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| borsky 2003-02-13, 11:50 am |
| I also found two incorrect answers about subnetting one of wich has been corrected. The other one is this:
Question A24: you plan to set up 4 subnets with up to 20 host on each. You'll use 192.168.0.0/24 private addr. Which subnet mask should you assign?
a. 255.255.255.192
b. 255.255.255.224
What is your suggestion? And why? | |
| Slinky 2003-02-13, 9:30 pm |
| I would to B, because 2^3-2=6 subnets which meets the minimum requirement of 4. | |
| borsky 2003-02-14, 3:54 am |
| I would choose a. because 2^2=4.
(When calculating subnets you don't subtract 2 only when claculating hosts.)
You need 2 bits from the 4. octet for creating 4 subnets.
The subnets:
192.168.0.0/26 (00)
192.168.0.64/26 (01)
192.168.0.128/26 (10)
192.168.0.192/26 (11)
Each subnet gives 62 hosts
The Transcender's answer was b. by the way
which is wrong IMHO. They probably calculated the number of hosts first and then the subnets. | |
| Slinky 2003-02-14, 10:31 am |
| Microsoft doesn't want you to consider the all 0s and all 1s subnets as valid, even though they are valid in the real networking world. So therefore you will need to subtract two from the number of subnets in order to get it right on the exam. | |
| borsky 2003-02-14, 11:18 am |
| I have done some exams and never seen what you are talking about. If you can create 4 subnets in reality from 2 bits then it must be valid in the exams too.
When you calculate subnets you don't subtract two only when calculate hosts.
All 0's in the host portion is the (sub)network address, 255 is the (sub)network loopback address, and 256=all 1's is the next (sub)network address. That's why you have to subtract 2. | |
| borsky 2003-02-14, 11:27 am |
| Almost forgot: there are no all 0's and all 1's subnets.
Or if you consider 192.168.1.0 as all zero than yes there are, but no need subtracting when creating subnets. | |
| Slinky 2003-02-14, 11:33 am |
| I suggest you review Freak's subnetting guide for more information. I'm not just making this stuff up for the sake of argument.  | |
| borsky 2003-02-14, 11:59 am |
| you actually did give any reason why I am wrong.
I give you another hint: you need only two subnets. According to your logic 2^1=2-2=0.
According to my logic 2^1=2 i.e you can create 2 subnets with two bits, I mean I can.
Can you? | |
| Slinky 2003-02-14, 12:10 pm |
| Sure you can create two subnets using 2 bits, but you cannot create 2 subnets using 1 bit. For example, 192.168.1.0 with a mask of 255.255.255.128. Available subnets are 192.168.1.0 and 192.168.1.128. The first one contains all 0s and is the network ID and the second one contains all 1s and is the broadcast ID. | |
| Slinky 2003-02-14, 12:22 pm |
| RFC 950 originally defined the all 0s and all 1s subents as invalid, because they could be confused between classful and classless routing. However, since routers can now support both types of routing there is no need to eliminate 2 of those subnets.
I'm pretty sure that M$ still wants you to subtract those 2 subnets just in case you administer an environment where CIDR is not supported. | |
| borsky 2003-02-14, 12:30 pm |
| No.
they are valid subnets and they don't hold all 1's and 0's.
the network portion of 192.168.1.0/24 is: 192.168.1 i.e 11000000 10101000 00000000
As you can see the network portion of the address has a number of 1's. Therefore whenever calculating networks subnetworks don't subtract. As for the host portion of this example the first available address is 00000001
In my previous thred I meant that I can create 2 subnets from 1 bit. | |
| borsky 2003-02-14, 12:32 pm |
| No.
they are valid subnets and they don't hold all 1's and 0's.
the network portion of 192.168.1.0/24 is: 192.168.1 i.e 11000000 10101000 00000001
As you can see the network portion of the address has a number of 1's. Therefore whenever calculating networks subnetworks don't subtract. As for the host portion of this example the first available address is 00000001
In my previous thred I meant that I can create 2 subnets from 1 bit. | |
| Slinky 2003-02-14, 12:52 pm |
| A class C address with two bits borrowed and an all 0s address will look like 11000000.10101000.00000001.00000000, and an address with all 1s will look like 11000000.10101000.00000001.11000000
If you borrow three bits then the all 1s will look like 11000000.10101000.00000001.11100000
Remember we are talking about the octet that we borrowed the bits from, and not the other stuff before that. | |
| borsky 2003-02-14, 12:56 pm |
| Ok, now I know what the misunderstanding is.
When you calculate the total number of subnets e.g. for an octet (it can;t hold all 0's and all 1's indeed) than you indeed have to subtract 2. But when you calculate to sub-divide a network segment and create e.g. 4 subnets then you don't have to take 2 from it. | |
| borsky 2003-02-14, 1:15 pm |
| An octet cannot look like this:
00000000 or 11111111
neither in the network nor in the host portion.
So when we calculate the total number of host or subnet for a network then have to subtruct 2. But because you don't calculate subnets without host portion you don't have to subtract.
See the original example:
You need 4 subnets and 20 hosts and you have
192.168.0.0 address. Now the total number of host is equal the total number of subnets that you can have which is 2^8-2=254. But because you don't have subnets without host therefore you never have to use the formula of 2^n-2 when calculating subnets.
4 subnets reguire 2 bits and the rest of 6 bits can give 2^6-2=62 hosts. | |
| borsky 2003-02-14, 7:58 pm |
| I saw that I was talking some rubbish too.
Obviously octets can hold all 0's and 1's
eg 192.255.0.1 is a valid IP address.
Also - now I checked - you never have to subtract 2 when calculating subnets.
There is a big misunderstanding about this and I noticed even some books are incorrect about it. Here are my reasons:
It is clear why we have to subtract 2 when calculating host numbers. It is because the host ID can't have all 0's and 1's. All 0's on the host ID equals the network ID (Net ID:192.168.1.0/24 where the first possible host ID is 192.168.1.1) and all 1's on the host ID is the loopback address of the (sub)network.
Now please tell why should you subtract to when calculating the network ID?
E.g you need 2 subnets, you borrow 1 bit. 192.168.0.0 is the Network addr. Two subnets will be: 192.168.0.0/25 (where the first host is 192.168.0.1 the last is 0.127) and 192.168.0.128/25 (where the first host is 192.168.0.129 and the last is 0.255)
Now if you can give me any satisfactory reason - apart from your book - why I am wrong I will be grateful.
By the wy if you check 70-216 Transcender question C20 it has a good explanation.
I have done 216 and spent some time with subnetting. Now I realized that sources are not always reliable including Transcender. | |
| borsky 2003-02-14, 8:12 pm |
| Final proof:
Check Microsoft:
http://support.microsoft.com/defaul...kb;EN-US;164015
Extract from the page:
Subnetting
A Class A, B, or C TCP/IP network can be further divided, or subnetted, by a system administrator. This becomes necessary as you reconcile the logical address scheme of the Internet (the abstract world of IP addresses and subnets) with the physical networks in use by the real world.
A system administrator who is allocated a block of IP addresses may be administering networks that are not organized in a way that easily fits these addresses. For example, you have a wide area network with 150 hosts on three networks (in different cities) that are connected by a TCP/IP router. Each of these three networks has 50 hosts. You are allocated the class C network 192.168.123.0. (For illustration, this address is actually from a range that is not allocated on the Internet.) This means that you can use the addresses 192.168.123.1 to 192.168.123.254 for your 150 hosts.
Two addresses that cannot be used in your example are 192.168.123.0 and 192.168.123.255 because binary addresses with a host portion of all ones and all zeros are invalid. The zero address is invalid because it is used to specify a network without specifying a host. The 255 address (in binary notation, a host address of all ones) is used to broadcast a message to every host on a network. Just remember that the first and last address in any network or subnet cannot be assigned to any individual host.
You should now be able to give IP addresses to 254 hosts. This works fine if all 150 computers are on a single network. However, your 150 computers are on three separate physical networks. Instead of requesting more address blocks for each network, you divide your network into subnets that enable you to use one block of addresses on multiple physical networks.
In this case, you divide your network into four subnets by using a subnet mask that makes the network address larger and the possible range of host addresses smaller. In other words, you are 'borrowing' some of the bits usually used for the host address, and using them for the network portion of the address. The subnet mask 255.255.255.192 gives you four networks of 62 hosts each. This works because in binary notation, 255.255.255.192 is the same as 1111111.11111111.1111111.11000000. The first two digits of the last octet become network addresses, so you get the additional networks 00000000 (0), 01000000 (64), 10000000 (128) and 11000000 (192). (Some administrators will only use two of the subnetworks using 255.255.255.192 as a subnet mask. For more information on this topic, see rfc 1878.) In these four networks, the last 6 binary digits can be used for host addresses.
Using a subnet mask of 255.255.255.192, your 192.168.123.0 network then becomes the four networks 192.168.123.0, 192.168.123.64, 192.168.123.128 and 192.168.123.192. These four networks would have as valid host addresses:
192.168.123.1-62
192.168.123.65-126
192.168.123.129-190
192.168.123.193-254
Remember, again, that binary host addresses with all ones or all zeros are invalid, so you cannot use addresses with the last octet of 0, 63, 64, 127, 128, 191, 192, or 255.
You can see how this works by looking at two host addresses, 192.168.123.71 and 192.168.123.133. If you used the default Class C subnet mask of 255.255.255.0, both addresses are on the 192.168.123.0 network. However, if you use the subnet mask of 255.255.255.192, they are on different networks; 192.168.123.71 is on the 192.168.123.64 network, 192.168.123.133 is on the 192.168.123.128 network.
Good Luck! | |
| The Light 2003-02-17, 5:34 am |
| I found the following text in EasyCert:
"Historically, subnets composed of all ones or all zeros were reserved. Therefore the minimum number of subnet bits allowed was two, which allotted for two usable subnets (01 and 10 in binary, subnets 1 and 2 in decimal) and two reserved subnet addresses (00 and 11 in binary, subnets 0 and 3 in decimal). However, rfc 1878 abolished this practice because the reserved subnets were generally not being used for any special purpose and was considered wasteful. Thus the minimum number of subnet bits is now one, allowing for two usable subnets (0 and 1 in binary, and decimal)."
Don't forget that in MS world, we're talking about CIDR. So, the formulas you should use in MS exams are:
2^x = number of subnets.
2^y-2 = number of valid hosts.
However, if you taking a Cisco exam, you should use the old method:
2^x-2 = valid subnets.
2^y-2 = valid hosts. | |
| borsky 2003-02-17, 5:48 am |
| Yes, thank you, it confirms what I have learnt.
I also referred to RFC1878 in this argument, - or was it another newsgroup? -
and recently found more info about this in the network infrastructure design manual (70-221) | |
| Teratoma 2003-02-18, 1:31 pm |
| Remember that we are talking about a M$ exam here. Everyone can quote rfc and anyone else for that matter, but what really counts is what M$ says...after all it is their exam. Review the MOC or Tech Net. I happen to have a copy of the MOC in my lap right now...Ch. 2 p. 35...IP Address Guidelines...
bullet #2 "The network ID and host ID cannot be '1's. If all bits are set to 1, the address is interpreted as a broadcast rather than a host ID."
bullet #3 "The network ID and host ID cannot be '0's. If all bits are set to 0, the address is interpreted to mean 'this network only."
I apologize for the length of the post, but I think this answers the question of whether or not to subtract 2 from the number of calculated networks. As far as M$ is concerned the answer is YES.
Remember...It's the Microsoft way or the highway!!
btw...excellent discussion, keep up the good posts!!! | |
| borsky 2003-02-18, 4:00 pm |
| I am afraid it is the other way round.
Precisely in MS exams you DON'T SUBTRACT!
Traditionally you had to subtract 2 when calculating subnets, as far as I know that is the Cisco way. But with Windows 2000 that uses OSPF and RIP, the network ID's can use all 1's and all 0's but not host ID's
Check Microsoft knowledge base, or designing network infrastracture (you can find it one of my previous thred in this topic) | |
| me? I dunno... 2003-02-18, 5:54 pm |
| In addition to what Teratoma and Slinky posted...
"this thread is turning into a bunch of 0's"
 | |
| borsky 2003-02-19, 4:59 am |
| This is a MS exam. I cannot add anymore to this argument.
Microsoft sources are clear about calculating subnets.
If on the exam there are questions about subnetting you have to make the decision.
Also you have to decide which sources you consider authentic. | |
| The Light 2003-02-19, 5:27 am |
| In many questions for 70-216 the addresses are given in a CIDR (Classless Inter-Domain Routing) format. With CIDR all "0" and all "1" networks are allowed. Also, there is a question about "rote sumarization" which is a CIDR term.
http://public.pacbell.net/dedicated/cidr.html
There is a difference between the two standards. So, for which one we're talking about? | |
| borsky 2003-02-19, 5:36 am |
| We are talkinabout the new standard.
The traditional way of calculating subnet did not allow all 1' and 0's in subnet ID's.
Cisco routers eg.
The latest protocols (OSPF, RIP)allow all 1's and 0's in subnet ID's but not in host ID's. | |
| Teratoma 2003-02-19, 12:05 pm |
| Referring back to the original question...there was no mention of which standard is being used or if communication with a classfull network is required. Using 2^x-2 will ALWAYS work regardless. 2^x may or may not work in every situation. Eliminating the first and last subnet is a safe bet.
borsky - you are right with what you are saying...it can be done. I know I can do it in a lab or test domain. Try to do it in the real world without losing connectivity.
More references from the Windows 2000 Server Resource Kit.
http://www.microsoft.com/windows200...bb_tcp_rlgr.asp
***RFC 950 forbade the use of the subnetted network IDs where the bits being used for subnetting are set to all 0's (the all-zeros subnet) and all 1's (the all-ones subnet). The all-zeros subnet caused problems for early routing protocols and the all-ones subnet conflicts with a special broadcast address called the all-subnets directed broadcast address.
However, rfc 1812 now permits the use of the all-zeros and all-ones subnets in a CIDR-compliant environment. CIDR-compliant environments use modern routing protocols that do not have a problem with the all-zeros subnet and the all-subnets directed broadcast is no longer relevant.
The all-zeros and all-ones subnets may cause problems for hosts or routers operating in a classful mode. Before you use the all-zeros and all-ones subnets, verify that they are supported by your hosts and routers. Windows 2000 and Windows NT support the use of the all-zeros and all-ones subnets.*** | |
| jwulf2000 2003-02-19, 6:13 pm |
| While you two are having your "discussion" just want to say thanks for the link on subnetting. Have always gotten confused over the whole issue. I let you guys get back to business.......... | |
| The Light 2003-02-20, 4:34 am |
| Sure, you can't use all 1's or all 0's for host ID as you still need to have a broadcast address for the subnet.
quote:
Originally posted by borsky
We are talkinabout the new standard.
The traditional way of calculating subnet did not allow all 1' and 0's in subnet ID's.
Cisco routers eg.
The latest protocols (OSPF, RIP)allow all 1's and 0's in subnet ID's but not in host ID's.
| |
| Teratoma 2003-02-20, 9:13 am |
| quote:
Originally posted by jwulf2000
While you two are having your "discussion" just want to say thanks for the link on subnetting. Have always gotten confused over the whole issue. I let you guys get back to business..........
That's the whole point of this, I think. I certainly have picked up more knowledge by getting involved. Everyone has given excellent input. This is certainly going to help me slay the beast. |
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