Using Solver to solve a minimization LP + interpretation of output BSAD 30 Dave Novak Source: Anderson et al., 2013 Quantitative Methods for Business 12.

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Using Solver to solve a minimization LP + interpretation of output BSAD 30 Dave Novak Source: Anderson et al., 2013 Quantitative Methods for Business 12 th edition – some slides are directly from J. Loucks © 2013 Cengage Learning

Overview Focus on section 7.5 Solving the minimization LP using MS Solver and interpreting output Surplus variables Standard formulation of LP

Setting up the problem Go to Excel  choose the “Data” tab and then click on Solver The Solver Screen pops up and we will need to reference the cells for our problem: Set Objective refers to the OF in B4 Click the Min Radio button as this is a minimization problem

Setting up the problem By Changing Variable Cells refers to our two variables (cells B1 and B2) Subject to the Constraints: refers to our constraints for the problem. We will add each one separately. Select “Add”  in the Add Constraint Menu choose cell B8 for the left-hand-side “Cell Reference” for constraint #1 and choose cell C8 for the right-hand- side “Cell Reference” for constraint #1

Setting up the problem Make sure the Make Unconstrained Variables non-negative check box is checked (this ensures that X1 and X2 are non-negative). We cannot produce a negative number of either decision variable And using the Pull-down menu make sure that Select Solving Method is Simplex LP Select SOLVE

Setting up the problem The cells in your spreadsheet model should change and cell B1 should now = 3.2, cell B2 should now = 0.8, and cell B4 should = 17.6 Using the Solver Results Menu that pops up on your screen highlight both “Answer” and “Sensitivity” on the right-hand-side under Reports and click OK

Minimization example from last class We solved this graphically ObjectiveFunctionObjectiveFunction “Regular”Constraints“Regular”Constraints Non-negativity Constraints ConstraintsNon-negativity Min 5x 1 + 2x 2 s.t. #1) 2x 1 + 5x 2 > 10 #2) 4x 1  x 2 > 12 #2) 4x 1  x 2 > 12 #3) x 1 + x 2 > 4 #3) x 1 + x 2 > 4 x 1, x 2 > 0 x 1, x 2 > 0

Minimization problem #2) 4x 1 - x 2 > 12 #1) 2x 1 + 5x 2 > 10 x1x1 x1x1 Feasible Region 654321654321 #3) x 1 + x 2 > 4 1 2 3 4 5 6 x2x2 x2x2 Objective function

Setting up the problem Download, save, and then open the Excel template file from the URL on the course schedule

Answer Report Excel Solver uses the term “slack” when referring to any non-binding ≥ or ≤ constraints Non-binding constraints do not restrict the feasible region at the optimal solution point

Slack and surplus variables Recall that slack and surplus variables are added to the standard form LP whenever there is an inequality Less-than or equal to constraints ≤ require a SLACK variable that is added to the LHS of the constraint so that we can set the LHS = the RHS

Slack and surplus variables Greater-than or equal to constraints ≥ require a SURPLUS variable that is subtracted from the LHS of the constraint so that we can set the LHS = the RHS Equality constraints do not require either a slack or surplus variable in standard form

Slack and surplus variables Surplus variables describe EXCESS quantity or the amount over the RHS of the constraint that is being used (associated with ≥) It is feasible to use less Slack variables describe IDLE resources or the amount of the RHS of the constraint that is not being used (associated with ≤) It is feasible to use more

Answer Report Constraint #1 is a non-binding constraint and has “surplus” = 0.4 (although Excel uses the term slack for both ≥ and ≤) Constraint #1 does not directly affect the optimal solution – it is a redundant, non- binding constraint Because this is a minimization problem, and the constraints are ≥, this lets us know that at the optimal point (3.2, 0.8) we exceed the RHS of constraint #1 by 0.4 units

Magnified view of extreme points #2) 4x 1 - x 2 > 12 #1) 2x 1 + 5x 2 > 10 #3) x 1 + x 2 > 4 Optimal  A (3.2, 0.8) This point is restricted or bound by constraints #2 and #3, but NOT by constraint #1 B (3.33, 0.67) Surplus in non-binding constraint #1 at optimal point

Answer Report Binding constraints let us know that all the resources associated with those specific constraints are used (there is no excess and there is no shortage) The constraint “binds” the problem, and all resources are used – there is no slack or surplus

Standard form of the LP Standard form of the LP involves adding slack or surplus variables to the mathematical model so the problem can be solved using strict equalities For each ≥ constraint, there should be a “surplus” variable (denoted by S i ) Surplus variables have a coefficient of 0 in the OF Surplus variables have a coefficient of 1 in the constraints

Standard form of the LP Reformulate our problem

Standard form of the LP Reformulate our problem

Slack or surplus? (I just randomly created this, so don’t solve it) ObjectiveFunctionObjectiveFunction “Regular”Constraints“Regular”Constraints Non-negativity Constraints ConstraintsNon-negativity Min 10x 1 + 25x 2 s.t. #1) 2x 1 + 5x 2 > 15 #2) 4x 1  x 2 > 20 #2) 4x 1  x 2 > 20 #3) x 1 + x 2 ≤ 7 #3) x 1 + x 2 ≤ 7 #4) 2x 1  15 #4) 2x 1  15 x 1, x 2 > 0 x 1, x 2 > 0

Standard Form?

Standard form of the LP You could think of slack/surplus variables as “placeholder” variables that hold the numeric difference between the LHS and RHS of any inequality In an equality constraint, the RHS MUST EQUAL the LHS, so there are no slack/surplus variables

Summary Solver setup for an LP Answer report summary Define surplus variable in context of any minimization problem Formulate Standard Form LP

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