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4/21


IP ADDRESS CLASSES

HOW TO CALCULATION FOR IP ADDRESS CLASSES

TASK 1: For a given IP Address, Determine Network Information

Given:

INDEX RESULT

Host IP Address 172.25.114.250

Network Mask 255.255.0.0 (/16)

Find:

Network Address 172.25.0.0

Network Broadcast Address 172.25.255.255

Total Number of Hosts Bits 65534Number of Hosts 2

16

Step 1: Determine the Network Address

Network Address = 172.25.0.0

Decimal Number

(IP Address)

128+32+8+4=172

16+8+1=25

0=0

0=0

: Just a coloured box that can only be added.

Step 2: Determine the Broadcast Address for the Network Address

Network Broadcast Address:

172.25.255.255

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 0 1 0 1 1 0 0

0 0 0 1 1 0 0 1

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Binary Number

Decimal Number

(IP Address)

128+32+8+4=172

16+8+1=25

128+64+32+16+8+4+2+1=255

128+64+32+16+8+4+2+1=255

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

1 0 1 0 1 1 0 0

0 0 0 1 1 0 0 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

Binary Number


5/21


How to Determine the Host ID and Host ID Number

Steps as follows:

a. Aware of Any IP Classes will we find host, the aim is to determine the amount of

net bits or id host id (a).

b. To know the subnet bits, can be determined by:

2n = Subnet

n = lots of bits Subnet

c. Now that you know a lot of bits used subnet, then we find a lot of bits used to

calculate the number of host id, by the way:

b = a-n

d. Formulas Total Host ID = 2b-2

Okay, after knowing the steps needed to find and count the number of hosts, it is time to

find the answer:

Q: Ip Address: 192.168.1.0 Subnet there are 4

Diket: Class C

JML bit host id = 8 bits

JML bit subnet: 2n = 4, n = 2 bits

JML bit of waste: 8-2 = 6, then b = 6

Number of Host Bits

10101100 : 00011001 : 00000000 : 00000000

8x2=2432-24=16

Host bits = 16

Total number of hosts:

2^16

= 65536

65536-2 = 65534 (Address that cannot use the all OS Address, Network Address, or the all

1st

Address, Broadcast Address).

Classes of IP Address

The following are the classes of IP addresses.

Class AThe first octet denotes the network address, and the last three octets are the host portion. AnyIP address whose first octet is between 1 and 126 is a Class A address. Note that 0 is reserved as apart of the default address, and 127 is reserved for internal loopback testing.

Class BThe first two octets denote the network address, and the last two octets are the host portion.Any address whose first octet is in the range 128 to 191 is a Class B address.

Class CThe first three octets denote the network address, and the last octet is the host portion. Thefirst octet range of 192 to 223 is a Class C address.

Class DUsed for multicast. Multicast IP addresses have their first octets in the range 224 to 239.

Class EReserved for future use and includes the range of addresses with a first octet from 240 to 255.


6/21


TASK 2: CHALLAGE

Problem 1

Host IP Address 172.30.1.33Network Mask 255.255.0.0



Total Number of Host Bits 65,536

Number of Hosts 216

Network Address = 172.30.0.0

Decimal Number

(IP Address)

128+32+8+4=172

16+8+4+2=30

0=0

0=0

Network Broadcast Address = 172.30.255.255

Number of Host Bits:

10101100 : 00011110 : 00000000 : 00000000

8x2=16

32-16=16

Host bits = 16

Total Number of Hosts:

2^16= 65536

65536-2 = 65534

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 0 1 0 1 1 0 0

0 0 0 1 1 1 1 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Binary Number

Decimal Number(IP Address)

128+32+8+4=172

16+8+4+2=30

128+64+32+16+8+4+2+1=255

128+64+32+16+8+4+2+1=255

2

7

2

6

2

5

2

4

2

3

2

2

2

1

2

0

128 64 32 16 8 4 2 1

1 0 1 0 1 1 0 0

0 0 0 1 1 0 0 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

Binary Number


7/21


Problem 2


Network Mask 255.255.255.0



Total Number of Host Bits 254

Number of Hosts 28

Network Address =172.30.1.0

Decimal Number

(IP Address)

128+32+8+4=172

16+8+4+2=30

1=1

0=0


Nu

mb

er of Host Bits:

10101100 : 00011110 : 00000001 : 00000000

8x3=24

32-24=8

Host bits = 8

Total Number of Host:2

^8=256

256-2=254

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 0 1 0 1 1 0 0

0 0 0 1 1 1 1 0

0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 0

Binary Number

Decimal Number

(IP Address)

128+32+8+4=172

16+8+4+2=301=1

128+64+32+16+8+4+2+1=255

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 0 1 0 1 1 0 0

0 0 0 1 1 1 1 0

0 0 0 0 0 0 0 1

1 1 1 1 1 1 1 1

Binary Number


8/21


Problem 3






Number of Hosts 28

Network Address =

192.168.10.0

Decimal Number

(IP Address)

128+64=192

128+32+8=168

8+2=10

0=0



11000000 : 10101000 : 00001010 : 00000000

8x3=24

32-24=8

Host bits = 8

Total Number of Host:

2^8=256

256-2=254

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0

1 0 1 0 1 0 0 0

0 0 0 0 1 0 1 0

0 0 0 0 0 0 0 0

Binary Number

Decimal Number

(IP Address)

128+64=192128+32+8=168

8+2=10

128+64+32+16+8+4+2+1=255

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0

1 0 1 0 1 0 0 0

0 0 0 0 1 0 1 0

1 1 1 1 1 1 1 1

Binary Number


9/21


Problem 4






Number of Hosts 216

Network Address =172.17.0.0

Decimal Number

(IP Address)

128+32+8+4=172

16+1=17

0=0

0=0


Nu

mb

er

of

Host Bits:

10101100 : 00010001 : 00000000 : 00000000

8x2=16

32-16=16

Host bits = 16


2^16= 65536

65536-2 = 65534

Problem 5


27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 0 1 0 1 1 0 0

0 0 0 1 0 0 0 1

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Binary Number

Decimal Number

(IP Address)

128+32+8+4=17216+1=17

128+64+32+16+8+4+2+1=255

128+64+32+16+8+4+2+1=255

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 0 1 0 1 1 0 0

0 0 0 1 0 0 0 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

Binary Number


10/21






Number of Hosts 216

Network Address =

192.168.0.0

Decimal Number

(IP Address)

128+64=192

128+32+8=168

0=0

0=0



11000000 : 10101000 : 00000000 : 00000000

8x2=16

32-16=16

Host bits = 16


2^16= 65536

65536-2 = 65534

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0

1 0 1 0 1 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Binary Number

Decimal Number

(IP Address)

128+64=192

128+32+8=168

128+64+32+16+8+4+2+1=255128+64+32+16+8+4+2+1=255

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0

1 0 1 0 1 0 0 0

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

Binary Number


11/21


Problem 6


Network Mask 255.255.255.224




Number of Hosts 25

Network Address =

192.168.3.192

Decimal Number

(IP Address)

128+64=192

128+32+8=168

2+1=3

128+64=192


Number of Host Bits:11000000 : 10101000 : 00000011 : 11000000

Subnet #6 = Subnet ID 110 / Host #0 = Host ID 00000

Host Bits = 5


2^5 = 32

32-2 = 30

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0

1 0 1 0 1 0 0 0

0 0 0 0 0 0 1 1

1 1 0 0 0 0 0 0

Binary Number

Decimal Number

(IP Address)

128+64=192

128+32+8=168

2+1=3

128+64+16+8+4+2+1=223

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0

1 0 1 0 1 0 0 0

0 0 0 0 0 0 1 1

1 1 0 1 1 1 1 1

Binary Number


12/21



13/21


TASK 2

TASK 1: For a given IP Address, Determine Network Information.

Given:

INDEX RESULT

Host IP Address 172.25.114.250Network Mask 255.255.0.0 (/16)

Subnet Mask 255.255.255.192 (/26)

Find:


Subnet Mask 255.255.255.192 (/26)

Number of Subnet Bits

Number of Subnet

10 Bits

210

= 1024 Subnet

Number of Host Bits per Subnet

Number of Usable Hosts per Subnet

6 Bits

= 2^6 = 64 -2 = 62 hosts per subnet

Subnet Address for this IP Address 172.25.114.192

IP Address of First Host on this Subnet 172.25.114.193

IP Address of Last Host on this Subnet 172.25.114.254

Broadcast Address for this Subnet 172.25.114.255

IP Address: 172.25.114.250

Subnet Mask: 255.255.255.192 = 11111111,11111111,11111111,11000000

NUMBER OF SUBNET BITS

Network = 11111111,11000000= 2^10 = 1024

: Host = 2^10 = 1024 Subnet

: 10 Bits

NUMBER OF HOST BITS PER SUBNET

: 11111111,11000000 = 2^6

: 6 Bits

NUMBER OF USABLE HOST PER SUBNET



14/21


Major Divide (M.D) Subnet Divide (S.D)

IP Address 172.25.114.250 10101100 00011001 01110010 11111010

Subnet

Mask

255.255.255.19211111111 11111111 11111111 11000000

Subnet

Address

172.25.114.19210101100 00011001 01110010 11000000

-Subnet

Counting Range-Host CountingRange

First Host 172.25.114.193 10101100 00011001 01110010 11000001

172 25 114 193

Last Host 172.25.114.254 10101100 00011001 01110010 11111110

172 25 114 254

Broadcast 172.25.114.255 10101100 00011001 01110010 11111111

172 25 114 255


15/21


16/21


Problem 2


Subnet Mask 255.255.255.252


Number of Subnet

14 Bits

= 2^14 = 16,384 Subnet



14 Bits

= 2^2 = 4 2 = 2 hosts per subnet





IP Address: 172.30.1.33

Subnet Mask: 255.255.255.252 = 11111111,11111111,11111111,00000011


Network = 11111111,00000011 = 2^14 = 16,384

: Host = 2^14 = 16,384 Subnet

: 14 Bits


: 2^2

: 2 Bits


= 2^2 = 4 2 = 2 hosts per subnet


IP Address 172.30.1.33 10101100 00011110 000000001 00100001

Subnet

Mask

255.255.255.25211111111 11111111 111111111 11111100

Subnet

Address

172.30.1.3210101100 00011110 00000001 00100000

-Subnet


First Host 172.30.1.33 10101100 00011110 00000001 00100001

172 30 1 33

Last Host 172.30.1.62 10101100 00011110 00000001 00111110

172 30 1 62

Broadcast 172.30.1.63 10101100 00011110 00000001 00111111

172 30 1 63


17/21


Problem 3


Subnet Mask 255.255.255.0


Number of Subnet

8 Bits

= 2^8 = 256 Subnet



8 Bits

= 2^8 = 256 2 = 254 hosts per subnet





IP Address: 192.192.10.234

Subnet Mask: 255.255.255.0 = 11111111,11111111,11111111,00000000


Network = 11111111,00000000 = 2^8 = 256

: Host = 2^8 = 256 Subnet

: 8 Bits


: 2^8

: 8 Bits


= 2^8 = 256 2 = 254 hosts per subnet


IP Address 192.192.10.234 11000000 11000000 00001010 11101010

Subnet

Mask

255.255.255.011111111 11111111 11111111 00000000

Subnet

Address

192.192.10.011000000 11000000 00001010 00000000

-Subnet


First Host 192.192.10.1 11000000 11000000 00001010 00000001

192 192 10 1

Last Host 192.192.10.254 11000000 11000000 00001010 11111110

192 192 10 254

Broadcast 192.192.10.255 11000000 11000000 00001010 11111111

192 192 10 255


18/21


Problem 4




Number of Subnet

0 Bits

= 2^0 = 1 Subnet



16 Bits

= 2^16 = 65536 2 = 65534 hosts per subnet





IP Address: 171.17.99.71

Subnet Mask: 255.255.0.0 = 11111111,11111111,00000000,00000000


Network =00000000,00000000 = 2^0 = 1

: Host = 2^0 = 1

: 0 Bits


: 2^16

: 16 Bits


= 2^16 = 65536 2 = 65534 hosts per subnet


IP Address 171.17.99.71 10101011 00010001 01100011 01000111

Subnet

Mask

255.255.0.011111111 11111111 00000000 00000000

Subnet

Address

171.17.0.010101011 00010001 00000000 00000000

-Subnet


First Host 171.17.0.1 10101011 00010001 00000000 00000001

171 17 0 1

Last Host 171.17.255.254 10101011 00010001 11111111 11111110

171 17 255 254

Broadcast 171.17.255.255 10101011 00010001 11111111 11111111

171 17 255 255


19/21


Problem 5




Number of Subnet

8 Bits

= 2^8 = 256 Subnet



8 Bits

= 2^8 = 256 2 = 254 hosts per subnet





IP Address: 192.168.3.219

Subnet Mask: 255.255.255.0 = 11111111,11111111,11111111,00000000


Network = 11111111,00000000 =2^8 = 256

: Host = 2^8 = 256 Subnet

: 8 Bits


: 2^8

: 8 Bits


= 2^8 = 256 2 = 254 hosts per subnet


IP Address 192.168.3.219 11000000 10101000 00000011 11011011

Subnet

Mask

255.255.255.011111111 11111111 11111111 00000000

Subnet

Address

192.168.3.011000000 10101000 00000011 00000000

-Subnet


First Host 192.168.3.1 11000000 10101000 00000011 00000001

192 168 3 1

Last Host 192.168.3.254 11000000 10101000 00000011 11111110

192 168 3 254

Broadcast 192.168.3.255 11000000 10101000 00000011 111111111

192 168 3 255


20/21


Problem 6


Subnet Mask 255.255.255.252


Number of Subnet

14 Bits

= 2^14 = 16,384 Subnet



2 Bits






IP Address: 192.168.3.219

Subnet Mask: 255.255.255.252 = 11111111,11111111,11111111,00000011


Network = 11111111,00000011 = 2^14 = 16,384

: Host = 2^14 = 16,384 Subnet

: 14 Bits


: 2^2

: 2 Bits




IP Address 192.168.3.219 11000000 10101000 00000011 11011011

Subnet

Mask

255.255.255.25211111111 11111111 11111111 11111100

Subnet

Address

192.168.3.21611000000 10101000 00000011 11011000

-Subnet


First Host 192.168.3.217 11000000 10101000 00000011 11011001

192 168 3 217

Last Host 192.168.3.222 11000000 10101000 00000011 11011110

192 168 3 222

Broadcast 192.168.3.223 11000000 10101000 00000011 11011111

192 168 3 223


21/21


NOTE:

Step One

Use the first octet of the IP address to determine the class of address (A, B, or

C).

169.199.109.137 = CLASS B255.255.255.192

Step Two

Use the class of the address to determine which octets are available for hosts.

CLASS B = Network. Network.Host. Host169.199.109.137255.255.255.192

Step Three

Look at the host octet(s) in the subnet mask. Use the"Possible Masks" charttodetermine 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.

169.199.109.137255.255.255.192 = 11111111 11000000 (host octets only)

Step Four

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"Powers of 2" chartif necessary.This is the number of potential subnets created by the mask. Two of thesepotential subnets are normally not usable.

11111111 11000000 = 10 ones. 210

= 1,024 - 2 = 1,022 usable subnets created.

Step Five

Count the total number of zeros in the host octet(s) of the subnet mask. Call thisnumber 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.
http://www.chabotcollege.edu/faculty/netacad/prot/sem2/reference/ipadd.dochttp://www.chabotcollege.edu/faculty/netacad/prot/sem2/reference/ipadd.dochttp://www.chabotcollege.edu/faculty/netacad/prot/sem2/reference/ipadd.dochttp://www.chabotcollege.edu/faculty/netacad/prot/sem2/reference/ipadd.dochttp://www.chabotcollege.edu/faculty/netacad/prot/sem2/reference/ipadd.dochttp://www.chabotcollege.edu/faculty/netacad/prot/sem2/reference/ipadd.dochttp://www.chabotcollege.edu/faculty/netacad/prot/sem2/reference/ipadd.dochttp://www.chabotcollege.edu/faculty/netacad/prot/sem2/reference/ipadd.doc

kfc 2044 network fundamental

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