 , x 2 , x 3 , S 1 , S 2 , S 3 > 0 Initial BFS is : x 1 = 0, x 2 = 0, x 3 =0, S 1 = 8, S 2 = 10, S 3 = 15 and Z=0.

Cont… Therefore, x 2 is the entering variable and S 1 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 x 3 S 1 S 2 S 3 Z 1 -3 -5 -4 0 0 0 0 S 1 0 2 3 0 1 0 0 8 8/3 S 2 0 0 2 5 0 1 0 10 5 S 3 0 3 2 4 0 0 1 15 15/2

Cont… Therefore, x 3 is the entering variable and S 2 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 x 3 S 1 S 2 S 3 Z 1 1/3 0 -4 5/3 0 0 40/3 x 2 0 2/3 1 0 1/3 0 0 8/3 - S 2 0 -4/3 0 5 -2/3 1 0 14/3 14/15 S 3 0 5/3 0 4 -2/3 0 1 29/3 29/12

Cont… Therefore, x 1 is the entering variable and S 3 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 x 3 S 1 S 2 S 3 Z 1 -11/15 0 0 17/15 4/5 0 256/15 x 2 0 2/3 1 0 1/3 0 0 8/3 4 x 3 0 -4/15 0 1 -2/15 1/5 0 14/15 - S 3 0 41/15 0 0 2/15 -4/5 1 89/15 89/41

Cont… Optimal Solution is : x 1 = 89/41, x 2 = 50/41, x 3 =62/41, Z= 765/41 Basic Variable Coefficients of: Sol. Z x 1 x 2 x 3 S 1 S 2 S 3 Z 1 0 0 0 45/41 24/41 11/41 765/41 x 2 0 0 1 0 15/41 8/41 -10/41 50/41 x 3 0 0 0 1 -6/41 5/41 4/41 62/41 x 1 0 1 0 0 -2/41 -12/41 15/41 89/41

Example Min.. Z = x 1 - 3x 2 + 2x 3 Subject to constraints: 3x 1 - x 2 + 3x 3 < 7 -2x 1 + 4x 2 < 12 -4x 1 + 3x 2 + 8x 3 < 10 x 1 , x 2 , x 3 > 0

Cont… Convert the problem into maximization problem Max.. Z’ = -x 1 + 3x 2 - 2x 3 where Z’= -Z Subject to constraints: 3x 1 - x 2 + 3x 3 < 7 -2x 1 + 4x 2 < 12 -4x 1 + 3x 2 + 8x 3 < 10 x 1 , x 2 , x 3 > 0

Cont… Let S 1 , S 2 and S 3 be three slack variables. Modified form is: Z’ + x 1 - 3x 2 + 2x 3 = 0 3x 1 - x 2 + 3x 3 +S 1 = 7 -2x 1 + 4x 2 + S 2 = 12 -4x 1 + 3x 2 + 8x 3 +S 3 = 10 x 1 , x 2 , x 3 > 0 Initial BFS is : x 1 = 0, x 2 = 0, x 3 =0, S 1 = 7, S 2 = 12, S 3 = 10 and Z=0.

Cont… Therefore, x 2 is the entering variable and S 2 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z’ x 1 x 2 x 3 S 1 S 2 S 3 Z’ 1 1 -3 2 0 0 0 0 S 1 0 3 -1 3 1 0 0 7 - S 2 0 -2 4 0 0 1 0 12 3 S 3 0 -4 3 8 0 0 1 10 10/3

Cont… Therefore, x 1 is the entering variable and S 1 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z’ x 1 x 2 x 3 S 1 S 2 S 3 Z’ 1 -1/2 0 2 0 3/4 0 9 S 1 0 5/2 0 3 1 1/4 0 10 4 x 2 0 -1/2 1 0 0 1/4 0 3 - S 3 0 -5/2 0 8 0 -3/4 1 1 -

Cont… ,[object Object],[object Object],Basic Variable Coefficients of: Sol. Z’ x 1 x 2 x 3 S 1 S 2 S 3 Z’ 1 0 0 13/5 1/5 8/10 0 11 x 1 0 1 0 6/5 2/5 1/10 0 4 x 2 0 0 1 3/5 1/5 3/10 0 5 S 3 0 0 0 11 1 -1/2 1 11

Example Max.. Z = 3x 1 + 4x 2 Subject to constraints: x 1 - x 2 < 1 -x 1 + x 2 < 2 x 1 , x 2 > 0

Cont… Let S 1 and S 2 be two slack variables . Modified form is: Z -3x 1 - 4x 2 = 0 x 1 - x 2 +S 1 = 1 -x 1 + x 2 +S 2 = 2 x 1 , x 2 , S 1 , S 2 > 0 Initial BFS is : x 1 = 0, x 2 = 0, S 1 = 1, S 2 = 2 and Z=0.

Cont… Therefore, x 2 is the entering variable and S 2 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 S 1 S 2 Z 1 -3 -4 0 0 0 S 1 0 1 -1 1 0 1 - S 2 0 -1 1 0 1 2 2

Cont… x 1 is the entering variable, but as in x 1 column every no. is less than equal to zero, ratio cannot be calculated. Therefore given problem is having a unbounded solution . Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 S 1 S 2 Z 1 -7 0 0 4 8 S 1 0 0 0 1 1 3 - x 2 0 -1 1 0 1 2 -

Simplex method

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