Duality. Primal program P: Maximize n. Dual program D: Minimize m. j=1 c jx j subject to n. j=1. i=1 b iy i subject to m. i=1

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1 Duality Primal program P: Maximize n j=1 c jx j subject to n a ij x j b i, i = 1, 2,..., m j=1 x j 0, j = 1, 2,..., n Dual program D: Minimize m i=1 b iy i subject to m a ij x j c j, j = 1, 2,..., n i=1 y j 0, i = 1, 2,..., m T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

2 The Duality Theorem Weak Duality Theorem If x is a feasible solution to P and y is a feasible solution to D then the value c T x is smaller than the value b T y. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

3 The Duality Theorem Weak Duality Theorem If x is a feasible solution to P and y is a feasible solution to D then the value c T x is smaller than the value b T y. (Strong) Duality Theorem If P has an optimal solution x then D has an optimal solution y and c T x = b T y. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

4 Finding Optimal Dual Solution from Primal Dictionary Primal program P: Maximize c T x under Ax b, x 0. Solve P using two phase simplex method, obtaining optimal solution x. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

5 Finding Optimal Dual Solution from Primal Dictionary Primal program P: Maximize c T x under Ax b, x 0. Solve P using two phase simplex method, obtaining optimal solution x. Last row of last dictionary: n+m z = z + c k x k. k=1 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

6 Finding Optimal Dual Solution from Primal Dictionary Primal program P: Maximize c T x under Ax b, x 0. Solve P using two phase simplex method, obtaining optimal solution x. Last row of last dictionary: Let y i = c n+i, i = 1, 2.,..., m. Then: n+m z = z + c k x k. k=1 1 y is a feasible solution to D: Minimize b T y under A T y c, y 0. 2 b T y = z. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

7 Explaining the magic... Primal P: Maximize 5x 1 + 4x 2 + 3x 3 Subject to 2x 1 + 3x 2 + x 3 5 4x 1 + x 2 + 2x x 1 + 4x 2 + 2x 3 8 x 1, x 2, x 3 0 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

8 Explaining the magic... Primal P: Maximize 5x 1 + 4x 2 + 3x 3 Subject to 2x 1 + 3x 2 + x 3 5 4x 1 + x 2 + 2x x 1 + 4x 2 + 2x 3 8 x 1, x 2, x 3 0 Dual D: Minimize 5y y 2 + 8y 3 Subject to 2y 1 + 4y y 1 + y 2 + 4y 3 4 y 1 + 2y 2 + 2y 3 3 y 1, y 2, y 3 0 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

9 Pivoting both primal and dual dictionary Primal: x 4 = 5 2x 1 3x 2 x 3 x 5 = 11 4x 1 x 2 2x 3 x 6 = 8 3x 1 4x 2 2x 3 z = 5x 1 + 4x 2 + 3x 3 Dual: (Not feasible!) y 4 = 5 + 2y 1 + 4y 2 + 3y 3 y 5 = 4 + 3y 1 + y 2 + 4y 3 y 6 = 3 + y 1 + 2y 2 + 2y 3 w = 5y y 2 + 8y 3 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

10 Pivoting x 1 and x 4 (and y 4 and y 1 ) Primal: x 1 = 5/2 1/2x 4 3/2x 2 1/2x 3 x 5 = 1 + 2x 4 + 5x 2 x 6 = 1/2 + 3/2x 4 + 1/2x 2 1/2x 3 z = 25/2 5/2x 4 7/2x 2 + 1/2x 3 Dual: (Still not feasible!) y 1 = 5/2 + 1/2y 4 2y 2 3/2y 3 y 5 = 7/2 + 3/2y 4 5y 2 1/2y 3 y 6 = 1/2 + 1/2y 4 + 1/2y 3 w = 25/2 + 5/2y 4 + y 2 + 1/2y 3 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

11 Pivoting x 3 and x 6 (and y 6 and y 3 ) Primal: Optimal! x 1 = 2 2x 4 2x 2 + x 6 x 5 = 1 + 2x 4 + 5x 2 x 3 = 1 + 3x 4 + x 2 2x 6 z = 13 x 4 3x 2 x 6 Dual: Feasible! (and optimal) y 1 = 1 + 2y 4 2y 2 3y 6 y 5 = 3 + 2y 4 5y 2 y 6 y 3 = 1 y 4 + 2y 6 w = y 4 + y 2 + y 6 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

12 Generalized Duality Theorem Primal program P: Maximize c 1 x 1 + c 2 x 2 + c 3 x 3 subject to y 1 : a 11 x 1 + a 12 x 2 + a 13 x 3 b 1 y 2 : a 21 x 1 + a 22 x 2 + a 23 x 3 b 2 y 3 : a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 x 1 0, x 2 0, x 3 UIS Dual program D: Minimize b 1 y 1 + b 2 y 2 + b 3 y 3 subject to x 1 : a 11 y 1 + a 21 y 2 + a 31 y 3 c 1 x 2 : a 12 y 1 + a 22 y 2 + a 32 y 3 c 2 x 3 : a 13 y 1 + a 23 y 2 + a 33 y 3 = c 3 y 1 0, y 2 0, y 3 UIS T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

13 Rules for Taking the Dual Constraints in primal corresponds to variables in dual (and vice versa) Coefficients of objective function in primal corresponds to constants in constraints in dual (and vice versa) Primal (Maximization) Dual (Minimization) for constraint 0 for variable for constraint 0 for variable = for constraint UIS for variable 0 for variable for constraint 0 for varaible for constraint UIS for variable = for constraint T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

14 Interpreting Duality A diet problem Buy food items in order to satisfy daily intake of energy, protein and calcium, minimizing the cost. Food Serving size Energy Protein Calcium Price Oatmeal Chicken Eggs Whole milk Cherry pie Pork with beans T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

15 Linear Programming Formulation Minimize 3x x x 3 + 9x x x 6 subject to 110x x x x x x x x x 3 + 8x 4 + 4x x x x x x x x x 1 0,..., x 6 0 Take the dual... T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

16 Linear Programming Formulation Minimize 3x x x 3 + 9x x x 6 subject to 110x x x x x x x x x 3 + 8x 4 + 4x x x x x x x x x 1 0,..., x 6 0 Take the dual... Maximize 2000y y y 3 subject to 110y 1 + 4y 2 + 2y y y y y y y y y y y 1 + 4y y y y y 3 19 y 1 0, y 2 0, y 3 0 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

17 Interpreting the Dual Maximize 2000y y y 3 subject to 110y 1 + 4y 2 + 2y y y y y y y y y y y 1 + 4y y y y y 3 19 y 1 0, y 2 0, y 3 0 How large can we price energy, protein and calcium? (Shadow prices) T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

18 Max Flow - Min Cut Let A be the vertex arc incidence matrix, i.e, 1 if e = (u, v) A u,e = 1 if e = (v, u) 0 otherwise Let d be the vector defined by d s = 1, d t = 1, and d i = 0 otherwise. Then we can express the max flow problem with following linear program. Maximize v subject to Af + dv = 0 f c f 0 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

19 Max Flow - Min Cut (II) Maximize v subject to Af + dv = 0 f c f 0 Take the dual... T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

20 Max Flow - Min Cut (II) Maximize v subject to Af + dv = 0 f c f 0 Take the dual... Minimize uv E g uvc(u, v) subject to p u p v + g uv 0 uv E p s + p t 1 g uv 0 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

21 Certifying Optimality Suppose we wish to prove to someone that a solution x to a linear program P is optimal. If we supply both the optimal solution x and an optimal solution y to the dual D, then one may easily verify that the solution x is fact optimal! T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

22 Certifying Optimality Suppose we wish to prove to someone that a solution x to a linear program P is optimal. If we supply both the optimal solution x and an optimal solution y to the dual D, then one may easily verify that the solution x is fact optimal! What if the linear program P is infeasible. How can we easily convince someone this is the case? T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

23 Certifying Optimality Suppose we wish to prove to someone that a solution x to a linear program P is optimal. If we supply both the optimal solution x and an optimal solution y to the dual D, then one may easily verify that the solution x is fact optimal! What if the linear program P is infeasible. How can we easily convince someone this is the case? Farkas Lemma Exactly one of the following is true: There exist x such that Ax b. There exist y 0 such that A T y = 0 but b T y < 0. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

24 Proof of Farkas Lemma Farkas Lemma Exactly one of the following is true: There exist x such that Ax b. There exist y 0 such that A T y = 0 but b T y < 0. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

25 Proof of Farkas Lemma Farkas Lemma Exactly one of the following is true: There exist x such that Ax b. There exist y 0 such that A T y = 0 but b T y < 0. Both cannot be true: 0 = 0 T x = (A T y) T x = y T Ax y T b = b T y < 0 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

26 Proof of Farkas Lemma Farkas Lemma Exactly one of the following is true: There exist x such that Ax b. There exist y 0 such that A T y = 0 but b T y < 0. Both cannot be true: 0 = 0 T x = (A T y) T x = y T Ax y T b = b T y < 0 We must show that both cannot be false. We do that be showing that if (1) is false then (2) is true. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

27 Proof of Farkas Lemma (II) Consider the program P and it s dual D: P: Maximize 0 subject to Ax b D: Minimize bt y subject to A T y = 0, y 0 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

28 Proof of Farkas Lemma (II) Consider the program P and it s dual D: P: Maximize 0 subject to Ax b D: Minimize bt y subject to A T y = 0, y 0 If P is infeasible then D is either infeasible or unbounded. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

29 Proof of Farkas Lemma (II) Consider the program P and it s dual D: P: Maximize 0 subject to Ax b D: Minimize bt y subject to A T y = 0, y 0 If P is infeasible then D is either infeasible or unbounded. D is always feasible. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

30 Proof of Farkas Lemma (II) Consider the program P and it s dual D: P: Maximize 0 subject to Ax b D: Minimize bt y subject to A T y = 0, y 0 If P is infeasible then D is either infeasible or unbounded. D is always feasible. Hence, if P is infeasible, then D is unbounded. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

31 Proof of Farkas Lemma (II) Consider the program P and it s dual D: P: Maximize 0 subject to Ax b D: Minimize bt y subject to A T y = 0, y 0 If P is infeasible then D is either infeasible or unbounded. D is always feasible. Hence, if P is infeasible, then D is unbounded. Thus we may find a solution y such that b T y < 0. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

32 Proof of Farkas Lemma (II) Consider the program P and it s dual D: P: Maximize 0 subject to Ax b D: Minimize bt y subject to A T y = 0, y 0 If P is infeasible then D is either infeasible or unbounded. D is always feasible. Hence, if P is infeasible, then D is unbounded. Thus we may find a solution y such that b T y < 0. Farkas Lemma follows from the Strong Duality Theorem. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

33 Proof of Farkas Lemma (II) Consider the program P and it s dual D: P: Maximize 0 subject to Ax b D: Minimize bt y subject to A T y = 0, y 0 If P is infeasible then D is either infeasible or unbounded. D is always feasible. Hence, if P is infeasible, then D is unbounded. Thus we may find a solution y such that b T y < 0. Farkas Lemma follows from the Strong Duality Theorem. We can also prove the Strong Duality Theorem from Farkas Lemma! T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

34 Proving Strong Duality from Farkas Lemma P: Maximize ct x subject to Ax b, x 0 (Strong) Duality Theorem D: Minimize bt y subject to A T y c, y 0 If P has an optimal solution x then D has an optimal solution y and c T x = b T y. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

35 Proving Strong Duality from Farkas Lemma P: Maximize ct x subject to Ax b, x 0 (Strong) Duality Theorem D: Minimize bt y subject to A T y c, y 0 If P has an optimal solution x then D has an optimal solution y and c T x = b T y. By weak duality we just have to provide feasible y such that b T y c T x. If no such y did exist then the following system is infeasible: A T c b T I y c x 0 T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

36 Proving Strong Duality from Farkas Lemma (II) If the following system is infeasible A T c b T I y c x 0 x by Farkas Lemma there exists λ 0 such that s [ A ] x b I λ = 0 s and [ c T c T x 0 ] x λ < 0 s T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

37 Proving Strong Duality from Farkas Lemma (III) Rewriting, and [ ] x A b I λ = 0 s [ c T c T x 0 ] x λ < 0 s we have x 0, λ 0, and s 0 such that Ax + λb Is = 0 and c T x + λc T x < 0. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

38 Proving Strong Duality from Farkas Lemma (III) Rewriting, and [ ] x A b I λ = 0 s [ c T c T x 0 ] x λ < 0 s we have x 0, λ 0, and s 0 such that Ax + λb Is = 0 and c T x + λc T x < 0. This implies: λc T x < c T x. Ax λb. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

39 Proving Strong Duality from Farkas Lemma (IV) We have x 0 and λ 0 such that λc T x < c T x. Ax λb. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

40 Proving Strong Duality from Farkas Lemma (IV) We have x 0 and λ 0 such that λc T x < c T x. Ax λb. Case 1: λ 0. Normalize x by λ. Then x is a better solution than x contradiction. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21

41 Proving Strong Duality from Farkas Lemma (IV) We have x 0 and λ 0 such that λc T x < c T x. Ax λb. Case 1: λ 0. Normalize x by λ. Then x is a better solution than x contradiction. Case 2: λ = 0. Then there exist x 0 such that Ax 0 and c T x > 0. Then we can improve x in the direction of x, and obtain a better solution than x contradiction. T. D. Hansen (Aarhus) Optimization, Lecture 8 February 18, / 21