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Question

Consider a machine with a byte addressable main memory of 2²⁴ bytes, block size is 32 bytes and 4 way set associative cache having 2¹⁵ cache blocks. Block size of cache is 32 byte then what is the set and tag address of memory (E4201F)₁₆ in hexadecimal?

A) 0100, 39

B) 0010, 37

C) 0110, 39

D) 0100, 36

Comments

option A)  ?

Answer 1

Block size  = 32 bytes ( 5 bit need to represent it)
Number of lines  = 2¹⁵ 
4 way set associative
Number of sets = 2¹³
24 bit will be represented as 6 ( TAG) + 13 (Number of set) + 5 (block)

Address given = (E4201F)₁₆ i.e (1110 0100 0010 0000 0001 1111)₂

First 6 bit will be use for tag i.e 1110 01 or 39 in hex
Next 13 bit will be used for set i.e 0100 in hex

option a is correct here

Comments

thanks

Answer 2

(E4201F)₁₆   ...

1110 01(39)    00 0010 0000 000(0100)     1 1111