4 Gig or ram- Only showing 3 gig

  • Thread starter Thread starter vyaw2003
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vyaw2003

Hello,
Just read the longest post on this issue with no result so let try
again.
If i have 4 gig, only 3 shows, 750mb is reservered for the system.
Doesn't that seem a little stupid? If i have 3 gig of ram, 3 gig
shows in vista, here it has not needed to reserve 750mb. So why does
the first scenario need 750 for the system, and the second does not?
This would mean that the second does not have any ram for its system,
same devices and os. I would have thought that both situations would
result in having 750 stripped for the system. My old box, XP 256 mb
ram all that was avaliable and did not need dedicated system ram.
Does anyone have a straight answer for this? Yes i could run 64 bit
os, but at the moment i want to get 4 gig running in my 32 bit
environment.

I tried the following switches with no change.
1. Reboot Vista to safe mode with command line
2. Run the command : bcdedit /set PAE ForceEnable
3. Run bcdedit again to verify the switch is added
4. Reboot the system and check whether the problem is fixed
Click to expand...

This is from the original post:

http://groups.google.com.au/group/m...+ram+enterprise&rnum=1&hl=en#37dbc2867738d289

"LOL, I was scratching my head over that one too. By definition, a 32-
bit CPU
can directly address up to 4GB of memory. But apparently nearly 1GB of
that
is reserved for devices. So the most you can possibly see is 3.12GB in
a
32-bit system. The way I read it, there is no way around it, no "fix"
if
you're using a 32-bit version of Vista.

To see any more than 3.12 GB you have to be using a 64-bit version of
Vista
plus meet all those other requirements (64-bit CPU instruction set,
chipset
with 8GB address space, BIOS that supports memory remapping). That's
the way
I read it anyway.

I think the DEP/ PAE thing is a whole different (but related) issue. "
 
also, my first reply on this:
someone posted that if you use PAE you could get the full 4 gig
showing, but it wont make any difference as the system will still need
to use 750mb for itself.
then why would a xp machine with 128 mb not fall over it would show
128mb, but need a fair wack for the system??
Sounds like to different outcomes?? which brings me to the original
point. The system cant hide the ram and use it for itself. XP and
Vista are the system and they use the ram that it can see and no more
"extra hidden ram".
how could a system address ram that it has not detected??
 
Try an experiment.
Turn off all virtual memory, page files, set it all to zero.
Reboot and see if you notice any performance difference in your 4 gb
machine.
 
Hi,

You aren't fully reading the explanations. It is the addresses that are
reserved for the system, and this is regardless of how much ram is
installed. In XP you didn't see it because the amount of ram installed was
nowhere near the 4GB of addressing space of this 32-bit OS. The /PAE switch
allows for more addressing space, so then you can see it.

--
Best of Luck,

Rick Rogers, aka "Nutcase" - Microsoft MVP

Windows help - www.rickrogers.org
My thoughts http://rick-mvp.blogspot.com
 
Hi,

You aren't fully reading the explanations. It is the addresses that are
reserved for the system, and this is regardless of how much ram is
installed. In XP you didn't see it because the amount of ram installed was
nowhere near the 4GB of addressing space of this 32-bit OS. The /PAE switch
allows for more addressing space, so then you can see it.
Click to expand...
Rick just to clarify, "so then you can see it". Even if i can see it
with a /PAE switch = 4gig, the machine will only ever use say 3 gig
because the other part is reservered for the system addresses. So the
max ram that can be used comes down to how many addresses it needs to
allocate for the system eg 750mb leaving the remaining addresses out
of 4096mb to be allocated to ram.

Long answer and short answer, is machines are now made for 64 bit ram
sizes, and 64 bit cpu's and dual cores, turn your Windows cd into
coasters and go and buy 64 bit OS's now!!
 
If i have 4 gig, only 3 shows, 750mb is reservered for the system.
Doesn't that seem a little stupid? If i have 3 gig of ram, 3 gig
shows in vista, here it has not needed to reserve 750mb.
Click to expand...

Yest it has reserved 750MB, itt's just that the 750MB is above the 3GB
so you don't see it. When you have 4GB RAM, the 750MB has to be
taken out of the total.

So why does
the first scenario need 750 for the system, and the second does not?
Click to expand...

Here's the diagram again:

System total:
|-------------------------4GB-------------------------------|

With 4GB:
|---------------3GB RAM-----------------|----System------|

With 3GB:
|---------------3GB RAM-----------------|----System------|

With 2GB:
|--------2GB RAM------------|--Unused--|----System------|

With 3GB or less, you won't see any difference between installed
RAM and amount available to Windows.

Tom Lake
 
You aren't fully reading the explanations. It is the addresses that are
reserved for the system, and this is regardless of how much ram is
installed. In XP you didn't see it because the amount of ram installed was
nowhere near the 4GB of addressing space of this 32-bit OS. The /PAE switch
allows for more addressing space, so then you can see it.
Click to expand...

This is really funny, like deja vu all over again.

Remember when IBM put all the ROMs at 640k because "no-one would ever
need that much RAM"?

Still, the system mappings have to do somewhere, I guess. Do they
fill the top part of the map, or is it a matter of 5M worth of stull
scattered from 3.12G upwards, breaking contiguous addressability?

Does addressability still need to be contiguous, in the post-286 age?


------------ ----- ---- --- -- - - - -
Click to expand...
The most accurate diagnostic instrument
in medicine is the Retrospectoscope
 
Hi Chris,
Remember when IBM put all the ROMs at 640k because "no-one would ever
need that much RAM"?
Click to expand...

Indeed, still talk about that with the old pc buffs.
Still, the system mappings have to do somewhere, I guess. Do they
fill the top part of the map, or is it a matter of 5M worth of stull
scattered from 3.12G upwards, breaking contiguous addressability?
Click to expand...

I honestly don't know on this one. I can't see where that much system
address space is required, but then again you never know. I think it's more
along the lines of an arbitrary point chosen as the cut off line, and
anything above it is marked as reserved, used or not.
Does addressability still need to be contiguous, in the post-286 age?
Click to expand...

I don't believe so, but again I don't know for sure.

--
Best of Luck,

Rick Rogers, aka "Nutcase" - Microsoft MVP

Windows help - www.rickrogers.org
My thoughts http://rick-mvp.blogspot.com
 
Addresses mapped
4GB of RAM installed
|---------------------|--------------|------|---|
1-2 GB for OS
2 GB for program "The hole"
| ~750MB |
Device
|------|--||
Video BIOS etc.

On your computer, you have a video card that has it's own memory. To make it
work faster, it is mapped as part of your RAM. You cannot have two areas of
memory with the same addresses, so the 750 MB of RAM above 3.12 GB is
"blocked", "re-routed" or "reserved" so that (in this case) the video card
can use those addresses for direct access to the processor. Other devices
_may_ do the same thing: Sound, BIOS, etc. So, the whole block of RAM is
"reserved." If you have a super-video card, you may even see less than 3GB.

When you only have 3GB installed, those addresses are not duplicated, so
they do not need to be "blocked" from program use and you see the whole 3GB.
When you have 4GB installed, those "reserved" addresses are blocked from
your use and available memory appears impaired.

The devices need unique addresses. 32-bit is limited to 4GB and the Vista
(32-bit) actually limits you to 3.12 GB because the "hole" must exist. So,
even with gimmicks, you can't get more because the devices need those
addresses. Instead, you can claim some of the OS operating space at a cost of
OS efficiency.
Gimmicks:
/PAE gives four more bits to the address bus (36-bits), but cannot be used
if you have less than 4GB of RAM installed:
http://msdn2.microsoft.com/en-us/library/aa366796.aspx
bcdedit /set increaseuserva (used to be /3GB) can be use to re-divide the
program/kernel memory usage:
http://msdn2.microsoft.com/en-us/library/bb613473.aspx
Use both with caution since they require certain setups to exist and impose
new limitations.

A 64-bit OS will allow you to use more memory, but the hole still exists (it
will vary in size) because those devices are _typically_ hard-coded to those
addresses.
 
With a 64Bit OS the OS addresses can start at 750 and can go to 4750 so you
don't loose any actual space because of the reserve for hardware requiring
system addresses.
 
[email protected] said:
Hello,
Just read the longest post on this issue with no result so let try
again.
If i have 4 gig, only 3 shows, 750mb is reservered for the system.
Doesn't that seem a little stupid? If i have 3 gig of ram, 3 gig
shows in vista, here it has not needed to reserve 750mb. So why does
the first scenario need 750 for the system, and the second does not?
This would mean that the second does not have any ram for its system,
same devices and os. I would have thought that both situations would
result in having 750 stripped for the system. My old box, XP 256 mb
ram all that was avaliable and did not need dedicated system ram.
Does anyone have a straight answer for this? Yes i could run 64 bit
os, but at the moment i want to get 4 gig running in my 32 bit
environment.

I tried the following switches with no change.

This is from the original post:

http://groups.google.com.au/group/m...+ram+enterprise&rnum=1&hl=en#37dbc2867738d289

"LOL, I was scratching my head over that one too. By definition, a 32-
bit CPU
can directly address up to 4GB of memory. But apparently nearly 1GB of
that
is reserved for devices. So the most you can possibly see is 3.12GB in
a
32-bit system. The way I read it, there is no way around it, no "fix"
if
you're using a 32-bit version of Vista.

To see any more than 3.12 GB you have to be using a 64-bit version of
Vista
plus meet all those other requirements (64-bit CPU instruction set,
chipset
with 8GB address space, BIOS that supports memory remapping). That's
the way
I read it anyway.

I think the DEP/ PAE thing is a whole different (but related) issue. "
Click to expand...
 
I'm showing 3.50 GB of RAM on a 4GB system.

[email protected] said:
Hello,
Just read the longest post on this issue with no result so let try
again.
If i have 4 gig, only 3 shows, 750mb is reservered for the system.
Doesn't that seem a little stupid? If i have 3 gig of ram, 3 gig
shows in vista, here it has not needed to reserve 750mb. So why does
the first scenario need 750 for the system, and the second does not?
This would mean that the second does not have any ram for its system,
same devices and os. I would have thought that both situations would
result in having 750 stripped for the system. My old box, XP 256 mb
ram all that was avaliable and did not need dedicated system ram.
Does anyone have a straight answer for this? Yes i could run 64 bit
os, but at the moment i want to get 4 gig running in my 32 bit
environment.

I tried the following switches with no change.

This is from the original post:

http://groups.google.com.au/group/m...+ram+enterprise&rnum=1&hl=en#37dbc2867738d289

"LOL, I was scratching my head over that one too. By definition, a 32-
bit CPU
can directly address up to 4GB of memory. But apparently nearly 1GB of
that
is reserved for devices. So the most you can possibly see is 3.12GB in
a
32-bit system. The way I read it, there is no way around it, no "fix"
if
you're using a 32-bit version of Vista.

To see any more than 3.12 GB you have to be using a 64-bit version of
Vista
plus meet all those other requirements (64-bit CPU instruction set,
chipset
with 8GB address space, BIOS that supports memory remapping). That's
the way
I read it anyway.

I think the DEP/ PAE thing is a whole different (but related) issue. "
Click to expand...
 
I have read thread and referenced KB articles. But what is unclear is whether
I am getting any performance benefit from 4gig. That is, is the 750 mb
reserved for addresses enhancing my performance in any way? Or did I just pay
for an extra gig that does me no good?
 
In message <[email protected]>
GaryReger said:
I have read thread and referenced KB articles. But what is unclear is whether
I am getting any performance benefit from 4gig. That is, is the 750 mb
reserved for addresses enhancing my performance in any way? Or did I just pay
for an extra gig that does me no good?
Click to expand...

There is no benefit in that last 750MB -- However, you are getting
benefit from the remainder of that gig.

More importantly though, that probably means you have 4x1GB sticks,
which probably means dual channel mode is available. This gives some
extra performance, so even though you aren't able to address or use the
last 750MB, you're better off with that stick in then without it.
 
If your Vista 64 GB system is showing that all 4GB are supported then you
are getting a lot of benefit especially if you run applications that use a
lot of memory such as video file conversions. In any case if all 4GB are
supported then all of it that is not currently in use will be used for disk
caching which can greatly increase the response time of reading files off of
disk such as video or photo files which have not been recently used or time
shifted video if you are using MC with TV programming.
 
I have read thread and referenced KB articles. But what is unclear is whether
I am getting any performance benefit from 4gig. That is, is the 750 mb
reserved for addresses enhancing my performance in any way? Or did I just pay
for an extra gig that does me no good?
Click to expand...
If you are using the 64 bit version of Vista then you ARE getting the
full benefit of the ful l4 gig. However if like most people you have
the 32bit version you spent money unnecessarily.
 
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