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The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory request are for read the rest for write. The hit ratio for the read access only is 0.9 and a write-through policy (where both main and cache memories are updated simultaneously) is used.
Determine the average time of the main memory.
Determine the average time of the main memory.
Correct answer is '180ns'. Can you explain this answer?
Verified Answer
The access times of the main memory and the Cache memory, in a compute...
Average memory access time = Time spend for read + Time spend for write= Read time when cache hit + Read time when cache miss+Write time when cache hit + Write time when cache miss= 0.8 ⨯ 0.9 ⨯ 50 + 0.8 ⨯ 0.1 ⨯ (500+50) (assuming hierarchical read from memory and cache as only simultaneous write is mentioned in question)+ 0.2 ⨯ 0.9 ⨯ 500 + 0.2 ⨯ 0.1 ⨯ 500 (simultaneous write mentioned in question)= 36 + 44 + 90 + 10 = 180 ns
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The access times of the main memory and the Cache memory, in a compute...
Question:
The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory requests are for read, and the rest are for write. The hit ratio for the read access only is 0.9, and a write-through policy (where both main and cache memories are updated simultaneously) is used. Determine the average time of the main memory.
Answer:
To determine the average time of the main memory, we need to consider the time spent for read and write operations separately.
Time spent for read:
The hit ratio for read access is given as 0.9, which means that 90% of the read requests will find the data in the cache memory. Therefore, the remaining 10% of read requests will result in cache misses and require access to the main memory.
For the 80% of read requests that hit the cache:
Time spent for read = Hit ratio * Cache access time = 0.9 * 50 n sec = 45 n sec
For the 10% of read requests that miss the cache:
Time spent for read = Miss ratio * (Cache access time + Main memory access time) = 0.1 * (50 n sec + 500 n sec) = 55 n sec
Time spent for write:
Since a write-through policy is used, both the main memory and cache memory are updated simultaneously for every write request. Therefore, the time spent for write operations is equal to the cache access time.
Time spent for write = Cache access time = 50 n sec
Average memory access time:
To calculate the average memory access time, we need to consider the proportion of read and write requests.
Average memory access time = (Proportion of read requests * Time spent for read) + (Proportion of write requests * Time spent for write)
Proportion of read requests = 80% = 0.8
Proportion of write requests = 20% = 0.2
Average memory access time = (0.8 * (45 n sec + 55 n sec)) + (0.2 * 50 n sec)
= (0.8 * 100 n sec) + (0.2 * 50 n sec)
= 80 n sec + 10 n sec
= 90 n sec
Therefore, the average time of the main memory is 90 n sec.
The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory requests are for read, and the rest are for write. The hit ratio for the read access only is 0.9, and a write-through policy (where both main and cache memories are updated simultaneously) is used. Determine the average time of the main memory.
Answer:
To determine the average time of the main memory, we need to consider the time spent for read and write operations separately.
Time spent for read:
The hit ratio for read access is given as 0.9, which means that 90% of the read requests will find the data in the cache memory. Therefore, the remaining 10% of read requests will result in cache misses and require access to the main memory.
For the 80% of read requests that hit the cache:
Time spent for read = Hit ratio * Cache access time = 0.9 * 50 n sec = 45 n sec
For the 10% of read requests that miss the cache:
Time spent for read = Miss ratio * (Cache access time + Main memory access time) = 0.1 * (50 n sec + 500 n sec) = 55 n sec
Time spent for write:
Since a write-through policy is used, both the main memory and cache memory are updated simultaneously for every write request. Therefore, the time spent for write operations is equal to the cache access time.
Time spent for write = Cache access time = 50 n sec
Average memory access time:
To calculate the average memory access time, we need to consider the proportion of read and write requests.
Average memory access time = (Proportion of read requests * Time spent for read) + (Proportion of write requests * Time spent for write)
Proportion of read requests = 80% = 0.8
Proportion of write requests = 20% = 0.2
Average memory access time = (0.8 * (45 n sec + 55 n sec)) + (0.2 * 50 n sec)
= (0.8 * 100 n sec) + (0.2 * 50 n sec)
= 80 n sec + 10 n sec
= 90 n sec
Therefore, the average time of the main memory is 90 n sec.
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The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory request are for read the rest for write. The hit ratio for the read access only is 0.9 and a write-through policy (where both main and cache memories are updated simultaneously) is used.Determine the average time of the main memory.Correct answer is '180ns'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 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 The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory request are for read the rest for write. The hit ratio for the read access only is 0.9 and a write-through policy (where both main and cache memories are updated simultaneously) is used.Determine the average time of the main memory.Correct answer is '180ns'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory request are for read the rest for write. The hit ratio for the read access only is 0.9 and a write-through policy (where both main and cache memories are updated simultaneously) is used.Determine the average time of the main memory.Correct answer is '180ns'. Can you explain this answer?.
The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory request are for read the rest for write. The hit ratio for the read access only is 0.9 and a write-through policy (where both main and cache memories are updated simultaneously) is used.Determine the average time of the main memory.Correct answer is '180ns'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 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 The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory request are for read the rest for write. The hit ratio for the read access only is 0.9 and a write-through policy (where both main and cache memories are updated simultaneously) is used.Determine the average time of the main memory.Correct answer is '180ns'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The access times of the main memory and the Cache memory, in a computer system, are 500 n sec and 50 n sec, respectively. It is estimated that 80% of the main memory request are for read the rest for write. The hit ratio for the read access only is 0.9 and a write-through policy (where both main and cache memories are updated simultaneously) is used.Determine the average time of the main memory.Correct answer is '180ns'. Can you explain this answer?.
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