Comments on: CCNA – Subnetting

By: KAUSHIK

KAUSHIK — Tue, 25 Nov 2014 07:32:13 +0000

Can anybody tell that in Q.5 The switch used is Intelligent(L3) or not? Since it divides the Broadcast Domain as well.
I ask the Administrator as well.

By: Daniel

Daniel — Wed, 20 Aug 2014 00:54:49 +0000

General question about subnetting… what is the usual average time to solve these type of questions?

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By: Anonymous

Anonymous — Sat, 05 Apr 2014 23:23:32 +0000

Netstack College has received a grant to wire all of its six buildings for 10/100 Ethernet. Four of the buildings contain classrooms and laboratories and will need to support 250 network connections each. One building is for administration and will need to support 180 network connections. The Administration Building also houses the Computer Center where the college servers and network backbone are located. The sixth building is the library, which must support 80 network connections.
You have been directed not to use a flat network of 1260 IP addresses assigned from a Class B network range. Instead, he would like to use multiple subnets.

Playing the role of the network administrator, what steps would you take to accomplish this setup using subnets instead of using a flat network of 1260 IP addresses assigned from a Class B network range? Determine and list the IP address ranges and masks associated with each building and explain the process you used to obtain the IP addresses and masks showing the obtained addresses and masks.

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By: qazi

qazi — Tue, 29 Oct 2013 05:57:17 +0000

Question 6)The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the increment is 256 / 8 = 32 please i need explanation on this.

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By: Haris Ques

Haris Ques — Sun, 27 Oct 2013 21:17:26 +0000

A national retail chain need to deisgn an ip addressing scheme to support a nationwide network.
The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet.
working with only one class B address, which of the following subnet masks will support an appropriate addressing scheme ?

i know that answer but i want to know about complete detail at this dump

By: moataz elshorbagy

moataz elshorbagy — Thu, 03 Oct 2013 17:52:18 +0000

do they let using calculators ?

By: Alex

Alex — Tue, 01 Oct 2013 14:52:20 +0000

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By: Kalyan

Kalyan — Fri, 27 Sep 2013 19:36:08 +0000

9tut is just awesome.But along with it you need to practice VCE files of Exam simulation which can be found at http://www.downloadexam.com/vce/cisco-640-802/

Thanks to 9tut I have passed by CCNA with 960

By: Bud

Bud — Fri, 27 Sep 2013 11:44:49 +0000

Atif,

Usable network addresses for the 115.64.4.0/22 network is

115.64.4.1 to 115.64.7.254

115.64.4.0 – is the network address
115.64.7.255 – is the broadcast address

7 is the last number of the 3 rd Octect
But the host part is in the forth octect can still change from 0 – 254 while 3rd octect is 7

Good Luck for your exam !

By: Atif

Atif — Fri, 27 Sep 2013 04:32:09 +0000

Q. 4 the answer B. 115.64.7.64, would it be the broadcast network ID based on the magic #4, network 0,4,8,12,16 etc, so 7 would be the broadcast IP. Please reply have exam tomorrow.

By: CiscoER

CiscoER — Thu, 26 Sep 2013 16:07:53 +0000

@ Nick Ques 1 Mask 224 gives block size of 32. You need to know this.
So for A – 15.234.118.63 The networks are 118.0 118.32 118.64 118.96 etc up to 224 and the broadcast are 118.31, 118.63, 118.95 etc Its explained there below the question. You cannot use 118.63 as a host cuz it is a broadcast address. Let’s try E…. E – 201.45.116.159 Networks are 116.128 116.160 116.192 up to 224….. (116.160 -1 ) = broadcast address which gives you 116.159. You cannot use E for a host. Its a broadcast address. Hope that helps but 9tut explanation is great.

By: Nick

Nick — Sun, 22 Sep 2013 16:12:40 +0000

hi, not really understand explanation for Q1. Anyone?

By: jasostrong

jasostrong — Wed, 11 Sep 2013 13:39:55 +0000

Thanks 9tut. I made it. I passed my CCNA 200-120 today. The sim is Access-list 1 , Access-list 2 & EIGRP. A lot of new questions like Netflow, Syslog, SNMP, VRRP, and GLBP.

By: Anonymous

Anonymous — Sat, 07 Sep 2013 04:58:38 +0000

great explanation for q6. csj0827.

By: @slesh

@slesh — Fri, 30 Aug 2013 13:52:51 +0000

hi 9tut i am also confuse with question of ndane

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ahmad saba — Thu, 29 Aug 2013 19:22:45 +0000

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By: ndane

ndane — Tue, 27 Aug 2013 15:00:14 +0000

Hi everyone!
Can anyone explain where did
The network administrator needs to address seven LANs. RIP version 1 is the only routing protocol in use on the network and subnet 0 is not being used. What is the maximum number of usable IP addresses that can be supported on each LAN if the organization is using one class C address block?

According to sekhar
Ans 14
Explanation
RIP version 1 does not support VLSM information, so all networks must have the same subnet mask. In the network above, there are a total of 12 networks (6 LANs and 6 different point to point WAN connections). Therefore, if each of the 12 networks use the 255.255.255.240 subnet mask, there will be a total of 16 networks with 14 usable hosts on each LAN.

Can anyone explain where the 6LANs and 6 WAN connection come from?

Please Im a bit confuse..

Thanks in advance

By: robert

robert — Tue, 27 Aug 2013 07:13:55 +0000

hi,

just want to ask if are the subnetting questions here the same with the actual ccna exam?

please reply. thanks.

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By: soll

soll — Wed, 21 Aug 2013 22:22:23 +0000

Hey guys I got a question on number 6 . check the following

The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the increment is 256 / 8 = 32 and its corresponding subnet mask is /19 (1111 1111.1111 1111.1110 0000).

By: please reply via my email address at laurantianhlapo@gmail.com

please reply via my email address at laurantianhlapo@gmail.com — Wed, 21 Aug 2013 09:50:28 +0000

An exam question read as follows:
You are required to provide IP addresses by subnetting the following network, 172.16.0.0/16.

The addresses are needed for the company’s 4 departments, the server room and for point to point WAN links.

The requirements of the network are as follow:

1) VF- 1000 host addresses

2) HR – 350 host addresses

3) FIN – 100 host addresses

4) LN – 50 host addresses

5) SERVER ROOM -20 host addresses

6) LINK 1,2,3,4 and 5 – 2 host addresses (these are point to point links, such as between 2 routers)

For EACH network show

a) How many bit/s did you use to provide the needed IP addresses? [6]

b) How many bits remained and how many networks would these bits create?[12]

c) What is the new subnet mask? [12]

d) For each network, generate the first and last valid IP addresses, and the broadcast addresses.(Do this in a table) [50]

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By: Bill_Ado

Bill_Ado — Wed, 24 Jul 2013 20:08:39 +0000

@dominic achira, the easy way to tackle with subnetting in class B is to find out first the subnet and the broadcast the address in which the address is for example, given the address 172.16.66.10/18, what subnet and broadcast is this address on? we all know that the default mask for a class B is a /16, then here we have 18 (18-16=2), 2 bits have been borrowed and we put them in the third octet and it will sit on 64 and 64 is our increment, we are working in the third octet, from our example 66 is between 0 and 64 and the next subnet is 128 since our increment is 64 and to find the broadcast for subnet 64 we take the next subnet which is 128-1 and we have 127, and the host addresses are any numbers between the subnet and broadcast, the most important thing to keep in mind is to know which octet you are working, it becomes easy once yoj know the number of bits borrowed,….my small contribution, i hope it makes sens to you

By: Bill_Ado

Bill_Ado — Wed, 24 Jul 2013 19:43:33 +0000

@Ruth, let me try to understand your question: you mean you’ve got 2 subnets:
192.168.64.0, and 192.168.65.0 and subnet of 255.255.254.0, given these 2 subnets, there’s only one posiible subnet mask we can work with and that is 255.255.255.0, and this is how we find our increment with this subnet mask: 256-256=1 and we will start adding 1 up to 255, but as you mention subnet mask 255.255.254.0, this mask gives us an increment of 2, which means our subnets will start, if subnet zero is being used from: 192.168.0.0, 192.168.2.0,……192.168.64.0, 192.168.66.0, we should use the increment(+2) in the third octet, as we are working with this subnet mask, that’s the way i try to understand your question, i hope it helps….

By: Rich

Rich — Wed, 24 Jul 2013 01:34:57 +0000

@drtedd

Since your default mask is /16 and you need 8 networks from the provided network you need to borrow however many bits will allow for 8 networks.
So…
2^1 = 1
2^2 = 4
2^3 = 8. There are your 8 networks. You already have a /16 and you borrowed 3 bits which now gives you /19 16+3.

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Vick — Sun, 21 Jul 2013 18:07:50 +0000

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By: drtedd

drtedd — Fri, 19 Jul 2013 19:24:12 +0000

Question 6??? – It’s obvious how “32″ came about dividing the network in 8 equal parts, however I do not understand how /19 is the corresponding subnet mask?

By: David

David — Thu, 18 Jul 2013 15:55:45 +0000

Will they let you use a calculator?

By: Syed Irfan Naseer

Syed Irfan Naseer — Thu, 18 Jul 2013 09:40:04 +0000

@ ibrahim
July 15th, 2013
find on youtube to understand subnetting
FREE URDU CCNA Lecture 11 Subnetting

By: Ruth

Ruth — Tue, 16 Jul 2013 19:58:44 +0000

Is it possible to use a network of 192.168.0.0 and have subnets 192.168.64.0 and 192.168.65.0 for one vlan and use S/M 255.255.254.0? Thank you for your response.

By: ibrahim

ibrahim — Mon, 15 Jul 2013 12:12:33 +0000

please can some one show me websites that i can read on subnetting basics to get the clear understanding.

By: RBK

RBK — Tue, 09 Jul 2013 14:32:09 +0000

hey guys plz tell me the ip addresses are same in exam? or not?

By: dominic achira

dominic achira — Mon, 01 Jul 2013 13:01:24 +0000

what are the easy ways to tackle with subnetting questions in class B.those guys who have mastered subnetting.

By: kim

kim — Mon, 01 Jul 2013 05:30:48 +0000

Rubel,, Letter “D” is belong to the first network w/c is 115.64.0.0 – 115.64.3.255/22.

By: Rubel

Rubel — Fri, 28 Jun 2013 20:24:04 +0000

Please someone describe question-4. I think D is correct.

By: mikeh

mikeh — Thu, 27 Jun 2013 09:33:24 +0000

@nicky they’ll only provide you w/ a erasble marker and plastic paper/board. no other materials used from outside.

By: Annoyn

Annoyn — Thu, 20 Jun 2013 16:35:45 +0000

Yes Nicky, They usually provide you with a pen and a pad of paper for the test. but restrict any outside materials that you might bring with you in case you have notes jotted in the back or other underhanded schemes people try to cheat on the tests (not that you would but some people do try)

By: nicky

nicky — Mon, 17 Jun 2013 23:01:25 +0000

I HAVE A QUESTION, FOR THE ONES WHO ALREADY TOOK THE EXAM, DOES THEY LET YOU TO USE ANY PENCIL OR PAPER, CU’Z TO SUBNETTING??
I CAN NOT DO EVERYTHING WITH MY MIND :s

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David Okeri — Wed, 05 Jun 2013 08:28:51 +0000

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By: Annoyn

Annoyn — Tue, 28 May 2013 16:22:56 +0000

You can use the 94.255 address in Q6, because the broadcast of that subnet is 95.255 it’s a tricky one

By: Rubie-sky

Rubie-sky — Sun, 26 May 2013 04:12:55 +0000

Q6 .. you cant use the .255 address it’s for broadcasts .. the 0 likewise is the network ID, hence why you -2 from the amount of available hosts you have.

By: Kris

Kris — Mon, 13 May 2013 13:58:29 +0000

Oh, on question 6, guess I should question of the .255 answer as well…

By: Kris

Kris — Mon, 13 May 2013 13:55:58 +0000

In question 6, is it a practice to assign a .0 address to a host? Is this correct? I realize the host would be in the middle of the subnet, so it isn’t the network nor is it a broadcast. I just don’t recall everything seeing a host with a .0 address.

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@Hes thank you

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By: Hes

Hes — Tue, 16 Apr 2013 13:08:42 +0000

@golu
Explanation is clear.
Network 172.25.0.0 is a Class B which means sunbet should be 255.255.0.0 and third part would be changed, i mean 255.255.X.0 ( X part of network or subnet )

Question say : divided into eight equal subnets which means : 255/8=32 and subnets will be : 0->32->64->96->128->160->196->224->256.

” ip subnet-zero ” command is configured which means you can use all subnets from 0 and third subnet would be :

1. 0–>32
2. 32–>64
*** 3. 64–>96 ***

It is Class B then in the network : 172.25.X.0 you need to use X part to calculate thefore third subnet would be :

1. 172.25.0.0 –> 172.25.31.255
2. 172.25.32.0 –> 172.25.63.255
*** 3. 172.25.64.0 –> 172.25.95.255 ***

As you look at the Answers : A C D are the correct.

By: golu

golu — Sat, 13 Apr 2013 13:30:23 +0000

can anyone explain me Q.6

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By: Yaser Arafat

Yaser Arafat — Sat, 23 Mar 2013 14:07:49 +0000

A network administrator is configuring ACLs on a Cisco router, to allow IP access form the 192.168.146.0/24,192.168.147.0/24,192.168.148.0/2,. and 192.168.149.0/24 networks only. Which two ACLs, when combined, should be used.

A.access-list 10 permit ip 192.168.146.0 0.0.0.255
B.access-list 10 permit ip 192.168.146.0 255 255.255.0
C.access-list 10 permit ip 192.168.147.0 0.0.255 255
D.access-list 10 permit ip 192.168.149.0 0.0.255.255.0
E.access-list 10 permit ip 192.168.148.0 0.0.1.255
F.access-list 10 permit ip 192.168.146.0 0.0.1.255

answer :E F
can anybody explain to me this one? Thanks!

By: Charles

Charles — Thu, 21 Mar 2013 19:53:40 +0000

Question 3:

B & E are correct!!!.. to compy with the first condition… 300 subnets.. you have to figure out how many subnet bits u need to rob from the 3rd octet and pssibly from the 4th of the class B network scheme and how many are left for the host bits.
Choose 2:

B. 255.255.255.128 = 11111111.11111111.11111111.10000000 = /25 => 2^9 subnets = 512 (300 subnets) and 2^7 hosts = 128 -2 = 126 (50)

E. 255.255.255.192 = 11111111.11111111.11111111.11000000 = /26 =>2^10 subnets = 1024 (300 subnets) and 2^6 hosts = 64 -2 = 62
Hope this helps for everyone!!! Regards!!

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By: Anonymous

Anonymous — Wed, 20 Mar 2013 02:31:25 +0000

@usman

Q1 – D. Issue the “show cdp neighbors detail” command on Router C.
this command with “detail” keyword will display all neighboring devices with their IP address.

Q2 – C. It is point-to-point only.

CSMA/CD is Ethernet protocol to detect collision in half-duplex. CSMA/CA is wireless protocol to avoid collision. Hub operate in half-duplex only.

Q3 – E & F
access-list 10 permit ip 192.168.148.0 0.0.1.255 will cover 192.168.148.0/24,. and 192.168.149.0/24 networks.
access-list 10 permit ip 192.168.146.0 0.0.1.255 will cover 192.168.146.0/24, 192.168.147.0/24 network.

Q4 – C. two

By: Mahesh V

Mahesh V — Sun, 17 Mar 2013 11:01:17 +0000

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By: El_Tarik

El_Tarik — Sat, 16 Mar 2013 17:50:43 +0000

@AN9
192.0.2.0/23 borrows one bit from the 3rd octet, the last bit (1st octet = 8 bits, 2nd octet = 16 bits, 7 bits of 3rd octet = 23 bits total).
The value of the bit is 2 (last bit is 1, second last 2, then 4, 8, 16, 32, 64 and 128).
So we know that each network address (subnetwork address) will be in multiples of 2 starting at 0, so the first answer is incorrect as it is a network address.
The third answer is 192.0.3.255. If you check the next network address it is 192.0.4.0. The broadcast address is always the next network (subnet) address minus 1, so in this case, 3.255 is the broadcast address so it can not be a host address.
The 4th answer is 192.0.4.0 which is the address for the subnet 192.0.4.0 (multiples of 2 in the 3rd octet).
The correct answer is the 2nd one, as it is 192.0.2.255. It is a trick question as when people see a 255 in an octet they assume it is a broadcast address but it is not so as this example proves. The next subnet address is 4.0 not 3.0, hence 2.255 is a valid host address.
If you translate it all into binary, you will see this very clearly and easily. Let me know if you have more questions mate, cheers!

By: daBeat

daBeat — Fri, 15 Mar 2013 01:30:35 +0000

@Question-5.. D is incorrect cuz it says Vlan 1… 172.16.1.128/25 is vlan 2

By: Question-05

Question-05 — Tue, 12 Mar 2013 14:44:37 +0000

@9tut
In Question-5
Option-D is also correct coz its in between
172.16.1.128 subnet which ranges from
172.16.1.128 to 172.16.1.254 so we can also assign 172.16.1.205 to any host, am i right..?

By: Question-05

Question-05 — Tue, 12 Mar 2013 14:18:09 +0000

In Q5 Option D is also correct, What should we do to that…???

By: mista-ug

mista-ug — Mon, 11 Mar 2013 16:45:48 +0000

the subnet mask here will be 255.255.255.254
so we have 256-254= 2subnets
so, 0,2,4,6,8 etc are the sunets so from here you work your way out if you can get the broadcast address of each
for example first network 192.0.2.0- 192.0.4.0
the broadcast is 192.0.2.3 host address 192.0.2.255

By: AN9

AN9 — Sat, 09 Mar 2013 18:18:42 +0000

HI THERE!
I HAVE BEEN GIVEN THIS NETWORK 19.0.2.0/23 AND I HAVE BEEN ASKED WHICH OPTION IS VALID IP ADDRESS THAT CAN BE ASSIGNED TO A HOST?
A) 192.0.2.0
B) 192.0.2.255
C)192.0.3.255
D) 192.0.4.0
I HAVE BEEN TOLD THAT THE ANSWER (B) BUT DO NOT UNDERSTAND HOW THEY GOT THIS.
PLEASE CAN SOMEONE EXPLAIN TO ME HOW THEY GOT THIS ANSWER ?
HERE’S MY EMAIL: an9one@gmail.com.
MANNY THANKS IN ADVANCE

By: Waqar

Waqar — Fri, 08 Mar 2013 02:11:45 +0000

Q3. E, F looks correct answer to me