linux-kernel - Re: [Announce] [patch] Modular Scheduler Core and Completely Fair Scheduler [CFS]

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Message-ID: <Pine.LNX.4.64.0704172141051.32323@alien.or.mcafeemobile.com>
Date:	Tue, 17 Apr 2007 21:42:34 -0700 (PDT)
From:	Davide Libenzi <davidel@...ilserver.org>
To:	William Lee Irwin III <wli@...omorphy.com>
cc:	Matt Mackall <mpm@...enic.com>, Ingo Molnar <mingo@...e.hu>,
	Nick Piggin <npiggin@...e.de>,
	Peter Williams <pwil3058@...pond.net.au>,
	Mike Galbraith <efault@....de>,
	Con Kolivas <kernel@...ivas.org>, ck list <ck@....kolivas.org>,
	Bill Huey <billh@...ppy.monkey.org>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
	Linus Torvalds <torvalds@...ux-foundation.org>,
	Andrew Morton <akpm@...ux-foundation.org>,
	Arjan van de Ven <arjan@...radead.org>,
	Thomas Gleixner <tglx@...utronix.de>
Subject: Re: [Announce] [patch] Modular Scheduler Core and Completely Fair
 Scheduler [CFS]

On Tue, 17 Apr 2007, William Lee Irwin III wrote:

> 100**(1/39.0) ~= 1.12534 may do if so, but it seems a little weak, and
> even 1000**(1/39.0) ~= 1.19378 still seems weak.
> 
> I suspect that in order to get low nice numbers strong enough without
> making high nice numbers too strong something sub-exponential may need
> to be used. Maybe just picking percentages outright as opposed to some
> particular function.
> 
> We may also be better off defining it in terms of a share weighting as
> opposed to two tasks in competition. In such a manner the extension to
> N tasks is more automatic. f(n) would be a univariate function of nice
> numbers and two tasks in competition with nice numbers n_1 and n_2
> would get shares f(n_1)/(f(n_1)+f(n_2)) and f(n_2)/(f(n_1)+f(n_2)). In
> the exponential case f(n) = K*e**(r*n) this ends up as
> 1/(1+e**(r*(n_2-n_1))) which is indeed a function of n_1-n_2 but for
> other choices it's not so. f(n) = n+K for K >= 20 results in a share
> weighting of (n_1+K,n_2+K)/(n_1+n_2+2*K), which is not entirely clear
> in its impact. My guess is that f(n)=1/(n+1) when n >= 0 and f(n)=1-n
> when n <= 0 is highly plausible. An exponent or an additive constant
> may be worthwhile to throw in. In this case, f(-19) = 20, f(20) = 1/21,
> and the ratio of shares is 420, which is still arithmeticaly feasible.
> -10 vs. 0 and 0 vs. 10 are both 10:1.

This makes more sense, and the ratio at the extremes is something 
reasonable.



- Davide


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