Homework 8. Replacement and Economic Life
1. Four years ago you bought a new machine for manufacturing semiconductors for
$1,000,000. Ever since you bought it, the machine has given you trouble. You have
spent approximately $100,000 a year to keep the machine working. Your income from
the sales of semiconductor products is $150,000 per year, so the machine is definitely
marginal. You are considering what to do about the machine. You have several options
as listed below. Whenever you “junk” the machine you will receive $50,000 in scrap
value. When annual payments are involved, use the end of year assumption.
A. The company who sold you the machine offers to fix it for $250,000. The operating
expenses will drop to $50,000 a year. The machine will last another 5 years and then
can be sold for $100,000.
B. You can contract to an outside supplier to produce the semiconductor products for you.
No investment is required, but the supplier will charge $115,000 per year. The contract
must be renewed after three years. In this case you will junk the machine you now own.
C. You can buy an entirely new machine for $600,000 and junk the old machine. This
machine should last you 10 years. Because you've had so much trouble, the company
selling the machine guarantees that the operating cost will be only $30,000 per year.
The full purchase price of the machine ($600,000) will be returned to you after 10
years. Again, if you choose this option, you will junk the machine you now own.
D. You can keep operating as you are now for one more year. You will junk the machine at
the end of the year.
Use any one of the methods discussed in class to determine the best option. Your
minimum acceptable rate of return is 20%.
Formula used for A
Must compare on an annual cost basis.
N A CA = (50 + 250)(A/P, .2, 5) + 50 - 100(A/F, .2. 5)
= 300*0.3344 + 50 - 100*0.1344
= 100.3 + 50 - 13.4 = 136.9
Formula used for B
N A CB = $115
There is no investment required in this case.
Number used for A in the
comparison
136.9 per year
Formula used for C
N A CC = 600(A/P, .2, 10) + 30 - 600(A/F, .2, 10)
= 600*0.2385 + 30 - 600*0.0385
= 143.1 + 30 - 23.1 = 150
Formula used for D
N A CD = (50 – 50)(A/P, .2, 1) + 100 + 50*.2
Or NACD = 50(A/P, .2, 1) + 100 –50 = 110
Number used for C in the
comparison
150
Number used for B in the
comparison
115
Number used for D in the
comparison
110
Note that the 150 annual revenue could have been
Decision (A, B or C)
added to the annual worths of all alternatives with D
not change in the result.
1
2. You work part-time as a pizza delivery driver. Four years ago you bought a car for this
purpose for $10,000. You spend approximately $1,000 a year to keep the car working.
Your income from the job is $3,000 per year. An acquaintance offers to buy the car
from you for $2,000. At this time, you identify the following options.
A. You can keep the car and continue to operate it as you are now. You expect that an
operating cost of $1,000 in the current year, and that the operating cost will increase by
$500 in subsequent years (i.e. the costs in the next three years are 1000, 1500, 2000).
If you continue to use the car as you are now for one more year, the salvage value will
be $1000. The salvage value will drop to $500 after two years. After that the car will be
worthless. You will keep the car for no more than three years.
B. You can fix the car for $3,000. The operating expenses will drop to $500 a year. The
economic life of the fixed car is 5 years. The salvage value after 5 years is $1,000.
C. You can buy a new Yugo for $6,000 and sell the old car. This car should last you 7
years. The operating cost will be only $300 per year. The car will be worthless after 7
years.
Your minimum acceptable rate of return is 20%.
Part a. What is the economic life for option A?
The investment in the defender is its market value 2000.
Salvage after one year is 1000. Salvage after 2 years is 500, and after 3 years is
zero.
1st Year NAWA = 3000 - 2000(A/P, .2, 1) - 1000 + 1000 = 600
2nd Year NAWA = 3000 - 2000(A/P, .2, 2) - 1000 - 500(A/G, .2, 2) +
500(A/F,.2,2) = 691.
3rd Year NAWA = 3000 - 2000(A/P, .2, 3) - 1000 - 500(A/G, .2, 3)= 611
The economic life of the defender is two years.
Economic Life = 2 years____________
Part b. What is the best option regarding the machine?
Formula used for A
N A WA = 3000 - 2000(A/P, .2, 2) - 1000 - 500(A/G,
.2, 2) + 500(A/F,.2,2)
Number used for A in the
comparison
690.07 per year
Formula used for B
The investment for B is 2000 + 3000 = 5000
N A WA = 3000 - 5000(A/P, .2, 5) - 500 + 1000(A/F,
.2. 5)
Number used for B in the
comparison
962.48
Formula used for C
N A WC = 3000 - 6000(A/P,.2,7) - 300 = 1035
or
Number used for C in the
comparison
1035
Decision (A, B or C): C
Note the 3000 income could have
been added to all alternatives
with no change in result.
2
3. In replacement analysis, we must assign an investment value to the asset being
considered for replacement (the defender). In each of the following situations, show the
value that is to be used for the investment in the defender.
The asset was purchased 3 years ago for $100,000. The life used for tax purposes
is five years. The tax salvage is zero. The SYD method is used for depreciation. At
present there is no market for the asset, however, we can sell its parts for scrap. The
total value of the parts is $10,000.
Situation
Investment in the Defender
a. We are doing a before tax analysis with a
PD = Opportunity cost = 10,000
MARR of 20%.
b. We are doing an after tax analysis. The
after tax MARR is 10%. The tax rate for
ordinary income and capital gains is 30%.
Compute depreciation for each
year.
SYD = 15. D1 = (5/15)(100) =
33.33
D2 = (4/15)100 = 26.67
D3 = (3/15)100 = 20
BV=100 - 33.33 - 26.67 -20 = 20.00
SV - BV = 10 - 20 = -10
Tax = (-10)*.3 = -3
PD = 10 + 3 = 3 or 13,000.
c. We are doing a before tax analysis with an
MARR of 20%. In addition to the
information above, we learn that if we keep
the defender we must spend $10,000 to
overhaul the machine for future use.
PD = Opportunity cost + Overhaul
cost = 10,000 + 10,000 = 20,000
4. You have been driving an old Cadillac automobile. Your father bought it in 1985 for
$25,000 with the thought that it would last ten years. He gave it to you in 1990. It is
now 1993. You really love the car, but the fuel and repair costs are becoming
burdensome. Because of the vintage condition of the car, you think you could sell it for
$8,000. For economic calculations, you estimate the future resale value and operating
costs as shown in the table below. The years are measured from today.
Year
Operating Cost
Resale Value
1
2700
6000
2
3000
5000
3
3300
3000
4
3600
1000
5
3900
0
If you do decide to get rid of the old beauty, you will be practical and buy a Plymouth
Shadow. It happens that this car only costs $8,000. With the money from the Cadillac,
you can move into the new car for no net out of pocket expense. The operating expense
is a low $2,500 per year. That amount will stay constant throughout the three years you
plan to keep the car. By that time you will have graduated and be able to buy a new
Cadillac of your own. After three years the Dodge will be worth about $1,000 in resale
value.
3
Find the economic life of the Cadillac. Should you sell the Cadillac and buy the
Plymouth? Base your decision on economics with your minimum acceptable rate of
return of 20%.
Challenger:
NAC = 8000(A/P, .2, 3) + 2500 - 1000(A/F, .2, 3)
NAC = 8000*0.4747 + 2500 - 1000*0. 2747
NAC = 6022.9
Defender: One year
NAC = 8000(A/P, .2, 1) + 2700 - 6000(A/F, .2, 1)
NAC = 8000*1.2000 + 2700 - 6000*1
NAC = 9600 + 2700 - 6000 = 6300
Defender: Two years
NAC = 8000(A/P, .2, 2) + 2700 + 300(A/G, .2, 2) - 5000(A/F, .2, 2)
NAC = 8000*0.6545 + 2700 + 300*0.455 - 5000*0.4545
NAC = 5236 + 2700 + 136.5- 2272.5 = 5800
Defender: Three years
NAC = 8000(A/P, .2, 3) + 2700 + 300(A/G, .2, 3) - 3000(A/F, .2, 3)
NAC = 8000*0.4747 + 2700 + 300*0.879 - 3000*0.2747
NAC = 3797.6 + 2700 + 263.7 - 824.1 = 5937.2
The economic life of the defender is 2 years. Since the annual cost
of the defender is 5800 and less than the cost of the challenger,
keep the Cadillac.
5. You are Alex Rogo and you have discovered that the robots in your plant are not as
good as you once thought. You paid $1,000,000 dollars for the robot three years ago,
but you are thinking of replacing it with several manual machines. The operating cost of
the robot is $20,000 per year. At the time of the purchase, the robots were projected to
last ten years with a salvage value of zero. We can sell the robot now for $500,000,
however, the selling price one year from now will be $400,000.
No initial investment is required for the manual machines, but the annual cost of
operating the machines is $200,000 per year.
We want to decide whether to replace the robot using a cost analysis. Both manual
machines and the robot give the same throughput. Evaluate the following situations. In
every case, we are considering keeping the robot for only one more year. Our minimum
acceptable rate of return on investments is 10%.
a. We must pay $100,000 to remove the robot from the plant. We must pay this whether
the robot is removed now or next year. Should we replace the robot?
b. Forget about the removal cost, but consider the effect of taxes. The robot is being
depreciated with the straight-line method with the depreciation equal to $100,000 per
year. With a tax rate of 50% and with an after tax MARR of 30%, should we replace
the robot?
4
Part a. Working in thousands.
The investment in the robot for analysis purposes is 500 - 100 = 400
The salvage value is 400 - 100 = 300
The cost for one more year is 400(1.1) + 20 - 300 = 440 + 20 - 300 = 160
The annual cost for one more year of operation is less than the cost of the
replacement (200). Do not replace the robot.
Part b.
Working in thousands.
For the robot
The book value of the robot is now $700. If we dispose of it there will be a tax loss
of 200. The tax savings will be 100. This must be added to the opportunity
cost of selling the machine. It will also be added next year, since there will still be
a capital gain of 200.
The investment in the robot for analysis purposes is 500 + 100 = 600
The salvage value is 400 + 100 = 500
The taxable income is -20-100 = -120. The tax then is (-120)*.5 = -60
The after tax cash flow is then -20 - (-60) = 40.
The cost for one more year is 600(1.3) - 40 - 500 = 780 - 40 - 500 = 240
For the manual machine
The tax is -200*.5 = -100
The net cost is 200 - 100 = 100
The manual machine has the smallest annual cost. Replace the robot.
6. Consider the following situation in a machine shop :
A drilling machine (machine A) is being considered for replacement. It was purchased
four years ago for $4000. It has since been deprecated by straight line depreciation.
An eight year useful life and zero salvage value have been assumed. The operating cost
for the machine is $2000 per year and is expected to stay constant for the rest of its life.
It could be sold now to a used equipment dealer for $1000 or be kept in service for
another four years. It would then have no resale value.
The new machine (replacement) would cost $4200, have a six year useful and
depreciable life, and no salvage value. For tax purposes, SOYD depreciation would be
used. This machine would have an operating cost of $500 per year that will be constant
over its life.
Assume a tax rate of 40% on operating income and capital gains.
a) Show the before and after tax cash flows for the existing machine and the new machine.
b ) If the after tax MARR is 10%, should the existing machine be replaced?
Existing machine
Year BTCF Depn TaxInc
Tax ATCF
0 -1000
1000* -400**
-1400
1 -2000
500 -2500 -1000 -1000
2 -2000
500 -2500 -1000 -1000
3 -2000
500 -2500 -1000 -1000
4 -2000
500 -2500 -1000 -1000
* Long term capital loss foregone by keeping machine:
= $2000 Book value - $1000 selling price = $1000 Capital loss.
** The $1000 long-term capital loss foregone would have offset the $1000 of
capital gains elsewhere in the firm. The result is that a tax saving of 40%(1000) =
400 is foregone
5
NPV = -4569
NAC = 1441.66
New machine
Year BTCF Depn TaxInc
Tax ATCF
0 -4200
-4200
1 -500
1200 -1700 -680 180
2 -500
1000 -1500 -600 100
3 -500
800 -1300 -520 20
4 -500
600 -1100 -440 -60
5 -500
400 -900 -360 -140
6 -500
200 -700 -280 -220
SOYD Deprec. Sum = (6*7/2) = 21
1st year SOYD = (6/21) (4200 - 0) = 1200
Annual decline = (1/21) (4200 - 0 ) = 200
NAC = 4200(A/P,10%,6) - 180 + 80(A/G, 10%, 6) = 962.24
Replace the machine
7. A bit for a machine tool costs $50. When new, the bit produces goods worth $100 per
week. However as the bit wears, the amount produced goes down in value by $5 for
each additional week of life. That is, the bit produces an income of $100 in the first
week, $95 in the second week, $90 in the third week, and so on decreasing by $5 per
week. How often should we replace the bit? There is no salvage value for the bit when
it is discarded. The MARR for the investment is 5% per week. Use time value of
money in answering this problem.
Write the Formula that you use to evaluate alternative lives.
Net Weekly worth = -50(A/P, .05, n) + 100 -5(A/G, .05, n)
n
1
2
3
4
5
6
7
NWW
47.5
70.7
76.8
78.7
78.9
78.3
77.3
What is the economic life?
The minimum NWC occurs at 5 weeks.
8. We bought a truck five years ago for $18,000. Its book value is now $3000.
We could sell the truck at this time for $5500. We estimate the resale value of the truck
will be $3500 next year, $2500 in two years and 0 after three years. The operating
costs for the next three years are: $1000, $1500, and $2000 respectively. The
depreciation for the truck will be $1000 per year in each of the next three years.
If we sell the truck, we will use a delivery service that will cost $3800 per year. There is
no investment in this option, and we assume that the cost occurs at the end of each year.
We can use the delivery service for any number of years.
6
Our minimum after tax acceptable rate of return is 12%. The tax rate on ordinary
income and capital gains is 40%.
Should we keep the truck or sell it and use the delivery service?
The challenger: NAC = 3800*.6 = 2280.
The defender is the truck now owned.
The investment in the defender is its after tax market value.
P = Market Value - Tax = MV - (MV - BV)*.4
P = 5500 - (5500 - 3000)*(.4) = $4500.
The after tax salvage in each of the next three years is:
S = MV - (MV - BV)*.4
S 1 = 3500 - (3500 - 2000)*(.4) = $2900
S 2 = 2500 - (2500 - 1000)*(.4) = $1900
S3 = 0 - 0 = 0
The after tax cash flow due to operating cost in the next three years is:
O = -OC - (-OC - Depr)*.4
O 1 = -1000 - (-2000)*.4 = -200.
O 2 = -1500 - (-2500)*.4 = -500.
O 3 = -2000 - (-3000)*.4 = -800.
For one year:
NAC(1) = 4500(A/P, .12, 1)+ 200 - 2900 = 4800*1.12 - 3000 = 2340
NAC(2) = 4500(A/P, .12, 2) + 200 + 300(A/G, .12, 2) - 1900(A/F. .12, 2)
NAC(2) = 4500 (.5917) + 200 + 300 (.4717) - 1900 (.4717) = 2107
This cost is less than challenger
The Defender wins.
7