| Author |
Examples of Class A Ranges, Little confused...
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| Ybentrepreneur 2002-11-05, 6:57 pm |
| Can someone please subnet this address for me: 15.0.0.0 I need to Make 15 Subnets from this......
Can someone give me a little list of about 6-7 Valid Address Ranges for this?
I came up with this, but forgot the Partern.
SubnetId Valid Range 15.0.0.0 15.0.0.0 -15.15.255.254
15.16.0.0 15.16.1.1-15.31.255.254
15.32.0.0 15.32.2.1-15.47.255.254
How would I continue with this? Im I doing this the Right Way?
In my Valid Range should the 3rd # Keep going up by 1? | |
| Ybentrepreneur 2002-11-05, 6:59 pm |
| sorry damn it didnt show the way i intended to show it on this Screen...
But can someone just give an example of some valid ranges for Class A Starting with
15.0.0.1-15.15.255.254
ETc
ETc | |
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| JPL333 2002-11-06, 6:39 am |
| maybee...
address15.0.0.0
mask 255.248.0.0 ie 5 bits for subnetting
that gives2**5-2=30 subnets
and 2**19-2 host addresses per subnet
therefore the pattern starts with
s/n 15.0.0.0
b/c 15.7.255.255
range 15.0.0.1-15.7.255.254
then
s/n 15.8.0.0
b/c 15.15.255.255
r 15.8.0.1-15.15.255.254
hope this is correct.... | |
| ANDRONDA 2002-11-06, 5:35 pm |
| I take this as a challenge to try to help.
First look at the requirement, which is to have at least 15 subnets.
Your network is 15.0.0.0 so you have to get your Subnet Mask.
Usually the question states that you cannot have a mask that sets aside more subnets than you absolutely need. So if you say that you need 15 subnets that means that you must reserve 16 bits because this is the closest power of two without going under the 15.
One thing you all have to realize is that Cisco does not use the formula of:
2*n –2 (where * means “raised to the power of” and “n” is the number of binary ones).
Unless the question specifically states that it is older equipment (such as unix). Believe me this is true. Ask anyone who has taken the test recently. Keep in mind, this applies to number of subnets not hosts. You still have to subtract out the network and broadcast addresses on each subnet.
Assuming that this the case (you are not using Unix or old equipment) in Binary that would be
11111111.11110000.00000000.00000000
That translates to 255.240.0.0.
Now find your different network addresses. Technically you are supposed to count your zeros and raise two to that power and subtract two. But this is an s-load of hosts: two raised to the power of 20, which I do not feel like calculating and converting. It works the other way too. Simply start with 15.0.0.0 and keep adding your 16 (derived from your four 1’s up top). See typically they ask you to do this with Class C addresses, which are much easier.
15.0.0.0
15.16.0.0
15.32.0.0
15.48.0.0
15.64.0.0
15.80.0.0
15.96.0.0
15.112.0.0
15.128.0.0
15.144.0.0
15.160.0.0
15.176.0.0
15.192.0.0
15.208.0.0
15.224.0.0
15.240.0.0
The broadcast addresses will be the ones preceding the next network address (for example 15.15.255.255 is the first one, 15.16.255.255 the second and so on.
I know some of you probably think this is wrong- you will say- Hey- you have to subtract two when you find subnets. You should have used a subnet mask of 255.248.0.0.
I argue otherwise. On the CCNA test they differentiate between older equipment and new equipment and if they ask how many subnets does a particular mask produce you had better not subtract the two or you will miss it. Believe me. I took the test four weeks ago and they did ask under two separate conditions and the answers were different depending on the scenarios. | |
| Ybentrepreneur 2002-11-06, 6:51 pm |
| so can u give me a list of what the valid ranges will look like? | |
| ANDRONDA 2002-11-07, 7:41 am |
| I did- look a tthe table provided.15.0.0.0
Your network addresses are:
15.0.0.0
15.16.0.0
15.32.0.0
15.48.0.0
15.64.0.0
15.80.0.0
15.96.0.0
15.112.0.0
15.128.0.0
15.144.0.0
15.160.0.0
15.176.0.0
15.192.0.0
15.208.0.0
15.224.0.0
15.240.0.0
add in the broadcast addresses (I did two for you- you can do the rest):
15.0.0.0 to 15.15.255.255
15.16.0.0 to 15.31.255.255
15.32.0.0 to 15.47.255.255
15.48.0.0 to 15.63.255.255
15.64.0.0 etc etc etc | |
| Ybentrepreneur 2002-11-07, 9:24 am |
| but anddronda, shoudnt it be this way?
15.0.0.0 15.1.0.0-15.15.255.254
15.16.0.0 15.2.0.0-15.31.255.254
15.32.0.0 15.3.0.0-15.47.255.254
ETC?
An addr 15.15.255.255 should be a broadcast, correct? | |
| ANDRONDA 2002-11-07, 10:25 am |
| We are talking apple and oranges here.
Ranges (in general) would be
15.0.0.0 to 15.15.255.255
15.16.0.0 to 15.31.255.255
15.32.0.0 to 15.47.255.255
15.48.0.0 to 15.63.255.255
Etc.
This includes network and broadcast addresses.
VALID HOST ID's would be
15.0.0.1 to 15.15.255.254
15.16.0.1 to 15.31.255.254
15.32.0.1 to 15.47.255.254
15.48.0.1 to 15.63.255.254
Etc.
Cisco likes to trick and ask for both. If they say VALID HOST ID's then be sure to take out the broadcast and network address but if they ask for ranges be sure to include the full ranges.
The important thing to remember is to include all ranges if they ask and do not subtract out the two as most subnetting tutorials teach. Cisco does not see it that way. | |
| Ybentrepreneur 2002-11-07, 5:00 pm |
| Hey ANDRONDA, do u have aol IM... I want to contact you if I have anymore cisco related questions, I am not going to bug you too much....
You can email me at ybentrepreneur@aol.com and let me know your Aol IM if you have or I can email you with some (But very little ) Questions...
Thanks a lot. | |
| ANDRONDA 2002-11-07, 5:55 pm |
| I will be happy to help if I can.
I do not use IM. You will have to leave me a private message on this board or just post publically. I must warn yu that I try to help, but certainly am not an expert. | |
| vschristopher 2002-11-11, 9:50 am |
| again the experts have given the judgement, i prefer not to lecture once again on the subnetting  |
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