Question

我使用nleqslv包来解决两个非线性方程组的系统

library(nleqslv)
read.table("C:\\Users\\Livia\\Desktop\\firm.txt",header=TRUE)
firm.txt<-"Q K r X.E t E0
"
firm<-read.table(textConnection(firm.txt),header=TRUE,stringsAsFactors=FALSE)
firm
#Loop with solver for 9 dates:
for(i in 1:9){
K<-firm[i,2]
r<-firm[i,3]
X.E<-firm[i,4]
t<-firm[i,5]
E0<-firm[i,6]
BS<-function(x){
f<-rep(NA,length(x))
f[1]<-(x[1]*pnorm((log(x[1]/K)+(r+(x[2]^2/2))*t)/(x[2]*sqrt(t)))-exp(-r*t)*K*pnorm((log(x[1]/K)+(r+(x[2]^2/2))*t)/(x[2]*sqrt(t))-x[2]*sqrt(t)))-E0
f[2]<-X.E*E0-pnorm((log(x[1]/K)+(r+(x[2]^2/2))*t)/(x[2]*sqrt(t)))*x[2]*x[1]
f
}
p0<-c((E0+K),X.E*E0/(E0+K))
print(p0)
ans<-nleqslv(p0,BS,control=list(trace=1,allowSingular=TRUE))
print(ans)
}

我真的很想了解它背后的逻辑。我有两个未知数，我根据

设置起始值

p0<-c((E0+K),X.E*E0/(E0+K))

迭代过程如何工作？我已经尝试从CRAN软件包“nleqslv”3.0 .pdf中获取它但是没有成功。

Answer 1

请参阅我上面的评论以获取警告。

话虽如此，这就是你运行这个功能时发生的事情：

> nleqslv
function (x, fn, jac = NULL, ..., method = c("Broyden", "Newton"), 
    global = c("dbldog", "pwldog", "cline", "qline", "gline", 
        "hook", "none"), xscalm = c("fixed", "auto"), jacobian = FALSE, 
    control = list()) 
{
    fn1 <- function(par) fn(par, ...)
    if (is.null(jac)) {
        jacfunc <- NULL
    }
    else {
        if (!is.function(jac)) 
            stop("argument 'jac' is not a function!")
        jacfunc <- function(par) jac(par, ...)
    }
    method <- match.arg(method)
    global <- match.arg(global)
    xscalm <- match.arg(xscalm)
    con <- list(ftol = 1e-08, xtol = 1e-08, btol = 0.001, stepmax = -1, 
        delta = "newton", sigma = 0.5, scalex = rep(1, length(x)), 
        maxit = 150, trace = 0, chkjac = FALSE, cndtol = 1e-12, 
        allowSingular = FALSE, dsub = -1L, dsuper = -1L)
    if (global == "none") 
        con$maxit = 20
    if (length(control)) {
        namc <- names(control)
        if (!is.list(control) || is.null(namc)) 
            stop("'control' argument must be a named list")
        if (!all(namc %in% names(con))) 
            stop("unknown names in control: ", paste(sQuote(namc[!(namc %in% 
                names(con))]), collapse = ", "))
        con[namc] <- control
    }
    tmp <- con[["delta"]]
    if (is.character(tmp)) {
        k <- match(tolower(tmp), c("cauchy", "newton"))
        con[["delta"]] <- as.numeric(-k)
    }
    else if (!is.numeric(tmp)) 
        stop("control[\"delta\"] should be either character or numeric")
    on.exit(.C("deactivatenleq", PACKAGE = "nleqslv"))
    out <- .Call("nleqslv", x, fn1, jacfunc, method, global, 
        xscalm, jacobian, con, new.env(), PACKAGE = "nleqslv")
    out
}
<environment: namespace:nleqslv>

如果你想看看其他功能是如何工作的，你可以用同样的方式探索它们（输入功能名称并按下回车键而不添加()）。

如果您想深入了解，请参阅完整的源代码：

https://cran.r-project.org/src/contrib/nleqslv_3.0.1.tar.gz

在这里：

https://github.com/cran/nleqslv

nleqslv理解迭代过程的逻辑

1 个答案: