Solution manual engineering economic analysis 9th edition ch06

Chapter 6: Annual Cash Flow Analysis6-1... While the projectdoes not meet the 8% interest rate criterion, it would be economically justified at a 4% interest rate.. Therefore the monthly

Chapter 6: Annual Cash Flow Analysis6-1

$200

$100

$200

$300

$200

APattern repeats

infinitely

$60,000 (A/P, 10%, 4) - $60,000 (A/F, 10%, 2)

This equals P*i = $60,000 (0.10) = $6,000

6-11

Prospective Cash Flow:

$30,000 (0.2229) = A + $35,000 (0.0729)

$6,687 = A + $2,551.50

6-12

This problem is much harder than it looks!

EUAC = {$600 (P/A, 8%, 5) + $100 (P/G, 8%, 5) + [$900 (P/A, 8%, 5) –

$60

0

$700

$800

$900

$30,000 = PW of all future scholarships

= $2,132.80 per semiannual period

Now, compute the equivalent uniform annual cost:

F = $1,000,000

= $25.13

6-18

(a) EUAC = $6,000 (A/P, 8%, 30) + $3,000 (labor) + $200 (material)

- 500 bales ($2.30/bale) – 12 ($200/mo trucker)

= $182.80Therefore, bailer is not economical

(b) The need to recycle materials is an important intangible consideration While the projectdoes not meet the 8% interest rate criterion, it would be economically justified at a 4% interest rate The bailer probably should be installed

$3,500

P

1 2 3 4 5 6 7 8 9

10

existing pipeline, it should be constructed.

Therefore the monthly deposit is $69.02.

Amount to deposit September 1:

= Future worth of 5 months deposits (May – Sep)

2 Quarterly interest payments to the savings account could

have an impact on the solution, but they do not in this

problem

3 The solution may be verified by computing the amount in thesavings account on Dec 1 just before making the payment (about $560.03) and the amount on April 1 after making thatpayment ($0)

Equivalent Uniform Monthly Cost = ($13,000 - $4,000) (A/P, 1%, 36) +

$4,000 (0.01)

= $338.80

(c) Lease with repurchase option= $360.00 - $500 (A/F, 1%, 36)

= $348.40Alternative (a) has the least equivalent monthly cost, but non-monetary considerations mightaffect the decision

Balance due at end of 10 years:

= $8,816 (P/A, 10%, 10) = $54,170making the unpaid loan at Year 0:

At year 10 this becomes:

= $25,830 (F/P, 10%, 10) = $67,000Note: The difference is due to four place accuracy in the compound interest tables The

A diagram is essential to properly see the timing of the 11 deposits:

These are beginning of period deposits, so the compound interest factors must be adjusted for this situation

-1 0 1 2 3 4 5 6 7 8 9 10

11

$500,000P

Around the Lake Under the Lake

Around the Lake

EUAC = $75,000 (A/P, 7%, 15) + $12,000 - $45,000 (A/F, 7%, 15)

= $75,000 (0.1098) + $12,000 - $45,000 (0.0398)

= $18,444

Under the Lake

EUAC = $125,000 (A/P, 7%, 15) + $7,000 - $25,000 (A/F, 7%, 15)

= $125,000 (0.1098) + $7,000 - $25,000 (0.0398)

= $19,730

Go around the lake

A’ A’ A’ A’ A’ A’ A’ A’

n = 8 quarterly periods

i = 1 ½% per quarter

A’ = $9,000 (A/F, 1.5%, 8)

= $9,000 (0.1186)

= $1,067.40

(c) In part (a) Bill Anderson’s monthly payment includes an interest payment on the loan The sum of his 24 monthly payments will exceed $9,000

In part (b) Doug James’ savings account monthly deposit earns interest for him that helps to accumulate the $9,000 The sum of Doug’s 24 monthly deposits will be less than $9,000

With neither input nor output fixed, maximize (EUAB – EUAC)

Continuous compounding capital recovery:

6-38

Annual Cost of Diesel Fuel = [$50,000km/(35 km/l)] x $0.48/l = $685.71Annual Cost of Gasoline = [$50,000km/(28 km/l)] x $0.51/l = $910.71

EUACdiesel = ($13,000 - $2,000) (A/P, 6%, 4) + $2,000 (0.06)

+ $685.71 fuel + $300 repairs + $500 insurance

= $11,000 (0.2886) + $120 + $1,485.71

= $4,780.31EUACgasoline = ($12,000 - $3,000) (A/P, 6%, 3) + $3,000 (0.06)

+ $910.71 fuel + $200 repairs + $500 insurance

= $5,157.61The diesel taxi is more economical

Check solution using NPW:

Note: The analysis period is seven years, hence one cannot compare three years of A vs

four years of B, If one does, the problem is constructed so he will get the wrong answer

= $5,400 (0.1627) + $60 + $800

= $1,739Select the electric motor

Additional Power Cost in last 30 years:

Use 20 year analysis period

Net Present Worth Approach

= -$1,000 - $10 (11.470) + $100 (0.3118)

= -$1,083Choose Masonite to save $52 on Present Worth of Cost

Equivalent Uniform Annual Cost Approach

EUACMas. = $20 + $250 (A/P, 6%, 4) - $10 (A/F, 6%, 4)

= $20 + $250 (0.2886) - $10 (0.2286)

= $90EUACBRK = $10 + $1,000 (A/P, 6%, 20) - $100 (A/F, 6%, 20)

= $10 + $1,000 (0.872) - $100 (0.0272)

= $94Choose Masonate to save $4 per year

6-46

Machine A

EUAB – EUAC = - First Cost (A/P, 12%, 7)

- Maintenance & Operating Costs+ Annual Benefit + Salvage Value (A/F, 12%, 7)

= -$15,000 (0.2191) - $1,600 + $8,000 + $3,000 (0.0991)

= $3,411

Machine B

EUAB – EUAC = - First Cost (A/P, 12%, 10)

- Maintenance & Operating Costs+ Annual Benefit + Salvage Value (A/F, 12%, 10)

= -$25,000 (0.1770) - $400 + $13,000 + $6,000 (0.0570)

= $8,517Choose Machine B to maximize (EUAB – EUAC)

Choose alternative with minimum EUAC

Buy the 36-month tire

DATA

9.00% Canadian conventional mortgage rate

0.7363% effectively monthly interest =((1+$A$2/2)^(1/6))-1

Ending Balance

DATA

9.00% Canadian conventional mortgage rate

Ending Balance

9.00% Canadian conventional mortgage rate

0.7363% effectively monthly interest =((1+$A$2/2)^(1/6))-1

9.00% Canadian conventional mortgage rate

Ending Balance

9.00% Canadian conventional mortgage rate

Ending Balance

6.00% Canadian conventional mortgage rate

0.4939% effectively monthly interest =((1+$A$2/2)^(1/6))-1

6.00% Canadian conventional mortgage rate

0.4939% effectively monthly interest =((1+$A$2/2)^(1/6))-1

6.00% Canadian conventional mortgage rate

0.4939% effectively monthly interest =((1+$A$2/2)^(1/6))-1

Additional Power Cost, yr

zero using solver