From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: from sc8-sf-mx1-b.sourceforge.net ([10.3.1.11] helo=sc8-sf-mx1.sourceforge.net) by sc8-sf-list1.sourceforge.net with esmtp (Exim 4.30) id 1DXhfC-0003y5-Rg for user-mode-linux-devel@lists.sourceforge.net; Mon, 16 May 2005 08:37:50 -0700 Received: from rproxy.gmail.com ([64.233.170.196]) by sc8-sf-mx1.sourceforge.net with esmtp (Exim 4.41) id 1DXhfB-0007go-W7 for user-mode-linux-devel@lists.sourceforge.net; Mon, 16 May 2005 08:37:50 -0700 Received: by rproxy.gmail.com with SMTP id j1so889804rnf for ; Mon, 16 May 2005 08:37:49 -0700 (PDT) Message-ID: <3524bf1f05051608377420ae37@mail.gmail.com> From: Young Koh Reply-To: Young Koh Subject: Re: [uml-devel] The UML scheduler In-Reply-To: <20050516150648.GB4977@ccure.user-mode-linux.org> Mime-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Content-Disposition: inline References: <3524bf1f05051607306756dcfc@mail.gmail.com> <20050516150648.GB4977@ccure.user-mode-linux.org> Sender: user-mode-linux-devel-admin@lists.sourceforge.net Errors-To: user-mode-linux-devel-admin@lists.sourceforge.net List-Unsubscribe: , List-Id: The user-mode Linux development list List-Post: List-Help: List-Subscribe: , List-Archive: Date: Mon, 16 May 2005 11:37:49 -0400 Content-Transfer-Encoding: quoted-printable To: user-mode-linux-devel@lists.sourceforge.net Thank you for your reply. I have one following question. in native Linux, a timer tick goes to the kernel, which has the previliege to save the current context and switch the running process. But UML kernel is just another user level process, even though it is tracing the application processes. in UML case, when a timer tick goes to the UML kernel, how does it stop the running process, which is another process? it seems it's a primitive quesion, but it will help me a lot to understand more about UML. Thank you! On 5/16/05, Jeff Dike wrote: > On Mon, May 16, 2005 at 10:30:33AM -0400, Young Koh wrote: > > suppose there are 2 processes on top of UML and 3 host processes. > > then, the host kernel sees 5 processes total, so, each of the process > > will get 20% of the CPU time? is that right? >=20 > In tt mode, there will be 5 processes, but it will be difficult for the 2= UML > processes to each get 20% of the time, since only one of them will be run= nable > at a time on the host. This is still unfair because each of them will ha= ve > a higher dynamic priority than the 3 host processes because they are each= only > runnable half the time. >=20 > In skas mode, there is only one host process for all of the UML processes. >=20 > > Or, UML is somehow viewed as a group of processes and UML get 25% of > > the CPU and each UML got 12.5% of the CPU time? if so, how is it > > possible? >=20 > In skas mode, this is true. >=20 > > one more question, when the UML kernel wants to schedule its > > processes, how does it preempt UML processes? >=20 > The same way as any other architecture. A timer tick comes in, and the > timer handler calls schedule(). >=20 > > this way of scheduling and preemption will be different from TT mode > > and SKAS mode? if so, how? >=20 > No, exactly the same. >=20 > Jeff > ------------------------------------------------------- This SF.Net email is sponsored by Oracle Space Sweepstakes Want to be the first software developer in space? Enter now for the Oracle Space Sweepstakes! http://ads.osdn.com/?ad_ids93&alloc_id=16281&op=CCk _______________________________________________ User-mode-linux-devel mailing list User-mode-linux-devel@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/user-mode-linux-devel