Task 2

Problem 1

Host IP Address 172.30.1.33Subnet Mask 255.255.0.0Number Of Subnet Bit 8Number Of Subnets 2^8=256 subnetsNumber Of Host Bits Per Subnet 8 bitNumber Of Usable Hosts Per Subnet 2^8=256-2=254 hosts per subnetSubnet Address For This IP Adress 172.30.1.0Ip Address Of First Host On This Subnet 172.30.1.1

Ip Address Of Last Host On Subnet 172.30.1.254Broadcast Address For This Subnet 172.30.1.255

How to get the binary number for IP Address

172 30 1 33172/2 = 0

86/2 = 0

43/2 = 1

21/2 = 1

10/2 = 0

5/2 = 1

2/2 = 0

1/2 = 1

30/2=0

15/2=1

7/2=1

3/2=1

1/2=1

0/2=0

0/2=0

0/2=0

1/2=1

0/2=0

0/2=0

0/2=0

0/2=0

0/2=0

0/2=0

0/2=0

33/2=1

16/2=0

8/2=0

4/2=0

2/2=0

1/2=1

0/2=0

0/2=0

The final answer in the binary number10101100 01111000 01110010 10000100

Host Ip Address

172 30 1 33

Ip Address 10101100 00011110 00000001 00100001

MultipleSubnet Mask

11111111 11111111 00000000 00000000

Subnet Address

10101100 00011110 00000001 00000000 Answer

Subnet Address For This IP Address

172 30 1 0

127.30.1.33

Ip address of first host:172.30.1.33

Ip address of last host:172.30.1.254

Broadcast address:172.30.1.255

0+254 hosts

Adress ip of host last calculation=254

Number of ip subnets and hosts

Look hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.

172.30.1.33 255.255.0.0

= 11111111 00000000

Calculate the amount of the resistance of the host subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.

11111111 00000000 = 8 ones. 2^8 = 256 - 2 = 254 usable subnets created.

Calculating the total number of zeros in octal host subnet mask. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for host addresses

11111111 00000000 = 8 zeros. 2^8 = 256 - 2 = 254 usable host addresses created

Number of hosts bit per subnet 8 bit

Number of Usable hosts per subnet 2^8=256-2=254 host per subnet

Number of subnet bit 8 bit

Number Of subnets 2^8=256 subnets

Problem 2

Host IP Address 172.30.1.33Subnet Mask 255.255.255.0Number Of Subnet Bit 14Number Of Subnets 2^14=16384 subnetsNumber Of Host Bits Per Subnet 4 bitNumber Of Usable Hosts Per Subnet 2^2=4-2=2 hosys per subnetSubnet Address For This IP Adress 172.30.1.32Ip Address Of First Host On This Subnet 172.30.1.33

Ip Address Of Last Host On Subnet 172.30.1.34Broadcast Address For This Subnet 172.30.1.35

How to get the binary number for IP Address

172 30 1 33172/2 = 0

86/2 = 0

43/2 = 1

21/2 = 1

10/2 = 0

5/2 = 1

2/2 = 0

1/2 = 1

30/2 = 0

15/2 = 1

7/2 = 1

3/2 = 1

½ = 1

0/2 = 0

0/2 = 0

0/2 = 0

½ = 1

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

33/2 = 1

16/2 = 0

8/2 = 0

4/2 = 0

2/2 = 0

½ = 1

0/2 = 0

0/2 = 0

The final answer in the binary number10101100 00011110 00000001 00100001

Host Ip Address

172 30 1 33

Ip Address 10101100 00011110 00000001 00100001

MultipleSubnet Mask

11111111 11111111 11111111 00000000

Subnet Address

10101100 00011110 00000001 00100000 Answer

Subnet Address For This IP Address

172 30 1 32

127.30.1.33

Ip address of first host:172.30.1.33

Ip address of last host:172.30.1.34

Broadcast address:172.30.1.255

32+2 hosts

Adress ip of host last calculation=34

Number of ip subnet and host

Look hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.

172.30.1.33 255.255.255.0

= 11111111 11111100

Calculate the amount of the resistance of the host subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.

11111111 11111100 = 14 ones. 2^14= 16384 usable subnets created

Number of subnet bit =14

Number of subnet 2^14=16384 subnets

Calculating the total number of zeros in octal host subnet mask. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for addressing hosts.

11111111 11111100 = 2 zeros. 2^2 = 4 - 2 = 2 usable host

Number of host bit per subnet =4 bit

Number of usable host per subnet 2^2=4-2=2 hosts per subnet

Problem 3

Host IP Address 192.168.10.234Subnet Mask 255.255.255.0Number Of Subnet Bit 8Number Of Subnets 2^8=256 subnetsNumber Of Host Bits Per Subnet 8 bitNumber Of Usable Hosts Per Subnet 2^8=256-2=254 hosts per subnetSubnet Address For This IP Adress 192.192.10.0Ip Address Of First Host On This Subnet 192.192.10.1

Ip Address Of Last Host On Subnet 192.192.10.254Broadcast Address For This Subnet 192.192.10.255

How to get the binary number for IP Address

192 168 10 234192/2 = 0

96/2 = 0

48/2 = 0

24/2 = 0

12/2 = 0

6/2 = 1

3/2 = 1

1/2 = 1

168/2 = 0

84/2 = 0

42/2 = 0

21/2 = 1

10/2 = 0

5/2 = 1

2/2 = 0

½ = 1

10/2 = 0

5/2 = 1

2/2 = 0

½ = 1

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

234/2 = 0

117/2 = 1

58/2 = 0

29/2 = 1

14/2 = 0

7/2 = 1

3/2 = 1

1/2 = 1

The final answer in the binary number11100000 10101000 00001010 11101010

Host Ip Address

192 168 10 234

Ip Address 11000000 11000000 00001010 11101010

MultipleSubnet Mask

11111111 11111111 11111111 00000000

Subnet Address

11000000 11000000 00001010 00000000 Answer

Subnet Address For This IP Address

192 192 10 0

192.168.10.234

Ip address of first host:192.168.10.234

Ip address of last host:192.168.10.235

Broadcast address:192.168.10.255

0+254 hosts

Adress ip of host last calculation=254

Number of ip subnets and host

Look at the hosts in the subnet mask. Use the "Mask of Possibilities" method to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.

192.168.10.234

255.255.255.0 = 11111111 00000000

Calculating the amount in octal host subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.

11111111 00000000 = 8 ones. 2^8= 256 usable subnets created

Number of subnet bit =8

Number of subnet 2^8=256 subnets

Calculating the total number of zeros in octal host subnet mask. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for addressing hosts.

11111111 00000000 = 8 zeros. 2^8 = 256 - 2 = 254 usable host

Number of host bit per subnet =4 bit

Number of usable host per subnet 2^8=256-2=254 hosts per subnet

Problem 4

Host IP Address 172.17.99.71Subnet Mask 255.255.0.0Number Of Subnet Bit 0Number Of Subnets 2^0=1 subnetsNumber Of Host Bits Per Subnet 16 bitNumber Of Usable Hosts Per Subnet 2^16=65536-2=65534 hosys per subnetSubnet Address For This IP Adress 172.17.0.0Ip Address Of First Host On This Subnet 172.17.0.1

Ip Address Of Last Host On Subnet 172.17.255.254Broadcast Address For This Subnet 172.17.255.255

How to get the binary number for IP Address

172 17 99 71172/2 = 0

86/2 = 0

43/2 = 1

21/2 = 1

10/2 = 0

5/2 = 1

2/2 = 0

1/2 = 1

17/2 = 1

8/2 = 0

4/2 = 0

2/2 = 0

1/2 = 1

0/2 = 0

0/2 = 0

0/2 = 0

99/2 = 1

49/2 = 1

24/2 = 0

12/2 = 0

6/2 = 0

3/2 = 1

1/2 = 1

0/2 = 0

71/2 = 1

35/2 = 1

17/2 = 1

8/2 = 0

4/2 = 0

2/2 = 0

1/2 = 1

0/2 = 0

The final answer in the binary number10101100 00010001 01100011 01000111

Host Ip Address

172 17 99 71

Ip Address 10101100 00010001 01100011 01000111

MultipleSubnet Mask

11111111 11111111 00000000 00000000

Subnet Address

10101100 00010001 00000001 00000000 Answer

Subnet Address For This IP Address

172 17 0 0

127.17.99.71

Ip address of first host:172.17.99.71

Ip address of last host:172.17.255.254

Broadcast address:192.17.99.255

0+254 hosts

Adress ip of host last calculation=254

Number of ip subnet and host

Look at the hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.

172.17.99.71

255.255.0.0 = 00000000 00000000

Calculating the amount in octal host subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.

00000000 00000000 = 0 ones. 2^0= 1 usable subnets created.

Number of Subnet bit 0

Number of Subnets 2^0=1 subnets

Calculating the total number of zeros in the subnet mask range. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for addressing hosts.

00000000 00000000 = 16 zeros. 2^16= 65536 - 2 = 65534 usable host

Number of Hosts Bit per Subnet 16 bit

Number of Usable Hosts per Subnet 2^16= 65536 - 2 = 65534 hosts per subnet

Problem 5

Host IP Address 192.168.3.219Subnet Mask 255.255.0.0Number Of Subnet Bit 8Number Of Subnets 2^8=256 subnetsNumber Of Host Bits Per Subnet 8 bitNumber Of Usable Hosts Per Subnet 2^8=256-2=254 hosys per subnetSubnet Address For This IP Adress 192.168.3.0Ip Address Of First Host On This Subnet 192.168.3.1

Ip Address Of Last Host On Subnet 192.168.3.254Broadcast Address For This Subnet 192.168.3.255

How to get the binary number for IP Address

192 168 3 219192/2 = 0

96/2 = 0

48/2 = 1

24/2 = 1

12/2 = 0

6/2 = 1

3/2 = 1

1/2 = 1

168/2 = 0

84/2 = 0

42/2 = 0

21/2 = 1

10/2 = 0

5/2 = 1

2/2 = 0

1/2 = 0

3/2 = 1

1/2 = 1

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

219/2 = 1

109/2 = 1

54/2 = 0

27/2 = 1

13/2 = 1

6/2 = 0

3/2 = 1

1/2 = 1

The final answer in the binary number11101100 00101000 00000011 11011011

Host Ip Address

192 168 3 219

Ip Address 11000000 10101000 00000011 11011011

MultipleSubnet Mask

11111111 11111111 00000000 00000000

Subnet Address

11000000 10101000 00000011 00000000 Answer

Subnet Address For This IP Address

192 168 3 0

192.168.3.219

Ip address of first host:192.168.3.219

Ip address of last host:192.168.3.254

Broadcast address:192.192.10.255

0+254 hosts

Adress ip of host last calculation=254

Number of ip subnet and host

Look at the hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.

192.168.3.219

255.255.0.0 = 11111111 00000000

Counting the number of people in various subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.

11111111 00000000 = 8 ones. 2^8= 256 usable subnets created

Number of subnet bit 8

Number of subnets 2^8=256 subnet

11111111 00000000 = 8 zeros. 2^8= 256 - 2 = 254 usable host

Number of host bit per subnet 8 bit

Number of Usable host per subnet 2^8=256-2=254 host per subnet

Problem 6

Host IP Address 192.168.3.219Subnet Mask 255.255.255.224Number Of Subnet Bit 14Number Of Subnets 2^14=16384 subnetsNumber Of Host Bits Per Subnet 2 bitNumber Of Usable Hosts Per Subnet 2^2=4-2=2 hosys per subnetSubnet Address For This IP Adress 192.168.3.216Ip Address Of First Host On This Subnet 192.168.3.217

Ip Address Of Last Host On Subnet 192.168.3.218Broadcast Address For This Subnet 192.168.3.255.219

How to get the binary number for IP Address

192 168 3 219192/2 = 0

96/2 = 0

48/2 = 1

24/2 = 1

12/2 = 0

6/2 = 1

3/2 = 1

1/2 = 1

168/2 = 0

84/2 = 0

42/2 = 0

21/2 = 1

10/2 = 0

5/2 = 1

2/2 = 0

1/2 = 0

3/2 = 1

1/2 = 1

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

0/2 = 0

219/2 = 1

109/2 = 1

54/2 = 0

27/2 = 1

13/2 = 1

6/2 = 0

3/2 = 1

1/2 = 1

The final answer in the binary number11101100 00101000 00000011 11011011

Host Ip Address

192 168 3 219

Ip Address 11000000 10101000 00000011 11011011

MultipleSubnet Mask

11111111 11111111 11111111 11111100

Subnet Address

11000000 10101000 00000011 11011000 Answer

Subnet Address For This IP Address

192 168 3 216

192.168.3.219

Ip address of first host:192.168.3.218

Ip address of last host:192.168.3.219

Broadcast address:192.192.10.255

216+2 hosts

Adress ip of host last calculation=218

Number of ip subnet and host

Look octet hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.

192.168.3.219

255.255.255.224 = 11111111 11111100

Counting the number of people in various subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of potential created by the subnet mask. Two of the subnet potential usually can not be used.

11111111 111111100 = 14 ones. 2^14= 16384 usable subnets created.

Number of subnet bit 14

Number of subnets 2^14=16384 subnet

Calculating the total number of zeros in the subnet mask range. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for addressing hosts.

11111111 11111100 = 2 zeros. 2^2= 4 - 2 = 2 usable host

Number of host bit per subnet 2 bit

Number of usable host per subnet 2^2=4-2=2 host per subnet