Ip Addressing and Subnetting

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Question 1

You are given a network address 192.168.10.0/27, show your calculation and answer the followings (20 Marks – 4 marks each)

|IP Address: 192.168.10.0 |Prefix Length - /27 |

Bits allocation:
|IP Address |128 |64 |32 |16 | |Prefix |11111111 |11111111 |11111111 |11100000 |

a) How many possible subnets?
|000 |001 |010 |011 |100 |101 |110 |111 |

b) How many hosts in each subnet?
= 2^5 – 2 = 32-2 = 30
c) List out all the subnet address with its subnet marks? Smallest number of /27 = 32. Therefore this has to be doubled 8 possible times till it reaches its maximum which is 224. All the subnet addresses with its subnet marks are listed below. |192.168.10.0 /27 |192.168.10.128 /27 | |192.168.10.32 /27 |192.168.10.160 /27 | |192.168.10.64 /27 |192.168.10.192 /27 | |192.168.10.96 /27 |192.168.10.224 /27 |

d) What is the broadcast address for each subnet?
Broadcast – last host before the first subnet starts Possible Hosts = 32
Usable Hosts = 30
Broadcast Host = 31 or 192.168.10.31 /27, because its the last host before the first subnet starts

e) What are the valid hosts (i.e. it can be allocated) in a subnet? Usable Hosts = 30
|Range: 1 – 30 |192.168.10.0 /27...192.168.10.30 /27 | |Broadcast: 31 | | |Range: 32 – 62 |192.168.10.32 /27... 192.168.10.62 /27 | |Broadcast: 63 | | |Range: 64 – 94 |192.168.10.64 /27... 192.168.10.94 /27 | |Broadcast: 95 | | |Range: 96 – 126 |192.168.10.96 /27... 192.168.10.126 /27 | |Broadcast: 127 | | |Range: 128 – 158 |192.168.10.128 /27... 192.168.10.158 /27 | |Broadcast: 159 | | |Range: 160 – 190 |192.168.10.160 /27... 192.168.10.190 /27 | |Broadcast: 191 | | |Range: 192 – 222 |192.168.10.192 /27... 192.168.10.222 /27 | |Broadcast: 223...
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