cache memory

Question

when 8-way set associative mapping  of cache memory is done and main memory size is 32 GB and Tag field has 10 bits. what will be the cache size ( consider, memory is byte addresable )

Comments

256 MB ??

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yes, this is right , but how ?

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@amrendra pal

same concept as in previous qsn.

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@joshi_nitish, can you tell that how many comparators are use here and what is the size of each comparator

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eight 10-bit comparators.

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@joshi_nitish, can you tell that how many multiplexers are used here and what will be the size of that multiplexers

Answer 1

Main memory size = 32 GB = 2^35 B (Byte addressbale) 

Size of physical address is = 35 bit 

Let x be the set offset, y be the word offset.

No. Of set = 2^x

No. of words in block = 2^y 

1 set = 8 blocks ( 8 way set associative ) 

2^x set contains = 2^x * 8 blocks = 2^(x+3) blocks

Cache size = No of blocks * No of words per blocks 

Cache size = 2^(x+3) * 2^y = 2^(x+y+3) 

Tag bits + x + y = 35 bit 

x + y = 35-10 = 25 bit 

Cache size = 2^(25+3) = 2^(28) B = 256 MB

Hence 256 MB is the answer

Comments

@ amrendra pal

can you tell me ....? what is the use of multiplexure here....??

i think...

if we talk about H/W configuration then  here in case of k-way set associative mapping we use......

k- comparator (size=tag bits) and 1 OR gate...??

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@Ashwani kumar 2,

can you tell that how many multiplexer are used here?

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@hs_yadav,

multiplexer is used here for selecting the sets ? but can you tell how many multiplexer is used and what is the size of that MUX

Answer 2

First Find the tag size in Direct mapped cache.In 8 way set associative it is of 10bits.So in direct mapped it is of 10-3=7bits.

This means main memory is 2⁷  Times bigger than cache memory.

So cache size = 32GB÷128 =256MB

Answer 3

LA=35bits

tag=10

set=p  let the block size is 2^x  then

word=x bits ...

total no blocks for MM=2^35-x

tag bits required=2^35-x  / 2^p =2^35-x-p  =35-x-p=10 ->x+p=25

now cache size=8 _* 2^x _* 2^p =2²⁸ ->256MB

Comments

This is also correct approach.

Answer 4

Main Memory is 32 GB which is 2⁵ x  2³⁰ bytes which would need 35 bits.

lets assume set needs x bits and word size needs y bits to represent.

TAG	SET	WORD SIZE
10	x	y

Where 10+x+y=35
hence x+y=25

now ,

Total Cache size = (total no of sets) x (no of blocks in one set)  x  (word size ).

Total cache size = 2^x x 8 x 2^y

Total cache size = 2^x+y x 8 = 2²⁵ x 8 = 256 MB.