Answer to Question #192971 in Computer Networks for danile

Question

, An organization has been assigned the prefix 212.1.1/24 (class C) and wants to form subnets for four departments, with hosts as follows:

a.      A 75 hosts

b.     B 35 hosts

c.      C 20 hosts

d.     D 18 hosts

           There are 148 hosts in all.

           a) Give a possible assignment of subnet masks to make this possible.

           b) Suggest what the organization might do if department D grows to 32 hosts.

Sol:

(a) Giving each department a single subnet, the nominal subnet sizes are 27 ,26 ,25 ,25

respectively; we obtain these by rounding up to the nearest power of 2. A possible

arrangement of subnet numbers is as follows. 

Answer 1

a)


212.1.1.0 	 11001000.00000001.00000001.00000000


Subnet A: 72 hosts  require 2^7 = 128, maximum number of hosts 2^7 – 2 = 126 valid IPs
11010100.00000001.00000001.00000000  212.1.1.0/25 (network address)
11111111.11111111.11111111.10000000  255.255.255.128/25 (subnet mask)


11010100.00000001.00000001.00000001  212.1.1.1 (the 1st host IP in subnet A)
11010100.00000001.00000001.01111110  212.1.1.126 (the last host IP in subnet A)
11010100.00000001.00000001.01111111  212.1.1.127 (broadcast address)


Subnet B: 35 hosts  require 2^6 = 64, maximum number of hosts 2^6 – 2 = 62 IPs
11010100.00000001.00000001.10000000  212.1.1.128/26 (network address)
11111111.11111111.11111111.11000000  255.255.255.192/26 (subnet mask)


11010100.00000001.00000001.10000001  212.1.1.129 (the 1st host IP in subnet B)
11010100.00000001.00000001.10111110  212.1.1.190 (the last host IP in subnet B)
11010100.00000001.00000001.10111111  212.1.1.191 (broadcast address)


Subnet C: 20 hosts  require 2^5 = 32, maximum number of hosts 2^5 – 2 = 30 IPs
11010100.00000001.00000001.11000000  212.1.1.192/27 (network address)
11111111.11111111.11111111.11100000  255.255.255.224/27 (subnet mask)


11010100.00000001.00000001.11000001  212.1.1.193 (the 1st host IP in subnet C)
11010100.00000001.00000001.11011110  212.1.1.222 (the last host IP in subnet C)
11010100.00000001.00000001.11011111  212.1.1.223 (broadcast address)




Subnet D: 18 hosts  require 2^5 = 32, maximum number of hosts 2^5 – 2 = 30 IPs
11010100.00000001.00000001.11100000  212.1.1.224/27 (network address)
11111111.11111111.11111111.11100000  255.255.255.224/27 (subnet mask)


11010100.00000001.00000001.11100001  212.1.1.225 (the 1st host IP in subnet D)
11010100.00000001.00000001.11111110  212.1.1.254 (the last host IP in subnet D)
11010100.00000001.00000001.11111111  212.1.1.255 (broadcast address)




b) Do away with the subnets or assign multiple subnets in a department