Consider a system with five processes P0 through P4 and three resource types A, B and C, with 7, 2 and 6 total instances respectively. At time T0 the system state is:

Process	Allocation (A,B,C)			Request (A,B,C)
P0	0	1	0	0	0	0
P1	2	0	0	2	0	2
P2	3	0	3	0	0	0
P3	2	1	1	1	0	0
P4	0	2	2	0	0	2

Available = (0, 0, 0). Applying the deadlock detection algorithm, the system is:

Explanation

Running the detection algorithm with Work = Available = (0,0,0): P0's request (0,0,0) ≤ Work, so it can finish, releasing its allocation and raising Work to (0,1,0). P2's request (0,0,0) ≤ (0,1,0) finishes next, raising Work to (3,1,3); then P3's request (1,0,0) ≤ (3,1,3) finishes, raising Work to (5,2,4). Now P4's request (0,0,2) ≤ (5,2,4) finishes, raising Work to (5,4,6), and finally P1's request (2,0,2) ≤ (5,4,6) also finishes. Since the sequence lets every process obtain its request and complete, no process is stuck waiting forever, so the system is not in a deadlock state.