L-2.3: First Come First Serve(FCFS) CPU Scheduling Algorithm with Example - YouTube

Hello Friends, welcome to Gate smashers

The topic is FCFS First come first serve scheduling algorithm

Let's understand with the numerical

So in numerical first of all we are given the process number P1, P2, P3, P4

Means 4 processes are here, along with arrival time... Means which process arrived at what time

Burst time... You can call it execution time also

This is time duration.. Means every process told how much time it'll take to get executed

There's no importance of its dimensions here if it's 2 mins, seconds or hours

It can be any value... So our main motive here is how FCFS works

So FCFS's working criteria is arrival time

Means it checks the arrival time, we'll serve the one which came first

And mode is non preemptive... means if a process arrived and I gave it to CPU

Means CPU started to execute that process, it won't stop in between, it'll get completely executed

And after execution it'll get terminated

So let's see how we'll solve this numerical

So first we'll check the arrival time of all the processes I.e. 0... 1... 5... 6

Now here I have to use gantt chart

This chart will help you in solving the numerical in the easiest way

So here time starts from 0

Now at 0, first you have to check which process has arrived

Means which process arrived at 0 from all these

So you can see clearly here, P1 arrived at 0 So there's no any other process left

Other than P1 there's no process which came at 0 So simply just bring it here

Bringing P1 here means we started executing P1 on the CPU

P1 says it'll take 2 unti of time to execute properly

So we'll run it for complete 2 units without any break

Non Preemptive means that we'll not stop the process in between

Now P1 got executed up to 2 and after that P1 left from here

Means you can say that P1 got executed at time 2

Now we are at time 2... means let's say the time is of 2'O clock

Now check again which process has already arrived

Let's say if you'll see P2 then P2 has arrived at 1

1 must be occurred between 0 and 2 Means P2 has already arrived in the ready queue

Is P3 arrived?.. No P3 will arrive at 5... but we are still at 2

Means P3 and P4 haven't arrived

But P2 has arrived and sitting in the ready queue

Now take out P2 from the ready queue and bring it here... Means execute it over the CPU

For how much time?... it said already that it'll take 2 unit of time,

so we'll execute it here for up to 2 unit of time

so here my time became 4 and P2 also got executed from here

Now check my time here, first we've to check the time here i.e. 4'O clock

Now at 4"o clock check which process has arrived

P3 will arrive at 5 and P4 will arrive at 6 And there's no process other than these two

So the important point here is that you can't do anything at 4

So it means, for one unit of time i.e. 4 - 5, CPU is idle

Idle means CPU don't have any process to execute So simply we made CPU Idle from 4 to 5

Now P3 arrived at 5'O clock SO we'll start executing P3

We executed P3 here for up to 3 time unit i.e. up to 8

Now P3 also left from here, now time is 8 And P4 arrived at 6, and 6 has already passed

Means P4 has already arrived in the ready queue So take out P4 from the ready queue and execute it

So here, simple... The process which came first, we have to take it out of the ready queue

So P4's time is 4... so 8 + 4 i.e. 12 So at time 12 all the processes got executed

P1, P2, P3, P4... All four have been executed

So many times they ask what is the time at which they all got executed... so you can write 12

Now we have to write completion time of all the processes

So here check when P1 got executed We have to check its right hand side

At 2... So P1 got completed at 2 P2 at 4... P3 at 8...

you have to check below in the right hand side

P4 got executed at 12... So simply write here, this is the time at which P1... P2... P3... P4 got executed

Now we've to find out TAT... i.e. Turn around time TAT is that for how much time a process stayed in the system

So it means... when it got completed and when it arrived, that's the TWT i.e. TWT = CT - AT

So you can find it easily... 2 - 0 i.e. 2 4 - 1 i.e. 3... 8 - 5 i.e. 3..... 12 - 6 i.e. 6

So this is the time turn around Now it's WT i.e. waiting time

WT is... how much was the useful time and how much time it had spent there

Now TWT means 2... Means P1 stayed for 2 unit of time

and how to know how much time it was supposed to stay.... Through burst time

Means useful time is the burst time but for how time it stayed... that's turnaround time

So TWT - BT... then it'll be waiting time So waiting time became 2 - 2 i.e. 0

3 - 2 = 1....... 3 - 3 = 0 6 - 4 = 2... so this is the waiting time

Means P1 stayed in the system for 2 unit time

And it was supposed to stay for 2 time, then obviously waiting time arrived 0

But if you'll see in case of P2... why 1 arrived?

P2 stayed for 3 time unit, but it was supposed to stay for 2 time only

So if you'd see carefully then P2 arrived at 1 but it git its turn at 2

So that's why it had to wait for its 1 unit time there

So next question they can ask from here is response time

Response time = time at which the process got the CPU first time - arrival time

Now When P1 got the CPU first time?... Now check in gantt chart when P1 got the CPU first

Because you can find out everything easily through gantt chart

So P1 got CPU at 0 ... now we've to check the left side, So P1 got CPU at 0 and it arrived at 0... So 0 - 0 = 0

Next is the P2... P2 got the CPU first time at 2 And P2 arrived at 1, so 2 - 1 i.e. 1

P3... P3 got the CPU first time at 5 And P3 came at 5 first... So 5 - 5 i.e. 0

P4 got the CPU first time at 8... but it arrived at 6, so 8 - 6 i.e. 2

Let me tell you a point here that in case of non preemptive... Because FCFS is non preemptive

so there's no benefit of calculating the response time in this case

,because response time will always come equal to waiting time

This question of response time will come more in case of round robin, in SJF...

Means in the preemptive case more But still we solved it here

No one will ask you more than this about different times

yeah but next they can ask what is the average turnaround time

So you can calculate average turnaround time... i.e.. 14 divide by total no. of processes 4

And how to calculate average waiting time... how would you calculate average waiting time

Just add all of that... 3 divided by no. of process i.e. 4

So it comes out to be 0.75 and it comes out to be 3.5

So this is the average turnaround time and waiting time

So this is all about the FCFS

Thank You!!