Alice is impatient
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Alice is impatient
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Re: Alice is impatient
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#6I find it much more inquisitive and visceral, to the extent that p99 now boggles my mind. 2N would be dreadful as an availability figure, yet for UX it's treated very different. So much so that my measurements corroborate exactly that; good UX requires the same many-nines reliability as e.g. DCs, not one or two.
I wonder if it's p90 and p99 to blame for the shoddy services we have, in a way. It's pretty hard to argue for fixing something when it's presented as only going wrong 0.5% or less of the time after all. Even if at scale that means most of your users are experiencing it weekly.
Re: Alice is impatient
#7This article contains very little substance. Show me the math!
Re: Alice is impatient
#8This article contains very little substance. Show me the math!
Yes I found this very hard to follow. I appreciate expressing ideas in math like E_a[X] as much as the next guy, but there is no definition or even description of what the heck E or E_a or Var(x) even mean, so how is anyone supposed to understand the reasoning here? All I get from this is a claim that experienced latency is different than the mean, which sounds important, but I still have no intuition as to why this…
I'm pretty sure what the author is saying is:
E(X) =:= \sum_t(t * P(X = t)) is the definition
another important note is P(X^2 = t^2) = P(X = t) - because it's the same distribution.
E_a(X) is a bit sloppy, but consider X_a aka Alice's latency "experience" distribution. The argument is:
P(X_a = t) = t * P(X = t) / \sum_u(u * P(X = u)) - i.e. scale the probability up by t but make it sum to 1.
Then
E(X_a) = \sum_t(t * P(X_a = t)) = \sum_t(t * t * P(X = t) / \sum_u(u * P(X = u))
aka
E(X^2) / E(X)
Then (from wikipedia)
Var(X) = E(X^2) - (E(X))^2
And we get
E(X_a) = (Var(X) + (E(X))^2) / E(X) = E(X) + Var(X) / E(X)
Re: Alice is impatient
#9Is the formula for E_a[X] trivial? I don't see it immediately...
Say that there are to different waiting times 1s and 3s, and they happen with probability 50% each. The average waiting time (1/2 1+1/2 3) is 2s. However, 75% of the time we are waiting on a 3s event and only 25% on a 1s event. The weighted average is 2.5s. E[X^2]=1/2 1+1/2 9=5(s^2) is not the right answer, it still has to be divided by E[X]=2(s) to get the correct answer.
Re: Alice is impatient
#10So.. apply that to Amazon design heuristics like author name search on books, and how Amazon return "in the style of" and "not a book but this guy called Charles Dickens makes jigsaws" as high order matches and consider how the end user experience weights to the pessimal yet Amazon can show on average they make more money doing this..
(Understood that engineers and AWS don't influence UX in the storefront or search)