Paging/Swapping

[Andreas Israel]

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."