1. Section 13.2 Loans

2. Example 8 Find the future value of each account at the end of 100 years if the initial balance is $1000 and the account earns: a) 7.5% simple interest. b) 7.5% compounded annually.

3. Example 8, cont’d P = 1000, t = 100, and r = 0.075. a) With simple interest, the future value is: F = 1000(1 +.075 100) = $8500 b) With annually compounded interest, the future value is: F = 1000(1 +.075/1) (1x100) = $1,383,000

4. Effective Annual Rate Effective Annual Rate (EAR) or Annual Percentage Yield (APY): the simple interest rate that would give the same result as a plan’s compounding interest in 1 year. Used for comparing different savings plans.  APY is used only for savings accounts.  EAR is used in any context.

5. Example 10 Find the effective annual rate by computing $100 over 1 year at 12% annual interest compounded every 3 months.

6. Example 10, cont’d The balance at the end of 1 year is: F = 100(1 +.12/4) (4x1) = $112.55 Since the account increased by $12.55 in 1 year, the EAR is 12.55%.  F = 100(1 +.1255 1) = $112.55

7. Effective Annual Rate, cont’d If r is the annual interest rate and m is the number of compounding periods per year, then the effective annual rate is: EAR = (1 + r/m) m – 1 The same formula is used for APY.

8. Example 11 A bank offers a savings account with an interest rate of 0.25% compounded daily, with a minimum deposit of $100. The same bank offers an 18-month CD with an interest rate of 2.13% compounded monthly, with deposits less than $10,000. Find the EAR for each option.

9. Example 11, cont’d The EAR for the savings account is: EAR = (1 + r/m) m – 1 = (1 +.0025/365) 365 – 1 =.00250312  The EAR for the account is about 0.2503%.

10. Example 11, cont’d The EAR for the CD is: EAR = (1 + r/m) m – 1 = (1 +.0213/12) 12 – 1 =.02150918  The EAR for the CD is about 2.1509%.

11. Simple Interest Loans Finance Charge: the simple interest charged per month  Charges are calculated using an average daily balance and a daily interest rate.

12. Example 1

13. Example 1, cont’d If the billing period is June 10 through July 9, determine: a) The average daily balance b) The daily percentage rate c) The finance charge d) The new balance

14. Example 1, cont’d a) The daily balances are shown below.

15. Example 1, cont’d a) The average daily balance is:

16. Example 1, cont’d b) The daily percentage rate is: c) The finance charge is the simple interest on the average daily balance at the daily rate: I = (250.72)(0.21/365)(30) = $4.33

17. Example 1, cont’d d) The new balance is the sum of the previous balance, new charges, and the finance charge, minus payments: 287.84 + 144.10 + 4.33 – 150.00 = 286.27  The new balance is $286.27.

18. Example 1, cont’d

19. Amortized Loans

20. Amortized loans: simple interest loans with equal periodic payments over the length of the loan. Important variables for an amortized loan are:  Principal  Interest rate  Term (length)  Monthly payment

21. Amortized Loans, cont’d Each payment pays the interest due since the last payment and an amount toward the balance.  The amount paid each month is constant, but the split between principal and interest changes.  The last payment may be slightly more or less than usual.

22. Example 2 Chart the history of an amortized loan of $1000 for 3 months at 12% interest with monthly payments of $340.

23. Example 2, cont’d Monthly payment #1:  The interest owed is I = (1000)(.12)(1/12) = $10  The payment toward the principal is $340 - $10 = $330  The new balance is $1000 - $330 = $670.

24. Example 2, cont’d Monthly payment #2:  The interest owed is I = (670)(.12)(1/12) = $6.70  The payment toward the principal is $340 - $6.70 = $333.30  New balance is $670 - $333.30 = $336.70

25. Example 2, cont’d Monthly payment #3:  The interest owed is I = (336.70)(.12)(1/12) = $3.37  The remaining balance plus the interest is: $336.70 + $3.37 = $340.07.  The third and final payment is $340.07.

26. Example 2, cont’d The amortization schedule is shown below.

27. Monthly Payments The monthly payments for an amortized loan can be determined in one of three ways:  Using an amortization table.  Using a formula.  Using financial software or an online calculator.

28. Amortization Table Amortization table: gives pre-calculated monthly payments for common loan rates and terms. An example of a table is shown on the next slide (and on pg. 824).

30. Example 3 A couple is buying a vehicle for $20,995. They pay $7000 down and finance the remainder at an annual interest rate of 4.5% for 48 months. Use the amortization table to determine their monthly payment.

31. Example 3, cont’d The amount being financed is: $20,995 – $7000 = $13,995. In the table, find the row corresponding to 4.5% and the column corresponding to 4 years.

32. Example 3, cont’d

33. The value 22.803486 means the couple will pay $22.803486 for each $1000 they borrowed. (22.803486)(13.995) = 319.13479 They will pay $319.14 per month.

34. Example 4 A couple borrowed $13,995 to buy a car and will pay the loan over 4 years. If their payments are $340.02, what interest rate are they being charged?

35. Example 4, cont’d They pay $340.02 a month for $13,995, or about $24.295820 per $1000.  Look at the amortization table to find the interest rate.

37. Example 4, cont’d The interest rate for the amortized loan, according to the table, is approximately 7.75%.

38. Homework Pg. 814  36 Pg. 828  6, 8, 22a, 36, 40