Given:
Time taken by non pipelined system = 200 ns = 200[tex]\times 10^{-9}[/tex] s
Clock cycle for 5- segment pipeline = 40 ns = 40[tex]\times 10^{-9}[/tex] s
Pipeline speed up ratio = 200
Formula used:
[tex](Speed-up)_{max} = \frac{System to process a task\times Speed-up ratio}{(speed-up ratio -1)clock cycle + segment}[/tex]
Solution:
Using formula:
[tex](Speed-up)_{max} = \frac{5\times 40\times 200}{((200 - 1)40 + 5}[/tex]
[tex](Speed-up)_{max}[/tex] = 5
