Exam
Name___________________________________
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Answer the question.
1)
Find the annual worth equivalent in actual–dollars from year 1 to 8 of an investment of
$28,000 now, if the inflation rate is 5.5% and the inflation–free interest rate is 15% per year.
1)
Answer:
$7588.00
Explanation:
f =5.5% per year; ir=15% per year
im=0.15 +0.055+ (0.15)(0.055)
=0.2133, or 21.33% per year
A =$28,000(A/P, 21.33%, 8)
=$28,000(0.2710)
=$7588.00
2)
Buckeye Manufacturing expects to generate additional revenue from its recently won
government contract. Buckeye forecasts that the revenue will be $40 million in the first
year, but will decline by $10.5 million every year for the next 3 years. What is the future
worth of total revenue at the end of 3 years in actual dollars if the real interest rate is 2%
per year and the average rate of inflation is 11.25% per year?
2)
Answer:
$103,988,036.03
Explanation:
f =11.25% per year; ir=2% per year
im=0.02 +0.1125 + (0.02)(0.1125)
=0.1348, or 13.48%
P =40,000,000 (P/A, 13.48%, 3) –10,500,000 (P/G, 13.48%, 3)
=71,156,450.00
F = P (F/P, 13.48%, 3)
=$71,156,450.00(1.4614)
=$103,988,036.03
1
3)
Aggie Satellite Corp. issued 10,000 debenture bonds with a face value of $32,000 each and a
bond interest rate of 14% per year, payable quarterly. The bonds have a maturity date of
13 years. If the inflation–free interest rate is 17% per year, and the inflation rate is 7.5%,
what is the present worth of one bond to a person who wants to purchase it?
3)
Answer:
$19,642.08
Explanation:
f =7.5% per year; ir=17% per year
im=0.17 +0.075+ (0.17)(0.075)
=0.2578, or 25.78% per year
From VN=C(P/F, i%, N) +rZ(P/A, i%, N)
Where (1 + i)4– 1 =0.2578
i =0.0590 or 5.90% per quarter
r= (14%)/4= (0.035%)
N =52
Thus, the present worth of the bond
=32,000(P/F, 5.90%, 52) +0.035(32,000)(P/A, 5.90%, 52)
=1622.40 +18,019.68
=$19,642.08
2
4)
A dentist is deciding between two X–ray machines for his new office. Estimated costs for
each machine are given below.
Machine A B
Installed cost $50,000 $52,000
Annual maintenance cost $1500 $1200
Market value at year 10 $5750 $6250
Life, years 10 10
Which machine should be recommended based on the annual worth method and an
actual–dollar analysis? Use an inflation–free MARR of 14%, an inflation rate of 6.5% per
year, and a study period of 10 years.
4)
Answer:
AW(A) = –$13,793.57
AW(B) = –$13,975.62
Therefore machine A should be recommended.
Explanation:
f =6.5% per year; ir=14% per year
im=0.14 +0.065+ (0.14)(0.065)
=0.2141, or 21.41%
AW(A) = –50,000(A/P, 21.41%, 10) –1500 +5750 (A/F, 21.41%, 10)
= –50,000(0.2500) –1500 +5750(0.0359)
= –13,793.57
AW(B) = –52,000(A/P, 21.41%, 10) –1200 +6250(A/F, 21.41%, 10)
= –52,000(0.2500) –1200 +6250(0.0359)
= –13,975.62
AW(A) > AW(B); therefore, machine A should be recommended.
3
5)
TransAtlantic Petroleum Corp. plans to seek two additional production licenses from the
Romanian government. The company estimates the costs to drill, acquire the seismic data,
and conduct technical studies for these new fields will be $14.2 billion, 7 years from now.
The company plans to establish a fund in a Romanian bank. The fund earns a rate of
return of 5% per year (a rate relative to the Romanian New Lei– RON). How much will the
company have to set aside now in U.S. dollars if it is estimated that U.S. dollars will be
devalued at an average of 2.25% per year and the present exchange rate is 2.57 RON per
U.S. dollar?
5)
Answer:
$8,603,780,000.00
Explanation:
ifm =5% per year; fe= –2.25% per year
Current exchange rate = $1 per 2.57 RON
ifm =0.05 =ius –0.0225 – (0.0225)ius
0.9775ius =0.0725,
ius =0.0742 or 7.42%,
P =$14,200,000,000(P/F, 7.42%, 7)
=$8,603,780,000.00
6)
Falcon Industries set aside $21,500 (now) in an investment account that earns 6.5% per
year, compounded monthly. Determine the future worth equivalent in real–dollars of this
investment 12 years from now, if the inflation rate is 3.5% per year.
6)
Answer:
$30,984.42
Explanation:
Effective interest rate =(1 +r/M)M=(1+ (0.065/12))12 –1
=0.0670 or 6.70% per year
FW in actual dollar =$21,500(F/P, 6.70%, 12)
=$21,500(2.1776)
=$46,818.40
If we assume the base year to be the present (b = 0), then:
FW in real dollars =$46,818.40(P/F, 3.5%, 12)
=$46,818.40(0.6618)
=$30,984.42
7)
Determine the real rate of return of an investment that yields 10.5% per year when the
inflation rate is 8.5% per year.
7)
Answer:
1.84%
Explanation:
im=10.5% per year; f =8.5% per year
im=0.105 =ir+0.085+ (0.085)ir
1.0850r =0.0200,
ir=0.0184, or 1.84% per year
4
8)
How much would equipment cost 7 years from now, if the cost today is $1000 and the
deflation rate is 14.75% per year?
8)
Answer:
$327.20
Explanation:
Price =$1000(1–0.1475)7
=$1000(0.3272)
=$327.20
9)
Two advanced thermal insulating and anti–condensation protection alternatives have been
proposed for new Antarctica marine vessels subject to the harsh marine environment. One
alternative must be selected. Estimated savings from reduced total installation and
maintenance costs over conventional insulation are the following:
Alternative Delta–TAlpha–B
Installed cost $44,000 $46,000
Annual savings $3750 $5250
Life, years 7 7
Which alternative should be recommended based on the present worth method and an
actual–dollar analysis? Use an inflation–free MARR of 8%, an inflation rate of 8% per
year, and a study period of 7 years. Assume negligible salvage values.
9)
Answer:
PW(Delta–T) = –$29,136.50
PW(Alpha–B) = –$25,191.10
Therefore Alpha–B should be recommended.
Explanation:
f =8% per year; ir=8% per year
im=0.08 +0.08 + (0.08)(0.08)
=0.1664, or 16.64%
PW(Delta–T) = –44,000 +3750(P/A, 16.64%, 7)
= –44,000 +3750(3.9636)
= –29,136.50
PW(Alpha–B) = –46,000 +5250(P/A, 16.64%, 7)
= –46,000 +5250(3.9636)
= –25,191.10
PW(Alpha–B) > PW(Delta–T); therefore, Alpha–B should be recommended.
10)
If the Chinese Yuan is worth 14.3045 Japanese Yen and the Yen trades for 0.0091 U.S.
dollars, determine how much $400 is worth in Chinese Yuan.
Answer:
3072.88 Chinese Yuan
Explanation:
(14.3045Yen/Yuan)($0.0091 US/1 Yen) =$0.1302 US/Yuan
Thus, $400 is worth 3072.88 Yuan.
5
11)
A Caribbean cruise line has purchased a new cruise ship for $35 million and expects to
realize a net revenue of $210,000 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 $134,000. Assume an effective
federal tax rate of 40%, a state income tax rate of 7% per year, and an after–tax
inflation–free MARR of 8% per year. Calculate the actual dollar present worth of ATCF if
straight–line depreciation is used and the average rate of inflation is 4% per year.
Answer:
–$25,709,971.00
Explanation:
SL: dk= (B–SVN)/N
= (35,000,000 –134,000)/10
=3,486,600.00
Effective tax rate (t) = state rate + federal rate (1–state rate)
=0.07 + (0.4)(1 –0.07)
=0.4420
Year BTCF Depreciation TI Taxes ATCF
0–35,000,000 – – – –35,000,000
1–10 210,000 3,486,600.00 –3,276,600.00 –1,448,257.20 1,658,257.20
10b 134,000 – – – 134,000
f =4% per year; ir=8% per year
im=0.08 +0.04 + (0.08)(0.04)
=0.1232, or 12.32%
PW (12.32%) = –35,000,000 +1,658,257.20 (P/A, 12.32%, 10) +134,000(P/F,
12.32%, 10) = –35,000,000 +1,658,257.20(5.5770) +134,000(0.3129)
= –$25,709,971.00
6
12)
An engineer wants to estimate the annual inflation–free cost of owning an aerobic digester
system with a capacity of 195 million gallons per day (MGD) for the first 8 years of
operation. Company records show that the cost of a similar system with a capacity of 75
MGD was $8 million five years ago. The equipment cost index has increased 26% per year
since then, and the future general inflation rate is forecast to be 2% per year. He estimates
that the annual operating expenses of the new system would be $440,000 per year for the
first two years and increase to $441,500 per year thereafter, due to the increase in
maintenance costs. Calculate the annual inflation–free cost of owning the 295–MGD
system for the first 8 years. Use a cost–capacity exponent of 0.14 for the system and a
market–based MARR of 8% per year.
Answer:
–$5,061,832.12
Explanation:
Using the power sizing technique with an adjustment for the price increase in
equipment costs, the initial cost of the system now is:
CA=CBSA
SB
x(1 + ec)8
= (8,000,000) 195
75 0.14 (1.26)5
=$29,042,833.25
PW(8%) = –29,042,833.25 –440,000(P/A, 8%, 2) –441,500(P/A, 8%, 6)(P/F, 8%, 2)
= –29,042,833.25 –440,000(1.7833) –441,500(4.6229)(0.8573)
= –29,042,833.25 –784,652.00 –1,749,758.17
= –$31,577,243.42
im=8% per year; f =2% per year
im=0.08 =ir+0.02 + (0.02)ir
1.0200r =0.0600,
ir=0.0588 or 5.88%,
If we assume the base year to be the present (b = 0), then the annual
inflation–free cost of owning the 195–MGD system for the first 8 years is:
AW(5.88%) = –$31,577,243.42 (A/P, 5.88%,8)
= –$31,577,243.42(0.1603)
= –$5,061,832.12 per year
13)
If the real interest rate is 7% per year and the inflation rate is 36.75% per year, what is the
inflation–adjusted interest rate?
Answer:
46.32%
Explanation:
f =36.75% per year; ir=7% per year
im=0.07 +0.3675 + (0.07)(0.3675)
=0.4632, or 46.32% per year
7
14)
Healthcare costs have risen an average of 85% per year over the past 5 years. What is the
average annual percentage increase over time? If the inflation rate was 7% per year, how
many percentage points over the inflation rate was the annual healthcare cost increase?
Answer:
13.09%; 6.09%
Explanation:
At a 85% increase, $1 would increase to $1.85.
Let x = average annual increase
1.85 =(1 + x)5
(1 + x) =1.850.20
x =0.1309 or 13.09%
Thus, percentage point over the inflation rate =13.09% –7%
=6.09%
15)
Calculate the implied annual inflation rate from a market interest of 15.75% per year and a
real interest rate of 15.67% per year.
Answer:
0.07%
Explanation:
im=15.75% per year; ir=15.67% per year
im=0.1575 =0.1567 + f + (0.1567)f
1.1567f =0.0008,
f =0.0007, or 0.07% per year
8
Answer Key
Testname: C8
1)
$7588.00
2)
$103,988,036.03
3)
$19,642.08
4)
AW(A) = –$13,793.57
AW(B) = –$13,975.62
Therefore machine A should be recommended.
5)
$8,603,780,000.00
6)
$30,984.42
7)
1.84%
8)
$327.20
9)
PW(Delta–T) = –$29,136.50
PW(Alpha–B) = –$25,191.10
Therefore Alpha–B should be recommended.
10)
3072.88 Chinese Yuan
11)
–$25,709,971.00
12)
–$5,061,832.12
13)
46.32%
14)
13.09%; 6.09%
15)
0.07%