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Option :

The correct answer is option 1, option 2, and option 4.

Concept:

The given data,

Cache Memory Size = 2 KB

Main Memory Size = 64 KB

Block Size = 64 B

Calculation:

Number of Lines =Cache Memory Size / Block Size

Number of Lines = 2K/ 64

Number of Lines = 211/26

Number of Lines = 25

16 bit 

Tag

5 bit

Number of lines

5 bit

Word Offset

6 bit

The addresses of the first bytes of P, Q, R, and S are,

P (A248)H 10100 01001 001000 9 th block
Q (C284)H 11000 01010 000100 10 th block
R (CA8A)H 11001 01010 001010  10 th block
S (A262)H 10100 01001 100010 9 th block

But P&S are from the same memory block (10100).

  • Every access of S is hit.
  • Once P is brought to the cache it is never evicted.
  • Every access to R evicts Q from the cache.

[ alt="F1 Savita Engineering 28-3-22 D10" src="//storage.googleapis.com/tb-img/production/22/03/F1_Savita_Engineering__28-3-22_D10.png" style="width: 185px; height: 161px;">

Hence the correct answer is option 1, option 2, and option 4.

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