Consider main memory of the size 64 kB with each word being 8 bits(one byte) only and a direct mapping Cache memory of size 4 kB also having data word size 8 bits.

Find the following :

    1. Find the size of tag and index fields of cache?
    1. In what location of cache hexadecimal address to main memory (AABB) (if exists in cache) will be located?
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    – D.W.
    Commented Aug 1, 2020 at 20:20

1 Answer 1


Answer to your first question is 4 tag bits and cache index bits should be 4 as it is byte-addressable so address bus size is 8 bits so the physical address is therefore of 8 bits and cache index = physical address - tag bits. Approach : tag bits = lg(main-memory size/ cache-memory size). Also clarifying the question will help, I don't think memory is byte-addressable if at all 2nd question has to make sense since AABB will take 16-bit address bus. Or according to the question, it won't exist.


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