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Re: Summary, was Re: Every time ..., was Re: General formula
At 01:17 AM 6/8/99 -0700, Ed Gerck wrote:
>
>
>Tony Bartoletti wrote:
>
>> The given formula "works" (assuming all component attributes have identical
>> lifetimes.)
>
>Tony:
>
>I can't understand the "works" -- do you find a mistake in the results?
>
>> I wanted to explore the more general case, where the component lifetimes
>> are unequal, and to help define "degrees of redundancy".
>>
>> A simple (?) example:
>>
>> Consider a certificate C over two attributes A and B with lifetimes given
>> by TA and TB, respectively.
>>
>> Assuming no redundancy, we allow 1/TC = 1/TA + 1/TB.
>
>You mean, there is no 100% redundancy? Because redundancy is NOT
>a problem in the equation -- as I showed in my e-mail anterior and
>you recognized when you wrote above "works" ;-))))
>
>Of course, you cannot just *repeat* the *same* attribute two times
>and expect that you have two attributes, in fact you would just one
>attribute, no?
At this point, we cannot tell from inspection whether A and B have any
components which are themselves 100% redundant. In any case, A and B
are not 100% redundant.
>> Suppose, upon closer inspection, we agree that attributes A and B are
>> themselves formed of components. Specifically:
>>
>> A = a1 and a2 and a3 (respective lifetimes Ta1, Ta2, Ta3)
>> B = b1 and b2 and b3 (respective lifetimes Tb1, Tb2, Tb3)
>>
>> Note: Attribute A is valid IFF all of a1, a2, a3 are valid, etc.
>>
>> Continuing under assumptions of no redundancy, we still find
>>
>> 1/TC = 1/TA + 1/TB
>> = 1/Ta1 + 1/Ta2 + 1/Ta3 + 1/Tb1 + 1/Tb2 + 1/Tb3
>
>You "forgot" to impose the conditions, derived directly by your
>hypothesis:
>
>1/TA = 1/Ta1 + 1/Ta2 + 1/Ta3
>1/TB = 1/Tb1 + 1/Tb2 + 1/Tb3
If I did not impose these conditions, how could I have made the
direct substitution in the equation for 1/TC above?
>> Suppose a "final investigation" reveals that the components a1 and a2 of A
>> are precisely the components b1 and b2 of B. Thus the certificate C
>> certifies exactly the components a1, a2, a3, b3.
>
>Your affirmation does not make mathematical sense, Tony. What do you mean by
>
>"the components a1 and a2 of A are precisely the components b1 and b2 of B."
>
>Do you mean they are equal? So what? You affirmed before that:
>
> "Assuming no redundancy, we allow 1/TC = 1/TA + 1/TB"
And I believe you maintain, in essense, that unless A and B are themselves
exactly redundant, "redundancy is not a problem in the equation."
Hence, I force A and B to be "partially redundant", and intentionally
drop the "Assuming no redundancy" to illustrate.
I mean that a1 is exactly redundant to b1, and a2 to b2. They ARE the
same attributes, just repeated. Although this does not make A and B
completely redundant, it DOES violate the original assumption, which
was exactly the point. The formula 1/TC = 1/TA + 1/TB will not provide
the right answer, even though A and B are "different" attributes.
Unless A and B reveal their structure to us, so that we may identify
and eliminate duplicated components, we cannot apply such a formula.
And if the duplicated components [ai,bi] for some of the i, all exhibit
different expected lifetimes, there is no opportunity for the "grouping"
you mentioned, which says nothing more than SUM[C] to n terms = n*C for
a constant C. They will be "groups of 1", and we are reduced to the
equation I gave:
1/TC = 1/TA + 1/TB - ( SUM[1/TCi] for ci in INTERSECT(A,B) )
The problem is in identifying this intersection, and not simply treating
attributes A and B as atoms.
___tony___
Tony Bartoletti LL
IOWA Center LL LL
Lawrence Livermore National Laboratory LL LL LL
PO Box 808, L - 303 LL LL LL
Livermore, CA 94551-9900 LL LL LLLLLLLL
phone: 925-422-3881 fax: 925-423-8002 LL LLLLLLLL
email: azb@llnl.gov LLLLLLLL