# How to calculate relative performance (non-finance) when lower is better?

Example: Car 1 goes from 0-60 in 6 seconds. Car 2 goes from 0-60 in 8 seconds. Car 3 goes from 0-60 in 9 seconds.

Now, Car 1 is obviously the fastest, so I want to make it my reference and express other cars' performance relative to it. If higher was better this would be a simple value/reference * 100, however, if I express Car 1 as 100% this formula makes the other two as 133% and 150% respectively, which is mathematically correct, but makes no sense on a scale of 1-100%. I've wasted the last two hours looking for an answer and only come across two:

1. (1 - value/reference) + 1 or 2 - value/reference. Problem with this is: if a car takes 12 seconds the result would be zero. Again no sense.

2. (1/value)/(1/reference) or reference/value. I'm not sure whether it's correct, but it seems to achieve what I want.

## 1 Answer

Your original try was using value/reference and you noticed it did the inverse of what you wished for. Therefore, taking the (multiplicative) inverse, which is reference/value, should indeed do the trick. Mathematically, this function is proportional to the reciprocal function $$f(x)=1/x$$.

Please note that this is one of many options for doing what you would like. For example, this function is not linear, so with a Low reference of $$r^L=6$$ it will not give you the same decrease in "performance" for a car with a value of 8 and another with a value of 10 (e.g. the score, 75%, is down 25% for a value of 8, but is 60%, so down a further 15% only, when going to a value of 10). This is perhaps a good thing since the decreases will be larger close to the reference, and thus it could enable more distinction between a few high-scoring cars for example; but it will tend to discriminate less well between cars that are further from the reference.

If you would like something linear, unfortunately you will have to specify where that line ends (otherwise you will start getting negative scores). So for example if you can give a maximum value that no car ever gets to, say a High reference $$r^H=60$$ seconds, you could in fact have a line be drawn using $$g(x)=\frac{r^H-x}{r^H-r^L}$$

• I'm glad that I took the time to write this question instead of wasting more time searching. It provided me with more insight than I had. My sincere gratitude. Commented Apr 21, 2023 at 16:37
• Happy you found this useful, and thanks for your kind words! Commented Apr 23, 2023 at 4:45