0 votes 0 votes psb asked Jan 11, 2017 psb 205 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 0 votes 0 votes no of main memory blocks = 400 no of cache blocks =20 no of sets = no of cache blocks / 2 = 20/2 =10 mod function realised during mapping of block no 232 is K MOD S= i where K is any block no S is any SET no .so 232 MOD 10=2 therefore value of Xis 2 focus _GATE answered Jan 11, 2017 • selected Jan 11, 2017 by Sushant Gokhale focus _GATE comment Share Follow See all 0 reply Please log in or register to add a comment.