Lecture 13

1. Lecture 13 • Chapter 7 • Duality and Sensitivity in Linear Programming • 7.1 Objective values usually can be interpreted as either minimize cost or maximize benefit. • 7.3 Page 301 < constraints usually restrict the supply of some commodity • 7.4 Page 301 > constraints usually require satisfaction of a demand of something

2. Activities • 7.4 Page 302 Decision variables usually imply the selection of the level of some activity. • Max 13x1+24x2+5x3+50x4 (1) • s.t. x1+3x2> 89 (2) • -3x3-5x4< -60 (3) • 10x1+6x2+8x3+2x4< 608 (4) • x1, x2, x3, x4> 0 (5) • Since (3) has RHS < 0, convert to > type constraint.

3. Activities • Max 13x1+24x2+5x3+50x4 (1) • s.t. x1+3x2> 89 (2) • 3x3+5x4> 60 (6) • 10x1+6x2+8x3+2x4< 608 (4) • x1, x2, x3, x4> 0 (5) • There are 4 activities. • How much do we undertake? (x1, x2, x3, x4 ) • There are demands for 2 commodities – const (2) & (6) • There is a restriction on the supply of some raw material – const (4)

4. Relaxation • What would we do to relax constraint (2) a little? • x1+3x2> 89 (2) • Goes to • x1+3x2>80 • What would we do to relax constraint (4) a little? • 10x1+6x2+8x3+2x4< 608 (4) • Goes to • 10x1+6x2+8x3+2x4<610

5. Relaxed Constraints • If we relax a constraint, then the optimal objective value either remains the same or improves!!!!!!!!!!! • If we are minimizing, then optimal cost either remains the same or is reduced. • Why?

6. Tightening And Relaxing • Tightening constraints reduces the feasible region. • Relaxing constraints increases the feasible region. Tightening Relaxing

7. Relaxed & Tightened • Max ______ • Original Relaxed Tightened Obj Value Obj Value Obj Value = A = B > A = C < A

8. Big Model On Page 305 • Min ____ • With x1< 75 • Solve for various RHS using x1< RHS slope = - infinity Opt Obj value slope = - 4.98 slope = - 3.38 slope = 0 75

9. Big Model On Page 305 • Min _____ • With 0.12x1 + 0.011x2 + 1.0x6> 10 (RHS) • Solve for various RHS slope=infinity slope=50.11 slope=0 slope=36.73 slope=8.75

10. Summary 7.12 Page 308 Adding constraints tightens. Dropping constraints relaxes

11. Adding & Dropping Columns • 7.17 page 315 • Adding columns to a min problem results in an optimal value < the value before the addition • Dropping columns of a min problem results in an optimal objective > the value before the drop.

12. The Dual Problem – Page 325 • Primal Dual • Max cx Min bv • s.t. Ax < b s.t. vA > c • x > 0 v > 0 • I call this form of the primal and dual the standard form.

13. Example 1 • Primal Dual • Max 5x1 + 6x2 Min 3v1 + 7v2 • s.t. x1 + 2x2< 3 (v1) s.t. v1 – 3v2> 5 (x1) • -3x1 + 4x2< 7 (v2) 2v1 + 4v2> 6 (x2) • x1 , x2> 0 v1, v2> 0 • In the primal there is a dual variable for each constraint. In the dual, there is a primal variable for each constraint.

14. Example 2 • Suppose that you are given the primal in the following form and asked to give the dual: • Min 3w1 + 7w2 • s. t. w1 – 3w2> 5 • 2w1 + 4w2> 6 • w1, w2> 0 • The 1st step is to place this problem in standard form.

15. Primal In Standard Form • Max –3w1 –7w2 • s.t. -w1 + 3w2< -5 (x1) • -2w1 – 4w2< -6 (x2) • w1, w2> 0 • Dual is • Min –5x1 – 6x2 Max 5x1 + 6x2 • s.t. -x1 – 2x2> -3 or s.t. x1 + 2x2< 3 • 3x1 – 4x2> -7 -3x1 + 4x2< 7 • x1, x2> 0 x1, x2> 0 This is standard form with dual var x1 & x2

16. Dual Of Dual Is Primal • Note from the previous slide that the dual of the dual problem is the primal. • How do you find the dual of any problem? Always place the problem in standard form for the primal and then apply the rules from slide 12. This will guarantee that you can produce the dual for any problem. Note: The exact presentation of the dual is not unique. On slide 15 both problems at the bottom are the dual problem.

17. Example 3 Exercise 7-12 Part A P 366 • Min 17x1 + 29x2 + 0x3 + 1x4 • s.t. 2x1 + 3x2 + 2x3 + 3x4< 40 • 4x1 + 4x2 + 0x3 + 1x4> 10 • 0x1 + 0x2 – 3x3 – x4 = 0 • x1, …, x4> 0 • Place in standard form for the primal.

18. Example 3 Standard Form • Max -17x1 - 29x2 - 0x3 - 1x4 Max cx • s.t. 2x1 + 3x2 + 2x3 + 3x4< 40 s.t. Ax < b • -4x1 - 4x2 - 0x3 - 1x4< -10 x > 0 • 0x1 + 0x2 – 3x3 – x4< 0 • 0x1 + 0x2 + 3x3 + x4< 0 • x1, …, x4> 0 • This is standard form for the primal.

19. Dual • Min 40w1 – 10w2 + 0w3 + 0w4 (1) • s.t. 2w1 – 4w2 + 0w3 + 0w4> -17 (2) • 3w1 – 4w2 + 0w3 + 0w4> -29 (3) • 2w1 – 0w2 –3w3 + 3w4> 0 (4) • 3w1 – 1w2 – 1w3 + 1w4> -1 (5) • w1, …,w4> 0 (6) • This is the dual, but it doesn’t look like the answer given by Rardin on page 892.

20. Dual Modified • Min 40w1 – 10w2 + 0w3 + 0w4 (1) • s.t. -2w1 + 4w2 - 0w3 - 0w4< +17 (7) = -(2) • -3w1 + 4w2 - 0w3 - 0w4< +29 (8) = -(3) • -2w1 + 0w2 +3w3 - 3w4< 0 (9) = -(4) • -3w1 + 1w2 + 1w3 - 1w4< +1 (10) = -(5) • w1, …,w4> 0 (6) • This is the dual, but it still doesn’t look like the answer given by Rardin on page 892.

21. Dual Modified Let v1 = -w1 • Min -40v1 – 10w2 + 0w3 + 0w4 (1) • s.t. +2v1 + 4w2 - 0w3 - 0w4< +17 (7) • +3v1 + 4w2 - 0w3 - 0w4< +29 (8) • +2v1 + 0w2 +3w3 - 3w4< 0 (9) • +3v1 + 1w2 + 1w3 - 1w4< +1 (10) • v1 < 0, w2, …,w4> 0 (6) • This is the dual, but it still doesn’t look like the answer given by Rardin on page 892.

22. Dual Modified Let v3 = w3-w4 • Max +40v1 + 10w2(11) = -(1) • s.t. +2v1 + 4w2< +17 (7) • +3v1 + 4w2< +29 (8) • +2v1 + 0w2 +3v3< 0 (9) • +3v1 + 1w2 +1v3< +1 (10) • v1 < 0, w2, w3> 0, v3 unrestricted • This is the dual, but it still doesn’t look like the answer given by Rardin on page 892 because Rardin made an error.

23. What do the dual variables mean? • 7.20 Page 316 The dual variables (1 per constraint) give the change in the objective value per unit change in the RHS. • Primal Dual Variable • 0.3x1 + 0.4x2< 2 v1 = 10 • Implies that the estimated change in the objective value per unit increase in the 2 is 10. That is one unit (2 to 3) is expected to improve the objective function by 10.

24. This Is An Estimate Only slope = 10 Can not be achieved due to slope change 2 3 2 3

25. Optimal Objective Values Match • Primal {max cx: Ax < b, x > 0} • Dual {min bv: vA > c, v > 0} • Let x* solve the primal and v* solve the dual. • Then cx* = bv*.

26. Example 4 • Max x1 + 2x2 • s. t. 2x1 + x2 < 5 • x1 + 3x2> 5 • x1, x2> 0 Opt = (0,5) obj value = 10

27. Example 4 Dual • Max x1 + 2x2 min 5v1 – 5v2 • s. t. 2x1 + x2 < 5 (v1) s.t. 2v1 – v2> 1 • -x1 - 3x2< -5 (v2) v1 – 3v2> 2 • x1, x2> 0 v1, v2> 0 Opt = (2,0) Obj = 10

28. Complementary Slackness • 7.26 • Primal Constraint Dual Variable • ax < b (v) • Or • ax + s = b (v) • Complementary Slackness Says (s)(v) = 0 • ----------------------------------------------------- • If s > 0, then v = 0. If v > 0, then s = 0. • Either the constraint is active or v = 0.

29. 7.27 On Page 323 • Dual Constraint Primal Variable • va > c (x) • Or • va – s = c (x) • Complementary Slackness Says (s)(x) = 0

30. Table 7.1 – Primal max cx • Primal Dual • ax > b v < 0 • ax < b v > 0 • ax = b v unrestricted • xj> 0 vaj> cj • xj< 0 vaj< cj • xj unrestricted vaj = cj

31. The Dual & The Primal • 7.30 Page 328 The dual of the dual is the primal. • 7.31 Primal min{cx:_______} • Dual max{vb: ______} • Let X be feasible for the Primal • Let V be feasible for the Dual • Then cX>Vb

32. The Dual & The Primal • 7.32 If either the Primal or Dual has an optimal solution, then both do and the objective functions are the same at optimality! Primal min{cx:___} optimal cx* = v*b Dual max{vb:___}

33. The Dual & The Primal • 7.33 Let x* = B-1b be an optimum for the Primal. • Then v* = cBB-1 is an optimum for the Dual. • Solving either the Primal or the Dual produces an optimum for the other problem!