Engineering Economic Analysis Solutions Part 5

4-86 (a) Bill’s monthly payment (b) = 2/3 ($4,200) (A/P, 0.75%, 36) = $2,800 (0.0318) = $89.04 Bill owed the October 1 payment plus the present worth of the 27 additional payments. Balance = $89.04 + $89.04 (P/A, 0.75%, 27) = $89.04 (1 + 24.360) = $2,258.05 4-87 Amount of each payment = $1,000 (A/P, 4.5%, 4) = $278.70 Effective interest rate = $1,000 (0.2787) = (1 + i)m – 1 = (1.045)4 – 1 = 0.19252 = 19.3% 4-88 Monthly Payment = $10,000 (A/P, 0.75%, 12) = $875.00 Total Interest Per Year = $10,000 (0.0875) = $875.00 x 12 - $10,000 = $500.00 Rule of 78s With early repayment: Interest Charge = ((12 + 11 + 10) / 78) ($500) = $211.54 Additional Sum (in addition to the 3rd $875.00 payment) Additional Sum = $10,000 + $211.54 interest – 3 ($875.00) = $7,586.54 Exact Method Additional Sum equals present worth of the nine future payments that would have been made: Additional Sum = $875.00 (P/A, 0.75%, 9) = $875.00 (8.672) = $7,588.00 4-89 P = $25,000 n = 60 months (a) A = $25,000 (A/P, 1.5%, 60) = $635 (b) P = $25,000 (0.98) = $24,500 $24,500 i = 18% per year = 1.5% per month = $635 (P/A, i%, 60) (P/A, i%, 60) = $24,500/$635 = 38.5827 Performing interpolation using interest tables: (P/A, i%, 60) 39.380 36.964 i 1.50% 1.75% i% = 0.015 + (0.0025) [(39.380 – 38.5827)/(39.380 – 36.964)] = 0.015 + 0.000825 = 0.015825 = 1.5825% per month ia = (1 + 0.015825)12 – 1 = 0.2073 = 20.72% 4-90 i = 18%/12 = 1.5% /month (a) A = $100 (A/P, 1.5%, 24) = $100 (0.0499) (b) P13 = $4.99 + $4.99 (P/A, 1.5%, 11) = (13th payment) + (PW of future 11 payments) = $4.99 + $4.99 (10.071) = $55.24 4-91 i = 6%/12 (a) P = ½% per month = $10 (P/A, 0.5%, 48) = $10 (42.580)= $425.80 = $4.99 (b) P24 = $10 (P/A, 0.5%, 24) = $10 (22.563)= $225.63 (c) P = $10 [(e(0.005)(48) – 1)/(e(0.005)(48)(e0.005 – 1))] = $10 (42.568) = $425.68 4-92 (a) P = $500,000 - $100,000 = $400,000 A i = r/m = 0.09/12 = 0.75% n = 360 = $400,000 (A/P, 0.75%, 360) = $3,220 = $400,000 (0.00805) (b) P = A (P/A, 0.75%, 240) = $357,887 = $3,220 (111.145) (c) A = $400,000 [(e(0.06/12)(360))(e(0.06/12) – 1)/(e(0.06/12)(360) – 1)] = $400,000 [(6.05)(0.005)/(5.05)] = $2,396 4-93 P = $3,000 + $280 (P/A, 1%, 60) = $15,587 = $3,000 + $280 (44.955) 4-94 5% compounded annually F = $5,000 (F/P, 5%, 3) = $5,000 (1.158) = $5,790 5% compounded continuously F = Pern = $5,000 (e0.05 (3)) = $5,809 = $5,000 (1.1618) 4-95 Compute effective interest rate for each alternative (a) 4.375% (b) (1 + 0.0425/4)4 – 1 = (1.0106)4 – 1 (c) ern – 1 = e0.04125 – 1 = 0.0421 = 0.0431 = 4.21% The 4 3/8% interest (a) has the highest effective interest rate. = 4.31% A=? 4-96 F = Pern = $100 e.0.04 (5) = $100 (1.2214) = $122.14 4-97 1 2 3 4 P P = F [(er – 1)/(r ern)] = $40,000 [(e0.07 – 1)/(0.07 e(0.07)(4))] = $40,000 (0.072508/0.092619) = $31,314.53 4-98 (a) Interest Rate per 6 months = $20,000/$500,000 = 0.0400 Effective Interest Rate per yr. = (1 + 0.04) – 1 = 8.16% 2 = 4% = 0.0816 (b) For continuous compounding: F = Pern $520,000 = $500,000 er(1) r = ln ($520,000/$500,000) = 3.92% per 6 months Nominal Interest Rate (per year) = 0.0392 = 3.92% (2) = 7.84% per year 4-99 P = $10,000 F = $30,000 F $30,000 =Pe = $10,000 e(0.05)n 0.05 n = ln ($30,000/$10,000) n i = 5% rn = 1.0986/0.05 = 21.97 years = 1.0986 n=? 4-100 (a) Effective Interest Rate = (1 + i)m – 1 = (1.025)4 – 1 = 0.1038 = 10.38% (b) Effective Interest Rate = 0.10516 = 10.52% = (1 + i)m – 1 = (1 + (0.10/365))365 – 1 (c) Effective Interest Rate = er – 1 = 10.52% 4-101 (a) P = Fe-rn = $8,000 e-(0.08)(4.5) = $5,581.41 (b) F F/P ln(F/P) = rn r = Pern = ern = (1/n) ln (F/P) = (1/4.5) ln($8,000/$5000) = 10.44% 4-102 (1) 11.98% compounded continuously F = $10,000 e(0.1198)(4) = $16,147.82 (2) 12% compounded daily F = $10,000 (1 + 0.12/365)365x4 = $16,159.47 (3) 12.01% compounded monthly F = $10,000 (1 + 0.1201/12)12x4 = $16,128.65 (4) 12.02% compounded quarterly F = $10,000 (1 + 0.1202/4)4x4 = $16,059.53 (5) 12.03% compounded yearly F = $10,000 (1 + 0.1203)4 = $15,752.06 Decision: Choose Alternative (2) = e0.10 – 1 = 0.10517 4-103 A = $1,000 …………… n = 54 P P 10/1/97 Continuous compounding Effective interest rate/ quarter year = 3.303% = e(0.13/4) – 1 = 0.03303 Solution One P10/1/97 = $1,000 + $1,000 (P/A, 3.303%, 53) = $1,000 + $1,000 [((1.03303)53 – 1))/(0.03303(1.03303)53)] = $25,866 Solution Two P10/1/97 = $1,000 (P/A, 3.303%, 54) (F/P, 3.303%, 1) = $1,000 [((1.03303)54 – 1)/(0.03303 (1.03303)54)] (1.03303) = $25,866 4-104 (a) Effective Interest Rate ia = (1 + r/m)m – 1 = (1 + 0.06/2)2 – 1 = 0.0609 = 6.09% Continuous Effective Interest Rate ia = er – 1 = e0.06 – 1 = 0.0618 = 6.18% (b) The future value of the loan, one period (6 months) before the first repayment: = $2,000 (F/P, 3%, 5) = $2,000 (1.159) = $2318 The uniform payment: = $2,318 (A/P, 3%, 4) = $2,318 (0.2690) = $623.54 every 6 months (c) Total interest paid: = 4 ($623.54) - $2,000 = $494.16 4-105 P = Fe-rn = $6,000 e-(0.12)(2.5) = $6,000 (0.7408) = $4,444.80 4-106 Nominal Interest Rate = (1.75%) 12 = 21% Effective Interest Rate = ern – 1 = e(0.21x1) – 1 = 0.2337 = 23.37% 4-107 P = $4,500 F i=? n = 1 six month interest period = P (1 + i) (1 + i) i F = $10,000 = F/P = $10,000/$4,500 = .0526 = 1.0526 = 5.26% Nominal Interest Rate = 5.26% (2) = 10.52% Effective Interest Rate = (1 + .0526)2 – 1 = 0.10797 = 10.80% 4-108 West Bank F = P (1 + i)n = $10,000 (1 + (0.065/365))365 = $10,671.53 East Bank F = P ern = $10,000 e(.065x1) = $10,671.59 Difference = $0.06 4-109 P = F e-rn = $10,000 e-(0.08)(0.5) = $9608 = $10,000 e-0.04 = $10,000 (0.9608) 4-110 (a) Continuous cash flow – continuous compounding (one period) F = P^ [(er – 1) (ern)/rer] = $1 x 109 [(e0.005 – 1) (e(0.005)(1))/(0.005 e0.005)] = $1 x 109 [(e0.005 – 1)/0.005] = $1 x 109 (0.00501252/0.005) = $1,002,504,000 Thus, the interest is $2,504,000. (b) Deposits of A = $250 x 106 occur four times a month Continuous compounding r = nominal interest rate per ¼ month = 0.005/4 = 0.00125 = 0.125% F = A [(ern – 1)/(er – 1)] = $250,000,000 [(e(0.00125)(4) – 1)/(e(0.00125) – 1)] = $250,000,000 [0.00501252/0.00125078] = $1,001,879,000 Here, the interest is $1,879,000. So it pays $625,000 a month to move quickly! 4-111 P = F^ [(er – 1)/rern] = $15,000 [(e0.08 – 1)/((0.08)(e(0.08)(6))] = $15,000 [0.083287/0.129286] = $9,663 4-112 P = $29,000 n = 3 years F=? (a) ia F = 0.13 = P (1 + i)n = $29,000 (1.13)3 = $41,844 (b) r F = 0.1275 = P ern = $29,000 e(0.1275)(3) = $29,000 (1.4659) = $42,511 We can see that although the interest rate was less with the continuous compounding, the future amount is greater because of the increased compounding periods (an infinite number of compounding periods). Thus, the correct choice for the company is to choose the 13% interest rate and discrete compounding. 4-113 A = $1,200 F r = 0.14/12 = 0.01167 n = 7 x 12 = 84 compounding periods = A [(ern – 1)/(er – 1)] = $1,200 [(e(0.01167)(84) – 1)/(e0.01167 – 1)] = $1,200 [1.66520/0.011738] = $170,237 4-114 First Bank- Continous Compounding Effective interest rate ia = er – 1 = 4.603% = e0.045 – 1 = 0.04603 Second Bank- Monthly Compounding Effective interest rate ia = (1 + r/m)m – 1 = 0.04698 = (1 + 0.046/12)12 – 1 = 4.698% No, Barry should have selected the Second Bank. 4-115 A = P (A/P, i%, 24) (A/P, i%, 24) = A/P = 499/10,000 = 0.499 From the compound interest tables we see that the interest rate per month is exactly 1.5%. 4-116 Common Stock Investment P = $1,000 F $1,307 (F/P, i%, 20) n = 20 quarters i=? = P (F/P, i%, n) = $1,000 (F/P, i%, 20) = $1,307/$1,000 = 1.307 Performing linear interpolation using interest tables: (P/A, i%, 20) 1.282 1.347 i i 1.25% 1.50% = 1.25% + 0.25% ((1.307 – 1.282)/(1.347 – 1.282)) = 1.25% + 0.10% = 1.35% Nominal Interest Rate = 4 quarters / year (1.35% / quarter) = 5.40% / year Effective Interest Rate = (1 + i)m – 1 = (1.0135)4 – 1 = 5.51% / year 4-117 F = P (1 + i)n i = (1.00/0.98) – 1 = 0.0204 = 2.04% ieff = (1 + i)m -1 = 0.4456 = 0.98F (1 + i)1 = (1.0204)365/20 – 1 = 44.6% 4-118 A = 0.05 P …………. n = 40 quarters P F = $1,307 P = 0.05 P (P/A, i%, 40) (P/A, i%, 40) = 1/0.05 = 20 From interest tables: i 3.5% 4.0% (P/A, i%, 40) 21.355 19.793 Performing linear interpolation: i = 3.5% + 0.5% ((21.355 – 20)/(21.355 – 19.793)) = 3.5% + 0.5% (1.355/1.562) = 3.93% per quarter year Effective rate of interest = (1 + i)m – 1 = (1.0393)4 – 1 = 0.1667 = 16.67% per year 4-119 $37 5 A = $93.41 ……………. $3575 n = 45 months i=? $3,575 = $375 + $93.41 (P/A, i%, 45) (P/A, i%, 45) = ($3,575 - $375)/$93.41 = 34.258 From compound interest tables, i = 1.25% per month For an $800 down payment, unpaid balance is $2775. P = $2,775 n = 45 months i = 1.25% A=? A = $2,775 (A/P, 1.25%, 45)* = $2,775 (0.0292) = $81.03 Effective interest rate = (1 + i)12 – 1 = (1.0125)12 – 1 = 0.161 = 16.1% per year * Note that no interpolation is required as (A/P, 1.25%, 45) = 1/(P/A, i%, 45) = 1/34.258 = 0.0292