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5.4 Multicast trees

It may be worth comparing the performance of a native multicast operation to the performance achieved by multicasting over a k-nomial tree to gauge how well the underlying communication layer implements multicasts. The following code records a wealth of data, varying the tree arity (i.e., k), the number of tasks receiving the multicast, and the message size. It provides a good demonstration of how to use the KNOMIAL_CHILDREN and KNOMIAL_CHILD functions.

# Test the performance of multicasting over various k-nomial trees
# By Scott Pakin <>

Require language version "1.5".

# Parse the command line.
minsize is "Min. message size (bytes)" and comes from "--minbytes" or "-n"
  with default 1.
maxsize is "Max. message size (bytes)" and comes from "--maxbytes" or "-x"
  with default 1M.
reps is "Repetitions to perform" and comes from "--reps" or "-r" with
  default 100.
maxarity is "Max. arity of the tree" and comes from "--maxarity" or "-a"
  with default 2.

Assert that "this program requires at least two processors" with

# Send messages from task 0 to 1, 2, 3, ... other tasks in a k-nomial tree.
For each arity in {2, ..., maxarity} {
  for each num_targets in {1, ..., num_tasks-1} {
    for each msgsize in {minsize, minsize*2, minsize*4, ..., maxsize} {
      task 0 outputs "Multicasting a " and msgsize and "-byte message to "
        and num_targets and " target(s) over a " and arity and
        "-nomial tree ..." then
      for reps repetitions {
        task 0 resets its counters then
        for each src in {0, ..., num_tasks}
          for each dstnum in {0, ..., knomial_children(src, arity,
            task src sends a msgsize byte message to task
              knomial_child(src, dstnum, arity) then
        all tasks synchronize then
        task 0 logs the arity as "k-nomial arity" and
                    the num_targets as "# of recipients" and
                    the msgsize as "Message size (bytes)" and
                    the median of (1E6/1M)*(msgsize/elapsed_usecs) as
                      "Incoming bandwidth (MB/s)" and
                    the median of (num_targets*msgsize/elapsed_usecs)*
                      (1E6/1M) as "Outgoing bandwidth (MB/s)"
      } then
      task 0 computes aggregates

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Scott Pakin,