ME FT: computer networks

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A large number of consecutive IP address are available starting at 198.16.0.0. Suppose that four organizations, A, B, C, and D, request 4000, 2000, 4000, and 8000 addresses, respectively, and in that order.

what is the correct mask for the organization castle?

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How did you get 198.16.32.0/20

Did you take 8000 first ?

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@Hemanth_13

no i have not taken 8000 first.

this is how i solved.

A: 198.16.0.0 – 198.16.15.255 written as 198.16.0.0/20

B: 198.16.16.0 – 198.16.23.255 written as 198.16.16.0/21

C: 198.16.32.0 – 198.16.47.255 written as 198.16.32.0/20

D: 198.16.64.0 – 198.16.95.255 written as 198.16.64.0/19

@Prateek Raghuvanshi

in your solution, tell me the network ID of B and C. if you find out, they come out to be the same. please check it once. and tell me is it possible to have two different subnets having the same network id?

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@aambazinga yeah you are absolutely right , i made mistake .we can't take from 24 because both B and C will have same NETid.so we have to take from 32.

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could someone explain the process in detail?

A: 198.16.0.0 – 198.16.15.255 written as 198.16.0.0/20 (198.16.00000000.0 to 198.16.00001111.255)

B: 198.16.16.0 – 198.16.23.255 written as 198.16.16.0/21 (198.16.00010000.0 to 198.16.00010111.255)

C: 198.16.32.0 – 198.16.47.255 written as 198.16.32.0/20 (198.16.00100000.0 to 198.16.00101111.255)

D: 198.16.64.0 – 198.16.95.255 written as 198.16.64.0/19 (198.16.01000000.0 to 198.16.01011111.255)

what is the correct mask for the organization castle?

This statement is not clear, kindly go through the addresses (bold are the network part and regular are the host part).

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