Question about ReadyBoot (Not ReadyBoost)

Everlong · Aug 18, 2008

Was having a poke around Vista, and like always I found something new (even after 10 months of using it, I'm finding new stuff). I came across ReadyBoot and looked around the properties, and it mentions something about the amount of RAM it uses to boot up to load processes and drivers etc. The default size is 20MB, so I was thinking, the more stuff you have to load at bootup, would increasing this size result in any benefit at all? As once this is full, I guess it uses the hard drive, which would be considerably slower.

motarola2 · Aug 18, 2008

The core idea of ReadyBoost is that a flash drive has a much faster seek time, allowing it to satisfy the requests fairly quickly compared to a hard drive when booting or reading certain system files.

Everlong · Aug 18, 2008

This is about ReadyBoot though :P

Barman58 · Aug 18, 2008

Microsofts explanation plus registry locations are shown here dont know if this will give you any further info

Everlong · Aug 18, 2008

Thanks, that helped

Barman58 · Aug 18, 2008

that's what we're here for :D

motarola2 · Aug 19, 2008

Everlong18 said:
This is about ReadyBoot though :P

hehe, oops, my bad :geek:

Question about ReadyBoot (Not ReadyBoost)

Everlong

To xor or not to xor

My Computer

System One

motarola2

My Computer

System One

Everlong

To xor or not to xor

My Computer

System One

Barman58

My Computers

System One System Two

Everlong

To xor or not to xor

My Computer

System One

Barman58

My Computers

System One System Two

motarola2

My Computer

System One