unacademy test series

Question

Answer 1

Given Physical address (Main memory address) is 36 bits

Physical address space (Main memory size) = $2^{36}$ Bytes

The Block Size is given as 128 Bytes = $2^{7}$ Bytes

The Total no of blocks in Main memory = $\frac{ 2^{36}B }{2^{7}B}$ = $2^{29}$ Blocks

Cache size is given to be 64 KB so the no of lines (Blocks) in Cache =  $\frac{ 64KB }{2^{7}B}$ = $\frac{ 2^{16}B }{2^{7}B}$

                                                                                                             =  $2^{9}$ Lines

The minimum tag bit size can be achieved when the cache uses Direct Mapping..

The maximum tag bit size can be achieved when the cache uses Fully Associative Mapping..

 

              

               

 

From the above example we can clearly visualize that the tag bits size gets reduced when we use

Direct mapping  as we map each block of main memory to a particular Cache line using Line bits

whereas in Fully associative...any block can be mapped to any Cache line as there are no Line bits…

Here in this question….

                                           

 

The Tag bits size in Fully associative = Main memory address bits – Block offset

                                                           = 36 – 7

                                                           = 29 bits

The Tag bits size in Direct Mapping = Main memory address bits – (Line bits + Block offset)

                                                           = 36 – (9 + 7)

                                                           = 20 bits

The Difference in the size = 29 – 20 = 9 bits