Question

XCOB 8 pts Question 1 The fill-in-the-blanks of this question are based on the problem data given below where 5 processes arrive in the order P1, P2, P3, p4 and P5, all at the time instant 0. In this problem priority increases with decreasing priority number, so the process with priority number 1 has the highest priority. Each of these blanks requires a single number to be written as answer, rounded off to one decimal place. For example, 5.3287 is 5.3, and 4.2675 is 4.3, after rounding off to one decimal place. Time R Attempt 1 Hour Process Burst time Priority 1 P P2 4 Pa 2 3 Pa 5 2 Ps 10 3 The average waiting time is when SJF scheduling is applied. The average turnaround time is when RR scheduling is applied with time quantum 2. When non-preemptive priority scheduling is applied, the average waiting time is When non-preemptive priority scheduling is applied, the average turnaround time is Question 2 6 pts

Answer 1

Process	Burst Time	Priority
P1	1	1
P2	1	4
P3	2	3
P4	5	2
P5	10	3

Turnaround time is the time a process spends upon its arrival in ready queue till its execution completes. Since the arrival time is not given in the problem, we will assume all process arrive at t = 0.

Waiting time is the time for which a process has to wait. Waiting time is given as: Turnaround Time - Burst time

A) In SJF (Shortest Job First) we select the process which has the least Burst Time and give it CPU execution until the process finishes. If two jobs have the same burst time, then we pick the process which arrived earlier.

The Gantt Chart for the above process will be as follows :

2 4. 0 1 P1 P2 P3 P3 P4 P5

Process	Arrival Time	Burst Time(BT)	Completion Time (CT)	Turnaround Time (TT = CT - AT)	Waiting Time( TT - BT)
P1	0	1	1	1	0
P2	0	1	2	2	1
P3	0	2	4	4	2
P4	0	5	9	9	4
P5	0	10	19	19	9
Total				35	16

Average Waiting Time = Total Waiting Time/5 = 3.2

B) In RR(Round Robin) scheduling we give each process a max n quanta CPU time. If the process finishes within this time or the time quanta expires, the control is transferred to next process in ready queue. If the process has remaining execution time left then the process is added to the tail of the ready queue.

The Gantt Chart is as follows:

q = 2 (r): remaining time 0 1 2 P1(0) P210) 10 P4(1) 12 13 P5(6) P4(0) P3(0) P4(3) P5(8) 19 P5(0)

Process

Process	Arrival Time	Burst Time(BT)	Completion Time (CT)	Turnaround Time (TT = CT - AT)	Waiting Time( TT - BT)
P1	0	1	1	1	0
P2	0	1	2	2	1
P3	0	2	4	4	2
P4	0	5	13	13	8
P5	0	10	19	19	9
Total				39	20

Average Turnaround Time = Total Turnaround Time/5 = 7.8

C) In Non Preemptive Priority Scheduling the process with highest priority is given CPU time until it completes it execution. If two processes have same priority, then process is chosen according to arrival time.

The Gantt chart is as follows:

1 18 19 0 P1 P4 P3 P5 PZ

Process	Arrival Time	Burst Time(BT)	Priority	Completion Time (CT)	Turnaround Time (TT = CT - AT)	Waiting Time( TT - BT)
P1	0	1	1	1	1	0
P2	0	1	4	19	19	18
P3	0	2	3	8	8	6
P4	0	5	2	6	6	1
P5	0	10	3	18	18	8
Total					52	33

Average Waiting Time = Total Waiting time / 5 = 6.6

D) Refer the above table

Average Turnaround Time = Total Turnaround Time/5 = 10.4