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Pesquisa Operacional Aplicada à Logística Prof. Fernando Augusto Silva Marins

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Sumário Introdução à Pesquisa Operacional (P.O.) Impacto da P.O. na Logística Modelagem e Softwares Exemplos Cases em Logística

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Pesquisa Operacional (P.O.) Operations Research Operational Research Management Sciences

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A P.O. e o Processo de Tomada de Decisão Tomar decisões é uma tarefa básica da gestão. Decidir: optar entre alternativas viáveis. Papel do Decisor : Identificar e Definir o Problema Formular objetivo (s) Analisar Limitações Avaliar Alternativas Escolher a melhor

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PROCESSO DE DECISÃO Abordagem Qualitativa: Problemas simples e experiência do decisor Abordagem Quantitativa: Problemas complexos, ótica científica e uso de métodos quantitativos.

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Pesquisa Operacional faz diferença no desempenho de organizações?

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Resultados - Finalistas do Prêmio Edelman INFORMS 2007

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FINALISTAS EDELMAN

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Como construir Modelos Matemáticos?

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Classification of Mathematical Models Classification by the model purpose – Optimization models – Prediction models Classification by the degree of certainty of the data in the model – Deterministic models – Probabilistic (stochastic) models

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Mathematical Modeling A constrained mathematical model consists of – An objective: Function to be optimised with one or more Control /Decision Variables Example: Max 2x – 3y; Min x + y – One or more constraints: Functions (,, =) with one or more Control /Decision Variables Examples: 3x + y 100; x - 4y 100; x + y 10;

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New Office Furniture Example Products Desks Chairs Molded Steel Profit $50 $30 $6 / pound Raw Steel Used 7 pounds (2.61 kg.) 3 pounds (1.12 kg.) 1.5 pounds (0.56 kg.) 1 pound (troy) = kg.

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Defining Control/Decision Variables Ask, Does the decision maker have the authority to decide the numerical value (amount) of the item? If the answer yes it is a control/decision variable. By very precise in the units (and if appropriate, the time frame) of each decision variable. D: amount of desks (number) C: amount of chairs (number) M: amount of molded steel (pound)

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Objective Function The objective of all optimization models, is to figure out how to do the best you can with what youve got. The best you can implies maximizing something (profit, efficiency...) or minimizing something (cost, time...). Total Profit =50 D + 30 C + 6 M Products Desks Chairs Molded Steel Profit $50 $30 $6 / pound D: amount of desks (number) C: amount of chairs (number) M: amount of molded steel (pound)

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Writing Constraints Create a limiting condition for each scarce resource : (amount of a resource required) (,, =) (resource availability) Make sure the units on the left side of the relation are the same as those on the right side. Use mathematical notation with known or estimated values for the parameters and the previously defined symbols for the decision/control variables. Rewrite the constraint, if necessary, so that all terms involving the decision variables are on the left side of the relationship, with only a constant value on the right side

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New Office Furniture Example If New Office has only 2000 pounds (746.5 kg) of raw steel available for production. 7 D + 3 C M2000 Products Desks Chairs Molded Steel Raw Steel Used 7 pounds (2.61 kg.) 3 pounds (1.12 kg.) 1.5 pounds (0.56 kg.) D: amount of desks (number) C: amount of chairs (number) M: amount of molded steel (pound)

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Special constraints or Variable Constraint Variable Constraint Non negativity constraint Lower bound constraint Upper bound constraint Integer constraint Binary constraint Mathematical Expression X 0 X L (number 0) X U (number 0) X = integer {0, 1, 2, 3, 4,…} X = 0 or 1 Writing Constraints

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No production can be negative; To satisfy contract commitments; at least 100 desks, and due to the availability of seat cushions, no more than 500 chairs must be produced. Quantities of desks and chairs produced during the production must be integer valued. New Office Furniture Example D 100, C 500 D, C integers D 0, C 0, M 0

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Example Mathematical Model MAXIMIZE Z = 50 D + 30 C + 6 M(Total Profit) SUBJECT TO: 7 D + 3 C M 2000 (Raw Steel) D 100(Contract) C 500(Cushions) D 0, C 0, M 0(Nonnegativity) D and C are integers Best or Optimal Solution: 100 Desks, 433 Chairs, 0.67 pounds Molded Steel Total Profit: $17,994 Ver Modelo no Lindo

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Delta Hardware Stores Problem Statement Delta Hardware Stores is a regional retailer with warehouses in three cities in California San Jose Fresno Azusa

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Delta Hardware Stores Problem Statement Each month, Delta restocks its warehouses with its own brand of paint. Delta has its own paint manufacturing plant in Phoenix, Arizona. San Jose Fresno Azusa Phoenix

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Although the plants production capacity is sometime inefficient to meet monthly demand, a recent feasibility study commissioned by Delta found that it was not cost effective to expand production capacity at this time. To meet demand, Delta subcontracts with a national paint manufacturer to produce paint under the Delta label and deliver it (at a higher cost) to any of its three California warehouses. Delta Hardware Stores Problem Statement

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Given that there is to be no expansion of plant capacity, the problem is to determine a least cost distribution scheme of paint produced at its manufacturing plant and shipments from the subcontractor to meet the demands of its California warehouses. Delta Hardware Stores Problem Statement

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Decision maker has no control over demand, production capacities, or unit costs. The decision maker is simply being asked, How much paint should be shipped this month (note the time frame) from the plant in Phoenix to San Jose, Fresno, and Asuza and How much extra should be purchased from the subcontractor and sent to each of the three cities to satisfy their orders? Delta Hardware Stores Variable Definition

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X 1 : amount of paint shipped this month from Phoenix to San Jose X 2 : amount of paint shipped this month from Phoenix to Fresno X 3 : amount of paint shipped this month from Phoenix to Azusa X 4 : amount of paint subcontracted this month for San Jose X 5 : amount of paint subcontracted this month for Fresno X 6 : amount of paint subcontracted this month for Azusa Decision/Control Variables

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National Subcontractor X4X4 X5X5 X6X6 X1X1 X2X2 X3X3 San Jose Fresno Azusa Phoenix Network Model and Decision Variables

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The objective is to minimize the total overall monthly costs of manufacturing, transporting and subcontracting paint, The constraints are (subject to): The Phoenix plant cannot operate beyond its capacity; The amount ordered from subcontractor cannot exceed a maximum limit; The orders for paint at each warehouse will be fulfilled. Model Structure

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To determine the overall costs: The manufacturing cost per 1000 gallons of paint at the plant in Phoenix - (M) The procurement cost per 1000 gallons of paint from National Subcontractor - (C) The respective truckload shipping costs form Phoenix to San Jose, Fresno, and Azusa - (T 1, T 2, T 3 ) The fixed purchase cost per 1000 gallons from the subcontractor to San Jose, Fresno, and Azusa (S 1, S 2, S 3 ) Costs

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MINIMIZE(M + T 1 ) X 1 + (M + T 2 ) X 2 + (M + T 3 ) X 3 + (C + S 1 ) X 4 + (C + S 2 ) X 5 + (C + S 3 ) X 6 Objective Function Where: Manufacturing cost at the plant in Phoenix: M Procurement cost from National Subcontractor: C Truckload shipping costs from Phoenix to San Jose, Fresno, and Azusa: T 1, T 2, T 3 Fixed purchase cost from the subcontractor to San Jose, Fresno, and Azusa: S 1, S 2, S 3 X 1 : amount of paint shipped this month from Phoenix to San Jose X 2 : amount of paint shipped this month from Phoenix to Fresno X 3 : amount of paint shipped this month from Phoenix to Azusa X 4 : amount of paint subcontracted this month for San Jose X 5 : amount of paint subcontracted this month for Fresno X 6 : amount of paint subcontracted this month for Azusa

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To write to constraints, we need to know: The capacity of the Phoenix plant (Q 1 ) The maximum number of gallons available from the subcontractor (Q 2 ) The respective orders for paint at the warehouses in San Jose, Fresno, and Azusa (R 1, R 2, R 3 ) Constraints

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The number of truckloads shipped out from Phoenix cannot exceed the plant capacity: X1 + X2 + X3 Q1 The number of thousands of gallons ordered from the subcontrator cannot exceed the order limit: X4 + X5 + X6 Q2 The number of thousands of gallons received at each warehouse equals the total orders of the warehouse: X1 + X4 = R1 X2 + X5 = R2 X3 + X6 = R3 All shipments must be nonnegative and integer: X1, X2, X3, X4, X5, X6 0 X1, X2, X3, X4, X5, X6 integer Constraints (Cont.)

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Respective Orders: R 1 = 4000, R 2 = 2000, R 3 = 5000 (gallons) Capacity: Q 1 = 8000, Q 2 = 5000 (gallons) Subcontractor price per 1000 gallons: C = $5000 Cost of production per 1000 gallons: M = $3000 Data Collection and Model Selection

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Transportation costs per 1000 gallons Subcontractor: S 1 = $1200;S 2 = $1400;S 3 = $1100 Phoenix Plant: T 1 = $1050;T 2 = $750;T 3 = $650 Data Collection and Model Selection (Cont.)

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Min ( )X 1 +( )X 2 +( )X 3 +( )X 4 +( )X 5 +( )X 6 Ou MIN 4050 X X X X X X 6 SUBJECT TO: X 1 + X 2 + X (Plant Capacity) X 4 + X 5 + X (Upper Bound - order from subcontracted) X 1 + X 4 = 4000 (Demand in San Jose) X 2 + X 5 = 2000 (Demand in Fresno) X 3 + X 6 = 5000 (Demand in Azusa) X 1, X 2, X 3, X 4, X 5, X 6 0 (non negativity) X 1, X 2, X 3, X 4, X 5, X 6 integer Delta Hardware Stores Operations Research Model

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X 1 = 1,000 gallons X 2 = 2,000 gallons X 3 = 5,000 gallons X 4 = 3,000 gallons X 5 = 0 X 6 = 0 Cost = $48,400 Delta Hardware Stores Solutions Ver Modelo no Excel

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CARLTON PHARMACEUTICALS Carlton Pharmaceuticals supplies drugs and other medical supplies. It has three plants in: Cleveland, Detroit, Greensboro. It has four distribution centers in: Boston, Richmond, Atlanta, St. Louis. Management at Carlton would like to ship cases of a certain vaccine as economically as possible.

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Data – Unit shipping cost, supply, and demand Assumptions – Unit shipping costs are constant. – All the shipping occurs simultaneously. – The only transportation considered is between sources and destinations. – Total supply equals total demand. To FromBostonRichmondAtlantaSt. LouisSupply Cleveland $ Detroit Greensboro Demand CARLTON PHARMACEUTICALS

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Boston Richmond Atlanta St.Louis Destinations Sources Cleveland Detroit Greensboro S 1 =1200 S 2 =1000 S 3 = 800 D 1 =1100 D 2 =400 D 3 =750 D 4 = CARLTON PHARMACEUTICALS Network presentation

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– The structure of the model is: Decision variables X ij = the number of cases shipped from plant i to warehouse j. where: i=1 (Cleveland), 2 (Detroit), 3 (Greensboro) j=1 (Boston), 2 (Richmond), 3 (Atlanta), 4(St.Louis) CARLTON PHARMACEUTICALS – Linear Programming Model Minimize Total Shipping Cost ST [Amount shipped from a source] [Supply at that source] [Amount received at a destination] = [Demand at that destination]

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Boston Richmond Atlanta St.Louis D 1 =1100 D 2 =400 D 3 =750 D 4 =750 Supply constraints Cleveland S 1 =1200 X11 X12 X13 X14 Supply from Cleveland X11+X12+X13+X Detroit S 2 =1000 X21 X22 X23 X24 Supply from Detroit X21+X22+X23+X Greensboro S 3 = 800 X31 X32 X33 X34 Supply from Greensboro X31+X32+X33+X34 800

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CARLTON PHARMACEUTICAL – The complete mathematical model Min 35X11+30X12+40X13+ 32X14 +37X21+40X22+42X23+25X24+ 40X31+15X32+20X33+38X34 ST Supply constraints: X11+X12+X13+X X21+X22+X23+X X31+X32+X33+X34800 Demand constraints: X11+X21+X X12+ X22+X32400 X13+ X23+X33750 X14+X24+X34750 All Xij are nonnegative = Total shipment out of a supply node cannot exceed the supply at the node. Total shipment received at a destination node, must equal the demand at that node.

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CARLTON PHARMACEUTICALS Spreadsheet =SUM(B7:E9) Drag to cells C11:E11 =SUMPRODUCT(B7:E9,B15:E17) =SUM(B7:E7) Drag to cells G8:G9

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MINIMIZE Total Cost SHIPMENTS Demands are met Supplies are not exceeded CARLTON PHARMACEUTICALS Spreadsheet

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CARLTON PHARMACEUTICALS Spreadsheet - solution

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CARLTON PHARMACEUTICALS - Sensitivity Report –Reduced costs The unit shipment cost between Cleveland and Atlanta must be reduced by at least $5, before it would become economically feasible to utilize it If this route is used, the total cost will increase by $5 for each case shipped between the two cities.

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CARLTON PHARMACEUTICALS - Sensitivity Report –Allowable Increase/Decrease This is the range of optimality. The unit shipment cost between Cleveland and Boston may increase up to $2 or decrease up to $5 with no change in the current optimal transportation plan.

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CARLTON PHARMACEUTICALS Sensitivity Report – Shadow prices For the plants, shadow prices convey the cost savings realized for each extra case of vaccine produced. For each additional unit available in Cleveland the total cost reduces by $2.

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CARLTON PHARMACEUTICALS Sensitivity Report –Shadow prices For the warehouses demand, shadow prices represent the cost savings for less cases being demanded. For each one unit decrease in demanded in Richmond, the total cost decreases by $32. –Allowable Increase/Decrease This is the range of feasibility. The total supply in Cleveland may increase up to $250, but doesn´t may decrease up, with no change in the current optimal transportation plan.

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– Cases may arise that require modifications to the basic model. Blocked routes - shipments along certain routes are prohibited. Remedies: – Assign a large objective coefficient to the route (C ij = 1,000,000) Modifications to the transportation problem

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– Cases may arise that require modifications to the basic model. Blocked routes - shipments along certain routes are prohibited. Remedies: – Assign a large objective coefficient to the blocked route (like as C ij = 1,000,000) – Add a constraint to Excel solver of the form X ij = 0 Modifications to the transportation problem Shipments on a Blocked Route = 0

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– Cases may arise that require modifications to the basic model. Blocked routes - shipments along certain routes are prohibited. Remedies: – Assign a large objective coefficient to the route (C ij = 1,000,000) – Add a constraint to Excel solver of the form X ij = 0 – Do not include the cell representing the rout in the Changing cells Modifications to the transportation problem Only Feasible Routes Included in Changing Cells Cell C9 is NOT Included Shipments from Greensboro to Cleveland are prohibited

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– Cases may arise that require modifications to the basic model. Minimum shipment - the amount shipped along a certain route must not fall below a pre-specified level. Maximum shipment - an upper limit is placed on the amount shipped along a certain route. Modifications to the transportation problem Remedy: Add a constraint to Excel of the form X ij B.

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The Capacitated Transshipment Model Sometimes shipments to destination nodes are made through transshipment nodes. Transshipment nodes may be – Independent intermediate nodes with no supply or demand – Supply or destination points themselves. Transportation on arcs may be bounded by given bounds

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The Linear Programming Model of this problem consists of: – Flow on arcs decision variables – Cost minimization objective function – Balance constraints on each node as follows: Supply node – net flow out does not exceed the supply Intermediate node – flow into the node is equal to the flow out Demand node – net flow into the node is equal to the demand – Bound constraints on each arc. Flow cannot exceed the capacity on the arc The Capacitated Transshipment Model

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DEPOT MAX A General Network Problem Depot Max has six stores located in the Washington D.C. area.

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5 6 DATA: The stores in Falls Church (FC) and Bethesda (BA) are running low on the model 5A Arcadia workstation. DEPOT MAX FC BA

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DATA: The stores in Alexandria (AA) and Chevy Chase (CC) have an access of 25 units. DEPOT MAX FC BA AA CC

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DATA: DEPOT MAX The stores in Fairfax and Georgetown are transshipment nodes with no access supply or demand of their own. FC BA AA CC FX GN Depot Max wishes to transport the available workstations to FC and BA at minimum total cost

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Data – There is a maximum limit for quantities shipped on various routes. – There are different unit transportation costs for different routes. DEPOT MAX

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FC BA DATA: The possible routes and the maximum flow limits are shown. DEPOT MAX FC BA AA CC FX GN

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FC BA DATA: The possible routes and the shipping unit costs are shown. DEPOT MAX FC BA AA CC FX GN 4 3

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–Supply nodes: Net flow out of the node] [Supply at the node] X 12 + X 13 + X 15 - X (Node 1) X 21 + X 24 - X 12 17(Node 2) –Intermediate transshipment nodes: [Total flow out of the node] = [Total flow into the node] X 34 +X 35 = X 13 (Node3) X 46 = X 24 + X 34 (Node 4) 7 –Demand nodes: [Net flow into the node] = [Demand for the node] X 15 + X 35 +X 65 - X 56 = 12 (Node 5) X 46 +X 56 - X 65 = 13 (Node 6) DEPOT MAX – Types of constraints

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DEPOT MAX The Complete mathematical model Min 5X X X X X X X X X X 65 S.T.X 12 +X 13 +X 15 –X X 12 +X 21 +X – X 13 +X 34 + X 35 = 0 – X 24 – X 34 +X 46 = 0 – X 15 – X 35 +X 56 - X 65 = -12 -X 46 – X 56 + X 65 = -13 X 12 3; X 15 6; X 21 7; X 24 10; X 34 8; X 35 8; X 46 17; X 56 7; X 65 5 All variables are non-negative

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DEPOT MAX - spreadsheet Usando o Template Network

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Uma empresa está planejando expandir suas atividades abrindo dois novos CDs, sendo que há três Locais sob estudo para a instalação destes CDs (Figura 1 adiante). Quatro Clientes devem ter atendidas suas Demandas (C i ): 50, 100, 150 e 200. As Capacidades de Armazenagem (A j ) em cada local são: 350, 300 e 200. Os Investimentos Iniciais em cada CD são: $50, $75 e $90. Os Custos Unitários de Operação em cada CD são: $5, $3 e $2. Admita que quaisquer dois locais são suficientes para atender toda a demanda existente, mas o Local 1 só pode atender Clientes 1, 2 e 4; o Local 3 pode atender Clientes 2, 3 e 4; enquanto o Local 2 pode atender todos os Clientes. Os Custos Unitários de Transporte do CD que pode ser construído no Local i ao Cliente j (C ij ) estão dados na Figura 1 do slide seguinte. Deseja-se selecionar os locais apropriados para a instalação dos CDs de forma a minimizar o custo total de investimento, operação e distribuição. Case em Logística Modelo de Pesquisa Operacional Expansão de Centros de Distribuição - CD

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Rede Logística, com Demandas (Clientes), Capacidades (Armazéns) e Custos de Transporte (Armazém-Cliente) A 1 =350 C 2 = 100 C 1 = 50 A 2 =300 C 3 =150 A 3 =200 C 4 =200 C 12 =9 C 14 =12 C 24 =4 C 34 =7 C 23 =11 C 33 =13 C 32 =2 C 22 =7 C 21 =10 C 11 =13 Figura 1

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Variáveis de Decisão/Controle: X ij = Quantidade enviada do CD i ao Cliente j L i é variável binária, i {1, 2, 3} sendo L i = 1, se o CD i for instalado 0, caso contrário

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Modelagem Função Objetivo: Minimizar CT = Custo Total de Investimento + Operação + Distribuição CT = 50L 1 + 5(X 11 + X 12 + X 14 ) + 13X X X L 2 + 3(X 21 +X 22 +X 23 +X 24 ) + 10X 21 +7X X 23 +4X L 3 + 2(X 32 + X 33 + X 34 ) + 2X X X 34 Cancelando os termos semelhantes, tem-se CT = 50L L L X X X X X X 23 +7X X X X 34

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Restrições: sujeito a X 11 + X 12 + X L 1 X 21 + X 22 + X 23 + X L 2 X 32 + X 33 + X L 3 L 1 + L 2 + L 3 = 2 Instalar 2 CDs X 11 + X 21 = 50 X 12 + X 22 + X 32 = 100 X 23 + X 33 = 150 X 14 + X 24 + X 34 = 200 X ij 0 L i {0, 1} Produção Demanda Não - Negatividade Integralidade Ver Modelo no Lindo

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Solução do Case em Logística OBJECTIVE FUNCTION VALUE = VARIABLE VALUE REDUCED COST L L L X X X X X X X X X X

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