Recap, and outline of Lecture 18

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1 Recap, and outline of Lecture 18 Previously Applications of duality: Farkas lemma (example of theorems of alternative) A geometric view of duality Degeneracy and multiple solutions: a duality connection Complementary basic primal and dual solutions Today Dual simplex: an alternative method sor solving standard form problems Economic interpretation of duality Sensitivity analysis of LPs IOE 510: Linear Programming I, Fall 2010 Outline of Lecture 18 Page 18 1 Dual simplex method for (primal) problem in standard form Consider a primal problem in standard form and its dual min c x s.t. Ax = b, x 0 and max p b s.t. p A c where A has linearly independent rows. Let B be a basis, and let B be the corresponding basis matrix Primal basic solution: x B = B 1 b, x N =0 Dual basic solution: use complementarity If x i is basic, set p A i = c i p B = c B Solution: p = c BB 1 ; c i =0(i.e.,ith dual constraint is active) for all basic i Note: basic solutions complementary by construction, and p b =(c BB 1 )b = c B(B 1 b)=c Bx B = c x. If we find a basis that is both dual-feasible and primal-feasible, then these basic solutions are optimal. In the dual simplex method We pick B so that c 0, or, equivalently, the vector p = c B B 1 is a BFS for the dual problem. We do not have B that is primal-feasible, i.e., B 1 b 0, until termination IOE 510: Linear Programming I, Fall 2010 Dual simplex method Page 18 2

2 The dual simplex method for problems in standard form min c x s.t. Ax = b, x 0 and max p b s.t. p A c Recall: need to find a basis B that is both primal- and dual-feasible Primal simplex method Works with primal-feasible (basic) solutions Travels from one primal BFS to another (adjacent) primal BFS, until dual feasibility is attained Each iteration consists of (i) selecting an entering non-basic variable (i.e., basic direction to move in), and then (ii) selecting leaving basic variable Dual simplex method Works with dual-feasible (basic) solutions Travels from one dual BFS to another (adjacent) dual BFS, until primal feasibility is attained Each iteration consists of (i) selecting leaving basic variable, and then (ii) selecting an entering non-basic variable (i.e., basic direction to move in) IOE 510: Linear Programming I, Fall 2010 Dual simplex method Page 18 3 Either method solves both the primal and dual problem. An iterations of the dual simplex method 1. A typical iteration starts with a basis matrix B such that all reduced costs are nonnegative. 2. If B 1 b 0, terminate we have an optimal basic feasible solution; otherwise, pick some l such that x B(l) < Consider the vector v n,whichisthelth row of the matrix B 1 A.Ifv 0, then the optimal dual cost is +, andthe algorithm terminates. 4. For each i such that v i < 0, compute the ratio c i / v i,andlet j be the index column that corresponds to the smallest ratio. The column A B(l) exist the basis, and is replaced by A j. 5. For every i, thenewvalue c i new c i new c = c i + v j i v j. of the reduced cost is IOE 510: Linear Programming I, Fall 2010 Dual simplex method Page 18 4

3 Convergence of the dual simplex method If x j is the entering variable, and c j > 0, then the dual cost increases when the change of basis is performed. If this is the case at every iteration, then the method is not going to repeat a basis, and will terminate by either finding an optimal solution, or establishing unboundedness of the dual/infeasibility of the primal. If at some iteration c j = 0, the above reasoning does not apply, and the algorithm may cycle. This can be avoided by using a suitable anticycling rule. IOE 510: Linear Programming I, Fall 2010 Dual simplex method Page 18 5 Duality and degeneracy Note: a dual basic solution p = c B B 1 is degenerate if and only if a non-basic primal variable has reduced cost 0. Consider the example: min 3x 1 + x 2 max 2p 1 s.t. x 1 + x 2 x 3 =2 s.t. p 1 +2p 2 3 2x 1 x 2 x 4 =0 p 1 p 2 1 x 1, x 2, x 3, x 4 0 p 1, p 2 0 If B =[A 3, A 4 ], primal BS x =(0, 0, 2, 0), dual BS p =(0, 0) If B =[A 1, A 3 ], primal BS x =(0, 0, 2, 0), dual BS p =(0, 1.5) If B =[A 2, A 3 ], primal BS x =(0, 0, 2, 0), dual BS p =(0, 1) IOE 510: Linear Programming I, Fall 2010 Dual simplex method Page 18 6

4 When should dual simplex be used? If a starting BFS is available for the dual, but not the primal If re-solving a previously solved problem with changed right-hand-sides If re-solving a previously solved problem with a new constraint added In general, dual simplex has been shown empirically to be very competitive, often more efficient than the primal simplex. IOE 510: Linear Programming I, Fall 2010 Dual simplex method Page 18 7 Optimal dual variables as marginal costs Consider a problem in standard form min c x s.t. Ax = b, x 0 where A has linearly independent rows. Let x be an optimal BFS, corresponding to basis B. Suppose x is non-degenerate. Then x B = B 1 b > 0. Suppose b is replaced with b + d. Then xb = B 1 (b + d) 0 if d is sufficiently small c is unaffected So, for small d, the current basis remains optimal, although the corresponding optimal BFS changes The change in optimal cost: c B x B c B x B = c B 1 ((b + d) b) =p d, where p = c B B 1 is the optimal solution to the dual problem. Hence, each component p i can be interpreted as the marginal cost, a.k.a.shadow price, per unit increase of the ith requirement b i. IOE 510: Linear Programming I, Fall 2010 Economic interpretation of dual variables Page 18 8

5 The Student and Pharmacist dual problems (P) min c x (D) max p b s.t. Ax b s.t. p A c x 0 p 0 Recall: x j is the amount of cereal j the student consumes p i is the price of nutrient/ingredient i Optimal solutions to either problem will have the same cost Let x and p be some optimal solutions of (P) and (D) If x j > 0, then A j p = c j : if the optimal diet includes cereal j, the pharmacist can charge the full price for its ingredients If a i x > b i,thenpi = 0. Interpretation? Suppose the student wants to increase his intake of i by a small amount: b i b i + δ. How would the cost of the diet change? Intuition: if δ is small, same constraint active/non-active at optimal primal/dual solutions of modified problems Then p stays optimal; cost: (p ) b + δp i Can view (D) as determining market value of cereal ingredients IOE 510: Linear Programming I, Fall 2010 Economic interpretation of dual variables Page 18 9

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