1. Chapter 21: Savings Models Lesson Plan Arithmetic Growth and Simple Interest Geometric Growth and Compound Interest A Limit to Compounding A Model for Saving Present Value and Inflation 1 Mathematical Literacy in Today’s World, 8th ed. For All Practical Purposes © 2009, W.H. Freeman and Company

2. Chapter 21: Savings Models Arithmetic Growth and Simple Interest 2 Principal – The initial balance of the savings account. Interest – Money earned on a savings account or a loan. Example: The amount of interest on 10% of the principal of $1000 is 10% × $1000 = 0.10 × $1000 = $100 Simple Interest  The method of paying interest only on the initial balance in an account, not on any accrued interest. Example: The following shows the simple interest of a savings account with a principal of $1000 and a 10% interest rate:  End of first year, you receive $100 interest.  The account total at the start of the second year is $1100.  End of second year, you receive again only $100, which is the interest from the original balance of $1000.  Account total at the beginning of the third year is $1200.  At the end of each year you receive just $100 in interest.

3. Chapter 21: Savings Models Arithmetic Growth and Simple Interest 3 Bonds  An obligation to repay a specified amount of money at the end of a fixed term, with simple interest usually paid annually.  Interest Rate Formula: I = Prt  The total amount accumulated: A = P(1 + rt ) Interest Rate Formula – I = Prt Where: I = Simple interest earned P = Principal amount r = Annual rate of interest t = Time in years Total Amount Accumulated A = P (1 + rt ) Arithmetic Growth – (linear growth) A = P (1 + rt ) Example: Say you bought a 10-year T-note (U.S. Treasure note) today. What would be the total amount accumulated in 10 years at 4.0% simple interest? Answer: P = $10,000, r = 4.0% = 0.40, and t = 10 yrs. Interest, I = Prt = (10,000)(0.04)(10) = $4000 Total A= P(1 + rt ) = 10,000(1 + (0.04)(10)) = 10,000(1.40) = $14,000

4. Chapter 21: Savings Models Geometric Growth and Compound Interest 4 Geometric Growth  Geometric growth is the growth proportional to the amount present (also called exponential growth). Compound Interest  Interest paid on both the principal and on the accumulated interest. Compound Interest Formula – A = P (1 + i ) nt Where: A = Amount earned after interest is made P = Principal amount i = Interest rate per compounding period, which is computed as i = r /n n = Number of compounding periods t = Time of the loan in years Rate per Compounding Period – For a nominal annual rate of interest r compounded n times per year, the rate per compounding period is: i = r / n

5. Chapter 21: Savings Models Geometric Growth and Compound Interest 5 Compounding Period  The amount of time elapsing before interest is paid. For the examples below (annual, quarterly, and monthly compounding), the amount earned increases when interest is paid more frequently. Example: Suppose the initial balance is $1000 (P = $1000) and the interest rate is 10% (r = 0.10). What is the amount earned in 10 years (t = 10) for the following compounding periods, n? To answer this problem you need to use the following equations: Rate per compounding period, i = r / n Compound Interest Formula, A = P (1 + i ) nt Annual compounding: i = 0.10, and nt = (1)10 years A = $1000(1 + 0.10) 10 = $1000(1.10) 10 = $2593.74 Quarterly compounding: i = 0.10/4 = 0.025, and nt = (4)(10) = 40 quarters A = $1000(1 + 0.025) 40 = $1000(1.025) 40 = $2685.06 Monthly compounding: i = 0.10/12 = 0.008333, and nt = (12)(10) = 120mo. A = $1000(1 + 0.10/12) 120 = $2707.04

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8. Chapter 21: Savings Models Geometric Growth and Compound Interest 8  The equation is obtained by manipulating the formula for finding the amount to be paid in the future after compound interest is earned.  With the compound interest formula A = P (1 + i ) nt, we can solve for the present value, P. Present Value Formula – P = Where: P = Present value (principal amount) A = Amount to be paid in the future after compounding interest is earned i = Interest rate per compounding period which is computed as i = r /n n = Number of compounding periods in a year t = The years of the loan A (1 + i ) nt Present Value  The present value P of an amount A to be paid in the future, after earning compound interest for n compounding periods at rate i per period is the future amount paid, A divided by (1 + i) nt.

9. Chapter 21: Savings Models Geometric Growth and Compound Interest Effective Rate  The effective rate of interest is: effective rate where i=r/m and n=mt. Annual Percentage Yield (APY)  The amount of interest earned in 1 year with a principal of $1.  The annual effective rate of interest. Example: With a nominal rate of 6% compounded monthly, what is the APY?  Solution: 9

10. Chapter 21: Savings Models Geometric Growth and Compound Interest 10 Compound Interest Compared to Simple Interest  The graph compares the growth of $1000 with compound interest and with simple interest.  The straight line explains why growth simple interest is also known as linear growth.  Example of geometric and arithmetic growth: Thomas Robert Malthus (1766– 1843), an English demographer and economist, claimed that human population grows geometrically but food supplies grow arithmetically—which he attributed to future problems.

11. Chapter 21: Savings Models A Limit to Compounding 11 A Limit to Compound Interest  The following table shows a trend: More frequent compounding yields more interest.  As the frequency of compounding increases, the interest tends to reach a limiting amount (shown in the right columns). Comparing Compound Interest The Value of $1000 at 10% Annual Interest, for Different Compounding Periods Compounded YearsYearlyQuarterlyMonthlyDailyContinuously 11100.001103.811104.711105.161105.17 51610.511638.621645.311648.611648.72 102593.742685.062707.042717.912718.28

12. Chapter 21: Savings Models A Limit to Compounding 12 Continuous Compounding  As n gets very large, (1+ 1 / n ) n approaches the constant e ≈ 2.71828.  For a principal P deposited in an account at a nominal annual rate r, compounded continuously, the balance after t years is: A = P e r t Example: For $1000 at an annual rate of 10%, compounded n times in the course of a single year, what is the balance at the end of the year? As the quantity gets closer and closer to $1000( e 0.1 ) = $1105.17. No matter how frequently interest is compounded, the original $1000 at the end of one year cannot grow beyond $1105.17. Yield of $1 at 100% Interest ( i = 1) Compounded n Times per Year n(1+ 1 / n ) n 12.0000000 52.4883200 102.5937424 502.6915880 1002.7048138 1,0002.7169239 10,0002.7181459 100,0002.7182682 1,000,0002.7182818 10,000,0002.7182818 It approaches e ≈ 2.71828 (which is the base of the natural logarithms).

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15. Chapter 21: Savings Models A Model for Saving 15 A Savings Plan  To have a specified amount of money in an account at a particular time in the future, you need to determine what size deposit you need to make regularly into an account with a fixed rate of interest. Savings Formula Where: A = Amount accumulated in the future after compounding interest is earned d = Uniform deposits (or payments made) i = Interest rate per compounding period which is computed as i = r /m n = Number of compounding, n=mt t = The years of the savings plan (or loan)  Savings Formula The amount A accumulated after a certain period of time can be calculated by stating a uniform deposit of d per compounding period (deposited at the end of the period) and using a certain interest rate i per compounding period.

16. Chapter 21: Savings Models A Model for Saving Payment Formula:  Solving for d in the Savings Formula so we can calculate how much our periodic payment should be in order to have $A in the future yields: Note: The periodic payment for a sinking fund may be calculated using the Payment Formula. An annuity is a specified number of (usually equal) periodic payments. A sinking fund is a savings plan to accumulate a fixed sum by a particular date, usually through equal periodic payments. 16

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19. Chapter 21: Savings Models Present Value and Inflation 19 Exponential Decay  Exponential decay is geometric growth with a negative rate of growth.  Present Value of a Dollar a Year from Now with Inflation Rate a.  Example: Suppose a 25% annual inflation rate from mid-2009 through mid-2013. What will be the value of a dollar in mid-2013 in constant mid-2009 dollars? Answer: a=0.25, so i=-a/(1+a)=-0.25/(1+.25)=-0.25/1.25=-0.20 and The quantity i=–a/(1+a) behaves like a negative interest rate so we can use the compound interest formula to find the present value of P dollars t years from now.

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22. Chapter 21: Savings Models Present Value and Inflation Depreciation Example Suppose you bought a car at the beginning of 2009 for $12,000 and its value in current dollars depreciates steadily at a rate of 15% per year. What will be its value at the beginning of 2012 in current dollars? Answer: Using the compound interest formula, P = $12,000, i = −0.15, and n = 3. The projected price is A = P (1 + r) n = $12,000 (1 - 0.15) 3 = $7369.50 22

23. Chapter 21: Savings Models Present Value and Inflation 23 Consumer Price Index  The official measure of inflation is the Consumer Price Index (CPI), prepared by the Bureau of Labor Statistics. This index represents all urban consumers (CPI-U) and covers about 80% of the U.S. population. This is the index of inflation that is referred to on television news broadcasts, in newspapers, and magazine articles. Each month, the Bureau of Labor Statistics determines the average cost of a “market basket” of goods, including food, housing, transportation, clothing, and other items. The base period used to construct the CPI-U is from 1982–1984 and is set to 100. CPI for other year cost of market basket in other year 100 cost of market basket in base year From this proportion calculation, you can also compute the cost of an item in dollars for one year to what it would cost in dollars in a different year. cost in year A CPI for year A cost in year B CPI for year B = =

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26. Chapter 21: Savings Models Present Value and Inflation 26 Real Growth  If your investments are growing at say 6% a year and inflation is growing at 3% a year, you cannot simply subtract the two to find the purchasing power of your investments. It’s not that simple. Where P is the initial investment principle and m is the price of goods. The investment rate is r and the inflation is a. q new = Here, you see the two influences on the investment (investment growth rate r and inflation a) have directly opposite effects. Fisher’s effect – Named after the economist Irving Fisher (1867–1947). To understand why you cannot simply find the difference between interest and inflation, you must realize that the gain itself is not in original dollars but in deflated dollars.  At the beginning of the year, your investment would buy the quantity: q old = P/m  At the end of the year, the investment would be: P(1 + r ) m(1 + a)

27. Chapter 21: Savings Models Present Value and Inflation Real Rate of Growth  The real (effective) annual rate of growth of an investment at annual interest rate r with annual inflation rate a is:  Example: Suppose you have an investment earning 8% per year and the annual inflation rate is 3% per year. What is your real rate of growth? Answer: r=.008 and a=0.03, so g=(0.08-0.03)/(1+0.03)=0.05/1.03=0.0485=4.85% 27

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31. Question 1 Solution 31

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33. Question 3 Solution 33