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Chapter 14 Applications of Linear Optimization Applications of Linear Optimization Building Linear Optimization Models Building optimization models is more of an art than a science ◦ Learning how to build models requires logical thought facilitated by studying examples and observing their characteristics Key issues: ◦ Formulation ◦ Spreadsheet implementation ◦ Interpreting results ◦ Scenario and sensitivity analysis ◦ Gaining insight for making good decisions Types of Constraints in Optimization Models Simple Bounds ◦ Constraints on values of a single variable Limitations ◦ Allocation of scarce resources Requirements ◦ Minimum levels of performance Proportional Relationships ◦ Requirements for mixtures or blends of materials or strategies Balance Constraints ◦ Ensure the flow of material or money is accounted for at locations or between time periods: input = output Process Selection Models Process selection models generally involve choosing among different types of processes to produce a good ◦ Example: make or buy decisions Example 14.1: Camm Textiles A mill that operates on a 24/7 basis produces three types of fabric on a make-to-order basis The key decision is which type of loom to use for each fabric type over the next 13 weeks The mill has dobbie looms and 15 regular looms If demand cannot be met, the fabric is outsourced Example 14.1 Continued Model Formulation Di = yards fabric i to produce on dobbie loom Ri = yards fabric i to produce on regular loom Pi = yards fabric i to outsource Objective Minimize total cost of milling and outsourcing Constraints Requirements: Total production and outsourcing of each fabric = demand Limitations: Production on each type of loom cannot exceed the available production time Nonnegativity Example 14.1 Continued Demand constraints Production + outsourcing = demand ◦ Fabric 1: D1 + P1 = 45,000 ◦ Fabric 2: D2 + R2 + P2 = 76,500 ◦ Fabric 3: D3 + R3 + P3 = 10,000 Example 14.1: Continued Loom capacity limitation constraints First convert yards/hour into hours/yard E.g., for fabric on the dobbie loom: hours/yard = 1/(4.7 yards/hour) = 0.213 hours/yard Capacity of the three dobbie looms: (24 hours/day)(7 days/week)(13 weeks)(3 looms) = 6,552 hours Constraint on available production time on dobbie looms: 0.213D1 + 0.192D2 0.227D3 ≤ 6,552 Constraint for regular looms: 0.192R2 + 0.227R3 ≤ 32,760 Example 14.1 Continued Full Model J&M Manufacturing Answer Report J&M Manufacturing Sensitivity Report Note that none of the bound constraints appear in the Constraints section Interpreting Reduced Costs For product B, the lower bound constraint is B ≥ How much more would the profit on B have to be in order for it to be economical to produce anything other than the minimum amount required? ◦ The answer is given by the reduced cost The unit profit on B would have to be reduced by at least - $1.905 (that is, increased by at least + $1.905) Product D is at its upper bound ◦ The reduced costs of $19.29 tells how much the unit profit have to be lowered before it is no longer economical to produce the maximum amount Interpreting Reduced Costs as Shadow Prices Increasing the right-hand-side value of the bound constraint, B ≥ 0, by unit will result in a profit reduction of $1.905 Increasing the right-hand side of the constraint D ≤ 1,000 by will increase the profit by $19.29 ◦ The reduced cost associated with a bounded variable is the same as the shadow price of the bound constraint However, we no longer have the allowable range over which we can change the constraint values Auxiliary Variables for Bound Constraints Auxiliary variables are a new set of cells for any decision variables that have upper- or lowerbound constraints created by referencing (not copying) the original changing cells In the Solver model, use auxiliary variable cells—not the changing variable cells as defined—to define the bound constraints Auxiliary variables allow easier interpretation of shadow prices for bounded variables Example 14.13: Using Auxiliary Variable Cells J&M Manufacturing model Define bound constraints using the auxiliary variable cells Example 14.13 Continued Sensitivity report Bound constraints not appear in the Constraints section Example 14.14: Walker Wines A production/marketing allocation problem Walker Wines buys grapes from local growers and blends the pressings to make two types of wine (Shiraz and Merlot) Grape costs are $1.60 per bottle Shiraz and $1.40 per bottle Merlot Their contract requires between 40% and 70% of their wine to be Shiraz Predicted demand for Shiraz is 1000 bottles but increases by bottles for each dollar spent on advertising; the base demand for merlot is 2,000 bottles and increases by bottles for each $1 spent on advertising Production should not exceed demand Shiraz sells to retail stores for $6.25 per bottle and merlot is sold for $5.25 per bottle $50,000 is available to purchase grapes and advertise products Example 14.14 Continued Model formulation Define ◦ ◦ ◦ ◦ S = number of bottles of Shiraz produced M = number of bottles of merlot produced As = dollar amount spent on advertising Shiraz Am = dollar amount spent on advertising merlot Maximize profit = ($6.25S + $5.25M) - ($1.60S + $1.40M + A + A ) = 4.65S + 3.85M - A - A s m s m Example 14.14 Continued Constraints Do not exceed budget ◦ $1.60S + $1.40M + As + Am ≤ $50,000 Meet contractual requirements ◦ 0.4 ≤ S/(S + M) ≤ 0.7, or expressed in linear form: Production must not exceed demand ◦ S ≤ 1,000 + 5As ◦ M ≤ 2,000 + 8Am Nonnegativity 0.6S - 0.4M ≥ and 0.3S - 0.7M ≤ Spreadsheet Implementation for Walker Wines Solver Model for Walker Wines Using Sensitivity Information Correctly One crucial assumption in interpreting sensitivity analysis information for changes in model parameters is that all other model parameters are held constant Example 14.15: Evaluating a Cost Increase for Walker Wines Suppose the cost of Shiraz grapes increase $0.05 per bottle A $0.05 increase in cost results in a drop in the unit profit of Shiraz from $4.65 to $4.60 In the Sensitivity report, however, the change in the profit coefficient is within the allowable decrease of 0.05328, thus concluding that no change in the optimal solution will result Example 14.15 Continued If the model is re-solved using the new cost parameter, the solution changes dramatically In this case, the unit cost is also reflected in the binding budget constraint When we change the cost parameter, the constraint also changes This violates the assumption that all other model parameters are held constant ... 15% to 14. 5% nd Scenario: Fiber limitation is raised from 14% to 14. 5% Optimal Cost per pound: $0 .148 if fat requirement lowered $0.152 if fiber limitation raised Portfolio Investment Models. .. Process Selection Models Process selection models generally involve choosing among different types of processes to produce a good ◦ Example: make or buy decisions Example 14. 1: Camm Textiles... Example 14. 1 Continued Demand constraints Production + outsourcing = demand ◦ Fabric 1: D1 + P1 = 45,000 ◦ Fabric 2: D2 + R2 + P2 = 76,500 ◦ Fabric 3: D3 + R3 + P3 = 10,000 Example 14. 1: Continued
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