It's important to remember your default mask for all 3 address ranges when
looking at a question like this. A class B address has a default mask of 16
bits, so if " You have an IP address of 172.16.4.58 with a 12-bit subnet
mask." you would add an additional 12 bits to the default mask and from
there determine your valid hosts. 16 bits (default mask) + 12 bits
(additional bits) = 28 = 255.255.255.240; then 256 - 240 = 16. You will have
valid ranges in multiples of 16, 1-16, 17-32, 33-48, 49-64, etc... you do
have to remember to leave out the high and low addresses (network and
broadcast) leaving 14 valid host id's per subnet. Using a 172.16.4.58/28
notation would accomplish the same thing and is just an easier way to say "I
have a class B address with a 12 bit subnet mask". I personally struggled
with subnetting before it finally clicked, it's just important to remember
that there are only just so many combinations available. I would suggest
approaching this from multiple different angles until you find the one that
clicks for you, once it does you'll be amazed at how simple it all seems.
Gragg Vaill
MCP CCNA
NOS Contractor
Sprint ION NOC
Kansas City, Ks.
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of
Timothy R Estes
Sent: Thursday, November 23, 2000 8:19 AM
To: [EMAIL PROTECTED]
Subject: Re: IP Address Calculation [1:1010]
Group,
That's pretty confusing. (IMHO). If we are going to refer to subnet masks
with the /30 notation, then we need to stick to it. How would we know if
someone meant with, or without the default mask included?
my $0.02
Timothy Estes
CCNA
Tampa FL
""Travis O'Hara"" wrote in message
news:[EMAIL PROTECTED]...
> Hi,
>
> It's important to understand the wording
>
> 172.16.4.58 is a Class B address. A subnet is the 'borrowed bits' from
the
> host portion of the address. In this case it is stated as being 12bits
>
> So by default it has a 16 bit network mask a Class a has 8 and a Class C
24
>
> 255 255 0 0
> 11111111 11111111 00000000 00000000 = 16Bit netmask
> |---------------| |---------------|
> network host
>
> 255 255 255 240
> 11111111 11111111 11111111 11110000 = 16bit netmask + 12bit subnet mask
+
> 4bit host portion
> |---------------| |-----------||--|
> network subnet host
>
>
> 172.16.4.58 255.255.255.240 is the ip address and network mask
>
> (anyone got a better way of explaining this bit?)
> As networks must start on a border of the subnet that they are divisible
by
> (anyone got a better way of explaining this bit?)
>
> we can figure out the the network address for this IP address with this
> subnet mask starts at 172.16.4.48 (this is not a host address it's a
> network/cable/segment address) first available host address is 172.16.4.49
> your network size is 16 IP addresses so counting from and including
> 172.16.4.48 you reach 172.16.4.63. The last address of the network is
> discarded as being a host address as well as this will be used as the
> broadcast address for the network so the available IP's assignable to
hosts
> are 172.16.4.49-62
>
> That explanation probably sounds as clear as mud but hopefully it's
helpful.
> It's important to read the questions carefully for the syntax they use and
> how it is applied to the subject.
>
> Trav.
>
> > -----Original Message-----
> > From: Mitsunori Sagae [mailto:[EMAIL PROTECTED]]
> > Sent: Thursday, 23 November 2000 3:46 PM
> > To: [EMAIL PROTECTED]
> > Subject: IP Address Calculation [1:1010]
> >
> >
> > Hi, and I am kinda lost with the following question on CCNA
> > sybex book.
> > It's on the Ch4 Review question, and says
> >
> > You have an IP address of 172.16.4.58 with a 12-bit subnet
> > mask. What are
> > your valid hosts?
> >
> > The answer to this question is 172.16.4.49 to 172.16.4.62
> > but I can never get this result,
> >
> > Can someone help me on this, explaining the logic behind it?
> >
> > Thanks
> >
> > mitzs
> >
> >
> >
> >
> > Message Posted at:
> > http://www.groupstudy.com/form/read.php?f=1&i=1010&t=1010
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>
>
>
>
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