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Q: "Internet Engineering" ( Answered,   0 Comments )
Question  
Subject: "Internet Engineering"
Category: Reference, Education and News > Teaching and Research
Asked by: sam7074-ga
List Price: $25.00
Posted: 19 Oct 2004 06:22 PDT
Expires: 18 Nov 2004 05:22 PST
Question ID: 416925
What is the total number of class B addresses available for assignment
to individual hosts
if no subnetting is used? If subnetting is allowed (assuming a
non-zero number of bits for the subnet ID),
what is the maximum number of hosts allowed? What is the value of the
subnet mask corresponding to the
point at which this maximum is reached?
Answer  
Subject: Re: "Internet Engineering"
Answered By: maniac-ga on 19 Oct 2004 17:39 PDT
 
Hello Sam7074,

Let's explain first what a Class B address is. Using
  http://www.geocities.com/SiliconValley/Vista/8672/network/ipaddr.html
as a guide (#20 through #24), a Class B address has two parts
 - Network ID - two bytes (up to 16384 possible networks)
 - Host ID - two bytes (up to 65534 possible hosts)
Note that the number of hosts is 2^16-2 to account for reserved values
  XXX.YYY.000.000
and
  XXX.XXX.255.255
the first one is sometimes used to refer to the "local host" (though
127.0.0.1 is typical as well) and the latter one is a directed
broadcast (#31). This is based on RFC 950 at
  http://www.faqs.org/rfcs/rfc950.html
which defines the Internet Standard Subnetting Procedure. Near the end
of section 2.1, it describes this assignment of these two values.

Other sites with basically the same answer include:
  http://linux.cudeso.be/tcpip1.php
(see also the calculator link at the end of the page)
  http://wks.uts.ohio-state.edu/sysadm_course/html/sysadm-341.html
or search with a phrase like
  class b maximum hosts

If subnetting is allowed, you basically allocate additional bits to
the Network ID and fewer to the Host ID. With two subnets (0 and 1),
the number of bits available to the Host ID is 15, or 32768 possible
values for each subnet. Again, if you reserve two addresses per subnet
(000.000, and 127.255 in this case), you get 32766 possible hosts per
subnet or 65532 total hosts. If you do more subnets, the number of
reserved addresses increases (two per subnet), so the maximum number
of subnetted hosts is when a single bit is used for the subnet.

The calculator at
  http://www.agt.net/public/sparkman/netcalc.htm
can be used to demonstrate that relationship. To use it, click on
Class B near the top, select 2 as the number of subnets and then click
on calculate to see the results below. Note this calculator gives you
the total number of addresses (assuming none are reserved). Follow
this link
  http://www.agt.net/public/sparkman/network/netcalchelp.htm#allzeroes
to get an explanation of how more modern protocols and routers may
allow you to use all possible values. Note in particular the caution
about answering questions that may be on exams to be aware of the
differences with classless networks.

There is also a good reference at
  http://www.redhat.com/advice/tips/ipnetworking.html
which has a table near the end showing six subnets allocated from a
class C address range with up to 30 hosts in each subnet.

The subnet mask is basically a bit pattern that can be used to extract
the Network ID from the IP address. Without subnetting, the class B
addresses has a subnet mask of
  255.255.0.0
With two class B subnets, the subnet mask is
  255.255.128.0
That represents the value at which the maximum number of subnetted
hosts is reached. This is sometimes referred to as a /17 subnet mask,
representing a 17 bit wide mask value.

To recap, a Class B without subnetting and with reserved addresses has
  65534
host addresses. With two subnets (0 and 1), the maximum number of hosts drops to
  65532
with half (32766) in each subnet. The subnet mask corresponding to this is
  255.255.128.0  (or /17)

If any part of this answer is unclear or you need a more complete
explanation, please make a clarification request. I would be glad to
help you on this kind of research.

  --Maniac

Request for Answer Clarification by sam7074-ga on 20 Oct 2004 08:36 PDT
I need a little bit of more explanation for the first part of my
question for which i am still confused.

As the address range for class B addresses is 128.0.0.0 to
191.255.255.255, so to calculate the total number of class B addresses
available for assignment to individual hosts, do we have to do the
calculation like this:
[191-128=64(including 128)]*[2^14=16384]*[2^16-2]= [2^20]*[2^16-2]

Please correct me if i am wrong for this part. Otherwise, your answer
for the other two parts is satisfactory.

Clarification of Answer by maniac-ga on 20 Oct 2004 16:03 PDT
Hello Sam7074,

If you need the total number of class B hosts (not the total number
within a class B network), the calculation is:
  16384 (# of class B networks) * 65534 (# of hosts per network) = 1073709056

You get 16384 for the number of networks from
  64 (values in first byte) * 256 (values in second byte) = 16384
The 64 in the equation you describe is not needed. To put it into the
form you used...
  [2^14]*[2^16-2]

  --Maniac
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