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7-2 (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 do not necessarily reflect what anyone will
The rate of return exceeds 60% so the interest tables are not useful. F = P (1 + i)n $25,000 = $5,000 (1 + i)3 (1 + i) = ($25,000/$5,000)1/3 = 1.71 i* = 0.71 Rate of Return = 71%
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%
7-20 Do nothing has a cash flow of zero, thus, the difference between alternatives is just
paid semi-annually and $1,000 is paid at the end of the 10th year (20th pay period).
NPW = 0 = +1000 – 34 (P/A, i, 20) – 1000 (P/F, i, 20) and interpolating
i = 59.518
3% (0.5%)59.518 14.192
= 3.404% (exact value = 3.400%),
r = (2) (3.404%) = 6.808%, and ia = (1+0.03404)2 – 1 = 0.06924 or 6.924%. (b) The fee is $1,000 x 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 interpolating
i = 67.018
3% (0.5%)67.018 6.692
= 3.4546% (exact value = 3.453%),
r = (2) (3.4546%) = 6.909%, and ia = (1 + 0.034546)2 – 1 = 0.07029 or 7.029%.
7-32 (a) NPW = 0 = -3118 + 10000 (P/F, i, 20), so, (P/F, i, 20) = 0.3118. Next you can
solve 20(1 ) 0.3118i for i or look in the tables to find i = 0.06 or 6.0%. Next, because it is paid annually, the effective annual interest rate is 6.0%. (b) The fee is $10,000 x 0.01 = $100. So ABC Corp. receives $3,118 - $100 =
$3,018. NPW = 0 = 3018 – 10000 (P/F, i, 20), so, (P/F, i, 20) = 0.3018. Next solve
20(1 ) 0.3018i and find i = 0.06173 or 6.173%. As above ia = 6.173%.
7-33
$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 − 1 = 0.144 = 14.4%
NPW = 0 = 6000 -250 – 166.67 (P/A, i, 36), so, (P/A, i, 36) = 34.50 . The tables don’t go to a low enough interest rate so must solve:
36
36
(1 ) 134.50
(1 )
i
i i
by trial and error or Excel using the IRR function. Excel
yields i = 0.00232, so, ia = (1 + 0.00232)12 – 1 = 0.0282 or 2.82%. (b) The fact that the dealer would accept $5,200 cash for the car indicates its true
worth so the extra $800 is a hidden finance charge. Your payments are still based on the original $6,000 cost but you only receive a car worth only $5,200!
NPW = 0 = 5200 -250 – 166.67 (P/A, i, 36), so, (P/A, i, 36) = 29.70 and
interpolating
i = 1% + (0.25%) 30.107 29.70
30.107 28.847
= 1.081% (exact value = 1.079%), so,
ia = (1 + 0.01081)12 – 1 = 13.77% (exact value = 13.75%).
(b) Worth of the car = $6,000 − $800 = $5,200 but the payments are determined by the actual cost to buyer, here $6,000. Thus, the payments are the same as above.
NPW = 0 = 5200 – 250 – 177.14 (P/A, i, 36), so, (P/A, i, 36) = 27.944 and
interpolating i = 1.25% + (0.25%) 28.847 27.944
28.847 27.661
= 1.440%, so,
r = (12) (1.440%) = 17.28% and ia = (1 + 0.01440)12 −1 = 0.1872 or 18.72%. (c) The actual value of the car seems to be the most important factor!
7-39 The 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 x 0.10 = $7,500, so, a 100% cash payment would be $75,000 − $7,500 = $67,500 (true value of equipment).
(b) The actual value received is $108,000, thus, to find the effective interest rate solve NPW = 0 = 108,000 – 660.46 (P/A, i, 360).
(P/A, i, 360) = 108000
660.46 = 163.522. Interpolating
imo = ½ % + (¼ %)[(163.522 – 166.792)/(124.282 – 166.792] = 0.51923% per month ia = (1 + 0.0051923)12 – 1 = 0.0641 or 6.41% (c) In ten years there are still 20 years left on the original loan, so, value of remaining loan at year ten = 660.46 (P/A, 0.5%, 240) = (660.46)(139.581) = $92,187.67 . To find the effective interest rate solve NPW = 108,000 – 660.46 (P/A, i, 120) – 92,187.67 (P/F, i, 120) . Interpolating imo = ½ % + (¼ %)[2156.62/(2156.62 + 18258.62] = 0.5264% (exact value 0.5236%) ia = (1 + 0.005264)12 – 1 = 0.0650 or 6.50% (exact value 6.467%)
7-45 (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%, NPW = $8,464 At i = 15%, NPW = −$6,816 IRR = 12% + (3%) [$8,464/($8,464 + $6,816)] = 13.7% (b) At 13.7% the apartment building is more attractive than the other options.
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%
7-48 (a) When n = ∞, i = A/P = $3,180/$100,000 = 3.18% (b) (A/P, i%, 100) = $3180/$100,000 = 0.318 From interest tables, i* = 3% (c) (A/P, i%, 50) = $3, 180/$100,000 = 0.318 From interest tables, i* = 2% 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.
7-51 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 2 years Benefit: $3,000 per year for 10 years 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%
Solving get i = 0.1318 or 13.18% and ia = (1 + 0.1318)12 – 1 = 3.418 or 342%. Unless the student is graduating in January or just doesn’t have the $100, it is clearly better to buy the permit a year at a time.
7-55 Details will vary by university, but is solved like Problem 7-54.
To solve for the monthly interest rate set the two PWs equal to each other, so, –65000 = –18000 – 18000 (P/A, i, 3) . Thus, (P/A, i, 3) = 2.611 and interpolating
i = 7% + (1%) 2.624 2.611
2.624 2.577
= 7.28%, so, r = 4 x 0.0728 = 0.2912 or 29.1% and
ia = (1 + 0.0728)4 – 1 = 0.3246 or 32.5%. This is a high rate of return, but some firms
7-57 –$65,000 = –$18,000( 1 + (P/A, i , 3)) The amount that the series of future payments is worth is: –65000 + 18000 = –47000 = –18000*(P/A, i , 3) Using the end-of-period designation (default) in RATE (Excel) yields: RATE(3,18000,-47000) = 7.2766% One could also solve with quarterly payments at the beginning of the period: RATE(4,18000,-65000,0,1) = 7.2766%
7-58 Insurance payments must be paid in advance, here on the first of the month or year.
To solve for the monthly interest rate set the PWs of the two cash flows equal to
each other. Thus, −1650 = −150 – 150 (P/A, i, 11), so, (P/A, i, 11) = 10.0. Interpolating
i = 1.5% + (0.25%) 10.071 10.0
10.071 9.928
= 1.624%. Next, ia = (1 + 0.01624)12 −1 = 0.2133
or 21.3%. This is a relatively high rate of return, but the student might prefer to pay monthly if there is a significant chance of wrecking the car before the year is up.
7-59 Details will vary by student, but solved like Problem 7-58.
IRR 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).
7-66
B A A- B First Cost $300,000 $615,000 $315,000 Maintenance & Operating Costs
$25,000 $10,000 −$15,000
Annual Benefit $92,000 $158,000 $66,000 Salvage Value −$5,000 $65,000 $70,000
NPW = −$315,000 + [$66,000 − (−$15,000)] (P/A, i*, 10) + $70,000 (P/F, i*, 10) = $0 Try i = 15% −$315,000 + [$66,000 − (−$15,000)] (5.019) + $70,000 (0.2472) = $108,840 ΔROR > MARR (15%) The higher cost alternative A is the more desirable alternative.
The 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.
7-68 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 − 1 = 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%.
($2,000 − $150) = $100 (P/A, i%, 20) (P/A, i%, 20) = $1,850/$100 = 18.5 I = ¾% per month The 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.
7-70 (a) Salvage = 0.15 x $380,000 = $57,000 and firm’s interest rate = 12%.
situation has changed. The interest rate on the borrowed amount is now well above the firm’s interest rate, so, buy the bulldozer. The rate of return for the bulldozer will clearly be largest for this cash flow and is given by
Note that the author has failed to give a practical scenario for how the $65,000 benefit can be realized if the bulldozer is purchased instead of leased!
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 by
Note that the author has failed to give a practical scenario for how the $80,000 benefit can be realized if the generator is purchased instead of leased!
7-72
Year A B A- B NPW at 7% NPW at 9% 0 −$9,200 −$5,000 −$4,200 −$4,200 −$4,200 1 +$1,850 +$1,750 +$100 +$93 +$92 2 +$1,850 +$1,750 +$100 +$87 +$84 3 +$1,850 +$1,750 +$100 +$82 +$77 4 +$1,850 +$1,750
7-76 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 +
For 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 deposit To solve, just vary the percent deposit to get $1M in savings for year 40. Amount saved is $1,000,132.92 in 40 years at 11.29%.
7-80 Details will vary by student, but solved like Problem 7-79.