KFC 2044 NETWORK FUNDEMENTAL

KFC 2044 NETWORK FUNDEMENTAL

Hazman Bin Abdul Rahim900515-14-6601

MOHD SHUKRI BIN MUHAMAD HUSIN


Hazman Bin Abdul Rahim

1.2 IP Address Classes

Task 1: For given IP address, Determine Network Information.

Given:

Host IP Address 172.25.114.250Network Mask 255.255.0.0(/16)

Find:

Network Address 172.25.0.0Network Broadcast Address 172.25.255.255Total Number of Host Bits 65534Number of Hosts 16

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

Convert the host IP address and network mask to binary:

IP Address : 172.25.114.250

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 0 1 0 1 1 0 0

128+32+8+4=172

IP Address : 172.25.114.250

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 1 1 0 0 1

16+8+1=25

IP Address : 172.25.114.250

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 1 1 1 0 0 1 0

64+32+16+2=114

IP Address : 172.25.114.250

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 0 1 0



128+64+32+16+8+2=250



Network Mask : 255.255.0.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0



Step 2: Determine THE NETWORK ADDRESS.

172 25 114 250

IP Address10101100 00011001 01110010

11111010

255 255 0 0

Subnet mask 11111111 11111111 00000000 00000000

Network Address 10101100 00011001 00000000 00000000

172 25 0 0

EXP:

172 25 114 25010101100 00011001 01110010 11111010

X255 255 0 011111111 11111111 00000000 00000000

172 25 0 010101100 00011001 00000000 00000000



Multiply

Answer

IP Address

Subnet mask

Network Address

Step 3:Determine the BROADCAST ADDRESS for the network address

172 25 0 0Network Add. 10101100 00011001 00000000 00000000

Mask 11111111 11111111 00000000 00000000

0x0 = 1 / 1x1 = 1 / 0x1 or 1x0 = 0

Broadcast. 10101100 00011001 11111111 11111111172 25 255 255

By counting the number of host bits, we can determine the total number of usable hosts for this network.

Look at the host octet(s) in the subnet mask.255.255.0.0 = 00000000 00000000 (host octets only)

Count the total number of zeros in the host octet(s) of the subnet mask.00000000 00000000 = 16 zeros.

Host bits: 16Total number of hosts:

216 = 65,536

65,536 - 2 = 65,534 (addresses that cannot use the all 0s address, network address, or the all 1s address, broadcast address.)

The answer:Host IP Address 172.25.114.250Network Mask 255.255.0.0 (/16)Network Address 172.25.0.0Network Broadcast Address 172.25.255.255Total Number of Host Bits 65,534Number of Hosts 16



Task 2: Challenge

For all problems:Create a Subnetting Worksheet to show and record all work for each problem.

Problem 1Host IP Address 172.30.1.33Network Mask 255.255.0.0Network Address Network Broadcast AddressTotal Number of Host BitsNumber of Hosts

IP Address : 172.30.1.33

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 0 1 0 1 1 0 0

128+32+8+4=172

IP Address : 172.30.1.33

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 1 1 1 1 0

16+8+4+2=30

IP Address : 172.30.1.33

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 1

=1

IP Address : 172.30.1.33

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 1 0 0 0 0 1

32+1=33




27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0



Step 2: Determine the NETWORK ADDRESS.

172 30 1 33

IP Address10101100 00011110 00000001

00100001

255 255 0 0

Subnet mask 11111111 11111111 00000000 00000000

Network Address 10101100 00011110 00000000 00000000

172 30 0 0

EXP:

172 30 1 3310101100 00011110 00000001 00100001

X255 255 0 011111111 11111111 00000000 00000000

172 30 0 010101100 00011110 00000000 00000000



Multiply

Answer

IP Address

Subnet mask

Network Address


172 30 0 0Network Add. 10101100 00011110 00000000 00000000Mask 11111111 11111111 00000000 00000000

0x0 = 1 / 1x1 = 1 / 0x1 or 1x0 = 0

Broadcast. 10101100 00011110 11111111 11111111172 30 255 255





216 = 65,536


The answer:Host IP Address 172.30.1.33Network Mask 255.255.0.0Network Address 172.30.0.0Network Broadcast Address 172.30.255.255Total Number of Host Bits 65,534Number of Hosts 216




IP Address : 172.30.1.33

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 0 1 0 1 1 0 0

128+32+8+4=172

IP Address : 172.30.1.33

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 1 1 1 1 0

16+8+4+2=30

IP Address : 172.30.1.33

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 1

=1

IP Address : 172.30.1.33

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 1 0 0 0 0 1

32+1=33




27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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

Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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

Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0




172 30 1 33

IP Address10101100 00011110 00000001

00100001

255 255 255 0

Subnet mask 11111111 11111111 11111111 00000000

Network Address 10101100 00011110 00000001 00000000

172 30 1 0

EXP:

172 30 1 3310101100 00011110 00000001 11111010

X255 255 255 011111111 11111111 11111111 00000000

172 30 1 010101100 00011110 00000001 00000000



Multiply

Answer

IP Address

Subnet mask

Network Address


172 30 1 0Network Add. 10101100 00011110 00000001 00000000

Mask 11111111 11111111 11111111 00000000

0x0 = 1 / 1x1 = 1 / 0x1 or 1x0 = 0

Broadcast. 10101100 00011110 00000001 11111111172 30 1 255





28 = 256

256 - 2 = 254 (addresses that cannot use the all 0s address, network address, or the all 1s address, broadcast address.)

The answer:Host IP Address 172.30.1.33Network Mask 255.255.255.0Network Address 172.30.1.0Network Broadcast Address 172.30.1.255Total Number of Host Bits 254Number of Hosts 28




IP Address : 192.168.10.234

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 0 0 0 0 0 0

128+64=192

IP Address : 192.168.10.234

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 0 1 0 1 0 0 0

128+32+8=168

IP Address : 192.168.10.234

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 1 0 1 0

8+2=10

IP Address : 192.168.10.234

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 0 1 0 1 0

128+64+32+8+2=234



Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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

Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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

Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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

Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0




192 168 10 234

IP Address11000000 10101000 00001010

11101010

255 255 255 0

Subnet mask 11111111 11111111 11111111 00000000

Network Address 11000000 10101000 00001010 00000000

192 168 10 0

EXP:

192 168 10 23411000000 10101000 00001010 11101010

X255 255 255 011111111 11111111 11111111 00000000

192 168 10 011000000 10101000 00001010 00000000



Multiply

Answer

IP Address

Subnet mask

Network Address


192 168 10 0Network Add. 11000000 10101000 00001010 00000000Mask 11111111 11111111 11111111 00000000

0x0 = 1 / 1x1 = 1 / 0x1 or 1x0 = 0

Broadcast. 11000000 10101000 00001010 11111111192 168 10 255





28 = 256






IP Address : 172.17.99.71

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 0 1 0 1 1 0 0

128+32+8+4=172

IP Address : 172.17.99.71

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 1 0 0 0 1

16+1=17

IP Address : 172.17.99.71

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 1 1 0 0 0 1 1

64+32+2+1=99

IP Address : 172.17.99.71

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 1 0 0 0 1 1 1

64+4+2+1=71




27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0




172 17 99 71

IP Address10101100 00010001 01100011

01000111

255 255 0 0

Subnet mask 11111111 11111111 00000000 00000000

Network Address 10101100 00010001 00000000 00000000

172 17 0 0

EXP:

172 17 99 7110101100 00010001 01100011 01000111

X255 255 0 011111111 11111111 00000000 00000000

172 17 0 010101100 00010001 00000000 00000000



Multiply

Answer

IP Address

Subnet mask

Network Address


172 17 0 0Network Add. 10101100 00010001 00000000 00000000Mask 11111111 11111111 00000000 00000000

0x0 = 1 / 1x1 = 1 / 0x1 or 1x0 = 0

Broadcast. 10101100 00010001 11111111 11111111172 17 255 255





216 = 65,536






IP Address : 192.168.3.219

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 0 0 0 0 0 0

128+64=192

IP Address : 192.168.3.219

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 0 1 0 1 0 0 0

128+32+8=168

IP Address : 192.168.3.219

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 1 1

2+1=3

IP Address : 192.168.3.219

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 0 1 1 0 1 1

128+64+16+8+2+1=219 / 27+26+24+23+21+20=219




27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0


27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 0 0




192 168 3 219

IP Address11000000 10101000 00000011

11011011

255 255 0 0

Subnet mask 11111111 11111111 00000000 00000000

Network Address 11000000 10101000 00000000 00000000

192 168 0 0

EXP:

192 168 3 21911000000 10101000 00000011 11011011

X255 255 0 011111111 11111111 00000000 00000000

192 168 0 011000000 10101000 00000000 00000000



Multiply

Answer

IP Address

Subnet mask

Network Address


192 168 0 0Network Add. 11000000 10101000 00000000 00000000

Mask 11111111 11111111 00000000 00000000

0x0 = 1 / 1x1 = 1 / 0x1 or 1x0 = 0

Broadcast. 11000000 10101000 11111111 11111111192 168 255 255





216 = 65,536






IP Address : 192.168.3.219

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 0 0 0 0 0 0

128+64=192

IP Address : 192.168.3.219

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 0 1 0 1 0 0 0

128+32+8=168

IP Address : 192.168.3.219

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 10 0 0 0 0 0 1 1

2+1=3

IP Address : 192.168.3.219

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 0 1 1 0 1 1

128+64+16+8+2+1=219



Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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

Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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

Network Mask : 255.255.255.0

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 1 1 1 1 1

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

Network Mask : 255.255.255.224

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 11 1 1 0 0 0 0 0

128+64+32=224




192 168 3 219

IP Address11000000 10101000 00000011

11011011

255 255 255 224

Subnet mask 11111111 11111111 11111111 11100000

Network Address 11000000 10101000 00000011 11000000

192 168 3 192

EXP:

192 168 3 21911000000 10101000 00000011 11011011

X255 255 255 22411111111 11111111 11111111 11100000

192 168 3 19211000000 10101000 00000011 11000000



Multiply

Answer

IP Address

Subnet mask

Network Address


192 168 3 192Network Add. 11000000 10101000 00000011 11000000Mask 11111111 11111111 11111111 11100000

0x0 = 1 / 1x1 = 1 / 0x1 or 1x0 = 0

Broadcast. 11000000 10101000 00000011 11011111192 168 3 223





25 = 32





TASK/EXERCISE :

Task 1: For a given IP Address and Subnet Mask, Determine Subnet Information.

Given :

Host IP Address 172.25.114.250Network Mask 255.255.0.00 (/16)Subnet Mask 255.255.255.192 (/26)

Find :

Number of Subnet BitsNumber of SubnetsNumber of Host Bits per SubnetNumber of Usable Hosts per SubnetSubnet Address for this IP AddressIP Address of First Host on this SubnetIP Address of Last Host on this SubnetBroadcast Address for this subnet



Step 1: Translate host IP address and subnet mask into binary notation.

Host IP Address 172 25 114 250

IP Address 10101100 00011001 01110010 11111010

Subnet Mask (x) 11111111 11111111 11111111 11000000

Subnet Address 10101100 00011001 01110010 11000000 = Subnet address for

this IP Address = 172 25 114 192 127.25.114.192

IP Address of First Host 172.25.114.193192+

62 Hosts

Adress IP of host last calculation

= 254

IP Address of Last Host 172.25.114.254

Broadcast Address 172.25.144.255



MULTIPLE

ANSWER

Step OneUse the first octetof the IP addressto determine the

class of address (A, B, or C).172.25.114.250

255.255.255.192

Step TwoUse the class of the address

to determine which octets are available for hosts.Network. Network. Host. Host

172.25.114.250255.255.255.192

Step ThreeLook 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 (host

Step FourCount 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" chart if necessary. This is the number of potential subnets created by the mask. Two of these potential subnets are normally not

usable.11111111 11000000 = 10 ones.

210 = 1,024 - 2 = 1,022 usable subnets created.

Step FiveCount 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 Hosts Bit per Subnet

Number of Usable Hosts per Subnet

6 bit

26 = 64 - 2 = 62 hosts per subnet



Final Answers

Host IP Address 172.25.114.250Subnet Mask 255.255.255.192 (/26)Number of subnet BitsNumber of Subnets

10 bits210 = 1024 subnets

Number of Host Bits per SubnetNumber of Usable Hosts per Subnet

6 bits28 – 2 = 64 – 2 = 62 hosts per subnet

Subnet Address for this IP Address 172.25.114.192IP Address of First Host on this Subnet 172.25.114.193IP Address of Last Host on this Subnet 172.25.114.254Broadcast Address for this Subnet 172.25.114.255



Problem 1

Host IP Address 172.30.1.33Subnet Mask 255.255.255.0Number of Subnet Bit 8Number of Subnets 28 = 256 subnetsNumber of Host Bits per Subnet 8 bitNumber of Usable Hosts per Subnet


Subnet Address for this IP Address

172.30.1.0

IP Address of First Host on this Subnet

172.30.1.1

IP Address of Last Host on Subnet

172.30.1.254

Broadcast Address for this Subnet

172.30.1.255


IP Address 10101100 00011110 00000001 00100001

Subnet Mask (x) 11111111 11111111 11111111 00000000


this IP Address = 172 30 1 0 127.25.1.0

IP Address of First Host 172.30.1.1 0+

254 Hosts


= 254





MULTIPLE

ANSWER

Determining the Number of IP 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 the step.

172.25.1.33255.255.255.0 = 11111111 00000000 (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 "Powers of 2" chart if necessary. This is the number of potential subnets created by the mask. Two of these potential subnets

are normally not usable.

11111111 00000000 = 8 ones. 28 = 256 - 2 = 254 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 Hosts Bit per Subnet


8 bit


Number of Subnet bit

Number of Subnets

(all 0s used , all 1s not used)

8

28 = 256 subnets



Problem 2

Host IP Address 172.30.1.33

Subnet Mask 255.255.255.252

Number of Subnet Bit 14

Number of Subnets 214 = 16384 subnets

Number of Host Bits per Subnet 4 bit

Number of Usable Hosts per Subnet 22 = 4 - 2 = 2 hosts per subnet

Subnet Address for this IP Address 172.30.1.32


172.30.1.33

IP Address of Last Host on Subnet 172.30.1.34

Broadcast Address for this Subnet 172.30.1.35


IP Address 10101100 00011110 00000001 00100001

Subnet Mask (x) 11111111 11111111 11111111 11111100


this IP Address = 172 30 1 32 127 . 30 . 1 . 32


2 Hosts


= 34





MULTIPLE

ANSWER




172.30.1.33255.255.255.252 = 11111111 11111100 (host

octets only)



11111111 11111100 = 14 ones. 214 = 16384 usable subnets created.Number of Subnet bit

Number of Subnets


14

214 = 16384 subnets



11111111 11111100 = 2 zeros. 22 = 4 - 2 = 2 usable host



4 bit




Problem 3

Host IP Address 192.192.10.234Subnet Mask 255.255.255.0Number of Subnet Bit 8Number of Subnets 28 = 256 subnetsNumber of Host Bits per Subnet 8 bitNumber of Usable Hosts per Subnet 28 = 256 - 2 = 254 hosts per subnetSubnet Address for this IP Address 192.192.10.0IP Address of First Host on this Subnet 192.192.10.1IP Address of Last Host on Subnet 192.192.10.254Broadcast Address for this Subnet 192.192.10.255


IP Address 11000000 11000000 00001010 11101010

Subnet Mask (x) 11111111 11111111 11111111 00000000

Subnet Address 11000000 11000000 00001010 00000000 =

Subnet address for this IP Address = 192 192 10 0

192 . 192 . 10 . 0


254 Hosts


= 254





MULTIPLE

ANSWER




192.192.10.230255.255.255.0 = 11111111 00000000 (host octets only)




Number of Subnets


8

28 = 256 subnets






8 bit




Problem 4

Host IP Address 172.17.99.71Subnet Mask 255.255.0.0Number of Subnet Bit 0Number of Subnets 20 = 1 subnetsNumber of Host Bits per Subnet 16 bitNumber of Usable Hosts per Subnet

216 = 65536 - 2 = 65534 hosts per subnet

Subnet Address for this IP Address

172.17.0.0


172.17.0.1

IP Address of Last Host on Subnet

172.17.255.254

Broadcast Address for this Subnet

172.17.255.255


IP Address 10101100 00010001 01100011 01000111

Subnet Mask (x) 11111111 11111111 00000000 00000000

Subnet Address 10101100 00010001 00000001 00000000


172 . 17 . 0 . 0


254 Hosts


= 254





MULTIPLE

ANSWER


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.

172.17.99.71255.255.0.0 = 00000000 00000000 (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 "Powers of 2" chart if

necessary. This is the number of potential subnets created by the mask. Two of these potential subnets are normally not usable.

00000000 00000000 = 0 ones. 20 = 1 usable subnets created.


Number of Subnets


0

20 = 1 subnets

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.




16 bit

216 = 65536 - 2 = 65534 hosts per subnet



Problem 5Host IP Address 192.168.3.219Subnet Mask 255.255.255.0



Number of Host Bits per Subnet 8 bitNumber of Usable Hosts per Subnet



IP Address of First Host on this Subnet 192.168.3.1




IP Address 11000000 10101000 00000011 11011011

Subnet Mask (x) 11111111 11111111 11111111 00000000

Subnet Address 11000000 10101000 00000011 00000000 =


192 . 168 . 3 . 0


254 Hosts


= 254





MULTIPLE

ANSWERANSWER


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.

192.168.3.219

255.255.255.0 = 11111111 00000000 (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 "Powers of 2" chart if

necessary. This is the number of potential subnets created by the mask. Two of these potential subnets are normally not usable.

11111111 00000000 = 8 ones. 28 = 256 usable subnets created.


Number of Subnets


8

28 = 256 subnets

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.




8 bit




Problem 6Host IP Address 192.168.3.219

Subnet Mask 255.255.255.252



Number of Host Bits per Subnet 2 bit

Number of Usable Hosts per Subnet 22 = 4 - 2 = 2 hosts per subnet


IP Address of First Host on this Subnet 192.168.3.217



Host IP Adress 192 168 3 219

IP Address 11000000 10101000 00000011 11011011

Subnet Mask (x) 11111111 11111111 11111111 11111100

Subnet Address 11000000 10101000 00000011 11011000 =


192 . 168 . 3 . 216


2 Hosts


= 218





ANSWER

MULTIPLE




192.168.3.219255.255.255.252 = 11111111 11111100 (host octets only)




Number of Subnets


14

214 = 16384 subnets






2 bit




KFC 2044 NETWORK FUNDEMENTAL

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