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10 December, 17:59

A four-set associative cache-memory system consists of 32 blocks of cache. Each block holds 256 bytes of data. Each block includes a control field with two LRU bits, a valid bit and a dirty bit. The address space is 4 gigabytes. Calculate the total size of the cache, including all control, data and tag bits.

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  1. 10 December, 19:31
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    Answer: total size of cache = 66, 336bits = 8,292B.

    Control bits = 4b.

    Offset=8bits

    Index=3bits

    Tag = 21bits

    Explanation: total $ size of cache = data + control bits + tag.

    For each block: control bit = LRU bit + valid bit + dirty bits = 2 + 1 + 1 = 4b.

    Address = tag + index + offset; 256 = 2 raised to power of 8, then offset = 8.

    Number of blocks in each direct mapped cache=32/4 = 2raised to power 3, index = 3.

    Tag = 32 - 8 - 3 = 21bits.

    Total size = 32 blocks * (4bits + 22bits + 256 bits * 8 bits) = 66,336bits = 8,292B.
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