For a non-pipelined system:
- Total number of instruction/task $(n)=1000$
- Total time required to perform a single task in pipelined processor $(T_{np})=130$ ns
For a pipelined system:
- Total number of stages $(k)=5$
- Total number of instruction/task $(n)=1000$
- Total time required to perform a single task in pipelined processor $(T_p)=30$ ns
$\because \text{Speedup ($S_k$)=$\frac{ET_{np}}{ET_p}$}$
$\implies S_k= \frac{(n*T_{np})}{(k+(n-1))T_p}$
$\implies S_k=\left[\frac{1000*130}{(5+(1000-1))*30}\right] $ ns
$\implies S_k=\left[\frac{130000}{(5+999)*30}\right]$ ns
$\implies S_k=\left[\frac{130000}{30120}\right]$ ns
$\implies S_k= 4.316$ ns
speedup =$4.316$ ns is the answer.