GATE CSE 2016 Set 1 | Question: 09

Question

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.

Answer 1

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 $= 31$

Comments

@HitechGa  what is the significance of the term at least ?

31 bits will be minimum to address 4GB and they themselves mentioned maximum memory  is 4GB, If 31 bits can do the job, any number > 31 will do the job i.e to adress 4GB memory.

I think to limit answer to 31 they used at least term.

 

 

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Why there is no significance of block size here? Does address bus refer to particular word or a particular block in physical memory? Is it possible that address bus can contain address of particular block on physical memory? Please answer these?

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i think this question should be in COA

Answer 2

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.

Answer 3

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.

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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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Answer 4

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.