GATE CSE 2017 | Set 2 | Question: 53

Question

Consider a machine with a byte addressable main memory of $2^{32}$ bytes divided into blocks of size $32$ bytes. Assume that a direct mapped cache having $512$ cache lines is used with this machine. The size of the tag field in bits is _______

Comments

A similar type of question is asked in GATE CSE 2021 Set 1 | Question: 22

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fill 18 in the blanks

 

Answer 1

No. of blocks of main Memory $= \dfrac{2^{32}}{2^5} = 2^{27}$

And there are $512 = 2^9$ lines in Cache Memory.

Tag bits tell us to how many blocks does $1$ line in Cache memory points to

$1$ cache line points to $ \large \dfrac{2^{27}}{2^9} = 2^{18}$ lines

So, $18$ bits are required as TAG bits.

Comments

Here they are asking for Size of tag field. SO , it should be 18*2^9 bit

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Tag field means Tag bits and $18*512$ would be size of complete tag memory.

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why are we not  taking address as 27 bit when it says main memory is divided into blocks of 32 bytes? why does it imply cache block is 32 bytes?

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Because the memory is byte addressable not block addressable.

Answer 2

$32 - 9 - 5 = 18$

Comments

The length of @Arjun Sir answer always varies with the complexity of the question.  $:)$

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This answer suits this phrase, “You reap what you sow“. This is the easiest question one can expect in CO for GATE.

Answer 3

Main memory = 32 bits

Cache memory = 512 lines = 9 bits + 5 bits for block offset = 14 bits

Tag bits = 32 - 14 = 18 bits

Answer 4

hope it might help.......

Answer 5

In Direct Mapped Cache

Tag bits = log(Main Memory size/ cache size) 

Cache size = No of blocks * Block size = 512*32 = 2^14

Tag bits = log(2^32/2^14) = log(2^18) = 18 bits

Answer 6

Answer is 18 bit

In the case of direct mapping Tag field is the ratio of main memory and cache memory.