Example 6 2 FixedCost Models Background Information n

Example 6. 2 Fixed-Cost Models

Background Information n The Great Threads Company is capable of manufacturing shirts, shorts and pants. n Each type of clothing requires that Great Threads have the appropriate type of machinery available. n The machinery needed to manufacture each type of clothing must be rented at the following rates: Shirt machinery $1500 per week; shorts machinery, $1200 per week; pants machinery, $1600 per week. n Each type of clothing requires the amounts of cloth and labor given in the following table. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Background Information – continued Data for Great Threads Example Labor Hours Cloth(Sq yd) Sales Price Unit Variable Cost Shirts 2. 0 3. 0 $35 $20 Shorts 1. 0 2. 5 $20 $10 Pants 6. 0 4. 0 $45 $25 n This table also shows the unit variable cost and selling price for each type of clothing. n There are 2000 hours and 2500 square yards of cloth available in a given week. n The company wants to find a solution that maximizes its weekly profit. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Solution n We first note that the cost of producing x shirts during a week is 0 if x=0, but it is 1500+20 x if x>0. n The cost structure violates the proportionality assumption that is needed for a linear model. n If proportionality were satisfied, then the cost of making, say, 10 shirts would be double the cost of making 5 shirts. n However, because of the fixed cost, the total cost of making 5 shirts is $1600, and the cost of making 10 shirts is only $1700. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Solution – continued n This violation of proportionality requires us to resort 01 variables to obtain a linear model. n To model the Great Threads problem, we need to keep track of the following: – Number of shirts, shorts, and pants produced – 0 -1 variable for each type of clothing that indicates whether any of that type of clothing is produced – Resource usage of labor and cloth – Total profit, which equals revenue from sales minus the cost of renting machines minus the variable cost of producing clothing n We must also ensure that if any of a given type of clothing is produced, then its 0 -1 variable equals 1. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

THREADS. XLS n The spreadsheet model is shown on the next slide. n This file can be used to complete the model. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Developing the Model n To formulate the model, follow these steps: – Inputs. Enter the given inputs in the shaded ranges. – 0 -1 values for shirts, shorts and pants. Enter any trial values for the 0 -1 variables for shirts, shorts, and pants in the Produce. Any range. For example, if you enter a 1 in cell C 16, you are implying that some shorts are produced. – Shirts, shorts and pants produced. Enter any trial values for the number of shirts, shorts, and pants produced in the Produced range. – Labor and cloth used. Calculate the total amount of labor hours in cell B 19 by entering the formula =SUMPRODUCT(Production, B 4: D 4). Then copy this to cell B 20 to calculate the amount of cloth used. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Developing the Model – continued – Effective capacities. Now we come to the tricky part of the formulation. We need to ensure that if any of the given type of clothing is produced, then its 0 -1 variable equals 1. This ensures that the model incurs the cost of renting a machine for this type of clothing. We could easily implement these constraints with IF statements. For example, to implement the constraint for shirts, we could enter the following formula in cell B 12: =IF(B 14>0, 1, 0). However Excel’s Solver is unable to deal with IF Functions accurately. Therefore we instead model the fixed cost constraint as follows: Shirts produced (Maximum number of shirts that could be produced) X (0 -1 variable for shirts). Of course, there are similar inequalities for shorts and pants. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Developing the Model – continued – The logic behind inequality. • If the 0 -1 variable for shirts is 0, then the right side of the inequality is 0, which means that the left side must be 0 – no shirts can be produced. That is, if the 0 -1 variable for shirts is 0, so that no fixed cost for shirts is incurred, then the inequality does not allow Great Threads to “cheat” and produce a positive number of shirts. • On the other hand, if the 0 -1 variable for shirts is 1, then the inequality is certainly true. It simply says that the number of shirts produced must be no greater than the maximum number that can be produced. • The inequality rules out the one case we want to rule out, namely, that Great Threads produces shirts but avoids the fixed cost. However, it will allow the 0 -1 variable to be 1 even if Great Threads plans to produce no shirts. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Developing the Model – continued – Fortunately, this is not a problem. When the Solver maximizes the total profit, it will never obtain such a solution because the total profit could be increased by setting the 0 -1 variable equal to 0 instead of 1. – To implement the inequality, we need an upper limit on the number of shirts that could be produced. To obtain this, observe that the number of shirts that could be produced is limited by the smaller of and 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Developing the Model – continued – Therefore, the smaller of these can be used as the maximum needed in the inequality. So in cell B 16, calculate the “effective capacity” for shirts with the formula =MIN($D$19/B 4, $D$20/B 5)*B 12. Then copy this formula to the range C 16: D 16 for shorts and pants. – Revenue and costs. Calculate the total sales revenue and the total valuable cost by entering the formula =SUMPRODUCT(Production, B 7: D 7) in cell B 23 and copying it to cell B 24. Then calculate the total fixed cost in the Tot. FCost cell with the formula =SUMPRODUCT(Produce. Any, Fixed. Costs). Note that this formula picks up the fixed costs only for those products with 0 -1 variables equal to 1. Finally, calculate the total profit in the Profit cell with the formula =Tot. Rev-Tot. VCost. Tot. FCost. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Using the Solver n The Solver dialog box is shown here. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Using the Solver – continued n We maximize the profit, subject to not using more hours or cloth than is available, and we ensure that production is no greater than effective capacity. n The key is that this effective capacity is 0 if we decide not to produce any of a given product. n Also, make sure to check the Assume Linear Model and Assume Non-Negative boxes under Solver options, and set the tolerance to 0%. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Solution n From the optimal solution shown before, we see that Great Threads should produce about 833 shirts, but no shorts or pants. n The total profit is $11, 000. Note that the 0 -1 variables for shorts and pants are both 0, which forces production of these products to be 0. n However, the 0 -1 variable for shirts, the product that is produced is 1. This ensures that the fixed cost of producing shirts is included in the total cost. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Solution – continued n It may be helpful to think of this solution as occurring in two stages. n In the first stage the Solver determines which products to produce – in this case, shirts only. n Then in the second stage, the Solver specifies how many shirts to produce. n Because each shirt is profitable, Great Threads makes as many shirts as possible, which is 833 because of the cloth constraint. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Solution – continued n Of course, these two stages are interrelated, and the Solver considers both of them in its solution process. n The Great Threads management might not be very excited about being a shirts-only shop. n Suppose the company wants to ensure that at least two types of clothing are produced at positive levels. n One approach is to add another constraint, namely, that the sum of the 0 -1 values in row 12 is greater than or equal to 2. n You can check, however, that when this constraint is added and the Solver is rerun, the 0 -1 variable for shorts becomes 1, but no shorts are produced! 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Solution – continued n Shirts are relatively more profitable than shorts, so only shirts are produced. n The new constraint forces Great Threads to rent an extra piece of machinery, but it doesn’t force the company to use it. n To force the company to produce some shorts, we would also need to add a constraint on the value in B 14, such as B 14>=100. n Any of these additional constraints will cost Great Threads money, but if they want to produce more than two types of clothing, this is their only option. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Sensitivity Analysis n Because the optimal solution currently calls for only shirts to be produced, an interesting sensitivity analysis is to see how much “incentive” is required for other products to be produced. n One way to check this is to increase unit revenues from shorts and pants simultaneously is a two-way Solver. Table. n We did this, keeping track of the binary variables for all three products, with the results shown on the following slide. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Sensitivity Analysis – continued n For clarity, we have shaded the cells that contain 1’s. These results are pretty much as expected. n As the revenues from other products increase, the company eventually shifts from producing shirts to producing shorts and/or pants. n The only possible surprise is in row 37. Here the unit revenue from shorts is still at its original value of $20. But when the unit revenue from pants increases to at least $60, Great Threads stops producing shirts and starts producing both shorts and pants. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7

Sensitivity Analysis – continued n On the other hand, when the unit revenue from the pants stay at $60 or $65 and the unit revenue from shorts increases sufficiently, the company stops producing pants and switches entirely to shorts. 6. 1 | 6. 3 | 6. 4 | 6. 5 | 6. 6 | 6. 7