70-216 - TCP/IP question... help!!

Author

TCP/IP question... help!!

l9nux

2002-09-18, 2:22 pm

I need some help understanding how the answer to this question is achieved...

You are the administrator of your company's network which consists of 400 computers equally distributed between two subnets. You have been assigned the network addresses 208.199.32.0 and 208.199.33.0, which you have used on Subnet A and Subnet B, respectivly. Your router supports BOOTP forwarding, classless interdomain routing (CIDR) and variable-length subnet masks (VLSM). The client computers on both subnets receive their IP configurations from a single DHCP server on Subnet A.

The users in the Publishing department routinely transfer very large graphics files between the computers in their department. Although the department has only eight computers, network analysis shows that they are using almost 50% of the bandwidth on Subnet B. To alleviate the network congestion being caused on Subnet B, you have decided to create a seperate segment, named Subnet C, for the Publishing department.

You do not have another Class C address to add to your network, so you have decided to segment a portion of the 208.199.33.0 address space for Subnet C and assign the remaining address space to the physical segment that contains Subnet B. The network address for Subnet C will be 208.199.33.240/28.

In order to provide the greatest number of available host addresses, which of the following network addresses should be assigned to the physical segment that contains Subnet B? (Select all choices that are correct.)

a. 208.199.33.0/24
b. 208.199.33.0/25
c. 208.199.33.0/26
d. 208.199.33.128/26
e. 208.199.33.192/26
f. 208.199.33.192/27
g. 208.199.33.224/28

Answer...

B, D, F and G

Can anyone explain this? I understand the concept of classless networks, and even how to calculate the number of hosts on a network, but I really don't know how to solve this one?

mikop

2002-09-18, 2:57 pm

put all possible /28 network in binary and start doing route summarization/supernetting.

240 is 1 1 1 1 0 0 0 0

so knwoing that, if we use /25 we have 2 network

one is

0 x x x x x x x

one is

1 x x x x x x x

can't be 2nd, so answer B is correct,

then we further do the granular supernetting of addresses that start with

1 x x x x x x x

till we fully utilize all avaialble addresses.

I am a bit too lazy to work it out to see if the answers given is correct, but this should clue you in as to where you should go to proceed to solve it.

let me clarify some,

the above, answer b includes all /28 networks from 0, 16, 32,48,64,80,96,112

now we have these /28 networks unaccounted for

128,144,160,176,192,208,224

so answer B got rid of half of them and left hte above for you to further supernet/summarize etc.

l9nux

2002-09-18, 4:09 pm

I've read the question over and over again, and this is how I have worked it out. Any suggestions or comments are very much welcome (I need the help!) and I'm still not 100% sure how I did it?

I've got a /24 address space and I've assigned a /28 mask:

11111111.11111111.11111111.11110000

The address assigned to Subnet C is 208.199.33.240/28 which is the last subnet in the /24 address space where a /28 mask is used.

11110000 = 208.199.33.240/28

With the original address 208.199.33.0/24 I just add an extra bit to the mask:

00000000(/24) + 1 bit = 10000000(/25)
208.199.33.0/25

This bit I'm not sure about, but the next available network is 208.199.33.128 so I add an extra bit to the mask:

10000000(/25) + 1 bit = 11000000(/26)
208.199.33.128/26

The next available network here is 208.199.33.192 so I add an extra bit to the mask again:

11000000(/26) + 1 bit = 11100000(/27)
208.199.33.192/27

The next available network is 208.199.33.224, so I add an extra bit to the mask:

11100000(/27) + 1 bit = 11110000(/28)
208.199.33.224/28

The next available network is 208.199.33.240 which is used for Subnet C. So the following are the correct answers:

208.199.33.0/25
208.199.33.128/26
208.199.33.192/27
208.199.33.224/28

Does this make sense??

mikop

2002-09-18, 5:13 pm

you are correct. this is how you worked it out.

I took separate pieces and aggregate, you count from 0-next biggest number.

let me clairfy your confusion as you indicated you are not sure how you come up wiht the 1st number.
this is how you work this through

1. we know anything 240 + is not avialable.

so starting from 0, what's the next biggest number we can get to while still keeping all the rule of subnetting valid?

1 bit, either on or off, either from 0-127 or 128-255, since 128-255 contain 240-255, its not valid, but 0-127 using 1 bit is valid

so answer one.

then you worked through the rest.

The thing that is most important is understand not just how, but WHY this is so. how is easy, its nothing but manipulating couple bits.

want to try it with the address use by subnet C being 208.199.33.64/28 ?

l9nux

2002-09-19, 2:59 pm

What I'm am confused about is that one of the answers:

208.199.33.0/25

contains a zero in the last octet? I didn't think that the last octet could contain a zero? It's just 1 - 254 right?

jeff_j_black

2002-09-19, 3:24 pm

Host addresses must be 1-254, 0 is the network ID and 255 or all 1's is the broadcast address. So when refering to a subnet you could say 208.199.33.0/25, with 208.199.33.0 as the network ID, hosts in the range of 208.199.33.1 to 208.199.33.126, with the broadcast address for that net being 209.199.33.127. You are getting it!

l9nux

2002-09-19, 5:01 pm

Thanks Jeff, I think I'm clear on the last octet now...

They were referring to Subnet C as 208.199.33.240/28 (the 240 would signify the network itself?)

mikop, you said try it with 208.199.33.64/28 (as Subnet C)?

If I'm thinking correctly, I would add an extra bit to the /24 subnet. Then assign the first available subnet and add another bit to the mask...

and do this until I reach the /28 mask...

208.199.33.0/25 (incorrect)
208.199.33.128/26
208.199.33.192/27
208.199.33.224/28

I am figuring that the first one is incorrect as it includes one of the networks from Subnet C (208.199.33.64) in the address space?

jeff_j_black

2002-09-19, 5:53 pm

First they state:

The network address for Subnet C will be 208.199.33.240/28

So you know they took the tail end of the Class C range you are working with. /28 in binary is 11110000 or 240, remember that when you look at a subnet mask, the 0's indicate the bits available for host addresses, so you are correct, the address 208.199.33.240/28 is the network ID, all the host bits are 0.

That network would have host addresses 208.199.33.241 - 208.199.33.254, with 208.199.33.255 being the broadcast address.

That leaves you to puzzle out the remaining pie slices from the available answers:

a. 208.199.33.0/24
b. 208.199.33.0/25
c. 208.199.33.0/26
d. 208.199.33.128/26
e. 208.199.33.192/26
f. 208.199.33.192/27
g. 208.199.33.224/28

a. is flat out wrong, as we already have taken a slice of pie so you cant have a whole pie left. To a degree, you are right that b. and c. collide, but b. is available. Remember we used 240 - 255 so if we start by taking the bigest remaining slice of pie, it's okay, I'm hungry!

b. works because it takes the addresses 0 - 127, which does not bother the 'Subnet C' that has already been taken. That leaves 128 - 239 for that smaller mouths. We have already eliminated c. as a possibility by selecting b.

The way I figure addressing is by the value of the significant bit. When you say /26, I say 64! so d. works because it starts at 128 (which is all 0's, by the way) and ends at 191 (the all 1's address or broadcast) This slice of pie is available so go for it!

Now, 208.199.33.192 + 63 = 208.199.33.255, we know that 'Subnet C' already has taken a slice of 240 - 255, so e. is wrong.

/27 has a significant valued bit of 32, so lets see if f. works: 208.199.33.192 + 31 = 208.199.33.223. So f. is a good slice of pie, starting where we left off with d. and still not cross 240 that is already gone.

Well now what is left of the pie? 224 - 239, thats what! Not much but pie is pie and as such should be consumed. Do we have a way to consume it? /28 significant valued bit is 16, so let's see: 208.199.33.224 + 15 = 208.199.33.239, so the last of the pie is gone.

Tallying our answers:
b, d, f, g are correct slices of pie to take after accounting for the slice that 'Subnet C' had taken. a was too big, c was already part of b, and e was already part of the 'Subnet C' slice. What we have done by selecting b over c is insured that the whole pie got eaten. Keep it up! let's talk more!

jocampo

2002-12-03, 10:30 pm

Ok, i think i got the idea now...but just tell me something about the question itself: it consist in find the valid Subnets from all the showed options, right?....because making calcs there is a total of 14 subnets!

so, if there are 14 subnets, why i have to make calcs from /24 till /28?

why i must use (1 1 1 1) 0 0 0 0 or all ones
and not (0 1 1 1) 0 0 0 0...for example....
or (1 0 1 1) 0 0 0 0

Damm! is not easy for me, sorry!!!!!!!!

mikop

2002-12-03, 11:22 pm

ok. let's go!

1st. they want you to use .240/28 for that network. /28 is 4 bit for subnets, 16. you are still using -2. if that's the case, 240/28 is not valid. therefor, subnet 0 is in use. so that's 16 valid subnets.

0
16
32
48
64
80
96
112
128
144
160
176
192
208
224
240<---the reserved.

now... what?! /24 /28 is subnet mask, if its not 11111111.11111111.11111111.00000000 or 11111111.11111111.11111111.11110000, what else can it be? 1111........10110000?

ok... I will prolly lose patience here so I will apologize in advance, but if I can't make it clear, someone will pick up the slack in the morning, I just want to get you started...

this will also be explained in great detail in the link I will provide later on but I will give it a shot.

the question as you to make use of all available ip address.

now. we have 16 networks. we use 1 for C, now we have 15 available for B right?
each network lose 2 ip address for network/broadcast right?
so we lose 30 ip address.

now that's not efficient right?... how do we overcome that waste?

route summarization / supernetting along with the use of VLSM and CIDR!!!! wheeee (now... really please do take the time, hours, a day or whatever and go through that PDF.)

the logic is... when we supernet, we can represent many networks as one single network. now let me explain.

in ordinary subnetting, we have class c network. we subnet to 8 subnet 32 (30 available) right? we lose 16 ip address. but if we do not subnet, we have 1 network with 256 (254) available, much more efficient right? losing only 2. This is the whole logic of supernetting.... to do the reverse and make better use of ip.

but for this question, we can't easily do that... because we CAN NOT include the .240/28 network... so taht 33.0/24 just would not work as that include 33.240/28... so we need to take what section we can summarize and do it in little chucks... but remmeber, even if we just combine 2 network into 1. that's a conservation of 2 ip address!!!

now... hopefully the concept is a little bit clearer... I would recommend you go over the previous posts... 2 methods there, each would work.

now. we are dealing with the MASK, not ip in binary, very important here. this is where you are confused...

lets. do this

33.0/28 is 00000000 right (ip in binary, mask is 11110000)
33.16/28 is 00010000 (mask again is 11110000)
.
.
.
33.112/28 is 0111000 (mask same)
33.128/28 is 10000000 (mask is still 11110000)
33.240/28 is 11110000 (mask is 11110000)

now... as we can see in binary, up from 0/28 to 112/28, the first bit is 0, (128 first bit is 1) now we supernet all that networks together and find a way to uniquely represent them. therefore

33.0 <-define the network, /25 <-subnet mask.

if we use /24, this include everything right? include 240/28. but if we use /25, this extend the mask, therefore, we have subnetted that 1 network into 2. and everything in that subnet is ok, the 2nd subnetted network we have problem as it now include the dreaded 240/28 network ip addresses...

but anyway, in that 1 single motion, we have instead of 8 subnets each waste 2 ip addresses each (total of 16), we now have 1 subnet wasting only 2 ip addresses!

so now. we have

33.0/25 (ip 33.0-33.127)
33.128/28 ip 33.128-33.143)
so on and so on
33.240/28 ip 33.240-33.255

now repeat that same process we did...

128/28 10000000
144/28 10010000
etc

I will just take this 2 (use for demonstration.

now let's see how we summarize this.

we see that the first 3 bits are the same right?

and that 1st is 128 and add 3 bits to /24

so we write 128/27

now for the actual question lets do this

128-10000000
144-10010000
160-10100000
176-10110000
192-11000000

now if we have to summarize this whole list, instead of just the 2 in the previous example, our summarized/supernetted network WILL be different. and hopefully now you will see

we count bits again!

now, 192 certainly don't fit in here so lets leave that.

now we deal with the 1st 4.

if we summarize the 1st 2, then the 2nd 2, we have 2 networks wasting 4 ips total right? how can we save more ip? summarize the 4 into 1 instead of 2 as my previous example.

common bits, the 1st 2.

so its now 128/26 this would uniquely identify these 4 networks... hopefully you will see now that in the previous example, by manipulating the subnet mask to /27, I uniquely identify only ip address from 128-159, and when using /26, I idetnify 128-191... this is the whole logic behind route summarization, to uniquely identify each network/address without duplicate while minimizing the number of networks.

now apply this and work throught the rest of the ip addresses.

there are easier shorter ways... but let's leave that alone and do it the RIGHT way so you understand... tricks and tips comes later... not when you need to learn it. (this is why I hate the *remmeber the subnet table* thingy... understand, not memorizing tricks.)

http://www.3com.com/other/pdfs/infr...n_US/501302.pdf

read that document, take 2 hours and read every word, do the exercise, read the history, the challenges, ... take a day ... whatever you need, that's pretty much EVERYTHING you really need to know about ip addressing etc... and would serve you VERY well for a very long time! (till we change to ipv6 of course ...)

I am not entirely convinced that you understand what's going on... and I dont mind explaining it as I sort of dig this stuff... (but I lack the patience and as you can tell, the more I write, the less patient I become

)

when you read that document, pay special attention to why we need VLSM, and what is accomplish when we do route summarization/supernetting (they are interchangeable).

jocampo

2002-12-04, 5:59 am

Mikop...

thanks a lot for your patient.

jocampo

2002-12-04, 7:03 am

Hey Miko, i'm reading the pdf document that you linked to me, and see some concepts about VLSM. Does the 216 cover this concepts too? Are VSLM a valid topic for 216?

mikop

2002-12-04, 7:34 am

prolly not... if it is in, not in detail... but what do I know... I don't remmeber a thing from any microsoft exam.

HOWEVER, that covers pretty much everything you will need to know regarding ip addressing, what any network administrator SHOULD know. if you intend to progress in your career, it will serve you will to know that document... atleast in theory.

a day spend studying that document will make your future study, if you intend, of cisco or any other networking cert so much easier...

jocampo

2002-12-04, 7:59 am

quote:
Originally posted by mikop
prolly not... if it is in, not in detail... but what do I know... I don't remmeber a thing from any microsoft exam.

HOWEVER, that covers pretty much everything you will need to know regarding ip addressing, what any network administrator SHOULD know. if you intend to progress in your career, it will serve you will to know that document... atleast in theory.

a day spend studying that document will make your future study, if you intend, of cisco or any other networking cert so much easier...

I know, and you're right, but IP is not easy some times; besides, english is my second language...so, imagine how can i feel sometimes.

I still do not understand "why" the Transcender questions have to be solve in that way...but i know HOW to do it. Give me some days before "eats" all the material.

What i really know well now is how to allocated a number of host of subnets: subnetting a common Network.

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