I am a Vista home premium user. I open my task manager and see that, the RAM usage is ~640M and that's annoying me... But while I calculate by adding up the RAM consumed by each application, I have a fas smaller number ~172M only
That means some 400M usage is not apparent to me, can anyone tell me where it goes or where I can see the remaining information, so that i can manage the usage of RAM better?
THanks!
Raymond
Full Version: Ram usage shows difference... Can any one tell me why?
Of course adding applications' RAM usage (assuming you're even checking the right column in task manager) isn't going to add up to your total mem usage -- it never has.
My copy of Vista (with no apps running at all) at bootup takes 365MB -- pretty close to the 400MB figure you've mentioned. That much RAM usage is perfectly normal.
BTW, RAM is cheap. You can get 2GB of DDR2 800MHz CL4 pretty much year-round for $30, or 4GB (2x2GB) of the same stuff for about $70...
My copy of Vista (with no apps running at all) at bootup takes 365MB -- pretty close to the 400MB figure you've mentioned. That much RAM usage is perfectly normal.
BTW, RAM is cheap. You can get 2GB of DDR2 800MHz CL4 pretty much year-round for $30, or 4GB (2x2GB) of the same stuff for about $70...
Vista and W2K8 work in the principle that "unused memory is wasted memory" - if you are not using it for applications then the Cache Manager will use it.
This does not mean that it is not available for use - it has not prevented anything from launching, or significantly slowed down the allocation of physical memory.
The system cache is a representation of data already on disk, so it speeds up the system when requests for that data are made - this is a good thing.
If the "free" physical memory is not enough to fullfil a request from a driver or process, then the Cache Manager can drop some of its data out of physical memory to allow the request to complete.
No data is lost as the cached data is still there in a non-volatile form on the disk, so if it is needed later then a disk I/O will fetch it back.
ReadyBoost is a mechanism to use flash memory storage purely for cache (it is not "extra RAM" or used for the page file at all), if you don't want to install extra memory.
This does not mean that it is not available for use - it has not prevented anything from launching, or significantly slowed down the allocation of physical memory.
The system cache is a representation of data already on disk, so it speeds up the system when requests for that data are made - this is a good thing.
If the "free" physical memory is not enough to fullfil a request from a driver or process, then the Cache Manager can drop some of its data out of physical memory to allow the request to complete.
No data is lost as the cached data is still there in a non-volatile form on the disk, so if it is needed later then a disk I/O will fetch it back.
ReadyBoost is a mechanism to use flash memory storage purely for cache (it is not "extra RAM" or used for the page file at all), if you don't want to install extra memory.
QUOTE (crahak @ Jul 14 2008, 03:55 PM) 
Of course adding applications' RAM usage (assuming you're even checking the right column in task manager) isn't going to add up to your total mem usage -- it never has.
My copy of Vista (with no apps running at all) at bootup takes 365MB -- pretty close to the 400MB figure you've mentioned. That much RAM usage is perfectly normal.
BTW, RAM is cheap. You can get 2GB of DDR2 800MHz CL4 pretty much year-round for $30, or 4GB (2x2GB) of the same stuff for about $70...
My copy of Vista (with no apps running at all) at bootup takes 365MB -- pretty close to the 400MB figure you've mentioned. That much RAM usage is perfectly normal.
BTW, RAM is cheap. You can get 2GB of DDR2 800MHz CL4 pretty much year-round for $30, or 4GB (2x2GB) of the same stuff for about $70...
QUOTE (Mr Snrub @ Jul 14 2008, 04:42 PM)
Vista and W2K8 work in the principle that "unused memory is wasted memory" - if you are not using it for applications then the Cache Manager will use it.
This does not mean that it is not available for use - it has not prevented anything from launching, or significantly slowed down the allocation of physical memory.
The system cache is a representation of data already on disk, so it speeds up the system when requests for that data are made - this is a good thing.
If the "free" physical memory is not enough to fullfil a request from a driver or process, then the Cache Manager can drop some of its data out of physical memory to allow the request to complete.
No data is lost as the cached data is still there in a non-volatile form on the disk, so if it is needed later then a disk I/O will fetch it back.
ReadyBoost is a mechanism to use flash memory storage purely for cache (it is not "extra RAM" or used for the page file at all), if you don't want to install extra memory.
This does not mean that it is not available for use - it has not prevented anything from launching, or significantly slowed down the allocation of physical memory.
The system cache is a representation of data already on disk, so it speeds up the system when requests for that data are made - this is a good thing.
If the "free" physical memory is not enough to fullfil a request from a driver or process, then the Cache Manager can drop some of its data out of physical memory to allow the request to complete.
No data is lost as the cached data is still there in a non-volatile form on the disk, so if it is needed later then a disk I/O will fetch it back.
ReadyBoost is a mechanism to use flash memory storage purely for cache (it is not "extra RAM" or used for the page file at all), if you don't want to install extra memory.
Thanks for both your reply!
Well, I was asking this because I see a far less difference in my XP desktop (around 100M) compare to Vista Laptop (around 400M).
Mr Snrub, as in your reply, where can I adjust the system cache? As I want to understand more about the use of RAM.
Thank you very much!
Raymond
You don't want to change the system cache - simply read from Task Manager / Performance / Physical Memory (MB):
"Cached + Free = available physical RAM"
"Cached + Free = available physical RAM"
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