IP Handouts

8/13/2019 IP Handouts

1/4

IP

ADDRESS

C

LASSES

CLASS

CLAS

S

ID*

START

IP

END

IP

A

0

1.

0.

0.

0

00000001.

00000000.

0000

0000.

00000000

126.25

5.255.255

01111111.

1111

1111.

11111111.

11111111

B

10

128.

0.

0.

0

10000000.

00000000.

0000

0000.

00000000

191.25

5.255.255

10111111.

1111

1111.

11111111.

11111111

C

110

192.

0.

0.

0

11000000.

00000000.

0000

0000.

00000000

223.25

5.255.255

11011111.

1111

1111.

11111111.

11111111

*Class

IDrefers

to

the

leading

bitsofa

lladdresses

in

th

eclass.

RESERvEd

(AKA

pRivAtE/unRou

tAblE)

AddRES

S

blocKS

CLASS

PREFIX

SIZE

*

START

IP

END

IP

A

8

Bi

t

10.

0.

0.

0

00001010.

00000000.

0000

0000.

00000000

010.25

5.255.255

00001010.

1111

1111.

11111111.

11111111

B

12

Bit

172.

16.

0.

0

10101100.

00010000.

0000

0000.

00000000

172.3

1.255.255

10111111.

0001

1111.

11111111.

11111111

C

16

Bit

192.168.

0.

0

11000000.

10101000.

0000

0000.

00000000

192.16

8.255.255

11000000.

1010

1000.

11111111.

11111111

lin

K-

locAl#

169.254.

0.

1

10101001.

11111110.

0000

0000.

00000001

169.25

4.254.255

10101001.

1111

1110.

11111110.

11111111

lo

opbAcK

127.

0.

0.

1

01111111.

00000000.

0000

0000.

00000001

127.25

5.255.254

01111111.

1111

1111.

11111111.

11111110

*Prefix

Sizerefers

to

the

leading

bitso

falladdresses

in

the

block.

#

Link

-localaddressesa

reselfassignedwh

ena

DHCPaddressi

sunavailable.

B

inary

Value

128

64

32

16

8

4

2

1

Sum

128

192

224

24

0

248

252

254

255

Pow

erof

2

28

27

26

2

5

24

23

22

21


2/4

SubnEt MASK bitS255.255.255.24011111111.11111111.11111111.11110000

^ ^Network Bits Host Bits

Network Bits (ones) are always next to each other, starting at the left.

Host Bits (zeroes) are always next to each other,

starting at the right.

2# Network Bits- 2 = Number of Available Subnets

2# Host Bits- 2 = Number of Available Host Addresses

(Remember, first and last arent legal)

incREMEnt oR blocK SizEThere are several ways to find the increment:

= 256 - (rightmost octet more than 0)= rightmost Network Bit (binary 1)= 2Host Bits

SubnEt id orSubnEt nuMbERYoull be given an IP and asked to find what subnet

it is in. Youll need the increment or subnet mask.

Find what class the IP is:

(first octet is: A=1-126; B=128-191; C=192-223)

Then write the Se Maskover the IP AddressIf the Se Maskis 255, copy the IP octet below

If the Se Maskis 0, copy the zero belowIf not 255 or 0, divide Maskoctet by IncrementRound down for your Subnet Number

Take Incrementtimes Subnet Numberand copy it foryour complete Subnet IP Address

BinaryValue 128 64 32 16 8 4 2 1Sum 128 192 224 240 248 252 254 255

Power of 2 27 26 25 24 23 22 21 20

255.255.255.24011111111.11111111.11111111.11110000

256-240=16V

^10000 = 16 Binary Decimal

^24=16=2x2x2x2

255.255.240. 0

177.168. 83.101 177.168.???. 0 83 / (256-240) 5.188 ^ 3rd IP Octet ^ Increment (its 16)

Youre in Subnet 5 ??? = 5

(Subnet)x16

(increment)=80

Subnet IP

177.168.80.0

SubnEt MASKinG


3/4

SubnEt MASKinG EXAMplESubnet Mask: 255.255.255.224IP License: 198.203. 18. 0Subnet Mask = 255.255.255.224; 224 in Binary = 11100000

Interval or Block Size = 2H; Count the Zeros = 5; 25= 32 -> Increment = 32

Usable Subnets = 2N- 2; Count the Ones = 3; 23= 8; 8 - 2 = 6 -> 6 usae SesUsable Hosts = Increment - 2; 32(Increment)- 2 = 30 -> 30 usae Hss per Se

the Firs and las Subne, and he Firs and las Hos on each Subne are no usabe.

SN# Subnet ID First Host las Hs braas Aress

1 198.203. 18. 0 198.203. 18. 1 198.203. 18. 30 198.203. 18. 31

2 198.203. 18. 32 198.203. 18. 33 198.203. 18. 62 198.203. 18. 63

3 198.203. 18. 64 198.203. 18. 65 198.203. 18. 94 198.203. 18. 95

4 198.203. 18. 96 198.203. 18. 97 198.203. 18.126 198.203. 18.127

5 198.203. 18.128 198.203. 18.129 198.203. 18.158 198.203. 18.158

6 198.203. 18.160 198.203. 18.161 198.203. 18.190 198.203. 18.191

7 198.203. 18.192 198.203. 18.193 198.203. 18.222 198.203. 18.223

8 198.203. 18.224 198.203. 18.225 198.203. 18.254 198.203. 18.255

Subnet ID + 1 Next Subnet - 2 Next Subnet - 1


4/4

vlSM:vARiAblE lEnGtH SubnEt MASKinGSubnet Mask: 255.255.255.224IP License: 198.203. 18. 0Standard subnetting provides a relatively small number of subnetworks, all with the same number of hosts.

VLSM allows us to split up these subnets into smaller and smaller groups, to use the addresses efficiently.

Say we were asked to subnet the 192.168.1.0/24 cidR k, and the following were the network sizes needed:

55, 2, 28, 2, 11, 2, 2. (Networks of 2 hosts are usually between routers.)

Now, if we used standard subnetting, we would have to use the 255.255.255.240 Se Maskto provide 14 subnets(because 6 isnt enough) but there are only 14 available hosts in each subnet! What do we do?

Lets start by sorting the necessary subnet sizes from largest to smallest: 55, 28, 11, 2, 2, 2, 2

Now, figure out what Subnet Mask is needed to fit 55 hosts (Check on your binary reference for powers of 2):

2H- 2 = Number of Available Hosts; So we need 2Hto be at least (55 + 2). The smallest power of 2 bigger

than 55 is 2

6

= 64. So that means we have 6 Host Bits (from 2

6

). That means the last octet has six zeros.11000000. Use your cheat sheet to find that means our overall subnet mask is 255.255.255.192.

Lets see whats happening in the diagram at the bottom:

We can take one of those two available subnets and give it to the office that needs 55 hosts.

Next, we need to fit 28. 25= 32, that will fit. That makes our last octet 11100000 or 255.255.255.224.

That split our subnet into two equal subnets with 32 hosts available. Lets do it again.

We need a subnet to fit 11 hosts now. 24= 16, that will fit. 11110000 is our new ending octet, or

255.255.255.240. That gives us two subnets with 16 hosts each. One goes to the office with 11 hosts, and

well split up the last one.

We need a Subnet Mask to support four groups of two hosts. So we split that last subnet twice instead of once. 23= 8... 22= 4. That means we use 11111100 as our last octet, or 255.255.255.252.

(Connections between two routers always use this subnet mask.)

1 2 3 4

nEtwoRKS in SubnEt MASK255.255.255.192:192.168.1.0(SN ID)

192.168.1.64 (fs 55 hss)

192.168.1.128

192.168.1.192 (bRoAdcASt)

nEtwoRKS in SubnEt MASK255.255.255.224:192.168.1.128 (fs 28 hss)

192.168.1.160


192.168.1.176


192.168.1.180 (fs 2 hss)

192.168.1.184 (fs 2 hss)

192.168.1.188 (fs 2 hss)

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