QUESTION 11 Assume a direct mapped cache with a block size of 64 2 to the...

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engineering Computer-Science

Answer 1

Answer #1

Block size = 64 B -> Block offset = log 64 = 6 bits

Total # of block inside cache = 1k -> index bits = log 1k = 10 bits

Let the memory size be x bits.

So tag = x - 10 - 6 = x - 16 bits.

Tag

(x - 16)

Index

(10 bits)

Block offset

(6 bits)

A3F2E9 -> 1010 0011 1111 0010 1110 1001

Index bits = 1111001011 = 971

Therefore the given address belong to 971th block of cache.

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Answer 2

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