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1 Dr. Lotfi K.GAAFAR Eng. Ahmed Salah RIFKY ENGR 345 Engineering Economy.

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Presentation on theme: "1 Dr. Lotfi K.GAAFAR Eng. Ahmed Salah RIFKY ENGR 345 Engineering Economy."— Presentation transcript:

1 1 Dr. Lotfi K.GAAFAR Eng. Ahmed Salah RIFKY ENGR 345 Engineering Economy

2 2 Q1: Depict the following transactions on a cash flow diagram: Depositing $4,000 today and withdrawing $2,000 after two years. The rest of the money is withdrawn after 4 years from today. If the interest rate is 10%, how much money will be available for the last withdrawal? Answer: $4,000 $2,000F i = 10% P = 4, *(P/F, 10%, 2) = 2,347.2 F= (F/P, 10%, 4) = 3,436.5  F = $3,436.5

3 3 Q2: How much should I deposit now in an investment that earns a profit of 10% in order to make 4 equal end-of-year withdrawals of $2,000 each starting one year from now? Show the cash flow diagram. Answer: P $2, i = 10% P = A*(P/A, i, n) P = 2,000*(P/A, 10%, 4)  P = $6,339.8

4 4 Q3: Which has a higher present value, 5 end-of-year payments that start at $500 at the end of year one and increase by $500 each year thereafter, or 6 end-of-year payments that start at $500 at the end of year one and increase by 25% each year thereafter? Assume that the interest rate is 10%. Answer: 500 2,500 2,000 1,500 1,000 P = 500*(P/G, 10%, 5)*(1.1) P = 500/ ,000/(1.1) 2 +1,500/(1.1) 3 +2,000/(1.1) 4 + 2,500/(1.1) 5 Or P = 500*(P/A, 10%, 5) + 500*(P/A,10%, 5)  P = $5, A) 500 B) 500*(0.25) 500*(0.25) 2 500*(0.25) 3 500*(0.25) 4 500*(0.25)  P = $3,844.32

5 5 Q4: A 5-year rental contract requires a down payment of $1,000 now (at time 0) and 5-end-of-year payments that start with $800 at the end of year 1 and increase by 15% every year thereafter. How much should be invested in an account that pays 12% to finance this contract? Answer: i = 12% g = 15% - P = 1, *[(1-(1.15/1.12) 5 )/( )]  P = $4, P 1, *(1.15) 800*(1.15) 2 800*(1.15) 3 800*(1.15)

6 6 Q5: What is the present equivalence of five end-of-year payments that start with $100 at the end of period one and increase by $100 every period thereafter? (Assume i = 10%) Answer: P  *(P/A,10%, 5)+100*(P/G,10%, 5) P =  P = 1,065.26

7 7 Q6: $4,000 were deposited at time 0 in an account that pays an interest of 10%. How much should be withdrawn at the end of period 2 to have $4,000 at the end of period 4 available in the account? Answer: 1234 $4,000 4,000*(1.1) 4 = 5,856.4 x *(1.1) 4 = 5, ,000 x = $1,856.4  Future Value $1,  Withdraw $4,000?

8 8 Q7: what is the nominal and the effective interest rates per year for an interest rate of 0.015% per day. Answer: Nominal i/year = * (365) =  5.48% Effective i= (1+i) * n –1 i= ( ) *  5.63%

9 9 Q8: What is the effective quarterly interest rate for a nominal rate of 12% compounded monthly? Answer: Solution 1: i = 1% i e = (1+0.01) 3 – 1  i e = 3.03% Solution 2: i Q = ? i = 12.68% (1+ i Q )4 – 1 = 12.68% (1+ i Q )4 =  i Q = 3.03%

10 10 Q9: Q: Calculate the capitalized cost of $60,000 in year 0 and uniform beginning-of-year rent payments of $25,000 for an infinite time using an interest rate of (a) 12% per year (b) 16% per year compounded monthly. Answer: (a) PW = - (60, ,000) /0.12  PW = -$293,333 (b) i/yr = (1+0.16/12) *12 -1= % PW = -85,000-25,000/  PW = -$230,121

11 11 Q10: I have $10,000 available for investment and the following four mutually exclusive investment opportunities. What is your advice? Assume a minimum attractive rate of return of 15%. ABCD Initial Investment($3,000)($4,000)($6,500)($10,000) Annual Income$700.00$800.00$1,750.00$2, Salvage Value$2, $3,200.00$4, Life6666 ROR19.97%14.17%22.07%19.37% Present Worth PW = - Initial Investment + Annual Income(P/A,15%,6)+Salvage Value (P/F,15%,6) PW A = $513.79, PW B = - $107.76, PW D = $1,  PW A = $1, Answer: $513.79($107.76) $1, $1, $1,506.29

12 12 Q11: Compare the five alternatives based on PW and choose the best one. Consider the following mutually exclusive investment alternatives (MARR=10%): PW = - Initial Investment + Annual Income(P/A,10%,4)+Salvage Value (P/F,10%,4) Answer: Alternative ABCDE Initial Investment-4,000-7,000-4,000-5,500-7,000 Annual Return8001, ,0001,400 Useful Life44444 Salvage Value2,2004,5002,5004,500 PW A = $38.52 PW B = $ PW C = $ PW D = $ PW E = $ B > E C > A PW = $38.52 $828.36$244.43$743.43$511.37

13 13 Q12: A bond with a face value of $5,000 pays interest of 8% per year. This bond will be redeemed at par value at the end of its 20-year life. The owner of the bond wants to sell it immediately after he received his fifth annual payment. How much should a prospective buyer pay to earn 10% per year? Answer: $400 P = face*interest*(P/A, i, n) + redemption*(P/F, i, n) P = 5,000*8%*(P/A, 10%, 15) + 5,000*(P/F, 10%, 15) P = 400*(7.606) + 5,000*(0.2394) P  $ $5,000

14 14 Q13: Develop an investment balance diagram for an investment that requires an initial payment of $5,000 in return for equal end-of-year returns of $2,000 for 5 years, and a salvage value of $1,000. Assume a 10% interest rate. Answer: PW = - Initial Investment + Annual Income(P/A,10%,5)+Salvage Value (P/F,10%,5) PW = - 5, ,000*(3.79)+1,000*(0.62)  PW = $3,200 $5,000 $2,000 $1,

15 15 Q14: A 5-year bond with a face value of $1,000 pays interest of 12% quarterly. A-If a buyer purchased the bond at $1,020 to yield a quarterly interest for 4% and sold it 3 years later after getting the 12th quarter payment, how much should he be asking for? B-What effective annual interest rate would the second buyer earn if he paid what the first buyer asked for, assuming that he keeps the bond until it matures and redeems it at par value? Answer: $1, $ F i = 4% F = 1,020 * (F/P, 4%, 12) – 30 * (F/A, 4%, 12)  F = $1, A) B) $ $1,000 $1, i = ?  i = 0.654%

16 16 Q15: How much should I deposit now in an account that pays 10% to be able to withdraw $500 every year for ever? Answer: ∞ P 1234 $500 P = 500 * { [i*(1+i) n ] / [(1+i) n -1] } n ===> ∞ P = 500/i = 500/0.1  P = $5,000

17 17 Q16: A particular product sells for a price of $70. If the fixed cost are $700/month and the variable cost is $35/unit, determine the breakeven quantity. Answer: Breakeven: Profit = 0 x+x Price x Quantity = Fixed Cost + Variable Cost x Quantity 70*Q = *Q 35*Q = 700  Q = 20

18 18 Q17: A utility vehicle is purchased for $30,000 and is expected to be sold for $7,500 at the end of its life. Determine the capital recovery cost at 10%. Answer: CR = P*(A/P, i, n) – F*(A/F, i, n) Assume n = 20 CR = 30,000*(A/P, 10%, 20) – 7,500*(A/F, 10%, 20)  CR = $3,393.3/year

19 19 Q18: How much should I deposit now in an account that pays an interest rate of 10% in order to be able to withdraw an amount every year forever that starts at 1000 at the end of the first year and increases by 5% every year thereafter. g = 5% i = 10% A= 1,000 P = A*(P/A, g, i, n) P = 1,000*(P/A, 5%, 10%, ∞) P = 1,000*[(1 – (1.05/1.1) ∞ )/(0.1 – 0.05)] P = 1,000*(1/0.05) = 20,000 P  $20,000 Answer:

20 20 Q19: Find the discounted payback period for the project that has the information given to the right. Is this a good project? Initial Investment($3,000) Annual Income$ Salvage Value$2, Life6 ROR19.97% MARR15% Answer: $3, $700 $2,000 3,000 = 700*(P/A, 15%, n) + 2,000*(P/F, 15%, n) Get n >>> trials and interpolation or NPER Excel function n  3.36

21 21 Q20: Consider a contractor who is considering purchasing a compressor for use during heavy demand periods. Local equipment rental firm will rent compressors at a cost of $50/day. Compressors can be purchased for $6000. The difference in operating and maintenance costs between owned and rented compressors is estimated to be $3000/year. Find X, the number of days a year that the compressor is required to justify buying over renting. Use a planning horizon of 5 years, zero salvage values, and 10% MARR. Answer: $6, x $3,000 6,000 = 50x * (P/A, 10%, 5) – 3,000 * (P/A, 10%, 5) x  days

22 22 Q21: A $5,000 loan at 10% is to be repaid over 5 years. How much is still owed immediately after the 3rd payment? Answer: A = 5,000 / A  $ F = 5,000 * (F/P, 10%, 3) – * (F/A, 10%, 3) F = 5,000 * (1.3310) – * (3.31) F  $2289.2

23 23 Q22: Calculate the B/C ratio for the following project that has an expected life of 20 years using a discount rate of 10% per year. Answer: B/C = 3,800,000 / [1,200, ,000 * (P/A, 10%, 20)] B/C = 3,800,000 / [1,200, ,000 * (8.5136)] B/C = 3,800,000 / 4,137,192 B/C  > 0.92 < 1 Bad Project ItemCash Flow PW of benefits$3,800,000 O&M costs per year$300,000 Other annual expenses$45,000 Initial Cost$1,200,000 B/C =

24 24 Q23: Compare the two alternatives below based on their B/C ratio assuming a life of 20 years and using a discount rate of 10% per year. Answer: B/C = Alternative AB PW of benefits$4,800,000$3,800,000 O&M costs per year$280,000$200,000 Other annual expenses$45,000$50,000 Initial Cost$1,200,000$1,000,000 Salvage Value$600,000 A – B $1,000,000 $80,000 - $5,000 $200,000 $ B/C =


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