Are we sure it's a bad thing when RAM usage goes up?

[ajac]

Originally Posted by Malatesta

Get an app like MemMaid to show CPU cycles.

There's no direct correlation between amount of RAM available and CPU usage. Just because RAM is occupied doesn't mean the CPU is doing any work, there's just no direct connection.

why not? For ram to be occupied it means the CPU has to be doing somethign ram just doesnt get used for no reason.

[krohnjw]

Originally Posted by ajac

why not? For ram to be occupied it means the CPU has to be doing somethign ram just doesnt get used for no reason.

Well, for the data set to get placed *into* ram the CPU must be used. However, it can remain in ram and if nothing is working with the data set then no CPU operations are required to keep it in RAM.

RAM gets used by any application in non sequential fashion. It is entirely possible that data in RAM is out dated or was placed there by a program that has since been closed or is sitting idle (which is why programs reading from an incorrect memory address exhibit sometimes unpredictable results).

However, having that data in ram doesn't require any further cpu operations.

whenever I see more ram used it also means more processor used.. and I wish we had somethign showing how much processor is being used like on windows vista/7

What you may see is that extra work is required validating that the current state of the data in memory is no longer needed so that it is then dirtied/swapped out to allow for newer data to be placed in the same memory location. This occurs when there is not enough free memory (the limit may be less than 100%) to place the required data set into memory or garbage collection occurs. This threshold is going to be determined by the memory manager (and 6.1's is really not very good).

So, with a large amount of used memory you may see extra CPU cycles used when new data has to be written into memory locations that are already occupied by old/stale data, but the actual storage of the current data (assuming it's not being used and nothing is being written to those locations) does not in itself cause any extra CPU usage.

[Malatesta]

Originally Posted by ajac

why not? For ram to be occupied it means the CPU has to be doing somethign ram just doesnt get used for no reason.

To add to krohnjw's excellent explanation, there is a whole other area of cache and allocation of that--in short I devote a lot of my RAM to cache (file system, file system filter, GDI/font, etc),which is variable to a certain extent. In turn I have a "faster" device but consume more RAM; however just because certain areas of the OS are pre-cached doesn't mean it is being acted on by the CPU when not being called up for a process.

Remember, RAM is just volatile memory--it's a temporary storage area for data to be accessed quickly but not written to storage. The RAM chips themselves, AFAIK, are powered by the system regardless of what is being "held" in memory at the time and it is only when that information is called or acted upon does the CPU need to be used.

At least that, in my non-engineering background, is how I understand it.

[1454]

Originally Posted by Malatesta

To add to krohnjw's excellent explanation, there is a whole other area of cache and allocation of that--in short I devote a lot of my RAM to cache (file system, file system filter, GDI/font, etc),which is variable to a certain extent. In turn I have a "faster" device but consume more RAM; however just because certain areas of the OS are pre-cached doesn't mean it is being acted on by the CPU when not being called up for a process.

Remember, RAM is just volatile memory--it's a temporary storage area for data to be accessed quickly but not written to storage. The RAM chips themselves, AFAIK, are powered by the system regardless of what is being "held" in memory at the time and it is only when that information is called or acted upon does the CPU need to be used.

At least that, in my non-engineering background, is how I understand it.

Is there a specific way in which one goes about this?