boinc/client/speed_stats.C

529 lines
16 KiB
C

// The contents of this file are subject to the Mozilla Public License
// Version 1.0 (the "License"); you may not use this file except in
// compliance with the License. You may obtain a copy of the License at
// http://www.mozilla.org/MPL/
//
// Software distributed under the License is distributed on an "AS IS"
// basis, WITHOUT WARRANTY OF ANY KIND, either express or implied. See the
// License for the specific language governing rights and limitations
// under the License.
//
// The Original Code is the Berkeley Open Infrastructure for Network Computing.
//
// The Initial Developer of the Original Code is the SETI@home project.
// Portions created by the SETI@home project are Copyright (C) 2002
// University of California at Berkeley. All Rights Reserved.
//
// Contributor(s):
//
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include "speed_stats.h"
#include "error_numbers.h"
// # of iterations of each test to run for initial timing purposes
#define D_FLOP_ITERS 100
#define I_OP_ITERS 100
#define BANDWIDTH_ITERS 5
//#define RUN_TEST
#ifdef RUN_TEST
int main(void) {
int cache_size;
cache_size = check_cache_size(CACHE_MAX);
run_test_suite(4);
return 0;
}
void run_test_suite(double num_secs_per_test) {
if (num_secs_per_test<0) {
fprintf(stderr, "error: run_test_suite: negative num_seconds_per_test\n");
}
printf(
"Running tests. This will take about %.1lf seconds.\n\n",
num_secs_per_test*3
);
printf(
"Speed: %.5lf million flops/sec\n\n",
run_double_prec_test(num_secs_per_test)/1000000
);
printf(
"Speed: %.5lf million integer ops/sec\n\n",
run_int_test(num_secs_per_test)/1000000
);
printf(
"Speed: %.5lf MB/sec\n\n",
12*sizeof(double)*run_mem_bandwidth_test(num_secs_per_test)/1000000
);
}
#endif
int check_cache_size(int mem_size) {
int i, n, index, stride, *memBlock, logStride, logCache;
double **results;
int steps, tsteps, csize, limit, temp, cind, sind;
clock_t total_sec, sec;
double secs, nanosecs, temp2;
int not_found;
if (mem_size<0) {
fprintf(stderr, "error: check_cache_size: negative mem_size\n");
return ERR_NEG;
}
logStride = (int)(log(STRIDE_MAX/STRIDE_MIN)/log(2))+1;
logCache = (int)(log(CACHE_MAX/CACHE_MIN)/log(2))+1;
printf("Test will take about %.2f seconds.\n", SECS_PER_RUN*logStride*logCache);
results = (double **)malloc(sizeof(double *)*logStride);
for (i=0;i<logStride;i++) {
results[i] = (double *)malloc(sizeof(double)*logCache);
for (n=0;n<logCache;n++) {
results[i][n] = 1.0;
}
}
printf("|");
for (i=0;i<logCache;i++) {
printf("-");
}
printf("|\n");
memBlock = (int *)malloc(sizeof(int)*mem_size);
printf(" ");
for (csize=CACHE_MIN,cind=0;csize<=CACHE_MAX;csize*=2,cind++) {
for (stride = STRIDE_MIN,sind=0; stride<=STRIDE_MAX; stride*=2,sind++) {
limit = csize - stride + 1; // cache size this loop
steps = 0;
sec = clock();
do { // repeat until collect 1 second
for (i = SAMPLE * stride; i != 0; i--) { // larger sample
for (index = 0; index < limit; index += stride) {
memBlock[index]++; // cache access
}
}
steps++; // count while loop iterations
} while (clock() < sec+(CLOCKS_PER_SEC*SECS_PER_RUN)); // until collect 1 second
total_sec = clock()-sec;
// Repeat empty loop to loop subtract overhead
tsteps = 0; // used to match no. while iterations
temp = 0;
sec = clock();
do { // repeat until same no. iterations as above
for (i = SAMPLE * stride; i != 0; i--) { // larger sample
for (index = 0; index < limit; index += stride) {
temp += index; // dummy code
}
}
tsteps++; // count while iterations
} while (tsteps < steps); // until = no. iterations
total_sec -= clock()-sec;
secs = ((double)total_sec) / CLOCKS_PER_SEC;
if (temp == 3) {
printf("Howdy\n");
}
nanosecs = (double) secs * 1e9 / (steps * SAMPLE * stride * ((limit - 1) / stride + 1));
results[sind][cind] = nanosecs;
//if (stride==STRIDE_MIN) printf("\n");
printf(
"Size (bytes): %7d Stride (bytes): %4d read+write: %4.0f ns, %d %d\n",
csize * sizeof (int), stride * sizeof(int), nanosecs, sind, cind
);
}
printf(".");
fflush(stdout);
}
printf("\n");
for (i=0;i<logStride;i++) {
for (n=0;n<logCache;n++) {
printf ("%4.0f ", results[i][n]);
}
printf("\n");
}
for (i=0;i<logStride;i++) {
for (n=logCache;n>0;n--) {
results[i][n] /= results[i][n-1];
}
}
for (i=0;i<logCache;i++) {
temp2 = 0;
for (n=0;n<logStride;n++) {
temp2 += results[n][i];
}
results[0][i] = temp2/logStride;
}
printf("\n");
for (i=0;i<logStride;i++) {
for (n=1;n<logCache;n++) {
printf ("%1.3f ", results[i][n]);
}
printf("\n");
}
csize=CACHE_MIN;
i = 1;
not_found = 2;
while(not_found && i < logCache) {
if (not_found == 1 && results[0][i] > 1.5) {
printf("Level 2 Data Cache is %d KB.\n", csize*sizeof(int)/CACHE_MIN);
not_found = 0;
}
if (not_found == 2 && results[0][i] > 1.5) {
printf("Level 1 Data Cache is %d KB.\n", csize*sizeof(int)/CACHE_MIN);
not_found = 1;
}
i++;
csize *= 2;
}
free(memBlock);
for (i=0;i<logStride;i++)
free(results[i]);
free(results);
return 0;
}
// Run the test of double precision math speed for about num_secs seconds
//
double run_double_prec_test(double num_secs) {
int df_test_time, df_iters;
double df_secs;
if (num_secs<0) {
fprintf(stderr, "error: run_double_prec_test: negatvie num_secs\n");
return ERR_NEG;
}
// Start by doing some quick timing tests for rough calibration
df_test_time = (int)double_flop_test(D_FLOP_ITERS, 0);
if (df_test_time <= 0) df_test_time = 1;
df_secs = (double)df_test_time/CLOCKS_PER_SEC;
// Calculate the # of iterations based on these tests
df_iters = (int)(D_FLOP_ITERS*num_secs/df_secs);
if (df_iters > D_FLOP_ITERS) { // no need to redo test if we already got enough
df_test_time = (int)double_flop_test(df_iters, 0);
} else {
df_iters = D_FLOP_ITERS;
}
df_secs = (double)df_test_time/CLOCKS_PER_SEC;
return 1000000*df_iters/df_secs;
}
// Run the test of integer math speed for about num_secs seconds
//
double run_int_test(double num_secs) {
int int_test_time, int_iters;
double int_secs;
if (num_secs<0) {
fprintf(stderr, "error: run_int_test: negative num_secs\n");
return ERR_NEG;
}
// Start by doing some quick timing tests for rough calibration
int_test_time = (int)int_op_test(I_OP_ITERS, 0);
if (int_test_time <= 0) int_test_time = 1;
int_secs = (double)int_test_time/CLOCKS_PER_SEC;
// Calculate the # of iterations based on these tests
int_iters = (int)(I_OP_ITERS*num_secs/int_secs);
if (int_iters > I_OP_ITERS) { // no need to redo test
int_test_time = (int)int_op_test(int_iters, 0);
} else {
int_iters = I_OP_ITERS;
}
int_secs = (double)int_test_time/CLOCKS_PER_SEC;
return 1000000*int_iters/int_secs;
}
// Run the test of memory bandwidth speed for about num_secs seconds
//
double run_mem_bandwidth_test(double num_secs) {
int bw_test_time;
double bw_secs;
int bw_iters;
if (num_secs<0) {
fprintf(stderr, "error: run_mem_bandwidth_test: negative num_secs\n");
return ERR_NEG;
}
// Start by doing some quick timing tests for rough calibration
bw_test_time = (int)bandwidth_test(BANDWIDTH_ITERS, 0);
if (bw_test_time <= 0) bw_test_time = 1;
bw_secs = (double)bw_test_time/CLOCKS_PER_SEC;
// Calculate the # of iterations based on these tests
bw_iters = (int)(BANDWIDTH_ITERS*num_secs/bw_secs);
if (bw_iters > BANDWIDTH_ITERS) { // no need to redo test
bw_test_time = (int)bandwidth_test(bw_iters, 0);
} else {
bw_iters = BANDWIDTH_ITERS;
}
bw_secs = (double)bw_test_time/CLOCKS_PER_SEC;
return 1000000*bw_iters/bw_secs;
}
// One iteration == D_LOOP_ITERS (1,000,000) floating point operations
// If time_total is negative, there was an error in the calculation,
// meaning there is probably something wrong with the CPU
clock_t double_flop_test(int iterations, int print_debug) {
double a[NUM_DOUBLES],t1,t2;
double temp;
clock_t time_start, time_total;
int i,j,k,calc_error;
if (iterations<0) {
fprintf(stderr, "error: double_flop_test: negative iterations\n");
return ERR_NEG;
}
// Initialize the array
a[0] = 1;
for (i=1;i<NUM_DOUBLES;i++)
a[i] = a[i-1]/2.0;
// Ideally, the array "a" will fit into cache, meaning this test doesn't
// really include memory accesses
time_start = clock();
for (i=0;i<iterations;i++) {
for (j=0;j<D_LOOP_ITERS;j+=((NUM_DOUBLES-1)*5)) {
temp = 1;
t1 = a[0];
// These tests do a pretty good job of preventing optimization
// since the result from all but one of the lines is required for the
// next line. At the end of each iteration through the for loop,
// the array should be the same as when it started
for (k=0;k<NUM_DOUBLES-1;k++) {
t2 = a[k+1];
t1 = t1 * t2; // 1st FLOP
temp = temp + temp; // 2nd FLOP
t1 = t1 * temp; // 3rd FLOP
t1 = t1 + t2; // 4th FLOP
t1 = t1 / 1.5; // 5th FLOP
a[k] = t1;
t1 = t2;
}
}
}
// Stop the clock
time_total = clock();
// Accomodate for the possibility of clock wraparound
if (time_total > time_start) {
time_total -= time_start;
} else {
time_total = 0; // this is just a kludge
}
calc_error = 0;
temp = 1;
// Check to make sure all the values are the same as when we started
for (i=0;i<NUM_DOUBLES;i++) {
if ((float)a[i] != (float)temp) {
calc_error = 1;
}
temp /= 2;
}
if (calc_error) {
time_total *= -1;
}
if (print_debug) {
for (i=0;i<NUM_DOUBLES;i++) {
printf("%3d: %.50f\n", i, a[i]);
}
}
return time_total;
}
// One iteration == 1,000,000 integer operations
// If time_total is negative, there was an error in the calculation,
// meaning there is probably something wrong with the CPU
clock_t int_op_test(int iterations, int print_debug) {
int a[NUM_INTS], temp;
clock_t time_start, time_total;
int i,j,k,calc_error;
if (iterations<0) {
fprintf(stderr, "error: int_op_test: negative iterations\n");
return ERR_NEG;
}
a[0] = 1;
for (i=1;i<NUM_INTS;i++) {
a[i] = 2*a[i-1];
}
time_start = clock();
for (i=0;i<iterations;i++) {
// The contents of the array "a" should be the same at the
// beginning and end of each loop iteration. Most compilers will
// partially unroll the individual loops within this one, so
// those integer operations (incrementing k) are not counted
for (j=0;j<I_LOOP_ITERS/(NUM_INTS*9);j++) {
for (k=0;k<NUM_INTS;k++) {
a[k] *= 3; // 1 int ops
}
for (k=NUM_INTS-1;k>=0;k--) {
a[k] += 6; // 2 int ops
}
for (k=0;k<NUM_INTS;k++) {
a[k] /= 3; // 3 int ops
}
for (k=NUM_INTS-1;k>=0;k--) {
a[k] -= 2; // 4 int ops
}
for (k=NUM_INTS-1;k>0;k--) {
a[k] -= a[k-1]; // 5 int ops
}
for (k=1;k<NUM_INTS;k++) {
a[k] = 2*a[k-1]; // 6 int ops
}
for (k=NUM_INTS-1;k>0;k--) {
if (a[k-1] != 0) // 7 int ops
a[k] /= a[k-1]; // 8 int ops
}
for (k=1;k<NUM_INTS;k++) {
a[k] = 2*a[k-1]; // 9 int ops
}
}
}
// Stop the clock
time_total = clock();
// Accomodate for the possibility of clock wraparound
if (time_total > time_start) {
time_total -= time_start;
} else {
time_total = 0; // this is just a kludge
}
calc_error = 0;
temp = 1;
// Check to make sure all the values are the same as when we started
for (i=0;i<NUM_INTS;i++) {
if (a[i] != temp) {
calc_error = 1;
}
temp *= 2;
}
if (calc_error) {
time_total *= -1;
}
if (print_debug) {
for (i=0;i<NUM_INTS;i++) {
printf("%3d: %d\n", i, a[i]);
}
}
return time_total;
}
// One iteration == Read of 6,000,000*sizeof(double), Write of 6,000,000*sizeof(double)
// If time_total is negative, there was an error in the copying,
// meaning there is probably something wrong with the CPU
clock_t bandwidth_test(int iterations, int print_debug) {
// a, b, and c are arrays of doubles we will copy around to test memory bandwidth
double *a, *b, *c;
// aVal and bVal are the values of all elements of a and b.
// These values use every other bit,
// so that if there is a HW problem it will easily manifest itself
double aVal, bVal;
// Start and stop times for the clock
clock_t time_start, time_total;
int i,j,copy_error;
if (iterations<0) {
fprintf(stderr, "error: bandwidth_test: negative iterations\n");
return ERR_NEG;
}
// These are doubles in order to make full use of bus and instruction bandwidth
a = (double *)malloc(MEM_SIZE * sizeof(double));
b = (double *)malloc(MEM_SIZE * sizeof(double));
c = (double *)malloc(MEM_SIZE * sizeof(double));
// These values use all the bits in a floating point number (Investigate these values)
aVal = (-2.0/3.0)*pow(2.0,-341.0);
bVal = (1.0/3.0)*pow(2.0,342.0);
// We add i to each value to prevent compiler optimizations of the copy
for (i=0;i<MEM_SIZE;i++) {
a[i] = aVal+i; b[i] = bVal+i; c[i] = 1.0;
}
// Start the clock
time_start = clock();
// 6 read, 6 write operations per iteration which will preserve a and b
for (i=0;i<iterations*2;i++) {
for (j=0;j<MEM_SIZE;j++) {
c[j] = a[j];
a[j] = b[j];
b[j] = c[j];
}
}
// Stop the clock
time_total = clock();
// Accomodate for the possibility of clock wraparound
if (time_total > time_start) {
time_total -= time_start;
} else {
time_total = 0; // this is just a kludge
}
copy_error = 0;
for (i=0;i<MEM_SIZE;i++) {
if (a[i] != aVal+i || b[i] != bVal+i) {
copy_error = 1;
}
}
if (copy_error) {
time_total *= -1;
}
free(a);
free(b);
free(c);
return time_total;
}