Will Any Current 100M Digit LL Tests Finish?

[James Heinrich]

Originally Posted by lfm

you still have costs of electricity but thats pretty cheap really too.

Assuming electricity costs around $0.114/kWh (which is within 5% of what it costs here), it costs $1/year per watt of power consumed. So figure out how much power in watts your old "free" computer consumes (typically 100W-200W under full load) and that's how much the system costs you per year in operating dollars.

[fivemack]

If you allow yourself two upgrades, then for large enough (greater than 6 MLT, where MLT is the time it takes for a computer to increase in speed by a factor E) jobs, you should

* wait a while (the value of 'a while' looks as if it involves the W-function)
* upgrade after 1 MLT
* upgrade again after 0.63212 MLT
* finish after 0.46853 more MLT

If you allow three upgrades, it only makes sense if the job takes more than 9 MLTs at the moment, and you should

* wait a while
* upgrade after 1 MLT
* upgrade again after 0.63212 MLT
* upgrade again after 0.46853 MLT
* finish after 0.37408 more MLT

If we assume MLT = 2 years, and have a job which takes 100 years if you started today, the recipe is

wait 3.6 years, buy a computer
run it for two years and replace it
run it for 1.26 years and replace it
run it for 0.92 years and replace it
finish the job at 8.49 years from today

[ATH]

Originally Posted by James Heinrich

Assuming electricity costs around $0.114/kWh (which is within 5% of what it costs here)

Here in Denmark electricity cost the equivalent of $0.354/kWh with the current dollar value So over 3 times as much.

[CRGreathouse]

Originally Posted by fivemack

If you allow yourself two upgrades, then for large enough (greater than 6 MLT, where MLT is the time it takes for a computer to increase in speed by a factor E) jobs, you should

What about including electric costs? Suppose a computer costs $A and consumes $B electricity per time unit. If you have a job that takes C time units now, when should you start it to compute it as quickly as possible without breaking the budget of $D dollars? Assume D > A so that a solution exists.

The current solutions are obtained by taking B = 0 and D = 2A, 3A, or so on; this generalizes to other amounts. Is a solution feasible, or are there too many variables?

[James Heinrich]

Originally Posted by ATH

Here in Denmark electricity cost the equivalent of $0.354/kWh with the current dollar value So over 3 times as much.

I was quoting Canadian dollars for my local cost, after all taxes, so ~US$0.091/kWh. So yours is nearly 4x more expensive, if that makes you feel any better

[James Heinrich]

Originally Posted by James Heinrich

As it relates to this thread, I will not be finishing M332203901, I'm just running through TF and P-1 (ETA: end of Jan 2009).

I have now finished TF 71-77, and P-1, and so I've released it for someone else to play with, if they want to run LL for several years :)

edit: hmm, looks like ANONYMOUS has already grabbed it... not sure who that is or if they'll actually complete anything.

[10metreh]

Originally Posted by James Heinrich

I have now finished TF 71-77, and P-1, and so I've released it for someone else to play with, if they want to run LL for several years :)

edit: hmm, looks like ANONYMOUS has already grabbed it... not sure who that is or if they'll actually complete anything.

Yes, ANONYMOUS seems to have the most computers of anybody contributing to GIMPS

[lavalamp]

Fighting scientology, finding primes, we salute you anonymous!

[Mr. P-1]

Originally Posted by lavalamp

Fighting scientology, finding primes, we salute you anonymous!

Not to mention being the most prolific author, poet, artist, and musician in history.

[Uncwilly]

Those who are LL in the 100M range, can you post your current % done please?

[JuanTutors]

Originally Posted by Uncwilly

Those who are LL in the 100M range, can you post your current % done please?

I'm a couple of days away from finishing my P-1 stage 2 test :)