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Back | Assign 1 | Assign 2  | Assign 3  | All  | [CNDS] 

Background

Duration: 1 hour

Format:

Section A: Multiple choice (based on material from Unit 5/6).
Section B: Fixed format answers.

Material covered:

Unit 5 (Network/IP)
Unit 6 (Transport/TCP)
Challenge 3: Network Configuration.
Challenge 4: Sockets and proxies.
Research 3: IP v6.

Ask a question from Unit5/6

If you want to ask a question, please complete the following:

Name:

Email:

Matriculation number:

Programme:

Question


Alternatively if you have MSN Messenger you can ask a question at:

cndsatnapier@hotmail.com.

Preparation

IP general:
IP classification.
Binary-to-dotted notation.
IP allocation.

IP/subnetting:
Subnet masks.
Subnet mask (255 subnet mask).
Subnet mask (240 subnet mask).
Number of subnets and hosts (for a given subnet mask).
Defining a subnet mask (for a required number of subnets).
Defining a subnet mask (for a required number of hosts).

TCP related:
TCP ports.

Just added ......:
Who Wants A Test for Test 2?.

 

Questions asked

Hi Bill Is the 4th Challenge, (sockets and proxies), part of the second test, (Ip & TCP), or is it part of a later test.

Tue 26/11/2002 11:19 AM

Answer: It's not as important as Challenge 3, but the concept is important so I would study it.

 

Hi Bill, For the test No 2 to undertake in december. Could you please tell me which sections of the Teaching Pack are going to be included? Are section 1 and 2 to be revised again??

Mon 25/11/2002 2:56 PM

Answer: It's only Unit 5 and Unit6, so ignore all other units.

 

MSN Messenger convseration:

" ... A good site for leaning to subnet is at www.learntosubnet.com. Please chould you forward it to everyone. Thanks."

Wed 27/11/2002 11am

Answer: Thanks

 

Hi Bill,

Just need a few points cleared up for the exam next Thursday.

Will we be allowed a calculator?
If I use scrap paper to do working out on will I need to hand that in as well?

Thu 11/28/2002 11:41 AM

Answer: Yes. You are allowed to use a calculator, and you hand in your extra work, but it goes in the bucket.

 

Can you help? What is the answer to this:

Q5.17.14 If a network has Class B IP address with 10 subnets, what will the subnet mask be?

Answer:

A Class B address has a default subnet mask of 255.255.0.0.

If we borrow one bit we can have 2 subnets (of which none would be useable).
If we borrow two bits we can have 4 subnets (of which 2 will be useable).
If we borrow three bits we can have 8 subnets (of which 6 will be useable).
If we borrow four bits we can have 16 subnets (of which 14 are useable).

This is enough. So we need to borrow 4 bits, thus the subnet mask will be:

255.255.1111 0000b. 0000 0000b

which is:

255.255.240.0

 

If a network has class C IP address with 50 subnets would the same formula that you used for the class B 10 subnets be used. i.e. in this case borrow 6 bits (of which 62 are usable)giving a subnet mask of 255.255.255.252?

Thu 11/28/2002

Answer:

Yup. A Class C address has a default subnet mask of 255.255.255.0.

If we borrow one bit we can have 2 subnets (of which none would be useable).
If we borrow two bits we can have 4 subnets (of which 2 will be useable).
If we borrow three bits we can have 8 subnets (of which 6 will be useable).
If we borrow four bits we can have 16 subnets (of which 14 are useable).
If we borrow five bits we can have 32 subnets (of which 30 are useable).
If we borrow six bits we can have 64 subnets (of which 62 are useable).

This is enough. So we need to borrow 6 bits, thus the subnet mask will be:

255.255.255.1111 1100b

which is:

255.255.255.252

 

Hi Bill,

Just a quick email concerning challenge 3.

At position B1 the IP address is 146.176.160.0, however at position G1 the IP address is the same.

Is this an error?

Answer: The actually address is 146.166.166.0 which is wrong as the whole network is 146.176.0.0, thus the 166 in the second field is wrong.

 

Hi Bill, Regarding challenge 4, Q3 asks about additional security checks for the proxy software. Could you give me a clue?

November 28, 2002 6:30 PM

Answer: Oh. There's lots, such as:

The only device that can talk to the other device is the one that uses a specific address, on a specific port (a bit like a firewall).

We could also provide a user ID and password for the proxy server.

We could encrypt the data over each of the links, such as creating public keys for each of the links.

and so on.

 

When calculating the subnet masks have we to presume that RFC 1812 is being used?

Answer: Good question. RFC1812 defines IPv4, so assume that we are using this specification.

 

When two computers set-up a connection using TCP is there 3 or 4 transmissions taking place. (would 4 be correct?)

Answer: Well. At the most it can be 4, but can be reduced to 3. So it'll depend on the context of the question.

 

Ip address is 61.x.y.z subnet mask is 255.255.240.0 Is the 1st subnet address 61.0.16.0 or 61.20.16.0 if answer is 61.20.16.0, could you tell me how to work out it? i think the answer is 61.0.16.0 but in your ip test2 flash the answer is 61.20.16.0 i am confused

Answer: It must just be a typo in the Flash movie. You're right it should be 61.0.16.0.

 

Trying to submit my Research for Section 3 IPv6 but it keeps saying can't find page, is it down just now or is it at my end. I have tried several times, could you let me know.

Answer: I've just tried and it seems to be okay, now. There have been a few problems with the Napier network, so if you cannot access any of the resources, then use the mirror sites at:

http://buchananweb.co.uk/cnds.html

or log onto Web CT at:

http://neo.napier.ac.uk

All the Flash movies will run quite happily from the cache of your computer, so check your history, and they should still be there.

 

hi bill, what do you mean with: "It's not as important as Challenge 3, but the concept is important so I would study it." do we have to do challenge 4 as a preparation or not?

Answer: Well. Proxies are not officially defined to be assessed in this test, and the concept of the WinSock code is important.

 

Hi Bill, Regarding challenge 4, Q2 asks about client server connection, modified so that it could act as a proxy server. Could you give me a clue?

Answer: If we assume Host A, Host B (the proxy) and Host C. Host A makes a connection with Host B, and Host B makes a connection with Host C. Anything received on Host B from A is send automatically to Host C, and anything received from Host C is sent to Host A.

 

Hello, 3 questions only :-)
1. In chapter 5 page 87 you say: Last address: 144.32.15.254 (143.32[00001 1111 1110]) I can't understand why 143 instead of 144 inside the brackets.
2. It's not very clear to me what is the difference between Broadcast & Multicast
3. What is the difference in using all 0's or all 1's in the host or subnet part of address since both of them describe the whole subnet or network?

Answer: It should be 144, of course. I'll fix the typo.

Multicast addresses are special address and use Class D addresses. Everyone listening on a certain address receives the data. Broadcast are sent to all the network on a subnet, and relate to the subnet that they are destined for (such as 146.176.150.255 for all the nodes on the 146.176.150 subnet). Broacasts can go over routers (if they are destined for other networks), but most routers block multicast data packets.

All zero's identify the network address, such as 146.176.150.0 represent the 146.176.150 subnet (assuming a 255.255.255.0 subnet mask). If there are all 1's in the host part it represents a broadcast to all the nodes on the subnet, such as 146.176.150.255.

 

Hi Bill, how do I calculate the broadcast address for a subnet of a differnt class of address?

Answer: Well. It depends on the network address and the subnet mask. For a Class B network of 131.10.0.0, and a subnet mask of 255.255.0.0, then the broadcast address for this network is:

131.10.255.255

If we subnetting the network, with a subnet mask of 255.255.240.0, then the first subnet will be at:

131.10.16.0

the broadcast address will thus be:

195.160.31 [0001 1111b]. 255 [1111 11111b]

For example if we have a Class C network of 195.160.66.0, and a subnet mask of 255.255.255.0, then the broadcast address for this network is:

195.160.66.255

If we subnetting the network, with a subnet mask of 255.255.255.240, then the first subnet will be at:

195.160.66.16

the broadcast address will thus be:

195.160.66.31 [0001 1111b]

 

Hi Bill, how many subnets can be created on a Class C network which uses a subnet mask of 255.255.255.192 and how many hosts can connect to each subnet?

Answer: 192 is 1100 0000,

Thus we can get four subnets (of which 2 are usable, 01 and 10).

We now have 6 bits left then we can have 2 to the power of 6 hosts which is 64 hosts on each subnet (of which 62 are usable).

 

MSN Messenger:

Hi in challenge 3 what do the red squares represent ?

cnds@napier says: Red squares are network address, as far as I can remember.

 

MSN Messenger:

> i can' t find any practical Ipv6 network
cnds@napier says: Search for: IPv6 university network


> do you expect us to be able to draw one
cnds@napier says: No. It's just for your own information, really. Try:


http://www.auburn.edu/its/i2/ipv6.html

or:

http://press.nokia.com/PR/200003/775048_5.html

where the quote is: 'allocate every device in the world an IP address' and '[benefits such as] network security, mobility and quality of service'

>for Ipv6 to Ipv4 tunelling?
cnds@napier says: IPv6 data packets are carried within IPv4 packets.

> so the format is still the same?
cnds@napier says: Yes. They are just contained within the IPv4 packet.

 

Hi Bill, Can you tell me whether or not bridges are required to have an IP address ?

Answer: No, as they operate at the Data Link Layer. Often though a switch or a bridge can have an IP address, so that it can be accessed remotely over the Internet. Typically they run a local WWW provision, which shows the statistics of the device.

 

MSN Messenger:

> In V.B. Does data Arrival identify the data has been received over the network?
cnds@napier says: Yes. It's an event, and calls up the winsockname_DataArrival routine. Within there, the event is serviced (such as reading the data from the incoming buffer).

 

MSN Messenger:

> Section 5 questions 26-32. ?!?! haven't a clue how to work them out.
cnds@napier says: Ignore them, as we don't cover that till a later unit.

 

MSN Messenger:

> When calculating the usable subnets and you convert the snm into binary does the counting go left to right in the form of 1 2 4 8 16 ect or 2 4 8 16 ?.

cnds@napier says: Start from 2, 4, 8 and so on.

> ok so there aint no 1

cnds@napier says: Nope. For 1 bit there are two cominations, a 0 or a 1. For 2 bits there are four combinations 00, 01, 10 and 11, and so on.

 

Hi Bill, What is the difference between a proxy server and router running NAT. They appear, essentially, to carry out the same task ie allowing a private network to connect to an internet? When would you deploy one rather than the other ?

Answer: NAT does address translations and operates on all the ports, whereas a proxy acts as an intermediate system for a transfer for specific ports.

 

MSN Messenger:

> ... So does that mean that the 3rd usable subnets for 146.176.y.z and a snm of 255.255.252.0 would be 146.176.8.0 because 252 in binary is 11111100 and the first value of 1 is in the value of 8 position?

cnds@napier says: I make it .12.0

cnds@napier says: The first is at .4.0.The next is at 8.0., and the next at 12.0.

> How did you get that?

cnds@napier says: Well. The address mask for the third field is 1111 1100. Thus we assign in sequence for the six bits on the left hand side. So we go: 0000 0100
0000 1000
0000 1100

4th will be:
0001 0000
and so on.

 


Hi Bill, I'm sweating over Q5.17.17 on page 100 of the workbook. \"If nodes connect to a network which has been allocated the address of 130.10.y.z and uses a subnet of 255.255.248.0. What will be the 1st allocatable address for a host on the 3rd subnet?\" I get 130.10.24.1 (Class B) Is this correct?

130.10.y.z
255.255.248.0

gives:
1st
2nd:
3rd:

.1111 1000b.0000 0000b
.0000 1000b.0000 0000b
.0001 0000b.0000 0000b
.0001 1000b.0000 0000b

first address will be:


130.10.0001 1000b.0000 0001b which is 130.10.24.1. So you were correct.

 

MSN Messenger:

> ... what's the answer to 5.17.26?

130.10.y.z with 14 subnets

Subnet mask for Class B is 255.255.0.0

1 bit
2 bits
3 bits
4 bits

2 subnets (0 useable)
4 subnets (2 useable)
8 subnets (6 useable)
16 subnets (8 useable)

Thus the subnet mask needs to borrow 4 bits. It will thus be:

255.255.1111 0000b.0000 0000b

which is 255.255.240

 

On one of the questions in your unit 5 test you ask for the subnet mask for an address with 64 subnets. Your answer is .252 Obviously correct but what i am concerned about is usable subnets. Although your answer does not ask for usable subnets will the actual exam be the same. Are we to watch out for the word usable? If usable is mentioned then we do not worry about "usable"? Because if the question I am reffering to had been usable subnets we would have had to borrow 1 more bit.

In an exam I would always make it clear whether it is total subnets, or usable. If the question was vague, I would accept both answers as correct.