Synergy User Manual and Tutorial. - THE CORE MEMORY
Synergy User Manual and Tutorial. - THE CORE MEMORY
Synergy User Manual and Tutorial. - THE CORE MEMORY
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<strong>Synergy</strong> <strong>User</strong> <strong>Manual</strong> <strong>and</strong> <strong>Tutorial</strong><br />
T<br />
n<br />
= Tc<br />
+ To<br />
+ T1<br />
− p<br />
Assume that the following values are valid for the matrix multiplication above:<br />
• Sequential run time T 1 120sec<br />
• Parallel computation time T c 60sec<br />
• Parallel overhead T o 20sec<br />
• Assume no non-parallizable code T 1-p 0sec<br />
Then speedup would be<br />
120sec<br />
T1 = 60sec,<br />
T2<br />
= Tc<br />
+ To<br />
+ T1<br />
− p<br />
= 60sec+<br />
20sec+<br />
0sec = 80sec, S = = 1.5 = 150%<br />
80sec<br />
This is somewhat less than the previous speedup.<br />
The cost C of a parallel system is calculated by multiplying the parallel run T n time <strong>and</strong><br />
the number of processors N p divided by the sequential run time T 1 :<br />
T<br />
C =<br />
n<br />
×<br />
T 1<br />
N<br />
p<br />
The values in the example above, ignoring overhead, would be.<br />
T<br />
C =<br />
n<br />
× N<br />
T<br />
1<br />
p<br />
60sec×<br />
2 120sec<br />
= = = 1<br />
120sec 120sec<br />
This equation is shows that the parallel system is optimal because the increase in speed is<br />
proportional with the number of processors added. Typically costs are not optimal.<br />
Considering the overhead in the example above, we have:<br />
T<br />
C =<br />
n<br />
× N<br />
T<br />
1<br />
p<br />
80sec×<br />
2 160sec<br />
= = = 1.333.<br />
120sec 120sec<br />
Timing Models<br />
111
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