Chapter 5, Problem 24E

Explanation of Solution

Explanation:

The time taken by the non-pipelined system to process the task is 100ns

The time taken by the five-segment pipeline to process the task using clock cycles is 20ns

The number of tasks present in the pipeline is 100

The number of segments present in the pipelined system is 5

Speedup:

The speedup gained can be calculated using the following formula

$\text{Actual}\text{speedup}\text{(S)}\text{=}\frac{\text{time}\text{taken}\text{by}\text{pipelined system to process a task}}{\text{time}\text{taken}\text{by}\text{non-pipelined system to process a task}}$

Substitute “${\text{nt}}_{\text{n}}$” for “the time taken by pipelined system” and “$(\text{k+}\text{n}-{\text{1)t}}_{\text{p}}$” for “the time taken by non-pipelined system”

$\begin{array}{c}\text{Actual}\text{speedup}(\text{S)}\text{=}\frac{{\text{nt}}_{\text{n}}}{(\text{k+}\text{n}-{\text{1)t}}_{\text{p}}}\\ =\frac{\text{100}\text{×}100}{\text{(5}\text{+}1\text{00}-\text{1)}\text{×}2\text{0}}\\ =\frac{10000}{(104)\times 20}\\ =\frac{10000}{2080}\\ =4.807\end{array}$

Therefore, the maximum speedup of the pipeline for 100 tasks is 4.807