4 4 votes Consider a n-way cache with 'x blocks of 64 words each. The main memory of the system is having 8 million words. Size of the tag field is 16 bits and additional memory required for tags is 1024 bytes. What will be the values of n and x respectively? Answer 256 512 CO & Architecture co-and-architecture cache-memory made-easy-test-series + – Ram Swaroop 2.1k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
Best answer 7 7 votes Memory required for tags = no. of blocks * tag size 1024B = x * 16-bit x = 512 Main memory size : 2^23 words can be represented in 23 bits Tag + Set + Word = 23 bit 16 + Set + 6 = 23-bit Set = 1bit No. of sets = 2 n = No.of blocks / No. of sets = 512 / 2 = 256 (x,n) = (512,256) balchandar reddy san answered Jan 30, 2019 • selected Jan 30, 2019 by Shaik Masthan balchandar reddy san comment Share Follow See all 2 Comments 2 2 Comments reply lokesh singh mahar commented Mar 22, 2020 reply Follow flag I think there is a correction required Tag + Set + Word = 23 bit 16 + P-Way + 6 = 23-bit P-way bit = 1bit so it is a 2-way set associative cache So answere must be :- (x,n) = (512, 2) 0 0 replyShare debasree88 commented Jun 17, 2020 reply Follow flag Main memory size : 2^23 words can be represented in 23 bits Tag + Set + Word = 23 bit why did you consider 23 bits as cache address length? main memory and cache memory length can not be same always...... i think more feasible is to take Tag + Set index = main memory tag and main memory tag represents main memory blocks.. pls clear 0 0 replyShare Please log in or register to add a comment.
