Download presentation

Published byJarred Scotton Modified over 6 years ago

1
Chapter 2, Problem 1. Find the correct numerical value for the following factors from the interest tables (F/P,8%,25) 2. (P/A,3%,8) (P/G,9%,20) (F/A,15%,18) (A/P,30%,15)

2
**Chapter 2, Solution 1. 1. (F/P,8%25) = 6. 8485 2. (P/A,3%,8) = 7**

Chapter 2, Solution 1. 1. (F/P,8%25) = (P/A,3%,8) = (P/G,9%,20) = (F/A,15%,18) = (A/P,30%,15) =

3
**Chapter 2, Problem 3. Pressure Systems, Inc**

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?

4
**Chapter 2, Solution 3. F = 200,000(F/P,10%,3) = 200,000(1**

5
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?

6
**Chapter 2, Solution 6. F = 150,000(F/P,18%,7) = 150,000(3**

7
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?

8
**Chapter 2, Solution 10. P = 162,000(P/F,12%,6) = 162,000(0**

9
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?

10
**Chapter 2, Solution 14. F = 65,000(F/P,4%,5) = 65,000(1**

11
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.

12
**Chapter 2, Solution 18. P = (280,000-90,000)(P/A,10%,5) = 190,000(3**

13
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)

14
Chapter 2, Solution 25. (a) 1. Interpolate between n = 32 and n = 34: /2 = x/ x = (P/F,18%,33) = – = Interpolate between n = 50 and n = 55: /5 = x/ x = (A/G,12%,54) = = (b) 1. (P/F,18%,33) = 1/(1+0.18)33 = (A/G,12%,54) = {(1/0.12) – 54/[(1+0.12)54 –1} =

15
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.

16
**Chapter 2, Solution 27 (a) G = $200 (b) CF8 = $1600 (c) n = 10**

17
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.

18
**Chapter 2, Solution 38. A = [4 + 0. 5(A/G,16%,5)] – [1 –0**

Chapter 2, Solution 38. A = [ (A/G,16%,5)] – [1 –0.1(A/G,16%,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

19
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?

20
**Chapter 2, Solution 40. For g = i, P = 60,000(0. 1)[15/(1 + 0**

21
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?

22
**Chapter 2, Solution 47. Find P and then convert to F**

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

23
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?

24
**Chapter 2, Solution 49. Simple: Total interest = P(0. 12)(15) = 1**

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= i = 7.11% (compounding rate)

25
Chapter 2, Problem 52. An investment of $600,000 increased to $1,000,000 over a 5-year period. What was the rate of return on the investment?

26
**Chapter 2, Solution 52. 1,000,000 = 600,000(F/P,i,5) (F/P,i,5) = 1**

Chapter 2, Solution 52. 1,000,000 = 600,000(F/P,i,5) (F/P,i,5) = i = 10.8% (Excel)

27
Chapter 2, Problem 55. 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?

28
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 spreadsheet: i = 50% (Excel)

29
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?

30
**Chapter 2, Solution 58. 2,000,000 = 100,000(P/A,4%,n) (P/A,4%,n) = 20**

Chapter 2, Solution 58. 2,000,000 = 100,000(P/A,4%,n) (P/A,4%,n) = From 4% table, n is between 40 and 45 years; by spreadsheet, 42 > n > 41 Therefore, n = 41 years

31
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?

32
**Chapter 2, Problem 61. 10A = A(F/A,10%,n) (F/A,10%,n) = 10**

Chapter 2, Problem A = A(F/A,10%,n) (F/A,10%,n) = From 10% table, n is between 7 and 8 years; therefore, n = 8 years

33
**Chapter 2, Solution 65. 160 = 235(P/F,i,5) (P/F,i,5) =0**

Chapter 2, Solution 65. 160 = 235(P/F,i,5) (P/F,i,5) = From tables, i = 8% Answer is (c)

34
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 $ million now. At what interest rate are the two options equivalent to each other? (a) 4% (b) 5% (c) 6% (d) 7%

35
**Chapter 2, Solution 67. 109. 355 = 7(P/A,i,25) (P/A,i,25) = 15**

Chapter 2, Solution 67. = 7(P/A,i,25) (P/A,i,25) = From tables, i = 4% Answer is (a)

36
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

37
Chapter 2, Solution 70. P = 8000(P/A,10%,10) + 500(P/G,10%,10) = 8000(6.1446) + 500( ) = $60, Answer is (a)

38
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

39
**Chapter 2, Solution 77. A = 100,000(A/F,12%,3) = 100,000(0**

Chapter 2, Solution 77. A = 100,000(A/F,12%,3) = 100,000( ) = $29,635 Answer is (c)

40
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

41
**Chapter 2, Solution 78. A = 10,000(F/A,12%,25) = 10,000(133**

Chapter 2, Solution 78. A = 10,000(F/A,12%,25) = 10,000( ) = $1,333,339 Answer is (d)

42
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

43
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 Answer is (d)

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google