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ISU CCEE CE 203 More Interest Formulas (EEA Chap 4)

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Presentation on theme: "ISU CCEE CE 203 More Interest Formulas (EEA Chap 4)"— Presentation transcript:

1 ISU CCEE CE 203 More Interest Formulas (EEA Chap 4)

2 ISU CCEE l Arithmetic Gradient: payments increase by a uniform AMOUNT each payment period l Geometric Gradient: payments increase by a uniform RATE each payment period Formulas for non-uniform payments

3 ISU CCEE Arithmetic Gradient Series AA P P A+GA+G A+2GA+2G A+3GA+3G A+(n-1)GA+(n-1)G = (can be divided into) Amount increases by “G” each period

4 ISU CCEE Arithmetic Gradient Series Components AA P’P’ P’P’ P ’’ GG AAAAAAAA 2G2G 3G3G (n-1)G(n-1)G + + 00

5 ISU CCEE Arithmetic Gradient Future Worth F ’’ GG 2G2G 3G3G (n-1)G(n-1)G F’’ = [ - n ] (1 + i) n - 1 i The Future Worth of an arithmetic gradient cash flow is given by (see text p. 99-100 for derivation of formula) 00 = G [ ] (1 + i) n – in - 1 i2i2 G i

6 ISU CCEE Arithmetic Gradient Uniform Series Factor A = G [ ] = G (A/G, i, n) (1 + i) n – in - 1 i {(1 + i) n - 1} If the arithmetic gradient factor is multiplied by the sinking fund factor, the result is called the Arithmetic Gradient Uniform Series Factor: (gives the uniform series cash flow payment equivalent to that of an arithmetic gradient cash flow) (see text p. 100 for derivation of formula)

7 ISU CCEE In-class Example 1 (arithmetic series) Your company just purchased a piece of equipment. Maintenance costs are estimated at $1200 for the first year and are expected to rise by $300 in each of the subsequent four years. How much should be set aside in a “maintenance account” now to cover these costs for the next five years? Assume payments are made at the end of each year and an interest rate of 6%.

8 ISU CCEE Geometric Gradient Series A1A1A1A1 A1A1A1A1 P P A 1 (1+g) - Amount changes at the uniform RATE, g - Useful for some types of problems such as those involving inflation A 1 (1+g) 2 A 1 (1+g) 3 A 1 (1+g) 4

9 ISU CCEE Present Worth for Geometric Gradient Series A1A1A1A1 A1A1A1A1 P P A 1 (1+g) for i ≠ g A 1 (1+g) 2 A 1 (1+g) 3 A 1 (1+g) 4 P = A 1 [ ] 1 - (1 + g) n (1 + i) -n i - g = A 1 (P/A, g, i, n) Can be very complex unless programmed on a computer

10 ISU CCEE In-class Example 2 (geometric series) You have just begun you first job as a civil engineer and decide to participate in the company’s retirement plan. You decide to invest the maximum allowed by the plan which is 6% of your salary. Your company has told you that you can expect a minimum 4% increase in salary each year assuming good performance and typical advancement within the company. 1) Choose a realistic starting salary 2) Assuming you stay with the company, the company matches your 6% investment in the retirement plan, expected minimum salary increases, and an interest rate of 10%, how much will you have in your retirement account after 40 years?

11 ISU CCEE Nominal vs. Effective interest rate Nominal interest rate, r: annual interest rate without considering the effect of any compounding at shorter intervals so that i (for use in equations) = r/m where m is number of compounding periods per year (this is what we have been doing)

12 ISU CCEE Nominal vs. Effective interest rate Effective interest rate, i a : annual interest rate taking into account the effect of any compounding at shorter intervals; also called “yield” An amount of $1, invested at r%, compounded m times per year, would be worth $1(1 + r/m) m and the effective interest 1(1 + r/m) m - 1 i a = (1 + i) m – 1 (see text p. 110, 9 th Ed for derivation)

13 ISU CCEE ExampleExample A bank pays 6% nominal interest rate. Calculate the effective interest with a) monthly, b) daily, c) hourly d) secondly compounding i a = (1 + i) m – 1 i a monthly = (1 +.06/12) 12 -1 = 6.1678 % i a daily = (1 +.06/365) 365 -1 = 6.183 % i a hourly = (1 +.06/8760) 8760 -1 = 6.1836 % i a secondly = (1 +.06/31.5M) 31.5M -1 = 6.18365 %

14 ISU CCEE Continuous Compounding The effective interest rate for continuous compounding (i.e., as the length of the compounding period → 0 and the number of periods → ∞) i a = e r – 1 (see text pp. 116-117 for derivation and equations involving continuous compounding) In our previous example, i a = e 0.06 – 1 = 6.183655%

15 ISU CCEE Nominal vs. Effective interest rate Comparison of nominal and effective interest rates for various APR values and compounding periods Nominal rateEffective interest rate, i a, for compounding … rYearly Semi- annuallyMonthlyDaily Contin- uously 55.00005.06255.11625.12685.1271 1010.000010.250010.471310.515610.5171 1515.000015.562516.075516.179816.1834 2020.000021.000021.939122.133622.1403

16 ISU CCEE l Cash flow period is not equal to compounding period l There is cash flow in only some of periods l Pattern does not exactly fit any of basic formulas l Etc. Adjustments may be needed because…

17 ISU CCEE Partial review l Single payment compound amount F = P(1+i) n = P(F/P,i,n) l Uniform series (sinking fund) l Arithmetic gradient series A = F [ ] = F (A/F, i, n) (1 + i) n - 1 i A = G [ ] = G (A/G, i, n) (1 + i) n – in - 1 i {(1 + i) n - 1}


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