- Uniform delay pipeline
In this type of pipeline, all the stages will take same time to complete an operation.
In uniform delay pipeline, Cycle Time (Tp) = Stage Delay
If buffers are included between the stages then, Cycle Time (Tp) = Stage Delay + Buffer Delay
- Non-Uniform delay pipeline
In this type of pipeline, different stages take different time to complete an operation.
In this type of pipeline, Cycle Time (Tp) = Maximum(Stage Delay)
For example, if there are 4 stages with delays, 1 ns, 2 ns, 3 ns, and 4 ns, then
Tp = Maximum(1 ns, 2 ns, 3 ns, 4 ns) = 4 ns
If buffers are included between the stages,
Tp = Maximum(Stage delay + Buffer delay)
Example : Consider a 4 segment pipeline with stage delays (2 ns, 8 ns, 3 ns, 10 ns). Find the time taken to execute 100 tasks in the above pipeline.
Solution : As the above pipeline is a non-linear pipeline,
Tp = max(2, 8, 3, 10) = 10 ns
We know that ETpipeline = (k + n – 1) Tp = (4 + 100 – 1) 10 ns = 1030 ns
NOTE: MIPS = Million instructions per second
Performance of pipeline with stalls
Speed Up (S) = Performancepipeline / Performancenon-pipeline => S = Average Execution Timenon-pipeline / Average Execution Timepipeline => S = CPInon-pipeline * Cycle Timenon-pipeline / CPIpipeline * Cycle Timepipeline
Ideal CPI of the pipelined processor is ‘1’. But due to stalls, it becomes greater than ‘1’.
S = CPInon-pipeline * Cycle Timenon-pipeline / (1 + Number of stalls per Instruction) * Cycle Timepipeline As Cycle Timenon-pipeline = Cycle Timepipeline, Speed Up (S) = CPInon-pipeline / (1 + Number of stalls per instruction)
Sources : goo.gl/J9KVNt
This article has been contributed by Saurabh Sharma.
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