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We have two tasks X and Y. The combination of them is called job Z. Thus Z= (Note the order in the sequence is the

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We have two tasks X and Y. The combination of them is called job Z. Thus Z= (Note the order in the sequence is the order of arrtival at t=0.) You are expected to give an answer that involves time in the unit of time utilized in this question. If you write 10 it is not clear whether it is 10 s, 10ms,10ns, or 10h or 10d or 10min. Task X is 3sCPU, followed by 4sI/O, followed by 1sCPU. Task Y is 1s CPU, followed by 4sI/O, followed by 3s CPU. For CPU utilization use integer fractions minized. Thus you do not write separately 4 and 8 to mean 4/8, but you use 1/2 minimized and you write in one box a 1 (not 1.0) and the other box a 2. (A). Serial Execution. (A1) What is the turnaround time of Z?Tr(Z)= (A2) What is the cpu utilization ? CPU Util(Z)= (B). Batch Multiprogramming. (B1) What is the turnaround time of Z ? Tr(Z)= (B2) What is the CPU utilization ? CPU Util (Z)= (C) Time sharing with timeslice 1s. Note: Time sharing affects only CPU allocation; it does not affect l/O. In I/O jobs are scheduled one after the other. Timeslicing is not affecting I/O execution. (C1) What is the turnaround time of Z ? Tr(Z)= (C2) What is the CPU utilization ? CPU Util (Z)= (D) Multitasking with quantum 1s. This problem is solved for you. (D1) What is the turnaround time of Z ? (D2) What is the CPU utilization ? The answer to question (D) follows with a pictorial proof. In a timesharing OS we have the following cpu timeline for two tasks X and Y. The timeslice is 1s. Both tasks are available in the system at the same time t=9:00:00.000 and order of arrival is the obvious X followed by Y. (The decimals reflect milliseconds if they showup in an indicated time reference.) There are no other processes (tasks) in the system other than X,Y. 1234567890 XYXYXYXY The time line 1 indicates that at t=9:00:00 s task X starts its execution and when t=9:00:01s is reached task Y takes over. The time line 1 indicates the 'first second' and time line 0 indicates the 'tenth second' above. Task Y completes its execution at t=9:00:10 s, the completion of the tenth second since X started its execution. Task X has completed its execution earlier. (a) What is the total number of context switches starting from prior to t=9:00:00 s (e.g. t=8:59:59.999) through the completion of $Y$? (b) What is the turnaround time for Y ? (c) What is the turnaround time for X ? (d) Tasks X,Y are processes and we utilize the five-state model of the textbook to describe their states. What is the state of X during the sixth second and just? Choose all states that are possible in your answer. (e) Tasks X,Y are processes and we utilize the five-state model as before. What is the state of Y during the sixth second (including just prior the sixth second)? Choose all states that are possible in your

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