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Inga [223]
3 years ago
15

One day your supervisor walks to you, saying: "here is the network ID 192.168.0.0/24, please create three separate networks or s

ubnets for a coffee shop: Sunny Cafe." The three separate subnets/networks are: One is for the office, one for the front desk and storage room, and one is for public use.
Required:
Your task is to list each subnet, network ID, subnet mask, Host ID Range, # of usable host IDs, and Broadcast ID
Computers and Technology
1 answer:
Harrizon [31]3 years ago
8 0

Answer:

See the explanation for the answer

Explanation:

Given, network address is: 192.168.0.0/24

The requirement is to create 3 subnets from this network.

3 is near to 4 in powers of 2 => 22

We need 2 bits for subnetting.

So, total number of network bits are: 24 + 2 = 26 bits

Number of host bits are: 32 - 26 = 6 bits [since total number of bits in a address is 32]

Total number of hosts in a subnet: 26 = 64 hosts

Total number of valid hosts: 64 - 2 = 62 hosts [2 => 1 is for network address and 1 is subnet mask]

    Subnet       Network ID                       Subnet Mask

       1              192.168.0.0/26               255.255.255.192

       2             192.168.0.64/26             255.255.255.192

       3             192.168.0.128/26            255.255.255.192

       4             192.168.0.192/26            255.255.255.192

Subnet Host ID range     #of usable host IDs      Broadcast ID

  1            192.168.0.1 -                    62                      192.168.0.63

               192.168.0.62                        

  2           192.168.0.65 -                 62                     192.168.0.127

               192.168.0.126

 3           192.168.0.129 -                 62                     192.168.0.191

               192.168.0.190

4             192.168.0.193 -                 62                    192.168.0.255

               192.168.0.254

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Answer:

  1.     public class Main {
  2.    public static void main(String[] args) {
  3.        int myArray[] = {3, 7, 2, 5, 9,11, 24, 6, 10, 12};
  4.        int myArray2 [] = {1, 2, 3, 4 ,5};
  5.        displayValue(myArray);
  6.        reverseDisplay(myArray);
  7.        displaySum(myArray);
  8.        displayLess(myArray, 10);
  9.        displayHighAvg(myArray);
  10.        displayBoth(myArray, myArray2);
  11.    }
  12.    public static void displayValue(int arr[]){
  13.        for(int i = 0; i < arr.length; i++){
  14.            System.out.print(arr[i] + " ");
  15.        }
  16.        System.out.println();
  17.    }
  18.    public static void reverseDisplay(int arr[]){
  19.        for(int i = arr.length-1; i >= 0; i--){
  20.            System.out.print(arr[i] + " ");
  21.        }
  22.        System.out.println();
  23.    }
  24.    public static void displaySum(int arr[]){
  25.        int sum = 0;
  26.        for(int i = 0; i < arr.length; i++){
  27.            sum += arr[i];
  28.        }
  29.        System.out.println(sum);
  30.    }
  31.    public static void displayLess(int arr[], int limit){
  32.        for(int i = 0; i < arr.length; i++){
  33.            if(arr[i] < limit){
  34.                System.out.print(arr[i] + " ");
  35.            }
  36.        }
  37.        System.out.println();
  38.    }
  39.    public static void displayHighAvg(int arr[]){
  40.        int sum = 0;
  41.        for(int i = 0; i < arr.length; i++){
  42.            sum += arr[i];
  43.        }
  44.        double avg = sum / arr.length;
  45.        for(int i = 0; i < arr.length; i++){
  46.            if(arr[i] > avg){
  47.                System.out.print(arr[i] + " ");
  48.            }
  49.        }
  50.        System.out.println();
  51.    }
  52.    public static void displayBoth(int arr1[], int arr2 []){
  53.        for(int i = 0; i < arr2.length; i++){
  54.            for(int j = 0; j < arr1.length; j++){
  55.                if(arr1[j] == arr2[i]){
  56.                    System.out.print(arr1[j] + " ");
  57.                }
  58.            }
  59.        }
  60.        System.out.println();
  61.    }
  62. }

Explanation:

There are five methods written to solve all the problems stated in the question.

Method 1 : displayValue    (Line 15 - 21)

This is the method that take one input array and use the print() method to display the all the elements in the array.

Method 2: reverseDisplay (Line 23 - 26)

This method will take one input array and print the value in the reverse order. We just need to start with the last index when running the for-loop to print the value.

Method 3: displaySum (Line 31 - 37)

This method will take one input array and use a for-loop to calculate the total of the values in the array.

Method 4: displayLess (Line 39 - 47)

This method will take one two inputs, a array and a limit. We use the limit as the condition to check if any value less than the limit, then the value will only be printed.

Method 5: displayHighAvg (Line 50 - 64)

This method will take one input array and calculate the average. The average will be used to check if any element in the array higher than it, then the value will only be printed.

Method 6: displayBoth  (Line 66 - 75)

This method will take two input arrays and compare both of them to find out if any value appears in both input arrays and print it out.

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4 years ago
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