Tag: First Come First Served

First Come First Serve | CPU Scheduling

Operating System

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

 

Also read- Various Times of Process

 

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

= 73.91%

 

To gain better understanding about FCFS Scheduling,

Watch this Video

 

Next Article- SJF Scheduling | SRTF Scheduling

 

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