SOLUTIONS TO CHAPTER 7 PROBLEMS
Consideration of Depreciation and Income Taxes
7–1. B = $40,000, Life = 10 years, SV10 = $4,000
(a) d5 = ($40,000-$4,000)/10 = $3,600
7-2. The cost basis for the machine is
B = $25,000 + $500 + $300 = $25,800
and the estimated salvage at N = 8 is
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7–3. B=$10,000 SV5 = $1,000
7-4. The depreciation charge per unit produced is
($25,000-$5,000)/100,000 = $0.20
Thus d3* = 60,000•$0.20 = $12,000
and BV3=$25,000 -$12,000=$13,000
d4 =10,000• $0.20 = $2000
BV4 = BV3 – d4 = $13,000 – $2,000 = $11,000
$10k
BV
($)
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7-5. R = 2/12 = 0.167
BV3 = $15,000(1- 0.167)3 = $8,700
d(straight line method) = 412
−
7-6. A s s u m i n g : ADR= 14 years, therefore MACRS recovery period = 7 years
Basis = $60,000, ES = $12,000. Find d3 and BV5
(a) d3 = dk 57.3428$
14
000,12000,60$ =
−
=
−
=N
ESB
)000,48($5
)(5 =−=
−
ESB
7–7. (a) Use MACRS, 5 year recovery period, no salvage allowance,
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7–8. (a) Assuming B = $150,000, purchased and put in use in 2005 and ADR = 9
years, then the MACRS recovery period would be 5 years.
(b) In 2008 there was a $17.280 depreciation deduction:
Depreciation End-of-Year
Deduction BookValue
2005 $150
,
000
(
0.2
)
$30
,
000 $120
,
000
7-9. B=$6,400 ADR= 9 years ES= $1,000 MV6 = $1,500
7-10. Assume the tank is placed in service in 2005, a MACRS recovery period of 5 years
and a cost basis =$99,500+$15,000(trade in) =$114,500
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7-11.
(1)
Year
(2)
Before-Tax
Cash Flow
(3)
Depreciation
for Tax
Purposes
(4)=(2)+(3) (5)=-(4)xRate
Taxable Cash Flow for
Income Income Taxes
(6)=(2)+(5)
After Tax
Cash Flow
7-12. Taxable income = gross income
– all expenses excluding capital expenditures
– depreciation deductions
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7-13. Before investment:
T.I. Rate Tax
$6,000 $40,000
15%
7-14. Income Tax $ 5,000 @ 25% = $1,250
$25,000 @ 34% = $8,500
Total $9,750
t = $9,750/$30,000 = 0.325
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Use PW Method
– assume alt. MACRS not used
– assume effective income tax rate = 40%
EOY ATCF
(P/F,15%,N)
0 -100,000.00
1.00
1 26,000.00
2 26,000.00
3 26,000.00
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7-15. MACHINE A:
YR BTCF Deprec. TI T ATCF
0 -$20,000 – – – $20,000
1-12 $12,000 $1,333.33 $10,666.67 –$4,266.67 $ 7,733.33
12 $ 4,000 – 0 0 $ 4,000
0.1468 0.0468
(a) AW = -20,000(A/P,10°%,12) + 7733.33 + 4,000(A/F,10%,12) = $4984.53
7-16. (a) Straight-line depreciation:
Method I
EOY BTCF Deprec. TI IT ATCF
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MACRS Depreciation
(b)
Method I
Assume that MV5 is $1,000.
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Method II
The MACRS class life is 7 years. Assume that MV10 = $5,000.
EOY BTCF Depre TI IT ATCF
0 -40,000 -40,000
1 -7,000 5,716 -12,716 5,086.40 -1,913.60
2 -7,000 9,796 -16,796 6,7113.4
0
-281.60
AW over the useful life of 10 years: AW(12%) = -51,1369.67 • (A/P,12%,10)
= -9,180.11
Thus, Method II is chosen also in this case.
7-17. Manufacturing designed for varying degrees of automation.
Deqree 1st Cost Annual Labor Cost
Annual Power &
Maintenance Cost
A $10,000 $9,000 $ 500
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Straight Line Depreciation: ($10,000-0)/5 = $2,000
(C)
EOY BTCF DEPN TI TAXES ATCF
0 –
$
20
,
000 –
$
20
,
000
1-5 -$6,000 $4,000 -$10,000 $4,000 -$2,000
Depreciation: ($20,000-0)/5 = $4,000
(a)
Annual Worth
.2983
AW A = -10, 000 (A/P,15%,5) – 4,900
AWA = -$7,883
.2983
AWB = –14,000(A/P,15%,5) – 3,860
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Select to automate to Degree A.
(c) IRR
Deqree
A B C D
First Cost 10,000 14,000 20,000 30,000
Annual Cost 4,900 3,860 2,000 300
B-A
-4,000 + 1,040(P/A,i’,5) = 0
i’ ≅ 9.431%
Since i'< 15% MARR, reject B
C-A
-10,000 + 2900(P/A,i’,5) = 0
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