Question

13. Calculate the total number of bits required to implement a 1024 KiB direct-mapped cache with 10-word blocks. Assume that the cache is byte addressable, and addresses and data words are both 32 bits. (1 KiB = 2¹⁰ bytes)

Answer 1

word size=32 bits=4B

cache memory size=1024KiB= Bytes= words

block size=10 words
There fore, number of blocks=ceil()=ceil(26214.4)=26215

There fore number of bits needed to index a block=15 bits.

Since block size=10 words, we need 4 bits for block offset.

The remaining bits are used for tag.

So number of tag bits=32-15-4=13 bits.

summarizing above,

Number of tag bits=13

Number of bits to index a block=15

Number of bits for block offset=4

Address format will be 13|15|4