Answer:
D
Explanation:
Not much of a tech person, but protocol then firewall seems the most reasonable.
Solution:
The process of transaction can guarantee the reliability of business applications. Locking resources is widely used in distributed transaction management (e.g; two phase commit, 2PC) to keep the system consistent. The locking mechanism, however, potentially results in various deadlocks. In service oriented architecture, the deadlock problem becomes even worse because multiple transactions try to lock shared resources in the unexpectable way due to the more randomicity of transaction requests, which has not been solved by existing research results. In this paper, we investigate how to prevent local deadlocks, caused by the resource competition among multiple sub-transactions of a gl obal transaction, and global deadlocks from the competition among different global transactions. We propose a replication based approach to avoid the local deadlocks, and a timestamp based approach to significantly mitigate the global deadlocks. A general algorithm is designed for both local and global deadlock prevention. The experimental results demonstrate the effectiveness and efficiency of our deadlock prevention approach. Further, it is also proved that our approach provides higher system performance than traditional resource allocation schemes.
This is the required answer.
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
