IE343_HW1_Solutions - HomeworkAssignment#1Solutions...

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Unformatted text preview: HomeworkAssignment#1Solutions PU R D U E U N I V E RSI T Y H om e w o r k A ssign m e n t #1 I E 343 : E n gi n e e r i n g E con om i cs A ssign e d : 27 A u g 2010 F a l l 2010 D u e : 3 Se p t 2010 I nst r u c to r : A . C a p pon i P r ob l e m 1 Textbook Problem 1-14 (10 points) a. The decision problem here is to decide which pizza to order while studying. Two options are available for pizza delivery and a decision has to be made based on suitable criteria, and in this case, least cost. b. Systematic application of the seven principles of engineering economy: 1. Alternatives: There are two alternatives, or pizza shops, from where to order the pizza: Pick 2. Differences: Type of pizza, round or square; Costs associated with each pizza (including delivery charges); Delivery time could be an issue. 3. The viewpoint in this particular case will be the students who want to order pizza. 4. The unit of measure to be used in this particular problem is the cost. Since a product is being purchased, the most suitable comparison measure is the cost of each product. 5. There are three people who will be sharing this particular pizza. So, the selection criteria 6. In this particular case, there is no included uncertainty, but the students buying the pizza may think there is variability in the quality of the pizzas, delivery time, and even cost. 7. Revisiting the decision: so that could have an impact on this decision. Perhaps the quality of the pizza from Pick Up Sticks was good the past two times, but before that it was just okay. c. Minimizing Costs per Unit of Volume Pick Up Sticks Pizza Cost = 15.00 + 0.05*(15.00) + 1.50 = $17.25 Volume of Pizza = 20*20*1.25 = 500 cubic inch Cost / unit of volume = $17.25/500 = $0.0345 / cubic inch Pizza Cost= 17.25 + 0.05*(17.25) = 18.11 Volume of Pizza = 3.14*10*10*1.75 = 549.5 cubic inch Cost / unit of volume = 18.11/549.5 = $0.0330 / cubic inch d. Other possible selection criteria: Simple overall cost, Quality of pizza, Personal taste preference, Time to delivery IE343 HomeworkAssignment#1Solutions Page1  HomeworkAssignment#1Solutions (10 points) P r ob l e m 2 Textbook Problem 2-13 a. 140 10D = 0 D* = 14 units per week. b. The profit at D* will be  IE343 HomeworkAssignment#1Solutions Page2  HomeworkAssignment#1Solutions P r ob l e m 3 Textbook Problem 2-15 a. Number of units that should be produced and sold to maximize profit. Total Cost = So, D* = 50 units/month b. Show that part (a) maximizes profit. So, D* = 50 is a point of maximum profit. IE343 HomeworkAssignment#1Solutions Page3  (10 points) HomeworkAssignment#1Solutions (10 points) P r ob l e m 4 Textbook Problem 2-18 Capacity = 4,100 pumps Fixed cost = $504,000 Variable cost = $166 per pump Sales Price = $328 Assume sales equal output volume. X = pumps produced Breakeven point  Breakeven point with 18% reduction of fixed cost and 6% reduction of variable cost    per month Percentage reduction = = 22.75% reduction in the breakeven point IE343 HomeworkAssignment#1Solutions Page4  HomeworkAssignment#1Solutions P r ob l e m 5 (10 points) A privately owned summer camp for youngsters has the following data for a 12-week session: Charge per camper: $120 per week Fixed Costs: $48,000 per session Variable cost per camper: $80 per week Capacity: 200 campers (a) Develop the mathematical relationships for total cost and total revenue. Total Cost = $48,000 + $80 * 12 X = $48,000 + $960 X Total Revenue = $120 * 12 x = $1440 X (b) What is the total number of campers that will allow the camp to breakeven? Total Cost = Total Revenue iff X = 100 $48,000 + $960 X = $1440 X 480 X = $48,000 X = 100 (c) What is the profit or loss for the 12-week session if the camp operates at 80% capacity? 80% capacity means X = 80% * 200 = 160 campers Total Profit (160) = Total Revenue (160) Total Cost (160) = $1440 * 160 ($48,000 + $960 * 160) = $28,800. IE343 HomeworkAssignment#1Solutions Page5  HomeworkAssignment#1Solutions P r ob l e m 6 (5 points) Two automatic systems for dispensing maps are being compared by the state highway department. The total cost for system 1 is y1 = 0.9 x + 1.0, and the total cost for system 2 is y2 = 0.1 x + 5. (a) What are the fixed and variable costs for system 1? Variable cost = 0.9 Fixed cost = 1 (b) What are the fixed and variable costs for system 2? Variable cost = 0.1 Fixed cost = 5 (c) What is the number of maps dispensed at which the two systems have equal annual costs? 0.9 x + 1.0 = 0.1 x + 5 0.8x = 4 x=5 (d) For what range of annual number of maps dispensed is system 1 recommended? 0,5 (e) For what range of annual number of maps dispensed is system 2 recommended? IE343 HomeworkAssignment#1Solutions Page6  HomeworkAssignment#1Solutions P r ob l e m 7 (15 points) A small company manufactures a certain product. Variable costs are $20 per unit and fixed costs are $10,875. Price-Demand relation is P = -0.25 D + 250, where P is the units sales price of the product and D the annual demand. (a) Develop equations for total cost and total revenue. Total Cost = Fixed cost + Variable cost = 10,875 + 20D Total Revenue = PD = (-0.25D + 250)D = -0.25D2 + 250D (b) Find the breakeven quantity (in terms of profit and loss) for the product. Breakeven, when Revenue = Cost OR Profit = 0 Profit = Total Revenue Total Cost = -0.25D2 + 250D (10,875 + 20D) = -0.25D2 + (250-20)D 10,875 Set Profit = 0 -0.25D2 + 230D 10,875 = 0 D = 50 or D = 870 B r e a k e v e n Q u a n t i t y = 50 o r 870 If you make less than 50 units, there will be a loss of money and if you produce more than 870 units, there will be a loss of money. (c) What profit would the company obtain by maximizing its total revenue? Total Revenue (TR) = PD = (250 Total Profit 0.25D)D = 250D 0.25D2 = -0.25D2 + 250D (10,875 + 20D) = -0.25*(500)2 + 250*500 (10,875 + 20*500) = $41,625 IE343 HomeworkAssignment#1Solutions Page7  HomeworkAssignment#1Solutions (d) Profit (Loss) = 250D 0.25D2 (10,875 + 20D) = -0.25D2 + (250-20)D 10,875 Set the first derivative of the profit equation to zero Solve for D* Check the second derivative at D* So D* maximizes the profit. Total Profit = -0.25D2 + 250D (10,875 + 20D) = -0.25*(460)2 + 250*460 (10,875 + 20*460) = $42,025 (e) Neatly graph the solutions from parts (a)-(d) on one graph where the x-axis is Demand and the y-axis is Revenue/Cost. IE343 HomeworkAssignment#1Solutions Page8  ...
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This note was uploaded on 01/20/2012 for the course INDUSTRIAL 343 taught by Professor Capponi during the Spring '11 term at Purdue University-West Lafayette.

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