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Processor Scheduling

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Assignment 3 Operating Systems Task 2 Introduction In this assignment we shall be looking at the four main memory management functions of an operating system. They are Process Management, Main-memory management, File and Disk management and Input/Output memory management. They shall be looked at further and in more detail within the task 3 section of the assignment. Process Management Scheduling Techniques There are three main states within the process management sub system. They are waiting, ready and running. Each process is started with the operation of system calls. Above can be seen a diagram of the scheduling operations within a system. The job queue is full of processes waiting to be entered into the ready queue, but must pass through admission to do so. When the job enters the ready queue it must wait for the processor to continue its processing method until it is time for the job to be dispatched. When dispatched it will enter the CPU where it will be executed. However, there are two errors that may occur. ...read more.


Looking at the figures above it comes as no surprise that this method is rarely used in a day to day environment. Shortest Job First The processor will execute each job in order of the completion time from lowest to highest. However; the first job received in the processor will be executed first even if it's bigger than any of the others in the queue. This method can reduce time and is far more efficient than FCFS. It can only work though if the CPU is aware if each processes burst time and then when this is achieved order can be obtained. This is a form of processing scheduling which has the lowest average waiting time using its algorithm. It's also a very clever processing technique. The processor cannot actually define the burst times individually so the operating system takes into account the previous burst times and then makes an average. This is a very clever feature. Process Arrival Time Burst Time P1 0 7 P2 2 4 P3 4 1 P4 5 4 0 7 8 12 16 Average waiting time = (0 + 6 +2 +7) ...read more.


When exceeded, the process shall be placed at the back of the queue and will be given an interrupt. It can be said that the round robin technique is a fair one. * If there are n processes in the ready queue and the time quantum is q then each process gets 1/nth of the CPU. * No process waits more than (n-1)q time units before receiving CPU allocation * It typically has a higher turnaround time than SRTF but a better average response time. The table below works on the basis of a time quantum of 4ms. Process Burst P1 24 P2 3 P3 3 0 4 7 10 14 18 22 26 30 Above can be seen a time quantum dictating the length of a process in the CPU. It can be said for definite that the process cannot exceed the quantum even if their are no other processes in the queue. We can also see that the P1 uses a whole time quantum; P2 and P3 however only use a part of it. P1 waits 0 + 6 = 6, P2 waits for 4 and P3 waits for 7 Average time = (6 + 4 + 7) / 3 = 5.66ms ...read more.

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