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7048523558570072303145454025Rate of Return Analysis00Rate of Return Analysis-914400-916305007-1$100 = $27 (P/A, i%, 10)(P/A, i%, 10) = 3.704Performing linear interpolation:(P/A, i%, 10)i4.19220%3.57125%Rate of Return= 20% + (5%) [(4.192 ? 3.704)/(4.912 ? 3.571)] = 23.9%7-3$175 ……….n = 12A = $12.64$175 ……….n = 12A = $12.64($175 ? $35) = $12.64 (P/A, i%, 12)(P/A, i%, 12) = $140/$12.64 = 11.08i = 1.25%Nominal interest rate = 12 (1.25%) = 15%7-5………….A = $325n = 36$3,000 $12,375………….A = $325n = 36$3,000 $12,375$9,375 = $325 (P/A, i%, 36)(P/A, i%, 36) = $9,375/$325 = 28.846From compound interest tables, i = 1.25%Nominal Interest Rate = 1.25 × 12 = 15%Effective Interest Rate = (1 + 0.0125)12 ? 1 = 16.08%7-7(F/A, i, 35) = = 172.414 and is very close to 8% from tables. (Exact = 8.003%)7-9YearCash Flow0$–1,0003+$1,094.606+$1,094.60$1,000 = $1,094 [(P/F, i%, 6) + (P/F, i%, 9)]Try i = 20%$1,094 [(0.5787) + (0.3349)] = $1,000Rate of Return = 20%7-11$3,000 = $119.67 (P/A, i%, 30)(P/A, i%, 30) = $3,000/$119.67 = 25.069Performing linear interpolation:(P/A, i%, 30)i25.8081%24.8891.25%i = 1% + (0.25%)((25.808?25.069)/(25.808?24.889)) = 1.201%(a) Nominal Interest Rate = 1.201 × 12 = 14.41%(b) Effective Interest Rate = (1 + 0.01201)12 ? 1 = 0.154 = 15.4%7-16The algebraic sum of the cash flows equals zero. Therefore, the rate of return is 0%.7-18 $80 $80 $80 $80 $80 $80$200 $200 $200 $80 $80 $80 $80 $80 $80$200 $200 $200The easiest solution is to solve one cycle of the repeating diagram:= $80 $80 $80$200$120= $80 $80 $80$200$120$120 = $80 (F/P, i%, 1)$120 = $80 (1 + i)(1 + i) = $120/$80 = 1.50i* = 0.50 = 50%Alternative Solution: Convert the outflows of $200 to their annual equivalent and equate to the inflows $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) = $79.99Therefore i* = 50%7-20………….n = 10$412Yr 0 $5,000 P’A = $1,000n = ∞………….n = 10$412Yr 0 $5,000 P’A = $1,000n = ∞At Year 0, PW of Cost = PW of Benefits$412 + $5,000 (P/F, i%, 10) = ($1,000/i) (P/F, i%, 10)Try i = 15%$412 + $5,000 (0.2472) = ($1,000/0.15) (0.2472)$1,648 = $1,648ROR = 15%7-21The one-time $2,000 life membership fee avoids the 40-year series of beginning-of-year membership dues that start at $200 and increase 3% annually.(a) The equation for determining the rate of return for the life membership is the difference of the present worth of the two cash flows set to zero:2000 – 200 – 206 (P/A, 3%, ROR, 39) = 0 (39 since beginning-of-year payments)(b) Use Excel where Result = and vary i = ROR until zero is obtained. ROR = 14.243%7-24$715…….A = $40$1,000n = 40 semiannual periods$715…….A = $40$1,000n = 40 semiannual periodsPW of Benefits – PW of Costs = 0$20 (P/A, i%, 40) + $1,000 (P/F, i%, 40) ? $715 = 0Try i = 3%$20 (23.115) + $1,000 (0.3066) ? $715 = $53.90 i is too lowTry i = 3.5%$20 (21.355) + $1,000 (0.2526) ? $715 = ?$35.30 i is too highPerforming linear interpolation:i* = 3% + (0.5%) [53.90/(53.90 ? (?35.30))] = 3.30%Nominal i* = 6.60%7-30(a) For the cash flow of the bond have i = = 3.4%, so (0.034) (1,000) = $34 is paid semi-annually and $1,000 is paid at the end of the 10th year (20th pay period).NPW = 0 = +1,000 – 34 (P/A, i, 20) – 1,000 (P/F, i, 20) and interpolatingi = = 3.404% r = (2) (3.404%) = 6.808%, and ia = (1+0.03404)2 – 1 = 0.06924 or 6.924%.(b) The fee is $1,000(0.0075) = $7.50. So ABC Corp. receives $1,000 – $7.50 = $992.50.NPW = 0 = 992.5 – 34 (P/A, i, 20) – 1000 (P/F, i, 20) and interpolatingi = = 3.4546% r = (2) (3.4546%) = 6.909%, and ia = (1 + 0.034546)2 – 1 = 0.07029 or 7.029%.7-32$3,500 – $1,200 = $2,300 ……….n = 24A = $110$3,500 – $1,200 = $2,300 ……….n = 24A = $110$2,300 = $110 (P/A, i%, 24)(P/A, i%, 24) = $2,300/$110 = 20.91From tables: 1 % < i < 1.25%On Financial Calculator: i = 1.13% per monthEffective annual interest rate = (1 + 0.0113)12 ? 1 = 0.144 = 14.4%7-34P = $1,845n = 4A = $50F = $2,242P = $1,845n = 4A = $50F = $2,242Set 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 > $1,845Try i = 8%450 (3.312) + $2,242 (0.7350) = $1,813 < $1,845Rate of Return= 7% + (1%) [($1,879 ? $1,845)/($1,879 ? $1,813)]= 7.52% for 6 monthsNominal annual rate of return = 2 (7.52%) = 15.0%Equivalent annual rate of return = (1 + 0.0752)2 ? 1 = 15.6%7-36(a) The foregone cash rebate is like a hidden finance charge. You pay $12,000 for the car but receive a car only worth $12,000 - $3,000 = $9,000. The monthly payments = = $250 for 48 months. NPW = 0 = 9,000 – 250 (P/A, i, 48), so, (P/A, i, 48) = 36.0 and interpolatingi = 1% + (0.25%) = 1.242%, so, r = (12) (1.242%) = 14.90% and ia = (1 + 0.01242)12 – 1 = 0.15965 or 15.97%.(b) Worth of car = Cost – Rebate = $18,000 – $3,000 = $15,000. The monthly payments = = $375 for 48 months.NPW = 0 = 15,000 – 375 (P/A, i, 48), so, (P/A, i, 48) = 40.0 and interpolatingi = 0.75% + (0.25%) = 0.771%, so, r = (12)(0.771%) = 9.65% and ia = (1 + 0.00771)12 – 1 = 0.0965 or 9.65%.(c) Worth of car = Cost – Rebate = $24,000 – $3,000 = $21,000. The monthly payments = = $500 for 48 months.NPW = 0 = 21,000 – 500 (P/A, i, 48), so, (P/A, i, 48) = 42.0 and interpolatingi = 0.50% + (0.25%) = 0.561%, so, r = (12)(0.561%) = 6.73% and ia = (1 + 0.00561)12 – 1 = 0.0694 or 6.94%.7-37The amount of cash paid will be $75,000 – $50,000 = $25,000 with $50,000 financed, so, the monthly payments will be 50000 (A/P, 8%, 4) = (50000) (0.3019) = $15,095. The reduction in cost if one pays entirely in cash is $75,000(0.10) = $7,500, so a 100% cash payment would be $75,000 ? $7,500 = $67,500 (true value of equipment).YearPay CashBorrow from ManufacturerIncremental Difference0–$67,500?$25,000?$42,5001——?15,09515,0952——?15,09515,0953——?15,09515,0954——?15,09515,095IRR = IRR (the (1) – (2) values for the Periods 0–4) = 15.69% per year7-39$2,000 = $91.05 (P/A, i*, 30)(P/A, i*, 30) = $2,000/$91.05 = 21.966(P/A, i%, 30)i22.3962%20.9302?%imo = 2% + (?%) [(22.396 ? 21.966)/(22.396 ? 20.930)] = 2.15% per monthNominal ROR received by finance company = 12 (2.15%) = 25.8%7-41$9,000A = $80$15,000$9,000A = $80$15,000PW of Benefits – PW of Cost = $0$15,000 (P/F, i%, 4) ? $9,000 ? $80 (P/A, i%, 4) = $0Try i = 12%$15,000 (0.6355) ? $9,000 ? $80 (3.037) = +$289.54Try i = 15%$15,000 (0.5718) ? $9,000 ? $80 (2.855) = ?$651.40Performing linear interpolation:i* = 12% + (3%) [289.54/(289.54 + 651.40)] = 12.92%7-42(a) Total Annual Revenues = $1,100 (12 months) (4 apt.) = $52,800Annual Revenues – Expenses = $52,800 ? $4,000 = $48,800To find Internal Rate of Return the Net Present Worth must be $0.NPW = $48,800 (P/A, i*, 5) + $1,200,000 (P/F, i*, 5) ? $960,000At i = 9%,NPW = $8,176At i = 10%,NPW = ?$31,420IRR = 9% + [$8,176/($8,176 + $31,420)] = 9.2%(b) At 9.2%, the apartment building is more attractive than the other options.7-43P = $20,000A1 = $1,100n = 20i = ?g = 10%P = $20,000A1 = $1,100n = 20i = ?g = 10%The payment schedule represents a geometric gradient.There are two possibilities: i ≠ g and i = gTry the easier i = g computation first:P = A1n (1 + i)?1 where g = i = 0.10$20,000 = $1,100 (20) (1.10)?1 = $20,000Rate of Return i* = g = 10%7-45$9,000$800$400$6,000$9,000$800$400$6,000YearCash Flow0$–9,0001–4+$8005–8+$4009+$6,000PW 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 < $9,000Try i = 2?%$400 (7.170) + $400 (3.762) + $6,000 (0.8007) = $9,177 > $9,000Rate of Return = 2?% + (1/2%) [($9,177 ? $9,000)/($9,177 ? $8,893)] = 2.81%7-47YearCase 1 (incl. deposit)0?$39,264.001+$599.002+$599.003+$599.004+$599.005+$599.006+$599.007+$599.008+$599.009+$599.0010+$599.0011+$599.0012+$599.00. . .+$599.0033+$599.0034+$599.0035+$599.0036+$27,854.00 ?$625.00 = +$27,229.00IRR = 0.86%Nominal IRR = 10.32%Effective IRR =10.83%7-50–$65,000 = –$18,000( 1 + (P/A, i, 3))The amount that the series of future payments is worth is:–65,000 + 18,000 = –47,000 = –18,000 (P/A, i , 3)Using the end-of-period designation (default) in RATE (Excel) yields:RATE(3, 18,000, –47,000) = 7.2766%One could also solve with quarterly payments at the beginning of the period:RATE(4, 18,000, –65,000, 0, 1) = 7.2766%7-54YearX Y(X – Y)0?$100?$50?$501+$35+$16.5+$18.52+$35+$16.5+$18.53+$35+$16.5+$18.54+$35+$16.5+$18.5Computed ROR15.0%12.1%17.8%The ΔROR on X – Y is greater than 10%. Therefore, the increment is desirable. Select X.7-55YearA B(B – A)0?$2,000?$2,800?$8001–10+$800+$1,100+$300Computed ROR9.7%8.7%6.1%ΔROR = 9.61% > MARR. Select A.7-56YearX Y(X – Y)0?$5,000?$5,000$01?$3,000+$2,000?$5,0002+$4,000+$2,000+$2,0003+$4,000+$2,000+$2,0004+$4,000+$2,000+$2,000Computed ROR16.9%21.9%9.7%Since X – Y difference between alternatives is desirable, select Alternative X.7-58Year0 12345678Alt. A-120001200Alt. B-3000-3000-3000-3000-3000-3000-3000-3000A-B-900030003000300030003000300030001200IRR of A ? B stream = IRR (the A – B values for the Years 0–8) = 27.90%Since ΔROR > MARR (15%), choose the higher initial cost alternative, A (purchasing the equipment).7-63(a)$2,000 ……….n = 20A = $100$150$2,000 ……….n = 20A = $100$150($2,000 ? $150) = $100 (P/A, i%, 20)(P/A, i%, 20) = $1,850/$100 = 18.5i = ?% per monthThe alternatives are equivalent at a nominal 9% annual interest.(b) Take Alt 1—the $2,000—and invest the money at a higher interest rate.7-64(a) Salvage = $50,000 and community’s interest rate = 8%.YearPurchaseLeasePurchase – Lease0–$480,000–$70,000–$410,00010–70,00070,00020–70,00070,00030–70,00070,00040–70,00070,00050–70,00070,00060–70,00070,00070–70,00070,00080–70,00070,00090–70,00070,0001050,000050,000NPW = 0 = ?410,000 +70,000 (P/A, IRR , 9) + 50,000 (P/A, IRR, 10) and interpolating IRR = 10% + (2%) = 10.74% (10.71% Excel). The IRR is above the community’s interest rate on the borrowed amount ($410,000) from leasing, so buy the generator.(b) The community spends $80,000 less on fuel and maintenance than it spends on buying power.YearPurchaseLeasePurchase – Lease0–$480,000–$70,000–$410,000180,000–70,000150,000280,000–70,000150,000380,000–70,000150,000480,000–70,000150,000580,000–70,000150,000680,000–70,000150,000780,000–70,000150,000880,000–70,000150,000980,000–70,000150,0001080,00050,0000130,000NPW = 0 = ?410,000 + 150,000 (P/A, IRR, 9) + 130,000 (P/F, IRR, 10) and interpolating IRR = 30% + (5%) = 34.66% (34.63% from Excel). The interest rate on the borrowed amount is now well above the firm’s interest rate, so buy the generator. The rate of return for the generator will clearly be largest for this cash flow and is given byPW = 0 = ?480,000 + 80,000 (P/A, ROR, 10) + 50,000 (P/F, ROR, 10) and interpolating ROR = 10% + (2%) = 11.44% (11.42% from Excel).7-66YearZappoKickoKicko – Zappo0?$56?$90?$341?$56$0+$562$0$0$0Compute the incremental rate of return on (Kicko ? Zappo)PW of Cost = PW of Benefit$34 = $56 (P/F, i%, 1)(P/F, i%, 1) = $34/$56 = 0.6071From interest tables, incremental rate of return > 60% (ΔROR = 64.7%), hence the increment of investment is desirable.Buy Kicko.7-68YearABA – B0?$150?$100?$501–10+$25+$22.25+$2.7511–15+$25$0+$2515+$20$0+$20Computed ROR14.8%18%11.6%Rate of Return (A – B):$50 = $2.75 (P/A, i%, 10) + $25 (P/A, i%, 5) (P/F, i%, 10) + $20 (P/F, i%, 15)Rate of Return = 11.65Select A.7-70$55,000income2.00%income gradient10.00%% deposit40horizon (years)5.00%savings rateYearSalaryDepositCumulative Savings1$55,000.00$5,500.00$5,500.00256,100.005,610.0011,385.00357,222.005,722.2017,676.45458,366.445,836.6424,396.92559,533.775,953.3831,570.14660,724.446,072.4439,221.09761,938.936,193.8947,376.04863,177.716,317.7756,062.61964,441.276,444.1365,309.871065,730.096,573.0175,148.371167,044.696,704.4785,610.261268,385.596,838.5696,729.331369,753.306,975.33108,541.131471,148.367,114.84121,083.021572,571.337,257.13134,394.301674,022.767,402.28148,516.301775,503.217,550.32163,492.431877,013.287,701.33179,368.381978,553.547,855.35196,192.152080,124.618,012.46214,014.222181,727.118,172.71232,887.652283,361.658,336.16252,868.192385,028.888,502.89274,014.492486,729.468,672.95296,388.162588,464.058,846.40320,053.972690,233.339,023.33345,080.012792,038.009,203.80371,537.812893,878.769,387.88399,502.572995,756.339,575.63429,053.333097,671.469,767.15460,273.153199,624.899,962.49493,249.2932101,617.3810,161.74528,073.4933103,649.7310,364.97564,842.1434105,722.7310,572.27603,656.5235107,837.1810,783.72644,623.0736109,993.9310,999.39687,853.6137112,193.8011,219.38733,465.6738114,437.6811,443.77781,582.7239116,726.4311,672.64832,334.5040119,060.9611,906.10885,857.33For any row: Salary = (1 + 0.02)(Previous year’s Salary)Deposit = (Percent Deposit)(Current year’s Salary)Savings = (1 + 0.05)(Previous year’s Savings) + Current year’s DepositAmount saved is $885,857.33 in 40 years.7-75Using Equivalent Uniform Annual Cost-EUACTh = ?$5 ? $20 (A/P, 12%, 3) = ?$5 ? $20 (0.4163) = ?$13.33EUACSL = ?$2 ? $40 (A/P, 12%, 5) = ?$2 ? $40 (0.2774) = ?$13.10You should choose slate over thatch to save $0.23/yr.To find incremental ROR, find i such that EUACSL ? EUACTH = 0.$0= ?$2 ? $40 (A/P, i*, 5) ? [?$5 ? $20 (A/P, i*, 3)]= $3 ? $40 (A/P, i*, 5) + $20 (A/P, i*, 3)At i = 12%$3 ? $40 (0.2774) + $20 (0.4163) = $0.23 > $0 12% too lowAt i = 15%$3 ? $40 (0.2983) + $20 (0.4380) = ?$0.172 < $0 15% too highUsing linear interpolation:ΔROR= 12 + 3[0.23/(0.23 ? (?0.172))] = 13.72%7-78YearGen. Dev.RJRRJR – Gen Dev.0?$480?$630?$1501?15+$94+$140+$4615+$1,000+$1,000$0Computed ROR21.0%22.8%30.0%Neither bond yields the desired 25% MARR? so do nothing.Note that simply examining the (RJR ? Gen Dev) increment might lead one to the wrong conclusion.7-82Incremental Rate of Return SolutionABCDC – DB – CA – CCost$1,000$800$600$500$100$200$400Uniform Annual Benefit$122$120$97$122?$25$23$25Salvage Value$750$500$500$0$500$0$250Compute Incremental Rate of Return10%< 0%1.8%The C?D increment is desirable. Reject D and retain C.The B?C increment is undesirable. Reject B and retain C.The A?C increment is undesirable. Reject A and retain C.Conclusion: Select alternative Present Worth Solution:Net Present Worth = Uniform Annual Benefit (P/A, 8%, 8) + Salvage Value (P/F, 8%, 8) – First CostNPWA= $122 (5.747) + $750 (0.5403) ? $1,000= +$106.36NPWB= $120 (5.747) + $500 (0.5403) ? $800 = +$159.79NPWC= $97 (5.747) + $500 (0.5403) ? $600 = +$227.61NPWD= $122 (5.747) ? $500 = +$201.137-85Monthly payment on new warehouse loan= $350,000 (A/P, 1.25%, 60)= $8,330MonthAlt. 1Alt. 2Alt. 1 – Alt. 2Alt. 3Alt. 1 – Alt. 30?$100,000?$100,000$0$0?$100,0001–60?$8,330+$2,500?$1,000?$8,300$0 $0$0 $0+$1,500?$2,700?$4,13060+$600,000+$600,000$0$0+$600,000DecisionBy inspection, this increment is desirable. Reject 2. Keep 1.Δ ROR = 1.34%/mo Nominal ROR = (1.34%)12 = 16.1% Effective ROR = (1 + 0.0134)12 ? 1 = 17.3%Being less desirable than Alternative 1, Alternative 2 may be rejected. The Alt. 1–Alt. 3 increment does not yield the required 20% MARR, so it is not desirable. Reject 1 and select 3 (continue as is).7-86Since there are alternatives with ROR greater than our 8% MARR, Alternative 3 may be immediately rejected as well as Alternative 5. Note also that Alternative 2 dominates Alternative 1 since its ROR > ROR Alt. 1. Thus ΔROR2–1 > 15%. So Alternative 1 can be rejected. This leaves Alternatives 2 and 4. Examine (4?2) increment.244 – 2Initial Investment$130.00$330.00$200.00Uniform Annual Benefit$38.78$91.55$52.77$200 = $52.77 (P/A, i%, 5)(P/A, i%, 5) = $200/$52.77 = 3.79ΔROR4–2 = 10%Since ΔROR4–2 > 8% MARR, select Alternative 4.7-88YearABB – ACC – ADD – CPW (D–C) at 8%0–1300–13000-13000-1300001+100+10–90+260+160+450+190+175.92+130+60–70+260+130+400+140+120.03+160+110–50+260+100+350+90+71.44+190+160–30+260+70+300+40+29.45+220+210–10+260+40+250–10–6.86+250+260+10+260+10+200–60–37.87+280+310+30+260–20+150–110–64.28+310+360+50+260–50+100–160–86.49+340+410+70+260–80+50–210–105.010+370+460+90+260–1100–260–120.40*+250**–350–23.9****By inspection, incremental rate of return = 0%. Reject B.**No positive value of i in this borrowing increment. The C?A increment is highly desirable. Reject A.***This borrowing increment has an i greater than 8%. This is not a desirable increment. Reject D.Decision: Choose Alternative C.7-89This problem is one of neither fixed input nor fixed output. When the estimated resale value equals the present total investment we have the special case where: A = Pi or i = A/PAlt.P*A*iΔPΔAΔiDecision?Sell Parking Lot000%Keep Parking Lot$200$2211.0%$200$2211.0% Keep Lot1 Storey Building$400$6015.0%$200$3819.0%1 storey2 Storey Building$555$7212.9%$155$127.7%1 storey3 Storey Building$750$10013.3%$350$4011.4%3 storey4 Storey Building$875$10512.0%$125$54.0%3 storey5 Storey Building$1,000$12012.0%$250$208.0%3 storey*All values in thousands.Conclusion: Build 3-storey building.7-91Lease: Pay $267 per month for 24 months.Purchase: A = $9,400 (A/P, 1%, 24) = $9,400 (0.0471) = $442.74 Salvage (resale) value = $4,700(a) Purchase Rather than LeaseΔMonthly payment = $442.71 ? $267= $175.74ΔSalvage value = $4,700 ? $0 = $4,700Δ Rate of ReturnPW of Cost = PW of Benefit$175.74 (P/A, i%, 24) = $4,700(P/A, i%, 24) = $4,700/$175.74 = 26.74i = 0.93% per monthThus, the additional monthly payment of $175.74 would yield an 11.2% rate of return. Leasing is therefore preferred at all interest rates above 11.2%.(b) Items that might make leasing more desirable:1. One does not have, or does not want to spend, the additional $175.74 per month.2. One can make more than 11.2% rate of return in other investment.3. One does not have to be concerned about the resale value of the car at the end of two years.From which we deduce that MIRR = 9.26%7-117YearA (Gas Station)B (Ice Cream Stand)(B – A)0?$80,000?$120,000?$40,0001?20+$8,000+$11,000+$3,000Computed ROR7.75%6.63%4.22%0?$80,000?$120,000?$40,000The rate of return in the incremental investment (B – A) is less than the desired 6%. In this situation the lower cost alternative (A) Gas Station should be selected. ................
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