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 have a 5-way SMP (A Symmetric Multi Processor computer with 5 identical CPUs that share the same main memory). We still have one printer. The five subtasks related to printing are fully parallelizable and thus can take advantage of the 5-way SMP machine's processors. The graphics task is also fully parallelizable and thus can keep all 5 CPU busy all the time by splitting evenly its work. Assume no overhead is incurred in such a splitting and processor reassignment. What is the turnaround time Tr(J) of J in s?Tr(J)= What is the processor utilization CPU(J)? Provide it in the form of a fractional CPU(J)=E/F=E/31 where F is given to you and thus you only need to provide E=