04-11-2010 11:57 PM - edited 03-06-2019 10:33 AM
This particular example problem asks for the smallest subnet number, and the zero subnet is still available (172.16.0.0/23, with broadcast address 172.16.1.255). If the question allowed the use of the zero subnet, the zero subnet (172.16.0.0/23) would be the correct answer. However, if the zero subnet was prohibited from use, the first four subnets listed in Table 5-4 would not be available, making the fifth subnet (172.16.6.0/23) the correct answer.
Pardon for this silly question I got from my ICND2 text book, to my understanding, the Zero subnet should 172.16.0.0 and NOT the first 4 subnet.
Pls correct me if I'm wrong.
Thanks!
Solved! Go to Solution.
04-12-2010 04:21 AM
macky1313 wrote:
Hi Jon,
Many thanks for the reply. Please find the Figure 5.2 attached.
To my understanding, if question allowed the use of the zero subnet, the zero subnet (172.16.0.0/23) would be the correct answer. But why if the zero subnet was prohibited from use, the first four subnets listed in Table 5-4 would not be available?
Andy
Andy
Because you cannot use a subnet that overlaps with one already in use in the network.
So
172.16.0.0/23 = 172.16.0.1 -> 172.16.1.254 broadcast 172.16.1.255
172.16.2.0/23 = 172.16.2.1 -> 172.16.3.254 broadcast 172.16.3.255
172.16.4.0/23 = 172.16.4.1 -> 172.16.5.254 broadcast 172.16.5.255
now if you look at figure 5-2 you can see addresses from 172.16.2.0/23 and 172.16.4.0/23 are already in use in the network. So you can't use these because the question is asking for the first available /23 subnet that does not overlap with any used addresses in figure 5-2.
So the first /23 subnet available that does not overlap with any used addresses in figure 5-2 is 172.16.6.0/23 which is
172.16.6.1 -> 172.16.7.254 broadcast 172.16.7.255
So if you look at figure 5-2 you can see that there are no addresses used in the above range so it is the first free /23 subnet that you can allocate that does not overlap with any addresses already in use.
Jon
04-12-2010 12:00 AM
Andy
Could you post the exact question from the book as it's not clear from your description what you mean.
Jon
04-12-2010 12:32 AM
I have attached the whole section of the exam:
Adding a New Subnet to an Existing Design
Another task required when working with VLSM-based internetworks is to choose a new subnet number for an existing internetwork. In particular, extra care must be taken when choosing new subnet numbers to avoid causing an overlap between the new subnet and any existing subnets. For example, consider the internetwork in Figure 5-2, with classful network 172.16.0.0. An exam question might suggest that a new subnet, with a /23 prefix length, needs to be added to the design. The question might also say “Pick the smallest subnet number that can be used for the new subnet.” So, the subnets must be identified, and a nonoverlapping subnet must be chosen.
To attack such a problem, you would essentially need to find all the subnet numbers that could be created in that classful network, using the stated or implied mask. Then, you would have to ensure that the new subnet did not overlap with any existing subnets. Specifically, you could use the following steps:
Step 1 If not already listed as part of the question, pick the subnet mask (prefix length) based on the design requirements, typically based on the number of hosts needed in the subnet.
Step 2 Calculate all possible subnet numbers of the classful network, using the mask determined at Step 1. (If the exam question asks for the numerically largest or smallest subnet number, you might choose to only do this math for the first few or last few subnets.)
Step 3 For the subnets found at Step 2, calculate the subnet broadcast address and range of addresses for each assumed subnet.
Step 4 Compare the lists of potential subnets and address ranges to the existing subnets and address ranges. Rule out any of the potential subnets that overlap with an existing subnet.
Step 5 Pick a subnet number from the list found at Step 2 that does not overlap with any existing subnets, noting whether the question asks for the smallest or largest subnet number.
Using this five-step process with the example started just before the step list with Figure 5-2, the question supplied the prefix length of /23 (Step 1). Table 5-4 lists the results for Steps 2 and 3, listing the subnet numbers, broadcast addresses, and range of addresses for the first five of the possible /23 subnets.
Step 4 compares the information in the table with the existing subnets. In this case, the second, third, and fifth subnets in Table 5-4 overlap with existing subnets in Figure 5-2.
Step 5 has more to do with the exam than with real networks. Multiple-choice questions sometimes need to force the question to have only a single answer, so asking for the numerically smallest or largest subnet solves the problem. This particular example problem asks for the smallest subnet number, and the zero subnet is still available (172.16.0.0/23, with broadcast address 172.16.1.255). If the question allowed the use of the zero subnet, the zero subnet (172.16.0.0/23) would be the correct answer.
However, if the zero subnet was prohibited from use, the first four subnets listed in Table 5-4 would not be available, making the fifth subnet (172.16.6.0/23) the correct answer.
04-12-2010 12:46 AM
Andy
Figure 5-2 might be useful. The subnet ero is 172.16.0.0/23. What the question seems to be saying is that if you can't use the subnet zero then the other subnets in table 5-4 are already in use in the network depicted in figure 5-2.
So you would need to then use the 5th subnet in table 5-4. It is not saying that subnet zero covers the first 4 subnets in table 4, it is saying that if you aren't allowed to use subnet zero then you must choose another of the subnets in table 4 and that they must not overlap with anything in use already. I'm assuming they are in use in figure 5-2.
Perhaps you could look at figure 5-2 and confirm or post it up here.
Jon
04-12-2010 02:22 AM
Hi Jon,
Many thanks for the reply. Please find the Figure 5.2 attached.
To my understanding, if question allowed the use of the zero subnet, the zero subnet (172.16.0.0/23) would be the correct answer. But why if the zero subnet was prohibited from use, the first four subnets listed in Table 5-4 would not be available?
Andy
04-12-2010 04:21 AM
macky1313 wrote:
Hi Jon,
Many thanks for the reply. Please find the Figure 5.2 attached.
To my understanding, if question allowed the use of the zero subnet, the zero subnet (172.16.0.0/23) would be the correct answer. But why if the zero subnet was prohibited from use, the first four subnets listed in Table 5-4 would not be available?
Andy
Andy
Because you cannot use a subnet that overlaps with one already in use in the network.
So
172.16.0.0/23 = 172.16.0.1 -> 172.16.1.254 broadcast 172.16.1.255
172.16.2.0/23 = 172.16.2.1 -> 172.16.3.254 broadcast 172.16.3.255
172.16.4.0/23 = 172.16.4.1 -> 172.16.5.254 broadcast 172.16.5.255
now if you look at figure 5-2 you can see addresses from 172.16.2.0/23 and 172.16.4.0/23 are already in use in the network. So you can't use these because the question is asking for the first available /23 subnet that does not overlap with any used addresses in figure 5-2.
So the first /23 subnet available that does not overlap with any used addresses in figure 5-2 is 172.16.6.0/23 which is
172.16.6.1 -> 172.16.7.254 broadcast 172.16.7.255
So if you look at figure 5-2 you can see that there are no addresses used in the above range so it is the first free /23 subnet that you can allocate that does not overlap with any addresses already in use.
Jon
04-12-2010 06:40 PM
Hi Jon,
Many thanks for your answer. I know this question sound weird and funny.
After reading the paragraph for a few times, I finally understand what is mean.
If subnet zero is allowed, we can use 172.16.0.0 as this subnet is available.
If subnet zero is not allowed, the next available is 172.16.6.0.
I got it.
Thanks again!
04-13-2010 04:25 AM
macky1313 wrote:
Hi Jon,
Many thanks for your answer. I know this question sound weird and funny.
After reading the paragraph for a few times, I finally understand what is mean.
If subnet zero is allowed, we can use 172.16.0.0 as this subnet is available.
If subnet zero is not allowed, the next available is 172.16.6.0.
I got it.
Thanks again!
Andy
Glad to have helped.
Jon
Find answers to your questions by entering keywords or phrases in the Search bar above. New here? Use these resources to familiarize yourself with the community: