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Mumbai University > Computer Engineering > Sem 8 > parallel and distributed systems
Marks: 5M
written 7.0 years ago by | modified 7.0 years ago by |
Mumbai University > Computer Engineering > Sem 8 > parallel and distributed systems
Marks: 5M
written 7.0 years ago by |
The ability to overlap stages of a sequential process for different input tasks (data or operations) results in an overall theoretical completion time of
$$T_pipe=m.P+(n-1).P$$
Here n is the number of input tasks, m is the number of stages in the pipeline, and P is the clock.
The three basic performance measures for the pipeline are as follows:
Speed up:
K-stage pipeline processes n tasks in k + (n-1) clock cycles: k cycles for the first task and n-1 cycles for the remaining n-1 tasks
Total time to process n tasks Tk = [k + (n-1)]
For the non-pipelined processor T1 = n k
Speed up factor is
$$speed up=(Time taken by non pipeline implemention)/(Time taken by pipelined implementation)$$
$$Speed up=nky/(K+(n-1))y=nk/(k+(n-1) )$$
$$n=α$$
$$speedup=nk/n=k$$
As the K stage reaches n and the value of n is approximated to infinity. The speedup factor is equivalent to K
Through put:
Throughput is the outputs produced per clock cycle and that throughput will be equal to 1, in case of ideal situation that means, when the pipeline is producing one output per clock cycle.
Formula is U(n)= m*f / n+(m-1)
Efficiency:
The efficiency of n stages in a pipeline is defined as ratio of the actual speedup to the maximum speed.
Formula is E(n)= m / n+m-1