ECOR 3800 - Assignment 2 Solutions

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Carleton University *

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3800

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Civil Engineering

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Apr 3, 2024

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2/26/24, 1:01 PM ECOR 3800 - Assignment 2 Solutions about:blank 1/10 Carleton University Department of Civil and Environmental Engineering Engineering Economics (ECOR 3800B) ASSIGNMENT # 2 Issued March 02, 2014 Due Date: March 14, 2014 at 12:00 Noon Drop off location: Filing cabinet near the entrance to the Civil and Environmental Engineering office. The cabinet located to the right of room 3424 ME. ============================================================= Question (1) (15 marks) (A) If your credit card calculates interest based on 12.5% APR, (a) What are your monthly interest rate and annual effective interest rate? (b) If your current outstanding balance is $2,000 and you skip payments for two months, what would be the total balance two months from now? Solution APR= 12.5% a) i/month =i m=r/12= 12.5%/12= 1.042% Annual effective interest rate : I a= (1+r/M) M -1= 13.2% b) F=2000(1+0.01042) 2 =2041.90 B) College Financial Sources, which makes small loans to college students, offers to lend $500. The borrower is required to pay $400 at the end of each week for 16 weeks. Find the interest rate per week. What is the nominal interest rate per year? What is the effective interest rate per year? Solution P = A(P/A, i,N) P/A = 500/40 = 12.5, From the table we can get P/A = 12.5 ,i= 3% Nominal = r=i w (52) week/year=156% I a = (1+i/M) M -1 =365.1%

2/26/24, 1:01 PM ECOR 3800 - Assignment 2 Solutions about:blank 2/10 (C) College Financial Sources, which makes small loans to college students, offers to lend $500. The borrower is required to pay $400 at the end of each week for 16 weeks. Find the interest rate per week. What is the nominal interest rate per year? What is the effective interest rate per year? Solution F= P(1+i) n F = 450, P = 400, n=1 Therefore i= 12.5% a) Nominal = i n = i x 52 weeks =( 0.125)52 = 650% b) Effective = i e = (1+6.5/52) 52 -1= 45702.2 a) AE= [ -10,000+ 3,000(P/A, 7% ,2) +4000 (P/A, 7%,2)(P/F,7%,2) +2000 (P/F, 7%, 5) + (2,000+ 200)(P/F , 7%, 6)] = [- 10,000 + 3,000 (1.808) + 4000 (1.808) (0.8734) + 2000(0.7130) + (2200) (0.6663)]. (0.2098) =4632.29 AE(7%)=4632.29(A/P,7%,6) =4632.29*0.2098 =$ 971.85 b) PROJECT A: Payback period P/A= $2500/$300 = 8.33 year Therefore project does not pay back PROJECT B: Payback period= 1+$1000/$1500 = 1.667 years PROJECT C: Payback period= 2+$1500/$2000 = 2.75 years PROJECT D Payback period=$4000/$5000 = 0.8 years, of initial payment Project D can be viewed as two separate projects, where the first investment is recovered at the end of year 1 and the investment that were made in year 2 and 3 will be recovered at the end of year 6.

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2/26/24, 1:01 PM ECOR 3800 - Assignment 2 Solutions about:blank 4/10 C) Consider the cash flow diagram. Compute the equivalent annual worth at = 11% i Solution A (P/A, 0.11, 6) = -{5500(P/A,0.11,2)+8500(P/F,0.11,3)+11500(P/F,11,4)+14500(P/F,0.11,5)+17500(P/F,0.11,6)} OR AE1 (11%) = +20500(A/P, 11%, 6) AE2 (11%) = -5000(P/A, 11%, 6) (A/P, 11%, 6) AE3 (11%) = -3000(P/G, 11%, 5) (P/F, 11%, 1) (A/P, 11%, 6) OR AE (9%) = +205000(A/P, 11%, 6)-5500-3000(P/G, 11%, 5) (P/F, 11%, 1)(A/P,11%, 6) Answer: P 1 = 20500 P 2 = -5500(P/A, 11%, 6) = -5500(4.233) Note = 10% ---4.355, 12% -----4.111, 11% ----4.233 = 23281.5 P 3 = -3000(P/G, 11%, 5) (P/F, 11%, 1) = -3000{(6.862+6.397)/2}{(.9091+.8929)/2} = -3000(6.624)(0.901) = 17904.67 AE 1 =20500(A/P, 11%, 6) = 20500{(0.2296+0.2432)/2}=20500(0.2364)=4846.2 AE 2 = -23281.5(0.2364) =5503.75 AE 3 = -17904.67(0.2364) = 4232.67 AE = 4846.2-5503.75-4232.67 = 4890.22

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