ReadyBoost options? SD or USB?

I tried a 4GB ReadyBoost USB device and I didn't see any difference in operation of my Vista Home Premium system.  I've quit using it.   If you do a google search you will see that it's questionable whether it really does help.  Unless you are constantly and continually accessing the hard drive it won't "wear out" in regular use.

You mention Google searches, basically internet access. However all your internet access or anything you use the internet explorer with will be stored on your disk “C” Under “temporary internet files”. That is a default and is built into the Windows, any version, operating system.   

I do not know if ReadyBoost will change the default address of the “temporary internet files” I personally find that difficult for many reasons; remember, internet explorer is nested into the windows O.S.

However, you can save all your work to the flash drive and save anything you like to into the flash drive. You can save anything from photos to college papers to work presentations, even data tables for Microsoft Access. That is what saves a tremendous amount  of  disk access.

Nevertheless ReadyBoost is just a memory increase of the main system  and it will be used after you run out of memory on the built in memory in your system. For example, if you have 2GB of  memory in your system  you will be accessing the flash drive very seldom unless you have a large number of applications and other data open and loaded in memory. Then, only when the memory gets overflowing is when your hard drive acts as a ”Virtual Memory” and starts “paginating”, there is when things really slow down. What the ready boost does is to take over for your hard drive and act as regular memory (solid state) saving your Hard Drive a ton of work and speeding up access to data. Remember, all that occurs after the regular memory goes into overflow and the HD would start the pagination prosess. And flash drives are slower than regular memory.

Please read this:

http://www.microsoft.com/windows/products/windowsvista/features/details/readyboost.mspx

mtabernig is not correct for the ReadyBoost perfromance. The flash drive or SD card is a mirror of the hard disk cache. It is not an add-on to system RAM.

 

The hard disk cache will contain all the cache data, nothing is missing. The ReadyBoost will also contain all or some of the hard drive cache, and will be used for faster transfer of data for some files. Use of ReadyBoost will not reduce the amount of hard drive activity used for cache operation.

 

Flash cache will read/write smaller files faster than the hard drive, while the hard drive could perform faster for larger files. It will all depend on the speed of the particular flash unit. Some flash contains a mix of fast and slow memory, some has more faster and some are just slow. If a flash is selected that just meets the minimum required by Windows, you might not see a great improvement. You should buy the fastest unit you can find.

 

Dimension 9100, Dual-Boot Win XP / Vista Home Premium, 3.0 GHz P4, 3 GB DDR2 533 MHz RAM, 2 GB SanDisk USB ReadyBoost, 160 GB SATA II Samsung (XP), 300 GB SATA II Seagate (Vista), 250 GB SimpleTech USB (WD Drive), Nvidia Go 6800 (425/825 MHz - XP, 325/600 MHz - Vista), Dell 1901 UltraSharp FP

Inspiron E1705, Win Vista Premium, T7200 Core 2 Duo (4MB, 2.0 GHz 667MHz), 2 GB DDR2 677 MHz RAM, 2 GB Transcend 150X SD ReadyBoost, 120 GB Samsung HD, Nvidia Go 7900 GS - 156.69 Driver, 17” Sharp UltraSharp TrueLife Wide-Screen WUXGA

This is what it says:

 

"The flash memory device serves as an additional memory cache—that is, memory that the computer can access much more quickly than it can access data on the hard drive. Windows ReadyBoost relies on the intelligent memory management of Windows SuperFetch and can significantly improve system responsiveness."

 

It says additional cache, not additional RAM for operation. Just what I said.

 

Try this MS link:

 

http://windowsvistablog.com/blogs/windowsvista/archive/2006/11/20/windows-readyboost.aspx

 

Is even more detailed and specifically declares that ReadyBoost IS NOT SYSTEM RAM.

 

Dimension 9100, Dual-Boot Win XP / Vista Home Premium, 3.0 GHz P4, 3 GB DDR2 533 MHz RAM, 2 GB SanDisk USB ReadyBoost, 160 GB SATA II Samsung (XP), 300 GB SATA II Seagate (Vista), 250 GB SimpleTech USB (WD Drive), Nvidia Go 6800 (425/825 MHz - XP, 325/600 MHz - Vista), Dell 1901 UltraSharp FP

Inspiron E1705, Win Vista Premium, T7200 Core 2 Duo (4MB, 2.0 GHz 667MHz), 2 GB DDR2 677 MHz RAM, 2 GB Transcend 150X SD ReadyBoost, 120 GB Samsung HD, Nvidia Go 7900 GS - 156.69 Driver, 17” Sharp UltraSharp TrueLife Wide-Screen WUXGA

Did you follow and read the Microsoft link? It do not mention anything about mirrowing anything. In fact, it supports my explanation. However, perhaps you can direct us to a site with a more comprehensive explenation.

This is a direct quote from the site you directed us

 

"I should be clear that while flash drives do contain memory, Windows ReadyBoost isn’t really using that memory to increase the main system RAM in your computer.  Instead, ReadyBoost uses the flash drive to store information that is being used by the memory manager.If you are running a lot of applications on a system that has limited memory, Windows ReadyBoost will use the flash drive to create a copy of -virtual memory - that is not quite as fast as RAM, but a whole lot faster than going to the hard disk. "

 

I think that is, in my own words, what my responce was. Perhaps I was not as eloquent.:smileywink:

Their use of the term Virtual Memory is not referring to memory that can be used to run actual programs. It is referring to CACHE. Cache is virtual memory. The cache can hold the image of a program that was in memory, but was moved to cache to make room for another program. The cache can then re-load the first program when needed and store the second till needed if there is not enough RAM to hold both. Thus, the speed of the cache can make it look like both programs are in RAM at the same time, when they are not. This saves time in not having to re-load a program that was pushed out of RAM to make room for something else. ReadyBoost is only a COPY of what is in the normal hard drive cache, nothing more. It can be faster when moving programs/data back into RAM than the actual hard disk cache, depending on file sizes. You wouldn't call a hard disk cache RAM, and ReadyBoost is not RAM and does not act like RAM. This is what Virtual Memory is - it is not real.

 

Adding ReadyBoost (or using a larger cache) will help speed up operations by minimizing the time to switch programs into and out of RAM. Increasing the amount of actual RAM is always more beneficial by having more room for the OS and applications. Cache is why you can run Windows with less than the amount of RAM actually needed to run it. However, the lower the RAM, the slower the operations because key parts of the OS are going in and out of cache. With lots of Apps running, low RAM systems tend to get constipated with all the in-out disk operations.

 

Adding an additional 1 GB RAM to a 1 GB Vista system will do much more for perfromance than adding a 1GB ReadyBoost flash drive.

 

Many people have the same misconceptions about ReadyBoost (and cache). Real RAM is better, and RAM costs are now about the same as usable flash drives or memory cards. Slow cards will not work, and most usable cards are not a whole lot faster than disk cache. Fast flash drives are considerably more expensive than RAM.

 

Dimension 9100, Dual-Boot Win XP / Vista Home Premium, 3.0 GHz P4, 3 GB DDR2 533 MHz RAM, 2 GB SanDisk USB ReadyBoost, 160 GB SATA II Samsung (XP), 300 GB SATA II Seagate (Vista), 250 GB SimpleTech USB (WD Drive), Nvidia Go 6800 (425/825 MHz - XP, 400/800 MHz - Vista, Vista Driver - 163.75), Dell 1901 UltraSharp FP

Inspiron E1705, Win Vista Premium, T7200 Core 2 Duo (4MB, 2.0 GHz 667MHz), 2 GB DDR2 677 MHz RAM, 2 GB Transcend 150X SD ReadyBoost, 120 GB Samsung HD, Nvidia Go 7900 GS - 156.69 Driver, 17” Sharp UltraSharp TrueLife Wide-Screen WUXGA

KirkD

 

That is what I was talking about, However, you will notice the difference when the goes into overrun and the computer starts to PAGINATE... when the cpu starts to send the small files to the HD then retrieving them....because there is no physical memory left....The HD acts as Virtual memory,,,, the process is very slow because of the access time of a mechanical shortcoming of the HD. RedyBoost redirects that HD access to the flashcard that is much faster than the HD.

As long that you have phisical memory (RAM) there will be no or very small diffirence on the pc's performance:smileyhappy: