1. INTEREST Simple Interest- this is where interest accumulates at a steady rate each period The formula for this is 1 +it Compound Interest is where interest is earned on interest. This process is known as compounding. The formula for this is (1+i) t..

2. Different components Principal is the original amount that was invested. i is the effective rate of interest per year. t is the time period in which the principal was invested. Accumulated Value is what your principal …

3. has grown to, denoted A(t). Therefore …. Interest = Accumulated Value-Principal Compound Interest is the most important to remember due to the fact that it is used mostly in situations. It has exponential growth whereas simple interest has linear growth.

4. Example – Someone borrows $1000 from the bank on January 1, 1996 at a 15% simple interest. How much does he owe on January 17, 1996? Solution – Exact simple interest would give you 1000[1+(.15)(16/365)]=1006.58. However…..

5. Banker’s rule uses 360 days, which gives a different result. Solution – 1000[1+(.15)(16/360)]=1006.67, which is slightly higher. Canada uses exact simple interest.

6. Example - Jessie borrows $1000 at 15% compound interest. How much does he owe after two years? Solution = 1000(1.15) 2 =1322.50.

7. Assuming a 3% rate of inflation $1 now will be worth 1.03 3 or $1.09 in three years. Example – How much was $1000 worth 4 years ago assuming a 3% inflation rate? Solution – It is worth 1000(1.03) -4, which is equal to $888.49.

8. Nominal rate of interest is a rate that is convertible other than once per year. i (m) is used to denote a nominal rate of interest convertible m times per year, which implies an effective rate of interest i (m) per mth a year, so the effective rate of interest is i=[1+ (i (m) /m)] m -1.

9. Example – Find the accumulated value of $1000 after three years at a rate of interest of 24% per year convertible monthly. Solution- i=[1+(.24/12)] 36 -1=.26824. So the answer to the problem is 1000(1.26824) 3 =2039.88.

10. Also, this is just something to remember. Suppose XXY credit card is offering 12% convertible monthly and Spragga Dap credit card is offering 12% convertible semi- annually, which has the best deal. Solution- XXY has an effective annual interest rate of [1+(.12/12)] 12 -1=.12683.

11. In the case of the Spragga Dap credit, the annual effective rate of interest is i=[1+(.12/2)] 2 -1=.1236, which is lower than the XXY credit card. So, the rule to remember is, given the same nominal rate, the effective annual rate of interest will be higher if it is compounded more.

12. Suppose we wanted to find a nominal rate of interest compounded continuously, which is the force of interest. There is a formula for this: ln(1+i). Example Suppose i was fixed at.12 and we wanted to find i (m), we would use the formula i=.12=[1+ (i (m) /m)] m -1 and solve for i (m). We will see that

13. i (2) =.1166 i (5) =.1146 i (10) =.1140 i (50) =.1135 …and if the nominal rate of interest is compounded continuously, then it would be ln(1.12)=.11333.

14. ANNUITIES An annuity is a stream of payments. The present value of a stream of payments of $1 is a n. The formula for a n is: (1-v n) /i……where v=(1/1+i) Suppose we were to take out a $50000 from the Spragga Dap bank. If the mortgage rate is 13% convertible semi-annually, what would the monthly payment be to pay off this mortgage in 20 years?

15. Solution: First, we find i, which is (1.065) (1/6) -1, then we proceed to set up the problem. 50000=X. a 240 A n =[1-(1/1.01055) 240 ]/.01055=87.1506 so… X=50000/87.1506=573.72

16. Here’s a tricky one! Suppose Haskell Inc. supplies you with a loan of $5000 that is supposed to be paid back in 60 monthly installments. If i=.18 and the first payment is not due until the end of the 9 th month, how much should each one of the 60 payments be?

17. Solution – first we convert i into a monthly rate, which is 1.18 (1/12) -1. Then we have to account for the fact that the $5000 earned interest in the 1 st 8 months. The new amount is 5000(1.013888) 8 which is 5583.29 so………. 5583.29=X. a 60

18. a 60 =[1-(1/1.013888) 60 ]/.013888=40.5299 Finally, 5583.3/40.5299=137.76 So we would need 60 payments of $137.76 to pay it off in 60 monthly installments. Note: If we were supposed to take out a loan which was repaid starting immediately, we would use a “double-dot” which is a n (1+i).

19. BONDS Investing in bonds is a good way to utilize your dollar. It is as simple as this. For a sum of money today, you will get interest annuity payments as well as another sum of money, known as redemption value, when the time period has elapsed.

20. There are a few key components to get familiar with when analyzing bonds. F is the face value or par value of the bond. r is the coupon rate per interest period. Normally, bonds are paid semi-annually. C is the redemption value of the bond. The phrase “redeemable at par” describes when F=C.

21. i is the yield rate per interest period n is the number of interest periods until the redemption date. P is the purchase price of the bond to obtain the yield rate i.

22. The price of the bond can be obtained by solving this formula: P=Fr. a n +C(1+i) -n Example – A bond of $500, redeemable at par in five years, pays interest at 13% per year convertible semi-annually. Find a price to yield an investor 8% effective per half a year.

23. Solution: F=C=500, r=.065, i=.08, n=10. So the price of this bond is: 32.5a 10 +500(1.08) -10 =449.67. Example: Spragga Dap Corporation decides to issue 15-year bonds, redeemable at par, with face amount of $1000 each. If interest payments are to be made at a rate of 10% convertible semi-annually,

24. And if the investor is happy with a yield of 8% convertible semi-annually, what should he pay for one of these bonds? F=C=1000, n=30, r=.05 and i=.04 so the price is 50. a 30 +1000(1.04) -30 =1172.92