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DUALITY THEORY

DUALITY THEORY. CONTENTS Defining the Dual Problem The Weak Duality Theorem The Strong Duality Theorem Simplex Algorithm and Dual Variables Complementary Slackness Conditions Dual Simplex Method Reference: Chapter 6 in Bazaraa, Jarvis and Sherali. Lower and Upper Bounds.

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DUALITY THEORY

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  1. DUALITY THEORY CONTENTS • Defining the Dual Problem • The Weak Duality Theorem • The Strong Duality Theorem • Simplex Algorithm and Dual Variables • Complementary Slackness Conditions • Dual Simplex Method Reference: Chapter 6 in Bazaraa, Jarvis and Sherali.

  2. Lower and Upper Bounds Maximize 4x1 + x2 + 3x3 subject to x1 + 4x2 1 3x1 - x2 + x3 3, x1, x2, x3 0 A feasible solution: x1 = 0, x2 = 0, x3 = 3, z = 9 • Each feasible solution gives a lower bound on the optimal objective function value. • We can also obtain an upper bound easily. 2 (x1 + 4x2 )  (1) 2 + 3 (3x1 - x2 + x3 )  (3) 3 11x1 + 5x2 + 3x3  11 • Observe that 4x1 + x2 + 3x3  11x1 + 5x2 + 3x3  11. Hence z 11

  3. Lower and Upper Bounds (contd.) Maximize 4x1 + x2 + 3x3 subject to x1 + 4x2 1 3x1 - x2 + x3 3, x1, x2, x3 0 4x1 + x2 + 3x3  11x1 + 5x2 + 3x3  11. Hence z 11. • How can we get the smallest possible upper bound? Find y1 and y2 such that 1 (x1 + 4x2 )  (1) 1 + 2 (3x1 - x2 + x3 )  (3) 2 (1+32)x1 + (41-2)x2 + (2)x3 1 + 32 1+ 32 4, 41- 2 1, 2  3, 1  0, 2  0 and 1+32 is minimum

  4. Defining the Dual Dual LP: Primal LP: Minimize 1 + 32 Maximize 4x1 + x2 + 3x3 subject to subject to 1 + 32 4 x1 + 4x2 1 41 - 2 1 3x1 - x2 + x3 3 2  3 x1, x2, x3 0 1  0, 2  0

  5. The Dual Problem PRIMAL LP: Maximize j=1,n cjxj subject to j=1,n aij xj bi for all i = 1, 2, …, m xj 0 for all j = 1, 2, … , n THE ASSOCIATED DUAL LP: Minimize i=1,m bii subject to i=1,mi aij cj for all j = 1, 2, … , n j 0 for all i = 1, 2, … , m

  6. The Dual Problem THE DUAL LP: - maximize i=1,m (-bi) i subject to i=1,m (-aij) i (-cj) for all j = 1, 2, …, n j 0 for all i = 1, 2, … , m DUAL OF THE DUAL LP: -minimize j=1,n (-cj)xj subject to j=1,n (-aij) xj (-bj) for all i = 1, 2, …, m xj 0 for all j = 1, 2, … , n which is the same as the primal problem.

  7. A Diet Problem • My diet consists of four items: Brownie, ice cream, soda, and cheesecake. • Each day, I must ingest at least 500 calories, 6 oz. of chocolate, 10 oz. of sugar, and 8 oz. of fat. • The nutritional contents of each type of per unit of food are given below. Find the minimum cost diet plan.

  8. A Pill Manufacturer’s Problem • A pill manufacturer is planning to manufacture pills for calories, chocolates, sugar, and fat. • He wants to set the prices for these pills which maximizes his revenue and at the same time the pills are more competitive than each of the items I consume (brownie, chocolate ice cream, cola, and pineapple cheese cake)

  9. The Weak Duality Theorem THEOREM: If (x1, x2, … , xn) is feasible for the primal problem and (1, 2, … , m) is feasible for the dual problem, then j=1,n cjxj  i=1,mi bi PROOF: j=1,n cjxj  j=1,n (i=1,mi aij) xj = j=1,ni=1,mi aij xj = i=1,m (j=1,n aijxj ) i  i=1,m bi i

  10. Primal Values Dual Values Duality Gap Primal Values Dual Values No Gap Implications of the Weak Duality Theorem THEOREM: If (x1, x2, … , xn) is feasible for the primal problem and (1, 2, … , m) is feasible for the dual problem, and j=1,n cjxj = i=1,m bi i, then x is optimal to the primal problem and  is optimal to the dual problem.

  11. Strong Duality Theorem THEOREM: If the primal problem has an optimal solution then the dual also has an optimal solution such that PROOF: To be proved using the simplex algorithm.

  12. Strong Duality Theorem (contd.) • The simplex algorithm maintains a set of dual variables  at each iteration. Simplex multipliers are these dual variables. • For the initial solution = 0. As it performs iterations and updates BFS, the associated set of dual variables changes. PRIMAL LP: Maximize j=1,n cjxj subject to j=1,n aij xj bi for all i = 1, 2, …, m xj 0 for all j = 1, 2, … , n or Maximize j=1,n cjxj +i=1,m 0wj subject to j=1,n aij xj + wi= bi for all i = 1, 2, …, m xj 0 for all j = 1, 2, … , n

  13. Strong Duality Theorem (contd.) Maximize j=1,n cjxj +i=n+1,n+m 0xj subject to j=1,n aij xj + xn+i= bi for all i = 1, 2, …, m xj 0 for all j = 1, 2, … , n+m Alternatively, Maximize j=1,n+m cjxj subject to j=1,n+m aij xj- bi = 0 for all i = 1, 2, …, m xj 0 for all j = 1, 2, … , n+m As the algorithm proceeds, we modify the objective function by performing elementary row operations - we multiply the ith constraint by i and add it to the objective function.

  14. Strong Duality Theorem (contd.) The modified objective function: z = i=1,m bi i +j=1,n+m (cj - i=1,m aij i) xj = i=1,m bi i -j=1,n+m xj Optimality criteria:  0, or i=1,m aiji  cj (These are the dual constraints.) The objective function value of the current primal solution: z = i=1,m bi i (This is also the value of dual obj.) Primal objective function value equals the dual objective function value, establishing the strong duality theorem.

  15. Complementary Slackness Theorem THEOREM: Suppose that x = (x1, x2, … , xn) is primal feasible and that  = (1, 2, … , n) is dual feasible. Let (w1, w2, … , wm) denote the corresponding primal slack variables, and let (z1, z2, … , zn) denote the corresponding dual slack variables. Then, x and y are optimal for their respective problems if and only if xjzj = 0 for j = 1, 2, … , n, (1) wi i = 0 for i = 1, 2, … , m. (2) PROOF: Show that j=1,n cjxj = j=1,m bi i if and only if (1) and (2) are satisfied.

  16. Complementary Slackness Theorem (contd.) PRIMAL LP: Maximize j=1,n cjxj subject to j=1,n aij xj bi for all i = 1, 2, …, m (dual var. = yi; slack = wi) xj 0 for all j = 1, 2, … , n DUAL LP: Minimize i=1,m bi i subject to i=1,mi aij cj for all j = 1, 2, …, n (primal var. = xj; slack = zj) j 0 for all i = 1, 2, … , m

  17. Dual for a Problem in General Form PRIMAL LP: Maximize j=1,n cjxj subject to j=1,n aij xj= bi for all i = 1, 2, …, m xj 0 for all j = 1, 2, … , n Equivalent Primal LP: Maximize j=1,n cjxj subject to j=1,n aij xj bi for all i = 1, 2, …, m - j=1,n aij xj -bi for all i = 1, 2, …, m xj 0 for all j = 1, 2, … , n

  18. Dual for a Problem in General Form (contd.) Primal LP: Maximize j=1,n cjxj subject to j=1,n aij xj bi for all i = 1, 2, …, m (Dual variable: ) - j=1,n aij xj -bi for all i = 1, 2, …, m (Dual variable: ) xj 0 for all j = 1, 2, … , n DUAL LP: Minimize i=1,m bi - i=1,m bi subject to i=1,m aij - i=1,m aij cj for all j = 1, 2, …, n ,  0 for all i = 1, 2, … , m

  19. Dual for a Problem in General Form (contd.) DUAL LP: Minimize Subject to for all for all EQUIVALENT DUAL LP: Minimize Subject to for all where

  20. Rules for Forming the Dual

  21. Dual Simplex Method • The dual simplex is an alternate method to solve linear programming problems. • The method starts with a BFS which satisfies optimality conditions (dual feasible) but violates non-negativity feasibility conditions (is not primal feasible). • The method performs a sequence of pivot operations where it maintains optimality conditions throughout and restores feasibility conditions. • For several problems, dual simplex method runs faster than the primal simplex method. • Dual simplex method is very useful for performing sensitivity analysis.

  22. Dual Simplex Method (contd.) • Initial simplex tableau: z - 60x1 - 30x2 - 20x3 = 0 8x1 + 6x2 + x3 + s1 = 48 4x1 + 2x2 + 1.5x3 + s2 = 20 2x1 + 1.5x2 + 0.5x3 + s3 = 8 • Optimal tableau: z + 5x2 + 10s2 + 10s3= 280 - 2x2 + s1 + 2s2 - 8s3 = 24 - 2x2 + x3 + 2s2 - 4s3 = 8 x1 + 1.25x2 - 0.5s2 + 1.5s3 = 2 • In our LP, 20 hours of finishing time is available. Suppose that we 30 hours of finishing time instead of 20 hours. • New updated RHS vector for this basis = B-1b = [44, 28, -3] which is primal infeasible.

  23. Dual Simplex Method (contd.) • Dual simplex method allows us to restore primal infeasibility of a solution while maintaining its dual infeasibility.

  24. Steps of the Dual Simplex Method Step 1. If the right-hand side of each constraint is nonnegative, an optimal solution has been found; otherwise go to Step 2. Step 2. Choose the most negative basic variable as the leaving variable. This row is the pivot row. Select the entering variable as the nonbasic variable which has a negative coefficient in the pivot row and for which the following ratio is minimum. Perform the pivot operation as usual. Step 3. If there is no negative coefficient in the pivot row, then the LP is infeasible.

  25. Uses of the Dual Simplex Method The dual simplex method is often used in the following situations: • Finding the new optimal solution after changing an LP’s right-hand side. • Solving a normal maximization or minimization problem. • When a new constraint is added.

  26. Solving a Normal LP by the Dual Simplex Method ORIGINAL LP: Minimize z = 4x1 + x2 subject to x1 + 4x2 1 3x1 - x2 3, x1, x2 0 LP with Surplus Variables: Minimize z = 4x1 + x2 subject to x1 + 4x2 - s1= 1 3x1 - x2 - s2= 3, x1, x2 0

  27. Solving a Normal LP by Dual Simplex Method (contd.) LP with Surplus Variables: Minimize z = 4x1 + x2 subject to - x1 - 4x2 + s1= -1 -3x1 + x2 + s2= -3, x1, x2 0 We can apply dual simplex method as this solution as the starting solution.

  28. Adding a New Constraint When a new constraint is added, one of the following three cases will occur: Case 1: The current solution satisfies the new constraint. The current solution is optimal for the new LP. Case 2: The current solution does not satisfy the new constraint, but the LP still has a feasible solution. We perform dual simplex method to reoptimize. Case 3: The additional constraint causes the LP to have no feasible solution. We perform dual simplex method to identify this possibility.

  29. Adding a New Constraint (contd.) Consider the following LP: z + 5x2 + 10s2 + 10s3= 280 - 2x2 + s1 + 2s2 - 8s3 = 24 - 2x2 + x3 + 2s2 - 4s3 = 8 x1 + 1.25x2 - 0.5s2 + 1.5s3 = 2 Additional Constraint: x1 + x2 12 Or, equivalently x1 + x2 – s4 = 12 or -x1 - x2 + s4 = -12

  30. Adding a New Constraint (contd.) • Add a row and make s4 as the basic variable of that row. • Obtain the new B-1 • Obtain the new B-1b • Perform dual pivot operations to restore the primal feasibility of the solution.

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