In a self-educational project I measure the bandwidth of the memory with help of the following code (here paraphrased, the whole code follows at the end of the question):
unsigned int doit(const std::vector<unsigned int> &mem){
const size_t BLOCK_SIZE=16;
size_t n = mem.size();
unsigned int result=0;
for(size_t i=0;i<n;i+=BLOCK_SIZE){
result+=mem[i];
}
return result;
}
//... initialize mem, result and so on
int NITER = 200;
//... measure time of
for(int i=0;i<NITER;i++)
resul+=doit(mem)
BLOCK_SIZE is choosen in such a way, that a whole 64byte cache line is fetched per single integer-addition. My machine (an Intel-Broadwell) needs about 0.35 nanosecond per integer-addion, so the code above could saturate a bandwith as high as 182GB/s (this value is just an upper bound and is probably quite off, what is important is the ratio of bandwidths for different sizes). The code is compiled with g++ and -O3.
Varying the size of the vector, I can observe expected bandwidths for L1(*)-, L2-, L3-caches and the RAM-memory:
However, there is an effect I’m really struggling to explain: the collapse of the measured bandwidth of L1-cache for sizes around 2 kB, here in somewhat higher resolution:
I could reproduce the results on all machines I have access to (which have Intel-Broadwell and Intel-Haswell processors).
My question: What is the reason for the performance-collapse for memory-sizes around 2 KB?
(*) I hope I understand correctly, that for L1-cache not 64 bytes but only 4 bytes per addition are read/transfered (there is no further faster cache where a cache line must be filled), so the plotted bandwidth for L1 is only the upper limit and not the badwidth itself.
Edit: When the step size in the inner for-loop is chosen to be
- 8 (instead of 16) the collapse happens for 1KB
- 4 (instead of 16) the collapse happens for 0.5KB
i.e. when the inner loop consists of about 31-35 steps/reads. That means the collapse isn’t due to the memory-size but due to the number of steps in the inner loop.
It can be explained with branch misses as shown in @user10605163’s great answer.
Listing for reproducing the results
bandwidth.cpp:
#include <vector>
#include <chrono>
#include <iostream>
#include <algorithm>
//returns minimal time needed for one execution in seconds:
template<typename Fun>
double timeit(Fun&& stmt, int repeat, int number)
{
std::vector<double> times;
for(int i=0;i<repeat;i++){
auto begin = std::chrono::high_resolution_clock::now();
for(int i=0;i<number;i++){
stmt();
}
auto end = std::chrono::high_resolution_clock::now();
double time = std::chrono::duration_cast<std::chrono::nanoseconds>(end-begin).count()/1e9/number;
times.push_back(time);
}
return *std::min_element(times.begin(), times.end());
}
const int NITER=200;
const int NTRIES=5;
const size_t BLOCK_SIZE=16;
struct Worker{
std::vector<unsigned int> &mem;
size_t n;
unsigned int result;
void operator()(){
for(size_t i=0;i<n;i+=BLOCK_SIZE){
result+=mem[i];
}
}
Worker(std::vector<unsigned int> &mem_):
mem(mem_), n(mem.size()), result(1)
{}
};
double PREVENT_OPTIMIZATION=0.0;
double get_size_in_kB(int SIZE){
return SIZE*sizeof(int)/(1024.0);
}
double get_speed_in_GB_per_sec(int SIZE){
std::vector<unsigned int> vals(SIZE, 42);
Worker worker(vals);
double time=timeit(worker, NTRIES, NITER);
PREVENT_OPTIMIZATION+=worker.result;
return get_size_in_kB(SIZE)/(1024*1024)/time;
}
int main(){
int size=BLOCK_SIZE*16;
std::cout<<"size(kB),bandwidth(GB/s)n";
while(size<10e3){
std::cout<<get_size_in_kB(size)<<","<<get_speed_in_GB_per_sec(size)<<"n";
size=(static_cast<int>(size+BLOCK_SIZE)/BLOCK_SIZE)*BLOCK_SIZE;
}
//ensure that nothing is optimized away:
std::cerr<<"Sum: "<<PREVENT_OPTIMIZATION<<"n";
}
create_report.py:
import sys import pandas as pd import matplotlib.pyplot as plt input_file=sys.argv[1] output_file=input_file[0:-3]+'png' data=pd.read_csv(input_file) labels=list(data) plt.plot(data[labels[0]], data[labels[1]], label="my laptop") plt.xlabel(labels[0]) plt.ylabel(labels[1]) plt.savefig(output_file) plt.close()
Building/running/creating report:
>>> g++ -O3 -std=c++11 bandwidth.cpp -o bandwidth >>> ./bandwidth > report.txt >>> python create_report.py report.txt # image is in report.png
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Answer
I changed the values slightly: NITER = 100000 and NTRIES=1 to get a less noisy result.
I don’t have a Broadwell available right now, however I tried your code on my Coffee-Lake and got a performance drop, not at 2KB, but around 4.5KB. In addition I find erratic behavior of the throughput slightly above 2KB.
The blue line in the graph corresponds to your measurement (left axis):
The red line here is the result from perf stat -e branch-instructions,branch-misses, giving the fraction of branches that were not correctly predicted (in percent, right axis). As you can see there is a clear anti-correlation between the two.
Looking into the more detailed perf report, I found that basically all of these branch mispredictions happen in the most inner loop in Worker::operator(). If the taken/non-taken pattern for the loop branch becomes too long the branch predictor will not be able to keep track of it and so the exit branch of the inner loop will be mispredicted, leading to the sharp drop in throughput. With further increasing number of iterations the impact of this single mispredict will become less significant leading to the slow recover of the throughput.
For further information on the erratic behavior before the drop see the comments made by @PeterCordes below.
In any case the best way to avoid branch mispredictions is to avoid branches and so I manually unrolled the loop in Worker::operator(), like e.g.:
void operator()(){
for(size_t i=0;i+3*BLOCK_SIZE<n;i+=BLOCK_SIZE*4){
result+=mem[i];
result+=mem[i+BLOCK_SIZE];
result+=mem[i+2*BLOCK_SIZE];
result+=mem[i+3*BLOCK_SIZE];
}
}
Unrolling 2, 3, 4, 6 or 8 iterations gives the results below. Note that I did not correct for the blocks at the end of the vector which were ignored due to the unrolling. Therefore the periodic peaks in the blue line should be ignored, the lower bound base line of the periodic pattern is the actual bandwidth.
As you can see the fraction of branch mispredictions didn’t really change, but because the total number of branches is reduced by the factor of unrolled iterations, they will not contribute strongly to the performance anymore.
There is also an additional benefit of the processor being more free to do the calculations out-of-order if the loop is unrolled.
If this is supposed to have practical application I would suggest to try to give the hot loop a compile-time fixed number of iteration or some guarantee on divisibility, so that (maybe with some extra hints) the compiler can decide on the optimal number of iterations to unroll.