Mi proyecto actual es una reprogramación de un modelo de plegamiento de proteínas que involucra la solución de miles de EDO en C ++. He estado deteniendo y comenzando a progresar mientras escribo el solucionador para que se ejecute totalmente en la GPU. Finalmente logré integrarlo, pero cuando trato de resolver dC / dt = -C usando la solución de quinto orden del algoritmo RKF45 con un tamaño de paso fijo de h = .001, obtengo la decisión del valor de la calculadora de e ^ {- t } del orden de 10 ^ {- 4} cuando se espera un error global del orden de 10 ^ {- 12}.
¿Es porque no estoy usando tamaños de pasos adaptativos y control de errores usando las soluciones de cuarto y quinto orden que esto está ocurriendo? Pensé que, dado que un algoritmo adaptativo analiza la diferencia entre las suposiciones de cuarto y quinto orden y básicamente trata el quinto orden como la "respuesta correcta", simplemente usaría esa respuesta para probar el integrador.
EDITAR:
Pensé que desde
Error global = (número de puntos) * (error local)
y
Error local = h ^ O
donde h es el tamaño del paso y O es el orden, calculo
O = ln (error global / número de puntos) / ln h
Haciendo un análisis de orden tanto como pensé que era correcto como sugerido en la respuesta, obtengo lo siguiente:
Number of Points h Global Error My Way The Answer's Way
10 0.1 2.89E-06 6.539E+00
20 0.05 7.09E-08 6.495E+00 5.350E+00
40 0.025 1.76E-07 5.216E+00 -1.315E+00
50 0.02 1.128E-07 5.089E+00 2.003E+00
80 0.0125 4.401E-08 4.866E+00 2.002E+00
100 0.01 2.816E-08 4.775E+00 2.001E+00
500 0.002 1.126E-09 4.316E+00 2.000E+00
1000 0.001 2.810E-10 4.184E+00 2.003E+00
10000 0.0001 3.000E-12 3.881E+00 1.972E+00Entonces, a pesar de mi avance casi anoche, donde realmente tuve un error de programación, el orden de mi solucionador aún está apagado. Sin embargo, el error antes mencionado asociado con el tamaño de paso .001 ha disminuido dramáticamente de O (10 ^ -4) a O (10 ^ -10).
2da EDICIÓN
Aquí hay un gráfico de LN (E) -LN (a) vs. LN (h). Lancé un ajuste lineal, así que esperaría que eso me revelara el orden. Lo que es realmente confuso ahora es que tengo tres cálculos para el orden de mi método. El gráfico me dice ~ 3, el método de Godric me dice casi exactamente 2 y mi método muestra que cambia de 6-> 4 para aumentar el tamaño del paso. Estoy más inclinado a creer mis cálculos porque el Error global es muy bajo (~ 10 ^ -10) y más cercano a lo que esperaría (10 ^ -15 * 10 ^ 3 = 10 ^ -12) para ese tamaño de paso (10 ^ -3).
EDITAR 3:
Mi código. Perdón por el novato.
#include <cuda.h>
#include <cuda_runtime.h>
#include <device_launch_parameters.h>
#include <stdio.h>
#include <iostream>
#include <iomanip> //display 2 decimal places
#include <math.h>
using namespace std;
/*
Things to do log:
1. Decide how adaptive step sizes is handled by the integrator loop. We can no longer say how many points to integrate if that is to be determined in the kernel
itself, only what the initial step size is.
2. Figure out how I'm going to graph all this stuff.
3. On paper solve for the step size.
4. Adaptive Step Size
a. arrMin kernel to get min of array
b. Decide how to implement tolerances
5. Use multiples of 256 for maxlength
Important things accomplished:
1. Can create 2D array for storing concentrations over time.
2. Can store a 1D array in 2D array.
3. Can calculate all the k-arrays given static step size.
4. Can take a step given static step size.
5. Can step through an integration from t0 to tf
6. Can accomodate fixed time-step
7. Calculated the current order of integrator using logs.
8. Learned how to profile the program.
9. Fixed the GPU timeout problem with WDDM.
Ideas:
1. Just use large fixed size array
*/
__global__ void rkf5(double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, int*, int*, size_t, double*, double*, double*);
__global__ void calcK(double*, double*, double*);
__global__ void k1(double*, double*, double*);
__global__ void k2(double*, double*, double*);
__global__ void k3(double*, double*, double*);
__global__ void k4(double*, double*, double*);
__global__ void k5(double*, double*, double*);
__global__ void k6(double*, double*, double*);
__global__ void arrAdd(double*, double*, double*);
__global__ void arrSub(double*, double*, double*);
__global__ void arrMult(double*, double*, double*);
__global__ void arrInit(double*, double);
__global__ void arrCopy(double*, double*);
__device__ void setup(double , double*, double*, double*, double*, int*);
__device__ double flux(int, double*) ;
__device__ double knowles_flux(int, double*);
__device__ void calcStepSize(double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, double*, int*);
__global__ void storeConcs(double*, size_t, double*, int);
__global__ void takeFourthOrderStep(double*, double*, double*, double*, double*, double*, double*);
__global__ void takeFifthOrderStep(double*, double*, double*, double*, double*, double*, double*, double*);
//Error checking that I don't understand yet.
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
//Main program.
int main(int argc, char** argv)
{
//std::cout << std::fixed; //display 2 decimal places
//std::cout << std::setprecision(12); //display 2 decimal places
const int maxlength = 1; //Number of discrete concentrations we are tracking.
double concs[maxlength]; //Meant to store the current concentrations
double temp1[maxlength]; //Used as a bin to store products of Butcher's tableau and k values.
double temp2[maxlength]; //Used as a bin to store products of Butcher's tableau and k values.
double tempsum[maxlength]; //Used as a bin to store cumulative sum of tableau and k values
double k1s[maxlength];
double k2s[maxlength];
double k3s[maxlength];
double k4s[maxlength];
double k5s[maxlength];
double k6s[maxlength];
const int numpoints = 40;
double to = 0;
double tf = 1;
//double dt = static_cast<double>(.5)/static_cast<double>(64);
double dt = (tf-to)/static_cast<double>(numpoints);
double mo = 1;
double concStorage[maxlength][numpoints]; //Stores concs vs. time
//Initialize all the arrays on the host to ensure arrays of 0's are sent to the device.
//Also, here is where we can seed the system.
std::cout<<dt;
std::cout<<"\n";
concs[0]=mo;
std::cout<<concs[0];
std::cout<<" ";
for (int i=0; i<maxlength; i++)
{
for (int j=0; j<numpoints; j++)
concStorage[i][j]=0;
concs[i]=0;
temp1[i]=0;
temp2[i]=0;
tempsum[i]=0;
k1s[i]=0;
k2s[i]=0;
k3s[i]=0;
k4s[i]=0;
k5s[i]=0;
k6s[i]=0;
std::cout<<concs[i];
std::cout<<" ";
}
concs[0]=mo;
std::cout<<"\n";
//Define all the pointers to device array memory addresses. These contain the on-GPU
//addresses of all the data we're generating/using.
double *d_concs;
double *d_temp1;
double *d_temp2;
double *d_tempsum;
double *d_k1s;
double *d_k2s;
double *d_k3s;
double *d_k4s;
double *d_k5s;
double *d_k6s;
double *d_dt;
int *d_maxlength;
int *d_numpoints;
double *d_to;
double *d_tf;
double *d_concStorage;
//Calculate all the sizes of the arrays in order to allocate the proper amount of memory on the GPU.
size_t size_concs = sizeof(concs);
size_t size_temp1 = sizeof(temp1);
size_t size_temp2 = sizeof(temp2);
size_t size_tempsum = sizeof(tempsum);
size_t size_ks = sizeof(k1s);
size_t size_maxlength = sizeof(maxlength);
size_t size_numpoints = sizeof(numpoints);
size_t size_dt = sizeof(dt);
size_t size_to = sizeof(to);
size_t size_tf = sizeof(tf);
size_t h_pitch = numpoints*sizeof(double);
size_t d_pitch;
//Calculate the "pitch" of the 2D array. The pitch is basically the length of a 2D array's row. IT's larger
//than the actual row full of data due to hadware issues. We thusly will use the pitch instead of the data
//size to traverse the array.
gpuErrchk(cudaMallocPitch( (void**)&d_concStorage, &d_pitch, numpoints * sizeof(double), maxlength));
//Allocate memory on the GPU for all the arrrays we're going to use in the integrator.
gpuErrchk(cudaMalloc((void**)&d_concs, size_concs));
gpuErrchk(cudaMalloc((void**)&d_temp1, size_temp1));
gpuErrchk(cudaMalloc((void**)&d_temp2, size_temp1));
gpuErrchk(cudaMalloc((void**)&d_tempsum, size_tempsum));
gpuErrchk(cudaMalloc((void**)&d_k1s, size_ks));
gpuErrchk(cudaMalloc((void**)&d_k2s, size_ks));
gpuErrchk(cudaMalloc((void**)&d_k3s, size_ks));
gpuErrchk(cudaMalloc((void**)&d_k4s, size_ks));
gpuErrchk(cudaMalloc((void**)&d_k5s, size_ks));
gpuErrchk(cudaMalloc((void**)&d_k6s, size_ks));
gpuErrchk(cudaMalloc((void**)&d_maxlength, size_maxlength));
gpuErrchk(cudaMalloc((void**)&d_numpoints, size_numpoints));
gpuErrchk(cudaMalloc((void**)&d_dt, size_dt));
gpuErrchk(cudaMalloc((void**)&d_to, size_to));
gpuErrchk(cudaMalloc((void**)&d_tf, size_tf));
//Copy all initial values of arrays to GPU.
gpuErrchk(cudaMemcpy2D(d_concStorage, d_pitch, concStorage, h_pitch, numpoints*sizeof(double), maxlength, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_concs, &concs, size_concs, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_temp1, &temp1, size_temp1, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_temp2, &temp2, size_temp2, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_tempsum, &tempsum, size_tempsum, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_k1s, &k1s, size_ks, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_k2s, &k2s, size_ks, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_k3s, &k3s, size_ks, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_k4s, &k4s, size_ks, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_k5s, &k5s, size_ks, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_k6s, &k6s, size_ks, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_maxlength, &maxlength, size_maxlength, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_numpoints, &numpoints, size_numpoints, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_dt, &dt, size_dt, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_to, &to, size_to, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_tf, &tf, size_tf, cudaMemcpyHostToDevice));
//Run the integrator.
rkf5<<<1,1>>>(d_concs, d_concStorage, d_temp1, d_temp2, d_tempsum, d_k1s, d_k2s, d_k3s, d_k4s, d_k5s, d_k6s, d_maxlength, d_numpoints, d_pitch, d_dt, d_to, d_tf);
gpuErrchk( cudaPeekAtLastError() );
gpuErrchk( cudaDeviceSynchronize() );
cudaDeviceSynchronize();
//Copy concentrations from GPU to Host. Almost defunct now that transferring the 2D array works.
cudaMemcpy(concs, d_concs, size_concs, cudaMemcpyDeviceToHost);
//Copy 2D array of concentrations vs. time from GPU to Host.
gpuErrchk( cudaMemcpy2D(concStorage, h_pitch, d_concStorage, d_pitch, numpoints*sizeof(double), maxlength, cudaMemcpyDeviceToHost) );
//Print concentrations after the integrator kernel runs. Used to test that data was transferring to and from GPU correctly.
std::cout << "\n";
for (int i=0; i<maxlength; i++)
{
std::cout<<concs[i];
std::cout<<" ";
}
double a[10];
double b[10];
double c[10];
for(int i = 0; i< 10; i++)
{
a[i]=0;
b[i]=0;
c[i]=0;
}
//Print out the concStorage array after the kernel runs. Used to test that the 2D array transferred correctly from host to GPU and back.
std::cout << "\n\n";
std::cout << "Calculated Array";
std::cout << "\n\n";
for (int i=0; i<maxlength; i++)
{
for(int j=0; j<numpoints; j++)
{
if (j%(numpoints/10)==0)
{
a[j/(numpoints/10)]=concStorage[i][j];
std::cout<<concStorage[i][j];
std::cout<<" ";
}
}
std::cout << "\n";
}
std::cout << "\n";
std::cout << "Exponential";
std::cout << "\n\n";
for (int i=0; i<10; i++)
{
b[i]=exp(-i*(tf-to)/10);
std::cout<<exp(-i*(tf-to)/10);
std::cout<<" ";
}
std::cout << "\n\n";
std::cout << "Error Array";
std::cout << "\n\n";
for (int i=0; i<10; i++)
{
c[i]=a[i]-b[i];
std::cout<<c[i];
std::cout<<" ";
}
std::cout << "\n\n";
cudaDeviceReset(); //Clean up all memory.
return 0;
}
//Main kernel. This is mean to be run as a master thread that calls all the other functions and thusly "runs" the integrator.
__global__ void rkf5(double* concs, double* concStorage, double* temp1, double* temp2, double* tempsum, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s, int* maxlength, int* numpoints, size_t pitch, double* dt, double* to, double* tf)
{
/*
axy variables represent the coefficients in the Butcher's tableau where x represents the order of the step and the y value corresponds to the ky value
the coefficient gets multiplied by. Have to cast them all as doubles, or the ratios evaluate as integers.
e.g. a21 -> a21 * k1
e.g. a31 -> a31 * k1 + a32 * k2
*/
double a21 = static_cast<double>(.25);
double a31 = static_cast<double>(3)/static_cast<double>(32);
double a32 = static_cast<double>(9)/static_cast<double>(32);
double a41 = static_cast<double>(1932)/static_cast<double>(2197);
double a42 = static_cast<double>(-7200)/static_cast<double>(2197);
double a43 = static_cast<double>(7296)/static_cast<double>(2197);
double a51 = static_cast<double>(439)/static_cast<double>(216);
double a52 = static_cast<double>(-8);
double a53 = static_cast<double>(3680)/static_cast<double>(513);
double a54 = static_cast<double>(-845)/static_cast<double>(4104);
double a61 = static_cast<double>(-8)/static_cast<double>(27);
double a62 = static_cast<double>(2);
double a63 = static_cast<double>(-3544)/static_cast<double>(2565);
double a64 = static_cast<double>(1859)/static_cast<double>(4104);
double a65 = static_cast<double>(-11)/static_cast<double>(40);
//for loop that integrates over the specified number of points. Actually, might have to make it a do-while loop for adaptive step sizes
for(int k = 0; k < *numpoints; k++)
{
if (k!=0)
{
arrCopy<<< 1, *maxlength >>>(concs, tempsum);
cudaDeviceSynchronize();
}
arrInit<<< 1, *maxlength >>>(tempsum, 0);
cudaDeviceSynchronize();
arrInit<<< 1, *maxlength >>>(temp1, 0);
cudaDeviceSynchronize();
arrInit<<< 1, *maxlength >>>(temp2, 0);
cudaDeviceSynchronize();
calcK<<< 1, *maxlength >>>(concs, k1s, dt); //k1 = dt * flux (concs)
cudaDeviceSynchronize(); //Sync here because kernel continues onto next line before k1 finished
setup(a21, temp1, tempsum, k1s, concs, maxlength); //tempsum = a21*k1
arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum); //tempsum = concs + a21*k1
cudaDeviceSynchronize();
calcK<<< 1, *maxlength >>>(tempsum, k2s, dt); //k2 = dt * flux (concs + a21*k1)
cudaDeviceSynchronize();
arrInit<<< 1, *maxlength >>>(tempsum, 0);
cudaDeviceSynchronize();
setup(a31, temp1, tempsum, k1s, concs, maxlength); //temp1sum = a31*k1
setup(a32, temp1, tempsum, k2s, concs, maxlength); //tempsum = a31*k1 + a32*k2
arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum); //tempsum = concs + a31*k1 + a32*k2
cudaDeviceSynchronize();
calcK<<< 1, *maxlength >>>(tempsum, k3s, dt); //k3 = dt * flux (concs + a31*k1 + a32*k2)
cudaDeviceSynchronize();
arrInit<<< 1, *maxlength >>>(tempsum, 0);
cudaDeviceSynchronize();
setup(a41, temp1, tempsum, k1s, concs, maxlength); //tempsum = a41*k1
setup(a42, temp1, tempsum, k2s, concs, maxlength); //tempsum = a41*k1 + a42*k2
setup(a43, temp1, tempsum, k3s, concs, maxlength); //tempsum = a41*k1 + a42*k2 + a43*k3
arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum); //tempsum = concs + a41*k1 + a42*k2 + a43*k3
cudaDeviceSynchronize();
calcK<<< 1, *maxlength >>>(tempsum, k4s, dt); //k4 = dt * flux (concs + a41*k1 + a42*k2 + a43*k3)
cudaDeviceSynchronize();
arrInit<<< 1, *maxlength >>>(tempsum, 0);
cudaDeviceSynchronize();
setup(a51, temp1, tempsum, k1s, concs, maxlength); //tempsum = a51*k1
setup(a52, temp1, tempsum, k2s, concs, maxlength); //tempsum = a51*k1 + a52*k2
setup(a53, temp1, tempsum, k3s, concs, maxlength); //tempsum = a51*k1 + a52*k2 + a53*k3
setup(a54, temp1, tempsum, k4s, concs, maxlength); //tempsum = a51*k1 + a52*k2 + a53*k3 + a54*k4
arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum); //tempsum = concs + a51*k1 + a52*k2 + a53*k3 + a54*k4
cudaDeviceSynchronize();
calcK<<< 1, *maxlength >>>(tempsum, k5s, dt); //k5 = dt * flux (concs + a51*k1 + a52*k2 + a53*k3 + a54*k4)
cudaDeviceSynchronize();
arrInit<<< 1, *maxlength >>>(tempsum, 0);
cudaDeviceSynchronize();
setup(a61, temp1, tempsum, k1s, concs, maxlength); //tempsum = a61*k1
setup(a62, temp1, tempsum, k2s, concs, maxlength); //tempsum = a61*k1 + a62*k2
setup(a63, temp1, tempsum, k3s, concs, maxlength); //tempsum = a61*k1 + a62*k2 + a63*k3
setup(a64, temp1, tempsum, k4s, concs, maxlength); //tempsum = a61*k1 + a62*k2 + a63*k3 + a64*k4
setup(a65, temp1, tempsum, **k4s**, concs, maxlength); //tempsum = a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5
arrAdd<<< 1, *maxlength >>>(concs, tempsum, tempsum); //tempsum = concs + a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5
cudaDeviceSynchronize();
calcK<<< 1, *maxlength >>>(tempsum, k6s, dt); //k6 = dt * flux (concs + a61*k1 + a62*k2 + a63*k3 + a64*k4 + a65*k5)
cudaDeviceSynchronize();
//At this point, temp1 and tempsum are maxlength dimension arrays that are able to be used for other things.
//Calculate acceptable step size before storing the concentrations.
calcStepSize(temp1, temp2, tempsum, concs, k1s, k2s, k3s, k4s, k5s, k6s, dt, maxlength); //temp1 = 4th Order guess, tempsum = 5th Order guess
cudaDeviceSynchronize();
//Store the initial conditions in the first column of the storage array.
if (k==0)
{
storeConcs<<< 1, *maxlength >>>(concStorage, pitch, concs, k); //Store this step's concentrations in 2D array
cudaDeviceSynchronize();
}
//Store future concentration in next column of storage array.
storeConcs<<< 1, *maxlength >>>(concStorage, pitch, tempsum, k+1); //Store this step's concentrations in 2D array
cudaDeviceSynchronize();
}
}
//calcStepSize will take in an error tolerance, the current concentrations and the k values and calculate the resulting step size according to the following equation
//e[n+1]=y4[n+1] - y5[n+1]
__device__ void calcStepSize(double* temp1, double*temp2, double* tempsum, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s, double* dt, int* maxlength)
{
//do
//{
takeFourthOrderStep<<< 1, *maxlength >>>(temp1, concs, k1s, k2s, k3s, k4s, k5s); //Store 4th order guess in temp1
takeFifthOrderStep<<< 1, *maxlength >>>(tempsum, concs, k1s, k2s, k3s, k4s, k5s, k6s); //Store 5th order guess in tempsum
cudaDeviceSynchronize();
arrSub<<< 1, *maxlength >>>(temp1, tempsum, temp2)
arrMin<<< 1, *maxlength >>>
//arrMult
//}
//while
}
//takeFourthOrderStep is going to overwrite the old temp1 array with the new array of concentrations that result from a 4th order step. This kernel is meant to be launched
// with as many threads as there are discrete concentrations to be tracked.
__global__ void takeFourthOrderStep(double* y4, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s)
{
double b41 = static_cast<double>(25)/static_cast<double>(216);
double b42 = static_cast<double>(0);
double b43 = static_cast<double>(1408)/static_cast<double>(2565);
double b44 = static_cast<double>(2197)/static_cast<double>(4104);
double b45 = static_cast<double>(-1)/static_cast<double>(5);
int idx = blockIdx.x * blockDim.x + threadIdx.x;
y4[idx] = concs[idx] + b41 * k1s[idx] + b42 * k2s[idx] + b43 * k3s[idx] + b44 * k4s[idx] + b45 * k5s[idx];
}
//takeFifthOrderStep is going to overwrite the old array of concentrations with the new array of concentrations. As of now, this will be the 5th order step. Another function can be d
//defined that will take a fourth order step if that is interesting for any reason. This kernel is meant to be launched with as many threads as there are discrete concentrations
//to be tracked.
//Store b values in register? Constants?
__global__ void takeFifthOrderStep(double* y5, double* concs, double* k1s, double* k2s, double* k3s, double* k4s, double* k5s, double* k6s)
{
double b51 = static_cast<double>(16)/static_cast<double>(135);
double b52 = static_cast<double>(0);
double b53 = static_cast<double>(6656)/static_cast<double>(12825);
double b54 = static_cast<double>(28561)/static_cast<double>(56430);
double b55 = static_cast<double>(-9)/static_cast<double>(50);
double b56 = static_cast<double>(2)/static_cast<double>(55);
int idx = blockIdx.x * blockDim.x + threadIdx.x;
y5[idx] = concs[idx] + b51 * k1s[idx] + b52 * k2s[idx] + b53 * k3s[idx] + b54 * k4s[idx] + b55 * k5s[idx] + b56 * k6s[idx];
}
//storeConcs takes the current array of concentrations and stores it in the cId'th column of the 2D concStorage array
//pitch = memory size of a row
__global__ void storeConcs(double* cS, size_t pitch, double* concs, int cId)
{
int tIdx = threadIdx.x;
//cS is basically the memory address of the first element of the flattened (1D) 2D array.
double* row = (double*)((char*)cS + tIdx * pitch);
row[cId] = concs[tIdx];
}
//Perhaps I can optimize by using shared memory to hold conc values.
__global__ void calcK(double* concs, double* ks, double* dt)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
ks[idx]=(*dt)*flux(idx, concs);
}
//Adds two arrays (a + b) element by element and stores the result in array c.
__global__ void arrAdd(double* a, double* b, double* c)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
c[idx]=a[idx]+b[idx];
}
//Subtracts two arrays (a - b) element by element and stores the result in array c.
__global__ void arrSub(double* a, double* b, double* c)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
c[idx]=a[idx]-b[idx];
}
//Multiplies two arrays (a * b) element by element and stores the result in array c.
__global__ void arrMult(double* a, double* b, double* c)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
c[idx]=a[idx]*b[idx];
}
//Will find the min of errors array.
__global__ void arrMin(double* errors)
{
//extern _shared_ double[7];
}
//Initializes an array a to double value b.
__global__ void arrInit(double* a, double b)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
a[idx]=b;
}
//Copies array b onto array a.
__global__ void arrCopy(double* a, double* b)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
a[idx]=b[idx];
}
//Placeholder function for the flux calculation. It will take the size of the oligomer and current concentrations as inputs.
__device__ double flux(int r, double *concs)
{
return -concs[r];
}
//This function multiplies a tableau value by the corresponding k array and adds the result to tempsum. Used to
//add all the a*k terms.
//e.g. setup(a21, temp1, tempsum, k1s, concs, maxlength) => tempsum = a21 * k1
__device__ void setup(double tableauValue, double *temp1, double *tempsum, double *ks, double *concs, int *maxlength)
{
//Sets tempsum to tabVal * k
arrInit<<< 1, *maxlength >>>(temp1, tableauValue); //Set [temp1] to tableau value, temp1 = a
cudaDeviceSynchronize();
arrMult<<< 1, *maxlength >>>(ks, temp1, temp1); //Multiply tableau value by appropriate [k], temp1 = a*k
cudaDeviceSynchronize();
arrAdd<<< 1, *maxlength >>>(tempsum, temp1, tempsum); //Move tabVal*k to [tempsum], tempsum = tempsum+temp1
cudaDeviceSynchronize();
//temp1 = tableauValue * kArray
//tempsum = current sum (tableauValue * kArray)
}EDICIÓN FINAL Encontré mi error. Fue un error tipográfico que calculaba los K6. Pongo asteriscos alrededor del término ofensivo "k4s" que debería ser un "k5s". Cualquier persona que quiera un integrador CUDA RKF45 puede usarlo como funciona ahora y es del orden correcto.
Number of Points h Global Error Calculated Order
10 0.1 3.59E-09
20 0.05 1.08E-10 5.06E+00
40 0.025 3.31E-12 5.03E+00
80 0.0125 1.03E-13 5.01E+00
160 0.00625 2.20E-15 5.55E+00
320 0.003125 1.20E-15 8.74E-01
640 0.0015625 1.40E-15 -2.22E-01
1280 0.00078125 2.00E-16 2.81E+00
ode
c++
time-integration
cuda
runge-kutta
Pelo de Slytherin
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Respuestas:
Supongo que está estableciendo la tolerancia de error en
1e-12. Tiene razón en que cuando un esquema adaptativo acepta el tamaño de paso actual, supone que el esquema de quinto orden fue, para todos los efectos, la respuesta "correcta". Sin embargo, esto es solo cuando acepta el paso actual. Si la diferencia entre el 4 y 5 pasos orden son demasiado grandes, que no acepta ya sea como suficientemente bueno. Para probar un solo algoritmo como este, debe ejecutarlo con múltiples pasos de tiempo y ver el orden de convergencia para verificarlo.Ahora, por otro lado, un paso
0.001con un esquema de quinto orden en su problema de prueba (supongo que fue 1D) debería ser más que suficiente para llevar el error a casi la máquina epsilon. Supongo que tiene un error tipográfico en algún lugar que reduce su integrador a primer orden. Como dije en el párrafo 1, ejecútelo con múltiples pasos de tiempo y verifique su orden de convergencia. Esto le dirá si el error que ve es razonable o si tiene un error en algún lugar de su integrador.EDITAR
Para calcular el orden de convergencia de un solucionador de odas, ejecútelo en un problema con una solución conocida para dos pasos de tiempo diferentes y . Su error global debería ser:hh h2
Tenga en cuenta que el 2 proviene de la relación entre los pasos de tiempo. Debería obtener para su esquema de quinto orden.n = 5
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