Computer-Organization
October 5, 2023NIC-NIELIT Scientist-B 2020
October 5, 2023Cache
| Question 25 |
A cache line is 64 bytes. The main memory has latency 32 ns and bandwidth 1 GBytes/s. The time required to fetch the entire cache line from the main memory is
| 32 ns | |
| 64 ns | |
| 96 ns | |
| 128 ns |
Question 25 Explanation:
For 1GBytes/s bandwidth → it takes 1 sec to load 109 bytes on line.
→ So, for 64 bytes it will take 64*1/109 = 64ns.
Main memory latency = 32
Total time required to place cache line is
64+32 = 96 ns
→ So, for 64 bytes it will take 64*1/109 = 64ns.
Main memory latency = 32
Total time required to place cache line is
64+32 = 96 ns
Correct Answer: C
Question 25 Explanation:
For 1GBytes/s bandwidth → it takes 1 sec to load 109 bytes on line.
→ So, for 64 bytes it will take 64*1/109 = 64ns.
Main memory latency = 32
Total time required to place cache line is
64+32 = 96 ns
→ So, for 64 bytes it will take 64*1/109 = 64ns.
Main memory latency = 32
Total time required to place cache line is
64+32 = 96 ns
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