Question

QUESTION 2 Suppose a computer using direct mapped cache has 216 bytes of byte-addressable main memory and a cache of 64 blocks, where each cache block contains 32 bytes. a. How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, (include field names and their sizes) c) To which cache block will the memory address (F8C916 map? What address in that block does it map to?

Answer 1

It is given that memory size = 2¹⁶ bytes

So memory address should be of size 16-bit because of byte-addressable main memory, each byte should be uniquely addressed.

a) Number of blocks of main memory

Since each block of main memory is mapped into a block in the cache.

The size of the main memory block is also the same as size of cache block.

Main memory size = 2¹⁶ bytes

Block size = 32 byte = 2⁵ bytes

Number of blocks of main memory = 2¹⁶ / 2⁵  = 2¹¹ = 2048 blocks.

b) Format of memory address seen by the cache.

The 16-bit memory address is seen by the cache is divided into 3 fields tag, block number, and offset(within a block).

Size there are 64 cache blocks, so the block number field is of size 6 bit. (6 bits are required to identify one block)

Size there one cache block is 32 bytes, so the offset field is of size 5 bit. (5 bits are required to identify one bye within a block)

Therefore the size of the tag field = 16 - 6 - 5 = 5 bits

Memory address =

tag(5 bit)

block(6 bit)

offset(5 bit)

c) Memory address (F8C9)₁₆

(F8C9)₁₆ = 1 1 1 1 1 0 0 0 1 1 0 0 1 0 0 1

tag(first 6 bits) = 1 1 1 1 1

block number(next 6 bits) = 0 0 0 1 1 0 = 6th block of the cache

offset(last 5 bits) = 0 1 0 0 1 = 9th byte within the block

So (F8C9)₁₆ maps to 6th block of the cache and it refers to 9th byte within that block.

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