Solution manual engineering economic analysis 9th edition ch07 rate of return analysis

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... Ther

Chapter 7: Rate of Return Analysis7-1

$125 = $10 (P/A, i%, 6) + $10 (P/G, i%, 6)

$120

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) = ($1,000/0.15) (0.2472)

$1,648 = $1,648ROR = 15%

(a) Nominal Interest Rate = 1.201 x 12 = 14.41%

(b) Effective Interest Rate = (1 + 0.01201)12 – 1 = 0.154 = 15.4%

7-15

$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 = 1.25 x 12 = 15%

Effective Interest Rate = (1 + 0.0125)12 – 1 = 16.08%

(F/P, 5%, 365) = (131.5)3 (23.84) = 0.542 X 108 (i too low)Performing linear interpolation:

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%

$20 (21.355) + $1,000 (0.2526) - $715 = -$35.30 i too highPerforming linear interpolation:

$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)

seventy-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,

$9,000

A = $80

$15,000

$200 - $12.90 = $12,90 (P/A, i%, 70)

(P/A, i%, 70) = $187.10/$12.90 = 14.506% < i* < 8%, By Calculator: i* = 6.83%

Nominal rate of return = 2 (3.75%) = 7.5%

$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%

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 yearsBenefit: $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

(Calculator Solution: i = 642.9%)

The payment schedule represents a geometric gradient

There are two possibilities:

From interest tables, i* = 2%

(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

Set PW of Cost = PW of Benefits

$1,845 = $50 (P/A, i%, 4) + $2,242 (P/F, i%, 4)

Equivalent annual rate of return = (1 + 0.0752)2 – 1 = 15.6%

(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

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

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)

($2,000 - $150) = $100 (P/A, i%, 20)

(P/A, i%, 20) = $1,850/$100 = 18.5

i = ¾% 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

The fixed output of +$17 may be obtained at a cost of either $50 or $53 The additional $3 for Alternative B does not increase the benefits Therefore it is not a desirable increment of investment

Select B

(b) Annual Cash Flow Analysis- Maximize (EUAB- EUAC)

(EUAB- EAUC)A = $746 - $2,500 (A/P, 8%, 5)

= $746 - $2,500 (0.2505)

= +$120(EUAB – EUAC)B = $1,664 - $6,000 (A/P, 8%, 5)

= +$161Select B

(c) Rate of Return Analysis: Compute the rate of return on the B- A increment of investmentand compare to 8% MARR

Maintenace & Operating Costs $25,000 $10,000 -$15,000

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 = 11.65

Select A