CCNA - Subnetting

Author

Subnetting

cybertechno

2002-12-17, 11:53 am

Hi,
Check out this question

The network IP is 199.141.27.0 with subnet mask 255.255.255.240
The question is to identify the valid host addressed

choices

A:199.141.27.33
B:199.141.27.112
C:199.141.27.119
D:199.141.27.126
E:199.141.27.175
F:199.141.27.208

I don't understand this question and how to solve it

Please help

davidbeecken

2002-12-17, 7:10 pm

www.ciscotrack.com/subnetting.html

this has helped some people, it covers what you want to know.

HOOLIGAN

2002-12-17, 7:51 pm

You need to figure out which ones are the network address and which are the broadcast addresses for each subnet address range. These can not be used for valid host addressing.

chodan

2002-12-17, 8:35 pm

ACD

nazeer_miya

2002-12-18, 3:42 am

there is no answer in the choices what u specified...
the valid hosts range for the above things r
199.141.27.1 to
199.141.27.16 and the broadcast for this is 199.141.27.15 hope i am right in my answer....

Hippo

2002-12-18, 4:58 am

OK here goes:

Given the Class C NETWORK address of:
199.141.27.0
and a subnet mask of:
255.255.255.240

the SUBNETS within the NETWORK address increase in units of 16. Get that from subtracting the last octet of the subnet mask from 256, ie 256 - 240 =16.

So the subnets are as follows:

Subnet 199.141.27.0
Range 199.141.27.1 - 199.141.27.14
B/Cast 199.141.27.15

Subnet 199.141.27.16
Range 199.141.27.17 - 199.141.27.30
B/Cast 199.141.27.31

Subnet 199.141.27.32
Range 199.141.27.33 - 199.141.27.46
B/Cast 199.141.27.47

and so on until

Subnet 199.141.27.240
Range 199.141.27.241 - 199.141.27.254
B/Cast 199.141.27.255

Looking at the options available, Chodan is right with ACD. Just write out the list I started, in full and see where the addresses fit in. It is that easy, honest.

Hippo

chodan

2002-12-18, 5:38 am

quote:
Originally posted by nazeer_miya

there is no answer in the choices what u specified...
the valid hosts range for the above things r
199.141.27.1 to
199.141.27.16 and the broadcast for this is 199.141.27.15 hope i am right in my answer....

Well your close
and the question is vague
it would be 0-15 16 starts the next network range but 15 is the broad cast.

I was assuming he was asking to give valid host addresses if you broke the entire /24 range into a series of /28 ranges although as you point out that wasn't specifically stated.

B.alkaff

2002-12-18, 5:56 am

Yes It is ACD.

another way to obtain the answer is:
256-240= 16
so subnet adresses will be multiplies of 16: 0, 16, 32, 48 etc.
Broadcast adresses will be the valid subnet adresses - 1 : 15, 31, 47.

Other Adresses will be valid host adresses.

so u can omit 112(16*7), 208(13*16) and 175(16*11-1). The others will be valid :33, 119, 126 i.e ACD.

Do You find that fast?!

cybertechno

2002-12-18, 1:13 pm

hi alkaff,
The answers given for the question are what you have told.so u r right.

But I don't understand how u did it?!?

Please be more clear.I will be grateful if there is more explanation.

Thanx

nazeer_miya

2002-12-18, 10:44 pm

Hi All I continue with Hippo method

Given the Class C NETWORK address of:
199.141.27.0
and a subnet mask of:
255.255.255.240

the SUBNETS within the NETWORK address increase in units of 16. Get that from subtracting the last octet of the subnet mask from 256, ie 256 - 240 =16.

So the subnets are as follows:

Subnet 199.141.27.0
Range 199.141.27.1 - 199.141.27.14
B/Cast 199.141.27.15

Subnet 199.141.27.16
Range 199.141.27.17 - 199.141.27.30
B/Cast 199.141.27.31

Subnet 199.141.27.32
Range 199.141.27.33 - 199.141.27.46
B/Cast 199.141.27.47

Subnet 199.141.27.48
Range 199.141.27.49 - 199.141.27.62
B/Cast 199.141.27.63

Subnet 199.141.27.64
Range 199.141.27.65 - 199.141.27.78
B/Cast 199.141.27.95

Subnet 199.141.27.96
Range 199.141.27.97 - 199.141.27.110
B/Cast 199.141.27.111

Subnet 199.141.27.112
Range 199.141.27.113 - 199.141.27.126
B/Cast 199.141.27.127

Subnet 199.141.27.128
Range 199.141.27.129 - 199.141.27.142
B/Cast 199.141.27.143

Subnet 199.141.27.144
Range 199.141.27.145 - 199.141.27.158
B/Cast 199.141.27.159

Subnet 199.141.27.160
Range 199.141.27.161 - 199.141.27.174
B/Cast 199.141.27.175

Subnet 199.141.27.176
Range 199.141.27.177 - 199.141.27.190
B/Cast 199.141.27.191

Subnet 199.141.27.192
Range 199.141.27.193 - 199.141.27.206
B/Cast 199.141.27.207

Subnet 199.141.27.208
Range 199.141.27.209 - 199.141.27.222
B/Cast 199.141.27.223

Subnet 199.141.27.224
Range 199.141.27.225 - 199.141.27.238
B/Cast 199.141.27.239

Subnet 199.141.27.240
Range 199.141.27.241 - 199.141.27.254
B/Cast 199.141.27.255

so cyber....
If u check at the answers other than ACD , BF are Network numbers and then E is an Broadcast IP which wont suit....

now To all my question if the said appears in exam means i think v can simply mention that there is no valid range for the given IP's.. is it necessary that v have to go through all the lengthy proceduer in short period of time.....

Cybertecho... hope u had understood the concept....

Nazeer

B.alkaff

2002-12-19, 1:17 am

Hippo & Nazeer Method is Right for Some one who want to Undersatnd the subject. Mine is for some one knows the concept and want quick way to answer the q using his
understanding of the concept.

Any way here is the explaination:
Since the Network Address is start with 199 then it is Class C Address. why? Cos class C Address starts with 110 (In binary) which equals to 192. 199 is 11000111.

Class C Network is Network has 256 Hosts or less. So the Part that will contains the Host address is the last part of Address A.B.C.D => D.

Default Subnet mask for class C network is 255.255.255.0 which means that there is no subnets so valid hosts 256 - 2= 254. why -2?
cos You have to reserve one Address for the Subnet as whole (The network itself in this case) which is the first no i.e 199.141.27.0.
The other Address will be the broadcast Address which used to broadcast to all hosts within the Network. It is the last Address i.e 199.141.27.255. The other Adresses from 199.141.27.1 to 199.141.27.254 are the host adrresses.

But here the Subnet mask is not the defalut. It is 255.255.255.240. Since last part is not 0 this means that this network was subnetted.

How can you know the No. of subnets & no of hosts per subnet? ok

You can use many ways.. here are 2 of:
1. 240 is = 11110000 (Binary)=> 4 ones => take 2 to the power 4 => 16 subnet. Now 256/16 (to obtain addresses per subnet)= 16 but you have to take 2 adresses again within each subnet for the subnet address itself (first Address in subnet) and one for Broadcast (last one in subnet). This means 16 - 2 = 14 hosts per subnet.

Now you can subnet: subnet 199.141.27.0 which had host addresses from 199.141.27.1 to 199.141.27.14 (14 hosts) and the broadcast address 199.141.27.15 (last one).

Second subnet is 199.141.27.16 (since it is the next Address after 199.141.27.15) hosts are from 199.141.27.17 to 199.141.27.30 (14 host) and broadcast address is 199.141.27.31. and You can continue to obtain the other subnets.

2. Also u can start with the No. of Addresses per subnet 256 - 240 = 16 Address per subnet 14 of them for hosts and the other 2 are for subnet address and broadcast. The No. of subnets are then 256/16=16 subnet. Same result & you can obtain the addresses now.

That is the concept. But to resolve the q You know that you have 16 addresses per subnet so each subnet will begin by a multipy of 16: 0, 16 , 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224 & 240. You can ignor any of these Addresses from the choices since that will be addresses of the subnets themselfes not Hosts within subnets.

Broadcast Address is last Address within a subnet.. so You can obtain it by subtract 1 from Subnet Address to know the last address of the previuos subnet (Broadcast). so You can ignor 15 (16-1), 31 (32-1), 47(48 - 1) and so on. Note that 0 - 1 = -1 which counts to 255.

Now in the q You must ignor any address that is Multiply of 16 cos it will be a subnet address and any address that is multiply of 16 subtracted by 1 (Broadcast). Here you find 112 (16 * 7) & 208 (16 * 13) as subnets addresses. So ignor them.

Also You find 175 (16 * 11 - 1) as broadcast address so ignor it. The Other will be the valid Hosts Addresses.

I hope that will help You

cybertechno

2002-12-19, 12:13 pm

Hi All,
Thanx I did get things right.I was wrong in my arithematic so had a few problems now I know.
But thanx all for ur explanation it did make me understand better.

Regards

Midas2

2002-12-19, 5:04 pm

Is this question from a CCNA Exam?

Sponsored Links