Question

1. Processes A, B, C, D, E, and F require service times of 3, 5, 2, 5, 3, and 5. Their arrival times are 0, 1, 3, 9, 10, and 12. What is the average turnaround time, waiting time, response time, and throughput when using SRJF, RR (q=2), and MLF (q=2²ⁱ⁺¹⁾. Show your work.

Answer 1

ANSWER:

Given ,processes A,B,C,D,E,F

we need to find average turn around time,waiting time ,response time and through put.

1) Shortest job first (SJF):It is the scheduling process that select the waiting process with smallest execution time to execute.

Turn around time:It is the interval from time of submission to time of completion i.e finished time -Arrival time.

waiting time: The total time spent by the process in ready waiting for CPU i.e turnaround time-service time.

Response time:The time when a job or process completes i.e process time-Arrival time.

throghput:It is the amount of work completed in a unit of time.

Gantt chart for the above processes are

  

Total process time for A-3

B-10

C-5

D-18

E-13

F-23

Average turn around time is calculated as

3+9+2+9+3+11/6    6.1667   

Average waiting time = 0+4+0+4+0+6/6   2.33

Average response time = (0+4+0+4+0+6)/6 2.33

throught put is calculated as overall efficiency of

CPU   ( 3+5+2+5+3+5)/6   3.866

ii) Round -Robin: Given Quantum =2

  

Response time is calculate as start time-Arrival time

2-0 = 2

3-1 2

5-3 2

9-9=0

11-10=1 ,13-12=1

Gantt chart :

   the average turn around time 3+15+4+13+9+11/6

   9.166

   The average response time =2+2-1 2+0+1.11/6

1.33

Average waiting time 5.33

   Throughput 3.866

iii) MLF (q = : MLF stands for multi level feed back.

Average turn around time = 7+13+2+9+5+10/6

7.66

Average response time =7+14+5+1+15+22/6

10.66

Average waiting time = 4+2+0+4+2+5/6

   2.833

through put =3 .866

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