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Example 6.4

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Example 6.4

Plant and Warehouse Location Models

- Huntco produces tomato sauce at five different plants.
- The capacity (in tons) of each plant is given in the following table.

- The tomato sauce is stored at one of three warehouses. The cost per ton of producing tomato sauce at each plant and shipping it to each warehouse is given in the table shown here.

- Huntco has four customers.
- The cost of shipping a ton of sauce from each warehouse to each customer is given in the table shown here.

- Each customer must receive the amount (in tons) of sauce given in the following table.

- The annual fixed cost of operating each plant and warehouse is listed in this table.

- Huntco’s goal is to minimize the annual cost of meeting customer demands.
- The company wants to determine which plants and warehouses to open, as well as the optimal shipping plan.

- To model Huntco’s situation we need to keep track of the following:
- The shipments from plants to warehouses
- The shipments from warehouses to customers
- The fixed costs of operating plants and warehouses
- The shipping and production costs from plants to warehouses
- The shipping costs from warehouses to customers
- The total amount shipped out of each plant

- We must also ensure that
- Huntco pays the fixed costs for all plants and warehouses that it uses.
- The amount shipped into each warehouse equals the amount received by each warehouse.
- Each customer receives the specified demand.

- The spreadsheet model is shown on the next slide.
- This file can be used to complete the model.

- To form the model, follow these steps:
- Inputs. Enter the given data in the shaded ranges.
- Shipments. Enter any trial values for the shipments from each plant to each warehouse in the Shipped1 range and any trial values for the shipments from each warehouse to each customer in the Shipped2 range.
- Binary fixed cost variables. Enter any trial 0-1 values for the plant fixed-cost variables in the UsePlants range and the warehouse fixed-cost variables in the UseWhses range. The fixed-cost variable for a plant equals 1 if the plant is used and 0 if the plant is not used. Similarly, the fixed-cost variable for a warehouse equals 1 if the warehouse is used and 0 if the warehouse is not used.

- Amount shipped out of each plant. Calculate the amounts shipped out of the plants as row sums in the ShippedOut1 range. Specifically, enter the formula =SUM(B30:D30) in cell E30 and copy it to the rest of the ShippedOut1 range.
- Upper limit on amount shipped out of each plant. For each plant we need a constraint of the form Total shipped out of plant Plant capacity * Fixed-cost variable for plantThis inequality ensures that if Huntco uses the plant, then this plant’s fixed-cost variable will equal 1 and the company will have to pay the plant’s operating cost. In this case the inequality states that the total shipped out of the plant is less than or equal to the plant’s capacity.

- We generate the right side of the inequality in the UpCounds1 range. Specifically, enter the formula =B21*H6 in cell G30 and copy it to the rest of the UpBounds1 range. Note that if a plant is not used, the Solver is free to make this plant’s fixed-cost variable 0, and no fixed cost for this plant will be incurred. Then the inequality will be satisfied trivially (0 0).
- Amount shipped into and out of each warehouse. For each warehouse, we need “flow balance” – that is, we need the following constraint:Total shipments into warehouse = Total shipments out of warehouseTo implement this equation, first calculate the left side as column sums in the ShippedIn1 range. That is, enter the formula =SUM(B30:B34) in cell B35 and copy it to the rest of the ShippedIn1 range.

- For the right side of the equality, first calculate total shipments out of warehouses as row sums in the ShippedOut2_Col column range. That is enter the formula =SUM(B42:E42) in cell F42 and copying it to the rest of the ShippedOut2_Col range, entering the formula totals in the ShippedOut2_Row row range by selecting this range, entering the formula =TRANSPOSE(ShippedOut2_Col) and pressing Ctrl-Shift-Enter. This allows us to compare a row with a row when we specify the equation in the Solver dialog box.
- Upper limit on amount shipped out of each warehouse. For each warehouse we need a constraint of the formTotal shipped out of warehouse UpperBound * Fixed-cost variable for warehouse

- Here UpperBound is an upper bound on the most that could possibly be shipped out of any warehouse. Several possibilities for UpperBound could be used. We use the smaller of the total demand for all customers and the total capacity for all plants. If a warehouse’s fixed-cost variable is 0, then the inequality ensures that this warehouse cannot be used, whereas if the fixed-cost variable is 1, then this inequality is satisfied automatically. To operationalize the inequality, note that we already have the left sides in the ShippedOut2_Col range. To calculate the right side, enter the formula =E21*MIN(SUM(Capacities),(SUM(Demands)) in cell H42 and copy it to the rest of the UpBounds2 range.

- Amount received by each customer. Calculate the total amounts received by the customers as column sums in the ShippedIn2 range. That is, enter the formula =SUM(B42:B44) in cell B45 and copy it to the rest of the ShippedIn2 range.
- Shipping costs. Calculate the total costs of shipping from plants to warehouses and from warehouses to customers in cells B50 and B51 with the formulas =SUMPRODUCT(UnitCosts1,Shipped1) and =SUMPRODUCT(UnitCosts2,Shipped2).
- Fixed costs. Calculate the annual fixed costs for operating plants and warehouses in cells B52 and B53 with the formulas =SUMPRODUCT(FCosts1,UsePlants) and =SUMPRODUCT(FCosts2,UseWhses).

- Total cost. Finally, calculate the total annual cost in the TotCost cell with the formula =SUM(ShipCosts,FixedCosts).

- The completed Solver dialog box is shown here.

- The following is the explanation of the setup of the previous dialog box.
- Objective. The objective to minimize is total annual cost.
- Changing cells. There are four sets of changing cells – two sets for amounts to ship and two sets of binary variables for which plants and warehouses to use.
- Plant upper bounds. The constraint ShippedOut1<=UpBounds1 operationalizes the first inequality.
- Warehouse upper bounds. The constraint ShippedOut2_Col<=UpBounds2 operationalizes the second inequality.

- Warehouse balance. The constraint ShippedIn1=ShippedOut2_Row operationalizes the equality.
- Demand constraints. The constraint ShippedIn2>=Demands ensures that each customer received the required amount.

- The optimal solution shown indicates that Huntco should use plants 2, 3, and 5 and warehouses 2 and 3.
- Of course, the optimal shipping plan, as specified in the Shipped1 and Shipped2 ranges, uses only these plants and warehouses.
- This solution incurs a total annual cost of $700,500.
- If you obtain an “optimal” solution with a total cost somewhat larger than this, check the Solver tolerance setting. If it is at its default level of 5%, the Solver might very well stop short of optimal. We obtained our solution by setting the tolerance to 0%.

- At this point, you might want to review the inputs for this problem and see whether the optimal solution appears reasonable from an economic point of view.
- For example, although plant 1 has a relatively small fixed cost, it has relatively large unit shipping costs.
- This is evidently the reason for not using plant 1.
- However, the situation is not so obvious for plant 4 or warehouse 1. We think you will agree that on logistics problems such as this – and this is not even a large problem – more than intuition is necessary!

- We will not report any specific sensitivity analyses for this model, but many are possible.
- For example, we might check whether adding larger capacities at plants 1 and 4 would induce Huntco to open them.
- Or we might see what would happen if all the fixed costs increases by some percentage.
- Or we might see what would happen if all customer demands increased by some percentage.
- SolverTable, after some slight model modifications, can easily analyze any of these situations.