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61 lines
2.0 KiB
61 lines
2.0 KiB
8 years ago
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#!/usr/bin/env python
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#
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# 20160920
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# Wirawan Purwanto
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#
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def est_hpl_timing(N, nprocs, proc_gflops, eff=0.8):
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"""Estimates the time it takes to do HPL calculation on
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an (N x N) problem, given `nprocs` processor cores which has
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`proc_gflops` GFLOPS.
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"""
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# Number of floating point operations
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# From HPL code, estimated to be
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# 2/3 N^3 - 1/2 N^2 flops for LU factorization + 2 N^2 flops for solve.
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assert N > 0
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assert nprocs >= 1
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assert proc_gflops > 0
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assert 0.0 < eff <= 1.0
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N = float(N)
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num_gflop = (2 * N**3 / 3 - 0.5 * N**2 + 2 * N**2) * 1e-9
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tot_proc_gflops = nprocs * proc_gflops
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est = num_gflop / tot_proc_gflops / eff
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proc_mem_gb = (N**2 * 1e-9 * 8) / nprocs
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#if verbose >= 1:
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return (est, num_gflop, tot_proc_gflops, proc_mem_gb)
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def est_hpl_timing2(proc_mem_gb, nprocs, proc_gflops, eff=0.8):
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"""Estimates the time it takes to do HPL calculation on
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a problem specified by `proc_mem_gb` RAM per core (in GB),
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`nprocs` processor cores, each having `proc_gflops` GFLOPS.
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We assume the matrix is evenly distributed across processors in
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a square fashion (i.e. P == Q for tile definition).
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"""
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from math import sqrt
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N1 = sqrt(proc_mem_gb * 1e9 / 8)
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N = float(int(N1 * sqrt(nprocs)))
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(est, num_gflop, tot_proc_flops, proc_mem_gb0) = \
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est_hpl_timing(N, nprocs, proc_gflops, eff)
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return est, num_gflop, tot_proc_flops, N
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def Test_64core_memscale(proc_mem_gb=[0.15, 0.25, 0.50, 0.780125, 1.25, 2.00, 3.00],
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proc_gflops=17.6, eff=0.8):
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"""[20160920]
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Test: keep at 64 cores, scale up memory and see how much time it takes."""
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nproc = 64
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from wpylib.text_tools import str_fmt_heading
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cols = ("N", "mem/proc", "nproc", "time", "numops_gf", "proc_gf")
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fmt = "%8d %8.3f %5d %6.0f %10.2f %10.2f"
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hfmt = str_fmt_heading(fmt)
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print(hfmt % cols)
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for pm1 in proc_mem_gb:
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(est_t, num_gflop, tot_proc_gflops, N) = \
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est_hpl_timing2(pm1, nproc, proc_gflops, eff)
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print(fmt % (N, pm1, nproc, est_t, num_gflop, tot_proc_gflops))
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