Solution manual engineering economic analysis 9th edition ch09 other analysis techniques

Alternative A may be considered if the investor is very short of cash and the short payback period is of importance to him... The Present Worth method requires common analysis period, wh

Chapter 9: Other Analysis Techniques9-1

i = 10%

i = 15%

A = $20,000Retirement

21 - xx

$48,500

21 years

Amount at Retirement = PW of needed retirement funds

F7 = $72,400 (F/A, 7%, 7)

= $626,550Compute the future worth of $626,550 in 3 years i.e at the end of year 10:

F10 = $626,550 (F/P, 7%, 3)

= $767,524Remaining two deposits = ($1,000,000 - $767,524) (A/F, 7%, 2)

- $7,500 = $417,940

(1) compute an equivalent twice a month savings account deposit, or(2) compute an equivalent monthly interest rate

Method (1)

Equivalent twice a month deposit (A) = $75 (A/F, i%, n)

= $75 [0.001875/((1 + 0.001875)2 – 1)]

= $37.4649Future Sum F1/1/15 = A (F/A, i%, 18 x 24)

= $37.4649 [((1 + 0.001875)432 – 1)/0.001875]

= $24,901

Method 2

Effective i per month (imonth) = (1 + 0.045/24)2 – 1 = 0.0037535

Future Sum F1/1/15 = A (F/A, imonth, 18 x 12)

Joe’s Plan

F = $40,000 (F/P, 3.5%, 31) = $40,000 (2.905) = $116,200Joe’s deposit will be insufficient He should deposit:

Required Value of GE stock = ($421,325 - $371,070)/600

= 9.818A

B/C Ratio = PW of Benefit/PW of

Incremental B/C Ratio Analysis

5 Stories Rather than 2 Stories

10 Stories Rather than 2 Stories

Δ PW of Cost $835,650 - $457,100

= $378,550

$2,114,200 - $457,100

= $1,657,100ΔPW of Benefit $1,030,890 - $687,260

= $343,630

$2,513,410 - $687,260

= $1,826,150ΔB/ΔC = ΔPW of

= 1.28

Rate of Return Solution

Compute B/C for each alternative

Form of computation used:

(PW of B)/(PW of C) = (UAB (P/A, 8%, 8))/(Cost – S (P/F, 8%, 8))

= (UAB (5.747))/(Cost – S (0.5403))B/CA = ($12.2 (5.747))/($100 - $75 (0.5403)) = 1.18

By inspection one can see that A, with its smaller cost and identical benefits, is preferred to

F in all situations, hence F may be immediately rejected Similarly, D, with greater benefits and identical cost, is preferred over B Hence B may be rejected Based on the B/C ratio for the remaining four alternatives, three exceed 1.0 and only C is less than 1.0 On this basis C may be rejected That leaves A, D, and E for incremental B/C analysis

Investment= $67,000

Annual Benefit = $26,000/yr for 2 years

Payback period = $67,000/$26,000q= 2.6 years

Do not buy because total benefits (2) ($26,000) < Cost

9-38

Payback Period = Cost/Annual Benefit = $3,800/(4*$400) = 2.4 years

$3,800= $400 (P/A, i%, 4) + $400 (P/A, i%, 4) (P/F, i%, 12)

+ $400 (P/A, i%, 4) (P/F, i%, 24) + $400 (P/A, i%, 4) (P/F, i%, 36)+ $400 (P/A, i%, 4) (P/F, i%, 48)

$3,800= $400 (P/A, i%, 4) [1 + (P/F, i%, 12) + (P/F, i%, 24) + (P/F, i%, 36)

2 more years

n = 60 months

= $3,798

So i 3.5% per month≈

Nominal Rate of Return = 12 (3.5%) = 42%

9-39

Costs = Benefits at end of year 8

Therefore, payback period = 8 years

PaybackB = 7 years (based on end of year cash flows)

(b) Equivalent Investment Cost

= $30 (F/P, 10%, 2) + $100 (F/P, 10%, 1) + $70

= $30 (1.210) + $100 (1.100) + $70

= $216.3 million

(c) Equivalent Uniform Annual Worth = EUAB – EUAC

EUAWA = $40 - $216.3 (A/P, 10%, 10) = $4.81 million

EUAWB = $32.5 - $216.3 (A/P, 10%, 20) = $7.08 million

Since the EUAW for the Alternative B is higher, this alternative should be selected Alternative A may be considered if the investor is very short of cash and the short payback period is of importance to him

(b) The key to solving this part of the problem is selecting a suitable analysis method

The Present Worth method requires common analysis period, which is virtually impossible for this problem The problem is easy to solve by Annual Cash Flow Analysis

EUACconventional- 20 yrs= $200 (A/P, 10%, 20) + $230 = $253.50

EUACsolar- n yrs = $1,400 (A/P, 10%, n) + $60

For equal EUAC:

To minimize Payback, select C

(a) Solve by Future Worth analysis In future worth analysis there must be a common

future time for all calculations In this case 12 years hence is a practical future time

NFWC = $39.6 (F/A, 12%, 12) - $110 (F/P, 12%, 4) - $110 (F/P, 12%, 8) –

$110 (F/P, 12%, 12)

= $39.6 (24.133) - $110 [1.574 + 2.476 + 3.896]

= +$81.61

Choose Alternative C because it maximizes Future Worth

(b) Solve by Benefit-Cost ratio analysis

With neither input nor output fixed, incremental analysis is required

Four years is a suitable analysis period for Alternatives C and A

For the increment C- A:

Twelve years is a suitable analysis period for Alternatives B and C

For the increment B- C

Ignoring the potential difficulties signaled by 3 sign changes in the B- C cash flow:

The increment is undesirable and therefore Alternative C is preferred over Alternative B

Alternative Analysis of the Increment B- C

An examination of the B- C cash flow suggests there is an external investment of money

at the end of Year 4 Using an external interest rate (say 12%) the +$110 at Year 4 becomes:

+$110 (F/P, 12%, 2) = $110 (1.254) = $137.94 at the end of Yr 6

The altered cash flow becomes:

To minimize the Payback Period, choose Alternative A

(d) Payback period is the time required to recover the investment Here we have

three alternatives that have rates of return varying from 10% to 16.4% Thus each generates uniform annual benefits in excess of the cost, during the life of the alternative From this is must follow that the alternative with a 2-year life has

a payback period less than 2 years The alternative with a 4-year life has a payback period less than 4 years, and the alternative with a 6-year life has a payback period less than 6 years

Thus we see that the shorter-lived asset automatically has an advantage over longer-lived alternatives in a situation like this While Alternative A takes the shortest amount of time to recover its investment, Alternative C is best for long-term economic efficiency

To minimize payback, select z

(c) No computations are needed The problem may be solved by inspection

Alternative x has a 0% rate of return (Total benefits = cost)

Alternative z dominates Alternative y (Both cost $50, but Alternative z yields more benefits)

Alternative z has a positive rate of return (actually 24.5%) and is obviously the best of the three mutually exclusive alternatives

Choose Alternative z

9-47

(a) Payback Period

Payback A = 4 + $150/$350 = Year 4.4

PaybackB = Year 4

PaybackC = 5 + $100/$200 Year 5.5

For shortest payback, choose Alternative B

(b) Net Future Worth

NFWA = $200 (F/A, 12%, 5) + [$50 (P/F, 12%, 5) - $400] (F/P, 12%, 5)

- $500 (F/P, 12%, 6)

= $200 (6.353) + [$50 (6.397) - $400] (1.762) - $500 (1.974)

= +$142.38NFWB = $350 (F/A, 12%, 5) + [-$50 (P/G, 12%, 5) - $300] (F/P, 12%, 5)

- $600 (F/P, 12%, 6)

= $350 (6.353) + [-$50 (6.397) - $300] (1.762) - $600 (1.974)

= -$53.03NFWC = $200 (F/A, 12%, 5) - $900 (F/P, 12%, 6)

= 1.36B/CC = ($100 (P/A, 10%, 4) + $100 (P/F, 10%, 1)/$300

∆B/∆CB-C = ($25 (P/A, 10%, 3)(P/F, 10%, 1) + $125 (P/F, 10%, 5))/$100

= 1.34This is a desirable increment Reject C

(b) For a 12% rate of return, the useful life must be 5.25 years.

(c) When n = ∞, rate of return = 26.7%

9-52

(EUAB – EUAC)A = $230 - $800 (A/P, 12%, 5) = +$8.08

Set (EUAB – EUAC)B = +$8.08 and solve for x

(EUAB – EUAC)B = $230 - $1,000 (A/P, 12%, x)= +$8.08

(A/P, 12%, x) = [$230 - $8.08]/$1,000 = 0.2219

A = $12

n = ?

P = $45

From the 12% compound interest table, x = 6.9 yrs

9-53

NPWA = $40 (P/A, 12%, 6) + $100 (P/F, 12%, 6) - $150 = +$65.10Set NPWB = NPWA

Alternative 1: Buy May 31 st

Alternative 2: Buy just before trip

Difference between alternatives

$1.71 = $1.38 + 0.1315 (treatment)Treatment = ($1.71 - $1.38)/0.1315 = $2.51

So, up to $2.51 could be paid for post treatment

The difference between the alternatives is that Plan A requires $20,000 extra now and Plan

B requires $40,000 extra n years hence

At breakeven:

$20,000 = $40,000 (P/F, 8%, n)

(P/F, 8%, n) = 0.5

From the 8% interest table, n = 9 years

(b) At 0% interest, from (a):

(P/A, 0%, n) = 5.33

Since (P/A, 0%, n) = n, the machines are equivalent at 5 1/3 years

9-63

(a) Payback Period

At first glance, payback would appear to be

$5,240/$1,000 = 5.24 yearsHowever, based on end-of-year benefits, as specified in the problem, the correct answer is:

Payback = 6 years

(b) Breakeven Point (in years)

Here interest is used in the computations For continuous compounding: