Question

由于许多机器学习算法依赖于矩阵乘法（或者至少可以使用矩阵乘法）来测试我的GPU，我计划创建矩阵a，b，将它们相乘并记录计算完成所需的时间。 / p>

以下代码将生成两个维度为300000,20000的矩阵并将它们相乘：

import tensorflow as tf
import numpy as np

init = tf.global_variables_initializer()
sess = tf.Session()
sess.run(init)


#a = np.array([[1, 2, 3], [4, 5, 6]])
#b = np.array([1, 2, 3])

a = np.random.rand(300000,20000)
b = np.random.rand(300000,20000)

println("Init complete");

result = tf.mul(a , b)
v = sess.run(result) 

print(v)

这是比较GPU性能的充分测试吗？我应该考虑哪些其他因素？

Answer 1

这里是example的matmul基准，避免了常见的陷阱，并与Titan X Pascal官方的11 TFLOP标记相匹配。

import os
import sys
os.environ["CUDA_VISIBLE_DEVICES"]="1"
import tensorflow as tf
import time

n = 8192
dtype = tf.float32
with tf.device("/gpu:0"):
    matrix1 = tf.Variable(tf.ones((n, n), dtype=dtype))
    matrix2 = tf.Variable(tf.ones((n, n), dtype=dtype))
    product = tf.matmul(matrix1, matrix2)


# avoid optimizing away redundant nodes
config = tf.ConfigProto(graph_options=tf.GraphOptions(optimizer_options=tf.OptimizerOptions(opt_level=tf.OptimizerOptions.L0)))
sess = tf.Session(config=config)

sess.run(tf.global_variables_initializer())
iters = 10

# pre-warming
sess.run(product.op)

start = time.time()
for i in range(iters):
  sess.run(product.op)
end = time.time()
ops = n**3 + (n-1)*n**2 # n^2*(n-1) additions, n^3 multiplications
elapsed = (end - start)
rate = iters*ops/elapsed/10**9
print('\n %d x %d matmul took: %.2f sec, %.2f G ops/sec' % (n, n,
                                                            elapsed/iters,
                                                            rate,))

使用张量流矩阵乘法测试GPU

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