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Why is Memory Increased by 2?

Posted on 2004-03-31
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Last Modified: 2010-04-26
Why is it taht memory increases, or doubles
like how we have 32ram, 64ram, 128ram, 256ram?
like why every time is it doubled? brief explanation would do :) thanks
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Question by:markyello
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InteraX earned 30 total points
ID: 10724032
Hi markyello,

This is due to the way memory and computers work. It is common sense to do everything as a power of 2. ie 2^4=16, 2^5=32 etc...

As everything in a computer is done in binary, ie. base2, then these are the easiest numbers to deal with.

Good luck.
;-)
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by:Huseyin1
Huseyin1 earned 20 total points
ID: 10724058
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by:juliangkeller
ID: 10724887
This tells you why and how they make the chips and name them the way they do:

http://internal.vusd.solanocoe.k12.ca.us/Buck/pc_tech/study/memory.htm
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by:multihull
ID: 10727456
Hi,
simply put, following on from Interax.

Its all to do with the addressing in binary. as you increase the number of bits in the address register you increase the amount of addressing available.

Working with the decimal system the increase is 10, 100, 1000, 10000. So when you reach 9 an extra it is required to make 10, when you reach 99 (still 2 bits) the next increment is 100 - extra bit aded.

The binary system is similar and the count needed has to include more than a single bit as base, so base 8 is used and is repesented by 4 single bits as:

   000 to 111 (7) then 1000 (8)   :  1111 (77) then 10000 (80) and so on in base 8 so the number doubles as it reaches the end of its count string and goes to the net bit.  (in decimal this would be 0 - 7, to  16. to 32). It appears complicated because we think in decimal and automatically add the extra bit to tak us from tens to hundreds. We do not consider that each bit can be from 0 to 9.

As computers count in binary the octal system is used to be able to get the address numers in easy arrays otherwise there would not be sufficient "holes" for each bit to fall in. And the decimal equivalents give the doubling effect without showing the need for the extra address bit required. Its all a matter of representation.

If thats enough I will finish as I will start to confuse myself.

Hope it helps

Multihull
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