Question

multiply hifhlighted number by 4 and please explain

Given: An 8-bit computer system with 4 MB of main memory. It has a 16K-words direct-mapped cache with a block size of 16 Bytes. Sought: Calculate the cache line number when Byte address 5678ten from main memory is mapped to cache. Consider that numbering of main memory blocks and cache lines starts from zero. Express your answer in hexadecimal. It is required to show ALL incremental steps of the solution.

Answer 1

Solution:

Given,

=>8-bit computer system

=>Size of main memory = 4 MB

=>Size of cache = 16 K words

=>Direct mapped cache

=>Block size = 16 bytes

=>Byte address = 5678 in decimal

=>Need to multiply byte address by 4.

Explanation:

Direct mapped cache:

Tag

Cache index

Block offset

8 bits                      10 bits                                      4 bits

Number of bits required for main memory address:

=>Number of bits required to express main memory address = log₂(main memory size)

=>Number of bits required to express main memory address = log₂(4 MB)

=>Number of bits required to express main memory address = log₂(4*2^20 B) as 1 MB = 2^20 B

=>Number of bits required to express main memory address = log₂(2^22 B)

=>Number of bits required to express main memory address = 22 bits

Number of bits required for cache index:

=>Size of cache = 16 K words

=>We know that system is 8-bits so 1 word = 8 bits

=>1 word = 1 byte as 1 byte = 8 bits

=>Size of cache = 16 KB

=>Number of blocks in the cache memory = Size of cache memory/size of cache block

=>Number of blocks in the cache memory = 16*1024 B/16 B as 1 KB = 1024 B

=>Number of blocks in the cache memory = 1024 blocks(0-1023)

=>Number of bits required for cache index = log₂(number of blocks)

=>Number of bits required for cache index = log₂(1024)

=>Number of bits required for cache index = log₂(2^10)

=>Number of bits required for cache index = 10 bits

Number of bits required for block offset:

=>Number of bits required for block offset = log₂(size of block)

=>Number of bits required for block offset = log₂(16 B)

=>Number of bits required for block offset = log₂(2^4 B)

=>Number of bits required for block offset = 4 bits

Number of bits required for tag:

=>Number of bits required for tag = 22 - 4 - 10

=>Number of bits required for tag = 8 bits

Calculating value of cache line number:

=>Mulitply the byte address by 4 = 4*5678

=>So calculated new address = 22712

=>In direct mapped cache byte address is mapped according to the relation: B mod P = i where B is the byte address of main memory, P is the number of blocks present in the cache memory and i the cache line number where the main memory block address is mapped.

=>22712 mod 1024 = 184 in decimal

=>Cache line number in hexadecimal = (B8)₁₆

I have explained each and every part with the help of statements attached to it.