How much RAM does Vista really take?

The task manager shows under the performance tab around 670 MB RAM
usage (on a 2gb computer), now I looked around in the processes tab,
clicked on the "show processes of all users" thingy, and looked how
much the processes take RAM. I added the usage of all the 45 processes
together, and they eat around 220 MB RAM up.

Why the big difference to the RAM Usage of the performance tab, and
the sum of the RAM usage of the processes in the processes tab?

How much do the processes really take?

It can't be the cache, or? Because the cache usage is shown in the
performance tab too, under "physical memory", and there is the cache
usage 1524 mb of 2029 mb, 51 mb is free.

Any RAM not actually used by programs and services is applied to the system
cache. This cache is populated by programs that, based upon you usage
history, you are likely to use again. By performing this caching, these
programs are already stored in the RAM and do not have to be read from the
hard drive when starting. This speeds up the system.

If you launch any program that need this RAM, sufficient space in the RAM is
cleared to make room for the program you are starting.

Most Vista systems will show a very low amount of "free" RAM. It is of no
concern. Mine usually show zero free RAM.

--


Regards,

Richard Urban
Microsoft MVP Windows Shell/User
(For email, remove the obvious from my address)



<> wrote in message
news: oups.com...
> The task manager shows under the performance tab around 670 MB RAM
> usage (on a 2gb computer), now I looked around in the processes tab,
> clicked on the "show processes of all users" thingy, and looked how
> much the processes take RAM. I added the usage of all the 45 processes
> together, and they eat around 220 MB RAM up.
>
> Why the big difference to the RAM Usage of the performance tab, and
> the sum of the RAM usage of the processes in the processes tab?
>
> How much do the processes really take?
>
> It can't be the cache, or? Because the cache usage is shown in the
> performance tab too, under "physical memory", and there is the cache
> usage 1524 mb of 2029 mb, 51 mb is free.
>

On Jul 7, 3:06 pm, "Richard Urban"
<richardurbanREMOVET...@hotmail.com> wrote:
> Any RAM not actually used by programs and services is applied to the system
> cache. This cache is populated by programs that, based upon you usage
> history, you are likely to use again. By performing this caching, these
> programs are already stored in the RAM and do not have to be read from the
> hard drive when starting. This speeds up the system.
>
> If you launch any program that need this RAM, sufficient space in the RAM is
> cleared to make room for the program you are starting.
>
> Most Vista systems will show a very low amount of "free" RAM. It is of no
> concern. Mine usually show zero free RAM.
>

I know. My Point was this: It shows, that the cache is using 1524 RAM,
I have 2029 RAM. It seems, that Vista filled around 600 MB with
"itself"... but, if I count up the usage of all the 45 processes, I
get only around 220 MB.

So: 600 - 220 = 380 (more 400, because I round here, the actual number
was a bit different). It seems 400 MB are "missing".

But, if I look at the memory usage number in the performance tab (the
green colored number) I get 600 MB used memory.

How can it use 600 MB, if the processes take up together 220 MB, where
does the rest go? It's not the cache, because that is shown
separetely! (around 1500 MB cache usage)


Just do the experiment: Open up the task manager, activate the show
all processes thingy, and count the usage of all the processes. You
will get a much lower number, than the RAM usage shows in the
performance tab. Since cache is separate shown (and is much higher),
this can't be the cache usage.

Open Resource Monitor and add up that which you find in the memory section.
Is it more accurate?

--


Regards,

Richard Urban
Microsoft MVP Windows Shell/User
(For email, remove the obvious from my address)



<> wrote in message
news: oups.com...
> On Jul 7, 3:06 pm, "Richard Urban"
> <richardurbanREMOVET...@hotmail.com> wrote:
>> Any RAM not actually used by programs and services is applied to the
>> system
>> cache. This cache is populated by programs that, based upon you usage
>> history, you are likely to use again. By performing this caching, these
>> programs are already stored in the RAM and do not have to be read from
>> the
>> hard drive when starting. This speeds up the system.
>>
>> If you launch any program that need this RAM, sufficient space in the RAM
>> is
>> cleared to make room for the program you are starting.
>>
>> Most Vista systems will show a very low amount of "free" RAM. It is of no
>> concern. Mine usually show zero free RAM.
>>
>
> I know. My Point was this: It shows, that the cache is using 1524 RAM,
> I have 2029 RAM. It seems, that Vista filled around 600 MB with
> "itself"... but, if I count up the usage of all the 45 processes, I
> get only around 220 MB.
>
> So: 600 - 220 = 380 (more 400, because I round here, the actual number
> was a bit different). It seems 400 MB are "missing".
>
> But, if I look at the memory usage number in the performance tab (the
> green colored number) I get 600 MB used memory.
>
> How can it use 600 MB, if the processes take up together 220 MB, where
> does the rest go? It's not the cache, because that is shown
> separetely! (around 1500 MB cache usage)
>
>
> Just do the experiment: Open up the task manager, activate the show
> all processes thingy, and count the usage of all the processes. You
> will get a much lower number, than the RAM usage shows in the
> performance tab. Since cache is separate shown (and is much higher),
> this can't be the cache usage.
>

On Jul 7, 3:45 pm, "Richard Urban"
<richardurbanREMOVET...@hotmail.com> wrote:
> Open Resource Monitor and add up that which you find in the memory section.
> Is it more accurate?
>
> --
>

Yes. Somewhat.

If I count up the RAM usage under "commit", I will get to the sum,
that the task manager shows under the performance tab.

If I count up the RAM usage under "private", I get a much smaller
number. The RAM usage that is shown in the task manager in the
processes tab is what in the resource monitor the tasks eat up under
"private".

What is the difference between "private" usage and "commit" usage? And
what is the 'real' usage?

Commit, I believe, is what is actually sent to RAM.

--


Regards,

Richard Urban
Microsoft MVP Windows Shell/User
(For email, remove the obvious from my address)



<> wrote in message
news: ups.com...
> On Jul 7, 3:45 pm, "Richard Urban"
> <richardurbanREMOVET...@hotmail.com> wrote:
>> Open Resource Monitor and add up that which you find in the memory
>> section.
>> Is it more accurate?
>>
>> --
>>
>
> Yes. Somewhat.
>
> If I count up the RAM usage under "commit", I will get to the sum,
> that the task manager shows under the performance tab.
>
> If I count up the RAM usage under "private", I get a much smaller
> number. The RAM usage that is shown in the task manager in the
> processes tab is what in the resource monitor the tasks eat up under
> "private".
>
> What is the difference between "private" usage and "commit" usage? And
> what is the 'real' usage?
>

Also, programs will usually commit more to RAM than they actually need to
operate. It is the programmers doing this.

--


Regards,

Richard Urban
Microsoft MVP Windows Shell/User
(For email, remove the obvious from my address)



<> wrote in message
news: ups.com...
> On Jul 7, 3:45 pm, "Richard Urban"
> <richardurbanREMOVET...@hotmail.com> wrote:
>> Open Resource Monitor and add up that which you find in the memory
>> section.
>> Is it more accurate?
>>
>> --
>>
>
> Yes. Somewhat.
>
> If I count up the RAM usage under "commit", I will get to the sum,
> that the task manager shows under the performance tab.
>
> If I count up the RAM usage under "private", I get a much smaller
> number. The RAM usage that is shown in the task manager in the
> processes tab is what in the resource monitor the tasks eat up under
> "private".
>
> What is the difference between "private" usage and "commit" usage? And
> what is the 'real' usage?
>

In task manager, select 'show processes from all users' and you will get the
system processes too.

<> wrote in message
news: ups.com...
> On Jul 7, 3:45 pm, "Richard Urban"
> <richardurbanREMOVET...@hotmail.com> wrote:
>> Open Resource Monitor and add up that which you find in the memory
>> section.
>> Is it more accurate?
>>
>> --
>>
>
> Yes. Somewhat.
>
> If I count up the RAM usage under "commit", I will get to the sum,
> that the task manager shows under the performance tab.
>
> If I count up the RAM usage under "private", I get a much smaller
> number. The RAM usage that is shown in the task manager in the
> processes tab is what in the resource monitor the tasks eat up under
> "private".
>
> What is the difference between "private" usage and "commit" usage? And
> what is the 'real' usage?
>