ip addresing games

8/22/2019 IP Addresing Games

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IP Addressing Games (They'll drive you crazy!)

1 Public or private?

2 Slash It! (An alternative subnet notation)

3 Bits on parade! (Identify the network, subnet and host bits in an IP address)-> Spot the network address

-> Spot the broadcast address

4 Calc those subnets! (How many subnets and how many hosts per subnet)

5 Reveal the mask - Part 1 (Calculate the subnet mask for a required number

6 Reveal the mask - Part 2 (Calculate the subnet mask for a required number

7 Spot the disaster(or collect your P45).

Future ideas for games

Binary weights ( for toning the mind ).

Convert those numbers!

m.t.stanhope

Dec-05

Updated : 17 Sept 06


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f subnets)

f subnets)


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Public or Private?

IP Address

Most significant byte

in binary Class

1 192.168.239.24 192= 11000000 C2 185.45.130.7 185= 10111001 B

3 10.220.34.188 10= 00001010 A

4 172.31.234.17 172= 10101100 B

5 192.168.253.230 192= 11000000 C

6 198.34.87.68

7 172.15.230.95 172= 10101100 B

8 172.23.34.2 172= 10101100 B

9

87.238.65.310 172.32.45.243 172= 10101100 B

Useful Reference

Class

Class Identification.

Most significant bit of IP

address starts with Private range in slash notation Private Addr

A 0 10.0.0.0/8

B 10 172.16.0.0/12C 110 192.168.0.0/16


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Public or Private?

ss Ranges in dotted decimal notation

10.0.0.0 - 10.255.255.255

172.16.0.0 - 172.31.255.255192.168.0.0 - 192.168.255.255


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Slash It! Complete the gaps in the table

Slash notation Dotted decimal Binary Notati

1

/24 255.255.255.0 1111 1111 1111 11112 /16

3 /8 1111 1111 0000 0000

4 /25

5 /17

6 1111 1111 1000 0000

7 /26

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9 /1010 1111 1111 1111 1111

11 /19

12 /11

13 255.255.255.240

14 /20

15 /12

16 255.255.255.248

17 /2118 1111 1111 1111 1000


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on

1111 1111 0000 0000

0000 0000 0000 0000

0000 0000 0000 0000

1111 1111 1110 0000

0000 0000 0000 0000


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Bits on parade (identify the network, subnet a

IP Address Subnet Mask Class

1

192.168.239.24 /24 C 1100 0000255.255.255.0 1111 1111

2 192.168.238.95 /27 C 1100 0000

255.255.255.224 1111 1111

3 192.168.237.48 /28 1100 0000

255.255.255.240 1111 1111

4 192.168.236.31 /29 1100 0000

5 192.168.236.13 /30 1100 0000

6 182.20.36.48 /16 1011 0110

7 182.20.63.255 /19 1011 0110

8 182.20.128.0 /19 1011 0110

9 84.34.68.7 /8 0101 0100

10 84.127.255.255 /11 0101 0100

Tips1. Write the subnet mask in binary. Colour to 'default' 1s part i

2. Convert the IP address into binary

3. Colour the bits of the IP address to match the colours used i

4. The IP structure is now as follows: the blue part is the netwo

and the pink part is the host number.

> A network address has all zeros in the host section of the IP

> A broadcast address has all 1s in the host section of the IP a


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> The host section of an IP address is that part lining up with z


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d host bits).

IP Address in binary What is it?

1010 1000 1110 1111 0001 1000Network host No. 24

1111 1111 1111 1111 0000 0000

1010 1000 1110 1110 01011111 Subnet 2, B/Cast

1111 1111 1111 1111 1110 0000

1010 1000

1111 1111 1111 1111 1111 0000

1010 1000 1110 1100 0001 1111

1010 1000 1110 1100 0000 1101

0001 0100 0010 0100 0011 0000

0001 0100 0111 1111 1111 1111

0001 0100 1000 0000 0000 0000

0010 0010 0100 0100 0000 0111

0111 111 1111 1111 1111 1111

blue and all other 1s in orange and the zeros in pink.

the subnet mask.

k id, the orange the subnet id

ddress.

dress.


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ros in the subnet mask.


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Calc those subnets! (How many subnets and ho

IP Address SNM ClassDefault number

of network bits

Default number

of network bits

1 192.168.239.0 /27 C 24 82 192.168.23.0 /28 C 24 8

3 192.168.10.0 /29

4 195.34.56.0 /26

5 190.34.0.0 /23 B 16 16

6 172.16.0.0 /25

7 86.0.0.0 /21

8 77.0.0.0 /16

9

10


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many hosts per subnet?).

s h

Usable

numberof

subnets

Usable

number ofhosts per

subnet

Number of

redefined

host bits,

i.e. the

subnet bits

Number of

remaining

hosts bits 2s

2h

- 2

27-24=3 8-3=52

3= 8 2

5- 2 = 30

28-24=4 8-4=4

23-16=7 16-7=9


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Reveal the mask - Part 1 (Determine the subnet mask for a re

ClassNo of subnets

required

Nearest power

of 2 to contain

the requirednumber Subnet mas

1 A 60 6 (as 26=64) 11111111 1111 1100

2 C 30 5 (as 25=32) 11111111 11111111

3 B 250

4 C 5

5 B 12

6 A 500 9 (as 29=512) 11111111 1111 1111

7 A 6

8 B 125 8 (as 28=128) 11111111 1111 1111

9 C 25

TipsWork out the power of 2.

Write down the default subnet mask for the class. Colour the 1s blu

Change a number of zeros (e.g. 6 in the first example) in the subnet


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uired number of subnets).

in binary

Resulting

number of

hosts persubnet

SNM in dotteddecimal

0000 0000 0000 0000 218

255.252.0.0

11111111 1111 1000 23

255.255.255.0

1000 0000 0000 0000 215

255.255.128.0

1111 1111 0000 0000 28

255.255.255.0

and zeros pink.

mask to ones and colour them orange.


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Reveal the mask - Part 1 (Determine the subnet mask for a requir

Class

Required

number of

hosts persubnet

Nearest power of 2

to contain therequired number Subnet mas

1 A 8000 13 (as 213

=8192) 11111111 1111 1111

2 C 25 5 (as 25=32) 11111111 11111111

3 B 500

4 C 12

5 B 700

6 A 32000 15 (as 215=32768) 11111111 1111 1111

7 A 15000

8 B 1000 10 (as 210

=1024) 11111111 1111 1111

9 C 50

TipsWork out the power of 2.

Write down the default subnet mask for the class. Colour the 1s blue a

Only keep the zeros required (e.g. 13 of them in example 1) then chan


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d number of hosts per subnet).

in binary

Resulting

number ofsubnets

SNM in dotteddecimal

1110 0000 0000 0000 211

255.255.224.0

11111111 1110 0000 23

255.255.255.224

1000 0000 0000 0000 29

255.255.128.0

1111 1100 0000 0000 26

255.255.252.0

d zeros pink.

e the other zeros to orange 1s.


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Spot the Disaster (or collect your P45).

Checking rules (in the order of the easiest to do first)

1 Check that the hosts and the default gateway all have the sam

2 Check that router interface addresses belong to different netw3 Check that hosts and the default gateway (i.e the router interfa

4 Check that the hosts and the default gateway have not been as

5 Check that the hosts and the default gateway have not been as

6 Check the default gateway setting on the hosts match the actu

Scenario diagram

Scenairo settings

Host 1 Host 2 Default Gateway (router)

1 192.168.239.65/24 192.168.239.66/24 192.168.239.1/24

2 192.168.238.65/27 192.168.239.66/27 192.168.239.254/27

3 192.168.239.65/27 192.168.239.93/27 192.168.239.129/27

4 192.168.239.18/24 192.168.239.60/24 192.168.239.254/24

5 192.168.239.65/27 192.168.239.66/28 192.168.239.254/27

6 192.168.239.65/24 192.168.239.66/24 192.168.238.1/24

7 192.168.239.66/27 192.168.239.65/27 192.168.239.98/27

8 192.168.239.65/24 192.168.239.255/24 192.168.239.1/24

9 192.168.239.0/24 192.168.239.66/24 192.168.239.254/24

10 192.168.238.65/27 192.168.239.64/27 192.168.239.90/27

11 192.168.238.92/27 192.168.239.91/27 192.168.239.65/27

12 172.28.16.61/26 172.28.16.63/26 172.28.16.62/26

ROUTER

SWITCH

HOSTNo.1

HOSTNo.2

ISP/Inte

DG


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subnet mask

rkse) are all on the same network

signed a network address

signed a broadcast address

l default gateway address on the router interface.

Other interface on the router Problem? (assume all subnets allowed)

192.168.240.2/24

not shown

not shown

192.168.239.253/24

not shown

192.168.240.2/24

not shown

192.168.240.2/24

192.168.240.2/24

not shown

not shown

Not shown

rnet

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