Talk:Virtual memory

swap file distribution across disks

I don't think RAID is required to distribute a Windows paging file across two disks, only a second available disk (partition) is required, whether RAID or not.Pat Palmer 10:14, 19 January 2008 (CST)

Another odd phenomenon: if the swap file is moved entirely to a second drive, say D:, a Windows machine may well boot up faster. I think that happens because the loading of boot programs from the primary partition (C: drive) can then go on in parallel with the loading of running program parts into the page file (D: drive), so overall, everything gets ready quicker. This is the kind of crap I learned from doing a lot of system administration and spending too many hours of my life watching Windows reboot (smile). It's typically 3 minutes per reboot (still) on practically any machine, because even though machines got faster, the amount of stuff that wants to load itself on startup has gotten correspondingly greater.Pat Palmer 10:19, 19 January 2008 (CST)

A RAID or stripe set is required. And no, moving the swap file to another drive doesn't make windows boot up faster, because the bandwidth across the IDE/SATA channel is slower between two discs. --Robert W King 11:03, 19 January 2008 (CST)

Robert, I not sure that I agree with this. I have the swap file on my D: drive (right here on my laptop) and it is certainly not a RAID. I know you think it's so, but would you please try it and make sure that you're correct?Pat Palmer 19:53, 19 January 2008 (CST)

To say that you're going to "distribute a paging file across disks" implies that you're allocating an amount of drive space on every drive for a portion of the swap, which requires a stripe or raid. If you're simply MOVING it to another drive all together, no you don't need a stripe then. But that is still a foolish thing to do because when you seperate your system files from the swap, it means now you have to transfer data between two hard drives over the relatively slow IDE/SATA channel and into RAM, versus right from the system disk into RAM. --Robert W King 20:36, 19 January 2008 (CST)