Solved The 32-bit RAM Issue

[skunksmash]

When the physical RAM that is installed on a computer equals the address space that is supported by the chipset, the total system memory that is available to the operating system is always less than the physical RAM that is installed.
For example, consider a computer that has an Intel 975X chipset that supports 8GB of address space.
If you install 8GB of RAM, the system memory that is available to the operating system will be reduced by the PCI configuration requirements.
In this scenario, PCI configuration requirements reduce the memory that is available to the operating system by an amount that is between approximately 200 MB and approximately 1 GB. The reduction depends on the configuration.

great discussion

 

[Fumz]

Fumz,

You still don't get it. Pay attention and read carefully now:

BIOS, I/O cards, NICs, video/graphics cards, etc. have their own address space that gets mapped into the 4GB address space of a 32 bit system. Do you under stand what that means?

Let me give you one example that should make this clear to you: My system has a video card installed with 640MB of RAM on it. That RAM is mapped into the 4GB address space of the system.

That means that at least 640MB of the 4GB of RAM installed is not mapped into the 4GB of address space. That means the system cannot address at least 640MB of the 4GB of RAM installed. In other words, that 640MB of RAM is wasted.

Please tell me you understand what I just told you and you now see how what you are saying is wrong!

S-

For example, if you have a video card that has 256 MB of onboard memory, that memory must be mapped within the first 4 GB of address space. If 4 GB of system memory is already installed, part of that address space must be reserved by the graphics memory mapping. Graphics memory mapping overwrites a part of the system memory. These conditions reduce the total amount of system memory that is available to the operating system.

The system memory that is reported in the System Information dialog box in Windows Vista is less than you expect if 4 GB of RAM is installed

As you may or may not know, 32-bit operating systems are limited to addressing a maximum memory size of 4,294,967,296 bytes, or "4GB" in more normal terms. However, this 4GB is shared between all MMIO (Memory-Mapped Input Output) devices - this includes graphics cards. Therefore, that nice, shiny new 512MB graphics card you've just bought limits your entire system memory use to just 3.5GB. What about the newest Nvidia GeForce 280 GTX with 1GB of GDDR3? That'll reduce your system memory to just 3GB.

bit-tech.net | Review - Is More Memory Better?

Microsoft disagrees with you, as do I. The quotes are clear. MMIO devices take a portion of system RAM. They "use" it. A 32-bit OS with 4GB's will also "reserve" a portion of system RAM to assure compatibility, and a portion of that RAM can be considered "wasted", but that portion is small and not nearly as big as the portion MMIO devices take.

If it's your understanding that your 640MB graphics card, an MMIO device, does not take system RAM, then your understanding is incorrect.

Perhaps this would go better if you ceased with the personal attacks? You know, it is possible to disagree and have a good discussion without attacking the person you're disagreeing with?

Again, my only point was that RAM, on the physical level, isn't being "wasted". A small portion is reserved, yes, but I don't consider an operating system reserving RAM to assure stability a "waste".

 

[sidewinder]

Fumz,

You just don't get it. The text you quote is saying exactly what I am saying. You just don't see it.

A 32-bit Vista has 4GB or address space. That address space is used to access system RAM, BIOS, I/O cards, NICs, video/graphics cards, etc. That 4GB of address space can be RAM, ROM, registers on APIC chips, PCI card memory (RAM or ROM), etc.

One way to look at this is that you have to plug the various devices you need to address into this 4GB address space. And the last item to get plugged into the address space is system RAM. On my 32-bit Vista system, I have about 1.2GB of address space used by devices. So when you get done plugging in all those devices, there is a 2.8GB spot left for system RAM. So only 2.8GB of system RAM is actually addressable by the system. The other 1.2GB cannot be addresses and, therefore, is wasted. If I had 3GB of RAM installed instead, only .2GB of RAM would be wasted.

You have to understand that other devices have their own memory of some type that is mapped into the 4GB address space. These device don't use system RAM for this as you seem to think.

S-

 

[Fumz]

The only problem here is the fundamental misunderstanding about what I said. There's only 1 question: is all 4GB's of RAM being used or not? I am not splitting hairs about how it's used, nor am I splitting hairs about what's using it. The only issue is whether or not that 4GB kit is being "used" by the machine in one way or another?

If 4 GB of system memory is already installed, part of that address space must be reserved by the graphics memory mapping. Graphics memory mapping overwrites a part of the system memory

That's the significant part and goes to the heart of what I said. You keep talking about the OS address space, and while what you say is correct, it has absolutely nothing to do with what I said and I never argued against it. You just aren't getting that I am not talking about the address space of a 32-bit OS.

... On my 32-bit Vista system, I have about 1.2GB of address space used by devices.

You said it yourself. 1.2GB is being used by MMIO devices. I don't take that to mean your MMIO devices are wasting RAM, I take that to mean they're using it.

 

[sidewinder]

Fumz,

No, not all 4GB of RAM is being used. All 4GB of address space is being used.

The 1.2GB of space used by devices is address space, not RAM.

You have to realize that devices are mapped into the address space and the last to get space allocated is RAM. If 1.2GB is allocated to devices other than system RAM, system RAM is mapped to what is left. In my system's case, system RAM is allocated 2.8GB of address space so 1.2GB of the 4GB installed is not addressable by the system and is wasted.

S-

 

[Fumz]

Is address space static or dynamic? By that I mean are things limited to a certain portion of physical RAM and never at any time are other address ranges on the modules used... all of course within the 4GB limit?

 

[sidewinder]

Fumz,

The 4GB address space itself is static. How the memory defined in the 4GB address space behaves depends on the type of memory it is. Clearly video card RAM and system RAM are dynamic. ROM would be static.

Let's be clear. If 2.8GB out of 4GB of system RAM are defined as part of the 4GB address space. That 2.8GB is constant and specific. The rest is wasted, not used, and not addressable.

S-

 

[Fumz]

My only problem with what you've said is what Microsoft now says:

For Windows Vista to use all 4 GB of memory on a computer that has 4 GB of memory installed, the computer must meet the following requirements:
...
The BIOS must support the memory remapping feature. The memory remapping feature allows for the segment of system memory that was previously overwritten by the Peripheral Component Interconnect (PCI) configuration space to be remapped above the 4 GB address line

The system memory that is reported in the System Information dialog box in Windows Vista is less than you expect if 4 GB of RAM is installed

They have apparently changed the way they map MMIO devices. They're now able to map them above the 4GB address line, meaning that RAM previously wasted is no longer wasted.

 

[sidewinder]

Fumz,

Oh come on! It would help if you read the whole section. Here it is:

WORKAROUND
For Windows Vista to use all 4 GB of memory on a computer that has 4 GB of memory installed, the computer must meet the following requirements:
• The chipset must support at least 8 GB of address space. Chipsets that have this capability include the following:
• Intel 975X
• Intel P965
• Intel 955X on Socket 775
• Chipsets that support AMD processors that use socket F, socket 940, socket 939, or socket AM2. These chipsets include any AMD socket and CPU combination in which the memory controller resides in the CPU.
• The CPU must support the x64 instruction set. The AMD64 CPU and the Intel EM64T CPU support this instruction set.
• The BIOS must support the memory remapping feature. The memory remapping feature allows for the segment of system memory that was previously overwritten by the Peripheral Component Interconnect (PCI) configuration space to be remapped above the 4 GB address line. This feature must be enabled in the BIOS configuration utility on the computer. View your computer product documentation for instructions that explain how to enable this feature. Many consumer-oriented computers may not support the memory remapping feature. No standard terminology is used in documentation or in BIOS configuration utilities for this feature. Therefore, you may have to read the descriptions of the various BIOS configuration settings that are available to determine whether any of the settings enable the memory remapping feature.
• An x64 (64-bit) version of Windows Vista must be used.

Note the last bullet item......

Jeez.....

S-

 

[Fumz]

Perhaps that was the wrong link to use in seeking clarification?

On computers that have a 32-bit operating system, more than 3 GB of system memory, and with a version of Windows that is earlier than Windows Vista SP1, users will see a larger difference in how much memory is reported as available to the operating system compared to how much physical memory is installed. This is because some physical address space must be reserved as I/O regions for memory mapped peripherals. These I/O regions are allocated between the 3 GB physical address and the 4 GB upper physical address limit.

Physical memory addresses that are mapped to these I/O regions cannot be used to address physical system memory. These addresses also cannot be used to prevent the operating system from using some physical memory that would ordinarily be accessed between the 3GB physical address and the 4GB upper physical address limit. The size of these I/O regions varies from system to system because they determine the type and configuration of the system’s peripherals.

Windows Vista SP1 includes reporting of Installed System Memory (RAM)
When the say that the physical address ranges between 3GB and 4GB are being used for I/O allocation, doesn't that mean that indeed, all 4GB are being used?

"that would ordinarily be accessed between the 3GB physical address and the 4GB upper physical address limit" That sentence is also confusing. It suggests, at least to me, that the operating system can, at least until prevented by MMIO allocation, access the RAM between the 3 and 4GB physical address ranges?

 

[sidewinder]

Fumz,

You can keep trying to save face, but you aren't going to do it. You are wrong and might as well admit it.

Let's take your bolded text:

These I/O regions are allocated between the 3 GB physical address and the 4 GB upper physical address limit.

What does it say, "physical address" or "physical RAM"?

Again, you are confusing address space with RAM.

S-

 

[Fumz]

wow... :eek:

I don't have a problem admitting when I'm wrong; if I am I am.

I've said what I've said based on what I've read and the language Microsoft has used; I even provided the links to back up my reasoning. If I'm getting it all wrong, that's ok; I love learning new things. However, ***REMOVED by dmex***

To me, physical address means just that, physical... ie.. RAM.

A memory address identifies a physical location in computer memory

Address space - Wikipedia, the free encyclopedia

In computing, a physical address, also real address, or binary address, is the memory address that is electronically (in the form of binary number) presented on the computer address bus circuitry in order to enable the data bus to access a particular storage cell of main memory.

Physical address - Wikipedia, the free encyclopedia

... "physical RAM" would be redundant, which is why they didn't say it.

 

[sidewinder]

Fumz,

You just don't know when to quit! I will use text you quoted against your argument:

On computers that have a 32-bit operating system, more than 3 GB of system memory, and with a version of Windows that is earlier than Windows Vista SP1, users will see a larger difference in how much memory is reported as available to the operating system compared to how much physical memory is installed. This is because some physical address space must be reserved as I/O regions for memory mapped peripherals. These I/O regions are allocated between the 3 GB physical address and the 4 GB upper physical address limit.

Clearly "physical memory" does not equal "physical address". Some or all of a systems "physical memory" is mapped into the "physical address" space. If the "physical address" space available is less than the "physical memory" available, some of the "physical memory" will not be mapped into the "physical address" space and will not be addressable by the system, thus wasted.

You claim that you are willing to learn but have shown no inclination to do so. This is a relatively simple concept and instead of attempting learn something, you are doing everything you can to prove that you are not wrong.

Who could blame me for getting short with you?????

S-

 

[Fumz]

I think you let your arrogance really get the best of you; that's unfortunate. My last two posts weren't arguments, they were questions. Good job recognizing that; even better job avoiding them.

I'm trying to get this, I really am. It's possible that I'm just a very stupid person and your level of smartness is just leaps and bounds beyond me; if so, please be patient with this ignorant mind.

Language is what prevents me from simply accepting what you say as fact (that and no links). Let's see what I mean, shall we?

On computers that have a 32-bit operating system, more than 3 GB of system memory, and with a version of Windows that is earlier than Windows Vista SP1, users will see a larger difference in how much memory is reported as available to the operating system compared to how much physical memory is installed.

This seems easy enough to understand. The operating system will report less memory than is actually installed. Here, "physical" is used in the material sense.

This is because some physical address space must be reserved as I/O regions for memory mapped peripherals.

Easy enough to get. Some of your RAM is used by some of your parts. Here again, "physical" is used in the material sense.

These I/O regions are allocated between the 3 GB physical address and the 4 GB upper physical address limit.

Maybe here is where I'm getting tied up? I assume that here again, "physical" is used as it has been used throughout the quote, in the material sense? When they say I/O mapping is reserved on physical address space between 3 and 4GB, I take them literally and assume they know what they mean when they use the word "physical".

If you're going to claim that physical does not mean physical, OK, support it with evidence; a link or two would be nice. If you're going to change the definition of a word, then you're going to have to back it up with more than just your mouth... arrogant as it is. This may come as a shock to you, but "supporting evidence" does not mean "because you say so".

Please find, and link an example of Microsoft using the term "physical address" where they DO NOT mean physical in the material sense.

Honestly, I tried to read that stuff you wrote about physical this and physical that... I got the sense you really didn't know what you were talking about. Real masters of a subject are able speak about that subject in simple terms; they're able to convey the subject to the novice in such a way as the novice can understand. A master on this you ain't.

 

[sidewinder]

Fumz,

You need to learn how to pose questions if those two posts were supposed to be anything other than rebuttals to what I am saying.

I'll take this down to a real basic level:

Pretend that the 4GB address space of the Vista 32-bit OS is a peg board with 4 billion peg holes. Let's also pretend that each device that fits into the Vista 32-bit OS 4GB address space has a peg for each byte of address space it takes up in that 4GB of address space.

Keep in mind that system memory (RAM) is the last device that gets allocated space in the 4GB of address space.

Let's again use my system as an example. My system's devices plug into 1.2 billion of those peg holes on the peg board. That leaves 2.8 billion peg holes available for the 4 billion pegs of system memory (RAM) that are installed in the system. So we can only plug in 2.8 billion system memory (RAM) pegs.

Now, for any of the pegs from any of the devices and system memory (RAM) to be accessed by the system, they have to be plugged into the peg board. Since only 2.8 billion system memory (RAM) pegs are in the peg board, only 2.8 billion system memory (RAM) pegs can be accessed by the system. The other 1.2 billion system memory (RAM) pegs are not plugged in and cannot be accessed.

Does that metaphor work for you?

S-

 

[Fumz]

When the say that the physical address ranges between 3GB and 4GB are being used for I/O allocation, doesn't that mean that indeed, all 4GB are being used?

That sentence is also confusing. It suggests, at least to me, that the operating system can, at least until prevented by MMIO allocation, access the RAM between the 3 and 4GB physical address ranges?

Are you telling me you didn't recognize these questions as questions? Even if you didn't like the wording, didn't the question marks give them away?

Your metaphor, while appreciated, doesn't work for me because it doesn't address the questions I asked. Microsoft said the I/O regions are in the 3GB to 4GB physical address ranges. Be patient, you're not going to like hearing this part... but to me, "physical" address range means physical. That's where I'm getting confused.

If I/O allocation is using the 3-4GB range of physical memory, then it is using memory not plugged into your peg board. How can this be?

 

[sidewinder]

Are you telling me you didn't recognize these questions as questions? Even if you didn't like the wording, didn't the question marks give them away?

Fumz,

Those questions, based on your history, looked completely rhetorical....you do understand that questions are not always posed with the intent of being answered, don't you? <---- An example of a rhetorical question

Your metaphor, while appreciated, doesn't work for me because it doesn't address the questions I asked. Microsoft said the I/O regions are in the 3GB to 4GB physical address ranges. Be patient, you're not going to like hearing this part... but to me, "physical" address range means physical. That's where I'm getting confused.

Of course the address range is physical. Are RAM sticks the only "physical" items in a computer? Address space is mapped to physical locations whether it is system RAM, RAM on a video card, ROM on a PCI card, BIOS, etc. We have been talking physical memory locations all along. That's why I used the peg board metaphor. Everything in the metaphor is physical.

How can parts be using pegs that aren't plugged into the board?

Huh? My example clearly states that only pegs plugged into the board can be accessed. That's why only 2.8GB of system RAM can be accessed.

S-

 

[skunksmash]

lol....... cant believe this debate is still going!!

im sure your both right ...if nothing else this should be sticky'd
for ''32-bit address space''

well done though lads you have managed to keep it civil....
im sure you love each other really...

 

[sidewinder]

skunksmash,

If Fumz was right, this thread would have concluded a long time ago....

S-

 

[skunksmash]

skunksmash,

If Fumz was right, this thread would have concluded a long time ago....

S-

this may shed some light

The 4GB Windows Memory Limit: What does it really mean? - From BrianMadden.com

im not even gonna attempt to jump in this debate ( b/c i don't know enough about it...lol) ....