Ubuntu 下 glpk 的安装及使用

作者：jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4156204.html

glpk是一个开源的求解线性规划的包。

添加源：

deb http://us.archive.ubuntu.com/ubuntu saucy main universe

更新源并安装：

sudo apt-get update

sudo apt-get install glpk

写入如下glpsolEx.mod 文件

 /* Variables */
 var x1 >= ;
 var x2 >= ;
 var x3 >= ;

 /* Object function */
 maximize z: x1 + *x2 + *x3;

 /* Constrains */
 s.t. con1: x1 + x2 + x3 <= ;
 s.t. con2: x1  <= ;
 s.t. con3: x3  <= ;
 s.t. con4: *x2 + x3  <= ;

 end;

运行 glpsol -m glpsolEx.mod -o glpsolEx.sol，输出到glpsolEx.sol文件中

结果为：

 Problem:    glpsolEx
 Rows:
 Columns:
 Non-zeros:
 Status:     OPTIMAL
 Objective:  z =  (MAXimum)

    No.   Row name   St   Activity     Lower bound   Upper bound    Marginal
 ------ ------------ -- ------------- ------------- ------------- -------------
       z            B
       con1         NU
       con2         B
       con3         B
       con4         NU                                                      

    No. Column name  St   Activity     Lower bound   Upper bound    Marginal
 ------ ------------ -- ------------- ------------- ------------- -------------
       x1           NL                                                    -
       x2           B
       x3           B                                          

 Karush-Kuhn-Tucker optimality conditions:

 KKT.PE: max.abs.err = 0.00e+00 on row
         max.rel.err = 0.00e+00 on row
         High quality

 KKT.PB: max.abs.err = 4.44e-16 on row
         max.rel.err = 1.11e-16 on row
         High quality

 KKT.DE: max.abs.err = 0.00e+00 on column
         max.rel.err = 0.00e+00 on column
         High quality

 KKT.DB: max.abs.err = 0.00e+00 on row
         max.rel.err = 0.00e+00 on row
         High quality

 End of output

帮助文档中一个求解八皇后的例子：

 /* QUEENS, a classic combinatorial optimization problem */

 /* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */

 /* The Queens Problem is to place as many queens as possible on the 8x8
    (or more generally, nxn) chess board in a way that they do not fight
    each other. This problem is probably as old as the chess game itself,
    and thus its origin is not known, but it is known that Gauss studied
    this problem. */

 param n, integer, > , default ;
 /* size of the chess board */

 var x{..n, ..n}, binary;
 /* x[i,j] = 1 means that a queen is placed in square [i,j] */

 s.t. a{i in ..n}: sum{j in ..n} x[i,j] <= ;
 /* at most one queen can be placed in each row */

 s.t. b{j in ..n}: sum{i in ..n} x[i,j] <= ;
 /* at most one queen can be placed in each column */

 s.t. c{k in -n..n-}: sum{i in ..n, j in ..n: i-j == k} x[i,j] <= ;
 /* at most one queen can be placed in each "\"-diagonal */

 s.t. d{k in ..n+n-}: sum{i in ..n, j in ..n: i+j == k} x[i,j] <= ;
 /* at most one queen can be placed in each "/"-diagonal */

 maximize obj: sum{i in ..n, j in ..n} x[i,j];
 /* objective is to place as many queens as possible */

 /* solve the problem */
 solve;

 /* and print its optimal solution */
 for {i in ..n}
 {  for {j in ..n} printf " %s", if x[i,j] then "Q" else ".";
    printf("\n");
 }

 end;