The Slide does not contain all the information and cannot be treated as a study material for Operating System. Please refer the text book for exams.

Transcription

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2 The Slide does not contain all the information and cannot be treated as a study material for Operating System. Please refer the text book for exams.

3 System Model Deadlock Characterization Methods of handling deadlock Deadlock prevention Deadlock Avoidance Deadlock Detection Recovery from deadlock

4 Deadlocks occur two or more process are each waiting for an event that will never occur, since it can only be generated by another process in that set Deadlock is one of the most difficult problems that OS designers face As we examine various approaches to deal with deadlocks, notice the tradeoff between, how well the approach solves the problem, and its performance/ OS overhead

5 Resource types R1, R2. Rm( CPU Cycles, memory space, I/O Devices Each resource type Ri has Wi instances Each process utilizes a resource as follows Request -> use -> Release Resources are Physical(tape drives, CPU cycles, printers, memory space) Logical(Files, Semaphores)

6 Mutual Exclusion Only one process can use a resource at a time. Hold and Wait A process are allowed to hold one resource and be waiting to acquire additional resources held by other processes No preemption Resources are released voluntarily; neither other process nor OS can force a process to release the resource Circular Wait: A set of waiting process must exist such that P0 is waiting for a process held by P1 and so on and Pn for P0

7 A deadlock condition can be modeled using a directed graph called a resource allocation graph 2 kinds of nodes: Boxes represent resources Instances of the resource are represented as dots within the box Circles represent threads / processes 2 kinds of (directed) edges: Request edge from process to resource - indicates the process has requested the resource, and is waiting to acquire it Assignment edge from resource instance to process indicates the process is holding the resource instance

8 When a request is made, a request edge is added When request is fulfilled, the request edge is transformed into an assignment edge When process releases the resource, the assignment edge is deleted

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12 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.

13 1. Ensure that the system will never enter a deadlock state. Deadlock prevention prevent deadlock from occurring by eliminating one of the 4 deadlock conditions Deadlock avoidance algorithms consider resources currently available, resources allocated to each thread, and possible future requests, and only fulfil requests that will not lead to deadlock 2. Allow the system to enter a deadlock state and detect it and then recover. Deadlock detection algorithm detect when deadlock occur, Deadlock recovery algorithms break the deadlock 3. Ignore the problem and pretend that deadlocks never occur in the system (The ostrich approach)

14 Basic idea: ensure that one of the 4 conditions for deadlock cannot hold Mutual exclusion if one process holds a resource, other processes requesting that resource must wait until the process releases it (only one can use it at a time) Hard to avoid mutual exclusion for nonsharable resources Printer & other I/O devices Files However, many resources are sharable, so deadlock can be avoided for those resources Read-only files How to avoid mutual exclusion for printer? through spooling then process won t have to wait on physical printer

15 Hold and wait processes are allowed to hold one (or more) resource and be waiting to acquire additional resources that are being held by other processes To avoid, ensure that whenever a process requests a resource, it doesn t hold any other resources Request all resources (at once) at beginning of process execution To get a new resource, release all current resources, and then request new ones Difficult to know what to request in advance Wasteful; ties up resources and reduces resource utilization Starvation is possible

16 No preemption resources are released voluntarily; neither another process nor the OS can force a process to release a resource To avoid, allow preemption If process A requests resources that aren t available, see who holds those resources If the holder (process B) is waiting on additional resources, preempt the resource requested by process A Otherwise, process A has to wait» While waiting, some of its current resources may be preempted» Can only wake up when it acquires the new resources plus any preempted resources If a process requests a resource that cannot be allocated to it, all resources held by that process are preempted Can only wake up when it can acquire all the requested resources Only works for resources whose state can be saved/restored(cpu registers and memory space, not printer or tape drives)

17 Circular wait there must exist a set of waiting processes such that P0 is waiting for a resource held by P1, P1 is waiting for a resource held by P2, Pn-1 is waiting for a resource held by Pn, and Pn is waiting for a resource held P0 To avoid, impose a total order on all resources, and require process to request resource in that order Order: disk drive, printer, CDROM Process A requests disk drive, then printer Process B requests disk drive, then printer Process B does not request printer, then disk drive, which could lead to deadlock Order should be in the logical sequence that the resources are usually acquired Allow process to release all resources, and start request sequence over Or force process to request total number of each resource in a single request

18 Requires that the system has some additional a priori information available about the use of resources by processes. 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.

19 A system is in safe if the system can allocate resources to each process(upto maximum) and still avoid deadlock safe sequence of all processes Sequence <P 1, P 2,, P n > is safe if for each P i, the resources that P i can still request can be satisfied by currently available resources + resources held by all the P j, with j<i. If P i resource needs are not immediately available, then P i can wait until all P j have finished. When P j is 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.

20 If a system is in safe state => no deadlocks If a system is in unsafe state => possibility of deadlock due to behavior of processes Avoidance => Ensure that system never enters an unsafe state

21 The behavior of process controls unsafe state 12 tape drives Process Max Current P P1 4 2 P2 9 2 What is the safe sequence? What if P2 requested 1 more tape drive at t1? Allocate more resources only if it leaves the system in safe state An unsafe state is not deadlocked, but no such scheduling order exists It may even work, if a process releases resources at the right time

22 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. And assignment edge when assigned. When a resource is released by a process, assignment edge reconverts to a claim edge. Resources must be claimed a priori in the system.

23 An algorithm for detecting a cycle in a graph requires order of n 2 operations A resource is not allocated if this action will create a cycle, so the claim edge will be left as request edge

24 Multiple instances of resources. 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.

25 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 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].

26 1. Let Work and Finish be vectors of length m and n, respectively. Initialize: Work = Available Finish [i] = false for i - 1,3,, n. 2. Find and i such that both: (a) Finish [i] = false (b) Need i Work (available) If no such i exists, go to step Work = Work + Allocation i Finish[i] = true go to step If Finish [i] == true for all i, then the system is in a safe state.

27 Request = 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 Pi. If unsafe Pi must wait, and the old resourceallocation state is restored

28 5 processes P 0 through P 4 ; 3 resource types A (10 instances) B (5 instances) C (7 instances). Snapshot at time T 0 : Allocation Max Available A B C A B C A B C P P P P P Is the system safe?

29 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 < P 1, P 3, P 4, P 2, P 0 > satisfies safety criteria.

30 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 P P P P P 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?

31 Allow system to enter deadlock state Detection algorithm Recovery scheme

32 Maintain wait-for graph Nodes are processes. P i P j if P i is waiting for P j. Periodically invoke an algorithm that searches for a cycle in the graph. An algorithm to detect a cycle in a graph requires an order of n 2 operations, where n is the number of vertices in the graph.

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34 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.

35 1. Let Work and Finish be vectors of length m and n, respectively 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 Work = Work + Allocation i Finish[i] = true go to step If Finish[i] == false, for some i, 1 i n, then the system is in deadlock state. Moreover, if Finish[i] == false, then P i is deadlocked. Algorithm requires an order of O(m x n 2) operations to detect whether the system is in deadlocked state.

36 Five processes P 0 through P 4 ; three resource types A (7 instances), B (2 instances), and C (6 instances). Snapshot at time T 0 : Allocation Request Available A B C A B C A B C P P P P P State of system? Can reclaim resources held by process P 0, but insufficient resources to fulfill other processes; requests. Deadlock exists, consisting of processes P 1, P 2, P 3, and P 4.

37 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 If detection algorithm is Invoke at each request we can identify deadlocked processes and specific process that caused the deadlock Incur considerable overhead Invoked arbitrarily - there may be many cycles in the resource graph and not be able to tell which of the many deadlocked processes caused the deadlock.

38 Whenever detection algorithm determines that deadlock exists several alternatives are available Two options of breaking the deadlock Process Termination Abort one or more process to break the circular wait Resource Preemption Preempt some resources from one or more of the deadlocked processes

39 1. Abort all deadlocked processes. great expense as processes must have computed for a long time and results must be discarded 2. Abort one process at a time until the deadlock cycle is eliminated Considerable overhead since after each process is aborted deadlock detection must be performed 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?

40 Preempt some resources from processes and give these to other process until deadlock cycle is broken Three issues needs to be addressed Selecting a victim chose the order of preemption to minimize cost. Rollback return to some safe state, restart process for that state or total rollback Starvation same process may always be picked as victim, include number of rollback in cost factor.

41 What method does your operating system use? Deadlock Prevention Deadlock Avoidance Deadlock Detection and Recovery Ignore the problem and pretend deadlock never occur The last solution is used in Unix and Windows and it is then up to the application developer to write programs that handle deadlock