operating systems and computer networks process scheduling … v7... · operating systems and...

Operating Systems and Computer Networks

Process Scheduling Part 2

Prof. Dr.-Ing. Axel HungerAlexander Maxeiner, M.Sc.

Institute of Computer EngineeringFaculty of Engineering

University Duisburg-Essen

Dr.-Ing. Pascal A. Klein & Alexander Maxeiner M.Sc.

[email protected]

Alexander Maxeiner, M.Sc.University Duisburg-Essen

2Prof. Dr.-Ing. Axel HungerInstitute of Computer Engineering OSCN – Process Scheduling

Scheduler Algorithms



Look back at last lecture:

Computer capable of multitasking need to organize and schedule tasks in advance.

Multitasking increases overall CPU efficiency according to the following formula:

CPU utilization = 1 – pn

Overall CPU utilization is shared among available tasks.

To maintain high efficiency tasks need to be observed and replaced should a memory access occur.

Summary of previous lecture I



Advantages of processing multiple tasks simultaneously:

Completion of multiple tasks takes less time than completion by sequential execution.

Reduced idle times within CPU.

Decrease of pipeline problems at the main bus. While one tasks is executed other tasks can be loaded into the cache.

Disadvantages of processing multiple tasks simultaneously:

Increased cache demands.

Individual tasks take longer until completion.

Summary of previous lecture II



Multiprogramming requires planning a sequence of tasks in advance.

We differentiate between long term and short term planning.

Long term planning involves the merger of several jobs into batches, and general sequence of complex procedures.

Short term planning handles the sequence of instructions at the lower level i.e. before a context switch.

A dispatcher executes the context switches and handles the change of references within the memory and CPU.

Scheduling



During the analysis of scheduling algorithm the turnaround time and the waiting time are most relevant.

Waiting time of a process is defined as:

𝑡𝑤𝑎𝑖𝑡 = time until the execution of a process is started.

Turnaround time of a process is defined as:

𝑡𝑡𝑢𝑟𝑛 = time until the execution of a process is finished.

The time it takes to execute a process will be defined as 𝑡𝑒𝑥 .

Execution times



The sequence of jobs is usually decided on the grounds of predefined algorithms.

On the higher complex operations within an operating system decide which jobs should be prioritized.

On the lower level the algorithms need to be simpler in structure to evade hardware and memory demands.

Examples include:

– First come, first serve

– Shortest job first

– Priority scheduling

– Round robin scheduling

Algorithms



This algorithm sorts the tasks according to its arrival time at the CPU.

In case of equal arrival times (i.e. a batch of processes) the tasks will be executed according to their entry in the process table.

New processes arriving at the CPU before the previous batch is finished will be added to the end of the queue.

Rescheduling may happen after each finished process or after a sequence of processes.

First come, first serve



First process that arrives the scheduler will be the next in line no Interrupts

Example:

• P1: 𝑡𝑒𝑥,1 = 300 ms waiting period 0 ms



Example: First Come, First Serve

Average Waiting Time:(0 + 300 + 340) / 3

= 640 / 3= 213.3 ms

Min Average Waiting Time:P3, P2, P1 (0 + 32 + 72) / 3 = 104 / 3= 34.67 ms

0

P1 (300ms)

P2(40ms)

300 340 372

P3(32ms)

P1 (300ms)

P2(40ms)

0 3727232

P3(32ms)



First process that arrives the scheduler will be the next in line no Interrupts

Example:




Example: First Come, First Serve

0

P1 (300ms)

P2(40ms)

300 340 372

P3(32ms)

P1 (300ms)

P2(40ms)

0 3727232

P3(32ms)

Average Turnaround Time:(300 + 340 + 372) / 3

= 1012 / 3= 337.3 ms

Min Average Turnaround Time:P3, P2, P1 (32 + 72 + 372) / 3 = 476 / 3= 158.67 ms



At fixed intervals all jobs in queue will be analyzed according their estimated runtime.

Then a sequence table is initialized sorting all by increasing runtimes.

New processes arriving at the CPU before the previous batch is finished will be added to the queue at the fixed intervals.

Any process with a shorter runtime than an already sequenced process will have a preferred position within the queue.

More complex jobs take longer if frequent rescheduling with shorter jobs happens.

Complications in Short-term planning, since next block computation time usually not known.

Shortest job first



Example: Shortest Job First I

Assumption: Time required for next block is known, arrival ~simultaneously

P1: 𝑡𝑒𝑥,1 = 8 ms

P2: 𝑡𝑒𝑥,2 = 4 ms

P3: 𝑡𝑒𝑥,3 = 9 ms

P4: 𝑡𝑒𝑥,4 = 5 ms

Average Waiting Time:[0 + 4 + 9 + 17] / 4

= 30 / 4= 7.5 ms

0 5 10 15 20 25 30

Arrival:(8)

(4)

(9)

(5)

P2 (4ms) P1 (8ms)P4 (5ms) P3 (9ms)

P1

P2

P3

P4



Example: Shortest Job First II

Assumption: Time required for next block is known, Interrupt at new arrival

P1: 𝑡𝑒𝑥,1 = 8 ms, arrival time: 0ms




Average Waiting Time:[0 + 0 + (5-3) + (10-1) + (17-2)] / 4

= 26 / 4= 6.5 ms

0 5 10 15 20 25 30

Arrival:

0

1

2

3

8

(8)

5

(4)

11

(9)

8

(5)

P1 (1ms) P2 (4ms) P1 (7ms)P4 (5ms) P3 (9ms)

P1

P2

P3

P4



Example: Shortest Job First III

Assumption: Time required for next block is known, non-preemptive





Average Waiting Time:[0 + (8-1) + (12-3) + (17-2)] / 4

= 26 / 4= 7.75 ms

FCFS: (0 + 7 + 10 + 18) / 4= 8.75 ms

0 5 10 15 20 25 30

Arrival:

0

1

2

3

8

(8)

5

(4)

11

(9)

8

(5)

P1

P2

P3

P4

P1 (8ms) P2 (4ms) P4 (5ms)P3 (9ms)FCFS

P1 (8ms) P2 (4ms) P4 (5ms) P3 (9ms)SJF



Sorts processes according to their priority starting with the highest priority

If another process with higher priority arrives Sending an interrupt to the CPU active task is replaced by the new one

Problem: Low-priority processes will wait a long time if constantly interrupted by higher priority processes

Solution: Increasing priority of waiting task or make the scheduling of a task list non-preemptive.

Scheduler: Priority Scheduling



Example: Priority scheduling

Assumption: Tasks with fixed priorities, interrupt at arrival, Priority 0 ≜ highest

P1: 𝑡𝑒𝑥,1 = 8 ms, 𝑡𝑎𝑟𝑟𝑖𝑣𝑎𝑙,1 = 0ms, Priority: 3




Average Waiting Time:[0 + 0 + 0 + (11-1) + (13-0) + (21-4)] / 4

= 40 / 4= 10 ms

0 5 10 15 20 25 30

Arrival:

0

1

2

3

8

(8)

5

(4)

11

(9)

8

(5)

P1

P2

P3

P4

P1 (1ms)

P2 (1ms)

P4 (5ms)P3 (9ms)Priority Scheduling

P2 (3ms) P1 (7ms)



Round robin scheduling algorithm gives every task a fixed quantum of time before swapping to another process.

Sequence of tasks depends on the sequence in the process table.

New tasks are added to the end of the table.

Context changes happen at the end of each quantum, no matter if the task is already finished.

Scheduler: Round Robin



Switches between tasks after fixed interval of time

Guarantees fair amount of time

Problems: time waste due to change of processes Completion time of long runtime processes (more active tasks, higher completion time)

Example: Round-robin

0 5 10 15 20 25 30

Arrival:

0

1

2

3

8

(8)

5

(4)

11

(9)

8

(5)

P1

P2

P3

P4

P1

P2

P1

P2

P1

P3

P2

P1

P3P4

P2

P1

P3

P4

P1

P3

P4

P1

P3

P4

P1

P3

P4

P3 P3 P3



Time sharing as background job (RR) plus Interrupt on demand based on priorities

Within a priority level, the allocation shall be made according to RR

After an appropriate waiting time processes can move up in priority – Aging

Scheduler: Combination of

several allocation levels



Two Processors: I & II

5 processes have to be handled

Scheduling: Priority Scheduling A running process will be interrupted if another process with higher priority arrives

If one of the CPUs is unoccupied and a higher priority task needs to be scheduled Free Core will be assigned to the task

Priorities and Arrival times for P1..P5:

• Priority 1: P2 (t=0, τ = 7 ms)

• Priority 2: P1 (t=0 , τ = 3 ms), P3 (t=1 , τ = 10 ms)

• Priority 3: P4 (t=5 , τ = 6 ms), P5 (t=6 , τ = 9 ms)

Example



At t=0ms P1, P2 arrive:

• Priority of P1 is higher, both processors are available => Assign P1 -> Processor I

• Assign P2 -> Processor II

Example

Priorities and Arrival times for P1..P5: •Priority 1: P2 (t=0, τ = 7 ms)•Priority 2: P1 (t=0 , τ = 3 ms), P3 (t=1 , τ = 10 ms)•Priority 3: P4 (t=5 , τ = 6 ms), P5 (t=6 , τ = 9 ms)

0 5 10 15 20 25 30

P1

P2

P3

P4

P5

Processor IP1

Processor IIP2




• Priority of P1 is higher, both processors are available => Assign P1 -> Processor I


At t=1ms P3 arrives:

• Priority P3 > P2

• P2 will be interrupted by P3 (on Processor II)

Example


0 5 10 15 20 25 30

P1

P2

P3

P4

P5

Processor IP1

Processor IIP2P3




• Priority of P1 is higher, both Processors are available => Assign P1 -> Processor I





Example


0 5 10 15 20 25 30

P1

P2

P3

P4

P5

Processor IP1

Processor IIP2P3









• P2 continues (on Processor I) after P1 finished

Example


0 5 10 15 20 25 30

P1

P2

P3

P4

P5

Processor IP1P2

Processor IIP2P3









• P2 continues (on Processor I) after P1 finished

Example

0 5 10 15 20 25 30

P4

P5


P1

P2

P3

Processor IP1P2

Processor IIP2P3



At t=5ms P4 arrives

• Priority P4 > P1 > P2


• P2 will be interrupted by P4 (on Processor I)

Example

0 5 10 15 20 25 30

P4

P5


P1

P2

P3

Processor IP1P2P4

Processor IIP2P3



At t=6ms P5 arrives

• P4, P5 > P1

• P3 > P2

• P1 > P2

• P3 will interrupted by P5 (on Processor II)

Example

0 5 10 15 20 25 30

P4

P5


P1

P2

P3

Processor IP1P2P4

Processor IIP2P3P5



At t=6ms P5 arrives

• P4, P5 > P1

• P3 > P2

• P1 > P2


Example

0 5 10 15 20 25 30

P4

P5


P1

P2

P3

Processor IP1P2P4

Processor IIP2P3P5



After P4 finished, P3 will be continued on Processor I

Example

0 5 10 15 20 25 30

P1

P2

P3

P4

P5


Processor IP1P2P4P3

Processor IIP2P3P5




Example

0 5 10 15 20 25 30

P1

P2

P3

P4

P5


Processor IP1P2P4P3

Processor IIP2P3P5




After P5 finished, P2 will be continued on Processor II

Example

0 5 10 15 20 25 30

P1

P2

P3

P4

P5


Processor IP1P2P4P3

Processor IIP2P3P5P2





P3 finished, Processor I unoccupied

Example

0 5 10 15 20 25 30

P1

P2

P3

P4

P5


Processor IP1P2P4P3







Example

0 5 10 15 20 25 30

P1

P2

P3

P4

P5


Processor IP1P2P4P3







P2 finished, Processor II unoccupied

Example

0 5 10 15 20 25 30

P1

P2

P3

P4

P5


Processor IP1P2P4P3




Questions?



Tanenbaum, Andrew S., “Modern Operating Systems”, 3rd edition, Pearson Education Inc, Amsterdam, Netherlands, 2008.

Tanenbaum, Andrew S., “Moderne Bertriebssysteme”, 3rd edition, Pearson Education Inc, Amsterdam, Netherlands, 2009.

Resources

operating systems and computer networks process scheduling … v7... · operating systems and...

Documents