Windows 7 not using all my ram?

OmegaZab said:

Well this is what it says now after what i did

Yes, the information is essentially the same.
The (3GB usable) is stating that 3GB is available for your use, bad terminology in my opinion.

The first time I saw that, it concerned me, so I posted a similar question and got a similar reply, confirmed several times on several sites.

The common misconception is that for some reason 1GB is 'not usable', when in fact it is already being used.

From Crucial website: Why doesn't my Windows® PC recognize the whole 4GB of memory I installed?

Installed memory(RAM): 4GB (2.94GB Usable)

If you go into the resource monitor ive got
hardware reserved 131MB my only explaination is that this is used to run just the operating system, thus making a more stable version of windows.
it all figures out if you take that away from 4096mb (4GB) you get 3.87GB

You can search the issue and find many view points. I find the 'It's being used by Windows' explanation for a 64bit OS as being the most likely.
You mileage may vary.....


Edit:
This has been clarified in this thread, the (3GB usable) is what's available to you computer.

Here is another explanation from a previous thread...

Solved 4GB (3.25 usabel of RAM)

Posted by Dzomlija

Dzomlija
Dark Lord of the Sith


Join Date: Oct 2008
Windows Vista x64 Ultimate
301 posts

devilmycry4

Help me Please.....???
i used Windows 7 64-bit with 4GB of My RAM, But In My Computer detect Is 4GB ( 3.25 GB usable)
can i use FuLLY of may RAM.....??????

Yes, you can. Go back to using Windows 7 64 Bit (or any other 64 Bit OS).

It all comes down to architecture, and hardware. Allow me to explain:

Computers are stupid machines, and "think" about numbers in Binary - 0's and 1's. What this means is the greater numbers of bits that can be processed at any given time, the greater the number that can be represented. Binary Numbers are read from right to left, with each successive digit representing a larger number. For example, the binary number "1111" is the number 15. It is calculated as follows:

1(8) 1(4) 1(2) 1(1)
8 + 4 + 2 + 1 = 15

Using various combinations of 0's and 1's, a 4 Bit number can represent the decimal numbers 0 to 15. You can calculate the possible number of deicmal numbers of any given "Bit Depth" using a simple math equation:

2 ^ [Bit Depth]

In the above example of a 4 Bit number, 2 ^ 4 = 16. With computers, when talking about 32 Bits or 64 Bits, we are actually referring to the maximum space addressable by memory. The maximum address space of 32-Bit can be calculated as follows

2 ^ 32 = 4,294,967,296 Bytes.

A common mistake to think of a KiloByte, MegaByte, Gigabyte, etc. in units of 1000, as we do with all decimal units of measurement. This is, however, incorrect. Think in units of 1024:

1 KiloByte (KB) = 1024 Bytes
1 Mega Byte (MB) = 1024 KiloBytes
1 GigaByte (GB) = 1024 MegaBytes
1 TeraByte (TB) = 1024 GigaBytes

4,294,967,296 Bytes = 4,194,304 KB
4,194,304 KB = 4,096 MB
4,096MB = 4GB

Some people don't understand it, but the "lost" memory is being used, just not for what you may think. Whether it is for a simple device such as a clock timer or you graphics card, all your hardware device drivers are using that memory. The memory is "lost" to the "usable" pool, which is defined as the memory that is not in use by hardware after all devices have been initialized. "Usable" memory is used for data and executable programs.

It is impossible for 32-Bit to exceed a 4GB address space, so the memory allocated to devices absolutely MUST fall within the 32-Bit range of 4GB. This is the the reason why on 64-Bit you could see the full 4GB, but you only get around 3.5 usable on 32-Bit. 64-Bit address is like this:

18,446,744,073,709,551,616 Bytes
18,014,398,509,481,984 KB
17,592,186,044,416 MB
17,179,869,184 GB
16,777,216 TB

As you can see, 64-Bit is fr, far larger than 32-Bit in terms of maximum addressable space. Heck, I don't even know how to pronounce the number in bytes!

FYI, if you're curious about it, HexaDecimal (Hex for short) is nothing more than a "human readable" form of Binary, with the numbers 0-9 and A-F representing the numbers from 0-9 and 10-15. It's easier to write down "0xFFFF" to represent 65535 instead of "1111 1111 1111 1111"

Hope that cleared things up for you...

Peter Alexander Dzomlija
Do you hear, huh? The Alpha and The Omega? Death and Rebirth? And as you die, so shall I be Reborn...

Hi, H2SO4

H2SO4 said:

Dave76 said:

OmegaZab said:

Well this is what it says now after what i did

Yes, the information is essentially the same.
The (3GB usable) is stating that 3GB is available for your use, bad terminology in my opinion.

The first time I saw that, it concerned me, so I posted a similar question and got a similar reply, confirmed several times on several sites.

The common misconception is that for some reason 1GB is 'not usable', when in fact it is already being used.

H2SO4 said:

With respect, no, that's not correct. As far as that UI is concerned, the terms mean something along these lines:

"Installed": if you pull all the RAM sticks out, this is the sum of their capacity.

Yes, thats what I'm trying to say.

H2SO4 said:

"Usable": This is the total amount available to Windows for its own needs and those of the apps, once the hardware has potentially "masked" certain ranges for itself. (It does not mean "what's left over for the apps when Windows takes what it needs".)

Looks like we agree on this.
Don't think I said "what's left over for the apps when Windows takes what it needs", Just re-read the post to be sure and didn't see it.
Are you saying the hardware and processes are using some of the RAM when you say: ""masked" certain ranges"?

H2SO4 said:

If there's a disparity between "installed" and "usable", that difference is effectively lost and inaccessible. Note that the "usable" value is reported only if it's different to "installed", and that's machine-specific.

Disclaimer: I'm not an expert in this area, just trying to explain what I've read, several different sites have explained it similarly to this pared down version of mine.

If the other explanations and therefore mine are incorrect I would like to find out the correct reason for the "installed" and "usable" difference. You said it is " effectively lost and inaccessible", what is the cause? Can you give possible incorrect settings that might be corrected to enable the use of all installed RAM. I would like to be able to correct this for myself and maybe pass it along to others.
Is it in the msconfig, Boot, Advanced Options, Maximum Memory setting?
I would appreciate your help.

H2SO4 said:

In this instance, the OP's configuration was for some reason artificially hobbled at startup with the "MAXMEM" parameter (that's what MSCONFIG exposes in its memory dialog). Windows was configured to behave as if there's 3GB of RAM in the box, not 4GB. That's normally used only for troubleshooting purposes, for example when trying to simulate a low resource situation in order to expose certain problems in memory management or drivers.

Your saying it's in the msconfig, Boot, Advanced Options, Maximum Memory setting?
So would it be correct to tick the box and put in the Installed memory for the computer?

Thanks for any clarification

Hello Dave76. Actually, stevecn70 did an excellent (rep-deserving) job of eloquently explaining the complex mechanisms involved, so all I can hope to do is to perhaps paraphrase some of the details.

The mix of 4.0GB of physical memory and 32-bit Windows leads to a curious anomaly. People are sometimes concerned that they paid for 4.0GB, while their computer registers 3.5GB or sometimes even less in Windows ("I got ripped off and received less than I paid for"). Prior to Vista SP1, the number they saw when they looked at their machine's properties, under Windows, was the equivalent of today's "usable" memory statistic.

On 32-bit systems, the disparity is caused by something called Memory Mapped I/O (MMIO). Many hardware devices and their drivers have a pact of sorts with Windows. They both agree that the hardware device will exchange data with the OS by presenting it as a memory range at an address that is specific to the device. For example, during the earliest boot phases, the standard IDE controller might declare its intention to use the memory address range between (say) 3.8GB and 4.0GB as a window of sorts into the disk. In that (over)simplified example, Windows reads disk-based data by first issuing a command to the disk controller, then subsequently accessing memory addresses starting at 3.8GB to read the raw bits and bytes which the controller found on the disk surface, as if 3.8-4.0GB directly corresponded to a 0.2GB chunk of the disk. To read any specific part of the disk, Windows tells the controller where to start, and then 3.8-4.0GB represents the intended disk section.

Back when 512MB and 1.0GB RAM was commonplace, Windows was already using the MMIO mechanism to exchange data with hardware controllers at memory address ranges corresponding to ~3.5GB and above. On such a machine, memory up to 1.0GB corresponded to "real" RAM, followed by a big blank, and then reading "memory" from around 3.8GB upwards redirected the read to the surface of the disk, courtesy of the disk controller. Kinda

With 2.0GB of installed RAM, there is still no overlap between the 0-2.0GB memory address range which is used for "real" RAM, and 3.8-4.0GB which maps to the surface of the disk. Now install 4.0GB in that same machine and there's a problem - the "real" RAM in the 3.8-4.0GB range cannot be accessed because reading from there constitutes MMIO to the disk. Add a few more hypothetical hardware devices, one doing MMIO from 3.7-3.8GB, and another from 3.5-3.7GB, and the result is a Windows machine which has 4.0GB of RAM "installed" but only 3.5GB "usable". The last 0.5GB of RAM is inaccessible, being masked/blanketed by the MMIO ranges. On a 32-bit system, math simply doesn't allow access to both 4.0GB of "real" RAM and 0.5GB worth of MMIO, or 4.5GB in total - it's impossible to describe numbers above 4.0GB with only 32 bits.

From Vista SP1 onwards, Windows differentiates between the "INSTALLED" value and the "USABLE" amount. With SP1 applied, Vista 32-bit users were reassured to see that their machine had the 4.0GB "installed" that they paid for, even if it wasn't immediately clear to them that some portion of that amount was lost to MMIO. Windows 7 presents info in the same way.

On a 64-bit system, owing to the vast address range, it's not a problem to accomodate both 0.5GB worth of MMIO and all 4.0GB of "real" RAM, or even far more of each. Hence, on a 64-bit system, the "installed" and "usable" values should normally be exactly the same, to the point where "usable" is not even displayed. One exception would be a (embedded) video card without its own memory, as stevecn70 mentioned. It would leech a portion of the "real" RAM for its own needs, thus depriving the software and causing a disparity between installed/usable.

In the OP's case, his Windows installation was for some unknown reason configured - through use of MSCONFIG - to artificially limit itself to only 3.0GB, which caused the difference between 4.0GB "installed" and 3.0GB "usable". That is NOT standard configuration however, and unticking the "maximum memory" tickbox in MSCONFIG removed the artificial limitation. The MSCONFIG functionality is provided only for troubleshooting purposes; normally, there's no good reason to limit the amount of RAM employed, and the tickbox should always be clear (empty).

I hope that serves to clear things up a little. Sorry about the verbosity :)

H2SO4 said:

Hello Dave76. Actually, stevecn70 did an excellent (rep-deserving) job of eloquently explaining the complex mechanisms involved, so all I can hope to do is to perhaps paraphrase some of the details.
...
From Vista SP1 onwards, Windows differentiates between the "INSTALLED" value and the "USABLE" amount. With SP1 applied, Vista 32-bit users were reassured to see that their machine had the 4.0GB "installed" that they paid for, even if it wasn't immediately clear to them that some portion of that amount was lost to MMIO. Windows 7 presents info in the same way. (*this is pertaining to 32bit OS)

On a 64-bit system, owing to the vast address range, it's not a problem to accomodate both 0.5GB worth of MMIO and all 4.0GB of "real" RAM, or even far more of each. Hence, on a 64-bit system, the "installed" and "usable" values should normally be exactly the same, to the point where "usable" is not even displayed. One exception would be a (embedded) video card without its own memory, as stevecn70 mentioned. It would leech a portion of the "real" RAM for its own needs, thus depriving the software and causing a disparity between installed/usable.

In the OP's case, his Windows installation was for some unknown reason configured - through use of MSCONFIG - to artificially limit itself to only 3.0GB, which caused the difference between 4.0GB "installed" and 3.0GB "usable". That is NOT standard configuration however, and unticking the "maximum memory" tickbox in MSCONFIG removed the artificial limitation. The MSCONFIG functionality is provided only for troubleshooting purposes; normally, there's no good reason to limit the amount of RAM employed, and the tickbox should always be clear (empty).

I hope that serves to clear things up a little. Sorry about the verbosity :)

Thank you for taking your time to explain this, it can be disconcerting reading so many contradicting opinions on issues like this. This explanation is clear on the subject.

This is why I enjoy this forum, you can learn and get clear affable replies. The prevailing characteristically gracious attitudes are appreciated by all.

Thanks to stevecn70 for his explanation.

Also, thanks to Antman for recently posting the correct fix on a similar thread, which was possibly referred to in this thread.