A job J consists of two tasks: printing of 5 pages plus some graphics work which is CPU intensive (also known as CPU bound). The task of printing the five pages is split into five subtasks indicated J1,J2,J3,J4 and J5 respectively. The graphics work is indicated as J. Start time is t=0 for all Js. J completes when J1,J2,J3,J4,J5,J6 get completed. Pages are printed one at a time.Preparing a single page to print takes 1s of instruction execution (service time). Printing it afterwards takes only printer time of 5s (printer service time). If interrupts are available interrupt handling also takes 1s (overhead time). The J6 graphics work requires service time of 10s and must start after all five pages are processed by the CPU but not necessarily printed by the printer. After the printer completes the printing of a page 1s is spent for interrupt handling to complete the corresponding job of printing the page (the job is not completed with the printing but with the conclusion of the interrupt associated with it). Interrupt handling has priority over other processor tasks. StartTime is t=0. CompletionTime or FinishTime is the time the job ends ( J is J1...J6). TurnaroundTime is CompletionTime minus StartTime. (Elapsed time can be used as a synonym to turnaround time by some books.) CPU utilization and Processor utilization mean the same thing (i.e. they are synonym) and give the percentage of time the CPU is busye doing useful things for say the job in question (i.e. preparing a page or graphics work). We do not have interrupts and thus Programmed I/O (also known as Blocked I/O) is to be employed. What is the turnaround time of J in s ? (We ask for time not just magnitude.) Tr(J)= What is the processor utilization CPU(J)? Use fractional answer CPU(J)=A/B , where A,B are integer in the simplest form, and then give A= and B= Note: A turnaround time would never be 2 or 2.0 s. Only 2 s would be marked as correct if it is indeed two seconds. If CPU utilization is 50% then the simplest form for A and B are A=1 and B=2. A pair A=50 and B=100 or similar pairs will not get you points. We do have interrupts and thus Interrupt-based I/O is to be employed. What is the turnaround time of J in s ? Tr(J)= What is the processor utilization CPU(J) ? Use a fractional answer CPU(J)=C/D= C/31 but provide only the numerator. The denominator is given to you as D=31