Operating Systems
1
A Case Study
On
Deadlock
4th
CSE ‘S’
Prepared by
Yash P. Dhol
Asimhusen N.Saiyad
Jay Pandya
Submitted ...
Operating Systems
2“Crises and deadlocks when they occur have at least this advantage that
they force us to think.”- Jawah...
Operating Systems
3
Explanation: There must exist a process that is holding a resource already
allocated to it while waiti...
Operating Systems
4
That means no Deadlock occurs..
Operating Systems
5
Ways of Handling Deadlock:
• Deadlock Prevention
• Deadlock Detection
• Deadlock Recovery
• Deadlock A...
Operating Systems
6
Removing the mutual exclusion condition means that no process
will have exclusive access to a resource...
Operating Systems
7
1. Process A holds R and wants S.
2. Process B holds nothing but wants T.
3. Process C holds nothing b...
Operating Systems
8
– choose process that can be rerun from the beginning
Deadlock Avoidance:
This approach to the deadloc...
Operating Systems
9
Unsafe State Consider what would happen if a request from B for one more
unit were grantedin above We ...
Operating Systems
10
The Banker’s Algoritham for Multiple Resources:
E = (6, 3, 4, 2) P = (5, 3, 2, 2) A = (1, 0, 2, 0)
No...
Operating Systems
11
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  1. 1. Operating Systems 1 A Case Study On Deadlock 4th CSE ‘S’ Prepared by Yash P. Dhol Asimhusen N.Saiyad Jay Pandya Submitted to Prof. Nikul Virapariya
  2. 2. Operating Systems 2“Crises and deadlocks when they occur have at least this advantage that they force us to think.”- Jawaharlal Nehru (1889 - 1964) Definition: “A set of processes is deadlocked if each process in the set is waiting for an event that only another process in the set can cause”. An example from US Kansas law: “When two trains approach each other at a crossing, both shall come to a full stop and neither shall start up again until the other has gone.” For another example, Two processes each want to record a scanned document on a CD.Process A requests permission to use the scanner and is granted it.Process B is programmed differently and requests the CD recorder first and is also granted it.Now A asks for the CD recorder, but the request is denied until B releases it.Unfortunately, instead of releasing the CD recorder B asks for the scanner.At this point both processes are blocked and will remain so forever. This situation is called a deadlock. Deadlocks can occur in a variety of situations besides requesting dedicated I/O devices. For example If process A locks record R1 and process B locks record R2, and then each process tries to lock the other one's record, we also have a deadlock. Thus deadlocks can occur on hardware resources or on software resources. Preemptable and Nonpreemptable Resources Resources come in two flavors: Preemptable and Nonpreemptable. A preemptable resource is one that can be taken away from the process with no ill effects. Memory is an example of a preemptable resource. On the other hand, a nonpreemptable resource is one that cannot be taken away from process (without causing ill effect). For example, CD resources are no preemptable at an arbitrary moment. Reallocating resources can resolve deadlocks that involve preemptable resources. Deadlocks that involve nonpreemptable resources are difficult to deal with. Necessary Conditions for Deadlock: Mutual Exclusion Condition: Explanation: At least one resource (thread) must be held in a non-shareable mode, that is, only one process at a time claims exclusive control of the resource.If another process requests that resource, the requesting process must be delayed until the resource has been released. Hold and Wait Condition:
  3. 3. Operating Systems 3 Explanation: There must exist a process that is holding a resource already allocated to it while waiting for additional resource that are currently being held by other processes. No-Preemptive Condition: Explanation: Resources cannot be removed from the processes are used to completion or released voluntarily by the process holding it. Circular Wait Condition: Explanation: There exists a set {T1, …, Tn} of waiting threads • T1 is waiting for a resource that is held by T2 • T2 is waiting for a resource that is held by T3 • … • Tn is waiting for a resource that is held by T1 T1 T2 R2 R1
  4. 4. Operating Systems 4 That means no Deadlock occurs..
  5. 5. Operating Systems 5 Ways of Handling Deadlock: • Deadlock Prevention • Deadlock Detection • Deadlock Recovery • Deadlock Avoidance Deadlock Prevention: Remove the possibility of deadlock occurring by denying one of the four necessary conditions: – Mutual Exclusion (Can we share everything?)
  6. 6. Operating Systems 6 Removing the mutual exclusion condition means that no process will have exclusive access to a resource. – Hold & Wait The hold & wait or resource holding conditions may be removed by requiring processes to request all the resources they will need before starting up. – No preemption The No preemption condition may also be difficult or impossible to avoid as a process has to be able to have a resource for a certain amount of time, or the processing outcome may be inconsistent. However, inability to enforce preemption may interfere with a priority algorithm. – Circular Wait The final condition is the circular wait condition. Approaches that avoid circular waits include disabling interrupts during critical sections. Deadlock Detection: Deadlock Detection with One Resource of Each Type
  7. 7. Operating Systems 7 1. Process A holds R and wants S. 2. Process B holds nothing but wants T. 3. Process C holds nothing but wants S. 4. Process D holds U and wants V. 5. Process E holds T and wants V. 6. Process F holds W and wants S. 7. Process G holds V wants U. Deadlock Recovery: • Recovery through preemption – take a resource from some other process • Recovery through rollback – checkpoint a process periodically – use this saved state – restart the process if it is found deadlocked • Recovery through killing processes – crudest but simplest way to break a deadlock – kill one of the processes in the deadlock cycle – the other processes get its resources
  8. 8. Operating Systems 8 – choose process that can be rerun from the beginning Deadlock Avoidance: This approach to the deadlock problem anticipates deadlock before it actually occurs. This method differs from deadlock prevention, which guarantees that deadlock cannot occur by denying one of the necessary conditions of deadlock. If the necessary conditions for a deadlock are in place, it is still possible to avoid deadlock by being careful when resources are allocated. Perhaps the most famous deadlock avoidance algorithm, due to Dijkstra [1965], is the Banker’s algorithm. So named because the process is analogous to that used by a banker in deciding if a loan can be safely made. Safe State The key to a state being safe is that there is at least one way for all users to finish. In other analogy, the state of figure 2 is safe because with 2 units left, the banker can delay any request except C's, thus letting C finish and release all four resources. With four units in hand, the banker can let either D or B have the necessary units and so on. Customers Used Max A 1 6 B 1 5 C 2 4 D 4 7 Available Units= 2
  9. 9. Operating Systems 9 Unsafe State Consider what would happen if a request from B for one more unit were grantedin above We would have following situation Available Units = 1 Customers Used Max A 1 6 B 2 5 C 2 4 D 4 7 This is an unsafe state. If all the customers namely A, B, C, and D asked for their maximum loans, then banker could not satisfy any of them and we would have a deadlock. • Three resource allocation states – safe – safe – unsafe
  10. 10. Operating Systems 10 The Banker’s Algoritham for Multiple Resources: E = (6, 3, 4, 2) P = (5, 3, 2, 2) A = (1, 0, 2, 0) Notice that A = E - P Going thru this algorithm with the foregoing data, we see that process D's requirements are smaller than A, so we virtually terminate D and add its resources back into the available pool: E = (6, 3, 4, 2) P = (5, 3, 2, 2) - (1, 1, 0, 1) = (4, 2, 2, 1) A = (1, 0, 2, 0) + (1, 1, 0, 1) = (2, 1, 2, 1) Now, A's requirements are less than A, so do the same thing with A: P = (4, 2, 2, 1) - (3, 0, 1, 1) = (1, 2, 1, 0) A = (2, 1, 2, 1) + (3, 0, 1, 1) = (5, 1, 3, 2) At this point, we see that there are no remaining processes that can't be satisfied from available resources, so the illustrated state is safe.
  11. 11. Operating Systems 11 Thank you

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