Parallel Programming in Fortran 95 using OpenMP - People

3.2. Other clauses 49
The 600 iterations are going to be distributed in different ways by changing the
value of chunk. The results are shown graphically in figure 3.6, where each block
represents a piece of work and the number inside it the thread identification number
processing the piece of work.
0
0
0
0
0
1
1
2
1
1
0
2
2
chunk = 150 chunk = 250 chunk = 300 default
50
100
150
200
250
300
350
400
450
500
550
600
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Figure 3.6: Graphical representation of the example explaining the general working principle
of the SCHEDULE(STATIC,chunk) clause for different values of the optional parameter chunk.
In the first case, chunk = 150, four blocks of work are needed to cover the complete
iteration space. Since the total number of iterations is exactly four times the value
of chunk, all blocks are exactly equal in size. To distribute the different blocks
among the existing threads, a round-robin procedure is used leading to the presented
distribution. The resulting parallel running code is not very efficient, since thread
0 is going to work while all the other threads are waiting in an idle state.
In the second case, chunk = 250, the number of blocks equals the number of threads,
which is something good. But this time the sizes of the blocks are not equal, since
the total number of iterations is not an exact multiply of chunk. Even though, this
value of the optional parameter leads to a faster code than in the previous case,
because the latency times are smaller 4 .
In the third case, chunk = 300, the number of blocks is smaller than the number of
available threads. This means that thread 2 is not going to work for the present
do-loop. The resulting efficiency is worse than in the previous case and equal to the
first case.
In the last case, which corresponds to do not specify the optional parameter chunk,
OpenMP creates a number of blocks equal to the number of available threads and
with equal block sizes. In general this option leads to the best performance, if all
iterations require the same computational time. This remark also applies to the
previous cases and explanations.
4 In the first case the fact, that thread 0 needs to compute in total 300 iterations, leads to a slower
code than in the second case, where the maximum number of iterations to compute by one thread is 250.