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 | | |最小平均延迟时间(MAL)为__________
(A) 3
(B) 2
(C) 1
(D) 4答案： (A)
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;
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