Chapter 7: Rate of Return Analysis 7-1 $12 $10 $20 $30 $40 $50 $60 $125 = $10 (P/A, i%, 6) + $10 (P/G, i%, 6) at 12%, $10 (4.111) + $10 (8.930) = $130.4 at 15%, $10 (3.784) + $10 (7.937) = $117.2 i* = 12% + (3%) ((130.4 – 125).(130.4-117.2)) = 13.23% 7-2 $80 $80 $80 $200 $200 $80 $80 $80 $200 The easiest solution is to solve one cycle of the repeating diagram: $80 $80 $80 = $20 $120 = $80 (F/P, i%, 1) $120 = $80 (1 + i) (1 + i) = $120/$80 = 1.50 $12 i* = 0.50 = 50% Alternative Solution: EUAB = EUAC $80 = [$200 (P/F, i%, 2) + $200 (P/F, i%, 4) + $200 (P/F, i%, 6)] (A/P, i%, 6) Try i = 50% $80 = [$200 (0.4444) + $200 (0.1975) + $200 (0.0878)] (0.5481) Therefore i* = 50% 7-3 $5 $10 $15 $20 $25 $42.55 $42.55 = $5 (P/A, i%, 5) + $5 (P/G, i%, 5) Try i = 15%, $5 (3.352) + $5 (5.775) = $45.64 > $42.55 Try i = 20%, $5 (2.991) + $5 (4.906) = $39.49 < $42.55 Rate of Return = 15% + (5%) [($45.64 - $42.55)/($45.64 - $39.49)] = 17.51% Exact Answer: 17.38% 7-4 For infinite series: A = Pi EUAC= EUAB $3,810 (i) = $250 + $250 (F/P, i%, 1) (A/F, i%, 2)* Try i = 10% $250 + $250 (1.10) (0.4762) = $381 $3,810 (0.10) = $381 i = 10% *Alternate Equations: $3,810 (i) = $250 + $250 (P/F, i%, 1) (A/P, i%, 2) $3,810 (j) = $500 - $250 (A/G, i%, 2) = $79.99 7-5 P’ A = $1,000 ………… Yr n = 10 $41 n=∞ $5,000 At Year 0, PW of Cost = PW of Benefits $412 + $5,000 (P/F, i%, 10) = ($1000/i) (P/F, i%, 10) Try i = 15% $412 + $5,000 (0.2472) $1,648 = ($1,000/0.15) (0.2472) = $1,648 ROR = 15% 7-6 The algebraic sum of the cash flows equals zero Therefore, the rate of return is 0% 7-7 A = $300 $1,000 Try i = 5% $1,000 =(?) Try i = 6% $1,000 =(?) $300 (3.546) (0.9524) =(?) $1,013.16 $300 (3.465) (0.9434) =(?) $980.66 Performing Linear Interpolation: i* = 5% + (1%) (($1,013.6 - $1,000)/($1,013.6 - $980.66)) = 5.4% 7-8 $400 = [$200 (P/A, i%, 4) - $50 (P/G, i%, 4)] (P/F, i%, 1) Try i = 7% [$200 (3.387) - $50 (4.795)] (0.9346) = 409.03 Try i = 8% [$200 (3.312) - $50 (4.650)] (0.9259) = $398.08 i* = 7% + (1%) [($409.03 - $400)/($409.03 - $398.04)] = 7.82% 7-9 $100 (P/A, i%, 10) = $27 (P/A, i%, 10) = 3.704 Performing Linear Interpolation: (P/A, i%, 10) i 4.192 20% 3.571 25% Rate of Return = 20% + (5%) [(4.192 – 3.704)/(4.912 – 3.571)] = 23.9% 7-10 Year Cash Flow -$500 -$100 +$300 +$300 +$400 +$500 $500 + $100 (P/F, i%, 1)= $300 (P/A, i%, 2) (P/F, i%, 1) + $400 (P/F, i%, 4) + $500 (P/F, i%, 5) Try i = 30% $500 + $100 (0.7692) = $576.92 $300 (1.361) (0.7692) + $400 (0.6501) + $500 (0.2693) = $588.75 ∆ = 11.83 Try i = 35% $500 + $100 (0.7407) = $574.07 $300 (1.289) (0.7407) + $400 (0.3011) + $500 (0.2230) = $518.37 ∆ = 55.70 Rate of Return = 30% + (5%) [11.83/55.70) = 31.06% Exact Answer: 30.81% 7-11 Year 10 Cash Flow -$223 -$223 -$223 -$223 -$223 -$223 +$1,000 +$1,000 +$1,000 +$1,000 +$1,000 The rate of return may be computed by any conventional means On closer inspection one observes that each $223 increases to $1,000 in five years $223 = $1,000 (P/F, i%, 5) (P/F, i%, 5) = $223/$1,000 = 0.2230 From interest tables, Rate of Return = 35% 7-12 Year Cash Flow -$640 40 +$100 +$200 +$300 +$300 $640 = $100 (P/G, i%, 4) + $300 (P/F, i%, 5) Try i = 9% $100 (4.511) + $300 (0.6499) = $646.07 > $640 Try i = 10% $100 (4.378) + $300 (0.6209) = $624.07 < $640 Rate of Return = 9% + (1%) [(%646.07 - $640)/($646.07 - $624.07)] = 9.28% 7-13 Since the rate of return exceeds 60%, the tables are useless F = P (1 + i)n $4,500 = $500 (1 + i)4 (1 + i)4 = $4,500/$500 =0 (1 + i) = 91/4 = 1.732 i* = 0.732 = 73.2% 7-14 $3,000 = $119.67 (P/A, i%, 30) (P/A, i%, 30) = $3,000/$119.67 = 25.069 Performing Linear Interpolation: (P/A, i%, 30) 25.808 24.889 i i 1% 1.25% = 1% + (0.25%)((25.808-25.069)/(25.808-24.889)) = 1.201% (a) Nominal Interest Rate = 1.201 x 12 = 14.41% (b) Effective Interest Rate = (1 + 0.01201)12 – = 0.154 = 15.4% 7-15 $3,000 A = $325 ………… n = 36 $12,375 $9,375 = $325 (P/A, i%, 36) (P/A, i%, 36) = $9,375/$325 = 28.846 From compound interest tables, i = 1.25% Nominal Interest Rate Effective Interest Rate = 1.25 x 12 = 15% = (1 + 0.0125)12 – = 16.08% 7-16 1991 – 1626 = 365 years = n F = P (1 + i)n 12 x 109 = 24 (1 + i)365 (1 + i)365 = 12 x 100/24 = 5.00 x 108 This may be immediately solved on most hand calculators: i* = 5.64% Solution based on compound interest tables: (F/P, i%, 365) = 5.00 x 108 = (F/P, i%, 100) (F/P, i%, 100) (F/P, i%, 100) (F/P, i%, 65) Try i = 6% (F/P, 6%, 365) = (339.3)3 (44.14) = 17.24 X 108 (i too high) Try i = 5% (F/P, 5%, 365) = (131.5)3 (23.84) = 0.542 X 108 (i too low) Performing linear interpolation: i* = 5% + (1%) [((5 – 0.54) (108))/((17.24 – 0.54) (108))] = 5% + 4.46/16.70 = 5.27% The linear interpolation is inaccurate 7-17 $1,000 A = $40 n = 10 $92 PW of Cost = PW of Benefits $925 = $40 (P/A, i%, 10) + $1,000 (P/F, i%, 10) Try i = 5% $925 = $40 (7.722) + $1,000 (0.6139) = $922.78 (i too high) Try i = 4.5% $925 = $40 (7.913) + $1,000 (0.6439) = $960.42 (i too low) i* ≈ 4.97% 7-18 $1,000 A = $40 …… n = 40 semiannual periods $715 PW of Benefits – PW of Costs = $20 (P/A, i%, 40) + $1,000 (P/F, i%, 40) - $715 = Try i = 3% $20 (23.115) + $1,000 (0.3066) - $715 = $53.90 i too low Try i = 3.5% $20 (21.355) + $1,000 (0.2526) - $715 = -$35.30 i too high Performing linear interpolation: i* = 3% + (0.5%) [53.90/(53.90 – (-35.30))] = 3.30% Nominal i* = 6.60% 7-19 $12,000 $6,000 $3,000 n = 10 n = 10 $28,000 PW of Cost = PW of Benefits n = 20 $28,000 = $3,000 (P/A, i%, 10) + $6,000 (P/A, i%, 10) (P/F, i%, 10) + $12,000 (P/A, i%, 20) (P/F, i%, 20) Try i = 12% $3,000 (5.650) + $6,000 (5.650) (0.3220) + $12,000 (7.469) (0.1037) = $37,160 > $28,000 Try i = 15% $3,000 (5.019) + $6,000 (5.019) (0.2472) + $12,000 (6.259) (0.0611) = $27,090 < $28,000 Performing Linear Interpolation: i* = 15% - (3%) [($28,000 - $27,090)/($37,160 - $27,090)] = 15% - (3%) (910/10,070) = 14.73% 7-20 $15,000 A = $80 $9,000 PW of Benefits – PW of Cost = $0 $15,000 (P/F, i%, 4) - $9,000 - $80 (P/A, i%, 4) = $0 Try i = 12% $15,000 (0.6355) - $9,000 - $80 (3.037) = +$289.54 Try i = 15% $15,000 (0.5718) - $9,000 - $80 (2.855) = -$651.40 Performing Linear Interpolation: i* = 12% + (3%) [289.54/(289.54 + 651.40)] = 12.92% 7-21 The problem requires an estimate for n- the expected life of the infant Seventy or seventyfive years might be the range of reasonable estimates Here we will use 71 years The purchase of a $200 life subscription avoids the series of beginning-of-year payments of $12.90 Based on 71 beginning-of-year payments, A = $12.90 ………… n = 70 $200 $200 - $12.90 (P/A, i%, 70) 6% < i* < 8%, = $12,90 (P/A, i%, 70) = $187.10/$12.90 = 14.50 By Calculator: i* = 6.83% 7-22 $1,000 A = $30 ……… n = 2(2001 – 1998) + = 27 $875 PW of Benefits – PW of Cost = $0 $30 (P/A, i%, 27) + $1,000 (P/F, i%, 27) - $875 = $0 Try i = ½% $30 (17.285) + $1,000 (0.3950) - $875 = $38.55 >$0 Try i = 4% $30 (16.330) + $1,000 (0.3468) - $875 = -$38.30 < $0 i* = 3.75% Nominal rate of return = (3.75%) = 7.5% 7-23 A = $110 ……… n = 24 $3,500 - $1,200 = $2,300 $2,300 = $110 (P/A, i%, 24) (P/A, i%, 24) = $2,300/$110 = 20.91 From tables: 1% < i < 1.25% On Financial Calculator: i = 1.13% per month Effective interest rate = (1 + 0.0113)12 – = 0.144 = 14.4% 7-24 A = $100 ……… n = 36 $3,168 PW of Cost = PW of Benefits $100 (P/A, i%, 36)= $3,168 (P/A, i%, 36) = $3,168/$100 = 31.68 Performing Linear Interpolation: (P/A, i%, 36) I 32.871 ẵ% 21.447 ắ% i* = (1/2%) + (1/4%) [(32.87 – 31.68)/(32.87 – 31.45)] = 0.71% Nominal Interest Rate = 12 (0.71%) = 8.5% 7-25 This is a thought-provoking problem for which there is no single answer Two possible solutions are provided below A Assuming the MS degree is obtained by attending graduate school at night while continuing with a full-time job: Cost: $1,500 per year for years Benefit: $3,000 per year for 10 years MS Degree A = $3,000 n = 10 $1,500 $1,500 Computation as of award of MS degree: $1,500 (F/A, i%, 2) = $3,000 (P/A, i%, 10) i* > 60 B Assuming the MS degree is obtained by one of year of full-time study Cost: Difference between working & going to school Whether working or at school there are living expenses The cost of the degree might be $24,000 Benefit: $3,000 per year for 10 years $24,000 = $3,000 (P/A, i%, 10) i* = 4.3% 7-26 $35 A = $12.64 ……… n = 12 $175 ($175 - $35) = $12.64 (P/A, i%, 12) (P/A, i%, 12) = $140/$12.64 = 11.08 i = ¼% Nominal interest rate = 12 (1 ¼%) = 15% 7-27 The rate of return exceeds 60% so the interest tables are not useful F $25,000 (1 + i) i* = P (1 + i)n = $5,000 (1 + i)3 = ($25,000/$5,000)1/3 = 1.71 = 0.71 Rate of Return = 71% 7-28 This is an unusual problem with an extremely high rate of return Available interest tables obviously are useless One may write: PW of Cost = PW of Benefits $0.5 = $3.5 (1 + i)-1 + $0.9 (1 + i)-2 + $3.9 (1 + i)-3 + $8.6 (1 + i)-4 + … For high interest rates only the first few terms of the series are significant: Try i = 650% PW of Benefits = $3.5/(1 + 6.5) + $0.9/(1 + 6.5)2 + $3.9/(1 + 6.5)3 + $8.6/(1 + 6.5)4 + … = 0.467 + 0.016 + 0.009 + 0.003 = 0.495 Try i = 640% PW of Benefits = $3.5/(1 + 6.4) + $0.9/(1 + 6.4)2 + $3.9/(1 + 6.4)3 + $8.6/(1 + 6.4)4 + … = 0.473 + 0.016 + 0.010 + 0.003 = 0.502 i* ≈ 642% (Calculator Solution: i = 642.9%) 7-29 g = 10% A1 = $1,100 n = 20 i=? P = $20,000 The payment schedule represents a geometric gradient There are two possibilities: i ≠ g and i = g Try the easier i = g computation first: P = A1n (1 + i)-1 where g = i = 0.10 $20,000 = $1,100 (20) (1.10)-1 = $20,000 Rate of Return i* = g = 10% 7-30 (a) Using Equation (4-39): F = Pern $4,000 = $2,000er(9) = er(9) 9r = ln = 0.693 r = 7.70% (b) Equation (4-34) ieff = er – = e0.077 – = 0.0800 = 8.00% 7-31 (a) When n = ∞, i = A/P = $3,180/$100,000 = 3.18% (b) (A/P, i%, 100) = $3180/$100,000 From interest tables, i* = 3% = 0.318 (c) (A/P, i%, 50) = $3,180/$100,000 From interest tables, i* = 2% = 0.318 (d) The saving in water truck expense is just a small part of the benefits of the pipeline Convenience, improved quality of life, increased value of the dwellings, etc., all are benefits Thus, the pipeline appears justified 7-32 F = $2,242 A = $50 n=4 P = $1,845 Set PW of Cost = PW of Benefits $1,845 = $50 (P/A, i%, 4) + $2,242 (P/F, i%, 4) Try i = 7% 450 (3.387) + $2,242 (0.7629) = $1,879 > $1,845 Try i = 8% 450 (3.312) + $2,242 (0.7350) = $1,813 < $1,845 Rate of Return = 7% + (1%) [($1,879 - $1,845)/($1,879 - $1,813)] = 7.52% for months Nominal annual rate of return = (7.52%) = 15.0% Equivalent annual rate of return = (1 + 0.0752)2 – = 15.6% 7-33 (a) F = $5 P = $1 n=5 F = P (1 + i)n $5 = $1 (1 + i)5 (1 + i) = 50.20 = 1.38 i* = 38% (b) For a 100% annual rate of return F = $1 (1 + 1.0)5 = $32, not $5! Note that the prices Diagonal charges not necessarily reflect what anyone will pay a collector for his/her stamps 7-34 $800 $400 $6,000 $9,000 Year 1- 5- Cash Flow -$9,000 +$800 +$400 +$6,000 PW of Cost = PW of Benefits $9,000 = $400 (P/A, i%, 8) + $400 (P/A, i%, 4) + $6,000 (P/F, i%, 9) Try i = 3% $400 (7.020) + $400 (3.717) + $6,000 (0.7664) = $8,893 < $9,000 Try i = ½% $400 (7.170) + $400 (3.762) + $6,000 (0.8007) = $9,177 > $9,000 Rate of Return = ½% + (1/2%) [($9,177 - $9,000)/($9,177 - $8,893)] = 2.81% 7-35 Year Cash Flow -$1,000 +$1,094.60 +$1,094.60 $1,000 = $1,094 [(P/F, i%, 6) + (P/F, i%, 9)] Try i = 20% $1,094 [(0.5787) + (0.3349)] = $1,000 Rate of Return = 20% 7-36 $65,000 $5,000 $240,000 $240,000 = $65,000 (P/A, i%, 13) - $5,000 (P/G, i%, 13) Try i = 15% $65,000 (5.583) -$5,000 (23.135) = $247,220 > $240,000 Try i = 18% $65,000 (4.910) -$5,000 (18.877) = $224,465 < $240,000 Rate of Return = 15% + 3% [($247,220 - $240,000)/($247,220 - $224,765)] = 15.96% 7-37 3,000 = 30 (P/A, i*, 120) (P/A, i*, 120) = 3,000/30 = 100 Performing Linear Interpolation: (P/A, i%, 120) 103.563 100 90.074 i* I ẳ% i* ẵ% = 0.0025 + 0.0025 [(103.562 – 100)/(103.562 – 90.074)] = 0.00316 per month Nominal Annual Rate = 12 (0.00316) = 0.03792 = 3.79% 7-38 (a) Total Annual Revenues = $500 (12 months) (4 apt.) = $24,000 Annual Revenues – Expenses = $24,000 - $8,000 = $16,000 To find Internal Rate of Return the Net Present Worth must be $0 NPW = $16,000 (P/A, i*, 5) + $160,000 (P/F, i*, 5) - $140,000 At i = 12%, At i = 15%, IRR NPW = $8,464 NPW = -$6,816 = 12% + (3%) [$8,464/($8,464 + $6,816)] = 13.7% (b) At 13.7% the apartment building is more attractive than the other options 7-39 NPW = -$300,000 + $20,000 (P/F, i*, 10) + ($67,000 - $3,000) (P/A, i*, 10) - $600 (P/G, i*, 10) Try i = 10% NPW= -$300,000 + $20,000 (0.3855) + ($64,000) (6.145) - $600 (22.891) = $87,255 > $0 The interest rate is too low Try i = 18% NPW = -$300,000 + $20,000 (0.1911) + ($64,000) (4.494) - $600 (14.352) = -$17,173 < $0 The interest rate is too high Try i = 15% NPW = -$300,000 + $20,000 (0.2472) + ($64,000) (5.019) - $600 (16.979) = $9,130 > $0 Thus, the rate of return (IRR) is between 15% and 18% By linear interpolation: i* = 15% + (3%) [$9,130/($9,130 - $17,173)] = 16.0% 7-40 (a) First payment = $2, Final payment = 132 x $2 = $264 Average Payment = ($264 + $2)/2 = $133 Total Amount = 132 payments x $133 average pmt = $17,556 Alternate Solution: The payments are same as sum-of-years digits form SUM = (n/2) (n + 1) = (132/2) (133) = 8,778 Total Amount = $2 (8778) = $17,556 (b) The bank will lend the present worth of the gradient series Loan (P) = $2 (P/G, 1%, 133) Note: n- = 132, so n = 133 By interpolation, (P/G, 1%, 133) = 3,334.11 + (13/120) (6,878.6 – 3,334.1) = 3,718.1 Loan (P) = $2 (3,718.1) = $7,436.20 7-41 Year 10 11 12 … 33 34 35 36 Case (incl Deposit) -$39,264.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$599.00 +$27,854.00 -$625.00 = +$27,229.00 IRR Nominal IRR Effective IRR = 0.86% = 10.32% = 10.83% 7-42 MARR = 5% P = $30,000 n = 35 years Alternative 1: Withdraw $15,000 today and lose $15,000 Alternative 2: Wait, leave your fund in the system until retirement Equivalency seeks to determine what future amount is equal to $15,000 now F = P (1 + i)n = $30,000 (1.05)35 = $30,000 (5.516015) = $165,480.46 Therefore: $15,000 = $165,480.46 (1 + i)-35 $15,000 (1 + i)35 = $165,480.46 (1 + i) = [(165,480.46/$15,000)]1/35 i = 1.071 – = 7.1002% > 5% Unless $15,000 can be invested with a return higher than 7.1%, it is better to wait for 35 years for the retirement fund $15,000 now is only equivalent to $165,480.46 35 years from now if the interest rate now is 7.1% instead of the quoted 5% 7-43 $2,000 = $91.05 (P/A, i*, 30) (P/A, i*, 30) = $2,000/$91.05 = 21.966 (P/A, i%, 30) i 22.396 20.930 2½ imo = 2% + (1/2%) [(22.396 – 21.966)/(22.396 – 20.930)] = 2.15% per month Nominal ROR received by finance company = 12 (2.15%) = 25.8% 7-44 $3,000 = $118.90 (P/A, i*, 36) (P/A, i*, 36) = $3,000/$118.90 = 26.771 (P/A, i%, 36) i 27.661 ẵ% 26.543 ắ% imo = ½% + ¼% [(27.661 – 26.771)/(27.661 – 26.543)] = 1.699% per month Nominal Annual ROR 7-45 (a) = 12 (1.699%) = 20.4% ... $1,813 < $1,845 Rate of Return = 7% + (1%) [($1,879 - $1,845)/($1,879 - $1,813)] = 7.52% for months Nominal annual rate of return = (7.52%) = 15.0% Equivalent annual rate of return = (1 + 0.0752)2... 10% $100 (4.378) + $300 (0.6209) = $624.07 < $640 Rate of Return = 9% + (1%) [(%646.07 - $640)/($646.07 - $624.07)] = 9.28% 7-13 Since the rate of return exceeds 60%, the tables are useless F =... rate = 12 (1 ¼%) = 15% 7-27 The rate of return exceeds 60% so the interest tables are not useful F $25,000 (1 + i) i* = P (1 + i)n = $5,000 (1 + i)3 = ($25,000/$5,000)1/3 = 1.71 = 0.71 Rate of