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我有我的 Black-Scholes 函数和用于看涨期权的二等分模型，其中包含来自 CSV 的数据。它似乎陷入了内循环，因为它保持在公差之上。我的 Black-Scholes 确实计算准确，我使用的是平均出价和要价，而不是期权的实际价格。经过几个小时的工作，也许我只是错过了一些明显的东西。

CSV 的链接在这里：http ://s000.tinyupload.com/?file_id=06213890949979926112

########################################################################
#Black-Scholes-Merton Call
bsmCall <- function(S, K, M, sig, r) {
  yrTime=(M/252)
  d1 <- (log(S/K)+(r+(sig^2/2))*(yrTime))/(sig*(sqrt(yrTime)))
  d2 <- d1-sig*(sqrt(yrTime))
  C <- (S*(pnorm(d1)))-((pnorm(d2))*K*(exp(-r*yrTime)))
  return(C)
}
########################################################################

myData = read.csv("09-26-16.csv", stringsAsFactors=FALSE)    #DATA
myData <- myData[,2:24]   #omit first column

####### start bisection method of CALLS and put IV in database #######
i <- 1    # reset counter
tol <- 0.000001   #tolerance

while(i <= nrow(myData)) {
  if((myData[i,5] != 0) & (myData[i,6] != 0)) {
    volLower <- .0001    #will need to reset with each iteration
    volUpper <- 1         #will need to reset with each iteration
    volMid <- (volLower + volUpper) / 2   #will need to reset with each iteration

    while(abs(bsmCall(as.numeric(as.character(myData[i,17])),as.numeric(as.character(myData[i,1])),as.numeric(as.character(myData[i,22])),volMid,as.numeric(as.character(myData[i,23])))-(as.numeric(as.character(myData[i,5])))) >= tol) {
      if((bsmCall(as.numeric(as.character(myData[i,17])),as.numeric(as.character(myData[i,1])),as.numeric(as.character(myData[i,22])),volMid,as.numeric(as.character(myData[i,23])))-(as.numeric(as.character(myData[i,5])))) < 0) {
        volLower <- volMid
        volMid <- (volUpper + volMid)/2
      } else {
        volUpper <- volMid
        volMid <- (volLower + volMid)/2
      }
    }
    myData[i,8] <- volMid
  } else { myData[i,8] <- 0 }
  i=i+1
}
4

2 回答 2

1

问题在这里：

while(abs(bsmCall(as.numeric(as.character(myData[i,17])),
                  as.numeric(as.character(myData[i,1])),
                  as.numeric(as.character(myData[i,22])),
                  volMid,
                  as.numeric(as.character(myData[i,23])))-(as.numeric(as.character(myData[i,5])))) >= tol)

您while在一个条件下使用循环，如果为真，则始终为真。这是一个无限循环。在您的第一行数据上遇到此问题。

如何修复此错误取决于您的用例，但如果您只是更改while为，if您将立即看到循环完成。

你问过二分法。包中有一些，这里有另一个：

bisect <- function(fn, lower, upper, tol=1.e-07, ...) {
f.lo <- fn(lower, ...)
f.hi <- fn(upper, ...)
feval <- 2

if (f.lo * f.hi > 0) stop("Root is not bracketed in the specified interval
\n")
chg <- upper - lower

while (abs(chg) > tol) {
        x.new <- (lower + upper) / 2
        f.new <- fn(x.new, ...)
        if (abs(f.new) <= tol) break
        if (f.lo * f.new < 0) upper <- x.new
        if (f.hi * f.new < 0) lower <- x.new
        chg <- upper - lower
        feval <- feval + 1
}
list(x = x.new, value = f.new, fevals=feval)
}

# An example
fn1 <- function(x, a) {
exp(-x) - a*x
}

bisect(fn1, 0, 2, a=1)

bisect(fn1, 0, 2, a=2)

递归版本：

bisectMatt <- function(fn, lo, hi, tol = 1e-7, ...) {

    flo <- fn(lo, ...)
    fhi <- fn(hi, ...)

    if(flo * fhi > 0)
        stop("root is not bracketed by lo and hi")

    mid <- (lo + hi) / 2
    fmid <- fn(mid, ...)
    if(abs(fmid) <= tol || abs(hi-lo) <= tol)
        return(mid)


    if(fmid * fhi > 0)
        return(bisectMatt(fn, lo, mid, tol, ...))

    return(bisectMatt(fn, mid, hi, tol, ...))
}
于 2016-09-27T03:47:23.773 回答
0

天哪，这是我迄今为止的第三次编辑......

让我们在何时重建while循环i=1并打印-在每次迭代后更新的条件volMid的唯一部分while

i <- 1
volLower <- .0001    #will need to reset with each iteration
volUpper <- 1         #will need to reset with each iteration
volMid <- (volLower + volUpper) / 2   #will need to reset with each iteration

j <- 1
while(abs(bsmCall(myData[i,17], myData[i,1], myData[i,22],volMid,myData[i,23])-myData[i,5]) >= tol & j < 30) {
  if(bsmCall(myData[i,17], myData[i,1], myData[i,22],volMid,myData[i,23])-myData[i,5] < 0) {
volLower <- volMid
volMid <- (volUpper + volMid)/2
  } else {
    print("pos")
    volUpper <- volMid
    volMid <- (volLower + volMid)/2
  }
  j <- j + 1
  print(volMid)
}

结果：

#[1] 0.750025
#[1] 0.8750125
#[1] 0.9375062
#[1] 0.9687531
#[1] 0.9843766
#[1] 0.9921883
#[1] 0.9960941
#[1] 0.9980471
#[1] 0.9990235
#[1] 0.9995118
#[1] 0.9997559
#[1] 0.9998779
#[1] 0.999939
#[1] 0.9999695
#[1] 0.9999847
#[1] 0.9999924
#[1] 0.9999962
#[1] 0.9999981
#[1] 0.999999
#[1] 0.9999995
#[1] 0.9999998
#[1] 0.9999999
#[1] 0.9999999
#[1] 1
#[1] 1
#[1] 1
#[1] 1
#[1] 1
#[1] 1

volMid1在不到 30 次迭代后收敛，从那里开始，它就卡住了。

于 2016-09-27T03:51:39.363 回答