Linear programming: Theory

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1 Division of the Humanities and Social Sciences Ec 181 KC Border Convex Analsis and Economic Theor Winter 2018 Topic 28: Linear programming: Theor 28.1 The saddlepoint theorem for linear programming The material for this chapter is based largel on the beautifull written book b David Gale [2]. A maximum linear program in standard inequalit form 1 is a constrained maximization problem of the form maximize x p x xa q (1) x 0 (2) where x and p belong to R n, q belongs to R m, and A is n m. The program is feasible if there is some x satisfing the constraints (1) and (2). Ever maximum linear program in standard inequalit form has a dual program, which is the minimization problem: q (3) A p (4) 0. (5) The original maximum linear program ma be called the primal program to distinguish it from the dual. Let us start b examining the Lagrangean for the primal program. Write (1) as q j (xa) j 0, j = 1,..., m, and let j denote the Lagrange multiplier for this constraint. Incorporate (2) b setting the domain X = R n +. The Lagrangean is then L(x, ) = p x + q xa. (6) 1 Gale [2] refers to this as the standard form of a linear program. However Dantzig [1] uses the term standard form in a different fashion. KC Border: for Ec 181, Winter 2018 src: LPTheor v ::15.57

2 KC Border Linear programming: Theor 28 2 Treating the dual as the problem of maximizing q A p 0, and using x to denote the vector of Lagrange multipliers, the Lagrangean for the dual is: q xa + x p, which is just (6) again. Consequentl, b the eas half of the Saddlepoint Theorem, if ( x, ȳ) is a saddlepoint of L(x, ) = p x + q xa over R n R m, then x is optimal for the primal program and ȳ is optimal (minimal) for the dual program. In particular, if there is a saddlepoint, then both programs are feasible. If we knew that both programs satisfied Slater s Condition, then the Saddlepoint Theorem would assert that an pair of optimal solutions would be a saddlepoint of the Lagrangean. Remarkabl, for the linear programming case, we do not need Slater s Condition Saddlepoint Theorem for Linear Programming The following are equivalent. 1. The Lagrangean has a saddlepoint over R n + R m The primal has an optimal solution. 3. The dual has an optimal solution. L(x, ) = p x + q xa 4. Both the primal and dual are feasible. Also, the following are equivalent. a. ( x, ȳ) is a saddlepoint of the Lagrangean. b. x is optimal for the primal and ȳ is optimal for the dual. c. x is feasible for the primal program and ȳ is feasible for the dual, and p x = q ȳ. The proof is broken down into a series of lemmas. The first of which is both eas and useful Lemma (Optimalit Criterion for LP) If x is feasible for the primal program and is feasible for its dual, then p x q. If in addition p x = q, then x is optimal and is optimal for the dual program, and p x = q = xa. Proof : Suppose x satisfies (1) and 0. Then xa q. Likewise if satisfies (4) and x 0, then xa x p. Combining these pieces proves the lemma. v ::15.57 src: LPTheor KC Border: for Ec 181, Winter 2018

3 KC Border Linear programming: Theor 28 3 If b some means we have found an optimum for the primal and the dual, then the Optimalit Criterion allows a simple wa to prove that it is an optimum. Most other optimization techniques do not include a proof that the outcome is optimal. The remaining gap is to show that if x is optimal, then the dual has an optimal solution ȳ and that p x = q ȳ (instead of p x < q ȳ). This brings us to the following Fundamental Dualit Theorem of LP If both a maximum linear program in standard inequalit form and its dual are feasible, then both have optimal solutions, and the values of the two programs are the same. If one of the programs is infeasible, neither has an optimum. Proof : (Gale [2]) Start b assuming both programs are feasible. We alread know that if x and are feasible for the primal and dual respectivel, then p x q. Thus it suffices to find a solution (x, ) 0 to the inequalities or, in matrix form xa q A p q x p 0, [x, ] A 0 p [q, p, 0]. (7) 0 A q Either these inequalities have a solution, or else b Farkas s Lemma , there is a nonnegative vector u 0, where u R m +, v R n +, and α R +, satisfing α A 0 p u A q α 0 (8) and [q, p, 0] u < 0. (9) α We shall show that this latter set of inequalities does not have a solution: Suppose b wa of contradiction that (8) and (9) have a nonnegative solution. Rewriting (8), we have Au αp (10) and va αq, (11) KC Border: for Ec 181, Winter 2018 src: LPTheor v ::15.57

4 KC Border Linear programming: Theor 28 4 while (9) becomes q u < p v. (12) Let x 0 be feasible for the primal, that is, xa q. Then xau q u (13) since u 0. Similarl let ȳ 0 be feasible for the dual, that is, Aȳ p. Then vaȳ v p (14) since v 0. We next show that α 0. For suppose α = 0. Then (10) becomes Au 0, which implies xau 0, since x 0. Also (11) implies vaȳ 0, since ȳ 0. Combining this with (13) and (14) ields q u xau 0 vaȳ v p, which contradicts (12). This shows that α > 0, so we ma without loss of generalit assume α = 1. In this case, (10) becomes Au p and (11) becomes va q, which impl that v is feasible for the primal program and u is feasible for the dual. Therefore, b Lemma , q u p v, which again contradicts (12). This contradiction shows that if both programs are feasible, then both have optimal solutions and both programs have the same value. If either program is infeasible, then certainl it cannot have an optimal solution. So suppose that the primal program is infeasible, but the dual is feasible. That is, xa q has no nonnegative solution, so b the theorem of the alternative again, there is a nonnegative satisfing A 0 and q < 0. Let z be an feasible nonnegative solution to the dual. Then z + α is feasible for an α 0, and q (z + α) = q z + αq as α. Therefore no optimal solution exists for the dual. A similar argument works if the dual is infeasible, but the primal is feasible. Proof of Theorem : B the eas half of the Saddlepoint Theorem, if ( x, ȳ) is a saddlepoint of the Lagrangean, then x is optimal for the primal and ȳ is optimal for the dual. Thus (1) = (2) & (3) and (a) = (b). B the Optimalit Criterion if x is optimal for the primal and ȳ is optimal for the dual, then ( x, ȳ) is a saddlepoint of the Lagrangean. It follows that (b) = (c) = (a) and (2) & (3) = (1). Clearl (2) & (3) = (4). The Fundamental Dualit Theorem implies (2) (3) and that (4) = (2) & (3). This finishes the proof of the Saddlepoint Theorem for Linear Programing. v ::15.57 src: LPTheor KC Border: for Ec 181, Winter 2018

5 KC Border Linear programming: Theor 28 5 Gale refers to the following result as the Equilibrium Theorem. It is also known as the Complementar Slackness Theorem. It is a simple consequence of the Optimalit Criterion (Lemma ) Complementar Slackness Theorem Suppose x and are feasible for the primal and dual respectivel. The are optimal if and onl if both (xa) j < q j = j = 0 (15) and (A) i > p i = x i = 0. (16) Proof : Suppose x and are feasible for the primal and dual respectivel. From xa q and 0, we have xa q with equalit if and onl if (15) holds. Similarl, (16) holds if and onl if xa = p x. The conclusion now follows from the Optimalit Criterion , which sas that x and are optimal if and onl if p x = q = xa. You ma suspect that it is possible to combine linear constraints with more general concave constraints that satisf Slater s Condition. This is indeed the case as Uzawa [4] has shown. (See also Moore [3].) 28.2 Other formulations Not ever linear program comes to us alread in standard inequalit form, nor is the inequalit form alwas the easiest to work with. There are other forms, some of which have names, and all of which can be translated into one another. In fact, we just translated a standard minimum inequalit form into a standard maximum inequalit form above. Each of these forms also has a dual, and the program and its dual satisf the Fundamental Dualit Theorem of LP That is, if both a linear program (in an form) and its dual are feasible, then both have optimal solutions, and the values of the two programs are the same. If one of the programs is infeasible, neither has a solution. Table summarizes these forms and their dual programs The primal is the dual of the dual Remember that the dual of a maximum linear program in standard inequalit form is a linear program of the form KC Border: for Ec 181, Winter 2018 src: LPTheor v ::15.57

6 KC Border Linear programming: Theor 28 6 q A p 0 where x and p belong to R n, q belongs to R m, and A is n m. Let us call this a minimum linear program in standard inequalit form. Now the dual program itself can be rewritten as the following maximum LP in standard inequalit form: maximize q ( A ) p 0, where A is the transpose of A. The dual of this program is: x p x A x p x 0, or maximize x p x xa p x 0, which is the primal. Thus the dual of a minimum LP in standard inequalit form is a maximum LP in standard inequalit form, and vice-versa. Moreover the dual of the dual is the primal The general form Let us start with a linear program in general maximum form, which allows for both linear inequalities and equations, and optional sign constraints on the components of x. v ::15.57 src: LPTheor KC Border: for Ec 181, Winter 2018

7 KC Border Linear programming: Theor 28 7 maximize x n p x = p j x j j=1 x A j q j j J L x A j = q j j J E x A j q j j J G x i 0 i I N x i free i I F x i 0 i I P where J L J E J G = {1,..., n}, and I N I F I P = {1,..., m}. We can translate this into standard inequalit maximum form as follows. Start b rewriting all the constraints as inequalities, x A j q j j J L x A j q j j J E x ( A j ) q j j J E x ( A j ) q j j J G. Next replace x b u v, where both 0 and v 0. This places no sign restrictions on the components of u v. To capture the requirement that x i 0 for i I N, we require u e i 0, where e i is the i th unit coordinate vector in R n. (Do ou see wh this works?) Similarl x i 0 corresponds to v e i 0. Thus our rewritten problem is maximize p (u v) u,v (u v) A j q j j J L (u v) A j q j j J E (u v) ( A j ) q j j J E (u v) ( A j ) q j j J G u e i 0 i I N v e i 0 i I P u 0 v 0. Or in matrix form maximize u,v (p, p) (u, v) KC Border: for Ec 181, Winter 2018 src: LPTheor v ::15.57

8 KC Border Linear programming: Theor 28 8 (u, v) AL A E A E A G I N 0 P (q L, q E, q E, q G, 0 N, 0 P ), A L A E A E A G 0 N I P where A L, A E, and A G are m-rowed matrices whose columns are A j for j J M, j J E, and j G respectivel, the columns of I N and I P are unit coordinate vectors e i for i N and i P respectivel, and q L, q E, and q P have components q j for j L, j E, and j G respectivel. The zeros are of the dimension the need to be. The dual of this maximum problem in standard inequalit form is thus the following: z,w z 0, w 0, and (q L, q E, q E, q G, 0 N, 0 P ) (z L, z E, w E, z G, z N, z P ) AL A E A E A G I N 0 P A L A E A E A G 0 N I P z L z E w E p, z G p z N z P where z L, z E, w E, z G, z N, z P are of the appropriate dimensions. Define R m b z j j J L j = z j w j j J E so that z j j J G and rewrite the dual as j 0 j J L j unsigned j J E j 0 j J G z 0, w 0, and A + IN 0 A 0 I P q z N z P p, p v ::15.57 src: LPTheor KC Border: for Ec 181, Winter 2018

9 KC Border Linear programming: Theor 28 9 where A is the m n matrix of columns A 1,..., A n. What this sas is A + z i e i p i N A z i e i p i P Since z 0, the dual can be written: q j 0 j J L j free j J E j 0 j J G A i p i i I N A i = p i i I F A i p i i I P Recall that the variables in the dual are the Lagrange multipliers for the primal. Thus we see that, the Lagrange multipliers associated with the equalit constraints (j J E ) are not a priori restricted in sign, while the multipliers for the inequalit constraints (i J L ) are nonnegative, and the multipliers for the inequalit constraints (i J G ) are nonpositive. Since the primal variables are the Lagrange multipliers for the dual program, the nonnegativit constraints (i I P ) on the primal correspond to inequalit constraints in the dual, the nonpositivit constraints (i I N ) on the primal correspond to inequalit constraints in the dual, and the unrestricted primal variable are associated with equalit constraints in the dual Canonical (equalit) form There is one more useful form for linear programs, the canonical or equalit form. 2 In it, all the constraints are equations, and all the variables are nonnegative. An LP is in canonical maximum form if it is written as: 2 The equalit form is what Gale [2] calls the canonical form. Dantzig uses the term canonical in a different fashion. KC Border: for Ec 181, Winter 2018 src: LPTheor v ::15.57

10 KC Border Linear programming: Theor maximize x p x xa = q x 0 To transform an inequalit form into the equalit form, introduce slack variables z R m and observe that xa q xa + z = q, z 0. It follows from the characterization of the dual to the general maximum problem that the dual program can be written as the decidedl non-equalit minimum problem q A p Note the lack of sign restrictions on. In the canonical maximization problem, we are looking to write q as linear combination of the rows of A. In discussing algorithms for solving linear programs it is useful to be able to restrict attention to solutions where q is a linear combination of a linearl independent set of rows. That is, we want to make sure that if an optimum exists, then there is an optimum x where {A i : x i > 0} is a linearl independent set, where A i denotes the i th row of A. We shall call such a x a basic solution to the linear program. To prove the existence of a basic solution we must first derive and prove the proper version of the Complementar Slackness Theorem for canonical programs Theorem (Complementar Slackness for Canonical From) Let x and be feasible for the canonical maximum problem and its dual. That is, x 0, xa = q, and A p. Then x and are optimal if and onl for i = 1,..., m, A i > p i = x i = 0. Similarl, let x and be feasible for the canonical minimum problem and its dual. Then x and are optimal if and onl for i = 1,..., m, A i < p i = x i = 0. Proof : I ll prove the result for the maximization problem. So let x 0, xa = q, and A p. v ::15.57 src: LPTheor KC Border: for Ec 181, Winter 2018

11 KC Border Linear programming: Theor ( = ) Assume that x and are optimal and that A i > p i. Since x and are optimal, we know b the Fundamental Dualit Theorem that p x = q. Since x 0 and A p, we have that On the other hand xa = q, so Combining these we get x A p x = q. m q = xa = x A = (A i )x i. i=1 xa = p x or m (A i p i )x i = 0. i=1 Since each x i 0 and each A i p i 0, the onl wa the sum can be zero is that each product term (A i p i )x i is zero. So if A i p i > 0, then x i = 0. which means if that can happens is ( = ) Assume that x i = 0 whenever A i > p i. Then since A p and x 0, we must have xa = p x. But xa = q so q = xa = p, which implies thatx and are both optimal Theorem If a canonical (equalit constraint) linear programming problem has an optimal solution, then it has a basic optimal solution. Proof : Let x and be an optimal solution to the primal and the dual. Put B = {i : x i > 0}. Then b the contrapositive of Complementar Slackness , A i = p i for all i B. B Lemma on nonnegative basic solutions there exists x satisfing xa = q such that B = {i : x i > 0} B and {A i : x i > 0} is linearl independent. Now for all i B B we also have A i = p i. That is, A i > p i implies x i = 0, so b Complementar Slackness again, x is also optimal Linear equations as LPs It is possible to recast the problem of solving linear equations and inequalities as LP problems. Consider the problem of finding a nonnegative solution to a sstem of equations. That is, find x such that xa = q x 0. KC Border: for Ec 181, Winter 2018 src: LPTheor v ::15.57

12 KC Border Linear programming: Theor Primal program Dual program General maximum form General minimum form maximize x p x q xa j q j j J L xa j = q j j J E xa j q j j J G x i 0 i I N x i free i I F x i 0 i I P j 0 j J L j free j J E j 0 j J G A i p i i I N A i = p i i I F A i p i i I P Standard maximum form Standard minimum form maximize x p x q xa q x 0 0 A p Canonical maximum form maximize x p x q xa = q x 0 A p free Canonical minimum form x p x maximize q Ax = q x 0 A p free Table v :: Selected forms src: LPTheor of linear programs KCand Border: theirfor duals. Ec 181, Winter 2018 The characterization of the general form implies the other results. The primal constraints ma also be expressed in terms of Ax rather than xa, in which case

13 KC Border Linear programming: Theor Consider the linear program in equalit minimum form: x,z 1 z xa + z = q [x, z] 0 Here 1 is the vector whose components are all 1. Without loss of generalit we ma assume q 0, for if q j < 0 we ma multipl A j and q j b 1 without affecting the solution set. Then note that this program is feasible, since x = 0, z = q is a nonnegative feasible solution. Since we require z 0, we have 1 z 0 and 1 z = 0 if and onl if z = 0, in which case xa = q. Thus, if this linear program has value 0 if and onl xa = q, x 0 has a solution, and an optimal (x, z) provides a nonnegative solution to the equation. At this point ou might be inclined to sa so what? In the next lecture, I describe the simplex algorithm, which is a special version of Gauss Jordan elimination, that is a reasonabl efficient and easil programmable method for solving linear programs. In other words, it also finds nonnegative solutions to linear equations when the exist. References [1] G. B. Dantzig Linear programming and extensions. Princeton: Princeton Universit Press. [2] D. Gale Theor of linear economic models. Chicago: Universit of Chicago Press. Reprint of the 1960 edition published b McGraw-Hill. [3] J. C. Moore Some extensions of the Kuhn Tucker results in concave programming. In J. P. Quirk and A. M. Zarle, eds., Papers in Quantitative Economics, 1, pages Lawrence, Kansas: Universit of Kansas Press. [4] H. Uzawa The Kuhn Tucker conditions in concave programming. In K. J. Arrow, L. Hurwicz, and H. Uzawa, eds., Studies in Linear and Nonlinear Programming, number 2 in Stanford Mathematical Studies in the Social Sciences, chapter 3, pages Stanford, California: Stanford Universit Press. KC Border: for Ec 181, Winter 2018 src: LPTheor v ::15.57

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