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Question 28 7 pts Consider a byte-addressable computer with 24-bit addresses, a cache capable of storing...

Question:

Question 28 7 pts Consider a byte-addressable computer with 24-bit addresses, a cache capable of storing a total of 64K bytes
Question 28 7 pts Consider a byte-addressable computer with 24-bit addresses, a cache capable of storing a total of 64K bytes of data, and blocks of 32 bytes. If the computer uses direct mapping, the format of the memory address is as follows: bits for the tag field, bits for the cache block number, and bits for the block offset.

Answers

Given byte-addressable computer with 24bit address.

Its meaning is that the computer has word size of 1 byte.

24 bit address means the main memory instructions size is of 24 bit address.

In direct mapping cache memory

Tag bit cache line offset bit block offset bit

<------------------------- 24 bits ------------------------------------------------------------------->

here given that computer has block offset of 32 bytes. means it will be represented by 5 bit of data for block offset.

Cache is capable of storing 64K (216) Bytes of data, it can be represented by 16 bit of the data

As there is only 3 fields and total bits is 24 bits . Therefore tag bit can be calculated as 24 - 16 - 5 = 3 bit. 3 bits assigned for the tag field.

Tag field (3 bit) Line offset (16 bit) Block offset (5 bit)

<------------------------------- 24 bits ----------------------------------------------------------->

For the directmapping, the format of the memory address is as follows

3 bits for the tag field

16 bits for the cache block number

and 5 bits for the block offset.

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