Question

Consider a machine with a byte addressable main memory of 2¹⁶ bytes and block size of 8 bytes. Assume that a direct mapped cache consisting of 32 lines used with this machine. How many bits will be there in Tag. line and word field of format of main memory addresses ?
1. 8, 5, 3
2. 8, 6, 2
3. 7, 5, 4
4. 7, 6, 3

Answer 1

Correct Answer - Option 1 : 8, 5, 3

The correct answer is option 1.

 

Direct Mapping: 

Given Data,

Main memory size =216 bytes

No of blocks in Main memory (M)=216  ÷ 23

Block Size (P)= 23 bytes

No of blocks in the cache memory (N) =25 bytes 

Tag=2⁸=Tag bits= log2(M/N)

line or cache block= 2⁵ =cache block  bits= log₂^N

word field or word offset= 2³=word offset bits=log2P^​

∴ Hence the correct answer is (8,5,3).

N = No of blocks in cache memory  =(cache memory size)/Block size

M=  No of blocks in main memory   = (main memory size)/Block size

P=Block size