Question

Task 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Five batch jobs, A through E, arrive at a computer at essentially at the same time.

They have an estimated running time of 15, 9, 3, 6 and 12 minutes, respectively.

Their externally defined priorities are 6, 3, 7, 9 and 4, respectively, with a lower value

corresponding to a higher priority. For each of the following scheduling algorithms,

determine the average turnaround time (TAT). Hint: First you should determine the

schedule, second you should determine the TAT of each job, and in the last step you

should determine the average TAT. Ignore process switching overhead. In the last 3

cases assume that only one job at a time runs until it finishes and that all jobs are

completely processor bound. Explain how you arrive at your answers.

2.1       Round robin with a time quantum of 1 minute (run in order A to E)                            (1)

2.2      Priority scheduling (1)

2.3      FCFS (run in order A to E) (1)

2.4      Shortest job first (1)

Answer 1

Note: Done accordingly. Please comment for any problem. Please Uprate. Thanks.

Round Robin
Process	Arrival Time	Burst Time	TAT	Waiting time (TAT-BT)
A	0	15	45	30
B	0	9	35	26
C	0	3	13	10
D	0	6	26	20
E	0	12	42	30

AVERAGE TAT =

32.2 min

Time	Queue
0	A
1	B
2	C
3	D
4	E
5	A
6	B
7	C
8	D
9	E
10	A
11	B
12     -->13	C	C OUT
13	D
14	E
15	A
16	B
17	D
18	E
19	A
20	B
21	D
22	E
23	A
24	B
25-->26	D	D OUT
26	E
27	A
28	B
29	E
30	A
31	B
32	E
33	A
34-->35	B	B OUT
35	E
36	A
37	E
38	A
39	E
40	A
41-->42	E	E OUT
42	A
43	A
44-->45	A	A OUT

Priority Scheduling
Process	Priority	Burst Time	TAT	Waiting time (TAT-BT)
A	6	15	36	36
B	3	9	9	9
C	7	3	39	39
D	9	6	45	45
E	4	12	21	21
QUEUE
0 --> 9	9 --> 21	21 --> 36	36-->39	39-->45	AVERAGE TAT =	(36+9+39+45+21)/5	30 min

FCFS
Process	Burst Time	TAT	Waiting time (TAT-BT)
A	15	15	0
B	9	24	15
C	3	27	24
D	6	33	27
E	12	45	33
QUEUE
0 --> 15	15 --> 24	24 --> 27	27-->33	33-->45	AVERAGE TAT =	(15+24+27+33+45)/5	28.8 min
A	B	C	D	E

Process	Burst Time	TAT	Waiting time (TAT-BT)
A	15	45	30
B	9	18	9
C	3	3	0
D	6	9	3
E	12	30	12
QUEUE
0-->3	3-->9	9 -->18	18-->30	30-->45	AVERAGE TAT =	(45+18+3+9+30)/5	21 min
C	D	B	E	A