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How can i make GPU process much faster than CPU process with CUDA 10.0 in Visual Studio 2017?
up vote
-1
down vote
favorite
Smart developer!
I am the beginner of CUDA programming and I have a big problem with my code.
Following code is a sample code from Nvidia and I changed a little bit for showing the GPU process much faster than from CPU process. However, after compiling this code, I got a unexpected result from that CPU process is much faster than GPU process.
This is my laptop gpu info.
This is my cuda code for Visual Studio 2017.
===========================================================================
#define N 10
This is add2 function() from GPU process
`___global____ void add2(int *a, int *b, int *c) {`
// GPU block from grid sector
//int tid = blockIdx.x; // checking the data of index = if you
insert min of N, you will get slow result from CPU. But if you put big number, this show much faster than CPU
// GPU thread
//int tid = threadIdx.x; // Same result as blockIdx.x
// GPU unexpected vector // Same result as above
int tid = threadIdx.x + blockIdx.x*blockDim.x;
if (tid < N) {
c[tid] = a[tid] + b[tid];
}
}
This is add function() from CPU process
`void add(int *a, int *b, int *c) {
int tid = 0;
while (tid < N) {
c[tid] = a[tid] + b[tid];
tid += 1;
}
}
This is Main function()
int main() {
// Values for time duration
LARGE_INTEGER tFreq, tStart, tEnd;
cudaEvent_t start, stop;
float tms, ms;
int a[N], b[N], c[N]; // CPU values
int *dev_a, *dev_b, *dev_c; // GPU values----------------------------------------------
// Creating alloc for GPU--------------------------------------------------------------
cudaMalloc((void**)&dev_a, N * sizeof(int));
cudaMalloc((void**)&dev_b, N * sizeof(int));
cudaMalloc((void**)&dev_c, N * sizeof(int));
// Fill 'a' and 'b' from CPU
for (int i = 0; i < N; i++) {
a[i] = -i;
b[i] = i * i;
}
// Copy values of CPU to GPU values----------------------------------------------------
cudaMemcpy(dev_a, a, N * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, N * sizeof(int), cudaMemcpyHostToDevice);
//////////////////////////////////////
QueryPerformanceFrequency(&tFreq); // Frequency set
QueryPerformanceCounter(&tStart); // Time count Start
// CPU operation
add(a, b, c);
//////////////////////////////////////
QueryPerformanceCounter(&tEnd); // TIme count End
tms = ((tEnd.QuadPart - tStart.QuadPart) / (float)tFreq.QuadPart) * 1000;
//////////////////////////////////////
// show result of CPU
cout << fixed;
cout.precision(10);
cout << "CPU Time=" << tms << endl << endl;
for (int i = 0; i < N; i++) {
printf("CPU calculate = %d + %d = %dn", a[i], b[i], c[i]);
}
cout << endl;
///////////////////////////////////////
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
// GPU operatinog---------------------------------------------------------------------
//add2 <<<N,1 >>> (dev_a, dev_b, dev_c); // block
//add2 << <1,N >> > (dev_a, dev_b, dev_c); // Thread
add2 << <N/32+1, 32 >> > (dev_a, dev_b, dev_c); // grid
///////////////////////////////////////
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&ms, start, stop);
///////////////////////////////////////
// show result of GPU
cudaMemcpy(c, dev_c, N * sizeof(int), cudaMemcpyDeviceToHost);
cout << fixed;
cout.precision(10);
cout << "GPU Time=" << ms << endl << endl;
for (int i = 0; i < N; i++) {
printf("GPU calculate = %d + %d = %dn", a[i], b[i], c[i]);
}
//Free GPU values
cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
return 0;
}
This is result of compiling this code.
I want to make GPU process much faster than CPU process.
c++ cuda gpu nvidia gpgpu
edited yesterday
Mike
1,6291421
asked yesterday
L SJin
33
New contributor
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
-1
down vote
favorite
Smart developer!
I am the beginner of CUDA programming and I have a big problem with my code.
Following code is a sample code from Nvidia and I changed a little bit for showing the GPU process much faster than from CPU process. However, after compiling this code, I got a unexpected result from that CPU process is much faster than GPU process.
This is my laptop gpu info.
This is my cuda code for Visual Studio 2017.
===========================================================================
#define N 10
This is add2 function() from GPU process
`___global____ void add2(int *a, int *b, int *c) {`
// GPU block from grid sector
//int tid = blockIdx.x; // checking the data of index = if you
insert min of N, you will get slow result from CPU. But if you put big number, this show much faster than CPU
// GPU thread
//int tid = threadIdx.x; // Same result as blockIdx.x
// GPU unexpected vector // Same result as above
int tid = threadIdx.x + blockIdx.x*blockDim.x;
if (tid < N) {
c[tid] = a[tid] + b[tid];
}
}
This is add function() from CPU process
`void add(int *a, int *b, int *c) {
int tid = 0;
while (tid < N) {
c[tid] = a[tid] + b[tid];
tid += 1;
}
}
This is Main function()
int main() {
// Values for time duration
LARGE_INTEGER tFreq, tStart, tEnd;
cudaEvent_t start, stop;
float tms, ms;
int a[N], b[N], c[N]; // CPU values
int *dev_a, *dev_b, *dev_c; // GPU values----------------------------------------------
// Creating alloc for GPU--------------------------------------------------------------
cudaMalloc((void**)&dev_a, N * sizeof(int));
cudaMalloc((void**)&dev_b, N * sizeof(int));
cudaMalloc((void**)&dev_c, N * sizeof(int));
// Fill 'a' and 'b' from CPU
for (int i = 0; i < N; i++) {
a[i] = -i;
b[i] = i * i;
}
// Copy values of CPU to GPU values----------------------------------------------------
cudaMemcpy(dev_a, a, N * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, N * sizeof(int), cudaMemcpyHostToDevice);
//////////////////////////////////////
QueryPerformanceFrequency(&tFreq); // Frequency set
QueryPerformanceCounter(&tStart); // Time count Start
// CPU operation
add(a, b, c);
//////////////////////////////////////
QueryPerformanceCounter(&tEnd); // TIme count End
tms = ((tEnd.QuadPart - tStart.QuadPart) / (float)tFreq.QuadPart) * 1000;
//////////////////////////////////////
// show result of CPU
cout << fixed;
cout.precision(10);
cout << "CPU Time=" << tms << endl << endl;
for (int i = 0; i < N; i++) {
printf("CPU calculate = %d + %d = %dn", a[i], b[i], c[i]);
}
cout << endl;
///////////////////////////////////////
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
// GPU operatinog---------------------------------------------------------------------
//add2 <<<N,1 >>> (dev_a, dev_b, dev_c); // block
//add2 << <1,N >> > (dev_a, dev_b, dev_c); // Thread
add2 << <N/32+1, 32 >> > (dev_a, dev_b, dev_c); // grid
///////////////////////////////////////
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&ms, start, stop);
///////////////////////////////////////
// show result of GPU
cudaMemcpy(c, dev_c, N * sizeof(int), cudaMemcpyDeviceToHost);
cout << fixed;
cout.precision(10);
cout << "GPU Time=" << ms << endl << endl;
for (int i = 0; i < N; i++) {
printf("GPU calculate = %d + %d = %dn", a[i], b[i], c[i]);
}
//Free GPU values
cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
return 0;
}
This is result of compiling this code.
I want to make GPU process much faster than CPU process.
c++ cuda gpu nvidia gpgpu
edited yesterday
Mike
1,6291421
asked yesterday
L SJin
33
New contributor
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
OP has acknowledged in cross posting here that they can get a different comparison between CPU and GPU by making some modifications to their code.
– Robert Crovella
yesterday
@RobertCrovella Thank you!
– L SJin
10 hours ago
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Smart developer!
I am the beginner of CUDA programming and I have a big problem with my code.
Following code is a sample code from Nvidia and I changed a little bit for showing the GPU process much faster than from CPU process. However, after compiling this code, I got a unexpected result from that CPU process is much faster than GPU process.
This is my laptop gpu info.
This is my cuda code for Visual Studio 2017.
===========================================================================
#define N 10
This is add2 function() from GPU process
`___global____ void add2(int *a, int *b, int *c) {`
// GPU block from grid sector
//int tid = blockIdx.x; // checking the data of index = if you
insert min of N, you will get slow result from CPU. But if you put big number, this show much faster than CPU
// GPU thread
//int tid = threadIdx.x; // Same result as blockIdx.x
// GPU unexpected vector // Same result as above
int tid = threadIdx.x + blockIdx.x*blockDim.x;
if (tid < N) {
c[tid] = a[tid] + b[tid];
}
}
This is add function() from CPU process
`void add(int *a, int *b, int *c) {
int tid = 0;
while (tid < N) {
c[tid] = a[tid] + b[tid];
tid += 1;
}
}
This is Main function()
int main() {
// Values for time duration
LARGE_INTEGER tFreq, tStart, tEnd;
cudaEvent_t start, stop;
float tms, ms;
int a[N], b[N], c[N]; // CPU values
int *dev_a, *dev_b, *dev_c; // GPU values----------------------------------------------
// Creating alloc for GPU--------------------------------------------------------------
cudaMalloc((void**)&dev_a, N * sizeof(int));
cudaMalloc((void**)&dev_b, N * sizeof(int));
cudaMalloc((void**)&dev_c, N * sizeof(int));
// Fill 'a' and 'b' from CPU
for (int i = 0; i < N; i++) {
a[i] = -i;
b[i] = i * i;
}
// Copy values of CPU to GPU values----------------------------------------------------
cudaMemcpy(dev_a, a, N * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, N * sizeof(int), cudaMemcpyHostToDevice);
//////////////////////////////////////
QueryPerformanceFrequency(&tFreq); // Frequency set
QueryPerformanceCounter(&tStart); // Time count Start
// CPU operation
add(a, b, c);
//////////////////////////////////////
QueryPerformanceCounter(&tEnd); // TIme count End
tms = ((tEnd.QuadPart - tStart.QuadPart) / (float)tFreq.QuadPart) * 1000;
//////////////////////////////////////
// show result of CPU
cout << fixed;
cout.precision(10);
cout << "CPU Time=" << tms << endl << endl;
for (int i = 0; i < N; i++) {
printf("CPU calculate = %d + %d = %dn", a[i], b[i], c[i]);
}
cout << endl;
///////////////////////////////////////
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
// GPU operatinog---------------------------------------------------------------------
//add2 <<<N,1 >>> (dev_a, dev_b, dev_c); // block
//add2 << <1,N >> > (dev_a, dev_b, dev_c); // Thread
add2 << <N/32+1, 32 >> > (dev_a, dev_b, dev_c); // grid
///////////////////////////////////////
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&ms, start, stop);
///////////////////////////////////////
// show result of GPU
cudaMemcpy(c, dev_c, N * sizeof(int), cudaMemcpyDeviceToHost);
cout << fixed;
cout.precision(10);
cout << "GPU Time=" << ms << endl << endl;
for (int i = 0; i < N; i++) {
printf("GPU calculate = %d + %d = %dn", a[i], b[i], c[i]);
}
//Free GPU values
cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
return 0;
}
This is result of compiling this code.
I want to make GPU process much faster than CPU process.
c++ cuda gpu nvidia gpgpu
edited yesterday
Mike
1,6291421
asked yesterday
L SJin
33
New contributor
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Smart developer!
I am the beginner of CUDA programming and I have a big problem with my code.
Following code is a sample code from Nvidia and I changed a little bit for showing the GPU process much faster than from CPU process. However, after compiling this code, I got a unexpected result from that CPU process is much faster than GPU process.
This is my laptop gpu info.
This is my cuda code for Visual Studio 2017.
===========================================================================
#define N 10
This is add2 function() from GPU process
`___global____ void add2(int *a, int *b, int *c) {`
// GPU block from grid sector
//int tid = blockIdx.x; // checking the data of index = if you
insert min of N, you will get slow result from CPU. But if you put big number, this show much faster than CPU
// GPU thread
//int tid = threadIdx.x; // Same result as blockIdx.x
// GPU unexpected vector // Same result as above
int tid = threadIdx.x + blockIdx.x*blockDim.x;
if (tid < N) {
c[tid] = a[tid] + b[tid];
}
}
This is add function() from CPU process
`void add(int *a, int *b, int *c) {
int tid = 0;
while (tid < N) {
c[tid] = a[tid] + b[tid];
tid += 1;
}
}
This is Main function()
int main() {
// Values for time duration
LARGE_INTEGER tFreq, tStart, tEnd;
cudaEvent_t start, stop;
float tms, ms;
int a[N], b[N], c[N]; // CPU values
int *dev_a, *dev_b, *dev_c; // GPU values----------------------------------------------
// Creating alloc for GPU--------------------------------------------------------------
cudaMalloc((void**)&dev_a, N * sizeof(int));
cudaMalloc((void**)&dev_b, N * sizeof(int));
cudaMalloc((void**)&dev_c, N * sizeof(int));
// Fill 'a' and 'b' from CPU
for (int i = 0; i < N; i++) {
a[i] = -i;
b[i] = i * i;
}
// Copy values of CPU to GPU values----------------------------------------------------
cudaMemcpy(dev_a, a, N * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, N * sizeof(int), cudaMemcpyHostToDevice);
//////////////////////////////////////
QueryPerformanceFrequency(&tFreq); // Frequency set
QueryPerformanceCounter(&tStart); // Time count Start
// CPU operation
add(a, b, c);
//////////////////////////////////////
QueryPerformanceCounter(&tEnd); // TIme count End
tms = ((tEnd.QuadPart - tStart.QuadPart) / (float)tFreq.QuadPart) * 1000;
//////////////////////////////////////
// show result of CPU
cout << fixed;
cout.precision(10);
cout << "CPU Time=" << tms << endl << endl;
for (int i = 0; i < N; i++) {
printf("CPU calculate = %d + %d = %dn", a[i], b[i], c[i]);
}
cout << endl;
///////////////////////////////////////
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
// GPU operatinog---------------------------------------------------------------------
//add2 <<<N,1 >>> (dev_a, dev_b, dev_c); // block
//add2 << <1,N >> > (dev_a, dev_b, dev_c); // Thread
add2 << <N/32+1, 32 >> > (dev_a, dev_b, dev_c); // grid
///////////////////////////////////////
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&ms, start, stop);
///////////////////////////////////////
// show result of GPU
cudaMemcpy(c, dev_c, N * sizeof(int), cudaMemcpyDeviceToHost);
cout << fixed;
cout.precision(10);
cout << "GPU Time=" << ms << endl << endl;
for (int i = 0; i < N; i++) {
printf("GPU calculate = %d + %d = %dn", a[i], b[i], c[i]);
}
//Free GPU values
cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
return 0;
}
This is result of compiling this code.
I want to make GPU process much faster than CPU process.
c++ cuda gpu nvidia gpgpu
c++ cuda gpu nvidia gpgpu
edited yesterday
Mike
1,6291421
asked yesterday
L SJin
33
New contributor
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Mike
1,6291421
asked yesterday
L SJin
33
New contributor
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Mike
1,6291421
edited yesterday
Mike
1,6291421
edited yesterday
Mike
1,6291421
1,6291421
asked yesterday
L SJin
33
New contributor
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
L SJin
33
asked yesterday
L SJin
33
33
New contributor
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
L SJin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
OP has acknowledged in cross posting here that they can get a different comparison between CPU and GPU by making some modifications to their code.
– Robert Crovella
yesterday
@RobertCrovella Thank you!
– L SJin
10 hours ago
add a comment |
1
OP has acknowledged in cross posting here that they can get a different comparison between CPU and GPU by making some modifications to their code.
– Robert Crovella
yesterday
@RobertCrovella Thank you!
– L SJin
10 hours ago
1
1
OP has acknowledged in cross posting here that they can get a different comparison between CPU and GPU by making some modifications to their code.
– Robert Crovella
yesterday
OP has acknowledged in cross posting here that they can get a different comparison between CPU and GPU by making some modifications to their code.
– Robert Crovella
yesterday
@RobertCrovella Thank you!
– L SJin
10 hours ago
@RobertCrovella Thank you!
– L SJin
10 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The GPU is generally actually slower than the CPU for running a single operation. Additionally it takes time to send data to the GPU and read it back again.
The advantage of the GPU is it can execute many operations in parallel.
As you have defined N to be 10 it probably takes longer to upload and download the data than to execute on the CPU. In order to see the advantage of the GPU increase your problem size to something much larger. Ideally you want to execute a minimum of a few operations on each GPU core before you start seeing some benefit. For example with your GPU's 1280 cores you would want to execute something like 4000 operations or more at once to get the benefit of the GPU.
answered yesterday
Alan Birtles
7,465733
You mean 400 000, not 4000. 4000 would like take the same time as 10...
– talonmies
yesterday
@talonmies it would take the same time as 10 to execute on the GPU but should take longer on the CPU. I agree that with GPU processing the more the merrier but I was giving an approximate minimum task size to start seeing some benefit.
– Alan Birtles
yesterday
@AlanBirtles Thank you so much to help me to understand what i don't know. By the way, I want to ask something else. Can I figure out what is minimum operations which can be faster than CPU? (This question is my curiosity to cuda programming. You can answer free or ignore it.)
– L SJin
10 hours ago
The only way to figure it out is by experimentation, depends on which operations you are performing, how much data you are using, whether your compiler optimises your cpu code to use SIMD instructions and of course how fast your CPU and GPU are
– Alan Birtles
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The GPU is generally actually slower than the CPU for running a single operation. Additionally it takes time to send data to the GPU and read it back again.
The advantage of the GPU is it can execute many operations in parallel.
As you have defined N to be 10 it probably takes longer to upload and download the data than to execute on the CPU. In order to see the advantage of the GPU increase your problem size to something much larger. Ideally you want to execute a minimum of a few operations on each GPU core before you start seeing some benefit. For example with your GPU's 1280 cores you would want to execute something like 4000 operations or more at once to get the benefit of the GPU.
answered yesterday
Alan Birtles
7,465733
You mean 400 000, not 4000. 4000 would like take the same time as 10...
– talonmies
yesterday
@talonmies it would take the same time as 10 to execute on the GPU but should take longer on the CPU. I agree that with GPU processing the more the merrier but I was giving an approximate minimum task size to start seeing some benefit.
– Alan Birtles
yesterday
@AlanBirtles Thank you so much to help me to understand what i don't know. By the way, I want to ask something else. Can I figure out what is minimum operations which can be faster than CPU? (This question is my curiosity to cuda programming. You can answer free or ignore it.)
– L SJin
10 hours ago
The only way to figure it out is by experimentation, depends on which operations you are performing, how much data you are using, whether your compiler optimises your cpu code to use SIMD instructions and of course how fast your CPU and GPU are
– Alan Birtles
8 hours ago
add a comment |
up vote
1
down vote
accepted
The GPU is generally actually slower than the CPU for running a single operation. Additionally it takes time to send data to the GPU and read it back again.
The advantage of the GPU is it can execute many operations in parallel.
As you have defined N to be 10 it probably takes longer to upload and download the data than to execute on the CPU. In order to see the advantage of the GPU increase your problem size to something much larger. Ideally you want to execute a minimum of a few operations on each GPU core before you start seeing some benefit. For example with your GPU's 1280 cores you would want to execute something like 4000 operations or more at once to get the benefit of the GPU.
answered yesterday
Alan Birtles
7,465733
You mean 400 000, not 4000. 4000 would like take the same time as 10...
– talonmies
yesterday
@talonmies it would take the same time as 10 to execute on the GPU but should take longer on the CPU. I agree that with GPU processing the more the merrier but I was giving an approximate minimum task size to start seeing some benefit.
– Alan Birtles
yesterday
@AlanBirtles Thank you so much to help me to understand what i don't know. By the way, I want to ask something else. Can I figure out what is minimum operations which can be faster than CPU? (This question is my curiosity to cuda programming. You can answer free or ignore it.)
– L SJin
10 hours ago
The only way to figure it out is by experimentation, depends on which operations you are performing, how much data you are using, whether your compiler optimises your cpu code to use SIMD instructions and of course how fast your CPU and GPU are
– Alan Birtles
8 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The GPU is generally actually slower than the CPU for running a single operation. Additionally it takes time to send data to the GPU and read it back again.
The advantage of the GPU is it can execute many operations in parallel.
As you have defined N to be 10 it probably takes longer to upload and download the data than to execute on the CPU. In order to see the advantage of the GPU increase your problem size to something much larger. Ideally you want to execute a minimum of a few operations on each GPU core before you start seeing some benefit. For example with your GPU's 1280 cores you would want to execute something like 4000 operations or more at once to get the benefit of the GPU.
answered yesterday
Alan Birtles
7,465733
The GPU is generally actually slower than the CPU for running a single operation. Additionally it takes time to send data to the GPU and read it back again.
The advantage of the GPU is it can execute many operations in parallel.
As you have defined N to be 10 it probably takes longer to upload and download the data than to execute on the CPU. In order to see the advantage of the GPU increase your problem size to something much larger. Ideally you want to execute a minimum of a few operations on each GPU core before you start seeing some benefit. For example with your GPU's 1280 cores you would want to execute something like 4000 operations or more at once to get the benefit of the GPU.
answered yesterday
Alan Birtles
7,465733
answered yesterday
Alan Birtles
7,465733
answered yesterday
Alan Birtles
7,465733
answered yesterday
Alan Birtles
7,465733
7,465733
You mean 400 000, not 4000. 4000 would like take the same time as 10...
– talonmies
yesterday
@talonmies it would take the same time as 10 to execute on the GPU but should take longer on the CPU. I agree that with GPU processing the more the merrier but I was giving an approximate minimum task size to start seeing some benefit.
– Alan Birtles
yesterday
@AlanBirtles Thank you so much to help me to understand what i don't know. By the way, I want to ask something else. Can I figure out what is minimum operations which can be faster than CPU? (This question is my curiosity to cuda programming. You can answer free or ignore it.)
– L SJin
10 hours ago
The only way to figure it out is by experimentation, depends on which operations you are performing, how much data you are using, whether your compiler optimises your cpu code to use SIMD instructions and of course how fast your CPU and GPU are
– Alan Birtles
8 hours ago
add a comment |
You mean 400 000, not 4000. 4000 would like take the same time as 10...
– talonmies
yesterday
@talonmies it would take the same time as 10 to execute on the GPU but should take longer on the CPU. I agree that with GPU processing the more the merrier but I was giving an approximate minimum task size to start seeing some benefit.
– Alan Birtles
yesterday
@AlanBirtles Thank you so much to help me to understand what i don't know. By the way, I want to ask something else. Can I figure out what is minimum operations which can be faster than CPU? (This question is my curiosity to cuda programming. You can answer free or ignore it.)
– L SJin
10 hours ago
The only way to figure it out is by experimentation, depends on which operations you are performing, how much data you are using, whether your compiler optimises your cpu code to use SIMD instructions and of course how fast your CPU and GPU are
– Alan Birtles
8 hours ago
You mean 400 000, not 4000. 4000 would like take the same time as 10...
– talonmies
yesterday
You mean 400 000, not 4000. 4000 would like take the same time as 10...
– talonmies
yesterday
@talonmies it would take the same time as 10 to execute on the GPU but should take longer on the CPU. I agree that with GPU processing the more the merrier but I was giving an approximate minimum task size to start seeing some benefit.
– Alan Birtles
yesterday
@talonmies it would take the same time as 10 to execute on the GPU but should take longer on the CPU. I agree that with GPU processing the more the merrier but I was giving an approximate minimum task size to start seeing some benefit.
– Alan Birtles
yesterday
@AlanBirtles Thank you so much to help me to understand what i don't know. By the way, I want to ask something else. Can I figure out what is minimum operations which can be faster than CPU? (This question is my curiosity to cuda programming. You can answer free or ignore it.)
– L SJin
10 hours ago
@AlanBirtles Thank you so much to help me to understand what i don't know. By the way, I want to ask something else. Can I figure out what is minimum operations which can be faster than CPU? (This question is my curiosity to cuda programming. You can answer free or ignore it.)
– L SJin
10 hours ago
The only way to figure it out is by experimentation, depends on which operations you are performing, how much data you are using, whether your compiler optimises your cpu code to use SIMD instructions and of course how fast your CPU and GPU are
– Alan Birtles
8 hours ago
The only way to figure it out is by experimentation, depends on which operations you are performing, how much data you are using, whether your compiler optimises your cpu code to use SIMD instructions and of course how fast your CPU and GPU are
– Alan Birtles
8 hours ago
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OP has acknowledged in cross posting here that they can get a different comparison between CPU and GPU by making some modifications to their code.
– Robert Crovella
yesterday
@RobertCrovella Thank you!
– L SJin
10 hours ago