Subnet and CIDR Activity

Solutions

Problem 1

An organization has a class C network 196.10.10 and wants to form subnets for five departments, which host as follows:

Hosts bits Net Address Address Range Mask
55 hosts. 6 bits 196.10.10.0 196.10.10.1 - 196.10.10.62 255.255.255.192
50 hosts 6 bits 196.10.10.64 196.10.10.65 - 196.10.10.126 255.255.255.192
45 hosts 6 bits 196.10.10.128 196.10.10.129 - 196.10.10.190 255.255.255.192
25 hosts 5 bits 196.10.10.192 196.10.10.193 - 196.10.10.222 255.255.255.224
20 hosts 5 bits 196.10.10.224 196.10.10.225 - 196.10.10.254 255.255.255.224

There are 195 hosts in all. Design a possible arrangement of subnets to make each department in a different subnet. For each subnet, give subnet mask and range of IP addresses.

Problem 2

1. You are the CTO of a growing startup and have to get IP addresses to connect 560 computers to the Internet. You can get IP addresses from two providers, IPMart and EastSideIP. IPMart sells classA, class B and class C blocks, while EastSideIP sells CIDR blocks. As the IPv4 address space is scarce, you want to save money and get the smallest number of addresses possible.

a. If you get one block from IPMart, which class do you have to get? What is the problem with that? [1 pt]

You have to get a class B block from IPMart, because a class C only has addresses for 254 computers. (The first address is the network number, and the last one is the broadcast address.) The problem with getting a class B block is that it allocates a space of 216 addresses, and you only need 560.

b. If you get one block from EastSideIP, how many bits are there in the mask (e.g., is it a /8, /22)? How many addresses are wasted?

A CIDR block has to have an integral power of two addresses. The smallest power of two larger than 560 is 210, which means that the network mask has 32 − 10 = 22 bits. The number of wasted addresses is 1024 − 2 − 560 = 462. (We also accepted 1024−560 = 464, if you didn’t take into account the first and the last addresses of the block.)

c. Suppose you can get two blocks from EastSideIP, and they can be of different sizes. How many bits are there in the masks for each of the blocks? How many addresses are wasted now?

If we can get two blocks, we should get a /23 block, which will be good for 512-2 = 510 computers. For the remaining 50 computers we need to get a /26 block, good for 26 − 2 = 62 addresses. The number of wasted addresses is 62 - 50 = 12. (We also accepted 16 as the number of wasted addresses, if you considered 512 addresses in the first block and 64 - 48 = 16).


Topic revision: r2 - 2015-11-28 - JimSkon
 
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