Consider a machine with a byte addressable main memory of 2²⁰ bytes, block size of 16 bytes and a direct mapped cache having 2¹² cache lines. Let the addresses of two consecutive bytes in main memory be (E201F)₁₆ and (E2020)₁₆. What are the tag and cache line address (in hex) for main memory address (E201F)₁₆?
(A) E, 201
(B) F, 201
(C) E, E20
(D) 2, 01F


Answer: (A)

Explanation:

Block Size = 16 bytes
Block Offset = 4 

No. of sets or cache lines = 212
Number of index bits = 12

Size of main memory = 220
Number of tag bits = 20 - 12 - 4 = 4 

Let us consider the hex address E201F
Tag lines = First 4 bits = E (in hex)
Cache lines = Next 12 bits  = 201 (In Hex) 

Refer http://virtual-labs.ac.in/labs/cse10/dmc.html

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