From: Anders Sandberg (nv91-asa@nada.kth.se)
Date: Thu Aug 14 1997 - 11:01:19 MDT
On Wed, 13 Aug 1997, Darren Reynolds wrote:
> Suppose you have two designs for a system. Both have the same amount of
> energy. System 1 duplicates all data, whereas system 2 does not. Clearly,
> system 2 will have twice as much capacity for data, although not
> necessarily twice as much usefulness. (Both systems would prioritize
> resource usage to store the most valuable data first, and therefore system
> 2's second half would have less value its first, which should be the same
> as system 1's total.)
>
> I'm no statistician, but you can figure it out mathematically. You want the
> system with the greatest total value over the systems' anticipated
> lifespan. The value would be a sum of the values of each datum stored
> successfully at each instant in time. If you can figure out the value of
> each datum, and the probability of it being destroyed at a given moment,
> then you can integrate to get the total value of the system.
OK, lets start by assuming all data is equally valuable (to make
things simple). The probability of a failure of a datum is p. System
one contains N datums, each with an extra copy, while system 2
contains 2N datums with no backup. Assume both systems have the same
lifespan t. In both systems memory would decay at rates
l*exp(-k*p^2*t) and l*exp(-k*p*t) respectively (l is a
proportionality constant).
Then the value of system one would be
W1=kN(1-exp(-l*p^2 t))/lp^2
and the value of system two would be
W2=2kN(1-exp(-l*p*t))/lp
In the long run, when t->infinity, W1 goes to kN/(lp^2) and W2 to
2kN/lp; for p<1/2 (which is the likely scenario, nobody uses
devices which fails more often than not) system 1 is more valuable
than system 2. Error correction is apparently worth its price.
-----------------------------------------------------------------------
Anders Sandberg Towards Ascension!
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