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Chapter 2, Solution 3. F = 200,000(F/P,10%,3) = 200,000(1.3310) = $266,200

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).

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Chapter 2, Solution 3. F = 200,000(F/P,10%,3) = 200,000(1.3310) = $266,200

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  1. 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)

  2. 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

  3. 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(1.3310) = $266,200

  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(3.1855) = $477,825

  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(0.5066) = $82,069

  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(1.2167) = $79,086

  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(3.7908) = $720,252

  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: 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

  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 –0.1(A/G,16%,5) = [4 + 0.5(1.7060)] – [1 –0.1(1.7060)] = $4,023,600or; 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 + 0.04)] = $86,538

  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.P = 3000{1 – [(1+0.05)4/(1+0.08)4}]}/(0.08 –0.05) = 3000{3.5522} = $10,657F = 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) = 1.8PCompound: 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)

  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) = 1.6667 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) = 20.000From 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) = 10.000From 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) =0.6809From 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 $109.355 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) = 15.6221From 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(22.8913) = $60,602.45Answer 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(0.29635) = $29,635Answer 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(133.3339) = $1,333,339Answer 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) = $36,502Answer is (d)

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