ISA decision-making

,

Here’s a situation that perhaps one or two of my readers may recognize: you have some debt and some surplus income, and need to decide whether to save the surplus, or to use it to pay down the debt.

Well, that’s easy, you say, suppose the relevant interest rates are \(d\) on the debt, and \(s\) on savings, and your returns are taxed at a rate of \(1−t\). We’ll assume that there are no complicating factors like a preference for liquidity. Then you pay down the debt if \(d > st\) and save otherwise.

However, in the UK the decision is complicated by the existence of Individual Saving Accounts (ISAs). Returns on investments in ISAs are not subject to tax, but there is a limit to the sum you can invest in each financial year. So if you fail to maximize your ISA allowance in some year (for example, by paying down debt instead), then you’ve lost the chance to receive a tax-free return on that sum for the remaining duration of the scheme. Since the scheme was started by John Major in 1990 and shows no signs of going away for some time yet, intuitively it seems that it might be costly to forego the allowance.

Under what circumstances is each decision right? Let’s work up a simplified model. Suppose that you have a sum \(x\) today and that in \(m\) years’ time you’ll have a sum \(y = x(1+d)^m\) over and above the ISA allowance for that future year (this value is chosen to make the sums easy), and you are deciding between these options:

  1. Pay down the debt by \(x\) today, and after \(m\) years invest the sum \(y\) in an ordinary (non-ISA) savings account.
  2. Save \(x\) in an ISA today, and after \(m\) years pay down the debt with the sum \(y\).

Then after \(m+n\) years you will have:

  1. \(y(1+st)^n = x(1+d)^m(1+st)^n\)
  2. \(x(1+s)^{m+n}\)

So paying down the debt is better if $$ (1+d)^m(1+st)^n > (1+s)^{m+n}. $$ What’s a good way to visualize this? Well, we can fix \(d\), \(s\) and \(t\), and look at the break-even point for \(m\) in terms of \(n\): $$ (1+d)^m(1+st)^n = (1+s)^{m+n}. $$ Then taking logarithms, $$ m \log(1+d) + n \log(1+st) = (m + n) \log(1+s), $$ and solving for \(m\), $$ m = n {\log(1+st) − \log(1+s) \over \log(1+s) − log(1+d)}. $$ So for a basic rate taxpayer (\(t = 80\%\)) we can plot the break-even ratio \(m\over n\) in terms of \(s\) and an assumption about how \(d\) varies with respect to \(s\). (In these plots the region below the line is the “invest in an ISA” area, and above the line is the “pay off the debt” area.)

Break-even ratio m/n for a basic rate taxpayer under various interest rate scenarios.

So for example if the saving rate is 2% and the borrowing rate is 4%, then the break-even \(m\over n\) is about 0.2. In other words, the if the time until you’ve paid off the debt (\(m\)) is a sixth or less of the time until you want to spend the proceeds (\(m+n\)), invest in the ISA now; if greater, pay off your debt now.

For example, under these assumptions (\(s = 2\%\); \(d = 4\%\); \(t = 80\%\)), if you plan to have paid off your debt in 5 years time (and so have a surplus to invest), and plan to spend the savings in 25 years time (\(m = 5\); \(m+n = 25\); \({m\over n} = 0.25\)), then it’s not worth investing in an ISA now (because 0.25 > 0.2).

For higher rate taxpayers (\(t = 60\%\)), ISAs are distinctly more attractive:

Break-even ratio m/n for a higher rate taxpayer under various interest rate scenarios.

I’m sure you can think of many problems with this simplified model. Let’s hear them! (The most obvious problem is the assumption of constant \(m\): in fact the time to pay off the debt depends on your decision about whether to pay it down now.)