NVS PGT CS 2017 Part-B
August 13, 2024ISRO CS 2009
August 13, 2024Memory-Management
| Question 16 |
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 __________.
| 1.68 | |
| 1.69 | |
| 1.70 | |
| 1.71 |
Question 16 Explanation:
Total instruction = 100 instruction fetch operation + 60 memory operand read operation + 40 memory operand write op
= 200 instructions (operations)
Time taken for fetching 100 instructions (equivalent to read) = 90*1ns + 10*5ns = 140ns
Memory operand Read operations = 90% (60)*1ns + 10% (60) × 5ns = 54ns + 30 ns = 84ns
Memory operands Write operations time = 90% (40)*2ns + 10% (40)*10ns
= 72ns + 40ns = 112ns
Total time taken for executing 200 instructions = 140 + 84 + 112 = 336ns
∴ Average memory access time = 336 ns/200 = 1.68ns
= 200 instructions (operations)
Time taken for fetching 100 instructions (equivalent to read) = 90*1ns + 10*5ns = 140ns
Memory operand Read operations = 90% (60)*1ns + 10% (60) × 5ns = 54ns + 30 ns = 84ns
Memory operands Write operations time = 90% (40)*2ns + 10% (40)*10ns
= 72ns + 40ns = 112ns
Total time taken for executing 200 instructions = 140 + 84 + 112 = 336ns
∴ Average memory access time = 336 ns/200 = 1.68ns
Correct Answer: A
Question 16 Explanation:
Total instruction = 100 instruction fetch operation + 60 memory operand read operation + 40 memory operand write op
= 200 instructions (operations)
Time taken for fetching 100 instructions (equivalent to read) = 90*1ns + 10*5ns = 140ns
Memory operand Read operations = 90% (60)*1ns + 10% (60) × 5ns = 54ns + 30 ns = 84ns
Memory operands Write operations time = 90% (40)*2ns + 10% (40)*10ns
= 72ns + 40ns = 112ns
Total time taken for executing 200 instructions = 140 + 84 + 112 = 336ns
∴ Average memory access time = 336 ns/200 = 1.68ns
= 200 instructions (operations)
Time taken for fetching 100 instructions (equivalent to read) = 90*1ns + 10*5ns = 140ns
Memory operand Read operations = 90% (60)*1ns + 10% (60) × 5ns = 54ns + 30 ns = 84ns
Memory operands Write operations time = 90% (40)*2ns + 10% (40)*10ns
= 72ns + 40ns = 112ns
Total time taken for executing 200 instructions = 140 + 84 + 112 = 336ns
∴ Average memory access time = 336 ns/200 = 1.68ns
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