FCFS Scheduling

In FCFS Scheduling,

• The process which arrives first in the ready queue is firstly assigned the CPU.
• In case of a tie, process with smaller process id is executed first.
• It is always non-preemptive in nature. 

Advantages- 

• It is simple and easy to understand.
• It can be easily implemented using queue data structure.
• It does not lead to starvation.

 Disadvantages- 

• It does not consider the priority or burst time of the processes.
• It suffers from convoy effect. 

Convoy Effect

In convoy effect, Consider processes with higher burst time arrived before the processes with smaller burst time. Then, smaller processes have to wait for a long time for longer processes to release the CPU.

 PRACTICE PROBLEMS BASED ON FCFS SCHEDULING-

 Problem-01:

 Consider the set of 5 processes whose arrival time and burst time are given below-

 

Process Id	Arrival time	Burst time
P1	3	4
P2	5	3
P3	0	2
P4	5	1
P5	4	3

 

If the CPU scheduling policy is FCFS, calculate the average waiting time and average turn around time. 

Solution-

 

Gantt Chart-

 

 

Here, black box represents the idle time of CPU.

Now, we know-

• Turn Around time = Exit time – Arrival time
• Waiting time = Turn Around time – Burst time 

Process Id	Exit time	Turn Around time	Waiting time
P1	7	7 – 3 = 4	4 – 4 = 0
P2	13	13 – 5 = 8	8 – 3 = 5
P3	2	2 – 0 = 2	2 – 2 = 0
P4	14	14 – 5 = 9	9 – 1 = 8
P5	10	10 – 4 = 6	6 – 3 = 3

 

Now,

• Average Turn Around time = (4 + 8 + 2 + 9 + 6) / 5 = 29 / 5 = 5.8 unit
• Average waiting time = (0 + 5 + 0 + 8 + 3) / 5 = 16 / 5 = 3.2 unit

 

Problem-02:

 

Consider the set of 3 processes whose arrival time and burst time are given below-

 

Process Id	Arrival time	Burst time
P1	0	2
P2	3	1
P3	5	6

 

If the CPU scheduling policy is FCFS, calculate the average waiting time and average turn around time. 

Solution-

 

Gantt Chart-

 

 

Here, black box represents the idle time of CPU.

 

Now, we know-

• Turn Around time = Exit time – Arrival time
• Waiting time = Turn Around time – Burst time

 

Process Id	Exit time	Turn Around time	Waiting time
P1	2	2 – 0 = 2	2 – 2 = 0
P2	4	4 – 3 = 1	1 – 1 = 0
P3	11	11- 5 = 6	6 – 6 = 0

 

Now,

• Average Turn Around time = (2 + 1 + 6) / 3 = 9 / 3 = 3 unit
• Average waiting time = (0 + 0 + 0) / 3 = 0 / 3 = 0 unit

 

Problem-03: 

Consider the set of 6 processes whose arrival time and burst time are given below-

 

Process Id	Arrival time	Burst time
P1	0	3
P2	1	2
P3	2	1
P4	3	4
P5	4	5
P6	5	2

 

If the CPU scheduling policy is FCFS and there is 1 unit of overhead in scheduling the processes, find the efficiency of the algorithm. 

Solution- 

Gantt Chart- 

 Here, δ denotes the context switching overhead. 

Now,

• Useless time / Wasted time = 6 x δ = 6 x 1 = 6 unit
• Total time = 23 unit
• Useful time = 23 unit – 6 unit = 17 unit 

Efficiency (η)

= Useful time / Total Total

= 17 unit / 23 unit

= 0.7391