Programming
April 5, 2025NTA UGC NET Dec 2024 Paper-1
April 6, 2025GATE 2008-IT
| Question 80 |
Consider a computer with a 4-ways set-associative mapped cache of the following characteristics: a total of 1 MB of main memory, a word size of 1 byte, a block size of 128 words and a cache size of 8 KB.
The number of bits in the TAG, SET and WORD fields, respectively are:
| 7, 6, 7 | |
| 8, 5, 7 | |
| 8, 6, 6 | |
| 9, 4, 7 |
Question 80 Explanation:
No. of cache blocks = 8KB/128 = 213/27 = 26
No. of sets in cache = 26/22 = 24
So no. of set bits = 4
Size of block = 128 words = 128 B = 27
So no. of word bit = 7
So tag bits = 20 – 4 – 7 = 9
No. of sets in cache = 26/22 = 24
So no. of set bits = 4
Size of block = 128 words = 128 B = 27
So no. of word bit = 7
So tag bits = 20 – 4 – 7 = 9
Correct Answer: D
Question 80 Explanation:
No. of cache blocks = 8KB/128 = 213/27 = 26
No. of sets in cache = 26/22 = 24
So no. of set bits = 4
Size of block = 128 words = 128 B = 27
So no. of word bit = 7
So tag bits = 20 – 4 – 7 = 9
No. of sets in cache = 26/22 = 24
So no. of set bits = 4
Size of block = 128 words = 128 B = 27
So no. of word bit = 7
So tag bits = 20 – 4 – 7 = 9