cache memory mapping

Question

Answer 1

Actually the significance of tag is 

How many blocks of main memory are mapped to same set of cache memory just like in case of direct mapping how many blocks of main memory gets mapped to a line(block) of cache memory..

So having said that ,

No of lines in cache given = 16

Associativity  i.e.  k  =  4

So no of sets   =  No of lines in cache / k

                      = 16 / 4

                      =  4

But there are 64 blocks of main memory in total..

Hence no of blocks that will map to same set  = No of blocks in main memory / No of sets

                                                                   =  64 / 4

                                                                   = 16

So no of tag bits as mentioned earlier the significance of tag  =  log₂16

                                                                                         =  4

Hence D) should be the correct option..

Comments

Thanks a lot Habib..got it now.very clear explanation ..:)

Answer 2

For direct mapped: TAG+ INDEX+  WORD-OFFSET = (2+4 (16 cache blocks)) 64 BLOCKS + X

Tag	INDEX	WR-OFF
2	4	x

for 4-way set associative

16/4 =4 sets 

Tag	index	wr-offset
4	2	x

so 4 is ans

Comments

@gabbar how are u getting word size and tag size?there is also no information given on cache and main memory sizes.

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pls reply.  @gabbar

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Here word offset or block size is not given...but main memory block size equal to cache memory block size I have taken as "x" bits for word offset I.e block size is takes as 2^x.

Now to find index cache blocks are given...log(16)=4 bit if we have used direct mapping...

Total address length = log(64) + x bits = 6+x bits...

Now tag = 6-4 =2 bits...

But here set associative mapping need to use...16/4 = 4 ...index bits are log(4)=2...

So tag =6-2 =4 bits ...