### Chapter 2, Problem 1. Find the correct numerical value for the

```Chapter 2, Problem 1.
Find the correct numerical value for
the following factors from the
interest tables.
1. (F/P,8%,25)
2. (P/A,3%,8)
3. (P/G,9%,20)
4. (F/A,15%,18)
5. (A/P,30%,15)
Chapter 2, Solution 1.
1. (F/P,8%25) = 6.8485
2. (P/A,3%,8) = 7.0197
3. (P/G,9%,20) = 61.7770
4. (F/A,15%,18) = 75.8364
5. (A/P,30%,15) = 0.30598
Chapter 2, Problem 3.
Pressure Systems, Inc., manufactures
high-accuracy liquid-level transducers. It
is investigating whether it should update
certain equipment now or wait to do it
later. If the cost now is \$200,000, what
will the equivalent amount be 3 years
from now at an interest rate of 10% per
year?
Chapter 2, Solution 3.
F = 200,000(F/P,10%,3)
= 200,000(1.3310)
= \$266,200
Chapter 2, Problem 6.
Thompson Mechanical Products is planning to
set aside \$150,000 now for possibly replacing
its large synchronous refiner motors whenever
it becomes necessary. If the replacement isn’t
needed for 7 years, how much will the
company have in its investment set-aside
account if it achieves a rate of return of 18%
per year?
Chapter 2, Solution 6.
F = 150,000(F/P,18%,7)
= 150,000(3.1855)
= \$477,825
Chapter 2, Problem 10.
What is the present worth of a
future cost of \$162,000 to
Corning, Inc., 6 years from now
at an interest rate of 12% per
year?
Chapter 2, Solution 10.
P = 162,000(P/F,12%,6)
= 162,000(0.5066)
= \$82,069
Chapter 2, Problem 14.
The current cost of liability
insurance for a certain consulting
firm is \$65,000. If the insurance cost
is expected to increase by 4% each
year, what will be the cost 5 years
from now?
Chapter 2, Solution 14.
F = 65,000(F/P,4%,5)
= 65,000(1.2167)
= \$79,086
Chapter 2, Problem 18.
How much money could RTT Environmental
Services borrow to finance a site reclamation
project if it expects revenues of \$280,000 per
year over a 5-year cleanup period? Expenses
associated with the project are expected to be
\$90,000 per year. Assume the interest rate is
10% per year.
Chapter 2, Solution 18.
P = (280,00090,000)(P/A,10%,5)
= 190,000(3.7908)
= \$720,252
Chapter 2, Problem 25.
Find the numerical value of the
following factors by (a) interpolation and
(b) using the appropriate formula.
1. (P/F,18%,33)
2. (A/G,12%,54)
Chapter 2, Solution 25.
(a) 1. Interpolate between n = 32 and n = 34:
1/2 = x/0.0014
x = 0.0007
(P/F,18%,33) = 0.0050 – 0.0007 = 0.0043
2. Interpolate between n = 50 and n = 55:
4/5 = x/0.0654
x = 0.05232
(A/G,12%,54) = 8.1597 + 0.05232 = 8.2120
(b) 1. (P/F,18%,33) = 1/(1+0.18)33 = 0.0042
2. (A/G,12%,54) = {(1/0.12) – 54/[(1+0.12)54 –1}
= 8.2143
Chapter 2, Problem 27.
A cash flow sequence starts in year 1 at
\$3000 and decreases by \$200 each year
through year 10. (a) Determine the value
of the gradient G; (b) determine the
amount of cash flow in year 8; and (c)
determine the value of n for the gradient.
Chapter 2, Solution 27
(a) G = \$200 (b) CF8 = \$1600
(c) n = 10
Chapter 2, Problem 38.
A start-up direct marketer of car parts expects
to spend \$1 million the first year for
\$100,000 each year. Income is expected to be
\$4 million the first year, increasing by
\$500,000 each year. Determine the equivalent
annual worth in years 1 through 5 of the
company’s net cash flow at an interest rate of
16% per year.
Chapter 2, Solution 38.
A = [4 + 0.5(A/G,16%,5)] – [1 –
0.1(A/G,16%,5)
= [4 + 0.5(1.7060)] – [1 –0.1(1.7060)]
= \$4,023,600
or; find the PW of inflows and Outflows then
calculate corresponding A as I demonstrated this
solution method in class
Chapter 2, Problem 40.
A chemical engineer planning for her
retirement will deposit 10% of her salary each
year into a high-technology stock fund. If her
salary this year is \$60,000 (i.e., end of year 1)
and she expects her salary to increase by 4%
each year, what will be the present worth of
the fund after 15 years if it earns 4% per year?
Chapter 2, Solution 40.
For
g = i,
P = 60,000(0.1)[15/(1 + 0.04)]
= \$86,538
Chapter 2, Problem 47.
A northern California consulting firm wants to
start saving money for replacement of
network servers. If the company invests
\$3000 at the end of year 1 and increases the
amount invested by 5% each year, how much
will be in the account 4 years from now if it
earns interest at a rate of 8% per year?
Chapter 2, Solution 47.
Find P and then convert to F.
P = 3000{1 – [(1+0.05)4/(1+0.08)4}]}/(0.08 –0.05)
= 3000{3.5522}
= \$10,657
F = 10,657(F/P,8%,4)
= 10,657(1.3605)
= \$14,498
Chapter 2, Problem 49.
What compound interest rate per
year is equivalent to a 12% per
year simple interest rate over a
15-year period?
Chapter 2, Solution 49.
Simple: Total interest = P(0.12)(15) = 1.8P
Compound: Total interest = P(1-i)^15-P
Total(simp.)=Total(comp.)
P(1-i)^15-P = 1.8P
(1-i)^15=2.8
i = 7.11%
(compounding rate)
Chapter 2, Problem 52.
An investment of \$600,000
increased to \$1,000,000 over a 5year period. What was the rate of
return on the investment?
Chapter 2, Solution 52.
1,000,000 = 600,000(F/P,i,5)
(F/P,i,5) = 1.6667
i = 10.8% (Excel)
Chapter 2, Problem 55.
A new company that makes mediumvoltage soft starters spent \$85,000 to
build a new website. Net income was
\$30,000 the first year, increasing by
\$15,000 each year. What rate of return
did the company make in its first 5
years?
Chapter 2, Solution 55.
85,000 = 30,000(P/A,i,5) +
15,000(P/G,i,5)
Solve for i by trial and error or
i = 50%
(Excel)
Chapter 2, Problem 58.
An engineer who invested very well
plans to retire now because she has
\$2,000,000 in her ORP account. How
long will she be able to withdraw
\$100,000 per year (beginning 1 year
from now) if her account earns interest at
a rate of 4% per year?
Chapter 2, Solution 58.
2,000,000 = 100,000(P/A,4%,n)
(P/A,4%,n) = 20.000
From 4% table, n is between 40 and
45 years; by spreadsheet, 42 > n > 41
Therefore, n = 41 years
Chapter 2, Problem 61.
How many years will it take for a
uniform annual deposit of size A
to accumulate to 10 times the size
of a single deposit if the rate of
return is 10% per year?
Chapter 2, Problem 61.
10A = A(F/A,10%,n)
(F/A,10%,n) = 10.000
From 10% table, n is between 7 and
8 years; therefore, n = 8 years
Chapter 2, Solution 65.
160 = 235(P/F,i,5)
(P/F,i,5) =0.6809
From tables, i = 8%
Chapter 2, Problem 67.
The winner of a multistate megamillions lottery
jackpot worth \$175 million was given the option of
taking payments of \$7 million per year for 25 years,
beginning 1 year now, or taking \$109.355 million
now. At what interest rate are the two options
equivalent to each other?
(a) 4%
(b) 5%
(c) 6%
(d) 7%
Chapter 2, Solution 67.
109.355 = 7(P/A,i,25)
(P/A,i,25) = 15.6221
From tables, i = 4%
Chapter 2, Problem 70.
An engineer deposits \$8000 in year 1, \$8500
in year 2, and amounts increasing by \$500 per
year through year 10. At an interest rate of
10% per year, the present worth in year 0 is
closest to
(a) \$60,600
(b) \$98,300
(c) \$157,200
(d) \$173,400
Chapter 2, Solution 70.
P = 8000(P/A,10%,10) +
500(P/G,10%,10)
= 8000(6.1446) + 500(22.8913)
= \$60,602.45
Chapter 2, Problem 77.
Simpson Electronics wants to have \$100,000
available in 3 years to replace a production line. The
amount of money that would have to be deposited
each year at an interest rate of 12% per year would
be closest to
(a) \$22,580
(b) \$23,380
(c) \$29,640
(d) Over \$30,000
Chapter 2, Solution 77.
A = 100,000(A/F,12%,3)
= 100,000(0.29635)
= \$29,635
Chapter 2, Problem 78.
A civil engineer deposits \$10,000 per year
into a retirement account that achieves a rate
of return of 12% per year. The amount of
money in the account at the end of 25 years is
closest to
(a) \$670,500
(b) \$902,800
(c) \$1,180,900
(d) \$1,333,300
Chapter 2, Solution 78.
A = 10,000(F/A,12%,25)
= 10,000(133.3339)
= \$1,333,339
Chapter 2, Problem 80.
Maintenance costs for a regenerative thermal
oxidizer have been increasing uniformly for 5
years. If the cost in year 1 was \$8000 and it
increased by \$900 per year through year 5, the
present worth of the costs at an interest rate of
10% per year is closest to
(a) \$31,670
(b) \$33,520
(c) \$34,140
(d) Over \$36,000
Chapter 2, Solution 80.
P = 8,000(P/A,10%,5) +
900(P/G,10%,5)
= 8,000(3.7908) + 900(6.8618)
= \$36,502