Tag: Bankers Algorithm

Banker’s Algorithm | Deadlock Avoidance

Operating System

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

 

Alloc Request
X Y Z X Y Z
P0 1 2 1 1 0 3
P1 2 0 1 0 1 2
P2 2 2 1 1 2 0

 

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.

Allocation Max
X Y Z X Y Z
P0 0 0 1 8 4 3
P1 3 2 0 6 2 0
P2 2 1 1 3 3 3

 

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-

 

Allocation Max Need
X Y Z X Y Z X Y Z
P0 0 0 1 8 4 3 8 4 2
P1 3 2 0 6 2 0 3 0 0
P2 2 1 1 3 3 3 1 2 2

 

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-

 

Allocation Max Need
X Y Z X Y Z X Y Z
P0 0 0 3 8 4 3 8 4 0
P1 3 2 0 6 2 0 3 0 0
P2 2 1 1 3 3 3 1 2 2

 

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-

 

Allocation Max Need
X Y Z X Y Z X Y Z
P0 0 0 1 8 4 3 8 4 2
P1 5 2 0 6 2 0 1 0 0
P2 2 1 1 3 3 3 1 2 2

 

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-

Allocated Maximum
A 1 0 2 1 1 1 1 2 1 3
B 2 0 1 1 0 2 2 2 1 0
C 1 1 0 1 1 2 1 3 1 1
D 1 1 1 1 0 1 1 2 2 0

 

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
A 0 1 0 0 2
B 0 2 1 0 0
C 1 0 3 0 0
D 0 0 1 1 0

 

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

 

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Banker’s Algorithm | Deadlock Avoidance

Operating System

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 Structure Definition Example
Available It 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.
Max It 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.
Allocation It 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.

 
Need It 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 Structure Definition Example
Work It 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.
Finish It 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

 

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