具有等式和不等式约束的R优化 [英] R optimization with equality and inequality constraints

问题描述

我正在尝试查找函数的局部最小值，并且参数具有固定的总和.例如，

I am trying to find the local minimum of a function, and the parameters have a fixed sum. For example,

Fx = 10-5x1 + 2x2-x3

，条件如下，

x1 + x2 + x3 = 15

(x1，x2，x3)> = 0

其中x1，x2和x3的总和具有已知值，并且都大于零.在R中，看起来像这样，

Where the sum of x1, x2, and x3 have a known value, and they are all greater than zero. In R, it would look something like this,

Fx = function(x) {10 - (5*x[1] + 2*x[2] + x[3])}
opt = optim(c(1,1,1), Fx, method = "L-BFGS-B", lower=c(0,0,0), upper=c(15,15,15))

我还尝试对constrOptim使用不等式来强制求和.我仍然认为这可能是可行的解决方法，但是我无法使其有效.这是实际问题的简化示例，但是任何帮助将不胜感激.

I also tried to use inequalities with constrOptim to force the sum to be fixed. I still think this may be a plausible work around, but I was unable to make it work. This is a simplified example of the real problem, but any help would be very appreciated.

推荐答案

在这种情况下，optim显然不起作用，因为您具有相等性约束. constrOptim出于相同的原因也不起作用(我尝试将等式转换为两个不等式，即大于和小于15，但这不适用于constrOptim).

On this occasion optim will not work obviously because you have equality constraints. constrOptim will not work either for the same reason (I tried converting the equality to two inequalities i.e. greater and less than 15 but this didn't work with constrOptim).

但是，有一个专门用于解决此类问题的程序包，它是Rsolnp.

However, there is a package dedicated to this kind of problem and that is Rsolnp.

您可以通过以下方式使用它:

You use it the following way:

#specify your function
opt_func <- function(x) {
  10 - 5*x[1] + 2 * x[2] - x[3]
}

#specify the equality function. The number 15 (to which the function is equal)
#is specified as an additional argument
equal <- function(x) {
  x[1] + x[2] + x[3] 
}

#the optimiser - minimises by default
solnp(c(5,5,5), #starting values (random - obviously need to be positive and sum to 15)
      opt_func, #function to optimise
      eqfun=equal, #equality function 
      eqB=15,   #the equality constraint
      LB=c(0,0,0), #lower bound for parameters i.e. greater than zero
      UB=c(100,100,100)) #upper bound for parameters (I just chose 100 randomly)

输出:

> solnp(c(5,5,5),
+       opt_func,
+       eqfun=equal,
+       eqB=15,
+       LB=c(0,0,0),
+       UB=c(100,100,100))

Iter: 1 fn: -65.0000     Pars:  14.99999993134  0.00000002235  0.00000004632
Iter: 2 fn: -65.0000     Pars:  14.999999973563  0.000000005745  0.000000020692
solnp--> Completed in 2 iterations
$pars
[1] 1.500000e+01 5.745236e-09 2.069192e-08

$convergence
[1] 0

$values
[1] -10 -65 -65

$lagrange
     [,1]
[1,]   -5

$hessian
          [,1]      [,2]      [,3]
[1,] 121313076 121313076 121313076
[2,] 121313076 121313076 121313076
[3,] 121313076 121313076 121313076

$ineqx0
NULL

$nfuneval
[1] 126

$outer.iter
[1] 2

$elapsed
Time difference of 0.1770101 secs

$vscale
[1] 6.5e+01 1.0e-08 1.0e+00 1.0e+00 1.0e+00

因此得出的最佳值为:

$pars
[1] 1.500000e+01 5.745236e-09 2.069192e-08

表示第一个参数为15，其余参数为零和零.这确实是函数中的全局最小值，因为x2正在添加到函数中，并且5 * x1对结果的影响(负)比x3大得多.选择15、0、0是解，并且根据约束选择函数的全局最小值.

which means that the first parameter is 15 and the rest zero and zero. This is indeed the global minimum in your function since the x2 is adding to the function and 5 * x1 has a much greater (negative) influence than x3 on the outcome. The choice of 15, 0, 0 is the solution and the global minimum to the function according to the constraints.

该功能很好用！

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