Tag: Deadlock Detection

Banker’s Algorithm | Deadlock Avoidance

Banker’s Algorithm-

 

Before you go through this article, make sure that you gone through the previous article on Banker’s Algorithm.

 

We have discussed-

  • Banker’s Algorithm is a deadlock avoidance algorithm.
  • It maintains a set of data using which it decides whether to entertain the request of any process or not.
  • It follows the safety algorithm to check whether the system is in a safe state or not.

 

 

Also read-Deadlock Handling Strategies

 

PRACTICE PROBLEMS BASED ON BANKER’S ALGORITHM-

 

Problem-01:

 

A single processor system has three resource types X, Y and Z, which are shared by three processes. There are 5 units of each resource type. Consider the following scenario, where the column alloc denotes the number of units of each resource type allocated to each process, and the column request denotes the number of units of each resource type requested by a process in order to complete execution. Which of these processes will finish LAST?

  1. P0
  2. P1
  3. P2
  4. None of the above since the system is in a deadlock

 

AllocRequest
XYZXYZ
P0121103
P1201012
P2221120

 

Solution-

 

According to question-

  • Total = [ X Y Z ] = [ 5 5 5 ]
  • Total _Alloc = [ X Y Z ] = [5 4 3]

 

Now,

Available

= Total – Total_Alloc

= [ 5 5 5 ] – [5 4 3]

= [ 0 1 2 ]

 

Step-01:

 

  • With the instances available currently, only the requirement of the process P1 can be satisfied.
  • So, process P1 is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then,

Available

= [ 0 1 2 ] + [ 2 0 1]

= [ 2 1 3 ]

 

Step-02:

 

  • With the instances available currently, only the requirement of the process P0 can be satisfied.
  • So, process P0 is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 2 1 3 ] + [ 1 2 1 ]

= [ 3 3 4 ]

 

Step-03:

 

  • With the instances available currently, the requirement of the process P2 can be satisfied.
  • So, process P2 is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 3 3 4 ] + [ 2 2 1 ]

= [ 5 5 5 ]

 

Thus,

  • There exists a safe sequence P1, P0, P2 in which all the processes can be executed.
  • So, the system is in a safe state.
  • Process P2 will be executed at last.

 

Thus, Option (C) is correct.

 

Problem-02:

 

An operating system uses the banker’s algorithm for deadlock avoidance when managing the allocation of three resource types X, Y and Z to three processes P0, P1 and P2. The table given below presents the current system state. Here, the Allocation matrix shows the current number of resources of each type allocated to each process and the Max matrix shows the maximum number of resources of each type required by each process during its execution.

AllocationMax
XYZXYZ
P0001843
P1320620
P2211333

 

There are 3 units of type X, 2 units of type Y and 2 units of type Z still available. The system is currently in safe state. Consider the following independent requests for additional resources in the current state-

 

REQ1: P0 requests 0 units of X, 0 units of Y and 2 units of Z

REQ2: P1 requests 2 units of X, 0 units of Y and 0 units of Z

 

Which of the following is TRUE?

  1. Only REQ1 can be permitted
  2. Only REQ2 can be permitted
  3. Both REQ1 and REQ2 can be permitted
  4. Neither REQ1 nor REQ2 can be permitted

 

Solution-

 

According to question,

Available = [ X Y Z ] = [ 3 2 2 ]

Now,

Need = Max – Allocation

So, we have-

 

AllocationMaxNeed
XYZXYZXYZ
P0001843842
P1320620300
P2211333122

 

Currently, the system is in safe state.

(It is given in question. If we want, we can check)

 

Checking Whether REQ1 Can Be Entertained-

 

  • Need of P0 = [ 0 0 2 ]
  • Available = [ 3 2 2 ]

 

Clearly,

  • With the instances available currently, the requirement of REQ1 can be satisfied.
  • So, banker’s algorithm assumes that the request REQ1 is entertained.
  • It then modifies its data structures as-

 

AllocationMaxNeed
XYZXYZXYZ
P0003843840
P1320620300
P2211333122

 

Available

= [ 3 2 2 ] – [ 0 0 2 ]

= [ 3 2 0 ]

 

  • Now, it follows the safety algorithm to check whether this resulting state is a safe state or not.
  • If it is a safe state, then REQ1 can be permitted otherwise not.

 

Step-01:

 

  • With the instances available currently, only the requirement of the process P1 can be satisfied.
  • So, process P1 is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 3 2 0 ] + [ 3 2 0 ]

= [ 6 4 0 ]

 

Now,

  • It is not possible to entertain any process.
  • The system has entered the deadlock state which is an unsafe state.
  • Thus, REQ1 will not be permitted.

 

Checking Whether REQ2 Can Be Entertained-

 

  • Need of P1 = [ 2 0 0 ]
  • Available = [ 3 2 2 ]

 

Clearly,

  • With the instances available currently, the requirement of REQ1 can be satisfied.
  • So, banker’s algorithm assumes the request REQ2 is entertained.
  • It then modifies its data structures as-

 

AllocationMaxNeed
XYZXYZXYZ
P0001843842
P1520620100
P2211333122

 

Available

= [ 3 2 2 ] – [ 2 0 0 ]

= [ 1 2 2 ]

 

  • Now, it follows the safety algorithm to check whether this resulting state is a safe state or not.
  • If it is a safe state, then REQ2 can be permitted otherwise not.

 

Step-01:

 

  • With the instances available currently, only the requirement of the process P1 can be satisfied.
  • So, process P1 is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 1 2 2 ] + [ 5 2 0 ]

= [ 6 4 2 ]

 

Step-02:

 

  • With the instances available currently, only the requirement of the process P2 can be satisfied.
  • So, process P2 is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 6 4 2 ] + [ 2 1 1 ]

= [ 8 5 3 ]

 

Step-03:

 

  • With the instances available currently, the requirement of the process P0 can be satisfied.
  • So, process P0 is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 8 5 3 ] + [ 0 0 1 ]

= [ 8 5 4 ]

 

Thus,

  • There exists a safe sequence P1, P2, P0 in which all the processes can be executed.
  • So, the system is in a safe state.
  • Thus, REQ2 can be permitted.

 

Thus, Correct Option is (B).

 

Problem-03:

 

A system has 4 processes and 5 allocatable resource. The current allocation and maximum needs are as follows-

AllocatedMaximum
A1021111213
B2011022210
C1101121311
D1111011220

 

If Available = [ 0 0 X 1 1 ], what is the smallest value of x for which this is a safe state?

 

Solution-

 

Let us calculate the additional instances of each resource type needed by each process.

We know,

Need = Maximum – Allocation

 

So, we have-

Need
A01002
B02100
C10300
D00110

 

Case-01: For X = 0

 

If X = 0, then-

Available

= [ 0 0 0 1 1 ]

 

  • With the instances available currently, the requirement of any process can not be satisfied.
  • So, for X = 0, system remains in a deadlock which is an unsafe state.

 

Case-02: For X = 1

 

If X = 1, then-

Available

= [ 0 0 1 1 1 ]

 

Step-01:

 

  • With the instances available currently, only the requirement of the process D can be satisfied.
  • So, process D is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 0 0 1 1 1 ] + [ 1 1 1 1 0 ]

= [ 1 1 2 2 1 ]

 

  • With the instances available currently, the requirement of any process can not be satisfied.
  • So, for X = 1, system remains in a deadlock which is an unsafe state.

 

Case-02: For X = 2

 

If X = 2, then-

Available

= [ 0 0 2 1 1 ]

 

Step-01:

 

  • With the instances available currently, only the requirement of the process D can be satisfied.
  • So, process D is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 0 0 2 1 1 ] + [ 1 1 1 1 0 ]

= [ 1 1 3 2 1 ]

 

Step-02:

 

  • With the instances available currently, only the requirement of the process C can be satisfied.
  • So, process C is allocated the requested resources.
  • It completes its execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 1 1 3 2 1 ] + [ 1 1 0 1 1 ]

= [ 2 2 3 3 2 ]

 

Step-03:

 

  • With the instances available currently, the requirement of both the processes A and B can be satisfied.
  • So, processes A and B are allocated the requested resources one by one.
  • They complete their execution and then free up the instances of resources held by it.

 

Then-

Available

= [ 2 2 3 3 2 ] + [ 1 0 2 1 1 ] + [ 2 0 1 1 0 ]

= [ 5 2 6 5 3 ]

 

Thus,

  • There exists a safe sequence in which all the processes can be executed.
  • So, the system is in a safe state.
  • Thus, minimum value of X that ensures system is in safe state = 2.

 

To watch video solutions and practice other problems,

Watch this Video Lecture

 

Next Article-Resource Allocation Graph

 

Get more notes and other study material of Operating System.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Banker’s Algorithm | Deadlock Avoidance

Deadlock Avoidance-

 

Before you go through this article, make sure that you have gone through the previous article on Deadlock and Handling Strategies.

 

We have discussed-

  • In a deadlock state, the execution of multiple processes is blocked.
  • Deadlock avoidance strategy involves maintaining a set of data.
  • The data is used to make a decision whether to entertain any new request or not.
  • If entertaining the new request causes the system to move in an unsafe state, then it is discarded.

 

Banker’s Algorithm-

 

  • Banker’s Algorithm is a deadlock avoidance strategy.
  • It is called so because it is used in banking systems to decide whether a loan can be granted or not.

 

Prerequisite-

 

Banker’s Algorithm requires-

  • Whenever a new process is created, it specifies the maximum number of instances of each resource type that it exactly needs.

 

Data Structures Used-

 

To implement banker’s algorithm, following four data structures are used-

 

 

  1. Available
  2. Max
  3. Allocation
  4. Need

 

Data StructureDefinitionExample
AvailableIt is a single dimensional array that specifies the number of instances of each resource type currently available.Available[R1] = K

It means K instances of resource type R1 are currently available.

MaxIt is a two dimensional array that specifies the maximum number of instances of each resource type that a process can request.Max[P1][R1] = K

It means process P1 is allowed to ask for maximum K instances of resource type R1.

AllocationIt is a two dimensional array that specifies the number of instances of each resource type that has been allocated to the process.Allocation[P1][R1] = K

It means K instances of resource type R1 have been allocated to the process P1.

 

NeedIt is a two dimensional array that specifies the number of instances of each resource type that a process requires for execution.Need[P1][R1] = K

It means process P1 requires K more instances of resource type R1 for execution.

 

Working-

 

  • Banker’s Algorithm is executed whenever any process puts forward the request for allocating the resources.
  • It involves the following steps-

 

Step-01:

 

  • Banker’s Algorithm checks whether the request made by the process is valid or not.
  • If the request is invalid, it aborts the request.
  • If the request is valid, it follows step-02.

 

Valid Request

A request is considered valid if and only if-

The number of requested instances of each resource type is less than the need declared by the process in the beginning.

 

Step-02:

 

  • Banker’s Algorithm checks if the number of requested instances of each resource type is less than the number of available instances of each type.
  • If the sufficient number of instances are not available, it asks the process to wait longer.
  • If the sufficient number of instances are available, it follows step-03.

 

Step-03:

 

  • Banker’s Algorithm makes an assumption that the requested resources have been allocated to the process.
  • Then, it modifies its data structures accordingly and moves from one state to the other state.

 

Available = Available - Request(i)
Allocation(i) = Allocation(i) + Request(i)
Need(i) = Need(i) - Request(i)

 

  • Now, Banker’s Algorithm follows the safety algorithm to check whether the resulting state it has entered in is a safe state or not.
  • If it is a safe state, then it allocates the requested resources to the process in actual.
  • If it is an unsafe state, then it rollbacks to its previous state and asks the process to wait longer.

 

Safe State

A system is said to be in safe state when-

All the processes can be executed in some arbitrary sequence with the available number of resources.

 

Safety Algorithm Data Structures-

 

To implement safety algorithm, following two data structures are used-

 

 

  1. Work
  2. Finish

 

Data StructureDefinitionExample
WorkIt is a single dimensional array that specifies the number of instances of each resource type currently available.Work[R1] = K

It means K instances of resource type R1 are currently available.

FinishIt is a single dimensional array that specifies whether the process has finished its execution or not.Finish[P1] = 0

It means process P1 is still left to execute.

 

Safety Algorithm-

 

Safety Algorithm is executed to check whether the resultant state after allocating the resources is safe or not.

 

Step-01:

 

Initially-

  • Number of instances of each resource type currently available = Available
  • All the processes are to be executed.
  • So, in Step-01, the data structures are initialized as-

 

Work = Available
Finish(i) = False for i = 0, 1, 2, ..., n-1

 

Step-02:

 

  • Safety Algorithm looks for an unfinished process whose need is less than or equal to work.
  • So, Step-02 finds an index i such that-

 

Finish[ i ] = False
Need(i) <= Work.

 

  • If such a process exists, then step-03 is followed otherwise step-05 is followed.

 

Step-03:

 

  • After finding the required process, safety algorithm assumes that the requested resources are allocated to the process.
  • The process runs, finishes its execution and the resources allocated to it gets free.
  • The resources are then added to the work and finish(i) of that process is set as true.

 

Work = Work + Allocation
Finish(i) = True

 

Step-04:

 

  • The loop of Step-02 and Step-03 is repeated.

 

Step-05:

 

  • If all the processes can be executed in some sequence, then the system is said to be in a safe state.
  • In other words, if Finish(i) becomes True for all i, then the system is in a safe state otherwise not.

 

To gain better understanding about Banker’s Algorithm,

Watch this Video Lecture

 

Next Article-Practice Problems On Banker’s Algorithm

 

Get more notes and other study material of Operating System.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Deadlock in OS | Deadlock Problems | Questions

Deadlock in OS-

 

Before you go through this article, make sure that you have gone through the previous article on Deadlock in OS.

 

We have discussed-

  • In a deadlock state, the execution of multiple processes is blocked.
  • There are 4 necessary conditions for the occurrence of deadlock.
  • Several Deadlock Handling Strategies exist to deal with the deadlock.

 

In this article, we will discuss practice problems based on deadlock.

 

Important Concept-

 

Consider there are n processes in the system P1, P2, P3, …… , Pn where-

  • Process P1 requires x1 units of resource R
  • Process P2 requires x2 units of resource R
  • Process P3 requires x3 units of resource R and so on.

 

In worst case,

The number of units that each process holds = One less than its maximum demand

 

So,

  • Process P1 holds x1 – 1 units of resource R
  • Process P2 holds x2 – 1 units of resource R
  • Process P3 holds x3 – 1 units of resource R and so on.

 

Now,

  • Had there been one more unit of resource R in the system, system could be ensured deadlock free.
  • This is because that unit would be allocated to one of the processes and it would get execute and then release its units.

 

From here, we have-

 

Maximum Number Of Units That Ensures Deadlock-

 

Maximum number of units of resource R that ensures deadlock

= (x1-1) + (x2-1) + (x3-1) + …. + (xn-1)

= ( x1 + x2 + x3 + …. + xn ) – n

= ∑xi – n

= Sum of max needs of all n processes – n

 

Minimum Number Of Units That Ensures No Deadlock-

 

Minimum number of units of resource R that ensures no deadlock

= One more than maximum number of units of resource R that ensures deadlock

= (∑xi – n) + 1

 

PRACTICE PROBLEMS BASED ON DEADLOCK IN OS-

 

Problem-01:

 

A system is having 3 user processes each requiring 2 units of resource R. The minimum number of units of R such that no deadlock will occur-

  1. 3
  2. 5
  3. 4
  4. 6

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 1 unit of resource R
  • Process P2 holds 1 unit of resource R
  • Process P3 holds 1 unit of resource R

 

Thus,

  • Maximum number of units of resource R that ensures deadlock = 1 + 1 + 1 = 3
  • Minimum number of units of resource R that ensures no deadlock = 3 + 1 = 4

 

Problem-02:

 

A system is having 10 user processes each requiring 3 units of resource R. The minimum number of units of R such that no deadlock will occur _____?

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 2 units of resource R
  • Process P2 holds 2 units of resource R
  • Process P3 holds 2 units of resource R and so on.
  • Process P10 holds 2 units of resource R

 

Thus,

  • Maximum number of units of resource R that ensures deadlock = 10 x 2 = 20
  • Minimum number of units of resource R that ensures no deadlock = 20 + 1 = 21

 

Problem-03:

 

A system is having 3 user processes P1, P2 and P3 where P1 requires 2 units of resource R, P2 requires 3 units of resource R, P3 requires 4 units of resource R. The minimum number of units of R that ensures no deadlock is _____?

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 1 unit of resource R
  • Process P2 holds 2 units of resource R
  • Process P3 holds 3 units of resource R

 

Thus,

  • Maximum number of units of resource R that ensures deadlock = 1 + 2 + 3 = 6
  • Minimum number of units of resource R that ensures no deadlock = 6 + 1 = 7

 

Problem-04:

 

A system is having 3 user processes P1, P2 and P3 where P1 requires 21 units of resource R, P2 requires 31 units of resource R, P3 requires 41 units of resource R. The minimum number of units of R that ensures no deadlock is _____?

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 20 units of resource R
  • Process P2 holds 30 units of resource R
  • Process P3 holds 40 units of resource R

 

Thus,

  • Maximum number of units of resource R that ensures deadlock = 20 + 30 + 40 = 90
  • Minimum number of units of resource R that ensures no deadlock = 90 + 1 = 91

 

Problem-05:

 

If there are 6 units of resource R in the system and each process in the system requires 2 units of resource R, then how many processes can be present at maximum so that no deadlock will occur?

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 1 unit of resource R
  • Process P2 holds 1 unit of resource R
  • Process P3 holds 1 unit of resource R
  • Process P4 holds 1 unit of resource R
  • Process P5 holds 1 unit of resource R
  • Process P6 holds 1 unit of resource R

 

Thus,

  • Minimum number of processes that ensures deadlock = 6
  • Maximum number of processes that ensures no deadlock = 6 – 1 = 5

 

Problem-06:

 

If there are 6 units of resource R in the system and each process in the system requires 3 units of resource R, then how many processes can be present at maximum so that no deadlock will occur?

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 2 units of resource R
  • Process P2 holds 2 units of resource R
  • Process P3 holds 2 units of resource R

 

Thus,

  • Minimum number of processes that ensures deadlock = 3
  • Maximum number of processes that ensures no deadlock = 3 – 1 = 2

 

Problem-07:

 

If there are 100 units of resource R in the system and each process in the system requires 2 units of resource R, then how many processes can be present at maximum so that no deadlock will occur?

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 1 unit of resource R
  • Process P2 holds 1 unit of resource R
  • Process P3 holds 1 unit of resource R and so on.
  • Process P100 holds 1 unit of resource R

 

Thus,

  • Minimum number of processes that ensures deadlock = 100
  • Maximum number of processes that ensures no deadlock = 100 – 1 = 99

 

Problem-08:

 

If there are 100 units of resource R in the system and each process in the system requires 4 units of resource R, then how many processes can be present at maximum so that no deadlock will occur?

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 3 units of resource R
  • Process P2 holds 3 units of resource R
  • Process P3 holds 3 units of resource R and so on.
  • Process P33 holds 3 units of resource R
  • Process P34 holds 1 unit of resource R

 

Thus,

  • Minimum number of processes that ensures deadlock = 34
  • Maximum number of processes that ensures no deadlock = 34 – 1 = 33

 

Problem-09:

 

If there are 100 units of resource R in the system and each process in the system requires 5 units of resource R, then how many processes can be present at maximum so that no deadlock will occur?

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process P1 holds 4 units of resource R
  • Process P2 holds 4 units of resource R
  • Process P3 holds 4 units of resource R and so on.
  • Process P25 holds 4 units of resource R

 

Thus,

  • Minimum number of processes that ensures deadlock = 25
  • Maximum number of processes that ensures no deadlock = 25 – 1 = 24

 

Problem-10:

 

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

  1. 2
  2. 3
  3. 4
  4. 1

 

Solution-

 

In worst case,

The number of tape drives that each process holds = One less than its maximum demand

So,

  • Process P1 holds 2 tape drives
  • Process P2 holds 2 tape drives
  • Process P3 holds 2 tape drives

 

Thus,

  • Minimum number of processes that ensures deadlock = 3
  • Maximum number of processes that ensures no deadlock = 3 – 1 = 2

 

Problem-11:

 

Consider a system having m resources of the same type. These resources are shared by 3 processes A, B and C which have peak demands of 3, 4 and 6 respectively. For what value of m, deadlock will not occur?

  1. 7
  2. 9
  3. 10
  4. 13

 

Solution-

 

In worst case,

The number of units that each process holds = One less than its maximum demand

So,

  • Process A holds 2 units of resource R
  • Process B holds 3 units of resource R
  • Process C holds 5 units of resource R

 

Thus,

  • Maximum number of units of resource R that ensures deadlock = 2 + 3 + 5 = 10
  • Minimum number of units of resource R that ensures no deadlock = 10 + 1 = 11

 

So, any number of units greater than 11 will ensure no deadlock.

Thus, Option (D) is correct.

 

Problem-12:

 

Consider a system having m resources of the same type being shared by n processes. Resources can be requested and released by processes only one at a time. The system is deadlock free if and only if-

  1. The sum of all max needs is < m+n
  2. The sum of all max needs is > m+n
  3. Both of above
  4. None of these

 

Solution-

 

We have derived above-

Maximum number of units of resource R that ensures deadlock = (∑xi – n)

 

Thus, For no deadlock occurrence,

Number of units of resource R must be > (∑xi – n)

i.e. m > (∑xi – n)

or ∑xi < m + n

Thus, Correct Option is (A).

 

Problem-13:

 

Consider the following snapshot of a system running n processes. Process i is holding xi instances of a resource R for 1<=i<=n. Currently, all instances of R are occupied. Further, for all i, process i has placed a request for an additional yi instances while holding the xi instances it already has. There are exactly two processes p and q such that yp = yq = 0. Which of the following can serve as a necessary condition to guarantee that the system is not approaching a deadlock?

  1. min(xp, xq) < maxk≠p,qyk
  2. xp + xq >= mink≠p,qyk
  3. min(xp, xq) < 1
  4. min(xp, xq) >1

 

Solution-

 

According to question, we have-

 

 

  • Clearly, processes Pp and Pq do not require any additional resource.
  • So they continue their execution.
  • After getting executed completely, they release the units allocated to them.
  • Thus, the total units that get free up = xp + xq

 

Now,

  • To ensure that other processes are executed without any deadlock, the total amount of units freely available currently ( xp + xq ) must be able to meet the requirements of some other process.
  • If available ( xp + xq ) units could not meet the requirement of any other process, then certainly there would be deadlock.

 

Thus, for no deadlock, the necessary condition is-

xp + xq >= min yk where k ≠ p, q

Thus, Correct Option is (B).

 

NOTE-

 

  • It is very important to note that the above condition is just a necessary condition and not at all a sufficient condition to avoid the deadlock.
  • The above condition just ensures that the system is able to proceed from the current state.
  • It does not guarantee that there won’t be a deadlock before all the other processes are finished.
  • The sufficient condition to avoid the deadlock would be either xp + xq >= ∑ yi or xp + xq >= max yk where k ≠ p, q.

 

To watch video solutions and practice other problems,

Watch this Video Lecture

 

Next Article-Banker’s Algorithm

 

Get more notes and other study material of Operating System.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Deadlock Detection | Deadlock Prevention

Deadlock Handling-

 

Before you go through this article, make sure that you have gone through the previous article on Deadlock in OS.

 

The various strategies for handling deadlock are-

 

 

  1. Deadlock Prevention
  2. Deadlock Avoidance
  3. Deadlock Detection and Recovery
  4. Deadlock Ignorance

 

Deadlock Prevention-

 

  • This strategy involves designing a system that violates one of the four necessary conditions required for the occurrence of deadlock.
  • This ensures that the system remains free from the deadlock.

 

The various conditions of deadlock occurrence may be violated as-

 

1. Mutual Exclusion-

 

  • To violate this condition, all the system resources must be such that they can be used in a shareable mode.
  • In a system, there are always some resources which are mutually exclusive by nature.
  • So, this condition can not be violated.

 

2. Hold and Wait-

 

This condition can be violated in the following ways-

 

Approach-01:

 

In this approach,

  • A process has to first request for all the resources it requires for execution.
  • Once it has acquired all the resources, only then it can start its execution.
  • This approach ensures that the process does not hold some resources and wait for other resources.

 

Drawbacks-

 

The drawbacks of this approach are-

  • It is less efficient.
  • It is not implementable since it is not possible to predict in advance which resources will be required during execution.

 

Approach-02:

 

In this approach,

  • A process is allowed to acquire the resources it desires at the current moment.
  • After acquiring the resources, it start its execution.
  • Now before making any new request, it has to compulsorily release all the resources that it holds currently.
  • This approach is efficient and implementable.

 

Approach-03:

 

In this approach,

  • A timer is set after the process acquires any resource.
  • After the timer expires, a process has to compulsorily release the resource.

 

3. No Preemption-

 

  • This condition can by violated by forceful preemption.
  • Consider a process is holding some resources and request other resources that can not be immediately allocated to it.
  • Then, by forcefully preempting the currently held resources, the condition can be violated.

 

A process is allowed to forcefully preempt the resources possessed by some other process only if-

  • It is a high priority process or a system process.
  • The victim process is in the waiting state.

 

4. Circular Wait-

 

  • This condition can be violated by not allowing the processes to wait for resources in a cyclic manner.
  • To violate this condition, the following approach is followed-

 

Approach-

 

  • A natural number is assigned to every resource.
  • Each process is allowed to request for the resources either in only increasing or only decreasing order of the resource number.
  • In case increasing order is followed, if a process requires a lesser number resource, then it must release all the resources having larger number and vice versa.
  • This approach is the most practical approach and implementable.
  • However, this approach may cause starvation but will never lead to deadlock.

 

Deadlock Avoidance-

 

  • This strategy involves maintaining a set of data using which a decision is made whether to entertain the new request or not.
  • If entertaining the new request causes the system to move in an unsafe state, then it is discarded.
  • This strategy requires that every process declares its maximum requirement of each resource type in the beginning.
  • The main challenge with this approach is predicting the requirement of the processes before execution.
  • Banker’s Algorithm is an example of a deadlock avoidance strategy.

 

Deadlock Detection and Recovery-

 

  • This strategy involves waiting until a deadlock occurs.
  • After deadlock occurs, the system state is recovered.
  • The main challenge with this approach is detecting the deadlock.

 

Deadlock Ignorance-

 

  • This strategy involves ignoring the concept of deadlock and assuming as if it does not exist.
  • This strategy helps to avoid the extra overhead of handling deadlock.
  • Windows and Linux use this strategy and it is the most widely used method.
  • It is also called as Ostrich approach.

 

To gain better understanding about Deadlock Handling Strategies,

Watch this Video Lecture

 

Next Article-Practice Problems On Deadlock

 

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Deadlock in OS | Conditions for Deadlock

Deadlock in OS-

 

Deadlock is a situation where-

  • The execution of two or more processes is blocked because each process holds some resource and waits for another resource held by some other process.

 

Example-

 

 

Here

  • Process P1 holds resource R1 and waits for resource R2 which is held by process P2.
  • Process P2 holds resource R2 and waits for resource R1 which is held by process P1.
  • None of the two processes can complete and release their resource.
  • Thus, both the processes keep waiting infinitely.

 

Conditions For Deadlock-

 

There are following 4 necessary conditions for the occurrence of deadlock-

  1. Mutual Exclusion
  2. Hold and Wait
  3. No preemption
  4. Circular wait

 

1. Mutual Exclusion-

 

By this condition,

  • There must exist at least one resource in the system which can be used by only one process at a time.
  • If there exists no such resource, then deadlock will never occur.
  • Printer is an example of a resource that can be used by only one process at a time.

 

2. Hold and Wait-

 

By this condition,

  • There must exist a process which holds some resource and waits for another resource held by some other process.

 

3. No Preemption-

 

By this condition,

  • Once the resource has been allocated to the process, it can not be preempted.
  • It means resource can not be snatched forcefully from one process and given to the other process.
  • The process must release the resource voluntarily by itself.

 

4. Circular Wait-

 

By this condition,

  • All the processes must wait for the resource in a cyclic manner where the last process waits for the resource held by the first process.

 

 

Here,

  • Process P1 waits for a resource held by process P2.
  • Process P2 waits for a resource held by process P3.
  • Process P3 waits for a resource held by process P4.
  • Process P4 waits for a resource held by process P1.

 

Important Note-

 

  • All these 4 conditions must hold simultaneously for the occurrence of deadlock.
  • If any of these conditions fail, then the system can be ensured deadlock free.

 

To gain better understanding about Deadlock in OS,

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Next Article-Deadlock Handling Strategies

 

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