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7th June 2006, 19:22 #1
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operating system numericals
Hi, this question is from Operating Systems By Galvin.
In this question according to solution manual I got the (a) part but for parts (b) & (c) I'm getting answers 6.86 and4.86 respectively and its not in the manual.
So , can someone help solving it.
6.4 Suppose that the following processes arrive for execution at the times indicated. Each
process will run the listed amount of time. In answering the questions, use nonpreemptive
scheduling and base all decisions on the information you have at the time the decision
must be made.
Process Arrival Time Burst Time
P1 0.0 8
P2 0.4 4
P3 1.0 1
a. What is the average turnaround time for these processes with the FCFS scheduling
algorithm?
b. What is the average turnaround time for these processes with the SJF scheduling algorithm?
c. The SJF algorithm is supposed to improve performance, but notice that we chose to
run process P1 at time 0 because we did not know that two shorter processes would
arrive soon. Compute what the average turnaround time will be if the CPU is left
idle for the first 1 unit and then SJF scheduling is used. Remember that processes P1
and P2 are waiting during this idle time, so their waiting time may increase. This
algorithm could be known as futureknowledge scheduling.
Answer:
a. 10.53
b. 9.53
c. 6.86

7th June 2006, 19:22

5th October 2011, 06:47 #2
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Re: operating system numericals
solution for b ) suppose the algorithm SJF í non preemptive in nature then only you ould get that asnwer 9.53.
In that case the first job arrives at 0.0 time and that is the shortest one having 8 and then it is non prrempmtive in nature so till s seconds you can't start another one. Meanwhile at 0.4 time and 1.0 time new job entered. So after 8 seconds shortest job is one whose time 1 sec so it willececute first . After 9 seconds the only job with 4 secinds left so it will exceute. so total time will be
( 8+ ( 9  1) + (13.4) ) /3 = 9.53 í the ănswer
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