Re: Summary, was Re: Every time ..., was Re: General formula

[Tony Bartoletti <azb@xxxxxxxx>]

At 12:32 AM 6/5/99 -0700, you wrote:

>Thus, the first formula is
>
>  To = collective attribute lifetime = x years
>  T1 = T2 = ... = T100 = individual lifetime = 1 year
>
>and, to begin, suppose there is no collective yet:
>
>   1/T = 1/T1 + ... + 1/T100 => T = 1/100 year
>
>but, as redundancy is increased to the next notch, n individuals
>are lost to the collective. Thus, the second formula has less attributes
>to sum up and we obtain the partial expression:
>
>  1/T = 1/To + 1/T1 + ... + 1/T(100 - n) => T =  x/((100-n)*x +1) years
>
>and, as redundancy is further increased, there is a time when all
>individuals are lost to the collective and n = 100, so that we only
>have the collective to take into account:
>
>  1/T = 1/To = x years
>
>which is *also* the result we obtain by making n =100 in the partial
>expression for n above:
>
> T =  x/((100-n)*x +1) years => x/((100-100)*x +1) = x years
>

The given formula "works" (assuming all component attributes have identical
lifetimes.)

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.

  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

  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.

  Comparing the result to the one that assumed no redundancy, I summarize

	1/TC = 1/TA + 1/TB - SUM(1/Tci) for ci in INTERSECT(A,B)

	     = 1/TA + 1/TB - [ 1/Tb1 + 1/Tb2 ]

	     < 1/TA + 1/TB

  Thus   TC > 1/( 1/TA + 1/TB )

Now this result is not really different than the one you give above, where
increasing redundancy gradually reduces the population of elements.  But it
demonstrates the difficulty in practice.

Since the individual components may have differing lifetimes, it may matter
greatly, exactly which components are removed from the intersection.  Saying
"two out of six components were redundant, so use only four" would not suffice.

And, most importantly, in those (many) cases where no one has the time or
know-how to correctly identify the component attributes of A and B, but
rather applies historical "lifetimes" exhibited by A-like and B-like
attributes, there is no opportunity to adjust proactively the formula

	1/TC = 1/TA + 1/TB

Instead, one would just apply it, and soon discover that observed TC
does not agree with its predicted value (and thus surmise that there
must be redundancy at work somewhere.)

Comments?

___tony___

Tony Bartoletti                                             LL
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