Question

operating systems question:

Below is a set of processes with CPU burst times listed in milliseconds.  Assume all processes arrive at time 0 in the  order shown.

Process            CPU burst

A                         10 ms

B                          3 ms

C                          6 ms

D                          8 ms

E                           5 ms

a) Create a Gantt chart for the Shortest Job First scheduling algorithm.  Assume no preemption.  Label the 
ending times of each process.  Use this format for your Gantt chart:

|  W     |     X     |     Y     |     Z     |
0         4             7            10          17

b) What are the wait times for each process, and the average wait time for SJF? Show your work.

c) Draw a Gantt chart for the Round Robin scheduling algorithm. Use a quantum of 5 ms.

d) What are the wait times for each process, and the average wait time for Round-Robin? Show your work.

Answer 1

Given

Process	CPU Burst
A	10
B	3
C	6
D	8
E	5

a) Gantt chart for Shortest Job First:

B	E	C	D	A
0	3	8	14	22	32

b)

Waiting Time for Process = Turn Around Time – Burst Time

Turn Around Time = Completion Time – Arrival Time

First we calculate Turn Around Time for each process,

Process	Turn Around Time
A	=32-0 = 32
B	=3 - 0 = 3
C	=14 - 0 = 14
D	=22 - 0 = 22
E	=8 - 0 = 8

Now Waiting time for each process,

Process	Waiting Time
A	= 32 - 10= 22
B	= 3 - 3 =0
C	=14 - 6 =8
D	= 22 - 8 =14
E	= 8 - 5 = 3

Average Waiting time = Total waiting time/ Number of process

Average Waiting time = (22+0+8+14+3)/5

Average Waiting time =47/5

Average Waiting time using SJF =9.4 ms

c) Gantt chart using Round Robbin CPU scheduling using time quantum 5ms

A	B	C	D	E	A	C	D
0	5	8	13	18	23	28	29	32

d)

Waiting Time for Process = Turn Around Time – Burst Time

Turn Around Time = Completion Time – Arrival Time

First we calculate Turn Around Time for each process,

Process	Turn Around Time
A	=28-0 = 28
B	=8 - 0 = 8
C	=29 - 0 = 29
D	=32 - 0 = 32
E	=23- 0 = 23

Now Waiting time for each process,

Process	Waiting Time
A	= 28- 10= 18
B	= 8- 3 =5
C	=29 - 6 =23
D	= 32- 8 =24
E	= 23 - 5 = 18

Average Waiting time = Total waiting time/ Number of process

Average Waiting time = (18+5+23+24+18)/5

Average Waiting time =88/5

Average Waiting time using Round Robbin =17.6 ms