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The memory access time is 1 nanosecond for a read operation with a hit in cache, 5 nanoseconds for a read operation with a miss in cache, 2 nanoseconds for a write operation with a hit in cache and 10 nanoseconds for a write operation with a miss in cache. Execution of a sequence of instructions involves 100 instruction fetch operations, 60 memory operand read
operations and 40 memory operand write operations. The cache hit-ratio is 0.9. The average memory access time (in nanoseconds) in executing the sequence of instructions is ______.
Correct answer is '1.68'. Can you explain this answer?
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The memory access time is 1 nanosecond for a read operation with a hit...
Fetch is also a memory read operation. Avg access time

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The memory access time is 1 nanosecond for a read operation with a hit...
Given information:
- Memory access time for a read operation with a hit in cache: 1 nanosecond
- Memory access time for a read operation with a miss in cache: 5 nanoseconds
- Memory access time for a write operation with a hit in cache: 2 nanoseconds
- Memory access time for a write operation with a miss in cache: 10 nanoseconds
- Number of instruction fetch operations: 100
- Number of memory operand read operations: 60
- Number of memory operand write operations: 40
- Cache hit-ratio: 0.9

Calculating average memory access time:
To calculate the average memory access time, we need to consider the time taken for each type of memory access operation (read with hit, read with miss, write with hit, write with miss) and the frequency of those operations.

1. Read operation with hit in cache:
- Time taken: 1 nanosecond
- Frequency: 100 instruction fetch operations
- Total time taken: 1 nanosecond * 100 = 100 nanoseconds

2. Read operation with miss in cache:
- Time taken: 5 nanoseconds
- Frequency: 60 memory operand read operations
- Total time taken: 5 nanoseconds * 60 = 300 nanoseconds

3. Write operation with hit in cache:
- Time taken: 2 nanoseconds
- Frequency: 40 memory operand write operations
- Total time taken: 2 nanoseconds * 40 = 80 nanoseconds

4. Write operation with miss in cache:
- Time taken: 10 nanoseconds
- Frequency: 40 memory operand write operations
- Total time taken: 10 nanoseconds * 40 = 400 nanoseconds

Calculating total time:
Total time = Total time for read operations + Total time for write operations
= (Time for read with hit + Time for read with miss) + (Time for write with hit + Time for write with miss)
= 100 nanoseconds + 300 nanoseconds + 80 nanoseconds + 400 nanoseconds
= 880 nanoseconds

Calculating average memory access time:
Average memory access time = Total time / Total number of memory operations

Total number of memory operations = Number of instruction fetch operations + Number of memory operand read operations + Number of memory operand write operations
= 100 + 60 + 40
= 200

Average memory access time = 880 nanoseconds / 200 = 4.4 nanoseconds

Considering cache hit-ratio:
The cache hit-ratio is given as 0.9, which means 90% of the memory operations are hits in the cache. Therefore, the remaining 10% are misses in the cache.

Average memory access time = (Cache hit-ratio * Time for hit) + ((1 - Cache hit-ratio) * Time for miss)
= (0.9 * 4.4 nanoseconds) + (0.1 * (100 nanoseconds + 300 nanoseconds + 80 nanoseconds + 400 nanoseconds))
= 3.96 nanoseconds + 88 nanoseconds
= 91.96 nanoseconds

Hence,
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The memory access time is 1 nanosecond for a read operation with a hit in cache, 5 nanoseconds for a read operation with a miss in cache, 2 nanoseconds for a write operation with a hit in cache and 10 nanoseconds for a write operation with a miss in cache. Execution of a sequence of instructions involves 100 instruction fetch operations, 60 memory operand readoperations and 40 memory operand write operations. The cache hit-ratio is 0.9. The average memory access time (innanoseconds) in executing the sequence of instructions is ______.Correct answer is '1.68'. Can you explain this answer?
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