1 / 37

390 likes | 548 Vues

Discounting Factors. Walesh (2000), p. 230. Decision economics requires placing costs on a comparable basis. Discounting factors are used to convert a set of discrete and continuous costs and revenues to a common point in time or to a common period.

Télécharger la présentation
## Discounting Factors

**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

**Discounting Factors**Walesh (2000), p. 230**Decision economics requires placing costs on a comparable**basis.**Discounting factors are used to convert a set of discrete**and continuous costs and revenues to a common point in time or to a common period.**Discounting factors account for the time value of money, and**allow you to compare apples to apples.**Example 1**Given: $10,000 is invested at 6% for four years. Earnings occur at simple interest with a single payment at the end of the four-year period. Find: Value of the single payment, that is, the original payment plus interest earned, at the end of four years. Solution: F = P(1+ni) = $10,000(1 + 4(0.06)) = $12,400**Total interest = ?**$2400 What is the difference between simple and compound interest? The investor does not earn interest on interest -- and the borrower does not pay interest.**Example 2**Given: $10,000 is invested at 6% for four years. Interest is compounded annually and a single payment is to be made at the end of four years. Find: Value of the single payment, i.e., original principal plus total interest earned (interest on interest) at the end of 4 years. Solution: F = P(1+i)n = 10,000(1.06)4 = $12,625**Total interest is $2625, or $225 more that the $2400**interest earned from the simple interest version of the example. What if the interest is compounded monthly? i = 0.06/12 = 0.005 & n = 48 F = 10,000(1.005)48 = $2705**Example 7**Given: As a result of a trust established for you, you will receive $75,000 in ten years. Find: Present value of your trust on the assumption that interest is 6%. Solution: P = F/(1+i)n = $75,000/ (1.06)10 = $41,880**$41,880 is the single-payment compound-amount that would**need to be invested at 6% to yield $75,000 in ten years.**The examples used up to now work with discrete (single)**expenses, costs, incomes, and revenues. Another class of transactions is those involving a series of equal costs or incomes. Annuities are transactions involving a series of equal disbursements and/or receipts.**Example 8**Given: Starting this year, you decide to put $2000 per year at the end of each year into a retirement plan earning 8%. Find: Value after 10 years and interest earned in 10 years. Solution: F = A((1+i)n – 1)/i = 2000(1.0810 – 1)/(0.08) = $28,973 Future value = $28,973, and interest = $28972 – (10)2000 = $8973.**Example 9**Given: A small manufacturing plant projects that it will need $1 million for construction of a plant expansion in 10 years and can invest funds at 6% interest. Find: The amount invested at the end of each of 10 years to fund the expansion. Solution: A = F (i)/((1+i)n – 1) = $1,000,000(0.06)/(0.7908) = $75,870**Example 10 (Series Present-Worth Example)**Given: Annual O&M costs for a left-handed wrench painter are $11,500 and the economic life is 20 years. Find: The present worth of the O&M costs with interest = 6%. Solution: P = A((1+i)n – 1)/((i)(1+i)n) = $11500(1.0620 – 1)/(0.1924) = $131,902 Why would you want to know this?**Example 11**Given: Your new car costs $25,000. You pay $2500 down and finance the remainder for 36 months at 10%. Find: Monthly payments and total interest paid. Solution: i = 0.10/12 = 0.00833 A = P((i)(1+i)n/((1+i)n – 1) = $22500(0.00833)1.0083336/(1.0083336 – 1) =$725.96 monthly payment Interest = (36)725.96 – 22500 = $3634.56 Do you pay equal amounts of interest each month?**Example 12 Gradient-Series Present-Worth**Given: For equipment being considered, the O&M cost is $1100 for the first year, and the cost per year increases at $100 per year. Find: Present worth of the O&M expenditures. Why would this be worth knowing? Can anyone give an example? Solution: Break the problem into an annuity and into a gradient series, as shown in Fig 8-15.**For the annual series (annuity)**P = A(P/A, i, n) = 1000((1.0610 – 1)/((0.06)(1.06)10) P = $7360 For the gradient series P = G(GSPWF) = $3696 What is another way to work this problem?**Page 239,**Example 13

More Related