1 / 14

CE 203 More Interest Formulas (EEA Chap 4)

CE 203 More Interest Formulas (EEA Chap 4). Example of a Uniform Series CFD. Cost Flow Diagram. F. P. A. A. A. A. A. A. 1. 2. 3. 4. 5. 6. 0. n = 6 for this example; i must “match” all amounts (arbitrarily) shown as positive. Uniform Series Formulas: Conventions.

abigail-puckett
Download Presentation

CE 203 More Interest Formulas (EEA Chap 4)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CE 203 More Interest Formulas(EEA Chap 4)

  2. Example of a Uniform Series CFD Cost Flow Diagram F P A A A A A A 1 2 3 4 5 6 0 n = 6 for this example; i must “match”all amounts (arbitrarily) shown as positive

  3. Uniform Series Formulas: Conventions • “A” is a payment that occurs at the end of each of a series of time periods • “P” occurs one payment period before the first “A” • “F” occurs at the same time as the last “A” and “n” periods after “P” • “i” is the interest rate per period • “n” is the number of periods

  4. Uniform Series Compound Amount Factor (e.g., many investment programs) F F = A [ ] = A (F/A, i, n) (1 + i)n - 1 i A A A A A A 1 2 3 4 5 6 0 (see text p. 87 for derivation of formula)

  5. Example 1: Series Compound Amount You deposit $1000 per year for 30 years in an account that earns 10%, compounded yearly. How much could you withdraw from the account at the end of 30 years? F = 1000 (F/A, 0.1, 30) = A[(1 + i)n - 1]/i = $1000 [(1 + 0.1)30 – 1] / 0.1 = $164,494.02 Or from text p. 578, F = 1000 (164.494)

  6. Example 2: Series Compound Amount You deposit $100 per month in a Roth IRA from age 25 to retirement at age 65 at 8%. How much would you have in the account at retirement? F = 100 (F/A, 0.08/12, 40 x 12) = $100 [(1 + 0.00667)480 – 1] / 0.00667 = $349,505.19

  7. Uniform Series Sinking Fund If the equation F = A [ ] = A (F/A, i, n) (1 + i)n - 1 i is solved for A A = F [ ] = F (A/F, i, n) i (1 + i)n - 1 Used to determine how much money (A) needs to be saved per period to obtain a given future amount (F)

  8. Example 1: Series Sinking Fund Your company plans to purchase a new instrument in 2 years at a guaranteed price of $60,000. Assuming your bank will pay 6% interest, compounded monthly, how much should your company put aside each month in order to purchase the instrument? A = 60,000 (A/F, 0.005, 24) = $60,000 (.005) / [(1 + 0.005)24 – 1] = $2,359.24 (per month) Note: $60,000/24 = $2,500

  9. Uniform Series Capital Recovery If the equations A = F []andF = P (1 + i)n i (1 + i)n - 1 are combined (and solved for A) A = P []= P (A/P, i, n) i (1 + i)n (1 + i)n - 1 Used to determine how much money (A) needs to be paid per period to repay or recover an initial amount

  10. Example 1: Series Capital Recovery You just purchased a home for $290,000. Your agreement is 10% down and a 20-year mortgage at 7%. What are your monthly payments going to be? P = .9 (290,000) = $261,000 i = .07/12 = .00583 n = 240 A = $261,000 [ ] = $2022.90 (Total payments = $485,496.86 + $29,000) 0.00583 (1 + 0.00583)240 (1 + 0.00583)240 - 1

  11. Uniform Series Present Worth If the equation A = P [ ] = P (A/P, i, n) i (1 + i)n (1 + i)n - 1 is solved for P (today’s value or Present Worth) P = A [ ] = A (P/A, i, n) (1 + i)n - 1 i (1 + i)n Used to calculate the Present Worth of a series of regular and equal future payments

  12. Example 1: Series Present Worth Jill just inherited a sum of money and has decided to put some of it into a savings account that can be used to pay her utility bills over the next three years. If her monthly utility bills are a uniform $180 per month, how much should she put into the savings account if it earns 6% per year, compounded monthly? P = $180 [ ] = $ 5916.78 (or P = 180 (P/A, .005, 36) = 180 (32.871) = $5916.78) (1 + 0.005)36 - 1 0.005 (1 + 0.005)36 Note: 36 x $180 = $6480

  13. In-class Example 2 (unknown i) A bank offers to pay $16,000 in 10 years for a deposit of $100 per month for that 10-year time period. A second bank promises a rate of return of 4.5% for the same payment plan. What interest rate is the first bank offering? Which is the better offer? (Assume monthly compounding for both.) Also can use EXCEL functions FV (trial and error) or IRATE F = FV (rate, nper, pmt, fv, type) = FV (trial, 120, -100,,) i = RATE (nper, pmt, pv, fv, type) = RATE (120, -100,,16000,,) This knowledge may be useful for your future spreadsheet homework

  14. In-class Example 3 Dave just bought a new car that came with an all inclusive (i.e., all maintenance included) 5-year warranty. Starting in year 6, Dave estimates that maintenance will be $800/year. He plans to keep the car for 8 years. Assuming he can earn 10% (compounded yearly) on his money, how much should he deposit now in order to cover the maintenance costs in years 6, 7, and 8? Answer: Calculate the “future present value” in Y5 = $1991 and convert to the present value in Y0 = $1236

More Related