How does Windows ReadyBoost improve Vista’s performance?

Vista site on MSDNOne of the Vista features I was intrigued by, and formerly a little confused by, is Windows ReadyBoost, which promises to improve system performance when you plug a flash drive (or even a fast enough USB 2.0 drive) into your Vista PC.

Jim Allchin explained ReadyBoost in the Vista Team blog:

"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.  What is very cool here is that there is nothing stored on this flash disk that isn’t also on the hard disk, so if you remove the flash drive, the memory manager sees the change and automatically goes to the hard disk.  While the performance gain from ReadyBoost is gone, you don’t lose any data and there is no interruption.  And because the Windows Readyboost cache on the flash drive is encrypted using AES-128, you don’t need to worry about exposing sensitive data if the flash drive is stolen or lost.  Also, the memory manager compresses the pages before writing them into the cache on the flash disk, which means you’ll get more mileage from each MB.

"So, if you just want your PC to run faster with Windows Vista -- it's pretty simple -- connect your flash drive through any USB 2.0 socket or PCI interface and when the autoplay interface comes up, choose 'Speed up my system using ReadyBoost.' " 

Read the rest of Jim's explanation here

Comments (1)

  1. Mark says:

    surely with Vista’s hardware specs being large as regards memory etc this facility is not really required as the system should already have enough memory to function.

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