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Consider three processes (process id 0, 1, 2 respectively) w

Consider three processes (process id 0, 1, 2 respectively) with compute time bursts 2, 4 and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF ties are broken by giving priority to the process with the lowest process id. What is the average turn around time ?

4Comments
Sanjay Sharma @sanjaysharma
17 Sep 2014 11:22 am

34/3=11.33 time units is the avg turn around time and 20/3=6.66 time units is the avg wt

 

KRISHN KUMAR GUPTA @kkgupta2015
21 Sep 2014 01:29 pm

avg TAT=39/3=13 time units

Vivek Vikram Singh @vivek14
21 Sep 2014 02:10 pm

LRTF, means the process which has remaining time largest,will run first and in case of same remaining time,lowest process wid will be given priority to run. 

First 4 sec ,P2 will run . then p2 remaining time =4 ,p1=4,p0=2. 

Now P1 will get chance to run for 1 sec. p2=4,p1=3,p0=2.

Now p2 will get chance to run for 1 sec, P2=3,p1=3,p0=2.

This way if u do carefully, TAT of P1 will be 14 sec,p1=13 sec and p0= 12 sec.

Total TAT = 14+13+12= 39

Avg TAT= 39/3 =13 sec.

Neha @neha30
18 Oct 2014 12:37 am

TAT of process P0= 12

TAT of process P1=13

TAT of process P2= 14

average TAT=12+!3+14/3=13