SCHEDULING POLICIES

B.Sc.IT325

ANKIT SURATI

TYPES OF SCHEDULER

1.LONG-TERM SCHEDULER:DETERMINE WHICH JOBS ARE ADMITTED TO THESYSTEM FOR PROCESSING.2.MEDIUM-TERM SCHEDULER:WHEN TO MANY PROCESSOR ARE COMPETING FOR MEMORY,DETERMINES WHICH GETS IN/OUT.3.SHORT –TERM SCHEDULER:DETERMINES WHICH PROCESS IN MEMORY GOES NEXT.

SCHEDULING PERFORMANCE

• TURN AROUND TIME:Time process present in system

• WAITING TIME:Time process present and not running

• RESPONSE RATIO:Factor by which processing rate reduces

• TURN AROUND TIME:T=[finish time]-[arrival time]

• WAITING TIME:W=T-[service time]

• RESPONSE RATIO:R= T/(service time)

Decision Mode

• NON-PREEMPTIVE:A PROCESS RUNS UNTIL ITS BOLCK ARE COMPLETES(runs until no longer ready)

• PREEMPTIVE:OS CAN INTERRUPT CURRENTLY RUNNING PROCESS

TO START ANOTHER ONE

FIRST-COME, FIRST-SERVED

• Criteria: BT-AT Where BT=burst time AT=arival time• Mode:non-preemptive• Scheduling criteria: Jobs are excuted on first cum first serve basis

• Advantage: Easy to understand and implement. • Disadvantage: Poor in performance as average wait time is high.• Example: Pno Burst time P1 24 P2 3 P3 3Result: p1 p2 p3 0 24 27 30

SHORTEST JOB FIRST

• Mode:non-preemptive• Scheduling criteria: It is assigned to the process that has smallest next cpu burst.• Disadvantage: SJF is non-preemptive like FCFS, therefore not useful in time sharing environment.

• Example: Pno BT P1 6 P2 8 P3 7 P4 3 Result: p4 p1 p3 p2 0 3 9 16 24

SHORTEST REMAINIG TIME FIRST

• Mode:pre-emptive• Scheduling criteria: The process which is having shortesr remaining time is excuted first and completed.• Advantage: Short process are handled quickly• Disadvantage: Estimation of remaining t UK me is not easy.

• Example: pno AT BT 1 0 7 2 1 6 3 2 3 4 3 1 5 4 2 6 5 1Result: p1 p1 p3 p4 p3 p3 p6 p5 p1 p2 0 1 2 3 4 5 6 7 9 14 20

Highest Response Ratio

• Highest Response Ratio Next(HRRN)Decision mode:non-preemptiveArbitration rule: schedules job having highest

response ratio and runs until completion.Priority function:p(a,r,t)= = (burst time+waiting time) _________________________ burst time

*Advantages:– It is better than SRTF & SJF

*Disadvantages:– Caculation of response ratio

*Example: Pno AT BT 1 0 3 2 2 6 3 4 4 4 6 5 5 8 2 Result: p1 p2 p3 p5 p4 0 3 9 13 15 20

ROUND ROBIN

• Criteria: AT+TQ where AT=arival time TQ=time quantum• Mode: pre-emptive• Scheduling criteria: It schedules job as per AT (takes from front of queue) & executes till TQ is completed (then puts back to request queue) or time is over.

Thank you