# Interest Factor Case Study

748 Words3 Pages
Interest factors Interest factors are multiplicative amounts computed from interest formulas. They are utilized to translate cash flow happening at distinctive times to a common time. The useful formats that are used to represent the factors are taken from ANSI Z94. Single payment Factors (F/P and P/F): The most important factor in engineering economy is the particular case that decides the amount of cash F aggregated after n periods from a single present worth P, with interest refers to interest compounded one time for every period. Recall the compound interest states to interest paid on the top of interest. Consequently, if the amount P is spent at time t = 0 the accrued F1 aggregated 1 year thus at an interest rate of i percent per year…show more content…
it can be used to simplify engineering economy problems. Example: An employee of a company got a bonus of \$10,000 that he will invest now. He needs to compute the equivalent worth after 24 years, when he plans to use the all preceding cash as the down payment on a trip. Assume a rate of return 8% per year for 24 years. Find total cash he can pay down, using the standard notation and the factor formula. Sol. Symbols and their values are given below P = \$ 10,000 F=? i= 8% per year n = 24 years Standard notation: determine F, using the F/P factor for 8% and 24 years. Above table provide the factor value. F = P (F/P, i, n) = 10,000(F/P, 8%, 24) = 10,000(6.3412) = \$ 63412 Factor formula: use Equation 2.2 to compute F. F = P (1+i) ^n = 10,000 (1+0.08) ^24 = 10,000(6.341181) = \$ 63411.81 This minor difference is due to round off error. An equivalence interpretation of this result is that \$10,000 today is worth \$63412 after 24 years of growth at 8% per