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

2. (Chapter-7) Deadlocks

3. Chapter 7: Deadlocks 1.The Deadlock Problem 2.Deadlock Characterization 3.Methods for Handling Deadlocks 3

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

5. The Deadlock Problem 5

6.  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. 6

7.  What is a deadlock problem?? o A set of blocked processes each holding a resource and waiting to acquire a resource held by another process in the set  Example o System has 2 disk drives o P 1 and P 2 each hold one disk drive and each need the other one n Example l 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 The Deadlock Problem (cont..) 7

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 8

9. System Model  The resources are partitioned into several types (R 1, R 2,..., R m ) o E.g.: CPU cycles, memory space, I/O devices  Each resource type R i has W i instances. o E.g :  If a system has (2) CPUs, then the resource type CPU has (2) instances.  The resource type printer may have (5) instances. 9

10. 10  Each process utilizes a resource as follows: 1. Request >> If the request cannot be granted immediately,then the requesting process must wait until it can acquire the resource. 2. Use 3. 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.

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. 11 Physical resources Logical resources e.g. printers, DVD drives, memory space, and CPU cycles e.g. files

12. The Deadlock Characterization 12

13. Deadlock Characterization 1. 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) 2. Hold and wait: a process holding at least one resource is waiting to acquire additional resources held by other processes Deadlock can arise if four conditions hold simultaneously: 13

14. Deadlock Characterization (cont..) 3. No preemption: a resource can be released only voluntarily by the process holding it, after that process has completed its task 4. Circular wait:  There exists a set {P 0, P 1, …, P n } of waiting processes such that : o P 0 is waiting for a resource that is held by P 1, o P 1 is waiting for a resource that is held by P 2, …, P n–1 is waiting for a resource that is held by P n, o P n is waiting for a resource that is held by P 0. 14 P0P0 P1P1 P2P2 PnPn

15. Resource-Allocation Graph 15

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. 16 V is partitioned into two types: E is partitioned into two types: P = {P 1, P 2, …, P n }, the set consisting of all the processes in the system Request edge – directed edge P i  R j R = {R 1, R 2, …, R m }, the set consisting of all resource types in the system Assignment edge – directed edge R j  P i

17. Resource-Allocation Graph (Cont.)  Process  Resource Type with 4 instances  P i requests instance of R j  P i is holding an instance of R j PiPi PiPi RjRj RjRj 17

18. Example of a Resource Allocation Graph n 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} n 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 18

19. Example of a Resource Allocation Graph n 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. 19

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. 20

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

22. l It ‘s a dead lock situation >>> Processes P1, P2, and P3 are deadlocked…….why??  There is no chance to break the cycle. Resource Allocation Graph With a Deadlock 22 n Example (2): l 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

23. Graph With A Cycle But No Deadlock l 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. 23 n Example (3): l The graph contains (1)cycle  may be a deadlock Cycle : P1 → R1 → P3 → R2 → P1

24. 24  To ensure that deadlocks never occur, the system can use either 1. a deadlock prevention or 2. a deadlock-avoidance scheme

25. 25 Deadlock preventionDeadlock 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.

26. 1-Deadlock Prevention 26

27. Deadlock Prevention 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 Remove the possibility of deadlock occurring by denying one of the four necessary conditions: The process must request all its resources before execution. 0r allow process to request resources only when the process has none 27

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 28

29. 29  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. 4- Circular Wait –

30. Deadlock Prevention (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 ). 4- Circular Wait – cont.. 30

31. Deadlock Prevention (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 4- Circular Wait – cont.. 31

32. 2-Deadlock Avoidance 32

33. Deadlock Avoidance  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 Requires that the system has some additional a priori information available 33

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 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 o That is: If Pi resource needs are not immediately available, then Pi can wait until all Pj have finished o When Pj is finished, Pi can obtain needed resources, execute, return allocated resources, and terminate o When Pi terminates, Pi +1 can obtain its needed resources, and so on 34

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. 35

36. Safe, Unsafe, Deadlock State 36

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 37

38. Resource-Allocation Graph Scheme  Claim edge P i → R j indicated that process P j may request resource R j ; 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 38

39. Resource-Allocation Graph 39

40. Unsafe State In Resource-Allocation Graph 40

41. 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 form of a cycle in the resource allocation graph 41

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 42

43. Banker’s Algorithm  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. o If it will, the resources are allocated; o otherwise, the process must wait until some other process releases enough resources. 43

44. 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 j to complete its task  Need [i][j] = Max [i][j] – Allocation [i][j] Let n = number of processes, and m = number of resources types. 44

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) 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 45

46. Example of Banker’s Algorithm  5 processes P 0 through P 4 ; 3 resource types: A (10 instances), B (7 instances), and C (5 instances) Snapshot at time T 0 : AllocationMaxAvailable A B C A B C A B C P 0 0 1 0 7 5 3 3 3 2 P 1 2 0 0 3 2 2 P 2 3 0 2 9 0 2 P 3 2 1 1 2 2 2 P 4 0 0 2 4 3 3 46

47. Example of Banker’s Algorithm Snapshot at time T 0 : AllocationMaxAvailable A B C A B C A B C P 0 0 1 0 7 5 3 3 3 2 P 1 2 0 0 3 2 2 P 2 3 0 2 9 0 2 P 3 2 1 1 2 2 2 P 4 0 0 2 4 3 3 47  The matrix Need is defined to be Max – Allocation A B C 7 4 3 1 2 2 6 0 0 0 1 1 4 3 1  The system is in a safe state since the sequence satisfies safety criteria Need

48. 48  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, Example of Banker’s Algorithm

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 0 0 1 0 7 4 3 2 3 0 P 1 3 0 2 0 2 0 P 2 3 0 2 6 0 0 P 3 2 1 1 0 1 1 P 4 0 0 2 4 3 1 49

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

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

52. Example: P 1 Request (1,0,2) cont. Executing safety algorithm shows that sequence satisfies safety requirement (cont.) o P 2 acquires 6 A more resources, achieving its maximum o The system now still has 1 A, 5 B, and 5 C resource available o P 2 terminates, returning 9 A, and 2 C resources to the system o The system now has 10 A, 5 B, and 7 C resources available o Because all processes were able to terminate, this state is safe  Can request for (3,3,0) by P 4 be granted?  Can request for (0,2,0) by P 0 be granted? 52

53. Methods for Handling Deadlocks 53

54. Methods for Handling Deadlocks  OS can deal with the deadlock problem in one of three ways 54 1-prevent or avoid deadlocks 2-Allow3-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

55. Thank you End of Chapter 7 Dr. Abeer Mahmoud 55