I have four different interfaces on a 2514 router (e0,e1,s0,s1), each
interface is on a different subnet, and the mask is 30 bits. Here's they
are:
int e0 = 172.16.10.4 /30
int e1 = 172.16.10.8 /30
int s0 = 172.16.10.12 /30
int s1 = 172.16.10.16 /30
( I only want two addresses per subnet)
I'm runn
It should be:
area 23 range 172.16.10.0 255.255.255.240
--
Date: Fri, 02 Jun 2000 23:49:40 -0700
From: Kurt <[EMAIL PROTECTED]>
Subject: VLSM Question
I have four different interfaces on a 2514 router (e0,e1,s0,s1), each
interface is on a different subnet, a
I have a guestion regarding VLSM
How many subnet addresses can be summarized by 172.108.168.0/21?
how could you calculate it?
thanks in advance.
iWon.com http://www.iwon.com why wouldn't you?
.255.240
--
Lou Nelson, CCNP, CCDA
- Original Message -
From: Kurt <[EMAIL PROTECTED]>
Newsgroups: groupstudy.cisco
To: <[EMAIL PROTECTED]>
Sent: Saturday, June 03, 2000 1:49 AM
Subject: VLSM Question
> I have four different interfaces on a 2514 router (e0,e1,s0,s1), eac
lt;[EMAIL PROTECTED]>; Kurt <[EMAIL PROTECTED]>
Sent: Saturday, June 03, 2000 2:36 PM
Subject: Re: VLSM Question
> Kurt,
> Let me clear... My OSPF is weak... Lotsa study/// ZERO Hands on... I
work
> with EIGRP, RIP etc... but will start some lab work soon with OSPF to
address wildcard-mask AREA area#
MENTAL NOTE
Proof read before you send!
Lou Nelson, CCNP, CCDA
- Original Message -
From: kurt <[EMAIL PROTECTED]>
To: Lou Nelson <[EMAIL PROTECTED]>
Sent: Saturday, June 03, 2000 4:47 AM
Subject: Re: VLSM Question
> Also, since this is an are
Summary address command in OSPF is for summarizing routes *INTO* the OSPF
domain. Area Range command is for summarizing routes from one area to
another. While summary address may work out of OSPF, this is a known bug
and may even stop working upon a reload of the router.
JOE
CCIE 5917
"Kurt"
The numbers you have might work, but they definately aren't the best to use.
I have a couple of comments inline below -
>I have four different interfaces on a 2514 router (e0,e1,s0,s1), each
>interface is on a different subnet, and the mask is 30 bits. Here's they
>are:
>int e0 = 172.16.10.4
Hello,
Your answer is'nt convincing to me. Please do not take it otherwise. Can you
explain your answer?
Thanks
- Original Message -
From: sumthin sumthin <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, June 05, 2000 9:34 AM
Subject: VLSM Question
> It
hey im just learning about vlsm, so theres a good chance this is wrong :)
anyways i'd say
172.108.168.0
x.x.169.0
x.x.170.0
x.x.171.0
x.x.172.0
x.x.173.0
x.x.174.0
x.x.175.0
Justin...
On Thu, 15 Jun 2000, jeongwoo park wrote:
> I have a guestion regarding VLSM
> How many subnet addresses can
You've got (the last) three bits on the third octet to play with, which
leaves you with 2^3 = 8 cont. numbers (the first one being 168) : 168,
169,..., 175.
You may also think of the /21 as 2 /22s, so 4 /23s, so 8 /24s.
mh
>hey im just learning about vlsm, so theres a good chance >this is wrong
>I have a guestion regarding VLSM
>How many subnet addresses can be summarized by 172.108.168.0/21?
>how could you calculate it?
>thanks in advance.
>
I'm not sure exactly what you are asking. Do you want to know the
total number of available host addresses in a /21? Or are you
looking for th
You could calculate it in this way,
for 172.108.168.0/24, it includes from 172.108.168.0 to 225
now for a mask of 21, 24-21=3
1000=248, the netmask is 255.255.248.0, it covers addreses of 8*256.
and for 172.108.168.0/21, it covers from 172.108.168.0 to 172.108.175.255.
> I have a guestion r
t;[EMAIL PROTECTED]>
>To: jeongwoo park <[EMAIL PROTECTED]>
>CC: [EMAIL PROTECTED]
>Subject: Re: VLSM Question
>Date: Fri, 16 Jun 2000 20:00:02 +1000 (EST)
>
>hey im just learning about vlsm, so theres a good chance this is wrong :)
>anyways i'd say
>172.108.16
taramani" <[EMAIL PROTECTED]>
>Reply-To: "Sriram Venkataramani" <[EMAIL PROTECTED]>
>To: "sumthin sumthin" <[EMAIL PROTECTED]>, <[EMAIL PROTECTED]>
>Subject: Re: VLSM Question
>Date: Mon, 5 Jun 2000 14:00:25 -0700
>
>Hello,
>
don't think you do much. ( If these 4
subnets were in the same area you accomplish that)
Cheers,Padhu
-Original Message-
From: sumthin sumthin [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, June 27, 2000 10:58 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: VLS
> > > I came up with summary-address 172.16.10.0 255.255.255.224
The summary-address command is only used on OSPF ASBRs. Use the area
range command to summarise routes on an
OSPF ABR.
The succinct difference lies in the type of LSA generated by these two
commands. The ABR generates Summary LS
As for VLSM, I found an example in Jeff Doyle (TCP/IP Vol 1) on p290 that I
don't understand.
192.168.50.0 /25, and it states that the reason it has /25 is because it
needs to have 100 hosts => so 2^7-2=126 hosts (as 2^6 would be too zmall),
so it makes sense.
What confuses me is that since 192.
A prep test I am using has a question for which I disagree with the answer.
Here is the questionÂ…
If I had a Class B address, what subnet mask would I use if I wanted to
split it into 8 class C addresses?
a.255.255.240.0
b.255.255.255.0
c.255.255.248.0
d.255.255.254.0
The answer from the
The 25 bit can be 0 or 1. It looks like the book chose 0. The CCNA books use
2^n - 2 for both hosts and subnets. But you also can use the
zero and one subnet so that would add 2. Then you would have 2^n for
subnets. So, 192.168.50.0 is one, and 192.168.50.128 is another subnet.
Using the first s
Sorry about the oversight:
192.168.50.0001 first host
192.168.50.0110 last host
should be:
192.168.50.0111 broadcast
Message Posted at:
http://www.groupstudy.com/form/read.php?f=7&i=35832&t=35827
--
FAQ, list archives, and subscrip
> As for VLSM, I found an example in Jeff Doyle (TCP/IP Vol 1) on
> p290 that I don't understand.
> 192.168.50.0 /25, and it states that the reason it has /25 is
> because it needs to have 100 hosts => so 2^7-2=126 hosts (as 2^6 would be
too small), so it makes sense.
>
> What confuses me is that
Don't think of the Class C at allif you have a /25, that means that, of
the 32 bits in the address, the most significant 25 represent the network
address. Dotted decimal notation is for human convenience, nothing more. The
address is a string of binary digits coming over the wire, and it is re
So does it mean on the first subent -> the host range is 192.168.50.1 -
192.168.50.126, and the second subnet host range is 192.168.50.129 -
192.168.50.254?
Best Regards,
Hunt Lee
""G Z"" wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> The 25 bit can be 0 or 1. It looks like th
, February 24, 2002 4:34 PM
To: [EMAIL PROTECTED]
Subject: Re: VLSM Question [7:35827]
So does it mean on the first subent -> the host range is 192.168.50.1 -
192.168.50.126, and the second subnet host range is 192.168.50.129 -
192.168.50.254?
Best Regards,
Hunt Lee
""G Z"&
The answer is correct, with the assumption "ip subnet-zero" was
configured.
Leo
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, December 04, 2002 3:52 PM
To: [EMAIL PROTECTED]
Subject: VLSM Question [7:58569]
A prep test I am u
Richard Burdette wrote:
>
> A prep test I am using has a question for which I disagree with
> the answer. Here is the questionÂ…
>
> If I had a Class B address, what subnet mask would I use if I
> wanted to split it into 8 class C addresses?
>
>
> a.255.255.240.0
> b.255.255.255.0
> c.255.255
""s vermill"" wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Richard Burdette wrote:
> >
> > A prep test I am using has a question for which I disagree with
> > the answer. Here is the question.
> >
> > If I had a Class B address, what subnet mask would I use if I
> > wanted to s
mes that subnet
zero and the all ones subnet are unusable. Therefore you have to create
16 subnets, resulting in 14 "usable" to get the required 8 subnets.
In the "real" world, 255.255.224.0 is correct.
BTW, what is the VLSM question here?
HTH,
Prof. Tom Lisa, CCAI
Commun
that subnet
> zero and the all ones subnet are unusable. Therefore you have to create
> 16 subnets, resulting in 14 "usable" to get the required 8 subnets.
>
> In the "real" world, 255.255.224.0 is correct.
> BTW, what is the VLSM question here?
>
> H
The Long and Winding Road wrote:
> > The question sucks and so do the answer choices. Eight
> *addresses* per
> > subnet or eight subnets? If the former, a mask of
> 255.255.255.248 would
> be
> > required - not 255.255.248.0. If the latter, I'd go with you
> (sorta). Or
> > perhaps a. if subne
""s vermill"" wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> The Long and Winding Road wrote:
> > > The question sucks and so do the answer choices. Eight
> > *addresses* per
> > > subnet or eight subnets? If the former, a mask of
> > 255.255.255.248 would
> > be
> > > required
: 12/05/02 09:48 AM
To: [EMAIL PROTECTED]
Subject: Re: VLSM Question [7:58569]
> you sure about that, Tom?
172.100..0
255.255.1110.0
subnet bits = 1.0
172.100.0.0 through 172.100.31.0 for /24's
these would be SUMMARIZED using the 224 mask in the third octet.
if you o
0 ! add 2 to both sides !
> n = 4 ! 2^4-2 = 14 !
>
> 128 64 32 16 8 4 2 1
> 1 1 1 1 0 0 0 0
>
> = 240, or answer A in the original post.
>
> BJ
>
>
>
> ---Original Message---
> From: The Long and Winding Road
> Sent: 12/05/02 09:48 A
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