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You are given a class B IP Address of 169.33.0.0 and you need to create a network with at least 20 subnets and at least 100 h
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Answer #1

IP address = 169.33.0.0

Number of subnets to be formed = atleast 20

Number of bits to be borrowed from host portion for subnets = ceil(log2(number of subnets)) = ceil(log220) = 5 bits

so, number of bits available for hosts = 16 - 5 = 11

1).Subnet mask:-

In subnet mask, all the bits that belongs to network portion are 1's and all other bits(host bits) are 0's,

subnet mask : - 11111111 11111111 11111000 00000000 = 255.255.248.0

2).

with 5 bits available number of subnets possible = 25 = 32

Subnet number Subnets IP address First host's IP address Last host's IP address Broadcast address
1 169.33.0.0 169.33.0.1 169.33.7.254 169.33.7.255
2 169.33.8.0 169.33.8.1 169.33.15.254 169.33.15.255
3 169.33.16.0 169.33.16.1 169.33.23.254 169.33.23.255
4 169.33.24.0 169.33.24.1 169.33.31.254 169.33.31.255
5 169.33.32.0 169.33.32.1 169.33.39.254 169.33.39.255
6 169.33.40.0 169.33.40.1 169.33.47.254 169.33.47.255
7 169.33.48.0 169.33.48.1 169.33.57.254 169.33.57.255
8 169.33.56.0 169.33.56.1 169.33.63.254 169.33.63.255
9 169.33.64.0 169.33.64.1 169.33.71.254 169.33.71.255
10 169.33.72.0 169.33.72.1 169.33.79.254 169.33.79.255
11 169.33.80.0 169.33.80.1 169.33.87.254 169.33.87.255
12 169.33.88.0 169.33.88.1 169.33.95.254 169.33.95.255
13 169.33.96.0 169.33.96.1 169.33.103.254 169.33.103.255
14 169.33.104.0 169.33.104.1 169.33.111.254 169.33.111.255
15 169.33.112.0 169.33.112.1 169.33.119.254 169.33.119.255
16 169.33.120.0 169.33.120.1 169.33.127.254 169.33.127.255
17 169.33.128.0 169.33.128.1 169.33.135.254 169.33.135.255
18 169.33.136.0 169.33.136.1 169.33.143.254 169.33.143.255
19 169.33.144.0 169.33.144.1 169.33.151.254 169.33.151.255
20 169.33.152.0 169.33.152.1 169.33.159.254 169.33.159.255
21 169.33.160.0 169.33.160.1 169.33.167.254 169.33.167.255
22 169.33.168.0 169.33.168.1 169.33.175.254 169.33.175.255
23 169.33.176.0 169.33.176.1 169.33.183.254 169.33.183.255
24 169.33.184.0 169.33.184.1 169.33.191.254 169.33.191.255
25 169.33.192.0 169.33.192.1 169.33.199.254 169.33.199.255
26 169.33.200.0 169.33.200.1 169.33.207.254 169.33.207.255
27 169.33.208.0 169.33.208.1 169.33.215.254 169.33.215.255
28 169.33.216.0 169.33.216.1 169.33.223.254 169.33.223.255
29 169.33.224.0 169.33.224.1 169.33.231.254 169.33.231.255
30 169.33.232.0 169.33.232.1 169.33.239.254 169.33.239.255
31 169.33.240.0 169.33.240.1 169.33.247.254 169.33.247.255
32 169.33.248.0 169.33.248.1 169.33.255.254 169.33.255.255

3).Number of IP address in 1 subnet = 2number of host bits = 211 = 2 * 1024 = 2048

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