retagged by
1,096 views
3 votes
3 votes
what will be size of main memory. when 4-way set associative mapping  of cache memory is done and cache size is 256 KB and Tag field has 7 bits( consider, memory is byte addresable )
retagged by

2 Answers

1 votes
1 votes

Just Find Total number of bits in tag when direct mapped cache.

7bit tag given in 4way set associative so in direct mapped cache it is of 7-2 =5bits tag.

So main memory is 25 times bigger than cache memory.

So main memory size = 32×256KB =8MB

If you get logic you can do directly without conversion.

1 votes
1 votes
tag set  block size
7 x y

cache size=no.of set*lines per set*size of set

256 KB=4*2y*2x

216  = 2y*2x

x+y=16

The number of bits in physical address=7+x+y=23

The main memort size =223 = 8 MB

No related questions found