Bigger Matrix Multiply problem
Now we have to multiply 2 matrices A[2000,2000] and B[2000,6000]. Our result will be a matrix C[2000,6000]. On bytes we have
Matrix A: 16 Mb
Matrix B/C: 4.8Mb
Basic our code multiply 10 times those matrices.
/*
Now we make the matrix much bigger
g++ -pg seq_matrix_big_mul.c -o seq_matrix_big_mul
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
int num_rows_A = 900; int num_rows_B = 900; int num_rows_C = 900;
int num_cols_A = 900; int num_cols_B = 600; int num_cols_C = 600;
// I'm forcing a malloc because I want to add the malloc time on the game
float *A = (float*) malloc(sizeof(float) * num_rows_A * num_cols_A);
float *B = (float*) malloc(sizeof(float) * num_rows_B * num_cols_B);
float *C = (float*) malloc(sizeof(float) * num_rows_C * num_cols_C);
void matrix_2d_mul_float(float *A, float *B, float *C, int num_rows_A, int num_cols_A, int num_cols_B) {
float sum = 0;
int num_rows_C = num_rows_A;
int num_cols_C = num_cols_B;
// Iterate on each row of A
//#pragma omp parallel for schedule(dynamic,1) collapse(2)
for(int i=0; i<num_rows_A; i++) {
// Iterate on each collumn of B
for (int k=0; k<num_cols_B; k++) {
sum = 0;
// Do the "multiply add between" row of A and collumn of B
for (int j=0; j<num_cols_A; j++){
// A[i][j] == A[i*num_cols_A+j]
// B[j][k] == B[j*num_cols_B+k]
//sum += A[i][j]*B[j][k];
sum += A[i*num_cols_A+j]*B[j*num_cols_B+k];
}
// C[i][k] == C[i*num_cols_C+k]
C[i*num_cols_C+k]=sum;
}
}
}
void fillRand(float *vec, int minValue, int maxValue, int sizeVec) {
srand(time(NULL));
for (int idx = 0; idx < sizeVec; idx++) {
vec[idx] = rand() % maxValue + minValue;
}
}
int main() {
// Get size in bytes for our vectors
int numBytesA = sizeof(float) * num_rows_A * num_cols_A;
int numBytesB = sizeof(float) * num_rows_B * num_cols_B;
printf("Size in bytes A: %d\n",numBytesA);
printf("Size in bytes B: %d\n",numBytesB);
// Fill arrays
fillRand(A, 1, 100, num_rows_A * num_cols_A);
fillRand(B, 1, 100, num_rows_B * num_cols_B);
// Call sequential function
//ProfilerStart("nameOfProfile.log");
for (int idxLoop=0; idxLoop < 10; idxLoop++) {
// Populate matricex on heap
matrix_2d_mul_float(A,B,C,num_rows_A,num_cols_A,num_cols_B);
printf("Matrix multiplication done %d\n",idxLoop);
}
// Free memory
free(A);free(B);free(C);
return 0;
}Executing sequentially
Executing this sequentially will be slow (30 seconds) to complete 10 matrix multiplications.
# Compile
g++ -g -O1 seq_matrix_big_mul.c -o seq_matrix_big_mul
# Check execution time
sudo perf stat ./seq_matrix_big_mul
Size in bytes A: 16000000
Size in bytes B: 4800000
Matrix multiplication done 0
Matrix multiplication done 1
Matrix multiplication done 2
Matrix multiplication done 3
Matrix multiplication done 4
Matrix multiplication done 5
Matrix multiplication done 6
Matrix multiplication done 7
Matrix multiplication done 8
Matrix multiplication done 9
Performance counter stats for './seq_matrix_big_mul':
30472.045987 task-clock (msec) # 1.000 CPUs utilized
82 context-switches # 0.003 K/sec
14 cpu-migrations # 0.000 K/sec
1,195 page-faults # 0.039 K/sec
91,004,382,307 cycles # 2.986 GHz
0 stalled-cycles-frontend # 0.00% frontend cycles idle
0 stalled-cycles-backend # 0.00% backend cycles idle
216,486,556,184 instructions # 2.38 insns per cycle
24,137,052,216 branches # 792.105 M/sec
12,256,039 branch-misses # 0.05% of all branches
30.473137262 seconds time elapsedTaking measurements with gprof each call takes 9.5 seconds
g++ -pg seq_matrix_big_mul.c -o seq_matrix_big_mul
./seq_matrix_big_mul
gprof ./seq_matrix_big_mul gmon.out > timeProf.txt
Flat profile:
Each sample counts as 0.01 seconds.
% cumulative self self total
time seconds seconds calls s/call s/call name
100.52 95.63 95.63 10 9.56 9.56 matrix_2d_mul_float(float*, float*, float*, int, int, int)
0.03 95.66 0.03 2 0.02 0.02 fillRand(float*, int, int, int)
0.00 95.66 0.00 1 0.00 0.00 _GLOBAL__sub_I_num_rows_A
0.00 95.66 0.00 1 0.00 0.00 __static_initialization_and_destruction_0(int, int)Executing on our CPU cores (7 cores)
Now we change our matrix2dmulfloat to run all rows in parallel with _#pragma omp parallel for schedule(dynamic,1) collapse(2). Now we have 2x speed-up. This is cool but could we done better on GPU?
# Compile
g++ -g -O1 -fopenmp seq_matrix_big_mul.c -o seq_matrix_big_mul
# Check execution time
sudo perf stat ./seq_matrix_big_mul
Size in bytes A: 16000000
Size in bytes B: 4800000
Matrix multiplication done 0
Matrix multiplication done 1
Matrix multiplication done 2
Matrix multiplication done 3
Matrix multiplication done 4
Matrix multiplication done 5
Matrix multiplication done 6
Matrix multiplication done 7
Matrix multiplication done 8
Matrix multiplication done 9
Performance counter stats for './seq_matrix_big_mul':
115167.655752 task-clock (msec) # 7.860 CPUs utilized
26,806 context-switches # 0.233 K/sec
11 cpu-migrations # 0.000 K/sec
1,903 page-faults # 0.017 K/sec
343,022,318,959 cycles # 2.978 GHz
0 stalled-cycles-frontend # 0.00% frontend cycles idle
0 stalled-cycles-backend # 0.00% backend cycles idle
265,276,470,662 instructions # 0.77 insns per cycle
24,270,361,665 branches # 210.739 M/sec
13,057,570 branch-misses # 0.05% of all branches
14.651747684 seconds time elapsedFirst check the Transfer time
Basically we need to know what is the time to send 20.8Mb (Matrices A and B) and receive 4.8Mb matrix (result matrix).
leo@leodev:~/work/learningOpenCl/chap1$ nvprof ./checkTransferMatMul 20800000 4800000
Checking GPU transfer...
Sending 20800000 bytes
Sending 4800000 bytes
==4451== NVPROF is profiling process 4451, command: ./checkTransferMatMul 20800000 4800000
==4451== Profiling application: ./checkTransferMatMul 20800000 4800000
==4451== Profiling result:
Time(%) Time Calls Avg Min Max Name
82.27% 6.7186ms 1 6.7186ms 6.7186ms 6.7186ms [CUDA memcpy HtoD]
17.73% 1.4480ms 1 1.4480ms 1.4480ms 1.4480ms [CUDA memcpy DtoH]
==4451== API calls:
Time(%) Time Calls Avg Min Max Name
90.92% 96.428ms 1 96.428ms 96.428ms 96.428ms cudaMalloc
8.04% 8.5233ms 2 4.2617ms 2.0535ms 6.4698ms cudaMemcpy
0.76% 805.41us 83 9.7030us 838ns 350.05us cuDeviceGetAttribute
0.19% 201.07us 1 201.07us 201.07us 201.07us cudaFree
0.05% 56.781us 1 56.781us 56.781us 56.781us cuDeviceTotalMem
0.04% 44.070us 1 44.070us 44.070us 44.070us cuDeviceGetName
0.00% 2.9340us 2 1.4670us 1.0480us 1.8860us cuDeviceGetCount
0.00% 1.9560us 2 978ns 908ns 1.0480us cuDeviceGetFrom the results the whole transfer takes 8ms. (On a system with a Quadro GPU). On a system with a Titan X (Maxwell) this transfer takes 2.6ms. So if our GPU calculate each matrix multiplication in less than 9 seconds (Sequential version) we have speed-ups.
Naive way on the GPU. (Using global memory)
So we have less than 9 seconds to multiply 2 matrices A[2000,2000] and B[2000,6000] on the GPU. Let's start by the Naive implementation. First let's take a look again on the CPU code.
void matrix_2d_mul_float(float *A, float *B, float *C, int num_rows_A, int num_cols_A, int num_cols_B) {
float sum = 0;
int num_rows_C = num_rows_A;
int num_cols_C = num_cols_B;
// Iterate on each row of A
for(int i=0; i<num_rows_A; i++) {
// Iterate on each collumn of B
for (int k=0; k<num_cols_B; k++) {
sum = 0;
// Do the "multiply add between" row of A and collumn of B
for (int j=0; j<num_cols_A; j++){
// A[i][j] == A[i*num_cols_A+j]
// B[j][k] == B[j*num_cols_B+k]
//sum += A[i][j]*B[j][k];
sum += A[i*num_cols_A+j]*B[j*num_cols_B+k];
}
// C[i][k] == C[i*num_cols_C+k]
C[i*num_cols_C+k]=sum;
}
}
}You can observe that there is no communication between the loop iterations. So each kernel would execute something like this:
sum = 0;
// Do the "multiply add between" row of A and collumn of B
for (int j=0; j<num_cols_A; j++){
// A[i][j] == A[i*num_cols_A+j]
// B[j][k] == B[j*num_cols_B+k]
//sum += A[i][j]*B[j][k];
sum += A[i*num_cols_A+j]*B[j*num_cols_B+k];
}
// C[i][k] == C[i*num_cols_C+k]
C[i*num_cols_C+k]=sum;As there is no communication between the threads the only issue is that each thread reads the global memory at least 2*j times. One for all the cols of A, one for all the rows of B) and one position for C. The issue is that accessing the global memory it's expensive in terms of time. Specially when you have multiple threads accessing at the same time, you start to have some congestion which make the kernel slower.
CUDA
__global__ void matrix_2d_mul_float_gpu(float *A, float *B, float *C, int num_rows_A, int num_cols_A, int num_cols_B) {
// Same code for all 2d kernel
int i = blockIdx.y * blockDim.y + threadIdx.y;
int k = blockIdx.x * blockDim.x + threadIdx.x;
if (i > num_rows_A || k > num_cols_B) return;
// Sum is on the register(local to each thread)
float sum = 0;
// This iterate a lot on the global memory 2*j times
for (int j=0; j<num_cols_A; j++){
// A[i][j] == A[i*num_cols_A+j]
// B[j][k] == B[j*num_cols_B+k]
//sum += A[i][j]*B[j][k];
sum += A[i*num_cols_A+j]*B[j*num_cols_B+k];
}
// And now one more time
C[i*num_cols_B+k]=sum;
}__kernel void matrix_2d_mul_float_gpu(__global float* A, __global float* B, __global float* C, int num_rows_A, int num_cols_A, int num_cols_B)
{
int i = get_global_id(0);
int k = get_global_id(1);
if (i > num_rows_A || k > num_cols_B) return;
// Sum is on the register(local to each thread)
float sum = 0;
// This iterate a lot on the global memory 2*j times
for (int j=0; j<num_cols_A; j++){
// A[i][j] == A[i*num_cols_A+j]
// B[j][k] == B[j*num_cols_B+k]
//sum += A[i][j]*B[j][k];
sum += A[i*num_cols_A+j]*B[j*num_cols_B+k];
}
// And now one more time
C[i*num_cols_B+k]=sum;
}On this case each thread create one entry on the matrix C(i,k), by sampling a row from A(i,j) and a col from B(j,k)
Ok let's check how the naive version performed.
==2543== Profiling application: ./matrix_big_mul
==2543== Profiling result:
Time(%) Time Calls Avg Min Max Name
89.60% 226.14ms 10 22.614ms 20.433ms 23.861ms matrix_2d_mul_float_gpu(float*, float*, float*, int, int, int)
8.20% 20.684ms 20 1.0342ms 456.25us 1.6279ms [CUDA memcpy HtoD]
2.20% 5.5601ms 10 556.01us 457.15us 982.74us [CUDA memcpy DtoH]Actually each matrix multiplication on the gpu took 23ms (kernel only) adding the time between GPU and CPU transfers we have now a total of 48 ms. This means an speed-up of .
So even with the naive implementation we got 199 times faster than the original sequential version.
Now let's compare with the multi-threaded cpu version (Took total of 14.6 seconds)
laraujo@lindev:~/work/learningOpenCl/chap2$ sudo perf stat ./matrix_big_mul
Size in bytes A: 16000000
Size in bytes B: 4800000
Matrix multiplication done 0
Matrix multiplication done 1
Matrix multiplication done 2
Matrix multiplication done 3
Matrix multiplication done 4
Matrix multiplication done 5
Matrix multiplication done 6
Matrix multiplication done 7
Matrix multiplication done 8
Matrix multiplication done 9
Performance counter stats for './matrix_big_mul':
348.548602 task-clock (msec) # 0.998 CPUs utilized
24 context-switches # 0.069 K/sec
2 cpu-migrations # 0.006 K/sec
3,387 page-faults # 0.010 M/sec
1,355,696,720 cycles # 3.890 GHz
<not supported> stalled-cycles-frontend
<not supported> stalled-cycles-backend
1,593,889,784 instructions # 1.18 insns per cycle
348,741,095 branches # 1000.552 M/sec
363,462 branch-misses # 0.10% of all branches
0.349369499 seconds time elapsedComparing with the multi-threaded version we had a speed up of 14.6/0.3, which is 48x faster.
This seems cool, also if we compare the execution timeline with Nvidia Visual Profiler. So the kernel execution time is bigger than the CPU/GPU transfer and we're still faster than the CPU.
So what's the problem with the naive implementation? Let's use the Nvidia visual profiler to also analyse the "Kernel performance Limiter" and the Kernel latency.
This shows that we're doing more memory operations than actually computing. Also on the title is saying "Kernel performance bounded by memory Bandwith"
Now let's check what is stalling our kernel.
Here our biggest "Stall Reason" is the "Execution Dependency" which means that our kernels are just waiting for data.
Making faster by Tiling
On the case of the matrix multiplication, we could divide our matrices in tiles and solve all the tiles in parallel. Also while doing this you bring the tiles to the shared memory which is faster than reading directly from the global memory (Kernel pointers, ex A[i,j]). And the tile C can be stored on the register memory (local variable). Which is really fast.
In the end we can just add the results.
Also inside the tiles the calculation remains the same.
This is the new version
__global__ void matrix_2d_mul_float_gpu(float *A, float *B, float *C, int num_rows_A, int num_cols_A, int num_cols_B) {
// Create shared variables (Available to all threads on the same block)
__shared__ float A_tile[N_THREADS][N_THREADS];
__shared__ float B_tile[N_THREADS][N_THREADS];
// Block index
int bx = blockIdx.x; int by = blockIdx.y;
// Thread index
int tx = threadIdx.x; int ty = threadIdx.y;
// Index of the first sub-matrix of A processed by the block
int aBegin = num_cols_A * N_THREADS * by;
// Index of the last sub-matrix of A processed by the block
int aEnd = aBegin + num_cols_A - 1;
// Index of the first sub-matrix of B processed by the block
int bBegin = N_THREADS * bx;
int bStep = N_THREADS * num_cols_B;
int aStep = N_THREADS;
float sum = 0;
for (int a = aBegin, b = bBegin;a <= aEnd;a += aStep, b += bStep) {
A_tile[ty][tx] = A[a + num_cols_A * ty + tx];
B_tile[tx][ty] = B[b + num_cols_B * tx + ty];
// Synchronize to make sure the matrices are loaded
__syncthreads();
for (int k = 0; k < N_THREADS; ++k)
sum += A_tile[ty][k] * B_tile[k][tx];
// Wait other threads to finish their sub-matrices
__syncthreads();
}
// Write the block sub-matrix to device memory;
// each thread writes one element
int c = num_cols_B * N_THREADS * by + N_THREADS * bx;
C[c + num_cols_B * ty + tx] = sum;
}Now if we compare the kernel execution time we're faster by a factor of ..... Now let's run again our profile to check the balance between computation and memory.
Now our kernel duration dropped to 5ms but if you see we're still bounded by memory bandwidth, but now it's not the L2 memory but the Shared memory, which is faster.
Get more information
There is an compile option -lineinfo that allows the nvidia profiler to annotate your source code to give an idea where you are loosing time (Or stalled due to memory).
nvcc matrix_big_mul_tile.cu -o matrix_big_mul_tile -lineinfoBellow we can see with "Kernel Profile PC Sampling" that the selected instruction (line 51) is using most of it's time doing "Execution dependency".
Now the line 44 it's spending most of it's time waiting for data.
The bad news is that this tool works only with Cuda. For more information on using the Nvidia Visual profiler you may refer here and refer to the code which shows a simple transpose algorithm been optimized.
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