矩阵矩阵乘法的 CPU 时间

Question

I am trying to decide wether several similar but independent problems should be dealt with simultaneously or sequentially (possibly in parallel on different computers). In order to decide, I need to compare the cpu times of the following operations :

• time_1 is the time for computing X(with shape (n,p)) @ b (with shape (p,1)).

• time_k is the time for computing X(with shape (n,p)) @ B (with shape (p,k)).

where X, b and B are random matrices. The difference between the two operations is the width of the second matrix.

Naively, we expect that time_k = kx time_1. With faster matrix multiplication algorithms (Strassen algorithm, Coppersmith–Winograd algorithm), time_k could be smaller than kx time_1 but the complexity of these algorithms remains much larger than what I observed in practice. Therefore my question is : How to explain the large difference in terms of cpu times for these two computations ?

The code I used is the following :

import time
import numpy as np
import matplotlib.pyplot as plt

p     = 100
width = np.concatenate([np.arange(1, 20), np.arange(20, 100, 10), np.arange(100, 4000, 100)]).astype(int)

mean_time = []
for nk, kk in enumerate(width):
    timings = []
    nb_tests = 10000 if kk <= 300 else 100
    for ni, ii in enumerate(range(nb_tests)):
        print('\r[', nk, '/', len(width), ', ',  ni, '/', nb_tests, ']', end = '')
        x     = np.random.randn(p).reshape((1, -1))
        coef  = np.random.randn(p, kk)
        d     = np.zeros((1, kk))
        start = time.time()
        d[:]  = x @ coef
        end   = time.time()
        timings.append(end - start)

    mean_time.append(np.mean(timings))

mean_time = np.array(mean_time)


fig, ax = plt.subplots(figsize =(14,8))
plt.plot(width, mean_time, label =  'mean(time\_k)')
plt.plot(width, width*mean_time[0], label = 'k*mean(time\_1)')
plt.legend()
plt.xlabel('k')
plt.ylabel('time (sec)')
plt.show()

Answer 1

You aren't only timing multiplication operation. time.time() takes time to complete.

>>> print(time.time() - time.time())
-9.53674316406e-07

When multiplied by the number of tries (10000) then the number of instances it becomes significant overhead, for n=100 you are in fact comparing what is 1.000.000 calls to time.time() to 100 regular numpy array multiplications.

For quick benchmarking, Python provides a dedicated module that doesn't have this problem : see timeit

Answer 2

This detail of the reason is very complex. You know that when PC run the X @ b , it will execute many other required instructions, maybe load data from RAM to cache and so on. In other words, the cost time contains two parts - the 'real calculate instructions' in CPU represented by Cost_A and 'other required instructions' represented by Cost_B . I have a idea, just my guess, that it's the Cost_B lead to time_k << kx time_1 .

For the shape of b is small (eg 1000 x 1), the 'other required instructions' cost relatively the most time. For the shape of b is huge (eg 1000 x 10000), it's relatively small. The following group of experiments could give a less rigorous proof. We can see that when the shape of b increases from (1000 x 1) to (1000 x ) the cost time increases very slowly.

import numpy as np
import time

X = np.random.random((1000, 1000))

b = np.random.random((1000, 1))
b3 = np.random.random((1000, 3))
b5 = np.random.random((1000, 5))
b7 = np.random.random((1000, 7))
b9 = np.random.random((1000, 9))
b10 = np.random.random((1000, 10))
b30 = np.random.random((1000, 30))
b60 = np.random.random((1000, 60))
b100 = np.random.random((1000, 100))
b1000 = np.random.random((1000, 1000))

def test_cost(X, b):
    begin = time.time()
    for i in range(100):
        _ = X @ b
    end = time.time()
    print((end-begin)/100.)

test_cost(X, b)
test_cost(X, b3)
test_cost(X, b5)
test_cost(X, b7)
test_cost(X, b9)
test_cost(X, b10)
test_cost(X, b30)
test_cost(X, b60) 
test_cost(X, b100)
test_cost(X, b1000)

output:
0.0003210139274597168
0.00040063619613647463
0.0002452659606933594
0.00026523590087890625
0.0002449488639831543
0.00024344682693481446
0.00040068864822387693
0.000691361427307129
0.0011700797080993653
0.009680757522583008

For more, I do a set of experiments with pref in linux. For the pref , the Cost_B maybe more big. I have 8 python files, the first one is as follows.

import numpy as np
import time
def broken2():
    mtx = np.random.random((1, 1000))
    c = None
    c = mtx ** 2

broken2()

I had process the output to table A, as follows.

I do a simple analysis that I divide the error of the number of operation (likes, cache-misses) in neighbor experiments by the error of time elapsed(seconds) . Then, I get the following table B. From the table, we can find that as the shape of b increasing the linear relation between of shape and cost time is more obvious. And maybe the main reason that lead to time_k << kx time_1 is cache misses (load data from RAM to cache), for it stabilized firstly .