cache size=4K word=212 words
Block size=64 words=26 words
So, number of blocks=212/26=26=64
Processor fetches 4352 words contiguously
So, number of blocks required to access all word=4352/64=68
so,(68-64)=4 blocks need to be replaced in each 9 accesses using LRU policy.Now consider the 1st time when blocks are being fetched in cache from MM.As LRU is used,so in the particular set(in which the block will be placed),the oldest block will be replaced.This will be done for all the 4 sets.Now,when 2nd time the sequence will be accessed,then the oldest block for the all 4 sets will not be found.So this block will be fetched from MM and it will be kept in the set by replacing the 2nd oldest block.So,for 2nd oldest block,miss will occur.Then this will be fetched and will be placed by replacing 3rd oldest block.This process will continue until the 5th block for the set is placed.So for all 5 blocks there will be miss.So total miss in each of the 9 accesses=(5*4)=20
so total cache misses=68+(9*20)=248
so,total MM word accesses for this miss=248*64=15872
total cache accesses=(9*(68-20)*64)=27648
if cache was not present,then total MM word access would be=4352*9=40788
so improvement factor=(40788*10*cache access time) / ((15872*10*cache access time)+(27648*cache access time))=2.1