in CO and Architecture edited by
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Block set associative cache consists of a total of 64blocks divided into 4blocks sets .The main memory contains 4096blocks ,each consisting of 128 words.

  1. how many bits for Main memory
  2. how many bits for TAG,SET,WORD .

solution:

MM=block size*words

         2^12 * 2^7=19 bits   

TAG=9

SET=4

WORD=6 

is this correct method or not please correct me

in CO and Architecture edited by
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2 Comments

1st part is correct.

2nd part is wrong

In cache memory block size is aligned with main memory block size. So it will contains 128 words within a block.

Now there are 64 blocks in cache which is 4 way set associative, means in each line is a set of 4 blocks, so totally there will be 16 lines, which required 4 bits and main memory require 12 bits apart from 7 bits for words. So out of 12 bits 4 bits will be used for $\text{SET OFFSET}$ and 8 remaining bits will be used for $\text{TAG OFFSET}$

$\text{TAG - 8 bits}$

$\text{SET - 4 bits}$

$\text{WORD- 7 bits}$
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thanks @!KARAN 

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