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Consider a small two-way set-associative cache memory, consisting of four blocks. For choosing the block to be replaced, use the least recently used (LRU) scheme. The number of cache misses for the following sequence of block addresses is:

$8, 12, 0, 12, 8$.

1. $2$
2. $3$
3. $4$
4. $5$

@akshay7797, it is a compulsary miss.
Cold miss → Cold miss → Cold miss → Hit → Conflict miss

correct.

We have $4$ blocks and $2$ blocks in a set

$\implies$ there are $2$ sets. So blocks will go to sets as follows:
$$\begin{array}{|c|c|} \hline \textbf {Set Number} & \textbf{Block Number} \\\hline \text{0} & \text{0,8,12} \\\hline\text{1} & \\\hline \end{array}$$
Since the lowest bit of block address is used for indexing into the set, so $8, 12$ and $0$ first miss in cache with $0$ replacing $8$ (there are two slots in each set due to $2-\text{way}$ set) and then $12$ hits in cache and $8$ again misses. So, totally $4$ misses.

Correct Answer: $C$
by

Cache is 2-way associative, and has total 4 blocks.

So, no. of sets $\frac{4}{2} = 2^1 \implies 1$ bit for deciding which set each block belongs to, which by standard is the least significant bit, which is $0$ for $\{0, 8, 12\}$

Each set can contain 2 blocks.

So, @`JEET set 2 should not contain anything.

sir why not 12 miss for the first time.

it is a  miss.

2 way set associative

4 blocks

Set size = # of blocks / # of sets

= 4/2 = 2

So Block addresses will be divided by 2 ... On the basis of their remainder we will place those blocks in the sets .. applying LRU approach ... answer will be 4 ..

Option C is the correct answer

total 4 cache misses are there.