computer networks ch-The network layer Q-30

Question

A large number of consecutive IP address are available at 198.16.0.0. Suppose that four organizations A, B, C and D request for 4000, 2000, 4000 and 8000 addresses respectively. For each of these, give the first IP address assigned, the last IP address assigned, and the mask in dotted decimal notation.

My answer is

A 198.16.0.0/20 to 198.16.15.255

B 198.16.16.0/21 to 198.16.23.255

C 198.16.24.0/20 to 198.16.39.255

D 198.16.40.0/19 to 198.16.71.255

Right or wrong??

Comments

Should I consider the given IP address as class C address?

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Your answer for Organization D is 

198.16.40.0/19 to 198.16.71.255

but it should be 

198.16.40.0/19 to 198.16.55.255

because /19 = 32 - 13 means we can manipulate only last 13 bits.

Explanation for D

198.16.40.0/19 

now 40.0 in binary notation is   00101000 00000000 

now 55.255 in binary notation is 00110111 11111111 (last 13 bits in bold letters are those which are changing)

hence for organization D it should be 

198.16.40.0/19 to 198.16.55.255

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Is it not our own choice how to give codes to these four organizations?

I mean like for org D, we need to fix 13 bits for host id so in third 8 bit segment of the IP address we get to have 3 bits to code. I gave it 198.16.00000000.00000000 to 198.16.00011111.11111111

Likewise,

org C- 198.16.10000000.00000000 to 198.16.10001111.11111111

org A - 198.16.11000000.00000000 to 198.16.11001111.11111111

org B - 198.16.11100000.00000000 to 198.16.11100111.11111111

The Bold represents the fixed bits.

The question is- Is there a standard way for giving code to the bits we need to fix by ourself? If not, there could be many ways for that, example : I can also give 198.16.01100000.00000000 as first IP address to organization D. I only need to fix first 3 bits of 3rd 8 bit segment of this IP address and I can do it any way possible.(unless the IP address clashes with any IP address given to other organisations.)

Answer 1

For organization A, which requested 4000 IP addresses, the number of bits for host portion would be at least 12, i.e., 2^12=4096. Therefore,

• its network mask should be 198.16.00000000/20, or 198.16.0.0/20
• its first IP address assigned is 198.16.0.1
• its last IP address assigned is 198.16.00001111.11111111 or 198.16.15.255