L13
Contributor
I think most people aren't understanding that you are not claiming an equality, but rather an upper bound. Note the "less" I bold underlined in my quote above.What I've a proof for is that spending a time T1 at depth D1 followed by a time T2 at depth D2 where D2 < D1 result in less load than spending T1+T2 at depth (T1*D1+T2*D2)/(T1+T2).