R中最优/ Synth中的卷积码

Question

我一直在使用Synth包(它使用了优化来进行优化)，当Synth不收敛时，我无法找到它是否会给出一个不收敛错误的答案。我可以在优化的帮助文件中看到，默认情况下，它不会保存失败(即那些不收敛的答案)，因此我猜测，在默认情况下，它不会将这些故障传递给Synth；但是，我想确定我得到的答案已经收敛。是否有一种方法可以通过卷积()设置或其他方法来确保这一点？

Answer 1

我看了一下代码，它没有为save.failures设置一个控制值，所以它应该使用optimx的默认行为，正如您已经知道的那样，这个行为是假的。下面是在optimx代码中对synth()的相关调用。

rgV.optim.1 <- optimx(par = SV1, fn = fn.V, gr = NULL, 
    hess = NULL, method = optimxmethod, itnmax = NULL, 
    hessian = FALSE, control = list(kkt = FALSE, starttests = FALSE, 
        dowarn = FALSE, all.methods = all.methods), X0.scaled = X0.scaled, 
    X1.scaled = X1.scaled, Z0 = Z0, Z1 = Z1, quadopt = quadopt, 
    margin.ipop = Margin.ipop, sigf.ipop = Sigf.ipop, 
    bound.ipop = Bound.ipop)

....

rgV.optim.2 <- optimx(par = SV2, fn = fn.V, gr = NULL, 
    hess = NULL, method = optimxmethod, itnmax = NULL, 
    hessian = FALSE, control = list(kkt = FALSE, 
      starttests = FALSE, dowarn = FALSE, all.methods = all.methods), 
    X0.scaled = X0.scaled, X1.scaled = X1.scaled, 
    Z0 = Z0, Z1 = Z1, quadopt = quadopt, margin.ipop = Margin.ipop, 
    sigf.ipop = Sigf.ipop, bound.ipop = Bound.ipop)