How to Make Things Faster

Chapter 89. Describing Performance Improvements

Once I saw an ad claiming that a product could improve response times by 1,000%. Sounds great. But what does that mean?

Here are two ways to express response time improvements:

What proportion of the time did you save?

p = (old − new)/old. It’s the improvement expressed as a proportion of the old value. You can multiply p by 100% if you want, to express it as a percentage.¹

How many times faster is it now?

That’s n = old/new. Don’t express this number as a percentage. It’s not helpful to the recipient of the information.

Here are some examples:

Before	After	“p% faster”	“n times faster”
(old)	(new)	(p = (old − new)/old × 100%)	(n = old/new)
60 sec	60 sec	0.0%	1
60 sec	59 sec	1.7%	1.017
60 sec	20 sec	66.7%	3
60 sec	1 sec	98.3%	60
60 sec	0 sec	100.0%	+∞

For throughput, just use the reciprocal of the n formula. For example, if you improve throughput from 1 thing/sec to 10 things/sec, then you’ve increased throughput by new/old = 10 times. As with n values, don’t express this value as a percentage.

Note that a p value can never be bigger than 1 (or 100%) without invoking reverse time travel.² So, when someone tells you that their thing “improves response time by 1,000%,” then you know they’re not telling you a p value. It’s probably an n value, meaning that their thing really makes your response time [merely] 10 times better than it was.

To say what sounds better, even though it’s misleading...is not your best pathway ...