Operating SystemsCMPSC 473Processes (contd.)

September 01, 2010 - Lecture 4Instructor: Bhuvan Urgaonkar

Process Switch• Suspend the current process and

resume a previously suspended process– Also called context switch or task switch

Process Switch

Process 0 Process 1

A1A2

B1B2

B3B4

Context_switch() {

}

Involuntary/Voluntary

Process Switch• What does the kernel need to save when

suspending a process?– Hint: The entire address space is already saved

(either in memory or on swap space). What else would the process need when it has to be resumed?

– CPU registers• This is called the hardware context of the

process• Execution context, PC, pointers to elements

within address space, page table, etc.

Context_switch() {

push R0, R1, … // save regs on its stack PCB[curr].SP = SP // save stack pointer PCB[curr].PT = PT // save ptr(s) to address space next = schedule() // find next process to run

PT = PCB[next].PT SP = PCB[next].SP pop Rn, … R0

return // NOTE: Ctrl returns to another process}

Overheads of Process Switch

• Direct– The time spent switching context

• Indirect– Cache pollution– TLB flush

CPU Scheduling

Ready

Waiting

Running

Disk

Lock

OS (scheduler)

Hmm .. Who shouldI pick to run?

Process Scheduling

When is the scheduler invoked?

• CPU scheduling decisions may take place when a process:1. Switches from running to waiting

state2. Switches from running to ready

state3. Switches from waiting to ready4. Terminates

• Scheduling only under 1 and 4: nonpreemptive scheduling– E.g., FCFS and SJF

• All other scheduling is preemptive

First-Come, First-Served Scheduling

(FCFS) Process Run TimeP1 24 P2 3 P3 3

• Suppose that the processes arrive in the order: P1 , P2 , P3 The Gantt Chart for the schedule is:

• Waiting time for P1 = 0; P2 = 24; P3 = 27• Average waiting time: (0 + 24 + 27)/3 = 17

P1 P2 P3

24 27 300

FCFS Scheduling (Cont.)

Suppose that the processes arrive in the order P2 , P3 , P1

• The Gantt chart for the schedule is:

• Waiting time for P1 = 6; P2 = 0; P3 = 3• Average waiting time: (6 + 0 + 3)/3 = 3• Much better than previous case• Convoy effect short process behind long process

P1P3P2

63 300

Choosing the Right Scheduling

Algorithm/Scheduling Criteria• CPU utilization – keep the CPU as busy as

possible• Throughput – # of processes that complete

their execution per time unit• Turnaround time – amount of time to execute

a particular process• Waiting time – amount of time a process has

been waiting in the ready queue• Response time – amount of time it takes from

when a request was submitted until the first response is produced, not output (for time-sharing environment)

• Fairness

Shortest-Job-First (SJF) Scheduling

• Associate with each process the length of its next CPU burst. Use these lengths to schedule the process with the shortest time

• SJF is optimal for avg. waiting time – gives minimum average waiting time for a given set of processes– In class: Compute average waiting time for the

previous example with SJF– Exercise: Prove the optimality claimed above

Why Pre-emption is Necessary

• To improve CPU utilization– Most processes are not ready at all times during their lifetimes– E.g., think of a text editor waiting for input from the keyboard– Also improves I/O utilization

• To improve responsiveness– Many processes would prefer “slow but steady progress” over

“long wait followed by fast process”

• Most modern CPU schedulers are pre-emptive

SJF: Variations on the theme• Non-preemptive: once CPU given to the process it

cannot be preempted until completes its CPU burst - the SJF we already saw

• Preemptive: if a new process arrives with CPU length less

than remaining time of current executing process, preempt. This scheme is know as Shortest-Remaining-Time-First (SRTF) Also called Shortest Remaining Processing Time (SRPT)

• Why SJF/SRTF may not be practical CPU requirement of a process rarely known in advance

Round Robin (RR)• Each process gets a small unit of CPU time (time

quantum), usually 10-100 milliseconds. After this time has elapsed, the process is preempted and added to the end of the ready queue.

• If there are n processes in the ready queue and the time quantum is q, then each process gets 1/n of the CPU time in chunks of at most q time units at once. No process waits more than (n-1)q time units.

• Performance– q large => FCFS– q small => q must be large with respect to

context switch, otherwise overhead is too high

Example of RR with Time Quantum = 20

Process CPU TimeP1 53 P2 17 P3 68 P4 24

• The Gantt chart is:

• Typically, higher average turnaround than SJF, but better response

P1 P2 P3 P4 P1 P3 P4 P1 P3 P3

0 20 37 57 77 97 117 121 134 154 162

Time Quantum and Context Switch Time

Turnaround Time Varies With Time Quantum

Proportional-Share Schedulers• A generalization of round robin

• Process Pi given a CPU weight wi > 0• The scheduler needs to ensure the following

– forall i, j, |Ti(t1, t2)/Tj(t1,t2) - wi/wj| ≤ e– Given Pi and Pj were backlogged during [t1,t2]

• Who chooses the weights and how?• Application modeling problem: non-trivial• Many PS schedulers developed in the 90s

– E.g., Start-time Fair Queueing (Qlinux UT-Austin/Umass-Amherst)

Lottery Scheduling[Carl Waldspurger, MIT,

~1995]• Perhaps the simplest proportional-share scheduler• Create lottery tickets equal to the sum of the weights of all

processes– What if the weights are non-integral?

• Draw a lottery ticket and schedule the process that owns that ticket– What if the process is not ready?

• Draw tickets only for ready processes– Exercise: Calculate the time/space complexity of the operations

Lottery scheduling will involve

Lottery Scheduling Example

1 4

2 5

3 6

7 10

8 11

9 12

13

14

15

9

P1=6 P2=9

Schedule P2

1 4

2 5

3 6

7 10

8 11

9 12

13

14

15

3

P1=6 P2=9

Schedule P1

Lottery Scheduling Example

1 4

2 5 11

6

7 10

8

3 9 12

13

14

15

11• As t ∞, processes will get their share (unless they were blocked a lot)• Problem with Lottery scheduling: Only probabilistic guarantee• What does the scheduler have to do

– When a new process arrives?– When a process terminates?

P1=6 P2=9

Schedule P2

Lottery Scheduling Example