Chapter 2, Problem 1. Find the correct numerical value for the following factors from the interest...

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


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?


F = 200,000(F/P,10%,3) = 200,000(1.3310)

= $266,200


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?


F = 150,000(F/P,18%,7) = 150,000(3.1855)

= $477,825


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?


P = 162,000(P/F,12%,6) = 162,000(0.5066)

= $82,069


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?


F = 65,000(F/P,4%,5)

= 65,000(1.2167) = $79,086


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.


P =

(280,000-90,000)(P/A,10%,5) = 190,000(3.7908)

= $720,252


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


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 advertising, with amounts decreasing by

$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.


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


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?


For

g = i, P = 60,000(0.1)[15/(1 + 0.04)]

= $86,538


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?


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


What compound interest rate per year is equivalent to a 12% per year simple interest rate over a

15-year period?


Simple: Total interest = P(0.12)(15) = 1.8P

Compound: Total interest = P(1-i)^15-PTotal(simp.)=Total(comp.)

P(1-i)^15-P = 1.8P(1-i)^15=2.8

i = 7.11% (compounding rate)


An investment of $600,000

increased to $1,000,000 over a 5-year period. What was the rate of

return on the investment?


1,000,000 = 600,000(F/P,i,5)

(F/P,i,5) = 1.6667 i = 10.8% (Excel)


A new company that makes medium-voltage 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?


85,000 = 30,000(P/A,i,5) +

15,000(P/G,i,5)Solve for i by trial and error or

spreadsheet:i = 50% (Excel)


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?


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


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?


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


160 = 235(P/F,i,5)(P/F,i,5) =0.6809

From tables, i = 8%Answer is (c)


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%


109.355 = 7(P/A,i,25)(P/A,i,25) = 15.6221From tables, i = 4%

Answer is (a)

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


P = 8000(P/A,10%,10) +

500(P/G,10%,10) = 8000(6.1446) + 500(22.8913)

= $60,602.45Answer is (a)


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


A = 100,000(A/F,12%,3)

= 100,000(0.29635) = $29,635Answer is (c)

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


A = 10,000(F/A,12%,25)

= 10,000(133.3339) = $1,333,339Answer is (d)

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


P = 8,000(P/A,10%,5) + 900(P/G,10%,5)

= 8,000(3.7908) + 900(6.8618) = $36,502

Answer is (d)

Chapter 2, Problem 1. Find the correct numerical value for the following factors from the interest...

Documents

Transcript of Chapter 2, Problem 1. Find the correct numerical value for the following factors from the interest...