1. You are thinking of opening a small copy shop. Renting a copier for a year costs $5000, and operating the copier costs $0.03 per copy. Other fixed costs of running the store will amount to $400 per month. You plan to charge an average of $0.10 per copy, and the store will be open 365 days per year. Each copier can make up to 100,000 copies per year. Find the annual profit for one to five copiers rented and the daily demands of 500, 1000, 1500, and 2000 copies daily. That is, find the yearly profit of each of these combinations of copiers rented and daily demand. If you rent three copiers, what daily demand for copies will allow you to break even? Graph profit as a function of the number of copiers for a daily demand of 500 copies and a daily demand of 2000 copies. Interpret your graphs.
1. A company manufactures and sells a product in the United States in England. The unit cost of manufacturing is $50. The current exchange rate (dollars per pound) is 1.221. The demand function, which indicates how many units the company can sell in England as a function of price (in pounds) is of the power type, with constant 27556759 and exponent −2.4 Develop a model for the company’s profit (in dollars) as a function of the price it charges (in pounds). Then, a data table will be used to find the profit-maximizing price for the nearest pound. If the exchange rate varies from its current value, does the profit-maximizing price increase or decrease? Does the maximum profit increase or decrease?
2.In PC Tech’s product mix problem, assume there is another PC model, the VXP, that the company can produce in addition to the Basics and XPs. Each VXP requires eight hours for assembling, three hours for testing, $275 for component parts, and sells for $560. At most 50 VXPs can be sold. Modify the spreadsheet model to include this new product, and use Solver to find the optimal product mix. You should find that the optimal solution is not integer-valued. If you round the values in the decision variable cells to the nearest integers, is the resulting solution still feasible? If not, how might you obtain a feasible solution that is at least close to optimal? Continuing the previous problem, perform a sensitivity analysis on the selling price of VXPs. Let this price vary from $500 to $650 in increments of $10, and keep track of the values in the decision variable cells and the objective cell. Discuss your findings. Again, continuing Problem 2, suppose that you want to force the optimal solution to be integers. Do this in Solver by adding a new constraint. Select the decision variable cells for the left side of the constraint, and in the middle dropdown list, select the “int” option. How does the optimal integer solution compare to the optimal noninteger solution in Problem 2? Are the decision variable cell values rounded versions of those in Problem 2? Is the objective value more or less than in Problem 2?If all of the inputs in PC Tech’s product mix problem are nonnegative (as they should be for any realistic version of the problem), are there any input values such that the resulting model has no feasible solutions? (Refer to the graphical solution.)There are five corner points in the feasible region for the product mix problem. We identified the coordinates of one of them: (560, 1200). Identify the coordinates of the others. Only one of these other corner points has positive values for both decision variable cells. Discuss the changes in the selling prices of either or both models that would be necessary to make this corner point optimal. Two of the other corner points have one decision variable cell value positive and the other zero. Discuss the changes in the selling prices of either or both models that would be necessary to make either of these corner points optimal.Using the graphical solution of the product mix model as a guide, suppose there are only 2800 testing hours available. How do the answers to the previous problem change? (Is the previous solution still optimal? Is it still feasible?)8.Again continuing Problem 2, perform a sensitivity analysis where the selling prices of Basics and XPs simultaneously change by the same percentage, but the selling price of VXPs remains at its original value. Let the percentage change vary from −25% to 50% in increments of 5%, and keep track of the values in the decision variable cells and the total profit. Discuss your findings.9.Consider the graphical solution to the product mix problem. Now imagine that another constraint—any constraint—is added. Which of the following three things are possible: (1) the feasible region shrinks; (2) the feasible region stays the same; (3) the feasible region expands? Which of the following three things are possible: (1) the optimal value in objective cell decreases; (2) the optimal value in objective cell stays the same; (3) the optimal value in objective cell increases? Explain your answers. Do they hold just for this particular model, or do they hold in general?
2. Suppose, as a matter of corporate policy, that General Flakes decides not to advertise on the “Rachael Ray” show. Modify the original advertising model appropriately and find the new optimal solution. How much has it cost the company to make this policy decision?
3. Five employees are available to perform four jobs. The time it takes each person to perform each job is given in the file P05_50.xlsx. Determine the assignment of employees to jobs that minimizes the total time required to perform the four jobs. (A blank indicates that a person cannot do that particular job. Also, assume that no person can do more than one job.)
3.NASA must determine how many of three types of objects to bring on board the space shuttle. The weight and benefit of each of the items are given in the file P06_42.xlsx. If the space shuttle can carry up to 2000 pounds of items 1 through 3, how many of each item should be taken on the space shuttle, assuming that at least one of each is necessary?
4.n the electricity pricing model in Example 7.4, the demand functions have positive and negative coefficients of prices. The negative coefficients indicate that as the price of a product increases, demand for that product decreases. The positive coefficients indicate that as the price of a product increases, demand for the other product increases.
Increase the magnitudes of the negative coefficients from −0.013 and −0.015 to −0.017 and −0.020, and then rerun Solver. Do the changes in the optimal solution go in the direction you would expect? Explain.Increase the magnitudes of the positive coefficients from 0.005 and 0.003 to 0.007 and 0.005, and then rerun Solver. Do the changes in the optimal solution go in the direction you would expect? Explain. Make the changes in parts a and b simultaneously, and then rerun Solver. What happens now?
4..In the truck-loading problem in Example 8.3, we assumed that any product could be loaded into any compartment. Suppose the following are not allowed: product 1 in compartment 2, product 2 in compartment 1, and product 3 in compartment 4. Modify the model appropriately, and then use Evolutionary Solver to find the new optimal solution. (Hint: Add a penalty to the objective for violating these new constraints.)
Part 1: Annual Profit for One to Five Copiers and Different Daily Demands
Let’s define the variables:
The total annual revenue is limited by the annual demand and the total capacity of the rented copiers (n×Cmax_copier). The number of copies made will be min(Dannual,n×Cmax_copier).
The annual cost of renting n copiers is n×Crent=5000n. The total number of copies made annually is Copiesannual=min(d×365,n×100000). The total operating cost is Copiesannual×Cop=0.03×Copiesannual. The total annual fixed costs are Fannual=4800.
The annual profit is given by: Profit=(Pcopy×Copiesannual)−(n×Crent)−(Cop×Copiesannual)−Fannual Profit=(0.10×Copiesannual)−5000n−(0.03×Copiesannual)−4800 Profit=0.07×Copiesannual−5000n−4800
Now, let’s calculate the annual profit for each combination:
Copiers (n) | Daily Demand (d) | Annual Demand (d×365) | Max Capacity (n×100000) | Annual Copies | Revenue (0.10×) | Rent Cost (5000n) | Operating Cost (0.03×) | Fixed Cost (4800) | Annual Profit |
---|---|---|---|---|---|---|---|---|---|
1 | 500 | 182,500 | 100,000 | 100,000 | 10,000 | 5,000 | 3,000 | 4,800 | -2,800 |
1 | 1000 | 365,000 | 100,000 | 100,000 | 10,000 | 5,000 | 3,000 | 4,800 | -2,800 |
1 | 1500 | 547,500 | 100,000 | 100,000 | 10,000 | 5,000 | 3,000 | 4,800 | -2,800 |
1 | 2000 | 730,000 | 100,000 | 100,000 | 10,000 | 5,000 | 3,000 | 4,800 | -2,800 |
2 | 500 | 182,500 | 200,000 | 182,500 | 18,250 | 10,000 | 5,475 | 4,800 | -2,025 |
2 | 1000 | 365,000 | 200,000 | 200,000 | 20,000 | 10,000 | 6,000 | 4,800 | -800 |
2 | 1500 | 547,500 | 200,000 | 200,000 | 20,000 | 10,000 | 6,000 | 4,800 | -800 |
2 | 2000 | 730,000 | 200,000 | 200,000 | 20,000 | 10,000 | 6,000 | 4,800 | -800 |
3 | 500 | 182,500 | 300,000 | 182,500 | 18,250 | 15,000 | 5,475 | 4,800 | -7,025 |
3 | 1000 | 365,000 | 300,000 | 300,000 | 30,000 | 15,000 | 9,000 | 4,800 | 1,200 |
3 | 1500 | 547,500 | 300,000 | 300,000 | 30,000 | 15,000 | 9,000 | 4,800 | 1,200 |
3 | 2000 | 730,000 | 300,000 | 300,000 | 30,000 | 15,000 | 9,000 | 4,800 | 1,200 |
4 | 500 | 182,500 | 400,000 | 182,500 | 18,250 | 20,000 | 5,475 | 4,800 | -12,025 |
4 | 1000 | 365,000 | 400,000 | 365,000 | 36,500 | 20,000 | 10,950 | 4,800 | 750 |
4 | 1500 | 547,500 | 400,000 | 400,000 | 40,000 | 20,000 | 12,000 | 4,800 | 3,200 |
4 | 2000 | 730,000 | 400,000 | 400,000 | 40,000 | 20,000 | 12,000 | 4,800 | 3,200 |
5 | 500 | 182,500 | 500,000 | 182,500 | 18,250 | 25,000 | 5,475 | 4,800 | -17,025 |
5 | 1000 | 365,000 | 500,000 | 365,000 | 36,500 | 25,000 | 10,950 | 4,800 | -4,250 |
5 | 1500 | 547,500 | 500,000 | 500,000 | 50,000 | 25,000 | 15,000 | 4,800 | 5,200 |
5 | 2000 | 730,000 | 500,000 | 500,000 | 50,000 | 25,000 | 15,000 | 4,800 | 5,200 |
Part 2: Break-Even Daily Demand for Three Copiers
For break-even, the profit must be zero. With three copiers (n=3), the annual rent cost is $3 \times 5000 = $15,000, and the annual fixed costs are $4800. So, the total annual fixed cost is $15000 + 4800 = $19,800.
Let the annual demand be Dannual=d×365. The number of copies made is Copiesannual=min(d×365,3×100000)=min(365d,300000).
The break-even condition is: 0.07×Copiesannual−19800=0 0.07×Copiesannual=19800 Copiesannual=0.0719800≈282857.14
Since the maximum capacity of three copiers is 300,000, this value is achievable. Now, we need to find the daily demand d such that 365d=282857.14: d=365282857.14≈774.95
So, a daily demand of approximately 775 copies will allow you to break even with three copiers.
Part 3: Graphing Profit as a Function of the Number of Copiers
We will plot the number of copiers (1 to 5) on the x-axis and the annual profit on the y-axis for two scenarios: daily demand of 500 copies and daily demand of 2000 copies.
Scenario 1: Daily Demand = 500 copies
Copiers (n) | Annual Copies | Annual Profit |
---|---|---|
1 | 100,000 | -2,800 |
2 | 182,500 | -2,025 |
3 | 182,500 | -7,025 |
4 | 182,500 | -12,025 |
5 | 182,500 | -17,025 |
Interpretation (Daily Demand = 500): With a low daily demand, increasing the number of copiers beyond one actually decreases the profit because the increased rental cost outweighs the marginal increase in revenue (which is limited by the demand). One copier can handle the annual demand, so renting more only adds to the fixed costs.
Scenario 2: Daily Demand = 2000 copies
Copiers (n) | Annual Copies | Annual Profit |
---|---|---|
1 | 100,000 | -2,800 |
2 | 200,000 | -800 |
3 | 300,000 | 1,200 |
4 | 400,000 | 3,200 |
5 | 500,000 | 5,200 |
Interpretation (Daily Demand = 2000): With a higher daily demand, increasing the number of copiers increases the annual profit up to the point where the demand is met by the capacity. Each additional copier allows for more copies to be sold, increasing revenue and profit. The profit increases linearly with the number of copiers until the capacity constraint due to demand is reached.
Graphs:
You would create two line graphs in Excel or another graphing tool:
Let:
The demand function is given by Q=27556759×P£−2.4.
The revenue in pounds is R£=Q×P£=(27556759×P£−2.4)×P£=27556759×P£−1.4.
To convert the revenue to dollars, we multiply by the exchange rate: RUSD=R£×ER=(27556759×P£−1.4)×1.221=33646713.399×P£−1.4.
The total cost in dollars is the unit cost times the quantity sold: Ctotal_USD=CUSD×Q=50×(27556759×P£−2.4)=1377837950×P£−2.4.
The profit in dollars is the total revenue in dollars minus the total cost in dollars: ProfitUSD(P£)=RUSD−Ctotal_USD ProfitUSD(P£)=33646713.399×P£−1.4−1377837950×P£−2.4
Data Table for Profit-Maximizing Price:
You would create a data table in Excel with a column for the price in pounds (P£) starting from a reasonable range (e.g., £1 to £100) and a column for the profit in dollars calculated using the formula above. By examining the profit column, you can find the price that yields the maximum profit to the nearest pound.
Impact of Exchange Rate Variation:
Let’s analyze the derivative of the profit function with respect to the price to understand the maximizing price. However, we can reason about the impact of the exchange rate.
If the exchange rate increases (more dollars per pound), the revenue in dollars for a given price in pounds will increase proportionally. The cost in dollars remains the same as it’s based on the quantity sold, which depends on the price in pounds and the demand function (independent of the exchange rate). Therefore, an increase in the exchange rate will likely lead to a higher profit. To maximize this higher profit, the company might be able to charge a slightly higher price in pounds without significantly decreasing demand, as the dollar revenue per pound is now greater. So, the profit-maximizing price might increase.