Hello Qweb88,
I will try to answer this in a couple ways
- describe how to get the answer
- provide the specific answers
There are a number of network calculators that are on line. A few
examples:
http://screamer.mobrien.com/net.shtml
which when you enter IP 122.125.120.200 and number of hosts per
network as 4094 (256*16-2) and press calculate to get the answers...
Class A subnetted as class A
Subnet Mask - 255.255.240.0 (or /20)
# of subnets - 4096
Network - 122.125.112.0, node # 8 200
Broadcast - 122.125.127.255
It also displays the IP address in a variety of formats - binary, hex,
and decimal (32 bit value).
http://screamer.mobrien.com/net.shtml
Lists ten different calculators - some on line, some as java
applications.
The first one provides the following...
Select class A (default), 12 subnet bits [calculates mask as your
problem described], enter IP address and it computes...
# of subnets - 4096
# of hosts - 4094
network - 122.125.112.0
broadcast - 122.125.127.255
range - 122.125.112.1 - 122.125.127.254 (for some reason the last #
not shown)
I think that hits the five values you asked for with slightly
different names from your question.
There are plenty of other sites available with a search such as
calculate subnet address
for more information.
Note that neither of the two calculators I used could directly take
the values that were provided in your question. You can adjust the
values requested to get the "right answer", but in general, you will
have to understand the specific technique to calculate network
addresses to use these effectively.
--Maniac |
Request for Answer Clarification by
qweb88-ga
on
13 Sep 2002 17:55 PDT
Please, could you just give me the exact anwsers to the qwustions.
Calclate the following subnetting problems,
1) major network number
2) Range of usable hosts
3) subnet address
4)subnet broadcast address
5)number of usable subnetworks
6)number of usable host per subnetwork
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Clarification of Answer by
maniac-ga
on
14 Sep 2002 07:16 PDT
Hello Qweb88,
I don't mind answering the specific query, but please note that
https://answers.google.com/answers/researcherguidelines.html#acceptablequestions
states that Google Answers discourages answers for homework.
The answer already provided gives you guidance on answering the
specific queries. The five specific values you originally requested
are listed in the original answer - the only thing I left out was
which one was which. If I answer the specific query [match the values
to the queries], and this is homework, you may be accused of cheating
if you fail to disclose the source of your answers.
Please advise if you want the values matched to the original queries.
Also note that some of the terms you have used (e.g., subnet address,
major network number) may have more than one definition. Please
include a definition of those terms if you want an accurate answer.
--Maniac
|
Request for Answer Clarification by
qweb88-ga
on
14 Sep 2002 09:59 PDT
Dear maniac,
Thank for your enfort so far, i just want some clarifications to
some of the answer you sent me. 1) the usabe number of host and
subnets.
2) range of uasable hosts
Thanks again for your help.
Just to clarify something you mention in your letter, Im a 45
years man and my reserach is mainly for personal use, ok.
|
Clarification of Answer by
maniac-ga
on
16 Sep 2002 17:47 PDT
Hello Qweb88,
The number of usable hosts and subnets is pretty straight forward. The
original answer states,
"number of hosts per network as 4094 ... number of subnets 4096"
This is because...
o there are 12 bits allocated for subnets based on the 20 bit mask
you provided (255.255.240.0) and because the IP address is in the
"class A" range (8 bits allocated for class A addresses). A class B or
C address range has fewer available subnets due to more bits for the
class number.
o there are 12 bits allocated for hosts, the difference between a 32
bits available and 20 bits already allocated.
Using A's for the class A portion, S's for the subnet portion, and H's
for the host portion, the bytes look something like this...
AAAAAAAA SSSSSSSS SSSSHHHH HHHHHHHH
01111010 01111101 01111000 11001000
122 125 120 200
in a 32 bit register (though may be byte swapped in a little endian
host).
In 12 bits, you can represent 4096 unique values (zero through 4095).
So there can be 4096 unique subnets in the class A address space
beginning with 122.
The reason you only get 4094 hosts in a subnet is that...
- the last address in the range is the "broadcast address"
- you don't get to use the first address either
So you only get 4096 (12 bits) minus 2 = 4094.
The range of host numbers for these 4094 addresses is
AAAAAAAA SSSSSSSS SSSSHHHH HHHHHHHH
01111010 01111101 01110000 00000001
122 125 112 001
through
AAAAAAAA SSSSSSSS SSSSHHHH HHHHHHHH
01111010 01111101 01111111 11111110
122 125 127 254
Not a problem w/ the clarification. I don't want to get a student in
trouble with their school.
--Maniac
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