GATE | Gate IT 2005 | Question 61

Consider a 2-way set associative cache memory with 4 sets and total 8 cache blocks (0-7) and a main memory with 128 blocks (0-127). What memory blocks will be present in the cache after the following sequence of memory block references if LRU policy is used for cache block replacement. Assuming that initially the cache did not have any memory block from the current job?
0 5 3 9 7 0 16 55

 
(A) 0 3 5 7 16 55
(B) 0 3 5 7 9 16 55
(C) 0 5 7 9 16 55
(D) 3 5 7 9 16 55


Answer: (C)

Explanation:

2-way set associative cache memory, .i.e K = 2.

No of sets is given as 4, i.e. S = 4 ( numbered 0 - 3 )

No of blocks in cache memory is given as 8, i.e. N =8 ( numbered from 0 -7)

Each set in cache memory contains 2 blocks.

The number of blocks in the main memory is 128, i.e  M = 128.  ( numbered from 0 -127)
A referred block numbered X of the main memory is placed in the 
set numbered ( X mod S ) of the the cache memory. In that set, the 
block can be placed at any location, but if the set has already become
 full, then the current referred block of the main memory should replace
 a block in that set according to some replacement policy. Here 
the replacement policy is LRU ( i.e. Least Recently Used block should 
be replaced with currently referred block).

X ( Referred block no ) and 
the corresponding Set values are as follows:

X-->set no ( X mod 4 )

0--->0   ( block 0 is placed in set 0, set 0 has 2 empty block locations,
              block 0 is placed in any one of them  )

5--->1   ( block 5 is placed in set 1, set 1 has 2 empty block locations,
              block 5 is placed in any one of them  )

3--->3  ( block 3 is placed in set 3, set 3 has 2 empty block locations,
             block 3 is placed in any one of them  )

9--->1  ( block 9 is placed in set 1, set 1 has currently 1 empty block location,
             block 9 is placed in that, now set 1 is full, and block 5 is the 
             least recently used block  )

7--->3  ( block 7 is placed in set 3, set 3 has 1 empty block location, 
             block 7 is placed in that, set 3 is full now, 
             and block 3 is the least recently used block)

0--->block 0 is referred again, and it is present in the cache memory in set 0,
            so no need to put again this block into the cache memory.

16--->0  ( block 16 is placed in set 0, set 0 has 1 empty block location, 
              block 0 is placed in that, set 0 is full now, and block 0 is the LRU one)

55--->3 ( block 55 should be placed in set 3, but set 3 is full with block 3 and 7, 
             hence need to replace one block with block 55, as block 3 is the least 
             recently used block in the set 3, it is replaced with block 55.

Hence the main memory blocks present in the cache memory are : 0, 5, 7, 9, 16, 55 .

(Note: block 3 is not present in the cache memory, it was replaced with block 55 )

 

Read the following articles to learn more related to the above question:
Cache Memory
Cache Organization | Introduction


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