Activity 1.1 Compound Interest and Accumulated Value

Activity 1.1 Compound Interest and Accumulated Value Remember that time is money. Ben Franklin, 1748 Reprinted by permission: Tribune Media Services Broom Hilda has discovered too late the power of compound interest. One dollar invested at an annual interest rate of 3% grows according to the table shown on the right. That is: Number of Value at the Years, n End of n th Year After one year, the dollar has accumulated $0.03 interest, so the investment has grown to a value of $1.03. In the second year, the entire $1.03 earns interest (not just the original $1.00 invested) and so the investment has grown to $1.03 plus the interest on $1.03. So the total value is $1.03 + (.03)($1.03). Applying the distributive law, we express this as $(1.03) 2. Each year the investment grows to 1.03 times its value at the end of the previous year, so the value at the end of three years is $(1.03) 3. In general, the value at the end of the n th year is $(1.03) n, so the value at the end of the 1 500 th year is $(1.03) 1 500. 1 2 3 1 500 1.03 (1.03) 2 (1.03) 3 (1.03) 1 500 The accumulated value of $10 at the end of the 1 500 th year would be 10(1.03) 1 500 or about $180 000 000 000 000 000 000. PART 2: ACTIVITY 1.1 COMPOUND INTEREST AND ACCUMULATED VALUE 33

Activity 1.2 Using Diagrams to Show How Money Grows It is often useful to use a diagram to relate the values of an investment at different times. The diagram below shows what the value of Broom Hilda s original $10 investment would have been at the end of various years if she hadn t bought cigars. Years after the original investment The example above shows that the accumulated value A to which an amount $P grows depends upon two factors, the annual rate of growth i and the number of years n for which it is invested. In the example above, the accumulated value was 10(1.03) 1 500, the principal was $10, the annual growth rate 3% and the period 1 500 years. Legend A = accumulated value P = principal i = annual growth rate n = period (years) Usually the accumulated amount and the principal are expressed in dollars, the growth rate as a percentage and the period in years. The accumulated value A is expressed in terms of P, i, and n in the formula: A = P (1 + i) n Accumulated Value of $1 000 growing at 6% per annum To find the accumulated value of $1 000 in 20 years growing at a rate of 6% per annum, we substitute P = 1 000, n = 20, and i = 0.06 into the formula above to obtain A = 1 000(1.06) 20 or 3 207.14. That is, $1 000 growing at 6% per annum becomes $3 207.14 at the end of 20 years. Years after the original investment This line graph shows, for a growth rate of 6%, the accumulated value of $1 000 at the end of every year for 20 years. 34 PART 2: ACTIVITY 1.2 USING DIAGRAMS TO SHOW HOW MONEY GROWS

Activity 1.3 How Money Grows at Different Interest Rates People who understand money often take some limited risks to gain an extra percentage point or two in the rate of return on their investments. In the following worked example, you will see how a small increase in an interest rate or stock yield can make a significant difference over a long period of time. Worked Example 1 A B What is the accumulated value of $1 000 at the end of each year for 30 years for each of the following annual rates of growth: 4%, 6%, 8%, 10%, and 12%? Display this data on a line graph for each growth rate. Does doubling the growth rate double the accumulated value? Solution A To display on a spreadsheet the accumulated value of $1 000 at the end of each year for 30 years for the given growth rates we enter the headings and formulas shown in the spreadsheet below. A B C D E F 1 Year @i=4% @i=6% @i=8% @i=10% @i=12% 2 1 =1000(1.04) =1000(1.06) =1000(1.08) =1000(1.10) =1000(1.12) 3 =$A2+1 =$B2*(1.04) =$C2*(1.06) =$D2*(1.08) =$E2*(1.10) =$F2*(1.12) 4 =$A3+1 =$B3*(1.04) =$C3*(1.06) =$D3*(1.08) =$E3*(1.10) =$F3*(1.12) 5 =$A4+1 =$B4*(1.04) =$C4*(1.06) =$D4*(1.08) =$E4*(1.10) =$F4*(1.12) We use the Fill Down command to extend these formulas to the 31st row of the spreadsheet (i.e., year 30). When we display the numbers (rather than the formulas), we obtain the display below. A B C D E F 1 2 3 4 5 Year 1 2 3 4 @i=4% $1040.00 $1081.60 $1124.86 $1169.86 @i=6% $1060.00 $1123.60 $1191.02 $1262.48 @i=8% $1080.00 $1166.40 $1259.71 $1360.49 @i=10% $1100.00 $1210.00 $1331.00 $1464.10 @i=12% $1120.00 $1254.40 $1404.93 $1573.52 Accumulated Value of $1 000 over 30 Years for Various Growth Rates B Using the Chart Menu, we obtain a graph like the one shown at right. We observe that $1 000 invested over 30 years at 6% grows to less than $6 000, while that same investment at 12% grows to about $30 000! In this case, doubling the growth rate from 6% to 12% increases the accumulated value about 5 times as much. In mathematical terms we say that accumulated value grows exponentially with the growth rate. PART 2: ACTIVITY 1.3 HOW MONEY GROWS AT DIFFERENT INTEREST RATES 35

Activity 1.4 Compounding Periods Less Than One Year Broom Hilda s investment, described in the cartoon, was compounded annually. However, many investments and most loans today are compounded semi-annually, monthly or even daily! When the compounding period is less than a year, the quoted annual interest rate (called the nominal rate), is less than the actual interest rate (called the effective rate). Definition: If interest is compounded m times per annum, then the effective interest rate for one compounding period is given by i /m where i is the nominal annual interest rate. The next example explores the relationship between i and the effective annual interest rate j. Worked Example 2 A A credit card charges a (nominal) interest rate of 15% on overdue balances. If the interest is compounded monthly, what is the effective annual interest rate? B A department store offers a computer system for $8 000 payable now or $8 800 payable six months from now. What nominal annual rate of interest compounded monthly would make the two prices equivalent (ignoring sales tax)? C A loan has a nominal interest rate i per annum, compounded m times per year. Write an expression for the effective annual rate j in terms of i and m. Express i in terms of j and m. Solution A The monthly interest rate is the nominal annual interest rate divided by 12, i.e. 0.15/12 = 0.0125. Therefore, the amount owing at the end of any month is (1.0125) times the amount owing at the beginning of that month (assuming no payments or new charges). At the end of 12 months, the amount owing is (1.0125) 12 1.16075 times the amount owing at the beginning of the year. That is, the amount of accrued interest is equivalent to an annual rate of 16.075%. The effective interest rate is 16.075%. B C If i denotes the nominal annual interest rate (expressed as a decimal number), then the accumulated value of $8 000 at the end of 6 months is $8 000 1+ 12 i 6. If this is equivalent to $8 800, then $8 000 1+ 12 i 6 = $8 800. That is 1+ 12 i 6 = 1.1. Solving for i yields i = 12 ( 6 1.1 1) 0.192. This shows that $8 000 at a nominal annual rate of 19.2% compounded monthly grows to $8 800 in 6 months. By definition, the effective interest rate over one compounding period is i/m. Over m compounding periods (1 year), $1 will have an accumulated value of $ 1+ i m m. The effective (actual) interest rate per annum j is the amount of interest accumulated on $1 by year end. That is, $ 1+ m i m 1 = j. Solving for i in terms of j and m, we obtain i = m ( m 1 + j 1). 36 PART 2: ACTIVITY 1.4 COMPOUNDING PERIODS LESS THAN ONE YEAR

Activity 1.5 Exploratory Activity A How Long Does It Take to Double Your Money? In Worked Example 1 of this activity, we saw how fast the value of an investment grows as the growth rate is increased. Another way to examine the effect of the growth rate on accumulated value is to compute for various growth rates the number of years it takes for an investment to double in value. Follow the procedure outlined in Worked Example 1 to create a spreadsheet for the accumulated value of $1 000 at the end of each year for 20 years, for growth rates of 4%, 6%, 8%, 10%, and 12%. Use the same formulas in the spreadsheet as shown in that worked example. The doubling time for each growth rate is the approximate number of years it takes an investment of $1 000 to have an accumulated value of $2 000. 1 Use your spreadsheet to help you complete the second row in the table below. Internet Exploration To explore the Rule of 72 further, check out this web site. www.moneychimp.com 2 Multiply each number in the first row by the number below it in the middle row and write the product in the bottom row. What pattern do you discover in the numbers in the bottom row? 3 Use the pattern you discovered in 2 to estimate the doubling time d corresponding to a growth rate of 7%. What value would you expect the expression (1.07) d to have if d is the exact doubling time for a 7% growth? 4 Calculate the exact doubling time corresponding to a growth of 18%. Does this doubling time follow the same pattern as you discovered for the other growth rates? Explain. 5 Use the pattern you discovered in 2 to estimate the number of years it would take an investment growing at 9% to quadruple in value. 6 An investment of $18 000 grows in value to $71 400 in a period of 16 years. Use the pattern you discovered above to estimate its average annual rate of growth. Explain how you obtained your estimate. Growth Rate as a Percent 4% 6% 8% 10% 12% Doubling Time in Years Growth Rate X Doubling Time Money invested at 12% doubles every 6 years. PART 2: ACTIVITY 1.5 EXPLORATORY ACTIVITY A 37

Activity 1.6 Exploratory Activity B And If You Have a Graphing Calculator In this activity and elsewhere, the keying sequences are for the TI-83 or TI-83 Plus graphing calculators because they are commonly accessible. However, if you have a different graphing calculator, follow the general instructions and apply the keying sequences appropriate to your calculator. 7 Create a table that shows the accumulated value of $1 at the end of every year for the first 30 years when growing at the rate of 12% annually. This can be accomplished by pressing Y= and entering the function Y 1 = (1.12)^X. 8 With Y 1 selected, press 2nd [TABLE] and scroll down the table display to check that the accumulated values for 12% growth match the values found in your spreadsheet display. 9 Set the window variables to 0 X 20; 0 Y 10 as shown in the display and press GRAPH to obtain the graph of the accumulated values of an investment of $1 at a growth rate of 12%. Growth Rate as a Percent Doubling Time in Years Growth Rate X Doubling Time 4% 6% 8% 10% 12% 10 Define Y 6 = 2 and graph Y 1 and Y 6. What is the significance of the point of intersection of these two curves? With these graphs displayed, press 2nd [CALC] 5 ENTER ENTER ENTER to obtain the point of intersection of these two graphs. Record the x-coordinate of the point of intersection in the appropriate cell of the table shown at left. 11 Repeat the procedure above for the functions corresponding to the other growth rates to complete the table. Is the Rule of 72 (that you discovered earlier) an exact relationship? Explain. 12 Graph the doubling time Y as a function of the growth rate X. First solve the equation (1 + X/100) Y = 2, for Y in terms of X. Then graph this function in the window 0 X 25; 0 Y 20. Graph also the function 72/X. Compare both graphs and describe the range of values of the growth rate for which the Rule of 72 is the most accurate. 38 PART 2: ACTIVITY 1.6 EXPLORATORY ACTIVITY B

Activity 2.1 Present Value Annual income twenty pounds, annual expenditure nineteen nineteen six, result happiness. Annual income twenty pounds, annual expenditure twenty pounds ought and six, result misery. Charles Dickens, 1851 To become a millionaire, you need to maximize your capital. Capital is not only money, but also assets, securities and inventory. You Can Be a Millionaire If You Start Soon Enough! How much do you think you would need to put into an RRSP (registered retirement savings plan) when you are 25 years old to have $1 000 000 when you reach 65 years of age? The answer is that it depends upon the rate at which your money grows. If your money or investment grows at the rate of 6% per annum, you would need to put almost $100 000 into your RRSP at age 25. However, if your investment grows at double that rate i.e., 12%, you would need only about $10 000! What a difference!? An understanding of the power of compound growth and how to maximize the rate of growth of an investment portfolio together constitute the cornerstone of wealth building. In this activity, you will learn how to calculate the amount of money P that must be invested now, called the present value, to grow to a particular amount A at some future time. In Activity 1, you discovered that $P invested now at an annual growth rate of i increases to $A in n years where A = P (1 + i) n. This formula allows us to calculate A for a given value of P. To find the present value P for a given future amount A, we solve the equation above for P to obtain P = A/(1 + i) n or, P = A (1+i) n To calculate the amount of money P that must be invested at 6% per annum now to grow to $1 000 000 in 40 years we substitute A =1 000 000, i = 0.06 and n = 40 into the formula above to obtain P = 1 000 000(1.06) 40 or 97 222.20. That is, $97 222.20 invested now at 6% per annum will grow to one million dollars in 40 years. This is called the present value of one million dollars when invested at 6% for 40 years. Use the formula above to calculate the present value of one million dollars when invested at 12% for 40 years. These examples show why growth rate is so important in long-term investments. 44 PART 2: ACTIVITY 2.1 PRESENT VALUE

Activity 2.2 Strip Bonds When a government or a company needs to borrow money, it issues a bond. A bond is a loan that the issuer promises to repay at a future time. When you purchase a bond, you are lending money to the bond issuer. The amount to be repaid to you in future is called the face value. The date when repayment is due is called the maturity date. The annual rate of interest paid to you is called the bond rate. The amount you pay for the bond is called the cost price. A Strip Bond Is Also Called a Zero Coupon Bond. Why? A typical bond such as a Government of Canada bond or a bond issued by a company has two parts: redeemable coupons that pay you, the bondholder, equal monthly or semi-annual installments of interest. the face value of the bond that is paid in cash at maturity. Brokerage firms sometimes separate these two parts and sell the coupons to investors who want regular payments, and the bond (without its coupons) to other investors who prefer to receive the face value at a future date. The bond stripped of its coupons is a special type of bond called a strip bond. The important question for you as an investor is: How much should I pay for a strip bond that pays $10 000, say, 20 years from now if I want a yield of 6% per annum? An equivalent question is: How much money $P invested now at 6% per annum will grow to an amount A = $10 000 in 20 years? Substitution of i = 0.06 and A = 10 000 into the present value formula on the previous page yields: P = 10 000(1.06) 20 or P = 3 118.05. That is, $3 118.05 should be paid now for a strip bond of face value $10 000 maturing 20 years from now to yield a 6% per annum return. Canadian financial author and former Minister of National Revenue, Garth Turner wrote in 1997 in his 1998 RRSP Guide: How to Build Your Wealth and Retire in Comfort (page 161): In May 1994, I bought a Government of Ontario strip bond that matures on February 18, 2002. The annual yield was good at the time, 8.86%, and on maturity day that bond will be worth $194 000 inside my RRSP. But the actual cost to buy it was $100 412, which worked out to be $51.75 for every $100 worth of bond. In effect, I was arranging to double my money in less than eight years without risk. PART 2: ACTIVITY 2.2 STRIP BONDS 45

Activity 2.3 How Much at Age 30 Becomes $1 000 000 at Age 60? How much money you need to invest at age 30 to yield $1 000 000 when you turn 60 depends on the annual growth rate of your investment. You learned in the previous activity how even a small increase in the annual growth rate over a long period of time yields substantially greater gains in the accumulated value. In this activity, you will investigate how the growth rate affects how much you need to invest to reach your goals. Worked Example 1 Use a spreadsheet to display the present value (cost) of the bond on January 2 of each year from 2000 to 2030 for each of the following annual yields: 4%, 6%, 8%, 10%, and 12%. Does doubling the yield rate reduce the cost on January 2, 2000 to half? Solution If the bond is to yield the investor a 4% annual return, then its value on January 2 for each year from 2000 to 2030 can be displayed on a time line as shown here. Why do we have negative exponents? To display on a spreadsheet these values on January 2 of each year for 30 years for the given yields we list the years from 2000 to 2030 A.D. by entering the formulas shown in column A below. We use the Fill Down command to extend these formulas to the 32nd row of the spreadsheet (i.e., year 2030). To enter the formulas in column B for the 4% yield, we observe from the time diagram above that the value of the 4% yield strip bond in year n is 1 000 000 (1.04) (2030 n) or 1 000 000(1.04) (n 2030). For the entry in column B of row 2 in the spreadsheet, n is the number in cell A2 so we write 1 000 000(1.04) (n 2030) as 10 6 *(1.04)^($A2 2030). We use the Fill Down command to extend these formulas to the 32 nd row of the spreadsheet (i.e., year 2030). Similarly we enter the corresponding formulas in columns C, D, E, and F (columns E and F are not shown here). A 1 Year 2 2000 3 =$A2+1 4 =$A3+1 5 =$A4+1 B @i=4% =10^6*(1.04) ^ ($A2 2030) =10^6*(1.04) ^ ($A3 2030) =10^6*(1.04) ^ ($A4 2030) =10^6*(1.04) ^ ($A5 2030) C @i=6% =10^6*(1.06) ^ ($A2 2030) =10^6*(1.06) ^ ($A3 2030) =10^6*(1.06) ^ ($A4 2030) =10^6*(1.06) ^ ($A5 2030) D @i=8% =10^6*(1.08) ^ ($A2 2030) =10^6*(1.08) ^ ($A3 2030) =10^6*(1.08) ^ ($A4 2030) =10^6*(1.08) ^ ($A5 2030) 46 PART 2: ACTIVITY 2.3 HOW MUCH AT AGE 30 BECOMES $1 000 000 AT AGE 60?

When we display the numbers (rather than the formulas), we obtain the display below. 1 2 3 4 5 A Year 2000 2001 2002 2003 B @i=4% $308318.67 $320651.41 $333477.47 $346816.57 C @i=6% $174110.13 $184556.74 $195630.14 $207367.95 D @i=8% $99377.33 $107327.52 $115913.72 $125186.82 On comparing columns B and D, we see that when the yield is 8%, the price of a bond in the year 2000 is less than one-third the price of a bond with a yield of 4%. Doubling the yield rate reduces the price in the year 2000 to much less than half! To be a millionaire at age 60, you must invest at age 30 either $174 110.13 in a bond with a 4% yield or only $33 377.92 in a bond with a 12% yield. You learned in Worked Example 1 that the yield rate of a bond has a significant effect on its price. When the yield rate is high, the bond is much cheaper. However, as the yield rate decreases, the price of the bond approaches its face value. The number of years to maturity also affects the price of a bond. The next example shows how a bond with a maturity date far in the future (called a long bond) has a significantly smaller price than a bond close to maturity. Worked Example 2 Using the spreadsheet for the strip bond described in Worked Example 1, display the present value (cost) of the bond on January 2 of each year from 2000 to 2030 for each of the following annual yields: 4%, 6%, 8%, 10%, and 12%. Describe how the price of the million dollar bond changes as the maturity date approaches. Solution Using the chart menu on the spreadsheet obtained in the solution of Worked Example 1, we obtain the graph shown at right. The graph reveals that the price of the strip bond grows exponentially as the maturity date approaches. Furthermore, we observe that the graphs converge as we approach the year 2030 indicating that as the maturity date draws near, the price of the bond does not vary much for different yield rates. In fact, the differences in the bond prices for various yield rates increase as the time to maturity is increased. Can you explain why this is? Price of a Million Dollar Strip Bond between 2000 and 2030 for Various Yield Rates. PART 2: ACTIVITY 2.3 HOW MUCH AT AGE 30 BECOMES $1 000 000 AT AGE 60? 47

Million Dollar Winners! The Stewart Family Revocable Trust of Bedford today claimed the top prize from the Texas Lottery scratch-off game Weekly Grand of $1 000 a week for the next 20 years. The total prize payout comes to $1 040 000.00. This makes the 44th winner for the Texas Lottery s Weekly Grand scratch-off game since its inception in 1995. (Austin TX) Activity 2.4 Exploratory Activity Did They Really Win a Million Dollars? A recent study revealed that a large number of Canadians expect that lottery winnings will constitute their major source of income for their retirement years. But if you consider the theory of probability, you ll find that this expectation is very unrealistic. In spite of the huge odds against winning the big prize, a small fraction of the world s population does hit the jackpot. The article to the left appeared in a press release by the Texas Lottery in April 2000. Grand Winner Claims Prize Worth $1 Million 1 Explain how the value of $1 040 000.00 was obtained. 2 Do you think that the prize was worth about the same as receiving a million dollars immediately? Explain why or why not. 3 If you won the prize now and you wanted to sell it for immediate cash, estimate roughly what you think you should receive in a lump sum payment. Show how you obtained your estimate. 48 PART 2: ACTIVITY 2.4 EXPLORATORY ACTIVITY

4 The weekly payments of $1 000 can be shown on a time line like the one below. Fill in the present value of each payment, assuming a yield rate of 5% per annum compounded weekly. The first one is done for you. 5 Use a spreadsheet or the formula for the sum of a geometric series to add the present values of all the 1 040 payments in Exercise 4. 6 Compare your answer in Exercise 5 with your rough estimate in Exercise 3. Are you surprised? Explain why or why not. 7 Why do you think the Texas Lottery Commission advertises the prize value as the sum of all the payments instead of the present value of all the payments? And If You Have a Graphing Calculator... You can verify our answer to Exercise 5 by using the TVM Solver (time-value-ofmoney) on the FINANCE menu of the TI-83 and TI-83 Plus. To activate press: APPS ENTER ENTER Then enter the number of payments, N = 1 040, and the interest rate I = 5%. Leave the present value PV unchanged and enter the payment value as 1 000 using the < > negative to show money paid out. Then enter the future value FV = 0 (value is zero after the last payment), the number of payments per year P/Y and the number of compoundings per year C/Y. Place the cursor opposite PV (payment) and press ALPHA [SOLVE]. Compare the present value you computed in Exercise 5 with the value opposite PV. Do they agree? Use the TVM menu to calculate the present value of a 20-year bond that has semiannual coupons of $1 000, and a face value of $10 000 if money is worth 5% compounded annually. N = 1 040 I% = 5 PV = PMT = -1 000 FV = 0 P/Y = 52 C/Y = 52 PMT: END BEGIN PART 2: ACTIVITY 2.4 EXPLORATORY ACTIVITY 49

Activity 2.5 Exercises Government of Canada Bond Yields Canada 2-year bond 6.25% Canada 10-year bond 6.19% Canada 30-year bond 5.81% 1 Explain why a cash payment of $1 000 now is worth more than a cash payment of $1 000 five years from now. 2 Determine the present value of each future amount if money is worth 6% per annum compounded monthly. a) $1 000 a month from now b)$5 000 a year from now c) $10 000 five years from now 3 The table to the left shows the yields for Government of Canada strip bonds on a particular day in May 2000. a) Calculate the cost of each of the bonds if each has a face value of $1 000. b) Which bond would you choose if you were to invest? Give reasons for your answer. 4 If money can earn 8% per annum compounded annually, what is worth more: $600 000 now or one million dollars ten years from now? Show how you obtain your answer. 5 How much money must be invested now at 6% per annum compounded monthly to grow into one million dollars: a) 15 years from now? b)25 years from now? c) 35 years from now? 50 PART 2: ACTIVITY 2.5 EXERCISES

Activity 3.1 Annuities & Installment Payments A Tale of Twins Amy and Amanda are identical twins at least in their external appearance. They have very different investment plans to provide for their retirement. Study both plans and then decide who you think will have more money in her RRSP when they both reach 65 years of age if their investments yield 9% per annum compounded annually. Amy Amanda Investing is all math. Harry Markowitz Nobel Prize for Economics, 1980 RRSP = Registered Retirement Savings Plan This is an account into which money can be placed so that it can compound freely without tax until withdrawn during retirement. I plan to put $1 000 each year into an RRSP account beginning on my birthday when I turn 20 years of age. When I turn 29, I will make my final contribution of $1 000. Then I will leave the money invested in my RRSP and let it grow, but I will make no further contributions. Q How much will Amy contribute to her RRSP between ages 20 and 30? I want to have some spending money while I m young, so I will not open an RRSP until I reach 30 years of age. However, when I turn 30 years of age, I will put $1 000 into an RRSP account each and every year until I turn 64. Then I will make my final contribution of $1 000. Q How much will Amanda contribute to her RRSP between ages 30 and 65? Both Amy and Amanda plan to make deposits of $1 000 into their RRSP accounts every year. Any such series of equal payments made at regular intervals of time is called an annuity. Annuities can be used to build assets or to pay off debt as in installment buying or the discharge of a mortgage. 56 PART 2: ACTIVITY 3.1 ANNUITIES & INSTALLMENT PAYMENTS

To calculate the value of Amy s RRSP at age 65, we cannot simply add her $1 000 contributions together, because the contributions are made at different times and therefore have grown into different amounts. To determine the value of Amy s RRSP at age 65, we must compute the accumulated value of each contribution on Amy s 65th birthday and then add them together, as shown in the time diagram. Amy s Plan To find the accumulated value of Amy s RRSP when she reaches age 65, we sum the accumulated values of all the contributions shown in the column under Age 65, using the fact that it is a geometric series of 10 terms with first term 1 000(1.09) 36 and common ratio (1.09). Accumulated value ($) = 1 000(1.09) 36 + 1 000(1.09) 37 + 1 000(1.09) 38 + + 1 000(1.09) 45 = 1 000[(1.09) 46 (1.09) 36 ] / 0.09 = 338 061.23 That is, Amy s RRSP will have an accumulated value of $338 061.23 on her 65 th birthday. Amanda s Plan To find the accumulated value of Amanda s RRSP when she reaches age 65, we sum the accumulated values of all the contributions, using the fact that it is a geometric series of 35 terms with first term 1 000(1.09) and common ratio (1.09). Accumulated value ($) = 1 000(1.09) + 1 000(1.09) 2 + 1 000(1.09) 3 + + 1 000(1.09) 35 = 1 000[(1.09) 36 (1.09)] / 0.09 = 235 124.72 That is, Amanda s RRSP will have an accumulated value of $235 124.72 on her 65 th birthday. Compare the contributions made by Amy and Amanda and the values of their RRSPs at age 65. Are you surprised? Explain why or why not. PART 2: ACTIVITY 3.1 ANNUITIES & INSTALLMENT PAYMENTS 57

Activity 3.2 Buying on the Installment Plan Using a Graphing Calculator to Verify your Answer We can verify our answer by using the TVM Solver (timevalue-of-money) on the FINANCE menu of the TI-83 Plus. To activate this, we press APPS ENTER ENTER. Then we enter the number of payments N, the interest rate I, the principal PV, the future value FV, the number of payments per year P/Y, and the number of compoundings per year C/Y, as in the display. We place the cursor opposite PMT(payment) and press ALPHA [SOLVE]. The display shows 444.88895 verifying that equal payments of $444.89 are required. N = 60 I% = 12 PV = 20 000 PMT = 0 FV = 0 P/Y = 12 C/Y = 12 PMT: END BEGIN Worked Example 1 Most people purchase cars on the installment plan. The lending institution advances a loan of $P called the principal. The borrower repays the loan in equal installments over a given number of years, n (called the term of the loan). Both parties agree to a particular interest rate i and compounding period (usually a month), and the lending institution informs the borrower what equal installment payments x are required to repay the principal. This example shows how to calculate the value of x. A car loan of $20 000 is offered at 12% per annum compounded monthly 1. What equal monthly payments are required to repay the loan over a term of 5 years? Solution To determine the amount of the monthly payments, we compute the present value of all the 60 payments made over the 5-year term. The sum of these present values should equal the amount of the loan, i.e., $20 000. Observe that when we calculate future value of money, we multiply by the growth factor, but when we calculate present value of future payments, we divide by the growth factor. The total present value of all the payments is given by: x (1.01) + x (1.01) +... x + 2 (1.01) + x 59 (1.01) 60 This is a geometric series with first term x (1.01) and common ratio (1.01) 1. The sum of this series is 100x [1 (1.01) 60 ]. Since the total present value of all the payments should be equal to the principal, i.e., $20 000, we write 100x [1 (1.01) 60 ] = 20 000. Solving this equation for x yields x = 444.89. That is, 60 monthly payments of $444.89 will repay the loan in 5 years. N = 60 I% = 12 PV = 20 000 PMT = -444.88895... 1 Note: 12% per annum compounded monthly means the interest is 1% per month and is calculated at the end of each month. 58 PART 2: ACTIVITY 3.2 BUYING ON THE INSTALLMENT PLAN

Activity 3.3 Calculating the Outstanding Balance on a Spreadsheet In Worked Example 1, we showed how to calculate the payment amounts in an annuity. However, when we have an installment loan, we already know the payment amount, but often wish to determine the balance owing at a particular time. The following worked example shows how we can use a spreadsheet to display the balance after each payment. Worked Example 2 For the $20 000 car loan in Worked Example 1: a) What is the balance of the loan after the 9th payment? b) How much of the 9th payment is allocated to interest, and how much to the principal? Solution To display on a spreadsheet the monthly balances of the $20 000 car loan described on the previous page, we enter the headings and formulas shown in the spreadsheet below. A B C D E 1 Payment # Initial Interest on Payment Final 2 Balance Balance Balance 3 1 20000 =$B3*0.01 444.89 =$B3+C3 D3 4 =$A3+1 =$E3 =$B4*0.01 444.89 =$B4+C4 D4 5 =$A4+1 =$E4 =$B5*0.01 444.89 =$B5+C5 D5 We use the Fill Down command to extend these formulas to the 11 th row of the spreadsheet (i.e., payment #9). When we display the numbers (rather than the formulas), we obtain the display below. A B C D E 1 Payment # Initial Interest on Payment Final 2 Balance Balance Balance 3 1 20000.00 200 444.89 19755.11 4 2 19755.11 197.55 444.89 19507.77 5 3 19507.77 195.08 444.89 19257.96 6 4 19257.96 192.58 444.89 19005.65 7 5 19005.65 190.06 444.89 18750.81 8 6 18750.81 187.51 444.89 18493.43 9 7 18493.43 184.93 444.89 18233.48 10 8 18233.48 182.33 444.89 17970.92 11 9 17970.92 179.71 444.89 17705.74 The 11 th row reveals that the balance before the 9 th payment is $17 970.92, but after the 9 th payment the (final) outstanding balance is $17 705.74. The difference, $17 970.92 $17 705.74 or $265.18, is the amount that has been allocated to the principal. Since the total payment is $444.89, then the interest included in that payment is $444.89 $265.18 or $179.71. Balance after the 9 th payment To verify that we have entered appropriate formulas into the spreadsheet, we can use the Fill Down command to extend the formulas to the 62 nd row (60 th payment). We then display the numbers and check that the balance after the 60 th payment is zero (or very close to it). PART 2: ACTIVITY 3.3 CALCULATING THE OUTSTANDING BALANCE ON A SPREADSHEET 59

Activity 3.4 Exploratory Activity Buy or Lease a Car Which Is the Cheaper Option? Many people debate whether it s cheaper to buy or to lease a car. When you buy a car, you negotiate with the dealer on a purchase price. Then you pay either the purchase price (plus tax, money levied by the government on income and sales) or a portion of the purchase price, called the down payment, and finance the rest. The part that you finance is the principal of an installment loan that is paid off in equal payments over the term of the loan (usually 3 to 5 years). When you lease a car, the dealer purchases the car on your behalf. The purchase price that you negotiate with the dealer is called the cap cost (short for capitalized cost). The lease agreement entitles you to the use of the car for a fixed term (usually for 2 or 3 years) and the option to buy the car at the end of the term at a pre-set price called the residual (value). In return for the use of the car during the term of the lease, you make monthly payments that are substantially less than if you were buying the car, but then you do not own the car at the end of the lease unless you exercise your option to buy. The mathematics of each financial arrangement are shown below. The Finances of Buying a Car 1 Reach agreement with the dealer on the purchase price of the car including optional extras. 2 Add the appropriate GST and PST to the purchase price. 3 Subtract the down payment from 2. 4 Agree upon an interest rate and term for an installment loan with principal obtained in 3. 5 Calculate the monthly payments required to pay off the principal of the loan in 4. The Finances of Leasing a Car 1 Reach agreement with the dealer on the purchase price of the car including optional extras. This is the cap cost. 2 Add to the cap cost the acquisition fees that the dealer charges for administering the lease. 3 Subtract the down payment from 2 to obtain the adjusted cap cost. 4 Subtract the residual from the adjusted cap cost to obtain the amount to be financed. 5 Agree to the terms of the payment schedule based on the amount obtained in 4. 6 Calculate the monthly payments required to pay off the principal of the loan in 5, and add the GST and applicable PST to each payment. 60 PART 2: ACTIVITY 3.4 EXPLORATORY ACTIVITY

Suppose you are interested in a car that sells for $18 995.00. You have saved a down payment of $6 000 that you want to put into a lease or purchase. A car dealer provides you with the purchase agreement and the lease agreement shown below. Read the terms of each offer and fill in the missing data. The Purchase Agreement Purchase Price: $18 995.00 Loan Information: 9% annually compounded monthly term of 3 years 1 The purchase price plus tax is: $ 2 The amount to be financed is: $ 3 The monthly payment is: $ 4 The accumulated value of all the monthly payments at the end of 3 years, assuming that they could have been invested at 3% per annum compounded monthly, is: The Lease Agreement Cap Cost: $18 995.00 Acquisition Fees: $5 286 Residual: $10 000.00 Lease Information: 9% annually compounded monthly term of 3 years 5 The cap cost is: $ 6 The adjusted cap cost is: $ 7 The amount to be financed is: $ 8 The monthly payment (including GST and PST) is: $ 9 The accumulated value of all the monthly payments at the end of 3 years, assuming that they could have been invested at 3% per annum compounded monthly, is: $ $ 10 Write a brief report comparing the costs of purchasing and leasing for the example given here. Discuss the advantages of each form of financing a car and describe the kinds of car use that would favour each method. In this activity, we have focussed on price only. Some people argue that leasing has the advantage that it doesn t tie up capital that could be invested elsewhere at higher rates of return. Tax considerations might also affect your decision whether to lease or buy. PART 2: ACTIVITY 3.4 EXPLORATORY ACTIVITY 61