Unless it says otherwise it's probably Vista 32bit. Vista was patched to show the total amount of physical RAM but it's only cosmetic. A 32bit OS can't address more than 4GB of total physical memory. Since the video card, sound card and other hardware devices also use physical memory the amount available for the system will be less than 4GB. As Riskyone said with XP the amount reported will depend on the hardware and will be different from one rig to another. It's all explained here : http://www.techsupportforum.com/f10/winxp-shows-less-ram-249007.html#post1484667ok thx then...it wuz 64bit, since its vista
XP manages the memory automatically. Say some program uses 10MB of RAM for now but may require up to 50MB. As long as there's enough free RAM XP may already allocate 50MB for that program. The result is that when the program needs more memory XP doesn't have to start to allocate it at that very moment, the address range is already available => gain of time. The point is not in having as much free ram as possible but in using it as efficiently as possible. You don't have to worry as long as the peak value under commit charge (= the amount of RAM + page file used currently) is lower than the total of physical memory. This means you have enough memory for all your apps and don't need to add more. If the commit charge was bigger than the physical memory the system would need to swap constantly between the page file and the physical memory, hence reducing the system's performances.1) Under the task manager performance tab, the Physical Memory section says i have:
So...my question is, why do I only have around 1.4 gigs of ram "availible" when all I got running really is Firefox, AV, APSyware, and just all standard services and processes related? How is my system already using 0.6 of a gig?
No, sorry. 1048576KB = 1024MB = 1GB. Dividing by 1 thousand (as if it was in base 10) usually gives a good enough approximation.2) Anyway to change the (K) to (MB) in the taskmanager? It's annoying trying to work it out ><