Chapter # 4 - Engineering Economy, 7 th edition , Leland Blank and Anthony Tarquin

Solutions to end-of-chapter problems Engineering Economy, 7th edition Leland Blank and Anthony Tarquin

Chapter 4 Nominal and Effective Interest Rates 4.1 t = one year; CP = one month; m = 12 4.2 t = one month; CP = one month; m = 1 4.3 (a) six times

(b) six times

(c) two times

4.4 (a) one time

(b) six times

(c) 18 times

4.5 (a) Quarter

(b) Semiannual

(c) Month

(d) Week

(e) Continuous

4.6 (a) Nominal; (b) Nominal; (c) Effective; (d) Nominal; (e) Effective; (f) Effective 4.7

1% per month = nominal 12% per year 3% per quarter = nominal 6% per six months 2% per quarter = nominal 8% per year 0.28% per week = nominal 3.36% per quarter 6.1% per six months = nominal 24.4% per two years

4.8 From interest statement, r = 11.5% per year is a nominal rate 4.9

i = 8/4 = 2% per quarter r = 2(2%) = 4% per six months

4.10

Hand solution:

i = (1 + 0.14/12)12 -1 = 14.93% per year

Spreadsheet solution: = EFFECT(14%,12) displays 14.93% 4.11 (a) Use Equation [4.4] i = (1 + 0.1587)1/4 – 1 = 0.0375 or 3.75% per quarter (b) r = 0.0375(4) = 15% per year (c) The function = NOMINAL(15.87%,4) displays 15% 1

4.12

i = (1 + 0.60)1/12 – 1 = 0.0399 or 3.99% per month

4.13

Hand solution:

i = (1 + 0.21/3)3 – 1 = 0.225 or 22.5% per year

Spreadsheet solution: = EFFECT(21%,3) displays 22.5% 4.14

8% per 6 months = 0.08/6 = 0.0133 per month i = (1 + 0.0133)3 – 1 = 0.0405 or 4.05% per quarter

4.15 (a) Use equation [4.4] for effective rate per month i = (1 + 0.04)1/3 – 1 = 0.0132 = 1.32% per month APR = 1.32(12) = 15.8% per year (b) Use Equation [4.3] for effective annual rate APY = (1 + 0.158/12)12 -1 = 17.0% 4.16 (a) Interest rate per month = (10/200)(100%) = 5% r = (5%)(12) = 60% per year (b) i = (1 + 0.60/12)12 – 1 = 0.796 or 79.6% per year 4.17 0.21/m = (1 + 0.2271)1/m – 1 By trial and error, m = 4; compounding is quarterly 4.18 (a) Interest rate per week = (10/100)(100%) = 10% r = (10%)(52) = 520% per year (b) i = (1 + 5.20/52)52 – 1 = 141.04 or 14,104% per year 4.19 (a) PP = one month; CP = six months (b) PP < CP since month is shorter than 6 months 2

4.20 (a) CP = years

(b) CP = quarters

(c) CP = months

4.21 i must be an effective rate per six months and n must be the number of semi-annual periods 4.22 F = 260,000(F/P,3%,12) = 260,000(1.4258) = $370,708 4.23 P = 1,700,000(P/F,1.5%,36) = 1,700,000(0.5851) = $994,670 4.24 P = 6(190,000)(P/F,7%,4) = 6(190,000)(0.7629) = $869,706 4.25 F = 5000(F/P,2%,48) + 7000(F/P,2%,28) = 5000(2.5871) + 7000(1.7410) = $25,123 4.26 In $1 million units, 28 = 12(F/P,3%,16) + x(F/P,3%,12) 28 = 12(1.6047) + x(1.4258) 1.4258x = 8.7436 x = $6.1324 ($6,132,400) 4.27 P = 21,000(P/F,5%,4) + 24,000(P/F,5%,6) + 10,000(P/F,5%,10) = 21,000(0.8227) + 24,000(0.7462) + 10,000(0.6139) = $41,325 4.28 P = 2,000,000(P/A,4%,20) = 2,000,000(13.5903) = $27,180,600 4.29 A = 7,000,000(A/P,6%,10) = 7,000,000(0.13587) = $951,090 4.30 926 = A(P/A,0.75%,60) 926 = A(48.1734) A = $19.22

3

4.31 A = 3,300,000(A/P,0.5%,240) + (200,000,000/1000)0.85 = 3,300,000(0.00716) + (200,000,000/1000)0.85 = 23,628 + 170,000 = $193,628 per month 4.32 First find savings at end of year 2011; use amount as an annual series for 10 years. Savings at end of year 2011 = 42,600(F/A,0.5%,5)(F/P,0.5%,3) = 42,600(5.0503)(1.0151) = $218,391 F = 218,391(F/A,0.5%,10) = 218,391(10.2280) = $2,233,708 4.33

A0% = 3199/12 = $266.58 per month A0.5% = 3199(A/P,0.5%,12) = 3199(0.08607) = $275.34 per month Savings = 275.34 – 266.58 = $8.76 per month

4.34 A = 28(F/A,1.5%,24)(A/P,1.5%,240) = 28(28.6335)(0.01543) = $12.3708 million per month 4.35 (a) Interest in payment = 5000(0.02) = $100 (b)

5000 = 110.25(P/A,2%,n) (P/A,2%,n) = 45.3515

From 2% interest table, n ≈ 120 months or 10 years 4.36 (a) Find the effective interest rate per month and calculate F after 12 months. Interest rate per month = (75/500)(100%) = 15% F = P(F/P,15%,12) = 500(5.3503) = $2675 (b) effective i = (1 + 0.15)12 – 1 = 4.35 or 435% per year 4

4.37

300 = A(P/A,1.5%,12) + [375 -10(12)](P/F,1.5%,12) 300 = A(10.9075) + [255](0.8364) 10.9075A = 86.72 A = $7.95 per month

4.38 F = 285,000(F/P,2%,60) = 285,000(3.2810) = $935,085 4.39 F = 3,600,000(F/P,6%,16) = 3,600,000(2.5404) = $9,145,440 4.40 First find F in year 5, then convert to A in years 1 through 5 using the effective annual i. F = 200,000(F/P,1.5%,48) + 350,000(F/P,1.5%,24) + 400,000 = 200,000(2.0435) + 350,000(1.4295) + 400,000 = $1,309,025 i = (1 + 0.18/12)12 - 1 = 19.56% per year A = 1,309,025(A/F,19.56%,5) Solve for A by interpolation between 18% and 20%, by formula, or use spreadsheet function. By spreadsheet function = PMT(19.56%,5,.-1309025) A = $177,435 per year 4.41 i = (1 + 0.12/12)12 - 1 = 12.68% per year F = 30(F/A,12.68%,9) + 20(F/A,12.68%,3) Find factor values by interpolation, formula, or spreadsheet. Figure 2-9 shows spreadsheet functions. F = 30(15.2077) + 20(3.3965) = $524.16 4.42 A = 480 + 20(A/G,0.25%,120) = 480 + 20(56.5084) = $1,610,168,000 per month

5

4.43 i = (1 + 0.10/4)4 - 1 = 10.38% per year Pg = 100,000{1 – [(1 + 0.04)/(1 + 0.1038)]5}/(0.1038 – 0.04) = 100,000(4.03556) = $403,556 4.44 A per quarter = 3(1000) = $3000 F = 3000(F/A,1.5%,20) = 3000(23.1237) = $69,371 4.45 Chemical cost = 11(30) = $3300 per month A = 2(950)(A/P,1%,36) + 3300 = 2(950)(0.03321) + 3300 = $3363.10 per month 4.46 A = 3000(3) = $9000 per quarter F = 9000(F/A,1.5%,10) = 9000(10.7027) = $96,324 4.47 A per 6 months = 900(6) = $5400 semiannually P = 5400(P/A,7%,6) = 5400(4.7665) = $25,739 4.48 Hand: r = 0.012(12) = 0.144 per year i = e0.144 - 1 = 15.49% per year Spreadsheet: = EFFECT(14.4%,10000) displays 15.49% 4.49 r = (0.016)(3) = 0.048% per quarter i = e0.048 - 1 = 4.92% per quarter 4.50

0.013 = er – 1 er = 1.013 r = ln 1.013 = 0.0129 or 1.29% per month 6

4.51

0.25 = er – 1 er = 1.25 r = ln 1.25 = 0.22.31 or 22.31% per year Nominal daily i = 22.31/365 = 0.061% per day

4.52 i = e0.12 – 1 = 0.1275 or 12.75% per year P = 13,000,000(P/F,12.75%,2) Find factor value by interpolation, formula, or spreadsheet. P = 13,000,000(0.7866) = $10,226,105 4.53 i = e0.10 – 1 = 0.10517 or 10.517% per year P = 150,000 + 200,000(P/F,10.517%,1) + 350,000(P/F,10.517%,2) Find factor values by interpolation, formula, or spreadsheet. P = 150,000 + 200,000(0.9048) + 350,000(0.8187) = $617,505 4.54 F = 300,000(F/P,1%,4)(F/P,1.25%,8) = 300,000(1.0406)(1.1045) = $344,803 4.55 Hand solution: F = 140,000(F/A,8%,3)(F/P,10%,2) + 140,000(F/A,10%,2) = 140,000(3.2464)(1.2100) + 140,000(2.1000) = $843,940 Spreadsheet solution: Use embedded FV functions = FV(10%,2,,FV(8%,3,140000)) + FV(10%,2,-140000) to display $843,940 4.56 In $1 million units P = 1.7(P/F,10%,1) + 2.1(P/F,12%,1)(P/F,10%,1) + 3.4(P/F,12%,2)(P/F,10%,1) = 1.7(0.9091) + 2.1(0.8929)(0.9091) + 3.4(0.7972) (0.9091) = $5,714,212

7

4.57 (a)

(b)

P = 100(P/A,10%,5) + 160(P/A,14%,3)(P/F,10%,5) = 100(3.7908) + 160(2.3216)(0.6209) = 100(3.7908) + 160(1.4415) = $609.72 609.72 = A(3.7908) + A(1.4415) A = 609.72/5.2323 = $116.53 per year

4.58 Answer is (b) 4.59 Answer is (d) 4.60 Answer is (c) 4.61 Answer is (b) 4.62 0.1268 = (1 + r/12)12 - 1 (1 + r/12)12 = 1.1268 12*log (1 + r/12) = log 1.1268 12*log (1 + r/12) = 0.05185 log (1 + r/12) = 0.00432 (1 + r/12) = 1.0100 r/12 = 0.0100 r = 0.12 r = 12% per year, compounded monthly = 1% per month Answer is (c) (Note: r = 12% per year, compounded monthly can be found in Table 4-3.) 4.63 i = (1 + 0.02)6 - 1 = 12.62% Answer is (d) 4.64 Answer is (b) 4.65 Answer is (c) 4.66 Answer is (c)

8

4.67 PP < CP; assume no interperiod compounding F = 1000(F/A,3%,50) = $112,796.90 Answer is (d) 4.68 Answer is (d) 4.69 Answer is (b) 4.70 Answer is (a) 4.71 Answer is (c) 4.72 A = 500,000(A/F,7%,12) = 500,000(0.05590) = $27,950 per 6 months Answer is (c)

9

Solution to Case Study, Chapter 4 There are not always definitive answers to case studies. The following are examples only.

IS OWNING A HOME A NET GAIN OR NET LOSS OVER TIME? 1. Summary of future worth values if sold at $363,000: A: 30-year, fixed rate plus investments, FA = $243,246 (from text) B: 15-year, fixed rate plus investments, FB = $246,010 (worked below) Rent-don’t buy: F = $109,199 (spreadsheet below) Conclusion: Select the 15-year loan Plan B analysis: 15-year fixed rate loan Amount of money required for closing costs: Down payment (10% of $330,000) Up-front fees (origination fee, attorney’s fee, survey, filing fee, etc.) Total

$33,000 3,000 $36,000

The amount of the loan is $297,000 and equivalent monthly principal and interest (P&I) is determined at 5.0%/12 = 0.4167% per month for 15(12) = 180 months. A = 297,000(A/P,0.4167%,180) = 297,000(0.00791) ≈ $2350 Add the T&I of $500 for a total monthly payment of PaymentB = $2850 per month The future worth of plan B is the sum of remainder of the $40,000 available for the closing costs (F1B); left over money from that available for monthly payments (F2B); and, increase in the house value when it is sold after 10 years (F3B). F1B = $7278 No money is available each month to invest after the mortgage payment of $2850. Therefore, F2B = $0 Net money from the sale in 10 years (F3B) is the difference in net selling price ($363,000) and remaining balance on the loan.

10

Loan balance = 297,000(F/P,0.4167%,120) - 2350(F/A,0.4167%,120) = 297,000(1.6471) - 2350(155.2856) = $124,268 F3B = 363,000 - 124,268 = $238,732 Total future worth of plan B is: FB = F1B + F2B + F3B = 7278 + 0 + 238,732 = $246,010 Rent-Don’t Buy Plan Analysis

2. Summary of future worth values if sold at $231,000: A: 30-year, fixed rate plus investments, FA = $111,246 F3A changes to 231,000 - 243,386 = $-12,386 (must pay purchasers to buy) Total future worth of plan A is: FA = F1A + F2A + F3A = 7278 + 116,354 - 12,386 = $111,246 B: 15-year, fixed rate plus investments, FB = $114,010 F3B changes to 231,000 - 124,286 = $106,714

11

Total future worth of plan B is: FB = F1B + F2B + F3B = 7278 + 0 + 106,714 = $113,992 Rent-don’t buy: F = $109,199 (same as above) Conclusion: Still select the 15-year loan, but the economic advantage is much less.

12