1. Simplex Method Adapting to Other Forms

2.  Until now, we have dealt with the standard form of the Simplex method  What if the model has a non-standard form?  Equality Constraints x 1 + x 2 = 8  Greater than Constraints x 1 + x 2 ≥ 8  Minimizing  How do we get the initial BF solution?

3. Original Form Maximize Z = 3x 1 + 5x 2 Subject to: x 1 ≤ 4 2x 2 ≤ 12 3x 1 + 2x 2 = 18 x 1 ≥ 0, x 2 ≥ 0 Augmented Form Maximize Z = 3x 1 + 5x 2 Subject to: Z - 3x 1 - 5x 2 = 0 x 1 + x 3 = 4 2x 2 + x 4 = 12 3x 1 + 2x 2 = 18 x 1, x 2, x 3, x 4 ≥ 0

4. Original Form Maximize Z = 3x 1 + 5x 2 Subject to: x 1 ≤ 4 2x 2 ≤ 12 3x 1 + 2x 2 = 18 x 1 ≥ 0, x 2 ≥ 0 Artificial Form (Big M Method) Maximize Z = 3x 1 + 5x 2 Subject to: Z - 3x 1 - 5x 2 + Mx 5 = 0 x 1 + x 3 = 4 2x 2 + x 4 = 12 3x 1 + 2x 2 + x 5 = 18 x 1, x 2, x 3, x 4, x 5 ≥ 0

5. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z1-3-500M0 x3x3 0101004 x4x4 00201012 x5x5 03200118 Z-3x 1 -5x 2 + Mx 5 = 0 x1x1 + x 3 = 4 x2x2 + x 4 = 12 3x 1 2x 2 + x 5 = 18

6. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z1 -3M-3-2M-5 000-18M x3x3 0101004 x4x4 00201012 x5x5 03200118 Select Initial Point nonbasic variables: x 1 and x 2 (origin) Initial BF solution: (x 1, x 2, x 3, x 4, x 5 ) = (0,0,4,12,18M) To get to initial point, remove x 5 coefficient (M) from Z - 3x 1 - 5x 2 + Mx 5 = 0 (-M) ( 3x 1 + 2x 2 + x 5 = 18) (-3M-3)x 1 + (-2M-5)x 2 = -18M

7. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z1 -3M-3-2M-5 000-18M x3x3 0101004 x4x4 00201012 x5x5 03200118 Optimality Test: Are all coefficients in row (0) ≥ 0? If yes, then STOP – optimal solution If no, then continue algorithm

8. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z1 -3M-3-2M-5 000-18M x3x3 0101004 x4x4 00201012 x5x5 03200118 Select Entering Basic Variable Choose variable with negative coefficient having largest absolute value Select Leaving Basic Variable 1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio 4/1 = 4 18/3 = 6

9. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z1 -3M-3-2M-5 000-18M x3x3 0101004 x4x4 00201012 x5x5 03200118 Z10 -2M-53M+3 00 -6M+12 x1x1 0101004 x4x4 00201012 x5x5 002-3016

10. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z10 -2M-53M+3 00 -6M+12 x1x1 0101004 x4x4 00201012 x5x5 002-3016 BF Solution: (x 1, x 2, x 3, x 4, x 5 ) = (4,0,0,12,6) Optimality Test: Are all coefficients in row (0) ≥ 0? If yes, then STOP – optimal solution If no, then continue algorithm

11. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z10 -2M-53M+3 00 -6M+12 x1x1 0101004 x4x4 00201012 x5x5 002-3016 Select Entering Basic Variable Choose variable with negative coefficient having largest absolute value Select Leaving Basic Variable 1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio 12/2 = 6 6/2 = 3

12. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z10 -2M-53M+3 00 -6M+12 x1x1 0101004 x4x4 00201012 x5x5 002-3016 Z100-9/20 M+5/2 27 x 1 0101004 x 4 000316 x 2 001-3/201/23

13. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z100-9/20 M+5/2 27 x 1 0101004 x 4 000316 x 2 001-3/201/23 BF Solution: (x 1, x 2, x 3, x 4, x 5 ) = (4,3,0,6,0) Optimality Test: Are all coefficients in row (0) ≥ 0? If yes, then STOP – optimal solution If no, then continue algorithm

14. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z100-9/20 M+5/2 27 x 1 0101004 x 4 000316 x 2 001-3/201/23 Select Entering Basic Variable Choose variable with negative coefficient having largest absolute value Select Leaving Basic Variable 1. Select coefficient in pivot column > 0 2. Divide Right Side value by this coefficient 3. Select row with smallest ratio 4/1 = 4 6/3 = 2

15. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z100-9/20 M+5/2 27 x 1 0101004 x 4 000316 x 2 001-3/201/23 Z10003/2 M+1 36 x 1 0100-1/31/32 x3 x3 0001 -1/32 x2 x2 00101/206

16. Basic Variable Coefficient of: Right Side Zx1x1 x2x2 x3x3 x4x4 x5x5 Z10003/2 M+1 36 x 1 0100-1/31/32 x3 x3 0001 -1/32 x2 x2 00101/206 BF Solution: (x 1, x 2, x 3, x 4, x 5 ) = (2,6,2,0,0) Optimality Test: Are all coefficients in row (0) ≥ 0? If yes, then STOP – optimal solution If no, then continue algorithm

17. Minimize Z = 3x 1 + 5x 2 Multiply by -1 Maximize -Z = -3x 1 - 5x 2

18. x 1 - x 2 ≤ -1 Multiply by -1 -x 1 + x 2 ≥ 1

19. x 1 - x 2 ≥ 1 x 1 - x 2 - x 5 ≥ 1 Change Inequality x 1 - x 2 - x 5 ≤ 1 x 1 - x 2 - x 5 + x 6 ≤ -1 Big M Augmented Form

20. Original Form Minimize Z = 4x 1 + 5x 2 Subject to: 3x 1 + x 2 ≤ 27 5x 1 + 5x 2 = 60 6x 1 + 4x 2 ≥ 60 x 1 ≥ 0, x 2 ≥ 0 Adaption Form Minimize Z = 4x 1 + 5x 2 Maximize -Z = -4x 1 - 5x 2 Subject to: -Z + 4x 1 + 5x 2 + Mx 4 + Mx 6 = 0 3x 1 + x 2 + x 3 = 27 5x 1 + 5x 2 + x 4 = 60 6x 1 + 4x 2 - x 5 + x 6 = 60