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 pyquickhelper.loghelper import run_cmd
from pandas import DataFrame, concat
from onnx_extended.ext_test_case import unit_test_going
from onnx_extended.validation.cpu._validation import (
    benchmark_cache,
    benchmark_cache_tree,
)

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]
 51%|#####     | 130/256 [00:00<00:00, 1291.94it/s]
100%|##########| 256/256 [00:00<00:00, 712.30it/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 1.609 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.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 \(2^10\) nodes.)

dfs = []
cols = []
drop = []
for n in tqdm(range(10)):
    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])
    if unit_test_going() and len(dfs) >= 2:
        break

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%|          | 0/10 [00:00<?, ?it/s]
 10%|#         | 1/10 [00:00<00:08,  1.12it/s]
 20%|##        | 2/10 [00:01<00:07,  1.13it/s]
 30%|###       | 3/10 [00:02<00:06,  1.14it/s]
 40%|####      | 4/10 [00:03<00:05,  1.15it/s]
 50%|#####     | 5/10 [00:04<00:04,  1.14it/s]
 60%|######    | 6/10 [00:05<00:03,  1.15it/s]
 70%|#######   | 7/10 [00:06<00:02,  1.13it/s]
 80%|########  | 8/10 [00:07<00:01,  1.13it/s]
 90%|######### | 9/10 [00:07<00:00,  1.13it/s]
100%|##########| 10/10 [00:08<00:00,  1.12it/s]
100%|##########| 10/10 [00:08<00:00,  1.13it/s]

	time0	time1	time2	time3	time4	time5	time6	time7	time8	time9	i	time_avg	time_med
0	0.028239	0.028142	0.028567	0.02891	0.028134	0.028104	0.027676	0.028184	0.028282	0.028113	0	0.028235	0.028163
1	0.028239	0.028142	0.028567	0.02891	0.028134	0.028104	0.027676	0.028184	0.028282	0.028113	1	0.028235	0.028163
2	0.028239	0.028142	0.028567	0.02891	0.028134	0.028104	0.027676	0.028184	0.028282	0.028113	2	0.028235	0.028163
3	0.028239	0.028142	0.028567	0.02891	0.028134	0.028104	0.027676	0.028184	0.028282	0.028113	3	0.028235	0.028163
4	0.028239	0.028142	0.028567	0.02891	0.028134	0.028104	0.027676	0.028184	0.028282	0.028113	4	0.028235	0.028163

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   768  0.027691  0.028317
1  1664  0.027801  0.027871
2   320  0.027830  0.028003
3  1920  0.027855  0.028854
4   576  0.027893  0.028547
5  1600  0.027905  0.027951
6   384  0.027924  0.027981
7   512  0.027926  0.028223
8  1856  0.027965  0.029282
9   704  0.027980  0.028077

One possible estimation

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

Estimation: 1048.3608045617461

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.savefig("plot_bench_cpu.png")