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
1 /* Variables */ 2 var x1 >= 0; 3 var x2 >= 0; 4 var x3 >= 0; 5 6 /* Object function */ 7 maximize z: x1 + 14*x2 + 6*x3; 8 9 /* Constrains */ 10 s.t. con1: x1 + x2 + x3 <= 4; 11 s.t. con2: x1 <= 2; 12 s.t. con3: x3 <= 3; 13 s.t. con4: 3*x2 + x3 <= 6; 14 15 end;
运行 glpsol -m glpsolEx.mod -o glpsolEx.sol，输出到glpsolEx.sol文件中
结果为：
1 Problem: glpsolEx 2 Rows: 5 3 Columns: 3 4 Non-zeros: 10 5 Status: OPTIMAL 6 Objective: z = 32 (MAXimum) 7 8 No. Row name St Activity Lower bound Upper bound Marginal 9 ------ ------------ -- ------------- ------------- ------------- ------------- 10 1 z B 32 11 2 con1 NU 4 4 2 12 3 con2 B 0 2 13 4 con3 B 3 3 14 5 con4 NU 6 6 4 15 16 No. Column name St Activity Lower bound Upper bound Marginal 17 ------ ------------ -- ------------- ------------- ------------- ------------- 18 1 x1 NL 0 0 -1 19 2 x2 B 1 0 20 3 x3 B 3 0 21 22 Karush-Kuhn-Tucker optimality conditions: 23 24 KKT.PE: max.abs.err = 0.00e+00 on row 0 25 max.rel.err = 0.00e+00 on row 0 26 High quality 27 28 KKT.PB: max.abs.err = 4.44e-16 on row 4 29 max.rel.err = 1.11e-16 on row 4 30 High quality 31 32 KKT.DE: max.abs.err = 0.00e+00 on column 0 33 max.rel.err = 0.00e+00 on column 0 34 High quality 35 36 KKT.DB: max.abs.err = 0.00e+00 on row 0 37 max.rel.err = 0.00e+00 on row 0 38 High quality 39 40 End of output
帮助文档中一个求解八皇后的例子：
1 /* QUEENS, a classic combinatorial optimization problem */ 2 3 /* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */ 4 5 /* The Queens Problem is to place as many queens as possible on the 8x8 6 (or more generally, nxn) chess board in a way that they do not fight 7 each other. This problem is probably as old as the chess game itself, 8 and thus its origin is not known, but it is known that Gauss studied 9 this problem. */ 10 11 param n, integer, > 0, default 8; 12 /* size of the chess board */ 13 14 var x{1..n, 1..n}, binary; 15 /* x[i,j] = 1 means that a queen is placed in square [i,j] */ 16 17 s.t. a{i in 1..n}: sum{j in 1..n} x[i,j] <= 1; 18 /* at most one queen can be placed in each row */ 19 20 s.t. b{j in 1..n}: sum{i in 1..n} x[i,j] <= 1; 21 /* at most one queen can be placed in each column */ 22 23 s.t. c{k in 2-n..n-2}: sum{i in 1..n, j in 1..n: i-j == k} x[i,j] <= 1; 24 /* at most one queen can be placed in each ""-diagonal */ 25 26 s.t. d{k in 3..n+n-1}: sum{i in 1..n, j in 1..n: i+j == k} x[i,j] <= 1; 27 /* at most one queen can be placed in each "/"-diagonal */ 28 29 maximize obj: sum{i in 1..n, j in 1..n} x[i,j]; 30 /* objective is to place as many queens as possible */ 31 32 /* solve the problem */ 33 solve; 34 35 /* and print its optimal solution */ 36 for {i in 1..n} 37 { for {j in 1..n} printf " %s", if x[i,j] then "Q" else "."; 38 printf(" "); 39 } 40 41 end;