Subnetting practice problems 1

[Cluebird]

Practice your subnetting skills with the following:

1. You have a router with 9 interfaces that need IP addresses. You are assigned 195.33.44.0. Construct a complete addressing plan for your router. Use FLSM only.

2. You have two hosts that are unable to connect with each other. On one host you see the ip address 192.168.100.33 255.255.255.240. On the other host you see 192.168.100.28 255.255.255.240. What is the problem?

3. Your administrator tells you to address a host with the following address: 172.16.2.255 255.255.252.0. Will this work? Why or why not?

4. What mask will give me 18 hosts with max number of networks if I start out with 192.168.255.0?

5. What class is the following mask: 255.255.255.0?
A. Class C
B. Class B
C. Class G
D. Not enough information to decide

6. Your server administrator wants to use the 172.168.200.127 255.255.255.128 address on his Exchange server. What is your response?
A. No problem
B. Won't work because that is a local broadcast
C. Won't work because that is a network address
D. Won't work because that is a directed broadcast
E. None of the above

Enjoy

 

[Cluebird]

bobby,

You're doing fine. I'm asking questions at the level the exam might without disclosing the exam. I haven't found any testprep questions that I recommend because they are too easy for the most part.

You've answered the question for the router from the reseller. Multicast has nothing to do with this one. You've correctly identified the 2811 with a WIC-2T will have four hardware interfaces: 2 FastEthernet and 2 Serial. That was the "dragoon". Once you know how many networks you need, all you have to do is make a mask for the 192.225.33.0 classC. The mask is 255.255.255.192 or /26 or
11111111.11111111.11111111.11000000

Keep working!

 

[bobbyforhire]

Given the address 208.23.24.249 255.255.255.192. Is this a network address, a host address, or a directed broadcast?


Ok due to my previous mistakes i'm going to start it off like this.

1) Is this address valid?
208 is a valid class C

2) What Subnet mask is given.
255.255.255.192 is a valid /26 subnet

3) What did the admin do before me to get here.
He decided that he needed a valid 62 hosts
per network.
128 64 32 16 8 4 2 1
1 1
= 192
His inc was 64



208.23.24.0 - 63
.64 - 127
.128 - 191
.192 - 255 <--249 fits in here and is useable.


249 is a Host address.

 

[Cluebird]

Correct!!! Always remember to do your first step!

I teach four steps

1. Know your starting point. (ClassA, ClassB, ClassC,ClassD,ClassE) If you get this wrong, you're toast.

2. Create new mask based on design criteria.

3. Fill in the details for the IP addresses based on block sizes.

4. Assign address space to networks.

 

[bobbyforhire]

You look at the output from ipconfig on an old Windows PC. You see the address 172.55.33.251 255.255.255.254. What is the network address and the directed broadcast for this host?

172.55.33.251
255.255.254.0.

172 - Class C Yes

11.11.11111110.00000000
= /23

172.55.33.0 - 172.55.33.255


I'm 100% sure i don't know how to do this problem.....HELP!

If i had to try i would say that the network is 33.0 and the broadcast is .255

 

[Cluebird]

Step 1: Know your starting point
This is a Class B address based on the 172 in the first octet.

Step 2: Create new mask based on design criteria
The mask is given therefore you already have the answer to the problem if you know how a network mask works. In this case the administrator has made 7 bits of additional networks or 128 new subnets. You've identified that correctly.

Step 3: Fill in details of ip addresses based on block size.
In this case the increment/blocksize/magic number occurs in the third octet based on one bit. That is a block size of two in the octet. The fourth octet is nothing but host bits so, as with an odometer, each network must spin through the bits in the fourth octet twice. The third and fourth octet increment as follows:
0.0-1.255
2.0-3.255
4.0-5.255
6.0-7.255
8.0-9.255
continuing this pattern until we reach
32.0-33.255
34.0-35.255

The address we're looking at is 172.55.33.251 so the network address for this host is 172.55.32.0/23 and the directed broadcast it 172.55.33.255/23

Step 4. Unnecessary since question is answered.

Cisco loves to work around the dot with /23 and /25.

 

[burtsbees]

You see the address 172.55.33.251 255.255.255.254."
No---he changed it to /23, not /31

Burt

 

[maczen]

Yeah, threw me off for a sec... was like that is easy since there are only two addresses.. lol

OK.. After that example I was playing around and got myself lost somewhere.. Would one of you guys tell me what I am doing wrong here... (Think my range should be 4 in stead of 16 but can not put it to binary to make sense...)

172.16.19.0 /22 Need 40 networks with 10 hosts per net
-------------------------------------------
Networks = borrow 6 bits for 64 subnets
Gives me /28 or 255.255.255.240 SNM 0.0.0.15 WCM
Hosts = that leaves 16-2= 14 hosts per net
Range =
172.16.19.0 /28
x.x.19.0 - 15
x.x.19.16 - 31
x.x.19.32 - 47
x.x.19.48 - 63
x.x.19.64 - 79
...
x.x.19.240 - 255
NOT ENOUGH.. LoL

OR

172.16.NNNNNNHH.HHHHHHHH(This is /22)
172.16.NNNNNNSS.SSSSHHHH (after your subnetting for /28)

THIS IS WHAT IS THROWING ME OFF.. THE TWO SUBNET BITS ARE ALREADY IN USE...
172.16.0001011.00000000 = 172.16.19.0

172.16.19.00010000 = 172.16.19.16
172.16.19.00100000 = 172.16.19.32
172.16.19.00110000 = 172.16.19.48
172.16.19.01000000 = 172.16.19.64
172.16.19.01010000 = 172.16.19.80
...
172.16.19.11110000 = 172.16.19.240 - 172.16.19.255
STILL WRONG.. LoL







B Haines
CCNA R&S, ETA FOI

 

[bobbyforhire]

172.55.33.251
255.255.254.0

172 = class B

255.255.0.0
255.255.254.0

11111111.11111111.00000000.00000000
11111111.11111111.11111110.00000000

INC = 2

32.0-33.255
34.0-35.255

-------------------------------------------
My own example:
10.8.254.249
255.254.0.0


10. = 255.0.0.0

11111111.00000000.00000000.00000000
11111111.11111110.00000000.00000000

INC = 2

10.0.254.0 - 10.1.254.255
10.2.254.0 - 10.3.254.255
.........
10.8.254.0 - 10.9.254.255

The broadcast for my 10.8.254.249
is 10.9.254.255 and my network is
10.8.254.0

I hope i'm getting better...

E.I.E.I.O

 

[bobbyforhire]

hey mac,

172.16.19.0 /22 Need 40 networks with 10 hosts per net


172.16.19.0
255.255.0.0


172.16.19.0
255.255.0.0
NEED 40 NET

128 64 32 16 8 4 2 1
0 0 1 0 1 0 0 0

Need 6 Bits


11111111.11111111.00000000.00000000

11111111.11111111.11111100.00000000
255.255.252.0 /22

Lowest INC = 4

 

[maczen]

Remember you begin at a /22 not at a /16... so you would end with a /28.. I am trying to determine what the range would be in that scenario...

172.16.19.0 /22
OR
172.16.00010011.00000000 /22

To get 40 networks/subnets
172.16.00010011.00000000 /28

So range would be...
16 right? But then you can not break out of the 172.16.19...

B Haines
CCNA R&S, ETA FOI

 

[maczen]

OK... My brain is now officially fried... LoL I don't know why this is giving me so much trouble but I have been looking at binary for too long today.. Will play with IOS for an hour then check back and try again.. hehe

B Haines
CCNA R&S, ETA FOI

 

[Cluebird]

bobby,

10.8.254.249
255.254.0.0

Based on this mask, your increment is 2 in the second octet.
10.0.0.0-10.1.255.255
10.2.0.0-10.3.255.255
....
10.8.0.0-10.9.255.255
10.10.0.0-10.11.255.255
10.12.0.0-10.13.255.255/15

mac,

You're try a VLSM problem. Trying not to go there is this thread.

Given: 172.16.19.0 /22 Need 40 networks with 10 hosts per net


You've already be given a Class B that has been subnetted. However, the starting address isn't valid for the block. You should have started with 172.16.16.0/22 (based on a block of 4 in the third octet) and then extended the mask. If your starting point was correct, then to get another 40 networks, you'd extend your mask another 6 bits to get /28.

Addresses would then look like this:
172.16.16.0-172.16.16.15
172.16.16.16-172.16.16.31
up to
172.16.19.0-172.16.19.255

You would actually have 64 networks on block of 16 with the extended VLSM mask.

Watch the bit boundaries with binary!!!!

 

[maczen]

172.16.19.0 /28

172.16.00010011.00000000 net addy
11111111.11111111.11111111.11110000 SNM
-----------------------------
172.16.00010011.00001111 broadcast addy for first subnet
or
172.16.19.15

Next...
172.16.00010011.11110000 net addy
11111111.11111111.11111111.11110000 SNM
------------------------------
172.16.00010011.11111111 broadcast addy for last subnet
or
172.16.19.255

WAIT...
I may have borrowed 6 bits but only gain 16 subnets since the first two were already utilized???

172.16.00010011.00000000
172.16.NNNNNNSS.SSSSHHHH
So 16 x 16 = 256.. DUH!!! Sorry guys!!!

See Bobby.. This is why it pays to write out the binary.. Anyone have any advice on how to avoid pitfalls like this?

B Haines
CCNA R&S, ETA FOI

 

[maczen]

Thanks Clue and sorry for posting that in here.. Was actually trying to help out but got myself lost there.. figured it out finally... Thanks for the help!

B Haines
CCNA R&S, ETA FOI

 

[bobbyforhire]

Ugh, ok so on my own example i failed beacuse i forgot that 0's are hosts and 1's are networks my very own problem i failed....kinda sad...Mac i just did the same thing..Got tired of subnets and went to play around with IOS.. come to find out my 2610 has got 32/8 and the ios i have doesn't support SSH.....got one that does but it wants 64/16 so i'm going to see if i can steal from peter to pay paul... but since im here why not test myself again.


172.1.0.0 need 500 networks

255.255.0.0

111.111.00000000.00000000

500 networks

512 256 128 64 32 16 8 4 2 1
0 1
9 BITS

11111111.11111111.11111111.1000000
255.255.255.128 = /28
Lowest INC is 128


172.1.0.0 - 172.1.0.127
172.1.0.128 - 172.1.0.255
172.1.1.0 - 172.1.1.127
172.1.1.128 - 172.1.1.255
172.1.2.0 - 172.1.2.127
172.1.2.128 0 172.1.2.255

How about now?
I did this because the last change that i made on my 1's and 0's were in the last octet.


Also Clue have i told you that i love you yet? Completely plutonic but at the same time i know it takes alot of WTF's when you look at my answers.

 

[Cluebird]

bobby,

The only issue is 255.255.255.128 is /25 not /28. All or nothing on exams.

 

[burtsbees]

Mac and bobby---not only is there RTFQ, but also another very important one...
KISS
lol
Clue---great attention to detail. You guys see how someone as anal as me can fatfinger things like his own pseudonym?

Clue also said for the /23..."In this case the increment/blocksize/magic number occurs in the third octet based on one bit."
Follow all the subnetting shortcut rules---every octet of course has 8 bits---the rules apply no matter what octet you're in! The third octet---the one bit=2, so the networks repeat every two, but in the THIRD octet.
KISS---that means don't complicate the $@%#! out of a simple question (though paying attention to detail is VITAL!

Burt Buirt

 

[maczen]

I did catch the Buirt earlier.. LoL

I have it now.. My biggest issue was that I learned to subnet via shortcuts for the CCNA all the way through the third course of the Network Academy then decided that I wanted to learn the binary method the last quarter... I grasped the concept rather quickly but the example that I threw out there earlier has haunted me a few times.. Did pose that question to my instructor but he was unable to explain in person what Clue did with a couple of sentences online. That is why I am quite ecstatic at this point.. LoL

Bobby, I would recommend that you do continue to hone your binary skills.. You may pass the CCNA with shortcuts but look at the above example again later when you are using VLSM and you will see why those shortcuts are sometimes ineffective...
172.16.4.0, 172.16.19.0 and 192.16.255.0 all /22 and you need to choose two that can be subnetted to provide 64 subnets with max hosts (14)... Can't really use the shortcuts for that..

You are doing the right thing by practicing the way you are.. I think that you will do very well.. In regard to subnetting.. develop a system.. In other words, do it the SAME WAY EVERY TIME... regardless of the question... Subnet for 30 networks with max yada yada, what is the broadcast address of the third yada yada, etc.. They can all be solved with the same formula/system with the only difference being that you sometimes work the problem out a little further..

B Haines
CCNA R&S, ETA FOI

 

[burtsbees]

Careful with that question...

172.16.4.0, 172.16.19.0 and 192.16.255.0 all /22 and you need to choose two that can be subnetted to provide 64 subnets with max hosts (14)... Can't really use the shortcuts for that..

There are such networks that supernet the 192.168.0.0/16...

Burt

 

[bobbyforhire]

ok, keeping it simple

192.168.0.0
255.255.255.0
800 hosts needed

1024 512 256 128 64 32 16 8 4 2 1
0 1
10 BITS REQUIRED

11.11.11111111.00000000
11111111.11111111.11111100.00000000
255.255.252.0 = 8 + 8 + 6 = /22
Lowest Inc = 4
Change was made from the default in the 3rd octet

192.168.0.0 - 192.168.3.255
0 - (255 - 2) = 253
1 - (255 - 2) = 253
2 - (255 - 2) = 253
3 - (255 - 2) = 253
----------------------------
253 x 4 = 1012 (GOOD)
253 x 3 = 759 (BAD)

Thus giving me 1 network with an avail 1012 hosts.


my hopeful rule of thumb. If i made a change in an octec that was not there by default that is where i make my inc change.