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| Nathaniel W. Cook 2003-09-29, 1:26 pm |
| Hello,
I am having trouble understanding an answer to a practice question. Here it
is:
If you had 22 bits used for subnetting, how many hosts are available per
subnet?
Answer? 2
This question is taken from the CCNA flashcards on the CD for Todd Lammle's
book. I usually use the following method to calculate hosts per subnet: 2
to the power of x (where x equals the # of bits not used in subnet mask)
minus 2. There are 22 bits used for subnetting leaving 10 bits for hosts.
2 to the 10th power gives 1024 minus 2 equals 1022. Obviously I am
miscalculating. Any help on this one?
Thanks,
Nathaniel
| |
| steve harris 2003-09-29, 1:26 pm |
| Nathaniel W. Cook wrote:
> Hello,
>
> I am having trouble understanding an answer to a practice question. Here it
> is:
>
> If you had 22 bits used for subnetting, how many hosts are available per
> subnet?
>
> Answer? 2
>
> This question is taken from the CCNA flashcards on the CD for Todd Lammle's
> book. I usually use the following method to calculate hosts per subnet: 2
> to the power of x (where x equals the # of bits not used in subnet mask)
> minus 2. There are 22 bits used for subnetting leaving 10 bits for hosts.
> 2 to the 10th power gives 1024 minus 2 equals 1022. Obviously I am
> miscalculating. Any help on this one?
>
> Thanks,
> Nathaniel
>
>
which book?
640-607
http://www.sybex.com/erratatracking...nForm&ISBN=4167
640-507
http://www.sybex.com/erratatracking...nForm&ISBN=2647
| |
| Nathaniel W. Cook 2003-09-29, 1:26 pm |
| Hi Steve.
It's the 607 book (3rd edition). Thanks for the heads up. Actually the
first thing I do now when I get a book is go to the errata page. This
flashcard thing is not listed on the respective errata page. So are you
saying that 1022 is possibly the correct answer?
Thanks,
Nathaniel
"steve harris" <steveharris1@hotmail.com> wrote in message
news:bl9nep$9l68n$1@ID-182617.news.uni-berlin.de...
> Nathaniel W. Cook wrote:
>
> > Hello,
> >
> > I am having trouble understanding an answer to a practice question.
Here it
> > is:
> >
> > If you had 22 bits used for subnetting, how many hosts are available per
> > subnet?
> >
> > Answer? 2
> >
> > This question is taken from the CCNA flashcards on the CD for Todd
Lammle's
> > book. I usually use the following method to calculate hosts per subnet:
2
> > to the power of x (where x equals the # of bits not used in subnet mask)
> > minus 2. There are 22 bits used for subnetting leaving 10 bits for
hosts.
> > 2 to the 10th power gives 1024 minus 2 equals 1022. Obviously I am
> > miscalculating. Any help on this one?
> >
> > Thanks,
> > Nathaniel
> >
> >
>
>
> which book?
> 640-607
> http://www.sybex.com/erratatracking...nForm&ISBN=4167
>
> 640-507
> http://www.sybex.com/erratatracking...nForm&ISBN=2647
>
| |
| Helmut 2003-09-29, 1:26 pm |
|
Użytkownik "Nathaniel W. Cook" <nospam_nathanielcook@comcast.net> napisał w
wiadomości news:F_Gdnf_4gYZ-wOWiU-KYgw@comcast.com...
> Hello,
>
> I am having trouble understanding an answer to a practice question. Here
it
> is:
>
> If you had 22 bits used for subnetting, how many hosts are available per
> subnet?
>
> Answer? 2
>
> This question is taken from the CCNA flashcards on the CD for Todd
Lammle's
> book. I usually use the following method to calculate hosts per subnet: 2
> to the power of x (where x equals the # of bits not used in subnet mask)
> minus 2. There are 22 bits used for subnetting leaving 10 bits for hosts.
> 2 to the 10th power gives 1024 minus 2 equals 1022. Obviously I am
> miscalculating. Any help on this one?
The answer is allright provided that you are considering
class A network; thus 8 + 22 bits gives 30 --> 2 hosts.
You have to assume (even if it is not explicitly given)
that it's class A. For B and C their "natural" mask plus
22 would result in subnet mask exceeding 32 bits.
HTH
Helmut
| |
| Nathaniel W. Cook 2003-09-29, 1:26 pm |
| Hi Helmut,
Oh ok. Well that makes sense. The only confusing part is that when they
said "22 bits for subnetting" I interpreted that as the shorthand /22 not as
class whatever PLUS 22. Would you all say this is or is not the type of
question that I might get on the official CCNA test?
"Helmut" <no_mail@no_spam.com> wrote in message
news:mDZdb.183186$hd6.2352988@news.chello.at...
>
> Użytkownik "Nathaniel W. Cook" <nospam_nathanielcook@comcast.net> napisał
w
> wiadomości news:F_Gdnf_4gYZ-wOWiU-KYgw@comcast.com...
> > Hello,
> >
> > I am having trouble understanding an answer to a practice question.
Here
> it
> > is:
> >
> > If you had 22 bits used for subnetting, how many hosts are available per
> > subnet?
> >
> > Answer? 2
> >
> > This question is taken from the CCNA flashcards on the CD for Todd
> Lammle's
> > book. I usually use the following method to calculate hosts per subnet:
2
> > to the power of x (where x equals the # of bits not used in subnet mask)
> > minus 2. There are 22 bits used for subnetting leaving 10 bits for
hosts.
> > 2 to the 10th power gives 1024 minus 2 equals 1022. Obviously I am
> > miscalculating. Any help on this one?
>
> The answer is allright provided that you are considering
> class A network; thus 8 + 22 bits gives 30 --> 2 hosts.
> You have to assume (even if it is not explicitly given)
> that it's class A. For B and C their "natural" mask plus
> 22 would result in subnet mask exceeding 32 bits.
>
> HTH
>
> Helmut
>
>
| |
| Mark Smythe 2003-09-29, 2:25 pm |
| With Cisco, I dont know why, but they like to train you to use "classfull" ip
addressing which means that subnet bits are the bits in addition to the
classfull network. So for 22 bits of subnetting they are probably referring to
a class A address with 22 bits added to the subnet mask. This would leave you
with 2 hosts available after removing the network and broadcast. However, your
calculation would be correct in the real world. When a network person calculates
hosts , the network bits are one in the same with subnet bits. i.e. 22 bits
refers to the first 22 bits like you did. And the remaining bits are the hosts.
But you will need to think classfully on your test . It is one of those Cisco
thangz.
"Nathaniel W. Cook" wrote:
> Hello,
>
> I am having trouble understanding an answer to a practice question. Here it
> is:
>
> If you had 22 bits used for subnetting, how many hosts are available per
> subnet?
>
> Answer? 2
>
> This question is taken from the CCNA flashcards on the CD for Todd Lammle's
> book. I usually use the following method to calculate hosts per subnet: 2
> to the power of x (where x equals the # of bits not used in subnet mask)
> minus 2. There are 22 bits used for subnetting leaving 10 bits for hosts.
> 2 to the 10th power gives 1024 minus 2 equals 1022. Obviously I am
> miscalculating. Any help on this one?
>
> Thanks,
> Nathaniel
| |
| Helmut 2003-09-29, 2:25 pm |
|
Użytkownik "Nathaniel W. Cook" <nospam_nathanielcook@comcast.net>
napisał w wiadomości news:h-6dnSExUuMX9eWiXTWJjw@comcast.com...
> Hi Helmut,
>
> Oh ok. Well that makes sense. The only confusing part is that when they
> said "22 bits for subnetting" I interpreted that as the shorthand /22 not
as
> class whatever PLUS 22.
Probably most of people would have their first thought like you.
>Would you all say this is or is not the type of
> question that I might get on the official CCNA test?
I hadn't question like this on my CCNA exam, however
AFAIR it was possible to meet a question about
"x bits for subnetting" in an initial configuration dialog
of a cisco box (sorry, don't remember platform/soft ver)
so I'd consider it as a "legal question" during a test.
cheers
Helmut
| |
| private 2003-09-29, 2:25 pm |
| I would calculate this as two hosts per subnet aswell. this is how I would
understand it.
22 bits used for sunbnetting, so I would interperet that as /22
giving the first 22 bits as the network portion of the ip address and the
rest to hosts.
255.255.252.0 subnet
i was taught take 256 combinations (0-255) subtract from 252 = 4 hosts for
that subnet
or 64 subnets, each containing 2 hosts per subnet
1 reserved for a network address and 1 for a broadcast address, this leave
4-2= 2 possible host for that 22bit subnet
"Nathaniel W. Cook" <nospam_nathanielcook@comcast.net> wrote in message
news:F_Gdnf_4gYZ-wOWiU-KYgw@comcast.com...
> Hello,
>
> I am having trouble understanding an answer to a practice question. Here
it
> is:
>
> If you had 22 bits used for subnetting, how many hosts are available per
> subnet?
>
> Answer? 2
>
> This question is taken from the CCNA flashcards on the CD for Todd
Lammle's
> book. I usually use the following method to calculate hosts per subnet: 2
> to the power of x (where x equals the # of bits not used in subnet mask)
> minus 2. There are 22 bits used for subnetting leaving 10 bits for hosts.
> 2 to the 10th power gives 1024 minus 2 equals 1022. Obviously I am
> miscalculating. Any help on this one?
>
> Thanks,
> Nathaniel
>
>
| |
| Helmut 2003-09-29, 2:25 pm |
|
Użytkownik "private" <tp007g2217@blueyonder.co.uk> napisał w wiadomości
news:k_Zdb.286$xG1.139@news-binary.blueyonder.co.uk...
> I would calculate this as two hosts per subnet aswell. this is how I would
> understand it.
> 22 bits used for sunbnetting, so I would interperet that as /22
> giving the first 22 bits as the network portion of the ip address and the
> rest to hosts.
> 255.255.252.0 subnet
sorry, you are wrong: if it's /22 , you have 10 bits left for a host part
> i was taught take 256 combinations (0-255) subtract from 252 = 4 hosts for
> that subnet
> or 64 subnets, each containing 2 hosts per subnet
> 1 reserved for a network address and 1 for a broadcast address, this leave
> 4-2= 2 possible host for that 22bit subnet
it's always good to see a background of the method rather than
just learning an algorithm by heart (as it can lead to wrong results)
cheers
Helmut
| |
| Nathaniel W. Cook 2003-09-29, 2:25 pm |
| wow, ok thanks for all the input guys.
Nathaniel
"Nathaniel W. Cook" <nospam_nathanielcook@comcast.net> wrote in message
news:F_Gdnf_4gYZ-wOWiU-KYgw@comcast.com...
> Hello,
>
> I am having trouble understanding an answer to a practice question. Here
it
> is:
>
> If you had 22 bits used for subnetting, how many hosts are available per
> subnet?
>
> Answer? 2
>
> This question is taken from the CCNA flashcards on the CD for Todd
Lammle's
> book. I usually use the following method to calculate hosts per subnet: 2
> to the power of x (where x equals the # of bits not used in subnet mask)
> minus 2. There are 22 bits used for subnetting leaving 10 bits for hosts.
> 2 to the 10th power gives 1024 minus 2 equals 1022. Obviously I am
> miscalculating. Any help on this one?
>
> Thanks,
> Nathaniel
>
>
| |
| private 2003-09-29, 3:25 pm |
|
"Helmut" <no_mail@no_spam.com> wrote in message
news:E2_db.183715$hd6.2361571@news.chello.at...
>
> Użytkownik "private" <tp007g2217@blueyonder.co.uk> napisał w wiadomości
> news:k_Zdb.286$xG1.139@news-binary.blueyonder.co.uk...
> > I would calculate this as two hosts per subnet aswell. this is how I
would
> > understand it.
> > 22 bits used for subnetting, so I would interperet that as /22
> > giving the first 22 bits as the network portion of the ip address and
the
> > rest to hosts.
> > 255.255.252.0 subnet
>
> sorry, you are wrong: if it's /22 , you have 10 bits left for a host part
Was the question posed not:
"If you had 22 bits used for subnetting, how many hosts are available per
subnet?"
Was my reply above not > > 22 bits used for subnetting, so I would
interperet that as /22
> > giving the first 22 bits as the network portion of the ip address and
the rest to hosts?
Therefore logic would dictate that 32 bit subnet mask - 22 bit network
portion and the rest of the 10 bits is left to????
HOSTS...............
But considering the question asked how many hosts per subnet?
which is 2
> > i was taught take 256 combinations (0-255) subtract from 252 = 4 hosts
for
> > that subnet
> > or 64 subnets, each containing 2 hosts per subnet
> > 1 reserved for a network address and 1 for a broadcast address, this
leave
> > 4-2= 2 possible host for that 22bit subnet
>
> it's always good to see a background of the method rather than
> just learning an algorithm by heart (as it can lead to wrong results)
>
> cheers
>
> Helmut
>
>
| |
|
| Sorry, pal, but when you have a subnet mask of /22 (255.255.252.0), you have
10 bits left for the host part (not 2 bits) because the router couldn't care
less about the decimal dot that separates octets. For the router, it's all a
long binary number. So, you get a subnet consisting of (2^10)-2 usable IP
addresses (1024 - 2 = 1022) per subnet. If this is a Class-B network, you
will get 2^6 = 64 subnets. If this is a class A network, you will get 2^14 =
16,384 subnets.
Grey
"private" <tp007g2217@blueyonder.co.uk> wrote in message
news:21%db.154$P02.153@news-binary.blueyonder.co.uk...
>
> "Helmut" <no_mail@no_spam.com> wrote in message
> news:E2_db.183715$hd6.2361571@news.chello.at...
> >
> > Użytkownik "private" <tp007g2217@blueyonder.co.uk> napisał w wiadomości
> > news:k_Zdb.286$xG1.139@news-binary.blueyonder.co.uk...
> > > I would calculate this as two hosts per subnet aswell. this is how I
> would
> > > understand it.
> > > 22 bits used for subnetting, so I would interperet that as /22
> > > giving the first 22 bits as the network portion of the ip address and
> the
> > > rest to hosts.
> > > 255.255.252.0 subnet
> >
> > sorry, you are wrong: if it's /22 , you have 10 bits left for a host
part
>
> Was the question posed not:
> "If you had 22 bits used for subnetting, how many hosts are available per
> subnet?"
> Was my reply above not > > 22 bits used for subnetting, so I would
> interperet that as / 22
> > > giving the first 22 bits as the network portion of the ip address and
> the rest to hosts?
> Therefore logic would dictate that 32 bit subnet mask - 22 bit network
> portion and the rest of the 10 bits is left to????
> HOSTS...............
> But considering the question asked how many hosts per subnet?
> which is 2
>
>
> > > i was taught take 256 combinations (0-255) subtract from 252 = 4 hosts
> for
> > > that subnet
> > > or 64 subnets, each containing 2 hosts per subnet
> > > 1 reserved for a network address and 1 for a broadcast address, this
> leave
> > > 4-2= 2 possible host for that 22bit subnet
> >
> > it's always good to see a background of the method rather than
> > just learning an algorithm by heart (as it can lead to wrong results)
> >
> > cheers
> >
> > Helmut
> >
> >
>
>
| |
| private 2003-09-29, 6:24 pm |
| Apologies Helmut and Grey. I am getting mixed up here. Had to sit down and
think about this, map it out . Just been doing access-lists with wildcard
masking, so I was getting mixed up here. A /30 bit subnet mask would give 2
hosts per subnet
A subnet mask forms a division between the network portion and host potion
of an ip address.
so if you an IP address of 136.20.40.2 has a subnet mask of 255.255.252.0 =
64 subnets each with 1024-2 =1022 hosts
SUBNET RANGE IS
136.20.0.0 - 136.20.3.0 where 136.20.0.0 is the network address and
136.20.3.255 is the broadcast address
thus 136.20.0.1 is the first host to 136.20.0.255 = 255 hosts (1st address
136.20.0.0 is the network address)
then 136.20.1.0. is the next host to 136.20.1.255 = 256 hosts
then 136.20.2.0. is the next host to 136.20.2.255 = 256 hosts
then 136.20.3.0. is the next host to 136.20.3.254 = 255 hosts (last address
is136.20.3.255 is the broadcast address for this subnet )
TOTAL HOSTS = 1022
136.20.4.0 - 136.20.4.0 where 136.20.4.0 is the network address and
136.20.7.255 is the broadcast address
136.20.8.0
136.20.12.0
136.20.16.0
136.20.20.0
136.20.24.0
136.20.28.0
136.20.32.0
136.20.36.0
136.20.40.0
"Grey" <bbb@ccc.com> wrote in message
news:E9KcnSwcC8VXGuWiXTWJig@co
mcast.com...
> Sorry, pal, but when you have a subnet mask of /22 (255.255.252.0), you
have
> 10 bits left for the host part (not 2 bits) because the router couldn't
care
> less about the decimal dot that separates octets. For the router, it's all
a
> long binary number. So, you get a subnet consisting of (2^10)-2 usable IP
> addresses (1024 - 2 = 1022) per subnet. If this is a Class-B network, you
> will get 2^6 = 64 subnets. If this is a class A network, you will get 2^14
=
> 16,384 subnets.
>
> Grey
>
> "private" <tp007g2217@blueyonder.co.uk> wrote in message
> news:21%db.154$P02.153@news-binary.blueyonder.co.uk...
> >
> > "Helmut" <no_mail@no_spam.com> wrote in message
> > news:E2_db.183715$hd6.2361571@news.chello.at...
> > >
> > > Użytkownik "private" <tp007g2217@blueyonder.co.uk> napisał w
wiadomości
> > > news:k_Zdb.286$xG1.139@news-binary.blueyonder.co.uk...
> > > > I would calculate this as two hosts per subnet aswell. this is how I
> > would
> > > > understand it.
> > > > 22 bits used for subnetting, so I would interperet that as /22
> > > > giving the first 22 bits as the network portion of the ip address
and
> > the
> > > > rest to hosts.
> > > > 255.255.252.0 subnet
> > >
> > > sorry, you are wrong: if it's /22 , you have 10 bits left for a host
> part
> >
> > Was the question posed not:
> > "If you had 22 bits used for subnetting, how many hosts are available
per
> > subnet?"
> > Was my reply above not > > 22 bits used for subnetting, so I would
> > interperet that as /22
> > > > giving the first 22 bits as the network portion of the ip address
and
> > the rest to hosts?
> > Therefore logic would dictate that 32 bit subnet mask - 22 bit network
> > portion and the rest of the 10 bits is left to????
> > HOSTS...............
> > But considering the question asked how many hosts per subnet?
> > which is 2
> >
> >
> > > > i was taught take 256 combinations (0-255) subtract from 252 = 4
hosts
> > for
> > > > that subnet
> > > > or 64 subnets, each containing 2 hosts per subnet
> > > > 1 reserved for a network address and 1 for a broadcast address, this
> > leave
> > > > 4-2= 2 possible host for that 22bit subnet
> > >
> > > it's always good to see a background of the method rather than
> > > just learning an algorithm by heart (as it can lead to wrong results)
> > >
> > > cheers
> > >
> > > Helmut
> > >
> > >
> >
> >
>
>
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