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i will try to Answer …...

In Set Associative Memory

       Tag Bit  

  Number Of Set

  Page Size

 

We Know Cache Size Is

Number Of Set  X  Number Of Page In Each Set   X Size Of Each Page     =  256KB

                Number Of Set  X  16   X Size Of Each Page                                   =  256KB

                 Number Of Set  X   Size Of Each Page                                            =  2^18 / 2^4

                 Number Of Set  X   Size Of Each Page                                            =   2^14

Number Of Bits ( Number Of Set  X   Size Of Each Page)                             =  14

 

  Total Size of Physical Address  = 36 bits

   Then Tag Size  =  36 – 14

                              =    22  bit

 

 

 

 

 

 

 

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Asim Abbas asked Jan 22, 2018
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The answer was given as 22 bits. I didn't get their method of solving this question. Kindly explain the solution. I even doubt their answer.