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author Chris Cannam
date Wed, 07 Sep 2016 10:40:32 +0100
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Chris@19 49 <a name="Load-balancing"></a>
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Chris@19 56
Chris@19 57 <h4 class="subsection">6.4.2 Load balancing</h4>
Chris@19 58
Chris@19 59 <p><a name="index-load-balancing-381"></a>
Chris@19 60 Ideally, when you parallelize a transform over some P
Chris@19 61 processes, each process should end up with work that takes equal time.
Chris@19 62 Otherwise, all of the processes end up waiting on whichever process is
Chris@19 63 slowest. This goal is known as &ldquo;load balancing.&rdquo; In this section,
Chris@19 64 we describe the circumstances under which FFTW is able to load-balance
Chris@19 65 well, and in particular how you should choose your transform size in
Chris@19 66 order to load balance.
Chris@19 67
Chris@19 68 <p>Load balancing is especially difficult when you are parallelizing over
Chris@19 69 heterogeneous machines; for example, if one of your processors is a
Chris@19 70 old 486 and another is a Pentium IV, obviously you should give the
Chris@19 71 Pentium more work to do than the 486 since the latter is much slower.
Chris@19 72 FFTW does not deal with this problem, however&mdash;it assumes that your
Chris@19 73 processes run on hardware of comparable speed, and that the goal is
Chris@19 74 therefore to divide the problem as equally as possible.
Chris@19 75
Chris@19 76 <p>For a multi-dimensional complex DFT, FFTW can divide the problem
Chris@19 77 equally among the processes if: (i) the <em>first</em> dimension
Chris@19 78 <code>n0</code> is divisible by P; and (ii), the <em>product</em> of
Chris@19 79 the subsequent dimensions is divisible by P. (For the advanced
Chris@19 80 interface, where you can specify multiple simultaneous transforms via
Chris@19 81 some &ldquo;vector&rdquo; length <code>howmany</code>, a factor of <code>howmany</code> is
Chris@19 82 included in the product of the subsequent dimensions.)
Chris@19 83
Chris@19 84 <p>For a one-dimensional complex DFT, the length <code>N</code> of the data
Chris@19 85 should be divisible by P <em>squared</em> to be able to divide
Chris@19 86 the problem equally among the processes.
Chris@19 87
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