3 ms·
log(2)÷log(1.1) ~= 7.27, so in principle sustained 10% improvements could double performance every 7 releases. But at some point we're bound to face diminishing
by Qem 3y ago
log(2)÷log(1.1) ~= 7.27, so in principle sustained 10% improvements could double performance every 7 releases. But at some point we're bound to face diminishing returns.
- hughdbrown 3y ago.9 * x = 0.5 x ln 0.9 = ln 0.5 x = ln 0.5 / ln 0.9 x = 6.5788 So decreasing runtime by 10% 6.5788 times results in the code running in half the original time.
- Dylan16807 3y agoI think their number is the right one. Ten percent faster is not a ten percent decrease in runtime, it's about a nine percent decrease.