Chapter 3Mathematics of FinanceMathematics of FinanceSection 2Compound and Continuous Compound InterestppCompound Interest Unlike simple interestcompound intereston an amountUnlike simple interest, compound intereston an amount accumulates at a faster rate than simple interest. The basic idea is that after the first interest period, the amount of interest is added to the principal amount and then the interest is computed on this higher principal. The latest computed interest is then added to the increased principal and then interest is calculated again. This process is 2gpcompleted over a certain number of compounding periods. The result is a much faster growth of money than simple interest would yield. Example Suppose a principal of $1.00 was invested in an account i 6% l i t t d d thl Hpaying 6% annual interest compounded monthly. How much would be in the account after one year? 3Solution Solution: Using the simple interest formula A = P (1 + rt) we obtain:we obtain: amount after one month after two months after three months ()()22230.061 (1) 1 1 0.005 1.005120.061.005(1 ) 1.005(1.005) 1.005120.061.005 1 1.005 1.005 1.00512+=+=+= =⎛⎞+= =⎜⎟⎝⎠4After 12 months, the amount is 1.00512 = 1.0616778.With simple interest, the amount after one year would be 1.06.The difference becomes more noticeable after several years.Graphical Illustration ofCompound Interest Growth of 1 00 compounded monthly at 6% annual interestGrowth of 1.00 compounded monthly at 6% annual interest over a 15 year period (Arrow indicates an increase in value of almost 2.5 times the original amount.)22.53Growth of 1.00 compounded monthly at 6% annual interest over a 15 year period 500.511.5General Formula In the previous example the amount ton060⎞⎛ This formula can be generalized to whereAis the future amountPis the principalris the interestIn the previous example, the amount to which 1.00 will grow after n months compounded monthly at 6% annual interest is 1mtrAPm⎛⎞=+⎜⎟⎝⎠nn)005.1(1206.01 =⎟⎠⎞⎜⎝⎛+6where Ais the future amount, Pis the principal, ris the interest rate as a decimal, m is the number of compounding periods in one year, and t is the total number of years. To simplify the formula, ()1nAPi=+rimnmt==Compound Interest General Formula ()1nAPi=+where i = r/mA = amount (future amount) at the end of n periodsP = principal (present value)r = annual nominal rate()7m = number of compounding periods per yeari = rate per compounding periodt = total number of compounding periodsExample Find the amount to which $1500 will grow if compounded gpquarterly at 6.75% interest for 10 years. 8Example Find the amount to which $1500 will grow if compounded quarterly at 6.75% interest for 10 years.  Solution: Use()1nAPi=+10(4)0.06751500 14A⎛⎞=+⎜⎟⎝⎠9 Helpful hint: Be sure to do the arithmetic using the rules for order of operations. 2929.50A =Same Problem Using Simple Interest Using the simple interest formula the amount to which $1500Using the simple interest formula, the amount to which $1500 will grow at an interest of 6.75% for 10 years is given byA = P (1 + rt)= 1500(1+0.0675(10)) = 2512.50which is more than $400 less than the amount earned using10which is more than $400 less than the amount earned using the compound interest formula. Changing the number of compounding periods per yearTo what amount will $1500 grow if compounded daily at 6 75%interest for 10 years?6.75%interest for 10 years? 11Changing the number of compounding periods per yearTo what amount will $1500 grow if compounded daily at 6 75%interest for 10 years?6.75%interest for 10 years? Solution: = 2945.87This is about $15.00 more than compounding $1500 quarterly at 6.75% interest. 10(365)0.06751500 1365A⎛⎞=+⎜⎟⎝⎠12Since there are 365 days in year (leap years excluded), the number of compounding periods is now 365. We divide the annual rate of interest by 365. Notice, too, that the number of compounding periods in 10 years is 10(365)= 3650.Effect of Increasing the Number of Compounding PeriodsIf the number of compounding periods per year isIf the number of compounding periods per year is increased while the principal, annual rate of interest and total number of years remain the same, the future amount of money will increase slightly. 13Computing the Inflation Rate Suppose a house that was worth $68,000 in 1987 is worth $,$104,000 in 2004. Assuming a constant rate of inflation from 1987 to 2004, what is the inflation rate?14Computing the Inflation RateSolution Suppose a house that was worth $68,000 in 1987 is worth Solution:$,$104,000 in 2004. Assuming a constant rate of inflation from 1987 to 2004, what is the inflation rate?1. Substitute in compound interest formula.Solution:()()171717104,000 68,000 1104,000168,000104,000(1 )rrr=+→=+ →=+→152. Divide both sides by 68,0003. Take the 17throot of both sides of equation 4. Subtract 1 from both sides to solve for r. 1717(1 )68,000104,0001 0.025368,000rr=+→−= =Computing the Inflation Rate(continued ) If the inflation rate te ato ateremains the same for the next 10 years, what will the house be worth in the year 2014? 16Computing the Inflation Rate(continued ) If the inflation rate  Solution: From 1987 to 2014 is If the inflation rate remains the same for the next 10 years, what will the house be worth in the year 2014? a period of 27 years. If the inflation rate stays the same over that period, r = 0.0253. Substituting into the compound interest formula, we have 172768,000(1 0.0253) 133,501A =+ =Growth Time of an Investment How long will it take for $5,000 g$,to grow to $15,000 if the money is invested at 8.5% compounded quarterly?18Growth Time of an InvestmentSolution How long will it take for $5,000 Solution:g$,to grow to $15,000 if the money is invested at 8.5% compounded quarterly?1. Substitute values in the compound interest formula. 2. divide both sides by 5,0003. Take the natural logarithm ofSolution:40.08515000 5000(1 )4t=+3 = 1.021254tln(3)=ln1 021254t193. Take the natural logarithm of both sides.4. Use the exponent property of logarithms 5. Solve for t.ln(3) ln1.02125ln(3) = 4t*ln1.02125ln(3)13.06174ln(1.02125)t ==Continuous Compound InterestPreviously we indicated that increasing the number of compounding periods while keeping the interest rate, principal, and time constant resulted in a somewhat higher compounded amount. What would happen to the amount if