Ip Address Classes

IP ADDRESS CLASSESTask 1: For a given IP address,Determine Network Information.Given :Host IP Address172.25.114.250

Network Mask255.255.0.0 (/16)

Find :

Network Address172.25.0.0

Network Broadcast Address172.25.255.255

Total Number of Host Bits65534

Number of Hosts16

Step 1 : Translate Host IP address and network mask into binary notation.FORMULA :2726252423222120

1286432168421

IP ADDRESS172 . 25 . 114 . 250Octet 1 Octet 2 Octet 3Octet 4Octet 12726252423222120

1286432168421

10101100

Let128+32+8+4= 172Octet 22726252423222120

1286432168421

00011001

Let16+8+1= 25

Octet 32726252423222120

1286432168421

01110010

Let64+32+16+2=144

Octet 42726252423222120

1286432168421

11111010

Let128+64+32+16+8+2=250

NETWORK MASK 255 . 255 . 0 . 0172 . 25 . 114 . 250Octet 1 Octet 2 Octet 3Octet 4Octet 1 2726252423222120

1286432168421

11111111

Let 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1= 255Octet 22726252423222120

1286432168421

11111111

Let128 + 64 + 32 + 16 + 4 + 2 + 1=255Octet 32726252423222120

1286432168421

00000000

LetThe value of network mask at octet 3 is 0 , so the binary number is still zero (0).

Octet 42726252423222120

1286432168421

00000000

LetThe value of network mask at octet 3 is 0 , so the binary number is still zero (0).

Step 2: Determine the network address. 17225114250IP Address10101100000110010111001011111010

Subnet Mask11111111111111110000000000000000

Network Address10101100000110010000000000000000

17225 0 0

EXP:IP Address10101100000110010111001011111010X MultiplySubnet Mask 11111111111111110000000000000000Network Address10101100000110010000000000000000 172 25 0 0 Answer

Step 3: Determine the broadcast address for the network addressThe network mask separates the network portion of the address from the host portion.The network address has all 0s in the host portion of the address and the broadcast address has all 1s in the host portion of the address. 172 25 0 0Network Add.10101100000110010000000000000000

Mask11111111111111110000000000000000

Broadcast.10101100000110011111111111111111

172 25 0 0

By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16`0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,5362n-2FORMULA = 65,536 2 = 65,536

Task 2: ChallengeFor all problem:Create a subnetting Worksheet to show and record all work for each problem.Problem 1Host Ip Address172.30.1.33

Network Mask255.255.0.0




Number of Hosts16

FORMULA :2726252423222120

1286432168421

IP ADDRESS172 . 30 . 1 . 33Octet 1 Octet 2 Octet 3Octet 4

Octet 12726252423222120

1286432168421

10101100

Let128+32+8+4=172

Octet 22726252423222120

1286432168421

00011110

Let16+8+4+2= 30Octet 32726252423222120

1286432168421

00000001

Let1= 1

Octet 42726252423222120

1286432168421

00100001

Let= 33

Step 2: Determine the network address. 17230 1 33IP Address10101100000111100000000100100001

Subnet Mask11111111111111110000000000000000

Network Address10101100000111100000000100000000

17230 0 0



Mask11111111111111110000000000000000

Broadcast.10101100000111101111111111111111

172 30 1 255

By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA = 65,536 2 = 65,536

Problem 2Host Ip Address172.30.1.33





Number of Hosts24

FORMULA :2726252423222120

1286432168421


1286432168421

10101100

Let128+32+8+4=172

Octet 22726252423222120

1286432168421

00011110

Let16+8+4+2= 30Octet 32726252423222120

1286432168421

00000001

Let1= 1Octet 42726252423222120

1286432168421

00100001

Let= 33


Subnet Mask11111111111111110000000000000000

Network Address10101100000111100000000000000000

17230 0 0



Mask11111111111111110000000000000000

Broadcast.10101100000111101111111111111111

172 25 0 0

By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA = 65,536 2 = 65,536






Number of Hosts28

FORMULA :2726252423222120

1286432168421


1286432168421

11000000

Let128+64= 192

Octet 22726252423222120

1286432168421

10101000

Let128+32+8= 168Octet 32726252423222120

1286432168421

00001010

Let8+2= 10Octet 42726252423222120

1286432168421

11111010

Let128+64+32+16+8+2= 234


Subnet Mask11111111111111110000000000000000

Network Address11000000101010000000101011111010

192168 0 0



Mask11111111111111110000000000000000

Broadcast.10101100000111101111111111111111

172 25 0 0By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA =65,536 2 = 65,536






Number of Hosts216

FORMULA :2726252423222120

1286432168421


1286432168421

10101100

Let128+32+8+4= 172

Octet 22726252423222120

1286432168421

00010001

Let16+1= 17

Octet 32726252423222120

1286432168421

01100011

Let64+32+2+1= 99Octet 42726252423222120

1286432168421

01000110

Let64+4+2= 71


Subnet Mask11111111111111110000000000000000

Network Address11000000000100010000101011111010

17217 0 0



Mask11111111111111110000000000000000

Broadcast.10101100000111101111111111111111

172 25 0 0By counting the number of host bits, we can determine the total number of usable host for this network.Host bits: 16 `0 number of mask in octet 3 and octet 4 Total number of host: 255.255.0.0 host216 = 65,536 FORMULA = 65,536 2 = 65,536






Number of Hosts216

FORMULA :2726252423222120

1286432168421


1286432168421

11000000

Let128+64= 192

Octet 22726252423222120

1286432168421

10101000

Let128+32+8= 168Octet 32726252423222120

1286432168421

00000011

Let2+1= 3Octet 42726252423222120

1286432168421

11011011

Let128+64+16+8+2+1= 219


Subnet Mask11111111111111110000000000000000

Network Address11000000101010000000101011111010

192168 0 0



Mask11111111111111110000000000000000

Broadcast.10101100000111101111111111111111







Number of Hosts216

FORMULA :2726252423222120

1286432168421


1286432168421

11000000

Let128+64= 192

Octet 22726252423222120

1286432168421

10101000

Let128+32+8= 168Octet 32726252423222120

1286432168421

00000011

Let2+1= 3Octet 42726252423222120

1286432168421

11011011

Let128+64+16+8+2+1= 219


Subnet Mask11111111111111110000000000000000

Network Address11000000101010000000101011111010

192168 0 0



Mask11111111111111110000000000000000

Broadcast.10101100000111101111111111111111


TASK/EXERCISETASK 1: For a Given IP Adress and Subnet Mask , Determine Subnet Information.Given:Host IP Address172.25.114.250

Network Mask255.255.0.0 (/16)

Subnet Mask255.255.255.192 (/26)

Find:Number of Subnet Bits10

Number of Subnets210=1,024 subnets

Number of Host Bits per Subnet6 bit

Number of Usable Hosts per Subnet26=64 - 2= 62 hosts per subnet

Subnet Address for this IP Address172.25.114.192

IP Address of First Host on this Subnet 172.25.114.193

IP Address of Last Host on this Subnet172.25.114.254

Broadcast Address for this Subnet172.25.114.255

Step 1 : Translate Host IP address and network mask into binary notation.FORMULA :2726252423222120

1286432168421

172 25 114 250IP Address10101100000110010111001011111010X MultiplySubnet Mask 11111111111111111111111111000000Subnet Address10101100000110010111001011000000Subnet address for 172 25 114 192 AnswerThis IP Address =IP Address of First Host 192 + 62 Hosts IP Address of Last Host

Broadcast Address172.25.144.255172.25.114.193 = 254 172.25.114.254

Determining the Number of IP Subnets and HostsStep 1-Use the first octet of the IP address to determine the class of address (A,B or C).172.25.114.250255.255.255.192Step 2Use the class of the address to determine which octets are available for hosts.Network. Network. Host. Host. 172 . 25 . 114 . 250 255 . 255 . 255 . 192Step 3Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed to step four.172.25.114.250255.255.255.192 = 11111111 11000000Step 4Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 11000000 = 10 ones.210 = 1,024 2 = 1,022 usable subnets created.Step 5Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 11000000 = 6 zeros. 26 = 64 2 = 62 usable host addresses created.-Number of Host Bit per Subnet - 6 bit Number of Usable Hosts per Subnet 26 = 64 2 = 62 hosts per subnet Problem 1Host IP Address172.30.1.33

Subnet Mask255.255.255.0

Number of Subnet Bits8

Number of Subnets28 = 256 subnets


Number of Usable Hosts per Subnet28 = 256 2 = 254 hosts per subnet


IP Address of First Host on this Subnet172.30.1.1



FORMULA :2726252423222120

1286432168421



Determining the Number of Subnets and Hosts

Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed to step four.172.25.114.250255.255.255.192 = 11111111 11000000

Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 11000000 = 10 ones.210 = 1,024 2 = 1,022 usable subnets created.

Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 00000000 = 8 zeros. 28 = 256 2 = 254 usable host addresses created.-Number of Host Bit per Subnet 8 bit Number of Usable Hosts per Subnet 28 = 256 2 = 254 hosts per subnet Number of Subnet bit 8 Number of Subnets 28= 256 subnets (all 0s used,all 1s not used)

Problem 2Host IP Address172.30.1.33

Subnet Mask255.255.255.252




Number of Usable Hosts per Subnet22 = 4 2 = 2hosts per subnet





FORMULA :2726252423222120

1286432168421




Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed to step four.172.30.1.33255.255.255.252 = 11111111 11111100 (host octets only)

Count the total number of ones in the host octet(s) of the subnet mask. Call this number X. Raise 2 to the power of X. Use the Power of 2 chart if necessary. This is the number of potential subnets created by mask. Two of these potential subnets are normally not usable.11111111 11000000 = 14 ones. 214 = 16384 usable subnets created.Number of Subnet bit14Number of Subnets1214 = 16384 subnets(all 0s used, all 1s not used)

Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 11111100 = 2 zeros. 22 = 4 2 = 2 usable host addresses created.-Number of Host Bit per Subnet 8 bit Number of Usable Hosts per Subnet 22 = 4 2 = 2 hosts per subnet











FORMULA :2726252423222120

1286432168421




Look at the host octet(s) in the subnet mask.Use the Possible Masks chart to determine which bits are set to one. If no bits are set to one, there are no subnets. If any bits are set to one, proceed the step .192.192.10.230255.255.255.0 = 11111111 00000000 (host octets only)


Count the total number of zeros in the host octet(s) of the subnet mask. Call this number Y. Raise 2 to the power of Y. Use the powers of 2 chart if necessary. This is the number of potential subnets created by the mask. Two of these numbers are never used to address hosts. 11111111 00000000= 8 zeros. 28 = 256 2 = 254 usable host addresses created.-Number of Host Bit per Subnet 8 bit Number of Usable Hosts per Subnet 28 = 256 2 = 254 hosts per subnet











FORMULA :2726252423222120

1286432168421

















FORMULA :2726252423222120

1286432168421








Subnet Mask255.255.255.252









FORMULA :2726252423222120

1286432168421







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Ip Address Classes