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e591e3622b
Double the number of samples that are used in the runs test. E.g. with 64 samples in regression the runs test will be tried over the 64 samples and up to 64 previous samples. The minimum number of samples in now 4. This improves the response with low-mid jitters by about 50%.
656 lines
17 KiB
C
656 lines
17 KiB
C
/*
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$Header: /cvs/src/chrony/regress.c,v 1.32 2003/09/22 21:22:30 richard Exp $
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=======================================================================
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chronyd/chronyc - Programs for keeping computer clocks accurate.
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**********************************************************************
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* Copyright (C) Richard P. Curnow 1997-2003
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*
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* This program is free software; you can redistribute it and/or modify
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* it under the terms of version 2 of the GNU General Public License as
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* published by the Free Software Foundation.
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*
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* This program is distributed in the hope that it will be useful, but
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* WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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* General Public License for more details.
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*
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* You should have received a copy of the GNU General Public License along
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* with this program; if not, write to the Free Software Foundation, Inc.,
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* 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA.
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*
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**********************************************************************
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=======================================================================
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Regression algorithms.
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*/
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#include <assert.h>
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#include <math.h>
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#include <stdio.h>
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#include <string.h>
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#include "regress.h"
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#include "logging.h"
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#include "util.h"
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#define MAX_POINTS 128
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void
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RGR_WeightedRegression
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(double *x, /* independent variable */
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double *y, /* measured data */
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double *w, /* weightings (large => data
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less reliable) */
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int n, /* number of data points */
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/* And now the results */
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double *b0, /* estimated y axis intercept */
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double *b1, /* estimated slope */
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double *s2, /* estimated variance of data points */
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double *sb0, /* estimated standard deviation of
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intercept */
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double *sb1 /* estimated standard deviation of
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slope */
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/* Could add correlation stuff later if required */
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)
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{
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double P, Q, U, V, W;
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double diff;
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double u, ui, aa;
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int i;
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assert(n >= 3);
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W = U = 0;
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for (i=0; i<n; i++) {
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U += x[i] / w[i];
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W += 1.0 / w[i];
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}
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u = U / W;
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/* Calculate statistics from data */
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P = Q = V = 0.0;
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for (i=0; i<n; i++) {
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ui = x[i] - u;
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P += y[i] / w[i];
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Q += y[i] * ui / w[i];
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V += ui * ui / w[i];
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}
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*b1 = Q / V;
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*b0 = (P / W) - (*b1) * u;
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*s2 = 0.0;
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for (i=0; i<n; i++) {
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diff = y[i] - *b0 - *b1*x[i];
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*s2 += diff*diff / w[i];
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}
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*s2 /= (double)(n-2);
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*sb1 = sqrt(*s2 / V);
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aa = u * (*sb1);
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*sb0 = sqrt(*s2 / W + aa * aa);
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*s2 *= (n / W); /* Giving weighted average of variances */
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return;
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}
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/* ================================================== */
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/* Get the coefficient to multiply the standard deviation by, to get a
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particular size of confidence interval (assuming a t-distribution) */
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double
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RGR_GetTCoef(int dof)
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{
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/* Assuming now the 99.95% quantile */
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static double coefs[] =
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{ 636.6, 31.6, 12.92, 8.61, 6.869,
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5.959, 5.408, 5.041, 4.781, 4.587,
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4.437, 4.318, 4.221, 4.140, 4.073,
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4.015, 3.965, 3.922, 3.883, 3.850,
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3.819, 3.792, 3.768, 3.745, 3.725,
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3.707, 3.690, 3.674, 3.659, 3.646,
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3.633, 3.622, 3.611, 3.601, 3.591,
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3.582, 3.574, 3.566, 3.558, 3.551};
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if (dof <= 40) {
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return coefs[dof-1];
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} else {
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return 3.5; /* Until I can be bothered to do something better */
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}
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}
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/* ================================================== */
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/* Get 90% quantile of chi-square distribution */
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double
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RGR_GetChi2Coef(int dof)
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{
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static double coefs[] = {
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2.706, 4.605, 6.251, 7.779, 9.236, 10.645, 12.017, 13.362,
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14.684, 15.987, 17.275, 18.549, 19.812, 21.064, 22.307, 23.542,
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24.769, 25.989, 27.204, 28.412, 29.615, 30.813, 32.007, 33.196,
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34.382, 35.563, 36.741, 37.916, 39.087, 40.256, 41.422, 42.585,
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43.745, 44.903, 46.059, 47.212, 48.363, 49.513, 50.660, 51.805,
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52.949, 54.090, 55.230, 56.369, 57.505, 58.641, 59.774, 60.907,
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62.038, 63.167, 64.295, 65.422, 66.548, 67.673, 68.796, 69.919,
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71.040, 72.160, 73.279, 74.397, 75.514, 76.630, 77.745, 78.860
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};
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if (dof <= 64) {
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return coefs[dof-1];
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} else {
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return 1.2 * dof; /* Until I can be bothered to do something better */
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}
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}
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/* ================================================== */
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/* Structure used for holding results of each regression */
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typedef struct {
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double variance;
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double slope_sd;
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double slope;
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double offset;
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double offset_sd;
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double K2; /* Variance / slope_var */
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int n; /* Number of points */
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int dof; /* Number of degrees of freedom */
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} RegressionResult;
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/* ================================================== */
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/* Critical value for number of runs of residuals with same sign. 10%
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critical region for now */
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static char critical_runs10[] = {
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0, 0, 0, 0, 0, 0, 0, 0, 3, 3,
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4, 4, 4, 5, 5, 6, 6, 6, 7, 7,
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8, 8, 9, 9, 9, 10, 10, 11, 11, 12,
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12, 12, 13, 13, 14, 14, 15, 15, 16, 16,
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17, 17, 17, 18, 18, 19, 19, 20, 20, 21,
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21, 21, 22, 22, 23, 23, 24, 24, 25, 25,
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26, 26, 27, 27, 27, 28, 28, 29, 29, 30,
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30, 31, 31, 32, 32, 32, 33, 33, 34, 34,
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35, 35, 36, 36, 37, 37, 38, 38, 39, 39,
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39, 40, 40, 41, 41, 42, 42, 43, 43, 44,
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44, 45, 45, 46, 46, 46, 47, 47, 48, 48,
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49, 49, 50, 50, 51, 51, 52, 52, 53, 53,
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54, 54, 54, 55, 55, 56, 56, 57, 57, 58
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};
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/* ================================================== */
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static int
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n_runs_from_residuals(double *resid, int n)
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{
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int nruns;
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int i;
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nruns = 1;
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for (i=1; i<n; i++) {
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if (((resid[i-1] < 0.0) && (resid[i] < 0.0)) ||
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((resid[i-1] > 0.0) && (resid[i] > 0.0))) {
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/* Nothing to do */
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} else {
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nruns++;
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}
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}
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return nruns;
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}
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/* ================================================== */
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/* Return a boolean indicating whether we had enough points for
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regression */
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#define MIN_SAMPLES_FOR_REGRESS 3
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int
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RGR_FindBestRegression
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(double *x, /* independent variable */
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double *y, /* measured data */
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double *w, /* weightings (large => data
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less reliable) */
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int n, /* number of data points */
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int m, /* number of extra samples in x and y arrays
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(negative index) which can be used to
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extend runs test */
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/* And now the results */
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double *b0, /* estimated y axis intercept */
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double *b1, /* estimated slope */
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double *s2, /* estimated variance of data points */
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double *sb0, /* estimated standard deviation of
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intercept */
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double *sb1, /* estimated standard deviation of
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slope */
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int *new_start, /* the new starting index to make the
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residuals pass the two tests */
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int *n_runs, /* number of runs amongst the residuals */
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int *dof /* degrees of freedom in statistics (needed
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to get confidence intervals later) */
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)
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{
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double P, Q, U, V, W; /* total */
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double resid[MAX_POINTS * REGRESS_RUNS_RATIO];
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double ss;
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double a, b, u, ui, aa;
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int start, resid_start, nruns, npoints;
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int i;
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assert(n <= MAX_POINTS);
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assert(n * REGRESS_RUNS_RATIO < sizeof (critical_runs10) / sizeof (critical_runs10[0]));
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if (n < MIN_SAMPLES_FOR_REGRESS) {
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return 0;
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}
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start = 0;
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do {
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W = U = 0;
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for (i=start; i<n; i++) {
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U += x[i] / w[i];
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W += 1.0 / w[i];
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}
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u = U / W;
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P = Q = V = 0.0;
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for (i=start; i<n; i++) {
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ui = x[i] - u;
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P += y[i] / w[i];
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Q += y[i] * ui / w[i];
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V += ui * ui / w[i];
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}
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b = Q / V;
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a = (P / W) - (b * u);
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/* Get residuals also for the extra samples before start */
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resid_start = n - (n - start) * REGRESS_RUNS_RATIO;
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if (resid_start < -m)
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resid_start = -m;
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for (i=resid_start; i<n; i++) {
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resid[i - resid_start] = y[i] - a - b*x[i];
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}
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/* Count number of runs */
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nruns = n_runs_from_residuals(resid, n - resid_start);
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if (nruns > critical_runs10[n - resid_start] || n - start <= MIN_SAMPLES_FOR_REGRESS) {
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break;
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} else {
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/* Try dropping one sample at a time until the runs test passes. */
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++start;
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}
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} while (1);
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/* Work out statistics from full dataset */
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*b1 = b;
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*b0 = a;
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ss = 0.0;
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for (i=start; i<n; i++) {
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ss += resid[i - resid_start]*resid[i - resid_start] / w[i];
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}
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npoints = n - start;
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ss /= (double)(npoints - 2);
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*sb1 = sqrt(ss / V);
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aa = u * (*sb1);
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*sb0 = sqrt((ss / W) + (aa * aa));
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*s2 = ss * (double) npoints / W;
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*new_start = start;
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*dof = npoints - 2;
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*n_runs = nruns;
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return 1;
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}
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/* ================================================== */
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#define EXCH(a,b) temp=(a); (a)=(b); (b)=temp
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/* ================================================== */
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/* Find the index'th biggest element in the array x of n elements.
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flags is an array where a 1 indicates that the corresponding entry
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in x is known to be sorted into its correct position and a 0
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indicates that the corresponding entry is not sorted. However, if
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flags[m] = flags[n] = 1 with m<n, then x[m] must be <= x[n] and for
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all i with m<i<n, x[m] <= x[i] <= x[n]. In practice, this means
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flags[] has to be the result of a previous call to this routine
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with the same array x, and is used to remember which parts of the
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x[] array we have already sorted.
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The approach used is a cut-down quicksort, where we only bother to
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keep sorting the partition that contains the index we are after.
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The approach comes from Numerical Recipes in C (ISBN
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0-521-43108-5). */
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static double
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find_ordered_entry_with_flags(double *x, int n, int index, int *flags)
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{
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int u, v, l, r;
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double temp;
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double piv;
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int pivind;
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assert(index >= 0);
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/* If this bit of the array is already sorted, simple! */
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if (flags[index]) {
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return x[index];
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}
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/* Find subrange to look at */
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u = v = index;
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while (u > 0 && !flags[u]) u--;
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if (flags[u]) u++;
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while (v < (n-1) && !flags[v]) v++;
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if (flags[v]) v--;
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do {
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if (v - u < 2) {
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if (x[v] < x[u]) {
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EXCH(x[v], x[u]);
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}
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flags[v] = flags[u] = 1;
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return x[index];
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} else {
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pivind = (u + v) >> 1;
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EXCH(x[u], x[pivind]);
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piv = x[u]; /* New value */
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l = u + 1;
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r = v;
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do {
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while (x[l] < piv && l < v) l++;
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while (x[r] > piv) r--;
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if (r <= l) break;
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EXCH(x[l], x[r]);
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l++;
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r--;
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} while (1);
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EXCH(x[u], x[r]);
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flags[r] = 1; /* Pivot now in correct place */
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if (index == r) {
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return x[r];
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} else if (index < r) {
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v = r - 1;
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} else if (index > r) {
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u = l;
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}
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}
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} while (1);
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}
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/* ================================================== */
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#if 0
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/* Not used, but this is how it can be done */
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static double
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find_ordered_entry(double *x, int n, int index)
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{
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int flags[MAX_POINTS];
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bzero(flags, n * sizeof(int));
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return find_ordered_entry_with_flags(x, n, index, flags);
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}
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#endif
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/* ================================================== */
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/* Find the median entry of an array x[] with n elements. */
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static double
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find_median(double *x, int n)
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{
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int k;
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int flags[MAX_POINTS];
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memset(flags, 0, n*sizeof(int));
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k = n>>1;
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if (n&1) {
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return find_ordered_entry_with_flags(x, n, k, flags);
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} else {
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return 0.5 * (find_ordered_entry_with_flags(x, n, k, flags) +
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find_ordered_entry_with_flags(x, n, k-1, flags));
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}
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}
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/* ================================================== */
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/* This function evaluates the equation
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\sum_{i=0}^{n-1} x_i sign(y_i - a - b x_i)
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and chooses the value of a that minimises the absolute value of the
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result. (See pp703-704 of Numerical Recipes in C). */
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static void
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eval_robust_residual
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(double *x, /* The independent points */
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double *y, /* The dependent points */
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int n, /* Number of points */
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double b, /* Slope */
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double *aa, /* Intercept giving smallest absolute
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value for the above equation */
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double *rr /* Corresponding value of equation */
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)
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{
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int i;
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double a, res, del;
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double d[MAX_POINTS];
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for (i=0; i<n; i++) {
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d[i] = y[i] - b * x[i];
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}
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a = find_median(d, n);
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res = 0.0;
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for (i=0; i<n; i++) {
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del = y[i] - a - b * x[i];
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if (del > 0.0) {
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res += x[i];
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} else if (del < 0.0) {
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res -= x[i];
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}
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}
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*aa = a;
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*rr = res;
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}
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/* ================================================== */
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/* This routine performs a 'robust' regression, i.e. one which has low
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susceptibility to outliers amongst the data. If one thinks of a
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normal (least squares) linear regression in 2D being analogous to
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the arithmetic mean in 1D, this algorithm in 2D is roughly
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analogous to the median in 1D. This algorithm seems to work quite
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well until the number of outliers is approximately half the number
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of data points.
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The return value is a status indicating whether there were enough
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data points to run the routine or not. */
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int
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RGR_FindBestRobustRegression
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(double *x, /* The independent axis points */
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double *y, /* The dependent axis points (which
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may contain outliers). */
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int n, /* The number of points */
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double tol, /* The tolerance required in
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determining the value of b1 */
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double *b0, /* The estimated Y-axis intercept */
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double *b1, /* The estimated slope */
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int *n_runs, /* The number of runs of residuals */
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int *best_start /* The best starting index */
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)
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{
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int i;
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int start;
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int n_points;
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double a, b;
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double P, U, V, W, X;
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double resid, resids[MAX_POINTS];
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double blo, bhi, bmid, rlo, rhi, rmid;
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double s2, sb, incr;
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double mx, dx, my, dy;
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int nruns = 0;
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assert(n < MAX_POINTS);
|
|
|
|
if (n < 2) {
|
|
return 0;
|
|
} else if (n == 2) {
|
|
/* Just a straight line fit (we need this for the manual mode) */
|
|
*b1 = (y[1] - y[0]) / (x[1] - x[0]);
|
|
*b0 = y[0] - (*b1) * x[0];
|
|
*n_runs = 0;
|
|
*best_start = 0;
|
|
return 1;
|
|
}
|
|
|
|
/* else at least 3 points, apply normal algorithm */
|
|
|
|
start = 0;
|
|
|
|
/* Loop to strip oldest points that cause the regression residuals
|
|
to fail the number of runs test */
|
|
do {
|
|
|
|
n_points = n - start;
|
|
|
|
/* Use standard least squares regression to get starting estimate */
|
|
|
|
P = U = 0.0;
|
|
for (i=start; i<n; i++) {
|
|
P += y[i];
|
|
U += x[i];
|
|
}
|
|
|
|
W = (double) n_points;
|
|
|
|
my = P/W;
|
|
mx = U/W;
|
|
|
|
X = V = 0.0;
|
|
for (i=start; i<n; i++) {
|
|
dy = y[i] - my;
|
|
dx = x[i] - mx;
|
|
X += dy * dx;
|
|
V += dx * dx;
|
|
}
|
|
|
|
b = X / V;
|
|
a = my - b*mx;
|
|
|
|
|
|
#if 0
|
|
printf("my=%20.12f mx=%20.12f a=%20.12f b=%20.12f\n", my, mx, a, b);
|
|
#endif
|
|
|
|
s2 = 0.0;
|
|
for (i=start; i<n; i++) {
|
|
resid = y[i] - a - b * x[i];
|
|
s2 += resid * resid;
|
|
}
|
|
|
|
/* Need to expand range of b to get a root in the interval.
|
|
Estimate standard deviation of b and expand range about b based
|
|
on that. */
|
|
sb = sqrt(s2 * W/V);
|
|
if (sb > 0.0) {
|
|
incr = 3.0 * sb;
|
|
} else {
|
|
incr = 3.0 * tol;
|
|
}
|
|
|
|
blo = b;
|
|
bhi = b;
|
|
|
|
do {
|
|
blo -= incr;
|
|
bhi += incr;
|
|
|
|
/* We don't want 'a' yet */
|
|
eval_robust_residual(x + start, y + start, n_points, blo, &a, &rlo);
|
|
eval_robust_residual(x + start, y + start, n_points, bhi, &a, &rhi);
|
|
|
|
} while (rlo * rhi > 0.0); /* fn vals have same sign, i.e. root not
|
|
in interval. */
|
|
|
|
/* OK, so the root for b lies in (blo, bhi). Start bisecting */
|
|
do {
|
|
bmid = 0.5 * (blo + bhi);
|
|
eval_robust_residual(x + start, y + start, n_points, bmid, &a, &rmid);
|
|
if (rmid == 0.0) {
|
|
break;
|
|
} else if (rmid * rlo > 0.0) {
|
|
blo = bmid;
|
|
rlo = rmid;
|
|
} else if (rmid * rhi > 0.0) {
|
|
bhi = bmid;
|
|
rhi = rmid;
|
|
} else {
|
|
assert(0);
|
|
}
|
|
} while ((bhi - blo) > tol);
|
|
|
|
*b0 = a;
|
|
*b1 = bmid;
|
|
|
|
/* Number of runs test, but not if we're already down to the
|
|
minimum number of points */
|
|
if (n_points == MIN_SAMPLES_FOR_REGRESS) {
|
|
break;
|
|
}
|
|
|
|
for (i=start; i<n; i++) {
|
|
resids[i] = y[i] - a - bmid * x[i];
|
|
}
|
|
|
|
nruns = n_runs_from_residuals(resids + start, n_points);
|
|
|
|
if (nruns > critical_runs10[n_points]) {
|
|
break;
|
|
} else {
|
|
start++;
|
|
}
|
|
|
|
} while (1);
|
|
|
|
*n_runs = nruns;
|
|
*best_start = start;
|
|
|
|
return 1;
|
|
|
|
}
|
|
|
|
/* ================================================== */
|