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9 Pages

### sec16_Notes

Course: ORIE 3300, Fall 2008
School: Cornell
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Word Count: 2493

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have 16 Duality We now seen in some detail how we can use the simplex method to solve linear programs, and how, at termination, the simplex method provides a proof of optimality. This proof, as we have seen, consists of a tableau equivalent to the system of equations defining the linear program, with no strictly positive reduced costs. In some contexts, we might be able to guess a good feasible solution for a...

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have 16 Duality We now seen in some detail how we can use the simplex method to solve linear programs, and how, at termination, the simplex method provides a proof of optimality. This proof, as we have seen, consists of a tableau equivalent to the system of equations defining the linear program, with no strictly positive reduced costs. In some contexts, we might be able to guess a good feasible solution for a linear program without using the simplex method. For example, we might have previously found an optimal solution of a related linear program, perhaps identical except for a slightly different objective function. How might we judge how good this solution is? If this solution was in fact optimal, how might we prove (or "certify") that fact? As a simple example, consider the linear program x1 - x2 + 7x3 maximize subject to 2x1 - x2 + x3 = 1 () x1 + x2 + 2x3 = 5 x1 , x 2 , x3 0. We might experiment by choosing feasible solutions: [2, 3, 0]T has value -1 [1, 2, 1]T has value 6 [0, 1, 2]T has value 13. Notice that any of these objective values is a lower bound on the optimal value of the linear program. In particular we conclude (16.1) optimal value 13. After some more experiments, we might guess that [0, 1, 2]T is a good feasible solution, and perhaps even optimal. To convince ourselves, we could try to discover upper bounds on the optimal value. As a simple example, notice that any feasible solution x satisfies the inequality x1 - x2 + 7x3 4(x1 + x2 + 2x3 ) = 20, as a consequence of the second constraint and the fact that each xj is nonnegative. We deduce optimal value 20. 79 By combining the constraints of the linear program in more complex ways, we can obtain other upper bounds. For example, we could multiply the second constraint by 3 and add the first constraint to deduce that any feasible x satisfies the inequality x1 - x2 + 7x3 (2x1 - x2 + x3 ) + 3(x1 + x2 + 2x3 ) = 16. Hence optimal value 16. This upper bound of 16 is smaller than the previous upper bound of 20, so provides more precise information about the true optimal value: the smaller the upper bound, the better. Experimenting in the same way, we might notice that any feasible x satisfies the inequality (16.2) so optimal value 13. But now we have solved out linear program! Combining this inequality with inequality (16.1), we learn that the optimal value is exactly 13, and furthermore that the feasible solution [0, 1, 2]T is optimal (since it attains the value 13). In this case we have managed to solve the linear program without using the simplex method. Furthermore, we could easily convince an observer that our solution is indeed optimal. We simply present them with the feasible solution [0, 1, 2]T and the "multipliers" 3 and 2 that we used to multiply the constraints to derive inequality (16.2). They then follow three simple steps: Verify the solution is feasible by substituting it into the constraints. Verify the objective value, a lower bound on the optimal value. Verify the inequality (16.2), an upper bound on the optimal value. The multipliers 3 and 2 comprise what we call a "certificate of optimality" for our feasible solution. In our example, we were, in some sense, lucky. First, we stumbled across what later turned out to be an optimal solution. Secondly, we were able to 80 x1 - x2 + 7x3 3(2x1 - x2 + x3 ) + 2(x1 + x2 + 2x3 ) = 13, certify that solution as optimal by guessing a certificate of optimality. For the second step, we could have taken a more systematic approach. We obtained all our upper bounds on the optimal value by taking multiples: y1 times the first constraint and y2 times the second constraint, where the numbers y1 , y2 are unrestricted in sign. We deduced that any feasible solution must satisfy the equation y1 (2x1 - x2 + x3 ) + y2 (x1 + x2 + 2x3 ) = y1 + 5y2 , or equivalently, (2y1 + y2 )x1 + (-y1 + y2 )x2 + (y1 + 2y2 )x3 = y1 + 5y2 . Any feasible solution x has all nonnegative components, so the corresponding objective value satisfies x1 - x2 + 7x3 (2y1 + y2 )x1 + (-y1 + y2 )x2 + (y1 + 2y2 )x3 = y1 + 5y2 , by comparing coefficients of each component xj , provided the following restrictions on y1 , y2 hold: 1 2y1 + y2 -1 -y1 + y2 7 y1 + 2y2 . In that case, we deduce the upper bound optimal value y1 + 5y2 . The smaller this upper bound is, the more precise the information on the true optimal value. To find the best possible upper bound, we therefore consider the following problem. y1 + 5y2 minimize subject to 2y1 + y2 1 () -y1 + y2 -1 y1 + 2y2 7. Notice that this problem is another linear program: we call it the "dual problem" for our original linear program (). Since by definition any feasible 81 solution of () has objective value an upper bound on the optimal value of (), we deduce (16.3) optimal value of () optimal value of (). We decided that the optimal value of () is 13. Our calculations above amount to observing that the vector y = [3, 2]T is feasible for (), with objective value 13. Hence in fact the optimal value of the dual problem () is also exactly 13. Observe how the dual problem () is related to the original problem (). The dual problem is a minimization problem, and the coefficients of the dual objective function are exactly the numbers on the right-hand side of the original constraints. Symmetrically, the numbers on the right-hand side of the dual constraints are exactly the coefficients of the original objective function. The matrix defining the dual constraints is just the transpose of the matrix defining the original constraints. Finally, all the dual variables are free. Motivated by this pattern, we make the following definition. The dual problem for a linear program in standard equality form, maximize cT x subject to Ax = b (P ) x 0, is the linear program (D) minimize bT y subject to AT y c. In this case, we call (P ) the primal problem. It is not hard to prove a general version of inequality (16.3). Theorem 16.4 (Weak Duality) Consider any linear program in standard equality form, and its dual problem. The primal objective value of any feasible solution for the primal problem is no larger than the dual objective value of any feasible solution for the dual problem. Consequently, primal optimal value dual optimal value. 82 Proof Suppose that the vector x is feasible for the primal problem (P ), ^ and that the vector y is feasible for the dual problem (D). For any index ^ j = 1, 2, . . . , n we know xj 0 and (AT y )j cj . ^ ^ Hence (AT y )j xj cj xj , ^ ^ ^ so we deduce (AT y )j xj ^ ^ j j cj xj . ^ Consequently, cT x (AT y )T x = (^T A)^ = y T (A^) = y T b = bT y , ^ ^ ^ y x ^ x ^ ^ as we wanted to prove. The last claim in the theorem follows. The following result is an easy consequence. Corollary 16.5 If a linear program in standard equality form is unbounded, then its dual problem must be infeasible. Proof If the dual problem (D) has a feasible solution y , then according to ^ the Weak Duality theorem above, we could deduce primal optimal value bT y , ^ contradicting the fact that the primal problem is unbounded. Hence (D) cannot have any feasible solutions. 2 By an almost identical proof, we deduce a corresponding result about the dual problem. Corollary 16.6 A linear program in standard equality must form be infeasible if its dual problem is unbounded. As in our first example, a certificate of optimality for a feasible solution x for the primal problem (P ) is a feasible solution for the dual problem (D) with the same objective value as the primal objective value of x. Our last corollary justifies the terminology. 83 2 Corollary 16.7 (Certificate of optimality) Consider a feasible solution x for a linear program in standard equality form, and a feasible solution y for its dual problem. If the primal objective value of x equals the dual objective value of y , then x is optimal for the primal, and y is optimal for the dual. Proof Suppose the primal and dual problems are (P ) and (D), so cT x = bT y . Any feasible solution x for (P ) satisfies c T x bT y , by the Weak Duality theorem, so it follows that x must be optimal for (P ). The argument that y is optimal for (D) is similar. 2 A certificate of optimality y for a feasible solution x is a powerful tool. It neatly allows us to persuade an observer of the optimality of x , simply by checking the validity of some equations and inequalities involving x and y : specifically, we just need to verify that x satisfies the primal constraints, that y satisfies the dual constraints, and that the two objective values are equal. The observer does not need to understand any subtleties of linear programming or the simplex method to be convinced by this proof. A natural question, therefore, is when certificates of optimality exist. Remarkably, the answer is very simple: corresponding to any optimal solution, there is always a certificate of optimality. This crucial result in the the theory of linear programming is called the Strong Duality theorem. Earlier in our study of linear programming, we have seen a systematic way to calculate and guarantee the optimal value of a linear program: the simplex method solves the linear program, and the final tableau provides the proof that final solution really is optimal. It should come as no surprise, therefore, that our proof of the Strong Duality theorem uses the simplex method. Theorem 16.8 (Strong Duality for standard equality form) If a linear program in standard equality form has an optimal solution, then so does its dual problem, and then the two optimal values are equal. Proof We consider the primal problem (P ), and suppose first that the columns of the constraint matrix A span. Since we are assuming that (P ) has an optimal solution, Theorem 13.1 (Existence of a basic feasible solution) guarantees that (P ) has a basic feasible solution. Starting from the 84 corresponding tableau, the simplex method with the smallest subscript rule must terminate, by Theorem 12.3 (Finite termination). Since (P ) is not unbounded, at termination we obtain an optimal tableau corresponding to a basic optimal solution x. By Theorem 12.2 (Tableau for a basis), we can ^ write the top equation of this tableau ^ ^ z - (cT - y T AN )xN = y T b, N where the vector y satisfies the equation ^ AT y = c B . B^ Since the tableau is optimal, we also know cT - y T AN 0, or in other words N AT y c N . N^ Putting these two equations together shows AT y c, so y is feasible for the ^ ^ dual problem (D). On the other hand, x is optimal for the primal problem ^ T T (P ), with objective value c x = b y , using the top row of the tableau again. ^ ^ Now Corollary 16.7 (Certificate of optimality) shows that y is optimal for ^ (D), and so we deduce furthermore that the primal and dual problems have equal optimal values. That completes the proof when the columns of the matrix A span. If, on the other hand, the columns of A don't span, then the rows of A are linear dependent. In other words, if we rewrite the problem (P ) in the form maximize cT x subject to aT x = bi (i = 1, 2, . . . , m) i x 0, then one of the vectors ai is a linear combination of the others. Since P has a feasible solution, the constraint involving that ai is redundant: it is a linear combination of the other equations. We could therefore delete that constraint to obtain an equivalent linear program with one less constraint. If necessary, we could repeat this process, until eventually we arrive at a linear program equivalent to (P ), maximize cT x subject to aT x = bi (i I) (P ) i x 0, 85 for some subset I {1, 2, . . . , m}, with the property that the set {ai : i I} is linear independent. Consider an optimal solution x for the original linear program (P ). Since the problem (P ) is equivalent, x must also be optimal for (P ). The columns of the constraint matrix for (P ) span, so we can apply the first half of our argument to deduce that the dual problem for (P ), namely minimize iI bi y i subject to iI yi ai c, also has an optimal solution, with components yi (for i I) say, and that the two optimal values are equal: bi yi = c T x . iI If we now define yi = 0 for every index i I, then the vector y is feasible for the dual problem (D), because m A y = i=1 T y i ai = iI yi ai c, and furthermore b y = T m bi yi = i=1 iI bi y i = c T x . Now Corollary 16.7 (Certificate of optimality) shows that y is optimal for (D), and so we deduce once again that the primal and dual problems have equal optimal values. 2 [A slight confession: there is one case in which we have not proved the Strong Duality theorem. Suppose the columns of A do not span, and in removing redundant rows we end up with nothing at all! This can only hold if A and b consist only of zeroes, and in this case it is not hard to prove the theorem directly.] As well as proving the Strong Duality theorem, the argument above gives some constructive insight into the dual problem. Suppose we use the revised 86 simplex method to solve a linear program in standard equality form (P ), where the columns of the constraint matrix A span. Suppose furthermore that we terminate with an optimal tableau corresponding to a basis B and solution x. In the final iteration, the algorithm computes the vector y as ^ usual by solving the equation AT y = c B . B But, as we observe in the proof, the solution of this equation is an optimal solution of the dual problem (D). We summarize in the following result. Corollary 16.9 (Dual solutions and revised simplex) If the revised simplex method terminates with an optimal solution, then the final vector y computed by the algorithm is optimal for the dual problem. We can interpret the earlier iterations in the revised simplex method in a similar vein. At each iteration, we have a current basis B and a corresponding basic feasible solution x. In Step 1 of that iteration, we compute a vector y ^ satisfying the equations AT y = c B B bT y = cT x. ^ We can interpret that computation as an attempt to find a feasible solution for the dual problem with the same objective value as the primal objective value of x. In particular, our chosen vector y satisfies each of the dual ^ constraints AT y ci (i B) i with equality. In Step 2 of the iteration, we examine whether y really is dual-feasible, by checking the inequalities AT y c j j (j N ). If all these inequalities hold too, we terminate, and y is a certificate of optimality for our current solution x. If, on the other hand, one of the inequalities ^ fails, we try to use that failure to improve the current solution. 87
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BEE 299 Lesson 6 March 2, 2008After reading the articles presented by Conway and Meadows, and analyzing their individual arguments, I came to the conclusion that my beliefs fall more inline with Meadows views on genetically modifies foods. Genetically mo
Cornell - BEE - 299
I need: Food: Money:I use/do thefollowing to meet need I use my own and my parents' income to purchase food in stores and on campus. Work and when necessary ask parentsShelter: I rent a room in a house with heating Clothing: For protection from the elem