Question

10. A 64 K cache has lines that are 128 bytes long, and is 4-way set associative. The cache is in a computer with a 32-bit address. Answer the following questions: A) How many lines are in the cache? B) How many sets are in the cache? C) How many tags are in the cache? D) How big is each tag? E) If the cache uses an LRU replacement algorithm, how many extra bits will be required to keep track of which set is least recently used?

Answer 1

Answer:-

(A) Cache is partitioned into lines.

Given: Each line is 128 bytes long.

Cache size=64 K=64*1024 bytes

Number of lines=Cache Size/Line size

=(64*1024)/128

=512

No. of lines in given Cache=512

(B) Given 4-way Associative Mapping

4 Lines form a Set

Total no. of Sets= (No. of Lines)/4 = 512/4 =128

(C) No. of Tags=No. of Lines=512

(D)

TAG

INDEX

OFFSET

MEMORY ADDRESS

Given 32 bit address

No. of Sets= 2⁷

Index=7 bits

Block Size=128=2⁷

Offset=7 bits

Tag Size=32-7-7=18 bits

(E) Each set has 4 Lines as it is 4-way Set Associative Cache

For LRU set replacement algorithm 4 extra bits are required.