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In a $k$-way set associative cache, the cache is divided into $v$ sets, each of which consists of $k$ lines. The lines of a set are placed in sequence one after another. The lines in set $s$ are sequenced before the lines in set $(s+1)$. The main memory blocks are numbered 0 onwards. The main memory block numbered $j$ must be mapped to any one of the cache lines from

1. $(j\text{ mod }v) * k \text{ to } (j \text{ mod } v) * k + (k-1)$
2. $(j \text{ mod } v) \text{ to } (j \text{ mod } v) + (k-1)$
3. $(j \text{ mod } k) \text{ to } (j \text{ mod } k) + (v-1)$
4. $(j \text{ mod } k) * v \text{ to } (j \text{ mod } k) * v + (v-1)$
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Number of sets in cache $= v$.

The question gives a sequencing for the cache lines. For set $0$, the cache lines are numbered $0, 1, .., k-1$. Now for set $1$, the cache lines are numbered $k, k+1,... k+k-1$ and so on.

So, main memory block $j$ will be mapped to set $(j \ \text{mod} \ v)$, which will be any one of the cache lines from $(j \ \text{mod } v) * k \ \text{ to } (j \ \text{mod } v) * k + (k-1)$.
(Associativity plays no role in mapping- $k$-way associativity means there are $k$ spaces for a block and hence reduces the chances of replacement.)
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Sir, how did you got the range..? I'm not able to understand how you got this (j mod v) * k + (k-1).

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why they multiply it with k in option (a) of range..
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That is to get the cache line number. I have added that in the answer. Now, it should be easy.
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k is block no and in this example there are 16 cache blocks

Block no. 13 rightly placed between 4 to 7

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ur architecture part is damn too good.....
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Simple and easy to understand!!
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nice explanation @Arjun Sir
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clear-cut explanation sir...
 set number block 0 0 0 1 0 2 0 3 1 0 1 1 1 2 1 3 2 0 2 1 2 2 2 3 3 0 3 1 3 2 3 3

In above exaple 16 blocks are there. and 4 sets.Each set contains 4 blocks.

Suppose memory address are like this : 0,4,8,14 these all can fit into set 0 in any order where it get place but mostly FIFO.

jth is memory block will be placed at j mod v = set number

Let j=14 the main memory block. 14 mod 4 = 2 we are at set number 2  but to read at set to we need to pass set0 and set1 each having k size .

so 2 * k =2*4 gives =8th cache block that is starting block of 3rd cache set.But each cache also having k way set associativy that means more k-1 block can be placed in that set.

so (j mod v) * k + (j mod v)*k +(k-1) should be the ans.

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Forget everything for a while and just see

The cache memory has k lines and v sets

Sets are built using lines.

Now there  are number of blocks but we shall consider block no j now

For mapping we know

To map block no J to set no V we use (j mod v)

Now we have number of lines as k

So multiply it by k

i. e (j mod v) *k

Now range part:

(J mod V) *K

And we  add k-1 for the numbee of lines - 1

So total is (j mod v) *k +k-1

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