Scheduling Algorithms | Gate Vidyalay

Tag: Scheduling Algorithms

CPU Scheduling | Practice Problems | Numericals

CPU Scheduling Algorithms-

Various CPU scheduling algorithms are-

PRACTICE PROBLEMS BASED ON CPU SCHEDULING ALGORITHMS-

Problem-01:

Consider three process, all arriving at time zero, with total execution time of 10, 20 and 30 units respectively. Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again. The operating system uses a shortest remaining compute time first scheduling algorithm and schedules a new process either when the running process gets blocked on I/O or when the running process finishes its compute burst. Assume that all I/O operations can be overlapped as much as possible. For what percentage of does the CPU remain idle?

0%
10.6%
30.0%
89.4%

Solution-

According to question, we have-

	Total Burst Time	I/O Burst	CPU Burst	I/O Burst
Process P1	10	2	7	1
Process P2	20	4	14	2
Process P3	30	6	21	3

The scheduling algorithm used is Shortest Remaining Time First.

Gantt Chart-

Percentage of time CPU remains idle

= (5 / 47) x 100

= 10.638%

Thus, Option (B) is correct.

Problem-02:

Consider the set of 4 processes whose arrival time and burst time are given below-

Process No.	Arrival Time	Burst Time
Process No.	Arrival Time	CPU Burst	I/O Burst	CPU Burst
P1	0	3	2	2
P2	0	2	4	1
P3	2	1	3	2
P4	5	2	2	1

If the CPU scheduling policy is Shortest Remaining Time First, calculate the average waiting time and average turn around time.

Solution-

Gantt Chart-

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	11	11 – 0 = 11	11 – (3+2) = 6
P2	7	7 – 0 = 7	7 – (2+1) = 4
P3	9	9 – 2 = 7	7 – (1+2) = 4
P4	16	16 – 5 = 11	11 – (2+1) = 8

Now,

Average Turn Around time = (11 + 7 + 7 + 11) / 4 = 36 / 4 = 9 units
Average waiting time = (6 + 4 + 4 + 8) / 4 = 22 / 5 = 4.4 units

Problem-03:

Consider the set of 4 processes whose arrival time and burst time are given below-

Process No.	Arrival Time	Priority	Burst Time
Process No.	Arrival Time	Priority	CPU Burst	I/O Burst	CPU Burst
P1	0	2	1	5	3
P2	2	3	3	3	1
P3	3	1	2	3	1

If the CPU scheduling policy is Priority Scheduling, calculate the average waiting time and average turn around time. (Lower number means higher priority)

Solution-

The scheduling algorithm used is Priority Scheduling.

Gantt Chart-

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	10	10 – 0 = 10	10 – (1+3) = 6
P2	15	15 – 2 = 13	13 – (3+1) = 9
P3	9	9 – 3 = 6	6 – (2+1) = 3

Now,

Average Turn Around time = (10 + 13 + 6) / 3 = 29 / 3 = 9.67 units
Average waiting time = (6 + 9 + 3) / 3 = 18 / 3 = 6 units

Next Article- Introduction to Process Synchronization

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HRRN Scheduling | CPU Scheduling

Operating System

HRRN Scheduling-

In HRRN Scheduling,

Out of all the available processes, CPU is assigned to the process having highest response ratio.
In case of a tie, it is broken by FCFS Scheduling.
It operates only in non-preemptive mode.

Calculating Response Ratio-

Response Ratio (RR) for any process is calculated by using the formula-

where-

W = Waiting time of the process so far
B = Burst time or Service time of the process

Advantages-

It performs better than SJF Scheduling.
It not only favors the shorter jobs but also limits the waiting time of longer jobs.

Disadvantages-

It can not be implemented practically.
This is because burst time of the processes can not be known in advance.

PRACTICE PROBLEM BASED ON HRRN SCHEDULING-

Problem-

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

Process Id	Arrival time	Burst time
P0	0	3
P1	2	6
P2	4	4
P3	6	5
P4	8	2

If the CPU scheduling policy is Highest Response Ratio Next, calculate the average waiting time and average turn around time.

Solution-

Step-01:

At t = 0, only the process P0 is available in the ready queue.
So, process P0 executes till its completion.

Step-02:

At t = 3, only the process P1 is available in the ready queue.
So, process P1 executes till its completion.

Step-03:

At t = 9, the processes P2, P3 and P4 are available in the ready queue.
The process having the highest response ratio will be executed next.

The response ratio are-

Response Ratio for process P2 = [(9-4) + 4] / 4 = 9 / 4 = 2.25
Response Ratio for process P3 = [(9-6) + 5] / 5 = 8 / 5 = 1.6
Response Ratio for process P4 = [(9-8) + 2] / 2 = 3 / 2 = 1.5

Since process P2 has the highest response ratio, so process P2 executes till completion.

Step-04:

At t = 13, the processes P3 and P4 are available in the ready queue.
The process having the highest response ratio will be executed next.

The response ratio are calculated as-

Response Ratio for process P3 = [(13-6) + 5] / 5 = 12 / 5 = 2.4
Response Ratio for process P4 = [(13-8) + 2] / 2 = 7 / 2 = 3.5

Since process P4 has the highest response ratio, so process P4 executes till completion.

Step-05:

At t = 15, only the process P3 is available in the ready queue.
So, process P3 executes till its completion.

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
P0	3	3 – 0 = 3	3 – 3 = 0
P1	9	9 – 2 = 7	7 – 6 = 1
P2	13	13 – 4 = 9	9 – 4 = 5
P3	20	20 – 6 = 14	14 – 5 = 9
P4	15	15 – 8 = 7	7 – 2 = 5

Now,

Average Turn Around time = (3 + 7 + 9 + 14 + 7) / 5 = 40 / 5 = 8 units
Average waiting time = (0 + 1 + 5 + 9 + 5) / 5 = 20 / 5 = 4 units

To gain better understanding about HRRN Scheduling,

Watch this Video Lecture

Next Article- Round Robin Scheduling

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Predicting Burst Time | SJF Scheduling

Operating System

SJF Scheduling-

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

In SJF Scheduling,

Out of all the available processes, CPU is assigned to the process having smallest burst time.
The main drawback of SJF Scheduling is that it can not be implemented practically.
This is because burst time of the processes can not be known in advance.

In this article, we will discuss techniques to predict burst time.

Techniques to Predict Burst Time-

There are several techniques which try to predict the burst time for the processes so that the algorithm can be implemented.

These techniques are-

Static Techniques-

There are two static techniques-

Based on process size
Based on process type

1. Based on Process Size-

This technique predicts the burst time for a process based on its size.
Burst time of the already executed process of similar size is taken as the burst time for the process to be executed.

Example-

Consider a process of size 200 KB took 20 units of time to complete its execution.
Then, burst time for any future process having size around 200 KB can be taken as 20 units.

NOTE

The predicted burst time may not always be right.
This is because the burst time of a process also depends on what kind of a process it is.

2. Based on Process Type-

This technique predicts the burst time for a process based on its type.
The following figure shows the burst time assumed for several kinds of processes.

Dynamic Techniques-

There are two dynamic techniques-

Based on simple averaging
Based on exponential averaging

1. Based on Simple Averaging-

Burst time for the process to be executed is taken as the average of all the processes that are executed till now.
Given n processes P₁, P₂, … , P_n and burst time of each process P_i as t_i, then predicted burst time for process P_n+1 is given as-

2. Based on Exponential Averaging-

Given n processes P₁, P₂, … , P_n and burst time of each process P_i as t_i. Then, predicted burst time for process P_n+1 is given as-

where-

α is called smoothening factor (0<= α <=1)
t_n = actual burst time of process P_n
T_n = Predicted burst time for process P_n

PRACTICE PROBLEM BASED ON PREDICTING BURST TIME-

Problem-

Calculate the predicted burst time using exponential averaging for the fifth process if the predicted burst time for the first process is 10 units and actual burst time of the first four processes is 4, 8, 6 and 7 units respectively. Given α = 0.5.

Solution-

Given-

Predicted burst time for 1^st process = 10 units
Actual burst time of the first four processes = 4, 8, 6, 7
α = 0.5

Predicted Burst Time for 2nd Process-

Predicted burst time for 2^nd process

= α x Actual burst time of 1^st process + (1-α) x Predicted burst time for 1^st process

= 0.5 x 4 + 0.5 x 10

= 2 + 5

= 7 units

Predicted Burst Time for 3rd Process-

Predicted burst time for 3^rd process

= α x Actual burst time of 2^nd process + (1-α) x Predicted burst time for 2^nd process

= 0.5 x 8 + 0.5 x 7

= 4 + 3.5

= 7.5 units

Predicted Burst Time for 4th Process-

Predicted burst time for 4^th process

= α x Actual burst time of 3^rd process + (1-α) x Predicted burst time for 3^rd process

= 0.5 x 6 + 0.5 x 7.5

= 3 + 3.75

= 6.75 units

Predicted Burst Time for 5th Process-

Predicted burst time for 5^th process

= α x Actual burst time of 4^th process + (1-α) x Predicted burst time for 4^th process

= 0.5 x 7 + 0.5 x 6.75

= 3.5 + 3.375

= 6.875 units

To gain better understanding about predicting burst time,

Watch this Video Lecture

Next Article- LJF Scheduling | LRTF Scheduling

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Watch video lectures by visiting our YouTube channel LearnVidFun.

Longest Job First Algorithm | LRTF Scheduling

Operating System

Longest Job First Algorithm-

In LJF Scheduling,

Out of all the available processes, CPU is assigned to the process having largest burst time.
In case of a tie, it is broken by FCFS Scheduling.

LJF Scheduling can be used in both preemptive and non-preemptive mode.
Preemptive mode of Longest Job First is called as Longest Remaining Time First (LRTF).

Advantages-

No process can complete until the longest job also reaches its completion.
All the processes approximately finishes at the same time.

Disadvantages-

The waiting time is high.
Processes with smaller burst time may starve for CPU.

PRACTICE PROBLEMS BASED ON LJF 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	0	3
P2	1	2
P3	2	4
P4	3	5
P5	4	6

If the CPU scheduling policy is LJF non-preemptive, calculate the average waiting time and average turn around time.

Solution-

Gantt Chart-

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	3	3 – 0 = 3	3 – 3 = 0
P2	20	20 – 1 = 19	19 – 2 = 17
P3	18	18 – 2 = 16	16 – 4 = 12
P4	8	8 – 3 = 5	5 – 5 = 0
P5	14	14 – 4 = 10	10 – 6 = 4

Now,

Average Turn Around time = (3 + 19 + 16 + 5 + 10) / 5 = 53 / 5 = 10.6 unit
Average waiting time = (0 + 17 + 12 + 0 + 4) / 5 = 33 / 5 = 6.6 unit

Problem-02:

Consider the set of 4 processes whose arrival time and burst time are given below-

Process Id	Arrival time	Burst time
P1	1	2
P2	2	4
P3	3	6
P4	4	8

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

Solution-

Gantt Chart-

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	18	18 – 1 = 17	17 – 2 = 15
P2	19	19 – 2 = 17	17 – 4 = 13
P3	20	20 – 3 = 17	17 – 6 = 11
P4	21	21 – 4 = 17	17 – 8 = 9

Now,

Average Turn Around time = (17 + 17 + 17 + 17) / 4 = 68 / 4 = 17 unit
Average waiting time = (15 + 13 + 11 + 9) / 4 = 48 / 4 = 12 unit

Problem-03:

Consider three processes (process id 0, 1, 2 respectively) with compute time bursts 2, 4 and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF, ties are broken by giving priority to the process with the lowest process id. The average turn around time is-