Deadlocks & Deadlock Detection Main Memory Management - PDF document

CSE 421/521 - Operating Systems Roadmap Fall 2011 • Deadlocks Lecture - XII – Resource Allocation Graphs Deadlocks & – Deadlock Detection Main Memory Management – Deadlock Prevention – Deadlock Avoidance – Deadlock Recovery • Main Memory Management Tevfik Ko ş ar University at Buffalo October 11 th , 2011 1 2 Deadlock Avoidance Deadlock Avoidance Deadlock Prevention: prevent deadlocks by restraining resources and making sure one of 4 necessary conditions • Simplest and most useful model requires that for a deadlock does not hold. (system design) each process declare the maximum number of --> possible side effect: low device utilization and reduced resources of each type that it may need. system throughput • The deadlock-avoidance algorithm dynamically Deadlock Avoidance: Requires that the system has some examines the resource-allocation state to ensure additional a priori information available. (dynamic request that there can never be a circular-wait condition. check) • Resource-allocation state is defined by the number of available and allocated resources, and i.e. request disk and then printer.. the maximum demands of the processes. or request at most n resources --> allows more concurrency • Similar to the difference between a traffic light and a police officer directing the traffic! 3 4 Safe State Example • A state is safe if the system can allocate resources to each process (upto its maximum) in some order and P2: P1: can still avoid a deadlock. Request Printer Request Disk Request Disk Request Printer • When a process requests an available resource, system must decide if immediate allocation leaves the system in a safe state. .... .... Release Disk Release Printer • System is in safe state if there exists a safe sequence Release Printer Release Disk of all processes. 5 6

Safe State Basic Facts • Sequence <P 1 , P 2 , …, P n > is safe if for each P i , the • If a system is in safe state ⇒ no resources that P i can still request can be satisfied by deadlocks. currently available resources + resources held by all the P j , with j<i. • If a system is in unsafe state ⇒ possibility – If P i resource needs are not immediately available, then P i of deadlock. can wait until all P j have finished. – When P j is finished, P i can obtain needed resources, • Avoidance ⇒ ensure that a system will execute, return allocated resources, and terminate. – When P i terminates, P i +1 can obtain its needed resources, never enter an unsafe state. and so on. • If no such sequence exists, the state is unsafe! 7 8 Safe, Unsafe , Deadlock State Example Consider a system with 3 processes and 12 disks. At t = t0; Maximum Needs Current Allocation P1 10 5 P2 4 2 P3 9 2 9 10 Example (cont.) Resource-Allocation Graph Algorithm Consider a system with 3 processes and 12 disks. • Claim edge P i → R j indicated that process P j At t = t1; may request resource R j ; represented by a Maximum Needs Current Allocation dashed line. P1 10 5 • Claim edge converts to request edge when a process requests a resource. P2 4 2 • When a resource is released by a process, assignment edge reconverts to a claim edge. P3 9 3 • Resources must be claimed a priori in the system. 11 12

Unsafe State In Resource-Allocation Graph Resource-Allocation Graph For Deadlock Avoidance 13 14 Data Structures for the Banker’s Algorithm Banker’s Algorithm Let n = number of processes, and m = number of resources types. • Works for multiple resource instances. • Available: Vector of length m . If available [ j ] = k , there are k instances of resource type R j • Each process declares maximum # of resources it available. may need. • Max: n x m matrix. If Max [ i,j ] = k , then process P i may request at most k instances of resource • When a process requests a resource, it may have type R j . to wait if this leads to an unsafe state. • Allocation: n x m matrix. If Allocation[ i,j ] = k • When a process gets all its resources it must then P i is currently allocated k instances of R j. return them in a finite amount of time. • 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 ]. 15 16 Resource-Request Algorithm for Process P i Safety Algorithm Let Request i be the request vector for process P i . 1. Let Work and Finish be vectors of length m and n , If Request i [ j ] = k then process P i wants k instances of respectively. Initialize: resource type R j . Work = Available Finish [ i ] = false for i = 1,2, …, n. 1. If Request i ≤ Need i go to step 2. Otherwise, raise error 2. Find an i such that both: condition, since process has exceeded its maximum claim. (a) Finish [ i ] = false 2. If Request i ≤ Available , go to step 3. Otherwise P i must (b) Need i ≤ Work wait, since resources are not available. If no such i exists, go to step 4. 3. Pretend to allocate requested resources to P i by modifying the state as follows: 3. Work = Work + Allocation i Available = Available - Request i ; Finish [ i ] = true Allocation i = Allocation i + Request i ; go to step 2. Need i = Need i – Request i ; 4. If Finish [ i ] == true for all i , then the system is in a  If safe ⇒ the resources are allocated to Pi. safe state.  If unsafe ⇒ Pi must wait, and the old resource-allocation state is restored 17 18

Example of Banker’s Algorithm Example of Banker’s Algorithm • The content of the matrix. Need is defined to be • 5 processes P 0 through P 4 ; 3 resource types: Max – Allocation. A (10 instances), B (5 instances), and C (7 instances). Need • Snapshot at time T 0 : A B C Allocation Max Available P 0 7 4 3 A B C A B C A B C P 1 1 2 2 P 0 0 1 0 7 5 3 3 3 2 P 2 6 0 0 P 1 2 0 0 3 2 2 P 3 0 1 1 P 2 3 0 2 9 0 2 P 4 4 3 1 P 3 2 1 1 2 2 2 P 4 0 0 2 4 3 3 19 20 Example of Banker’s Algorithm Example of Banker’s Algorithm • Snapshot at time T 0 : • Snapshot at time T 0 : Allocation Max Available Need Allocation Max Available Need A B C A B C A B C A B C A B C A B C A B C A B C P 0 0 1 0 7 5 3 3 3 2 7 4 3 P 0 0 1 0 7 5 3 3 3 2 7 4 3 1 2 2 1 2 2 P 1 2 0 0 3 2 2 P 1 2 0 0 3 2 2 6 0 0 6 0 0 P 2 3 0 2 9 0 2 P 2 3 0 2 9 0 2 0 1 1 0 1 1 P 3 2 1 1 2 2 2 P 3 2 1 1 2 2 2 4 3 1 4 3 1 P 4 0 0 2 4 3 3 P 4 0 0 2 4 3 3 • The system is in a safe state since the sequence < P 1 , P 3 , P 4 , P 2 , P 0 > satisfies safety criteria. 21 22 Example: P 1 Requests (1,0,2) Recovery from Deadlock: Process Termination • Check that Request ≤ Available (that is, (1,0,2) ≤ (3,3,2) ⇒ true . • Abort all deadlocked processes. --> expensive Allocation Need Available • Abort one process at a time until the deadlock cycle is A B C A B C A B C eliminated. --> overhead of deadlock detection alg. P 0 0 1 0 7 4 3 2 3 0 P 1 3 0 2 0 2 0 • In which order should we choose to abort? P 2 3 0 1 6 0 0 – Priority of the process. – How long process has computed, and how much longer to P 3 2 1 1 0 1 1 completion. P 4 0 0 2 4 3 1 – Resources the process has used. – Resources process needs to complete. • Executing safety algorithm shows that sequence – How many processes will need to be terminated. <P1, P3, P4, P0, P2> satisfies safety requirement. – Is process interactive or batch? • Can request for (3,3,0) by P4 be granted? • Can request for (0,2,0) by P0 be granted? 23 24