## Disk Scheduling-

Before you go through this article, make sure that you have gone through the previous article on Magnetic Disk.

 Disk scheduling is a technique used by the operating system to schedule multiple requests for accessing the disk.

## Disk Scheduling Algorithms-

• The algorithms used for disk scheduling are called as disk scheduling algorithms.
• The purpose of disk scheduling algorithms is to reduce the total seek time.

Various disk scheduling algorithms are-

1. FCFS Algorithm
2. SSTF Algorithm
3. SCAN Algorithm
4. C-SCAN Algorithm
5. LOOK Algorithm
6. C-LOOK Algorithm

## FCFS Disk Scheduling Algorithm-

• As the name suggests, this algorithm entertains requests in the order they arrive in the disk queue.
• It is the simplest disk scheduling algorithm.

• It is simple, easy to understand and implement.
• It does not cause starvation to any request.

• It results in increased total seek time.
• It is inefficient.

## Problem-

Consider a disk queue with requests for I/O to blocks on cylinders 98, 183, 41, 122, 14, 124, 65, 67. The FCFS scheduling algorithm is used. The head is initially at cylinder number 53. The cylinders are numbered from 0 to 199. The total head movement (in number of cylinders) incurred while servicing these requests is _______.

## Solution-

Total head movements incurred while servicing these requests

= (98 – 53) + (183 – 98) + (183 – 41) + (122 – 41) + (122 – 14) + (124 – 14) + (124 – 65) + (67 – 65)

= 45 + 85 + 142 + 81 + 108 + 110 + 59 + 2

= 632

To gain better understanding about FCFS Disk Scheduling Algorithm,

Watch this Video Lecture

Next Article- SSTF Disk Scheduling Algorithm

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## 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.

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

• 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.

## 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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