Classical

Operating System - Deadlocks

1:  

With a single resource, deadlock occurs

A.

if there are more than two processes competing for that resource

B.

if there are only two processes competing for that resource

C.

if there is a single process competing for that resource

D.

none of these

 
 

Option: D

Explanation :

Deadlock doesn't occurs with a single resource

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Nikhil said: (11:33pm on Thursday 6th June 2013)
why option D?with two or more process accessing one resource may leads to deadlock or not?
Yatendra Dagur said: (7:33am on Monday 10th June 2013)
option A is right because deadlock occur when more than one process requesting single resource.
chandru said: (10:06pm on Wednesday 26th June 2013)
correct answer since there is no cycle formation deadloclk not occur
chandan kumar said: (4:22pm on Friday 5th July 2013)
it correct answer because single resource can not forme cycle
Priyanka said: (2:02pm on Friday 21st March 2014)
Yes,no cycle formation is necessary and sufficient condition for no deadlock.So,option d is correct.
aj said: (1:39am on Tuesday 8th November 2016)
Deadlock not occur when single resource is available.It occurs only when more than one resources available

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2:  

A state is safe if the system can allocate resources to each process (up to its maximum) in some order and still avoid deadlock. Then

A.

deadlocked state is unsafe

B.

unsafe state may lead to a deadlock situation

C.

deadlocked state is a subset of unsafe state

D.

all of these

 
 

Option: D

Explanation :

Safe State

  • A system is in a safe state only if there exists a safe sequence of processes P1, P2, P3, Pn where:

    • For each Pi, the resources that Pi can still request can be satisfied by the currently available resources plus the resources help by all Pj, j<i.

  • If a system is in safe state, there is no deadlock.

  • If the system is deadlocked, it is in an unsafe state.

  • If a system is in unsafe state, there is a possibility for a deadlock.

  • Avoidance: making sure the system will not enter an unsafe state.

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3:  

A computer system has 6 tape drives, with 'n' processes competing for them. Each process may need 3 tape drives. The maximum value of 'n' for which the system is guaranteed to be deadlock free is

A.

4

B.

3

C.

2

D.

1

 
 

Option: C

Explanation :

Two processes can never lead to deadlock as the peak time demand of 6 (3 + 3) tape drives can be satisfied. But 3 processes can lead to a deadlock if each hold 2 drives and then demand one more.

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4:  

'm' processes share 'n' resources of the same type. The maximum need of each process doesn't exceed 'n' and the sum all the their maximum needs is always less than m + n. In this set up

A.

deadlock can never occur

B.

deadlock may occur

C.

deadlock has to occur

D.

none of these

 
 

Option: A

Explanation :

Using Banker's algorithm, one can show that one process has to acquire all its needed resources. This process, after completing its task, will release all its resources, thereby avoiding any possible deadlock.

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5:  

Consider a system having 'm' resources of the same type. These resources are shared by 3 processes A, B, C, which have peak time demands of 3, 4, 6 respectively. The minimum value of 'm' that ensures that deadlock will never occur is

A.

11

B.

12

C.

13

D.

14

 
 

Option: A

Explanation :

Having 11 resources ensures that atleast 1 process will have no pending request. This process after using will release the resources and so deadlock can never occur.

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mushi,oscar said: (12:18pm on Monday 20th May 2013)
option c having 13 resources all process will be satisfied,hence no deadlock occur
mushi,oscar said: (12:20pm on Monday 20th May 2013)
option c having 13 resources all process will be satisfied,hence no deadlock occur
chandan kumar said: (4:35pm on Friday 5th July 2013)
correct answer because for deadlock every procees must be wait for at least one other procees so that it can be able to form cycle hence answer 11
Amit Ahlawat said: (4:05am on Thursday 11th July 2013)
if 2,3, and 5 resources allocated to each process have to wait.. after getting 1 more atleast 1 process satisfied so minimam 11
balvant said: (12:08am on Monday 12th October 2015)
the deadlock occurs for 11 because a and b have 3 and 4 hence c have to wait to release a or b

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Syllabus covered in this section is-

  • Main functions of operating systems
  • Multiprogramming, multiprocessing and multitasking
  • Memory Management- Virtual memory, paging, fragmentation
  • Concurrent Processing: Mutual exclusion, Critical regions, locks and unlock.
  • Scheduling: CPU scheduling. I/O scheduling, Resource scheduling
  • Deadlock and scheduling algorithms.
  • Banker's algorithm for deadlock handling.

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