1. Operating Systems (CS 340 D)
Princess Nora University
Faculty of Computer & Information Systems
Computer science Department
Operating Systems (CS 340 D)
Dr. Abeer Mahmoud

2. (Chapter-7) Deadlocks

3. Chapter 7: Deadlocks The Deadlock Problem Deadlock Characterization
Methods for Handling Deadlocks
Dr. Abeer Mahmoud

4. To present methods for Handling Deadlocks
OBJECTIVES:
To develop a description of deadlocks, which prevent sets of concurrent processes from completing their tasks
To present methods for Handling Deadlocks
Dr. Abeer Mahmoud

5. The Deadlock Problem
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6. The Deadlock Problem
In a multiprogramming environment, several processes may compete for a finite number of resources.
A process requests resources; if the resources are not available at that time, the process enters a waiting state.
Sometimes, a waiting process is never again able to change state, because the resources it has requested are held by other waiting processes.
This situation is called a DEADLOCK.
Dr. Abeer Mahmoud

7. The Deadlock Problem (cont..)
What is a deadlock problem??
A set of blocked processes each holding a resource and waiting to acquire a resource held by another process in the set
Example
System has 2 disk drives
P1 and P2 each hold one disk drive and each need the other one
semaphores A and B, initialized to 1
P0 P1
wait (A); wait(B)
wait (B); wait(A)
Note :– Most OSs do not prevent or deal with deadlocks
Dr. Abeer Mahmoud

8. 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
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9. System Model
The resources are partitioned into several types (R1, R2, . . ., Rm )
E.g.: CPU cycles, memory space, I/O devices
Each resource type Ri has Wi instances.
E.g :
If a system has (2) CPUs, then the resource type CPU has (2) instances.
The resource type printer may have (5) instances.
Dr. Abeer Mahmoud

10. Each process utilizes a resource as follows: Request
>> If the request cannot be granted immediately ,then the requesting process must wait until it can acquire the resource.
Use
Release
A set of processes is in a deadlocked state when every process in the set is waiting for an event that can be caused only by another process in the set.
Dr. Abeer Mahmoud

11. System Model (cont.) Resource forms: Example:
consider a system with one printer and one DVD drive. Suppose that process Pi is holding the DVD and process Pj is holding the printer. If Pi requests the printer and Pj requests the DVD drive, a deadlock occurs.
Physical resources
Logical resources
e.g. printers, DVD drives, memory space, and CPU cycles
e.g. files
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12. The Deadlock Characterization
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13. Deadlock Characterization
Deadlock can arise if four conditions hold simultaneously:
Mutual exclusion: the resources are sharable or non sharable  mutual exclusion happen when only one process at a time can use a resource…(i.e. No resource sharing)
Hold and wait: a process holding at least one resource is waiting to acquire additional resources held by other processes
Dr. Abeer Mahmoud

14. Deadlock Characterization (cont..)
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, …, Pn} 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,
Pn is waiting for a resource that is held by P0. P0 P1 P2 Pn
Dr. Abeer Mahmoud

15. Resource-Allocation Graph
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16. Resource-Allocation Graph
Deadlocks can be described in terms of a directed graph called a system resource-allocation graph.
This graph consists of a set of vertices V and a set of edges E.
V is partitioned into two types:
E is partitioned into two types:
P = {P1, P2, …, Pn}, the set consisting of all the processes in the system
Request edge – directed edge
Pi  Rj
R = {R1, R2, …, Rm}, the set consisting of all resource types in the system
Assignment edge – directed edge
Rj  Pi
Dr. Abeer Mahmoud

17. Resource-Allocation Graph (Cont.)
Process
Resource Type with 4 instances
Pi requests instance of Rj
Pi is holding an instance of Rj Pi Rj Pi Rj
Dr. Abeer Mahmoud

18. Example of a Resource Allocation Graph
The sets P, R, and E:
P = {P1, P2, P3}
R = {R1, R2, R3, R4}
E = {P1 → R1, P2 → R3, R1 → P2, R2 → P2, R2 → P1, R3 → P3}
Resource instances:
(1) instance of resource type R1
(2) instances of resource type R2
(1) instance of resource type R3
(3) instances of resource type R4
Dr. Abeer Mahmoud

19. Example of a Resource Allocation Graph
Process states:
Process P1 is holding an instance of resource type R2 and is waiting for an instance of resource type R1.
Process P2 is holding an instance of R1 and an instance of R2 and is waiting for an instance of R3.
Process P3 is holding an instance of R3.
Dr. Abeer Mahmoud

20. Basic Facts
How we recognize deadlock by using a Resource Allocation Graph??
If graph contains no cycles  no deadlock
If graph contains a cycle 
If only one instance per resource type, then there is a deadlock
If several instances per resource type, possibility of deadlock .
Dr. Abeer Mahmoud

21. Resource Allocation Graph With NO Deadlock
Example (1):
The graph contains NO cycles  NO deadlock is occurred.
Dr. Abeer Mahmoud

22. Resource Allocation Graph With a Deadlock
Example (2):
The graph contains (2)cycles  may be a deadlock is occurred.
Cycle (1) : P1 → R1 → P2 → R3 → P3 → R2 → P1
Cycle (2) : P2 → R3 → P3 → R2 → P2
It ‘s a dead lock situation >>> Processes P1, P2, and P3 are deadlocked…….why??
There is no chance to break the cycle.
Dr. Abeer Mahmoud

23. Graph With A Cycle But No Deadlock
Example (3):
The graph contains (1)cycle  may be a deadlock
Cycle : P1 → R1 → P3 → R2 → P1
NO dead lock situation .....WHY????
process P4 may release its instance of resource type R2…… That resource can then be allocated to P3, breaking the cycle.
Dr. Abeer Mahmoud

24. To ensure that deadlocks never occur, the system can use either
a deadlock prevention or
a deadlock-avoidance scheme
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25. Deadlock prevention Deadlock avoidance
provides a set of methods to ensure that at least one of the necessary 4 conditions (explained before ) cannot hold.
These methods prevent deadlocks by constraining how requests for resources can be made.
requires that the operating system be given additional information in advance concerning which resources a process will request and use during its lifetime.
With this additional knowledge, the operating system can decide for each request whether or not the process should wait.
Dr. Abeer Mahmoud

26. 1-Deadlock Prevention
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27. Deadlock Prevention
Remove the possibility of deadlock occurring by denying one of the four necessary conditions:
1- Mutual Exclusion – not required for sharable resources; must hold for non-sharable resources.
That is, at least one resource must be nonsharable. Sharable resources, in contrast, do not require mutually exclusive access and thus cannot be involved in a deadlock
2- Hold and Wait – must guarantee that whenever a process requests a resource, it does not hold any other resources by using one of the following protocols:
Problems-> Low resource utilization; starvation possible
The process must request all its resources before execution.
0r allow process to request resources only when the process has none *

28. Deadlock Prevention (Cont.)
3- 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 released
Process will be restarted only when it can regain its old resources, as well as the new ones that it is requesting
It is applied to resources whose state easily saved and restored later, such as CPU registers and memory space
The main Advantages is more better resource utilization
Problems
The cost of removing a process's resources
starvation possible *

29. 4- Circular Wait –
One way to ensure that this condition never holds is to impose a total ordering of all resource types and to require that each process requests resources in an increasing order of enumeration.
let R = {R1, R2, ..., Rm} be the set of resource types. We assign to each resource type a unique integer number, which allows us to compare two resources and to determine whether one precedes another in our ordering.
Formally, we define a one-to-one function F: R→N, where N is the set of natural numbers.
Dr. Abeer Mahmoud

30. Deadlock Prevention (Cont.)
4- Circular Wait – cont..
Example : if the set of resource types R includes tape drives, disk drives,
and printers,
then the function F might be defined as follows:
F(tape drive) = 1
F(disk drive) = 5
F(printer) = 12
We can now consider the following protocol to prevent deadlocks:
Each process can request resources only in an increasing order of enumeration.
That is, a process can initially request any number of instances of a resource type —say, Ri . After that, the process can request instances of resource type Rj if and only if F(Rj ) > F(Ri ). *

31. Deadlock Prevention (Cont.)
4- Circular Wait – cont..
EX: a process that wants to use the tape drive and printer at the same time must first request the tape drive and then request the printer.
Alternatively, we can require that a process requesting an instance of resource type Rj must have released any resources Ri such that F(Ri ) ≥ F(Rj ).
Problems
Resources must be requested in ascending order of resource number rather than as needed *

32. 2-Deadlock Avoidance
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33. 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 *

34. 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 sequence <P1, P2, …, Pn> of ALL the processes in the systems such that for each Pi, the resources that Pi can still 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 *

35. 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. *

36. Safe, Unsafe , Deadlock State*

37. Avoidance algorithms
The avoidance algorithms ensure that the system will always remain in a safe state.
Single instance of a resource type
Use a resource-allocation graph
Multiple instances of a resource type
Use the banker’s algorithm *

38. 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 *

39. Resource-Allocation Graph*

40. Unsafe State In Resource-Allocation Graph*

41. Resource-Allocation Graph Algorithm
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 form of a cycle in the resource allocation graph *

42. Banker’s Algorithm Multiple instances
Each process must a priori claim maximum use, When a new process enters a system, it must declare the maximum number of instances of each resource type that may not exceed the total number of resources in the system
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
When a new process enters the system, it must declare the maximum number of instances of each resource type that it may need. This number may not exceed the total number of esources in the system. When a user requests a set of resources, the system must determine whether the allocation of these
resources will leave the system in a safe state. If it will, the resources are
allocated; otherwise, the process must wait until some other process releases
enough resources. *

43. Banker’s Algorithm
When a new process enters the system, it must declare the maximum number of instances of each resource type that it may need. This number may not exceed the total number of resources in the system.
When a user requests a set of resources, the system must determine whether the allocation of these resources will leave the system in a safe state.
If it will, the resources are allocated;
otherwise, the process must wait until some other process releases enough resources. *

44. 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] *

45. 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 index 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 *

46. Example of Banker’s Algorithm
5 processes P0 through P4;
3 resource types:
A (10 instances), B (5 instances), and C (7 instances)
Snapshot at time T0:
Allocation Max Available
A B C A B C A B C P P P P P *

47. Example (Cont.)
The content of the matrix Need is defined to be Max – Allocation
Need
A B C P P P P P
The system is in a safe state since the sequence < P1, P3, P4, P2, P0> satisfies safety criteria *

48. Example of Banker’s Algorithm
Suppose now that process P1 requests one additional instance of resource type A and two instances of resource type C, so Request1 = (1,0,2). To decide whether this request can be immediately granted,
Dr. Abeer Mahmoud

49. Example: P1 Request (1,0,2)
1- Check that Request ≤ Need (that is, (1,0,2) ≤ (1,2,2) ⇒ true
2- Check that Request ≤ Available (that is, (1,0,2) ≤ (3,3,2) ⇒ true
3- Available = Available – Request => (3,3,2) - (1,0,2) = (2,3,0)
Allocation = Allocation + Request => (2,0,0) - (1,0,2) = (3,0,2)
Need = Need – Request => (1,2,2) - (1,0,2) = (0,2,0)
Allocation Need Available
A B C A B C A B C P P P P P *

50. Example: P1 Request (1,0,2) cont.
Executing safety algorithm shows that sequence < P1, P3, P4, P0, P2> satisfies safety requirement
P1 acquires 2 B more resources, achieving its maximum
The system now still has 2 A, 1 B, and no C resource available
P1 terminates, returning 3 A, 3 B, 2 C resources to the system
The system now has 5 A, 3 B, and 2 C resources available
P3 acquires 1B and 1C more resources, achieving its maximum
The system now still has 5 A, 2 B, and 1 C resource available
P3 terminates, returning 2 A, 2 B, 2 C resources to the system
The system now has 7 A, 4 B, and 3 C resources available *

51. Example: P1 Request (1,0,2) cont.
Executing safety algorithm shows that sequence < P1, P3, P4, P0, P2> satisfies safety requirement
P4 acquires 4 A, 3 B AND 1 C more resources, achieving its maximum
The system now still has 3 A, 1 B, and 2 C resource available
P4 terminates, returning 4 A, 3 B, 3 C resources to the system
The system now has 7 A, 4 B, and 5 C resources available
P0 acquires 7 A, 4 B AND 3 C more resources, achieving its maximum
The system now still has no A, no B, and 2 C resource available
P0 terminates, returning 7 A, 5 B, 3 C resources to the system
The system now has 7 A, 5 B, and 5 C resources available *

52. Example: P1 Request (1,0,2) cont.
Executing safety algorithm shows that sequence < P1, P3, P4, P0, P2> satisfies safety requirement (cont.)
P2 acquires 6 A more resources, achieving its maximum
The system now still has 1 A, 5 B, and 5 C resource available
P2 terminates, returning 9 A, and 2 C resources to the system
The system now has 10 A, 5 B, and 7 C resources available
Because all processes were able to terminate, this state is safe
Can request for (3,3,0) by P4 be granted?
Can request for (0,2,0) by P0 be granted? *

53. Methods for Handling Deadlocks
Dr. Abeer Mahmoud

54. Methods for Handling Deadlocks
OS can deal with the deadlock problem in one of three ways
1-prevent or avoid deadlocks
2-Allow
3-Ignore
ensuring that the system will never enter a deadlocked state.
Allow the system to enter a deadlocked state, detect it, and recover.
Ignore the problem
Used by most operating systems, including Unix & windows
This method is cheaper than 1 or 2 .
Solution: The system must be restarted manually
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55. Thank you End of Chapter 7
Dr. Abeer Mahmoud