Question: Comparison between FCFS and SJF scheduling algorithms.
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Mumbai University > Information Technology > Sem 5 > Operating System

Marks: 6M

Year: May 2016

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modified 2.8 years ago  • written 2.8 years ago by gravatar for Juilee Juilee2.4k
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A Process Scheduler schedules different processes to be assigned to the CPU based on particular scheduling algorithms.

There are six popular process scheduling algorithms which we are going to discuss in this chapter −

  • First-Come, First-Served (FCFS) Scheduling
  • Shortest-Job-Next (SJN) Scheduling
  • Priority Scheduling
  • Shortest Remaining Time
  • Round Robin(RR) Scheduling
  • Multiple-Level Queues Scheduling

These algorithms are either non-preemptive or preemptive. Non-preemptive algorithms are designed so that once a process enters the running state, it cannot be preempted until it completes its allotted time, whereas the preemptive scheduling is based on priority where a scheduler may preempt a low priority running process anytime when a high priority process enters into a ready state.

First Come First Serve (FCFS)

  • Jobs are executed on first come, first serve basis.
  • It is a non-preemptive scheduling algorithm.
  • Easy to understand and implement.
  • Its implementation is based on FIFO queue.
  • Poor in performance as average wait time is high.

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Wait time of each process is as follows −

Process Wait Time : Service Time - Arrival Time
P0 0 - 0 = 0
P1 5 - 1 = 4
P2 8 - 2 = 6
P3 16 - 3 = 13

Average Wait Time: (0+4+6+13) / 4 = 5.75

Shortest Job Next (SJN)

  • This is also known as shortest job first, or SJF
  • This is a non-preemptive scheduling algorithm.
  • Best approach to minimize waiting time.
  • Easy to implement in Batch systems where required CPU time is known in advance.
  • Impossible to implement in interactive systems where required CPU time is not known.
  • The processer should know in advance how much time process will take

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Wait time of each process is as follows −

Process Wait Time : Service Time - Arrival Time
P0 3 - 0 = 3
P1 0 - 0 = 0
P2 16 - 2 = 14
P3 8 - 3 = 5

Average Wait Time: (3+0+14+5) / 4 = 5.50

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written 2.8 years ago by gravatar for Juilee Juilee2.4k
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