Solution manual engineering economic analysis 9th edition ch13

Chapter 13: Replacement Analysis13-1For the Replacement Analysis Decision Map, the appropriate analysis method is a function of the cash flows and assumptions made regarding the defender

Chapter 13: Replacement Analysis13-1

For the Replacement Analysis Decision Map, the appropriate analysis method is a function

of the cash flows and assumptions made regarding the defender and challenger assets.Thus, the answer would be the last it depends on the data and the assumptions

13-2

The replacement decision is a function of both the defender and the challenger

The statement is false

The answer is one year

13-5

The EUAC of installed cost will decline as the service life increases The EUAC of

maintenance is constant Thus total EUAC is declining over time

Answer: For minimum EUAC, keep the bottling machine indefinitely

13-6

The value to use is the present market value of the defender equipment (The book

indicates that trade-in value may be purposely inflated as a selling strategy, hence it may or may not represent market value.)

13-7

(a) Expected good performance, productivity, energy efficiency, safety, long service life

Retraining in operation and maintenance may be required High comfort of

operation High purchase price May not be immediately available Sales taxes to

be paid Can be depreciated Supplier warranty and spare parts backup available

(b) All as in (a) except for lower price and probably faster delivery

(c) All as in (a) except for still lower cost, lost production during the rebuild period, and

that the rebuild costs can be expensed, at least partially No sales tax applies (d) Performance and productivity may not be as good as in option (c) Retraining in

operation and maintenance is not required Production will be lost during the

rebuilding period Cost may be substantially lower than in previous options The rebuild costs can be expensed No sales tax applies

(e) Performance, productivity, service life, energy efficiency, safety, reliability may be

significantly lower than in the other options Retraining in operation and

maintenance may be required if the new unit is different from the previous one Costmay be only 20-50% of the new equipment Immediate delivery is a possibility The sales tax applies Equipment can be depreciated

For a 1-year useful life

For a 2-year useful life

$2,000

$500

EUAC = $2,000 (1 + 0.15)1 + $500 = $2,800

Alternate computation of maintenance in any year N:

EUACN = A = $2,000 (A/P, 15%, N) + $500 + $500 (A/G, 15%, N)

(a) Total EUAC = $2,250 + EUAC of Maintenance

Therefore, to minimize Total EUAC, choose the alternative with minimum EUAC of maintenance

Economical life = 3 years

(b) The stainless steel tank will always be compared with the best available replacement (the challenger) If the challenger is superior, then the defender tank probably will be replaced

It will cost a substantial amount of money to remove the existing tank from the plant, sell

it to someone else, and then buy and install another one As a practical matter, it seems unlikely that this will be economical

+ $2,100 (F/P, 15%, 1) + $4,500] (A/F, 15%, 7)

= $1,074

EUAC of Installed Cost

Year (P – S) (A/P, i%, n) + (S) (i) = EUAC of

Year EUAC of Installed

Cost + EUAC of Maintenance = Total EUAC

Useful Life- 1 year

Useful Life- 2 years

$12,000

EUAC = $12,000 (A/P, 10%, 2) = $6,914

$12,000

EUAC = $12,000 (F/P, 10%, 1) = $13,200

Useful Life- 3 years

Useful Life- 4 years

Useful Life- 5 years

Useful Life- 6 years

Maintenance & Operating Cost = $235,000

Maintenance & Operating Gradient = $75,000

MARR = 10%

EUAB – EAUC = $1,050,000 (A/P, 10%, n) + $225,000 (A/F, 10%, n)

- $235,000 - $75,000 (A/G, 10%, n)Try n = 4 years:

Maintenance for the Year

Sum of Decline in Value +

MaintenanceNew $11,200

Find: NPWOVERHAUL and NPWREPLACE

Note: All costs which occur before today are sunk costs and are irrelevant

NPWOVERHAUL = -$1,800 - $800 (P/A, 5%, 2)

= -$1,800 - $800 (1.859) = -$3,287

= +$1,500 - $2,800 (1.859) = -$3,705Since the PW of Cost of the overhaul is less than the PW of Cost of the replacement car, thedecision is to overhaul the 1988 auto

By linear interpolation, the incremental before-tax rate of return is 12.7%

The 12.7% rate of return on the increment is unsatisfactory, so reject the increment and recondition the old tank car

(a) Before-Tax Analysis

Year New Machine

BTCF Existing Machine BTCF New Machine rather than Existing Machine BTCF

40% IncomeTaxes

* Long term capital loss foregone by keeping machine:

$2,000 Book Value - $1,000 Selling price = $1,000 capital loss

** The $1,000 long term capital loss foregone would have offset $1,000 of long term

capital gains elsewhere in the firm The result is a tax saving of 20% ($1,000) =

$200 is foregone

New Machine rather than Existing Machine

Year New Tool ATCF Existing

Tool ATCF

New- Existing ATCF

Compute Equivalent Uniform Annual Cost (EUAC)

Initial Cost: $196,000 (A/P, 9%, 8) = $196,000 (0.1807) = $35,453

Less Salvage Value: (6 x $750) (A/F, 9%, 8)= $4,500 (0.0907) = -$408

EUAC = $26,645

Alternative II: Keep 4 old machines and buy 3 new ones

Initial Cost:

Value of 4 old machines 4 x $2,000 $8,000

3 new machines at $32,000 each$96,000

Training Program at 3 x $700 $2,100

Annual Maintenance= 4 old x $1,500 + 3 new x $600 = $7,800 per year

Salvage Value 8 years hence= 4 old x $500 + 3 new x $750 = $4,250

Compute Equivalent Uniform Annual Cost (EUAC)

Initial Cost: $106,100 (A/P, 9%, 8) = $106,100 (0.1807) = $19,172

Less Salvage Value: ($4,250) (A/F, 9%, 8) = $4,250 (0.0907) = -$385

EUAC = $26,587Decision: Choose Alternative II with its slightly lower EUAC

Loss in

MV (n)

Annual Costs (n)

Lost Interest

in (n)

Total Marg

13-19

For this problem we have marginal cost data for the defender, so we will check to see if that data is strictly increasing

Defender:

Current Market Value = $25,000 (0.70)5 = $4,202

Year Time Market Loss in Annual Lost Total

Line Value (n) MV (n) Costs

(n)

Interest in (n)

(b) In looking at the table above one can see that the marginal cost data of the defender is strictly increasing over the next five year period Thus the Replacement Decision Analysis Map would suggest that we use Replacement Analysis Technique #1 We compare the defender marginal cost data against the challenger’s minimum EUAC

We would keep the defender asset for two more years and then replace it with the new automated shearing equipment After two years the MC (def) > Min EUAC (chal):

$113,000 > $110,000

In this case we first compute the total marginal costs of the defender asset From Figure

13-1 the marginal cost data is available, and it is not strictly increasing (see Total MC column inthe table below) Thus, we use Replacement Analysis Technique #2, comparing minimumEUAC defender against minimum EUAC of challenger In the table below the minimumEUAC is at year 5 for the old paver (five years from today), the value is $59,703 Wecompare this value to the minimum EUAC for the challenger of $62,000 Thus, werecommend keeping the defender for at least one more year and reviewing the data forchanges

MARR% 20%

First

Cost

120000

YEAR OPER MAIN

T MV in LostMV LostInt. TotalMC NPW EUAC(n) Cost Cost (n) (n) (n) (n) (1 >n) (1 >n)

(a) The minimum cost life is where the EUAC of ownership is minimized for the number

of years held This would occur at 4 years for the defender where EUAC = $4,400.(b) The minimum cost life of the challenger is 5 years where the EUAC = $6,200

(c) Using Replacement Analysis Technique #3: Assuming that the defender and

challenger costs do not change over the next 4 years we should keep the defender for four years and then reevaluate the costs with challengers at that time Here we are comparing the min EUAC (def) vs min EUAC (challenger) and $4,400 < $6,200thus we keep the defender

13-23

(a) The minimum cost life is where the EUAC of ownership is minimized for the number

of years held This would occur at 1 year for the defender, where EUAC = $4,000.(b) The minimum cost life of the challenger is 4 years where the EUAC = $3,300 (c) Using Replacement Analysis Technique #3: Given these costs for the defender and

challenger we should replace the defender with the challenger asset now This is because the min EUAC (def) > min EUAC (challenger): $4,000 > $3,300

Here we use Replacement Analysis Technique #3 Because the remaining life of the defender and the life of the challenger are both 10 years we can use either the “opportunity cost” or “cash flow” approach to setting the first cost of each option (keep defender or replace with challenger) Let’s show each solution:

Opportunity Cost Approach

EUAC (def) = 4 ($600) (A/P, 25%, 10) = $672

EUAC (chal) = $5,000 (A/P, 25%, 10) - $10,000 (0.075) = $650

Cash Flow Cost Approach

EUAC (def) = $0.00

EUAC (chal) = ($5,000 - $2,400) (A/P, 255, 10) - $10,000 (0.075)= -$22

In either case we recommend that the new high efficiency machine be implemented today

13-25

From the facts stated, we see that if the old forklift is retained the EUAC is minimum for a one year useful life The problem says the challenger economic life is 10 years (Using the data provided this fact could be verified, but that is not part of the problem.)

Annual Cash-Flow Analysis:

Keep Old Forklift Another Year

Income

40% IncomeTaxes

ATCF

EUAC for one more year with old forklift = $240

Buy New Forklift

Year BTCF SL Deprec Taxable

Income

40% IncomeTaxes

13-26

The problem, with a 7-year analysis period, may be solved in a variety of ways A first step

is to compute an after-tax cash flow for each alternative

Alternative A

Income

40% IncomeTaxes

40% IncomeTaxes

ATCF

Alternative E (Do Nothing)

Income

40% IncomeTaxes

Choose the solution that maximizes NPW Choose Alternative C

Rate of Return Solution

Alternative A rather than Alternative E (Do nothing)

Year Alt A ATCF Alt E ATCF (A- E) ATCF

∆ROR = 6.4%

Reject Alternative A

Alternative D rather than Alternative E

Year Alt D ATCF Alt E ATCF (D- E) ATCF

∆ROR = 12.8%

Reject Alternative E

Alternative C rather than Alternative D

Year Alt C ATCF Alt D ATCF (C- D) ATCF

= $16,500

Machine A annual depreciation = (P – S)/n = ($54,000 - $0)/12 = $4,500

Machine B annual depreciation = (P – S)/n = ($42,000 - $0)/12 = $3,500

Alternate 1: Keep A for 12 more years

Year BTCF SL Deprec Taxable

Income

40% IncomeTaxes

* If A were sold the Year 0 entries would be:

Year BTCF SL Deprec Taxable

Income

40% IncomeTaxes

ATCF

If A is kept, the entries are just the reverse

After-Tax Annual Cost

Alternate 2: Buy Machine B

Year BTCF SL Deprec Taxable

Income

40% IncomeTaxes

SOYD Deprec Book Value

∆ Tax Income

+$2,560

* Foregone recaptured depreciation is $7,000 – BV5 = $1,000

** Loss is $1,600 – BV5 = -$2,000(b)

Year BTCF

CCA Depreciation (30% Income ∆ Tax ∆ Tax (40%) (SHSS) ATCF (Sonar) ATCF

Here we use the Opportunity Cost Approach for finding the first costs

(a) Problem as given

Defender: SL Depr = ($50,000 - $15,000)/10 = $3,500 per year

(c) Defender Market Value = $18,000

Defender: SL Depr = ($50,000 - $15,000)/10 = $3,500 per year

1 )(

2 1 (

2 )

1 )(

2 1 (

1 )

2 (

1 2 1

d P UCC

n for d

d Pd CCA

n for

d P CCA

n n

n n

Capital

Cost CCA

Book Value

AT Salvage

O&M cash flow

Taxable Income

PW Sum O&M - tax EAC