MATH 111 Worksheet 1 Replacement Partial Compounding Periods Key Questions: I. XYZ Corporation issues promissory notes in $1,000 denominations under the following terms. You give them $1,000 now, and eight years later they will give $,000 back to you. You bought $8,000 worth of these notes 4 years ago and wish to sell them today to a friend who will cash them in for $16,000 4 years from now. How much money should your friend give you to make the deal fair? II. You have a bacteria colony that triples its population every hour. You start with 10 million bacteria and come back 0 minutes later and then 0 minutes after that. How many bacteria should you expect to see at each of these two times? We ll start with Key Question I. 1. Your friend says that, since the investment will be earning $8,000 in interest over the next 8 years, you re earning $1,000 in interest each year. (a) With this approach, fill in the following table: Time Supposed Value of Investment 4 years ago $8,000 3 years ago years ago 1 year ago $11,000 now $1,000 1 year from now years from now 3 years from now 4 years from now $16,000 (b) With this approach, you will pay $8,000 and receive $4,000 in interest in four years. Your friend will pay $1,000 and receive $4,000 in interest in four years. Is this deal fair to both you and your friend?. You say that, since the investment will be earning 100% interest over the next 8 years and 100 8 = 1.5, you re earning 1.5% interest each year. (a) With this approach, fill in the following table: 1
Time Supposed Value of Investment 4 years ago $8,000 3 years ago $8,000+8,000(0.15)=$9,000 years ago $9,000+9,000(0.15)=$10,15 1 year ago now $1,814 1 year from now years from now 3 years from now 4 years from now (b) The supposed values of the investment in the second column of the table form a multiplicative sequence. Give the values of the multiplier and the proportionate change. (See Worksheet 18, #0, for the relationship between the multiplier and proportionate change.) (c) Use the last entry in the table to explain why you were wrong to assume that 100% over 8 years was the same as 1.5% per year for 8 years. 3. Your Math 111 TA tells you that the way to make the deal fair is to make sure that you both earn in interest the same percentage of your respective investments. Let p be that percentage, expressed as a decimal. Suppose the amount your friend will pay you is $X. (a) You will pay $8,000 and receive $X from your friend. The amount X should include your original investment plus (p 100)% interest. That is, X = 8, 000 + 8, 000p OR X = 8, 000(1 + p). Similarly, your friend will pay $X and receive $16,000. But this $16,000 will also be your friend s original investment plus (p 100)% interest. That is, 16, 000 = X + Xp OR 16, 000 = X(1 + p). This gives two equations in two unknowns. Find the values of X and p. (b) How much will you earn in interest (in dollars) with this approach? How much will your friend earn? (c) Does this deal seem fair to both you and your friend? 4. We know from # that 100% over 8 years does not mean 1.5% per year for 8 years. But there is what s called an effective annual interest rate, a rate that, if applied each year for 8 years, would result in a 100% gain in the investment. Let r be this effective annual interest rate, expressed as a decimal, and let A(k) be the value of the investment k years after the notes were purchased. Then, A(0) = 8, 000 and A(8) = 16, 000, for example.
(a) Write an equation that relates A(1) to A(0). (Your equation should contain an r.) (b) Write an equation that relates A() to A(1). Then use your answer to part (a) to write an equation that relates A() to A(0). (c) Write an equation that relates A(3) to A(). Then use your answer to part (b) to write an equation that relates A(3) to A(0). (d) Your answers to parts (a), (b), and (c) should be revealing a pattern: the numbers A(k) form a multiplicative sequence. What are its multiplier and proportionate change? (e) Write an equation that relates A(k) to A(0). (f) We know that A(0) = 8, 000 and A(8) = 16, 000. Use your formula from part (e) to determine the value of r, the effective annual interest rate. Express this rate as a percentage. 5. Suppose you place $1,000 into a bank account that earns 6% interest annually. After 6 months, you withdraw the entire balance, which is your $1,000 plus some interest. In the following problem, you ll determine this balance. (a) Let A(k) be the balance after k years. What is A(0)? If you left the money in the account, what would be the value of A(1)? (b) The 6% annual interest corresponds to some rate of interest r (expressed as a decimal) per six months. Use your answers to part (b) to show that r does not equal 3% (6% ). (HINT: Show what A(1) would be if you earned 3% every six months for one year.) (c) Let X be the balance in the account after 6 months. Then, X = A(0) + A(0)r OR X = A(0)(1 + r) and A(1) = X + Xr OR A(1) = X(1 + r). Combining these, we get that A(1) = A(0)(1 + r). Use your answers to part (a) to solve for r. (d) If you did part (c) correctly, you should have passed through a step that said that 1 + r = 1.06 or 1 + r = (1.06) 1. Recall that, if X is the balance after six months, then X = A(0)(1 + r). This can be re-written as X = A(0)(1.06) 1. Compute the value of X, the balance in the account after 6 months ( 1 a year). In Worksheet #19, we introduced the Compound Amount Formula: A(k) = P (1+i) k. P =principal i =interest rate, expressed as a decimal k =number of times compounding has taken place A(k) =balance after k compoundings 3
One limitation with this formula, as derived in Worksheet #19, was that k must be a whole number. We re now ready to extend the CAF to allow for fractional values of k. In the last problem, you showed that you can compute the balance in the account after 6 months ( 1 a year) by using the CAF: ( ) 1 A = 1, 000(1 + 0.06) 1. That is, instead of thinking of k as the number of compoundings that have occurred, we can think of k as the number of compounding periods (years in this case) that have passed (and this number need not be a whole number). 6. Suppose you deposit $17,000 into an account that offers 5.5% annual interest. Use the CAF to compute the balance in the account after 6.5 years. 7. The same types of computations could be done to answer Key Question II. (a) The population triples every hour. So, we can think of the compounding period as being one hour. How many compounding periods have passed after 0 minutes? After 40 minutes? (b) Let B(k) be the number of bacteria present after k hours. Write an equation that relates: i. B(1) to B(0) ii. B() to B(0) iii. B(3) to B(0) iv. B(k) to B(0). (c) Your answer to part (b)iv gives a formula for B(k) that you should believe works as long as k is a whole number of hours. However, you could do work similar to what we did with the bank accounts to show that this formula also works if k is not a whole number. Use this fact to answer Key Question II. 8. Here are some routine exercises to give you practice in applying the formulas with fractional periods. (a) Find the value of $10,000 after 4.5 years, if it carries interest of 15% per year. (b) A bacteria strain multiplies by a factor of 1.5 every hour. How many bacteria are in a colony after 10 minutes, if the colony started with 5 million bacteria? (c) Money gains interest at a rate of 50% every 4 years. What is the value of $5,000 after 6 months? after 7 years? (d) An account gains interest at a rate of 10% per annum. How much must you have now in order to have $4,000.5 years from now? (e) Bacteria double every 30 minutes. How much must you have now in order to have 5 million bacteria in 45 minutes? 4
MATH 111 Worksheet 1 Replacement Answers 1 (b) No, this doesn t seem fair. Your friend invests more money than you do and should receive more interest. (b) multiplier=1.15, proportionate change = multiplier 1 = 0.15 (c) If we were earning 1.5% per year for 8 years, then we would get $0,56.8 at the end of the 8 years. But we only get $16,000. 3 (a) p = 1 = 0.414136, X = $11, 313.71 (b) You ll earn $3,317.71 in interest; your friend will earn $4686.9 in interest. (c) You re each earning 4.14136% on your respective investments. 4 (a) A(1) = A(0) + A(0) r OR A(1) = A(0) (1 + r) (b) A() = A(1) (1 + r); A() = A(0) (1 + r) (c) A(3) = A() (1 + r); A(3) = A(0) (1 + r) 3 (d) multiplier=1 + r; proportionate change = multiplier 1 = r (e) A(k) = A(0) (1 + r) k (f) r = 8 1 = 0.0905077. The effective annual interest rate is 9.05077%. 5 (a) A(0) = $1, 000; A(1) = $1, 060 (b) If r = 0.03, then A(1) = A(0) (1 + 0.03) = 1, 000 (1.03) = $1060.90. (c) r = 1.06 1 = 0.09563 (d) X = 1, 000(1.06) 1/ = $1, 09.56 6 A(6.5) = 17, 000 (1 + 0.055) 6.5 = $3, 756.19 7 (a) 0 minutes = 1 of an hour; 40 minutes = of an hour 3 3 (b) i. B(1) = 3 B(0) ii. B() = 3 B(0) iii. B(3) = 3 3 B(0) iv. B(k) = 3 k B(0) (c) B ( 1 3 8 (a) $18,756.00; ) = 3 1/3 10 million = 14.4 million; B ( 3) = 3 /3 10 million = 0.801 million (b) 0.668 million; (c) $559.95; $10,165.5; (d) $3151.94; (e) 1.768 million. 5