Ip Addressing and Subnetting

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