Subnetting question

5uu

What is the first valid host on the subnetwork that the node 172.27.103.132 255.255.255.128 belongs to?

plz help me by givin me a simple explanaton for this question...
thanku

ZZOmega

First, you need to calculate how many subnets the 255.255.255.128 mask will create for the 172.27.103.XXX network;

A subnet mask of 255.255.255.128 equals 25 network bits, and 7 host bits, or to represent it in a different way, this is a /25 mask (CIDR).

This means you have 2 subnets, with 128 addresses each. Of course you know by now that with 128 possible addresses, only 126 are valid host addresses.

Second, find out which subnet the host in question resides in. Here we see that the question is asking for the first valid host address on the same network as the node 172.27.103.132

There are two networks created with the /25 mask;

Subnet 0 ) 172.27.103.0 through 172.27.103.127
Subnet 1 ) 172.27.103.128 through 172.27.103.255

Quickly checking our given information, we can conclude that the network in question is Subnet 1. Now, the first address in the subnet, 172.27.103.128, is the network ID address, making it an invalid host address.

The second possible address in any subnet, or in this case 172.27.103.129, is known as the first valid host address, since it is the first address that is able to be assigned to any host(valid).

Answer: 172.27.103.129

If you need any more help, or if my explanation is confusing, please let me know!

-Devon

5uu

"This means you have 2 subnets, with 128 addresses each."

Could u plz explain this???

ZZOmega

The address for this question, 172.27.103.132, is a Class B address, with a /25 subnet mask(255.255.255.12. For the sake of clarity, let's say that it originally had a Class C default subnet mask, or a /24 mask(255.255.255.0). Class C addresses are easier to explain and work with, and once you get the hang of these, Class B addresses are just the final stretch.

So, with a Class C subnet mask of 255.255.255.0, you know that the first 3 octets are the Network portion (24 "N-Bits"), and the last octet is the host portion (8 "H-Bits"). A Class C subnet mask allows for 256 addresses (254 valid hosts) on one network;

172.27.103.0 - 255

By borrowing another N-Bit, you have changed your /24 mask to a /25 mask, and created subnetworks within the original 172.27.103.0 network. I don't remember which way is recommended for calculating subnets, but I'll show you mine!

As the question states;

What is the first valid host on the subnetwork that the node 172.27.103.132 255.255.255.128 belongs to?

Step One (Identify Number of Subnets)

All you need to do for Step One, is pay attention to the subnet mask. More importantly, the octet of the mask that is neither 255 nor 0. This brings us to the fourth octet, which is 128.

To find the number of subnetworks that exist from a subnet mask, take the interesting number(128 in this case), and convert to binary.

1000 0000

Wherever there are 1's, means "This part is the network"
whereas 0's, mean "This part is the host"

At step one, we're trying to figure out information about the network, so pay attention to the 1's that are there. Count how many 1's you see, which in this case is one, then raise 2 to that power.

So, 2^1 is 2, resulting in 2 subnetworks.

Step Two (Identify Subnetwork Range / Number of Hosts per Subnet)

Step two isn't the final step, but it's the end to all the headache and information gathering! Before, we calculated how many subnets were created from the subnet mask. Here, we are going to see the range of each subnet, or to put it differently, the number of hosts per subnet.

Take the number we got from Step One, which in this case would be 2. Now divide 256 by that number;

256 / 2 = 128

The reason we are dividing 256 by the number from Step One, is because there were 256 possible addresses in the network by default(before subnetting occurred). So, if have 2 subnetworks that share the same pool of the possible 256 addresses, they have 128 addresses each! Easy right?

The subnetwork range is in terms of 128, or there are 128 hosts per subnet. This fact brings us to the third and final step of the subnetting process!

Step Three (Which Subnetwork is the right one?)

Luckily we have easy numbers to work with! As concluded by steps One and Two, we have 2 ranges with 128 addresses each to work with:

Subnet 0 ) 172.27.103.0 through 172.27.103.127
Subnet 1 ) 172.27.103.128 through 172.27.103.255

Our address, 172.27.103.132, resides in the range of Subnet 1. As stated before, the first and last possible address in each range are not valid host addresses, meaning that addresses 172.27.103.128 and 172.27.103.255 are invalid host addresses. The question asks for the first valid host address, which would be the next up from .128!

Answer: 172.27.103.129

---

I basically just walked you through my thought process on this problem, the steps and process here isn't official, as there's no "official" way of picking apart questions like these. I just prefer my method

If you are looking for an in-depth binary explanation, post a reply saying so and I shall try to help!

5uu

(said with tears of joy)... i understood very wll... n i tried another question n got the correct answer... thanku

ZZOmega

5uu wrote: »

(said with tears of joy)... i understood very wll... n i tried another question n got the correct answer... thanku

You're very welcome, glad I could help!

TonyCCNAontheway

I have a subnetting question with CCNA Network Fundamentals. With a subnet using 172.16.0.0/22 address block I have created a /23 subnet with 510 hosts per subnet. I have to configure the DNS server, default gateway and a few other items and am stuck. Any ideas? Do I need to give you more details?

Thanks,
Tony

alan2308

TonyCCNAontheway wrote: »

I have a subnetting question with CCNA Network Fundamentals. With a subnet using 172.16.0.0/22 address block I have created a /23 subnet with 510 hosts per subnet. I have to configure the DNS server, default gateway and a few other items and am stuck. Any ideas? Do I need to give you more details?

Thanks,
Tony

Where exactly are you stuck?

TonyCCNAontheway

Here is what I need to find
I have a network block 127.16.0.0/22
I need 7 subnets with the biggest being 400 hosts so I created the subnet
127.16.0.0/23 which gives me 510 usable hosts per subnet.

Using packet tracer I have a server, 2 routers, 1 switch and and 2 pc's in the network.

I have to configure each device with a static address, subnet, default gateway and setup the server for dns.

I have have of the network configured but still need the IP addresses for the two pc's. ( the book told me to use the lowest 2 ip addresses in the 1st subnet.

I need to get the default gateway and domain name to work on the server. The Fa 0/0 port still need the IP address and subnet mast.

Router 1 & 2 need IP's and subnet masks
Router 1 needs IP, subnet mask, serial 0/0/0 IP and subnet mask
and Links for Router 1 and 2 need IP's and subnet masks and static routes.

Hope I explained it right.

Tony

Here are the 7 subnets using VLSM

Subnetwork First Host Last Host Broadcast

172.16.0.0 172.16.0.1 172.16.1.254 172.16.1.255
172.16.2.0 172.16.2.1 172.16.3.254 172.16.3.255
172.16.2.192 172.16.2.193 172.16.2.254 172.16.2.255
172.16.3.0 172.16.3.1 172.16.3.30 172.16.3.31
172.16.3.32 172.16.8.1 172.16.3.34 172.16.3.35
172.16.3.36 172.16.10.1 172.16.3.38 172.16.3.39
172.16.3.40 172.16.12.1 172.16.3.42 172.16.3.43

alan2308

You're working on "Skills Integration Challenge: Data Link Layer Issues(7.6.1.3)," aren't you? I've still got my paper with the breakdown of the subnets stuffed in my Lab book.


The first problem I see is your subnets. They should always go from biggest subnet to smallest subnet. And your second subnet is WAY too big for 180 hosts. Here's what I had:

1 - 172.16.0.0/23 (0.0 - 1.2.55)
2 - 172.16.2.0/24 (2.0 - 2.255)
3 - 172.16.3.0/26 (3.0 - 3.63)
4 - 172.16.3.64/27
5 - 172.16.3.96/30
6 - 172.16.3.100/30
7 - 172.16.3.104/30

The two PC's are in the first subnet (172.16.0.0/23) and need the first two IP addresses in that range. R2's fa0/0 interface is also in that range and should have the last address:

PC1 - 172.16.0.1 255.255.254.0
PC2 - 172.16.0.2 255.255.254.0
R2 fa0/0 - 172.16.1.254 255.255.254.0

The serial link between R1 and R2 is the 5th subnet. There's 2 usable devices and 2 possible host addresses. You should be able to figure this one out.

The Ethernet link between R2 and Eagle Server is the third subnet. R1 gets The highest address (172.16.3.62) goes to the router, the next highest goes to the server.

For the configuration of the PC's, the default gateway is the interface of R2 on that subnet, the DNS server is the IP address of Eagle Server.

2URGSE

OP:

172.27.103.132
255.255.255.128

So the basics are:

255 + (any #) = that #

0 + (any #) = 0


Last octet is the "interesting" one.

magic # is 256-128 = 128

so then you count 0....128...and you know you need to get to
132 so 128 is the closest without passing 132 so...

172.127.103.128 is your network #.

from here, just add a 1 to it and that's the first

usable host address of 172.127.103.129



If subnetting isn't your stronghold, I HIGHLY recommend the video mentor from Wendel Odom, he makes very easy it's funny. If you're doing this in binary that's cool to understand since the machine does it like that, but on the exam you need to be able to do it in decimal but the good news that doing this in decimal is much faster and easier.

TonyCCNAontheway

Who this is time consuming but I think I have it. I am only missing one thing.

I have to put two static routes in for the serial ports on R1 - R2 and reverse. The static on R2 is fine I used the IP address from the serial on R2 and it is ok I show 0.0.0.0/0: 172.16.3.98.

On the R1-ISP router I can not seem to make it work. It shows 172.16.0.0/23: 172.16.3.97. I entered them both the same way.

Network 0.0.0.0
Mask 0.0.0.0
Next hop 172.16.3.97

I don't get it?

Any help would make me complete.

Tony

alan2308

TonyCCNAontheway wrote: »

Who this is time consuming but I think I have it. I am only missing one thing.

I have to put two static routes in for the serial ports on R1 - R2 and reverse. The static on R2 is fine I used the IP address from the serial on R2 and it is ok I show 0.0.0.0/0: 172.16.3.98.

On the R1-ISP router I can not seem to make it work. It shows 172.16.0.0/23: 172.16.3.97. I entered them both the same way.

Network 0.0.0.0
Mask 0.0.0.0
Next hop 172.16.3.97

I don't get it?

Any help would make me complete.

Tony

Is there any error message when you enter the static route? Is the 172.16.3.96/30 network in the routing table for R1?

TonyCCNAontheway

No the only thing I notice is that the numbers that I enter
0.0.0.0
0.0.0.0
172.16.3.98

Stay in the top portion. Before they disappeared and where put in the lower box now they stay in both places? I made sure the interfaces are one.

TonyCCNAontheway

Also the clock rate is being set by that interface. I even tried it with the clock set off.

alan2308

TonyCCNAontheway wrote: »

No the only thing I notice is that the numbers that I enter
0.0.0.0
0.0.0.0
172.16.3.98

Stay in the top portion. Before they disappeared and where put in the lower box now they stay in both places? I made sure the interfaces are one.

What do you mean by top portion and lower box?

TonyCCNAontheway

In the configuration tab on packet tracer.

alan2308

TonyCCNAontheway wrote: »

In the configuration tab on packet tracer.

Interesting, I don't remember seeing all that in the configuration tab.

Make sure that the port status on the serial interface is "ON," double check that the IP address and subnet mask are correct and then do the same for the other router. If they're both configured correctly, then try restarting packet tracer. What you're entering is correct so I'm at a loss unless you're still using version 5.2 which was full of bugs. No matter what I enter for a static route, it takes it no problem.

And one end of the serial connection does need to have a clock rate set. You'll get more into that in semester 2 so I won't bore you with the gory details now. I usually just set a clock rate on both ends and it will be ignored on the end that doesn't need it.

TonyCCNAontheway

I will look at it more later tonight and let you know if I find out what I did wrong. Thanks for all you help you should teach the class I have right now.

alan2308

TonyCCNAontheway wrote: »

I will look at it more later tonight and let you know if I find out what I did wrong. Thanks for all you help you should teach the class I have right now.

Well, I'm sure I'll have to wait at least until I take the CCNA myself before I can do that.

TonyCCNAontheway

I just submitted without the static route was so slammed today that I just got on line now. Thanks for all your help.

alan2308

That's what we're all here for.

I still don't see anything wrong with what you were doing, nor can I reproduce what you are seeing still. I just really wish I had my completed copy of that activity to look at now. I know I saved them all when they were done but I can't find them now.

The only other thing that I can think of is it may be a version issue. When I took the classes, we were using 4.2 and a lot of the activities had issues since 4.2 was buggy and they were all written with 4.1. And I haven't used 4.3 enough to say that all the problems are fixed.

xxfrozen

This board proved very helpful to me just to make sure I was doing the assignment for class correctly.

The reason why the static routers may have not worked in the previous posts, is that perhaps you didn't configure the default gateway's which would be the routers in each network segment, once I completed that I was able to do everything else fairly easily.

Taddow1

Thank you so much mate, especially for taking the time out to write that up in detail
for others to benefit! Much appreciated

kurosaki00

Didnt want to make a whole new thread for just one question, so Im Hi-jacking this since its about subnetting question

Was practicing classless subnetting, can you guys tell me if I did good?

boredgamelad

Looks fine to me, nice job.

On future network diagrams you can omit listing the range and just list the subnet ID and mask but it's not bad practice to write them out like this. Helps make sure you're doing it right.

vtiniall

I have this question and I don't seem to be getting the correct answer. Can anyone help here with the correct step to get the correct answer, please.

What is the first valid host on the subnetwork that the node 172.25.209.197 255.255.255.224 belongs to?

Welly_59

X.x.x.193

vtiniall

Hi,

Thanks for the input, would you have steps to get to that answer, please.

rob42

vtiniall wrote: »

Hi,

Thanks for the input, would you have steps to get to that answer, please.

If I may...

I created a 'Subnetting **** Sheet' in this thread... http://www.techexams.net/forums/ccna-ccent/125960-subnetting-****-sheet.html



CIDR
SUBNET /bits
/1
/9
/17
/25
/2
/10
/18
/26
/3
/11
/19
/27
/4
/12
/20
/28
/5
/13
/21
/29
/6
/14
/22
/30
/7
/15
/23
/31
/8
/16
/24
/32


Block Size
128
64
32
16
8
4
2
1


Mask
128
192
224
240
248
252
254
255


HOST bits
31
23
15
7
30
22
14
6
29
21
13
5
28
20
12
4
27
19
11
3
26
18
10
2
25
17
9
1
24
16
8
0



So to answer your question

What is the first valid host on the subnetwork that the node 172.25.209.197 255.255.255.224 belongs to?

Understand that the Octet that is of interest is the 4th Octet (224 in your mask, /27 in CIDR Notation). This equates to a Block Size of 32.

So, the Subnets are 0, 32, 64, 96, 128, 160, 192, 224.

172.25.209.197 is in the 192 Subnet.

Valid Host range 172.25.209.193 -> 172.25.209.222
B/Cast 172.25.209.223

Is this of help?

vtiniall

Thanks for taking the time to write out the solution. It does help alot