13)
Aggie Research Laboratory purchased a new High Performance Liquid Chromatography
(HPLC) unit for $39,000. The unit has a functional life of 5 years and a salvage value of
10% of the purchased price. Using declining balance depreciation at a rate of 1.5 times the
straight line rate, determine the annual depreciation at the end of year 2.
13)
Answer:
d2=$8190.00
Explanation:
From R=1.5
N, dk=B(1 –R)k–1(R) and BVk=B(1 –R)k
R =1.5/5 =0.30
d2=39,000(1 –0.30)1(0.30)
=8190.00
14)
Seminole Lighting, a specialty lamps and specialty light sources manufacturer, had the
following information on its annual tax returns. Determine Seminole’s taxable income
and calculate the federal income tax for the year.
Sales $650,000
Interest Revenues $8500
Operating Expense $53,000
Depreciation $6500
14)
Answer:
TI =$599,000.00
Income tax =$203,660.00
Explanation:
TI = GI – Expense – Depreciation
=650,000 +8500 –53,000 –6500
=599,000.00
From Table 7.5, $335,000 < TI < $10,000,000
Income tax = 113,900 + (0.34)(599,000.00 – 335,000)
=203,660.00
11
15)
A plantation has purchased an automated propagation system for $26,000. The declining
balance depreciation at a rate of 2 times the straight line rate and a $2022 salvage value are
used to write off the capital investment. The company expects to realize net revenue of
$8000 each year for the next 5 years. Assume an effective federal tax of 39%, state income
tax of 12.25% per year, and an after–tax MARR of 4% per year. Calculate the present worth
of this investment.
15)
Answer:
PW =$5008.01
Explanation:
BTCF = – Capital Investment + GI – Expense
TI = GI – Expense – Depreciation
DB: R=2
N, dk=B(1 –R)k–1(R) and BVk=B(1 –R)k
R =2/5 =0.40
BV5=26,000(0.60)5=2022; therefore, Depreciation Recapture = 0
Effective tax rate (t) = state rate + federal rate (1– state rate)
=0.1225 + (0.39)(1 –0.1225)
=0.4647 or 46.47%
Taxes = TI(t)
ATCF = BTCF – Taxes
Year BTCF Depreciation TI Taxes ATCF
0–26,000 – – – – 26,000
18000 10,400.00 –2400.00 –1115.28 9115.28
28000 6240.00 1760.00 817.87 7182.13
38000 3744.00 4256.00 1977.76 6022.24
48000 2246.40 5753.60 2673.70 5326.30
58000 1347.84 6652.16 3091.26 4908.74
5b 2022 –0 0 2022
PW (4%) = – 26,000 +9115.28 (P/F, 4%, 1) +7182.13 (P/F, 4%, 2) +…+6930.74 (P/F,
4%, 5)
= – 26,000 +9115.28 (0.9615) +7182.13 (0.9246) +…+6930.74(0.8219)
=5008.01
16)
An inspecting and profiling web controller that costs $40,000 has a life of 8 years with a
$5000 salvage value. The estimated annual operating and maintenance cost is $3700 per
year. Use classical straight line depreciation to determine the annual depreciation at the
end of year 5.
16)
Answer:
d5=$4375.00
Explanation:
From dk= (B–SVN)/N
d5= (40,000 –5000)/8
=4375.00
12
17)
A Caribbean cruise line has purchased a new cruise ship for $670,000 and expects to realize
a net revenue of $190,000.00 each year for the next 10 years. The estimated salvage value of
the ship at the end of its useful life of 10 years is $52,000. Assume an effective federal tax
of 40%, state income tax of 10.75% per year, and an after–tax MARR of 14% per year.
Calculate the present worth of ATCF if a straight–line depreciation method is used.
17)
Answer:
PW =$17,956.05
Explanation:
BTCF = – Capital Investment + GI – Expense + Salvage Value
TI = GI – Expense – Depreciation
SL: dk= (B–SVN)/N
= (670,000 –52,000)/10
=61,800.00
Effective tax rate (t) = state rate + federal rate (1– state rate)
=0.1075 + (0.4)(1 –0.1075)
=0.4645
Taxes = TI (t)
ATCF = BTCF – Taxes
Year BTCF Depreciation TI Taxes ATCF
0–670,000.00 – – – – 670,000.00
1–9190,000.00 61,800.00 128,200.00 59,548.90 130,451.10
10 242,000.00 61,800.00 180,200.00 83,702.90 158,297.10
PW (14%) = – 670,000.00 +130,451.10 (P/A, 14%, 9) +158,297.10(P/F, 14%, 10)
= – 670,000.00 +130,451.10(4.9464) +158,297.10(0.2697)
=17,956.05
13
18)
An piece of automated assembly equipment has an initial cost of $64,000 and generates net
annual benefits of $150,000 per year. The equipment is expected to have zero salvage
value at the end of its useful life of 5 years. Using straight–line depreciation, an after–tax
MARR of 2%, a federal tax rate of 39%, and a state tax rate of 9%, determine if the
investment in this equipment is economically justifiable on the basis of the present worth
of the EVA estimates.
18)
Answer:
PW(2)% =355,303.51
The investment in the equipment is justifiable.
Explanation:
Depreciation = (64,000)/5 =12,800
BTCF = – Capital Investment + GI – Expense
TI = GI – Expense – Depreciation
Taxes = TI (t)
ATCF= BTCF – Taxes
NOPATk= TI –Taxes
EVAk=NOPATk– i BVk–1
BTCF DtTI Taxes ATCF NOPAT BV E
–64,000 – – – –64,000 064,000
150,000 12,800 137,200 61,040.28 88,959.72 76,159.72 51,200.00 74,87
150,000 12,800 137,200 61,040.28 88,959.72 76,159.72 38,400.00 75,13
150,000 12,800 137,200 61,040.28 88,959.72 76,159.72 25,600.00 75,39
150,000 12,800 137,200 61,040.28 88,959.72 76,159.72 12,800.00 75,64
150,000 12,800 137,200 61,040.28 88,959.72 76,159.72 075,90
PW (2%) =74,879.72 (P/F, 2%, 1) +75,135.72 (P/F, 2%, 2) +75,391.72 (P/F, 2%, 3) +
75,647.72 (P/F, 2%, 4) +75,903.72 (P/F, 2%, 5)
=74,879.72 (0.9804) +75,135.72 (0.9612) +75,391.72 (0.9423) +75,647.72
(0.9238) +
75,903.72 (0.9057)
=355,303.51
The investment in the equipment is justifiable.
14
19)
An engineer–to–order manufacturer is considering purchasing new equipment for its film
adhesive assembly. The initial cost of the equipment is $12,100 and annual maintenance
costs are estimated to be $185,000 per year. Annual operating costs will be in direct
proportion to the hours of use at $12 per hour. The expected annual revenue is $220,000 per
year. The equipment has a salvage value of $500 at the end of 5 years. What is the
maximum annual hours of use for which the equipment is economically justified? Use
straight–line depreciation, an effective tax rate of 39%, and an after–tax MARR of 16% per
year.
19)
Answer:
X =$2545.38
Explanation:
Depreciation = (12,100 –500)/5 =2320.00
BV =500
BTCF = – Capital Investment + GI – Expense
TI = GI – Expense – Depreciation
Taxes = TI (t)
ATCF= BTCF – Taxes
Let annual hours = X hours per year
Year BTCF DtTI Taxes ATCF
0–12,100 – – – – 12,100
1–535,000 –12X2320.00 32,680.00 –12X
12,745.20 –
4.68X22,254.80 –7.32
5500 – – – 500
AW(16%) = 0 = – 12,100(A/P, 16%, 5) +22,254.80 –7.32X +500(A/F, 16%, 5)
= – 12,100(0.3054) +22,254.80 –7.32X+500(0.1454)
=18,632.16 –7.32X
X =2545.38
20)
New spray coating equipment costs $24,000 is to be used in an oversea shipyard. It has a
12–year life and an estimated salvage value of $4750. If the global naval shipbuilding
company uses MACRS with an ADS recovery period to calculate the depreciation and
book value for the new coating equipment, determine the annual depreciation and book
value at the end of year 7.
20)
Answer:
d7=$2000.00
BV7=$11,000.00
Explanation:
MACRS with a 12–year ADS recovery period should be used.
Depreciation with SL =24,000/12 =2000.00
Therefore, depreciation at the end of year 7=2000.00
BV7=24,000 – [(0.5(2000.00)) + (6)(2000.00)]
=11,000.00
15
21)
Bulldog Shipping, Inc. has purchased new cargo containers for $500,000. MACRS with a
5–year recovery period and an estimated salvage value of $96,000 is to be used to write off
the capital investment. The company expects to realize net revenue of $170,000 each year
for the next 6 years. Assume an effective federal tax of 38%, state income tax of 10.5% per
year, and an after–tax MARR of 13% per year. Calculate the present worth of this
investment.
21)
Answer:
PW =$63,264.58
Explanation:
BTCF = – Capital Investment + GI – Expense + Salvage Value
TI = GI – Expense – Depreciation
MACRS: from Table 7.3 and dk=rkB
Effective tax rate (t) = state rate + federal rate (1 – state rate)
=0.105 + (0.13)(1 –0.105)
=0.4451 or 44.51%
Taxes = TI(t)
ATCF = BTCF – Taxes
Year BTCF rkDepreciation TI Taxes ATCF
0–500,000 – – – – – 500,000.00
1170,000 0.2 100,000.00 70,000.00 31,157.00 138,843.00
2170,000 0.32 160,000.00 10,000.00 4451.00 165,549.00
3170,000 0.192 96,000.00 74,000.00 32,937.40 137,062.60
4170,000 0.1152 57,600.00 112,400.00 50,029.24 119,970.76
5170,000 0.1152 57,600.00 112,400.00 50,029.24 119,970.76
6266,000 0.0576 28,800.00 237,200.00 105,577.72 160,422.28
PW (13%) = – 500,000 +138,843.00(P/F, 13%, 1) +165,549.00(P/F, 13%, 2) +…+
160,422.28(P/F, 13%, 6)
= – 500,000 +138,843.00(0.885) +165,549.00(0.7831) +…+160,422.28(
0.4803)
=63,264.58
22)
A nuclear power plant is planning to replace the outdated equipment with more
environmental–friendly equipment. The new equipment has an initial cost of $410,000.
The equipment is expected to yield an annual savings of $190,000 each year for the first 4
years and $191,200 each year thereafter. The MACRS with a 15–year recovery period is to
be used for tax purposes. Should the equipment be purchased if the equipment will be
sold for $148,263.00 at the end of year 10? Assume an effective tax of 38% and a before–tax
MARR of 16.13% per year.
22)
Answer:
PW =$432,628.96
PW (10%) > 0; therefore, the equipment should be purchased.
16
Explanation:
BTCF = – Capital Investment + GI – Expense + Salvage Value
From Table 7.3 , dk=rkB and BVk=B –k
i–1
di
BV10 =410,000 –
(0.05+0.095+0.0855+0.077+0.0693+0.0623+0.059+0.059+0.0591+0.0590/2)
(410,000)
=145,263.00
MV =148,263.00
Therefore, depreciation recapture =148,263.00 –145,263.00 = $3,000
TI = GI – Expense – Depreciation +Depreciation Recapture
MACRS: from Table 7.3 and dk=rkB
Taxes = TI (t)
ATCF = BTCF – Taxes
After–tax MARR = Before–tax MARR (1– Effective income tax rate)
=0.1613(1 –0.38) = 0.10 or 10%
Year BTCF rkDepreciation TI Taxes ATCF
0–410,000 – – – – – 410,000
1190,000 0.05 20,500.00 169,500.00 64,410.00 125,590.00
2190,000 0.095 38,950.00 151,050.00 57,399.00 132,601.00
… … … … … … …
10a 191,200 0.0590/2 12,095.00 179,105.00 68,059.90 123,140.10
10b 148,263.00 3000 1140.00 147,123.00
PW(10%) = – 410,000 +125,590.00 (P/F, 10%, 1) +132,601.00 (P/F, 10%, 2) + … +
[123,140.10 +147,123.00] (P/F, 10%, 10)
= – 410,000 +125,590.00 (0.9091) +132,601.00 (0.8264) + … +
[123,140.10 +147,123.00](0.3855)
=432,628.96
23)
A logistics company is deciding between two models of semi–trailer trucks to add to its
fleet. The manager has prepared the following information for the economic evaluation.
The new trucks are to be used for 7 years and sold for the estimated salvage value. The
before–tax MARR is 16.39% per year and the effective tax rate is 39%. Select a machine on
the basis of after–tax annual worth analysis using MACRS with a 5–year recovery period.
Alternative P1 P2
First costs $245,000 $230,000
Net annual benefits $155,000 $140,000
Salvage value $44,500 $37,500
Life, years 10 10
23)
Answer:
AWP1 (10%) =$62,261.12
AWP2 (10%) =$54,813.17
P1 should be selected.
17
Explanation:
BTCF = – Capital Investment + GI – Expense
MACRS: from Table 7.3 and dk=rkB
BV7= 0; therefore, Depreciation recapture = salvage value
TI = GI – Expense – Depreciation + Depreciation recapture
Taxes = TI (t)
ATCF = BTCF – Taxes
After–tax MARR = Before–tax MARR (1– Effective income tax rate)
=0.16391 –0.39) = 0.10 or 10%
Alternative P1:
Year BTCF rkDepreciation TI Taxes ATCF
0–245,000 – – – – – 245,000
1155,000 0.2 49,000.00 106,000.00 41,340.00 113,660
2155,000 0.32 78,400.00 76,600.00 29,874.00 125,126
… … … … … … …
6155,000 0.0576 14,112.00 140,888.00 54,946.32 100,053.68
7199,500 – – 199,500 77,805.00 121,695
AWP1 (10%) = [–245,000 +113,660 (P/F, 10%, 1) +125,126 (P/F, 10%, 2) + …
+100,053.68 (P/F, 10%, 6) +121,695 (P/F, 10%, 7)] (A/P, 10%, 7)
= [–245,000 +113,660 (0.9091) +125,126 (0.8264) +…+121,695
(0.5132)](0.2054)
=62,261.12
Alternative P2:
Year BTCF rkDepreciation TI Taxes ATCF
0–230,000 – – – – – 230,000
1140,000 0.2 46,000.00 94,000.00 36,660.00 103,340
2140,000 0.32 73,600.00 66,400.00 25,896.00 114,104
… … … … … … …
6140,000 0.0576 13,248.00 126,752.00 49,433.28 90,566.72
7177,500 – – 177,500 69,225.00 108,275.00
AWP2 (10%) = [–230,000 +103,340 (P/F, 10%, 1) +114,104 (P/F, 10%, 2) + …
+108,275.00 (P/F, 10%, 7)] (A/P, 10%, 7)
= [–230,000 +103,340 (0.9091) +114,104 (0.8264) + …+108,275.00
(0.5132)](0.2054)
=54,813.17
AWP1 >AWP2 > 0; therefore, P1 should be selected.
18