1. 1 Advanced Operating Systems - Spring 2009 Lecture 10 – February 11, 2009 Dan C. Marinescu Email: dcm@cs.ucf.edudcm@cs.ucf.edu Office: HEC 439 B. Office hours: M, Wd 3 – 4:30 PM. TA: Chen Yu Email: yuchen@cs.ucf.edu@cs.ucf.edu Office: HEC 354. Office hours: M, Wd 1.00 – 3:00 PM. Warning: A student missing classes takes a serious risks as the problems at exams require understanding of fine points discussed in class and problem analyzed in class.

2. 2 Last, Current, Next Lecture Last time: Deadlocks Today Problem solving More about deadlocks Consumable resources Elements of queuing theory Arrival and service processes Little’s law Problem Next time: M/M/1 systems

3. 3 Problem Write a procedure consisting of a single executable instruction that could cause the process calling this procedure to deadlock.

4. 4 Problem You have a system consisting of n processes and one non- sharable resource with m units. Show that a sufficient condition for the system to be deadlock-free is that the sum of the maximum amount of resources required by all processes is less than (m+n) units.

5. 5 Problem Two processes P1 and P2 share a resource with two units. Each process can be in one of five states: S0  no unit S1  no unit, request one unit S2  hold one unit S3  hold one unit request a second one S4  hold two units The following transitions may occur: T1  request one unit T2  acquire one unit T3  release one unit Construct a state diagram of the system and identify the states Sij with i,j in the range (0,4) when the system is deadlocked.

6. 6 Problem Prove that a necessary condition for deadlock is that a cycle exists in the resource allocation graph.

7. 7 Deadlocks and graph theory If a directed graph does not contain a cycle  there exists a linear ordering of nodes such that when there exits a path from node i to node j then node i appears in this ordering before node j

8. 8 Banker’s algorithm Multiple resource instances. Each process must a priori claim maximum use. When a process requests a resource it may have to wait gets all its resources it must return them in a finite amount of time.

9. 9 Data structures for banker’s algorithm Available: Vector of length m: Available [j] = k  there are k instances of resource type R j available. Max: n x m matrix: Max [i,j] = k  P i may request at most k instances of resource type R j. Allocation: n x m matrix: Allocation[i,j] = k  P i is currently allocated k instances of R j. Need: n x m matrix: Need[i,j] = k  P i may need k more instances of R j to complete its task. Need [i,j] = Max[i,j] – Allocation [i,j]. n  # of processes; m  # of resources types.

10. 10 Safety algorithm 1. Work and Finish are vectors of length m and n, respectively. Initialize: Work = Available Finish [i] = false for i = 0, 1, …, n- 1. 2.Find i such that: (a) Finish [i] = false (b) Need i  Work If no such i exists, go to step 4. 3. Work i = Work i + 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.

11. 11 Resource request algorithm for process P i If request vector Request i [j] = k then P i wants k instances of resource type j (R j. ) 1.If Request i  Need i go to step 2. Otherwise  error (process has exceeded its maximum claim). 2.If Request i  Available, go to step 3. Otherwise  P i must wait (resources are not available). 3.Pretend to allocate requested resources to P i by modifying the state as follows: Available = Available – Request; Allocation i = Allocation i + Request i ; Need i = Need i – Request i ; l If safe  the resources are allocated to P i l If unsafe  P i must wait, and the old resource- allocation state is restored

12. 12 Example 5 processes P 0 through P 4 ; 3 resource types: A (10 instances), B (5instances), and C (7 instances). Snapshot at time T 0 : AllocationMaxAvailable A B CA B C A B C P 0 0 1 07 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 24 3 3

13. 13 Example (cont’d) The content of the matrix Need is defined to be Max – Allocation. Need A B C P 0 7 4 3 P 1 1 2 2 P 2 6 0 0 P 3 0 1 1 P 4 4 3 1 The system is in a safe state since the sequence satisfies safety criteria.

14. 14 Example: P 1 Request (1,0,2) Check that Request  Available (that is, (1,0,2)  (3,3,2)  true. AllocationNeedAvailable A B CA B CA B C P 0 0 1 0 7 4 3 2 3 0 P 1 3 0 20 2 0 P 2 3 0 1 6 0 0 P 3 2 1 1 0 1 1 P 4 0 0 2 4 3 1 Executing safety algorithm shows that sequence satisfies safety requirement. Can request for (3,3,0) by P 4 be granted? Can request for (0,2,0) by P 0 be granted?

15. 15 Deadlock detection Allow system to enter deadlock state Detection algorithm Recovery scheme

16. 16 Single instance of each resource type 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. If there is a cycle, there exists a deadlock. 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.

17. 17 Resource allocation and wait-for graphs Resource-Allocation Graph Corresponding wait-for graph

18. 18 Multiple 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.

19. 19 Detection algorithm 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 4.

20. 20 Detection algorithm (cont’d) 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 state. Moreover, if Finish[i] == false, then P i is deadlocked. The algorithm requires O(m x n 2) operations to detect whether the system is in deadlocked state.

21. 21 Example 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 : AllocationRequestAvailable A B C A B C A B C P 0 0 1 0 0 0 0 0 0 0 P 1 2 0 0 2 0 2 P 2 3 0 30 0 0 P 3 2 1 1 1 0 0 P 4 0 0 2 0 0 2 Sequence will result in Finish[i] = true for all i.

22. 22 Example (Cont’d) P 2 requests an additional instance of type C. Request A B C P 0 0 0 0 P 1 2 0 1 P 2 0 0 1 P 3 1 0 0 P 4 0 0 2 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.

23. 23 Detection algorithm 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 invoked arbitrarily, there may be many cycles in the resource graph and so we would not be able to tell which of the many deadlocked processes “caused” the deadlock.

24. 24 Recovery from deadlock: 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 complete. How many processes will need to be terminated. Is process interactive or batch?

25. 25 Recovery from deadlock: resource 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.