Rolf wrote:

> So, this thread has been alive for long enough to capture my interest 
> even though I do not need the solution myself. Here's how I would get 
> 67M counts ot of 1024 bytes of EEPROM without exceeding 100K writes per 
> cell (in fact, keeping about 30% lower than the limit).

If it had captured your interest a bit earlier (Jan 20), you probably would
have read the message where I show how I get 102M (1022 * 100k) writes :)
Quite similar to your description.

> With some more thought, and a more complicated count mechanism, I am 
> sure I could get close to 100M count in 1024 bytes, but would be much 
> more complex (I think).

The thought involved is not so much more, and the mechanism is only a bit
more complex. Instead of 2-byte counters you'd use 3-byte counters and
count each one to 100k (or a bit less, if you want to discount the 2nd and
3rd byte writes).

My question was what the exact way is that Shawn calculates his 50M, and
whether there's some advantage to his algorithm that I don't yet see
(because I don't know it :). These 50M is a number that doesn't make much
sense to me in this context. You just showed that with a 2-byte counter
(the probably simplest algorithm) you get some 67M, with a 3-byte counter
you get 100M+.

Gerhard

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