Need help on a practice question

helms20 · February 2007

I am studying using practice tests from examforce and need to try and understand the explanation given for this question. I will post the question and the possible answers and then the explanation given. Sorry it will be long.

Q: You are the IP addressing architect for your company which has purchased 191.14.10.0, 191.14.11.0, and 191.14.12.0 IP networks from an ISP. The network is designed as shown in the diagram below:

(the diagram is one bar on either side of a firewall router research on one and sales & marketing on the other.)

The research department has been moved to its own isolated portion of the network separate from sales and marketing. A firewall router separates the two portions of the network. The research department has 195 hosts that require the assignment of an IP address while the sales and marketing departments have a combined total of 380 hosts that require an assigned IP address.

Select the Network ID and subnet mask for the sales and marketing segment.

A 191.14.11.0
B 191.14.10.0
C 191.14.12.0
D 255.255.255.0
E 255.255.255.128
F 255.255.254.0
G 255.255.252.0

Now for the explanation.

The correct answer is:

Sales & Marketing: 191.14.10.0 /23 or 255.255.254.0

The scenario presented must be quickly identified as a scenario in which supernetting is required. Notice that all of the purchased IP networks are by default class C networks which only provide 254 hosts per network and use a default subnet mask of 255.255.255.0 and a CIDR notation of /24. However the sales and marketing subnet has 380 hosts. Therefore 2 of the IP networkds must be combined into a single network to accommodate the sales and marketing subnet. Research only has 195 and that can be accommodated with a single class C network.

Because of restrictions in the binary ocnversions of the IP networks the 191.14.10.0 and the 191.15.11.0 networks should be the two that get combined for the sales and marketing subnet while the 191.14.12.0 IP network is used for research. There are two methods for determining the new IP network that results from the combination, either solution will work you do not need to perform both:

The rest of the explanation tells you the two methods on how to determine how many hosts are on a subnetted network. What my question is why are they saying you have to combine the networks? Would you not just change the subnet mask of one of the networks to the 255.255.254.0 to allow for the extra hosts?

sprkymrk · February 2007

When you change the subnet mask, you ARE combining the networks. If you didn't own the 191.14.11.0 network, but still changed the netmask on your 191.14.10.0 network to 255.255.254.0 you wouldn't be gaining anything.

helms20 · February 2007

Yes but isn't a 191 address range a class B range? They would hold more than the required 380 hosts.

sprkymrk · February 2007

Hey, good catch. Yes, they worded the question wrong by saying "by default" they are all class C networks. However, in the question it states that you have 191.14.10.0, 191.14.11.0 and 191.14.12.0. If you had purchased a full class B it would have read 191.14.0.0 with a mask of 255.255.0.0.

Did you notice in the explanation where it mentioned the CIDR notation of /24? That means the network portion of the IP Address matches the first 24 bits of the binary number 191.14.10.0 which is all of the first 3 octets. I am going to let one of the Cisco guru's explain it beyond that, because while I can do it and know it, I usually mess up trying to explain it.

helms20 · February 2007

Thanks sprkymrk as supernetting is one of the few things I have left to get down before I take the test. That and WSUS

This whole supernetting thing gets me a little though. Like why do you always use the lower IP address for your new address? I understand the basic concept of combining networks to create more hosts possibilities at a physical location, but I am still a little lost as to what is previous stated

Anyways thanks for all of your help so far.

helms20 · February 2007

Ok so while searching around google, looking for something to increase my knowledge on the current subject, I found this.

In order to supernet eight Class C addresses, they must beconsecutive, and the third octet of the first address must be divisible by 8. For example,131.103.15.0 and 131.103.16.0 cannot be combined into a supernet, but 131.103.18.0 and131.103.19.0 may.

Now according to what I "think" I know the description above is correct except that 18 is not divisible by 8 so this really would not work it would need to be 16 or 24. Am I correct or way off

techster79 · February 2007

Lets see if I can explain this. Supernetting 8 Class C addresses:

In order to supernet eight Class C addresses, they must beconsecutive, and the third octet of the first address must be divisible by 8. For example,131.103.15.0 and 131.103.16.0 cannot be combined into a supernet, but 131.103.18.0 and131.103.19.0 may.

To get 8 supernetted class C addresses the subnet mask would have to be 255.255.248.0 or /21.

Now to figure out the network portion of 131.103.15.0 and 131.103.16.0 we AND the IP and subnet mask in binary. Since we know the first two octets are gonna be network I'll focus on the third.

15 into binary is:
00001111
248 in binary is:
11111000
AND them together and you get 00001000 or the number 8 in the third octet.
Therefore the network portion for 131.103.15.0 is 131.103.8.0

Now lets do 19(16 would be too easy):
19 in binary is:
00010011 AND'd with
11111000 equals
00010000 which is 16, if the mask is /21 or 3 bits borrowed which is 2^3=8 networks, all addresses combined must be within a multiple of 8.
131.103.0.0
131.103.8.0-contains .15
131.103.16.0-contains .16, .18, and .19
131.103.24.0
131.103.32.0
131.103.40.0
and so on...its been a little while since I have subnetted so someone correct me if I'm wrong.

helms20 · February 2007

Awesome thanks alot techster, and for everyone else I found a link to help elaborate on techsters post as well. http://www.sethenol.com/supernetex.htm Hopefully that links

So from my understanding, and correct me if I am wrong, when you supernet you need to work out the network address and the subnet mask to binary first. Then you have to AND them together and then divide the last network octect by the number of networks you are looking to make. If you do not get a whole number then the supernet is not possible. If the third octect is divisible by the number of networks that you want to create then you can supernet them. Is that the basics of it?? Thanks for all the help

Need help on a practice question

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