Priority Scheduling

 In Priority Scheduling,

• Out of all the available processes, CPU is assigned to the process having the highest priority.
• In case of a tie, it is broken by FCFS Scheduling. 

 

• Priority Scheduling can be used in both preemptive and non-preemptive mode.

 

Advantages- 

• It considers the priority of the processes and allows the important processes to run first.
• Priority scheduling in preemptive mode is best suited for real time operating system.

 

Disadvantages- 

• Processes with lesser priority may starve for CPU.
• There is no idea of response time and waiting time.

 

Important Notes-

 

Note-01:

 

• The waiting time for the process having the highest priority will always be zero in preemptive mode.
• The waiting time for the process having the highest priority may not be zero in non-preemptive mode.

 

Note-02:

 

Priority scheduling in preemptive and non-preemptive mode behaves exactly same under following conditions-

• The arrival time of all the processes is same
• All the processes become available

 

PRACTICE PROBLEMS BASED ON PRIORITY SCHEDULING-

 

Problem-01:

 

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

 

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

 

If the CPU scheduling policy is priority non-preemptive, calculate the average waiting time and average turn around time. (Higher number represents higher priority)

 

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	4	4 – 0 = 4	4 – 4 = 0
P2	15	15 – 1 = 14	14 – 3 = 11
P3	12	12 – 2 = 10	10 – 1 = 9
P4	9	9 – 3 = 6	6 – 5 = 1
P5	11	11 – 4 = 7	7 – 2 = 5

 

Now,

• Average Turn Around time = (4 + 14 + 10 + 6 + 7) / 5 = 41 / 5 = 8.2 unit
• Average waiting time = (0 + 11 + 9 + 1 + 5) / 5 = 26 / 5 = 5.2 unit

 

Problem-02:

 

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

 

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

 

If the CPU scheduling policy is priority preemptive, calculate the average waiting time and average turn around time. (Higher number represents higher priority) 

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	15	15 – 0 = 15	15 – 4 = 11
P2	12	12 – 1 = 11	11 – 3 = 8
P3	3	3 – 2 = 1	1 – 1 = 0
P4	8	8 – 3 = 5	5 – 5 = 0
P5	10	10 – 4 = 6	6 – 2 = 4

 

Now,

• Average Turn Around time = (15 + 11 + 1 + 5 + 6) / 5 = 38 / 5 = 7.6 unit
• Average waiting time = (11 + 8 + 0 + 0 + 4) / 5 = 23 / 5 = 4.6 unit