Chapter 5, Problem 23E

Explanation of Solution

Explanation:

The time taken by the non-pipelined system to process the task is 200ns

The time taken by the five-segment pipeline to process the task using clock cycles is 40ns

The number of tasks present in the pipeline is 200

Using five-segment pipeline:

Assume a k-stage pipeline which includes clock cycle time of ${\text{t}}_{\text{p}}$ with n number of tasks to process.

The time required by the task 1 to complete the process is $\text{k}\text{×}{\text{t}}_{\text{p}}$

The time required to process the remaining tasks is $\text{(n}-{\text{1)t}}_{\text{p}}$

The time required to complete n tasks using a k-stage pipeline can be calculated as follows

$\begin{array}{c}\text{=}\text{k}\text{×}{\text{t}}_{\text{p}}\text{+}\text{(n}-\text{1)}\text{×}{\text{t}}_{\text{p}}\\ \text{=}\text{(k}\text{+}\text{n}-\text{1)}\text{×}{\text{t}}_{\text{p}}\end{array}$

Thus, the time taken to complete n tasks using k-stage pipeline is $\text{=}\text{(k}\text{+}\text{n}-\text{1)}\text{×}{\text{t}}_{\text{p}}$

Without using pipeline:

The time required by the task 1 to complete the process without using pipeline is ${\text{t}}_{\text{n}}\text{=}\text{k}\text{×}{\text{t}}_{\text{p}}$

The number of cycles required for n tasks is $\text{n}\text{×}{\text{t}}_{\text{n}}$

Speedup:

The speedup gained can be calculated using the following formula

$\text{Ac}$