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Example 5.5 Non-logistics Network Models

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Background Information n The city of Spring View is taking bids from six companies on the right routes that must be driven in the surrounding school district. n Each company enter a bid on how much it will charge to drive selected routes, although not all companies bid on all routes. n The data appears in the table on the next slide.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Background Information -- continued n Blank cells indicates routes on which a company does not bid.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Background Information -- continued n The city needs to select which companies to assign to which routes with the specifications that –If a company does not bid on a route, it cannot be assigned to that route. –Exactly one company must be assigned to each route. –A company can be assigned to at most two routes. n The objective is to minimize the total cost of covering all routes.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Solution n We formulate this model in the “network” way. n You can imagine nodes for the bus companies on the left, nodes for the routes on the right, and all arrows going from left to right. n All flows are 0 or 1 – a company is either assigned to a route or it isn’t. n The constraint that a bus company can be assigned to at most two routes is handled by constraining the outflow from any company node to be at most 2. n To ensure that each route is covered by some company, we constrain the inflow to each route node to be at least 1.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a BUSROUTES.XLS n The completed model and corresponding Solver dialog box are shown on the next slides. n This file can be used to create the model. n Because this model is so similar to the MidWest Electric transportation model, we will not repeat all of the details here.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Developing the Model n The key steps are as follows: –Arc lists. The list of arcs (company – route pairs) in rows 16-46 correspond to nonblank cells only in the CostMatrix range. There is no point in including arcs that are infeasible. –Costs and flows. The costs in the Costs range are found with the usual method. The formula in cell C16 is =INDEX(CostMatrix,A16,B16) which is then copied down column C. Then enter any values in the Flows range. –Inflows and Outflows. Because no arcs point into company nodes and no arcs point out of route nodes, we require only outflows for company nodes and inflows for route nodes. To calculate these, enter the formulas =SUMIF(Origins,H16,Flows) and =SUMIF(Destinations,H24,Flows) in cells I16 and I24, and copy them down their respective ranges.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Developing the Model -- continued –Total cost. Calculate the total cost to the city in the TotalCost cell with the formula =SUMPRODUCT(Costs,Flows).

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Solution -- continued n The optimal solution shown on the next slide indicates the following assignments: company 1 covers route 3, company 2 covers routes 6 and 7, company 3 covers route 2, company 5 covers routes 4 and 8, and company 6 covers routes 1 and 5. n The total cost to the city of this assignment is $40,300. n After all, there is no constraint that every company must be assigned to at least one route, and company 4 is evidently underbid by at least one company for all routes.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Solution -- continued n If we wanted to require that all companies be assigned to at least one route, we would simply put a lower bound of 1 on all of the outflows from the company nodes (in rows 16-21). n Of course, this would increase the total cost to the city.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Sensitivity Analysis n One interesting sensitivity analysis is to see what effect the maximum routes upper bound constraints has on the total cost. n Presumably, if we allow more routes per bus company (assuming this is physically possible for the companies), the companies who tends to bid lowest will be assigned to the bulk of the routes, and the total cost will decrease. n The analysis itself is straightforward, with no modifications of the model necessary.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Sensitivity Analysis -- continued n We specify the MaxRoutes cell as the single input cell, allow it to vary from 1 to 7, say, in increments of 1, and keep track of total cost. n The resulting output appears here.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Sensitivity Analysis -- continued n We see first that if each company can be assigned to only one route, there is no feasible solution. But this is clear, there are eight routes to cover and only six companies! n For larger values of the MaxRoutes value, the total cost begins to decrease, but only until MaxRoutes reaches 3. n From that point, the city achieves no added flexibility by allowing companies to travel more routes.

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5.15.1 | 5.2 | 5.3 | 5.4 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a5.25.35.45.65.75.85.95.105.10a Sensitivity Analysis -- continued n Evidently, there is no single company or pair of companies who are consistently underbidding all others. n The following model is considerably more challenging, but we believe it is also much more interesting. n As with many problems that can be formulated as assignment models, the trick here is to recognize that it indeed involves appropriate assignments.

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