Ace Test Series: CO & Architecture - Cache Memory

Question

 

Comments

@manisha11 tag bits will obviously be needed and for block we will take as byte addressable and so 64 bits  = 64/8 = 8 bytes = 2^3 , so min value of x = 16+3 = 19 

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oh yes 64/8 thanks

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I am also getting 19 as answer Any one please Verifiy….

Answer 1

To get minimun value of X , you should consider that it is block addressable where block size should be 64 bits.

So minimun number of bits will be only for tag bits.

So x=16.

Comments

how you got it, please explain little more ?

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In full associative cache

PA is divided in two part - TAG and WO

where WO is word offset to represent which block in the word.

Now if system is word addressable i.e. we access whole word at a time , we dont need WO.

So PA will only consist of TAG bits.

Thus minimun PA = 16 bits.

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What is block addressable, you can assume byte addressable giving total answer of 19 bits or word addressable giving answer of 16 bits (assuming one word of 8 Bytes). But default we always we take byte addressable i went with 19 bits.

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@Rishabh Agrawal

does this mean here one word is of 64bits ?

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Actually somewhere it is Block addressable while somewhere it is word addressable.Just different terminology.

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@kman30

Yes