Measuring CPU performance

Processor caches must be taken into account when writing an algorithm, see Memory part 2: CPU caches from Ulrich Drepper.

Cache Performance

from tqdm import tqdm
import matplotlib.pyplot as plt
from pandas import DataFrame, concat
from sphinx_runpython.runpython import run_cmd
from onnx_extended.ext_test_case import unit_test_going
from onnx_extended.validation.cpu._validation import (
    benchmark_cache,
    benchmark_cache_tree,
)

Code of benchmark_cache.

obs = []
step = 2**12
for i in tqdm(range(step, 2**20 + step, step)):
    res = min(
        [
            benchmark_cache(i, False),
            benchmark_cache(i, False),
            benchmark_cache(i, False),
        ]
    )
    if res < 0:
        # overflow
        continue
    obs.append(dict(size=i, perf=res))

df = DataFrame(obs)
mean = df.perf.mean()
lag = 32
for i in range(2, df.shape[0]):
    df.loc[i, "smooth"] = df.loc[i - 8 : i + 8, "perf"].median()
    if i > lag and i < df.shape[0] - lag:
        df.loc[i, "delta"] = (
            mean
            + df.loc[i : i + lag, "perf"].mean()
            - df.loc[i - lag + 1 : i + 1, "perf"]
        ).mean()

  0%|          | 0/256 [00:00<?, ?it/s]
 54%|█████▎    | 137/256 [00:00<00:00, 1352.31it/s]
100%|██████████| 256/256 [00:00<00:00, 708.71it/s]

Cache size estimator

cache_size_index = int(df.delta.argmax())
cache_size = df.loc[cache_size_index, "size"] * 2
print(f"L2 cache size estimation is {cache_size / 2 ** 20:1.3f} Mb.")

L2 cache size estimation is 1.695 Mb.

Verification

try:
    out, err = run_cmd("lscpu", wait=True)
    print("\n".join(_ for _ in out.split("\n") if "cache:" in _))
except Exception as e:
    print(f"failed due to {e}")

df = df.set_index("size")
fig, ax = plt.subplots(1, 1, figsize=(12, 4))
df.plot(ax=ax, title="Cache Performance time/size", logy=True)
fig.tight_layout()
fig.savefig("plot_benchmark_cpu_array.png")

L1d cache:                       128 KiB (4 instances)
L1i cache:                       128 KiB (4 instances)
L2 cache:                        1 MiB (4 instances)
L3 cache:                        8 MiB (1 instance)

TreeEnsemble Performance

We simulate the computation of a TreeEnsemble of 50 features, 100 trees and depth of 10 (so nodes.) The code of benchmark_cache_tree

dfs = []
cols = []
drop = []
for n in tqdm(range(2 if unit_test_going() else 5)):
    res = benchmark_cache_tree(
        n_rows=2000,
        n_features=50,
        n_trees=100,
        tree_size=1024,
        max_depth=10,
        search_step=64,
    )
    res = [[max(r.row, i), r.time] for i, r in enumerate(res)]
    df = DataFrame(res)
    df.columns = [f"i{n}", f"time{n}"]
    dfs.append(df)
    cols.append(df.columns[-1])
    drop.append(df.columns[0])

df = concat(dfs, axis=1).reset_index(drop=True)
df["i"] = df["i0"]
df = df.drop(drop, axis=1)
df["time_avg"] = df[cols].mean(axis=1)
df["time_med"] = df[cols].median(axis=1)

df.head()

  0%|          | 0/5 [00:00<?, ?it/s]
 20%|██        | 1/5 [00:00<00:03,  1.03it/s]
 40%|████      | 2/5 [00:01<00:02,  1.07it/s]
 60%|██████    | 3/5 [00:02<00:01,  1.10it/s]
 80%|████████  | 4/5 [00:03<00:00,  1.12it/s]
100%|██████████| 5/5 [00:04<00:00,  1.12it/s]
100%|██████████| 5/5 [00:04<00:00,  1.11it/s]

	time0	time1	time2	time3	time4	i	time_avg	time_med
0	0.028126	0.038091	0.028161	0.028261	0.028189	0	0.030166	0.028189
1	0.028126	0.038091	0.028161	0.028261	0.028189	1	0.030166	0.028189
2	0.028126	0.038091	0.028161	0.028261	0.028189	2	0.030166	0.028189
3	0.028126	0.038091	0.028161	0.028261	0.028189	3	0.030166	0.028189
4	0.028126	0.038091	0.028161	0.028261	0.028189	4	0.030166	0.028189

Estimation

print("Optimal batch size is among:")
dfi = df[["time_med", "i"]].groupby("time_med").min()
dfi_min = set(dfi["i"])
dfsub = df[df["i"].isin(dfi_min)]
dfs = dfsub.sort_values("time_med").reset_index()
print(dfs[["i", "time_med", "time_avg"]].head(10))

Optimal batch size is among:
      i  time_med  time_avg
0   448  0.027966  0.028792
1    64  0.028016  0.028505
2   192  0.028045  0.028000
3   896  0.028057  0.028381
4  1856  0.028061  0.028154
5  1920  0.028076  0.027879
6  1216  0.028079  0.029016
7   128  0.028087  0.028028
8   384  0.028093  0.028341
9   704  0.028118  0.028209

One possible estimation

subdfs = dfs[:20]
avg = (subdfs["i"] / subdfs["time_avg"]).sum() / (subdfs["time_avg"] ** (-1)).sum()
print(f"Estimation: {avg}")

Estimation: 938.7128273422819

Plots.

cols_time = ["time_avg", "time_med"]
fig, ax = plt.subplots(2, 1, figsize=(12, 6))
df.set_index("i").drop(cols_time, axis=1).plot(
    ax=ax[0], title="TreeEnsemble Performance time per row", logy=True, linewidth=0.2
)
df.set_index("i")[cols_time].plot(ax=ax[1], linewidth=1.0, logy=True)
fig.tight_layout()
fig.savefig("plot_bench_cpu.png")