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Komok [63]
3 years ago
6

Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first

acknowledgement is lost, but the second acknowledgement arrives before the timer for the first segment expires. True or False.
Computers and Technology
1 answer:
irinina [24]3 years ago
3 0

Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first acknowledgement is lost, but the second acknowledgement arrives before the timer for the first segment expires is True.

True

<u>Explanation:</u>

In network packet loss is considered as connectivity loss. In this scenario host A send two segment to host B and acknowledgement from host B Is awaiting at host A.

Since first acknowledgement is lost it is marked as packet lost. Since in network packet waiting for acknowledgement is keep continues process and waiting or trying to accept acknowledgement for certain period of time, once period limits cross then it is declared as packet loss.

Meanwhile second comes acknowledged is success. For end user assumes second segments comes first before first segment. But any how first segment expires.

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alina1380 [7]

Answer:

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5 0
2 years ago
Write a program to sort an array of 100,000 random elements using quicksort as follows: Sort the arrays using pivot as the middl
Stels [109]

Answer:

Check the explanation

Explanation:

#include<iostream.h>

#include<algorithm.h>

#include<climits.h>

#include<bits/stdc++.h>

#include<cstring.h>

using namespace std;

int partition(int arr[], int l, int r, int k);

int kthSmallest(int arr[], int l, int r, int k);

void quickSort(int arr[], int l, int h)

{

if (l < h)

{

// Find size of current subarray

int n = h-l+1;

 

// Find median of arr[].

int med = kthSmallest(arr, l, h, n/2);

 

// Partition the array around median

int p = partition(arr, l, h, med);

 

// Recur for left and right of partition

quickSort(arr, l, p - 1);

quickSort(arr, p + 1, h);

}

int findMedian(int arr[], int n)

{

sort(arr, arr+n); // Sort the array

return arr[n/2]; // Return middle element

}

int kthSmallest(int arr[], int l, int r, int k)

{

// If k is smaller than number of elements in array

if (k > 0 && k <= r - l + 1)

{

int n = r-l+1; // Number of elements in arr[l..r]

 

// Divide arr[] in groups of size 5, calculate median

// of every group and store it in median[] array.

int i, median[(n+4)/5]; // There will be floor((n+4)/5) groups;

for (i=0; i<n/5; i++)

median[i] = findMedian(arr+l+i*5, 5);

if (i*5 < n) //For last group with less than 5 elements

{

median[i] = findMedian(arr+l+i*5, n%5);

i++;

}

int medOfMed = (i == 1)? median[i-1]:

kthSmallest(median, 0, i-1, i/2);

int pos = partition(arr, l, r, medOfMed);

if (pos-l == k-1)

return arr[pos];

if (pos-l > k-1) // If position is more, recur for left

return kthSmallest(arr, l, pos-1, k);

return kthSmallest(arr, pos+1, r, k-pos+l-1);

}

return INT_MAX;

}

void swap(int *a, int *b)

{

int temp = *a;

*a = *b;

*b = temp;

}

int partition(int arr[], int l, int r, int x)

{

// Search for x in arr[l..r] and move it to end

int i;

for (i=l; i<r; i++)

if (arr[i] == x)

break;

swap(&arr[i], &arr[r]);

 

// Standard partition algorithm

i = l;

for (int j = l; j <= r - 1; j++)

{

if (arr[j] <= x)

{

swap(&arr[i], &arr[j]);

i++;

}

}

swap(&arr[i], &arr[r]);

return i;

}

 

/* Function to print an array */

void printArray(int arr[], int size)

{

int i;

for (i=0; i < size; i++)

cout << arr[i] << " ";

cout << endl;

}

 

// Driver program to test above functions

int main()

{

float a;

clock_t time_req;

int arr[] = {1000, 10, 7, 8, 9, 30, 900, 1, 5, 6, 20};

int n = sizeof(arr)/sizeof(arr[0]);

quickSort(arr, 0, n-1);

cout << "Sorted array is\n";

printArray(arr, n);

time_req = clock();

for(int i=0; i<200000; i++)

{

a = log(i*i*i*i);

}

time_req = clock()- time_req;

cout << "Processor time taken for multiplication: "

<< (float)time_req/CLOCKS_PER_SEC << " seconds" << endl;

 

// Using pow function

time_req = clock();

for(int i=0; i<200000; i++)

{

a = log(pow(i, 4));

}

time_req = clock() - time_req;

cout << "Processor time taken in pow function: "

<< (float)time_req/CLOCKS_PER_S

return 0;

}

..................................................................................................................................................................................................................................................................................................................................

OR

.......................

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

 

// Swap utility

void swap(long int* a, long int* b)

{

int tmp = *a;

*a = *b;

*b = tmp;

}

 

// Bubble sort

void bubbleSort(long int a[], long int n)

{

for (long int i = 0; i < n - 1; i++) {

for (long int j = 0; j < n - 1 - i; j++) {

if (a[j] > a[j + 1]) {

swap(&a[j], &a[j + 1]);

}

}

}

}

 

// Insertion sort

void insertionSort(long int arr[], long int n)

{

long int i, key, j;

for (i = 1; i < n; i++) {

key = arr[i];

j = i - 1;

 

// Move elements of arr[0..i-1], that are

// greater than key, to one position ahead

// of their current position

while (j >= 0 && arr[j] > key) {

arr[j + 1] = arr[j];

j = j - 1;

}

arr[j + 1] = key;

}

}

 

// Selection sort

void selectionSort(long int arr[], long int n)

{

long int i, j, midx;

 

for (i = 0; i < n - 1; i++) {

 

// Find the minimum element in unsorted array

midx = i;

 

for (j = i + 1; j < n; j++)

if (arr[j] < arr[min_idx])

midx = j;

 

// for plotting graph with integer values

printf("%li, %li, %li, %li\n",

n,

(long int)tim1[it],

(long int)tim2[it],

(long int)tim3[it]);

 

// increases the size of array by 10000

n += 10000;

}

 

return 0;

}

8 0
4 years ago
Read 2 more answers
A user is implementing security on a small office network. Which two actions would provide the minimum security requirements for
yarga [219]

Answer:

implementing a firewall

installing antivirus software

Explanation:

implementing a firewall: implementing this you could prevent the inbound and outbound not permitted connections to your internal network.

installing antivirus software: We need to install a antivirus software in the endpoints to detect and delete possible infections.

8 0
4 years ago
Helpppp pleaseee thanksss​
Archy [21]

Answer:

B!

Explanation:

i think... Here is an example

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3 years ago
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Romashka-Z-Leto [24]

Answer:

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The Flat file database is a database developed by IBM and it is the primary type of database for storing data in text files such as in Microsoft Excel

Individual database records are stored in a line of plain text and are separated from other records by delimiters including commas and tabs

The advantages of a Flat file database are;

1) The records of the database are stored in a single place

2) A Flat file is easier to set up with Microsoft Excel or other office applications

3) The Flat file database is easier to comprehend and understand

4) The records of the database can be simply stored

5) Simple criteria can be used for viewing or extracting data from a Flat file database

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