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我有 N 个股票的时间序列数据。

sample.data<-replicate(10,rnorm(1000))，其中每列显示不同股票随时间的回报。

我正在尝试构建一个投资组合权重向量以最小化收益的方差。

目标函数：

min w^{T}\sum w
s.t. e_{n}^{T}w=1
\left \| w \right \|\leq C

其中 w 是权重向量，\sum是协方差矩阵，e_{n}^{T}是 1 的向量，C是常数。其中第二个约束 ( \left \| w \right \|) 是不等式约束。

我使用以下代码来解决这个问题：

library(Rsolnp) 
 gamma<-1  
fn<-function(x) {cov.Rt<-cov(sample.data); return(t(x)%*%cov.Rt%*%x)}   #OBJECTIVE FUNCTION TO MINIMIZE 
eqn<-function(x){one.vec<-matrix(1,ncol=10,nrow=1); return(one.vec%*%x)} #EQUALITY CONSTRAINT
constraints<-1   #EQUALITY CONSTRAINT
ineq<-function(x){one.vec<-matrix(1,ncol=10,nrow=1); #INEQUALITY CONSTRAINT
z1<-one.vec%*%abs(x)
return(z1)  
    }   

uh<-gamma #UPPER BOUND
lb<-0 #LOWER BOUND
x0<-matrix(0,10,1) #STARTING PARAMETER VECTOR (NOT SURE WHAT STARTING VALUES TO PUT HERE)

sol1<-solnp(x0,fun=fn,eqfun=eqn,eqB=constraints, ineqfun=ineq,ineqUB=gamma,ineqLB=lb)

运行此代码时，我收到以下错误消息：

Error in solve.default(cz,tol = 1e-25) : system is computationally singular: reciprocal condition number = 0 In addition: There were 50 warnings (use warnings() to see the first 50)

warnings()
1: In cbind(temp, funv): number of rows of result is not a multiple of vector length

有什么想法我可能做错了吗？起始参数向量有问题x0吗？

4

1 回答 1

4

我稍微更改了您的代码：

library(Rsolnp)
set.seed(1)
sample.data <- matrix(rnorm(10*1000),ncol=10)
gamma <- 1

#OBJECTIVE FUNCTION TO MINIMIZE 
fn <- function(x){
  cov.Rt <- cov(sample.data); 
  as.numeric(t(x) %*% cov.Rt %*% x) 
} 
#EQUALITY CONSTRAINT
eqn <- function(x){
  one.vec <- matrix(1, ncol=10, nrow=1)
  as.numeric(one.vec %*% x) 
} 
constraints <- 1
#INEQUALITY CONSTRAINT
ineq <- function(x){ 
  one.vec <- matrix(1, ncol=10, nrow=1); 
  z1<-one.vec %*% abs(x)
  as.numeric(z1)  
}   

uh <- gamma #UPPER BOUND
lb <- 0 #LOWER BOUND
x0 <- matrix(1, 10, 1) #STARTING PARAMETER VECTOR (NOT SURE WHAT STARTING VALUES TO PUT HERE)

sol1 <- solnp(x0, fun=fn, eqfun=eqn, eqB=constraints)

当我们运行上面的代码时，我们得到了一个解决方案：

Iter: 1 fn: 0.09624      Pars:  0.08355 0.08307 0.10154 0.09108 0.11745 0.12076 0.09020 0.09435 0.10884 0.10918
Iter: 2 fn: 0.09624      Pars:  0.08354 0.08308 0.10153 0.09107 0.11746 0.12078 0.09021 0.09434 0.10883 0.10918
solnp--> Completed in 2 iterations

但是当我们将不等式限制添加到优化问题中时，我们就会遇到问题：

> sol2 <- solnp(x0, fun=fn, eqfun=eqn, eqB=constraints, 
          ineqfun=ineq, ineqUB=gamma, ineqLB=lb)

Iter: 1 fn: 0.09624      Pars:  0.08356 0.08305 0.10153 0.09111 0.11748 0.12078 0.09021 0.09431 0.10881 0.10916
solnp-->The linearized problem has no feasible
solnp-->solution.  The problem may not be feasible.

Iter: 2 fn: 272.5459     Pars:  4.44541 4.42066 5.40272 4.84595 6.25082 6.42718 4.80020 5.02004 5.79138 5.81029
solnp-->The linearized problem has no feasible
solnp-->solution.  The problem may not be feasible.

Iter: 3 fn: 272.5459     Pars:  4.44547 4.42070 5.40274 4.84596 6.25078 6.42712 4.80023 5.02006 5.79138 5.81023
Iter: 4 fn: 0.09624      Pars:  0.08357 0.08304 0.10157 0.09106 0.11744 0.12074 0.09021 0.09432 0.10886 0.10918
Iter: 5 fn: 0.09625      Pars:  0.08354 0.08308 0.10153 0.09107 0.11747 0.12078 0.09021 0.09434 0.10883 0.10919
Iter: 6 fn: 0.09717      Pars:  0.08394 0.08347 0.10201 0.09150 0.11803 0.12135 0.09064 0.09479 0.10935 0.10971
Iter: 7 fn: 0.09624      Pars:  0.08353 0.08307 0.10153 0.09106 0.11747 0.12078 0.09020 0.09433 0.10883 0.10919
Iter: 8 fn: 0.09624      Pars:  0.08353 0.08307 0.10153 0.09106 0.11747 0.12078 0.09020 0.09433 0.10883 0.10919
solnp--> Solution not reliable....Problem Inverting Hessian.
Warning message:
In p0 * vscale[(neq + 2):(nc + np + 1)] :
  longer object length is not a multiple of shorter object length

让我们尝试稍微改变一下 gamma： gamma <- 1.01

> sol2 <- solnp(x0, fun=fn, eqfun=eqn, eqB=constraints, 
          ineqfun=ineq, ineqUB=gamma, ineqLB=lb)
Iter: 1 fn: 0.09624      Pars:  0.08355 0.08307 0.10153 0.09108 0.11745 0.12076 0.09020 0.09435 0.10884 0.10918
Iter: 2 fn: 0.09624      Pars:  0.08354 0.08308 0.10153 0.09107 0.11746 0.12078 0.09021 0.09434 0.10883 0.10918
solnp--> Completed in 2 iterations

所以你的不等式约束似乎完全围绕着等式约束。还把这两个约束放在一起看，我觉得有点奇怪。我的猜测是，您可能错误地指定了您的不等式约束，并且只是想要卖空约束之类的东西，例如。权重介于 0 和 1 之间。这可以按照您的方法通过以下方式归档：

ineq <- function(x){ return(x) }   
uh <- rep(gamma, 10) #UPPER BOUND
lb <- rep(0, 10) #LOWER BOUND
sol3 <- solnp(x0, fun=fn, eqfun=eqn, eqB=constraints, ineqfun=ineq, ineqUB=uh, ineqLB=lb)
于 2014-11-19T21:00:00.177 回答