標題:
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Hard Maths
發問:
Mr. Chan is 25 years old and has $60M in the bank. Bank interest is 1% p.a., if Mr. Chan spends $40,000 per month ( he draws this amount from the bank every month) and inflation is 4% per year, can he survive till 85 years old with this money? If not, then to what age?
最佳解答:
assumptions for calculation below as listed: 1. bank interest is compound yearly 2. draws the amount with inflation yearly ($480,000 + inflation) 3. draws the amount at each end of month 1st year left = 60M(1.01) - 480000(1.04) 2nd year left = [60M(1.01) - 480000(1.04)](1.01) - 480000(1.04)^2 = 60M(1.01)^2 - 480000(1.04)(1.01) - 480000(1.04)^2 = 60M(1.01)^2 - 480000[(1.04)(1.01) + (1.04)^2] 3rd year left = {60M(1.01)^2 - 480000[(1.04)(1.01) + (1.04)^2]}(1.01) - 480000(1.04)^3 = 60M(1.01)^3 - 480000[(1.04)(1.01)^2 + (1.04)^2(1.01) + (1.04)^3] ... 60th year left = 60M(1.01)^60 - 480000[(1.04)(1.01)^59 + (1.04)^2(1.01)^58 + ... + (1.04)^58(1.01)^2 + (1.04)^59(1.01) + (1.04)^60] [(1.04)(1.01)^59 + (1.04)^2(1.01)^58 + ... + (1.04)^58(1.01)^2 + (1.04)^59(1.01) + (1.04)^60] is a sum of G.P. with R=1.04/1.01, n=60, a=(1.04)(1.01)^59 Sum of G.P. = a(R^n - 1) / (R-1) = (1.04)(1.01)^59 [(1.04/1.01)^60 - 1] / (1.04/1.01 - 1) = 301.7 60th year left = 60,000,000(1+1%)^60 - 480000(301.7) = 109,001,802 - 144,816,000 480000 [a(R^n - 1) / (R-1)] 125 (1.01)^n > (1.04)(1.01)^(n-1) [(1.04/1.01)^n - 1] / (1.04/1.01 - 1) 125 (1.01) > (1.04) [(1.04/1.01)^n - 1] / (1.04/1.01 - 1) 125 (0.03) / 1.04 > (1.04/1.01)^n - 1 [125 (0.03) / 1.04] +1 > (1.04/1.01)^n log (4.79/1.04) > n log (1.04/1.01) n > 52.2 2011-11-22 13:30:43 補充： n is years to go, not the age of Mr. Chan
其他解答:A9A3995907B431A4
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