An 8KB direct-mapped write-back cache is organized as multiple blocks, each…
2011
An 8KB direct-mapped write-back cache is organized as multiple blocks, each size of 32-bytes. The processor generates 32-bit addresses. The cache controller contains the tag information for each cache block comprising of the following.
1 valid bit
1 modified bit
As many bits as the minimum needed to identify the memory block mapped in the cache.
What is the total size of memory needed at the cache controller to store meta-data (tags) for the cache?
- A.
4864 bits
- B.
6144 bits
- C.
6656 bits
- D.
5376 bits
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Correct answer: D
Solution:
Compute sizes: Cache size = 8 KB = 8192 bytes. Block size = 32 bytes.
Number of blocks (lines) = 8192 / 32 = 256.
Offset bits = log2(block size) = log2(32) = 5 bits.
Index bits = log2(number of lines) = log2(256) = 8 bits.
Tag bits = address bits - index bits - offset bits = 32 - 8 - 5 = 19 bits.
Metadata per block = tag bits + valid bit + modified bit = 19 + 1 + 1 = 21 bits.
Total metadata = number of blocks × metadata per block = 256 × 21 = 5376 bits.
Answer: 5376 bits
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