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

1) Main memory contains = 4096 blocks = 2¹² blocks

each block size = 128 words = 2⁷ bits

Hence totals number of bits in main memory = 2¹² x 2⁷ = 2¹⁹ bits.

2) as number of blocks = 64 = 2⁶

=> word field will require 6+1 = 7 bits

one set contains 4 blocks, so for 64 blocks we require 16 sets = 2⁴ set

Hence set field will require 4 bits.

Since total memory is 2¹⁹ so TAG field will require,

19-(7+4) =  8 bits.

Hence number of bits in TAG, SET and word field are 8,4,7

Answer 2

It depends on the bits in the memory address of a computer that will decide the maximum number of bytes.

For example, if you using 16 bit computer then it allow 16 bits for the address.

1 byte=8 bit

128 words is equal to 128*16=2048 bits  

Totally, 2048 bits are there in the main memory, so 2048 bits is the right answer.