Networking fun...... -.-

[Kailea]

I have a problem with a lab I am doing for my CCNA class, if there is anyone here on the forums that can help me, for the love of god please do ;p

it wants to use the IP range of 192.168.8.0/22, and that should be 192.168.8.01 - 192.168.11.254

now it seems that the lab wants 6 subnets...but I keep getting that I can only make 4, not sure what I am doing wrong here.

to even being the lab, I have to split the range of IPs in that range of addresses into 6 subnets.

[Mhurron]

Re: Networking fun...... -.-

What the question is asking is turn the /22 into 6 subnets. You're leaving out an important detail of the question, being how many hosts on a subnet they need, which will determine how many bits from the host octet you'll take.

CCNA questions are often worded in a tricky way, you probably should post the question word for word.

[Kailea]

Re: Networking fun...... -.-

What the question is asking is turn the /22 into 6 subnets. You're leaving out an important detail of the question, being how many hosts on a subnet they need, which will determine how many bits from the host octet you'll take.

CCNA questions are often worded in a tricky way, you probably should post the question word for word.

This is the whole 1st task of the lab, I hate that I missed class last night, or I would not have this problem, I like examples and being shown things, it is how I learn ;p

--------------------------------------------------------------------------------
Determine the number of subnets needed.
Determine the number of hosts needed for each LAN.
Design a hierarchical addressing scheme.
Assign addresses and subnet masks to device interfaces and hosts.
Examine the use of the available network address space.
Introduction:

In this lab, you have been given the IP addresses included in 192.168.8.0/22. Use this range of IP addresses to subnet and provide the IP addressing for the network. The network has the following addressing requirements:

The R2 LAN will require 120 host IP addresses.
The R3 LAN will require 120 host IP addresses.
The R4 LAN will require 120 host IP addresses.
The WAN links between each of the routers will require an IP address for each end of the link.
Task 1: Examine the network requirements.

Examine the network requirements and answer the questions below. Keep in mind that IP addresses will be needed for each LAN interface on each router.

How many subnets are needed?
What is the maximum number of IP addresses that are needed for a single subnet?
How many IP addresses are needed for each of the LANs?
How many IP addresses are needed for all of the connections between routers?
What is the total number of IP addresses that are needed?
This activity has multiple windows of instructions. Select the ">" button to view the next portion of instructions.


EDIT: going to look at it more when I get home from work, my brain is shot right now and I have work in 30 mins ;p

I thank you in advanced for any help you can give me *bows*

[Mhurron]

Re: Networking fun...... -.-

Your first problem is that /22 is not a classful Class C mask. This means that 192.168.8.0/22 is a single network comprising of 192.168.8.01 - 192.168.11.254. The 4 subnets you think you see (192.168.8.0, 192.168.9.0 192.168.10.0 and 192.168.11.0) are actually all addresses on the one /22 supernet. They would only be separate subnets if you had a /24.

For this to make sense, you have to convert to binary. A /22 (255.255.252.0) looks like this:
(s for subnet bit, h for host bit)
ssssssss.ssssssss.sssssshh.hhhhhhhh

The answer to the question doesn't make much sense in decimal, so work in binary.

A /22 has taken the last two bits from the third octet and transferred them to the host portion of the address. To turn the /22 - which is one network - into smaller ones, you start taking them back.

Now in your question I only see 4 networks required, one for the LAN off of R2, one for the LAN off of R3, one for the LAN off of r4 and one for the WAN between the three routers. We're going to go on that, since I don't see where you got 6 from.

The WAN is easy, it only needs 5 IP's (one for each interface, one broadcast and one network address.) So that doesn't really help in determining the subnet mask. The other three have to have 120 usable addresses, so this is what we go on.

You have the host portion of the /22 is 1111111111 in binary or 1023 in decimal giving you 1022 usable host addresses. (1023 + 1 because the network address would be all 0's giving you 1024 addresses, but you have to take 2 away for the network and broadcast leaving 1022 usable.)

Obviously the first thing to try is taking one host bit away (/23), but that only gives you at most 2 subnets because that bit would either be 1 or 0. So we try taking another one (/24). Now we have 4 subnets to work with ( 00, 01, 10 , 11) each with 256 addresses, or 254 addresses. Giving you the /24 subnet mask (255.255.255.0) with the network addresses of 192.168.8.0, 192.168.9.0 192.168.10.0 and 192.168.11.0

If you did leave something out and do need 6 networks, you would continue on like this, (your next test would be to use a /25 by taking the most significant bit from the fourth octet leaving 7 bits for the host address) and seeing if that gave you enough networks with enough usable addresses on them.

[Kailea]

Re: Networking fun...... -.-

Thanks for the reply,vwill look at t morewhen I get home (on phone rgt now) from work.

[Kailea]

Re: Networking fun...... -.-

ok I think I see where I am getting confused here.....

at first it tells me this.....
----------------------------------------------------------
The R2 LAN will require 120 host IP addresses.
The R3 LAN will require 120 host IP addresses.
The R4 LAN will require 120 host IP addresses.
The WAN links between each of the routers will require an IP address for each end of the link.
---------------------------------------------------------

but then it tells me this......
---------------------------------------------------------
When assigning the subnets, keep in mind that routing will need to occur to allow information to be sent throughout the network.

The subnets will be assigned to the networks to allow for route summarization on each of the routers.

Assign subnet 0 to the R1 to R2 WAN.
Assign subnet 1 to the R2 LAN.
Assign subnet 2 to the R1 to R3 WAN.
Assign subnet 3 to the R3 LAN.
Assign subnet 4 to the R1 to R4 WAN.
Assign subnet 5 to the R4 LAN.
---------------------------------------------------------
I need at least 120 address per subnet.... but then it wants me to have 6 subnets...... from what I have learned so far, I do not see how to do this -.-

[Mhurron]

Re: Networking fun...... -.-

No, still just 4 subnets listed. One off R2, one off R3, one off R4 and one between the routers.

Like this -


I still do not see where you are getting 6 subnets from.

BTW, you could have 8 subnets with a /25 and each subnet would have 126 usable addresses.

[Kailea]

Re: Networking fun...... -.-

yeah that is what I need /25 then, I had tried using /25 though and I seemed to only get 2 subnets..... I have a feeling my math is all wrong though, I am not good with math -.-

EDIT: oh wait I think I got it..... silly me was trying to use the C class calculations on a B class....... *facepalm* I think that is where all my confusion is coming from.

[Mhurron]

Re: Networking fun...... -.-

/25 gives you 8 subnets with the whole thing looking like this in binary

(n for network, s for subnet, h for host)

nnnnnnnn.nnnnnnnn.nnnnnnss.shhhhhhh

so you have networks -
192.168.8.0/25 (255.255.255.128)
192.168.8.128/25 (255.255.255.128)
192.168.9.0/25 (255.255.255.128)
192.168.9.128/25 (255.255.255.128)
192.168.10.0/25 (255.255.255.128)
192.168.10.128/25 (255.255.255.128)
192.168.11.0/25 (255.255.255.128)
192.168.11.128/25 (255.255.255.128)

It's just bit flipping and makes much more sense in binary.

---------- Post added at 01:10 PM ---------- Previous post was at 01:07 PM ----------

See, you left off the details about how the WAN link worked. Thats why I couldn't see where you were coming up with 6 subnets.

[Kailea]

Re: Networking fun...... -.-

my bad ;p thanks for the help though I think I do finaly get it, now to finish this lab, so I can do other things I need to do ^.^

Now that I have this, I have to use Packet Tracer to build the network with these addresses.

[Aeni]

Re: Networking fun...... -.-

Wow ... blast from the past ... studying for MS, Intel and Cisco certs.

[Kailea]

Re: Networking fun...... -.-

yeah ;p I want to get out of the Dell on site work, and get into working with networks, but my brain likes to be stubborn most of the time, and my math skills are not the greatest.

[Aeni]

Re: Networking fun...... -.-

yeah ;p I want to get out of the Dell on site work, and get into working with networks, but my brain likes to be stubborn most of the time, and my math skills are not the greatest.

Ask anyone in the field and they'll tell you that some diagnostic tool (Hey, there's an app for that!) will do most of the work. You just have to worry about the whole problem solving end of the equation. But yeah, unfortunately, the certification test won't allow you to *ahem* circumvent that bit of issue. In my opinion it should be about how to fix something broken and not about how much shit you can remember by rote and answer multiple choice questions. Fucking testing industry fraud. No I'm not bitter at all.

[Kailea]

Re: Networking fun...... -.-

lol that is pretty much what our teacher said, if I can just get past the stupid test, I will be ok, but I really learn from doing, so until I can start getting my hands "dirty" my learning has not even started.

[Aeni]

Re: Networking fun...... -.-

but I really learn from doing, so until I can start getting my hands "dirty" my learning has not even started.

This is actually the experience when you start to work. Most of the stuff you learn from school won't even be applied to the work you do. All it does is make you a more informed and well-rounded individual in society.