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Ready Boost.. Can This Break The 3gb Ram Ceiling For 32 Bit O/s?


CosmicSurfer

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I'm still a newbie to Vista, and I've just run into this... ReadyBoost....

http://en.wikipedia.org/wiki/ReadyBoost

From what I'm reading, I can connect a Top Quality 4GB USB Thumb drive to my computer and RAISE my RAM from 3 GB internal that I run now to an amazing potential of 11GB... yes... that is 11 GB !!!!

I'm running Vista Ultimate... 32 Bit (not 64)

I now have 2x1 GB plus 2x 512 MB = 3GB RAM... (I don't want to get into any discussion here about whether Vista sees the extra 1 GB.. had this discussion many times before....)

Acording to what I read on the Wiki page above, it seems that the 11 GB is achievable... But does Vista actually see and use the additional 8 GB possible?

Or is the traditional 2 or 3 GB ceiling still operative.... ????

BTW... according to the Wiki page, "Microsoft has stated that a 2:1 compression ratio is typical, so that a 4 GB cache could contain upwards of 8 GB of data"

8 + 3 = 11 ... RAM Heaven.

Am I interpreting this correctly????

CS

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Guest Reimar
I'm still a newbie to Vista, and I've just run into this... ReadyBoost....

http://en.wikipedia.org/wiki/ReadyBoost

From what I'm reading, I can connect a Top Quality 4GB USB Thumb drive to my computer and RAISE my RAM from 3 GB internal that I run now to an amazing potential of 11GB... yes... that is 11 GB !!!!

I'm running Vista Ultimate... 32 Bit (not 64)

I now have 2x1 GB plus 2x 512 MB = 3GB RAM... (I don't want to get into any discussion here about whether Vista sees the extra 1 GB.. had this discussion many times before....)

Acording to what I read on the Wiki page above, it seems that the 11 GB is achievable... But does Vista actually see and use the additional 8 GB possible?

Or is the traditional 2 or 3 GB ceiling still operative.... ????

BTW... according to the Wiki page, "Microsoft has stated that a 2:1 compression ratio is typical, so that a 4 GB cache could contain upwards of 8 GB of data"

8 + 3 = 11 ... RAM Heaven.

Am I interpreting this correctly????

CS

Er.. looks good but will it work?

First at all, I tested several Flash cards (CF, SD pp.) and the only card which was recognized as Readyboost able was an Flash-Stick from Sandisk with an readspeed of 25 MB/s! But I wasn't able to use that Stick because of the to slow Write speed of 15 MB/s!

The 3/4 GB barricade of the 32 Bit OS's until now are NOT breakable as far as I know. The only way for to use Sticks pp. for to boost is to install them as kind Ram-Drive.

If you like to bypass that barricade, install Vista 64 Bit and you can use up to 32 GB of memory.

Cheers.

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Er.. looks good but will it work?

First at all, I tested several Flash cards (CF, SD pp.) and the only card which was recognized as Readyboost able was an Flash-Stick from Sandisk with an readspeed of 25 MB/s! But I wasn't able to use that Stick because of the to slow Write speed of 15 MB/s!

The 3/4 GB barricade of the 32 Bit OS's until now are NOT breakable as far as I know. The only way for to use Sticks pp. for to boost is to install them as kind Ram-Drive.

If you like to bypass that barricade, install Vista 64 Bit and you can use up to 32 GB of memory.

Cheers.

Sawadee Bee Mai Reimar... Merry Merry & Happy Happy

Hmmm that is the question... isn't it.

Unfortunately 64 Bit comes with too many downsides and program issues to make it good for me.

According to the Wikipedia page I linked to in my OP these are the required Specs for a Flash drive to work properly:

According to Jim Allchin, for future releases of Windows, ReadyBoost will be able to use spare RAM on other networked Windows Vista PCs.[4]

For a device to be compatible and useful it must conform to the following requirements:

The capacity of the removable media must be at least 256 MB (250 after formatting)

Devices larger than 4 GB will have only 4 GB used for ReadyBoost

The device should have an access time of 1 ms or less

The device must be capable of 2.5 MB/s read speeds for 4 KB random reads spread uniformly across the entire device and 1.75 MB/s write speeds for 512 KB random writes spread uniformly across the device

The device must have at least 235 MB of free space

NTFS, FAT16 and FAT32 are supported (Vista Service Pack 1 and Windows 7 also support the new exFAT filesystem)

The initial release of ReadyBoost supports one device

The recommended amount of flash memory to use for Windows ReadyBoost acceleration is one to three times the amount of random access memory (RAM) installed in your computer

Depending on the brand, wear and tear due to read-write cycles, and size of the flash memory, the ability to format as NTFS may not be available. Enabling write caching on the flash drive by selecting Optimize for performance in Device Manager will allow formatting as NTFS. [5]

It may be worth the investment if I can find the Flash that meets these specs... If it doesn't work, I can always use the Thumb Drive in more traditional ways.

I guess the next question is how will I ever Know if it's working?

Isn't computing Fun?

Sawadee Bee Mai to All

CS

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Readyboost is disk caching memory. It can't be used to run applications in.

OK.. Does this mean that it will relieve any caching currently being done by the RAM...

and theerby free RAM up for Applications???

Or is RAM not involved with caching at all?

Slightly confused,

CS

Edited by CosmicSurfer
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I'm not completely sure but I think it just adds to ram caching rather than replacing it. As far as the 32 bit thing goes, 2 raised to the power 32 gives 4GB which is the maximum address space that Win32 can use. It uses some of that for its own purposes so you finish up with a maximum of 3GBish for app use.

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I'm still a newbie to Vista, and I've just run into this... ReadyBoost....

http://en.wikipedia.org/wiki/ReadyBoost

From what I'm reading, I can connect a Top Quality 4GB USB Thumb drive to my computer and RAISE my RAM from 3 GB internal that I run now to an amazing potential of 11GB... yes... that is 11 GB !!!!

I'm running Vista Ultimate... 32 Bit (not 64)

I now have 2x1 GB plus 2x 512 MB = 3GB RAM... (I don't want to get into any discussion here about whether Vista sees the extra 1 GB.. had this discussion many times before....)

Acording to what I read on the Wiki page above, it seems that the 11 GB is achievable... But does Vista actually see and use the additional 8 GB possible?

Or is the traditional 2 or 3 GB ceiling still operative.... ????

BTW... according to the Wiki page, "Microsoft has stated that a 2:1 compression ratio is typical, so that a 4 GB cache could contain upwards of 8 GB of data"

8 + 3 = 11 ... RAM Heaven.

Am I interpreting this correctly????

CS

Er.. looks good but will it work?

First at all, I tested several Flash cards (CF, SD pp.) and the only card which was recognized as Readyboost able was an Flash-Stick from Sandisk with an readspeed of 25 MB/s! But I wasn't able to use that Stick because of the to slow Write speed of 15 MB/s!

The 3/4 GB barricade of the 32 Bit OS's until now are NOT breakable as far as I know. The only way for to use Sticks pp. for to boost is to install them as kind Ram-Drive.

If you like to bypass that barricade, install Vista 64 Bit and you can use up to 32 GB of memory.

Cheers.

As pointed out earlier, Windows thinks of the Ready Boost device as more of a disk cache than RAM. I.E., rather than paging out the file/program to your HDD, it puts it on the Ready Boost device. Since these are just small files/programs, you don't need crazy throughput. However, having the sub 1 ms access time is a boon since the data can be accessed MUCH more quickly (7x times as fast as a 10k RPM drive!).

It's better to get more RAM, since the access time is that much better than even those flash drives (90 nsec for Opteron or 0.00009 ms versus the 1ms required for Ready Boost). Add in the bandwidth (some 10 GB/s for Opteron or 10 000 MB/s versus the ~25 MB/s that you usually get over USB) and there's no contest.

Under XP you used to be able to use the PAE kernel, worked ok but not as well as using native 64 bit. Also everything I've read indicates that Vista64 is MUCH improved over XP64. If you want to try out PAE on your 32 bit install, follow Microsoft's instructions.

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As a point aside, I cant think that fetching data over a USB bus can be particularly efficient way of increasing your RAM, even for paged data. Consider the speed of reading/writing data to a thumb drive, even with USB 2.0, to disk IO. Now consider the speed of disk IO to RAM IO. On that premise I couldn't see any possible benefits from configuring your system like this.

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As a point aside, I cant think that fetching data over a USB bus can be particularly efficient way of increasing your RAM, even for paged data. Consider the speed of reading/writing data to a thumb drive, even with USB 2.0, to disk IO. Now consider the speed of disk IO to RAM IO. On that premise I couldn't see any possible benefits from configuring your system like this.

The important section from Wikipedia:

This caching is applied to all disk content, not just the page file or system DLLs. Flash devices are typically slower than the hard drive for sequential I/O, so to maximize performance, ReadyBoost includes logic to recognize large, sequential read requests and then allows these requests to be serviced by the hard drive

All the small files that would normally be paged to your swap file are written instead to the Ready Boost device. The swifter access time allows the system to quickly find them, and since they're readily available you can have your disk serving up something else at the same time instead of thrashing about trying to find all those little swapped files while working on a large file.

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I'm not completely sure but I think it just adds to ram caching rather than replacing it. As far as the 32 bit thing goes, 2 raised to the power 32 gives 4GB which is the maximum address space that Win32 can use. It uses some of that for its own purposes so you finish up with a maximum of 3GBish for app use.

Even if you were able to get 32 bit Windows to note all 4GB, Windows LIMITS APPLICATIONS TO 2GB. Documented behavior, well discussed.

Technically 32-bit systems can very well handle 4GB RAM. If you remember primary school sums, it went like this. Right-most column counts ones (1-9), next to the left counts tens, then hundreds and so on. (Ten to the power 0, 1, 2 etc.). Counting in two's (computers can only count in two's because electronic states are nominally on or off) it will look like this from right to left; 8 4 2 1. and also like this; 0000, 0001, 0010, 0011, 0110, 0111, 1000, 1+1=2 which causes the carry. With four bits like here you can only count to fifteen decimal: 8+4+2+1=15. When you get to the 32nd place (32-bit) you will be able to count to 4,294,967,295 bits. This will leave 294,967,295 addresses since 4GB is 4,000,000,000 bits.

The reason why a system is limited to 3GB most likely comes from an operational decision since not even Microsoft can change the physical realities of the machines they support.

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