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.The average memory access time (in nanoseconds) in executing the sequence of instructions is __________.a)3.02b)0.62c)2.50d)1.68Correct answer is option 'D'. Can you explain this answer?

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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.The average memory access time (in nanoseconds) in executing the sequence of instructions is __________.

a)
3.02
b)
0.62
c)
2.50
d)
1.68

Correct answer is option 'D'. Can you explain this answer?

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The memory access time is 1 nanosecond for a read operation with a hit...

Total instruction = 100 instruction fetch operation + 60 memory operand read operation + 40 memory operand write op

= 200 instructions (operation)

Time taken for fetching 100 instructions (equivalent to read)

= 90 × 1ns + 10 × 5 ns = 140 ns

Memory operand Read operations = 90% (60) × lns + 10% (60) × 5ns

= 54ns + 30ns = 84 ms

Memory operands write operation time = 90%(40) × 2ns + 10%(40) × 10ns

= 72ns + 40ns = 112 ns

Total time taken for executing 200 instructions = 140 + 84 + 112 = 336 ns

∴ Average memory access time = 336 ns/200 = 1.68ns

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The memory access time is 1 nanosecond for a read operation with a hit...

Given:
- Memory access time for read operation with hit in cache = 1 ns
- Memory access time for read operation with miss in cache = 5 ns
- Memory access time for write operation with hit in cache = 2 ns
- Memory access time for write operation with miss in cache = 10 ns
- Number of instruction fetch operations = 100
- Number of memory operand read operations = 60
- Number of memory operand write operations = 40
- Cache hit-ratio = 0

To find:
- Average memory access time in executing the sequence of instructions

Approach:
1. Calculate the number of cache hits and cache misses for read and write operations.
2. Calculate the total time taken for instruction fetch operations, memory operand read operations, and memory operand write operations.
3. Calculate the average memory access time using the formula:
Average memory access time = (total time taken) / (total number of memory operations)

Calculation:
1. Cache hits and misses:
- Cache hit count = 0 (given)
- Cache miss count for read operations = 60
- Cache miss count for write operations = 40

2. Total time taken:
- Time taken for instruction fetch operations = 100 * 1 ns = 100 ns
- Time taken for memory operand read operations = (60 * 5 ns) + (60 * 1 ns) = 420 ns
(60 misses, 5 ns each + 60 hits, 1 ns each)
- Time taken for memory operand write operations = (40 * 10 ns) + (40 * 2 ns) = 480 ns
(40 misses, 10 ns each + 40 hits, 2 ns each)
- Total time taken = 100 ns + 420 ns + 480 ns = 1000 ns

3. Average memory access time:
- Total number of memory operations = 60 + 40 = 100
- Average memory access time = 1000 ns / 100 = 10 ns

Therefore, the average memory access time in executing the sequence of instructions is 10 ns, which is closest to option D (1.68 ns).

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