254 views

1 Answer

0 votes
0 votes
The total number of blocks is 8192.

Since it is a 4-way set associative cache, each set can contain 4 blocks.

Therefore, total number of sets is $\frac{8192}{4} = 2048 = 2^{11}$.

Hence, number of bits required will be $log_2(2^{11}) = 11$

Also, size of one block = $32 \, \text{words} \Rightarrow 32 \times 64 \, \text{bits} = 2^{11} \, \text{bits}$

Now since the addresses are word addressable, the number of words is $\frac{2^{11}}{2^6} = 2^5$
Therefore, word offset is $log_2(2^{5}) = 5 \, \text{bits} $

Now we know that $$\text{Number of bits in physical address = Tag + Set + Word Offset}$$

Plugging in the values, we get the value of Tag as $48$ bits.

If this is not the correct answer, do let me know. I tried to answer it to the best of my knowledge.
edited by

Related questions

3 votes
3 votes
1 answer
2
Rakesh K asked Aug 26, 2016
2,200 views
"Assuming a cache of 4K blocks, a 4-word block size, and a 32‐bit address, find the total number of sets and the total number of tag bits for caches that are direct map...
0 votes
0 votes
1 answer
4
rahuldb asked Nov 17, 2016
3,576 views
Consider a memory of 8 words per block.If 2 clock cycle are required to transfer address from CPU to main memory, 6 clock cycle to access the first word, 3 clock cycles e...