==================================================================
The gateway between this list and the sci.stat.edu newsgroup will
be disabled on June 9. This list will be discontinued on June 21.
Subscribe to the new list EDSTAT-L at Penn State using the web
interface at http://lists.psu.edu/archives/edstat-l.html.
==================================================================
.
In article <ITSnetNOTcom/[EMAIL PROTECTED]>,
Virgil <ITSnetNOTcom/[EMAIL PROTECTED]> wrote:
>In article <[EMAIL PROTECTED]>,
> [EMAIL PROTECTED] (ZHANG Yan) wrote:
>> Thanks for the comments. And sorry for the inclear question. I will
>> re-state the problem in the following.
>> Suppose that d is positive integer, 1 <= d <= D. A function is defined
>> as follows C(d)=C1(d)+C2(d).
>> And for any d in 1 to D-1, we have
>> C1(d) >= C1(d+1)
>> C1(d) <= C2(d+1)
(I assume this was supposed to be C2(d) <= C2(d+1), since in the previous
article you said C2 is increasing)
>> Then, I have to determine that C(d) has minimum value with an optimal
>> value d_{op}, and to find d_{op}. I am wondering whether these
>> conditions are sufficient and whether I should think about or add more
>> conditions?
>Since there are only finitely many, D, values of C(d), there must be a
>minimum value, but I do not see any better way of finding it that by
>testing all values of C(d).
Indeed. The condition C(d) = C1(d) + C2(d) with C1(d) >= C1(d+1)
and C2(d) <= C2(d+1) tells us nothing at all, since any real function
C(d) on the natural numbers can be written this way. Namely, you could
define C1(1) = C(1), C2(1) = 0,
C2(d+1) = max(C2(d), C(d+1) - C1(d))
C1(d+1) = C(d+1) - C2(d+1) = min(C1(d), C(d+1) - C2(d))
Robert Israel [EMAIL PROTECTED]
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
