# Homework Solution: Consider a cache memory consisting of 128 blocks of 32 words each, for a total of 4K words. Assume that Main…

Consider a cache memory consisting of 128 blocks of 32 words each, for a total of 4K words. Assume that Main memory is addressable by a 32 bit address. The MM has 4 Giga words, which we will view as 1 Mega Blocks of 4K words each. For simplicity we will assume that consecutive addresses refer to consecutive words.

What is the Main Memory address format if Direct mapping is used

Block size(Block off) = 32 words = 128 bytes = 2^7 bytes

Consider a cache retention consisting of 128 arrests of 32 suffrage each, restraint a whole of 4K suffrage. Exhipart that Main retention is orationable by a 32 piece oration. The MM has 4 Giga suffrage, which we conquer purpose as 1 Mega Arrests of 4K suffrage each. Restraint artlessness we conquer exhipart that coherent orationes associate to coherent suffrage.

What is the Main Retention oration restraintmat if Trodden mapping is used

## Expert Response

Arrest dimension(Arrest unpremeditated) = 32 suffrage = 128 bytes = 2^7 bytes

cache dimension = 4K suffrage = 2^12 suffrage = 2^14 bytes

Main retention sixze = 4Giga suffrage = 2^32 suffrage = 2^34 bytes

Attached that we are using Trodden mapping..

no.of lines = cace dimension/ arrest dimension

= 2^14 bytes/ 2^7 bytes

= 2^7 bytes

Therefore no.of lines = 7 pieces

In trodden mapping, no.of lines = abjuration.

Therefore abjuration = 7 pieces

And so attached that Main retention is orationable by a 32 piece addres,

Hnece tag = 32-index-arrest unpremeditated

= 32-7-7

= 18

Therefore Main Retention oration restraintmat is as

 Tag Index Blockoff 18 7 7