8/20/2019 Assignment 2 Subnetting

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8/20/2019 Assignment 2 Subnetting

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Problem 2

Host IP Address 172.30.1.33

Subnet Mask 255.255.255.252

Number of Subnets Bits 14 bits

Number of Subnets 214

 = 16384

Number of Host Bits per Subnet 4

Number of Usable Hosts per Subnet 4 – 2 = 2 usable hosts

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 172.30.1.34Broadcast Address for this Subnet 172.30.1.35

1.  Bilangan Biner :

172.30.1.33 = 10101100 00011110 00000001 00100001

255.255.255.252 = 11111111 11111111 11111111 11111100

2. 

Subnet Address untuk IP Address :

IP = 10101100 00011110 00000001 00100001 Subnet Mask = 11111111 11111111 11111111 11111100 

Subnet Add = 10101100 00011110 00000001 00100000 

Subnet 14bits  host

Subnet = 172 30 1 32

3.  Tentukan First Host , Last Host dan Broadcast :

First Host = 10101100 00011110 00000001  00100001 

First Host IP = 172 30 1 33

Last Host = 10101100 00011110 00000001  00100010 Last Host IP = 172 30 1 34

Broadcast = 10101100 00011110 00000001  00100011 

Broadcast IP = 172 30 1 35

4. 

Number of Subnet Bits and Usable Host Bits :

Subnet = 14 bits

Number of Subnets = 214

 = 16384

Host Bits = 22 = 4

Usable Host Bits = 4 – 2 = 2 Usable Host Bits

8/20/2019 Assignment 2 Subnetting

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Problem 3

Host IP Address 192.192.10.234

Subnet Mask 255.255.255.0

Number of Subnets Bits 8 bits

Number of Subnets 28 = 256

Number of Host Bits per Subnet 256

Number of Usable Hosts per Subnet 256 – 2 = 254 usable hosts

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

1.  Bilangan Biner :

192.192.10.234 = 11000000 11000000 00001010 11101010

255.255.255.0 = 11111111 11111111 11111111 00000000

2. 

Subnet Address untuk IP Address :

IP = 11000000 11000000 00001010  11101010 Subnet Mask = 11111111 11111111 11111111  00000000 

Subnet Add = 11000000 11000000 00001010  00000000 Subnet 8bits  host 8bits

Subnet = 192 192 10 0

3. 

Tentukan First Host , Last Host dan Broadcast :

First Host = 11000000 11000000 00001010 00000001 

First Host IP = 192 192 10 1

Last Host = 11000000 11000000 00001010 11111110 Last Host IP = 192 192 10 254

Broadcast = 11000000 11000000 00001010 11111111 

Broadcast IP = 192 192 10 255

4. 

Number of Subnet Bits and Usable Host Bits :

Subnet = 8 bits

Number of Subnets = 28 = 256

Host Bits = 28 = 256

Usable Host Bits = 256 – 2 = 254 Usable Host Bits

8/20/2019 Assignment 2 Subnetting

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Problem 4

Host IP Address 172.17.99.71

Subnet Mask 255.255.0.0

Number of Subnets Bits 0 bits

Number of Subnets 2 = 2Number of Host Bits per Subnet 65536

Number of Usable Hosts per Subnet 65536 – 2 = 65534 Usable Host Bits

Subnet 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

1. 

Bilangan Biner :

172.17.99.71 = 10101100 00010001 01100011 01000111

255.255.0.0 = 11111111 11111111 00000000 00000000

2. 

Subnet Address untuk IP Address :

IP = 10101100 00010001 01100011 01000111 

Subnet Mask = 11111111 11111111 00000000 00000000 

Subnet Add = 10101100 00010001 00000000 00000000

Host 16 bits

Subnet = 172 17 0 0

3.  Tentukan First Host , Last Host dan Broadcast :

First Host = 10101100 00010001 00000000  00000001 

First Host IP = 172 17 0 1

Last Host = 10101100 00010001 11111111  11111110 

Last Host IP = 172 17 255 254

Broadcast = 10101100 00010001 11111111 11111111 

Broadcast IP = 172 17 255 255

4. 

Number of Subnet Bits and Usable Host Bits :

Subnet = 0 bitsNumber of Subnets = 2 = 2

Host Bits = 216

 = 65536

Usable Host Bits = 65536 – 2 = 65534 Usable Host Bits

8/20/2019 Assignment 2 Subnetting

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Problem 5

Host IP Address 192.168.3.219

Subnet Mask 255.255.255.0

Number of Subnets Bits 8 bits

Number of Subnets 28

 = 256Number of Host Bits per Subnet 256

Number of Usable Hosts per Subnet 256 – 2 = 254 Usable Host Bits

Subnet Address for this IP Address 192.168.3.0

IP 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

1. 

Bilangan Biner :

192.168.3.219 = 11000000 10101000 00000011 11011011

255.255.255.0 = 11111111 11111111 11111111 00000000

2.  Subnet Address untuk IP Address :

IP = 11000000 10101000 00000011  11011011 

Subnet Mask = 11111111 11111111 11111111  00000000 

Subnet Add = 11000000 10101000 00000011  00000000 Subnet 8bits Host 8bits 

Subnet = 192 168 3 0

3. 

Tentukan First Host , Last Host dan Broadcast :

First Host = 11000000 10101000 00000011  00000001 

First Host IP = 192 168 3 1

Last Host = 11000000 10101000 00000011  11111110 

Last Host IP = 192 168 3 254

Broadcast = 11000000 10101000 00000011  11111111 

Broadcast IP = 192 168 3 255

4.  Number of Subnet Bits and Usable Host Bits :

Subnet = 8 bits

Number of Subnets = 28 = 256

Host Bits = 28 = 256

Usable Host Bits = 256 – 2 = 254 Usable Host Bits

8/20/2019 Assignment 2 Subnetting

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Problem 6

Host IP Address 192.168.3.129

Subnet Mask 255.255.255.252

Number of Subnets Bits 14 bits

Number of Subnets 214

 = 16384Number of Host Bits per Subnet 4

Number of Usable Hosts per Subnet 4 – 2 = 2 Usable Host Bits

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

1. 

Bilangan Biner :

192.168.3.129 = 11000000 10101000 00000011 11011011

255.255.255.252 = 11111111 11111111 11111111 11111100

2.  Subnet Address Untuk IP Address :

IP = 11000000 10101000 00000011  11011011 

Subnet Mask = 11111111 11111111 11111111  11111100 

Subnet Add = 11000000 10101000 00000011  11011000 

Subnet 14bits Host

Subnet = 192 168 3 216

3. 

Tentukan First Host , Last Host dan Broadcast :

First Host = 11000000 10101000 00000011  11011001 

First Host IP = 192 168 3 217

Last Host = 11000000 10101000 00000011  11011010 

Last Host IP = 192 168 3 218

Broadcast = 11000000 10101000 00000011  11011011 

Broadcast IP = 192 168 3 219

4. 

Number of Subnet Bits and Usable Host Bits :

Subnet = 14 bits

Number of Subnets = 214

= 16384 

Host Bits = 22 = 4

Usable Host Bits = 4 – 2 = 2 Usable Host Bits

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