A non-pipeline system takes 50ns to process a task. The same task can be processed in six-segment pipeline with a clockcycle of 10ns. Determine approximately the speedup ratio of the pipeline for 500 tasks. 

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UGC NET Computer Science (Paper 2) 2020 Official Paper
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  1. 6
  2. 4.95
  3. 5.7
  4. 5.5

Answer (Detailed Solution Below)

Option 2 : 4.95
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UGC NET Paper 1: Held on 21st August 2024 Shift 1
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Detailed Solution

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Concept:

Speed up factor is defined as the ratio of time required for non-pipelined execution to that of time received for pipelined execution.

Data:

Time for non-pipelined execution per task = t= 50 ns

Time for pipelined execution per task =  t= 10 ns

Number of stages in the pipeline = k = 6

Number of tasks = 500

Formula

\(S = \frac{{{T_n}}}{{{T_p}}}\)

S = speed up factor

Calculation:

Time for non-pipelined = Tn = tn × Number of tasks

Time for non-pipelined = Tn =  50 × 500

Time for pipelined = Tp​ = 1st task × k × tp + (All Remaining Tasks (k - 1)) × tp

Time for pipelined = Tp​ =  1 × 6 × 10 + (500 - 1) × 10

\(S = \frac{{T_n}}{{T_p}} = 4.95\)

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