Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN: 9780133594140
Author: James Kurose, Keith Ross
Publisher: PEARSON
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Table 1. 4 processes arriving shown in Operating System Scheduling PDF for Shortest Job Next, non-
preemptive
Process Arrival Times Execution Times Service Times sequence Wait Times calculations
PO
0
12
P1
21
9
P2
41
6
P3
6
3
Total wait time = ?
Here the 4 processes of the PDF are repeated in Table 1, with arrival times changed to 0, 2, 4, and 6
respectively. Also, the execution times are changed to 12, 9, 6, and 3.
You will be asked to fill table 1 with the numbers like the service times (when is each process serviced), the
sequence of executions of these 4 processes in part (a) and part (b) below. Also, you will compute the total wait
time.
Note for FCFS, it is 1, 2, 3, 4 for P0, P1, P2, and P3;
• for the SJN or Shortest Job Next with non-preemptive, PO is #1, P1 may not be #2;
for SRT, or shortest remaining time, the preemptive version of SJN, PO is #1, but after 2 time units when
P1 comes, P1 kicks out PO since its execution time is 4 < 9- 2 = 7; and P1 may be preempted by P2 later
on, etc. You may use a time line to help computing when which process takes over.
(a)
Fill table T1 and compute total wait times if the queue is SJN or Shortest Job Next (note in case
of a tie, the process arriving first is executed first). This is non-preemptive, so PO will be executed to the
completion first
(b)
Repeat part (a) if the queue is SRT, which is the preemptive version of SJN. (so PO is kicked
out by P1 at time 2, and P1 is kicked out by P2 at time 4, etc. Please make a new table T2 for this part
(b).
Note you may do this question manually or by program if you have a computer program, or any other methods
you know.
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Transcribed Image Text:Table 1. 4 processes arriving shown in Operating System Scheduling PDF for Shortest Job Next, non- preemptive Process Arrival Times Execution Times Service Times sequence Wait Times calculations PO 0 12 P1 21 9 P2 41 6 P3 6 3 Total wait time = ? Here the 4 processes of the PDF are repeated in Table 1, with arrival times changed to 0, 2, 4, and 6 respectively. Also, the execution times are changed to 12, 9, 6, and 3. You will be asked to fill table 1 with the numbers like the service times (when is each process serviced), the sequence of executions of these 4 processes in part (a) and part (b) below. Also, you will compute the total wait time. Note for FCFS, it is 1, 2, 3, 4 for P0, P1, P2, and P3; • for the SJN or Shortest Job Next with non-preemptive, PO is #1, P1 may not be #2; for SRT, or shortest remaining time, the preemptive version of SJN, PO is #1, but after 2 time units when P1 comes, P1 kicks out PO since its execution time is 4 < 9- 2 = 7; and P1 may be preempted by P2 later on, etc. You may use a time line to help computing when which process takes over. (a) Fill table T1 and compute total wait times if the queue is SJN or Shortest Job Next (note in case of a tie, the process arriving first is executed first). This is non-preemptive, so PO will be executed to the completion first (b) Repeat part (a) if the queue is SRT, which is the preemptive version of SJN. (so PO is kicked out by P1 at time 2, and P1 is kicked out by P2 at time 4, etc. Please make a new table T2 for this part (b). Note you may do this question manually or by program if you have a computer program, or any other methods you know.
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