Let us consider your example scenario #1:
Quote:
Originally Posted by cheesehead
Suppose that P1 is choosing its own B1/B2 limits, and it's choosing among three alternatives:
combination X (B1=590000,B2=590000  so only Stage 1) has a 1.0% chance of finding a factor,
combination Y (B1=460000,B2=2100000) has a 1.3% chance of finding a factor, and
combination Z (B1=410000,B2=3900000) has a 1.5% chance of finding a factor.
Suppose that if "Available Memory" is 512M, combination X takes 10 hours, combination Y takes 15 hours, and combination Z takes 16 hours.

Let
be the LL cost (depending on the context, this could be 1 LL test or 2 LL tests).
Code:
P1 level  Expected cost  Break even pt
++
no P1  t  N/A
combo X  0.990t + 10  t > 10/0.010 = 1000.00 hrs
combo Y  0.987t + 15  t > 15/0.013 = 1153.85 hrs
combo Z  0.985t + 16  t > 16/0.015 = 1066.67 hrs
The table gives the break even pts for each of the P1 levels as specified.
Now, the thing to note is that, given a sufficiently large
, a combo with a lower coefficient for
/will/ be better. So let's calculate how large is sufficiently large.
Break even pt between Combo X & Y
or
i.e, for tests longer than 1666.67 hrs, combo Y is preferred to X.
Similarly, X vs Z:
And, Y vs Z:
To summarize, if
<= 1000, no P1. For 1000 <
<= 1200, combo X. For
> 1200, Combo Z

The same calculation can be repeated for the other scenarios.