如何在lpsolve中公式化x！= y [英] How to formulate x != y in lpsolve?

问题描述

我试图表述变量x，y，z必须全部不同，并且它们仅接受值1、2或3(当然，这是一个玩具示例):

I'm trying to formulate that the variables x,y,z must all be different and that they accept only the values 1, 2 or 3 (this is, of course, a toy example):

min: x+y+z;

1 <= x <= 3;
1 <= y <= 3;
1 <= z <= 3;

但是要完成这项工作，我仍然需要访问布尔运算符或！=运算符，它们似乎在lpsolve中不存在！我该怎么办?我想这样做:

but to make this work I still need either access to boolean operators or to a != operator, which don't seem to exist in lpsolve! How can I go about this? I wanted to do this:

x != y;
x != z;
y != z;

谢谢

这是我当前的代码:

/* Objective function */
min: 1;

/* Variable bounds */
1 <= x1 <= 4;
1 <= x2 <= 4;
1 <= x3 <= 4;
1 <= x4 <= 4;


x1 + x2 + x3 + x4 = 10;

x1 < x2;
x2 < x3;
x3 < x4;

int x1;
int x2;
int x3;
int x4;

lpsolve给我的结果是

lpsolve is giving me as result:

x1 = 1
x2 = 3
x3 = 3
x4 = 3

这是错误的.为什么?

推荐答案

总的来说，我会同意迈克尔·拉法格(Michael Laffargue)的观点，不可能在lpSolve中使用像a < b这样的实数a，b.但是对于整数表达式则有些不同.

In general I would agree with Michael Laffargue that it is not possible to have something like a < b for real a,b in lpSolve. But for integer expressions that is a bit different.

也许我们可以从一个更简单的问题开始. 让我们考虑两个整数变量x和y以及一个常数M，使得 1 <= x <= M and 1<=y<=M. 如果x和y不相等，则x> y或y> x.但是由于这两个整数都是仅以下一个不等式成立

Maybe we can start with a more simplified problem. Let us think of two integer variables x and y and a constant M such that 1 <= x <= M and 1<=y<=M. If x and y may not be equal then x>y or y>x. But as both are integers only one of the following inequalities hold

x+1 <= y
y+1 <= x

我们可以通过引入二进制变量r来强制上述不等式成立，这样对于x，y，r的以下不等式都都:

We can enforce that only one of the inequalities above holds by introducing a binary variable r such that for x,y, r the following inequalities hold both :

(i)   x+1 <= y+ Mr
(ii)  y+1 <= x+M-Mr

因为如果r=0则(i) x+1 <=y和(ii)是微不足道的，但是如果r=1则(ii) y+1 <= x和(i)则微不足道.

because if r=0 then (i) x+1 <=y and (ii) is trivial, but if r=1 then (ii) y+1 <= x and (i) is trivial.

现在我们从上方将解决方案应用于OP问题时，我们可以针对OP问题和M=4的所有变量对构建具有不等式(i)和(ii)的线性程序:

When we now apply the solution from above to the problem of the OP, then we can build a linear program with inequalities (i) and (ii) for all pairs of variables from the OP's problem and M=4:

/* Objective function */
min: 1;

/* Variable bounds */
1 <= x1 <= 4;
1 <= x2 <= 4;
1 <= x3 <= 4;
1 <= x4 <= 4;
r_12 <= 1;
r_13 <= 1;
r_14 <= 1;
r_23 <= 1;
r_24 <= 1;
r_34 <= 1;

/* This is done automatically because all x1,..,x4 are different
   x1 + x2 + x3 + x4 = 10;
*/

/*  Apply (i) and (ii) to all pairs of x1,..x4
(i)  x+1 <= y + Mr
(ii) y+1 <= x + M-Mr
*/

x1 + 1 <= x2 + 4 r_12;
x2 + 1 <= x1 + 4 - 4 r_12;

x1 + 1 <= x3 + 4 r_13;
x3 + 1 <= x1 + 4 - 4 r_13;

x1 + 1 <= x4 + 4 r_14;
x4 + 1 <= x4 + 4 - 4 r_14;

x2 + 1 <= x3 +  4 r_23;
x3 + 1 <= x2 + 4 - 4 r_23;

x2 + 1 <= x4 + 4 r_24;
x4 + 1 <= x2 + 4 - 4 r_24;

x3 + 1 <= x4 + 4 r_34;
x4 + 1 <= x3 + 4 - 4 r_34;

/*
x1 < x2;
x2 < x3;
x3 < x4;
*/

int r_12;
int r_13;
int r_14;
int r_23;
int r_24;
int r_34;
int x1;
int x2;
int x3;
int x4;

解决上述MILP会提供一个解决方案，其中所有x1,..,x4均不同:

Solving the MILP above renders a solution with all x1,..,x4 different:

x1=3
x2=1
x3=2
x4=4

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