ECOR 3800 - Assignment 2 Solutions

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Carleton University *

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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
2/26/24, 1:01 PM ECOR 3800 - Assignment 2 Solutions about:blank 5/10 Question (3) (15 marks) (A) Compute the future worth of the cash flows with different interest rates specified. The cash flows occur at the end of each year over four years. Solution i a1 = 9% daily = (1+r/M) M -1=(1+0.09/365) 365 -1 For years 3 and 4 i= 9% continuous I a2 = e r/k -1 =e 0.09/1 -1=0.09417 F= 400(F/P,9.416,2)(F/P,9.417,2)+250(F/P,9.416%,1)(F/P,9.417%,2)+100(F/P,9.417%,2)+100(F/P,9.417%,1 ) + 250 =400(1.1972)(1.1972)+250(1.0942)(1.1972)+100(1.1972)+100(1.0942)+250 =573.32+327.49+119.72+109.42+250 = $1379.95
2/26/24, 1:01 PM ECOR 3800 - Assignment 2 Solutions about:blank 6/10 This is a preview Do you want full access? Go Premium and unlock all 10 pages Access to all documents Get Unlimited Downloads Improve your grades Already Premium? Free Trial Get 30 days of free Premium Upload Share your documents to unlock Log in
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2/26/24, 1:01 PM ECOR 3800 - Assignment 2 Solutions about:blank 7/10 (B) A large food-processing corporation is considering using laser technology to speed up and eliminate waste in the potato-peeling process. To implement the system, the company anticipates needing $2.5 million to purchase the industrial-strength lasers. The systems will save $1,200,000 per year in labor and materials. However, it will require an additional operating and maintenance cost of $275,000. Annual income taxes will also increase by $145,000. The system is expected to have a 10-year service life and will have a salvage value of about $225,000. If the company’s MARR is 10%, justify the economics of the project based on: (a) PE method (b) FE method (c) AE method Solution PE(10%) =-2.5X10 6 +1.2X10 6 -275000-145000)(P/A, 10%, 10) + 225000 (P/F, 10%, 10) = $ 2379525.50 FE (10%) = PE(10%) (P/F, 10%, 10) = $ 6171775.29 AE (10%) = FE(10%) (A/P,10%,10) = $ 387148.80 Question (5) (20 Marks) (A) Consider the following project balances for a typical investment project with a service life of 4 years (a) Construct the original cash flows of the project. (b) Determine the interest rate in computing the project balance. (c) At =15%, would this project be acceptable? i Solution At year 3 cost of funds = -800 i -500= [460+800i]+(-800)= 0.20=20% A 1 = -1100+1000(1.2) = $100 A 2 = -800+1100(1.20) = $520 A 3 = 0+500(1.2) = $600 C) FE(15%)= -1000(F/P, 15%, 4)+ 100(F/P, 15%, 3) + 520(F/P, 15%, 2)+460(F/P,15%,1)+ 600 = $ 219.78 FE(15%) > 0 the project is acceptable
2/26/24, 1:01 PM ECOR 3800 - Assignment 2 Solutions about:blank 8/10 Question (5) (20 Marks) Solution PW(24%) = -1000+400(P/F, 24%, 1) +800(P/F, 24%, 2)+ X(P/F, 24%, 3) X = $ 299.58 PW(23%) = -1000+300(P/F, 23%, 1) +800(P/F, 23%, 2)+ Y(P/F, 23%, 2) Y=$493.49 FW(24%) = -1000(1.24) 3 + 400(1.24) 2 +800(1.24) 2 +299.49 = 0 INTEREST i for Project 400(1+i 1 ) -1 +800 (1+i 1 ) -2 +229.58((1+i 1 ) -3 =300((1+i 1 ) -1 +493.49(1+i 1 ) -2 +800((1+i 1 ) -3 100((1+i 1 ) -2 +30651((1+i 1 )-500=0 i=0.1783 OR 17.83%
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2/26/24, 1:01 PM ECOR 3800 - Assignment 2 Solutions about:blank 10/10 (B) Consider the following investment projects a) Compute the future worth at the end of life for each project i=12% b) Determine the acceptability of each project Solution FW(12%) A = -1800(F/P, 12%, 5)-500 (F/P, 12%, 4)+900((F/P, 12%, 3)+1300((F/P, 12%, 2)+ 2200(F/P, 12%, 1)-700 =$700.18 FW(12%) B =-5200(F/P, 12%, 5)+2500 (F/P, 12%, 4)-4000((F/P, 12%, 3)+5000((F/P, 12%, 2)+ 6000(F/P, 12%, 1)+3000 =5141.19 FW (12%) C =-3800(F/P, 12%, 5)+4000 (F/P, 12%, 2)+7000((F/P, 12%, 1)+12000 =$18160.7 FW (12%) D =-4000(F/P, 12%, 5)+500 (F/P, 12%, 4)+2000((F/P, 12%, 3)+3000((F/P, 12%, 2)+ 4000(F/P, 12%, 1)+1250 =$6040.45 FW (12%) D =-6500(F/P, 12%, 3)+1000 (F/P, 12%, 2)+3600((F/P, 12%, 1)+2400 =$-1446.5 All projects are acceptable except D

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