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我正在尝试使用solve.QP来求解一个有三个约束的二次程序：

  • 所有总和为 1（这工作正常）
  • 全部必须大于 0（这也可以）
  • 所有都必须小于 0.30（这是行不通的）

一、测试数据：

n <- 6
r <- array(rnorm(n^2, mean=1), c(100,n)) # the 'returns'
colnames(r) <- c("A", "B", "C", "D", "E", "F")

我的功能：

eff.frontier <- function (returns, short="no", max.allocation=NULL, risk.premium.up=.5, risk.increment=.005){     
  covariance <- cov(returns)
  n <- ncol(covariance)
  Amat <- matrix (1, nrow=n)
  bvec <- 1
  meq <- 1
  if(short=="no"){
    Amat <- cbind(1, diag(n))
    bvec <- c(bvec, rep(0, n))
  }
  if(!is.null(max.allocation)){
    if(max.allocation > 1 | max.allocation <0){
      stop("max.allocation must be greater than 0 and less than 1")
    }
    Amat <- cbind(Amat, -1*diag(n))
    # This seems to be the issue area....
    bvec <- c(bvec, rep(-max.allocation, n))
    print(bvec)
  }
  loops <- risk.premium.up / risk.increment + 1
  loop <- 1
  #Initialize a matrix to contain allocation and statistics
  eff <- matrix(nrow=loops, ncol=n+3)
  # Now I need to give the matrix column names
  colnames(eff) <- c(colnames(returns), "Std.Dev", "Exp.Return", "sharpe")
  # Loop through the quadratic program solver
  for (i in seq(from=0, to=risk.premium.up, by=risk.increment)){
    dvec <- colMeans(returns) * i
    sol <- solve.QP(covariance, dvec=dvec, Amat=Amat, bvec=bvec, meq=meq)
    eff[loop,"Std.Dev"] <- sqrt(sum(sol$solution *colSums((covariance * sol$solution))))
    eff[loop,"Exp.Return"] <- as.numeric(sol$solution %*% colMeans(returns))
    eff[loop,"sharpe"] <- eff[loop,"Exp.Return"] / eff[loop,"Std.Dev"]
    eff[loop,1:n] <- sol$solution
    loop <- loop+1
  }
  return(as.data.frame(eff))
}

当我不尝试设置分配最大值时，这可以正常工作。但无论我尝试将什么设置为 max.allocation，它总是会失败。

我是否错误地设置了 Amat 或 bvec？

这是错误：

> eff <- eff.frontier(returns=r, short="yes", risk.premium.up=.5, risk.increment=.001, max.allocation=.15)
     Error in solve.QP(covariance, dvec = dvec, Amat = Amat, bvec = bvec, meq = meq) : 
  constraints are inconsistent, no solution!

这是阿马特：

   [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]    1   -1    0    0    0    0    0
[2,]    1    0   -1    0    0    0    0
[3,]    1    0    0   -1    0    0    0
[4,]    1    0    0    0   -1    0    0
[5,]    1    0    0    0    0   -1    0
[6,]    1    0    0    0    0    0   -1

这是 bvec：

[1]  1.00 -0.15 -0.15 -0.15 -0.15 -0.15 -0.15

这些怎么不一致？？

谢谢，

安德鲁

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