Scheduling Deadlock

8/12/2019 Scheduling Deadlock

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C 5: P S


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C 5: P

S

B

T CS P P S S H

S C P S

M S E A A

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O T ,

T

T

T

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B P

M ,

C

I :S . W

counter . I, counter 0. I .

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Synchronization, Deadlock andScheduling


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Pwhile (true) {

/* produce an item in next produced */

while (counter == BUFFER_SIZE) ;

/* do nothing */

buffer in = next roduced_

in = (in + 1) % BUFFER_SIZE;

counter++;

}

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Cwhile (true) {

while (counter == 0); /* do nothing */

next_consumed = buffer[out];

= _

counter--;

/* consume the item in next consumed */

}

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R C counter++

register1 = counterregister1 = register1 + 1counter = register1

counter--

register2 = counter -

counter = register2 C = 5:

S0: register1 = counter 1 = 5S1: register1 = register1 + 1 1 = 6S2: register2 = counter 2 = 5S3: register2 = register2 1 2 = 4S4: counter = register1 = 6 S5: counter = register2 = 4

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C S P C , ,

E P , , , W ,

E , ,

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C S

G

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S CS

P1. I ,

2. I ,

3. A A

N

T , , CPU

E

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P

S G

T

A load store ; ,

T : int turn; Boolean flag[2]

T turn

T flag . flag[i] = true Pi !

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A P Pdo {

flag[i] = true;turn = j;

while (flag[j] && turn == j);

flag[i] = false;remainder section

} while (true);

P 1. M

2. P

3. B

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S H M

A P

U

C G O

M

= E O

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do {

acquire lock

critical section

release lock

S C P U L

remainder section} while (TRUE);

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I

D:

*_ _

{boolean rv = *target;

*target = TRUE;

return rv:

}

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S ()

S , FALSE

S:

do {

while (test_and_set(&lock))

; /* do nothing *//* critical section */

lock = false;

/* remainder section */

} while (true);

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OS Concepts 053006 -



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I

D:

, , _

int temp = *value;if (*value == expected)

*value = new_value;

return temp;

}

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S

S B FALSE;

E B S:

do {while (compare_and_swap(&lock, 0, 1) != 0)

; /* do nothing */

/* critical section */

lock = 0;


} while (true);

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B M E

do {

waiting[i] = true;

key = true;while (waiting[i] && key)

key = test_and_set(&lock);

waiting[i] = false;

/* critical section */

j = (i + 1) % n;

while ((j != i) && !waiting[j])

j = (j + 1) % n;

if (j == i)

lock = false;

else

waiting[j] = false;


} while (true);

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M L

P

OS

S P acquire() release()

B

U

B

T

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() ()

acquire() {

while (!available)

; /* busy wait */available = false;;

}

release() {

available = true;

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}

do {

acquire lock

critical section

release lock

remainder section

} while (true);


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S

S

S

T : wait() signal() O P() V()

L

C ()

wa t S

while (S


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S U

0 1

T

C

C

C

P1:S1;

signal(synch);

P2:

wait(synch);

S2;

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S I

M wait() signal()

T,

B L

N

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S I

B

W

E : ( )

T :

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S I

B (C.)

typedef struct{

int value;

struct process *list;} semaphore;

wait(semaphore *S) {

S->value--;

if (S->value < 0) { -

block();}

}

signal(semaphore *S) {

S->value++;

if (S->value list;

wakeup(P);

}

}

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D S

L S Q 10 1

wait(S); wait(Q);

wait(Q); wait(S);. .

signal(S); signal(Q);

signal(Q); signal(S);

A

S S

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0 1

wait(S); wait(Q);wait(Q); wait(S);

signal(S); signal(Q);signal(Q); signal(S);

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C P

S

C

R W P

DP P

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BB P

,

Smutex 1

S full 0

S empty

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B B P (C.)

T

do {...

/* produce an item in next_produced */

...

wait(mutex);...

/* add next produced to the buffer */

...

signal(mutex);

signal(full);

} while (true);

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B B P (C.)

T

do {

wait(full);

wait(mutex);

...

_

...signal(mutex);

signal(empty);

...

/* consume the item in next consumed */

...

} while (true);

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RW P

A R ;

W

P O

S

S D

D S rw_mutex 1 Smutex 1 I read_count 0

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RW P (C.)

T

do {wait(rw_mutex);

...* *

...

signal(rw_mutex);

} while (true);

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RW P (C.)

T

do {

wait(mutex);read count++;if (read_count == 1)

wait(rw_mutex);

signal(mutex);

...

...

wait(mutex);read count--;if (read_count == 0)

signal(rw_mutex);

signal(mutex);

} while (true);

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R W P


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RW P

V

,

B

P

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DP P

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DP P

P

D , 2 ( )

N , I 5

S

B ( ) S 5 1

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D P P


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DP P

A T P:

do {wait (chopstick[i] );wait (chopStick[ (i + 1) % 5] );

// eat

signal (chopstick[i] );signal (chopstick[ (i + 1) % 5] );

// think

} while (TRUE);

W ?

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P S

I :

() . ()

() ()

O () () ( )

D

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M

A

,

O B

monitor monitor-name

// shared variable declarationsprocedure P1 () { . }

procedure Pn () {}

Initialization code () { }}

}

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S M

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C V

condition x, y;

T :x.wait()

x.signal()

x.signal() ( ) x.wait()

I x.wait() ,

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M C


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M C

V

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C V C

I P x.signal(), Q x.wait() , ?

I Q , P

O

Q P

B

M C P P , Q

I M, C#, J

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S D P

monitor DiningPhilosophers{enum { THINKING; HUNGRY, EATING) state [5] ;

condition self [5];

void pickup (int i) {state[i] = HUNGRY;test(i);if (state[i] != EATING) self [i].wait;

}

void putdown (int i) {state[i] = THINKING;

// test left and right neighborstest((i + 4) % 5);

test((i + 1) % 5);}

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S D P


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S D P

(C.)

void test (int i) {

if ( (state[(i + 4) % 5] != EATING) &&(state[i] == HUNGRY) &&(state[(i + 1) % 5] != EATING) ) {

state[i] = EATING ;se .s gna ;

}}

initialization_code() {for (int i = 0; i < 5; i++)

state[i] = THINKING;}

}

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S l i Di i Phil h (C )


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E pickup() putdown() :

DiningPhilosophers.pickup(i);

EAT

Solution to Dining Philosophers (Cont.)

DiningPhilosophers.putdown(i);

N ,

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M I U S

V

semaphore mutex; // (initially = 1)semaphore next; // (initially = 0)

int next_count = 0; E

wait(mutex);

body of F;

if (next_count > 0)signal(next)

elsesignal(mutex);

M

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M I C V

F , :

semaphore x_sem; // (initially = 0)

int x_count = 0; T . :

x_count++;if (next_count > 0)

signal(next);else

signal(mutex);wait(x_sem);

x_count--;

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M I (C )


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M I (C.)

T x.signal :

if (x_count > 0) {

next_count++;

signal(x_sem);

wait(next);

next_count--;

}

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R P


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M I ,

.() , ?

.()W

P ( )

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A M A S


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Rmonitor ResourceAllocator

{

boolean busy;condition x;

void acquire(int time) {

if (busy)

x.wait(time);

busy = TRUE;

}void release() {

busy = FALSE;

x.signal();

}

initialization code() {

busy = FALSE;}

}

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A T


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A T

S M L R

C C A T

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S M


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S M

A , ,

R C

H T T ( ) ( )

A

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T S M


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T S M

V

E: ,

N I

E:

S I

N , RAID

Goal is to assure transaction atomicity where failures cause loss of information on volatile storage

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L B R


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LB R

R

M L ,

, T

D O

N

T

T L

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L B R A


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LB R A

U ,

() T() T

U(T) (T) M

I ,

I , () I , ()

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C T


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C T

M

C

I,

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S


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S

C A B

C T T0 T1 E T0, T1

E

A

F N , N!

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S 1: T T


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S 1: T0 T1

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N S


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N S

R

C S, O, O ,

I O, O

& O O T S OO S

I S S S

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S 2: C S S


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S 2: C S S

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L P


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L P

E F

L T (S) Q, T Q

Q

T (X) Q, T

Q R Q I ,

S

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T L P


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T L P

G

E

G

S

D

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T P


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T P

S

T T TS(T) T

T T < T T T TTS

T

I TS(T) < TS(T), T T

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T P I


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D Q W(Q)

(Q) R(Q) (Q)U (Q) (Q)

S T ()I TS(T) < W(Q), T Q

T I TS(T) W(Q) , R(Q) (R(Q), TS(T))

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T P


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S T (Q)I TS(T) < R(Q), Q T

T , T

I TS(T) < W(Q), T Q

T O,

A T

A

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S P U T P


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E C 5


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Synchronization, Deadlock and

Scheduling


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C 6: CPU S

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Scheduling

72

C 6: CPU S


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B C

S C S A

MP S

RT CPU S

O S E A E

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Scheduling

O


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T CPU ,

T CPU

T CPU

T

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Scheduling

B C


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M CPU

CPUI/O B C P

I/O

/

CPU

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Scheduling

H CPU T


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Scheduling

CPU S


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, CPU

Q CPU :

1. S

.

3. S

4. T

S 1 4

A C

C

C OS

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Scheduling

D


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D CPU ;

:

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Scheduling

S C


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CPU

#

, ( )

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Scheduling

S A O C


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M CPU

M M

M M

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Scheduling

FC, FS (FCFS) S


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P B T1 24

2 33 3 S : 1, 2, 3

W 1 = 0; 2 = 24; 3 = 27 A : (0 + 24 + 27)/3 = 17

P1 P2 P3

24 27 300

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Scheduling

FCFS S (C.)


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S :

2, 3, 1 T G :

P1P3P2

W 1 6 2= 0 3 3 A : (6 + 0 + 3)/3 = 3 M

C CPU I/O

63 300

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Scheduling

SJF (SJF)


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S A

CPU U

SJF T CPU

C

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Scheduling

E SJF


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PA T B T

1 0.0 6

2 2.0 83 4.0 7

4 5.0 3

SJF

A = (3 + 16 + 9 + 0) / 4 = 7

P4 P3P1

3 160 9

P2

24

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Scheduling

D L N CPU


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B C

T CPU

C CPU ,

burstCPUnexttheforvaluepredicted2.

burstCPUoflengthactual1.

=

=

+ 1n

thn nt

C, P

:Define4.

10,3.

( ) .11 nnn t +==

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Scheduling

P L

N CPU B


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Scheduling

E E


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A =0

+1=

R =1

+1= O CPU

I , :+1= +(1)1 +

+(1 )+ +(1 )+1 0

S (1 ) 1,

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Scheduling

E S


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N

PA TT B T

1 0 82 1 4

3 2 9

4

SJF G C

A = (101)+(11)+(172)+53)/4 = 26/4 = 6.5

P1 P1P2

1 170 10

P3

265

P4

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Scheduling

P S


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A ()

T CPU ( ) P

SJF CPU

P

S

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Scheduling

E P S


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PA B TT P

1 10 3

2 1 13 2 4

4 1 5

5

P G C

A = 8.2

P2 P3P5

1 180 16

P4

196

P1

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Scheduling

R R (RR)


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E CPU ( ), 10100 . A , .

I , 1/ CPU . N 1 .

T

P FIFO ,

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Scheduling

E RR T

Q 4


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Q = 4P B T

1 24

2 3

3 3

T, SJF, 10 100, < 10

P1 P2 P3 P1 P1 P1 P1 P1

0 4 7 10 14 18 22 26 30

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Scheduling

T Q C S T


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Scheduling

T T V W

T T Q


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80% of CPU bursts shouldbe shorter than q

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Scheduling

M Q


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R , : ()

()

P

E : RR

FCFS S :

F ; (.., ). P .

T CPU ; .., 80% RR 20% FCFS

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Scheduling

M Q S


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Scheduling

M F Q


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A ;

M

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Scheduling

E M

F Q


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F Q T :

0 RR 8 1 RR 16

2 FCFS

S

A 0 FCFS W CPU, 8

I 8 ,

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OS Concepts 053006 -Synchronization, Deadlock and

Scheduling

T S


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D

W , ,

M , LWP

K ()

T

K CPU

()

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Scheduling

MP S


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CPU CPU

,

() , , C,

V

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Scheduling

NUMA CPU S


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Note that memory-placement algorithms can alsoconsider affinity

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Scheduling

MP S L B


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I SMP, CPU

, CPU CPU

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Scheduling

M P


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R

F

M

T

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Scheduling

M M

S


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S

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Scheduling

L F


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=

=

=

,

=

V

F , 7 ,

14 , = 2

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Scheduling

S


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Q

P

G D

R

D T

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Scheduling

E CPU S

S


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Scheduling

E C 6


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Scheduling


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C 7: D

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Scheduling

109

C 7: D


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M H D

D P D A

D D

R D

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Scheduling

C O


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T ,

T

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Scheduling

S M


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S .

R 1, 2, . . ., , , /

E

.

E :

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Scheduling

D C

Deadlock can arise if fo r conditions hold sim ltaneo sl


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Deadlock can arise if four conditions hold simultaneously.

:

:

: ,

: 0, 1, , 0 1, 1 2, , 1 , 0.

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Scheduling

RA G

A set of vertices Vand a set of edges E.


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V :=

1

, 2

, ,

,

= 1, 2, , ,

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RA G

(C.)

P


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P

R T 4

Pi

Pi

Rj

Rj

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E R A G


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Q?


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03 P 04 R.

R1 01 , R2 02 , R3 01 , R4 03 .

P1 01 R2 01 R1

P2 01 R1, 01 R2 01 R3

P3 01 R3 01 R2

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117

R A G W A D


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G W A C B N

D


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B F

I


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I

I

,

,

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M H

D

E


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E

A

I ;

, UNIX

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D P

;Restrain the ways request can be made


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;

,

R ,

L ;

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D P (C.)


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I ,

P

P ,

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Scheduling

D A

S

Requires that the system has some additional a priori informationavailable


124/153

S

T

R ,

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Scheduling

S S

W ,


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W ,

.

S ALL P,

+ , <

T : I P ,

W , , , ,

W , +1 ,

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B F

I


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I

I

A

.

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S, U, D S


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A

S


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S

U

M

U

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B A

M


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M

E W

W

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D S B A

Let n= number of processes, and m = number of resources types.


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V . I = ,

. I , = ,

. I A, = . I , =,

,= , ,

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S A

1. L , .I:


131/153

I: = = = 0, 1, , 1

2. F : =

() I , 4

3. = + = 2

4. I == ,

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RR A P

= . I =


132/153

1. I 2. O,

, 2. I , 3. O ,

3. P

: = =+ ;=

I

I

,

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E B A

5 0 4;3


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3 :

(10 ), (5), (7 )

S 0:

0 0 1 0 7 5 3 3 3 21 2 0 0 3 2 2

2 3 0 2 9 0 2

3 2 1 1 2 2 2

4 0 0 2 4 3 3

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E (C.)

T


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0 7 4 3

1 1 2 2

2 6 0 03 0 1 1

4 4 3 1

T < 1, 3, 4, 2, 0>

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E: 1R (1,0,2)

C R A ( , (1,0,2) (3,3,2)


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0 0 1 0 7 4 3 2 3 01 3 0 2 0 2 0

2

3 2 1 1 0 1 1

4 0 0 2 4 3 1 E < 1, 3, 4, 0, 2>

C (3,3,0) 4 ?

C (0,2,0) 0 ?

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RA G

S


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;

C

R

W ,

R

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RA G


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U S I RA G


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RA G

A

S


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T

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D D

A


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D

R

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S I E

R T

M


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N

P . I ,

A 2 ,

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RA G W G


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Resource-Allocation Graph Corresponding wait-for graph

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S I

R T

A


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A

A

. I = ,

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D A

1. L , I:


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() =

() F = 1,2, ,, 0, ; , =

2. F :() ==

()

I , 4

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D A (C.)

3.= + =


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=

2

4.I , , 1 , . M, ==,

Algorithm requires an order of O(m x n2

) operations to detect whether the system is in deadlocked state

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E D A

F 0 4; A (7 ), (2 ), (6 )


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S 0:

0 0 1 0 0 0 0 0 0 0

1 2 0 0 2 0 22 3 0 3 0 0 0

3 2 1 1 1 0 0

4 0 0 2 0 0 2

S

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E (C.)

2


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0 0 0 0

1 2 0 2

2 0 0 1

3

1 0 0

4 0 0 2

S ? C 0,

;

D , 1, 2, 3, 4

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DA U

W, , :

H ?


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H ?

H ?

I ,

.

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R D:

P T

A

A


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A

I ?.

2. H ,

3. R 4. R

5. H

6. I ?

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R D:

R P


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,

,

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E

A system with 05 process.a.Prove that system is Safe?b.P1 Request (0, 4, 3, 0) ?


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b.P1 Request (0, 4, 3, 0) ?

c.P1 Request (0, 4, 1, 0) ?

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151

E


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All process came at 0.0.

Calculate average roundabout time?Calculate average waiting time?

For each method of scheduling.

E C 7


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Documents

Scheduling Deadlock