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Unformatted text preview: Space IOE 202: lecture 3 outline Announcements Last time... Optimization problems and optimization models Another example of a Linear Programming model: Monet picture frames Linear Programming (LP) optimization models — brief history Solving small LP models graphically and interpreting solutions IOE 202: Operations Modeling, Fall 2009 Page 1 Space Last time Extension of the EOQ model for inventory management: quantity discounts Pet food supplier’s inventory problem
Does not ﬁt EOQ framework; diﬀerent model needed Identiﬁed
Decisions that need to be made Performance measure used to evaluate solutions Restrictions/constraints on decisions Simplifying assumptions made in the modeling process Started formulating a model of the problem IOE 202: Operations Modeling, Fall 2009 Page 2 Space Another inventory management problem1
• You are the Michigan distributor of Nature’s Peak, a highend brand of frozen dog food. You have (prepaid) contracts with local “boutique” pet stores to deliver, in each of the next 4 months, respectively, 50, 65, 100, and 70 lb of food, and these orders must be ﬁlled on time. • You obtain the food from the manufacturer at wholesale prices which vary month to month. In the next four months, unit prices are $5, $8, $4, and $7 per pound, respectively, and you can buy at most 80 lb each month. • Food needs to be delivered to the stores at the end of each month. You place your order with the manufacturer in the beginning of each month, receive your order at the end of the month, and immediately deliver food to the local stores. • If you have food remaining after the demand has been satisﬁed, you can keep some of in your wearhouse at a cost of $2 per pound per month until the next delivery, and donate the rest to the Humane Society. • In 4 months, Nature’s Peak is planning to change the recipe and packaging for this food. If you have any food left at that time, you can sell it to discount pet stores in the area for $6 per pound. (Until then, the company wants to maintain the product’s highend image.) • How should you manage your inventory for the next 4 months?
1 Section 4.12 of Denardo describes a problem in which such inventory management issues are part of the decisions
IOE 202: Operations Modeling, Fall 2009 Page 3 Space Operational decisions in the pet food distributor’s problem What decisions do you need to make for the coming 4 months?
How many pounds of food to order in each of the next 4 months How many pounds of food to store in each of the next 3 months, and how much to sell after 4th month How many pounds of food to donate in each of the next 4 months What performance measure would you use to compare diﬀerent decisions?
Net cost (Ordering and holding costs, minus the revenue from resale at the end) — to be minimized Ordering capacity of 80 lb each month Demand must be met each month We assumed all parameters given are known with certainty We assumed that there are no setup ordering costs, and no cost or ﬁnancial beneﬁt associated with donations
Page 4 What constraints (restrictions) must your decisions satisfy?
What assumptions are being made?
IOE 202: Operations Modeling, Fall 2009 Space Formulation of a mathematical model for the pet food distributor’s problem Decision variables: represent decisions by variables. x1 , x2 , x3 , x4 : lbs of food ordered in each of the 4 months s1 , s2 , s3 , s4 : lbs of food stored/sold at the end of each of the 4 months y1 , y2 , y3 , y4 : lbs of food donated in each of the 4 months Constraints: express all (explicit and implicit) constraints and restrictions on the values of the decision variables. IOE 202: Operations Modeling, Fall 2009 Page 5 Space Formulation of a model for the pet food distributor’s – cont. Objective function: express the performance criterion in terms of the decision variables; should it be minimized of maximized?
Minimize 5x1 + 8x2 + 4x3 + 7x4 + 2(s1 + s2 + s3 ) − 6s4 IOE 202: Operations Modeling, Fall 2009 Page 6 Space Mathematical model for pet food distributor’s problem Decision variables: x1 , x2 , x3 , x4 : lbs of food ordered in each of the 4 months s1 , s2 , s3 , s4 : lbs of food stored/sold at the end of each of the 4 months y1 , y2 , y3 , y4 : lbs of food donated in each of the 4 months Mathematical model: minimize 5x1 + 8x2 + 4x3 + 7x4 +2(s1 + s2 + s3 ) − 6s4 (Net) cost objective subject to x1 = 50 + s1 + y1 Inventory balance in month 1 constraint x2 + s1 = 65 + s2 + y2 Inventory balance in month 2 constraint x3 + s2 = 100 + s3 + y3 Inventory balance in month 3 constraint x4 + s3 = 70 + s4 + y4 Inventory balance in month 4 constraint x1 , x2 , x3 , x4 ≤ 80 Ordering capacity constraint(s) x1 , x2 , x 3 , x4 , Nonnegativity constraint(s) s1 , s2 , s3 , s4 , y1 , y2 , y 3 , y4 ≥ 0
IOE 202: Operations Modeling, Fall 2009 Page 7 Space Diﬀerent, but equivalent, mathematical model for pet food problem Decision variables: x1 , x2 , x3 , x4 : lbs of food ordered in each of the 4 months s1 , s2 , s3 , s4 : lbs of food stored/sold at the end of each of the 4 months Mathematical model: minimize 5x1 + 8x2 + 4x3 + 7x4 +2(s1 + s2 + s3 ) − 6s4 (Net) cost objective subject to x1 ≥ 50 + s1 Inventory balance in month 1 constraint x2 + s1 ≥ 65 + s2 Inventory balance in month 2 constraint x3 + s2 ≥ 100 + s3 Inventory balance in month 3 constraint x4 + s3 ≥ 70 + s4 Inventory balance in month 4 constraint x1 , x2 , x3 , x4 ≤ 80 Ordering capacity constraint(s) x1 , x2 , x3 , x4 ≤ 80 Ordering capacity constraint(s) x1 , x2 , x3 , x4 ≤ 80 Ordering capacity constraint(s) x1 , x2 , x 3 , x4 , Nonnegativity constraint(s) s1 , s2 , s3 , s4 ≥ 0
IOE 202: Operations Modeling, Fall 2009 Page 8 Space Using a spreadsheet and Excel solver
We will use a spreadsheet to obtain the solution of this model. Components of such a spreadsheet:
Inputs The data needed to form the objective and constraints Changing cells Use designated cells whose values will play the role of the decision variables. In particular, the component of Excel that ﬁnds the optimal (best) values of variables will change the contents of these cells. Target (objective) cell This cell will contain the formula for computing the value of objective function, referencing the changing cells. Constraints Will be speciﬁed in the Solver dialog box. Cells with expressions for left and right hand sides need to be prepared in the spreadsheet referencing the changing cells. Nonnegativity constraints Can be speciﬁed by checking a box in the Solver dialog (Also, check “Assume Linear Model”) Suggestion: use references to input cells, rather than numbers, as much as possible in all formulas.
IOE 202: Operations Modeling, Fall 2009 Page 9 Space Optimal solution of pet food distributor’s problem Optimal solution: Optimal (net) cost: IOE 202: Operations Modeling, Fall 2009 Page 10 Space Optimization problems
Optimization problems: problems involving deciding which actions to select among all feasible ones to achieve the best (or optimal) performance, as measured by a speciﬁed performance criterion.
Pet food distributor’s problem Deciding on the size of a production batch in TV speakers problem Choose the leastcost daily diet among all those that satisfy dietary (and taste!) requirements. Choose the shortestdistance route from home to school. In a production facility, choose the most proﬁtable combination of products to manufacture from the available raw materials. Others? Optimization models: prescriptive mathematical models of optimization problems IOE 202: Operations Modeling, Fall 2009 Page 11 Space Terminology of optimization models – I Decision variables: the quantities that can vary; we often call them simply variables
Represent quantitive decisions that need to be made Objective function: the expression that is being minimized of maximized
Represents the performance measure Constraints: equations and inequalities that the decision variables must satisfy
Represent restrictions on decisions being made IOE 202: Operations Modeling, Fall 2009 Page 12 Space Linear functions and constraints
Examples of linear functions: 17xA − 57xB + 91xC or A − 3.4B + 2C . The variables of the ﬁrst expression are xA , xB , xC , and of the second — A, B , C . Both functions depend on their respective variables linearly. A linear constraint requires a linear function to be equal, greaterthanorequalto, or lessthanorequalto, a number: 2A − 3B = 6 and A − 3.4B + 2C ≤ −2 and C ≥ 0. (We do not consider the inequality 17xA − 57xB + 91xC > 12 to be a correct form of a constraint!) A linear programming model is an optimization model with a linear objective function and linear constraints
Page 13 IOE 202: Operations Modeling, Fall 2009 Space Terminology of optimization models – II Solution: an assignment of values to the decision variables Feasible solution: an assignment of values to the variables that satisﬁes all of the constraints
x1 = x2 = x3 = x4 = 80, s1 = s2 = s3 = s4 = 0 is a solution, but it is not feasible x1 = x2 = x3 = x4 = 80, s1 = 30, s2 = 45, s3 = 25, s4 = 35 is a feasible solution; objective value $1,910.00 Optimal solution: feasible solution with the best objective value among all feasible solutions IOE 202: Operations Modeling, Fall 2009 Page 14 Space Example: problem of optimal resource allocation2 • The Monet company produces four types of picture frames, labeled A, B, C, and D. The table below lists the unit selling price Monet charges for each type of frame. • Each type requires a certain amount of skilled labor, metal, and glass, as shown in the table. For production during the coming week, Monet can purchase up to 4000 hours of skilled labor, 6000 ounces of metal, and 10,000 ounces of glass. The unit costs are also indicated in the table. • Also, market constraints are such that it is impossible to sell more than 1000 typeA frames, 2000 typeB frames, 500 typeC frames, and 1000 typeD frames. • How many frames of each type should Monet produce during the coming week to maximize its proﬁt?
2 Another problem of this type: Recreational Vehicle problem on p. 19 of Denardo.
IOE 202: Operations Modeling, Fall 2009 Page 15 Space Data (inputs) for the Monet production problem Frame type Frame A Frame B Frame C Frame D Max. amount of resource Resource Unit prices Skilled labor 2 1 3 2 4000 hours $8.00 per hour Metal 4 2 1 2 6000 oz $0.50 per 1 oz Glass 6 2 1 2 10,000 oz $0.75 per 1 oz Selling price $28.50 $12.50 $29.25 $21.50 Maximal production 1000 2000 500 1000 IOE 202: Operations Modeling, Fall 2009 Page 16 Space Operational decisions at Monet
What decisions do managers at Monet need to make for the coming week? What performance measure is the company using to compare diﬀerent decisions? What constraints (restrictions) must their decisions satisfy? IOE 202: Operations Modeling, Fall 2009 Page 17 Space Formulation of a mathematical model for Monet Decision variables: represent decisions by variables. Objective function: express the performance criterion in terms of the decision variables; should it be minimized of maximized. IOE 202: Operations Modeling, Fall 2009 Page 18 Space Formulation of a model for Monet – cont. Constraints: express all (explicit and implicit) constraints and restrictions on the values of the decision variables. IOE 202: Operations Modeling, Fall 2009 Page 19 Space Mathematical model for Monet: Decision variables: xA , xB , xC , xD denote the number of frames A, B, C, and D to produce, respectively. Mathematical model: maximize 6xA + 2xB + 4xC + 3xD subject to 2xA + xB + 3xC + 2xD ≤ 4000 4xA + 2xB + xC + 2xD ≤ 6000 6xA + 2xB + xC + 2xD ≤ 10000 xA ≤ 1000 xB ≤ 2000 xC ≤ 500 xD ≤ 1000 xA , x B , xC , x D ≥ 0 Proﬁt objective Labor constraint Metal constraint Glass constraint Frame A sales constraint Frame B sales constraint Frame C sales constraint Frame D sales constraint Nonnegativity constraint This model is also a linear programming model, since all constraints are linear, and the objective function is a linear function.
IOE 202: Operations Modeling, Fall 2009 Page 20 Space Some comments An alternative (but equivalent) formulation can be constructed by including variables to represent the amount of labor, metal, and glass purchased. Objective function and constraints could be expressed in terms of this variables. Both integer and noninteger levels of production were allowed in our formulation. (Is it a reasonable assumption?) Next week, we will discuss the changes in the models/solution methods when the last assumption is not reasonable; for now, let us allow variables to be noninteger. IOE 202: Operations Modeling, Fall 2009 Page 21 Space Linear Programming (LP) history
Linear Programming (LP) models constitute a special class of optimization models. Some history: • Theoretical tools for solving systems of linear equations and inequalities, too calculationintense • 1930’s  early 1940’s — ﬁrst applications of LP models to speciﬁc problems in production planning (L. Kantorovich, Soviet Union) and transportation planning (T. Koopmans, Netherlands/USA) (“programming” means “planning”) • Mid1940’s — development of an algorithm (the Simplex Method) capable of ﬁnding an optimal solution of any LP model by G. Dantzig coincides with development of computers, enabling broad practical applications. • Late1940’s  today: LP models are increasingly used in many applications; simultaneously, new algorithms for solving LPs, and software implementing these algorithms, are developed, as larger models need to be solved.
IOE 202: Operations Modeling, Fall 2009 Page 22 Space LP today Used in all types of organizations, with applications in: eﬃcient resource allocation, military operations planning, production and inventory planning, capacity expansion, manufacturing process design, staﬀ scheduling, location planning, traﬃc routing, supply chain management, economic game theory, airline crew and plane scheduling, telecommunication capacity allocation and network design, medical treatment planning, image reconstruction, publishing (typesetting), ﬁnance (asset allocation), mathematics (as a proof technique and computational method), data analysis, pattern classiﬁcation, optimal control, mechanical structure design, electromagnetic antennae design, etc. IOE 202: Operations Modeling, Fall 2009 Page 23 Space Optimal solution of the Monet problem Optimal solution: Optimal proﬁt: Which inequality constraints are “tight” (i.e., hold as equalities) at the optimal solution? An alternative way to characterize the optimal solution: “Do not make any frames of type D. Produce as many frames of type A as you can sell. Use up all available labor and metal.” IOE 202: Operations Modeling, Fall 2009 Page 24 ...
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 Fall '09
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