Question

1. Suppose a computer using direct mapped cache has 2³² bytes of main memory and a cache of 1024 blocks, where each block contains 32 bytes.
   1. [2] How many blocks of main memory does this computer have?
   2. [4] Show the format of a memory address as seen by cache; be sure to include the field names as well as their sizes.
   3. [3] Given the memory address 0x00001328, to which cache block will this address map? (Give you answer in decimal.)
2. A fully associative mapped cache has 8 blocks, with 16 bytes per block. Main memory is 128K bytes.
   1. [3] Show the format for a main memory address, assuming memory is byte addressable. Include the field names as well as their sizes.
   2. [3] Show the format for a main memory address, assuming memory is word addressable, where a word is 32 bits. Include the field names as well as their sizes.

Answer 1

Main Memory Size = 2³² Bytes

Number of Cache Blocks = 1024 = 2¹⁰

Block Size = 32 Bytes = 2⁵

Answer a

Number of blocks in Main memory

= Main Memory Size / Block Size

= 2³² / 2⁵ = 2²⁷

=  134,217,728 blocks

Answer b

Cache is Direct Mapped. So

Block Offset = 10 bits (to access the blocks, number of blocks = 2¹⁰)

Byte Offset = 5 Bits (to access each byte in selected block, block size = 2⁵)

Tag bits = 32 - 10 - 5 = 17 bits

Main Memory Address = Tag(17) Block(10) Byte(5)

Answer c

0x00001328 = 00000000000000000 0010011001 01000

Block offset =  0010011001 = 153 block

Fully Associative Cache

Number of blocks in Cache = 8 = 2³

Blocks Size = 16 Bytes

Main Memory Size = 128 K bytes = 2¹⁷ which means main memory address = 17 bit

Answer a

Byte offset = 4 bits

Tag Field = 17-4 = 13 bits

Main memory address = Tag(13) Byte(4)

Answer b

Memory is word addressable which means block size = 16 bytes = 4 words where 1 word = 32 bits (4 bytes)

Word offset = 2 bits  

Tag Field = 17-2 = 15 bits

Main memory address = Tag(15) Word(2)