Math of Finance: Solving for Time

Math 103 Math of Finance: solving for time Three ways to compute future value Simple interest A = P (1 + rt) Compound interest A = P (1 + i)n Continuous compounded interest A = P ert These formulas can also be used to compute the present value required to attain a given future value. Example: What present value P is required for a future value F of $4,000? Interest is compounded semiannually for 5 years at a rate of 8%. Solve the equation for P : 4000 = = P = = P (1 + .08/2)10 P (1.48024) 4000/1.48024 2702.26 Summary: A present value of $2702.26 is required for a future value of $4000 if interest is compounded semiannually for 5 years at a rate of 8%. 1 Solving the future/present value formula for time t Example: Use the graph to solve the equation for the number of years t: 3000 = 1000e(.10)t Future Value 4000 3500 3000 2500 2000 1500 1000 500 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Years 2 Solving the future/present value formula for time t Use logarithms graph to solve the equation for the number of years t: 3000 = 1000e(.10)t ! e(.10)t = 3 ln e(.10)t " = ln(3) (.10)t = ln(3) t = ln(3)/.10 = 10.98612 Summary: It takes 10.98612 years for a present value of $1000 to grow to a future value of $3000 at a rate of 10% compounded continuously 3 Solving the future/present value formula for time t Use logarithms graph to solve the equation for the number of years t: 5000 = 1200e(.08)t Future Value 6000 5500 5000 4500 4000 3500 3000 2500 2000 1500 1000 500 2 4 6 8 10 12 14 16 18 20 Years 4 Solving the future/present value formula for time t Use logarithms graph to solve the equation for the number of years t: 2500 = 1000e(.09)t 3600 = 1000e(.05)t 5 Two facts about the natural logarithm, ln: ln (ex) = x (1) x ln (a ) = x ln(a) Fact (1): ! ln e(.03)t ! ln e(.09)t ! ln e(.06)t ! ln e(.10)t " " " " (2) Fact (2): = (.03)t ln ((1.02)n) = n ln(1.02) = (.09)t ln ((1.10)n) = n ln(1.10) = (0.0198026)n = (0.0953102)n ln ((1.045)n) = n ln(1.045) = = (0.0440169)n ln ((1.01)n) = n ln(1.01) = (.10)t = 6 How to use Fact (2): Compound Interest Formula: A = P (1 + i)n Problem: Deposit $100 into an account earning 4.5% interest compounded annually. How many years will it take to have a future value of $200? Solve for n: 200 = 100(1 + .045)n. (1.045)n = 2 ln ((1.045)n) = ln(2) n ln(1.045) = ln(2) n = ln(2)/ ln(1.045) n = = 15.747302 Summary: It takes 15.75 years to have a future value of $200 if a present value of $100 earns 4.5% interest compounded annually. 7 How to use Fact (2): Compound Interest Formula: A = P (1 + i)n Problem: Deposit $100 into an account earning 9% interest compounded semiannually. How many years will it take to have a future value of $200? Solve for n: 200 = 100(1 + .045)n. This is the same equation as in the previous slide. The answer is still n = 15.747302, but it must be interpreted differently. n is the number of compounding periods. n isn’t always the number of years. In this example, interest is compounded semiannually. So 15.747302 periods is 15.747302/2 = 7.873651 years. Summary: It takes 7.87 years to have a future value of $200 if a present value of $100 earns 9% interest compounded semiannually. 8 Example: The present value is $200. Interest is compounded quarterly at a rate of 10%. How many years does it take for a future value of $500? Compound interest formula: A = P (1 + i)n i = 0.10/4 = 0.025 Solve for n: 200(1.025)n = 500 (1.025)n = 2.5 ln ((1.025)n) = ln(2.5) n ln(1.025) = ln(2.5) n = ln(2.5)/ ln(1.025) n = 37.107890 There are 37.108 periods. Each period is a quarter (of a year). So that’s 37.108/4 = 9.277 years. Summary: It takes 9.277 years to have a future value of $500 if a present value of $200 earns 10% compounded quarterly. 9