Deadlock Detection Algorithm in Operating System
If a system does not employ either a deadlock prevention or deadlock avoidance algorithm then a deadlock situation may occur. In this case-
- Apply an algorithm to examine state of system to determine whether deadlock has occurred or not.
- Apply an algorithm to recover from the deadlock. For more refer- Deadlock Recovery
Deadlock Avoidance Algorithm/ Bankers Algorithm:
The algorithm employs several times varying data structures:
- Available –
A vector of length m indicates the number of available resources of each type. - Allocation –
An n*m matrix defines the number of resources of each type currently allocated to a process. Column represents resource and rows represent process. - Request –
An n*m matrix indicates the current request of each process. If request[i][j] equals k then process Pi is requesting k more instances of resource type Rj.
This algorithm has already been discussed here
Now, Bankers algorithm includes a Safety Algorithm / Deadlock Detection Algorithm
The algorithm for finding out whether a system is in a safe state can be described as follows:
Steps of Algorithm:
- Let Work and Finish be vectors of length m and n respectively. Initialize Work= Available. For i=0, 1, …., n-1, if Requesti = 0, then Finish[i] = true; otherwise, Finish[i]= false.
- Find an index i such that both
a) Finish[i] == false
b) Requesti <= Work
If no such i exists go to step 4. - Work= Work+ Allocationi
Finish[i]= true
Go to Step 2. - If Finish[i]== false for some i, 0<=i<n, then the system is in a deadlocked state. Moreover, if Finish[i]==false the process Pi is deadlocked.
For example,
- In this, Work = [0, 0, 0] &
Finish = [false, false, false, false, false]
- i=0 is selected as both Finish[0] = false and [0, 0, 0]<=[0, 0, 0].
- Work =[0, 0, 0]+[0, 1, 0] =>[0, 1, 0] &
Finish = [true, false, false, false, false].
- i=2 is selected as both Finish[2] = false and [0, 0, 0]<=[0, 1, 0].
- Work =[0, 1, 0]+[3, 0, 3] =>[3, 1, 3] &
Finish = [true, false, true, false, false].
- i=1 is selected as both Finish[1] = false and [2, 0, 2]<=[3, 1, 3].
- Work =[3, 1, 3]+[2, 0, 0] =>[5, 1, 3] &
Finish = [true, true, true, false, false].
- i=3 is selected as both Finish[3] = false and [1, 0, 0]<=[5, 1, 3].
- Work =[5, 1, 3]+[2, 1, 1] =>[7, 2, 4] &
Finish = [true, true, true, true, false].
- i=4 is selected as both Finish[4] = false and [0, 0, 2]<=[7, 2, 4].
- Work =[7, 2, 4]+[0, 0, 2] =>[7, 2, 6] &
Finish = [true, true, true, true, true].
- Since Finish is a vector of all true it means there is no deadlock in this example.
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