Subnetting - WAN Links

JJBladester

You have been assigned the address block 10.255.255.224/28 to create the network addresses for point-to-point WAN links. How many of these WANs can you support with this address block?

A. 1
B. 4 (correct answer)
C. 7
D. 14

My question is: "Why is B correct?"

I assume that they expect you to see 10.255.255.244/28 and realize that you have 28 network bits and 4 host bits (adding up to a total of 32 bits).

I also assume that they expect you to make the WAN links take up the least amount of addresses, thus requiring you to use VLSM. This would mean you would have to use a different slash format like /29, /30, or /31.

If you use /30, you get:

---

Network

---

Assignable Host Addresses----
WAN1 10.255.255.224 /30 10.255.255.225 - 10.255.255.226
WAN2 10.255.255.228 /30 10.255.255.229 - 10.255.255.230
WAN3 10.255.255.232 /30 10.255.255.233 - 10.255.255.234
WAN4 10.255.255.236 /30 10.255.255.237 - 10.255.255.238

However, is it possible (assuming you don't care about wasting IP addresses) to have more than 4 WANs?

Lastly, do you assign WAN links a network address (like 10.255.255.234) or a host address (like 10.255.255.235)? I ask because my Cisco book says "Each WAN link is a network. You create subnets for the WAN that interconnect different geographic locations."

Zartanasaurus

/30 only gets you 4 subnets in that range. 10.255.255.224/28 ends at 10.255.255.239. Yes, there are more valid IPs after that, but they aren't in the range specified.

Forsaken_GA

tehcnically speaking, you can get 8, you can use /31 masks to address point to point links, though not all ios versions support it. However, of the choices available, B is correct.

Forsaken_GA

JJBladester wrote: »

However, is it possible (assuming you don't care about wasting IP addresses) to have more than 4 WANs?

Given that the question specifies point to point links, you'd be abslutely foolish to assign anything longer than the smallest possible mask to the links. There will only ever be two points on that link, so there's no need for expansion.

The difference between WAN and LAN is one of scope. WAN's tend to be global, international, or at the very least, regional, and they tend to be internetwork links, as opposed to your local LAN, which are intranetwork links. LAN links are your intrastate highways, WAN links are your interstate highways. They're both roads, it's just in how they're used that's different.

Lastly, do you assign WAN links a network address (like 10.255.255.234) or a host address (like 10.255.255.235)? I ask because my Cisco book says "Each WAN link is a network. You create subnets for the WAN that interconnect different geographic locations."

WAN links are like any other layer 3 link, you address them with a host address. It's no different than your layer 3 routed LAN links. Each layer 3 LAN also represents a seperate and distinct network. Go ahead and try to assign a router 2 IP's in the same subnet to different interfaces, see what happens

JJBladester

Forsaken_GA wrote: »

Go ahead and try to assign a router 2 IP's in the same subnet to different interfaces, see what happens

Out of curiosity, what would happen?

lon21

Zartanasaurus wrote: »

/30 only gets you 4 subnets in that range. 10.255.255.224/28 ends at 10.255.255.239. Yes, there are more valid IPs after that, but they aren't in the range specified.

Sorry would you mind explain in more detail?

I make out that you have a total of 14 host which will give 7 WLAN Links. from 224-239.

Thanks

Eildor

The block you are given is 10.255.255.224/28.

The 10.255.255.224/28 block gives you a range of 10.255.255.255.224 - 10.255.255.255.240.

If you subnet that with a /30 you get the following subnets:

10.255.255.224 - 10.255.255.227
10.255.255.228 - 10.255.255.231
10.255.255.232 - 10.255.255.235
10.255.255.236 - 10.255.255.239

If you were to subnet it further, it would leave the 10.255.255.255.224 - 10.255.255.255.240 range.

lon21

Eildor wrote: »

The block you are given is 10.255.255.224/28.

The 10.255.255.224/28 block gives you a range of 10.255.255.255.224 - 10.255.255.255.240.

If you subnet that with a /30 you get the following subnets:

10.255.255.224 - 10.255.255.227
10.255.255.228 - 10.255.255.231
10.255.255.232 - 10.255.255.235
10.255.255.236 - 10.255.255.239

If you were to subnet it further, it would leave the 10.255.255.255.224 - 10.255.255.255.240 range.

Yes thanks,

previously I would have said 7, but I guess its asking you to create WLAN links from a certain block of ip address.

JJBladester

Forsaken_GA wrote: »

tehcnically speaking, you can get 8, you can use /31 masks to address point to point links, though not all ios versions support it.

So, is there something taboo about /31 masks? Should I assume that any Cisco Press books will automatically assume /31 masks?

I did find an RFC entitled Using 31-Bit Prefixes on IPv4 Point-to-Point Links which made the use of /31 sound a bit out of the ordinary.

Eildor

JJBladester wrote: »

So, is there something taboo about /31 masks? Should I assume that any Cisco Press books will automatically assume /31 masks?

I did find an RFC entitled Using 31-Bit Prefixes on IPv4 Point-to-Point Links which made the use of /31 sound a bit out of the ordinary.

As far as CCNA is concerned you use a /30 for a WAN link. I don't remember reading about using a /31 in any of the CCNA literature I've read, and I'm not even sure how that would work to be honest (but I trust Forsaken_GA that it's possible).

Forsaken_GA

lon21 wrote: »

Sorry would you mind explain in more detail?

I make out that you have a total of 14 host which will give 7 WLAN Links. from 224-239.

Thanks

The fact that you're getting an odd number off binary math should tell you that your math is wrong. When dealing with subnets and ranges, it's all about powers of 2, everything is divisible by 2.

It's pretty simple math. If you start with a 28 bit mask, how many subnets do you create if you borrow one bit?

2.

You get 2 /29's instead of 1 /28.

So now you can borrow a bit from each of the /29's creating 2 pairs of /30's. So your /28 just became 4 /30's by borrowing 2 bits. Convention is that you cannot borrow any deeper than that.

The normal rules say you must have a network address and a broadcast address, so if you subnet down to a /31, you have no IP's left for host addressing, so a /30 is as deep as you can go.

and 224 from 239 is 15, not 14, but you have to remember that 0 is a valid number when it comes to IP addressing, so you actually have 16. Remember that 8 bit binary range is 0 to 255, which is 256 total values.

If you doubt me, then count:

224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239

There are 16 distinct numbers there.

So we have 16 IP's, which assuming 2 IP's to each link, would give 8 WAN links. Which is true if you're using /31 masks. However, that's not considered cannon, so you have to account for network and broadcast addresses, which means each WAN link consumes 4 IP's, not 2. So 4 IP's per link, 16 total IP's, 16 / 4 = 4

Forsaken_GA

JJBladester wrote: »

Out of curiosity, what would happen?

You get an error message complaining about overlapping IP space. A router cannot have 2 layer 3 ports that reside in the same subnet.

Forsaken_GA

JJBladester wrote: »

So, is there something taboo about /31 masks? Should I assume that any Cisco Press books will automatically assume /31 masks?

I did find an RFC entitled Using 31-Bit Prefixes on IPv4 Point-to-Point Links which made the use of /31 sound a bit out of the ordinary.

No, most everything official from Cisco says that /30 links are considered canon. /31's aren't supported on all platforms, or IOS images. In practical usage, if I'm peering with a service provider on a point to point link, even if my end supports it, their end might not. So you can ask, but you'll probably be told no. The only time I'd ever use a /31 is for internal addressing between platforms that I knew supported it.

Forsaken_GA

Eildor wrote: »

As far as CCNA is concerned you use a /30 for a WAN link. I don't remember reading about using a /31 in any of the CCNA literature I've read, and I'm not even sure how that would work to be honest (but I trust Forsaken_GA that it's possible).

correct, for a CCNA, you should ignore the existence of /31 links. It's one of those things they don't like to teach you about early on because it breaks the rules of the subnet that they're trying to impart. If you'll continue your studies, you'll find there are several issues where the CCNA material is being less than truthful (for example, when I was studying for the CCNA, it was God's Own Truth that subnet-zero and the all ones subnet could Never Be Used. Imagine my surprise when I got to CCNP level material and found out that wasn't exactly true)

As far as /31's go, Cisco does have official literature on it:

Using 31-Bit Prefixes on IPv4 Point-to-Point Links [Cisco IOS Software Releases 12.2 T] - Cisco Systems

But again, for CCNA studies, you should ignore it's existence. Just take advantage of it in the real world if the opportunity presents itself.

Eildor

Forsaken_GA wrote: »

correct, for a CCNA, you should ignore the existence of /31 links. It's one of those things they don't like to teach you about early on because it breaks the rules of the subnet that they're trying to impart. If you'll continue your studies, you'll find there are several issues where the CCNA material is being less than truthful (for example, when I was studying for the CCNA, it was God's Own Truth that subnet-zero and the all ones subnet could Never Be Used. Imagine my surprise when I got to CCNP level material and found out that wasn't exactly true)

As far as /31's go, Cisco does have official literature on it:

Using 31-Bit Prefixes on IPv4 Point-to-Point Links [Cisco IOS Software Releases 12.2 T] - Cisco Systems

But again, for CCNA studies, you should ignore it's existence. Just take advantage of it in the real world if the opportunity presents itself.

Thanks, I'll have a read of that link

JJBladester

Forsaken_GA wrote: »

The fact that you're getting an odd number off binary math should tell you that your math is wrong. When dealing with subnets and ranges, it's all about powers of 2, everything is divisible by 2.

It's pretty simple math. If you start with a 28 bit mask, how many subnets do you create if you borrow one bit?

2.

You get 2 /29's instead of 1 /28.

So now you can borrow a bit from each of the /29's creating 2 pairs of /30's. So your /28 just became 4 /30's by borrowing 2 bits. Convention is that you cannot borrow any deeper than that.

Since CCNA exams are probably timed, I need to know how to answer the original question faster than the 20 minutes it took me to write down the original IP address block + subnet mask and do all the logical ANDing necessary to show the four different subnets (and their net/host/b-cast addresses) created by a /30 mask.

So, here's how I think I would answer a qustion like this going forward:

Each WAN link must be on a different subnet. WAN links are assumed to have /30 masks, per industry standards.

Therefore, the real question is how many subnets can we get out of a /28 block of IP addresses by borrowing two bits, making the new subnets all /30 format?

The math is:

2^(# of borrowed bits) = number of subnets available after borrowing bits

In our case:

2^2 = 4

The longer answer would be to recognize that the two extra bits can only be represented by the following combinations:

00 (.224) (with 4 addy's in this subnet including network and brodacast addresses b/c you have 2 host bits with a possible combination of 00, 01, 10, 11)

Since the host bits are to the far-right of the /30 network, they represent the numbers:

00 = 0
01 = 1
10 = 2
11 = 3

Adding these numbers to the original .224 yields:

.224 + 0 = .224 (network)
.224 + 1 = .225 (host)
.224 + 2 = .226 (host)
.224 + 3 = .227 (broadcast)

I won't do this math for the remaining subnets because you get the point.

01 ( .220 ) (with 4 addy's in this subnet including network and brodacast addresses b/c you have 2 host bits with a possible combination of 00, 01, 10, 11)
10 ( .232 ) (with 4 addy's in this subnet including network and brodacast addresses b/c you have 2 host bits with a possible combination of 00, 01, 10, 11)
11 ( .236 ) (with 4 addy's in this subnet including network and brodacast addresses b/c you have 2 host bits with a possible combination of 00, 01, 10, 11)

So, Forsaken_GA... Given my "quick answer" (not the math one I drew up above simply for clarity), am I on the right path to just do
2^(# of borrowed bits) = number of subnets available after borrowing bits = number of available point-to-point WAN links

Forsaken_GA

JJBladester wrote: »

So, Forsaken_GA... Given my "quick answer" (not the math one I drew up above simply for clarity), am I on the right path to just do
2^(# of borrowed bits) = number of subnets available after borrowing bits = number of available point-to-point WAN links

Yes. If you're given the base mask, and asked how many possible subnets you can create from it, the answer is always going to be 2 to the power of # of bits borrowed, you just have to recognize that you have to leave at least 2 bits for host addressing.

Now, the questions can be presented in other ways. for example, if I give you the mask 172.16.12.0/23, and I ask you how many bits I would need to borrow to create at least 5 subnets that can support at least 50 hosts each, you have to understand the correlation between how many bits you borrow effects the number of subnets you have, and how that effects the number of addresses you have left for addressing hosts.

To answer my own example -

Ok, i need 5 subnets. If I borrow one bit, I get 2. If I borrow 2 bits, I get 4. Neither of these is good enough. If I borrow 3 bits, I get 8. This covers my needs, gives me more than I need, but it's the smallest number of bits I can possibly borrow in order to fulfill my needs.

So that means I'm using 26 bits for masking, 32 - 26 = 6. So I have 6 bits left for host addressing. 2^6 = 64, so each of my subnets can address 64 hosts (really 62 once you account for broadcast and network ID). Since I need 50 hosts per subnet, this meets my requirements. I borrow 3 bits for subnetting, and use the additional 6 bits for host addressing.

JJBladester

Forsaken... Thanks so much for the clarity of your response. I find this Cisco Press book to ask ambiguous questions sometimes. I'm feeling a lot more comfortable now with IPv4 addressing.

Eildor

This question shouldn't take longer than 2 minutes.

10.255.255.224 is the network address. /28 gives you an increment of 16, which gives you the block range. /30 gives you an increment of 4. Keep incrementing until you reach the end of the block range.

If you're not sure about what I just said then let me know and I'll try post a more thorough explanation of how I personally would go about answering the question.

A good website to practice your subnetting is subnettingquestions.com

cpartin

I second subnettingquestions.com. I studied for my CCENT by drilling problems there a few minutes a day until it was second nature. That helped immensely come exam time and I cannot recommend that site enough.

Forsaken_GA

cpartin wrote: »

I second subnettingquestions.com. I studied for my CCENT by drilling problems there a few minutes a day until it was second nature. That helped immensely come exam time and I cannot recommend that site enough.

I agree it's a valuable website, my only caveat is to be careful of something like this. It can teach you the mechanics of subnetting, but it's important to understand the why of it, and how it all works, and the effects it has. Maybe not for the CCNA, but definitely for your career.

Sadly, this is becoming something of a lost art, given the number of software solutions available for managing IP address space.

JJBladester

Eildor wrote: »

This question shouldn't take longer than 2 minutes.

10.255.255.224 is the network address. /28 gives you an increment of 16, which gives you the block range. /30 gives you an increment of 4. Keep incrementing until you reach the end of the block range.

If you're not sure about what I just said then let me know and I'll try post a more thorough explanation of how I personally would go about answering the question.

A good website to practice your subnetting is subnettingquestions.com

Before an hour ago, I'd have no clue what you meant by "increment" but I understand it now thanks to your example and the CBT Nugget video I just watched where the guy talks about finding the increment which is "the lowest network bit converted back to a decimal number."

Thanks for the reply and the awesome site reference for subnetting.