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Posted on 2008-10-22
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Suppose that instead of using 16 bits for the network part of a class B address originally, 20 bits had been used.  How many class B networks would there have been?
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Question by:ashp87
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by:sullivanr6
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your subnetting down, to increase the number of total networks, while decreasing the total number of hosts you can support. If you orignally had a /16 bit mask then you'd have a total of 65536 networks, if you take host bits (which is what you're doing when you subnet down) away and give them to the network you'd have 2^20 which would give you over a million networks (1048576), and you'd be able to support 4094 hosts on each network.
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by:larsga
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The historical "class B" address range was 128.0.0.0 to 191.255.255.255.

That is, all addresses in the range:
01000000.00000000.00000000.00000000
01011111.11111111.11111111.11111111

This was historically spilt so that the 16 first bits were used for network address, and the remaining 16 bits were used for host addresses.

That is, the mask used to split network address and host address was:
11111111.11111111.00000000.00000000

This resulted in:
00011111.11111111  = 2^14 = 16384 subnets

If one instead had used /20 instead of /16, the mask would have been:
11111111.11111111.11110000.00000000

Which would have given:
00011111.11111111.1111 = 2^18 = 262144 subnets.


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by:sullivanr6
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i don't understand how you've reached this conclusion???

"This resulted in:
00011111.11111111  = 2^14 = 16384 subnets"

why do you have the first three bits off?

J
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by:larsga
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The range for Class B was 128.0.0.0-191.255.255.255. The three first bits do not change.
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larsga earned 500 total points
ID: 22780095
Oops. Sorry. Managed to get that wrong.

The address range of class B was:
10000000.00000000.00000000.00000000
10111111.11111111.11111111.11111111

With 16-bit netmask, that gives (n for network bits, H for host bits):
10nnnnnn.nnnnnnnn.HHHHHHHH.HHHHHHHH
Giving 14 bits for network (2^14 = 16.384) and 16 bits for host (2^16 = 65.536).

If they had used 20-bit netmask instead, it would have been:
10nnnnnn.nnnnnnnn.nnnnHHHH.HHHHHHHH
Giving 18 bits for network (2^18= 262.144) and 12 bits for host (2^12 = 4.096)
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by:sullivanr6
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yea, i keep wanna to think everything is classless. but i guess in the "real world" your class B address ranges actaully mean somthing.
thanx, for clearing that up.

J
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by:larsga
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Note that I said "was" above. The traditional "classful" A/B/C/D/E were introduced in 1981. Routers were very short on CPU and RAM in those days, and one useful feature of "classful" was that the router only had to look at the first 3 bits of the address to determine the netmask.

It did have some serious problems, though. It was very inflexible, and the "C class" was very inefficient. Classful was abandoned around 1993, and was replaced by netmasks/prefixes (cidr). So what "class" an address is in has absolutely no practical meaning today in terms of inter-/intra-network routing - we use explicit netmasks instead.

The question was very explicit in what it asked for (I suppose he is cramming for something like CCDA/CCNA, the question is designed to show if you have an understanding of how the historical "classfull" division of the 32bit IP address space was done).

ashp87, the following should be useful reading for you:
http://en.wikipedia.org/wiki/Classful_network
http://en.wikipedia.org/wiki/CIDR
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by:ashp87
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The answer was actually 4094 i believe.
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by:larsga
ID: 22818073
Yes, you have to subtract 2 for the number of usable host addresses. The top and bottom host address have special meaning (broadcast and network respectively), so those two can't be used for actual hosts.

So the number of usable host addresses in a subnet is 2^n -2 (n being the number of bits used for the host part of the address).

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