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October 26, 2023Sorting
October 26, 2023GATE 2011
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Question 32
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An 8KB direct-mapped write-back cache is organized as multiple blocks, each of size 32-bytes. The processor generates 32-bit addresses. The cache controller maintains the tag information for each cache block comprising of the following.
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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?
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4864 bits
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6144 bits
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6656 bits
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5376 bits
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Question 32 Explanation:
No. of cache blocks = cache size/size of a block = 8KB/32B = 256
So we need 8 bits for indexing the 256 blocks in the cache. And since a block is 32 bytes we need 5 word bits to address each byte.
So out of remaining (32 – 8 – 5), 19 bits should be tag bits.
So tag entry size = 19 + 1 (valid bit) + 1 (modified bit) = 21 bits
∴ Total size of metadata = 21 × Number blocks = 21 × 256 = 5376 bits
So we need 8 bits for indexing the 256 blocks in the cache. And since a block is 32 bytes we need 5 word bits to address each byte.
So out of remaining (32 – 8 – 5), 19 bits should be tag bits.
So tag entry size = 19 + 1 (valid bit) + 1 (modified bit) = 21 bits
∴ Total size of metadata = 21 × Number blocks = 21 × 256 = 5376 bits
Correct Answer: D
Question 32 Explanation:
No. of cache blocks = cache size/size of a block = 8KB/32B = 256
So we need 8 bits for indexing the 256 blocks in the cache. And since a block is 32 bytes we need 5 word bits to address each byte.
So out of remaining (32 – 8 – 5), 19 bits should be tag bits.
So tag entry size = 19 + 1 (valid bit) + 1 (modified bit) = 21 bits
∴ Total size of metadata = 21 × Number blocks = 21 × 256 = 5376 bits
So we need 8 bits for indexing the 256 blocks in the cache. And since a block is 32 bytes we need 5 word bits to address each byte.
So out of remaining (32 – 8 – 5), 19 bits should be tag bits.
So tag entry size = 19 + 1 (valid bit) + 1 (modified bit) = 21 bits
∴ Total size of metadata = 21 × Number blocks = 21 × 256 = 5376 bits
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