Missing 1mb of ram???

Q_Ball

I ran the diagnostic tool on B & W and it tells me i have 127mb of RAM?

How did i lose the 1mb and how do i get it back?

WinME btw
PC100

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It's a 106 miles to Chicago. We've got a full tank of gas, half a pack of cigarettes, it's dark and we're wearing sunglasses.

HIT IT!

Red Moose

You haven't lost it. It's all based on old ways of what a Kilobyte is and how manufacturers quote easy numbers (same for hard disks).

Data is made up of "bits" - 0 or 1

A byte is 8 bits

A kilobyte is 1024 bytes (not 1000, but remember this no-one ever said it's based on a metric system).

A megabyte therefore is 1024 * 1024 bytes. Or maybe it's 1000 of the 1024 bytes. That's where the confusion lies.

1024 bytes * 1024 = 1,048,576 bytes in a megabyte

Multiply by 128 (because you have "128MB of RAM") and you get 134,217,728 bytes in this memory amount.

Now the other way would be to say 128MB of RAM is 128,000,000 bytes (1,000,000 bytes is 1 megabyte).

Dividing it down by 1024 to give the amount of kilobytes is: 125,000. But why not divide by 1000 if you take 128MB to be 128,000,000 and so on.

Basically I've lost track a little but for example, I have 128MB RAM in my system. Windows reads this as 130,576KB of RAM, which if you divide by 1024 you get 127.51MB of RAM, which I guess B&W's installer reads as 127MB because that's what it told me as well.

Basically it doesn't matter. I read an explanation of the confusion a while ago; it dates back a long time, and I suppose a quick Google search on "Origin of 1024 kilobyte" might reveal something.

It's more likely to cause consumer confusion in hard disks, where the multiples make much more of a difference when talking about 40GB of data.

This is the advertised amount, but it really means 1024 * 1024 for 1 MB. Then * by 1024 for 1 GB, then * by 40 for 40 GB, giving a number that isn't 40GB if you see all the bytes at once and so on.

Good to make use of the Windows calculator though.

SickBoy

I would say its more down to the newer intel i8xx chipsets. They for some reason steal 1 meg of RAM for their intergrated graphics chip at boot. These chipsets(i810/i815/i820...) have only 512k of real vid memory, the rest is system RAM. The 512k is very fast memory though and is used as a cache basically. I've seen this on loads of machines in work especially when booting NT4, it says "Windows NT4 workstation, 1 CPU, 127megs RAM" Nothing to worry about...

Q_Ball

Cheers lads

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It's a 106 miles to Chicago. We've got a full tank of gas, half a pack of cigarettes, it's dark and we're wearing sunglasses.

HIT IT!

_CreeD_

And 2 more....

Some BIOS's "deduct" your conventional Ram allocated at start up from the displayed amount. It's purely cosmetic, as all systems do this, most just list the full installed amount anyway.

The original version of Win95 (and Win95a I think) used to deduct any ram allocated to SmartDrv from it's own ram count.
Ah, all those nice Gateway customers demanding theire 2mb of ram back....was music to my ears...

Samson

I know this off thread but I'm curious:

"It's a 106 miles to Chicago. We've got a full tank of gas, half a pack of cigarettes, it's dark and we're wearing sunglasses."


Where is this quote from Q_Ball ?

gandalf

I'll answer this for Q_Ball its from the The Blues Brothers just after the Big Ballroom gig

Gandalf.

po0k

There was also something in PCGamer aw hilembcak about the Kilobuyte thing

according to LAtin, or greek or english or maths or wahtever......2 to the power of 20 is a kibibyte i think. if we were to take the 1bit * 8 * 1024 * 1024 * 1024 = 1 GB then we should prounce it and spell it GibiByte. or so I'm told.
Macs also use the 1bit * 8 * 1000 * 1000 * 1000 = 1GB. Its strange. But the things said about BIOSes and chipsets stealing RAM, though possibly (and probably knowing my track record on these boards) correct, don't sound quite as convincing as teh confusion around the calculation of the bytes simply because these usnits ahve bee arounf a hell of a lot longer then the i810/i810e/i815/i820 and they are found on my old hard drives here beside me. There for I conclude taht it is the calculation explanation which is true.
Now I'll sit back and be scorned for giving my 2 cents. or maybe not as this isn't teh CS boards :P