Suppose you are given n tasks J = {j1, j2, · · · jn}.
Each task ji has two parts - a preprocessing phase which takes pi units and a main phase which
takes fi units of time. There are n machines that can execute the main phases of the jobs in parallel.
However, the preprocessing phases need to be executed sequentially on a special machine. The
completion time of any schedule is the earliest time when all tasks have finished execution. Design
a greedy
need to give formal proof of correctness. Your proof should not exceed one side of an A4 sheet.
Anything more than that will not be considered by the evaluator. Provide only pseudocode.
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- CPU scheduler will decide which process should be given the CPU for its execution. For this it uses different algorithm to choose among the process. One among that algorithm is FCFS algorithm. In this algorithm the process which arrives first is given the CPU after finishing its request only it will allow the CPU to execute another process. Write a c program to implement the first come first serve CPU scheduling algorithm Sample output: Enter number of processes:3 Enter the burst times: Gantt chartPi 92 P3 Process Burst Time Waiting Time Turnaround Time 8 5 P1 P2 P3 0 13 Average Waiting Time is 7.000000 Average Turn around Time is 14.333333 8 22arrow_forwardConsider a system of three independent, periodic real-time tasks: T1= (4, 3), T2 = (12,7), and T3 = (8, 5). Construct the schedule for these tasks in the interval [0, 24) with q=1 on twoprocessors with PF. Note that there are many PFair algorithms, while PF is one of them. PF prioritize subtasks by deadlines, and break ties by inspecting future subtask deadlines. Also, if a currently-executing job ties with a newly- released job, then continue executing the currently-executing job; break any other ties by task ID (T1 = highest, T3 = lowest). a. Show all sub jobs’ releases and deadlines for (the first job of) each task. Note that q=1 so each sub job should have execution requirement of 1b. Show the PF schedule.c. Will this task set be schedulable with q=2? Please explain.arrow_forwardSuppose that an operating system supports two kinds of sequential processes: high-priority interactive processes, and low-priority non-interactive processes. The behavior of the high-priority processes is to alternate between periods of computation of duration Tc and periods of blocking (waiting for input) of duration Tb. The behavior of the low-priority processes is to compute constantly, with no blocking. The operating system’s scheduling policy is round-robin with a quantum q, where Tc < q. Scheduling decisions are made only when a quantum expires, or when the running process blocks. The scheduler selects a low-priority process to run only if no high-priority processes are ready. Suppose there is one high-priority process and one low-priority process in the system, and that both processes will run for a long time. For what fraction of the time does the low-priority process run?arrow_forward
- Consider a computer environment in which there are 4 magnetic drives, 2 printers, 3 scanners, and 1 optical drive. Three processes P1, P2, P3 are using these resources. At any point in time, the allocations and requests for these resources are as follows. Allocations: Magnetic drive Printer Scanner Optical drive P1 1 P2 2 1 P3 1 Requests: Magnetic drive 2 Printer Scanner Optical drive P1 1 P2 1 1 P3 2 1 Is there any way that the processes can complete without deadlock? If yes, explain how. If not, explain why not.arrow_forwardGiven the list of processes, their CPU burst times, arrival times and priorities implement SJF,Priority and Round Robin scheduling algorithms on the processes with preemption. For each ofthe scheduling policies, compute and print the completion Time(CT), Turnaround Time(TAT),and Waiting Time(WT) for each process using C Programming.Waiting time: Processes need to wait in the process queue before execution starts and inexecution while they get preempted. Turnaround time: Time elapsed by each process to get completely served. (Difference betweensubmission time and completion time).arrow_forwardFor the following function, select the most appropriate time complexity in terms of n: def rate(n): i = 0 total = 0 while in: j= n while j > 1: total + j j // = 2 i += 1 return total IMPORTANT: The penalty scheme for this question is 50, 100%. Select one: O a. Time complexity = O(n) O b. Time complexity = O(n^2) O c. Time complexity = O(2n) O d. Time complexity = O(log n) e. Time complexity = O(n log n) Checkarrow_forward
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