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 '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'. Can you explain this answer?

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This discussion on 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 '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'. Can you explain this answer? is done on EduRev Study Group by Computer Science Engineering (CSE) Students. The Questions and Answers of 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 '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'. Can you explain this answer? are solved by group of students and teacher of Computer Science Engineering (CSE), which is also the largest student community of Computer Science Engineering (CSE). If the answer is not available please wait for a while and a community member will probably answer this soon. You can study other questions, MCQs, videos and tests for Computer Science Engineering (CSE) on EduRev and even discuss your questions like 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 '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'. Can you explain this answer? over here on EduRev! Apart from being the largest Computer Science Engineering (CSE) community, EduRev has the largest solved Question bank for Computer Science Engineering (CSE).
This discussion on 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 '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'. Can you explain this answer? is done on EduRev Study Group by Computer Science Engineering (CSE) Students. The Questions and Answers of 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 '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'. Can you explain this answer? are solved by group of students and teacher of Computer Science Engineering (CSE), which is also the largest student community of Computer Science Engineering (CSE). If the answer is not available please wait for a while and a community member will probably answer this soon. You can study other questions, MCQs, videos and tests for Computer Science Engineering (CSE) on EduRev and even discuss your questions like 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 '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'. Can you explain this answer? over here on EduRev! Apart from being the largest Computer Science Engineering (CSE) community, EduRev has the largest solved Question bank for Computer Science Engineering (CSE).