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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).
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