# 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)

## Presentation on theme: "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)"— Presentation transcript:

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)

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) =

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?

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

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

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

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

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 = (280,000-90,000)(P/A,10%,5) = 190,000(3

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: /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} =

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 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.

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

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

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
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
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)

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?

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)

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?

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)

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
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

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
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

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)

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%

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)

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( ) = \$60, Answer is (a)

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

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)

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 Answer is (d)

Similar presentations