Transcript Lecture 1: Overview - City University of New York

Lecture 9: Deadlocks
Advanced Operating System
Fall 2009
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Contents






What is deadlock
Conditions for deadlock
Methods for Handling Deadlocks
Deadlock Prevention
Deadlock Avoidance
Deadlock Detection and Recovery
2
What is deadlock?


A set of blocked processes each holding a resource
and waiting to acquire a resource held by another
process in the set.
Example:
Process P
…
request A
…
request B
…
release A
…
release B
Process Q
…
request B
…
request A
…
release B
…
release A
3
Bridge Crossing Example





Traffic only in one direction.
Each section of a bridge can be viewed as a resource.
If a deadlock occurs, it can be resolved if one car
backs up (preempt resources and rollback).
Several cars may have to be backed up if a deadlock
occurs.
Starvation is possible.
4
Reusable Resources



One that can be safely used by only one process at a
time and is not depleted by that use.
Serially Reusable
Example:







Memory (main & secondary)
Processors
I/O channels
Semaphores
Files
Databases
Printer etc.
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Example 1
P
Request disk drive
Lock disk drive
Request tape drive
Lock tape drive
… perform function
Unlock disk drive
Unlock tape drive
Q
Request tape drive
Lock tape drive
Request disk drive
Lock disk drive
… perform function
Unlock tape drive
Unlock disk drive
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Example 2

Suppose we have a pool of 200 kbytes of
memory for allocation:
P
…
Request 80 Kbytes
…
Request 60 Kbytes
Q
…
Request 70 Kbytes
…
Request 80 Kbytes
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Consumable Resources


One that can be created and destroyed
Example:

Message passed among processes
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Conditions for deadlock




Mutual exclusion: only one process at a time can use a
resource.
Hold and wait: a process holding at least one resource is
waiting to acquire additional resources held by other processes.
No preemption: A resource can be released only voluntarily
by the process holding it, after that process has completed its
task. No resources can be forcibly removed from a process
holding it.
Circular wait: there exists a set {P0, P1, …, P0} of waiting
processes such that P0 is waiting for a resource that is held by
P1, P1 is waiting for a resource that is held by
P2, …, Pn–1 is waiting for a resource that is held by
Pn, and P0 is waiting for a resource that is held by P0.
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Methods for Handling Deadlocks




Deadlock prevention: Design policies so that deadlock
can’t occur.
Deadlock avoidance: The OS will monitor the progress of
processes so that deadlock can always be avoided.
Deadlock detection and recovery: Do nothing and allow
deadlock occur, if deadlock are detected, then recover at
a minimal cost.
Ignore the problem and pretend that deadlocks never
occur in the system; used by most operating systems,
including UNIX.
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Deadlock Prevention

Design policies so that deadlock can’t occur:


Mutual Exclusion – usually cannot be disallowed
Hold and wait




No preemption – preemption is allowed.


require each process to release all resources held before making a new
request.
a process may request resources ahead of time.
Drawback: inefficient utilization of resources.
A good example is the example of the memory pool. If we get into a
deadlock, move the allocated memory to secondary memory
(preemption) and restore later.
Circular Wait – Assign a linear ordering of resource types.


R1<R2<R3…
If a process has been allocated resources of type i (Ri), then it may
request resources of type higher than i. If it wants resources of type
lower than i, then it must release resources of type i.
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Deadlock Avoidance

When a process requests resources, a decision is
made whether the current request will, if granted,
potentially lead to a deadlock. Deadlock avoidance
thus requires knowledge of future processes resource
request.


Do not start a process if its demands might lead to deadlock
Do not grant an incremental resource request to a process if
this allocation might lead to deadlock.
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Deadlock Avoidance(cont.)

Resource-allocation state is defined by the number of available
and allocated resources, and the maximum demands of the
processes.
resource  ( R1, R2 ,..., Rm )
Total amount of each resource in the system
Available  (V1 ,V2 ,...,Vm ) Total amount of each resource not allocated to
a process
 C11 C12

C
C22
claim   21
 ...
...

 Cn1 Cn 2
 A11

A
Allocation   21
 ...

 An1
... C1m 

... C1m  Requirement of each process of each resource
(declared by the processes ahead of time)
... ... 

... Cnm 
A12
A22
...
An 2
A1m 

... A1m 
Current allocation

... ...

... Anm 
...
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Example
4 processes, 3 resources in the system
n=4, m=3
R  (9,3,6)
Total amount of each resource in the system
V  (0,1,1)
Total amount of each resource not allocated to
a process
3

6
claim  
3

4
2
1
1
2
1

6
Allocation  
2

0
2

3
4

2
0
1
1
0
Requirement of each process of each resource
(declared by the processes ahead of time)
0

2
1

2
Current allocation
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Safe State



The state of the system is simply the current
allocation of resources to processes.
A safe state is one in which there is at least
one sequence that does not result in a
deadlock
An unsafe state is a state that is not safe.
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Is this example a safe state?
n=4, m=3
4 processes, 3 resources
R  (9,3,6)
Total amount of each
resource in the system
V  (0,1,1)
Total amount of each
resource not allocated to
a process
3

6

claim 
3

4
2
1
1
2
1

6
Allocation  
2

0
2

3  Requirement of
each process of

4 each resource

2
0
1
1
0
0

2  Current
allocation
1

2
P1 cannot proceed. In the worst case, P1 will request 2
units of R1, 2 units of R2 and 2 units of R3.
Impossible to satisfy
P2 can proceed. In the worst case, P2 will request 1 more
units of R3. Possible to satisfy.
When it completes, it will release all resources.
V  (0,1,1)  (6,1, 2)  (6, 2,3)
Now P1 can proceed.
V  (6, 2,3)  (1, 0, 0)  (7, 2,3)
Now P3 can proceed.
V  (7, 2,3)  (2,1,1)  (9,3, 4)
Now P4 can proceed.
V  (9,3, 4)  (0, 0, 2)  (9,3, 6)
It is a safe sate.
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Basic Facts


If a system is in safe state  no deadlocks.
If a system is in unsafe state  possibility of
deadlock.
Avoidance  ensure that a system will never enter
an unsafe state.


In deadlock avoidance, each time a process requests
resource, the OS will determine if it will, after granting the
resources to the process, lead it to an unsafe state. At the
beginning, the system is in a safe sate, at each step, the
system is always in a safe state.
Avoidance algorithm: Banker’s Algorithm
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Banker’s Algorithm
Checking to see if a state is safe or not
Algorithm:


1.
2.
3.
4.
Find a process Pk such that Claim(k,*)Allocation(k,*)Available(*) for *. If not found,
go to 4.
Available(*) <- Available(*)+Allocation(k,*) for
*.
Goto 1
If there is no process left, the state is safe.
Otherwise, it is not safe.
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Example 1
n=4, m=3
4 processes, 3 resources
R  (9,3,6)
Total amount of each
resource in the system
V  (0,1,1)
Total amount of each
resource not allocated to
a process
3

6
claim  
3

4
2
1
1
2
1

6
Allocation  
2

0
Suppose P1 makes a request of 1 unit of R2. Should the OS
Grants the request? Well, the OS should if it lead it to a safe
State.
V  (0, 0,1)
3

6
claim  
3

4
2

3  Requirement of
each process of
4  each resource

2
0
1
1
0
This is a safe state.
0

2  Current
allocation
1

2
2
1
1
2
1

6
Allocation  
2

0
2

3
4

2
1
1
1
0
0

2
1

2
Is this a safe state?
Yes.
So the OS will grant this request.
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Example 2
n=4, m=3
4 processes, 3 resources
R  (9,3,6)
Total amount of each
resource in the system
V  (0,1,1)
Total amount of each
resource not allocated to
a process
3

6
claim  
3

4
2
1
1
2
1

6
Allocation  
2

0
Suppose P3 asks for one unit of R3. Should the OS grants
the request?
V  (0,1, 0)
3

6
claim  
3

4
2

3  Requirement of
each process of
4  each resource

2
0
1
1
0
This is a safe state.
0

2  Current
allocation
1

2
2
1
1
2
1

6
Allocation  
2

0
2

3
4

2
0
1
1
0
0

2
2

2
Is this a safe state?
No.
So the OS will block this request.
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Deadlock Detection and Recovery


Allow deadlocks to occur. If and when it occurs,
recover from it.
The system checks periodically whether it is in
deadlock. If a deadlock occurs, recovery procedure
will be called.
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Deadlock Detection Algorithm
Detection Algorithm:
Available  (V1 ,V2 ,...,Vm )
 A11

A
Allocation   21
 ...

 An1
A12
A22
...
An 2
 Q11 Q12

Q
Q22
Request   21
 ...
...

 Qn1 Qn 2
A1m 

... A1m 
... ... 

... Anm 
...
... Q1m 

... Q1m 
... ... 

... Qnm 
1. Initialize a temporary vector w to equal to Available
vector. w(*)=V(*)
2. Mark each process that has a row in the Allocation
matrix of all zeros.
3. Find an index k such that process k is currently
unmarked and the kth row of Q  w. That is,
Q(k,*) w(*) for *. If no such row is found,
goto step 5
4. Mark process k and add the corresponding row of
the allocation matrix to w.
That is, w(*)=w(*)+A(k,*). Go to step 3
5. If every process is marked, then there is no deadlock.
Otherwise, the unmarked process are deadlocked.
Are we in a deadlock state?
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Example 1
V  (0,1,1)
1

6
Allocation  
2

0
0

0
Request  
1

3
1
1
0
2
0
1
1
0
1

0
1

3
0

2
1

2
w=(0,1,1)
P1’s request  w
w=(0,1,1)+(1,0,0)=(1,1,1)
P2’s request  w
w=(1,1,1)+(6,1,2)=(7,2,3)
P3’s request  w
w=(7,2,3)+(2,1,1)=(9,3,4)
P4’s request  w
w=(9,3,4)+(0,0,2)=(9,3,6)
Every process is marked. No deadlock.
Is there a deadlock?
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Example 2
V  (0,1,1)
1

6
Allocation  
2

0
4

0
Request  
2

0
1
2
0
1
0
1
1
0
0

2
1

2
w=(0,1,1)
P4’s request  w
w=(0,1,1)+(0,0,2)=(0,1,3)
P1,P2,P3 are involved in a deadlock.
0

1
0

1
Is there a deadlock?
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Recovery

Two methods:
1.
2.
Abort all deadlocked processes.
Find a subset of deadlocked processes
with the minimum cost that can break the
deadlock.
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Example for method 2
V  (0,1,3)
1 0 0


Allocation   6 1 2 
2 1 1


Aborting P2 will resolve the deadlock,
the cost = 9
Aborting P1 will not resolve the deadlock
Aborting P3 will resolve the deadlock,
the cost = 4
 4 1 0


Request   0 2 1 
 2 0 0


P3 is the optimal choice.
Cost of aborting P1 = c1 = 3
Cost of aborting P2 = c2 = 9
Cost of aborting P3 = c3 = 4
The problem is NP-Hard.
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End of lecture 9
Thank you!
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