True. I was curious as to what happens when I am time sharing the CPU.

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Sent from my iPhone

On May 8, 2012, at 3:11 AM, TERRY DONTJE <terry.dontje@oracle.com> wrote:

On 5/7/2012 8:40 PM, Jeff Squyres (jsquyres) wrote:
On May 7, 2012, at 8:31 PM, Jingcha Joba wrote:

So in the above stated example, end-start will be: <whatever the solver took> + 20ms ?
 
(time slice of P2 + P3 = 20ms)
More or less (there's nonzero amount of time required for the kernel scheduler, and the time quantum for each of P2 and P3 is likely not *exactly* 10ms).  But you're over thinking this.  

The elapsed wall-clock time is simply (end-start).

To kind of add to what Jeff is saying, the case you are describing sounds like oversubscription.  If you really need to find the "pure" performance of the code you should be running on a dedicated cluster otherwise you'll be battling other issues in addition to timeslicing. 

--
Terry D. Dontje | Principal Software Engineer
Developer Tools Engineering | +1.781.442.2631
Oracle - Performance Technologies
95 Network Drive, Burlington, MA 01803
Email terry.dontje@oracle.com



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