優化R中的maxDrawdown（）

Question

所以我有一個交易策略，平均有一個大於20％的最大回撤。我想找到一種方法來優化maxDrawdown()。不過考慮一下，我發現了很好的文章，通過DEoptim()來展示如何通過這個文章，但只展示了投資組合優化的例子。我不知道這是否可行？

#Choose the Adjusted Close of a Symbol 
stock <- Ad(TNA) 

# I want to create a table with all possible combinations from the ranges below 
i = c(3:45) 
k = c(3:45) 
j = c(3:45) 

# stores possible combinations into z 
z <- expand.grid(i,k,j) 
z <- z[z[,1]<z[,2], ] 
colnames(z)<- c("one","two","three")    
row.names(z)<- c(paste(z[,1],z[,2],z[,3],sep=",")) 



# Function that will be used 

getStratRet <- function(nFast, nSlow, nSig, stock, stockret) { 
x <- MACD((stock), nFast=nFast, nSlow=nSlow, nSig=nSig, maType="EMA") 
x <- na.omit(x) 
sig <- Lag(ifelse((x$macd <= x$signal),-1, 0)) + Lag(ifelse((x$macd >= x$signal),1, 0)) 
return(na.omit(stockret * sig)) 
} 

# time elapsed for 3:45 combinations: 479.886 seconds, 
system.time(
Returns <- do.call(merge, mapply(FUN = getStratRet, nFast = z[,1], nSlow = z[,2], nSig  
= z[,3], MoreArgs = list(stock = stock, stockret = stockret), SIMPLIFY = TRUE)) 
) 


#My strategy returns a matrix of returns 
View(Returns[1:10,1:5]) 

row.names  3,4,3   3,5,3   4,5,3   3,6,3   4,6,3 
2011-01-11 -0.0035308990  NA    NA    NA    NA 
2011-01-12 0.0090226176 0.0090226176 0.0090226176  NA    NA 
2011-01-13 -0.0016647249 -0.0016647249 -0.0016647249 -0.0016647249 -0.0016647249 
2011-01-14 0.0072214466 0.0072214466 0.0072214466 0.0072214466 0.0072214466 
2011-01-18 0.0017353225 0.0017353225 0.0017353225 0.0017353225 0.0017353225 
2011-01-19 -0.0098735504 -0.0098735504 -0.0098735504 -0.0098735504 -0.0098735504 
2011-01-20 0.0013350023 0.0013350023 0.0013350023 0.0013350023 0.0013350023 
2011-01-21 -0.0022517836 -0.0022517836 -0.0022517836 -0.0022517836 -0.0022517836 
2011-01-24 -0.0056487939 -0.0056487939 -0.0056487939 -0.0056487939 -0.0056487939 
2011-01-25 0.0005796862 0.0005796862 0.0005796862 0.0005796862 0.0005796862 

我已經創建了一個功能通過DEoptim()調用，以優化此MAXDD函數計算maxDrawdown（從PerformanceAnalytics）爲每個返回。

# MAX DrawDown 
MAXDD <- function(ret) { 
ret <- na.omit(ret) 
maxdd<- maxDrawdown(ret) 
return (maxdd) 
} 

# MAX DRAWDOWN 
system.time(
MaxDraw <- sapply(Returns, FUN = MAXDD) 
) 

然後我用DEoptim()的，我希望儘量減少功能：MAXDD

library(DEoptim) 
lower <- c(-0.10, 0) # I think this sets the max drawdown that will be acceptable 
upper<- -lower  # I think this will set what the maximum return acceptable 
res <- DEoptim(MAXDD,lower,upper, control=list(NP=2000, itermax=10)) 
# I set the iteration to 5, to print them here: 
Iteration: 1 bestvalit: -0.000000 bestmemit: 0.051143 0.000000 
Iteration: 2 bestvalit: -0.000000 bestmemit: 0.034017 0.000000 
Iteration: 3 bestvalit: -0.000000 bestmemit: 0.020190 0.000000 
Iteration: 4 bestvalit: -0.000000 bestmemit: 0.028910 0.000000 
Iteration: 5 bestvalit: -0.000000 bestmemit: 0.038250 0.000000 

這是我堅持，因爲我不知道我是否做正確嗎？我不知道如何翻譯這個以用於我的目標。

Answer 1

由於您的代碼已經計算了所有策略的回報，因此您可以計算所有策略的回撤併取最佳回報。 這是一個詳盡的搜索：你不需要DEoptim。

z[ which.min(apply(Returns, 2, maxDrawdown)), ] 

要使用DEoptim，您需要提供一個功能 ，其參數是你的戰略的參數， 和返回縮編。 lower和upper參數是您的策略參數的下限和上限 。

f <- function(u) 
    maxDrawdown(getStratRet( 
    u[1], u[2], u[3], stock=stock, stockret=stockret 
)) 
r <- DEoptim(f, c(3,3,3), c(45,45,45)) 
r$optim$bestmem 
#  par1  par2  par3 
# 3.486908 40.528064 13.813589