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Still Hard for me!!!!
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Rajan
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Hi Guys,
give me solution in detail please.
find subnet, broad cast,valid host range
1.
176.16.10.33 255.255.255.224
176.16.10.65 255.255.255.192
176.16.10.17 255.255.255.252
176.16.10.33 255.255.255,240
192.168.100.25 255.255.255.252
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 01-07-01 11:44 PM
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Azam
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Hey Rajan, don't feel bad about this, I'm really stuck on subnet ?'s also. I'm sure people will help us out on these one.
Good Luck
Azam
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01-08-01 12:01 AM
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Silkysmoothe1
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Guys, may want to try going here. www.mcsefreak.com
good luck,,,,,,silky,,,,
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01-08-01 12:16 AM
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dmaftei
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First here's the answers (subnet, broadcast, host range):
176.16.10.33 255.255.255.224
176.16.10.32 176.16.10.63 176.16.10.33-176.16.10.62
176.16.10.65 255.255.255.192
176.16.10.64 176.16.10.127 176.16.10.65-176.16.10.126
176.16.10.17 255.255.255.252
176.16.10.16 176.16.10.19 176.16.10.17-176.16.10.18
176.16.10.33 255.255.255.240
176.16.10.32 176.16.10.47 176.16.10.33-176.16.10.46
192.168.100.25 255.255.255.252
192.168.100.24 192.168.100.27 192.168.100.25-192.168.100.26
Let's work on the first example to see how to do it: 176.16.10.33 255.255.255.224
- Determine the natural class of your address. If first byte is 1-126, that's a class A address, with default mask 255.0.0.0. If first byte is 128-191, that's a class B address, with default mask 255.255.0.0. If first byte is 192-223, that's a class C address, with default mask 255.255.255.0. Don't worry about the rest (first byte 0 and 224-255). Your address is 176.16.10.33. This is a class B address, with default mask 255.255.0.0.
- Now look at the default mask (255.255.0.0) and your mask (255.255.255.224). Count how many bits are "1" in the host portion of your mask, and how many bits are "0". For class B, the last 2 bytes are the host portion. The host portion of your mask is "255.224". There are 8 "1" bits in 255, and 3 "1" bits in 224 (255 is binary 11111111, while 224 is binary 11100000). The total number of "1" bits is 11, the total number of "0" bits is 5.
- The total number of subnets is 2^11 = 2048. The total number of hosts per subnet is 2^5 = 32. This "total number of hosts" is the "block size" you'll use to find the subnet addresses.
- Now start at the natural network (in your case, that's 176.16.0.0), and keep adding the block size. You do this like in decimal arithmetic:
176.16.0.0
176.16.0.32
176.16.0.64
176.16.0.96
176.16.0.128
176.16.0.160
176.16.0.192
176.16.0.224
176.16.1.32 (here you carry to the third byte, 'cause 224 + 32 = 256, which is too big)
176.16.1.64
...........
176.16.10.0
176.16.10.32
176.16.10.64
............
176.16.255.160
176.16.255.192
176.16.255.224
Note that you have a total of 2048 subnets in this list (you may count them, if you don't believe me  )
- Now look at your address and see in what subnet it falls. Your address is 176.16.10.33. The subnet for your address is 176.16.10.32. You just got the first answer.
- To find the broadcast for your subnet, find the next subnet, and substract 1. The next subnet is 176.16.10.64. When you substract 1, you get 176.16.10.63. This is the second answer.
- To find the host range, add 1 to the subnet address, and substract 1 from the broadcast address. You get 176.16.10.33 through 176.16.10.62. You're done!
This example is the most difficult, because you borrow a whole host octet, and a few bits from the next host octet. Here are a few more hard examples:
- 10.10.10.1 255.255.128.0 (class A, borrow byte 2 and 1 bit from byte 3)
- 10.10.10.1 255.255.255.192 (class A, borrow bytes 2 and 3, and 2 bits from byte 4)
- your example (class B, borrow byte 3 and 3 bits from byte 4).
You have easy subnetting when you borrow bits only from the first host octet:
- 10.10.10.1 255.128.0.0
- 176.16.10.1 255.255.192.0
- your last example
There are a few points I'm not going to touch now (the 2^n - 2 rule, shortcuts for hard subnetting). Get the basics first, and come back for more later.
Is it easier now?
P.S.
Double check all the answers; I didn't spend too much time checking...
[This message has been edited by dmaftei (edited 01-07-2001).]
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01-08-01 12:45 AM
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Azam
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Hey dmaftei, thanks for your reply.
If I have an IP Adress of...
209.202.42.155 , I know this is a Class C address, but what will the MASK be?
Like in your example you have...
176.16.10.33 255.255.255.224 <- How did you get this
176.16.10.32 176.16.10.63 176.16.10.33-176.16.10.62
Thanks in advance
Azam
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01-08-01 01:16 AM
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dmaftei
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quote:
Originally posted by Azam:
Like in your example you have...
176.16.10.33 255.255.255.224 <- How did you get this
It was given in the text of the problem.  (take a look at the first post in this thread...)
If a mask is not specified I would say it's safe to assume you're talking "classful" (so you'll have the default mask). You cannot subnet without a mask...
Cheers!
[This message has been edited by dmaftei (edited 01-07-2001).]
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01-08-01 01:38 AM
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Marker
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I have some worksheets that will get you started in subnetting. If anyone would like them E-mail me at:
mark-er@marksmail.com
--Mark
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01-08-01 01:38 AM
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Azam
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Hey,
Ok if my IP Adress was..
209.202.42.34 and my Mask was 255.255.255.240
Then these will be the subnet, broadcast, host ranges...
256-240 = 16, 16+16 = 32, 32+16 = 48
Subnet = 32
Broadcast = 47
Host Ranges = 33-46
Is all this correct?
Thanks in advance
Azam
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01-08-01 04:00 AM
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Azam
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01-08-01 04:21 AM
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dmaftei
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Yup! But I would advise you (if you don't mind) to get into the habit of writing the whole address:
Subnet = 209.202.42.32
Broadcast = 209.202.42.47
Host Range = 209.202.42.33-209.202.42.46
It may avoid potential confusion, and, you know, better safe than sorry...
Cheers!
[This message has been edited by dmaftei (edited 01-09-2001).]
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01-08-01 04:22 AM
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