Consider a cache memory of 16 words. Each block consists of 4 words. Size of the main memory is 256 bytes. Draw associative mapping and calculate TAG, and WORD size.

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written 6.7 years ago by

teamques10 ★ 70k

• modified 3.9 years ago

computer organization and architecture

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4.4 years ago by teamques10 ★ 70k

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written 3.9 years ago by

teamques10 ★ 70k

Given Data -

Main Memory Size = 256 bytes = $2^8 bytes$

Cache Memory Size = 16 Words = $2^4 bytes$

Block Size = 4 Words = $2^2 bytes$

To Find -

TAG Field Size = ?

WORD Field Size = ?

Associative Mapping Structure = ?

Solution -

Physical Address = 8 bits obtained from the size of Main Memory ($2^8$)

WORD Field Size = 2 bits obtained from the size of Cache Block Size ($2^2$)

TAG Field Size = Number of bits in the Physical Address - Number of bits in the WORD Field

= 8 - 2 = 6 bits

Associative Mapping Structure looks like as follows:

Associative Mapping

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