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.

2 Comments

  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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