Scheduling Algorithms
First-Come, First-Served (FCFS) Scheduling
Process Burst Time
P1 24
P2 3
P3 3
Suppose that the processes arrive in the order: P1 , P2 , P3
The Gantt Chart for the schedule is:
P1
|
P2
|
P3
|
0 24 27 30
Waiting time for P1 = 0; P2 = 24; P3 = 27
Average waiting time: (0 + 24 + 27)/3 = 17
Suppose that the processes arrive in the order
P2 , P3 , P1 .
The Gantt chart for the schedule is:
P2
|
P3
|
P1
|
0 3 6 30
Waiting time for P1 = 6; P2 = 0; P3 = 3
Average waiting time: (6 + 0 + 3)/3 = 3
Much better than previous case.
Convoy effect short process behind long process
Shortest-Job-First (SJF) Scheduling
Associate with each process the length of its next CPU burst. Use these lengths to schedule the process with the shortest time.
Two schemes:
1. Non pre- emptive – once CPU given to the process it cannot be preempted until completes its CPU burst.
2. Preemptive – Preemption takes place when a new process arrives with CPU burst length less than remaining time of current executing process. This scheme is also known as the Shortest-Remaining-Time-Next (SRTN) Scheduling.
SJF is optimal – gives minimum average waiting time for a given set of processes.
Process Arrival Time Burst Time
P1 0.0 7
P2 2.0 4
P3 4.0 1
P4 5.0 4
SJF (non-preemptive)
P1
|
P3
|
P2
|
P4
|
0 7 8 12 16
Average waiting time = [0 +(8-2)+(7-4) +(12-5)] /4 =4
Example of Preemptive SJF
Proces Arrival Time Burst Time
P1 0.0 7
P2 2.0 4
P3 4.0 1
P4 5.0 4
SJF (preemptive)
P1
|
P2
|
P3
|
P2
|
P4
|
P1
|
0 2 4 5 7 11 16
Average waiting time = (9 + 1 + 0 +2)/4 =3
Determining length of cpu burst is possiblee by using the length of previous CPU bursts, using exponential averaging though it's a very complex calculation to carry out.
Priority Scheduling
A priority number (integer) is associated with each process
The CPU is allocated to the process with the highest priority (smallest integer ≡ highest priority).
1. Preemptive
2. nonpreemptive
SJF is a priority scheduling where priority is the predicted next CPU burst time.
Problem ≡ Starvation – low priority processes may never execute.
Solution ≡ Aging – as time progresses increase the priority of the process.
Round Robin (RR)
Each process gets a small unit of CPU time (time quantum), usually 10-100 milliseconds. After this time has elapsed, the process is preempted and added to the end of the ready queue.
If there are n processes in the ready queue and the time quantum is q, then each process gets 1/n of the
CPU time in chunks of at most q time units at once. No process waits more than (n-1)q time units.
Performance
1. q large _ FIFO
2. q small _ q must be large with respect to context switch, otherwise overhead is too high.
Example of RR with Time Quantum = 4
Process Burst Time
P1 24
P2 3
P3 3
The Gantt chart is:
P1
|
P2
|
P3
|
P1
|
P1
|
P1
|
P1
|
P1
|
0 4 7 10 14 18 22 26 30
Average waiting time = ((30-24)+4+7)/3 = 17/3 =5.66
Multilevel Queue
Ready queue is partitioned into separate queues:
foreground (interactive)
background (batch)
Each queue has its own scheduling algorithm,
foreground – RR
background – FCFS
Scheduling must be done between the queues.
1. Fixed priority scheduling; (i.e., serve all from foreground then from background). Possibility of starvation.
2. Time slice – each queue gets a certain amount of CPU time
which it can schedule amongst its processes; i.e., 80% to foreground in RR
1. 20% to background in FCFS
Multilevel Queue Scheduling
Multilevel Feedback Queue
A process can move between the various queues; aging can be implemented this way.
Multilevel-feedback-queue scheduler defined by the following parameters:
1. number of queues
2. scheduling algorithms for each queue
3. method used to determine when to upgrade a process
4. method used to determine when to degrade a process
5. method used to determine which queue a process will enter when that process needs service
Three queues:
1. Q0 – time quantum 8 milliseconds
2. Q1 – time quantum 16 milliseconds
3. Q2 – FCFS
Scheduling
1. A new job enters queue Q0 which is served FCFS . When it gains CPU, job receives 8 milliseconds.
If it does not finish in 8 milliseconds, job is moved to queue Q1.
2. At Q1 job is again served FCFS and receives 16 additional milliseconds. If it still does not complete,
it is preempted and moved to queue Q2.
No comments:
Post a Comment