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Address bus - patterns - HELP!

Discussion in 'A+' started by tony_baduk, Nov 14, 2007.

  1. tony_baduk

    tony_baduk Bit Poster

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    Hi all,
    Currently studying the A+ course - 601 & 602 path.

    Just a quick one really. On page 67 of Mike Meyers Sixth Edition, it mentions if you have 20 wires, you would have 1,048, 576 combinations - how did he calculate this.

    Maths was always my low point so would appreciate your help.

    Thanks.
     
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  2. greenbrucelee
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    greenbrucelee Zettabyte Poster

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    The address Bus is basically a set of wires that connect the the CPU and memory, so lets say a CPU has a 32 bit address bus(32 wires) which means the calculation is 2^32 which is 4GB so that means the 32bit address can access 4GB of memory.

    Hope this helps, maths aint my strong point either
     
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  3. greenbrucelee
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    greenbrucelee Zettabyte Poster

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    just to add when it was older CPUs it would have been 2^16 or even 2^8
     
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  4. tony_baduk

    tony_baduk Bit Poster

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    Cheers Greenbrucelee for your reply. But the question is HOW did you get that answer? What's the method?
     
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  5. greenbrucelee
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    greenbrucelee Zettabyte Poster

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    There are 1024 bytes per kilobyte.

    There are 1024 kilobytes per megabyte or 1,048,576 bytes.

    There are 1024 megabytes per gigabyte or 1,073,741,824 bytes per gigabyte.

    The conversion on the address bus calculation is done to the power of ^ so when Working the calculation of lets say a 16 bit data bus (16 bits wide) = 65536 bytes / 1024 = 64KB this is in the Mike Myeres AIO 6th edit if you have it

    / means devide
     
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  6. tony_baduk

    tony_baduk Bit Poster

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    Thanks again. I kind of get it.
     
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  7. greenbrucelee
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    What study materials are you using?
     
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  8. greenbrucelee
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    When doing to the power of calculations I do it in a way that suits me so 2 to the power of 16, is 16 time by 2 11 times which is 65536 devide by 1024 is 64.
     
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  9. Fergal1982

    Fergal1982 Petabyte Poster

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    actually, to get 65536 with your method you have to multiply 16 by 2 only 11 times. 2^16 is actually
    2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2
     
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  10. greenbrucelee
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  11. Fergal1982

    Fergal1982 Petabyte Poster

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    The thing to remember with computing is that a lot of it is based on binary. Binary is a base 2 counting system, hence the 2^x. Think about it like this:

    Every line can either be on or off, 1 or 0. Binary counting (base 2 remember) works like that so if only the first lane is on, then the address is 1. if only the second lane is one, then the address is 2, if both are on, the address is 3.

    So if every lane has two possible states, and there are 20 lanes, then there are 2^20 (2x2x2x2x2x2x2x2x2x2... - 20 2's in total - or 20 lines with two possible states) combinations (1048576).
     
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  12. Mathematix

    Mathematix Megabyte Poster

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  13. BosonMichael
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    He's got it, guys... rep provided.
     
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  14. greenbrucelee
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    greenbrucelee Zettabyte Poster

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    I knew someone could explain it better than I.

    Explain Maths aint my strong point, I can do it but find it hard to explain it well.
     
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