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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 ?

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 ? Correct Answer 8, 5, 3

The correct answer is option 1.

 Key Points

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= log2N

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

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

Important Points

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

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