如何在约束优化中将参数'sum设置为1

Question

这是代码(如果它太久了,我很抱歉,但这是我的第一个例子); 我正在使用CreditMetricsA. Wittmann和DEoptim求解器包中的CVaR示例进行优化:

library(CreditMetrics)
library(DEoptim)

N <- 3
n <- 100000
r <- 0.003
ead <- rep(1/N,N)
rc <- c("AAA", "AA", "A", "BBB", "BB", "B", "CCC", "D")
lgd <- 0.99
rating <- c("BBB", "AA", "B")   
firmnames <- c("firm 1", "firm 2", "firm 3")
alpha <- 0.99

# correlation matrix
rho <- matrix(c(  1, 0.4, 0.6,
                  0.4,   1, 0.5,
                  0.6, 0.5,   1), 3, 3, dimnames = list(firmnames, firmnames),
              byrow = TRUE)

# one year empirical migration matrix from standard&poors website
rc <- c("AAA", "AA", "A", "BBB", "BB", "B", "CCC", "D")
M <- matrix(c(90.81,  8.33,  0.68,  0.06,  0.08,  0.02,  0.01,   0.01,
              0.70, 90.65,  7.79,  0.64,  0.06,  0.13,  0.02,   0.01,
              0.09,  2.27, 91.05,  5.52,  0.74,  0.26,  0.01,   0.06,
              0.02,  0.33,  5.95, 85.93,  5.30,  1.17,  1.12,   0.18,
              0.03,  0.14,  0.67,  7.73, 80.53,  8.84,  1.00,   1.06,
              0.01,  0.11,  0.24,  0.43,  6.48, 83.46,  4.07,   5.20,
              0.21,     0,  0.22,  1.30,  2.38, 11.24, 64.86,  19.79,
              0,     0,     0,     0,     0,     0,     0, 100
)/100, 8, 8, dimnames = list(rc, rc), byrow = TRUE)

cm.CVaR(M, lgd, ead, N, n, r, rho, alpha, rating)

y <- cm.cs(M, lgd)[which(names(cm.cs(M, lgd)) == rating)]

现在我写我的功能......

fun <- function(w) {
  # ... 
  - (t(w) %*% y - r) / cm.CVaR(M, lgd, ead = w, N, n, r, 
                           rho, alpha, rating)
}

......我想优化它:

DEoptim(fn = fun, lower = rep(0, N), upper = rep(1, N), 
        control = DEoptim.control())

你能告诉我什么才是我要插入# ...,使sum(w) = 1优化过程中？

下面我根据flodel的提示向您展示优化结果:

# The first trick is to include B as large number to force the algorithm to put sum(w) = 1

fun <- function(w) {
  - (t(w) %*% y - r) / cm.CVaR(M, lgd, ead = w, N, n, r, rho, alpha, rating) + 
    abs(10000 * (sum(w) - 1))
}

DEoptim(fn = fun, lower = rep(0, N), upper = rep(1, N), 
        control = DEoptim.control())

$optim$bestval
[1] -0.05326055

$optim$bestmem
par1        par2        par3 
0.005046258 0.000201286 0.994752456

parsB <- c(0.005046258, 0.000201286, 0.994752456)

> fun(parsB)
            [,1]
[1,] -0.05326089

...和...

正如你所看到的,第一个技巧效果更好,因为他发现的结果小于第二个结果.不幸的是,他需要更长时间.

# The second trick needs you use w <- w / sum(w) in the function itself

fun <- function(w) {
  w <- w / sum(w)
  - (t(w) %*% y - r) / cm.CVaR(M, lgd, ead = w, N, n, r, rho, alpha, rating) #+ 
    #abs(10000 * (sum(w) - 1))
}

DEoptim(fn = fun, lower = rep(0, N), upper = rep(1, N), 
        control = DEoptim.control())

$optim$bestval
[1] -0.0532794

$optim$bestmem
par1         par2         par3 
1.306302e-15 2.586823e-15 9.307001e-01

parsC <- c(1.306302e-15, 2.586823e-15, 9.307001e-01)
parC <- parsC / sum(parsC)

> fun(parC)
           [,1]
[1,] -0.0532794

任何意见？

我是否应该增加迭代次数,因为"太随机"的待优化函数？

Answer 1

尝试:

w <- w / sum(w)

如果DEoptim为您提供最佳解决方案w*,sum(w*) != 1那么 w*/sum(w*)应该是您的最佳解决方案.

另一种方法是解决所有变量而不是一个.我们知道1 - sum(w)函数体中最后一个变量的值必须如此:

w <- c(w, 1-sum(w))

并对返回的最佳解决方案执行相同的操作DEoptim:w* <- c(w*, 1-sum(w*))

这两种解决方案都要求您将问题重新制定为无约束(不计入变量边界)优化,以便DEoptim可以使用; 这迫使你在外面做一些额外的工作DEoptim来恢复原始问题的解决方案.

在回答您的意见,如果你想DEoptim给你正确的答案马上(即无需后转化),你也可以尝试包括惩罚成本,以你的目标函数:例如加B * abs(sum(w)-1),其中B一些武断大量的人sum(w)将被迫1.