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A CPU has a cache with block size 64 bytes. The main memory has k blocks, each block being c bytes wide. Consecutive c-byte chunks are mapped on consecutive blocks with warp-around. All the k blocks can be accessed in parallel, but two accesses to the same block must be serialized. A cache block access may involve multiple iterations of parallel block accesses depending on the amount of data obtained by accessing all the k-blocks in parallel. Each iteration requires decoding the block numbers to be accessed in parallel and this takes k/2 ns.
The latency of one block access is 80 ns. If c = 2 and k = 24, then latency of retrieving a cache block starting at address zero from main memory is
  • a)
    92 ns
  • b)
    104 ns
  • c)
    172 ns
  • d)
    184 ns
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A CPU has a cache with block size 64 bytes. The main memory has k bloc...
Cache block size = 64 bytes
Main memory has K banks or k = 24
Each bank is 2 byte long because c = 2
Total time for one parallel access

Total latency time = CT
= 2 x 92 = 184 ns
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Most Upvoted Answer
A CPU has a cache with block size 64 bytes. The main memory has k bloc...
Given Information:
- Cache block size: 64 bytes
- Main memory block size: c bytes
- Number of main memory blocks: k = 24
- Latency of one block access: 80 ns
- Time to decode block numbers: k/2 ns
- Consecutive c-byte chunks mapped to consecutive blocks with wrap-around

Calculating Total Latency:
- Time to access all k blocks in parallel: k/2 ns
- Total time for one cache block access = 80 ns + k/2 ns
- Substituting k = 24, total latency = 80 ns + 24/2 ns = 80 ns + 12 ns = 92 ns

Calculating Iterations:
- Number of cache blocks to access = 64 bytes / c bytes = 64 / 2 = 32 blocks
- Number of iterations = 32 / k = 32 / 24 = 4 iterations
- Each iteration takes k/2 ns to decode block numbers
- Total time for all iterations = 4 * k/2 ns = 4 * 12 ns = 48 ns

Final Calculation:
- Total latency of retrieving a cache block = Total time for one cache block access + Total time for all iterations
- Total latency = 92 ns + 48 ns = 140 ns
Therefore, the correct answer is option 'D' with a latency of 184 ns.
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A CPU has a cache with block size 64 bytes. The main memory has k blocks, each block being c bytes wide. Consecutive c-byte chunks are mapped on consecutive blocks with warp-around. All the k blocks can be accessed in parallel, but two accesses to the same block must be serialized. A cache block access may involve multiple iterations of parallel block accesses depending on the amount of data obtained by accessing all the k-blocks in parallel. Each iteration requires decoding the block numbers to be accessed in parallel and this takes k/2 ns.The latency of one block access is 80 ns. If c = 2 and k = 24, then latency of retrieving a cache block starting at address zero from main memory isa)92 nsb)104 nsc)172 nsd)184 nsCorrect answer is option 'D'. Can you explain this answer?
Question Description
A CPU has a cache with block size 64 bytes. The main memory has k blocks, each block being c bytes wide. Consecutive c-byte chunks are mapped on consecutive blocks with warp-around. All the k blocks can be accessed in parallel, but two accesses to the same block must be serialized. A cache block access may involve multiple iterations of parallel block accesses depending on the amount of data obtained by accessing all the k-blocks in parallel. Each iteration requires decoding the block numbers to be accessed in parallel and this takes k/2 ns.The latency of one block access is 80 ns. If c = 2 and k = 24, then latency of retrieving a cache block starting at address zero from main memory isa)92 nsb)104 nsc)172 nsd)184 nsCorrect 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 A CPU has a cache with block size 64 bytes. The main memory has k blocks, each block being c bytes wide. Consecutive c-byte chunks are mapped on consecutive blocks with warp-around. All the k blocks can be accessed in parallel, but two accesses to the same block must be serialized. A cache block access may involve multiple iterations of parallel block accesses depending on the amount of data obtained by accessing all the k-blocks in parallel. Each iteration requires decoding the block numbers to be accessed in parallel and this takes k/2 ns.The latency of one block access is 80 ns. If c = 2 and k = 24, then latency of retrieving a cache block starting at address zero from main memory isa)92 nsb)104 nsc)172 nsd)184 nsCorrect 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 A CPU has a cache with block size 64 bytes. The main memory has k blocks, each block being c bytes wide. Consecutive c-byte chunks are mapped on consecutive blocks with warp-around. All the k blocks can be accessed in parallel, but two accesses to the same block must be serialized. A cache block access may involve multiple iterations of parallel block accesses depending on the amount of data obtained by accessing all the k-blocks in parallel. Each iteration requires decoding the block numbers to be accessed in parallel and this takes k/2 ns.The latency of one block access is 80 ns. If c = 2 and k = 24, then latency of retrieving a cache block starting at address zero from main memory isa)92 nsb)104 nsc)172 nsd)184 nsCorrect answer is option 'D'. Can you explain this answer?.
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