What is Shortest Remaining Time Next Algorithm ?
This is the preemptive version of Shortest job first algorithm .This permits the process
that enters the ready list to preempt the running process if the time for the new process
is less than the remaining time fro the running process.
Let us understand this with the help of an example:
Consider the set of four process arrived as per timings described in the table
Process Arrival Time Processing Time
P1 0 5
P2 1 2
P3 2 5
P4 3 3
At time 0 only process P1 has entered the system, so its is the process that executes.
At time 1, process P2 arrives. At that time Process P1 has 4 units left to execute.
At this juncture process 2's processing time is less than P1. so P2 starts executing
at time 1.At time 3 P3 enters the system with the processing time 5 units.Process
P2 continues executing as it has minimum processing time compared to P1 and P3
both.At time 4 process p2 terminates and process P4 enters the system with time 3
units. Since process P4 has minimum time duration than P1 and P2 process P4 starts
executing. when process P1 terminates at time 10, process P3 enters the system.
The Gantt chart is explained below:
P1 P2 P4 P1 P3
0 1 3 6 10 15
Turn around Time for each of the process is given below:
P1 = 10 - 0 = 10
P2 = 3 - 1= 2
P3 = 15 - 2 = 13
P4 = 6 - 3 = 3
Average Turn around Time = 10+2+13+3/4 = 7
The waiting time fro each of the process is given as follows:
P1= 10 -5 =5
P2 = 2 - 2 =0
P3 =13 - 5 =8
P4 = 3 - 3 = 0
Average waiting Time = 5+0+8+0/4 = 3.25milliseconds
Four jobs executed in 15 time units, so throughput is 15/4 = 3.75 time units/job
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