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?

  1. 0%
  2. 10.6%
  3. 30.0%
  4. 89.4%

 

Solution-

 

According to question, we have-

 

Total Burst TimeI/O BurstCPU BurstI/O Burst
Process P110271
Process P2204142
Process P3306213

 

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 TimeBurst Time
CPU BurstI/O BurstCPU Burst
P10322
P20241
P32132
P45221

 

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 IdExit timeTurn Around timeWaiting time
P11111 – 0 = 1111 – (3+2) = 6
P277 – 0 = 77 – (2+1) = 4
P399 – 2 = 77 – (1+2) = 4
P41616 – 5 = 1111 – (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 TimePriority Burst Time
CPU BurstI/O BurstCPU Burst
P102153
P223331
P331231

 

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 IdExit timeTurn Around timeWaiting time
P11010 – 0 = 1010 – (1+3) = 6
P21515 – 2 = 1313 – (3+1) = 9
P399 – 3 = 66 – (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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Summary
CPU Scheduling | Practice Problems | Numericals
Article Name
CPU Scheduling | Practice Problems | Numericals
Description
Practice Problems on CPU Scheduling Algorithms. Various CPU Scheduling Algorithms are- FCFS Scheduling, SJF Scheduling, SRTF Scheduling, Round Robin Scheduling, Priority Scheduling.
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Gate Vidyalay
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