2009 Mid-term Solution

Student Name:_______________________

Student Number:_________________

Instructor Use Serial Number

UNIVERSITY OF WINDSOR FACULTY OF ENGINEERING

Saturday, November 7th, 2009 Closed Book

85-313-(01 & 02) Engineering Economy MIDTERM EXAMINATION Dr. Fouzia Baki & Dr. Amr Shabaka Total Marks: 100

Student Name: Student ID.: Section No.: Instructor’s Name: Notes: * The time allocated for this examination is 2:00 hours * Attempt ALL Questions * Draw Cash Flow Diagrams Whenever Applicable * Be Neat & Precise * Show All Computations and/or Assumptions * There are 10 pages in total. * If you are running out of space for a specific problem than write at the back of the sheet containing that problem (an extra blank sheet (page 9)has been provided for extra space if needed) Total Mark

___________________________________________________________________ Question 1

Marks

Remarks /10 Pa ge 1 of 15

Student Name:_______________________ 2

/15

3

/25

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Total

/ 100

Student Number:_________________

QUESTION 1 - 10marks For each of the following problems indicate on the figure at which point in time you are doing the equivalence P 01234n $50 $1000 $50

(a) What is the value of n that makes the disbursements equivalent to the receipts, based on a 3% interest rate? Solution P = A (P/A, 3%, n) $1,000 = $50 (P/A, 3%, n) (P/A, 3%, n) = 20 From the 3% interest table, n = 31

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Student Name:_______________________ Student Number:_________________ (b) Assuming a 12% interest rate, find the value of J that makes the disbursements equivalent to 5J 6012345 $ J3 2500 46J JJJ

$500

Solution 0P $500 G $500 6G 5G 4G 3G 2G

Present Worth P of the two $500 amounts (at year -1): P = $500 (P/F, 12%, 2) + $500 (P/F, 12%, 3) = $500 (0.7972) + $500 (0.7118) = $754.50 Also: P $754.50

= G (P/G, 12%, 7) = G (P/G, 12%, 7) Pa ge 3 of 15

Student Name:_______________________ = G (11.644) G

Student Number:_________________

= $754.50/11.644 = $64.80

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Student Name:_______________________

Student Number:_________________

QUESTION 2 - 15 marks John bought a car at a local used car lot. Including tax and insurance, the total price was $3000. He is to pay for the car in 12 equal monthly payments using a Nominal interest of 12% compounded monthly on the loan, with the first payment due immediately. After six payments, he decides to sell the car. A buyer agrees to pay a cash amount to payoff the loan in full at the time the next payment is due and also to pay John $1000. If there are no penalty charges for this early payment of the loan, how much will the car cost the new buyer? [Cash flow diagram  3Marks]

Solution 10 11 0123456789 Pay A = off $3,000 ? loan

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Student Name:_______________________

Student Number:_________________

Compute monthly payment: $3,000 = A + A (P/A, 1%, 11) = A + A (10.368) = 11.368 A A = $3,000/11.368 = $263.90 Car will cost new buyer: = $1,000 + 263.90 + 263.90 (P/A, 1%, 5) = $1263.90 + 263.90 (4.853) = $2,544.61

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Student Name:_______________________

Student Number:_________________

QUESTION 3 - 25 marks Chrysler wants to build a spare parts storage facility in Windsor. A plant engineer has identified four different location options. Initial cost and annual net cash flow estimates are detailed in the Table below. If the MARR is 15%, use the IRR analysis to select the economically best location. Estimates for Four Alternative Building Locations A B C D Initial cost, $ -130,000 -150,000 -170,000 -100,000 Annual cash flow, $ +25,000 +30,000 +33,000 +20,000 Life, years 15 15 15 15 P 0123415 $100Project 000 $20 000 (D)

Solution Sort projects in order of least initial cost DABC Compare D to Do Nothing (DN) 100,000 = 20,000(P/A, i*%,15) (P/A, i*%,15) = 5 if i=20%  (P/A,20%,15)= 4.6755 if i=15%  (P/A,15%,15)= 5.8474 P 0123415 $30Project 000 $5 000 (A-D)

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Student Name:_______________________ 15%< i* < 20% D is current best because i* >MARR -----------------------Incremental Comparison from DA A is Challenger

Student Number:_________________

30,000 = 5,000(P/A, i*%,15) (P/A, i*%,15) = 6 if i=14%  (P/A,14%,15)= 6.1422 if i=15%  (P/A,15%,15)= 5.8474 P 0123415 $50Project 000 $10 000 (B-D)

because i* <MARR ------------------------Incremental Comparison from DB B is Challenger 50,000 = 10,000(P/A, i*%,15) (P/A, i*%,15) = 5 if i=20%  (P/A,20%,15)= 4.6755 if i=15%  (P/A,15%,15)= 5.8474 15%< i* < 20% B is current best because i* >MARR ------------------------------------P 0123415 $20Project 000 $3 000 (C-B)

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Student Name:_______________________ C is Challenger 20,000 = 3,000(P/A, i*%,15) (P/A, i*%,15) = 6.67 if i=12%  (P/A,12%,15)= 6.8109 if i=13%  (P/A,13%,15)= 6.4624 12%< i* < 13% C fails test & B is still current best because i* <MARR

Student Number:_________________

B is the Best Alternative

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Student Name:_______________________

Student Number:_________________

QUESTION 4 - 20 marks 11 234567890110 $1500 $3000 $1000 $2000 $4000 Years 1$500 $5000 2

PW P 31 5

+

A mechanical engineering student is going to deposit money into her bank account for the following five years using the pattern shown in figure. She is then planning to withdraw part of her money for the following 7 years as shown. If she is investing her money at an interest rate of 7% per year, find net present worth and equivalent annual worth for the shown cash flow diagram.

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Student Name:_______________________

Student Number:_________________

11 234567890110 $1000 $5000 $2000 $2500 $4000 $3500 $3000 Years 12

P1 = present worth of deposits = - [2 000 + 500 (A/G, 7%,5)] (P/A, 7%,5) = - [2 000 + 500 (1.86505)] (4.1002)= - $12,023.84 Add 5000 +ve and 5000 –ve annuity from end of year 6 until end of year 10 Let PW5= The PW at end of year 5 for all payments from end of year 6 until end of year 10 Pa ge 11 of 15

Student Name:_______________________ PW5

Student Number:_________________

= [5 000 – 1 000 (A/G, 7%,5)] (P/A, 7%,5) = [5 000 + 1000 (1.86505)] (4.1002) =12 853.9

P2 = present worth of PW5 = PW5(P/F, 7%, 5) = 12 853.9 (0.71299) = $9 164.72 P3 = present worth for payments at end of year 11 and 12 = 1000 (P/A, 7%,2) (P/F, 7%, 10) = 1000 (1.8080) (0.50835) = $919.15 Total PW P = P1 + P2 + P3 = -12,023.84 + 9 164.72 + 919.15 = - $1 939.97 -------------------------------------AW =P(A/P,7%,12)= 919 (0.12590) = -244.24

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Student Name:_______________________ QUESTION 5 - 15 marks ABC Automotive has $100 000 to invest in internal projects. The choices are:

Project A. Line Improvements B. New Manual Tester C. New Automated Tester C. Overhauling Press

Student Number:_________________

Cost $20 000 $30 000 $60 000 $50 000

Only one tester may be bought and the press will not need overhauling if the line improvements are not made. What mutually exclusive project combinations are available if ABC Auto will invest in at least one project? [For each infeasible alternative: indicate the reason why you think that the alternative is infeasible]

Solution

Possible combinations

Testers

Press

Total Cost

Feasible?

Do nothing A B C D AB AC AD BC BD CD ABC ABD ACD BCD ABCD

OK OK OK OK OK OK OK OK >1 OK OK >1 OK OK >1 >1

OK OK OK OK No Need to OH OK OK OK OK No Need to OH No Need to OH OK OK OK No Need to OH OK

0 OK OK OK OK OK OK OK OK OK >100 000 >100 000 OK >100 000 >100 000 >100 000

No Yes Yes Yes Yes Yes Yes

Yes

There are 7 mutually exclusive projects: {A, B, C, AB, AC, AD, ABD}

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Student Name:_______________________ QUESTION 6 - 10 marks

Student Number:_________________

(a) What is the difference in annual worth between an investment of $100, 000 per year for 100 years and an investment of $100 000 per year forever at an interest rate of 10% per year? Solution: For the infinit life project PW∞= A/i = [100 000]/0.1= $1 000 000 The equivelant Annuity over 100 years is: AW∞ for 100 yrs= 1000 000 (A/P,10%,100) =1000000 i (1 + i )

(1 + i ) N

N

−1

= 1000000

0.1(1 + 0.1)

100

(1 + 0.1) 100 − 1

= 100007.26

The differnce in Annual worth is $7.26 Other way For 100 year project PW100 =100 000 (P/A,10%,100)=100 000

(1 + i ) N − 1 = 100000(1 + 0.1) 100 − 1 = 999927.43 N 100 i (1 + i ) 0.1(1 + 0.1) Equivelent Annual worth for an infinit project AW100 for ∞ yrs= PWi= *0.1= 99992.74 999927.43 Difference = 100 000 -99992.74 = $7.26 (b) Consider alternatives A & B: A B Initial cost, $ -60,000 -65,000 Annual cash flow, $ +25,000 +25,000 Life, years 8 8 If the two projects are mutually exclusive and the MARR = 5% which alternative should be taken.

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Student Name:_______________________ P 012348

Student Number:_________________

$65 000 $60 $25A B000

Solution

Since both investments are of equal life and bring an equal +ve cash, than the project with the smaller investment is the desired one (Select A) By Calculation PW(A)= -60 000 +25 000(P/A, 5%, 8) = -60 000 + 25 000 (6.4632) = 101 580 PW(B)= -65 000 +25 000(P/A, 5%, 8) = -65 000 + 25 000 (6.4632)= 96580 Select A

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