Longest Remaining Time First (LRTF) or Preemptive Longest Job First CPU Scheduling Algorithm

Longest Remaining Time First (LRTF) is a preemptive version of Longest Job First (LJF) scheduling algorithm. In this scheduling algorithm, we find the process with the maximum remaining time and then process it, i.e. check for the maximum remaining time after some interval of time(say 1 unit each) to check if another process having more Burst Time arrived up to that time. 

Characteristics of Longest Remaining Time First (LRTF) 

• Among all the processes waiting in a waiting queue, CPU is always assigned to the process having largest burst time.
• If two processes have the same burst time then the tie is broken using FCFS i.e. the process that arrived first is processed first. 
• LJF CPU Scheduling can be of both preemptive and non-preemptive type.

Advantages of Longest Remaining Time First (LRTF) 

• No other process can execute until the longest job or process executes completely.
• All the jobs or processes finishes at the same time approximately.

Disadvantages of Longest Remaining Time First (LRTF) 

• This algorithm gives very high average waiting time and average turn-around time for a given set of processes.
• This may lead to convoy effect.
• It may happen that a short process may never get executed and the system keeps on executing the longer processes.
• It reduces the processing speed and thus reduces the efficiency and utilization of the system.

Longest Remaining Time First (LRTF) CPU Scheduling Algorithm

• Step-1: First, sort the processes in increasing order of their Arrival Time. 
• Step-2: Choose the process having least arrival time but with most Burst Time. 
• Step-3: Then process it for 1 unit. Check if any other process arrives upto that time of execution or not. 
• Step-4: Repeat the above both steps until execute all the processes. 
  

Examples to show working of Preemptive Longest Job First CPU Scheduling Algorithm:

Example-1: Consider the following table of arrival time and burst time for four processes P1, P2, P3 and P4. 

Processes	Arrival time	Burst Time
P1	1 ms	2 ms
P2	2 ms	4 ms
P3	3 ms	6 ms
P4	4 ms	8 ms

The Longest Remaining Time First CPU Scheduling Algorithm will work on the basis of steps as mentioned below:

At time = 1, 

• Available Process : P1. So, select P1 and execute 1 ms. 

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
1-2ms	P1	1ms		1ms	2ms	1ms

At time = 2, 

• Available Process : P1, P2. 
• So, select P2 and execute 1 ms (since B.T(P1) = 1 which is less than B.T(P2) = 4) 

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
2-3ms	P1	1ms	P1	0ms	1ms	1ms
	P2	2ms		1ms	4ms	3ms

At time = 3,

• Available Process : P1, P2, P3. 
• So, select P3 and execute 1 ms (since, B.T(P1) = 1 , B.T(P2) = 3 , B.T(P3) = 6). 

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
3-4ms	P1	1ms	P1, P2	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		1ms	6ms	5ms

At time = 4,

• Available processes: P1, P2, P3, P4. 
• So, select P4 (as burst time of P4 is largest) and execute 1 ms (since, B.T(P1) = 1 , B.T(P2) = 3 , B.T(P3) = 6, B.T(P4) = 8). 

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
4-5ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		0ms	5ms	5ms
	P4	4ms		1ms	8ms	7ms

At time = 5,

• Available processes: P1, P2, P3, P4, 
• Process P4 will continue its execution as no other process has burst time larger than the Process P4

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
5-7ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		0ms	5ms	5ms
	P4	4ms		2ms	7ms	5ms

At time = 7,

• The processes P3 and P4 have same remaining burst time, 
• hence If two processes have the same burst time then the tie is broken using FCFS i.e. the process that arrived first is processed first. 
• Therefore P3 will get executed for 1ms

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
7-8ms	P1	1ms	P1, P2, P4	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		1ms	5ms	4ms
	P4	4ms		0ms	5ms	5ms

At time = 8,

• Available processes: P1, P2, P3, P4, 
• Process P4 will again continue its execution as no other process has burst time larger than the Process P4

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
8-9ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		0ms	4ms	4ms
	P4	4ms		1ms	5ms	4ms

At time = 9,

• Available processes: P1, P2, P3, P4, 
• Process P3 continue its execution on the basis of FCFS rule.

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
9-10ms	P1	1ms	P1, P2, P4	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		1ms	4ms	3ms
	P4	4ms		0ms	4ms	4ms

At time = 10,

• Available processes: P1, P2, P3, P4, 
• Now again the burst time of P4 is largest, thus it will execute further.

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
10-11ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		0ms	3ms	3ms
	P4	4ms		1ms	4ms	3ms

At time = 11,

• Available processes: P1, P2, P3, P4, 
• Process P2 will continue its execution as the burst time of P2, P3, P4 is same 
• Thus, in this case the further execution will be decided on the basis of FCFS i.e. the process that arrived first is processed first. 

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
11-12ms	P1	1ms	P1, P3, P4	0ms	1ms	1ms
	P2	2ms		1ms	3ms	2ms
	P3	3ms		0ms	3ms	3ms
	P4	4ms		0ms	3ms	3ms

At time = 12, 

• Available processes: P1, P2, P3, P4, 
• Process P3 continue its execution on the basis of above explanation.

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
12-13ms	P1	1ms	P1, P2, P4	0ms	1ms	1ms
	P2	2ms		0ms	2ms	2ms
	P3	3ms		1ms	3ms	2ms
	P4	4ms		0ms	3ms	3ms

At time = 13,

• Available processes: P1, P2, P3, P4, 
• Now again the burst time of P4 is largest, thus it will execute further.

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
13-14ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	2ms	2ms
	P3	3ms		0ms	2ms	2ms
	P4	4ms		1ms	3ms	2ms

At time = 14,

• Available processes: P1, P2, P3, P4 
• Now, the process P2 will again begin to execute first among all

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
14-15ms	P1	1ms	P1, P3, P4	0ms	1ms	1ms
	P2	2ms		1ms	2ms	1ms
	P3	3ms		0ms	2ms	2ms
	P4	4ms		0ms	2ms	2ms

At time = 15,

• Available processes: P1, P2, P3, P4, now P3 will execute

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
15-16ms	P1	1ms	P1, P2, P4	0ms	1ms	1ms
	P2	2ms		0ms	1ms	1ms
	P3	3ms		1ms	2ms	1ms
	P4	4ms		0ms	2ms	2ms

At time = 16,

• Available processes: P1, P2, P3, P4, 
• here, P4 will execute as it has the largest Burst time among all

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
16-17ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	1ms	1ms
	P3	3ms		0ms	1ms	1ms
	P4	4ms		1ms	2ms	1ms

At time = 17,

• Available processes: P1, P2, P3, P4, 
• Process P1 will execute here on the basis of above explanation

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
17-18ms	P1	1ms	P2, P3, P4	1ms	1ms	0ms
	P2	2ms		0ms	1ms	1ms
	P3	3ms		0ms	1ms	1ms
	P4	4ms		0ms	1ms	1ms

At time = 18,

• Available processes: P2, P3, P4,  
• Process P2 will execute.

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
18-19ms	P2	2ms	P3, P4	1ms	1ms	0ms
	P3	3ms		0ms	1ms	1ms
	P4	4ms		0ms	1ms	1ms

At time = 19,

• Available processes: P3, P4,  
• Process P3 will execute.

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
19-20ms	P3	3ms	P4	1ms	1ms	0ms
	P4	4ms		0ms	1ms	1ms

At time = 20,

• Process P4 will execute at the end.

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
20-21ms	P4	4ms		1ms	1ms	0ms

At time = 22,

• Process P4 will finish its execution.

The overall execution of the processes will be as shown below

Time Instance	Process	Arrival Time	Waiting Table	Execution Time	Initial Burst Time	Remaining Burst  Time
1-2ms	P1	1ms		1ms	2ms	1ms
2-3ms	P1	1ms	P1	0ms	1ms	1ms
	P2	2ms		1ms	4ms	3ms
3-4ms	P1	1ms	P1, P2	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		1ms	6ms	5ms
4-5ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		0ms	5ms	5ms
	P4	4ms		1ms	8ms	7ms
5-7ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		0ms	5ms	5ms
	P4	4ms		2ms	7ms	5ms
7-8ms	P1	1ms	P1, P2, P4	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		1ms	5ms	4ms
	P4	4ms		0ms	7ms	5ms
8-9ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		0ms	4ms	4ms
	P4	4ms		1ms	5ms	4ms
9-10ms	P1	1ms	P1, P2, P4	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		1ms	4ms	3ms
	P4	4ms		0ms	4ms	4ms
10-11ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	3ms	3ms
	P3	3ms		0ms	3ms	3ms
	P4	4ms		1ms	4ms	3ms
11-12ms	P1	1ms	P1, P3, P4	0ms	1ms	1ms
	P2	2ms		1ms	3ms	2ms
	P3	3ms		0ms	3ms	3ms
	P4	4ms		0ms	3ms	3ms
12-13ms	P1	1ms	P1, P2, P4	0ms	1ms	1ms
	P2	2ms		0ms	2ms	2ms
	P3	3ms		1ms	3ms	2ms
	P4	4ms		0ms	3ms	3ms
13-14ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	2ms	2ms
	P3	3ms		0ms	2ms	2ms
	P4	4ms		1ms	3ms	2ms
14-15ms	P1	1ms	P1, P3, P4	0ms	1ms	1ms
	P2	2ms		1ms	2ms	1ms
	P3	3ms		0ms	2ms	2ms
	P4	4ms		0ms	2ms	2ms
15-16ms	P1	1ms	P1, P2, P4	0ms	1ms	1ms
	P2	2ms		0ms	1ms	1ms
	P3	3ms		1ms	2ms	1ms
	P4	4ms		0ms	2ms	2ms
16-17ms	P1	1ms	P1, P2, P3	0ms	1ms	1ms
	P2	2ms		0ms	1ms	1ms
	P3	3ms		0ms	1ms	1ms
	P4	4ms		1ms	2ms	1ms
17-18ms	P1	1ms	P2, P3, P4	1ms	1ms	0ms
	P2	2ms		0ms	1ms	1ms
	P3	3ms		0ms	1ms	1ms
	P4	4ms		0ms	1ms	1ms
18-19ms	P2	2ms	P3, P4	1ms	1ms	0ms
	P3	3ms		0ms	1ms	1ms
	P4	4ms		0ms	1ms	1ms
19-20ms	P3	3ms	P4	1ms	1ms	0ms
	P4	4ms		0ms	1ms	1ms
20-21ms	P4	4ms		1ms	1ms	0ms

Note: CPU will be idle for 0 to 1 unit time since there is no process available in the given interval. 

Gantt chart will be as following below: 

Since, completion time (C.T) can be directly determined by Gantt chart, and 

Turn Around Time (TAT)
= (Completion Time) – (Arrival Time)

Also, Waiting Time (WT)
= (Turn Around Time) – (Burst Time) 

Therefore, final table look like, 

Total Turn Around Time = 68 ms
So, Average Turn Around Time = 68/4 = 17.00 ms

And, Total Waiting Time = 48 ms
So Average Waiting Time = 48/4 = 12.00 ms 

Example-2: Consider the following table of arrival time and burst time for four processes P1, P2, P3,P4 and P5. 

Processes	Arrival Time	Burst Time
P1	0ms	2ms
P2	0ms	3ms
P3	2ms	2ms
P4	3ms	5ms
P5	4ms	4ms

Similarly example-1, Gantt chart for this example, 

Since, completion time (CT) can be directly determined by Gantt chart, and 

Turn Around Time (TAT)
= (Completion Time) – (Arrival Time)

Also, Waiting Time (WT)
= (Turn Around Time) – (Burst Time) 

Therefore, final table look like, 

Total Turn Around Time = 61 ms
So, Average Turn Around Time = 61/5 = 12.20 ms

And, Total Waiting Time = 45 ms
So, Average Waiting Time = 45/5 = 9.00 ms