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A processor can support a maximum memory of $4\;\textsf{GB}$, where the memory is word-addressable (a word consists of two bytes). The size of address bus of the processor is at least _________bits.
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Size of Memory = No of words (Addresses)  $\times$ No of bits per word

$2^{32}\;\textsf{B} =$  No of words (Addresses)  $\times \;2\;\textsf{B}$

No of words (Addresses)  $= 2^{31}$

Number of Address lines $= 31$

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Max memory $= 4\,GB = 2^{32}\, Bytes$

since $1\,word=2B$.. Total number of words$=\dfrac{2^{32}}{2}=2^{31}$ ..

so to address these words we need minimum $31$-$bit$ address bus.

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Memory Size = $4GB/2B=2G \ words=2^{31} \ words$

If the processor has to refer to any word in this memory, it's address would contain 31 bits. Hence, the address bus will have a minimum size of 31 bits.


Size of a word is the smallest unit of data that can be fetched or stored in a basic operation. So, the data bus will have a minimum size of 16 bits (Size of a word)


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Maximum Memory = 4GB = 232 bytes
Size of a word = 2 bytes
Therefore, Number of words = 232 / 2 = 231
So, we require 31 bits for the address bus of the processor.
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