Consider a system that uses Round-Robin scheduling and with a scheduling overhead of duration o=0.1ms. We represent the execution timeline of a job mix on this system by indicating 1ms of computation for a process via an uppercase letter. For example, the following string: AAAoBoCoAAoDDD means that process A executes for 3ms, then process B for 1ms, then process C for 1ms, then process A again for 2ms, then process D for 3ms, with appropriate context-switches. This example execution lasts 10ms + 4*0.1ms = 10.4ms (10ms of actual process execution, .4ms of context-switch overhead).

Question #1: Consider the following execution: AAAAAoBBoCoDDDDDoAAAAAoCoDDDDDoAAAoCoDDDDDoAAAAAoBBB Knowing the there is at least one CPU-intensive job in this mix, what is the tightest lower bound on the time quantum used by the scheduler? Why? Which process seems to have the longest I/O burst? Why? Which process would you say is interactive (e.g., a text editor in which a user is typing text)? Why? What fraction of the time is spent doing context-switch overhead in this execution?

Question #2 : Assume a job mix with 5 I/O-bound tasks and 1 fully CPU-bound task. The fully CPU-bound task is just one infinite CPU burst. Each I/O-bound task has a CPU burst of 1ms and then an I/O burst of 4ms (i.e., it does nothing on the CPU for 4ms and waits for the completion of whatever I/O operation it issues at the end of its CPU burst), and repeats this burst cycle ad infinitum. Assuming a Round-Robin scheduler with a time quantum of 1ms, we reason on steady-state execution (i.e., consider the behavior of the workload in the long run). Write the ASCII representation of the execution of the job mix as in Question #1, using I to denote 1ms of execution of any of the I/O-bound tasks, and C to denote 1ms of execution of the CPU-bound task, and o to denote a 0.1ms context-switch overhead. Even though the execution is infinite, you should identify a repeating pattern, which should be your answer. Your answer should look like this "IoIICoCCoIoIoC" (this is a very wrong answer though).