Question

Answer ALL Questions 1. For the processes listed in Table 1, draw a chart illustrating their execution using: a. First Come First Serve (FCFS) b. Shortest Job First Shortest Remaining Time d. Round Robin (Quantum =2) e. Round Robin (Quantum =1) Table 1 Process Arrival Time Burst Time Calculate: a. Average turnaround time b. Average waiting time

Answer 1

Turn around time = Finish time - Arrival time .

Waiting Time= Turn around time - Burst time.

Finish time is determined from the chart .

For the question 1)

part A) FCFS :

In fcfs , the processes are executed in the order of their arrival. If the process enters into execution phase , it completes its execution .

the image attached shows the gantt chart and average turnaround time and average waiting time.

Please Note: the image contains average turnaround time and average waiting time.

PILI detecte Perovery from Suchessor and (a) first come first serve (FCFS) u Turn Around W aiting Process Arrival Baust Time fuiswenn Time Time! J Time Time une Finish - Arrival) (TAT-Bursa- 4 دل که 4 www too bolo to 10 10 15 12 5:20 12:35 TA BT To Since A first comes, A is executed first then B and as c and come together we execute by its sequence given. Average Turn Around Time Sum of Turu Around the (1 35 8.75 No: 8 pears 4 Ang. TAT = 8.75 " Average waiting time sum of waiting Time 90 - | 5a Arg. Wote of procese

Part B) Shortest Job first.

Shortest job first can be considered as a non preemptive job scheduling as there is no context switching .

In this , the job with shortest burst time is served first and completes executes.

Now the image attached shows the chart as well as average turnaround time and average waiting time.

Please Note: the image contains average turnaround time and average waiting time.

Locommunication OS m Communicatie Shortest Job Fir STF) Process | Arrival Time Burst ! Finish Time Time 44 4 715 Turn Around Wouting TimelF1-ATA E-Bicine 4 13 O د Role د 225 3 2:10 Now stock is maintained and arrival M D -2 by 4, all process are in stack c-20 Grantt Chart : TA o 4 6 8 15 А ТствB Now 6C is chosen as it is shortest. Average Turn 95 Around Time sum of Turn Around time No. of process. 6.25: Ang TAI of waiting time No. of process. Average waiting time - Sum - 10 = 12.5 = Aug WT

Part c) Shortest remaining time first .

In this , context switching is allowed in this method .

It is similar to Shortest job first as the one with shortest burst time is served but switching is allowed in between the processes.

Now the image attached shows the chart as well as average turnaround time and average waiting time.

Please Note: the image contains average turnaround time and average waiting time.

(3) Shortest Remaining time in Process / Arrival (Burst / Finish / Turn Around I waiting Time Time Time ume TAT) - AFT-AT = TAT-BT) Time (wt 4 0 13 lo lote JA Stack 3 En 25 =10 Confucte ubitene kan doa rotto 2 30 64 6 8 A LALAC TD BT Aty 21 (1 Average Turn Around time sum of TAT No. of Process. = 25 2 6.25 · AugTATI (2) Average waiting time Sum of waiting time. No. of process 10105Avg. or

Part D) Round Robin scheduling .

In the round robin method , we give each process a slot of time and then the queue is shifted to next process.

By doing this , we can prevent the processes from starvation.

Now the image attached shows the chart as well as average turnaround time and average waiting time.

Please Note: the image contains average turnaround time and average waiting time.

d Round Robin Arrival / Burst Time Time finish Process quantom - 2) Turn Around Waiting Time Time LAT-AT) (TAT-BT) Time Time ho leo blete woor 2 31 2-16 Bolita queue is maintained - (TQ = 2) DITABO DB As the process finishes, Petr - it moves out of queue. B75 A Gantt Chart : A B AC DB Average Turn Around time sum of TAT No.of peocens 2 3 = 7.75 Avg TAT] No. of process Average Waiting Time - Sum of WT = 16 - 4 - Avg. WT

Part E) Round robin

This part is same as the above but with time quantom = 1.

Now the image attached shows the chart as well as average turnaround time and average waiting time.

Please Note: the image contains average turnaround time and average waiting time.

(e) Round Robin (Time quantom Process) Arrival / Burst I finish! Turn Around Time Time Time time CET-AT) waiting I Time ICTAT-BT) A / و 4 7 نه le Roc در دنا {=19 Queue: AA8A4884488 Stacki Tot8X GANTS Chart: Hax FB 760 1 2 3 4 5 6 7 8 9 10 TAYSTATABACO BLACOB 15 enters at 2 A finishes. I finishes (a) Average Turn Around time sum Dfinishes - No of Process = 34 18.5 = Aug TAT (6) Average waiting time i sume 14.453 Aug w Process

Thank you.

I hope i was able to provide you the desired answer.

if you have any queries please be comfortable to ask in the comments i will be glad to help.

:)