7/24/2019 TASK 1(Noradibah)

1/17

ASSIGNMENT

Calculating the Network Address:

The network address is the logical AND of the respective bits in the binary representation of the IP

address and network mask. Align the bits in both addresses, and perform a logical AND on each pair

of the respective bits. Then convert the individual octets of the result back to decimal.

TASK 2:

PROBLEM 1

THE IP ADDRESS: 172.30.1.33

THENETWORK MASK: 255.255.0.0

Host IP Address:

The host IP address is:

172.30.1.33 in binary is 10101100 . 00011110 . 00000001 . 00100001

1. 172.30.1.33

2 172

2 86

2 43

2 21

2 10

2 5

2 2

2 1 =10101100

INPUT OUTPUT

A B A AND B

0 0 0

0 1 0

1 0 0

1 1 1

172 30 1 3310101100 00011110 00000001 00100001

0

0

1

1

0

1

0

1

. Mencari Network Address, dengan

cara menjumlahkan angka biner dari

IP Address dan Network Mask

menggunakan operator AND.

Catatan; 1 AND 1 = 1, sedangkan jika

ada 0 di AND kan maka hasilnya 0.

7/24/2019 TASK 1(Noradibah)

2/17

2 30

2 15

2 7

2 3

2 1 = 00011110

2 1

2 33

2 16

2 8

2 4

2 2 = 00100001

2 1

2.

255.255.0.0

= 11111111

0

1

1

1

1

1

1

0

0

0

0

1

2

2

2

2

2

2

2

2

255

127

63

31

15

7

3

1

1

1

1

1

1

1

1

1

7/24/2019 TASK 1(Noradibah)

3/17

= 11111111

= 00000000

= 00000000

Network Mask:

The network mask is:

Network Address:

> [Logical AND]

172.30.0.0

As you can see, the network address of 172.30.1.33/116 is 172.30.0.0

255.255.0.0in binary is 11111111 , 11111111 , 00000000 , 00000000

255 255 0 0

11111111 11111111 00000000 00000000

172.30.1.33in binary is 10101100,00011110,00000001,00100001

255.255.0.0in binary is 11111111,11111111,00000000,00000000

10101100,00011110,00000000,00000000

2

2

2

2

2

2

2

2

255

127

63

31

15

7

3

1

1

1

1

1

1

1

1

1

2 0 0

2 0 0

7/24/2019 TASK 1(Noradibah)

4/17

Network Broadcast Address:

The broadcast address converts all host bits to 1s...

Remember that our IP address in decimal is:

172.30.1.33 in binary is 10101100,00011110,00000001,00100001

To calculate the broadcast address, we force all host bits to be 1s :

> Force host bits

172.30.255.255

Mencari Broadcast Address dengan cara mengganti seluruh angka 0 pada bilangan host menjadi

angka 1.

Total Number Of Host Bits:

This means our host bits are the last 11 bits of the IP address, because we find the host mask by

inverting the network mask:

Host bits mask 2^16 = 65, 536

Total Number Of Host Bits : 65, 536

To find the number of host bitsis : (Total Number Of Host Bits[ 65 , 536 ] 2)

Number of Host Bits : 65, 534

PROBLEM 2

THE IP ADDRESS: 172.30.1.33

THENETWORK MASK: 255.255.255.0

Host IP Address: The host IP address is:

172.30.1.33 in binary is 10101100,00011110,00000001,00100001

172.30.1.33 in binary is 10101100,00011110,00000001,00100001

Host bits mask 11111111,11111111,11111111,11111111

10101100,00011110,11111111,11111111

7/24/2019 TASK 1(Noradibah)

5/17

1. 172.30.1.33

2 172

2 86

2 43

2 21

2 10

2 5

2 2

2 1 =10101100

= 11111111

= 00000000

Network Mask:

The network mask is:

255.255.255.0in binary is 11111111,11111111,11111111,00000000

172 30 1 3310101100 00011110 00000001 00100001

255 255 255 0

11111111 11111111 11111111 00000000

0

0

1

1

0

1

0

1

2

2

2

2

2

2

2

2

255

127

63

31

15

7

3

1

1

1

1

1

1

1

1

1

2 0 0

7/24/2019 TASK 1(Noradibah)

6/17

Network Address:

172.30.1.33in binary is 10101100,00011110,00000001,00100001

255.255.0.0in binary is 11111111,11111111,11111111,00000000

10101100,00011110,00000001,00000000

As you can see, the network address of 172.30.1.33/116 is 172.30.0.0

Network Broadcast Address:

The broadcast address converts all host bits to 1s...

Remember that our IP address in decimal is:

172.30.1.33 in binary is 10101100,00011110,00000001,00100001

To calculate the broadcast address, we force all host bits to be 1s :

> Force host bits

172.30.255.255

Total Number Of Host Bits:

This means our host bits are the last 11 bits of the IP address, because we find the host mask by

inverting the network mask:

Host bits mask 2^8 = 256

Total Number Of Host Bits : 256

To find the number of host bitsis : (Total Number Of Host Bits [ 256 ]2)

Number of Host Bits: 25

172.30.1.33 in binary is 10101100,00011110,00000001,00100001

Host bits mask 11111111,11111111,11111111,11111111

10101100,00011110,00000001,11111111

7/24/2019 TASK 1(Noradibah)

7/17

PROBLEM 3

THE IP ADDRESS: 192.168.10.234

THENETWORK MASK: 255.255.255.0

Host IP Address:

The host IP address is:

192.168.10.234 in binary is 11000000,10101000,00001010,11101010

1. 192.168.10.234

= 11000000

= 10101000

192 168 10 234

11000000 10101000 00001010 11101010

2

2

2

2

2

2

2

2

255

127

63

31

15

7

3

1

0

0

0

0

0

0

1

1

2

2

2

2

2

2

2

2

168

84

42

21

10

5

2

1

0

0

0

1

0

1

0

1

7/24/2019 TASK 1(Noradibah)

8/17

= 00001010

= 11101010

Network Mask:

The network mask is:

255.255.255.0in binary is 11111111,11111111,11111111,00000000

Network Address:

192.168.10.234 in binary is 11000000,10101000,00001010,11101010

255.255.255.0in binary is 11111111,11111111,11111111,00000000

11000000,10101000, 00001010,00000000

As you can see, the network address of 192.168.10.234 is 192.168.10.0

255 255 255 0

11111111 11111111 11111111 00000000

2

2

2

2

10

5

2

1

0

1

0

1

2

2

2

2

2

2

2

2

234

117

58

29

14

7

3

1

0

1

0

1

0

1

1

1

7/24/2019 TASK 1(Noradibah)

9/17

Network Broadcast Address:

The broadcast address converts all host bits to 1s...

Remember that our IP address in decimal is:

192.168.10.234 in binary is 11000000,10101000,00001010,11101010

To calculate the broadcast address, we force all host bits to be 1s :

> Force host bits

192.268.10.0

Total Number Of Host Bits:

This means our host bits are the last 11 bits of the IP address, because we find the host mask by

inverting the network mask:

Host bits mask 2^8 = 256

Total Number Of Host Bits : 256

To find the number of host bitsis : (Total Number Of Host Bits [ 256 ]

2)

Number of Host Bits : 254

PROBLEM 4

THE IP ADDRESS: 172.12.99.71

THENETWORK MASK: 255.255.0.0

Host IP Address:

The host IP address is:

172.17.99.71 in binary is ------- 10101100 00010001 01100011 01000111

1.

172.17.99.71

192.168.10.234 in binary is 11000000,10101000,00001010,11101010

Host bits mask 11111111,11111111,11111111,00000000

11000000,10101000,00001011,11111111

172 17 99 7110101100 00010001 01100011 01000111

7/24/2019 TASK 1(Noradibah)

10/17

= 10101100

= 00010001

= 01100011

2

2

2

2

2

2

2

2

172

86

43

21

10

5

2

1

0

0

1

1

0

1

0

1

2

2

2

2

2

17

8

4

2

1

1

0

0

0

1

2

2

2

2

2

2

2

99

49

24

12

6

3

1

1

1

0

0

0

1

1

7/24/2019 TASK 1(Noradibah)

11/17

= 01000111

Network Mask:The network mask is:

255.255.0.0in binary is 11111111,11111111,00000000,00000000

Network Address:172.17.99.71 in binary is 10101100,00010001,01100011,01000111

255.255.0.0in binary is 11111111,11111111,00000000,00000000

10101100,00010001,00000000,00000000

As you can see, the network address of 172.17.99.71 is 172.17.0.0

Network Broadcast Address:The broadcast address converts all host bits to 1s...

Remember that our IP address in decimal is:

172.17.99.71 in binary is 10101100,00010001,01100011,01000111

255 255 0 0

11111111 11111111 00000000 00000000

2

2

2

2

2

2

2

71

35

17

8

4

2

1

1

1

1

0

0

0

1

7/24/2019 TASK 1(Noradibah)

12/17

To calculate the broadcast address, we force all host bits to be 1s :

> Force host bits

192.268.255.255

Total Number Of Host Bits:

This means our host bits are the last 11 bits of the IP address, because we find the host mask by

inverting the network mask:

Host bits mask 2^16 = 65536

Total Number Of Host Bits : 65536

To find the number of host bitsis : (Total Number Of Host Bits [ 65536 ]2)

Number of Host Bits : 65534

PROBLEM 5

THE IP ADDRESS: 192.168.3.219

THENETWORK MASK: 255.255.0.0

Host IP Address:The host IP address is:

192.168.3.219 in binary is 11000000,10101000,00000011,11011011

1. 192.168.3.219

172.17.99.71 in binary is 10101100,00010001,01100011,01000111

Host bits mask 11111111,11111111,00000000,00000000

10101100,00010001,11111111,11111111

192 168 3 21911000000 10101000 00000011 11011011

7/24/2019 TASK 1(Noradibah)

13/17

= 11000000

= 10101000

= 00000011

= 11011011

2

2

2

2

2

2

2

2

192

96

48

24

12

6

3

1

0

0

0

0

0

0

1

1

2

2

2

2

2

2

2

2

168

84

42

21

10

5

2

1

0

0

0

1

0

1

0

1

2

2

3

1

1

1

2

2

2

2

2

2

2

2

219

109

54

27

13

6

3

1

1

1

0

1

1

0

1

1

7/24/2019 TASK 1(Noradibah)

14/17

Network Mask:

The network mask is:

255.255.0.0in binary is 11111111,11111111,00000000,00000000

Network Address:

192.168.3.219 in binary is 11000000,10101000,00000011,11011011

255.255.0.0in binary is 11111111,11111111,00000000,00000000

10101100,00010001,00000000,00000000

As you can see, the network address of 192.168.3.219 is 192.168.0.0

Network Broadcast Address:

The broadcast address converts all host bits to 1s...

Remember that our IP address in decimal is:

192.168.3.219 in binary is 11000000,10101000,00000011,11011011

To calculate the broadcast address, we force all host bits to be 1s :

> Force host bits

192.168.255.255

Total Number Of Host Bits:

This means our host bits are the last 11 bits of the IP address, because we find the host mask by

inverting the network mask:

255 255 0 0

11111111 11111111 00000000 00000000

192.168.3.219 in binary is 11000000,10101000,00000011,11011011

Host bits mask 11111111,11111111,00000000,00000000

11000000,10101000, 11111111,11111111

7/24/2019 TASK 1(Noradibah)

15/17

Host bits mask 2^16 = 65536

Total Number Of Host Bits : 65536

To find the number of host bitsis : (Total Number Of Host Bits [ 65536 ]2)

Number of Host Bits : 65534

PROBLEM 6

THE IP ADDRESS: 192.168.3.219

THENETWORK MASK: 255.255.255.254

Host IP Address:

The host IP address is:

192.168.3.219 in binary is 11000000,10101000,00000011,11011011

= 11000000

192 168 3 21911000000 10101000 00000011 11011011

2

2

2

2

2

2

2

2

192

96

48

24

12

6

3

1

0

0

0

0

0

0

1

1

7/24/2019 TASK 1(Noradibah)

16/17

= 10101000

= 00000011

= 11100000

Network Mask:

The network mask is:

255.255.0.0in binary is 11111111,11111111,00000000,00000000

255 255 255 224

11111111 11111111 11111111 11100000

2

2

2

2

2

2

2

2

168

84

42

21

10

5

2

1

0

0

0

1

0

1

0

1

2

2

3

1

1

1

2

2

2

2

2

2

2

2

224

112

56

28

14

7

3

1

0

0

0

0

0

1

1

1

7/24/2019 TASK 1(Noradibah)

17/17

Network Address:

192.168.3.219 in binary is 11000000,10101000,00000011,11011011

255.255.255.224in binary is--

11111111,11111111,11111111,11100000

11000000,10101000,00000011,11011111

As you can see, the network address of 192.168.3.219 is 192.168.3.224

Network Broadcast Address:

The broadcast address converts all host bits to 1s...

Remember that our IP address in decimal is:

192.168.3.219 in binary is 11000000,10101000,00000011,11011011

To calculate the broadcast address, we force all host bits to be 1s :

> Force host bits

192.168.255.255

Total Number Of Host Bits:

This means our host bits are the last 11 bits of the IP address, because we find the host mask by

inverting the network mask:

Host bits mask 2^16 = 65536

Total Number Of Host Bits : 65536

To find the number of host bitsis : (Total Number Of Host Bits [ 65536 ]2)

Number of Host Bits : 65534

192.168.3.219 in binary is 11000000,10101000,00000011,11011011

Host bits mask 11111111,11111111,00000000,00000000

11000000,10101000,11111111,11111111