1. DUALITY

4. Dual Problem Primal LP : Max z = c 1 x 1 + c 2 x 2 + ... + c n x n subject to: a 11 x 1 + a 12 x 2 + ... + a 1n x n ≤ b 1 a 21 x 1 + a 22 x 2 + ... + a 2n x n ≤ b 2 : a m1 x 1 + a m2 x 2 + ... + a mn x n ≤ b m x 1 ≥ 0, x 2 ≥ 0,…….x j ≥ 0,……., x n ≥ 0. Associated Dual LP : Min. z = b 1 y 1 + b 2 y 2 + ... + b m y m subject to: a 11 y 1 + a 21 y 2 + ... + a m1 y m ≥ c 1 a 12 y 1 + a 22 y 2 + ... + a m2 y m ≥ c 2 : a 1n y 1 + a 2n y 2 + ... + a mn y m ≥ c n y 1 ≥ 0, y 2 ≥ 0,…….y j ≥ 0,……., y m ≥ 0.

5. Example Primal Max. Z = 3x 1 +5x 2 Subject to constraints: x 1 < 4 y 1 2x 2 < 12 y 2 3x 1 +2x 2 < 18 y 3 x 1 , x 2 > 0 The Primal has: 2 variables and 3 constraints. So the Dual has: 3 variables and 2 constraints Dual Min. Z’ = 4y 1 +12y 2 +18y 3 Subject to constraints: y 1 + 3y 3 > 3 2y 2 +2y 3 > 5 y 1 , y 2 , y 3 > 0 We define one dual variable for each primal constraint.

6. Example Primal Min.. Z = 10x 1 +15x 2 Subject to constraints: 5x 1 + 7x 2 > 80 6x 1 + 11x 2 > 100 x 1 , x 2 > 0

7. Solution Dual Max.. Z’ = 80y 1 +100y 2 Subject to constraints: 5y 1 + 6y 2 < 10 7y 1 + 11y 2 < 15 y 1 , y 2 > 0

8. Example Primal Max. Z = 12x 1 + 4x 2 Subject to constraints: 4x 1 + 7x 2 < 56 2x 1 + 5x 2 > 20 5x 1 + 4x 2 = 40 x 1 , x 2 > 0

10. Cont… The primal problem can now take the following standard form: Max. Z = 12x 1 + 4x 2 Subject to constraints: 4x 1 + 7x 2 < 56 -2x 1 - 5x 2 < -20 5x 1 + 4x 2 < 40 -5x 1 - 4x 2 < -40 x 1 , x 2 > 0

11. Cont… Min. Z’ = 56y 1 -20y 2 + 40y 3 – 40y 4 Subject to constraints: 4y 1 – 2y 2 + 5y 3 – 5y 4 > 12 7y 1 - 5y 2 + 4y 3 – 4y 4 > 4 y 1 , y 2 , y 3 , y 4 > 0 The dual of this problem can now be obtained as follows:

12. Example Primal Min.. Z = 2x 2 + 5x 3 Subject to constraints: x 1 + x 2 > 2 2x 1 + x 2 +6x 3 < 6 x 1 - x 2 +3x 3 = 4 x 1 , x 2 , x 3 > 0

13. Solution Primal in standard form : Max.. Z = -2x 2 - 5x 3 Subject to constraints: -x 1 - x 2 < -2 2x 1 + x 2 +6x 3 < 6 x 1 - x 2 +3x 3 < 4 - x 1 + x 2 - 3x 3 < -4 x 1 , x 2 , x 3 > 0

14. Cont… Dual Min. Z’ = -2y 1 + 6y 2 + 4y 3 – 4y 4 Subject to constraints: -y 1 + 2y 2 + y 3 – y 4 > 0 -y 1 + y 2 - y 3 + y 4 > -2 6y 2 + 3y 3 - 3y 4 > -5 y 1 , y 2 , y 3 , y 4 > 0

15. DUAL SIMPLEX METHOD

16. Introduction Suppose a “basic solution” satisfies the optimality condition but not feasible, then we apply dual simplex method. In regular Simplex method, we start with a Basic Feasible solution (which is not optimal) and move towards optimality always retaining feasibility. In the dual simplex method, the exact opposite occurs. We start with a “optimal” solution (which is not feasible) and move towards feasibility always retaining optimality condition.The algorithm ends once we obtain feasibility.

18. Example Min. Z = 3x 1 + 2x 2 Subject to constraints: 3x 1 + x 2 > 3 4x 1 + 3x 2 > 6 x 1 + x 2 < 3 x 1 , x 2 > 0

19. Cont… Step I: The first two inequalities are multiplied by –1 to convert them to < constraints and convert the objective function into maximization function. Max. Z’ = -3x 1 - 2x 2 where Z’= -Z Subject to constraints: -3x 1 - x 2 < -3 -4x 1 - 3x 2 < -6 x 1 + x 2 < 3 x 1 , x 2 > 0

20. Cont… Let S 1 , S 2 , S 3 be three slack variables Model can rewritten as: Z’ + 3x 1 + 2x 2 = 0 -3x 1 - x 2 +S 1 = -3 -4x 1 - 3x 2 +S 2 = -6 x 1 + x 2 +S 3 = 3 Initial BS is : x 1 = 0, x 2 = 0, S 1 = -3, S 2 = -6, S 3 = 3 and Z=0.

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

23. Cont… Optimal Solution is : x 1 = 3/5, x 2 = 6/5, Z= 21/5 6/5 1 2/5 -1/5 0 0 0 S 3 6/5 0 -3/5 4/5 1 0 0 x 2 3/5 0 1/5 -3/5 0 1 0 x 1 21/5 0 3/5 1/5 0 0 1 Z Sol. S 3 S 2 S 1 x 2 x 1 Z Coefficients of: Basic Variable

24. Example Max. Z = -x 1 - x 2 Subject to constraints: x 1 + x 2 < 8 x 2 > 3 -x 1 + x 2 < 2 x 1 , x 2 > 0

25. Cont… Let S 1 , S 2 , S 3 be three slack variables Model can rewritten as: Z + x 1 + x 2 = 0 x 1 + x 2 + S 1 = 8 -x 2 + S 2 = -3 -x 1 + x 2 + S 3 = 2 x 1 , x 2 > 0 Initial BS is : x 1 = 0, x 2 = 0, S 1 = 8, S 2 = -3, S 3 = 2 and Z=0.

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

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

28. Cont… Optimal Solution is : x 1 = 1, x 2 = 3, Z= -4 1 -1 -1 0 0 1 0 x 1 3 0 -1 0 1 0 0 x 2 4 0 2 1 0 0 0 S 1 -4 1 2 0 0 0 1 Z Sol. S 3 S 2 S 1 x 2 x 1 Z Coefficients of: Basic Variable