5 ms·
5.5% compounded over 5 years is a bit over 30%: not a huge amount but an easily noticeable speed-up. What were you thinking of when you typed “significantly fas
by chalst 3y ago
5.5% compounded over 5 years is a bit over 30%: not a huge amount but an easily noticeable speed-up. What were you thinking of when you typed “significantly faster”?
- aktenlage 3y agoCompunding a decrease works differently than an increase. If something gets 10% faster twice it actually got 19% faster. In other words, the runtime is 90% of 90%, i.e. 81%.
- BeetleB 3y ago> If something gets 10% faster twice it actually got 19% faster 21%, not 19%. It is 1.1 * 1.1 = 1.21 You are right in the opposite direction. If it got 10% slower, then it is 0.9 * 0.9 = 0.81 = 19% slower
- readams 3y agoIt's easy to see there must be something wrong with this since if you get 10% faster 8 times, we can be sure that this doesn't mean you're 114% faster (1.1^8 = 2.14). You can't get more than 100% faster! When you say something is 10% faster, what you mean is it took 10% less time to finish. So 19% is correct.
- nuancebydefault 3y ago100 percent faster means doubling the speed,just like a 100 percent salary increase means doubling the salary, or 100 percent car speed increase means doubling its speed. Unless one uses a bit more esoteric definition of speed, in which a 50 percent of car speed increase, makes it go from 100 km/h to 200 km/h, such that it arrives in half the time.
- ummonk 3y agoNot if “faster” refers to computation rate rather than runtime, in which case it becomes 100/81 i.e. 23% faster.
- moomin 3y agoYes, but I’ve literally never heard anyone say that.
- kaashif 3y agoIt's not really possible to tell if someone saying "50% faster" means a 50% speed increase or a 50% time decrease. Even in other contexts it can be ambiguous. Yesterday I drove 60mph, today I drove 50% faster. Yesterday I got there in 1 hour, today I got there 50% faster. It's not really possible to tell them apart without looking at the numbers.
- ummonk 3y agoYou've never seen anyone say "x% faster" where x is a number larger than 100? I find that hard to believe.
- squeaky-clean 3y agoNot for a programming language because it's extremely rare for the computation rate to increase, rather than the work being done to compute something decrease. If you've rewritten something to better use cachelines, removed saturating memory bandwidth, etc then sure you've increased The computation rate. But that's rarely how these language specific optimizations occur.
- chalst 3y agoI have, in technical descriptions of compilers.
- chalst 3y agoIn a certain sense, decreases are reciprocals of increases. You can calculate the reciprocal of each “faster”, or you can work with the figures as given and what you want is the reciprocal of the final result. This follows from the elementary fact that division is the same as multiplication by the reciprocal and we are only treating multiplication and division.