Annuities With Inflation
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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