Re[2]: Summary, was Re: Every time ..., was Re: General form

[veikko.punkka@xxxxxxxxxxx]

I think we need a simple example to clarify the question:

Suppose I have two different business cards, one of them (A) says:

Veikko Punkka
Symbian Ltd.

while the other one (B) says:

Veikko Punkka
Software Engineer

Now, each of these business cards has a lifetime that is given by the now 
well known Gerck formula:

1/TA = 1/Ta1 + 1/Ta2 (1)

1/TB = 1/Tb1 + 1/Tb2 (2)

Now, suppose I print another business card that has card A on one side and 
card B on the other. Naively thinking you might expect that the lifetime of 
the new card (C) would be given by the formula (after all A and B are not 
100% redundant)

1/TC = 1/TA + 1/TB (3)

but this is not true since a1 and b1 are 100% redundant and therefore

1/TC = 1/Ta1 + 1/Ta2 + 1/Tb2 (4)

Now, my interpretation of the question is: How can we be sure we are not 
making false conclusions like (3)?

Cheers,
    Veikko Punkka


______________________________ Reply Separator _________________________________
Subject: Re: Summary, was Re: Every time ..., was Re: General formula
Author:  Ed Gerck <egerck@mcg.org.br> at symb-internet
Date:    08/06/99 01:17




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?

>   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

>   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"

so we know we are not dealing here with the *same* attributes that were
just repeated, like:

Attribute1 = 'tony'
Attribute2 = 'tony'

So,  your :"results" do not apply. But, it is a simple confusion. Calculate
again, taking into account that:

1/TA = 1/Ta1 + 1/Ta2 + 1/Ta3
1/TB = 1/Tb1 + 1/Tb2 + 1/Tb3

and you will see that it "works".

BTW, the case you want to deal with above, by extending my example
of your suggestion to the case where the lifetimes are different can be
dealt with by regrouping a different number of terms of like lifetimes in
the equation and then simply assigning them to a different set of attributes
of unlike lifetimes. The results will remain the same.

Cheers,

Ed Gerck