70-291 MS Self-Paced Kit - Page 13-8

I'm having a difficult time understanding the answer to the following question.

Your company has leased the network 138.16.0.0/21. You want to subdivide this network into 10 evenly sized subnets. No single subnet will ever contain more than 100 users. You do not use NAT or private network addresses. The default gateway is always the lowest possible user address on the subnet. You use DHCP for your client PCs, but your servers are configured manually. The 10 lowest IP addresses on each subnet are excluded from the DHCP scope so they can be used for manual configuration. Which of the following configurations is a valid server?

A. IP Address: 138.16.1.5
Subnet Mask: 255.255.255.0
Gateway: 138.16.1.1

B. IP Address: 138.16.8.5
Subnet Mask: 255.255.255.128
Gateway: 138.16.8.1

C. IP Address: 138.16.3.133
Subnet Mask: 255.255.255.128
Gateway: 138.16.3.129

D. IP Address: 138.16.1.134
Subnet Mask: 255.255.255.128
Gateway: 138.16.1.128

Correct Answser: C
The 138.16.3.128/25 network supports an address range of 138.16.3.129 through 138.16.3.255. The default gateway is therefore 138.16.3.129. The IP address 138.16.3.133 is one of the lowest 10 IP addresses in the subnet.

Thanks in advace.

JLuna

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Richardxs

That does look a tough one!!

well we know its not A since thats a single subnet

I dont think its D because (using a subnet calc to help me) the gateway is outside the address range

And from what i can see its cannot be B because it gives an Ip address range of 1-129 which contains more than 100

I think thats right but hell I used a subnet calc and thats a tough one to get your head around..

deneb829

You can just eyeball this one with having to do any of the math:

Well right away, we can eliminate A, using the /24 subnet will not give us enough networks.
I eliminated B since it is outside the scope of the DHCP server.
Looking at the remaining two answers, the default gateway gives it away. C is 138.16.1.129 and D is 138.16.1.128. Based on the Subnet Mask 138.16.1.128 is the network address, so by process of elimination, it has to be C.
Hope my logic makes sense

davenport

Unlike deneb829 I had to do the math, cause I'm not that good at subnetting. But here is the way I see it. If you do the math and lay the ranges out, C is the only one that could be correct. Someone correct me if I'm wrong here but I think the range would fall like.
(this is including the 0 address at the start, and the broadcast at the end. it's easier for me to see that way.)

138.16.0.0 - 138.16.0.127
138.16.0.128 - 138.16.0.255
138.16.1.0 - 138.16.1.127
138.16.1.128 - 138.16.1.255
138.16.2.0 - 138.16.2.127
138.16.2.128 - 138.16.2.255
138.16.3.0 - 138.16.3.127
138.16.3.128 - 138.16.3.255
138.16.4.0 - 138.16.4.127
138.16.4.128 - 138.16.4.255

Ok if I laid that out right.
A is obviously wrong cause the subnet is wrong
B is wrong cause it's out of range
C is right cause the 138.16.3.129 is the first address (the lowest) and 138.16.3.133 is within the first 10
D is wrong cause the gateway is the 0 address.

Sound about right?? Someone jump in if I'm wrong here.

JLL

Thanks guys.

I was able to choose the right answer as well by process of elimination, but would like someone to actually show me the work.

My thoughts are that since the 138.16.0.0/21 leased network is "fixed", I would need to start subnetting from there.

Since I need to accomodate 10 subnets, 4 bits would need to be used from the host portion which changes the /21 to /25 (255.255.255.12

. This will allow me to have 16 subnets with 126 hosts per subnet.

The problem I have is defining the available subnets. Once I know what the subnets are, I can easily figure out the ranges within each subnet.

I would really appreciate someone showing the work for this question.

Thanks again.

JLuna

davenport

jluna wrote:

Thanks guys.

I was able to choose the right answer as well by process of elimination, but would like someone to actually show me the work.

My thoughts are that since the 138.16.0.0/21 leased network is "fixed", I would need to start subnetting from there.

Since I need to accomodate 10 subnets, 4 bits would need to be used from the host portion which changes the /21 to /25 (255.255.255.12. This will allow me to have 16 subnets with 126 hosts per subnet.

The problem I have is defining the available subnets. Once I know what the subnets are, I can easily figure out the ranges within each subnet.

I would really appreciate someone showing the work for this question.

Thanks again.

JLuna

I'm not sure I get what you are asking. You're right about starting from the fixed address of /21. Are you asking how (or why) you get the /25 subnet mask? Sorry for the misunderstanding.

JLL

Well, since one of the requirements is to have 10 evenly sized subnets; I would need to borrow 4 bits from the host portion of the 138.16.0.0/21 network address. I am basically extending the string of 1-bits to configure the subnet mask that would reserve enough bits to configure enough subnets (10). This would still leave 7 bits for host addresses per subnet (128-2=126).

Am I not approaching this correctly?

JL

turbo

Edit

davenport

This is why I'm not a teacher. I'm not very good at explaining things. lol

Maybe someone else will jump in that can explain it better than me. But until then here goes, the way I've always looked at it is this. I’m sure you know the majority (if not all) of this, but anyway, one of the main reasons to subnet a network was so we wouldn't waste IP addresses. So the way I start any problem like that one is just meet the minimum amount of hosts with the least amount of wasted addresses and start the subnet there. You could have subnetted that network with a different mask but it would have been more wasteful of IP addresses. So what you should do is start working from the host side of the mask until you meet your required hosts. When I first started subnetting I would always start from the network side of the mask and normally end up with a solution that would in theory work but it was wasting addresses. I would end up with something that would function, yet was always wrong on a test question. When I stopped and thought about it from different standing point, not being so wasteful, I started on the other side (host side) of the mask it’s been easier to understand. Like I said, I’m not sure that is what you are wanting but maybe someone else can jump in for you. Normally it doesn’t take but for one thing to “click” and you’ll understand it all. Good luck with it!

JLL

davenport,

I definitely agree about not wasting addresses. From that perspective, I would still need 7 bits reserved in the host portion for host addresses; leaving me with 4 bits for subnets. Still end up in the same delima.

Thanks.

JL

davenport

Right, you need 7 bits for the hosts. Being that it was a class B address that leaves 9 bits for the network.

That leaves you with 512 networks and 126 hosts per subnet.

JLL

I assumed that since it is a leased network address (138.16.0.0/21), the organization (ISP, ARIN, etc.) that leased it to me, has it configured on one of their routers that will route external traffic to my network.

Since the leased network address is fixed at /21, I can't begin subnetting it as a default class B network (/16 or 255.255.0.0). So, that leaves me with only 11 bits to play with and not 16. This is where the fun starts. In order to fulfill the 10 subnets with a maximum of 100 hosts per subnet requirement, I need to steal 4 bits out of the 11. With those 4 bits, I can create 16 subnets. With the remaining 7 bits, I can have 128-2=126 host addresses per subnet.

Now, to create the subnet mask, I would need to translate:

11111111 11111111 11111111 10000000 to decimal format which gives me:

255.255.255.128

And this is about as far as I get

Help!

JL

davenport

You're right, in the midst of all this I forgot the orginal question said the subnet was fixed at /21. I was thinking of it as an internal address that we assigned so we could use the default class B mask.

Check your PMs jluna

eurotrash

I'm not really sure what your question is here, but this is how I approached the problem:

100 IPs per subnet: needs 7 bits (2⁷-2 = 126) = /25 mask.
10 subnets: well, if the rightmost 7 bits are going for IPs, let's see how many potential subnets we're left with: 255.255.11111xxx.x0000000. That looks like 4 bits (from the original /21 mask to a /25 mask).
That gives us 16 (or 14, if you subtract two) subnets, meeting the requirements.

Now all of these subnets' IP addresses must start at .129, as the /25 mask = 255.255.255.128.

A: Wrong subnet.
B: Violates the network range we were given
xxxxxxxx.xxxxxxxx.00000 (original /21 network - can't touch this portion)
xxxxxxxx.xxxxxxxx.00001 (B: modifies it)
C: Gateway and mask look right, .3 doesn't modify the /21 portion, .133 is within the excluded range as this is a static address (right option).

IP address is good. Mask is good. Lowest usable IP is .129, gateway is incorrect.

JLL

OK, you say that you can get 16 subnets out of the newly configured mask (/25). Let me see what the first 5 subnets look like?

The part that is throwing me off is after extending the string of 1-bits from /21 to /25, the affected subnet mask octet is no longer the third. It is now the fourth octet that is affected. So instead of 255.255.248.0 (/21), you have 255.255.255.128 (/25). Now what? I want to see the subnets now.

Sorry for the confusion.

JL

davenport

Look at my first post, that is how the network would lay out. The address range is there, then the subnet you alreay figured out is /25.

You have the hard part figured out, I think you know how to do this and you're just screwing with us. lol J/K

The first one would be

zero address: 138.16.0.0
First IP address: 138.16.0.1
Last IP address: 138.16.0.126
Broadcast address: 138.16.0.127
subnet mask: 255.255.255.128

Next subnet
zero address: 138.16.0.128
first ip address: 138.16.0.129
last ip address: 138.16.0.254
broadcast address: 138.16.0.255
subnet mask: 255.255.255.128

etc.

JLL

OK, now back to the answer:

C. IP Address: 138.16.3.133
Subnet Mask: 255.255.255.128
Gateway: 138.16.3.129

Where did 138.16.3 come from??? That wouldn't be one of the subnets. Or would it????

JL

davenport

jluna wrote:

OK, now back to the answer:
C. IP Address: 138.16.3.133
Subnet Mask: 255.255.255.128
Gateway: 138.16.3.129

Where did 138.16.3 come from??? That wouldn't be one of the subnets. Or would it????
JL

The 7th and 8th network are both 138.16.3

zero 138.16.3.0
first ip 138.16.3.1
last ip 138.16.3.126
broadbast 138.16.3.127
subnet 255.255.255.128

and

zero 138.16.3.128
first ip 138.16.3.129
last ip 138.16.3.254
broadcast 138.16.3.255
subnet 255.255.255.128

so yes the 138.16.3.133 address is the 5th address in the 8th subnet

JLL

OK, the reason why this question is giving me some trouble is because the Microsoft 70-291 Self-Paced Kit does not explain the subnetting of a Class B network address whereby the subnet mask values extend to the fourth octet; making it look like a Class C mask (e.g. 172.16.0.0/25)

I do make it a habit, when studying for an exam, to read as much as possible from as many different sources as possible. This will hopefully allow me to have an extensive and thorough understanding of the subject matter. The mistake I made was to skip over the subnetting sections of all other sources except the Microsoft 70-291 Self-Paced book. This was mainly because after reading the subnetting section in the Self-Paced book, I felt very comfortable subnetting. I was able to answer all of the practice questions and scenarios provided at the end of the lesson.

Anyhow, I would like to suggest an excellent document that explained it to me. It is a sample chapter of the Sybex CCNA study guide that is available for downloading at no cost. Just do a Google Search for:

Sybex IP Subnetting Sample Chapter

JL