Knowing the formula Speedupoverall $=1/((1-$ Fractionenhanced$)+($Fractionenhanced $/$ Speedupenhanced$))$ or Speedup$=1/((1$P$)+($P$/$N$))$

As indicated in the text, when an information$-$processing task is parallelized on a computing system, the ideal speedup occurs when the speedup is equal to the number of processors the system has. In practice, though, that$$s not achieved because, as noted in the text, there are communication overheads involved with sharing$/$synchronizing the processors and not all of the task can be parallelized. As also noted there, Amdahl$$s Law focuses on this last point. Considering all that and what you have read in the text and we have discussed in the lecture, answer the following.

$($a$)$ Assuming that only $75\%$ of a task is parallelizable, compute what the speedup would be if there were$4$ processors. Ignore communication cost and assume that the $4$ processors are equally utilized.

$($b$)$ Assuming that only $75\%$ of a task is parallelizable, compute what the speedup would be if there were $10$ processors. Ignore communication cost and assume that the $10$ processors are equally utilized.

$($c$)$ Remember, the results in $($a$)$ and $($b$)$ are optimal results. They are optimal in the sense that they haven$$t considered communication overhead$$communication overhead would be expected to reduce speedup below what you computed. Re$-$compute the speedup for running a task that is $75\%$ parallelizable on $10$ processors assuming that these two things also both apply : the original, non$-$parallelized task ran in $10$ seconds run$-$time and when the parallelized part of the task ran on the $10$ processor environment it spent $1$ sec more of run$-$time in communication following the end of the parallelized part.