Program for Deadlock free condition in Operating System

Given: A system has R identical resources, P processes competing for them and N is the maximum need of each process. The task is to find the minimum number of Resources required So that deadlock will never occur.

Formula:

R >= P * (N - 1) + 1 

Examples:

Input : P = 3, N = 4
Output : R >= 10

Input : P = 7, N = 2
Output : R >= 8 

Approach:

Consider, 3 process A, B and C.
Let, Need of each process is 4
Therefore, The maximum resources require will be 3 * 4 = 12 i.e, Give 4 resources to each Process.
And, The minimum resources required will be 3 * (4 – 1) + 1 = 10.
i.e, Give 3 Resources to each of the Process, and we are left out with 1 Resource.
That 1 resource will be given to any of the Process A, B or C.
So that after using that resource by any one of the Process, It left the resources and that resources will be used by any other Process and thus Deadlock will Never Occur.

C++

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// C++ implementation of above program.
#include <bits/stdc++.h>
using namespace std;
  
// function that calculates
// the minimum no. of resources
int Resources(int process, int need)
{
    int minResources = 0;
  
    // Condition so that deadlock
    // will not occuur
    minResources = process * (need - 1) + 1;
  
    return minResources;
}
  
// Driver code
int main()
{
    int process = 3, need = 4;
  
    cout << "R >= " << Resources(process, need);
    return 0;
}

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Java

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// Java implementation of above program
  
class GFG 
  
// function that calculates 
// the minimum no. of resources 
static int Resources(int process, int need)
    int minResources = 0
  
    // Condition so that deadlock 
    // will not occuur 
    minResources = process * (need - 1) + 1
  
    return minResources; 
  
// Driver Code 
public static void main(String args[]) 
    int process = 3, need = 4;
      
    System.out.print("R >= "); 
    System.out.print(Resources(process, need)); 

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Python3

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# Python 3 implementation of 
# above program
   
# function that calculates 
# the minimum no. of resources 
def Resources(process, need):
  
    minResources = 0
  
    # Condition so that deadlock 
    # will not occuur 
    minResources = process * (need - 1) + 1
  
    return minResources 
  
# Driver Code
if __name__ == "__main__"
  
    process, need = 3, 4
  
    print("R >=", Resources(process, need))
  
# This Code is Contributed 
# by Naman_Garg

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C#

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// C# implementation of above program
using System; 
  
class GFG 
  
// function that calculates 
// the minimum no. of resources 
static int Resources(int process, int need)
    int minResources = 0; 
  
    // Condition so that deadlock 
    // will not occuur 
    minResources = process * (need - 1) + 1; 
  
    return minResources; 
  
// Driver Code 
public static void Main() 
    int process = 3, need = 4;
      
    Console.Write("R >= "); 
    Console.Write(Resources(process, need)); 
  
// This code is contributed 
// by Sanjit_Prasad 

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Output:

R >= 10


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