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CPU scheduling is a process used by the operating system to decide which task or process gets to use the CPU at a particular time. This is important because a CPU can only handle one task at a time, but there are usually many tasks that need to be processed. The following are different purposes of a CPU scheduling time.
CPU scheduling is the process of deciding which process will own the CPU to use while another process is suspended. The main function of CPU scheduling is to ensure that whenever the CPU remains idle, the OS has at least selected one of the processes available in the ready queue.
Turn Around Time = Completion Time – Arrival Time
Waiting Time = Turn Around Time – Burst Time
Different CPU Scheduling algorithms have different structures and the choice of a particular algorithm depends on a variety of factors.
There are mainly two types of scheduling methods:
Please refer Preemptive vs Non-Preemptive Scheduling for details.
Let us now learn about these CPU scheduling algorithms in operating systems one by one:
Here is a brief comparison between different CPU scheduling algorithms:
| Algorithm | Allocation | Complexity | Average waiting time (AWT) | Preemption | Starvation | Performance |
|---|---|---|---|---|---|---|
| FCFS | According to the arrival time of the processes, the CPU is allocated. | Simple and easy to implement | Large. | No | No | Slow performance |
| SJF | Based on the lowest CPU burst time (BT). | More complex than FCFS | Smaller than FCFS | No | Yes | Minimum Average Waiting Time |
| SRTF | Same as SJF the allocation of the CPU is based on the lowest CPU burst time (BT). But it is preemptive. | More complex than FCFS | Depending on some measures e.g., arrival time, process size, etc | Yes | Yes | The preference is given to the short jobs |
| RR | According to the order of the process arrives with fixed time quantum (TQ) | The complexity depends on Time Quantum size | Large as compared to SJF and Priority scheduling. | Yes | No | Each process has given a fairly fixed time |
| Priority Pre-emptive | According to the priority. The bigger priority task executes first | This type is less complex | Smaller than FCFS | Yes | Yes | Well performance but contain a starvation problem |
| Priority non-preemptive | According to the priority with monitoring the new incoming higher priority jobs | This type is less complex than Priority preemptive | Preemptive Smaller than FCFS | No | Yes | Most beneficial with batch systems |
| MLQ | According to the process that resides in the bigger queue priority | More complex than the priority scheduling algorithms | Smaller than FCFS | No | Yes | Good performance but contain a starvation problem |
| MFLQ | According to the process of a bigger priority queue. | It is the most Complex but its complexity rate depends on the TQ size | Smaller than all scheduling types in many cases | No | No | Good performance |
Read Related Article - Starvation and Aging
S1: It causes minimum average waiting time
S2: It can cause starvation
(A) Only S1
(B) Only S2
(C) Both S1 and S2
(D) Neither S1 nor S2
Answer: (D) S1 is true SJF will always give minimum average waiting time. S2 is true SJF can cause starvation.
| Process | Arrival time | Burst Time |
| P0 | 0 ms | 9 ms |
| P1 | 1 ms | 4 ms |
| P2 | 2 ms | 9 ms |
The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?
(A) 5.0 ms
(B) 4.33 ms
(C) 6.33
(D) 7.33
Solution: (A)
Process P0 is allocated processor at 0 ms as there is no other process in the ready queue. P0 is preempted after 1 ms as P1 arrives at 1 ms and burst time for P1 is less than remaining time of P0. P1 runs for 4ms. P2 arrived at 2 ms but P1 continued as burst time of P2 is longer than P1. After P1 completes, P0 is scheduled again as the remaining time for P0 is less than the burst time of P2. P0 waits for 4 ms, P1 waits for 0 ms and P2 waits for 11 ms. So average waiting time is (0+4+11)/3 = 5.
| Process | Arrival time | Burst Time |
| P1 | 0 ms | 5 ms |
| P2 | 1 ms | 3 ms |
| P3 | 2 ms | 3 ms |
| P4 | 4 ms | 1 ms |
What is the average turnaround time for these processes with the preemptive Shortest Remaining Processing Time First algorithm ?
(A) 5.50
(B) 5.75
(C) 6.00
(D) 6.25
Answer: (A)
Solution: The following is Gantt Chart of execution
| P1 | P2 | P4 | P3 | P1 |
| 1 | 4 | 5 | 8 | 12 |
Turn Around Time = Completion Time – Arrival Time Avg Turn Around Time = (12 + 3 + 6+ 1)/4 = 5.50
| Process | Burst Time | Arrival Time |
| P1 | 20 ms | 0 ms |
| P2 | 25 ms | 15 ms |
| P3 | 10 ms | 30 ms |
| P4 | 15 ms | 45 ms |
(A) 5
(B) 15
(C) 40
(D) 55
Answer (B)
Solution: At time 0, P1 is the only process, P1 runs for 15 time units. At time 15, P2 arrives, but P1 has the shortest remaining time. So P1 continues for 5 more time units. At time 20, P2 is the only process. So it runs for 10 time units At time 30, P3 is the shortest remaining time process. So it runs for 10 time units At time 40, P2 runs as it is the only process. P2 runs for 5 time units. At time 45, P3 arrives, but P2 has the shortest remaining time. So P2 continues for 10 more time units. P2 completes its execution at time 55.
Total waiting time for P2
= Completion time - (Arrival time + Execution time)
= 55 - (15 + 25)
= 15
