GATE 2023
August 11, 2023GATE 2019
August 11, 2023GATE 2023
| Question 64 |
An 8-way set associative cache of size 64 KB (1 KB = 1024 bytes) is used in a system with 32-bit address. The address is sub-divided into TAG, INDEX, and BLOCK OFFSET.
The number of bits in the TAG is ____?
The number of bits in the TAG is ____?
| 19 |
Question 64 Explanation:
Let block offset be x bits
Block size = 2x bytes
Number of blocks per cache = 64*1024 / 2x = 216-x
Number of blocks per set = 8
Thus, Number of set = 216-x/8 = 213-x
Address = 32 bits = index + tag + block offset
32 = 13-x + tag + x
x = 19 bits
Block size = 2x bytes
Number of blocks per cache = 64*1024 / 2x = 216-x
Number of blocks per set = 8
Thus, Number of set = 216-x/8 = 213-x
Address = 32 bits = index + tag + block offset
32 = 13-x + tag + x
x = 19 bits
Correct Answer: A
Question 64 Explanation:
Let block offset be x bits
Block size = 2x bytes
Number of blocks per cache = 64*1024 / 2x = 216-x
Number of blocks per set = 8
Thus, Number of set = 216-x/8 = 213-x
Address = 32 bits = index + tag + block offset
32 = 13-x + tag + x
x = 19 bits
Block size = 2x bytes
Number of blocks per cache = 64*1024 / 2x = 216-x
Number of blocks per set = 8
Thus, Number of set = 216-x/8 = 213-x
Address = 32 bits = index + tag + block offset
32 = 13-x + tag + x
x = 19 bits
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