Hi Ravi,

Thank you for your reply and please excuse my late response.

Plugging w2’ = k/w1’ from (A) into (B) yields

(C) f(w1') = (w1-w1’)^2 + (w2-k/w1’)^2

The partial derivative wrt w1' is

(D) df(w1')/ dw1' = -2(w1-w1’) + 2(w2-k/w1’)*k/(w1')^2

in order for this to be a minimum the f.o.c. df(w1')/ dw1' = 0 and the
s.o.c. d^2f(w1')/ d(w1')^2 >0 must be satisfied.

I follow you on substituting the constraint into the distance from the
iterate to (w1', w2') and then minimizing this distance, but I'm not quite
sure how to turn this into a projection function. Any suggestions?

Regards,

Kristian

2011/9/8 Ravi Varadhan <rvaradhan@jhmi.edu>

>  Hi Kristian,****
>
> ** **
>
> The idea behind projection is that you take an iterate that violates the
> constraints and project it onto a point such that it is the nearest point
> that satisfies the constraints.  ****
>
> ** **
>
> Suppose you have an iterate (w1, w4) that does not satisfy the constraint
> that w1 * w4 != (1 + eps)/2.  Our goal is to find a (w1’, w2’), given (w1,
> w2), such that****
>
> ** **
>
> **(A)   **w1’ * w2’ = (1+eps)/2 = k****
>
> **(B)   **(w1-w1’)^2 + (w2-w2’)^2 is minimum.  ****
>
> ** **
>
> This is quite easy to solve.  We know (w1, w2).  You plug in w2’ = k/w1’
> from (A) into (B) and minimize the function of w1’.  This is a simple
> calculus exercise, and I will leave this as a homework problem for you to
> solve!****
>
> ** **
>
> Best,****
>
> Ravi.****
>
> ** **
>
> -------------------------------------------------------****
>
> Ravi Varadhan, Ph.D.****
>
> Assistant Professor,****
>
> Division of Geriatric Medicine and Gerontology School of Medicine Johns
> Hopkins University****
>
> ** **
>
> Ph. (410) 502-2619****
>
> email: rvaradhan@jhmi.edu****
>
> ** **
>

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