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67 Section 8.4 –Compound Interest Objective #1: Using the compound interest formula. If an account earns interest during a period of time, the interest is then added to the principal. During the next cycle, the new principal is the old principal plus the interest earned during the last cycle. Thus, at the end of the next cycle, the account not only earns interest on the original principal but also the interest earned in the last cycle. We say that the interest is compounded. Thus, compound interest is the interest earned on the principal and the interest earned previously. How often the interest is calculated depends on the length of the cycle or payment period. Below are some common payment periods: Interest is compounded … Means Interest in calculated once a year. annually Interest in calculated twice a year or every six months. semiannually Interest in calculated 4 times a year or every three months. quarterly Interest in calculated 12 times a year or every month. monthly Interest in calculated 52 times a year or every week. weekly Interest in calculated 360* times a year or every day. daily * Banks will define 1 month = 30 days and 1 year = 360 days. Solve the following: Ex. 1 Maria invests $3000 in an account that pays an annual interest rate of 5% compounded quarterly. How much does she have in the account: a) after 1 year. b) after 12 years. Solution: In both parts, P = $3000, r = 0.05 a) In one year's time, the interest will be calculated four times. st 1 quarter: A = P(1 + rt) = 3000(1 + (0.05)(1/4)) = $3037.50 nd 2 quarter: A = P(1 + rt) = 3037.50(1 + (0.05)(1/4)) = $3075.47 rd 3 quarter: A = P(1 + rt) = 3075.47(1 + (0.05)(1/4)) = $3113.91 th 4 quarter: A = P(1 + rt) = 3113.91(1 + (0.05)(1/4)) = $3152.84* * - if you do not round any of your calculations until the end, you get 3152.8360… which rounds to $3152.84 instead of $3152.83. 68 b) In twelve years time, the interest will be calculate 12•4 = 48 times. Thus, Beginning Ending Balance Quarter Balance P A = P(1 + rt) st 1 $3,000.00 $3,037.50 nd 2 $3,037.50 $3,075.47 rd 3 $3,075.47 $3,113.91 th 4 $3,113.91 $3,152.84 : : : : : : th 48 $5,378.83 $5,446.06 Part b of the last example illustrates that we need a better way to compute compound interest since our current method is unrealistic to use for most situations. We will need to derive a formula for compound interest: If we start with principal P, then the total at the end of the 1st cycle is A1 = P(1 + rt) If we start with principal A1, then the total at the end of the 2nd cycle is A2 = A1(1 + rt) Since A1 is equal to the amount in the account at the end of the first cycle, then we can replace A1 by P(1 + rt). So, A2 = A1(1 + rt) = P(1 + rt)(1 + rt) = P(1 + rt)2 If we start with principal A2, then the total at the end of the 3rd cycle is A3 = A2(1 + rt) Since A2 is equal to the amount in the account at the end of the second cycle, then we can replace A2 by P(1 + rt)2. So, A3 = A2(1 + rt) = P(1 + rt)2(1 + rt) = P(1 + rt)3 Continuing this pattern, the amount after k cycles is Ak = P(1 + rt)k Let n = the number of times interest is compounded per year. The length of time t for each payment period is 1/n and the number of cycles k is the time t in years multiplied by the number of times interest is compounded per ( year n. Replacing t by 1/n, k by nt, and Ak with A to get: A = P 1 + r n nt ) Compound Interest Formula If the principal P (present value) is invested at an annual interest rate r compounded n times a year for t years, the amount A (future value) will be: nt € r A=P 1+ where r has been converted to a decimal. ( € n ) 69 We can now rework example 1 using the compound interest formula with n = 4: a) ( r n ( r A=P 1+ nt ) ( 0.05 4 ( 0.05 = 3000 1 + nt ) 4(1) ) = 3000(1.0125)4 ≈ $3152.84 4(12) ) A=P 1+ = 3000 1 + = 3000(1.0125)48 ≈ $5446.06 n 4 € € Find the following: Ex. 2 Leroy invests $10000 in an account that pays an annual € interest rate € of 4%. How much does he have in the account after 1 year if interest is compounded: a) annually? b) semiannually? c) quarterly? d) monthly? Solution: In all four parts, P = $10,000, r = 0.04 and t = 1 year. a) Compounded annually means n = 1: b) ( A=P 1+ r n nt ) ( 0.04 1 = 10000 1 + 1(1) ) = 10000(1.04) = $10,400 Interest = 10400 – 10000 = $400 b) Compounded semiannually means n = 2: nt 2(1) r 0.04 € € A=P 1+ = 10000 1 + = 10000(1.02)2 = $10,404 ( n ) ( 2 ) Interest = 10404 – 10000 = $404 c) Compounded quarterly means n = 4: nt 4(1) r 0.04 € A€= P 1 + = 10000 1+ = 10000(1.01)4 ( n ) ( = $10,406.04 d) 4 ) Interest = 10406.04 – 10000 = $406.04 Compounded monthly means n = 12: nt 12(1) r 0.04 € € A=P 1+ = 10000 1 + = 10000(1.00333…)12 ( n ) ( = $10,407.42 12 ) Interest = 10407.42 – 10000 = $407.42 Increasing the number of times interest is compounded from one to 12 netted us $7.42 As we increasing the number of times € in additional interest. € interest is compounded per year, the amount increases so the question becomes if we increase the number of times interest is compounded without bound, how much more money can we earn in interest? To find out, consider the following: n r 1 n Let h = which implies that = and nt = •rt = h•rt r € n € € h r € 70 nt ( A = P( A€= P[( A=P ) ) )] r r 1+ (replace n n hrt 1 1+ (rewrite as h h rt 1 1+ € € h by 1 h and nt by h•rt) a power of rt) n Now, consider the expression inside the square brackets. Since h = , as r € we increase n without bound, h increases without bound. This gives us a € special number that occurs in nature. It is the natural base e. The number e is an irrational number that is approximately 2.71828182846.... It comes h € 1 from examining the behavior of 1 + as h → ∞ (increases without ( bound). It may seem that 1 + ( h) 1 h should approach 1, but if you plug in h) some large values for h, you will see this is not the case. For example, if h h € 1 = 10,000, then 1 + ≈ 2.71814592682... h € h (1 + 1/h)h h (1 + 1/h)h 1 2 1000 2.716923932 € 10 2.59374246 10,000 2.718145927 100 2.704813829 1,000,000 2.718280469 ( ) [( Thus, as h increases without bound, A = P 1 + 1 h h rt )] becomes A = Pert. It is important to be able to evaluate expressions involving e. On your calculator, you should have an ex key. On the TI-30XS, you will hit the 2nd key € hit enter. and hit ln key. Then type the exponent and Evaluate the following: Ex. 3a e4.3 Solution: a) e4.3 ≈ 73.6997937 Ex. 3b 2e3 b) 2e3 ≈ 40.17107385 For the previous example, if number of times interest is compounded increases without bound, the amount in the account will be: A = 10000e0.04(1) ≈ $10408.11 This means there is a limit to how much more interest can be gained. When we have an "infinite" number of times interest is compounded, we say that interest is compounded continuously. 71 Interest Compounded Continuously If the principal P (present value) is invested at an annual interest rate r compounded continuously for t years, the amount A (future value) will be: A = Pert where r has been converted to a decimal. Find the following: Ex. 4 Maria invests $3000 in an account that pays an annual interest rate of 5% compounded continuously. How much does she have in the account after 12 years? Solution: Plug in P = $3000, r = 0.05 and t = 12: A = Pert = 3000e0.05(12) = 3000e0.6 ≈ $5466.36 Ex. 5 Alissa has $5000 that she can invest for 6 years in one of two accounts. The first account pays 8% interest compounded quarterly and the second pays 7.92% interest compounded continuously. Which is the better investment? Solution: First, plug in P = $5000, r = 0.08, n = 4, and t = 6 in the compound interest formula: ( A=P 1+ r n nt ) ( = 5000 1 + 0.08 4 4(6) ) = 5000(1.02)24 ≈ $8042.19 Interest = $8042.19 – 5000 = $3042.19 Now, plug in P = 5000, r = 0.0792, and t = 6 in the compounded continuously formula: € A€= Pert = 5000e0.0792(6) = 5000e0.4752 ≈ $8041.68 Interest = $8041.68 – 5000 = $3041.68 The first investment paying 8% compounded quarterly is the better investment. Objective #2: Determining the Present Value of a Lump Sum. You can think of the present value as how much money that you have to put in an account today so that it grows to some predetermined amount in the future. Thus, if you want to have $30,000 in an account 20 years from now, the present value would be how much you would put into the account today so it grows to $30,000 in 20 years. Starting with our compound interest formulas, we need to solve for P to find the present value: 72 Compounded n times per year: ( A=P 1+ ( r n A 1+ – nt nt ) (multiply both sides by 1 + =P (rewrite the formula) ) ( P€= A 1 + € r n r n – nt ) ( = A (1 + r n – nt ) ) (book uses the € second formula) r nt ) n Present Value Formulas € A be the balance wanted in the future and r be the annual interest rate Let (written as€a decimal) compounded n times a year for t years. Then the present value P of the account is: ( P=A 1+ r n – nt ) = A (1 + r nt ) n Solve the following: Ex. 6 € Find the amount that Juanita should invest in an account paying € 8% annual interest so that after 20 years, she will have $50,000 if the interest is compounded quarterly. Solution: If interest is compounded quarterly, then n = 4, A = $50,000, r = 0.08, and t = 20. Thus, ( P=A 1+ r n – nt ) ( = 50000 1 + 0.08 4 – 4•20 ) = 50000(1.02)– 80 ≈ 50000(0.2051097282) ≈ $10,255.49. Objective #3: € Calculating Effective Rates of Return. € When comparing several different accounts, it is useful to determine what the equivalent interest rate would be if the money was invested an account paying simple interest for one year. This equivalent interest rate is known as the effective rate of interest. Effective Rate of Interest The effective rate of interest Y of an investment earning an interest rate r ( compounded n times a year is Y = 1 + decimal. € r n n ) – 1 where r is converted to a 73 Solve the following: Ex. 7 Suppose Bank A pays 4.52% annual interest compounded annually while Bank B pays 4.5% annual interest compounded quarterly and Bank C pays 4.47% annual interest compounded daily. Which one is the best deal? Solution: Bank A Bank B Bank C r = 0.0452; n = 1 r = 0.045; n = 4 r = 0.0447; n = 360 ( ( = 1+ € n ) r –1 n 1 0.0452 – 1 Y= 1+ ) ( ( 1 = (1.0452) – 1 € = 0.0452 = 4.52% Bank B is the best deal. = 1+ n ) ( r –1 n 4 0.045 –1 4 Y= 1+ 4 ) ( = 1+ = (1.01125) – 1 € = 0.04576… ≈ 4.58% € n ) ) r –1 n 360 0.0447 –1 360 Y= 1+ = 0.04571… € ≈ 4.57% € When investing money, you will want to look for the account with the highest effective rate of interest. When borrowing money, you will want to look for the loan that has the lowest effective rate of return. The effective rate of return in borrowing money is also called the annual percentage rate or APR.