ie 2324 chap 13 homework with answers
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Economics PracitceTRANSCRIPT
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