Operating System Test 4

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Operating System Test 4

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Weekly Quiz Competition

Question 1
1 / -0

Consider a system having 'N' resources of the same type. X, Y and Z are the three processes that shared the resource. The peak demand of X, Y and Z are 5, 11, and 9 respectively. What is the maximum value of 'N' to ensures the system is in deadlock?

A
20

B
21

C
22

D
23

Solution

Data:
number of processes = 3
The available instance of Resources = N
Max needs of process (X, Y, Z) = (5, 11, 9)
Calculation
Deadlock can occur: needed (requested) resource > available resources
Max resource per process to be in deadlock = needed – 1.
N_max= (5 - 1) + (11 - 1) + (9 - 1) = 4 + 10 + 8 = 22
Question 2
1 / -0

The total number of instances of a resource is 11. Four processes A, B, C, and D are present in the system. The allocation table of the resources is given below.
Process
Max. Need
Current Allocation
D
9
1
C
5
4
B
10
3
A 7 2

Which of the following orders of process execution forms a deadlock-free sequence?

A
<C, A, D, B>

B
<A, C, B, D>

C
<C, A, B, D>

D
<C, B, A, D>

Solution

Process
Max. Need
Current Allocation
Need = Max – current allocation
D
9
1
8
C
5
4
1
B
10
3
7
A 7 2 5

Total resources that are currently allocated = 1 + 4 + 3 + 2 = 10
Total available = 11
Current available = 11 – 10 = 1 resources
With 1 resource, we can satisfy the need of C.
After this, resources allocated to C added to current available.
New current available = 1 + 4 = 5
With 5 resources we can fulfil the need of A
new current available = 5 + 2 = 7,
With 7 resources need of B can be easily satisfied.
new current available = 7 + 3 = 10,
With 10 resources need of D can be easily satisfied.
So, the correct order is <C, A, B, D>
Question 3
1 / -0
Which of the below-given statement is/are true?
1. Deadlock is waiting for a resource from a process that will never release it resource.
2. Banker’s Algorithm is used to avoid deadlock in a system.
A
Only I

B
Only II

C
Both I and II

D
None of these
Solution
- Deadlock is infinite waiting in which process is waiting for an event that will never occur
- Banker’s algorithm detects the unsafe state and hence prevent the system in going to deadlock state
Question 4
1 / -0

Consider a system in which there are P processes. A₁, A₂, … A_P is the processes that share K identical resource units, which can be reserved and release at a time. The peak requirement per process is X_i. Which of the following condition is necessary to say that the system is in deadlock state?(Assume X_i > 0)

A
\(\mathop \sum \limits_{i = 1}^P {X_i} + 1 \ge K + P\)

B
\(\mathop \sum \limits_{i = 1}^P {X_i} \le K - P\)

C
\(\mathop \sum \limits_{i = 1}^P {X_i} \ge K + P\)

D
\(\mathop \sum \limits_{i = 1}^P {X_i} \le K + P\)

Solution

Process
A1
A2
A3
A4
…
..
Ap
Maximum need
X1
X2
X3
X4
..
…
Xn
Total resource = K
For the system to be in a deadlock
(X1 – 1) + (X2 – 1) + (X3 – 1) + (X4 – 1)... (Xp – 1) ≥ K
\(\mathop \sum \limits_{i = 1}^P {X_i} - P \ge K\)
\(\mathop \sum \limits_{i = 1}^P {X_i} \ge K + P\)
Question 5
1 / -0

An operating system uses the Banker’s algorithm for deadlock avoidance when managing the allocation of three resource types A, B, and C to four processes P0, P1, P2, and P3.
Process
Allocation
Max
Available

A
B
C
A
B
C
A
B
C
P0
4
3
3
5
4
4
3
2
2
P1
1
2
0
3
2
0
P2
3
0
1
6
0
2
P3
1
1
0
2
1
1

Which of the following best describes the current state of the system?

A
Safe, Deadlocked

B
Safe, Not deadlocked

C
Not Safe, Deadlocked

D
Not Safe, Not deadlocked

Solution

Need = Max – Allocation
Process
Need

A
B
C
P0
1
1
1
P1
2
0
0
P2
3
0
1
P3
1
0
1
Here, Need < available. Hence, the system is in the safe state and not deadlocked.
Question 6
1 / -0

Which of the following statements are True?
I. All unsafe states are deadlocks.
II. A deadlock-avoidance algorithm dynamically examines the resource-allocation state to ensure that a circular-wait condition can never exist.
III. To eliminate deadlock if a process and resources are selected for preemption on the basis of cost leads to starvation.

A
I

B
I and II

C
II and III

D
All of the above

Solution

1. False: A deadlocked state is an unsafe state. Not all unsafe states are deadlocks, however an unsafe state may lead to a deadlock.
2. True: A method for avoiding deadlocks, requires that the operating system have a priori information about how each process will utilize system resources. The resource allocation state is deﬁned by the number of available and allocated resources and the maximum demands of the processes.
3. True: In a system that selects victims i.e. processes and resources for rollback primarily on the basis of cost factors, starvation may occur, and the selected process can never complete its designated task.
Question 7
1 / -0

In a computer system, 28 processes are present and they share R resources of the same type. Each process can request a maximum of 5 resources. What is the minimum value of R such that no deadlock occurs?

A
113-113

B
1134-1131

C
1137-1136

D
1139-1132

Solution

Data:
number of process = n = 28
The available instance of Resources = R
Max needs per process = 5
Concepts:
Deadlock can occur: needed (requested) resource > available resources
Max resource per process to be in deadlock = needed – 1 = 5 – 1 = 4
For n process, max resource to be in deadlock = 4 × n = 4n
If one instance is added, any of the n processes can take it a finish its execution leaving behind the allocated resource, and hence the system will be deadlock-free.
Formula:
For n process, min resource for the system not to be in deadlock state = 4n + 1 ≤ R
Calculation:
4×(28) + 1 ≤ R
113 ≤ R
Tips and Tricks:
n × (m - 1) + 1 ≤ R
where n is the number of processes
m is the maximum resource need per process
R is the available system resources.
Question 8
1 / -0

An operating system uses the banker's algorithm for deadlock avoidance to manage the allocation of four resources A, B, C, and D. The table given below represents the current system state.

Allocation
Max

A
B
C
D
A
B
C
D
P1
1
2
1
0
1
3
1
2
P2
3
1
2
1
5
3
2
1
P3
4
2
3
0
7
2
3
4
P4
0
0
1
5
1
1
3
6

There are 3 units of type B, 2 units of type D still available. Which of the following sequence is a safe sequence?

A
P3 → P2 → P1 → P4

B
P1 → P4 → P2 → P3

C
P2 → P3 → P1 → P4

D
No safe sequence available

Solution

Available = {0, 3, 0, 2}

Allocation
Max
Needed

A
B
C
D
A
B
C
D
A
B
C
D
P1
1
2
1
0
1
3
1
2
0
1
0
2
P2
3
1
2
1
5
3
2
1
2
2
0
0
P3
4
2
3
0
7
2
3
4
3
0
0
4
P4
0
0
1
5
1
1
3
6
1
1
2
1

The given system is not in a safe state.
Question 9
1 / -0

Which of the following is/are not true about the deadlock scheme?
I. In deadlock avoidance, number the resource uniquely and never requests a lower-numbered resource
II. In deadlock avoidance, the request for the resources is always granted if the resulting state is safe.
III. In deadlock prevention, never request a resource after releasing any resource.

A
Only II

B
Only II and III

C
Only I and III

D
Only I, II and III

Solution

Statement I: not true (False)
In deadlock prevention, number the resource uniquely and never requests a lower-numbered resource
Statement II: True
In deadlock avoidance, the request for the resources is always granted if the resulting state is safe.
Statement IIi: not true (False)
In deadlock prevention, the request can be made after releasing the resource
Statement I and III are not true

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Process	Max. Need	Current Allocation	Need = Max – current allocation
D	9	1	8
C	5	4	1
B	10	3	7
A	7	2	5

Process	Allocation			Max			Available
	A	B	C	A	B	C	A	B	C
P0	4	3	3	5	4	4	3	2	2
P1	1	2	0	3	2	0
P2	3	0	1	6	0	2
P3	1	1	0	2	1	1

	Allocation				Max				Needed
	A	B	C	D	A	B	C	D	A	B	C	D
P1	1	2	1	0	1	3	1	2	0	1	0	2
P2	3	1	2	1	5	3	2	1	2	2	0	0
P3	4	2	3	0	7	2	3	4	3	0	0	4
P4	0	0	1	5	1	1	3	6	1	1	2	1

Process	A1	A2	A3	A4	…	..	Ap
Maximum need	X1	X2	X3	X4	..	…	Xn

Operating System Test 4

Result

Operating System Test 4

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SHARING IS CARING

Question 1

20

21

22

23

Solution

Question 2

<C, A, D, B>

<A, C, B, D>

<C, A, B, D>

<C, B, A, D>

Solution

Question 3

Only I

Only II

Both I and II

None of these

Solution

Question 4

\(\mathop \sum \limits_{i = 1}^P {X_i} + 1 \ge K + P\)

\(\mathop \sum \limits_{i = 1}^P {X_i} \le K - P\)

\(\mathop \sum \limits_{i = 1}^P {X_i} \ge K + P\)

\(\mathop \sum \limits_{i = 1}^P {X_i} \le K + P\)

Solution

Question 5

Safe, Deadlocked

Safe, Not deadlocked

Not Safe, Deadlocked

Not Safe, Not deadlocked

Solution

Question 6

I

I and II

II and III

All of the above

Solution

Question 7

113-113

1134-1131

1137-1136

1139-1132

Solution

Question 8

P3 → P2 → P1 → P4

P1 → P4 → P2 → P3

P2 → P3 → P1 → P4

No safe sequence available

Solution

Question 9

Only II

Only II and III

Only I and III

Only I, II and III

Solution

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