Saving for retirement - how much?

I'm working through a problem in the book "An Undergraduate Introduction to Financial Mathematics" and there is an example I can follow.

The problem is: Suppose you want to save for retirement. The savings account is compounded monthly at a rate of 10%. You are 25 years and you plan to retire in 40 years. You plan on being retired for 30 years and you plan on receiving a monthly payment of \$1500 as retired. How much (fixed amount) should you save monthly? I can't follow the computation in the book so I tried to solve it my own way. What am I doing wrong? Let's call the monthly savings amount$x$. After 480 months (40 years) the account statement will read:$x + x(1+0.1/12) + ... + x(1+0.1/12)^480 = x(1 - r^{481})/(1-r) =: P$where I write$r = (1+0.1/12)$for short. Next we will start subtracting \$1500 every month, but we should not forget that we still have interest: 30 years = 360 months. Let $P_n$ be the amount we have in our account after $n$ withdrawals.

$P_0 = P$

$P_1 = P_0\times r - 1500$

...

$P_n = P_{n-1}\times r - 1500$

Finally we wish that $P_{360} = 0$. Thus $P = r^{-360}\times 1500\left(\frac{1-r^{360}}{1-r}\right)$

Solving for $x$ gives me the answer $x \approx 26.8$. The correct answer should be $x \approx 27.03$. Is this just a rounding error?

1 Answer

Your payments-in are made at the end of the months. This information is not given explicit in the text.

The equation is

$x\cdot \frac{1-(1+\frac{0.1}{12})^{480}}{-\frac{0.1}{12}}=1500\cdot\frac{1-(1+\frac{0.1}{12})^{360}}{-\frac{0.1}{12}}\cdot \frac{1}{(1+\frac{0.1}{12})^{360}}$

This gives $x\approx 27.03$