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umka21 [38]
3 years ago
11

Instructions

Computers and Technology
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

See explaination

Explanation:

dateType.h

#ifndef dateType_H

#define dateType_H

class dateType

{

public:

void setDate(int month, int day, int year);

//Function to set the date.

//The member variables dMonth, dDay, and dYear are set

//according to the parameters.

//Postcondition: dMonth = month; dDay = day;

// dYear = year

int getDay() const;

//Function to return the day.

//Postcondition: The value of dDay is returned.

int getMonth() const;

//Function to return the month.

//Postcondition: The value of dMonth is returned.

int getYear() const;

//Function to return the year.

//Postcondition: The value of dYear is returned.

void printDate() const;

//Function to output the date in the form mm-dd-yyyy.

void isLeapYear() const;

dateType(int month = 1, int day = 1, int year = 1900);

//Constructor to set the date

//The member variables dMonth, dDay, and dYear are set

//according to the parameters.

//Postcondition: dMonth = month; dDay = day; dYear = year;

// If no values are specified, the default

// values are used to initialize the member

// variables.

private:

int dMonth; //variable to store the month

int dDay; //variable to store the day

int dYear; //variable to store the year

};

#endif

dateType.cpp

#include <iostream>

#include "dateType.h"

using namespace std;

void dateType::setDate(int month, int day, int year)

{

// Checking month is valid

while(month<1 || month>12)

{

cout << "Enterd month "<<month<< " is wrong"<<endl;

cout << "Enter correct month"<<endl;

cin>>month;

}

dMonth = month;

// Checking date is valid

while(day<1 || day>31)

{

cout << "Enterd date "<<day<<" is wrong"<<endl;

cout<<"Enter correct date"<<endl;

cin>>day;

}

dDay = day;

int count_digits = 0;

int flag=0;

int year1;

// Counting number of digits in year

while(flag==0)

{

year1=year;

count_digits=0;

while (year) {

year /= 10;

count_digits++;

}

if(count_digits != 4)

{

cout << "Enterd year "<<year1<<" is wrong"<<endl;

cout<<"Enter correct year"<<endl;

cin>>year;

flag=0;

}

else

flag=1;

}

dYear = year1;

}

int dateType::getDay() const

{

return dDay;

}

int dateType::getMonth() const

{

return dMonth;

}

int dateType::getYear() const

{

return dYear;

}

void dateType::printDate() const

{

cout << dMonth << "-" << dDay << "-" << dYear;

}

void dateType::isLeapYear() const

{

if ( dYear%400 == 0)

cout<<endl<<dYear<< " is leap year.\n";

else if ( dYear%100 == 0)

cout<<endl<<dYear<< " is leap year.\n";

else if ( dYear%4 == 0 )

cout<<endl<<dYear<< " is leap year.\n";

else

cout<<endl<<dYear<< " is not leap year.\n";

}

//Constructor with parameters

dateType::dateType(int month, int day, int year)

{

// Checking month is valid

while(month<1 || month>12)

{

cout << "Enterd month "<<month<< " is wrong"<<endl;

cout << "Enter correct month"<<endl;

cin>>month;

}

dMonth = month;

// Checking date is valid

while(day<1 || day>31)

{

cout << "Enterd date "<<day<<" is wrong"<<endl;

cout<<"Enter correct date"<<endl;

cin>>day;

}

dDay = day;

int count_digits = 0;

int flag=0;

int year1;

// Counting number of digits in year

while(flag==0)

{

year1=year;

count_digits=0;

while (year) {

year /= 10;

count_digits++;

}

if(count_digits != 4)

{

cout << "Enterd year "<<year1<<" is wrong"<<endl;

cout<<"Enter correct year"<<endl;

cin>>year;

flag=0;

}

else

flag=1;

}

dYear = year1;

}

main.cpp

#include<iostream>

#include "dateType.h"

using namespace std;

int main()

{

dateType *dt1=new dateType();

cout<<"Date is "<<endl;

dt1->printDate();

cout<<endl;

dt1->isLeapYear();

cout<<endl;

dateType *dt2=new dateType(11,14,2019);

cout<<"Date is "<<endl;

dt2->printDate();

cout<<endl;

dt2->isLeapYear();

cout<<endl;

dt2->setDate(13,32,2016);

cout<<"Date is "<<endl;

dt2->printDate();

cout<<endl;

dt2->isLeapYear();

cout<<endl;

dt1->setDate(10,10,198);

cout<<"Date is "<<endl;

dt1->printDate();

cout<<endl;

dt1->isLeapYear();

cout<<endl;

system("pause");

return 0;

}

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Assume that a large number of consecutive IP addresses are available starting at 198.16.0.0 and suppose that two organizations,
Fiesta28 [93]

Answer & Explanation:

An IP version 4 address is of the form w.x.y.z/s

where s = subnet mask

w = first 8 bit field, x = 2nd 8 bit field, y = 3rd 8 bit field, and z = 4th 8 bit field

each field has 256 decimal equivalent. that is

binary                                        denary or decimal

11111111      =        2⁸      =             256

w.x.y.z represents

in binary

11111111.11111111.11111111.11111111

in denary

255.255.255.255

note that 255 = 2⁸ - 1 = no of valid hosts/addresses

there are classes of addresses, that is

class A = w.0.0.0 example 10.0.0.0

class B = w.x.0.0 example 172.16.0.0

class C = w.x.y.0 example 198.16.8.1

where w, x, y, z could take numbers from 1 to 255

Now in the question

we were given the ip address : 198.16.0.0 (class B)

address of quantity 4000, 2000, 8000 is possible with a subnet mask of type

255.255.0.0 (denary) or

11111111.11111111.00000000.00000000(binary) where /s =  /16 That is no of 1s

In a VLSM (Variable Length Subnet Mask)

Step 1

we convert the number of host/addresses for company A to binary

4000 = 111110100000 = 12 bit

step 2 (subnet mask)

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

<u><em>subnet mask: 255.255.</em></u><u><em>16.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1      </em></u><u><em> 1 </em></u><u><em>      1     1     1     1</em></u>

<u><em>128  64     32    </em></u><u><em>16</em></u><u><em>    8    4     2    1</em></u>

step 3

in the ip network address: 198.16.0.0/19 <em>(subnet representation)</em> we increment this using 16

that is 16 is added to the 3rd field as follows

That means the ist Valid Ip address starts from

          Ist valid Ip add: 198.16.0.1 - 198.16.15.255(last valid IP address)

Company B starts<u><em>+16: 198.16.</em></u><u><em>16</em></u><u><em>.0 - 198.16.31.255</em></u>

<u><em>                   +16: 198.16.</em></u><u><em>32</em></u><u><em>.0- 198.16.47.255 et</em></u>c

we repeat the steps for other companies as follows

Company B

Step 1

we convert the number of host/addresses for company B to binary

2000 = 11111010000 = 11 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 11 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11111000.000000                /21

now we have added 5 1s in the third field to reserve 11 0s

<u><em>subnet mask: 255.255.</em></u><u><em>8.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1       1       </em></u><u><em>1 </em></u><u><em>    1     1     1</em></u>

<u><em>128  64     32    16    </em></u><u><em>8 </em></u><u><em>   4     2    1</em></u>

Step 3

Starting from after the last valid Ip address for company A

in the ip network address: 198.16.16.0/21 (<em>subnet representation</em>) we increment this using 8

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.16.1 - 198.16.23.255(last valid IP address)

Company C starts <u><em>+16: 198.16.</em></u><u><em>24</em></u><u><em>.0- 198.16.31.255</em></u>

<em>                             </em><u><em> +16: 198.16.</em></u><u><em>32</em></u><u><em>.0- 198.16.112.255 et</em></u>c

Company C

Step 1

we convert the number of host/addresses for company C to binary

4000 = 111110100000 = 12 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

<u><em>subnet mask: 255.255.</em></u><u><em>16.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1       1       1       1     1     1     1</em></u>

<u><em>128  64     32    16    8    4     2    1</em></u>

Step 3

Starting from after the last valid ip address for company B

in the ip network address: 198.16.24.0/20 (subnet representation) we increment this using 16

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.24.1 - 198.16.39.255(last valid IP address)

Company C starts <u><em>+16: 198.16.40.0- 198.16.55.255</em></u>

<em>                          </em><u><em>    +16: 198.16.56.0- 198.16.71.255 et</em></u>c

Company D

Step 1

we convert the number of host/addresses for company D to binary

8000 = 1111101000000 = 13 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 13 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11100000.000000                /19

now we have added 3 1s in the 3rd field to reserve 13 0s

<u><em>subnet mask: 255.255.</em></u><u><em>32.</em></u><u><em>0 (where the 1s in each field represent a denary number as follows)</em></u>

<u><em>1st 1 = 128, 2nd 1 = 64 as follows</em></u>

<u><em>1        1      </em></u><u><em> 1 </em></u><u><em>      1       1     1     1     1</em></u>

<u><em>128  64     </em></u><u><em>32  </em></u><u><em>  16    8    4     2    1</em></u>

Step 3

Starting from after the last valid ip address for company C

in the ip network address: 198.16.40.0/20 (subnet representation) we increment this using 32

That means the ist Valid Ip address starts from

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Company C starts <u><em>+16: 198.16.72.0- 198.16.103.255</em></u>

<em>                          </em><u><em>    +16: 198.16.104.0- 198.16.136.255 et</em></u>c

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