Where's my 4GB of RAM? (Windows XP 32-bit question)

I upgraded RAM and since it's kind of cheap now (DDR2 at least) I decided to go all the way to the 4GB max specified by my laptop's instruction manual. Had I read the *, I'd have realized that over 3 GB may not show up on 32-bit Windows OS (I have 32-bit Windows XP installed), and in fact that last GB doesn't register.

I googled around a bit, and it appears a year ago the answer was to stick with 3 GB, and that was that. However, there are other claims out there; maybe your machine can use that last GB even if it doesn't show (if the computer is new enough), or maybe you can get it to show up, maybe something has changed over the past year now that this has become a more common problem as DDR2 SDRAM prices have fallen, so is anyone here on top of this topic? Best approach/solution?

 

The maximum amount of RAM that can be addressed by a 32bit operating system is 3.2gb. It is a limitation of the 32bit architecture and no Windows tweak or registry hack is going to change it. The remaining three-quarters of a gig of RAM does not get used or addressed, and just kind of sits there in limbo. This topic has been discussed in-depth in the Hardware forum, and the math is explained in detail if you're interested. Go to the hardware forum and search it for "RAM limits" or maybe "32bit OS RAM" or something similar. Apparently, there is some "rumor" going around that you can 'fix' the 3.2gb limit by modifying the registry, or that any amount of RAM over 2gb requires certain Windows tweaks; both of these are untrue. I'm running 3gb of DDR2 and never had to do anything other than buying the memory, and installing it on my motherboard.

I hope this helps!

[dlb]
:major

 

Thanks for the reply. That confirms what I've heard elsewhere.

FWIW, if I had to do it again I'd probably still buy 4 GB instead of 3 GB because a) it's probably cheaper given the way they sell it in pairs, and b) in case the 32-bit Windows gets "fixed," or the 64-bit version of Vista becomes worth installing.