考虑以下具有三个阶段 S1、S2 和 S3 的管道的预留表。
Time -->
-----------------------------
1 2 3 4 5
-----------------------------
S1 | X | | | | X |
S2 | | X | | X | |
S3 | | | X | | |最小平均延迟 (MAL) 是 __________
(一) 3
(乙) 2
(三) 1
(四) 4答案：(一)
解释：
S1 | X | Y | | | X | Y | X | Y | | | X | Y |
S2 | | X | Y | X | Y | | | X | Y | X | Y | |
S3 | | | X | Y | | | | | X | Y | | |
We can interleave instructions like the above
pattern.
Latency between X and Y is 1.
Latency between fist and second X is 5.
The pattern repeats after that.
So, MAL is (1 + 5)/2;
Quiz of this Question
Learn all GATE CS concepts with Free Live Classes on our youtube channel.