Suppose you invest $ at 4% annual interest. How much will you have at the end of two years?

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1 Example 1 Suppose you invest $ at 4% annual interest. How much will you have at the end of two years? Paul Koester () MA 111, Simple Interest September 19, / 13

2 Example 1 Suppose you invest $ at 4% annual interest. How much will you have at the end of two years? Solution 1: (Simple interest) Last class we studied simple interest, which case the answer could be found by ( ) = (1.08) = $ (Effectively, this says 4% at two years is 8%.) Paul Koester () MA 111, Simple Interest September 19, / 13

3 Example 1 Suppose you invest $ at 4% annual interest. How much will you have at the end of two years? Solution 1: (Simple interest) Last class we studied simple interest, which case the answer could be found by ( ) = (1.08) = $ (Effectively, this says 4% at two years is 8%.) This is NOT how banks typically compute interest! Paul Koester () MA 111, Simple Interest September 19, / 13

4 Example 1 Suppose you invest $ at 4% annual interest. How much will you have at the end of two years? Paul Koester () MA 111, Simple Interest September 19, / 13

5 Example 1 Suppose you invest $ at 4% annual interest. How much will you have at the end of two years? Solution 2: (How a bank MIGHT compute this) If we invest $ for only one year, we would have at the end of the year (1.04) = $ Paul Koester () MA 111, Simple Interest September 19, / 13

6 Example 1 Suppose you invest $ at 4% annual interest. How much will you have at the end of two years? Solution 2: (How a bank MIGHT compute this) If we invest $ for only one year, we would have (1.04) = $ at the end of the year. Now, if we invest the $ for an additional year, we will have at the end of the second year (1.04) = $ Paul Koester () MA 111, Simple Interest September 19, / 13

7 Why different answers? The second method of solution resulted in an additional $1.60. Where did this extra money come from? Paul Koester () MA 111, Simple Interest September 19, / 13

8 Why different answers? The second method of solution resulted in an additional $1.60. Where did this extra money come from? In this first solution, the interest was only applied to the initial investment (sometimes called the principal). In the second solution, one year s worth of interest was earned at the end of the first year. During the second year, interest was being earned not just on the initial $ , but also on the additional $4.00 that was earned in the first year. Compound Interest: earlier earned interest is allowed to earn interest There are different notions of compound interest. We are currently looking at Annual Compounding. We ll study other notions of Compound interest next class. Paul Koester () MA 111, Simple Interest September 19, / 13

9 Another Look at the Compound Interest Example Suppose you invest $ at 4% annual interest. How much will you have at the end of 2 years? Paul Koester () MA 111, Simple Interest September 19, / 13

10 Another Look at the Compound Interest Example Suppose you invest $ at 4% annual interest. How much will you have at the end of 2 years? We argued that after one year we ll have (1.04) = $ and at the end of the second year we ll have (1.04) = $ But notice the bottom line could be rewritten as = ( ) (1.04) = (1.04) 2 This suggests a general formula Paul Koester () MA 111, Simple Interest September 19, / 13

11 (Annual) Compound Interest Formula P is the Present Value (i.e., Principal, Initial Investment) F is the Future Value t is the time of the investment, measured in years R is the Annual Percentage Rate (APR) r = R/100 Then for annual compound interest: F = P (1 + r) t Paul Koester () MA 111, Simple Interest September 19, / 13

12 An unknown P You can invest money at 3.5% APR, compounded annually. How much money must you invest now in order to have $20, for a down payment on a new home 5 years from now? Paul Koester () MA 111, Simple Interest September 19, / 13

13 An unknown P You can invest money at 3.5% APR, compounded annually. How much money must you invest now in order to have $20, for a down payment on a new home 5 years from now? The present value is unknown. F = 20, , r = and t = 5 20, = P (1.035) 5 Dividing both sides by (1.035) 5 yields P = (1.035) 5 = $ Paul Koester () MA 111, Simple Interest September 19, / 13

14 An unknown P You can invest money at 3.5% APR, compounded annually. How much money must you invest now in order to have $20, for a down payment on a new home 5 years from now? The present value is unknown. F = 20, , r = and t = 5 20, = P (1.035) 5 Dividing both sides by (1.035) 5 yields P = (1.035) 5 = $ Caution: Be careful with rounding! Your answer should be correct to the penny. Small errors in intermediate calculations can lead to significant errors in the final answer. For example, if you say (1.035) then your P would be $ , which is off by about $4. Paul Koester () MA 111, Simple Interest September 19, / 13

15 Interest Earned You invest $P at % APR compounded annually. The total interest earned in 10 years in $ How much was P? Paul Koester () MA 111, Simple Interest September 19, / 13

16 Interest Earned You invest $P at % APR compounded annually. The total interest earned in 10 years in $ How much was P? Solution: F = P (1.0525) 10 The earned interest is F P = P (1.0525) 10 P = P ((1.0525) 10 1 ) So Solving for P: = P ((1.0525) 10 1 ) P = (1.0525) 10 1 = $ Paul Koester () MA 111, Simple Interest September 19, / 13

17 The Case of Fractional Years Invest $ at 2.5% APR, compounded annually. How much will you have at the end of 7 years and 9 months? Paul Koester () MA 111, Simple Interest September 19, / 13

18 The Case of Fractional Years Invest $ at 2.5% APR, compounded annually. How much will you have at the end of 7 years and 9 months? Solution: If interest is compounded annually, then you have to wait until the end of a given year before you earn that year s interest. Thus, we only earn interest for the 7 years (1.025) 7 = $ Paul Koester () MA 111, Simple Interest September 19, / 13

19 Multiple Investments 3.6% APR, compounded annually. Suppose $2500 is invested. Two years later, $ is withdrawn. Three years after that, $ is added to the account. What is the balance 10 years from time the original $ was invested? Paul Koester () MA 111, Simple Interest September 19, / 13

20 Multiple Investments 3.6% APR, compounded annually. Suppose $2500 is invested. Two years later, $ is withdrawn. Three years after that, $ is added to the account. What is the balance 10 years from time the original $ was invested? Solution 1: The initial $ left alone, would have grown to (1.036) 10 = The $ that was withdrawn after two years would have grown to (1.036) 8 = The $ was added in 5 years from the start, so earns interest for 10 5 = 5 years, and grows to 500 (1.036) 5 = The total balance after 10 years is therefore = Rounding to the nearest penny, $ Paul Koester () MA 111, Simple Interest September 19, / 13

21 Multiple Investments 3.6% APR, compounded annually. Suppose $2500 is invested. Two years later, $ is withdrawn. Three years after that, $ is added to the account. What is the balance 10 years from time the original $ was invested? Paul Koester () MA 111, Simple Interest September 19, / 13

22 Multiple Investments 3.6% APR, compounded annually. Suppose $2500 is invested. Two years later, $ is withdrawn. Three years after that, $ is added to the account. What is the balance 10 years from time the original $ was invested? Solution 2: The initial $ earns interest for 2 years: (1.036) 2 = $ is deducted from that ( = ) and this new balance earns interest for 3 years (1.036) 3 = Add in (to get $ ) and this new balance earns interest for 5 years: (1.036) 5 = Rounding to the nearest penny, $ Paul Koester () MA 111, Simple Interest September 19, / 13

23 In solution 1, we found the future value of each transaction, then added/subtracted those future values. In solution 2, we recomputed the balance at each time when a withdrawal/deposit was made. Theoretically, these two methods should give the same answer. You should carry out the intermediate calculations to many digits. If you round too liberally in the intermediate calculations, your final answer may be off by a few cents. Paul Koester () MA 111, Simple Interest September 19, / 13

24 Doubling Time Suppose is invested at 2.4% APR, compounded annually. How long will it take to in order for your investment to double in value? Paul Koester () MA 111, Simple Interest September 19, / 13

25 Doubling Time Suppose is invested at 2.4% APR, compounded annually. How long will it take to in order for your investment to double in value? Solution We want F to be at least $ There are a few ways we can find t Mehtod 1: Trial and Error We make an educated guess of t = 20: F = 1800 (1.024) 20 = This is too small, so increase t, say to t = 40. F = 1800 (1.024) 40 = So t should be between 20 and 40. Try t = 30. F = 1800 (1.024) 30 = This is just a little more than twice. Check if it has double by t = 29? No, so it takes 30 years to double. F = 1800 (1.024) 29 = Paul Koester () MA 111, Simple Interest September 19, / 13

26 Doubling Time Suppose is invested at 2.4% APR, compounded annually. How long will it take to in order for your investment to double in value? Mehtod 2: Algebraic Approach Lets try to find t so that Divide both sides by 1800 : 3600 = 1800 (1.024) t 2 = (1.024) t We can get t by itself using logarithms: ln (2) = ln ( (1.024) t) = t ln (1.024) Divide both sides by ln (1.024) to get t = ln 2 ln (1.024) = Now, we have to wait until the end of a year to get that year s interest. Thus, we need to wait 30 years to double the value. If you re not comfortable with algebraic properties of logarithms, you may be better off using method 1 or method 3. Paul Koester () MA 111, Simple Interest September 19, / 13

27 Doubling Time Suppose is invested at 2.4% APR, compounded annually. How long will it take to in order for your investment to double in value? Mehtod 3: Banker s Rule of 70 This is a rule of thumb used in practice. The time to double is approximately t = i.e., we take 70 and divide by the APR. If you use this method, you should plug back in and check. It is a rule of thumb, so it usually gives a good ballpark estimate, but could be a little off. The rule of 70 works because ln (2) 0.70 and ln (1 + r) r and so ln (2) ln (1.024) = Paul Koester () MA 111, Simple Interest September 19, / 13