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In a two level memory hierarchy, the access time of the cache memory is 12 nsec and the access time of the main memory is 1.5 msec. The hit ratio is 0.98. What is the average access time of the two level memory system?
  • a)
    13.5 nsec
  • b)
    42 nsec
  • c)
    7.56 nsec
  • d)
    41.76 nsec
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
In a two level memory hierarchy, the access time of the cache memory i...
Suppose 100 instructions are accessed from cache.
So, 98 instructions will take 12 ns � 98 and 2 instructions will take 1500 ns � 2.
So, average access time = (12 � 98 + 3000)/100 = 41.76 ns

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Most Upvoted Answer
In a two level memory hierarchy, the access time of the cache memory i...
Explanation:

To calculate the average access time of the two-level memory system, we need to consider the hit ratio and the access time of each level of memory.

Given information:
- Access time of cache memory (Level 1): 12 nsec
- Access time of main memory (Level 2): 1.5 msec
- Hit ratio: 0.98

Step 1: Calculate the miss ratio (miss rate) using the hit ratio:
Miss ratio = 1 - Hit ratio
Miss ratio = 1 - 0.98
Miss ratio = 0.02

Step 2: Calculate the average access time:
Average access time = Hit time + Miss rate * Miss penalty

Step 3: Calculate the miss penalty:
Miss penalty = Access time of main memory - Access time of cache memory
Miss penalty = 1.5 msec - 12 nsec

Step 4: Convert the miss penalty to the same unit as the average access time (nanoseconds):
1.5 msec = 1.5 * 10^6 nsec

Step 5: Substitute the values into the formula:
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 12 nsec + 0.02 * (1.5 * 10^6 nsec - 12 nsec)
Average access time = 41.76 nsec

Therefore, the average access time of the two-level memory system is 41.76 nsec, which corresponds to option D.
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In a two level memory hierarchy, the access time of the cache memory is 12 nsec and the access time of the main memory is 1.5 msec. The hit ratio is 0.98. What is the average access time of the two level memory system?a)13.5 nsecb)42 nsecc)7.56 nsecd)41.76 nsecCorrect answer is option 'D'. Can you explain this answer?
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