IE 3301 ENGINEERING ECONOMIC ANALYSIS

Ch.13 Problems - Solutions

13.1

XYZ Beverage Company just bought a brewing machine for $35,000 with an estimated

useful life of 5 years. It is estimated that the annual operating and maintenance cost for

the brewing machine will be $1000 until the 3rd year and then this cost will increase by

$700 per year from the 4th

year on. The brewing machine belongs to the MACRS

property class of 3 years for depreciation purposes. Find the marginal cost for this

machine assuming that the brewing machine could be sold for its current book value at

the end of any given year. Use MARR = 10%

Solution:

The market value after depreciation using MACRS 3 year property class is given below:

Year

MACRS

rt Cost basis Depreciation Market Value

1 33.33% $35,000.00 $11,665.50 $23,334.50

2 44.45% $35,000.00 $15,557.50 $7,777.00

3 14.81% $35,000.00 $5,183.50 $2,593.50

4 7.41% $35,000.00 $2,593.50 $0.00

5 0.00% $35,000.00 $0.00 $0.00

The Marginal Cost for the brewing machinery (year to year cost of ownership) over its 5

year useful life is given below:

Year (n)

Loss in

market Value

(depreciation) Interest O&M

Marginal

Cost

1 $11,665.50 $3,500.00 $1,000.00 $16,165.50

2 $15,557.50 $2,333.45 $1,000.00 $18,890.95

3 $5,183.50 $777.70 $1,000.00 $6,961.20

4 $2,593.50 $259.35 $1,700.00 $4,552.85

5 $0.00 $0.00 $2,400.00 $2,400.00

13.2

Find the most economic life for a machine with the following information:

a. First Cost = $1,000,000

b. Salvage Value = $225,000 (if sold in any year)

c. Maintenance and operating cost = $200,000 for the first year, with a

$70,000 gradient

d. MARR = 10%

Useful life = 8 years

Solution:

EUAB – EAUC = -$1,000,000 (A/P, 10%, n) + $225,000 (A/F, 10%, n) − $200,000 −

$70,000 (A/G, 10%, n)

Substituting the different values of n, we get the following result:

year n EUAB-EUAC

1 -$1,075,000.00

2 -$702,375.00

3 -$599,717.50

4 -$563,682.50

5 -$553,645.00

6 -$556,120.00

7 -$565,225.00

8 -$578,015.00

The most economic life is 5 years.

13.3

A water pump has a first cost of $45,000. The market value declines by 10% every year,

as the water pump erodes with every year in use due to the chemical reaction of pump

material with minerals present in the water. This chemical reaction is also believed to be

the cause for pipe joint leakage, which costs $800 for the first year and then climbs by

$800 per year. Assume the MARR is 10%. Find the minimum EUAC for this water pump

and the most economic life of the water pump.

Solution:

First Cost = $15,000

Market value decline % = 10%

Leakage cost = $800

MARR = 10%

year (n) Cost Salvage (P/F,10%,n) PW (A/P,10%,n) EUAC

0 45,000

1 800 40500 0.9091 -8,908.73 1.1 $9,799.60

2 1600 36450 0.8264 -16927.24 0.5762 $9,753.48

3 2400 32805 0.7513 -24206.24 0.4021 $9,733.33

4 3200 29524.5 0.683 -30873.00 0.3155 $9,740.43

5 4000 26572.05 0.6209 37023.25 0.2638 $9,766.73

Thus, the minimum EUAC = $9,733 and the Minimum Cost Life = 3 years

13.4

Jack installed industrial air-conditioner equipment for $25,000, 5 years ago (at the

beginning of 2005) with the marginal costs given in the table below. There is new

equipment available now that has a first cost of $27,900, no yearly costs, zero salvage

value, and a maximum life of 5 years. Using an 8% MARR:

a) Which replacement analysis technique should Jack use and why?

b) When should he replace the existing equipment with the new equipment?

End of

Year (n)

Year Marginal Cost

1 2005 $5,750

2 2006 $5,800

3 2007 $5,895

4 2008 $6,031

5 2009 $6,202

6 2010 $6,407

7 2011 $6,641

8 2012 $6,902

9 2013 $7,187

10 2014 $7,493

Solution:

a) The marginal cost data is strictly increasing; hence we use replacement analysis

technique I.

b) In order to make a decision of when to replace the defender (existing equipment) with

the challenger (new equipment), we must find the minimum EUAC for challenger.

Since there is no annual cost gradient, only a capital recovery function is at work and the

minimum EUAC for challengers occurs at 5 years: EUAC = $27,900 (A/P, 8%, 5) =

$27,900 (A/P, 8%, 5) =$6,989

From this we would recommend that we keep the Defender for three more years and then

replace it with the Challenger at the end of year 8 (beginning of year 9). This is because

after three years the marginal costs of the Defender become greater than the min. EUAC

of the Challenger.

End of

Year (n)

Marginal Cost of

Defender

Min. EUAC of

the Challenger

Result

6 $6,407 $6,989 Keep

7 $6,641 $6,989 Keep

8 $6,902 $6,989 Keep

9 $7,187 $6,989 Replace defender

with challenger

at the start of this

year

10 $7,493 $6,989

13.5

B&B construction is considering the replacing its existing mid-size excavator, which was

purchased 3 years ago for $200,000. The current market price of the excavator is

$120,000 and annual maintenance and operating cost are given in the table below. Data

for a new excavator were analyzed and it was found that the most economic life is 8

Years with the minimum annual cost of $62,000. When should the excavator be

replaced? Use MARR = 20%.

Year Operating cost Maintenance Cost Market Value

1 $15,000 $9000 $85,000

2 $15,000 $10,000 $65,000

3 $17,000 $12.000 $50,000

4 $20,000 $18,000 $40,000

5 $25,000 $20,000 $35,000

6 $30,000 $25,000 $30,000

7 $35,000 $30,000 $25,000

Solution:

First the marginal cost for the defender (existing excavator) has to be found out to

determine which replacement method is to be used. The marginal cost for the defender is

calculated in the table below: year Operating

Cost

Maintenance

cost

Market

Value(MV)

Loss in

interest

Loss in

MV

Total Marginal

Cost

1 15000 9000 85000 24000 35000 83000

2 15000 10000 65000 17000 20000 62000

3 17000 12000 50000 13000 15000 57000

4 20000 18000 40000 10000 10000 58000

5 25000 20000 35000 8000 5000 58000

6 30000 25000 30000 7000 5000 67000

7 35000 30000 25000 6000 5000 76000

The marginal cost data in the table above are not strictly increasing; therefore, use

replacement analysis technique II. We compare this value to the minimum EUAC for the

challenger of $62,000. We note that Marginal Cost def < MC Challenger for years 1-5.

Therefore, the two obvious choices are: 1) replace now or 2) replace after year 5.

Calculate the NPV of both choices and compare (below).

Replace now Replace after year 5

1 62,000 83000

2 62,000 62000

3 62,000 57000

4 62,000 58000

5 62,000 58000

6 62,000 62,000

7 62,000 62,000

$223,484.69 $234,554.65

Choose to replace now!

13.6

One of the three annealing machines at a heat treatment plant is being considered for

replacement. The salvage value and the maintenance cost of the machine are given

below, along with the data for the new annealing machine being considered. It is believed

that new and old annealing machine have similar productivity and energy costs. If the

MARR is 10%:

a) Which replacement analysis technique would you consider and why?

b) When should the defender be replaced?

New Annealing Machine Old Annealing Machine

Year Salvage value at

end of year ($) Maintenance

Cost ($) Salvage Value at

end of year ($) Maintenance Cost ($)

0 80000 - 20000 -

1 75000 0 17000 9500

2 70000 0 14000 9600

3 66000 1000 11000 9700

4 62000 3000 7000 9800

Solution:

Marginal Cost of the Defender (Old Machine):

Year

Salvage Value

at end of year Maintenance Cost

Loss in market

Value Interest in year (n) Marginal Cost

0 20000 - - - -

1 17000 9500 3000 2000 14500

2 14000 9600 3000 1700 14300

3 11000 9700 3000 1400 14100

4 7000 9800 4000 1100 14900

The marginal cost for defender is not strictly increasing; therefore, we use Replacement

Analysis technique II. We could use the same approach as in Problem 5 or we can

compare the EUAC of the defender for n years or ownership to the minimum EUAC of

the challenger.

EUAC for Defender for year 1 = 20000(A/P,10%,n) - 17000(A/F,10%,n) + 9500 +

100(A/G,10%,n) = $14500.

Other calculations are shown in the table below:

Year

Salvage Value

at end of year Maintenance Cost EUAC

0 20000 -

1 17000 9500 14500

2 14000 9600 14404.8

3 11000 9700 14312.6

4 7000 9800 14439.6

Minimum EUAC for defender = $14312.6 at year 3.

EUAC for Challenger:

EUAC for year 1 = 80,000(A/P,10%,1)-75,000(A/F,10%,1) =$13,000

EUAC for year 2 =80,000(A/P,10%,2)-70,000(A/F,10%,2) =$12762

EUAC for year 3 = 80,000(A/P,10%,3)-66,000(A/F,10%,3)+1000(A/F,10%,3) =

$12531.5

EUAC for year 4 = 80,000(A/P,10%,4)-62,000(A/F,10%,4)+1000(P/F,10%,3)

(A/P,10%4) + 3000(A/F,10%,4) = $12762.55

EUAC for Challenger in tabulation

Year

Salvage Value

at end of year Maintenance Cost EUAC

0 80000 -

1 75000 0 13000

2 70000 0 12762

3 66000 1000 12531.5

4 62000 3000 12762.55

Minimum EUAC for challenger is $12531.5 at year 3

From above calculations, minimum EUAC of defender > minimum EUAC for challenger

so, we must replace it immediately.

13.7

You are considering purchasing new equipment to replace old equipment. The old

equipment has a current market value of $2400. The new equipment will cost $4500. The

use of the new equipment is expected to produce energy savings of $500 per year. It is

estimated that both the new and old equipment have a remaining useful life of 10 years

with zero salvage value. Should you replace the old equipment today with the new one?

Use 12% MARR

Solution:

Here we use Replacement Analysis technique 3 because we do not have defender

marginal cost data. In this case, we simply compare defender’s EUAC over its stated

useful life and the challenger’s minimum EUAC.

EUAC(defender) = 2400 (A/P,12%,10) = 2400*0.1770 = $424.8

EUAC(challenger) = 4500 (A/P, 12%, 10) – 500 = 4500*0.1770 -500 = $296.5

EUAC(defender) > EUAC(challenger), Choose Challenger (New Equipment).

13.8

XYZ Inc has decided to build a natural gas production plant with equipment purchased

for $135000. The annual O&M cost is expected to be $2000 in year 1 and then increasing

by $1000 per year for every year after that. It is estimated that the plant would last for up

to 10 years. The equipment’s market value will be 85,000 at the end of the first year,

decreasing by $3500 annually. MACRS depreciation method will be used with MACRS

GDS property class of 7 years. The combined federal and state tax rate is estimated to be

40%. Assuming the after-tax MARR is 10%, calculate the after-tax marginal cost of the

equipment for each of the 10 years.

Solution:

ATCF for O&M and Depreciation

Year O&M MACRS (rt) Macrs dt Taxable Income Tax Saving

(40%)

O&M

+depreciation

ATCF

1 -$2,000.00 14.29% $19,291.50 -$21,291.50 $8,516.60 $6,516.60

2 -$3,000.00 24.49% $33,061.50 -$36,061.50 $14,424.60 $11,424.60

3 -$4,000.00 17.49% $23,611.50 -$27,611.50 $11,044.60 $7,044.60

4 -$5,000.00 12.49% $16,861.50 -$21,861.50 $8,744.60 $3,744.60

5 -$6,000.00 8.93% $12,055.50 -$18,055.50 $7,222.20 $1,222.20

6 -$7,000.00 8.92% $12,042.00 -$19,042.00 $7,616.80 $616.80

7 -$8,000.00 8.93% $12,055.50 -$20,055.50 $8,022.20 $22.20

8 -$9,000.00 4.46% $6,021.00 -$15,021.00 $6,008.4 -$2,291.600

9 -$10,000.00 0.00% $0.00 -$10,000.00 $4,000.00 -$6,000.00

10 -$11,000.00 0.00% $0.00 -$11,000.00 $4,400.00 -$6,600.00

ATCF in Year of Disposal

Year Market

Value Book value

Depreciation

Recapture or

Loss

Gain/Loss

Tax at 40%*

AT MV end of

year

AT MV beg of year

AT Marginal

Cost

1 $85,000.00 $115,708.50 -$30,708.50 $12,283.40 $97,283.40 $135,000.00 $44,700.00

2 $81,500.00 $82,647.00 -$1,147.00 $458.80 $81,958.80 $97,283.40 $13,628.34

3 $78,000.00 $59,035.50 $18,964.50 -$7,585.80 $70,414.20 $81,958.80 $12,695.88

4 $74,500.00 $42,174.00 $32,326.00 -$12,930.40 $61,569.60 $70,414.20 $12,141.42

5 $71,000.00 $30,118.50 $40,881.50 -$16,352.60 $54,647.40 $61,569.60 $11,856.96

6 $67,500.00 $18,076.50 $49,423.50 -$19,769.40 $47,730.60 $54,647.40 $11,764.74

7 $64,000.00 $6,021.00 $57,979.00 -$23,191.60 $40,808.40 $47,730.60 $11,673.06

8 $60,500.00 $0.00 $60,500.00 -$24,200.00 $36,300.00 $40,808.40 $11,580.84

9 $57,000.00 $0.00 $57,000.00 -$22,800.00 $34,200.00 $36,300.00 $11,730.00

10 $53,500.00 $0.00 $53,500.00 -$21,400.00 $32,100.00 $34,200.00 $12,120.00

*Tax savings from this depreciation Loss is added to market value (salvage) to determine

ATCF and vice versa.

Partial Answers

13.1 On your own

13.2 Economic life = 5 years

13.3 Min EUAC = $9733 and Min cost life = 3 years

13.4 a) on your own b) replace defender at the end of year 8 (Year 2012)

13.5 Replace the defender immediately

13.6 replacement analysis technique II b) immediately

13.7 EUAC(defender) > EUAC(challenger), Choose Challenger (New Equipment).

13.8 On your own