Question

For a direct-mapped cache memory, the following data is given.         

          Main memory                                       Cache memory

           Size = 64KB                                               Size = 128B

          Block size = 8Bytes                                    Block size = 4Bytes

Calculate the following:

1. Number of blocks created in main memory.
2. Number of blocks created in cache memory.
3. The distribution of the address fields in the system.

Q5. For a direct-mapped cache memory, the following data is given. Main memory Cache memory Size = 64KB Size = 128B Block size = Bytes Block size = 4Bytes Calculate the following: 1. Number of blocks created in main memory. 2. Number of blocks created in cache memory. 3. The distribution of the address fields in the system. Tag Block Word

Answer 1

1. Number of blocks in main memory=Main memory size / Block size

1 KB = 2^10 Bytes

Number of blocks in main memory= 64 * 2^10 / 8

Number of blocks in main memory= 2^13 Blocks

2. Number of blocks created in cache memory =Cache memory size / Block size

Number of blocks in cache memory = 128 / 4

Number of blocks in cache memory = 32 = 2^5

3.Distribution of the address fields in the system -

It is given that system follows direct mapping

address field have 16 bits ( 64KB-->2^16 B )

These 16 bits can be visualised as distributed in TAG, BLOCK, WORD.

Word is nothing but Block offset

since Block size = 8 B = 2^3 B

so Word require 3 bits

The Block for a direct mapped cache is the number of blocks in the cache (5 bits in this case, because 2^5 = 32 )

Then the tag is all the bits that are left

Tag = Total address bits- Block -Word

Tag = 16-5-3 = 8 bits

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