Consider a state of the system with the Allocation matrix as shown below, and in which 3 instances of E and 3 instances of F are the only resources available.AllocationMaxEFGEFGP0101431P1112214P2103133P3200541From the perspective of deadlock avoidance, which one of the following is true?

The system is in a safe state.Need = Max − Allocation gives P0 (3, 3, 0), P1 (1, 0, 2), P2 (0, 3, 0) and P3 (3, 4, 1), with (3, 3, 0) available. The banker's safety check finds an order in which every process can finish:P0 needs (3, 3, 0), which is available. It finishes and releases (1, 0, 1): available becomes (4, 3, 1).P2 needs (0, 3, 0). It finishes and releases (1, 0, 3): available becomes (5, 3, 4).P1 needs (1, 0, 2). It finishes and releases (1, 1, 2): available becomes (6, 4, 6).P3 needs (3, 4, 1). It finishes.The safe sequence P0, P2, P1, P3 exists, so the state is safe and none of the options claiming it is unsafe can be right, whichever extra resource they add.

Deadlocks Questions and Answers

Q: Let m[0] ...... m[4] be mutexes (binary semaphores) and P[0] ........ P[4] be processes. Suppose each process P[i] executes the following :wait (m[i]); wait (m[(i+1) mod 4]);.......................release (m[i]); release (m[(i+1) mod 4]);This could be :

This could lead to Deadlock.Process P[i] locks m[i] and then m[(i+1) mod 4]. For P[0] to P[3] this gives a circular chain: P[0] holds m[0] and wants m[1], P[1] holds m[1] and wants m[2], P[2] holds m[2] and wants m[3], and P[3] holds m[3] and wants m[0]. If all four acquire their first mutex before any gets its second, each waits for the next forever: hold-and-wait with circular wait, a deadlock.Thrashing is excessive paging and has nothing to do with mutexes, and since a deadlock is possible, 'starvation but not deadlock' and 'none of the above' are wrong.

Q: Consider a system with seven processes A through G and six resources R through W. Resource ownership is as follows :Process A holds R and wants TProcess B holds nothing but wants TProcess C holds nothing but wants SProcess D holds U and wants S and TProcess E holds T and wants VProcess F holds W and wants SProcess G holds V and wants UAre there any processes deadlocked ?

Yes, D, E and G are deadlocked.Follow who waits for whom. D holds U and wants T; T is held by E. E wants V, held by G. G wants U, held by D. This gives the cycle D, E, G, D, and since each resource has a single instance, the three are deadlocked.A and B also want T, so they are blocked waiting for E, but they are not part of the cycle. C and F want S, which nobody holds, so they can proceed. So the processes involved in the deadlock are D, E and G, which rules out 'no process' and the sets containing A, B, C or F.

Q: An operating system contains 3 user processes each requiring 2 unit of resources 'R'. The minimum number of units of 'r' such that no deadlocks will ever arise is :

The minimum is 4 units.With n processes each needing at most k units, deadlock is impossible if there are at least \(n(k - 1) + 1\) units. In the worst case each process holds k - 1 units and waits for one more; one extra unit lets one process finish and release its resources.\[ 3(2 - 1) + 1 = 4 \]With 3 units, each process can hold one and wait forever for a second, which is a deadlock. Five or six units also avoid deadlock but are not the minimum.

Direction for [ Question No: 1 To 1 ] :

In a system, there are three types of resources: E, F and G. Four processes P₀, P₁, P₂ and P₃ execute concurrently. At the outset, the processes have declared their maximum resource requirements using a matrix named Max as given below. For example, Max[P₂,F] is the maximum number of instances of F that P₂ would require. The number of instances of the resources allocated to the various processes at any given state is given by a matrix named Allocation.

Consider a state of the system with the Allocation matrix as shown below, and in which 3 instances of E and 3 instances of F are the only resources available.

	Allocation			Max
	E	F	G	E	F	G
P₀	1	0	1	4	3	1
P₁	1	1	2	2	1	4
P₂	1	0	3	1	3	3
P₃	2	0	0	5	4	1

From the perspective of deadlock avoidance, which one of the following is true?

A.The system is in safe state.
B.The system is not in safe state, but would be safe if one more instance of E were available
C.The system is not in safe state, but would be safe if one more instance of F were available
D.The system is not in safe state, but would be safe if one more instance of G were available

Exam Year(s)

GATE CS - 2018,

Answer: A

The system is in a safe state.

Need = Max − Allocation gives P0 (3, 3, 0), P1 (1, 0, 2), P2 (0, 3, 0) and P3 (3, 4, 1), with (3, 3, 0) available. The banker's safety check finds an order in which every process can finish:

P0 needs (3, 3, 0), which is available. It finishes and releases (1, 0, 1): available becomes (4, 3, 1).
P2 needs (0, 3, 0). It finishes and releases (1, 0, 3): available becomes (5, 3, 4).
P1 needs (1, 0, 2). It finishes and releases (1, 1, 2): available becomes (6, 4, 6).
P3 needs (3, 4, 1). It finishes.

The safe sequence P0, P2, P1, P3 exists, so the state is safe and none of the options claiming it is unsafe can be right, whichever extra resource they add.

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Let m[0] ...... m[4] be mutexes (binary semaphores) and P[0] ........ P[4] be processes. Suppose each process P[i] executes the following :

wait (m[i]); wait (m[(i+1) mod 4]);
.......................
release (m[i]); release (m[(i+1) mod 4]);

This could be :

A.Thrashing
B.Deadlock
C.Starvation but not deadlock
D.None of the above

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

This could lead to Deadlock.

Process P[i] locks m[i] and then m[(i+1) mod 4]. For P[0] to P[3] this gives a circular chain: P[0] holds m[0] and wants m[1], P[1] holds m[1] and wants m[2], P[2] holds m[2] and wants m[3], and P[3] holds m[3] and wants m[0]. If all four acquire their first mutex before any gets its second, each waits for the next forever: hold-and-wait with circular wait, a deadlock.

Thrashing is excessive paging and has nothing to do with mutexes, and since a deadlock is possible, 'starvation but not deadlock' and 'none of the above' are wrong.

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Consider a system with seven processes A through G and six resources R through W. Resource ownership is as follows :

Process A holds R and wants T
Process B holds nothing but wants T
Process C holds nothing but wants S
Process D holds U and wants S and T
Process E holds T and wants V
Process F holds W and wants S
Process G holds V and wants U

Are there any processes deadlocked ?

A.No process
B.Yes, A, B, C
C.Yes, D, E, G
D.Yes, A, B, F

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

Yes, D, E and G are deadlocked.

Follow who waits for whom. D holds U and wants T; T is held by E. E wants V, held by G. G wants U, held by D. This gives the cycle D, E, G, D, and since each resource has a single instance, the three are deadlocked.

A and B also want T, so they are blocked waiting for E, but they are not part of the cycle. C and F want S, which nobody holds, so they can proceed. So the processes involved in the deadlock are D, E and G, which rules out 'no process' and the sets containing A, B, C or F.

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An operating system contains 3 user processes each requiring 2 unit of resources 'R'. The minimum number of units of 'r' such that no deadlocks will ever arise is :

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The minimum is 4 units.

With n processes each needing at most k units, deadlock is impossible if there are at least \(n(k - 1) + 1\) units. In the worst case each process holds k - 1 units and waits for one more; one extra unit lets one process finish and release its resources.

\[ 3(2 - 1) + 1 = 4 \]

With 3 units, each process can hold one and wait forever for a second, which is a deadlock. Five or six units also avoid deadlock but are not the minimum.

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