Lec 1 Exercise

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3360U

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Economics

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Feb 20, 2024

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ENGR 3360U, ENGINEERING ECONOMICS SET of PROBLEMS N. 2 / WINTER 2024 / SOLUTIONS Assigned on 22 nd of January 2023 / SOLUTIONS out 29 th of January 2024 Please USE these problems to practice and above all REFLECT and THINK to test your understanding of concepts. Note that some of these “reflections” can have a different outcome depending on the contest. 1. PROBLEM / CHAPT. 1 Which of the following problems is most suitable for analysis by engineering economic analysis? (a) Some 45¢ chocolate bars are on sale at 12 bars for $3. Sandy, who eats a couple of chocolate bars a week, must decide whether to buy a dozen at the lower price. (b) A woman has $150,000 in a bank chequing account that pays no interest. She can either invest it immediately at a desirable interest rate or wait a week and obtain an interest rate that is 0.15% higher. (c) Joe backed his car into a tree, damaging the fender. He has car insurance that will pay for the fender repair. But if he files a claim for payment, they may change his "good driver" rating downward and charge him more for car insurance in the future. Solution) Of the three alternatives, the $150,000 investment problem is most suitable for economic analysis. There are not enough data to figure out how to proceed, but if the “desirable interest rate” were 9%, then foregoing it for one week would mean an immediate loss of: 1/52 (0.09) = 0.0017= 0.17% It would take over a year at 0.15% more to equal the 0.17% foregone now. The candy bar problem is suitable for economic analysis. Compared to the investment problem it is, of course, trivial. Joe’s problem is a real problem with serious economic consequences. The difficulty may be in figuring out what one gains if he pays for the fender damage, instead of having the insurance company pay for it. 2. PROBLEM / CHAPT. 1

Maria, a university student, is getting ready for three final examinations. Between now and the start of exams, she has 15 hours of study time available. She would like to get the highest possible overall grade average based on her grades in her math, physics, and engineering economy classes. She feels she must study at least two hours for each course and, if necessary, will settle for the low grade that the limited study would yield. How much time should Maria devote to each class if she estimates her grade in each subject as follows: Mathematics Physics Engineering Economy Study Hours Grade Study Hours Grade Study Hours Grade 2 25 2 35 2 50 3 35 3 41 3 61 4 44 4 49 4 71 5 52 5 59 5 79 6 59 6 68 6 86 7 65 7 77 7 92 8 70 8 85 8 96 Solution) The fundamental concept here is that we will trade an hour of study in one subject for an hour of study in another subject so long as we are improving the total results. The stated criterion is to “get as high an average grade as possible in the combined classes.” (This is the same as saying “get the highest combined total score.”) Since the data in the problem indicate that additional study always increases the grade, the question is how to apportion the available 15 hours of study among the courses. One might begin, for example, assuming five hours of study on each course. The combined total score would be 190. Decreasing the study of mathematics one hour reduces the math grade by 8 points (from 52 to 44). This hour could be used to increase the physics grade by 9 points (from 59 to 68). The result would be: Math 4 hours 44 Physics 6 hours 68 Engr. Econ. 5 hours 79

Total 15 hours 191 Further study would show that the best use of the time is: Math 4 hours 44 Physics 7 hours 77 Engr. Econ. 4 hours 71 Total 15 hours 192 3. PROBLEM / CHAPT. 1 / COSTS One of your firm's suppliers discounts prices for larger quantities. The first 1 , 000 parts are $13 each. The next 2,000 are $12 each. All parts in excess of 3,000 cost $11 each. What are the average and marginal costs per part for each of thefollowingquantities? (a) 500 (b) 1,500 (c) 2 , 500 (d) 3 , 500 Solution) (a) 500 parts Average cost = $13 Marginal cost = $13 (b) 1500 parts Average cost = ((1000)($13) + (500)($12)) / 1500 = $ 12.67 Marginal cost = $12 (c) 2500 parts Average cost = ((1000)($13) + (1500)($12)) / 2500 = $12.40 Marginal cost = $12 (d) 3500 parts Average cost = ((1000)($13) + (2000)($12) + 500($11)) / 3500 = $12.14 Marginal cost = $11 4. PROBLEM / CHAPT. 1 / COSTS Two automatic systems for dispensing maps are being compared by a provincial highway department. The accompanying break-even chart of the

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comparison of these systems (System I vs System II) shows total yearly costs for the number of maps dispensed per year for both alternatives. Answer the following questions. (a) What is the fixed cost for System I? (b) What is the fixed cost for System II? (c) What is the variable cost per map dispensed for System I? (d) What is the variable cost per map dispensed for System II? (e) What is the break-even point in terms of maps dispensed at which the two systems have equal annual costs? (f) For what range of annual number of maps dispensed is System I recommended? (g) For what range of annual number of maps dispensed is System II recommended? (h) At 3,000 maps a year, what are the marginal and average map costs for each system? Solution) x = number of maps dispensed per year (a) Fixed Cost (I) = $1,000 (b) Fixed Cost (II) = $5,000 (c) Variable Costs (I) = 0.900 (d) Variable Costs (II) = 0.100 (e) Set Total Cost (I) = Total Cost (II) $1,000 + 0.90 x = $5,000 + 0.10x thus x = 5,000 maps dispensed per year. The student can visually verify this from the figure. (f) System I is recommended if the annual need for maps is <5,000 (g) System II is recommended if the annual need for maps is >5,000 (h) Average Cost @ 3,000 maps: TC(I) = (0.9) (3.0) + 1.0 = 3.7/3.0 = $1.23 per map TC(II) = (0.1) (3.0) + 5.0 = 5.3/3.0 = $1.77 per map Marginal Cost is the variable cost for each alternative, thus:

Marginal Cost (I) = $0.90 per map Marginal Cost (II) = $0.10 per map 5. PROBLEM / CHAPT. 1 / COSTS Mr Sam Spade, the president ofAjax, recently read in a report that a competitor named Bendix has thefollowing relationship between cost and production quantity: C = $3,000,000 -‐ $18,000 Q+$75 Q 2 where C =total manufacturing cost peryear and Q = number ofunitsproduced peryear. A newly hired employee, who previously worked for Bendix,tells Mr Spadethat Bendix is now producing 110 units a year. If the selling price remains unchanged, Mr Spade wonders if Bendix is likelyto increase the number of units produced per year in the near future. He asks you to look at the information andtell him what you are able to deduce from it. Solution) C = $3,000,000 − $18,000Q + $75Q 2 Where C = Total cost per year Q = Number of units produced per year Set the first derivative equal to zero and solve for Q. dC/dQ = −$18,000 + $150 Q = 0 Q = $18,000/$150 = 120 Therefore total cost is a minimum at Q equal to 120. This indicates that production below 120 units per year is most undesirable, as it costs more to produce 110 units than to produce 120 units. Check the sign of the second derivative: d 2 C/dQ 2 = +$150 The + indicates the curve is concave upward, ensuring that Q = 120 is the point of a minimum. Average unit cost at Q = 120/year: = [$3,000,000 – $18,000 (120) + $75 (120) 2 ]/120 = $16,000 Average unit cost at Q = 110/year:

= [$3,000,000 – $18,000 (110) + $75 (120) 2 ]/110 = $17,523 One must note, of course, that 120 units per year is not necessarily the optimal level of production. Economists would remind us that the optimum point is where Marginal Cost = Marginal Revenue, and Marginal Cost is increasing. Since we do not know the Selling Price, we cannot know Marginal Revenue, and hence we cannot compute the optimum level of output. We can say, however, that if the firm is profitable at the 110 units/year level, then it will be much more profitable at levels greater than 120 units. 6. PROBLEM / CHAPT. 1 / COSTS Bonka Toys isconsidering arobot that will cost $20,000.Aftersevenyears its Salvage valuewill be $2,000. An overhaul costing $5,000 will be needed in year4 . O&M costswillbe $2,500 per year.Draw the cash flow diagram. Solution) Year Capital Costs O & M Overhaul 0.00 –20 0 0 1.00 0 –2.5 0 2.00 0 –2.5 0 3.00 0 –2.5 0 4.00 0 –2.5 –5 5.00 0 –2.5 0 6.00 0 –2.5 0 7.00 2 –2.5 0

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-20 -15 -10 -5 0 5 10 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 Year Cash Flow ($1,000) Overhaul O & M Capital Costs

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