Report for Uni

[Lkane]

Hello forum, i just made this account today in order to gain some help with some uni reports and projects. ( Not against the rules, i checked and seeking help and discussion is encouraged.)

Anyway i have a report due for next Friday, and i have hit a stumbling block on Sub netting.

As i have came from a tech support background which is a new qualification being offered in the computing world in scotland at colleges , I have felt at quite a loss this year when entering directly into a 3rd year computer networking course at university. My problem has came from being taught subnetting a number of diffrent ways from diffrent lecturers using diffrent methods and now everything is sort of jumbled in my head and am at a complete brick wall with it. especially after undertaking a cisco ccna 1 module recently.

Below is an extract of the task in the report on which i am stuck on.

You will be given an IP address block 183.200.100.0/24. Your task is to devise a feasible network addressing scheme for a large company with a number of geographically distributed offices around the country. There are five locations, with the number of computers at each site 47, 9, 25, 15 and 30 respectively. The location with 47 hosts is the head office, to which every other branch has a wide area network connection through a serial link.

i have took this to mean that there must be 5 subnets at least, for each of the offices, but i cannot find a way to subnet the network enough as to accomadate the appropriate number of hosts.

If someone could show me a subnetting scheme in order to do this correctly and briefly explain how they have done it i would greatly appreciate it !

 

[frowertr]

It would work according to my coffee deprived brain. You just net to calculate how many hosts you need per subnet from that class 3 block you were given.

62 host capable for the first subnet (/26) gives you more than 47, 14 host capable for the 2nd subnet gives you more than 9 (/28), 30 host capable for the 3rd subnet gives you more than 25 (/27), etc...

 

[Lkane]

It would work according to my coffee deprived brain. 62 host capable for the first subnet (/26) of 47, 14 host capable for the 2nd subnet of 9 (/28), 30 host capable for the 3rd subnet of 25 (/27), etc...

this is the orignal way i thought of it and was going to do it but as im going through the cnna1 qualifcation i hit the subnetting module and its just confused everything i thought it knew.

is there anything wrong with doing it like this ? (having each segment a diffrent subnet mask due to the various prefix's /25/26/27 etc etc) and would these all still be connected over a WAN through a serial link with no issues ?

 

[frowertr]

Well that becomes a network question and not really related to subnetting. Sure, assuming the routing is setup correctly, this is not an issue.

I assumed from the question that every site wanted it's own subnet. If you just wanted to clump them all together you could just use a /24 mask.

 

[Lkane]

The task is to subnet that i.p Address block down between 5 offices that are connected through a serial WAN link.

I am now further confused however after going through Cisco's cnna1 module again and finding this situation (very similar to my own problem) within the course material.



"Determine the Number and Size of the Networks

Next, consider the number of subnets required and the number of host addresses needed on each subnet. Based on the network topology consisting of 5 LAN segments and 4 inter network connections between routers, 9 subnets are required. The largest subnet requires 40 hosts. When designing an addressing scheme, you should anticipate growth in both the number of subnets and the hosts per subnet. "

This is practically the same as what i have been asked to do although my largest subnet would require 47 hosts.

This apparently also states that separate subnets are required for the inter network links?

 

[frowertr]

Oh, you also need the subnets for the PtP links for the routers. That is why they are saying 9 is required. So you need the orignal subnets for the networks in addition to the PtP links. You only need two working IP addresses for PtP links so you would use a /30 mask for those.

It's been so long since I took my CCNA I've forgotten much of the coursework.

 

[Pheran]

For the scenario you describe I would use 2 /26 (47, 30), 2 /27 (25, 15) and 1 /28 (9). That still leaves enough room for the 4 /30 WAN links. There's no problem with using different masks, that's what VLSM (variable length subnet masking) is.

Don't try to use a /27 for 30 computers, otherwise have fun figuring out what address you're going to use for the gateway.

 

[Lkane]

For the scenario you describe I would use 2 /26 (47, 30), 2 /27 (25, 15) and 1 /28 (9). That still leaves enough room for the 4 /30 WAN links. There's no problem with using different masks, that's what VLSM (variable length subnet masking) is.

Don't try to use a /27 for 30 computers, otherwise have fun figuring out what address you're going to use for the gateway.

taking on-board what you said i have done this.

Office 1 (HQ - 47 Hosts required) 183.200.100.0/26
Host Address Range (62 available hosts) 183.200.100.193 – 183.200.100.254
Subnet Mask 255.255.255.192
Network address 183.200.100.0
Broadcast address 183.200.100.255
Host bits 6 host bits available
Network Bits 26 network bits used.

Office 2 (9 Hosts required) 183.200.100.0/28
Host Address Range (14 available hosts) 183.200.100.241 – 183.200.100.254
Subnet Mask 255.255.255.240
Network address 183.200.100.0
Broadcast address 183.200.100.255
Host bits 4 host bits available
Network Bits 28 network bits used.




Office 3 (25 Hosts required) 183.200.100.0/27
Host Address Range (30 available hosts) 183.200.100.225 – 183.200.100.254
Subnet Mask 255.255.255.224
Network address 183.200.100.0
Broadcast address 183.200.100.255
Host bits 5 host bits available
Network Bits 27 network bits used.

Office 4 (15 Hosts required) 183.200.100.0/27
Host Address Range (30 available hosts) 183.200.100.225 – 183.200.100.254
Subnet Mask 255.255.255.224
Network address 183.200.100.0
Broadcast address 183.200.100.255
Host bits 5 host bits available
Network Bits 27 network bits used.

Office 5 (30 Hosts required) 183.200.100.0/26
Host Address Range (62 available hosts) 183.200.100.193 – 183.200.100.254
Subnet Mask 255.255.255.192
Network address 183.200.100.0
Broadcast address 183.200.100.255
Host bits 6 host bits available
Network Bits 26 network bits used.

Interconnections (WAN serial links) (2 Hosts required) 183.200.100.0/30
Host Address Range (2 available hosts) 183.200.100.253 – 183.200.100.254
Subnet Mask 255.255.255.252
Network address 183.200.100.0
Broadcast address 183.200.100.255
Host bits 2
Network Bits 30 network bits used.


sorry if thats a bit hard to read its taken directly from my report.

Is my i.p address ranges fine? thats the bit thats throwing me off right now

 

[Pheran]

Unfortunately there are lots of errors in there - nearly all the network and broadcast addresses are incorrect, and you've got overlapping subnets (for example office 2 is stomping all over office 1). I don't know why all of your host address ranges look like they are trying to use the top of the /24.

Here's one example of a /26 network that is correct:

183.200.100.64/26
Network Address: 183.200.100.64
Subnet Mask: 255.255.255.192
Broadcast Address: 183.200.100.127
Host Address Range: 183.200.100.65 - 183.200.100.126

 

[Pheran]

By the way you might want to check out these helpful Kevin Wallace videos:

Part 1: Binary Numbering Review

Part 2: Basic Subnetting

Part 3: Advanced Subnetting

 

[Lkane]

absolutely hopeless for me i'm afraid, having been put straight into a 3rd year networking program has messed up every basic understanding i had of networking.

would both /26 networks have the same details then ? or would they be different. i just cannot understand this its driving me mad.

 

[Lkane]

By the way you might want to check out these helpful Kevin Wallace videos:

Part 1: Binary Numbering Review

Part 2: Basic Subnetting

Part 3: Advanced Subnetting

giving them a watch now, thanks for the help !

 

[Pheran]

The two /26 networks would be completely different. You have this /24 which is a chunk of 256 addresses. A /26 is going to use up 64 of those addresses, which cannot be used again. So every single subnet is going to have different network and broadcast addresses, and different host ranges, none of which overlap.

 

[Lkane]

The two /26 networks would be completely different. You have this /24 which is a chunk of 256 addresses. A /26 is going to use up 64 of those addresses, which cannot be used again. So every single subnet is going to have different network and broadcast addresses, and different host ranges, none of which overlap.

That makes sense and was the way i was thinking about it logically until i done the cisco stuff, however in the example you gave if the first subnet you were to create was a /26 why would you not start at the at the beginning of the I.P block ?

for example

183.200.100.0/26
Network Address: 183.200.100.0
Subnet Mask: 255.255.255.192
Broadcast Address: 183.200.100.64
Host Address Range: 183.200.100.1 - 183.200.100.63

would that be correct?

 

[Pheran]

That makes sense and was the way i was thinking about it logically until i done the cisco stuff, however in the example you gave if the first subnet you were to create was a /26 why would you not start at the at the beginning of the I.P block ?



for example



183.200.100.0/26

Network Address: 183.200.100.0

Subnet Mask: 255.255.255.192

Broadcast Address: 183.200.100.64

Host Address Range: 183.200.100.1 - 183.200.100.63



would that be correct?

I would start at the beginning of the block, that was just an example. What you just posted is close to correct but your broadcast is off by 1. In fact it overlaps my example which can't happen.

 

[Lkane]

I would start at the beginning of the block, that was just an example. What you just posted is close to correct but your broadcast is off by 1. In fact it overlaps my example which can't happen.

ah i see also just realized a mistake i have been making and not counting the ".0" address as an actual number so 64 addresses from the 183.200.100.0/26 would only go up to 183.200.100.63 - and there for that would be the broadcast address?

You have brought back a bit of sanity and understanding here.

so just as a continuation of the previous part of work, If i was to add another subnet on to that /26 network, say a /28 network it would look like this;


183.200.100.0/28

Network Address: 183.200.100.64

Subnet Mask: 255.255.255.240

Broadcast Address: 183.200.100.80

Host Address Range: 183.200.100.65 - 183.200.100.79

 

[Pheran]

ah i see also just realized a mistake i have been making and not counting the ".0" address as an actual number so 64 addresses from the 183.200.100.0/26 would only go up to 183.200.100.63 - and there for that would be the broadcast address?



You have brought back a bit of sanity and understanding here.



so just as a continuation of the previous part of work, If i was to add another subnet on to that /26 network, say a /28 network it would look like this;





183.200.100.0/28



Network Address: 183.200.100.64



Subnet Mask: 255.255.255.240



Broadcast Address: 183.200.100.80



Host Address Range: 183.200.100.65 - 183.200.100.79

Close, but your broadcast is off again; count the addresses - you've used 17. Also you should write that as 183.200.100.64/28. Writing .0 is talking about a different network.

 

[mv2devnull]

No.

The 183.200.100.0/28 says that the network address is 183.200.100.0
You want network address 183.200.100.64, and thus you should write 183.200.100.64/28.

The broadcast address ... didn't you just notice with the first /26 subnet that the broadcast goes to odd number?

 

[Lkane]

Close, but your broadcast is off again; count the addresses - you've used 17. Also you should write that as 183.200.100.64/28. Writing .0 is talking about a different network.

Was about to ask you to explain where i'm going wrong with the broadcast but just listed it without doing it in my head and I've found the problem, i am adding the full next set of hosts to the network address without taking one off for that network address.

so do this ?

/28 network. 16 hosts. 14 useable.

so

1. 183.200.100.64 - Network address
2. 183.200.100.65 - Usable host
3. 183.200.100.66 - Usable host
4. 183.200.100.67 - Usable host
5. 183.200.100.68 - Usable host
6. 183.200.100.69 - Usable host
7. 183.200.100.70 - Usable host
8. 183.200.100.71 - Usable host
9. 183.200.100.72 - Usable host
10. 183.200.100.73 - Usable host
11. 183.200.100.74 - Usable host
12. 183.200.100.75 - Usable host
13. 183.200.100.76 - Usable host
14. 183.200.100.77 - Usable host
15. 183.200.100.78 - Usable host
16. 183.200.100.79 - Broadcast

currently re-doing my network table and will post the corrected version in 10 mins or so, hopefully you are still awake to show me any errors in it !

the helps appreciated.

 

[Lkane]

Removed for proofing reasons !

 

[Pheran]

Getting better - the first 2 subnets are correct, but after that you began to go wrong. Because of the way subnet masks actually work in binary, the beginning network address of any subnet is not arbitrary. When dealing with the common case of splitting up a /24, the end result is that the final octet of the first IP must be a multiple of the subnet size. For example, the only valid starting points for /26 networks are multiples of 64. In other words, they must begin with either .0, .64, .128, or .192.

Thus, 183.200.100.80/27 is an illegal subnet because 80 cannot be divided by 32. Most likely you have more of these problems but I didn't check them all.

 

[Lkane]

Getting better - the first 2 subnets are correct, but after that you began to go wrong. Because of the way subnet masks actually work in binary, the beginning network address of any subnet is not arbitrary. When dealing with the common case of splitting up a /24, the end result is that the final octet of the first IP must be a multiple of the subnet size. For example, the only valid starting points for /26 networks are multiples of 64. In other words, they must begin with either .0, .64, .128, or .192.

Thus, 183.200.100.80/27 is an illegal subnet because 80 cannot be divided by 32. Most likely you have more of these problems but I didn't check them all.

Thanks for all the help today, i have submitted it as was, the overall subnetting details are only worth 30% of the mark and im confident enough is done to pass the report. just wanted it out of the way as it was driving me insane. also what you have just said would have confused me further . Thank you !

 

[JackMDS]

Hmm.

Forum's members doing/helping in personal Educational work is Not Kosher.

Thread Closed.