Difference between SRJF and LRJF CPU scheduling algorithms

1. Shortest remaining job first (SRJF) :
Shortest remaining job first also called the shortest remaining time first is the preemptive version of the shortest job first scheduling algorithm. 
In the shortest remaining job first, the process with the smallest runtime to complete (i.e remaining time) is scheduled to run next, In SRJF, a running process will be preempted if a new process arrives in the CPU scheduler which requires less burst time for execution. 

2. Longest remaining job first (LRJF) :
Longest remaining job first also called longest remaining time first is exactly opposite of SRJF. In this type of CPU scheduling, the process which has the longest processing time is scheduled to run next and a running process will be preempted if a new process with longer burst time enters the queue. 

Difference table : 

Shortest Remaining Job First (SRJF)	Longest Remaining Job First(LRJF)
short processes are executed first and at  any instant if a process with shorter time  arrives it will be executed first.	Long processes are executed first and at  any instant of time if a long process  appears it will be executed first.
It does not have large average turn around  time therefore it is more effective than LFJT	It has a very large average turn around  time and waiting time therefore  reduces the effectiveness of the  operating system
It does not lead to convoy effect	It will lead to convoy effect.
More process are executed in less amount of time	Less process are executed in certain amount of time
SRJF is a “preemptive” scheduling algorithm.	LRJF is a “non-preemptive” scheduling algorithm.

Let’s solve one problem for each:
LJFT : 

Processes	Arrival Time	Burst Time
P1	0	2
P2	0	4
P3	0	8

Longest remaining job first:
Gantt chart: 

Therefore the final table will be: 
LJFT : 

Processes	Arrival Time	Burst Time	Completion Time	Turn Around Time	Waiting Time
P1	0	2	12	12	10
P2	0	4	13	13	9
P3	0	8	14	14	6

Turn around time = Completion time - Arrival time 

Average turn around time, 
= [(12-0) + (13-0) + (14-0)]/3
= (12 + 13 + 14)/3
= 13


Waiting time = Turn around time - Burst time

Average waiting time,
= [(12-2) + (13-4) + (14-8)]/3
= (10 + 9 + 6)/3
= 8.34 

Problem two:

LJFT 

LJFT 

Processes	Arrival Time	Burst Time
P1	0	20
P2	15	25
P3	30	10
P4	45	15

Shortest remaining time first:
Gantt chart: 

Therefore the final table will be: 
SRFT 

Processes	Arrival Time	Burst Time	Completion Time	Turn Around Time	Waiting Time
P1	0	20	20	20	0
P2	15	25	55	40	15
P3	30	10	40	10	0
P4	45	15	70	25	10

Turn around time = Completion time - Arrival time 

Average turn around time, 
= [(20-0) + (55-15) + (40-10) + (70-45)]/4
= (20 + 40 + 30 + 25)/4
= 28.75


Waiting time = Turn around time - Burst time

Average waiting time,     
= [(20-20) + (40-25) + (10-10) + (25-15)] / 4
= (0 + 15 + 0 + 10) / 4
= 6.25 

My Personal Notes arrow_drop_up

Save