[1] | 1 | /* glpios07.c (mixed cover cut generator) */ |
---|
| 2 | |
---|
| 3 | /*********************************************************************** |
---|
| 4 | * This code is part of GLPK (GNU Linear Programming Kit). |
---|
| 5 | * |
---|
| 6 | * Copyright (C) 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, |
---|
| 7 | * 2009, 2010 Andrew Makhorin, Department for Applied Informatics, |
---|
| 8 | * Moscow Aviation Institute, Moscow, Russia. All rights reserved. |
---|
| 9 | * E-mail: <mao@gnu.org>. |
---|
| 10 | * |
---|
| 11 | * GLPK is free software: you can redistribute it and/or modify it |
---|
| 12 | * under the terms of the GNU General Public License as published by |
---|
| 13 | * the Free Software Foundation, either version 3 of the License, or |
---|
| 14 | * (at your option) any later version. |
---|
| 15 | * |
---|
| 16 | * GLPK is distributed in the hope that it will be useful, but WITHOUT |
---|
| 17 | * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY |
---|
| 18 | * or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public |
---|
| 19 | * License for more details. |
---|
| 20 | * |
---|
| 21 | * You should have received a copy of the GNU General Public License |
---|
| 22 | * along with GLPK. If not, see <http://www.gnu.org/licenses/>. |
---|
| 23 | ***********************************************************************/ |
---|
| 24 | |
---|
| 25 | #include "glpios.h" |
---|
| 26 | |
---|
| 27 | /*---------------------------------------------------------------------- |
---|
| 28 | -- COVER INEQUALITIES |
---|
| 29 | -- |
---|
| 30 | -- Consider the set of feasible solutions to 0-1 knapsack problem: |
---|
| 31 | -- |
---|
| 32 | -- sum a[j]*x[j] <= b, (1) |
---|
| 33 | -- j in J |
---|
| 34 | -- |
---|
| 35 | -- x[j] is binary, (2) |
---|
| 36 | -- |
---|
| 37 | -- where, wlog, we assume that a[j] > 0 (since 0-1 variables can be |
---|
| 38 | -- complemented) and a[j] <= b (since a[j] > b implies x[j] = 0). |
---|
| 39 | -- |
---|
| 40 | -- A set C within J is called a cover if |
---|
| 41 | -- |
---|
| 42 | -- sum a[j] > b. (3) |
---|
| 43 | -- j in C |
---|
| 44 | -- |
---|
| 45 | -- For any cover C the inequality |
---|
| 46 | -- |
---|
| 47 | -- sum x[j] <= |C| - 1 (4) |
---|
| 48 | -- j in C |
---|
| 49 | -- |
---|
| 50 | -- is called a cover inequality and is valid for (1)-(2). |
---|
| 51 | -- |
---|
| 52 | -- MIXED COVER INEQUALITIES |
---|
| 53 | -- |
---|
| 54 | -- Consider the set of feasible solutions to mixed knapsack problem: |
---|
| 55 | -- |
---|
| 56 | -- sum a[j]*x[j] + y <= b, (5) |
---|
| 57 | -- j in J |
---|
| 58 | -- |
---|
| 59 | -- x[j] is binary, (6) |
---|
| 60 | -- |
---|
| 61 | -- 0 <= y <= u is continuous, (7) |
---|
| 62 | -- |
---|
| 63 | -- where again we assume that a[j] > 0. |
---|
| 64 | -- |
---|
| 65 | -- Let C within J be some set. From (1)-(4) it follows that |
---|
| 66 | -- |
---|
| 67 | -- sum a[j] > b - y (8) |
---|
| 68 | -- j in C |
---|
| 69 | -- |
---|
| 70 | -- implies |
---|
| 71 | -- |
---|
| 72 | -- sum x[j] <= |C| - 1. (9) |
---|
| 73 | -- j in C |
---|
| 74 | -- |
---|
| 75 | -- Thus, we need to modify the inequality (9) in such a way that it be |
---|
| 76 | -- a constraint only if the condition (8) is satisfied. |
---|
| 77 | -- |
---|
| 78 | -- Consider the following inequality: |
---|
| 79 | -- |
---|
| 80 | -- sum x[j] <= |C| - t. (10) |
---|
| 81 | -- j in C |
---|
| 82 | -- |
---|
| 83 | -- If 0 < t <= 1, then (10) is equivalent to (9), because all x[j] are |
---|
| 84 | -- binary variables. On the other hand, if t <= 0, (10) being satisfied |
---|
| 85 | -- for any values of x[j] is not a constraint. |
---|
| 86 | -- |
---|
| 87 | -- Let |
---|
| 88 | -- |
---|
| 89 | -- t' = sum a[j] + y - b. (11) |
---|
| 90 | -- j in C |
---|
| 91 | -- |
---|
| 92 | -- It is understood that the condition t' > 0 is equivalent to (8). |
---|
| 93 | -- Besides, from (6)-(7) it follows that t' has an implied upper bound: |
---|
| 94 | -- |
---|
| 95 | -- t'max = sum a[j] + u - b. (12) |
---|
| 96 | -- j in C |
---|
| 97 | -- |
---|
| 98 | -- This allows to express the parameter t having desired properties: |
---|
| 99 | -- |
---|
| 100 | -- t = t' / t'max. (13) |
---|
| 101 | -- |
---|
| 102 | -- In fact, t <= 1 by definition, and t > 0 being equivalent to t' > 0 |
---|
| 103 | -- is equivalent to (8). |
---|
| 104 | -- |
---|
| 105 | -- Thus, the inequality (10), where t is given by formula (13) is valid |
---|
| 106 | -- for (5)-(7). |
---|
| 107 | -- |
---|
| 108 | -- Note that if u = 0, then y = 0, so t = 1, and the conditions (8) and |
---|
| 109 | -- (10) is transformed to the conditions (3) and (4). |
---|
| 110 | -- |
---|
| 111 | -- GENERATING MIXED COVER CUTS |
---|
| 112 | -- |
---|
| 113 | -- To generate a mixed cover cut in the form (10) we need to find such |
---|
| 114 | -- set C which satisfies to the inequality (8) and for which, in turn, |
---|
| 115 | -- the inequality (10) is violated in the current point. |
---|
| 116 | -- |
---|
| 117 | -- Substituting t from (13) to (10) gives: |
---|
| 118 | -- |
---|
| 119 | -- 1 |
---|
| 120 | -- sum x[j] <= |C| - ----- (sum a[j] + y - b), (14) |
---|
| 121 | -- j in C t'max j in C |
---|
| 122 | -- |
---|
| 123 | -- and finally we have the cut inequality in the standard form: |
---|
| 124 | -- |
---|
| 125 | -- sum x[j] + alfa * y <= beta, (15) |
---|
| 126 | -- j in C |
---|
| 127 | -- |
---|
| 128 | -- where: |
---|
| 129 | -- |
---|
| 130 | -- alfa = 1 / t'max, (16) |
---|
| 131 | -- |
---|
| 132 | -- beta = |C| - alfa * (sum a[j] - b). (17) |
---|
| 133 | -- j in C */ |
---|
| 134 | |
---|
| 135 | #if 1 |
---|
| 136 | #define MAXTRY 1000 |
---|
| 137 | #else |
---|
| 138 | #define MAXTRY 10000 |
---|
| 139 | #endif |
---|
| 140 | |
---|
| 141 | static int cover2(int n, double a[], double b, double u, double x[], |
---|
| 142 | double y, int cov[], double *_alfa, double *_beta) |
---|
| 143 | { /* try to generate mixed cover cut using two-element cover */ |
---|
| 144 | int i, j, try = 0, ret = 0; |
---|
| 145 | double eps, alfa, beta, temp, rmax = 0.001; |
---|
| 146 | eps = 0.001 * (1.0 + fabs(b)); |
---|
| 147 | for (i = 0+1; i <= n; i++) |
---|
| 148 | for (j = i+1; j <= n; j++) |
---|
| 149 | { /* C = {i, j} */ |
---|
| 150 | try++; |
---|
| 151 | if (try > MAXTRY) goto done; |
---|
| 152 | /* check if condition (8) is satisfied */ |
---|
| 153 | if (a[i] + a[j] + y > b + eps) |
---|
| 154 | { /* compute parameters for inequality (15) */ |
---|
| 155 | temp = a[i] + a[j] - b; |
---|
| 156 | alfa = 1.0 / (temp + u); |
---|
| 157 | beta = 2.0 - alfa * temp; |
---|
| 158 | /* compute violation of inequality (15) */ |
---|
| 159 | temp = x[i] + x[j] + alfa * y - beta; |
---|
| 160 | /* choose C providing maximum violation */ |
---|
| 161 | if (rmax < temp) |
---|
| 162 | { rmax = temp; |
---|
| 163 | cov[1] = i; |
---|
| 164 | cov[2] = j; |
---|
| 165 | *_alfa = alfa; |
---|
| 166 | *_beta = beta; |
---|
| 167 | ret = 1; |
---|
| 168 | } |
---|
| 169 | } |
---|
| 170 | } |
---|
| 171 | done: return ret; |
---|
| 172 | } |
---|
| 173 | |
---|
| 174 | static int cover3(int n, double a[], double b, double u, double x[], |
---|
| 175 | double y, int cov[], double *_alfa, double *_beta) |
---|
| 176 | { /* try to generate mixed cover cut using three-element cover */ |
---|
| 177 | int i, j, k, try = 0, ret = 0; |
---|
| 178 | double eps, alfa, beta, temp, rmax = 0.001; |
---|
| 179 | eps = 0.001 * (1.0 + fabs(b)); |
---|
| 180 | for (i = 0+1; i <= n; i++) |
---|
| 181 | for (j = i+1; j <= n; j++) |
---|
| 182 | for (k = j+1; k <= n; k++) |
---|
| 183 | { /* C = {i, j, k} */ |
---|
| 184 | try++; |
---|
| 185 | if (try > MAXTRY) goto done; |
---|
| 186 | /* check if condition (8) is satisfied */ |
---|
| 187 | if (a[i] + a[j] + a[k] + y > b + eps) |
---|
| 188 | { /* compute parameters for inequality (15) */ |
---|
| 189 | temp = a[i] + a[j] + a[k] - b; |
---|
| 190 | alfa = 1.0 / (temp + u); |
---|
| 191 | beta = 3.0 - alfa * temp; |
---|
| 192 | /* compute violation of inequality (15) */ |
---|
| 193 | temp = x[i] + x[j] + x[k] + alfa * y - beta; |
---|
| 194 | /* choose C providing maximum violation */ |
---|
| 195 | if (rmax < temp) |
---|
| 196 | { rmax = temp; |
---|
| 197 | cov[1] = i; |
---|
| 198 | cov[2] = j; |
---|
| 199 | cov[3] = k; |
---|
| 200 | *_alfa = alfa; |
---|
| 201 | *_beta = beta; |
---|
| 202 | ret = 1; |
---|
| 203 | } |
---|
| 204 | } |
---|
| 205 | } |
---|
| 206 | done: return ret; |
---|
| 207 | } |
---|
| 208 | |
---|
| 209 | static int cover4(int n, double a[], double b, double u, double x[], |
---|
| 210 | double y, int cov[], double *_alfa, double *_beta) |
---|
| 211 | { /* try to generate mixed cover cut using four-element cover */ |
---|
| 212 | int i, j, k, l, try = 0, ret = 0; |
---|
| 213 | double eps, alfa, beta, temp, rmax = 0.001; |
---|
| 214 | eps = 0.001 * (1.0 + fabs(b)); |
---|
| 215 | for (i = 0+1; i <= n; i++) |
---|
| 216 | for (j = i+1; j <= n; j++) |
---|
| 217 | for (k = j+1; k <= n; k++) |
---|
| 218 | for (l = k+1; l <= n; l++) |
---|
| 219 | { /* C = {i, j, k, l} */ |
---|
| 220 | try++; |
---|
| 221 | if (try > MAXTRY) goto done; |
---|
| 222 | /* check if condition (8) is satisfied */ |
---|
| 223 | if (a[i] + a[j] + a[k] + a[l] + y > b + eps) |
---|
| 224 | { /* compute parameters for inequality (15) */ |
---|
| 225 | temp = a[i] + a[j] + a[k] + a[l] - b; |
---|
| 226 | alfa = 1.0 / (temp + u); |
---|
| 227 | beta = 4.0 - alfa * temp; |
---|
| 228 | /* compute violation of inequality (15) */ |
---|
| 229 | temp = x[i] + x[j] + x[k] + x[l] + alfa * y - beta; |
---|
| 230 | /* choose C providing maximum violation */ |
---|
| 231 | if (rmax < temp) |
---|
| 232 | { rmax = temp; |
---|
| 233 | cov[1] = i; |
---|
| 234 | cov[2] = j; |
---|
| 235 | cov[3] = k; |
---|
| 236 | cov[4] = l; |
---|
| 237 | *_alfa = alfa; |
---|
| 238 | *_beta = beta; |
---|
| 239 | ret = 1; |
---|
| 240 | } |
---|
| 241 | } |
---|
| 242 | } |
---|
| 243 | done: return ret; |
---|
| 244 | } |
---|
| 245 | |
---|
| 246 | static int cover(int n, double a[], double b, double u, double x[], |
---|
| 247 | double y, int cov[], double *alfa, double *beta) |
---|
| 248 | { /* try to generate mixed cover cut; |
---|
| 249 | input (see (5)): |
---|
| 250 | n is the number of binary variables; |
---|
| 251 | a[1:n] are coefficients at binary variables; |
---|
| 252 | b is the right-hand side; |
---|
| 253 | u is upper bound of continuous variable; |
---|
| 254 | x[1:n] are values of binary variables at current point; |
---|
| 255 | y is value of continuous variable at current point; |
---|
| 256 | output (see (15), (16), (17)): |
---|
| 257 | cov[1:r] are indices of binary variables included in cover C, |
---|
| 258 | where r is the set cardinality returned on exit; |
---|
| 259 | alfa coefficient at continuous variable; |
---|
| 260 | beta is the right-hand side; */ |
---|
| 261 | int j; |
---|
| 262 | /* perform some sanity checks */ |
---|
| 263 | xassert(n >= 2); |
---|
| 264 | for (j = 1; j <= n; j++) xassert(a[j] > 0.0); |
---|
| 265 | #if 1 /* ??? */ |
---|
| 266 | xassert(b > -1e-5); |
---|
| 267 | #else |
---|
| 268 | xassert(b > 0.0); |
---|
| 269 | #endif |
---|
| 270 | xassert(u >= 0.0); |
---|
| 271 | for (j = 1; j <= n; j++) xassert(0.0 <= x[j] && x[j] <= 1.0); |
---|
| 272 | xassert(0.0 <= y && y <= u); |
---|
| 273 | /* try to generate mixed cover cut */ |
---|
| 274 | if (cover2(n, a, b, u, x, y, cov, alfa, beta)) return 2; |
---|
| 275 | if (cover3(n, a, b, u, x, y, cov, alfa, beta)) return 3; |
---|
| 276 | if (cover4(n, a, b, u, x, y, cov, alfa, beta)) return 4; |
---|
| 277 | return 0; |
---|
| 278 | } |
---|
| 279 | |
---|
| 280 | /*---------------------------------------------------------------------- |
---|
| 281 | -- lpx_cover_cut - generate mixed cover cut. |
---|
| 282 | -- |
---|
| 283 | -- SYNOPSIS |
---|
| 284 | -- |
---|
| 285 | -- #include "glplpx.h" |
---|
| 286 | -- int lpx_cover_cut(LPX *lp, int len, int ind[], double val[], |
---|
| 287 | -- double work[]); |
---|
| 288 | -- |
---|
| 289 | -- DESCRIPTION |
---|
| 290 | -- |
---|
| 291 | -- The routine lpx_cover_cut generates a mixed cover cut for a given |
---|
| 292 | -- row of the MIP problem. |
---|
| 293 | -- |
---|
| 294 | -- The given row of the MIP problem should be explicitly specified in |
---|
| 295 | -- the form: |
---|
| 296 | -- |
---|
| 297 | -- sum{j in J} a[j]*x[j] <= b. (1) |
---|
| 298 | -- |
---|
| 299 | -- On entry indices (ordinal numbers) of structural variables, which |
---|
| 300 | -- have non-zero constraint coefficients, should be placed in locations |
---|
| 301 | -- ind[1], ..., ind[len], and corresponding constraint coefficients |
---|
| 302 | -- should be placed in locations val[1], ..., val[len]. The right-hand |
---|
| 303 | -- side b should be stored in location val[0]. |
---|
| 304 | -- |
---|
| 305 | -- The working array work should have at least nb locations, where nb |
---|
| 306 | -- is the number of binary variables in (1). |
---|
| 307 | -- |
---|
| 308 | -- The routine generates a mixed cover cut in the same form as (1) and |
---|
| 309 | -- stores the cut coefficients and right-hand side in the same way as |
---|
| 310 | -- just described above. |
---|
| 311 | -- |
---|
| 312 | -- RETURNS |
---|
| 313 | -- |
---|
| 314 | -- If the cutting plane has been successfully generated, the routine |
---|
| 315 | -- returns 1 <= len' <= n, which is the number of non-zero coefficients |
---|
| 316 | -- in the inequality constraint. Otherwise, the routine returns zero. */ |
---|
| 317 | |
---|
| 318 | static int lpx_cover_cut(LPX *lp, int len, int ind[], double val[], |
---|
| 319 | double work[]) |
---|
| 320 | { int cov[1+4], j, k, nb, newlen, r; |
---|
| 321 | double f_min, f_max, alfa, beta, u, *x = work, y; |
---|
| 322 | /* substitute and remove fixed variables */ |
---|
| 323 | newlen = 0; |
---|
| 324 | for (k = 1; k <= len; k++) |
---|
| 325 | { j = ind[k]; |
---|
| 326 | if (lpx_get_col_type(lp, j) == LPX_FX) |
---|
| 327 | val[0] -= val[k] * lpx_get_col_lb(lp, j); |
---|
| 328 | else |
---|
| 329 | { newlen++; |
---|
| 330 | ind[newlen] = ind[k]; |
---|
| 331 | val[newlen] = val[k]; |
---|
| 332 | } |
---|
| 333 | } |
---|
| 334 | len = newlen; |
---|
| 335 | /* move binary variables to the beginning of the list so that |
---|
| 336 | elements 1, 2, ..., nb correspond to binary variables, and |
---|
| 337 | elements nb+1, nb+2, ..., len correspond to rest variables */ |
---|
| 338 | nb = 0; |
---|
| 339 | for (k = 1; k <= len; k++) |
---|
| 340 | { j = ind[k]; |
---|
| 341 | if (lpx_get_col_kind(lp, j) == LPX_IV && |
---|
| 342 | lpx_get_col_type(lp, j) == LPX_DB && |
---|
| 343 | lpx_get_col_lb(lp, j) == 0.0 && |
---|
| 344 | lpx_get_col_ub(lp, j) == 1.0) |
---|
| 345 | { /* binary variable */ |
---|
| 346 | int ind_k; |
---|
| 347 | double val_k; |
---|
| 348 | nb++; |
---|
| 349 | ind_k = ind[nb], val_k = val[nb]; |
---|
| 350 | ind[nb] = ind[k], val[nb] = val[k]; |
---|
| 351 | ind[k] = ind_k, val[k] = val_k; |
---|
| 352 | } |
---|
| 353 | } |
---|
| 354 | /* now the specified row has the form: |
---|
| 355 | sum a[j]*x[j] + sum a[j]*y[j] <= b, |
---|
| 356 | where x[j] are binary variables, y[j] are rest variables */ |
---|
| 357 | /* at least two binary variables are needed */ |
---|
| 358 | if (nb < 2) return 0; |
---|
| 359 | /* compute implied lower and upper bounds for sum a[j]*y[j] */ |
---|
| 360 | f_min = f_max = 0.0; |
---|
| 361 | for (k = nb+1; k <= len; k++) |
---|
| 362 | { j = ind[k]; |
---|
| 363 | /* both bounds must be finite */ |
---|
| 364 | if (lpx_get_col_type(lp, j) != LPX_DB) return 0; |
---|
| 365 | if (val[k] > 0.0) |
---|
| 366 | { f_min += val[k] * lpx_get_col_lb(lp, j); |
---|
| 367 | f_max += val[k] * lpx_get_col_ub(lp, j); |
---|
| 368 | } |
---|
| 369 | else |
---|
| 370 | { f_min += val[k] * lpx_get_col_ub(lp, j); |
---|
| 371 | f_max += val[k] * lpx_get_col_lb(lp, j); |
---|
| 372 | } |
---|
| 373 | } |
---|
| 374 | /* sum a[j]*x[j] + sum a[j]*y[j] <= b ===> |
---|
| 375 | sum a[j]*x[j] + (sum a[j]*y[j] - f_min) <= b - f_min ===> |
---|
| 376 | sum a[j]*x[j] + y <= b - f_min, |
---|
| 377 | where y = sum a[j]*y[j] - f_min; |
---|
| 378 | note that 0 <= y <= u, u = f_max - f_min */ |
---|
| 379 | /* determine upper bound of y */ |
---|
| 380 | u = f_max - f_min; |
---|
| 381 | /* determine value of y at the current point */ |
---|
| 382 | y = 0.0; |
---|
| 383 | for (k = nb+1; k <= len; k++) |
---|
| 384 | { j = ind[k]; |
---|
| 385 | y += val[k] * lpx_get_col_prim(lp, j); |
---|
| 386 | } |
---|
| 387 | y -= f_min; |
---|
| 388 | if (y < 0.0) y = 0.0; |
---|
| 389 | if (y > u) y = u; |
---|
| 390 | /* modify the right-hand side b */ |
---|
| 391 | val[0] -= f_min; |
---|
| 392 | /* now the transformed row has the form: |
---|
| 393 | sum a[j]*x[j] + y <= b, where 0 <= y <= u */ |
---|
| 394 | /* determine values of x[j] at the current point */ |
---|
| 395 | for (k = 1; k <= nb; k++) |
---|
| 396 | { j = ind[k]; |
---|
| 397 | x[k] = lpx_get_col_prim(lp, j); |
---|
| 398 | if (x[k] < 0.0) x[k] = 0.0; |
---|
| 399 | if (x[k] > 1.0) x[k] = 1.0; |
---|
| 400 | } |
---|
| 401 | /* if a[j] < 0, replace x[j] by its complement 1 - x'[j] */ |
---|
| 402 | for (k = 1; k <= nb; k++) |
---|
| 403 | { if (val[k] < 0.0) |
---|
| 404 | { ind[k] = - ind[k]; |
---|
| 405 | val[k] = - val[k]; |
---|
| 406 | val[0] += val[k]; |
---|
| 407 | x[k] = 1.0 - x[k]; |
---|
| 408 | } |
---|
| 409 | } |
---|
| 410 | /* try to generate a mixed cover cut for the transformed row */ |
---|
| 411 | r = cover(nb, val, val[0], u, x, y, cov, &alfa, &beta); |
---|
| 412 | if (r == 0) return 0; |
---|
| 413 | xassert(2 <= r && r <= 4); |
---|
| 414 | /* now the cut is in the form: |
---|
| 415 | sum{j in C} x[j] + alfa * y <= beta */ |
---|
| 416 | /* store the right-hand side beta */ |
---|
| 417 | ind[0] = 0, val[0] = beta; |
---|
| 418 | /* restore the original ordinal numbers of x[j] */ |
---|
| 419 | for (j = 1; j <= r; j++) cov[j] = ind[cov[j]]; |
---|
| 420 | /* store cut coefficients at binary variables complementing back |
---|
| 421 | the variables having negative row coefficients */ |
---|
| 422 | xassert(r <= nb); |
---|
| 423 | for (k = 1; k <= r; k++) |
---|
| 424 | { if (cov[k] > 0) |
---|
| 425 | { ind[k] = +cov[k]; |
---|
| 426 | val[k] = +1.0; |
---|
| 427 | } |
---|
| 428 | else |
---|
| 429 | { ind[k] = -cov[k]; |
---|
| 430 | val[k] = -1.0; |
---|
| 431 | val[0] -= 1.0; |
---|
| 432 | } |
---|
| 433 | } |
---|
| 434 | /* substitute y = sum a[j]*y[j] - f_min */ |
---|
| 435 | for (k = nb+1; k <= len; k++) |
---|
| 436 | { r++; |
---|
| 437 | ind[r] = ind[k]; |
---|
| 438 | val[r] = alfa * val[k]; |
---|
| 439 | } |
---|
| 440 | val[0] += alfa * f_min; |
---|
| 441 | xassert(r <= len); |
---|
| 442 | len = r; |
---|
| 443 | return len; |
---|
| 444 | } |
---|
| 445 | |
---|
| 446 | /*---------------------------------------------------------------------- |
---|
| 447 | -- lpx_eval_row - compute explictily specified row. |
---|
| 448 | -- |
---|
| 449 | -- SYNOPSIS |
---|
| 450 | -- |
---|
| 451 | -- #include "glplpx.h" |
---|
| 452 | -- double lpx_eval_row(LPX *lp, int len, int ind[], double val[]); |
---|
| 453 | -- |
---|
| 454 | -- DESCRIPTION |
---|
| 455 | -- |
---|
| 456 | -- The routine lpx_eval_row computes the primal value of an explicitly |
---|
| 457 | -- specified row using current values of structural variables. |
---|
| 458 | -- |
---|
| 459 | -- The explicitly specified row may be thought as a linear form: |
---|
| 460 | -- |
---|
| 461 | -- y = a[1]*x[m+1] + a[2]*x[m+2] + ... + a[n]*x[m+n], |
---|
| 462 | -- |
---|
| 463 | -- where y is an auxiliary variable for this row, a[j] are coefficients |
---|
| 464 | -- of the linear form, x[m+j] are structural variables. |
---|
| 465 | -- |
---|
| 466 | -- On entry column indices and numerical values of non-zero elements of |
---|
| 467 | -- the row should be stored in locations ind[1], ..., ind[len] and |
---|
| 468 | -- val[1], ..., val[len], where len is the number of non-zero elements. |
---|
| 469 | -- The array ind and val are not changed on exit. |
---|
| 470 | -- |
---|
| 471 | -- RETURNS |
---|
| 472 | -- |
---|
| 473 | -- The routine returns a computed value of y, the auxiliary variable of |
---|
| 474 | -- the specified row. */ |
---|
| 475 | |
---|
| 476 | static double lpx_eval_row(LPX *lp, int len, int ind[], double val[]) |
---|
| 477 | { int n = lpx_get_num_cols(lp); |
---|
| 478 | int j, k; |
---|
| 479 | double sum = 0.0; |
---|
| 480 | if (len < 0) |
---|
| 481 | xerror("lpx_eval_row: len = %d; invalid row length\n", len); |
---|
| 482 | for (k = 1; k <= len; k++) |
---|
| 483 | { j = ind[k]; |
---|
| 484 | if (!(1 <= j && j <= n)) |
---|
| 485 | xerror("lpx_eval_row: j = %d; column number out of range\n", |
---|
| 486 | j); |
---|
| 487 | sum += val[k] * lpx_get_col_prim(lp, j); |
---|
| 488 | } |
---|
| 489 | return sum; |
---|
| 490 | } |
---|
| 491 | |
---|
| 492 | /*********************************************************************** |
---|
| 493 | * NAME |
---|
| 494 | * |
---|
| 495 | * ios_cov_gen - generate mixed cover cuts |
---|
| 496 | * |
---|
| 497 | * SYNOPSIS |
---|
| 498 | * |
---|
| 499 | * #include "glpios.h" |
---|
| 500 | * void ios_cov_gen(glp_tree *tree); |
---|
| 501 | * |
---|
| 502 | * DESCRIPTION |
---|
| 503 | * |
---|
| 504 | * The routine ios_cov_gen generates mixed cover cuts for the current |
---|
| 505 | * point and adds them to the cut pool. */ |
---|
| 506 | |
---|
| 507 | void ios_cov_gen(glp_tree *tree) |
---|
| 508 | { glp_prob *prob = tree->mip; |
---|
| 509 | int m = lpx_get_num_rows(prob); |
---|
| 510 | int n = lpx_get_num_cols(prob); |
---|
| 511 | int i, k, type, kase, len, *ind; |
---|
| 512 | double r, *val, *work; |
---|
| 513 | xassert(lpx_get_status(prob) == LPX_OPT); |
---|
| 514 | /* allocate working arrays */ |
---|
| 515 | ind = xcalloc(1+n, sizeof(int)); |
---|
| 516 | val = xcalloc(1+n, sizeof(double)); |
---|
| 517 | work = xcalloc(1+n, sizeof(double)); |
---|
| 518 | /* look through all rows */ |
---|
| 519 | for (i = 1; i <= m; i++) |
---|
| 520 | for (kase = 1; kase <= 2; kase++) |
---|
| 521 | { type = lpx_get_row_type(prob, i); |
---|
| 522 | if (kase == 1) |
---|
| 523 | { /* consider rows of '<=' type */ |
---|
| 524 | if (!(type == LPX_UP || type == LPX_DB)) continue; |
---|
| 525 | len = lpx_get_mat_row(prob, i, ind, val); |
---|
| 526 | val[0] = lpx_get_row_ub(prob, i); |
---|
| 527 | } |
---|
| 528 | else |
---|
| 529 | { /* consider rows of '>=' type */ |
---|
| 530 | if (!(type == LPX_LO || type == LPX_DB)) continue; |
---|
| 531 | len = lpx_get_mat_row(prob, i, ind, val); |
---|
| 532 | for (k = 1; k <= len; k++) val[k] = - val[k]; |
---|
| 533 | val[0] = - lpx_get_row_lb(prob, i); |
---|
| 534 | } |
---|
| 535 | /* generate mixed cover cut: |
---|
| 536 | sum{j in J} a[j] * x[j] <= b */ |
---|
| 537 | len = lpx_cover_cut(prob, len, ind, val, work); |
---|
| 538 | if (len == 0) continue; |
---|
| 539 | /* at the current point the cut inequality is violated, i.e. |
---|
| 540 | sum{j in J} a[j] * x[j] - b > 0 */ |
---|
| 541 | r = lpx_eval_row(prob, len, ind, val) - val[0]; |
---|
| 542 | if (r < 1e-3) continue; |
---|
| 543 | /* add the cut to the cut pool */ |
---|
| 544 | glp_ios_add_row(tree, NULL, GLP_RF_COV, 0, len, ind, val, |
---|
| 545 | GLP_UP, val[0]); |
---|
| 546 | } |
---|
| 547 | /* free working arrays */ |
---|
| 548 | xfree(ind); |
---|
| 549 | xfree(val); |
---|
| 550 | xfree(work); |
---|
| 551 | return; |
---|
| 552 | } |
---|
| 553 | |
---|
| 554 | /* eof */ |
---|