Suppose we have a certain amount of savings p at age 65, invested in a fund that yields 10% annual return, and we wish to make withdrawals in amounts that increase at 4% each year (to keep up with inflation) in such a way that our savings are exhausted at age 85. What should be the size of our withdrawals? The precise answer depends on how we compound the interest/inflation, and how we schedule our withdrawals, but the general method is the same, regardless. Suppose we want to do our calculations on an annual basis. An inflation rate of 4% means our withdrawals must increase by a factor of I = 1.04 each year. Also, an investment yield of 10% means our principle would increase by a factor of Y = 1.10 each year, if we made no withdrawals. Beginning with a principal amount of p[0] we would start by with- drawing w[0] dollars, and then let the remaining principal gather interest for 1 year. At the end of the first year our principle is p[1] = (p[0]-w[0])*Y (1) At that point we withdraw w[1] dollars, and then let the remaining principle gather interest for a year, at the end of which time we have p[2] = (p[1]-w[1])*Y (2) Of course, to keep up with inflation, we have w[1] = I*w[0]. Also, from equation (1) we have w[0] = p[0] - p[1]/Y (3) Therefore, w[1] equals I*(p[0] - p[1]/Y), so we can substitute into equation (2) to give p[2] = Y*(p[1] - I*(p[0] - p[1]/Y)) (4) which simplifies to p[2] = (Y+I)*p[1] - (YI)*p[0] (5) Obviously this same relation applies to each successive step, so we have the recurrence p[k+2] = (Y+I) p[k+1] - (YI) p[k] (6) The roots of the characteristic polynomial are just Y and I, so the general term is of the form p[n] = A Y^n + B I^n (7) where the coefficients A and B are determined by the initial conditions. We require A + B = p[0] YA + IB = p[1] = (p[0]-w[0])*Y Solving for A and B gives Ip[0] - Y(p[0]-w[0]) -Yw[0] A = -------------------- B = --------- I - Y I - Y so the general formula for the principle after n years is (Y-I)p[0] - Yw[0] Yw[0] p[n] = ----------------- Y^n + --------- I^n (8) Y - I Y - I This value reaches zero when [ Yw[0] - (Y-I)p[0] ] Y^n = [ Yw[0] ] I^n so we have (Y-I)p[0] (I/Y)^n = 1 - ----------- (9) Yw[0] Solving for w[0] gives the result 1 - (I/Y) w[0] = p[0] ------------ (10) 1 - (I/Y)^n As an example, suppose our initial principal is p[0] = $200,000, the inflation rate is I = 1.04, the investment yield is Y = 1.10, and we intend to run out of money in exactly 20 years. The above equation says our initial withdrawal should be $16,178.31 (increased by 4% every year thereafter). Conversely, we might ask how many years our money would last if we start by withdrawing $20,000 the first year. Solving (9) for n gives / / Y-I \ p[0] \ ln ( 1 - ( ----- ) ---- ) \ \ Y / w[0] / n = --------------------------- ln(I/Y) which indicates we would run out of money in 14.06 years. There's actually a much simpler way to derive this formula. Let I=1.04 be the inflation rate, and Y=1.10 be the investment yield. If x is our initial payout, then it's present value is obviously just x. The payout one year from now will be Ix, which has a present investment value of x(I/Y). In general, the present value of the payout k years from now is x(I/Y)^k. If we want to run out of money in exactly n=20 years, we want the sum of the present values of the 20 payouts at year 0 through 19 to equal our initial principle P. Thus, P = x [ 1 + (I/Y) + (I/Y)^2 + ... + (I/Y)^(n-1) ] 1 - (I/Y)^n = x ------------- 1 - (I/Y) which is identical to the result given previously.

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