Consider a pipe-lined process with s stages. Assuming n tasks are being processed by this pipe-line Both s and n are positive integers. In each of the scenarios listed below in (2.1), (2.2) and (2.3), compute formulas giving the following quantities: C(n, s): the total number of clock-cycles during which the pipe-lined process runs R(n, s): the speedup ratio with respect a non-pipe-lined serial execution, F(n, s): the fill time, that is, the total number of clock-cycles before the pipe-line runs at full occupancy (that is, before all pipe-line stages are used), no longer runs at full occupano which all pipe-line stages are used) D(n, s): the drain time, that is, the total number of clock-cycles remaining after the pipe-line O(n, s): the percentage of time during which the pipe-line runs at full occupancy (that is, during The three considered scenarios are the following ones 2.1 Each stage runs within the same amount of time. 2.2 Each stage, but the first one, runs within t units of time (say pico-seconds) meanwhile the first stage runs within r t units of time where r is a constant greater than one, thus, the first stage is slower than the other ones. You can assume t-1 in order to simplify calculations. Keeping t non-specialized (so as a symbol instead of letting be 1) yields a bonus of 5 marks 2.3 Each stage, but the last one, runs within t units of time (say pico-seconds) meanwhile the last stage runs within rt units of time where r is a constant greater than one, thus the last stage is slower than the other ones. You can assume t - 1 in order to simplify calculations In each of the scenarios (2.1), (2.2) and (2.3), you can assume that every stage runs within one clock-cycle. Moreover, you can assume that this clock-cycle is equal in time to the longest stage of the pipeline