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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?
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
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
An 8KB direct-mapped write-back cache is organized as multiple blocks,...
Number of cache blocks = cache size / size of a block
=8 KB/32 B
=256
So, we need 8 bits for indexing the 256 blocks of the cache. And since a block is 32 bytes we need 5 WORD bits to address
each byte. So, out of the remaining 19 bits (32 - 8 - 5) should be tag bits.
So, a tag entry size = 19 + 1(valid bit) + 1(modified bit) = 21 bits.
Total size of metadata = 21 * Number of cache blocks
= 21 * 256
= 5376 bits
=8 KB/32 B
=256
So, we need 8 bits for indexing the 256 blocks of the cache. And since a block is 32 bytes we need 5 WORD bits to address
each byte. So, out of the remaining 19 bits (32 - 8 - 5) should be tag bits.
So, a tag entry size = 19 + 1(valid bit) + 1(modified bit) = 21 bits.
Total size of metadata = 21 * Number of cache blocks
= 21 * 256
= 5376 bits
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Most Upvoted Answer
An 8KB direct-mapped write-back cache is organized as multiple blocks,...
The total size of memory needed at the cache controller to store meta-data (tags) for the cache can be calculated by considering the number of blocks in the cache and the number of bits required for each block's tag information.
Given information:
- Cache size: 8KB
- Block size: 32 bytes
To find the number of blocks in the cache, we need to divide the cache size by the block size:
Number of blocks = Cache size / Block size = 8KB / 32 bytes = 256 blocks
Now, let's calculate the number of bits required for each block's tag information:
- 1 valid bit: This bit is used to indicate whether the block contains valid data or not. As there are 256 blocks, we need 256 valid bits.
- 1 modified bit: This bit is used to indicate whether the block has been modified or not. Again, we need 256 modified bits.
- Number of bits required to identify the memory block mapped in the cache: Since the cache is direct-mapped, we only need enough bits to uniquely identify each memory block. Since there are 256 blocks, we need log2(256) = 8 bits.
Total number of bits required for each block's tag information = 1 (valid bit) + 1 (modified bit) + 8 (bits to identify the memory block) = 10 bits
Finally, to calculate the total size of memory needed for storing the tag information, we multiply the number of blocks by the number of bits required for each block's tag information:
Total size of memory needed = Number of blocks * Number of bits per block = 256 blocks * 10 bits = 2560 bits
Converting the total size to bytes, we get:
Total size of memory needed = 2560 bits / 8 bits/byte = 320 bytes
Since the options given in the question are in bits, we can convert the total size to bits:
Total size of memory needed = 320 bytes * 8 bits/byte = 2560 bits
Therefore, the total size of memory needed at the cache controller to store meta-data (tags) for the cache is 2560 bits, which corresponds to option 'D'.
Given information:
- Cache size: 8KB
- Block size: 32 bytes
To find the number of blocks in the cache, we need to divide the cache size by the block size:
Number of blocks = Cache size / Block size = 8KB / 32 bytes = 256 blocks
Now, let's calculate the number of bits required for each block's tag information:
- 1 valid bit: This bit is used to indicate whether the block contains valid data or not. As there are 256 blocks, we need 256 valid bits.
- 1 modified bit: This bit is used to indicate whether the block has been modified or not. Again, we need 256 modified bits.
- Number of bits required to identify the memory block mapped in the cache: Since the cache is direct-mapped, we only need enough bits to uniquely identify each memory block. Since there are 256 blocks, we need log2(256) = 8 bits.
Total number of bits required for each block's tag information = 1 (valid bit) + 1 (modified bit) + 8 (bits to identify the memory block) = 10 bits
Finally, to calculate the total size of memory needed for storing the tag information, we multiply the number of blocks by the number of bits required for each block's tag information:
Total size of memory needed = Number of blocks * Number of bits per block = 256 blocks * 10 bits = 2560 bits
Converting the total size to bytes, we get:
Total size of memory needed = 2560 bits / 8 bits/byte = 320 bytes
Since the options given in the question are in bits, we can convert the total size to bits:
Total size of memory needed = 320 bytes * 8 bits/byte = 2560 bits
Therefore, the total size of memory needed at the cache controller to store meta-data (tags) for the cache is 2560 bits, which corresponds to option 'D'.
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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 bit1 modified bitAs 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 bitsb)6144 bitsc)6656 bitsd)5376 bitsCorrect answer is option 'D'. Can you explain this answer?
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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 bit1 modified bitAs 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 bitsb)6144 bitsc)6656 bitsd)5376 bitsCorrect answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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 bit1 modified bitAs 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 bitsb)6144 bitsc)6656 bitsd)5376 bitsCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 bit1 modified bitAs 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 bitsb)6144 bitsc)6656 bitsd)5376 bitsCorrect answer is option 'D'. Can you explain this answer?.
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 bit1 modified bitAs 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 bitsb)6144 bitsc)6656 bitsd)5376 bitsCorrect answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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 bit1 modified bitAs 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 bitsb)6144 bitsc)6656 bitsd)5376 bitsCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 bit1 modified bitAs 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 bitsb)6144 bitsc)6656 bitsd)5376 bitsCorrect answer is option 'D'. Can you explain this answer?.
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