Why do results from optim() depend on initial valu

问题:

In R, I am using the function optim() to find the minimum of an objective function of two variables. The real objective functions I'm working with are quite complex, so I tried to familiarize myself with the a simpler objective function. The simplest way to run optim() is optim(par,function) where par is a vector of initial values for the algorithm. I find that the answer I get depends heavily on the initial values I input. However, the function I used is so simple, I'm worried that I am misunderstanding either the input or output of optim().

The objective function I am using is:

f <- function(x){
    abs(x[1])*abs(x[2])
}

When I run optim(c(-1.2,1),f) and optim(c(-1.2,10),f) I get drastically different output for the optimal arguments (par) and the minimum (value). Does anyone have an idea why this would be so?

回答1:

In this case your objective function has infinitely many optimal points (not necessarily just different local maxima). Anywhere one of the parameters is zero 0 is just as good as any other point where a parameter is near 0. I'm not sure if you were expecting (0,0), but (0,34) has the same value and can also be considered optimal.

回答2:

An associated function is:

g <- function(x, y) abs(x)*abs(y)

We can visualize the levels of the graph with contour and plot the points given:

A reasonable field given your initial and final conditions (noted from running optim):

x <- seq(-1.5, 0, by=.1)
y <- seq(0, 11, by=1)

A matrix of the values in g:

m <- outer(x, y, g)

Plot the results, including the results of optim. Note that the values at x==0 or y==0 are optimal.

contour(x, y, m)
o1 <- optim(c(-1.2,1),f)$par
o2 <- optim(c(-1.2,10),f)$par
segments(-1.2, 1, o1[1], o1[2], col='red')
segments(-1.2, 10, o2[1], o2[2], col='red')
# Add zero lines as contour does not draw them
segments(0, 11, 0, 0)
segments(-1.5, 0, 0, 0)

This shows a straight line from the initial condition (left side) to the zero of the function (right side). Note that the optimization does not follow a straight line, but this shows that it is reasonable that quite different results will be achieved.