10/09/2007ecs150 Fall 20071 Operating System ecs150 Fall 2007 : Operating System #2: Scheduling and...

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ecs150 Fall 2007:Operating SystemOperating System#2: Scheduling and Mutual Exclusion(chapter 4)

Dr. S. Felix Wu

Computer Science Department

University of California, Davishttp://www.cs.ucdavis.edu/~wu/

[email protected]

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Kernel and User SpaceKernel and User Space

Process FOOFOOMemoryspace for thisprocess

System call(or trap into the kernel)

program

System Call

conceptually

Kernel Resources(disk or IO devices)

Process FOOFOOin the Kernel

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States of a ProcessStates of a Process

Running, Blocked, and Ready

Running

Waiting Ready

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Running

Blocked Ready

Running

Blocked Ready

Running

Blocked Ready

Running

Blocked Ready

Scheduling &Context Switching

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Basic ConceptsBasic Concepts

Maximum CPU utilization obtained with multiprogramming

CPU–I/O Burst Cycle – Process execution consists of a cycle of CPU execution and I/O wait.

CPU burst distribution Processes classified as CPU bound or I/O

bound

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CPU SchedulerCPU Scheduler Selects from among the processes in memory that

are ready to execute, and allocates the CPU to one of them.

CPU scheduling decisions may take place when a process:

1.Switches from running to waiting state.

2.Switches from running to ready state.

3.Switches from waiting to ready.

4.Terminates. Scheduling under 1 and 4 is nonpreemptive. All other scheduling is preemptive.

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Preemptive vs.Preemptive vs.NonpreemptiveNonpreemptive

Preemptive:

Nonpreemptive:

Pros and Cons…..

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User P Kernel Process Kernel P Kernel Process

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Context SwitchingContext Switching

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DispatcherDispatcher

Dispatcher module gives control of the CPU to the process selected by the short-term scheduler; this involves:– switching context– switching to user mode– jumping to the proper location in the user

program to restart that program Dispatch latency – time it takes for the dispatcher

to stop one process and start another running.

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Preemptive SchedulingPreemptive Scheduling

fixed time window timer/clock interrupt– scheduling decision– “bill” the process– context switching might or might not happen

Priority Fairness …

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Example: two schedulesExample: two schedules

100 msec slotwith around 5 mseccontext switchingoverhead….

Which one is better?And Why?

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Optimization CriteriaOptimization Criteria

Max CPU utilization Max throughput Min turnaround time Min waiting time Min response time

– but obviously, we have some trade-off…

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First-Come, First-Served (FCFS)First-Come, First-Served (FCFS)

Process Burst Time

P1 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

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Example 1Example 1

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Shortest-Job-First (SJF)Shortest-Job-First (SJF) Associate with each process the length of its next CPU burst.

Use these lengths to schedule the process with the shortest time. Two schemes:

– nonpreemptive – once CPU given to the process it cannot be preempted until completes its CPU burst.

– preemptive – if a new process arrives with CPU burst length less than remaining time of current executing process, preempt. This scheme is know as the Shortest-Remaining-Time-First (SRTF).

SJF is optimal – gives minimum average waiting time for a given set of processes.

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Process Arrival Time Burst Time

P1 0.0 7

P2 2.0 4

P3 4.0 1

P4 5.0 4

SJF (non-preemptive)

Average waiting time = (0 + 6 + 3 + 7)/4 - 4

ExampleExample

P1 P3 P2

73 160

P4

8 12

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Issues for SJFIssues for SJF

???

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CPU BurstsCPU Bursts

CPU–I/O Burst Cycle – Process execution consists of a cycle of CPU execution and I/O wait.

CPU burst distribution Processes classified as CPU bound or

I/O bound

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Determining Length of Next CPU Determining Length of Next CPU BurstBurst

Can only estimate the length. Can be done by using the length of previous CPU

bursts, using exponential averaging.

:Define 4.

10 , 3.

burst CPU next the for value predicted 2.

burst CPU of lenght actual 1.

≤≤=

=

+

αατ 1n

thn nt

( ) .1 1 nnn t ταατ −+=+

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Priority SchedulingPriority Scheduling A priority number (integer) is associated with each

process The CPU is allocated to the process with the highest

priority (smallest integer highest priority).

– Preemptive

– Non-preemptive

SJF is a priority scheduling where priority is the predicted next CPU burst time.

FCFS is a priority scheduling where priority is the arrival time.

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““Fixed” PriorityFixed” Priority

What is it?– The process sticks with the origin assigned priority.

A good or bad idea? What other possible policy?

– Dynamic policy. Problem Starvation – low priority processes may never

execute. Solution Aging – as time progresses increase the priority of

the process.

– Hybrid policy.

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Round Robin (RR)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 FIFO

– q small q must be large with respect to context switch, otherwise overhead is too high.

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Process Burst Time

P1 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

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I/O system calls

1

2

3

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Past CPU UsagePast CPU Usage

Round-Robin How many CPU cycles so far

– fairness How about #3?

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Past CPU UsagePast CPU Usage

Round-Robin How many CPU cycles so far

– fairness How many CPU cycles you have used

lately– aging factor: 4 cycles

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BSD SchedulingBSD Scheduling

For each tick (10 msec):– Pi

estcpu = Piestcpu + Tick(Pi)

For each decaying period (1 second):– Pi

estcpu = Piestcpu*(2*load)/(2*load+1)+Pi

nice load: # of processes in the ready queue

Example: 1 process in the queue– Pi

estcpu = Piestcpu*0.66 + Pi

nice

– in 4 seconds, (0.66)4 = 0.001296

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LinuxLinux

Linux provides two process scheduling algorithms– Time sharing for fairness

– Real-time, where priorities are more important than fairness

Linux allows only user-mode processes to be preempted– kernel mode processes may not be interrupted

Formula – a balance between software real-time and fairness

– Picredit = Pi

credit / 2 + Priorityi

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SchedulingScheduling

Fairness: Round-Robin Responsiveness: SJF Utilization: Aging, promoting System Calls. Fairness in Inter-process communication

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BSD SchedulingBSD Scheduling

For each tick (10 msec):– Pi

estcpu = Piestcpu + Tick(Pi)

For each decaying period (1 second):– Pi

estcpu = Piestcpu*(2*load)/(2*load+1)+Pi

nice load: # of processes in the ready queue

Example: 1 process in the queue– Pi

estcpu = Piestcpu*0.66 + Pi

nice

– in 4 seconds, (0.66)4 = 0.001296

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Load = 1Load = 1

2Load/(2Load+1) ~ 0.66 Pi

estcpu = Piestcpu*(0.66)1 – 20

Piestcpu = Pi

estcpu*(0.66)2 – 20

Piestcpu = Pi

estcpu*(0.66)3 – 20

Piestcpu = Pi

estcpu*(0.66)4 – 20

– (0.66)4 = 0.001296

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Load = 50Load = 50

2Load/(2Load+1) ~ 0.99 Pi

estcpu = Piestcpu*(0.99)1 – 20

Piestcpu = Pi

estcpu*(0.99)2 – 20

Piestcpu = Pi

estcpu*(0.99)3 – 20

Piestcpu = Pi

estcpu*(0.99)4 – 20

– (0.99)4 = 0.9606

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1

0

0

1

0

1

::.

256 different priorities64 scheduling classes

RR

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1

0

0

1

0

1

::. 256 different priorities

64 scheduling classes

RR

0~63 bottom-half kernel (interrupt)64~127 top-half kernel128~159 real-time user160~223 timeshare224~255 idle

kg_estcpu

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static voidschedcpu(void *arg){ ... FOREACH_PROC_IN_SYSTEM(p) { FOREACH_KSEGRP_IN_PROC(p, kg) { awake = 0; FOREACH_KSE_IN_GROUP(kg, ke) { ... } /* end of kse loop */

kg->kg_estcpu = decay_cpu(loadfac, kg->kg_estcpu); resetpriority(kg); FOREACH_THREAD_IN_GROUP(kg, td) { if (td->td_priority >= PUSER) { sched_prio(td, kg->kg_user_pri); } } } /* end of ksegrp loop */ } /* end of process loop */}

/usr/src/sys/kern/sched_4bsd.c

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static voidresetpriority(struct ksegrp *kg){ register unsigned int newpriority; struct thread *td;

if (kg->kg_pri_class == PRI_TIMESHARE) { newpriority = PUSER + kg->kg_estcpu / INVERSE_ESTCPU_WEIGHT + NICE_WEIGHT * (kg->kg_nice - PRIO_MIN); newpriority = min(max(newpriority, PRI_MIN_TIMESHARE), PRI_MAX_TIMESHARE); kg->kg_user_pri = newpriority; }

FOREACH_THREAD_IN_GROUP(kg, td) { maybe_resched(td); /* XXXKSE silly */ }}


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struct kse *sched_choose(void){

struct kse *ke;

ke = runq_choose(&runq);

if (ke != NULL) {runq_remove(&runq, ke);ke->ke_state = KES_THREAD;

KASSERT((ke->ke_thread != NULL), ("runq_choose: No thread on KSE"));KASSERT((ke->ke_thread->td_kse != NULL), ("runq_choose: No KSE on thread"));KASSERT(ke->ke_proc->p_sflag & PS_INMEM, ("runq_choose: process swapped out"));

}return (ke);

}


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#define loadfactor(loadav) (2 * (loadav))#define decay_cpu(loadfac, cpu) (((loadfac) * (cpu)) / ((loadfac) + FSCALE))

static fixpt_t ccpu = 0.95122942450071400909 * FSCALE; #define CCPU_SHIFT 11


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/* * Compute a tenex style load average of a quantity on * 1, 5 and 15 minute intervals. * XXXKSE Needs complete rewrite when correct info is * available. * Completely Bogus.. * only works with 1:1 (but compiles ok now :-) */static voidloadav(void *arg){

}

/usr/src/sys/kern/kern_synch.c

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Lottery SchedulingLottery Scheduling(a dollar and a dream)(a dollar and a dream)

A process P(I) has L(I) lottery tickets. Totally, we have N tickets among all processes. P(I) has L(I)/N probability to get CPU cycles. Difference between Priority and Tickets:

– Priority: which process is more important? The importance is “infinitely” big!!

– Tickets: how much more important? Quantify the importance.

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J, Pri(J) = 1 I, Pri(I) = 120

J,80% I J I

I J J J I J J J J J J J I J J J J I J J

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Lottery SchedulingLottery Scheduling

At each scheduling decision point:– Ticket assignment:

For each ticket in each process, produce a unique random number (we can not have more than one process sharing the prize).

– Ticket drawing: Produce another random number until a winner is

identified.

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Efficient ImplementationEfficient Implementation

Maybe only the # of tickets matters…

One rand( ) could produce 32 random bits.

0-16

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ExampleExample422

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static voidschedcpu(void *arg){ ... FOREACH_PROC_IN_SYSTEM(p) { mtx_lock_spin(&sched_lock); FOREACH_KSEGRP_IN_PROC(p, kg) { awake = 0; FOREACH_KSE_IN_GROUP(kg, ke) { ... ke->ke_sched->ske_cpticks = 0; } /* end of kse loop */ /*

* If there are ANY running threads in this KSEGRP, * then don't count it as sleeping. */


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if (awake) { ... kg->kg_slptime = 0; } else { kg->kg_slptime++; } if (kg->kg_slptime > 1) continue; kg->kg_estcpu = decay_cpu(loadfac, kg->kg_estcpu); resetpriority(kg); FOREACH_THREAD_IN_GROUP(kg, td) { if (td->td_priority >= PUSER) { sched_prio(td, kg->kg_user_pri); } } } /* end of ksegrp loop */


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voidsched_clock(struct kse *ke){

struct ksegrp *kg;struct thread *td;

mtx_assert(&sched_lock, MA_OWNED);kg = ke->ke_ksegrp;td = ke->ke_thread;

ke->ke_sched->ske_cpticks++;kg->kg_estcpu = ESTCPULIM(kg->kg_estcpu + 1);if ((kg->kg_estcpu % INVERSE_ESTCPU_WEIGHT) == 0) {

resetpriority(kg);if (td->td_priority >= PUSER)

td->td_priority = kg->kg_user_pri;}

}


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avg = &averunnable;sx_slock(&allproc_lock);nrun = 0;FOREACH_PROC_IN_SYSTEM(p) {

FOREACH_THREAD_IN_PROC(p, td) {switch (td->td_state) {case TDS_RUNQ:case TDS_RUNNING:

if ((p->p_flag & P_NOLOAD) != 0)goto nextproc;

nrun++; /* XXXKSE */default:

break;}

nextproc: continue;}

}sx_sunlock(&allproc_lock);for (i = 0; i < 3; i++)

avg->ldavg[i] = (cexp[i] * avg->ldavg[i] + nrun * FSCALE * (FSCALE - cexp[i])) >> FSHIFT;

/* * Schedule the next update to occur after 5 seconds, but add a * random variation to avoid synchronisation with processes that * run at regular intervals. */callout_reset(&loadav_callout, hz * 4 + (int)(random() % (hz * 2 + 1)), loadav, NULL);

/usr/src/sys/kern/kern_synch.c

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static voidschedcpu(void *arg){ ... FOREACH_PROC_IN_SYSTEM(p) { mtx_lock_spin(&sched_lock); FOREACH_KSEGRP_IN_PROC(p, kg) { awake = 0; FOREACH_KSE_IN_GROUP(kg, ke) { ... ke->ke_sched->ske_cpticks = 0; } /* end of kse loop */ /*

* If there are ANY running threads in this KSEGRP, * then don't count it as sleeping. */


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Mutual ExclusionMutual Exclusion Consistent access to the information. Critical sections. Race conditions.

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Critical SectionsCritical Sections No two processes may be simultaneously inside

their critical regions/sections– safety condition

No assumptions may be made about speeds or the number of CPUs

No process running outside of its critical region may block other processes

No process should have to wait forever to enter its critical region/section– liveness condition

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Race ConditionRace Condition

4

2

2

1

Withdraw $2 from X:

Read balance X; (1)If X < 2 then exit;Write new X = X - 2;(4)Release $2;

Withdraw $3 from X:

Read balance X;(2)If X < 3 then exit;Write new X = X - 3;(3)Release $3;

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How to Prevent?How to Prevent?

Disable Interrupts Busy Waiting and TSL Strict Alternation Protected Critical Sections

Hardware versus software solution….

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Let’s try some software solutions!!Let’s try some software solutions!!

Assuming we have only two processes. The solutions should be able to extend to

cover more general cases.

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Solution #1Solution #1Process 0 Process 1…. …..while (turn != 0); while (turn != 1);<critical section>; <critical section>;turn = 1; turn = 0;…. …...

What is the problem??

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ProblemProblemProcess 0 Process 1…. …..while (turn != 0);<critical section>;turn = 1;…. …...while (turn != 0);…...

while (turn != 1);<critical section>turn = 0;

…... <critical section>;turn = 1;

Maybe a very long period of time…

Blocked…

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Solution #2Solution #2

Process 0 Process 1…. …..while (flag[1] = = true); while (flag[0] = = true);flag[0] = true; flag[1] = true;<critical section>; <critical section>;flag[0] = false; flag[1] = false;…. …...


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ProblemProblem

Process 0 Process 1flag[0] = false; flag[1] = false;…. …..while (flag[1] = = true);

while (flag[0] = = true);flag[0] = true;

flag[1] = true;<critical section>;

<critical section>;flag[1] = false;

flag[0] = false;

…. …...

NOTSAFE!!

00011011

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Two Properties for METwo Properties for ME

Safe: no two may enter Live: eventually one will win

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Process 0 Process 1…. …..flag[0] = true; flag[1] = true; while (flag[1] = = true); while (flag[0] = = true);<critical section>; <critical section>;flag[0] = false; flag[1] = false;…. …...


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ProblemProblem

Process 0 Process 1…. …..flag[0] = true;

flag[1] = true; while (flag[1] = = true);

while (flag[0] = = true);

…. …...

Nobody can progress!! (Liveness)

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Process 0 Process 1…. …..flag[0] = true; flag[1] = true; while (flag[1] = = true) while (flag[0] = = true){ { flag[0] = false; flag[1] = false; <delay a while> <delay a while> flag[0] = true; flag[1] = true;} }<critical section>; <critical section>;flag[0] = false; flag[1] = false;…. …...

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ideaideaProcess 0 Process 1….

…..flag[0] = true; flag[1] = true; while (flag[1] = = true && f2 = = true) while (flag[0] = = true && f2 == false){ { flag[0] = false; /*mnbn */ flag[1] = false; /*mnbn*/ flag[2] = ~flag[2]; flag[2] = ~flag[2]; flag[0] = true; /*mnbn */ flag[1] = true; /*mnbn */} }<critical section>; <critical section>;flag[0] = false; flag[1] = false;…. …...

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ProblemsProblems

Delay for how long? A different problem: Starvation with a small

but non-zero probability.

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Dekker’s AlgorithmDekker’s AlgorithmProcess 0 Process 1…. …..flag[0] = true; flag[1] = true; while (flag[1] = = true) while (flag[0] = = true){ { if (turn = = 1) if (turn = = 0) { {

flag[0] = false; flag[1] = false;while (turn = = 1); while (turn = = 0);flag[0] = true; flag[1] = true;

} }} }<critical section>; <critical section>;turn = 1; turn = 0;flag[0] = false; flag[1] = false;…. …...

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Dekker’sDekker’s

Three shared variables to solve the problem:– flag[0], flag[1], and turn

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Peterson’s AlgorithmPeterson’s Algorithm

Process 0 Process 1…. …..flag[0] = true; flag[1] = true; turn = 0; turn = 1;while (flag[1] && (turn = = 1)); while (flag[0] && (turn = = 0));<critical section>; <critical section>;flag[0] = false; flag[1] = false;…. …...

Comparing Dekker’s with Peterson’s!!

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flag[1] = true; turn = 0;while (flag[1] && (turn = = 1));

turn = 1;while (flag[0] && (turn = = 0));

<critical section>; <critical section>;flag[0] = false; flag[1] = false;…. …...

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WWW stands for WWW stands for What Went WrongWhat Went Wrong??


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WWW stands for WWW stands for What Went WrongWhat Went Wrong??


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Previous Exam Question?Previous Exam Question?

Process 0 Process 1…. …..turn = 0; turn = 1;flag[0] = true; flag[1] = true; while (flag[1] && (turn = = 0)); while (flag[0] && (turn = = 1));<critical section>; <critical section>;flag[0] = false; flag[1] = false;…. …...

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turn = 1;<critical section>;

while (flag[0] && (turn = = 1));flag[0] = false;

<critical section>;flag[1] = false;

…. …...

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Process 0 Process 1…. …..flag[0] = true;turn = 0;while (flag[1] && (turn = = 0));<critical section>;


flag[0] = false;<critical section>;flag[1] = false;

…. …...

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Process 0 Process 1…. …..flag[0] = true;turn = 0;

flag[1] = true; turn = 1;

while (flag[1] && (turn = = 0));while (flag[0] && (turn = = 1));<critical section>;flag[1] = false;

<critical section>;flag[0] = false;…. …...

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Safety and Liveliness propertiesSafety and Liveliness properties

Process 0 Process 1…. …..flag[0] = true; flag[1] = true; turn = 0; turn = 1;while (flag[1] && (turn = = 0)); while (flag[0] && (turn = = 1));

At the moment when both are in the CS:flag[0] = 1; flag[1] = 1; turn = 0 or 1;


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Process 0 Process 1…. …..flag[0] = true; flag[1] = true; turn = 0;

turn = 1;while (flag[1] && (turn = = 0)); while (flag[0] && (turn = = 1));

flag[0] = 1; flag[1] = 1; turn = 0 or 1;case 1: 0 1 process 1 would not enter!


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Process 0 Process 1…. …..flag[0] = true; flag[1] = true; turn = 0;

turn = 1;while (flag[1] && (turn = = 0)); while (flag[0] && (turn = = 1));

flag[0] = 1; flag[1] = 1; turn = 0 or 1;case 1: 0 1 process 1 would not enter!case 2: 1 0 process 0 would not enter!


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Monitor

Multiple Threaded Kernel

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TSLTSLtsl register, lock

Do the following jobs atomically (they can not be broken apart):

copies lock to register sets lock to 1

BEFORE: Lock = 1AFTER: Reg = 1 Lock = 1


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Critical Section with TSLCritical Section with TSL

Enter??

Leave??

tsl register, lock



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Critical Section with TSLCritical Section with TSL

Enter??

loop: tsl register, lock cmp register, #0 jne loop

Leave??

move lock, #0

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CMPXCHGCMPXCHG

Compare and Exchange

CMPXCHG reg, op1, op2

if (reg == op1)

op1 op2;

else

reg op1;

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SemaphoresSemaphores

A “non-busy-waiting” approach. General and Binary Semaphore.

Waiting Queuefor resources

I can only holdN (n=4) units ofresources

not reallynot reallya queue!!!a queue!!!

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Binary SemaphoreBinary SemaphorewaitB(s):if (s.value == 1)

s.value = 0;else

place the process in s.queue;block the process;

signalB(s):if (s.queue is empty)

s.value = 1;else

remove P from s.queue;place P in the ready queue;

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General SemaphoreGeneral SemaphorewaitG(s):s.count --;if (s.count < 0)

place P in s.queue;block this process;

signalG(s):s.count ++;if (s.count <= 0)

remove P from s.queue;place P in the ready queue;

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Binary SemaphoreBinary Semaphore

1. How to do Mutual Exclusion with Semaphores?2. How to implement Semaphores in Software?

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Mutual ExclusionMutual Exclusion

mutex_lock

mutex_unlock

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ME using SemaphoreME using Semaphore….waitB(semA)<enter critical section>signalB(semA)…..

Waiting Queuefor resources

I can only holdN (n=1) units ofresources

not reallynot reallya queue!!!a queue!!!

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MutexesMutexes

A Mutex (Mutual Exclusion) is a data element that allows multiple threads to synchronize their access to shared resources

Like a binary semaphore, a mutex has two states, locked and unlocked

Only one thread can lock a mutex Once a mutex is locked, other threads will block

when they try to lock the same mutex, until the locking mutex unlocks the mutex, at which point one of the waiting thread’s lock will succeed, and the process begins again

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Lock and TryLockLock and TryLock

What is the difference?

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Mutex vs. CondMutex vs. Cond

mutex_lock

mutex_unlock

cond_wait

cond_signal

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Pthread ImplementationPthread Implementation

in pthread_mutex_lock (mutex.c):SET_KERNEL_FLAG;semaphore implementation…CLEAR_KERNEL_FLAG;

in sighandler (signal.c):if (!is_in_kernel)

pthread_schedule_wrapper();…else

sigaddset(…);

10/09/2007 ecs150 Fall 2007 98


10/09/2007 ecs150 Fall 2007 99

HW#2HW#2

2 test programs each team LT kernel

– Please tell us the amount of tickets to set

Contest/test suties– 2xN + M– Is the lottery scehduling working clearly? (e.g., very,

moderate, not really)

– You are encouraged to do cross-team testing before your submission.

10/09/2007ecs150 Fall 20071 Operating System ecs150 Fall 2007 : Operating System #2: Scheduling and...

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