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Old 2007-05-27, 05:50   #10
axn
 
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Jun 2003

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Let us consider your example scenario #1:
Quote:
Originally Posted by cheesehead View Post
Suppose that P-1 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 t be the LL cost (depending on the context, this could be 1 LL test or 2 LL tests).
Code:
P-1 level | Expected cost | Break even pt
----------+---------------+------------------------
no P-1    | 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 P-1 levels as specified.

Now, the thing to note is that, given a sufficiently large t, a combo with a lower coefficient for t /will/ be better. So let's calculate how large is sufficiently large.

Break even pt between Combo X & Y

0.990t + 10 > 0.987t + 15

or t > (15-10)/(0.990-0.987) = 1666.67 hrs

i.e, for tests longer than 1666.67 hrs, combo Y is preferred to X.


Similarly, X vs Z: t > (16-10)/(0.990-0.985) = 1200 hrs

And, Y vs Z: t > (16-15)/(0.987-0.985) = 500 hrs


To summarize, if t <= 1000, no P-1. For 1000 < t <= 1200, combo X. For t > 1200, Combo Z

--------------
The same calculation can be repeated for the other scenarios.
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