| Author |
Broadcast Addresses
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| Hello,
I am currently being taught subnetting by a MSCE. I take CCNA at night. I am almost certain he is incorrect about the broadcast addresses for the subnets. Does anyone have a site I can reference specifically were broadcasts numbers are listed next to their subnets? | |
| harbyma 2002-04-10, 12:44 pm |
| The broadcast address is the last address of the subnet eg 192.168.56.0 255.255.255.0 the broadcast would be 192.168.56.255 | |
| strikeattack 2002-04-10, 1:48 pm |
| quote:
The broadcast address is the last address of the subnet eg 192.168.56.0 255.255.255.0 the broadcast would be 192.168.56.255
More appropriately, the broadcast address is the address in which the host portion of the address is set to all binary 1s. The reason I point this out is because with bit-level subnetting, one of the octets might not have a value of 255. And yes, the example is correct.
Oh, and out of curiosity, I would like to know what a fellow MCSE is saying about the broadcast address... Would you tell us? | |
|
| It is the number right before the next subnet. For example if you have the following subnets 32,64,96 etc it should be like this:
subnet 32 64 96
First host 33 65
Last host 62 94
Broadcast 63 95
and so on, the following link might give you a better idea
http://www.certnotes.com/microsoft/...ySubnetting.htm
HTH | |
| strikeattack 2002-04-10, 3:37 pm |
| Hmm... I still have to think that my explanation is the clearest...
<slight sarcasm>
Someone else care to say the same thing with different words?
</slight sarcasm> | |
|
| You guys are the most awsome.
Thanks!!
My teacher was right I was backassward, but willing to change my ways.
Where I am confused again is about this:
given 10.15.17.22 /13
What is the last usable IP address for the first subnet.
I'm unclear but I believe the first usable IP maybe 10.8.0.1 and I'm very unclear on the last IP in the same subnet is something like 10.14.255.254 or 10.15....... | |
| harbyma 2002-04-10, 4:04 pm |
| If the host portion is all binary 1s is that not the last address of the subnet?
I was trying to keep it simple, like me. | |
|
| Unfortunately, I'm not simple, I strive to complicate most things. There is more truth to that then I care to admit.
Really I dont get it. | |
|
| so what is the answer here? I'm thinking 10.15.255.254 but I'm a bit confused too. | |
|
| I reposted the newer question | |
| MadChef 2002-04-10, 8:15 pm |
| quote:
Originally posted by strikeattack
Hmm... I still have to think that my explanation is the clearest...
And you'd be right. 
MadChef | |
| guitarjim 2002-04-11, 6:13 am |
|  Hmmm...
00001010.00001111.00010001.00010110
11111111.11111000.00000000.00000000
-----------------------------------
00001010.00001000.00000000.00000001
00001010.00001111.11111111.11111110
That works for me | |
| strikeattack 2002-04-11, 8:54 am |
| quote:
00001010.00001111.11111111.11111110
Guitarjim,
Look at the very last bit in the fourth line. A mistake? | |
|
| Ok Kid
"""given 10.15.17.22 /13
What is the last usable IP address for the first subnet. """
Now let's see this
That's having 5 bits in the mask for subnetting
Fo the first Subnet is :
10.0.0.0 -- 10.31.255.255
** Here the first usable Host IP address is: 10.0.0.1
The last subnet is
10.224.0.0 -- 10.255.255.255
Here the last usable Ip address 10.255.255.254
Cheers
LordV | |
| guitarjim 2002-04-11, 11:33 am |
| No. That is the last usable host for the first subnet. The broadcast would be all ones in the 4th octet. | |
| strikeattack 2002-04-11, 11:46 am |
| Given the title of this thread, I thought that you were showing the broadcast address. | |
| adam salam 2002-04-11, 11:49 am |
| 
For ip address 10.15.17.22/13
Here 13 mains 13 bits for subnetting that's 255.248.0.0 witch is mains
11111111.11111000.00000000.00000000
Here we used all bits in first octet and only 5 bits in the second
octet, we have no thing to do with the first octet now if we analyze
second octet we got this:
128, 64, 32, 16, 8 and here the most important thing the last bit we
used is "8" so our subnet going 8 by 8 that mains first subnet should
be 8 the second is 16 the third is 24 and so on.
if we apply this we got our subnets like that:
first subnet 10.8.0.0
second subnet 10.16.0.0
third subnet 10.24.0.0
forth subnet 10.32.0.0
-and so on
-
and here's the answer to your Q:
first subnet 10.8.0.0
first host 10.8.0.1
- 10.8.0.2
- 10.8.0.3
- blah
- blah
- blah
- 10.8.255.254
- 10.8.255.255
- 10.9.0.0
- 10.9.0.1
- 10.9.0.2
- blah
- blah
- blah
- 10.15.255.252
- 10.15.255.253
last host 10.15.255.254
broadcast 10.15.255.255
and now start the second subnet
second subnet 10.16.0.0
if you follow the process you end with the last subnet:
last subnet 10.240.0.0
first host 10.240.0.1
last host 10.247.255.254
& broadcast 10.247.255.255
now you reach the limit for the nodes
255.248.0.0
and you can't go ferther
I hope that can help | |
|
| adam salam, Well done. Explained in detail perfectly! | |
| dmaftei 2002-04-11, 12:28 pm |
| quote:
Originally posted by guitarjim
The broadcast would be all ones in the 4th octet.
So, would 192.168.0.127/25 (the 4th octet is 01111111) be the broadcast address for subnet 192.168.0.0/25, or not? | |
| adam salam 2002-04-11, 1:17 pm |
| dmaftei
you can't use 192.168.0.0 as a subnet because it's all 0's this is only a network address and can't be used.
any way for 192.168.0.0/25 the first and the only subnet well be 192.168.0.128
please reply | |
|
| adam salam's got it right but nearly
quote:
last subnet 10.240.0.0
first host 10.240.0.1
last host 10.247.255.254
& broadcast 10.247.255.255
now you reach the limit for the nodes
255.248.0.0
and you can't go ferther
No the last subnet here is
10.248.0.0
First Host here is 10.248.0.1
Last host is : 10.255.255.254
And Broadcast is 10.255.255.255
dmaftei
quote:
So, would 192.168.0.127/25 (the 4th octet is 01111111) be the broadcast address for subnet 192.168.0.0/25, or not?
Hmmmm the boradcast address for the first subnet here is
192.168.0.127
Network Address for the first is
192.168.0.0
(rest in between are usable hosts from 1 to 126)
Second subnet is
192.168.0.128 --to-- 192.168.0.255
LordV | |
| marathoner 2002-04-11, 1:24 pm |
| quote:
Originally posted by drc1
You guys are the most awsome.
Thanks!!
My teacher was right I was backassward, but willing to change my ways.
Where I am confused again is about this:
given 10.15.17.22 /13
What is the last usable IP address for the first subnet.
I'm unclear but I believe the first usable IP maybe 10.8.0.1 and I'm very unclear on the last IP in the same subnet is something like 10.14.255.254 or 10.15.......
--------
OK I'll take a crack at this. The key is to understand what Cisco means by subnet. As I understand it, they mean "whatever portion of host address space from the classic A, B, C scheme you have reallocated to make network space that you administer internally. " 10.x.x.x is class A, (normally /8 ) You have decided to take over 5 more bits (/13) so you can have 2 ^ 5 = 32 subnets. <b>Both 00000 and 11111 are valid subnets -- reserving the all zero and all-1 for special purposes applies to HOSTS not subnets </b> [See Odom p. 266] If we add the lower three bits as zeros, this octet can have values from 0 [00000000] to 248 [11111000]
The subnets would go as follows:
10.0.0.0 (0 lowest)
10.8.0.0 (1)
10.16.0.0 (2)
10.24.0.0 (3)
10.32.0.0 (4)
10.40.0.0 (5)
...<br>
10.240.0.0 (30)
10.248.0.0 (31 highest)
See the pattern?
I would say, therefore, that
10.0.0.1 is the FIRST usable IP address on the FIRST subnet
10.7.255.254 is the LAST usable IP address on the FIRST subnet
the example given is on the SECOND subnet, which is 10.8.0.0
10.248.0.1 is the FIRST usable IP on the LAST subnet
10.255.255.254 is the LAST usable IP on the LAST subnet.
Please if I am wrong correct me, because if this thinking is not right I am in DEEP trouble!!
 | |
| adam salam 2002-04-11, 1:49 pm |
| quote:
Originally posted by LordV
adam salam's got it right but nearly
No the last subnet here is
10.248.0.0
First Host here is 10.248.0.1
Last host is : 10.255.255.254
And Broadcast is 10.255.255.255
LordV
you can't go further after 10.248.0.0
this is the whole network broadcast address
the 10.255.255.255 is the default broadcast address but here you have a limit by 13 bits
pleas verify
 | |
| adam salam 2002-04-11, 2:03 pm |
| <B>marathoner</B>
one thing I like to explain is that you don't have to use the first ip address witch is the whole NETWORK address so the first valid subnet is 10.8.0.0
and go from here forth ..... >>>>>>>>>>>> | |
|
| Yes but the 13th bit is 248.. so we have that subnet
see
0 is the first subnet
8 is the second (8x2=16-8 = 8)
...
...
...
...
80 is the 11th subnet (8x11=88-8=80)
...
...
...
...
So what is the 32nd subnet??
(8x32=256-8=248)
similarly how many hosts? say just for 2nd ocklet is 8 per subnet
i.e 8X32=256 so if you do not take the hosts from 248 to 255 then you lose exponentially large number of hosts this being a class A address ....
I hope you understand
Think
LordV
Per subnet = | |
| adam salam 2002-04-11, 3:02 pm |
| LordV
Let’s take it easy
To calculate the available subnets or hosts there is the formulae:
2n(2 to the power of n)-2= the number of available address
n= number of bits
the number of subnets bits we can play with here is 5 bits : 248 = 11111000
now lets substitute 25(2 to the power of 5)-2= 30 subnet address
witch are : 8, 16, 24, 32, 40, 48, 56, 64,72, 80,
88, 96, 104, 112,120, 128,136, 144, 152, 160,
168, 176, 184, 192, 200, 208, 216, 224, 232, and 240 the last valid subnet witch cast is 247
now if you count you got exactly 30 subnets
if you have no idea about this formulae you should know it because it ease your work.
and that’s it. | |
|
| kid what u fail to realise is that the formule you quoted is true for hosts when subnetting and not subnets...
That was the old concept when some the routers/ network hardware did not support all ones and all zero subnets.
LordV | |
| marathoner 2002-04-11, 8:38 pm |
| quote:
Originally posted by LordV
kid what u fail to realise is that the formule you quoted is true for hosts when subnetting and not subnets...
That was the old concept when some the routers/ network hardware did not support all ones and all zero subnets.
I'm with LordV here, ditto Odom. The point of contention is [B]what is the first subnet? which keys off the standard what is the maximum # of subnets you can have? A lot of study material that say if you have n subnet bits you can have 2^n-2 subnets and if you have m host bits you can have 2^m-2 hosts. I believe they are wrong about the subnets.
Odom says on p. 266 that you no longer subtract 2 from 2^n to get the number of subnets, i.e. a subnet of all 0's or all 1's is fine. LordV agrees -- he/she just passed the exam. (Congrats.)
I have never seen any one in the real world implement a subnet of 0's, probably due to having been taught that your first subnet always begins with 1. In the real world I would probably not implement a subnet of all 0's or 1's either just because it would confuse people. However right now I want to know what the Cisco exam setters think.
 | |
| dmaftei 2002-04-11, 8:44 pm |
| quote:
I have never seen any one in the real world implement a subnet of 0's...
That's why there's a shortage of IP addresses nowadays... [grin]
Seriously now... Last time I checked the CCNA exam wanted you to drop the first and last subnets, so I guess you can use the (in)famous 2^2-2 for subnets too... | |
| LordV 2002-04-11, 11:46 pm |
| I checked ICND and pg 236 has a chart and it seems that Cisco does indeed want you to subtract 2 bits from the total usable set.. But what the book does not specify is if the decuction if for the all ones and the all zero subnet (must be).
But could someone check some other book for CCNA and revert the same. with Page number and details would appreciate.
(I may seem that CCNP or other Ciso courses may have explained the comcept of using the all ones and all zero subnets with the use of higher level Routers/Switches. I will check and keep you informed)
LordV | |
| guitarjim 2002-04-12, 11:13 am |
| In regards to how many subnets:
CCNA exam does not include ip subnet-zero.
In CCNP you can use subnet zero. I don't beleive that cisco recommends using subnet broadcast (all ones in subnet). | |
| MadChef 2002-04-12, 7:04 pm |
| quote:
Originally posted by marathoner
I have never seen any one in the real world implement a subnet of 0's,
I make a regular practice of doing so. I nearly always use the zero subnet for my internetwork links. I also like to use the all ones subnet for my loopbacks. Just like Yankee said, everyone should be so lucky to have my networks.
MadChef | |
| Clangashe 2002-04-15, 8:25 am |
| Just for a bit of fun.
MCT use a different method for subnetting than CCAI. They are both correct methods, it is up to you to find what suits you.
Adam_Salam gave a good version of the break down.
Or Take a look at this thread
http://www.examnotes.net/forums/sho...&threadid=26339
Hope it sorts things out a little bit | |
| tvyoda 2002-04-15, 12:36 pm |
| quote:
Originally posted by drc1
You guys are the most awsome.
Thanks!!
My teacher was right I was backassward, but willing to change my ways.
Where I am confused again is about this:
given 10.15.17.22 /13
What is the last usable IP address for the first subnet.
I'm unclear but I believe the first usable IP maybe 10.8.0.1 and I'm very unclear on the last IP in the same subnet is something like 10.14.255.254 or 10.15.......
Hi,
10.15.17.22 /13 is really a class A network Sub-netted.
Orignal class A subnet mask is 255.0.0.0 i.e 8 bits (or /8)
/13 suggests next 5 bits (13-8 = 5)are now also part of the subnet. The new subnet mask is 255.248.0.0.
Part of the second octet (from the left) i.e 5 bits now are network bits and the last 3 bits are now host bits.
The first network is therefore 10.8.0.0
The second network is 10.16.0.0
The broadcast address for first network is 10.15.255.255.
The last usable host address is one less i.e 10.15.255.254
Hope this helps | |
| GilGraber 2002-04-15, 1:05 pm |
| 1st) 192.168.0.0 - 192.168.255.255 is a private address range.
2nd) using bit mask of /25 is sugesting that you are using VLSM (Variable Lenght Subnet Mask) and thus studying MCSE, since CCNA does not touch VLSM. CCNA only covers Classfull addressing and routing with RIPv1 and IGRP. CCNP is when VLSM is deffined and implemented with EIGRP, OSPF (somehow cisco forget to look at RIPv2 which supports VLSM, so make sure you learn it at one point).
Say you are using VLSM on your net and you are using Private IP address 192.168.0.0/24.
And you need 1 subnet for 100 hosts and 2 subnets for 60 hosts each and have only one class C block of addresses, You will need to deploy VLSM and CIRD (Classless Inter-Domain Routing) supporting protocols like RIPv2,EIGRP,OSPF,IS-IS. Here is a breake down of this problem:
major net. addr.: 192.168.0.0/24
1st create the largest 2 subnets:
192.168.0.0/25 = 192.168.0.0/25
192.168.0.128/25
(leave subnet 192.168.0.0/25 alone for 1 subnet of 100 hosts. This will give you:
1st usable host 192.168.0.1/25
last usable host 192.168.0.126/25
subnet broadcast 192.168.0.127/25)
2nd breake down the second subnet into two:
192.168.0.128/26 = 192.168.0.128/26
192.168.0.192/26
subnet address 192.198.0.128/26
1st usable host 192.168.0.129/26
last usable host 192.168.0.190/26
subnet broadcast 192.168.0.191/26
subnet address 192.168.0.192/26
1st usable host 192.168.0.193/26
last usable host 192.168.0.254/26
subnet broadcast 192.168.0.255/26
(the problem here is that all network broadcast is also a last subnet directed broadcast address, you must make sure that you do not need to use directed broadcasts between subnets and/or disable or filter them. Cisco routers will not propagate all net broadcasts and directed broadcasts are disabled by default on latest IOS.)
Nothing wrong here any CIDR protocol will recognize it and use it. To implement it you will use "ip subnet-zero" command in global config mode on cisco router. Each VLSM protocol will have different way for you to to control the advertisements of these subnets and thus you are required to know these protocols inside out, othervise you can loose connectivity to some of your subnets. | |
| Carebear 2002-04-16, 9:22 am |
| It's really quite simple...if you draw it out on paper...you've figured out that the network ID is using the first 13 bits which takes you to the fifth bit in the second octet...the one that has a value of 8...the remaining 19 bits signify your host bits...in order to calculate the last available IP, you have to turn all the host bits on (value of 1) except for the last one (all host bits cannot be on or off at the same time)...so, your second octet value for network is 8 and with the host bits turned on, you have 8+4+2+1 giving a value in the second octet of 15...the third octet, all bits are on, giving the value of 255...and finaly the last octet, all bits are on except for the last one, giving a value of 254...so your final IP in the first network is 10.15.255.254...draw it out and it becomes much clearer...good luck | |
| thonguyen 2002-04-16, 1:07 pm |
| Originally posted by drc1
You guys are the most awsome.
............................................
Where I am confused again is about this:
given 10.15.17.22 /13
What is the last usable IP address for the first subnet.
.............................................
Hi guys,
The 1st usable ip address of the 1st subnet is 10.8.0.1. The last ip of the 1st subnet is 10.15.255.254. The broadcast ip address of the 1st subnet is 10.15.255.255.
Here is how to.
10.15.17.22 /13 is class A with the subnet mask host bits of 13. This will give us a subnet mask of 255.248.0.0. Therefore, we know that the 1st octet of a given ip address, in this case it's 10, will always be the same. The KEY here is the 248, the 2nd octet of the subnet mask. This is where is all begins.
Let's draw a chart for 248.
248 in binary: 11111000
Subnet mask: N/A 192 224 240 248 252 254 255.
Beginning range os network id: N/A 64 32 16 8 4 2 1
Number of subnets: N/A 2 6 14 30 62 126 254
By looking at the above chart, 248 will have the 1st network ip address beginning with 8; 2nd network ip address beginning with 16(multiple of 8); 3rd network ip address beginning with 24 and so on........and it will have total of 30 subnets. Why 30? Using 2^n-2, we will have 2^5-2=30. Now let's put them in order.
1st subnet ip address: 10.8.0.0
1st usable ip address of 1st subnet: 10.8.0.1
2nd usable ip address of 1st subnet: 10.8.0.2...and so on.....
The last ip address of 1st subnet: 10.15.255.254
The broadcast ip address of 1st subnet: 10.15.255.255
2nd subnet ip address: 10.16.0.0
1st usable ip address of 2nd subnet: 10.16.0.1
2nd usable ip address of 2nd subnet: 10.16.0.2....and so on.....
The last ip address of 2nd subnet: 10.16.255.254
The broadcast ip address of 2nd subnet: 10.16.255.255
And on and on we go........until........
The last(30th)subnet ip address: 10.240.0.0. Why 240? 8(range)x30(subnets)= 240.
1st ip address of last subnet: 10.240.0.1
2nd ip address of last subnet: 10.240.0.2....and so on........
The last ip address of the last subnet: 10.240.255.254
The broadcast ip address of last subnet: 10.240.255.255
There you go. Hope this helps.
TN | |
| pshalley 2002-04-16, 3:32 pm |
| ok matey, subnetting is not straight and forward, you can start going into variable lenght subnetting also, just to confuse you a little also. However remember that a broadcast address is required and also a network address is required for every subnet that you have. when you see the /30 or /24 etc I always use the following which may help you. if you use 24, thats 8 8 8 bits which is relevant to 255.255.255.0 and if you use /30 that has 6 bits left over which is 128 + 64 + 32 + 16 + 8 + 4 which is 252
which only allows you to have a mask of 255.255.255.252 that will give you one network address and only one broadcast address, although it does give you 4 ip address only 2 are useable. But then if your using RIP it does not support VLSM, so you have to think what your going to be using the routing protocol. Well mate hopefully i have not confused you, the thing to remember which is most important before you subnet, how many ip addresses are required and is there any future growth. and subnet accordingly, never just subnet for the sake of it, use your ip numbers sparingly as a bad plan will create extra work at a later stage. On your links use /30 or ipunumbered will free up some addresses for you, but use ipunumbered sparingly, if you want to use snmp etc on the interface
well thats if from me mate, | |
| Wookieasd 2002-04-22, 10:20 am |
|
1) Go www.download.com
2) search with word "subnet" in the top right corner
3) Download IP Subnet Calculator Ver 3.0 (400Kb)
It's all Free
4) Check out how many donkeys are around : )
5) See the REAL solution I'm too lazy to explain ; )
Ciao | |
| Wookieasd 2002-04-22, 10:46 am |
| Sorry I can't attach the file as 400Kb seems to be too much to attach here . . . | |
| va1encia 2002-04-27, 1:44 pm |
| I don't mean to be a pest but....
All page 4 submits have restated previous posts. There have been several great explanation of the same thing.
Although there are some very talented people who are posting great info. We must use what I like to call K.I.S.S. method.
1) This is the CCNA forum (no VLSM). (strikeattack and guitarjim)
2) Original post requested answer from teching in open forum. (included VLSM)(GilGraber)
3) Both parties whom explained the use of subnetting and supernetting are correct but on the CCNA test NO VLSM, NO all 1's or all 0's can be included in subnetting (for this test).
4) Beyond CCNA VLSM allows all 1's and all 0's
By the way, KISS method is Keep It Simple Stupid method for those not so informed.
Awesome info from a wealth of information. Thanks for the added info from all. |
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