question for top ppl?

Question

when you were 40, and you want to retire. You have one million in the bank, the interest rate is 5%. You take out 75,000 every year. How long can you survive?

so far, no body gets the right answer,
t = 21.9722

dQ/dt = rate in - rate out
rate in is the intertest, the rate out is 75,000, Q is the one million
dQ/dt = 0.05Q - 75,000
dQ = (0.05Q - 75,000)dt
dQ/(0.05Q - 75,000) = dt
intergrade both sides
20ln(0.05Q - 75,000) = t + c
ln(0.05Q - 75,000) = 0.05t + c
0.05Q - 75,000 = e^(0.05t + c)
0.05Q - 75,000 = ce^0.05t
(0.05Q = ce^0.05t + 75,000) x 20
Q = ce^0.05t + 1,500,000
solve for c,
Q(0)=1,000,000
1,000,000 = ce^0.05(0) + 1,500,000
1,000,000 = c + 1,500,000
1,000,000 - 1,500,000 = c
c = -500,000
Q(t) = -500,000e^0.05t + 1,500,000
plug in the number
0 = -500,000e^0.05t + 1,500,000
-1,500,000/-500,000 = e^0.05t
3 = e^0.05t
ln3 = 0.05t
20ln3 = t
t = 21.9722

Answer 1

about 22 years

Answer 2

Let f(n) denote the amount of money you have at the end of the nth year. We assume that interest is compounded annually and that money is withdrawn all at once, after the interest has been calculated. Then:

f(0) = 1,000,000
f(n) = f(n-1)*1.05 - 75,000

Solving this recurrence relation gives us:

f(n) = 1,000,000 * 1.05^n - 75,000 * (1.05^n - 1)/.05

We wish to know when this will equal zero. Thus we set:

1,000,000 * 1.05^n - 75,000 * (1.05^n - 1)/.05 = 0
1,000,000 * (21/20)^n - 1,500,000 * ((21/20)^n - 1) = 0
-500,000 * (21/20)^n + 1,500,000 = 0
-(21/20)^n + 3 = 0
3 = (21/20)^n
ln 3 = n ln (21/20)
n = ln 3/ln (21/20) ≈ 22.51708530

So you could live about 22.5 years on that income.

Answer 3

22 years. You would run out of money somewhere in the middle of the 23rd year.

My answer is based on the assumption that the first payment of $75,000 was at time t = 0.

Answer 4

Your additional details are incorrect. You are assuming a continuous solution to a discrete problem, which is good for an approximation, but you can do better. Here's a website which can help you. Cheers!

Answer 5

Amount left at the end of the 1st year
= (1,000,000 - 75,000)*105%
= 925,000*1.05

Amount left at the end of the 2nd year = (925,000*1.05 - 75,000)*1.05
= 925,000*1.05² - 75,000*1.05

Amount left at the end of the 3rd year = (925,000*1.05² - 75,000*1.05 - 75,000)*1.05
= 925,000*1.05³ - 75,000*1.05² - 75,000*1.05

Amount left at the end of the nth year
= 925,000*1.05^n - 75,000*1.05^(n - 1) - 75,000*1.05^(n - 2) - .... - 75,000*1.05² - 75,000*1.05

= 0 when all is spent

So 925,000*1.05^n = 75,000*1.05(1 + 1.05 + 1.05² + ... + 1.05^(n - 2))

= 75,000*1.05(1.05^(n - 1) - 1)/(1.05 - 1)

= 75,000(1.05^n - 1.05)/0.05

So 925,000 *0.05/75000= 1 - 1.05^(1 - n)

ie 1.05^(1 - n) = 1 - 37/60 = 23/60

So (1 - n) log 1.05 = log (23/60)

So 1 - n = log (23/60)/log 1.05
So n = 1 - log (23/60)/log 1.05
≈ 20.6525

ie it would last just over 20 years

Answer 6

i can survive for 14 months with that amount of money