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