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Difference between SRJF and LRJF CPU scheduling algorithms
  • Difficulty Level : Easy
  • Last Updated : 26 May, 2021

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 effectIt will lead to convoy effect.
More process are executed in less amount of timeLess process are executed in certain amount of time

Let’s solve one problem for each: 



LJFT : 

 

ProcessesArrival TimeBurst Time
P102
P204
P308

Longest remaining job first: 
Gantt chart: 

 

Therefore the final table will be: 

LJFT : 

 



ProcessesArrival TimeBurst TimeCompletion TimeTurn Around TimeWating Time
P102121210
P20413139
P30814146

 

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 

 

ProcessesArrival TimeBurst Time
P1020
P21525
P33010
P44515

Shortest remaining time first: 
Gantt chart: 

 

Therefore the final table will be: 

SRFT 

 

ProcessesArrival TimeBurst TimeCompletion TimeTurn Around TimeWating Time
P102020200
P21525554015
P3301040100
P44515702510

 

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 

 

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