γ = speed doubling time | ||

t = time | ||

t_{0} = year processing was started
| ||

T = processing time with constant cpu speed starting at t=0 |

solving gives This is minimal for

The duration of the calculation will be

Or, when

This has the following result:

When however the calculation is started at t=0, running at ever increasing CPU speeds: The # operations per second performed as function of t

Where C is the current number of ops/sec

The total number of operations needed to complete the calculation is T*C

So for determine minimal ending time So This is minimal for t