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Math 212 β Week 2 β Finance Quick Reference Card This Quick Reference Card (QRC) contains information about the concepts presented in Math 212 Week 2 (Chapter 3 of the course textbook). For more information about any of these concepts, visit the Center for Math Excellence (CME). Even more information about Math 212 Week 2 concepts can be found on the Web at sites such as www.JoleneMorris.com and www.youtube.com. As soon as your Math 212 class begins, you will also have access to FREE tutoring 24-hours a day at the CME. Ask for tutoring in any of the Week 2 concepts that are not clear to you. The Simple Interest Formula The simple interest formula is: π° = π·ππ where I = interest, P = principal, r = annual simple interest rate (written as a decimal), and t = time in years. EXAMPLE: Find the interest on a loan of $1,000 at 7.5% for 6 months. πΌ = πππ‘ = (1000)(0.075)(0.5) = $37.50 To find the present value of an investment or the total amount due on a loan, use this formula (based on the simple interest formula--the sum of the principal and the interest): π¨ = π· + π·ππ = π·(π + ππ) EXAMPLE: Find the total amount due on a car loan of $9,500 at 4% simple interest at the end of 9 months. π΄ = π(1 + ππ‘) = 9500(1 + [0.04][0.75]) = 9500(1 + 0.03) = 9500(1.03) = $97.85 Compound Interest Compound interest is where interest is paid at the end of a period (monthly, quarterly, etc.). Then the next time interest is paid, it is paid on the principal and the interest earned. Thus, interest is compounded each period. To find compound π ππ interest, we use this formula: π¨ = π· οΏ½π + οΏ½ where A = π amount at the end of time, P = principal amount, r = interest rate expressed as a decimal number, m is the number of compounding periods per year, and t is the time expressed in years. EXAMPLE: Find the amount to which $1500 will grow if compounded quarterly at 6.75% interest for 10 years. π¨ = π·(π + π)π = 1500 οΏ½1 + 10(4) 0.0675 οΏ½ 4 Growth and Time = $2,929.50 How much should you invest now to have a given amount at a future date? What annual rate of return have your investments earned? How long will it take your investment to double in value? The formulas for compound interest and continuous compound interest can be used to answer such questions. EXAMPLE: How much should you invest now at 10% to have $8,000 toward the purchase of a car in 5 years if interest is (a) compounded quarterly? (b) compounded continuously? (π) π΄ = π(1 + π)π (π) π΄ = ππ ππ‘ 8000 β π= = $4,882.17 (1 + 0.025)20 8000 β π = 0.10(5) = $4,852.25 π Simple Interest and Investments To find the total value of an investment, use the same formula: π¨ = π·(π + ππ) EXAMPLE: Find the total amount due on a loan of $600 at 16% interest at the end of 15 months. π΄ = π(1 + ππ‘) = 600οΏ½1 + 0.16(1.25)οΏ½ = $720 To find the interest rate, use the above formula but solve it for r: π΄βπ π= ππ‘ EXAMPLE: What is the annual interest rate earned by a 33-day T- bill with a maturity value of $1,000 that sells for $996.16? NOTE: We normally use 360 days for a financial year π= π΄ β π 1000 β 996.16 = = 0.042 = 4.2% 33 ππ‘ (996.16) οΏ½ οΏ½ 360 Continuous Compound Interest As the number of compounding periods per year increases without bound, the compounded amount approaches a limiting value. This value is given by the following formula: π¨ = π·πππ where r = annual interest rate compounded continuously and t = number of years. EXAMPLE: What amount will an account have after 10 years if $1,500 is invested at an annual rate of 6.75% compounded continuously? A = 1500e(0.0675)(10) = $2,946.05 Caution: Do not use the approximation 2.7183 for e; it is not accurate enough to compute the correct amount to the nearest cent. Instead, use your calculatorβs built-in π π₯ . In order to solve for rate or time, you will need to take the natural logarithm of both sides of the equation formula. Annual Percentage Yield The simple interest rate that will produce the same amount as a given compound interest rate in 1 year is called the annual percentage yield (APY). This is also called the effective rate π¨π·π = οΏ½π + π π οΏ½ βπ π where r = rate of interest and m = number of periods in a year. EXAMPLE: What is the annual percentage yield for money that is invested at 6% compounded monthly? π΄ππ = οΏ½1 + π π 0.06 12 οΏ½ β 1 = οΏ½1 + οΏ½ β 1 = 0.06168 π 12 This Quick Reference Card was prepared by Jolene M. Morris ([email protected]) using the course textbook. Future Value of an Annuity An annuity is any sequence of equal periodic payments. The future value of an annuity is found by using the following formula ππ½ = π·π΄π» (π+π)πβπ π where FV = future value, PMT = periodic payment, i = interest rate per period (r/m), and n = total number of periods (mt). EXAMPLE: Suppose a $1000 payment is made at the end of each quarter and the money in the account is compounded quarterly at 6.5% interest for 15 years. How much is in the account after the 15 year period? οΏ½1 + (1 + π)π β 1 πΉπ = πππ = 1000 οΏ½ π 0.065 4(15) οΏ½ β1 4 οΏ½ = 100,336.68 0.065 4 Approximating Interest Rates Mr. Ray has deposited $150 per month into an annuity. After 14 years, the annuity is worth $85,000. What annual rate has this annuity earned (compounded monthly) during the 14 year period? Solution: Use the FV formula; substitute these values into the formula; solution is approximated by graphing each side of the equation separately and finding the point of intersection. (1 + π)π β 1 (1 + π‘)14(12) β 1 πΉπ = πππ β 85,000 = 150 οΏ½ οΏ½ π π (1 + π)168 β 1 (1 + π)168 β 1 85,000 =οΏ½ οΏ½ β 566.67 = οΏ½ οΏ½ 150 π π Sinking Funds Any account that is established for accumulating funds to meet future obligations or debts is called a sinking fund. To derive the sinking fund payment formula, we use algebraic techniques to rewrite the formula for the future value of an annuity and solve for the variable PMT: π π·π΄π» = ππ½ οΏ½ οΏ½ (π + π)π β π EXAMPLE: How much must Harry save each month in order to buy a new car for $12,000 in three years if the interest rate is 6% compounded monthly? π πππ = πΉπ οΏ½ οΏ½ = 12000 οΏ½ (1 + π)π β 1 0.06 12 οΏ½ = $305.06 0.06 36 οΏ½1 + οΏ½ β1 12 Present Value of an Annuity π β (π + π)βπ π·π½ = π·π΄π» οΏ½ οΏ½ π where PV = present value of all payments, PMT = periodic payment, i = interest rate per period, and n = number of periods. EXAMPLE: How much money must you deposit now at 6% interest compounded quarterly in order to be able to withdraw $3,000 at the end of each quarter year for two years? 1 β (1 + π)βπ ππ = πππ οΏ½ οΏ½ π 1 β (1.015)β8 = 3000 οΏ½ οΏ½ 0.015 = $22,457.78 The present value of all payments is $22,457.78. The total amount of money withdrawn over two years is 3000(4)(2)=$24,000. Thus, the accrued interest is the difference between the two amounts: Monthly interest rate β1.253%. Annual interest rate β 15%. $24,000 β $22,457.78 =$1,542.22. Amortization Amortization Schedules Problem: A bank loans a customer $50,000 at 4.5% interest per year to purchase a house. The customer agrees to make monthly payments for the next 15 years for a total of 180 payments. How much should the monthly payment be? If you borrow $500 that you agree to repay in six equal monthly payments at 1% interest per month on the unpaid balance, how much of each monthly payment is used for interest and how much is used to reduce the unpaid balance? Use the formula for present value of an annuity and solve for PMT: π 0.01 πππ = ππ οΏ½ οΏ½ = 500 οΏ½ οΏ½ = $86.27 1 β (1 + π)βπ 1 β (1.01)β6 π 1 β (1 + π)βπ οΏ½ οΏ½ β π·π΄π» = π·π½ οΏ½ π π β (π + π)βπ 0.045 12 = 50,000 οΏ½ οΏ½ = $382.50 0.045 β180 1 β οΏ½1 + οΏ½ 12 If the customer makes a monthly payment of $382.50 to the bank for 180 payments, then the total amount paid to the bank is the product of $382.50 and 180 = $68,850. Thus, the interest earned by the bank is the difference between $68,850 and $50,000 (original loan) = $18,850. ππ = πππ οΏ½ In reality, the last payment would be increased by $0.03, so that the balance is zero. General Problem-Solving Strategy Step 1: Determine whether the problem involves a single payment or a sequence of equal periodic payments. Simple and compound interest problems involve a single present value and a single future value. Ordinary annuities may be concerned with a present value or a future value but always involve a sequence of equal periodic payments. Step 2: If a single payment is involved, determine whether simple or compound interest is used. Often simple interest is used for durations of a year or less and compound interest for longer periods. Step 3: If a sequence of periodic payments is involved, determine whether the payments are being made into an account that is increasing in value--a future value problem--or the payments are being made out of an account that is decreasing in value--a present value problem. Remember that amortization problems always involve the present value of an ordinary annuity.