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Replacement Analysis Should we replace an asset that we own now or later? Reasons for replacing an asset • Physical Impairment • Altered Requirements • Technology The replacement of assets often represents economic opportunity for the firm. We compare the two alternatives: • The asset that we own: The Defender • The Asset that we might buy to replace it: The Challenger Factors to consider (or ignore) • Sunk Costs • Existing Asset Value and the outsider viewpoint • Income Tax Considerations • Economic Life of the challenger and the defender An asset has various types of lives • Useful Life • Tax Life • Economic Life The Economic Life of an asset is • the period of time that minimizes the net annual cost (NAC) for the investment (when it primarily consists of costs) or • the period of time that maximizes the net annual worth (NAW) for the investment (when it consists of costs and revenues) Example 1 • Investment in a machine: $15,000 • Useful life 10 years • Salvage decreasing with time like the book value using SYD method. Assume salvage after 10 years is zero • Operating cost is $500 in first year, but increases by 40% in each subsequent year • What is the economic life? The MARR is 18%
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Computation of the Economic Life Tim e
O perating Cost
Salvage Value
500 700 980 1372 1921 2689 3765 5271 7379 10331
15000 12273 9818 7636 5727 4091 2727 1636 818 273 0
0 1 2 3 4 5 6 7 8 9 10
Let A(n) = cost in year n S(n) = Salvage in year n P = Investment at time 0
NAC
5927 5669 5462 5307 5207 5162 5178 5256 5404 5627
PA = present worth of annual costs = A(1)(P/F, i, 1) + A(2)(P/F, i, 2) + … + A(n)(P/F, i, n) NAC = (P + PA)(A/P, i, n) – S(n)(A/F, i, n)
The Economic Life Minimizes the NAC Find NAC(1), NAC(2),…, NAC(10). The economic life is the life that minimizes NAC. In this case the economic life = 6 years. Economic Life 16000 14000 12000 10000
Operating Cost
8000
Salvage Value
6000
NAC
4000 2000 0 0
1
2
3
4
5
6
7
8
9
10
Economic Life Summary: • For many situations, the economic life and the useful life are the same. • When the economic life is different from the useful life, use the economic life when determining the advisability of an investment. • When comparing two alternatives, compute and use the economic life of each alternative in the comparison.
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Replacement Analysis: Example 2 It is 2003. Should we buy a new car to replace the old wreck? Your minimum acceptable rate of return is 12%. Old Car: Initial cost was $4,500 in 1995. The car was supposed to last for 8 years with a trade-in value $500 at the end of 8 years. Maintenance costs next year will be $800, and are expected to go up by $400 a year in the coming years (800, 1200, 1600, . . .). The car is now a death trap, not worth more than $250. To trade it in, you're going to have to clean it up for a cost of $50. In the future, net salvage value is also expected to be $200 at any time.
New car: Cost is $8,250, less trade-in allowance of $250. This car is supposed to last for 10 years with a trade-in value of $750 at the end of that time. If you sell it before the end of its useful life, you expect the trade-in value to be the same as the book value computed with straight-line depreciation. Maintenance will be $100 per year for the first three years and $300 per year thereafter.
The Defender Investment
The Challenger Investment
Operating Costs and Revenues
Operating Costs and Revenues
Salvage Value
Salvage Value
Economic Life
Economic Life
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Investment for the Defender The opportunity cost for the defender is the money that you give up by not disposing of it. You must also add any costs at time 0 to make it equivalent to the challenger. The investment consists of • the current market value for the defender, • less costs necessary for its disposal, • less taxes on the capital gain (when taxes are considered), • plus any real costs at time 0 necessary to keep the defender. Beware Don't use any defender characteristics to compute the challenger investment, costs, or salvage. Don't use any challenger characteristics to compute the defender investment, costs, or salvage. The Defender
The Challenger
Investment: PD = $200. Operating Cost: AD(n) = $800 + $400(n-1) Salvage Value: SD(n) = $200
Investment: PC = $8,250 Operating Cost: AC(n) = $100, $100, $100, $300, $300, $300, ... Salvage Value: SC(n) = $8,250 –$750n
NACD(n) = PD(A/P, i, n) + $800 + $400(A/G, i, n) – 200(A/F, i, n)
NACC(n) = PC(A/P, i, n) + 100 –SC(n)(A/F, i, n) for n = 1, 2, 3 NACC(n) = PC(A/P, i, n) + 300 – 200(P/A, i, 3) (A/P, i, n) – SC(n)(A/F, i, n) for n = 4,…,10
Defender Age, n 0 1 2 3
A(n) 800 1200 1600
Challenger S(n) 200 200 200 200
NAC 824 1013 1194
Age, n 0 1 2 3 4 5 6 7 8 9 10
A(n) 100 100 100 300 300 300 300 300 300 300
S(n) 8250 7500 6750 6000 5250 4500 3750 3000 2250 1500 750
NAC 1840 1798 1757 1760 1747 1728 1705 1681 1657 1632
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Which do we select? • Defender cost for one more year: $824 • Challenger cost per year: $1632 Notes • All parameters of the defender and challenger are independent. • Use economic lives in the analysis. • The economic life of the defender is often one year. • The economic life of the challenger is often its useful life. • Choose the winner on the basis of minimum NAC or maximum NAW Replacement Analysis Examples • Do we replace now or later? • Case 1: When the useful lives of the defender and the challenger are known and the same. • Case 2: When the useful lives of the defender and the challenger are not known or are not the same. Example 3: Known and Equal Useful Lives Defender: Existing Pump A Capital investment when purchased 5 years ago: $17,000 Useful life:
another 9 years
Depreciation:
SL with half-year convention
Tax Life:
9 yrs
Annual Expenses Replacement of impeller and bearings = Operating and maintenance = Taxes and insurance = $17,000 × 2% = Total
$1,750 $3,250 $340 $5,340
Present Market Value $750 Estimated market value at the end of the 9 years = $200 Current book value = $8500 Challenger: Replacement Pump B
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Capital investment = Useful life = Depreciation: Tax Life: Estimated market value at the end of 9 years = Annual Expenses Operating and maintenance = Taxes and insurance = $16,000 × 2% = Total
$16,000 9 years MACRS 5 years $3,200 $3,000 $320 $3,320
Effective income tax (capital gains) rate = 40% MARR (before taxes) = 10%, MARR (after taxes) = 6% Before Tax Analysis: 9 years study period Defender Investment Opportunity Cost = Current Market Value = Salvage Value = Yearly Total Expenses = NAC(9) of Defender = Challenger Investment Initial Investment = Salvage Value = Yearly Total Expenses = NAC(9) of Challenger =
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After Tax Replacement Analysis Before-tax analysis is often not valid because of • the effect of depreciation • the effect of any significant gain or loss on disposal on income taxes. Therefore, an after-tax analysis is often necessary to evaluate the benefit of replacement. Example 4 Computing the investment in the defender General Case BTMV = Market value of the Defender when we are considering disposal
Existing Pump A (defender) BTMV =750
Before tax cash flow is –BTMV
BTCF = –750
ATMV = After tax market value of the Defender • ATMV = BTMV – Tax • BV = Book value of the Defender assuming we do dispose of it • Taxable income if we dispose of the defender = BTMV – BV • Tax = (BTMV – BV) × (Tax Rate)
If we sell the asset at the end of the 5th year of its tax life, then Depreciation: $944 in year 1 $1,889 in years 2 through 5 (assuming full depr. in year 5) Accum. Depr. = $944 + 4 × $1,889 = $8,500 BV = $17,000 – $8,500 = $8,500 Taxable income = $750 – $8,500 = –$7,750 Tax = –$7,750 × 0.4 = –$3,100 ATMV = $750 – (–$3,100) = $3,850
After tax opportunity cost is the ATMV. Use –ATMV for ATCF at time 0.
ATCF = –$3,850
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Example 4: In the table below for the defender, row 0 is the opportunity cost of not disposing of pump A (what we would have received had we sold it). Rows 1 through 9 assume that we do not dispose of the defender. Cash flows and depreciation are as if we keep the defender for the 9 year horizon. ATCF for the Defender End of Year k 0
Deprec.
Taxable income -$7750
Income taxes $3100
BTCF -$750
ATCF -$3850
1-4
-$5,340
$1,889
-7229
-2892
-2448
5
-$5,340
$944
-6284
-2513
-2826
6-9
-$5,340
$0
-5,340
-2136
-3204
200
80
120
9 Salvage
$200
BV = $0
Using 6% After Tax Present Cost = –$22,669 After Tax Net Annual Cost = – $3,333 ATCF for the Challenger Construct the challenger cash flow independently of the defender End of Year k 0 1 2 3 4 5 6 7-8 9
BTCF -$16,000 -$3,320 -$3,320 -$3,320 -$3,320 -$3,320 -$3,320 -$3,320 -$120
MACRS deprec.
Taxable income
Income taxes
$3,200 $5,120 $3,072 $1,843 $1,843 $922 $0 $0
-$6,520 -$8,440 -$6,392 -$5,163 -$5,163 -$4,242 -$3,320 -$120
-$2,608 -$3,376 -$2,557 -$2,065 -$2,065 -$1,697 -$1,328 -$48
After Tax NAC using 6% =$3,375
ATCF -$16,000 -$712 +$56 -$763 -$1,255 -$1,255 -$1,623 -$1,992 -$72
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Lessons from Example 4 • Before-Tax and After-Tax Analysis can yield different results. When taxes play a role in cash flows, an after-tax analysis should be performed. • The after-tax NAC of the challenger and the defender are very close ($3,375 vs. $3,33s). In such cases, other factors (such as the improved reliability of the new pump or productivity loss due to training) can be considered. Example 5: Comparing Projects with Unknown Lives Challenger: New Forklift Truck Capital investment = $20,000 For the next five years, we have Year 1 2 3 4 5
Estimated Decline in MV $5000 $3750 $2750 $2000 $1750
Estimated MV $15,000 $11,250 $8,500 $6,500 $4,750
Annual Expenses $2,000 $3,000 $4,620 $8,000 $12,000
Effective income tax rate = 40% MARR (before taxes) = 10%, MARR (after taxes) = 6% Economic Life Recall that NAC(k) = (MV(0) + End of Year k 0 1 2 3 4 5
k
∑ l=1
A(l)(P/F, i, l) – MV(k)(P/F, i, k)) (A/P, i, k)
MV $20,000 $15,000 $11,250 $8,500 $6,500 $4,750
Annual Expenses
NAC(k)
$2,000 $3,000 $4,620 $8,000 $12,000
$9,000 $8,643 $8,598 $9,083 $9,954
The minimum NAC is achieved if we keep the asset 3 years. Compare against Defender
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Defender: Current Forklift Capital investment = $13,000 two years ago; for the next five years, we have Annual Expenses Estimated MV Year 0 $5,000 1 $4,000 $5,500 2 $3,000 $6,600 3 $2,000 $7,800 4 $1,000 $8,800 MARR (before taxes) = 10% Economic Life NAC(1) = $5,000 (A/P, 0.1, 1) + $5,500 – $4,000(A/F, 0.1, 1) = $5,000 (1+0.1) + $5,500 + $4,000 = $1,500 + $5,500 = $7,000
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End of Year k 0 1 2 3 4
MV $5,000 $4,000 $3,000 $2,000 $1,000
Annual expenses
NAC(k)
$5,500 $6,600 $7,800 $8,800
$7,000 $7,476 $7,967 $8,405
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The minimum NAC is achieved if we keep the asset one more year. Lessons from Example 5 • Keep the old truck at least one more year. • We could have stopped computing after one year. For this economic life the defender has a smaller NAC than the challenger. • We may keep the defender longer than one year. This analysis does not tell how long to keep it. One could compute marginal costs to get some idea on how long to keep the defender.
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Replacement Analysis Summary The opportunity cost for the defender is the money that you give up by not disposing of the defender. The opportunity cost is the current market value for the defender, less costs necessary for its disposal, less taxes on the capital gain. The investment in the defender is the opportunity cost plus any real costs that must be expended at time zero necessary to keep the defender. The investment in the challenger is just the cost of purchasing the challenger. Don't use any defender characteristics to compute the challenger investment, costs, or salvage. Don't use any challenger characteristics to compute the defender investment, costs, or salvage. Use the economic lives for both the defender and challenger. The economic life minimizes the NAC of operation (or maximizes NAW when benefits are considered). Don't forget to use the proper investment and salvage values for the calculations. In cases where the operating cost of the defender is increasing, the economic life of the defender is often one year. To simplify, compute the NAC for one year. If the defender wins on the basis of one year, keep it for one more year. Do the analysis again next year. In cases where the operating costs of the challenger are constant and the salvage value is not decreasing too rapidly, the economic life of the challenger is often the same as the useful life. When lives of the defender or challenger are different (the usual case), use NAC or NAW to make the decision. When considering taxes, use the after tax cash flows in the analysis. Note that the investment in the defender may involve taxes. There is a tax effect if the book value of the defender does not equal its net market value.
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Economic Life and Replacement Problems 1a.
The tables below show the operating cost and salvage value for a machine that was purchased for $50,000 and has a useful live of 3 years. Find its economic life using an MARR of 10%. (a) Year 1 2 3
Operating cost Salvage value $10,000 0 $40,000 0 $70,000 0
Year 1 2 3
Operating cost $10,000 $10,000 $10,000
(b) Salvage value $30,000 $20,000 0
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2. A piece of equipment was purchased one year ago for $100,000. The annual cost of operating the equipment is $20,000. We expect it to last another 5 years and this operating cost will remain constant during that time. During the year, a remarkable new process was discovered that uses waste material to run the equipment. The result is that the same output can be obtained with zero operating cost. The cost of purchasing and installing the new equipment is $200,000, and is expected to last 10 years. At that time, it will have zero salvage value. Note that under these conditions the economic life of the new equipment is 10 years. You must have either the new or old equipment. Both options perform the function with equal quality. You are now considering selling the old equipment and replacing it with the new one. There is a buyer that is willing to take the old equipment off your hands for $30,000. If you don't make the deal now, you doubt that you will find a buyer for any price in the future. After this year, you will have to pay $10,000 to get rid of the old piece of equipment. The minimum acceptable rate of return is 10%. a. What is the Net Annual Cost of the challenger? b. What is the economic life of the defender? Economic life
1
c. What is the best decision?
2
3
4
5
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3. Your company purchased a machine for $14,000 with a 6-year tax life. The sum-ofthe-years digits method is used for depreciation and the tax salvage value is zero. a. After the third year of use, the machine is sold for $10,000. How much does the company get from the sale after taxes assuming the tax rate on capital gains is 40%%. b. Neglect taxes in this part. After the third year of life, the company is thinking about replacing the machine with a new one. It can be sold now for $10,000. Next year it will only be worth $6,000 and in two years, only $4,000. Three years from now the machine will have no resale value. The operating cost of the machine is expected to be constant for the next three years at $1,000 per annum. The new machine has a life of 10 years with a NAC of $5,000. Should the old machine be replaced with the new one if the company’s MARR is 10%?
4. You are considering replacing your car with a new one. After much bickering, the dealer offers you the new car for $10,000 with your car as a trade-in or $12,000 without your car. An acquaintance offers you $3,500 for your car if you fix the air conditioner. That will cost you $1,000. The expected maintenance cost of your car for the next year is $1200 not including the air conditioner. You think you can sell it for $1,000 at this time next year without fixing the air conditioner. Assume that the life of the new car is 10 years with annual maintenance cost of $800 per year. The new car will have $2,000 salvage value at the end of the 10 years. Show the cash flows you would use in an analysis. Your MARR is 10%. What is your most economic action?
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5. A company faces the following equipment replacement problem. An existing piece of equipment is 4 years old and was originally bought for $8,000. The economic life of this equipment is 8 years and it has no salvage at the end of its useful life. The machine has been depreciated by the straight-line method and its current market value is equal to its book value. If the machine is replaced, $500 must be spent to remove it and ready it for sale. Neglect any tax effects associated with the sale or removal costs. The net annual before tax benefit of $2,000 from the existing equipment is expected to continue for another four years. Use four years as the economic life, and assume the salvage value is zero at the end of four years. The new machine being considered as a replacement costs $10,000. It has an economic life of 10 years at the end of which it can be sold for $2,000. The net annual before tax benefit expected from this machine is $3,000. Again the straightline method is used for depreciation using a tax salvage of $2,000. The tax rate is given to be 50%. The after tax MARR is 10%. Should you replace the existing machine with a new one? Show all calculations. Existing Machine (Defender) Year 0 1-4
BTCF
Depr.
TI
Tax
ATCF
BTCF
Depr.
TI
Tax
ATCF
Challenger Year 0 1-9 10
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6. A milling machine (machine A) in your company’s shop has a current market value of $30,000. It was bought nine years ago for $54,000 and has since been depreciated by the straight-line method assuming a 12-year tax life. If the decision is made to keep the machine at this point it time it can be expected to last another 12 years (measured from today). At the end of the 12 years it will be worthless. The operating costs of this machine are $7,500 per year and are not expected to change for its remaining life. Alternatively, machine A can be replaced by a smaller machine B which costs $42,000 and is expected to last 12 years. Its operating costs are $5,000 per year and would be depreciated by the straight-line method over the 12-year period with no salvage value expected. Both income and capital gains are taxed at 40%. Compare the after-tax equivalent uniform annual costs of the two machines and decide whether machine A should be retained or replaced by machine B. Use a 10% after-tax MARR in your calculations. Machine A: Year 0 1-3 4-12
BTCF
Depr.
Taxable Income
Income tax
ATCF
Depr.
Taxable income
Income tax
ATCF
Machine B: Year 0 1-12
BTCF
131
3/18/2003
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