Irwin/McGraw-Hill Network Optimization Problem. Distribution Unlimited Co. Problem The Distribution...

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Irwin/McGraw-Hill Network Optimization Problem

Transcript of Irwin/McGraw-Hill Network Optimization Problem. Distribution Unlimited Co. Problem The Distribution...

Table of Contents Chapter 6 (Network Optimization Problems)

Network Optimization ProblemIrwin/McGraw-HillDistribution Unlimited Co. ProblemThe Distribution Unlimited Co. has two factories producing a product that needs to be shipped to two warehousesFactory 1 produces 80 units.Factory 2 produces 70 units.Warehouse 1 needs 60 units.Warehouse 2 needs 90 units.There are rail links directly from Factory 1 to Warehouse 1 and Factory 2 to Warehouse 2.Independent truckers are available to ship up to 50 units from each factory to the distribution center, and then 50 units from the distribution center to each warehouse.Question: How many units (truckloads) should be shipped along each shipping lane?6-2The Distribution Network

6-33Figure 6.1 The distribution network for the Distribution Unlimited Co. problem, where each feasible shipping lane is represented by an arrow.Data for Distribution Network6-4

4Figure 6.2 The data for the distribution network for the Distribution Unlimited Co. problem.A Network Model6-5

5Figure 6.3 A network model for the Distribution Unlimited Co. problem as a minimum-cost flow problem.The Optimal Solution

6-66Figure 6.4 The optimal solution for the Distribution Unlimited Co. problem, where the shipping amounts are shown in parentheses over the arrows.Terminology for Minimum-Cost Flow ProblemsThe model for any minimum-cost flow problem is represented by a network with flow passing through it.The circles in the network are called nodes.Each node where the net amount of flow generated (outflow minus inflow) is a fixed positive number is a supply node.Each node where the net amount of flow generated is a fixed negative number is a demand node.Any node where the net amount of flow generated is fixed at zero is a transshipment node. Having the amount of flow out of the node equal the amount of flow into the node is referred to as conservation of flow.The arrows in the network are called arcs.The maximum amount of flow allowed through an arc is referred to as the capacity of that arc.6-7Assumptions of a Minimum-Cost Flow ProblemAt least one of the nodes is a supply node.At least one of the other nodes is a demand node.All the remaining nodes are transshipment nodes.Flow through an arc is only allowed in the direction indicated by the arrowhead, where the maximum amount of flow is given by the capacity of that arc. (If flow can occur in both directions, this would be represented by a pair of arcs pointing in opposite directions.)The network has enough arcs with sufficient capacity to enable all the flow generated at the supply nodes to reach all the demand nodes.The cost of the flow through each arc is proportional to the amount of that flow, where the cost per unit flow is known.The objective is to minimize the total cost of sending the available supply through the network to satisfy the given demand. (An alternative objective is to maximize the total profit from doing this.)6-8Properties of Minimum-Cost Flow ProblemsThe Feasible Solutions Property: Under the previous assumptions, a minimum-cost flow problem will have feasible solutions if and only if the sum of the supplies from its supply nodes equals the sum of the demands at its demand nodes.The Integer Solutions Property: As long as all the supplies, demands, and arc capacities have integer values, any minimum-cost flow problem with feasible solutions is guaranteed to have an optimal solution with integer values for all its flow quantities.6-9Spreadsheet Model

6-1010Figure 6.5 A spreadsheet model for the Distribution Unlimited Co. minimum-cost flow problem, including the target cell Total Cost (D11). The changing cells Ship (D4:D9) show the optimal shipping quantities through the distribution network obtained by the Solver.The SUMIF FunctionThe SUMIF formula can be used to simplify the node flow constraints.=SUMIF(Range A, x, Range B)For each quantity in (Range A) that equals x, SUMIF sums the corresponding entries in (Range B).The net outflow (flow out flow in) from node x is then=SUMIF(From labels, x, Flow) SUMIF(To labels, x, Flow)6-11Typical Applications of Minimum-Cost Flow ProblemsKind ofApplicationSupplyNodesTransshipment NodesDemandNodesOperation of a distribution networkSources of goodsIntermediate storage facilitiesCustomersSolid waste managementSources of solid wasteProcessing facilitiesLandfill locationsOperation of a supply networkVendorsIntermediate warehousesProcessing facilitiesCoordinating product mixes at plantsPlantsProduction of a specific productMarket for a specific productCash flow managementSources of cash at a specific timeShort-term investment optionsNeeds for cash at a specific time6-1212Table 6.1 Typical kinds of applications of minimum-cost flow problems.The BMZ Maximum Flow ProblemThe BMZ Company is a European manufacturer of luxury automobiles. Its exports to the United States are particularly important.BMZ cars are becoming especially popular in California, so it is particularly important to keep the Los Angeles center well supplied with replacement parts for repairing these cars.BMZ needs to execute a plan quickly for shipping as much as possible from the main factory in Stuttgart, Germany to the distribution center in Los Angeles over the next month.The limiting factor on how much can be shipped is the limited capacity of the companys distribution network.Question: How many units should be sent through each shipping lane to maximize the total units flowing from Stuttgart to Los Angeles?6-13The BMZ Distribution Network

6-1414Figure 6.6 The BMZ Co. distribution network from its main factory in Stuttgart, Germany, to a distribution center in Los Angeles.A Network Model for BMZ

6-1515Figure 6.7 A network model for the BMZ Co. problem as a maximum flow problem, where the number in square brackets below each arc is the capacity of that arc.Spreadsheet Model for BMZ

6-1616Figure 6.8 A spreadsheet model for the BMZ Co. maximum flow problem.Assumptions of Maximum Flow ProblemsAll flow through the network originates at one node, called the source, and terminates at one other node, called the sink. (The source and sink in the BMZ problem are the factory and the distribution center, respectively.)All the remaining nodes are transshipment nodes.Flow through an arc is only allowed in the direction indicated by the arrowhead, where the maximum amount of flow is given by the capacity of that arc. At the source, all arcs point away from the node. At the sink, all arcs point into the node.The objective is to maximize the total amount of flow from the source to the sink. This amount is measured in either of two equivalent ways, namely, either the amount leaving the source or the amount entering the sink.6-17BMZ with Multiple Supply and Demand PointsBMZ has a second, smaller factory in Berlin.The distribution center in Seattle has the capability of supplying parts to the customers of the distribution center in Los Angeles when shortages occur at the latter center.Question: How many units should be sent through each shipping lane to maximize the total units flowing from Stuttgart and Berlin to Los Angeles and Seattle?6-18Network Model for The Expanded BMZ Problem

6-1919Figure 6.9 A network model for the expanded BMZ Co. problem as a maximum flow problem, where the number in square brackets below each arc is the capacity of that arc.Spreadsheet Model

6-2020Figure 6.10 A spreadsheet model for the expanded BMZ Co. problem as a variant of a maximum flow problem with sources in both Stuttgart and Berlin and sinks in both Los Angeles and Seattle. Using the target cell Maximum Flow (D21) to maximize the total flow from the two sources to the two sinks, the Solver yields the optimal shipping plan shown in the changing cells Ship (D4:D19).Distribution UnlimitedDistribution Unlimited Co. Minimum Cost Flow ProblemFromToShipCapacityUnit CostNodesNet FlowSupply/DemandRange NameCellsF1W130$700F180=80CapacityF5:F8F1DC50