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.

Question

Bosco

Computers & Technology

17 April, 06:59

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.

+4

Answers (1)

Know the Answer?

Karter Singh · Answer 1

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.