4.7 Compound Interest 2011 November 11, 2011 4.7 Compound Interest Objective: Determine the future value of a lump sum of money. 1 4.7 Compound Interest 2011 November 11, 2011 Simple Interest Formula: te In t s e r I = Prt l pa i c rin P a e y rs in e m ti te a r t s e ter in 2 4.7 Compound Interest 2011 November 11, 2011 A credit union pays interest of 8% interest compounded quarterly on a savings account. If $1,000 is deposited and the interest is left to accumulate, how much is in the account after 1 year? I = Prt 1 I = ($1000)(0.08)( ) 4 I = $20 3 4.7 Compound Interest 2011 November 11, 2011 Therefore, at the end of one quarter we have earned $20 in interest. My principal is now $1020. So at the end of the second quarter, how much do I earn? 1 I = ($1020)(0.08)( ) 4 I = $20.40 Adding that interest to the money in my account, I now have $1040.40 six months into the year. 4 4.7 Compound Interest 2011 November 11, 2011 At the end of the third quarter, how much do I have? 1 I = ($1040.40)(0.08)( ) 4 I = $20.81 This brings my principal to $1,061.21. 5 4.7 Compound Interest 2011 November 11, 2011 At the end of the fourth quarter, or at the end of the year what do I have? 1 I = ($1061.21)(0.08)( ) 4 I = $21.22 This brings the total in my bank account to $1,082.43 just by collecting interest. 6 4.7 Compound Interest 2011 November 11, 2011 This process can be very lengthy especially if the interest were being compounded daily. This leads to a more general formula for calculating interest that is compounded. Compounded interest formula: r nt A = P(1 + ) n n = the number of compounds per year 7 4.7 Compound Interest 2011 November 11, 2011 Compounded interest is interest that is paid on principal and previously earned interest. Common forms of compounding are: annually: once per year semiannually: twice per year quarterly: four times per year monthly: 12 times per year daily: 365 times per year 8 4.7 Compound Interest 2011 November 11, 2011 Lets compare investments using different compounding periods. nt r A = P(1 + ) n n = the number of compounds per year Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years. (1 3) 0.04 A = (3000)(1 + ) 1 3 A = (3000)(1 + 0.04) 3 A = (3000)(1.04) A = $3,374.59 9 4.7 Compound Interest 2011 November 11, 2011 Lets compare investments using different compounding periods. nt r A = P(1 + ) n n = the number of compounds per year Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years. 0.04 (2 3) A = (3000)(1 + ) 2 A = (3000)(1 + 0.02) 6 6 A = (3000)(1.02) A = $3,378.49 10 4.7 Compound Interest 2011 November 11, 2011 Lets compare investments using different compounding periods. nt r A = P(1 + ) n n = the number of compounds per year Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years. 0.04 (4 3) A = (3000)(1 + ) 4 12 A = (3000)(1 + 0.01) 12 A = (3000)(1.01) A = $3,380.48 11 4.7 Compound Interest 2011 November 11, 2011 Lets compare investments using different compounding periods. nt r A = P(1 + ) n n = the number of compounds per year Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years. 0.04 (12 3) A = (3000)(1 + ) 12 A = (3000)(1 + 0.0033333333)36 A = (3000)(1.0033333333)36 A = $3,381.82 12 4.7 Compound Interest 2011 November 11, 2011 Lets compare investments using different compounding periods. nt r A = P(1 + ) n n = the number of compounds per year Find the resulting amount of an investment of $3000 when it is compounded annually, semiannually, quarterly, monthly and daily at a rate of 4% for a period of 3 years. (365 3) 0.04 A = (3000)(1 + ) 365 A = (3000)(1 + 0.0001095890411)1095 1095 A = (3000)(1.0001095890411) A = $3,382.47 13 4.7 Compound Interest 2011 November 11, 2011 What if the investment is compounded more often than daily? Compounding Interest Continuously: rt A = Pe Find the resulting amount of an investment of $3000 when it is compounded continuously at a rate of 4% for a period of 3 years. A = Pe rt A = (3000)e A = (3000)e (0.04 3) 0.12 A = $3,382.49 14 4.7 Compound Interest 2011 November 11, 2011 How much principal is needed now to have $4000 after 2 years at 6% compounded quarterly? nt r A = P(1 + ) n n = the number of compounds per year 0.06 (4 2) 4000 = P(1 + ) 4 4000 = P(1 + 0.015) 8 4000 = P(1.015)8 4000 = P(1.126492587) P = $3,550.84 15 4.7 Compound Interest 2011 November 11, 2011 How long will it take for an investment of $250 to reach $675 in value if it earns 9% compounded monthly? nt r A = P(1 + ) n n = the number of compounds per year 0.09 675 = (250)(1 + ) 12 (12 t) 675 = (250)(1 + 0.0075) 12t 675 = (250)(1.0075) 12t 2.7 = (1.0075) 12t log1.00752.7 = 12t log 2.7 = 12t log 1.0075 132.9295772 = 12t t ≈ 11.08 years 16 4.7 Compound Interest 2011 November 11, 2011 How long will it take for an investment to double in value if it earns 5% compounded continuously? A = Pert 2P = Pe0.05t 2 = e0.05t ln 2 = ln e0.05t ln 2 = 0.05t ln e ln 2 = 0.05t t = 13.86 years 17 4.7 Compound Interest 2011 November 11, 2011 Homework page 322 (4, 5, 8, 9, 12, 14, 16, 18, 19, 28, 32, 36, 37, 48) 18