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WOOD 492 MODELLING FOR DECISION SUPPORT Lecture 12 Duality Theory

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Quiz 3 A multi period production plan –Maximize revenue –Limited production capacity –Limited storage capacity –Seasonal demand –Storage costs Oct 1, 2012Wood 492 - Saba Vahid2 Quiz3 matrix

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The Dual model Each LP problem, has another LP problem associated with it, called “the dual problem” Relationship between the original LP (Primal) and the dual model is very useful in solving large models and performing sensitivity analysis Use the example from lecture 9 to understand the dual problem –Produce 5 products using 3 moulders, 2 sanders and labor –Factory works 2 shifts, 8 hours per shift, 6 days / week –Each product takes 20 hrs of labor, with 8 workers working 48hrs/wk Oct 1, 2012Wood 492 - Saba Vahid3

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The primal model Find the unit value of each resource to minimize the total value of the resources used to produce the optimal solution (of the primal model) Therefore, our objective is to: minimize total value of resources Unit value of a resource: its contribution to the profits Our resources (i.e. decision variables): –moulder = contribution of each hour of moulding to total profits (Y1) –sander = contribution of each hour of sanding to total profits (Y2) –labor = contribution of each hour of labor to total profits (Y3) Oct 1, 2012Wood 492 - Saba Vahid4

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The objective function: Min Total value of resources = ∑ (amount of allocated resource * unit value of the resource) Min Z = total moulding hrs * Y1+ total sanding hrs * Y2+ total labor hrs * Y3 Min Z = 288 Y1+ 192 Y2+ 384 Y3 Oct 1, 2012Wood 492 - Saba Vahid5

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The constraints Each unit of Prod 1 uses 12 hours of moulding, 10 hours of sanding, and 20 hours of labor and contributes $550 to profits This mix of resources (moulder, sander and labor hrs) used for making Prod1 may also be used in other ways, as long as they generate at least $550 (otherwise we would use them to make Prod1) So the total contributions of this mix of resources to profits should be at least $550 Total contributions to profit => 550 12 Y1 + 10 Y2 + 20 Y3 =>550 Use the same logic to write down the remaining four constraints Oct 1, 2012Wood 492 - Saba Vahid6

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The dual problem Y1, Y2, and Y3 : contribution of each resource to profits Min Z = 288 Y1+ 192 Y2+ 384 Y3 S.t. 12 Y1 + 10 Y2 + 20 Y3 =>550 20 Y1 + 8 Y2 + 20 Y3 =>600 16 Y2 + 20 Y3 =>350 25 Y1 + 20 Y3 =>400 15Y1 + 20 Y3 =>200 All contributions to profits should be =>0, so all Y i =>0 Oct 1, 2012Wood 492 - Saba Vahid7

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Solution to the dual model Oct 1, 2012Wood 492 - Saba Vahid8 Y1Y2Y3 1.The final contribution to profits for this mix of resources is the same as the price of Prod1 and Prod2, so the resource are allocated to producing Prod1 and Prod2 2.The final contribution of this mix of resources is higher than the prices of the products, so they are already being used in a more profitable way and should not be assigned to produce Prod3, Prod4, and Prod5. 1212

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Constructing the dual problem To generate the dual model from the primal model (in standard form): –Variables in primal model will be represented by constraints in the dual model and vice versa –Change Maximization to Minimization and “ “ –The Obj. Fn. Coefficients will be the RHS in the dual problem –The RHS parameters will be the Obj. Fn. Coefficients –Coefficients related to each variable in the primal model will be used to construct the corresponding dual constraint Oct 3, 20129Wood 492 - Saba Vahid

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Example: primal model (from Lecture 9) Primal objective : maximization → dual objective : minimization Primal model has 3 constraints → dual model has 3 variables Dual objective coefficients are the RHS parameters of the primal model: 288, 192, and 384 –Primal objective: Max Z =550 Prod1 + 600 Prod2 + 350 Prod3 + 400 Prod4 + 200 Prod5 –Dual objective: Min Z = 288 Moulder + 192 Sander + 384 Labor Oct 1, 2012Wood 492 - Saba Vahid10 Max Z Moulder Sander Labor

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Example: primal-dual model (from Lecture 9) Dual objective: Min Z = 288 Moulder + 192 Sander+ 384 Labor Primal model has 5 variables→ dual model has 5 constraints Dual RHS values are the objective coefficients of the primal model: 550, 600, 350, 400 and 200 Each variable in primal model is represented by a constraint in dual model: –Constraint related to Prod1: 12 moulder + 10 sander + 20 labor => 550 –Constraint related to Prod2: 20 moulder + 8 sander + 20 labor => 600 Oct 1, 2012Wood 492 - Saba Vahid11 Max Z

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The dual model properties Important : Strong Duality property Optimal solution of the dual model= Optimal solution of the primal model Symmetry property: For any primal and its dual problem, all relationships between them must be symmetric because the dual of a dual problem, is the primal problem – therefore, it doesn’t matter which problem you call primal or dual Duality Theorem (relationship between dual and primal) –If one problem has an optimal solution, so does the other –If one problem has feasible solutions with an “unbounded” objective function (no optimal value), then the other problem has no feasible solution –If one problem has no feasible solution, the other problem has either no feasible solutions, or an unbounded objective function Oct 1, 2012Wood 492 - Saba Vahid12

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Applications of the dual model In solving LPs with the Simplex method, the more variables and constraints we have, the more computational effort is required –However, the number of constraints has a stronger impact on the required computational effort (compared to the number of variables) –If: number of constraints> number of variables, solving the dual model with Simplex reduces the computational effort and results in the same optimal solution In an algorithm for solving LP problems called the “dual simplex” method which is very useful in re-optimization and sensitivity analysis In generating insight into the primal model through economic interpretations of the dual model Oct 1, 2012Wood 492 - Saba Vahid13

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Next Class Economic interpretations of duality 14Wood 492 - Saba VahidOct 1, 2012

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