20061013, 20:02  #1  
Sep 2006
Brussels, Belgium
3^{3}·61 Posts 
optimal memory settings for the P1 stage
I did a search through the FAQs, the help file and the forum and I still have some questions about the optimal amount of memory for the P1 stage. In the help file there is a table with some obsolete values (it gives three values : for 6M, 10M and 33M.)
There is a thread where the optimum amount is said to be 5.5 times the number of millions of the exponent but I suppose this is not actual anymore because in the current readme files the text is missing : Quote:


20061013, 23:08  #2  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Quote:
When reading about P1 stage 2 memory allocations, it's sometimes important to distinguish between two different cases because the behavior differs between the two. But this is often difficult because most users aren't even aware that more than one case exists! First case (the one to which most users, including wackeyh in your linked quote as far as I can tell, refer in their posts): the worktodo.ini line command is Test=nnnnnnnn... or Pfactor=nnnnnnnn... In this case, prime95 calculates an optimum combination of B1 and B2 bounds, taking into account several factors including the Available Memory setting. Example (where I'm guessing at roughly realistic values  don't quote these as authoritative): For a given exponent such as 33500153, if Available Memory is the 8M default or other small value < 30M, prime95 may assign B1 = B2 = 505000, so that only Stage 1 is run. If Avail. Mem. = 60M, prime95 may decide on B1 = 380000, B2 = 9500000. For Avail. Mem. > 150M, prime95 may choose B1 = 325000, B2 = 21775000. The first choice (<30M) will take the shortest time, and has the smallest probability of finding a factor. The second choice (60M) will take longer, but has a noticeably higher chance of finding a factor. The third case (>150M) will take the longest time, but also has the highest chance of finding a factor. Thus we can see why wackeyh writes that increasing the Available Memory will increase the time. What wackeyh leaves out, however, is that prime95 has calculated that the increased time is justified by the increased chance of finding a factor. (Perhaps wackeyh is even unaware that increasing the memory allows prime95 to have a greater chance of finding a factor in P1.) My advice is to trust prime95's algorithm for choosing B1/B2 limits  the more memory you give it, the better your result will be. Prime95's B1/B2 choosing algorithm takes more factors into account than you know of (unless you read the source code or my past posting where I list them :), and its choice will optimize GIMPS throughput! Second case: The worktodo.ini command line is Pminus1=nnnnnnnn... Here, the user explicitly specifies the B1 and B2 bounds. Then the only effect of the Available Memory figure is to limit the number of auxiliary work areas that prime95 can allocate during Stage 2 to speed up execution. Here, increasing Available Memory will decrease running time, the opposite of the effect it had in the first case! And if you give it a whole lot of memory (like a GB or more), it may even shift into a "higher gear" (The BrentSuyama Extension) that increases its chance of finding a factor! Last fiddled with by cheesehead on 20061013 at 23:16 

20070525, 12:03  #3 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Update: "P1 RAM Vs ????" at http://mersenneforum.org/showthread.php?t=8251, where I explain another possible reason for misunderstanding.
Note that my previous post here doesn't mention the possibility of system thrashing if "Available Memory" is set too high. Last fiddled with by cheesehead on 20070525 at 12:09 
20070525, 17:41  #4 
Sep 2002
296_{10} Posts 
wasnt there something else too
like p1 with 512 ram you had lets say 50% chance of finding a factor (i know its much lower) but with 1 gig you only had like a 60% and with 2 gigs you only had like a 65% chance something like that 
20070526, 04:53  #5  
"Richard B. Woods"
Aug 2002
Wisconsin USA
1111000001100_{2} Posts 
Quote:
Increasing memory does not increase the chance of factorfinding in _linear proportion_. ( to be continued in a few minutes ) Last fiddled with by cheesehead on 20070526 at 05:21 

20070526, 06:06  #6 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
P1 Stage 2 uses "Available Memory" to set up work areas for storing certain repeatedlyused intermediate computed values so that when they are needed again, they need not be recomputed from scratch but can be simply copied from a work area. So the firstorder effect of increased memory is to speed up calculations. One specifies larger "Available Memory" in order to reach a given B2 bound more rapidly.
Or ... (ways it's more commonly thought of) ... One can specify larger "Available Memory" in order to reach higher B2 bounds in the same amount of time ... or to reach even higher B2 bounds in an amount of time that is larger, but not as much larger as it would take at a lower "Available Memory".   When Prime95 is choosing the B1/B2 combination that optimizes GIMPS throughput, it is only amounts of LL time saved per time of P1 run that are optimized, not space used to perform the P1 calculation. Example: (Note: In this example, I made up the limit numbers, probabilities, and run times, to approximate realistic actual numbers I've seen in P1 runs. Yeah, I chose numbers so that the results came out the way I wanted, but I think they're close enough to reality for illustration purposes.) 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. The efficiency of each choice is proportional to the ratio (chance of finding a factor)/(time to find a factor), so that ratio calculation is the one I use in the following: Prime95 will prefer combination X because 1.0%/10 (.010/10) is greater than 1.3%/15 (.013/15) or 1.5%/16 (.015/16). Suppose that if "Available Memory" is 1024M, then combination X takes 10 hours, combination Y takes 11 hours, and combination Z takes 15 hours. Prime95 will prefer combination Y because .013/11 is greater than .010/10 or .015/15. Suppose that if "Available Memory" is 2048M, then combination X takes 10 hours, combination Y takes 10 hours, and combination Z takes 11 hours. Prime95 will prefer combination Z because .015/11 is greater than .010/10 or .013/10. So with 512M, P1 has a 1.0% chance of finding a factor, with 1024M it has a 1.3% chance, and with 2048M it has a 1.5% chance. Last fiddled with by cheesehead on 20070526 at 06:42 
20070526, 09:42  #7  
Jun 2003
2×3^{2}×269 Posts 
Quote:
Quote:


20070526, 20:32  #8  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Quote:
Please explain what's wrong. Quote:
I was treating the cost of an LL test as an implicit multiplication constant (constant for a specific exponent, that is  so that it would multiply each of the calculations in my example by exactly the same number and thus need not be explicitly included  hence my "proportional" remark). How would you rewrite my example, or at least part of it? (It's quite possible that when I see your explanation, I'll think, "Oh, how could I have forgotten?" or "There it is, right in the middle of my earlier posting in this thread, but I got mixedup!" I eagerly await your explanation.)    Added: Possible source of problem: I have used the term "efficiency" privately in certain calculations for a certain concept. It's possible that: (a) "Efficiency" has a formal meaning in the context of real, public P1 discussion that conflicts with my private definition, or (b) I may have a "fuzzy" private definition of "efficiency" and have gonewrong in using it here, or (c) I may have two separate private definitions of "efficiency" for two different contexts and mixed them up here. My private usage: One of my private definitions of "efficiency" in this context is: (total LL time saved by finding a factor) / (time spent in P1 effort to find that factor) Another (apparently equivalent, to me) is: (Total LL time expected to be saved by finding factors of a given set of Mnumbers with P1, on average given fixed B1/B2/"Available Memory" for all Mnumbers) / (Total time spent in P1 effort to find factors of a given set of Mnumbers with P1, on average given fixed B1/B2/"Available Memory" for all Mnumbers) Last fiddled with by cheesehead on 20070526 at 21:02 

20070527, 05:08  #9  
Jun 2003
2·3^{2}·269 Posts 
Quote:
However, as per the expression I have given, the correct metric is: (total LL time saved by finding a factor)  (time spent in P1 effort to find that factor) i.e. the net savings in the execution of an LL test. This is what we need to maximize. Here, one cannot discount the LL cost. To illustrate: (contd in next post) 

20070527, 05:50  #10  
Jun 2003
2·3^{2}·269 Posts 
Let us consider your example scenario #1:
Quote:
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 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. 

20070527, 10:33  #11  
"Richard B. Woods"
Aug 2002
Wisconsin USA
1E0C_{16} Posts 
Quote:
It's not choosing an exponent (or choosing an LL cost) to best utilize combination X, or Y, or Z. If one wants to analyze that way, one can, but that's not what the Prime95 software is doing. When Prime95 is given the task of choosing a B1/B2 combination for doing P1 on some exponent, it's selecting the combination that's most efficient to use on the given exponent (while also taking into consideration the specified "Available Memory", the CPU type, and the bitlevel to which the exponent has been previously TFed). So the LL cost is not a variable whose value can be chosen; it's a constant during any one search for an optimal B1/B2 combination. In my example, combinations X, Y, and Z are just representing three out of hundreds or thousands of possible combinations that Prime95 examines each time it's choosing P1 bounds for one single exponent. Last fiddled with by cheesehead on 20070527 at 11:17 

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