GATE|GATE-CS-2014 PAPER-03|Question5
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 operation and 40 memory operand write operations.The cache hit-ratio is 0.9.The average memory access time in executing the sequence of instructions is
Answer
Total instruction=100 instruction fetch operation+60 memory operand read operation+40 memory operand write operation=200 instruction.
Time taken for fetching 100 instructions(equivalent to read)=90%(60)*1ns+10%(60)*5ns=54ns+30ns=84ns.
Memory operands write operation time=90%(40)*2ns+10%(40)*10ns=72ns+40ns=112ns.
Total time taken for executing 200 instruction=140+84+112=336ns.
Average memory access time=336ns/200=1.68ns.
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