Os module 2 d

Objectives
 To develop a description of deadlocks,
which prevent sets of concurrent
processes from completing their tasks
 To present a number of different
methods for preventing or avoiding
deadlocks in a computer system.

Deadlock
 Defined as the permanent blocking of a
set of processes that compete for a
system resource, including database
records and communication lines

 A situation wherein each of a collection of
processes is waiting for something from
other processes in the collection. Since all
are waiting, none can provide any of the
things being waited for.

Deadlock
 A situation wherein a process is
requesting a resource which is withheld
by another process and never get hold
of the resource.

Example
 System has 2 tape drives
 P1 and P2 each hold one tape drive and
each needs another one.

 Semaphores A and B initialized to 1
 P0 P1
 Wait(A); Wait(B);
 Wait(B); Wait(A);

Deadlock
 Deadlock occurs when a set of
processes are in a wait state, because
each process is waiting for a resource
that is held by some other waiting
process. Therefore, all deadlocks
involve conflicting resource that are
needed by two or more processes.

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.

Deadlock
 Unlike some other problems in
multiprogramming systems, there is no
efficient solution to the deadlock
problem in the general case. It is one
area where there is a strong theory but it
is completely ignored in practice
because solutions to deadlock are
expensive.

Resource
 A hardware device (e.g. tape drive,
memory)

 A piece of information (e.g. a locked
record in a database)

General Categories of Resources
 Reusable – something that can be
safely used by one process at a time
and is not depleted by that use.
Processes obtain resources that they
later release for reuse by others.
 Example:
 CPU, Memory, specific I/O devices or files.

General Categories of Resources
 Consumable – these can be created and
destroyed. When a resource is acquired
by a process, the resource ceases to
exist.
 Examples:
 Interrupts, signals, messages.

Types of Resources
 Preemptable resource – a resource that
can be taken away from the process
owning it with no ill effect (needs
save/restore)

 Example: Memory or CPU

Types of Resources
 Non-preemptable – a resource that
cannot be taken away from its current
owner without causing the computation
to fail.

 Example: Printer or floppy disk.

Note
 Deadlocks occur when sharing reusable
and non-preemptable resources

Pop Exercise

 Classify the following as either sharable
or non-sharable
 A. Hard disk
 B. file
 C. keyboard

System Model
 System consists of a finite number of
resources and a finite number of
processes competing for resources.
 Resources are partitioned into several
types, each of which consists of some
number of identical instances

System Model
 When a process requests an instance of
a resource type, the allocation of any
instance of the type will satisfy the
request (otherwise, resource types have
not been defined properly).

Protocol for resource utilization
 Request – the process will request for a
certain resource. If the request cannot
be granted immediately, then the
requesting process must wait until it can
acquire the resource.
 Use – the process can operate the
resource
 Release – the process relinquishes the
resources.

Note
 A set of processes is deadlocked when
every process in the set is waiting for an
event that can be caused only by
another process in the set.

Deadlock Characterization
Deadlock can arise if four conditions hold simultaneously:
 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.
 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.

Additional characteristic
 PROCESS IRREVERSIBLE – Unable to
reset to an earlier state where resources
are not held.

Systems avoiding deadlocks
 Systems involving only shared
resources cannot deadlock. – negates
mutual exclusion
 System that aborts processes which is
requesting a resource but currently
being used. – negates hold and wait
 Preemption may be possible if a process
does not use its resources until it has
acquired all it needs – negates no
preemption

Systems…
 Transaction processing systems provide
checkpoints so that processes may back
out of a transaction. – negates
irreversible process
 System that detects or avoids deadlocks
– prevents cycle.

Resource-Allocation Graph (RAG)
 Consists of a set of vertices V and a set of edges
E. Devised by Holt in 1972.

 V is partitioned into two types:
 P = {P1, P2, …, Pn}, the set consisting of all the processes
in the system.

 R = {R1, R2, …, Rm}, the set consisting of all resource
types in the system.
 request edge – directed edge P1 → Rj
 assignment edge – directed edge Rj → Pi

Resource-Allocation Graph (RAG)

 Process

 Resource Type with 4 instances

 Pi requests instance of Rj Pi
Rj

 Pi is holding an instance of Rj Pi
Rj

Example
 P = P1, P2, P3
 R = R1, R2, R3, R4
 Instances: R1=1, R2=2, R3=1, R4=3

 E={ P1 -> R1, P2 -> R3, R1 -> P2, R2 ->
P1, R2 -> P2, R3 -> P3,

 Question: is there a deadlock?

Resource Allocation Graph With A
Deadlock

Graph With A Cycle But No Deadlock

Basic Facts
 If graph contains no cycles ⇒ no
deadlock.

 If graph contains a cycle ⇒
 if only one instance per resource type, then
deadlock.
 if several instances per resource type,
possibility of deadlock.

Methods for handling deadlocks
 3 strategies:

1. Ostrich Solution: Ignore the problem
and pretend deadlocks never occur in
the system. Used by most OS, including
Unix

2. Ensure that the system will never enter a
deadlock state.

 Deadlock prevention – Design a system
in such a way that the 4 conditions
leading to deadlock will never occur.
 Deadlock avoidance – make a decision
dynamically checking whether the request
will, if granted, potentially lead to a
deadlock or not.

3. Allow the system to enter a deadlock
state and recover

 Deadlock detection and recovery

Deadlock Prevention
 Preventing the occurrence of one of the
necessary conditions leading to
deadlock. Deadlock prevention strategy
is very conservative. It solves the
problem of deadlock by limiting access
to resources and by imposing
restrictions on processes.

Mutual Exclusion
 In general, this condition cannot be
required for non-sharable resources but
is a must for sharable resources.

Hold and Wait
 Must guarantee that whenever a process requests a
resource, it does not hold any other resources. A
process is blocked until all requests can be granted
simultaneously.

 Require process to request and be allocated all its
resources before it begins execution or allow
process to request resources only when the process
has none.

 Low resource utilization; starvation possible

No preemption
 If a process that is holding some
resource requests another resource that
cannot be immediately allocated to it,
then all resources currently being held
are released.
 Preempted resources are added to the
list of resources for which the process is
waiting.
 Process will be restarted only when it
can regain its old resources, as well as
new ones that it is requesting.

Circular Wait
 Impose a total ordering of all resource
types, and require that each process
requests resources in an increasing
order of enumeration.

Deadlock Avoidance
• Requires that the system has some additional a priori
information available.

 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.

Ways to avoid deadlock by careful
resource allocation
1. Resource trajectories
2. Safe/Unsafe states
3. Dijkstra’s Banker’s Algorithm

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.

Avoidance algorithms
 Single instance of a resource type. Use
a resource-allocation graph

 Multiple instances of a resource type.
Use the banker’s algorithm

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.

Unsafe State In Resource-Allocation Graph

Sample Algorithm
 Given the RAG of the system (Initially on
a safe state, grant a resource allocation
request if and only if doing so will not
introduce a cycle in the graph. (taking
into consideration all assignment,
request and claim edges). Otherwise,
the requesting process has to wait.

Sample Algorithm
1. Operation: P2 requests R3
Action: Request granted
System State: SAFE

2. Operation: P3 requests R2
Action: REQUEST DENIED!!!
System State: Leads to deadlock

Sample 2
 Consider a system with 15 identical instances of a resource

Process Max Need Curr Allo Remaining
P0 14 5
P1 10 5
P2 3 2
 Currently available resource: 15 – (5+5+2) = 3
 Is it safe?
 Yes with a safe sequence of <P2, P1, P0>
 What if P0 requests two more resources, will it be granted?
 No, OS cannot satisfy remaining needs of all processes. Unsafe.

Sample 3
 Total instances: 8

P0 5 3
P1 8 1
P2 7 2

 Is it safe or unsafe? What is the safe sequence?

Sample 4
 Total instances: 8

P0 5 3
P1 8 1
P2 7 2

 Is it safe or unsafe? What is the safe sequence?

Banker’s Algorithm
 Multiple instances.

 Each process must a priori claim maximum use.

 When a process requests a resource it may have
to wait.

 When a process gets all its resources it must
return them in a finite amount of time.

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].

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.

Example of 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
ABC ABC ABC
P0 010 753 332
P1 200 322
P2 302 902
P3 211 222
P4 002 433

Example (Cont.)
 The content of the matrix Need is defined to be Max – Allocation.
Need
ABC
P0 743
P1 122
P2 600
P3 011
P4 431

 The system is in a safe state since the sequence < P1, P3, P4, P2,
P0> satisfies safety criteria.

Example: P1 Request (1,0,2)
 Check that Request ≤ Available (that is, (1,0,2) ≤ (3,3,2) ⇒
true.
Allocation Need Available
ABC ABC ABC
P0 010 743 230
P1 302 020
P2 302 600
P3 211 011
P4 002 431
 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?

Deadlock detection and recovery
 Implies no conscious effort to
prevent/avoid deadlocks
 At certain times during processing, the
system executes an algorithm to
determine whether or not a deadlock
has occurred.
 If so, an attempt to recover from the
deadlock is made.

Detection Algorithm
Single Instance of 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.

Resource-Allocation Graph and Wait-for
Graph

Resource-Allocation Graph Corresponding wait-for graph

Several Instances of a 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.

Detection Algorithm
1. Let Work and Finish be vectors of length m and n, respectively
Initialize:
(a) Work = Available
n 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.

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.

Note:
The algorithm requires an order of O(m
x n2) operations to detect whether
the system is in deadlocked state.

Example of Detection Algorithm
 Five processes P0 through P4; three resource types
A (7 instances), B (2 instances), and C (6 instances).
Allocation Request Available
ABC ABC ABC
P0 010 000 000
P1 200 202
P2 303 000
P3 211 100
P4 002 002
 Sequence <P0, P2, P3, P1, P4> will result in Finish[i] = true for
all i. Therefore, no deadlock

Example (cont.)
 P2 requests an additional instance of type C.
Request
ABC
P0 000
P1 202
P2 001
P3 100
P4 002
 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.

Detection Algorithm Usage
 When and how often, to invoke deadlock detection?

 Factors to consider:
 How often a deadlock is likely to occur
 How many process will need to be rolled back
 Approaches
 Execute the algorithm at predetermined time interval
 Execute the algorithm whenever a resource request cannot
be granted immediatetly

Recovery from deadlock: Process
Termination
 Who’s responsibility?
 Inform the operator and let him deal with the
deadlock manually
 Allow the system to recover from deadlock
automatically.

Approaches: I. 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 process
 How long process has computed and
how much longer to completion
 Resources the process has used
 Resources process needs to complete
 How many process will need to be
terminated?
 Is process interactive or batch?

II. Process 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

Quiz
Consider the following and answer the questions that follow:
P = P1, P2, P3, P4, P5
R = R1, R2, R3, R4
Instances: R1=2, R2=3, R3=1, R4=1
E={ R3 -> P1, P1 -> R2, R2 -> P2, R2 -> P4, R1 -> P2, R1 -> P3,
P4 -> R4}
2 points each:
 Question 1: Is there a cycle? (Y/N)
 Question 2: Is there a deadlock? (Y/N)
 Question 3: Will P1’s request for R2 be granted? (Y/N)
 Question 4: Will P4’s request for R4 be granted? (Y/N)
 Question 5: How many available instances are there for R2?

Quiz
 8 processes P0 through P7; 4 resource types:
A (10 instances), B (5), C (11) and D (6).
Allocation Max
ABCD ABCD
P0 0121 7492
P1 2100 3220
P2 0020 9072
P3 2102 2243
P4 0020 4363
P5 2020 2132
P6 2111 5221
P7 1020 7485
Question 6: Is the system safe?
Question 7: If yes, what is the safe sequence? If no, what

Os module 2 d

More Related Content

What's hot (20)

Viewers also liked (20)

Similar to Os module 2 d (20)

More from Gichelle Amon (20)

Recently uploaded (20)

Os module 2 d