- Throughput: It is the ratio of total number of
processes executed to a given amount of time.
- Latency: This factor can be sub – divided in
to two namely response time and the turnaround time. Response time is the
time taken from the submission of the process till its output is produced
by the processor. The latter i.e., the turnaround time is the time period
elapsed between the process submission and its completion.
- Waiting/ fairness time: This is the equal CPU
time given to each process or we can say that the time is allocated as
per the priority of the processes. The time for which the processes wait
in the ready queue is also counted in this.
Thursday, May 2, 2013
What is a CPU Scheduler?
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5/02/2013 05:28:00 PM
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Labels: Algorithm, Communication, CPU, CPU Scheduling, Data, Latency, Memory, Methods, Multi-tasking, Operating System, Resources, Scheduling, System, Tasks, Throughput, Transmit, Types, Waiting time
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Tuesday, April 23, 2013
What is Throughput, Turnaround time, waiting time and Response time?
What is Throughput?
What is Turnaround Time?
What is Waiting Time?
What is Response Time?
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Sunflower
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4/23/2013 06:57:00 PM
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Labels: Communication, Complete, CPU, Data, digital, Factors, Network, Operating System, Packets, Performance, Processes, Response time, Submit, Tasks, Thread, Throughput, Turnaround time, Waiting time
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Wednesday, August 26, 2009
Shortest-Job-First (SJF) Scheduling
Shortest-Job-First (SJF) is a non-preemptive discipline in which waiting job (or process) with the smallest estimated run-time-to-completion, is run next. In other words, when CPU is available, it is assigned to the process that has smallest next CPU burst. The SJF scheduling is especially appropriate for batch jobs for which the run times are known in advance. Since the SJF scheduling algorithm gives the minimum average time for a given set of processes, it is probably optimal. The SJF algorithm favors short jobs (or processors) at the expense of longer ones.
Example :
Process Burst time Arrival
P1 6 0
P2 8 0
P3 7 0
P4 3 0
Gantt chart: Order P1, P2, P3, P4
| P4 | P1 | P3 | P2 |
0 3 9 16 24
Average waiting time: (0+3+16+9)/4 = 7
With FCFS: (0+6+(6+8)+(6+8+7))/4 = 10.25
Problem: SJF minimizes the average wait time because it services small processes before it services large ones. While it minimizes average wiat time, it may penalize processes with high service time requests. If the ready list is saturated, then processes with large service times tend to be left in the ready list while small processes receive service. In extreme case, where the system has little idle time, processes with large service times will never be served. This total starvation of large processes may be a serious liability of this algorithm.
Posted by
Sunflower
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8/26/2009 09:12:00 PM
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Labels: Process, Scheduling algorithm, Shortest-Job-First, SJF, Waiting time
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