如何在R中使用GenSA函数放置数学约束 [英] How to put mathematical constraints with GenSA function in R

问题描述

我目前正在尝试使用模拟退火程序包GenSA，以最小化以下功能:

I am currently trying to use Simulated Annealing package GenSA in order to minimize the function below :

efficientFunction <- function(v) {
  t(v)  %*% Cov_Mat   %*%  v
 }

其中Cov_Mat是从4种资产获得的协方差矩阵，而v是维度4的权重向量.

Where Cov_Mat is a covariance matrix obtained from 4 assets and v is a weight vector of dimension 4.

我正在尝试以这种方式解决Markowitz资产分配方法，并且我想知道如何引入数学约束，例如所有系数之和必须等于1:

I'm trying to solve the Markowitz asset allocation approach this way and I would like to know how I could introduce mathematical constraint such as the sum of all coefficients have to equal 1 :

sum(v) = 1

此外，由于我打算依靠GenSA功能，因此我想在约束条件下使用类似的东西:

Moreover since I intend to rely on the GenSA function, I would like to use something like this with the constraint :

v <- c(0.25, 0.25, 0.25, 0.25)
dimension <- 4
lower <- rep(0, dimension)
upper <- rep(1, dimension)

out <- GenSA(v, lower = lower, upper = upper, fn = efficientFunction)

我在本文中发现:

I have found in this paper : http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.97.6091&rep=rep1&type=pdf how to handle such constraint within the Simulated Annealing Algorithm but I don't know how I could implement it in R.

对于任何建议，我将不胜感激.这是我第一次使用SO，所以请不要犹豫告诉我我的提问方式是否错误.

I'd be very grateful for any advice. It is my first time using SO so don't hesitate to tell me if I have the wrong approach in the way I ask question.

推荐答案

一种可行的方法是利用所谓的拉格朗日乘数(参见，

A possible approach would be to make use of so-called Lagrange multipliers (cf., http://en.wikipedia.org/wiki/Lagrange_multiplier). For example, set

efficientFunction <- function(v) {
  lambda <- 100
  t(v)  %*% Cov_Mat   %*%  v + lambda * abs( sum(v) - 1 )
}

，因此为了最小化目标函数efficientFunction，生成的参数也将惩罚项lambda * abs( sum(v) - 1 )最小化.拉格朗日乘数lambda设置为任意但足够高的水平.

, so that in order to minimize the objective function efficientFunction the resulting parameter also minimize the penalty term lambda * abs( sum(v) - 1 ). The Lagrange multiplier lambda is set to an arbitrary but sufficiently high level.

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