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Video RAM takes total system RAM amount down?

post #1 of 7
1/15/09 at 7:51pm
Thread Starter 
Hi guys, first off this post is in no way related to anything regarding integrated graphics cards and shared memory, etc. We're talking dedicated video cards only.

So...
From what I know/understand, a 32-bit OS will cap out at 3.25Gb of memory utilization. Anything above that, it can "see" but not use.

So I'm about to buy a new laptop, and was going to get 4Gb of RAM with it, but I know the OS will only use 3.25Gb (or whatever the number is). But what about the video memory on the 512mb 9800M GTS? That 512mb of video memory still has to be mapped, correct?

So in real world: I have 4gb of RAM installed, I can use 3.25gb on a 32-bit install of Vista, but then that video memory has to be mapped as well, which takes away from the total address pool the system has (I know, my wording is terrible, I don't know exactly how to explain this). So the system had 3.25gb of memory to use, minus the 512mb graphics memory = 2.75gb of memory for the system to use.

Meaning I have an entire gigabyte of RAM not being used whatsoever (aka waste of money).

So is this the case? Or is video memory considered completely different, and if so, how? How does the 32-bit OS utilize its maximum address pool of 3.25Gb PLUS another 512mb that is (in theory) out of the boundaries to what a 32-bit OS can address?

Thank you very much in advance
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post #2 of 7
1/15/09 at 7:58pm
Quote:
Originally Posted by Sapphire View Post
Hi guys, first off this post is in no way related to anything regarding integrated graphics cards and shared memory, etc. We're talking dedicated video cards only.

So...
From what I know/understand, a 32-bit OS will cap out at 3.25Gb of memory utilization. Anything above that, it can "see" but not use.

So I'm about to buy a new laptop, and was going to get 4Gb of RAM with it, but I know the OS will only use 3.25Gb (or whatever the number is). But what about the video memory on the 512mb 9800M GTS? That 512mb of video memory still has to be mapped, correct?

So in real world: I have 4gb of RAM installed, I can use 3.25gb on a 32-bit install of Vista, but then that video memory has to be mapped as well, which takes away from the total address pool the system has (I know, my wording is terrible, I don't know exactly how to explain this). So the system had 3.25gb of memory to use, minus the 512mb graphics memory = 2.75gb of memory for the system to use.

Meaning I have an entire gigabyte of RAM not being used whatsoever (aka waste of money).

So is this the case? Or is video memory considered completely different, and if so, how? How does the 32-bit OS utilize its maximum address pool of 3.25Gb PLUS another 512mb that is (in theory) out of the boundaries to what a 32-bit OS can address?

Thank you very much in advance
You've got the right idea, but you've gone a step further than you need to. 32-bit operating systems can utilize a total of 4GB total memory of any kind. So in the case you mentioned, 512MB are dedicated to video memory, which means the OS can only utilize 3.5GB or so for the system. Then subtract out the cache memory on the CPU and a couple other things, and it works out to around 3.25GB of usable system memory. However, that number isn't an exact one, it's just a general approximation. If, for instance, you had a video card with 1GB of dedicated video memory, then you'd have less than 3GB RAM that your system can use with a 32-bit OS.
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post #3 of 7
1/15/09 at 8:16pm
Thread Starter 
Oh

So when its all said and done, if i buy 4gb of RAM with a 32-bit OS, I'm only actually using 3.25Gb of that 4gb. And that 3.25gb (or whatever) has already accounted for a subtraction of 512mb for the graphics memory. Is that correct?

In other words, we have 4Gb total to work with for anything/everything. Subtract 512mb for graphics = 3.5Gb. Subtract all the other minutia and you're left with the 3.25Gb of RAM for the OS to use like you said. MEANING at the end of all this memory mapping is that there is 768mb of RAM that is sitting there doing nothing...yes?
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post #4 of 7
1/15/09 at 8:30pm
Quote:
Originally Posted by Sapphire View Post
Oh

So when its all said and done, if i buy 4gb of RAM with a 32-bit OS, I'm only actually using 3.25Gb of that 4gb. And that 3.25gb (or whatever) has already accounted for a subtraction of 512mb for the graphics memory. Is that correct?

In other words, we have 4Gb total to work with for anything/everything. Subtract 512mb for graphics = 3.5Gb. Subtract all the other minutia and you're left with the 3.25Gb of RAM for the OS to use like you said. MEANING at the end of all this memory mapping is that there is 768mb of RAM that is sitting there doing nothing...yes?
correct. and that's currently the biggest reason to use a 64-bit operating system.
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post #5 of 7
1/15/09 at 8:32pm
Thread Starter 
Perfect. Thank you so much buddy for clearing that up for me.
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post #6 of 7
1/15/09 at 8:40pm
no prob
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post #7 of 7
1/21/09 at 8:40am
Yeah 64 bit is becoming the way to go. I thought that it didn't matter and that all OSes running 32 bit could only access 3.25Gb, not 4gb. So I'm totally bogusly wrong? (I've never ran higher than 3Gbs so yet to test never felt a need.)
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