Question

QUESTION 1

1. Consider the following set of processes, with the length of the CPU burst

   time given in milliseconds:

   Process            Burst Time      Priority

   P1                    7                      5

   P2                    2                      4

   P3                    11                    3

   P4                    9                      1

   P5                    5                      3

   The processes are assumed to have arrived in the order P1,P2, P3, P4, P5,

   all at time 0.

   a. Draw four Gantt charts that illustrate the execution of these

   processes using the following scheduling algorithms: FCFS, SJF,

   nonpreemptive priority (a smaller priority number implies a higher

   priority), and RR (quantum = 2).

b. What is the turnaround time of each process for each of the

scheduling algorithms in part a?

X	FCFS	SJF	PRI	RR
P1
P2
P3
P4
P5

c. What is the waiting time of each process for each of these scheduling

algorithms?

x	FCFS	SJF	PRI	RR
P1
P2
P3
P4
P5

d. Which of the algorithms results in the minimum average waiting

time (over all processes)?

Answer 1

FCFS: Since process have arrived in order P1, P2, P3, P4, PS, they will be scheduled in same ordee. rocess Waiting Arrival Time Bust Time Conupletion Time Time * 7 0 7 o 0 P3 Pu Ps 0 o o 7 = on 1 q 5 20 29 34 20 29 34 0 0 P P2 P3 P4 Ps | 7 9 20 29 34. o Turn Around - Completion Time - Arrival Time Waiting Time - Turn Around Time - Burst Time = 7+9+20+29 - 13

SJF: Process with minimum assi burst time will be scheduled first onto CPU. Anival Burst Completion Twen Around Waiting tion Ture Time Process Trocess Time Time Time 1 3 it 14 y 0 lé ar an a Po 0 1 34 Puoa 23 Ps o 5 7 - 0 1 O 2 7 14 23 34 :: Average waiting time = 4+23+14+2 = 9.2.

Non-Preemptive Priority Process with higher priority will be executed first till its completion. In case of tie, give priority to process with less burst time. Process 1. Arrival Time Waiting Burst - Time completion Time Turn Around Tive Tinting - 27. a P Pe Pa P4 o o 0 0 7 34 27 25 a 14 34 a7 25 25 14. a 5 a õ a 14 a5 27 34 Average waiting Time = 27+25+14+9=15

given RRCTQ-2): Each process is time sto slice of 2 units to run on CPU. Process Anival Burst Completion Trusu Around waiting Time Time Time Time P. Pa - P3 P4 Ps 0 0 o 0 0 #48% 26 Ko 4 Kart 34 A$AX 33 Stře 25 26 19 4 2. 34 23. 33 24. 25 20. Re: B.YZBB%839.8%B7B2%BB,图 12 | 13 | Ral 6 1 10 e : 12 12 14 16 17 18 ! 0 2 4 6 8 10 12 14 16 18 20 22 24 25 26 28 28 30 32 33 34 ime - 19+2 +23+24+20 :11.6 ed

Among all the scheduling algorithm, Shortest Job First(SJF) has got least average waiting time

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