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GATE | GATE-CS-2015 (Set 3) | Question 61

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  • Last Updated : 07 Dec, 2021
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Consider the following reservation table for a pipeline having three stages S1, S2 and S3.

     Time -->
-----------------------------
      1    2   3    4     5
-----------------------------
S1  | X  |   |   |    |  X |    
S2  |    | X |   | X  |    |
S3  |    |   | X |    |    |

The minimum average latency (MAL) is __________
(A) 3
(B) 2
(C) 1
(D) 4


Answer: (A)

Explanation:

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 first and second X is 5.

The pattern repeats after that.
So, MAL is (1 + 5)/2;


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