Help... Given the following information: JOb A Arrival 00 CPU Cycle 10 Job B Arr
ID: 3677143 • Letter: H
Question
Help...
Given the following information:
JOb A Arrival 00 CPU Cycle 10
Job B Arrival 02 CPU Cycle 12
JOb C Arrival 03 CPU Cycle 03
Job D Arrival 06 CPU Cycle 01
Job E Arrival 09 CPU Cycle 15
a. Calculate which jobs will have arrived ready for processing by the time
the first job is finished or interrupted for the following algorithms.
FCFS, SJN, SRT ,Round Robin (quantum=5, ignore time required for context switching and natural wait)
b. Calculate the start and finish time for each job for the following processing algorithms.
FCFS, SJN, SRT ,Round Robin (quantum=5, ignore time required for context switching and natural wait)
c. Calculate the turnaround time for every job for each of the following algorithms.
FCFS, SJN, SRT ,Round Robin (quantum=5)
Please show work and I will rate answers.
Example:
Chapter 4 : T=turnaround time F=finish time A=arrival time AVGW=average wait time W=wait time C=CPU cycles LF=last finish time AVGT=average tunr around time FORMULAS: turnaround time = finish time - arrival time average wait time = total wait time/# jobs average turn around time = total turnaround time/# jobs wait = turnaround time - CPU cycles Four Jobs arrive in quick succession. #8 d Job Arrival CPU Finish (calc) Turnaround (calc) Wait ROUND Number Time Cycles Time F-A Time T-C Time ROBIN AVGW A 0 10 B 2 12 C 3 3 AVGT D 6 1 E 9 15 TOTAL Find where Job finished (on graph) ROUND ROBIN #7 dExplanation / Answer
Given information is :
Job A Arrival 00 CPU Cycle 10
Job B Arrival 02 CPU Cycle 12
Job C Arrival 03 CPU Cycle 03
Job D Arrival 06 CPU Cycle 01
Job E Arrival 09 CPU Cycle 15
Answers:
a) Answer :
FCFS
All of the jobs will have arrived : A , B , C , D , E
SJN
All of the jobs will have arrived : A , B , C , D , E
SRT
Jobs B & C will have arrived at time 2 , but job A is not interrupted . Job C arrives at time 3 and interrupts job A
Round Robin
Job A is interrupted at time 5 by which time Jobs B & C have arrived for processing.
b) Answer :
FCFS
Job
A
B
C
D
E
Arrival Time
00
02
03
06
09
Finish Time
10
22
25
26
41
A
B
C
D
E
0 10 22 25 26 41
SJN
Job
A
B
C
D
E
Arrival Time
00
02
03
06
09
Finish Time
10
26
14
11
41
A
D
C
B
E
0 10 11 14 26 41
SRT
SRT is a preemptive and interrupts a job when another job arrives that has a shorter remaining CPU time required to finish. Therefore job A begins , and when job B arrives , the two jobs are compared , by this time require 8 more cycles to finish , while job B requires 12 , so job A is allowed to continue cycle 3 when job c arrives .
Job
A
B
C
D
E
Arrival Time
00
02
03
06
09
Finish Time
14
26
6
7
41
A
C
D
A
B
E
0 3 6 7 14 26 41
ROUND ROBIN
Job
A
B
C
D
E
Arrival Time
00
02
03
06
09
Finish Time
24
36
13
14
41
A1
B1
C
D
E1
A2
B2
E2
B3
E3
0 5 10 13 14 19 24 29 34 36 41
c) Answer :
Turn Around Time = Finish Time – Arrival Time
FCFS
Job
A
B
C
D
E
Turn Around Time
10
20
22
20
32
SJN
Job
A
B
C
D
E
Turn Around Time
10
24
11
5
32
SRT
Job
A
B
C
D
E
Turn Around Time
14
24
3
1
32
ROUND ROBIN
Job
A
B
C
D
E
Turn Around Time
24
34
10
8
32
Job
A
B
C
D
E
Arrival Time
00
02
03
06
09
Finish Time
10
22
25
26
41
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