WOOD 492 MODELLING FOR DECISION SUPPORT Lecture 13 Duality Theory

The Dual Problem Oct 3, 20122Wood Saba Vahid Another way of looking at a linear program Has the same optimal value for the objective function as the primal model The shadow prices of the dual model, are the optimal values of the decision variables in the primal model and vice versa –Extremely helpful when we can solve the dual model but not the primal one (e.g. large primal model which are computationally intensive)

Re-visiting the last example Oct 3, 20123Wood Saba Vahid

The constraints What does this constraint mean? 12 Y Y Y3 =>550 Y1 is the marginal value of resource 1 (moulding) when the optimum profits are being produced Y2 and Y3 are defined similarly (12 Y Y Y3) = total marginal value of producing 1 unit of product 1 $550 = marginal profit of 1 unit of product 1 If the total value of resources used and the total profits are the same, it is worth producing product 1, other wise it’s not worth it Oct 3, 20124Wood Saba Vahid

Oct 3, 2012Wood Saba Vahid5 Y1Y2Y3 16 Y Y3 is valued at $475 (with the current valuation as shown in the answer row) Product 3 produces $350 profits, so it doesn’t make sense to use these resources for making product 3 (for example, you are better off selling these resources in a “market” at their current value)

Shadow prices and reduced costs The results of the dual model are the shadow prices for the primal model and vice versa Shadow prices show a form of internal pricing/valuation for the resources at the optimal level: –How much is a resource worth for producing the optimal result –E.g. an hour of labor is worth $23.75 to the firm Reduced cost is calculated from the dual model as: Reduced cost = RHS (in the dual problem) – actual constraint value e.g. reduced cost of product 3 = 350 – 475 = -125 –So the actual value of the constraint (e.g. 475) shows the price that will make it feasible to produce product 3 Oct 3, 2012Wood Saba Vahid6

Dual-primal relationship Oct 3, 2012Wood Saba Vahid7 Primal (Maximize) Dual (Minimize) ith constraint 0 ith constraint =>ith variable <= 0 ith constraint = ith variable unrestricted j th variable =>0j th constraint => j th variable <= 0 j th constraint <= j th variable unrestricted j th constraint = ith constraint is bindingith variable is non-zero jth variable is non-zerojth constraint is binding

Lab 4 preview Harvest scheduling –3 stands with different ages and species –Different harvest regimes (based on the period of first cut, first cut can happen in any period*) * Note this is different from what we discussed in the class. There is now no restrictions on when the first cut can happen –Considering the total forest inventory (standing tree volume) –Even-flow restrictions –Beginning and ending inventory limits –Growth and yield volumes (how much trees grow in each period) Our objective is to maximize the total harvest (over 8 decades) Our decision variable is the total area (ha) in each stand to manage with each harvest regime –e.g. how many ha of stand 1 should we harvest in the first period (S1_P1) –e.g. how many ha of stand 3 should we leave uncut (S3_uncut) Oct 3, 2012Wood Saba Vahid8

Data tables The first table is related to existing (natural) stands Shows the available timber volume in the stands in each decade (decade 1 shows the current age of the stands) Oct 3, 2012Wood Saba Vahid9 The second table is related to regenerated stands (natural stands after being harvested once) Shows the available timber volume in the stands after different periods of growth

Data tables Yield tables show the harvest volumes per ha in each decade, depending on the period in which the first cut happens (the column headings correspond to our decision variables) e.g. for stand 1, if we choose to enter in the first period (first column), we will harvest 490 m3/ha (value extracted from the natural stands table), and then we will have to wait three decades before being able to harvest again, in which case we will get 364 m3/ha (value extracted from the regenerated stands table) Oct 3, 2012Wood Saba Vahid10 Fill the rest of the data table

Data tables Ending inventory tables, show the remaining volume of timber in the stands (m3/ha) at the end of each period, depending on when the first cut happens E.g. when we enter stand 1 in the first period, the remaining volume at the end of decade 1 is “0” because we have just harvested it. At the end of decade2 and 3, there is a growing inventory of standing timber (42 and 224 m3/ha, values extracted from the regenerated stands table), and since we harvest again in decade 4, the inventory drops to “0” again. Oct 3, 2012Wood Saba Vahid11 Fill the rest of the data table Lab 4 matrix

Next Class Economic interpretations of duality Oct 3, Wood Saba Vahid