8/11/2019 assignment kfc 2044 - networking fundamental

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INSTITUT KEMAHIRAN MARA LUMUT

ASSIGNMENT

KFC 2044

NETWORK FUNDAMENTAL

TOPIC

COMPUTER NETWORKING FUNDAMENTALS

CERTIFICATE IN COMPUTER ENGINEERING TECHNOLOGY

(NETWORKING)

NAME : NURUL ISWANI BINTI AHMAD SHAWARI

STUDENT ID : 1401114

NO. IC : 940525-08-6426

CLASS : SKN 2A

SESSION : JULDIS 2014

LECTURER : EN. MOHD SHUKRI BIN MUHAMAD HUSIN

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IP ADDRESS CLASES

CALCULATION OF IP ADDRESS CLASSES

TASK 1 : For a given IP Address, determine network information

Given 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 Host Bit > 66534

Number of Host > 16

STEP 1

TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION

IP ADDRESS

OCTET 1 OCTET 2 OCTET 3 OCTET 4

IP ADDRESS 172 25 114 250

FORMULA

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 0 1 0 1 1 0 0

OCTET 2 0 0 0 1 1 0 0 1

OCTET 3 0 1 1 1 0 0 1 0

OCTET 4 1 1 1 1 1 0 1 0

BINARY NUMBER

Let

= 172= 25

= 114

= 250

OCTET 1 128 + 32 + 8 + 4OCTET 2 16 + 8 + 1

OCTET 3 64 + 32 + 16 + 2

OCTET 4 128 + 64 + 32 + 16 + 8 + 2

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NETWORK MASK

OCTET 1 OCTET 2 OCTET 3 OCTET 4

NETWORK MASK 255 255 0 0

FORMULA

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 1 1 1 1 1 1

OCTET 2 1 1 1 1 1 1 1 1

OCTET 3 0 0 0 0 0 0 0 0

OCTET 4 0 0 0 0 0 0 0 0

BINARY NUMBER

Let

= 255

= 255

= 0

= 0

STEP 2

DETERMINE THE NETWORK ADDRESS

172 . 25 . 0 . 0

172 25 114 250

IP ADDRESS 10101100 00011001 01110010 11111010

SUBNET MASK 11111111 11111111 00000000 00000000

255 255 0 0

Let the binary notation of IP Address times with binary notation of subnet mask

( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )

10101100 00011001 01110010 11111010

x 11111111 11111111 00000000 00000000

________________________________________

10101100 00011001 00000000 00000000

________________________________________

So

172 255 0 0

OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 3 0

OCTET 4 0

NETWORK

ADDRESS10101100 00011001 00000000 00000000

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STEP 3

DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS

172 . 25 . 255 . 255

172 25 0 0

NETWORK ADD. 10101100 00011001 00000000 00000000

SUBNET MASK 11111111 11111111 00000000 00000000

BROADCAST ADD. 10101100 00011001 11111111 11111111

172 25 255 255

Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask

= 16

Total Number of Host Bits : 2n2 use this formula to calculate

= 2162

= 65 5362

= 65 534

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TASK

PROBLEM

Host IP Address 172.30.1.33Network Mask 255.255.0.0 ( /16 )

Network Address 172.30.0.0

Network Broadcast Address 172.30.255.255

Total Number of Host Bits 65534

Number of Host 16

TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION

OCTET 1 OCTET 2 OCTET 3 OCTET 4

IP ADDRESS 172 30 1 33

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 0 1 0 1 1 0 0

OCTET 2 0 0 0 1 1 1 1 0

OCTET 3 0 0 0 0 0 0 0 1

OCTET 4 0 0 1 0 0 0 0 1

Let

= 172

= 30

= 1

= 33

OCTET 1 OCTET 2 OCTET 3 OCTET 4

NETWORK MASK 255 255 0 0

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 1 1 1 1 1 1

OCTET 2 1 1 1 1 1 1 1 1

OCTET 3 0 0 0 0 0 0 0 0

OCTET 4 0 0 0 0 0 0 0 0

Let

= 255

= 255

= 0

= 0

OCTET 1 128 + 32 + 8 + 4

OCTET 2 16 + 8 + 4 + 2

OCTET 3 1

OCTET 4 32 + 1

OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 3 0

OCTET 4 0

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DETERMINE THE NETWORK ADDRESS

172 . 30 . 0 . 0

172 30 0 33

IP ADDRESS 10101100 00011110 00000000 00100001

SUBNET MASK 11111111 11111111 00000000 00000000

255 255 0 0

Let the binary notation of IP Address times with binary notation of subnet mask

( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )

10101100 00011110 00000000 00100001

x 11111111 11111111 00000000 00000000

________________________________________

10101100 00011110 00000000 00000000

________________________________________

So

172 30 0 0

DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS

172 . 30 . 255 . 255

172 30 0 0

NETWORK ADD. 10101100 00011001 00000000 00000000

SUBNET MASK 11111111 11111111 00000000 00000000

BROADCAST ADD. 10101100 00011001 11111111 11111111

172 30 255 255

Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask

= 16

Total Number of Host Bits : 2n2 use this formula to calculate

= 216

2

= 65 5362

= 65 534

NETWORK

ADDRESS10101100 00011110 00000000 00000000

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TASK

PROBLEM 2

Host IP Address 172.30.1.33Network Mask 255.255.255.0 ( /24 )

Network Address 172.30.1.0

Network Broadcast Address 172.30.1.255

Total Number of Host Bits 254

Number of Host 8

TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION

OCTET 1 OCTET 2 OCTET 3 OCTET 4

IP ADDRESS 172 30 1 33

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 0 1 0 1 1 0 0

OCTET 2 0 0 0 1 1 1 1 0

OCTET 3 0 0 0 0 0 0 0 1

OCTET 4 0 0 1 0 0 0 0 1

Let

= 172

= 30

= 1

= 33

OCTET 1 OCTET 2 OCTET 3 OCTET 4

NETWORK MASK 255 255 255 0

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 1 1 1 1 1 1

OCTET 2 1 1 1 1 1 1 1 1

OCTET 3 1 1 1 1 1 1 1 1

OCTET 4 0 0 0 0 0 0 0 0

Let

= 255

= 255

= 255

= 0

OCTET 1 128 + 32 + 8 + 4

OCTET 2 16 + 8 + 4 + 2

OCTET 3 1

OCTET 4 32 + 1

OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 3 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 4 0

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DETERMINE THE NETWORK ADDRESS

172 . 30 . 1 . 0

172 30 1 33

IP ADDRESS 10101100 00011110 00000001 00100001

SUBNET MASK 11111111 11111111 11111111 00000000

255 255 255 0

Let the binary notation of IP Address times with binary notation of subnet mask

( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )

10101100 00011001 00000001 00100001

x 11111111 11111111 11111111 00000000

________________________________________

10101100 00011001 00000001 00000000

________________________________________

So

172 30 1 0

DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS

172 . 30 . 1 . 255

172 30 1 0

NETWORK ADD. 10101100 00011001 00000001 00000000

SUBNET MASK 11111111 11111111 11111111 00000000BROADCAST ADD. 10101100 00011001 00000001 11111111

172 30 1 255

Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask

= 8

Total Number of Host Bits : 2n2 use this formula to calculate

= 282

= 2562= 254

NETWORK

ADDRESS10101100 00011001 00000001 00000000

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TASK

PROBLEM 3

Host IP Address 192.168.10.234Network Mask 255.255.255.0 ( /24 )

Network Address 192.168.10.0

Network Broadcast Address 192.168.10.255

Total Number of Host Bits 254

Number of Host 8

TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION

OCTET 1 OCTET 2 OCTET 3 OCTET 4

IP ADDRESS 192 168 10 234

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 0 0 0 0 0 0

OCTET 2 1 0 1 0 1 0 0 0

OCTET 3 0 0 0 0 1 0 1 0

OCTET 4 1 1 1 0 1 0 1 0

Let

= 192

= 168

= 10

= 234

OCTET 1 OCTET 2 OCTET 3 OCTET 4

NETWORK MASK 255 255 255 0

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 1 1 1 1 1 1

OCTET 2 1 1 1 1 1 1 1 1

OCTET 3 1 1 1 1 1 1 1 1

OCTET 4 0 0 0 0 0 0 0 0

Let

= 255

= 255

= 255

= 0

OCTET 1 128 + 64

OCTET 2 128 + 32 + 8

OCTET 3 8 + 2

OCTET 4 128 + 64 + 32 + 8 + 2

OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 3 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 4 0

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DETERMINE THE NETWORK ADDRESS

192 . 168 . 10 . 0

192 168 10 234

IP ADDRESS 11000000 10101000 00001010 11101010

SUBNET MASK 11111111 11111111 11111111 00000000

255 255 255 0

Let the binary notation of IP Address times with binary notation of subnet mask

( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )

11000000 10101000 00001010 11101010

x 11111111 11111111 11111111 00000000

________________________________________

11000000 10101000 00001010 00000000

________________________________________

So

192 168 10 0

DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS

192 . 168 . 10 . 255

192 168 10 0

NETWORK ADD. 11000000 10101000 00001010 00000000

SUBNET MASK 11111111 11111111 11111111 00000000

BROADCAST ADD. 11000000 10101000 00001010 11111111

192 168 10 255

Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask

= 8

Total Number of Host Bits : 2n2 use this formula to calculate

= 282

= 2562

= 254

NETWORK

ADDRESS

11000000 10101000 00001010 00000000

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TASK

PROBLEM 4

Host IP Address 172.17.99.71Network Mask 255.255.0.0 ( /16 )

Network Address 172.17.0.0

Network Broadcast Address 172.17.255.255

Total Number of Host Bits 65534

Number of Host 16

TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION

OCTET 1 OCTET 2 OCTET 3 OCTET 4

IP ADDRESS 172 17 99 71

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 0 1 0 1 1 0 0

OCTET 2 0 0 0 1 0 0 0 1

OCTET 3 0 1 1 0 0 0 1 1

OCTET 4 0 1 0 0 0 1 1 1

Let

= 172

= 17

= 99

= 71

OCTET 1 OCTET 2 OCTET 3 OCTET 4

NETWORK MASK 255 255 0 0

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 1 1 1 1 1 1

OCTET 2 1 1 1 1 1 1 1 1

OCTET 3 0 0 0 0 0 0 0 0

OCTET 4 0 0 0 0 0 0 0 0

Let

= 255

= 255

= 0

= 0

OCTET 1 128 + 32 + 8 + 4

OCTET 2 16 + 1

OCTET 3 64 + 32 + 2 + 1

OCTET 4 64 + 4 + 2 + 1

OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 3 0

OCTET 4 0

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DETERMINE THE NETWORK ADDRESS

172 . 17 . 0 . 0

172 17 99 71

IP ADDRESS 10101100 00010001 01100011 01000111

SUBNET MASK 11111111 11111111 00000000 00000000

255 255 0 0

Let the binary notation of IP Address times with binary notation of subnet mask

( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )

10101100 00010001 01100011 01000111

x 11111111 11111111 00000000 00000000

________________________________________

10101100 00010001 00000000 00000000

________________________________________

So

172 17 0 0

DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS

172 . 17 . 255 . 255

172 17 0 0

NETWORK ADD. 10101100 00010001 00000000 00000000

SUBNET MASK 11111111 11111111 00000000 00000000

BROADCAST ADD. 10101100 00010001 11111111 11111111

172 17 255 255

Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask

= 16

Total Number of Host Bits : 2n2 use this formula to calculate

= 216

2

= 655362

= 65534

NETWORK

ADDRESS10101100 00010001 00000000 00000000

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TASK

PROBLEM 5

Host IP Address 192.168.3.219Network Mask 255.255.0.0 ( /16 )

Network Address 192.168.0.0

Network Broadcast Address 192.168.255.255

Total Number of Host Bits 65534

Number of Host 16

TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION

OCTET 1 OCTET 2 OCTET 3 OCTET 4

IP ADDRESS 172 17 99 71

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 0 1 0 1 1 0 0

OCTET 2 0 0 0 1 0 0 0 1

OCTET 3 0 1 1 0 0 0 1 1

OCTET 4 0 1 0 0 0 1 1 1

Let

= 172

= 17

= 99

= 71

OCTET 1 OCTET 2 OCTET 3 OCTET 4

NETWORK MASK 255 255 0 0

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 1 1 1 1 1 1

OCTET 2 1 1 1 1 1 1 1 1

OCTET 3 0 0 0 0 0 0 0 0

OCTET 4 0 0 0 0 0 0 0 0

Let

= 255

= 255

= 0

= 0

OCTET 1 128 + 32 + 8 + 4

OCTET 2 16 + 1

OCTET 3 64 + 32 + 2 + 1

OCTET 4 64 + 4 + 2 + 1

OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 3 0

OCTET 4 0

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DETERMINE THE NETWORK ADDRESS

172 . 17 . 0 . 0

172 17 99 71

IP ADDRESS 10101100 00010001 01100011 01000111

SUBNET MASK 11111111 11111111 00000000 00000000

255 255 0 0

Let the binary notation of IP Address times with binary notation of subnet mask

( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )

10101100 00010001 01100011 01000111

x 11111111 11111111 00000000 00000000

________________________________________

10101100 00010001 00000000 00000000

________________________________________

So

172 17 0 0

DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS

172 . 17 . 255 . 255

172 17 0 0

NETWORK ADD. 10101100 00010001 00000000 00000000

SUBNET MASK 11111111 11111111 00000000 00000000

BROADCAST ADD. 10101100 00010001 11111111 11111111

172 17 255 255

Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask

= 16

Total Number of Host Bits : 2n2 use this formula to calculate

= 216

2

= 655362

= 65534

NETWORK

ADDRESS10101100 00010001 00000000 00000000

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TASK

PROBLEM 6

Host IP Address 192.168.3.219Network Mask 255.255.255.224 ( /27 )

Network Address 192.168.3.192

Network Broadcast Address 192.168.3.223

Total Number of Host Bits 30

Number of Host 5

TRANSLATE IP ADDRESS AND NETWORK MASK INTO BINARY NOTATION

OCTET 1 OCTET 2 OCTET 3 OCTET 4

IP ADDRESS 192 168 3 219

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 0 0 0 0 0 0

OCTET 2 1 0 1 0 1 0 0 0

OCTET 3 0 0 0 0 0 0 1 1

OCTET 4 1 1 0 1 1 0 1 1

Let

= 192

= 168

= 3

= 219

OCTET 1 OCTET 2 OCTET 3 OCTET 4

NETWORK MASK 255 255 255 224

27 2

6 2

5 2

4 2

3 2

2 2

1 2

0

128 64 32 16 8 4 2 1

OCTET 1 1 1 1 1 1 1 1 1

OCTET 2 1 1 1 1 1 1 1 1

OCTET 3 1 1 1 1 1 1 1 1

OCTET 4 1 1 1 0 0 0 0 0

Let

= 255

= 255

= 255

= 224

OCTET 1 128 + 64

OCTET 2 128 + 32 + 8

OCTET 3 2 + 1

OCTET 4 128 + 64 + 16 + 8 + 2 + 1

OCTET 1 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 2 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 3 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

OCTET 4 128 + 64 + 32

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DETERMINE THE NETWORK ADDRESS

192 . 168 . 3 . 192

192 168 3 219

IP ADDRESS 11000000 10101000 00000011 11011011

SUBNET MASK 11111111 11111111 11111111 11100000

255 255 255 224

Let the binary notation of IP Address times with binary notation of subnet mask

( note : 1 and 1 = 1 ; 0 and 1 or 0 = 0 )

11000000 10101000 00000011 11011011

x 11111111 11111111 11111111 11100000

________________________________________

11000000 10101000 00000011 11000000

________________________________________

So

192 168 3 192

DETERMINE THE BROADCAST ADDRESS FOR THE NETWORK ADDRESS

192 . 168 . 3 . 223

192 168 3 192

NETWORK ADD. 11000000 10101000 00000011 11000000

SUBNET MASK 11111111 11111111 11111111 11100000BROADCAST ADD. 11000000 10101000 00000011 11011111

192 168 3 223

Number of Host (n) : `0 number in octet 3 and octet 4 of subnet mask

= 5

Total Number of Host Bits : 2n2 use this formula to calculate

= 252

= 322= 30

NETWORK

ADDRESS

11000000 10101000 00000011 11000000

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STEP OF SUBNETING CALCULATION

TASK 1 : For a Given IP Address and Subnet Mask, Determine Subnet Information.

Given

Host IP Address 172.25.114.250

Network Mask 255.255.0.0(/16)Subnet Mask 255.255.255.192(/26)

Find

Number of Subnet Bits 10

Number of Subnets 210

=1 024 subnets

Number of Host Bits per Subnet 6 bit

Number of Useable Host per Subnet 26

= 642 =62 hosts per subnet

Subnet Address for this IP Address 172.25.114.192

IP Address of First Host on this Subnet 172.25.114.193IP Address of Last Host on this Subnet 172.25.114.254

Broadcast Address for this Subnet 172.25.114.255

TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION

172 25 114 250

IP ADDRESS 10101100 00011001 01110010 11111010SUBNET MASK 11111111 11111111 11111111 11000000

255 255 255 192

DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS

Binary notation of IP address X Binary notation of subnet mask( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )

10101100 00011001 01110010 11111010

X 11111111 11111111 11111111 11000000

_____________________________________________

10101100 00011001 01110010 11000000

_____________________________________________

Subnet Address > 172.25.114.192

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DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION

AND WHICH CONTAIN HOST INFORMATION

Subnet Address

Network . Network . Host . Host

172 . 25 . 114 . 192

DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST

To find broadcast address, use subnet mask and subnet address

SUBNET MASK 255

11111111

255

11111111

255

11111111

192

11000000

SUBNET ADDRESS 172

10101100

25

00011001

114

01110010

192

11000000

BROADCAST

ADDRESS

10101100

172

00011001

25

01110010

114

11111111

255

To find IP Address of first Host, the subnet address is add with 1

172 . 25 . 114 . 192

+ 1

________________

172 . 25 . 114 . 193 IP Address of First Host

________________

To find IP Address of last Host, the broadcast address is subtract with 1

172 . 25 . 114 . 255

- 1

________________

172 . 25 . 114 . 254 IP Address of Last Host

________________

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DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS

ON THIS SUBNET

M.D S.D

IP ADDRESS10101100 00011001 01110010 11 111010

SUBNET MASK 11111111 11111111 11111111 11 000000

SUBNET ADDRESS10101100 00011001 01110010 11 000000

subneting host

counting counting

range range

FIRST HOST 10101100

172

00011001

25

01110010

114

11 000001

193

LAST HOST 10101100

172

00011001

25

01110010

114

11 111110

254BROADCAST 10101100

172

00011001

25

01110010

114

11 111111

255

DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS

Firstly, use the first octet of the IP address to determine the class of address and to determine which

octet are available for host.

Network.network.host.host

IP Address

172.25.114.250

255.255.255.255

Subnet Mask

Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask 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 count the number of host bits.

172.25.114.250

255.255.255.255 = 11111111

11000000 (host)

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Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to

count the number of subnets.

11111111 11000000 = 10 ones of subnet bits

2n2

= 210

2

= 10242

= 1022 useable subnet

Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to

calculate the number of host bits.

11111111 11000000 = 6 zeros of host bits

2n2

= 262

= 642

= 62 useable host addresses

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TASK 2

CHALLENGE

PROBLEM

Given

Host IP Address 172.30.1.33

Subnet Mask 255.255.255.0

Find

Number of Subnet Bits 8

Number of Subnets 28=256 subnets

Number of Host Bits per Subnet 8 bit

Number of Useable Host per Subnet 28

= 2562 =254 hosts per subnetSubnet 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 this Subnet 172.30.1.254

Broadcast Address for this Subnet 172.30.1.255

TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION

172 30 1 33

IP ADDRESS 10101100 00011110 00000001 00100001

SUBNET MASK 11111111 11111111 11111111 00000000

255 255 255 0

DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS

Binary notation of IP address X Binary notation of subnet mask

( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )

10101100 00011110 00000001 00100001

X 11111111 11111111 11111111 00000000

_____________________________________________

10101100 00011110 00000001 00000000

_____________________________________________

Subnet Address > 172.30.1.0

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DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION

AND WHICH CONTAIN HOST INFORMATION

Subnet Address

Network . Network . Host . Host

172 . 30 . 1 . 192

DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST

To find broadcast address, use subnet mask and subnet address

SUBNET MASK 255

11111111

255

11111111

255

11111111

0

00000000

SUBNET ADDRESS 172

10101100

30

00011110

1

00000001

0

00000000

BROADCAST

ADDRESS

10101100

172

00011110

30

00000001

1

11111111

255

To find IP Address of first Host, the subnet address is add with 1

172 . 30 . 1 . 0

+ 1

________________

172 . 30 . 1 . 1 IP Address of First Host

________________

To find IP Address of last Host, the broadcast address is subtract with 1

172 . 30 . 1 . 255

- 1

________________

172 . 30 . 1 . 254 IP Address of Last Host

________________

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DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS

ON THIS SUBNET

M.D S.D

IP ADDRESS10101100 00011110 00000001 00 100001

SUBNET MASK 11111111 11111111 11111111 00 000000

SUBNET ADDRESS10101100 00011110 00000001 00 000000

subneting host

counting counting

range range

FIRST HOST 10101100

172

00011110

30

00000001

1

00 000001

1

LAST HOST 10101100

172

00011110

30

00000001

1

11 111110

254BROADCAST 10101100

172

00011110

30

00000001

1

11 111111

255

DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS

Firstly, use the first octet of the IP address to determine the class of address and to determine which

octet are available for host.

Network.network.host.host

IP Address

172.30.1.33

255.255.255.0

Subnet Mask

Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask 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 count the number of host bits.

172.30.1.33

255.255.255.0 = 11111111

00000000 (host)

8/11/2019 assignment kfc 2044 - networking fundamental

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Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to

count the number of subnets.

11111111 00000000 = 8 ones of subnet bits

2n2

= 282

= 2562

= 254 useable subnet

Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to

calculate the number of host bits.

11111111 00000000 = 8 zeros of host bits

2n2

= 282

= 2562

= 254 useable host addresses

8/11/2019 assignment kfc 2044 - networking fundamental

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TASK 2

CHALLENGE

PROBLEM 2

Given

Host IP Address 172.30.1.33

Subnet Mask 255.255.255.252

Find

Number of Subnet Bits 14

Number of Subnets 214

=16834 subnets

Number of Host Bits per Subnet 2 bit

Number of Useable Host per Subnet 22 = 4 2 =2 hosts per subnet

Subnet Address for this IP Address 172.30.1.32

IP Address of First Host on this Subnet 172.30.1.33

IP Address of Last Host on this Subnet 172.30.1.34

Broadcast Address for this Subnet 172.30.1.35

TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION

172 30 1 33

IP ADDRESS 10101100 00011110 00000001 00100001

SUBNET MASK 11111111 11111111 11111111 11111100

255 255 255 252

DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS

Binary notation of IP address X Binary notation of subnet mask

( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )

10101100 00011110 00000001 00100001

X 11111111 11111111 11111111 11111100

_____________________________________________

10101100 00011110 00000001 00100000

_____________________________________________

Subnet Address > 172.30.1.32

8/11/2019 assignment kfc 2044 - networking fundamental

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DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION

AND WHICH CONTAIN HOST INFORMATION

Subnet Address

Network . Network . Host . Host

172 . 30 . 1 . 32

DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST

To find broadcast address, use subnet mask and subnet address

SUBNET MASK 255

11111111

255

11111111

255

11111111

252

11000000

SUBNET ADDRESS 172

10101100

30

00011110

1

00000001

32

00100000

BROADCAST

ADDRESS

10101100

172

00011110

30

00000001

1

00100011

35

To find IP Address of first Host, the subnet address is add with 1

172 . 30 . 1 . 32

+ 1

________________

172 . 30 . 1 . 33 IP Address of First Host

________________

To find IP Address of last Host, the broadcast address is subtract with 1

172 . 30 . 1 . 35

- 1

________________

172 . 30 . 1 . 34 IP Address of Last Host

________________

8/11/2019 assignment kfc 2044 - networking fundamental

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DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS

ON THIS SUBNET

M.D S.D

IP ADDRESS10101100 00011110 00000001 00 100001

SUBNET MASK 11111111 11111111 11111111 11 111100

SUBNET ADDRESS10101100 00011110 00000001 00 100000

subneting host

counting counting

range range

FIRST HOST 10101100

172

00011110

30

00000001

1

00 100001

33

LAST HOST 10101100

172

00011110

30

00000001

1

00 100010

34BROADCAST 10101100

172

00011110

30

00000001

1

00 100011

35

DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS

Firstly, use the first octet of the IP address to determine the class of address and to determine which

octet are available for host.

Network.network.host.host

IP Address

172.30.1.33

255.255.255.252

Subnet Mask

Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask 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 count the number of host bits.

172.30.1.33

255.255.255.252 = 11111111

11111100 (host)

8/11/2019 assignment kfc 2044 - networking fundamental

28/44

Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to

count the number of subnets.

11111111 11111100 = 14 ones of subnet bits

2n2

= 214

2

= 163842

= 16382 useable subnet

Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to

calculate the number of host bits.

11111111 11111100 = 2 zeros of host bits

2n2

= 222

= 42

= 2 useable host addresses

8/11/2019 assignment kfc 2044 - networking fundamental

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TASK 2

CHALLENGE

PROBLEM 3

Given

Host IP Address 192.192.10.234

Subnet Mask 255.255.255.0

Find

Number of Subnet Bits 8

Number of Subnets 28=256 subnets

Number of Host Bits per Subnet 8 bit

Number of Useable Host per Subnet 28 = 2562 =254 hosts per subnet

Subnet Address for this IP Address 192.192.10.0

IP Address of First Host on this Subnet 192.192.10.1

IP Address of Last Host on this Subnet 192.192.10.254

Broadcast Address for this Subnet 192.192.10.255

TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION

192 192 10 234

IP ADDRESS 11000000 11000000 00001010 11101010

SUBNET MASK 11111111 11111111 11111111 00000000

255 255 255 0

DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS

Binary notation of IP address X Binary notation of subnet mask

( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )

11000000 11000000 00001010 11101010

X 11111111 11111111 11111111 00000000

_____________________________________________

11000000 11000000 00001010 00000000

_____________________________________________

Subnet Address > 192.192.10.0

8/11/2019 assignment kfc 2044 - networking fundamental

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DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION

AND WHICH CONTAIN HOST INFORMATION

Subnet Address

Network . Network . Host . Host

192 . 192 . 10 . 0

DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST

To find broadcast address, use subnet mask and subnet address

SUBNET MASK 255

11111111

255

11111111

255

11111111

0

00000000

SUBNET ADDRESS 192

11000000

192

11000000

10

00001010

0

00000000

BROADCAST

ADDRESS

11000000

192

11000000

192

00001010

10

11111111

255

To find IP Address of first Host, the subnet address is add with 1

192 . 192 . 10 . 0

+ 1

________________

192 . 192 . 10 . 1 IP Address of First Host

________________

To find IP Address of last Host, the broadcast address is subtract with 1

192 . 192 . 10 . 255

- 1

________________

192 . 192 . 10 . 254 IP Address of Last Host

________________

8/11/2019 assignment kfc 2044 - networking fundamental

31/44

DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS

ON THIS SUBNET

M.D S.D

IP ADDRESS11000000 11000000 00001010 11 101010

SUBNET MASK 11111111 11111111 11111111 00 000000

SUBNET ADDRESS11000000 11000000 00001010 00 000000

subneting host

counting counting

range range

FIRST HOST 11000000

192

11000000

192

00001010

10

00 000001

1

LAST HOST 11000000

192

11000000

192

00001010

10

11 111110

254BROADCAST 11000000

192

11000000

192

00001010

10

11 111111

255

DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS

Firstly, use the first octet of the IP address to determine the class of address and to determine which

octet are available for host.

Network.network.host.host

IP Address

192.192.10.234

255.255.255.0

Subnet Mask

Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask 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 count the number of host bits.

172.25.10.234

255.255.255.0 = 11111111

00000000 (host)

8/11/2019 assignment kfc 2044 - networking fundamental

32/44

Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to

count the number of subnets.

11111111 00000000 = 8 ones of subnet bits

2n2

= 282

= 2562

= 254 useable subnet

Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to

calculate the number of host bits.

11111111 00000000 = 8 zeros of host bits

2n2

= 282

= 2562

= 254 useable host addresses

8/11/2019 assignment kfc 2044 - networking fundamental

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TASK 2

CHALLENGE

PROBLEM 4

Given

Host IP Address 172.17.99.71

Subnet Mask 255.255.0.0

Find

Number of Subnet Bits 0

Number of Subnets 20=1 subnets

Number of Host Bits per Subnet 16 bit

Number of Useable Host per Subnet 216

= 6553 -2 =65534 hosts per subnetSubnet Address for this IP Address 172.17.0.0

IP Address of First Host on this Subnet 172.17.0.1

IP Address of Last Host on this Subnet 172.17.255.254

Broadcast Address for this Subnet 172.17.255.255

TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION

172 17 99 71

IP ADDRESS 10101100 00010001 01100011 01000111

SUBNET MASK 11111111 11111111 00000000 00000000

255 255 0 0

DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS

Binary notation of IP address X Binary notation of subnet mask

( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )

10101100 00010001 01100011 01000111

X 11111111 11111111 00000000 00000000

_____________________________________________

10101100 00010001 00000000 00000000

_____________________________________________

Subnet Address > 172.17.0.0

8/11/2019 assignment kfc 2044 - networking fundamental

34/44

DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION

AND WHICH CONTAIN HOST INFORMATION

Subnet Address

Network . Network . Host . Host

172 . 17 . 0 . 0

DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST

To find broadcast address, use subnet mask and subnet address

SUBNET MASK 255

11111111

255

11111111

0

00000000

0

00000000

SUBNET ADDRESS 172

10101100

17

00010001

0

00000000

0

00000000

BROADCAST

ADDRESS

10101100

172

00010001

17

11111111

255

11111111

255

To find IP Address of first Host, the subnet address is add with 1

172 . 17 . 0 . 0

+ 1

________________

172 . 17 . 0 . 1 IP Address of First Host

________________

To find IP Address of last Host, the broadcast address is subtract with 1

172 . 17 . 255 . 255

- 1

________________

172 . 17 . 255 . 254 IP Address of Last Host

________________

8/11/2019 assignment kfc 2044 - networking fundamental

35/44

DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS

ON THIS SUBNET

M.D S.D

IP ADDRESS10101100 00010001 011100011 11 111010

SUBNET MASK 11111111 11111111 11111111 11 000000

SUBNET ADDRESS10101100 00010001 00000000 00 000000

subneting host

counting counting

range range

FIRST HOST 10101100

172

00010001

17

00000000

0

00 000001

1

LAST HOST 10101100

172

00010001

17

11111111

255

11 111110

254BROADCAST 10101100

172

00010001

17

11111111

255

11 111111

255

DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS

Firstly, use the first octet of the IP address to determine the class of address and to determine which

octet are available for host.

Network.network.host.host

IP Address

172.17.99.71

255.255.0.0

Subnet Mask

Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask 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 count the number of host bits.

172.17.99.71

255.255.0.0 = 00000000

00000000 (host)

8/11/2019 assignment kfc 2044 - networking fundamental

36/44

Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to

count the number of subnets.

00000000 00000000 = 0 ones of subnet bits

2n

= 20

= 1 useable subnet

Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to

calculate the number of host bits.

00000000 00000000 = 16 zeros of host bits

2n2

= 216

2

= 655362

= 65534 useable host addresses

8/11/2019 assignment kfc 2044 - networking fundamental

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TASK 2

CHALLENGE

PROBLEM 5

Given

Host IP Address 192.168.3.219

Subnet Mask 255.255.255.0

Find

Number of Subnet Bits 8

Number of Subnets 28=256 subnets

Number of Host Bits per Subnet 8 bit

Number of Useable Host per Subnet 28

= 2562 =254 hosts per subnet

Subnet Address for this IP Address 192.168.3.0IP Address of First Host on this Subnet 192.168.3.1

IP Address of Last Host on this Subnet 192.168.3.254

Broadcast Address for this Subnet 192.168.3.255

TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION

192 168 3 219

IP ADDRESS 11000000 10101000 00000011 11011011

SUBNET MASK 11111111 11111111 11111111 00000000

255 255 255 0

DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS

Binary notation of IP address X Binary notation of subnet mask

( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )

11000000 10101000 00000011 11011011

X 11111111 11111111 11111111 00000000

_____________________________________________

11000000 10101000 00000011 00000000

_____________________________________________

Subnet Address > 192.168.3.0

DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION

8/11/2019 assignment kfc 2044 - networking fundamental

38/44

AND WHICH CONTAIN HOST INFORMATION

Subnet Address

Network . Network . Host . Host

192 . 168 . 3 . 0

DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST

To find broadcast address, use subnet mask and subnet address

SUBNET MASK 255

11111111

255

11111111

255

11111111

0

00000000

SUBNET ADDRESS 192

11000000

168

10101000

3

00000011

0

00000000

BROADCAST

ADDRESS

11000000

192

10101000

168

00000011

3

11111111

255

To find IP Address of first Host, the subnet address is add with 1

192 . 168 . 3 . 0+ 1

________________

192 . 168 . 3 . 1 IP Address of First Host

________________

To find IP Address of last Host, the broadcast address is subtract with 1

192 . 168 . 3 . 255

- 1

________________

192 . 168 . 3 . 254 IP Address of Last Host

________________

DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS

ON THIS SUBNET

8/11/2019 assignment kfc 2044 - networking fundamental

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M.D S.D

IP ADDRESS11000000 10101000 00000011 11 011011

SUBNET MASK11111111 11111111 11111111 00 000000

SUBNET ADDRESS 10101100 10101000 00000011 00 000000

subneting host

counting counting

range range

FIRST HOST 11000000

192

10101000

168

00000011

3

00 000001

1

LAST HOST 11000000

192

10101000

168

00000011

3

11 111110

254

BROADCAST 11000000

192

10101000

168

00000011

3

11 111111

255

DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS

Firstly, use the first octet of the IP address to determine the class of address and to determine which

octet are available for host.

Network.network.host.host

IP Address192.168.3.219

255.255.255.0

Subnet Mask

Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask 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 count the number of host bits.

192.168.3.219

255.255.255.0 = 11111111

00000000 (host)

8/11/2019 assignment kfc 2044 - networking fundamental

40/44

Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to

count the number of subnets.

11111111 00000000 = 8 ones of subnet bits

2n2

= 282

= 2562

= 254 useable subnet

Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to

calculate the number of host bits.

11111111 00000000 = 8 zeros of host bits

2n2

= 282

= 2562

= 254 useable host addresses

8/11/2019 assignment kfc 2044 - networking fundamental

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TASK 2

CHALLENGE

PROBLEM 6

Given

Host IP Address 192.168.3.219

Subnet Mask 255.255.255.252

Find

Number of Subnet Bits 14

Number of Subnets 214

=16384 subnets

Number of Host Bits per Subnet 2 bit

Number of Useable Host per Subnet 22 = 4 2 =2 hosts per subnet

Subnet Address for this IP Address 192.168.3.216

IP Address of First Host on this Subnet 192.168.3.217

IP Address of Last Host on this Subnet 192.168.3.218

Broadcast Address for this Subnet 192.168.3.219

TRANSLATE HOST IP ADDRESS AND SUBNET MASK INTO BINARY NOTATION

192 168 3 219

IP ADDRESS 11000000 10101000 00000011 11011011

SUBNET MASK 11111111 11111111 11111111 11111100

255 255 255 252

DETERMINE THE NETWORK(OR SUBNET) WHERE THIS HOST ADDRESS BELONGS

Binary notation of IP address X Binary notation of subnet mask

( 1 X 1 = 1 0 X 0 = 0 0 X 1 = 0 1 X 0 = 0 )

11000000 10101000 00000011 11011011

X 11111111 11111111 11111111 11111100

_____________________________________________

11000000 10101000 00000011 11011000

_____________________________________________

Subnet Address > 192.168.3.216

8/11/2019 assignment kfc 2044 - networking fundamental

42/44

DETERMINE WHICH BITS IN THE ADDRESS CONTAIN NETWORK INFORMATION

AND WHICH CONTAIN HOST INFORMATION

Subnet Address

Network . Network . Host . Host

192 . 168 . 3 . 216

DETERMINE THE BROADCAST ADDRESS AND IP ADDRESS OF FIRST HOST AND LAST HOST

To find broadcast address, use subnet mask and subnet address

SUBNET MASK 255

11111111

255

11111111

255

11111111

252

11111100

SUBNET ADDRESS 192

11000000

168

10101000

3

00000011

216

11011000

BROADCAST

ADDRESS

11000000

192

10101000

168

00000011

3

11011011

219

To find IP Address of first Host, the subnet address is add with 1

192 . 168 . 3 . 216

+ 1

________________

192 . 168 . 3 . 217 IP Address of First Host

________________

To find IP Address of last Host, the broadcast address is subtract with 1

192 . 168 . 3 . 219

- 1

________________

192 . 168 . 3 . 218 IP Address of Last Host

________________

8/11/2019 assignment kfc 2044 - networking fundamental

43/44

DETERMINE THE RANGE OF HOST ADDRESSES AVAILABLE ON THIS SUBNET AND THE BROADCAST ADDRESS

ON THIS SUBNET

M.D S.D

IP ADDRESS11000000 10101000 00000011 11 011011

SUBNET MASK 11111111 11111111 11111111 11 111100

SUBNET ADDRESS11000000 10101000 00000011 11 011000

subneting host

counting counting

range range

FIRST HOST 11000000

192

10101000

168

00000011

3

11 011001

217

LAST HOST 11000000

192

10101000

168

00000011

3

11 011010

218BROADCAST 11000000

192

10101000

168

00000011

3

11 011011

219

DETERMININING THE NUMBER OF IP SUBNETS AND HOSTS

Firstly, use the first octet of the IP address to determine the class of address and to determine which

octet are available for host.

Network.network.host.host

IP Address

192.168.3.219

255.255.255.252

Subnet Mask

Look at the host octet (octet 3 and octet 4) in the subnet mask. Use the possible mask 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 count the number of host bits.

192.168.3.219

255.255.255.252 = 11111111

11111100 (host)

8/11/2019 assignment kfc 2044 - networking fundamental

44/44

Count the total number of ones in the host octet(s) of the subnet mask and use formula 2n2 to

count the number of subnets.

11111111 11111100 = 14 ones of subnet bits

2n2

= 214

2

= 163842

= 16382 useable subnet

Count the total number of zeros in the host octet(s) of the subnet mask and use formula 2n2 to

calculate the number of host bits.

11111111 11111100 = 2 zeros of host bits

2n2

= 222

= 42

= 2 useable host addresses