set associative (carl hamacher)

Question

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

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