Reading diagrams to find subnetwork numbers

funkwoo-10

Hi all, I am in need of some direction as to where I can learn more about reading network diagram IP's (like below) so I can find out more about how many subnets exists based on the diagrams provided? My current books do not have details info about this.

Darril

I'm trying to understand what your goal is here. What are you studying? What books are you using now? Do you have any idea of how how many subnets exist in the diagram?

funkwoo-10

Hi Darril, my goal is to learn how many subnets exist and how many additional hosts can be support by each LAN.

I am reading the CompTIA Network+ book, but It does not cover this in depth.

Thanks!

malcybood

Sounds like you need to learn subnetting.

Try some youtube videos like this:

subnetting tutorial - Bing Videos

also try doing binary conversions using the 128 64 32 16 8 4 2 1 method etc. I learned subnetting in around 1999 and still find myself writing this out sometimes to work out subnets.

funkwoo-10

Thanks Malcybood, I will watch that video tonight. I have some basic subnetting before but it is the diagram that confuses me, I do not know where to start with it!

Thanks.

malcybood

funkwoo-10 wrote: »

Thanks Malcybood, I will watch that video tonight. I have some basic subnetting before but it is the diagram that confuses me, I do not know where to start with it!

Thanks.

The key is getting familiar with what subnet mask ties to which "/number"

So in basic terms an ip address can be up to 32 bits (8 bits per octet)

8bits.8bits.8bits.8bits - 32 bits

A subnet mask defines the network element of the address so to take the network on the left, that is a /26 mask where the network begins at 192.168.100.64

The servers to the left are using the first and second usable addresses for the 192.168.100.64 subnet of .65 & .66

The fact that the subnet is a /26 means that the subnet mask is 255(8 bits).255(8bits).255(8 bits).192(2 bits) = 26 bits = a "/26" network

The answer to how many addresses is below but I'll leave it to you to work out how to get there as I'll probably just end up confusing things if I try to explain

Remember this for each octet (8 binary bits):

128 64 32 16 8 4 2 1

Network element = .64
Host element = 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127

Usable address range = 192.168.100.65 - 126 (.64 is network and .127 is broadcast)

LAN 2 has the same amount of hosts in it as it's also a /26 mask.

Practice practice practice

funkwoo-10

Thanks for taking the time to write that out Malcybood, so, first up from the diagram I can see that 3 subnetworks are shown?:

Left:
192.168.100.64/26

Center Link:
192.168.100.128/30

Right:
192.168.100.0/26

Webmaster

That's correct. A shortcut to determine the number of subnets in your diagram, without looking at the IP addresses (which you still should to verify), is simply to count the unique router segments. Especially since it makes no mention of VLANs, the 2 routers basically divide the network into 3 subnets.

funkwoo-10

Hi Malcybood,

Just re-re-re reading over this

I wanted to check I am understanding this, LAN 1 would support 62 hosts? I would then minus 2 from 62 as the dns and web server are already fitted. Leaving 60 additional hosts remaining in LAN 1?

Thanks.

malcybood

Hi mate, yes that's correct. I was always taught to subtract 2 addresses for the network ie .64 and the broadcast which I still believe is the best way to teach this topic until a student is comfortable to subnet "in their head".

There are ways that you can use the subnet so you only subtract 1 but I don't seethe point in ever doing that in a production network as there are millions of private addresses available.

If you are interested in this topic look over ip subnet zero configuration on Cisco routers and this will explain it.

As already confirmed above yes there are 3 segments where two are /26 with 64 addresses, 62 usable and 60 available after servers used the first couple of addresses.

A /30 network gives 4 addresses, with two being usable hence why it is used in a point to point link. This is best practice for point to point links so if you ever get a look at an MPLS wan router it's most likely this will be how it is setup from te customer site to the service provider edge router.

Hope this is helpful.

Cheers

schmalz2

This video is how I learned. CCENT 19 - SUBNETTING PART 1 - (CBT NUGGETS) - (xxiisnip3diixx) - YouTubeIt's long but worth the watch. The video series is for the CCNA but this video pertains to what you are asking.

funkwoo-10

Thanks again Malcybood, then with LAN 2 we have a total of 62 usable hosts then 62 minus the DHCP server would leave me with 61 additional hosts as PC0 and PC1 get IP's from the DHCP pool?

Thanks

malcybood

funkwoo-10 wrote: »

Thanks again Malcybood, then with LAN 2 we have a total of 62 usable hosts then 62 minus the DHCP server would leave me with 61 additional hosts as PC0 and PC1 get IP's from the DHCP pool?

Thanks

Well yes and no.

Technically you could use 61 addresses for dhcp on that segment but you also need to assign an address for the default gateway from the subnet.

Typically in most production networks you use the first usable or the last usable in the range as the default gateway (black line side of the router)

Also typically (not always though) you would keep a number of addresses spare for static devices (such as the dhcp server) then start the dhcp pool after that so could look something like this for LAN 2

.1 - .10 = static addresses
.11 - .61 = client dhcp scope
.62 = default gateway (router inside)

My preference is to configure static devices at the start of te range and give the default gateway the first usable ie .1 and this is the most common deployment that I've seen but there's many ways to skin a cat as they say.

The other /26 segment would follow the same format

funkwoo-10

Hmm so rather then having 61 additional hosts I would need to subrract another 1 address for the default gateway making it 60 additional hosts supported by LAN 2. Taking in to account the hosts already fitted in the diagram..

Thanks again!