> 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.
Just an additional note:
This assumes the system lives until long after the last useful data
has vanished, which isn't that likely. For lifetimes short compared
to 1/p, then the non-error correcting system has an advantage, but it
vanishes in time.
-----------------------------------------------------------------------
Anders Sandberg Towards Ascension!
nv91-asa@nada.kth.se http://www.nada.kth.se/~nv91-asa/main.html
GCS/M/S/O d++ -p+ c++++ !l u+ e++ m++ s+/+ n--- h+/* f+ g+ w++ t+ r+ !y