Reputation: 1
I'm a little stumped here. I'm supposed to calculate turnaround time for SFJ (non-preemptive) and Round Robin quantums 1 and 10. I'm trying to go over some notes from the web, but I'm not really seem to grasping it. Can anyone easily explain how I can find turnaround time for SJF and RR Q 1 and 10 for the following table?
Process Arrival Time Burst Time
P1 3.0 3.0
P2 2.0 7.0
P3 1.0 2.0
Upvotes: 0
Views: 3488
Reputation: 176
we find the Gantt chart first.
| P3 | P1 | P2 |
1-----3-----6---------13
at 2.0 (millisecond i guess...) P2 came and P1 came at 3.0 ms .It's non preemptive so CPU was not preempted while executing any process , then the process with smallest burst time is chosen.
Average waiting time = [(1-1)+(6-2)+(3-3)]/3 = 1.33 ms
Average Turnaround time = [(0+2) + (7+4) + (3+0)]/3 = 5.33 ms
Upvotes: 1
Reputation: 815
Arrival time is the response time, in other words the time it takes to start the process. You should explain what you mean by Burst time... When counting scheduler times, I know of response/arrival, turnaround time, and wait time. Never heard of "Burst".
Upvotes: 0