1. There are three methods to handle deadlocks: prevention, avoidance, and detection with recovery.
2. Deadlock prevention ensures that at least one of the necessary conditions for deadlock cannot occur. Deadlock avoidance requires processes to declare maximum resource needs upfront.
3. The Banker's algorithm is a deadlock avoidance technique that dynamically checks the resource allocation state to ensure it remains safe and no circular wait can occur.
Gives an overview about Process, PCB, Process States, Process Operations, Scheduling, Schedulers, Interprocess communication, shared memory and message passing systems
A brief introduction to Process synchronization in Operating Systems with classical examples and solutions using semaphores. A good starting tutorial for beginners.
Goals of Protection
Principles of Protection
Domain of Protection
Access Matrix
Implementation of Access Matrix
Access Control
Revocation of Access Rights
Capability-Based Systems
Language-Based Protection
Gives an overview about Process, PCB, Process States, Process Operations, Scheduling, Schedulers, Interprocess communication, shared memory and message passing systems
A brief introduction to Process synchronization in Operating Systems with classical examples and solutions using semaphores. A good starting tutorial for beginners.
Goals of Protection
Principles of Protection
Domain of Protection
Access Matrix
Implementation of Access Matrix
Access Control
Revocation of Access Rights
Capability-Based Systems
Language-Based Protection
The Deadlock Problem
System Model
Deadlock Characterization
Methods for Handling Deadlocks
Deadlock Prevention
Deadlock Avoidance
Deadlock Detection
Recovery from Deadlock
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1. Unit 4 : Deadlocks
Prepared By: Mr. Sangram A. Patil
Assistant Professor PVPIT,Budhgaon
2. Methods for handling deadlocks
We can deal with deadlock in one of three ways
1. we can use protocol to prevent or avoid deadlock, ensure that system will never
enters into deadlock state .
2. Allow system to enter in deadlock state, detect it and recover.
3. We ignore the problem and behaves like deadlock never occurs.
To ensure deadlock never occurs system can use either a deadlock prevention or
deadlock avoidance scheme.
Deadlock prevention scheme ensure that at least one of necessary condition can
not hold.
Deadlock avoidance requires that the operating system be given in advance
additional information concerning which resources a process will request and use
during its lifetime.
3. Deadlock Prevention
Deadlock to occur each of four necessary conditions must hold.
If one of these condition not hold it prevents deadlock from occur.
1. Mutual Exclusion: ME condition must hold for non-sharable resources such as
printers.
also for sharable resources
2. Hold and wait : to ensure that hold and wait condition never occurs in the system
we must guarantee that whenever process request a resource it does not hold any
other resource.
Example of disk drive, file and printer
4. 3. No preemption: there be no preemption of resources that have already been
allocated . Process is holding some resources and request another resource that
cannot be immediately allocated to it ,then all resources the process is currently
holding are preempted.
4. Circular wait: to ensure this condition never holds is to impose a total ordering of
all resource types and to require that each process request resources in increasing
order of enumeration.
R={R1,R2,….Rm} resource types assign to each unique number, which allow us to
compare two resource and determine whether one precedes another in our
Define one to one function F: R N
F(tape drives)=1
F(disk drive)=5
F(printers)=12
5. Deadlock Avoidance
Require additional information how resources are to be requested.
Simplest and most useful model requires that each process declare the
maximum number of resources of each type that it may need.
The deadlock-avoidance algorithm dynamically examines the resource-
allocation state to ensure that there can never be a circular-wait condition.
Resource-allocation state is defined by the number of available and
allocated resources, and the maximum demands of the processes.
6. Safe state
When a process requests an available resource, system must decide if
immediate allocation leaves the system in a safe state.
System is in safe state if there exists a safe sequence. A sequence of processes
<P1, P2, …,Pn> is safe sequence for current allocation state if, of ALL the processes
in the systems such that for each Pi, the resources requestes that Pi can still make
request can be satisfied by currently available resources + resources held by all the
Pj, with j < i.
That is:
If Pi resource needs are not immediately available, then Pi can wait until all Pj have
finished.
When Pj is finished, Pi can obtain needed resources, execute, return allocated
resources, and terminate.
When Pi terminates, Pi +1 can obtain its needed resources, and so on.
7. 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.
11. Resource-Allocation Graph Scheme
Claim edge Pi Rj indicated that process Pj may request resource Rj;
represented by a dashed line.
Claim edge converts to request edge when a process requests a resource.
Request edge converted to an assignment edge when the resource is allocated
to the process.
When a resource is released by a process, assignment edge reconverts to a claim
edge.
Resources must be claimed a priori in the system
13. Suppose that process Pi requests a resource Rj
The request can be granted only if converting the request edge to an assignment
edge does not result in the formation of a cycle in the resource allocation graph
14. Banker’s Algorithm
Multiple instances.
Each process must claim maximum need in advance.
When a process requests a resource it may have to wait.
When a process gets all its resources it execute successfully and must return them
in a finite amount of time.
15. Data Structures for the Banker’s
Algorithm
Let n = number of processes, and m = number of resources types.
Available : Vector of length m. If available [j] = k, there are
k instances of resource type Rj available.
Max: n x m matrix. If Max [i][j] = k, then process Pi may request at most k
instances of resource type Rj.
Allocation: n x m matrix. If Allocation [i][j] = k then Pi is currently allocated
k instances of Rj.
Need: n x m matrix. If Need [i][j] = k, then Pi may need k more instances of Rj to
complete its task.
Need [i][j] = Max [i][j] – Allocation [i][j]
16. Safety Algorithm
1. Let Work and Finish be vectors of length m and n, respectively. Initialize:
Work = Available
Finish [i] = false for i = 0, 1, …, n- 1.
2. Find and i such that both:
(a) Finish [i] = false
(b) Needi Work
If no such i exists, go to step 4.
3. Work = Work + Allocationi Finish[i] = true
go to step 2.
4. If Finish [i] == true for all i, then the system is in a safe state.
17. Resource-Request Algorithm for Process
Request = request vector for process Pi.
If Requesti [j] = k then process Pi wants k instances of resource type Rj.
When request for resource made by process Pi then following actions are taken
1. If Requesti Needi go to step 2. Otherwise, raise error condition, since process has
exceeded its maximum claim.
2. If Requesti Available, go to step 3. Otherwise Pi must wait, since resources are
not available.
3. Pretend to allocate requested resources to Pi by modifying the state as follows:
Available = Available – Request; Allocationi = Allocationi + Requesti;
Needi = Needi – Requesti;
If safe the resources are allocated to Pi.
If unsafe Pi must wait, and the old resource-allocation state is restored
18. Exampleof Banker’s Algorithm
5 processes P0 through P4; 3 resource types:
A (10 instances), B (5instances), and C (7 instances).
Snapshot at time T0:
Allocation Max Available
A B C A B C A B C
P0 0 1 0 7 5 3 3 3 2
P1 2 0 0 3 2 2
P2 3 0 2 9 0 2
P3 2 1 1 2 2 2
P4 0 0 2 4 3 3
19. Example(Cont.)
The content of the matrix Need is defined to be Max – Allocation.
Need
A B C
P0 7 4 3
P1 1 2 2
P2 6 0 0
P3 0 1 1
P4 4 3 1
The system is in a safe state since the sequence < P1, P3, P4, P2, P0> satisfies
safety criteria.
20. Example: P1 Request (1,0,2)
Check that Request Available (that is, (1,0,2) (3,3,2) true.
Allocation Need Available
A B C A B C A B C
P0 0 1 0 7 4 3 2 3 0
P1 3 0 2 0 2 0
P2 3 0 1 6 0 0
P3 2 1 1 0 1 1
P4 0 0 2 4 3 1
Executing safety algorithm shows that sequence < P1, P3, P4, P0, P2>
satisfies safety requirement.
Can request for (3,3,0) by P4 be granted?
Can request for (0,2,0) by P0 be granted?
24. Single Instanceof Each Resource Type
Maintain wait-for graph
Nodes are processes.
Pi Pj if Pi is waiting for Pj.
Periodically invoke an algorithm that searches for a cycle in the graph. If there is a
cycle, there exists a deadlock.
An algorithm to detect a cycle in a graph requires an order of n2 operations,
where n is the number of vertices in the graph.
26. Several Instances of a each Resource
type
Available : A vector of length m indicates the number of available resources
of each type.
Allocation: An n x m matrix defines the number of resources of each type
currently allocated to each process.
Request: An n x m matrix indicates the current request of each process. If
Request [ij] = k, then process Pi is requesting k more instances of resource type. Rj.
27. Detection Algorithm
1. Let Work and Finish be vectors of length m and n, respectively Initialize:
(a) Work = Available
(b) For i = 1,2, …, n, if Allocationi 0, then
Finish[i] = false;otherwise, Finish[i] = true.
2. Find an index i such that both:
(a) Finish[i] == false
(b) Requesti Work
If no such i exists, go to step 4.
28. Detection Algorithm (Cont.)
3. Work = Work + Allocationi Finish[i] =
true
go to step 2.
4. If Finish[i] == false, for some i, 1 i n, then the system is in deadlock
state. Moreover, if Finish[i] == false, then Pi is deadlocked.
Algorithm requires an order of O( m x n2) operations to detect whether
the system is in deadlocked state.
29. Exampleof Detection Algorithm
Five processes P0 through P4; three resource types A (7 instances), B (2
instances), and C (6 instances).
Snapshot at time T0:
0 0 0
0 1 0
3 1 3
5 2 4
7 2 4
7 2 6
A B C
30. Example(Cont.)
P2 requests an additional instance of type C.
Request
A B C
P0 0 0 0
P1 2 0 1
P2 0 0 1
P3 1 0 0
P4 0 0 2
State of system?
Can reclaim resources held by process P0, but insufficient resources to fulfill other
processes; requests.
Deadlock exists, consisting of processes P1, P2, P3, and P4.
31. Recovery from Deadlock:
Process Termination
Abort all deadlocked processes.
Abort one process at a time until the deadlock cycle is eliminated. In which order
should we choose to abort?
Priority of the process.
How long process has computed, and how much longer to completion.
Resources the process has used.
Resources process needs to complete.
How many processes will need to be terminated.
Is process interactive or batch?
32. Recovery from Deadlock: Resource
Preemption
Selecting a victim – minimize cost.
Rollback – return to some safe state, restart process for that state.
Starvation – same process may always be picked as victim, include number
of rollback in cost factor.