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Optimization Models Module 9

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MODEL OUTPUT EXTERNAL INPUTS DECISION INPUTS Optimization models answer the question, “What decision values give the best outputs?” OPTIMIZATION MODELS

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DECISION INPUTS Optimization models are necessary because resources are always limited. SCARCITY CONSTRAINTS

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Linear programming is a special type of optimization model that sets up constraints as linear equations and solves them simultaneously while maximizing an objective function. LINEAR PROGRAMMING CONSTRAINTS 2x + 6y < 150 y DECISION INPUTS x x +.8y < 40 x + 1.5y < 50 MODEL (objective function) 10x + 18y OUTPUT

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EXAMPLE 1: BOB’S MUFFLER SHOP Bob manufactures 2 types of mufflers, family and sport. Each family muffler nets him $10 and each sport muffler nets him $18. These numbers are known as the products’ ‘contribution margin’. His profit is determined by the function $10f + $18s, where f and s are the total number of each type of muffler sold. This is his objective function. His decision variables are the number of family and sport mufflers to manufacture each month. In an unconstrained situation, he would want to sell as many sport mufflers as because of their higher contribution. However…

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EXAMPLE 1: BACKGROUND In order to create a muffler, Bob must use brackets, labor hours, and alloy. The resources needed for each type are: FamilySports Brackets2.006.00 Labor hours1.000.80 Max Alloy11.5 Quantity Brackets150 Labor hours40 Max Alloy50 In any given month, he is limited in all three resources to these amounts: These are the problem constraints.

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EXAMPLE 1: MODELING CONSTRAINTS AS LINEAR EQUATIONS Bob also has an outstanding contract that forces him to manufacture 5 sports style mufflers each month and knows, from historical performance, that demand for family mufflers in a given month will not exceed 35. These constraints can be modeled as: These constraints can be modeled as a system of linear equations: 2f + 6s ≤ 150 (brackets) f +.8s ≤ 40 (labor) 50f + 33.33s ≤ 50 (alloy) f ≤ 35 (maximum demand) s ≥ 5 (contractual obligation) The objective function (profit) is: 10f + 18s

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EXAMPLE 1: GRAPHING A CONSTRAINT This situation can be represented graphically. Consider the constraint imposed by brackets: 2f + 6s ≤ 150. If zero family mufflers are manufactured, the equation changes to (0)f + 6s ≤ 150, which reduces to s ≤ 25. Thus a maximum of 25 sports mufflers can be made if no family mufflers are made. Similarly, 75 family mufflers can be made if there are no sports mufflers manufactured. Connecting these points gives a graphical representation of this constraint.

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EXAMPLE 1: RANGE OF FEASIBILITY Points on the line are scenarios in which all available brackets are used. The area under the line is called the Range of Feasibility because it covers all product mixes that the constraint allows. Any point above the line is infeasible because Bob does not have enough brackets to create that combination of mufflers. Range of Feasibility Infeasible Point: (30 sport, 40 family)

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EXAMPLE 1: GRAPHING ALL CONSTRAINTS The rest of the constraints can then be added in a similar fashion. 2f + 6s ≤ 150 (brackets) f +.8s ≤ 40 (labor) f + 1.5s ≤ 50 (alloy) f ≤ 35 (maximum demand) s ≥ 5 (contractual obligation)

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EXAMPLE 1: THE COMPOSITE RANGE OF FEASIBILITY The Range of Feasibility for the model is the area enclosed by all lines. 2f + 6s ≤ 150 (brackets) f +.8s ≤ 40 (labor) f + 1.5s ≤ 50 (alloy) f ≤ 35 (maximum demand) s ≥ 5 (contractual obligation) Range of Feasibility

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EXAMPLE 1: POTENTIAL PROFIT LINES The profit function will be a line with the slope of -5/9, the ratio of the contribution margin of family mufflers to sports mufflers. Each line corresponds to a particular level of total profit. The further out the line is, the greater the amount of profit. $100 $190 $280 $370 $460 $550 Total Profit:

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EXAMPLE 1: FEASIBLE PROFIT LINES By combining the profit lines with the range of feasibility, you can identify the highest profit line that touches a feasible point.

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EXAMPLE 1: OPTIMAL SOLUTION VIA GRAPHICAL METHOD The optimal point is a mix of 25 family mufflers and 16.67 sport mufflers. This combination will net Bob $550. Optimal point

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EXAMPLE 1: SPREADSHEET MODEL The situation can be modeled in Excel and an optimal solution found with Solver.

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EXAMPLE 1: SPREADSHEET MODEL Profit Available resources Contribution Margins Resources needed for each type of muffler Decisions: How many of each muffler to make Modeled constraints Contract Max demand

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EXAMPLE 1: SOLVER INPUTS Select the profit cell and choose solver from the tools menu.

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EXAMPLE 1: SOLVER INPUTS Objective function (profit) Decision variables Constraints Click

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EXAMPLE 1: SOLVER RESULTS Solver finds a solution. Select the answer and sensitivity reports for further analysis.

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EXAMPLE 1: OPTIMAL SOLUTION Solver found the same optimal solution as the graphical method.

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EXAMPLE 1: ANSWER REPORT All available brackets are used in the optimal solution, making it a binding constraint. There are 1.67 labor hours left over, or slack. ‘Binding’ means all available resources are used Slack is the amount of leftover resources The total resources used in the optimal solution Brackets: 2f + 6s ≤ 150 Labor: f +.8s ≤ 40

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EXAMPLE 1: SENSITIVITY ANALYSIS The shadow price is the amount of profit that an additional unit of available resources would yield. A non-binding constraint will have a shadow price of 0 because there is already an excess of the resource. An additional available bracket (a total of 151) will yield $1 of profit. An additional labor hour will yield no profit because there is already an excess. The allowable increase is the amount of resources that would have to be available to change the optimal solution. If 50 additional brackets are available, the optimal product mix will change. The allowable decrease is the amount of fewer resources that will change the optimal mix.

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EXAMPLE 1: SENSITIVITY ANALYSIS If an additional unit of resource can be obtained for less than the shadow price, then the difference will be profit. The allowable increase and decrease tells you how sensitive the decision inputs of the optimal solution (final value column) are to the right hand sides of each constraint.

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Profit: 10 f + 18 s EXAMPLE 1: RANGE OF OPTIMALITY The range of optimality is the range of contribution margins (coefficients in the objective function) in which a particular solution is the optimal solution. The contribution margin for each type of muffler may be uncertain.

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Profit: 10 f + 18 s EXAMPLE 1: RANGE OF OPTIMALITY The range of optimality tells you how sensitive the decision inputs for the optimal solution are to variation in the coefficients of the objective function. 8 ≤ contribution margin ≤ 1415 ≤ contribution margin ≤ 30 $18 + $12 = $30$18 - $3 = $15Sports mufflers range:

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