The main memory of a computer has 2 cm blocks while the cache has 2 c blocks. If the cache uses the set associative mapping scheme with 2 blocks per set, then block 'k' of the main memory maps to the set.

(k m o d 2 c m) o f t h e c a c h e

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(k m o d c) o f t h e c a c h e

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(k m o d m) o f t h e c a c h e

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(k m o d 2 c) o f t h e c a c h e

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Solution

The correct option is B $(k m o d c) o f t h e c a c h e$

# of set in set associative cache

$= \frac{# of blocks in cache}{# of blocks in one set}$

$= \frac{2 c}{2} = c$

# of sets in cache $= c$

Therefore, the block 'k' of the main memory maps to the set $k m o d c$ of the cache.

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