A. FCFS Answer - by the time the first job A is finished, then next job will be
ID: 644821 • Letter: A
Question
A. FCFS
Answer - by the time the first job A is finished, then next job will be B,C,D,E.
B. SJN
Answer - by the time the first job E is finished, then next job will be B,D, C, A.
C. SRT
Answer - by the time the first job E is finished, then next job will be B,D,C,A,C,D,A,A.
D. Round Robin
Answer - by the time the first job A is finished, then next job will be B,C,D,E,A,C,D,A.
9. Using the same information from the previous exercise, calculate the start time and finish time for each of the five jobs using each of the following scheduling algorithms. It may help to draw the timeline. a. FCFS b. SJN c. SRT d. Round Robin (use a time quantum of 5, but ignore the time required for context switching and natural wait) 10. Using the same information given for Exercise 8, compute the turnaround time for every job for each of the following scheduling algorithms (ignore context switching overhead times). It may help to draw the timeline. a. FCFS b. 94 c. SRT d. Round Robin (using a time quantum of 5) 8. Given the following information:Explanation / Answer
9)a
10a)
First Come First Served Scheduling
The process that requests the cpu first is allocated the CPU first.
Managed with a FIFO queue when a process enters the ready queue, its PCB is linked to the tail of the queue and when the cpu is free, it is allocated to the process at the head of the queue.
Process
Burst time
Arrival Time
Starting Time
Waiting Time
Finishing Time
Turnaround Time
P1
10
0
0
0
10
10
P2
29
0
10
10
39
39
P3
3
0
39
39
42
42
P4
7
0
42
42
49
49
P5
12
0
49
49
61
61
P1
P2
P3
P4
P5
Gantt Chart:
0 10 39 42 49 61
Avg waiting time=(0+10+39+42+49)/5=28
Avg Turnaround time=(10+39+42+49+61)/5=40.2
9) b
10)b)
Shortest Job Next(SJN)
It is also known as Shortest Job First(SJF) or Shortest Process Next(SPN) .
It selects the waiting process with the smallest execution time to execute next.SJN is non preemptive algorithm. it minimizes the average amount of time each process has to wait until its execution is complete
It has the potential for process starvation which will require a long time to complete if short processes are continually added. Highest response ration next is similar but provides a solution to this problem.
Another disadvantage of using shortest job next is that the total execution time of a job must be known before execution. While it is not possible to perfectly predict execution time, several methods can be used to estimate the execution time for a job, such as a weighted average of previous execution times.
Shortest job next can be effectively used with interactive processes which generally follow a pattern of alternating between waiting for a command and executing it. If the execution burst of a process is regarded as a separate "job", past behaviour can indicate which process to run next, based on an estimate of its running time.
Shortest job next is used in specialized environments where accurate estimates of running time are available. Estimating the running time of queued processes is sometimes done using a technique called aging..
The Gantt chart is:
P2
P1
P3
P4
P0
0 10 30 50 100 180
Average waiting time calculation:
Waiting time for process
p0 = 0 sec.
p1 =10 sec
p2 = 30 sec.
p3 = 50 sec.
p4 = 100 sec.
Therefore, the average waiting time is = (0+10+30+50+100)/5=38
c) Shortest remaining time first(SRT)
It selects the process for executing which has the smallest amount of time remaining until completion. SRT algorithm may lead to starvation.If the short processes are continually added to the cpu scheduler than the currently running process will never be able to execute, hence SRT is not starvation free.
Process
Burst time
Arrival time
Start time
Wait time
Finish time
Turnaround time
1
8
0
0
9
17
17
2
4
1
1
0
5
4
3
9
2
17
15
26
24
4
5
3
5
2
10
7
Gantt Chart
P1
P2
P4
P1
P3
0 1 5 10 17 26
9) d
10) d) Round Robin
(q = 10):
Process
Burst Time
Arrival
Start
Wait
Finish
TA
1
10
0
0
0
10
10
2
29
0
10
32
61
61
3
3
0
20
20
23
23
4
7
0
23
23
30
30
5
12
0
30
40
52
52
Gantt chart:
average waiting time: (0+32+20+23+40)/5 = 23
average turnaround time: (10+39+42+49+61)/5 = 35.2
Process
Burst time
Arrival Time
Starting Time
Waiting Time
Finishing Time
Turnaround Time
P1
10
0
0
0
10
10
P2
29
0
10
10
39
39
P3
3
0
39
39
42
42
P4
7
0
42
42
49
49
P5
12
0
49
49
61
61
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