Interest incrementally

I was looking at bank accounts and interest rates the other day — such as the Virgin Super page that lists every single fund returning a negative percentage, except for cash and one of the funds available to over-60s. It seems really hard to compare percentage rates, for example if you gain 10% on year, then have a rate of -10% the next year, you’re not actually square, you end up worse off. But to really work it out, you have to multiply it out — “1.10*0.9=0.99, oh I’m down a percent, damn”.

At some point it crossed my mind that working with exponents would be much more sensible — rather than multiplying, you’re just adding which is a lot easier to do in your head, and compounding just falls out naturally, rather than being horribly confusing. So creating a new unit, “i%”, an incremental percentage improvement, where “1 i%” is the same as a 1% return, and “2 i%” is the same as a 1% return on top of a 1% return (ie, 1.01*1.01=1.0201, so 2 i%=2.01%). The formula for going from an i% to a percentage interest rate is straightforward, it’s n i% = 1.01n%. Unfortunately the formula for getting the i% in the first place is more complicated, it’s r% = log(1+r/100)/log(1.01) i%.

Some particular values:

100% = 69.661 i%
(what it takes to double your money)
13% = 12.283 i%
(the low end of current credit card rates)
6% = 5.856 i%
(current savings interest rate, if you’re lucky)
5.25% = 5.142 i%
(current Reserve Bank policy rate)
0% = 0 i%
(what happens if you don’t get anything)
-0.9901% = -1 i%
(the negative interest rate that exactly cancels out a prior 1% profit)
-13.36% = -14.412 i%
(12 month performance on Virgin’s 100% growth agressive fund)
-50% = -69.661 i%
(what it takes to halve your money)

That, to me, seems like it makes comparisons a lot easier. If you’re getting a flat interest rate of 5% is it better or worse to change to a 2.4% interest rate compounded twice? 5%=4.903i%, 2.4%=2.383i%. Double the latter because you get it twice, and you’re at 4.766i%, which is worse off. 2.5% on the other hand would be 2.482i% which doubles to 4.964i%. If you get 6% for three years, then -13.36%, what’s that cumulatively? 5.856 + 5.856 + 5.856 – 14.412 = 3.156 i% (or a 3.19% improvement). What’s that as an annual rate over four years? 3.156 / 4 = 0.789 i% (or a 0.788% pa average). If you want to work out how long it’ll take you to double your money at 6% interest per annum? 69.661/5.856 = 11.9 years.

Anyway, that seemed like an interesting (and better) way of comparing things to me than what people usually put up with, YMMV.


  1. Why not just use straight logs of the multiplier and skip the percent?

    Percentages seem like a bad idea anyway. They do nothing that can’t be done another way and just serve to confuse people.

  2. aj says:

    It’s just a scaling factor — if you had it as log(x) and e^x, a 1% rate would be a rate of 0.00995…, doubling your money would be a rate of 0.693…, etc, which just seems unnecessarily awkward to me.

    Doing it with base 2 might work okay, doubling your money is then a rate of 1, and a 1% rate is 0.0143…, with the formulas being log(x)/log(2) and 2^x.

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