Question

Q1.

a. A block-set-associative cache consists of a total 64 blocks divided into 4 - block sets. The main memory contains 4096 blocks, each consisting of 128 words. Find: i. How many bits are there in the main memory address? Ii. How many bits are there in tag ,set and word field

Answer 1

i)The main memory size = 4096 blocks

= 2¹²

Each block consist of 128 words = 2⁷ words

Hence total size of main memory in word lengths

= 2¹² x 2⁷

= 2¹⁹

Hence there will be 19 bits in the main memory

ii)The main memory size = 4096 blocks

= 2¹²

Hence the tag and set field combine will have 12 bits.i.e.

TAG + SET = 12

Number of sets = 64/4 = 16

=> The cache is divided into 16( =  2⁴) sets  

SET = 4

=> TAG = 12-4 = 8

Each block consist of 128 words = 2⁷ words

Hence the word field length will be 7 bits

Hence there are 8,4,7 bits in tag set and word field respectively.

Answer 2

Ans. main memory size is = 4096 blocks = 2¹²

Every block consist of 128 words = 2⁷ words

Therefore the total size of main memory in length of words = 2¹² x 2⁷

= 2¹⁹

So, there is at least 19 bits in the main memory.

ii) Size of the main memory = 4096 blocks = 2¹²

Therefore, the tag and set the field in a combined manner will consist of 12 bits.i.e. TAG + SET = 12

So, the Number of sets is = 64/4 = 16

=> The cache is then divided into 16(= 2⁴) sets  

SET = 4

=> TAG = 12-4 = 8

Every block contains of 128 words = 2⁷ words

Hence the word field length will be 7 bits. 8, 4, 7 bits respectively are in the tag set and word field.