The old "why do i only have "blank" memory question?

lion149

lion149

Member
My computer currently has 512mb of RAM. Yet the computer says it only has 504mb. I understand that integrated functions take memory but when i installed a new standalone graphics card the number remained the same.

I would reallllly like to know the answer to this...anyone know?
 
If the onboard video is still configured to take some RAM then it will take it regardless of whether you use a discrete card or not. You can try disabling onboard in the BIOS but there may not be an option for it.
 
With two 512s you would see 1,048mb total while 1,024mb available on any system since the system even without onboard video will grab some as well as what Windows does for background services.
 
Interesting...well i booted up my old extra dell computer to check out this theory and i found something interesting. It also has 512mb of RAM. It says under properties (523,832 KB)

so this divided by (1,000) would equal 523.832 mb....so what gives???
not only does it appear to have no loss but an actual gain!!!?

first machine in question (win xp OS)
second machine (windows 2000)
 
Actually you are using two systems here. Windows think 1024MB=1GB, not 1000MB=1GB, so 1024KB=1MB therefore you should have divided by 1024 NOT by 100. 523,832/1024 = is 511.55MB so if you round that off the to nearest number, it will be 512MB. Hard drives demonstrates this in a very dramatic way. On the Box it says 80GB, how come in Windows its only 73.2GB? Where has my 6.8GB gone? Its just the way its read. Manufacturers round off to 1000 to make it all simple for the consumer.
 
Gee? My new 500gb sata drive after partitioning says 465gb! :eek: where did those gigs go? :confused: In actuality the 512mb dimm sees 524mb until reduced by the system to what the OS reports as 512mb total. 2x524=1,048mb less memory reserved by the system plus what is used up with the OS loaded into ram(random access memory).

Hard drive space is calculated by cylinders, sectors, and other things. When you boot up with an old 98 startup floppy to fdisk a small chuck out of a 250gb ide drive and use the SYS C: command to transfer dos system files onto the new partition(any size will do) and type C: while at the cd rom D:> prompt to change to the hard drive type in either dir or dir/w and press enter. You will then see a figure larger then 250,000,000 bytes. Funny how that works when Windows reports 238gb total!

And 1000K=1mb not 512,xxx mb!
 
In actuality the 512mb dimm sees 524mb until reduced by the system to what the OS reports as 512mb total. 2x524=1,048mb less memory reserved by the syste
Click to expand...
No. RAM is counted in binary (ie 1MB = 1024 bytes). OS reserved address space is in the 2.5-4GB range (depending on the mother board).
 
If you look on the post screen you will see the total memory figure exceed the usual amount listed when looking at the 1.024mb amount seen in Windows. The 1,024mb is actually the retailers rounded figure rather then the amount you don't see taken and reserved by the board itself for things like caches and basic hardware processes performed by the ide and other controllers.

Windows then takes some more for it's own background services as well as reserving a certain amount of memory addresses into higher memory. Vista does like a little more then XP there for the latest whatevers. The conversion from bytes into kilobytes and then into megabytes does lose something to end up with the rounded 512mb figure. To really see how this is done you need a math calculator like the one seen at http://familyinternet.about.com/gi/...=http://www.onlineconversion.com/computer.htm

1024 Byte = 0.0009766 Megabyte (MB)
512 Megabyte (MB) = 524,288 Kilobyte (KB)
1024 Megabyte (MB) = 1,048,576 Kilobyte (KB)
2048 Megabyte (MB) = 2,097,152 Kilobyte (KB)
2 Gigabyte (GB) = 2,097,152 Kilobyte (KB)
2 Gigabyte (GB) = 2,048 Megabyte (MB)
2 Gigabyte (GB) = 2,147,483,648 Byte
 
I understand how binary math works. I can do base 2 math in my sleep (it was a long boring easy class I had to take in school). The RAM taken in backround services is allocated RAM, it still shows up in my computer, as seen below. RAM taken for device addressing is taken off the top end of the memory addressing range.
 

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Are you kidding? That was the "line drive" of the season there. "way back.. way back... gggone! for the home run!" :P It kind of reminds you of a tv show running on CBS called "numbers".
 
I would be lying if i said i wasnt still a little confused but i understand my obvious first mistake. Thanks for clearing that up. I will continue to read and if i have more questions. Well i know who to go to, to say the least. haha
 
Even after years of working on systems and being an E tech one statement by a 30yr. computer tech holds true when he stated: "I know I can always go back to the book." meaning referring to techical references. Somehow you'll get there eventually. :P
 
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