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2. ISE 211 Engineering Economy More Interest Formulas (Chapter 4)

3. Uniform Series Compound Interest Formulas  Many times we will find situations where there are a uniform series of receipts or disbursement.  Examples: Automobile loan, house payments, etc.  The series A is defined as: An end-of-period cash receipt or disbursement in a uniform series, continuing for n periods – the entire series equivalent to P or F at interest rate i.

4. Uniform Series Compound Interest Formulas (cont’d)  The uniform series is equivalent to a future worth value, F, at interest rate i, and can be expressed as follows:  In functional notation:  The term ___________________ is called the uniform series compound amount factor.  Or we can say that the term _________________ is called the uniform series sinking fund factor.

5. Example 1 A man deposits $500 in a credit union at the end of each year for five years. The credit union pays 5% interest, compounded annually. At the end of five years, immediately following his fifth deposit, how much will he have in his account? Solution:

6. Example 2 Jim Hayes read that in western United States, a ten- acre parcel of land could be purchased for $1000 cash. Jim decided to save a uniform amount at the end of each month so that he would have the required $1000 at the end of one year. The local credit union pays 6% interest, compounded monthly. How much would Jim have to deposit each month? Solution:

7. How can we find P if we know A?  The factor __________________ is called the uniform series capital recovery factor.  The factor __________________ is called the uniform series present worth factor.

8. Example 1 On January 1 st a man deposits $5000 in a credit union that pays 8% interest, compounded annually. He wishes to withdraw all the money in five equal end-of-year sums, beginning December 31 st of the first year. How much should he withdraw each year? Solution:

9. Example 2 An investor holds a time payment purchase contract on some machine tools. The contract calls for the payment of $140 at the end of each month for a five year period. The first payment is due in one month. He offers to sell you the contract for $6800 cash today. If you otherwise can make 1% per month on your money, would you accept or reject the investor’s offer? Solution:

10. Example 3 Suppose we decided to pay the $6,800 for the time purchase contract in Example 2. What monthly rate of return would we obtain on our investment? Solution:

11. Example 4 Using a 15% interest rate, compute the value of F in the following cash flow: YearCashflow 1+100 2+100 3+100 40 5F Solution:

12. Example 5 Every month starting today, you save $50 for 12 months. If your account earns 3% interest, compounded monthly, how much will you have just after your last deposit? Solution:

13. Example 6 Consider the following situation, and find the value of P. Solution: P=? 20 30 20 0 i = 15%

14. Relationships Between Compound Interest Factors  Single Payment:  Compound amount factor = 1 / Present worth factor (F/P, i, n) = 1 / (P/F, i, n)  Uniform Series:  Capital recovery factor = 1 / Present worth factor (A/P, i,n) = 1 / (P/A, i, n)  Compound amount factor = 1 / Sinking fund factor (F/A, i, n) = 1 / (A/F, i, n)

15. Miscellaneous Relationships Miscellaneous Relationships    i

16. Arithmetic Gradient Arithmetic Gradient  In an arithmetic series, a constant amount is added each period.  Suppose the amount added = G.  Examples: 2, 4, 6, 8, ….(G=2).  : 225, 250, 275, 300, … (G=25)  The cash flow diagram for such a situation is shown as:

17. Arithmetic Gradient (cont’d) Arithmetic Gradient (cont’d)  In an arithmetic series, a constant amount is added each period.  This can be expressed as follows:

18. Arithmetic Gradient (cont’d) Arithmetic Gradient (cont’d)  The arithmetic gradient present worth factor is given as follows:  The arithmetic gradient uniform series factor:

19. Example 1 Example 1 A man purchased a new automobile. He wishes to set aside enough money in a bank account to pay the maintenance on the car for the first five years. It has been estimated that the maintenance cost of an automobile is as follows: Year Maintenance Cost 1 $120 2150 3180 4210 5240 Assume the maintenance costs occur at the end of each year and that the bank pays i=5%. How much should he deposit in the bank now?

20. Example 2 Example 2 On a certain piece of machinery, it is estimated that the maintenance expense will be as follows: Year Maintenance Cost 1 $100 2200 3300 4400 What is the equivalent uniform annual maintenance cost for the machinery if 6% interest is used?

21. Example 3 Example 3 Consider the following cash flows YearAmount 1$24,000 2 18,000 3 12,000 4 6,000 Assuming 10% interest, what is the equivalent uniform annual series over 4 years?

22. Example 4 Example 4 Compute the value of P in the diagram below, using 10% interest rate. P 000 50 150 100

23. Geometric Gradient  This illustrates the situation when the period-by-period change is a uniform rate, g.  Example: The maintenance costs for an automobile are $100 the first year and increasing at a uniform rate, g, of 10% per year. Illustration:

24. Geometric Gradient (cont’d)  From the previous table, we can calculate the maintenance in any given year as follows: $100(1+g) n-1  Which can be written in general as follows: A n = A 1 (1+g) n-1 Whereg = geometric gradient -- uniform rate of cash flow increase/decrease from period to period A 1 = value of cash flow at year 1 A n = value of cash flow at any year n.

25. Geometric Gradient (cont’d)  What if we are interested in obtaining the present worth of the geometric series shown below? P= ??  This can be done using the following equation:

26. Geometric Gradient (cont’d)  This can be expressed as follows in functional notation:P = A 1 (P/A, g, i, n) i ≠ g  The term A 1 (P/A, g, i, n) is the geometric series present worth factor where i  g.  In the special case where i = g, the equation becomes: P = A 1 n (1 + i) -1

27. Example 1 The first year maintenance for a new automobile is estimated to be $100, and it increases at a uniform rate of 10% per year. What is the present worth of cost of the first five years of maintenance in this situation, using an 8% interest rate? Solution:

28. Nominal and Effective Interest Example: Consider a situation of a person depositing $100 into a bank account that pays 5% interest compounded semi-annually. How much would be in the savings account at the end of one year? Solution:  What interest rate compounded annually, results in this value of F = ______? Solution:

29. Nominal and Effective Interest (cont’d) Definitions: r = nominal interest rate per interest period (usually one year) – interest rate per interest period without considering the compounding effect. i = effective interest rate per interest period (e.g., year) – the interest rate taking into account the effecting of compounding per interest period. i a = effective interest rate per year (annum) m= # of compounding sub-periods per time period

30.  The effective interest rate per year is given as follows: i a = (1 + r/m) m – 1 i a = (1 + i) m – 1  Example: If a savings bank pays 1.5% interest every three months, what are the nominal and effective interest rates per year? Nominal and Effective Interest (cont’d)

32. Example You go to a check cashing store to get a $50 advance on your pay check. They are willing to do so provided you write a check for $60 dated for 1 week from now. a) What nominal interest rate per year are you paying? b) What effective interest rate per year are you paying? c) How much would you have to write the check for if your pay back period is 1 year instead of 1 week? Solution:

33. What if compounding interval  cash flow period? Example: A bank pays 8% nominal annual interest per year, compounded quarterly. A person deposits $5000 now (at time 0). After the end of each of 5 years, she wants to withdraw an equal amount of money. What is this amount? Solution # 1: