israel@saturn.informatik.tu-chemnitz.de (Andreas Israel) (06/26/91)
I remember from some erlier UNIX kernel 'hacking', that the maximum amount of memory that can be given to all processes is computed as the maximum max(phys_memory, swap_space). This was done to avoid deadlocks, because every page in memory needs potentially one page on swap space, and if one was allocated, that swap page was released only on exit (if I understood the sources properly). Now my question: If you have a computer with, for instance 48 MByte RAM, what amount of swap space is needed? Is any swap space size less than 48 MByte of much use (or of none, if the formula above is still right). Any ideas how the amount of swap space should be computed and how UNIX kernels compute upper limits (esp. for HP-UX and SunOS) Thanks -- Name: Andreas Israel | Chemnitz University of Technology | *************** Nick: Easy | Dept. of Computer Science | * empty * Phone: +37 71 668 361 | Germany, O-9010 Chemnitz, PSF 964 | *************** "Think of your family tonight. Try to crawl home after the computer crashes."
pd@x.co.uk (Paul Davey) (06/28/91)
>>>>> On 26 Jun 91 08:46:54 GMT, israel@saturn.informatik.tu-chemnitz.de (Andreas Israel) said:
Andreas> Now my question: If you have a computer with, for instance 48 MByte RAM,
Andreas> what amount of swap space is needed? Is any swap space size less than
Andreas> 48 MByte of much use (or of none, if the formula above is still right).
Andreas> Any ideas how the amount of swap space should be computed and how
Andreas> UNIX kernels compute upper limits (esp. for HP-UX and SunOS)
I can't help with the kernal computations but you should have more
swap than physical RAM, normally a factor of 1.5 to 2 times, more if
you have memory heavy applications, eg lots of gnu-emacs's, CAD,
expert systems etc.
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