Newsgroups: comp.parallel.pvm
From: Graham Edward Fagg <sssfagg@reading.ac.uk>
Subject: Re: Timing PVM programs
Organization: University of Reading, U.K.
Date: Wed, 8 Mar 1995 12:00:46 +0000
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII
Message-ID: <Pine.SUN.3.91.950308114329.22181C-100000@sssscsc1.reading.ac.uk>

On 6 Mar 1995, Niall Adams wrote:
> Hi,
> 	I am interested in timing my PVM application (yes, I know this
> can be a mine field!). 
You said it!
> Anyway, if I use the UNIX time command, I get 
> something like;
>   % time pvm_prog.out
> 460.509u 1.792s 18:03.06 42.6% 130+344k 4+66io 4pf+0w
> So the wall clock time for execution is about 18 minutes. The sum of 
> cpu and system time comes to less than eight minutes. Is this eight
> minutes a valid estimate of the throughput time for this job? Is there
> a more meaningful method of measurement?
> Thanks,
> Niall Adams
> cmsnadam@kryten.cms.livjm.ac.uk
Well yes, the master probably only spent 8 minutes doing any real work. 
When it was waiting for results on pvm_recv calls it would be swapped 
out, and hense not take any cpu time. So you have two solutions:
(1) the 18 minutes is the time that is really important as its the time 
that users have to wait for their results..
(2) use something like :
	getrusage(RUSAGE_SELF,&usage) where usage.ru_utime and usage.ru_stime
	contain user and system time so-far
And do this for all the slaves and the master and hense work out the *total*
CPU+SYSTEM time used. 

If your on a Sun machine try using the /usr/5bin/time command as its alot 
earier to read (and also parse).

Graham.
===============================================================================
Graham Edward Fagg ||| *** Cluster Computing Lab. ***  ||| e-mail me some time
 Computer Science ||  Software Engineering Subject Grp.  ||  G.E.Fagg@rdg.ac.uk
0734-875123 7626 | http://www.cs.reading.ac.uk/people/gef/ | PVM/MPI/LINDA/VMD
===============================================================================


