CS307 Operating Systems Deadlocks Fan Wu Department of Computer Science and Engineering Shanghai Jiao Tong University Spring 2020

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CS307 Operating Systems

Deadlocks

Fan Wu

Department of Computer Science and Engineering

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Bridge Crossing Example

 Traffic only in one direction

 Each section of a bridge can be viewed as a resource

 A deadlock occurs when two cars get on the bridge from different directions at the same time

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The Problem of Deadlock

 Example

 System has 2 disk drives

 P₁ and P₂ each hold one disk drive and each needs another one

 Example

 semaphores S and Q, initialized to 1

P₀ P₁

① wait (S); ② wait (Q);

③ wait (Q); ④ wait (S);

 Deadlock: A set of blocked processes each holding some resources and

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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, …, P_n} of waiting processes such that P₀is waiting for a resource that is held by P₁, P₁ is waiting for a resource that is held by P₂, …, P_n–1 is waiting for a resource that is held by P_n, and P_n is waiting for a resource that is held by P₀.

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System Model

 Processes P₁, P₂, …, P_n

 Resource types R₁, R₂, ..., R_m

e.g., CPU, memory space, I/O devices

 Each resource type R_i has W_i instances.

 Each process utilizes a resource as follows:

 request

 use

 release

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Resource-Allocation Graph

 Deadlocks can be identified with system resource- allocation graph.

 A set of vertices V and a set of edges E.

 V is partitioned into two types:

 P = {P₁, P₂, …, P_n}, the set consisting of all the processes in the system

 R = {R₁, R₂, …, R_m}, the set consisting of all resource types in the system

 E has two types:

 request edge – directed edge P_i R_j

 assignment edge – directed edge R_j  P_i

P_i

R_j

P_i

R_j

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Example of a Resource Allocation Graph



P = {P

₁

, P

₂

, P

₃

}



R = {R

₁

, R

₂

, R

₃

, R

₄

}



Resource instances:



W

₁

=W

₃

=1



W

₂

=2



W

₄

=3

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Resource Allocation Graph With A Deadlock

 A circle



P

₁

R

₁

P

₂

R

₃

P

₃



R

₂

P

₁

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Resource Allocation Graph With A Deadlock

 Two circles



P

₁

R

₁

P

₂

R

₃

P

₃

 R

₂

P

₁



P

₂

R

₃

P

₃

R

₂

P

₂

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Graph With A Cycle But No Deadlock

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Basic Facts

 If graph contains no circle  no deadlock

 If graph contains a circle 

 if only one instance per resource type, then deadlock

 if several instances per resource type, possibility of deadlock

 Question:

 Can you find a way to determine whether there is a deadlock, given a resource allocation graph with several instances per resource type?

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Methods for Handling Deadlocks

 Ensure that the system will never enter a deadlock state

 Deadlock prevention

 Deadlock avoidance

 Allow the system to enter a deadlock state and then recover

 Deadlock detection

 Deadlock recovery

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Deadlock Prevention

 Mutual Exclusion – not required for sharable resources; must hold for non-sharable resources

 Hold and Wait – must guarantee that whenever a process requests a resource, it does not hold any other resources

 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 (has released all its resources)

 Low resource utilization; starvation possible Restrain the ways request can be made

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Deadlock Prevention (Cont.)

 No Preemption

 If a process that is holding some resources requests another resource that cannot be immediately allocated to it, then all resources currently being held are preempted

 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 the 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

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Deadlock Avoidance

 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 Requires that the system has some additional a priori information

available

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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 <P₁, P₂, …, P_n> of ALL the processes in the systems such that for each P_i, the resources that P_ican still request can be satisfied by currently available resources +

resources held by all the P_j, with j < i

 That is:

 If P_i’s resource needs are not immediately available, then P_i can wait until all P_j have finished

 When all P_j are finished, P_i can obtain needed resources, execute, return allocated resources, and terminate

 When P_i terminates, P_{i +1} can obtain its needed resources, and so on

 Otherwise, system is in unsafe state

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Safe, Unsafe, Deadlock State

 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.

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 5 5

P₁ 4 2 2

P₂ 9 2 7

Available 3

Safe sequence: ?

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 5 5

P₁ 4 4 0

P₂ 9 2 7

Available 1

Safe sequence: P₁

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 5 5

P₁ 4 -- --

P₂ 9 2 7

Available 5

Safe sequence: P₁

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 10 0

P₁ 4 -- --

P₂ 9 2 7

Available 0

Safe sequence: P₁  P₀

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 -- --

P₁ 4 -- --

P₂ 9 2 7

Available 10

Safe sequence: P₁  P₀

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 -- --

P₁ 4 -- --

P₂ 9 9 0

Available 3

Safe sequence: P₁  P₀  P₂

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 -- --

P₁ 4 -- --

P₂ 9 -- --

Available 12

Safe sequence: P₁  P₀  P₂

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 5 5

P₁ 4 2 2

P₂ 9 3 6

Available 2

Safe sequence: ?

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Safe & Unsafe States

Maximum

Needs Holds Needs

P₀ 10 5 5

P₁ 4 -- --

P₂ 9 3 6

Available 4

Safe sequence: P₁  ?

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Avoidance Algorithms

 Avoidance algorithms ensure that the system will never deadlock.

 Whenever a process requests a resource, the request is granted only if the allocation leaves the system in a safe state.

 Two avoidance algorithms

 Single instance of a resource type

 Use a resource-allocation graph

 Multiple instances of a resource type

 Use the banker’s algorithm

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Resource-Allocation-Graph Algorithm

 Claim edge P_i  R_j indicates that process P_j may request resource R_j; represented by a directed dashed line

 Resources must be claimed a priori in the system

 Claim edge converts to request edge when a process requests a resource

 Request edge converts 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 (the edge is removed if the process finishes)

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Resource-Allocation Graph Algorithm

 Suppose that process P_i requests a resource R_j

 The request can be granted only if converting the request edge to an

assignment edge does not result in the formation of a circle in the resource allocation graph

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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

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Data Structures for the Banker’s Algorithm

 Available: Vector of length m. If available[j] = k, there are k instances of resource type R_j available

 Max: n x m matrix. If Max[i,j] = k, then process P_i may request at most k instances of resource type R_j

 Allocation: n x m matrix. If Allocation[i,j] = k then P_i is currently allocated k instances of R_j

 Need: n x m matrix. If Need[i,j] = k, then P_i may need k more instances of R_jto complete its task

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

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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 an i such that both:

(a) Finish [i] = false (b) Need_i  Work

If no such i exists, go to step 4 3. Work = Work + Allocation_i

Finish[i] = true go to step 2

4. If Finish [i] == true for all i, then the system is in a safe state

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Resource-Request Algorithm for Process P

_i

Request_i = request vector for process P_i. If Request_i [j] = k then process P_i wants k instances of resource type R_j

1. If Request_i  Need_i, go to step 2. Otherwise, raise error condition, since process has exceeded its maximum claim

2. If Request_i  Available, go to step 3. Otherwise P_i must wait, since resources are not available

3. Pretend to allocate requested resources to P_i by modifying the state as follows:

Available = Available – Request_i; Allocation_i= Allocation_i + Request_i; Need_i = Need_i – Request_i;

 If safe  the resources are allocated to P_i

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Example of Banker’s Algorithm

 5 processes P₀through P₄; 3 resource types:

A (10 instances), B (5 instances), and C (7 instances) Snapshot at time T₀:

Max Allocation Need Available

A B C A B C A B C A B C

P₀ 7 5 3 0 1 0 7 4 3 3 3 2 P₁ 3 2 2 2 0 0 1 2 2

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1 Is the system in safe state?

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Applying Safety Algorithm

P₀ 7 5 3 0 1 0 7 4 3 3 3 2 P₁ 3 2 2 2 0 0 1 2 2

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1

P₀ 7 5 3 0 1 0 7 4 3 2 1 1 P₁ 3 2 2 3 2 2 0 0 0

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1

P₀ 7 5 3 0 1 0 7 4 3 5 3 2

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1

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Applying Safety Algorithm

P₀ 7 5 3 0 1 0 7 4 3 5 3 2

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1 Safe sequence: P₁  P₃

P₀ 7 5 3 0 1 0 7 4 3 5 2 1

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 2 2 0 0 0

P₄ 4 3 3 0 0 2 4 3 1

P₀ 7 5 3 0 1 0 7 4 3 7 4 3

P₂ 9 0 2 3 0 2 6 0 0

P₄ 4 3 3 0 0 2 4 3 1

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Applying Safety Algorithm

P₀ 7 5 3 0 1 0 7 4 3 7 4 3

P₂ 9 0 2 3 0 2 6 0 0

P₄ 4 3 3 0 0 2 4 3 1

P₀ 7 5 3 7 5 3 0 0 0 0 0 0

P₂ 9 0 2 3 0 2 6 0 0

P₄ 4 3 3 0 0 2 4 3 1

7 5 3

P₂ 9 0 2 3 0 2 6 0 0

P₄ 4 3 3 0 0 2 4 3 1

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Applying Safety Algorithm

7 5 3

P₂ 9 0 2 3 0 2 6 0 0

P₄ 4 3 3 0 0 2 4 3 1 Safe sequence: P₁  P₃  P₀  P₂

1 5 3

P₂ 9 0 2 9 0 2 0 0 0

P₄ 4 3 3 0 0 2 4 3 1

10 5 5

P₄ 4 3 3 0 0 2 4 3 1

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Applying Safety Algorithm

10 5 5

P₄ 4 3 3 0 0 2 4 3 1

6 2 4

P₄ 4 3 3 4 3 3 0 0 0

10 5 7

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P₀ 7 5 3 0 1 0 7 4 3 3 3 2 P₁ 3 2 2 2 0 0 1 2 2

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1

Example: P ₁ Request (1,0,2)

 Check that Request  Available (that is, (1,0,2)  (3,3,2)  true)

 Executing safety algorithm shows that sequence < P₁, P₃, P₀, P₂, P₄>

satisfies safety requirement

P₀ 7 5 3 0 1 0 7 4 3 2 3 0 P₁ 3 2 2 3 0 2 0 2 0

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1

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Example: P ₀ Request (0,2,0)

 Check that Request  Available (that is, (0,2,0)  (2,3,0)  true)

 Does there a safe sequence exist?

P₀ 7 5 3 0 1 0 7 4 3 2 3 0 P₁ 3 2 2 3 0 2 0 2 0

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1

P₀ 7 5 3 0 3 0 7 2 3 2 1 0 P₁ 3 2 2 3 0 2 0 2 0

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1

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Pop Quiz

 5 processes P₀through P₄; 3 resource types:

A (10 instances), B (5 instances), and C (7 instances) Snapshot at time T₀:

P₀ 7 5 3 0 1 0 7 4 3 3 3 2 P₁ 3 2 2 2 0 0 1 2 2

P₂ 9 0 2 3 0 2 6 0 0

P₃ 2 2 2 2 1 1 0 1 1

P₄ 4 3 3 0 0 2 4 3 1

 Can P4’s request (2, 1, 0) be granted?

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Deadlock Detection

 Allow system to enter deadlock state

 Detection algorithm

 Recovery scheme

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Single Instance of Each Resource Type

 Maintain wait-for graph

 Nodes are processes

 P_i  P_jif P_i is waiting for P_j

 Periodically invoke an algorithm that searches for a cycle in the graph. If there is a cycle, there exists a deadlock

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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[i][j] = k, then process P_i is requesting k more instances of resource type R_j.

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Detection Algorithm

1. Let Work and Finish be vectors of length m and n, and initialize:

(a) Work = Available

(b) For i = 1,2, …, n, if Allocation_i  0, then Finish[i] = false; otherwise, Finish[i] = true

2. Find an index i such that both:

(a) Finish[i] == false (b) Request_i  Work

If no such i exists, go to step 4 3. Work = Work + Allocation_i

Finish[i] = true go to step 2

4. If Finish[i] == false, for some i, 1  i  n, then the system is in deadlock

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Example of Detection Algorithm

 Five processes P₀ through P₄; three resource types A (7 instances), B (2 instances), and C (6 instances)

 Snapshot at time T₀:

Allocation Request Available

A B C A B C A B C

P₀ 0 1 0 0 0 0 0 0 0 P₁ 2 0 0 2 0 2

P₂ 3 0 3 0 0 0 P₃ 2 1 1 1 0 0

P 0 0 2 0 0 2

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Example (Cont.)

 P₂ requests an additional instance of type C

 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 Allocation Request Available

A B C A B C A B C

P₀ 0 1 0 0 0 0 0 0 0 P₁ 2 0 0 2 0 2

P₂ 3 0 3 0 0 1 P₃ 2 1 1 1 0 0 P₄ 0 0 2 0 0 2

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Detection-Algorithm Usage

 When, and how often, to invoke depends on:

 How often a deadlock is likely to occur?

 How many processes will need to be rolled back?

 one for each disjoint cycle

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Recovery from Deadlock

 Process Termination

 abort one or more processes to break the circular wait

 Resource Preemption

 preempt some resources from one or more of the deadlocked processes

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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 compete

 How many processes will need to be terminated

 Is process interactive or batch?

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Resource Preemption

 Selecting a victim – minimize cost

 Rollback – return to some safe state, restart process from that state

 Starvation – same process may always be picked as victim, include number of rollback in cost factor

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Homework

 Reading

 Chapter 7

 Exercise

 See course website