I udnerstand part a of this question. I just do not know how to find the 5 for part b. can somebody explain it ?

2. (4 points) A nonpipelined system takes 300ns to process a task. The same task can be processed in a 5-segment pipeline with a clock cycle of 60ns. a) Determine the speedup ratio of the pipeline for 100 tasks. N*Ketp = 100 * 5x 6Ons = 5(100) - 500 (n+K - tp 100+5 -1 ) bons 104 inatuotioks ne katr. K+K-1) te b) What is the maximum speedup that could be achieved with the pipeline unit over the nonpipelined unit? ( 5 ) K 0o> kys (n?k -15tr SU