Mon 01/21/2002 5:01 PM

The data link layer is responsible for the point-to-point transmission of the data packet as it travels over interconnected network. There is no guarentee that the error checking will occur. If there was an error than the data frame transmitted would be lost. It would then be up to the TCP layer to detect that the data packet had not arrived (as there would be a gap in the sequence of delivered). TCP thus works at a higher-layer, and is the only layer that asks for a receipt for the delivery. Data link does not do this.

Mon 01/21/2002 5:01 PM

It depends on the context of the question, as it all depends on the number of potential nodes and networks, as some addresses are used for other purposes.

In a Class A, there is the potential for 2^7 (128) different network addresses, but the all 1's network address is reserved, thus it gives 127 useable network addresses. The maximum number of hosts on a Class A network is 2^24 (16,777,216), but again there would be restrictions in the number of usable addresses.

In a Class B, there is the potential for 2^14 (16,384) different network addresses. The maximum number of hosts on a Class B network is 2^16 (65,536), but again there would be restrictions in the number of usable addresses.

In a Class C, there is the potential for 2^21 (2,097,152) different network addresses. The maximum number of hosts on a Class B network is 2^8 (256), but again there would be restrictions in the number of usable addresses.

Friday, January 12, 2001 at 10:39:02

Routers react to ARP requests in the same way that any other host does. They will only respond to an ARP request if one of their ports has the IP address which is contained in the ARP request. They listen to all the ARP replies, though, and will update their ARP table whenever they hear an ARP reply. If they do not know the MAC address of a host on the network then they will send out an ARP request to find it. If it cannot find the host, it will discard the incoming message (as the host either doesn't exist or has went off-line).

Friday, January 12, 2001 at 10:39:02

Monday, January 15, 2001 at 16:31:05

A hub operates a Level 1 of the OSI model and basically just provides a basic network connection. Thus it still has collision between the ports of the hub. A switch operates at Layer 2 of the OSI model, and allows simulataneous communications between pairs of ports (and thus does not have any collisions). Thus a 10Mbps switch can support multiples of this bandwidth if two or more nodes are communicating at a time. The layers are really:

Layer 1	Hub, which basically just creates a network connection, and propagates collisions)
Layer 2	Switch, which operates on the MAC address, and switches data between connected ports. Switches can either be cut-through, where the data frame is forwarded immediately as the destination MAC address is read, or store-and-forward where the data frame is stored in memory before it is forwarded.
Layer 3	Router, which uses the network address to route data. These will operate at a slower rate than switches as they must look inside the data packet before it can be forwarded to another port.

Thursday, January 18, 2001 at 23:56:38

The data packet is sent up from the data link layer, and it obviously has a certain size, thus if the length of the header is known, then the length of the data part can be simply determined (total length minus the header length).

Thursday, January 18, 2001 at 23:02:21

Yup, you're totally correct (and no, you're not daft!), 2^32 is 4 billion (4,294,967,296). It's a little typo. I'll change it in the teaching pack for next year. Of course, it's a US billion (which is one thousand million), and not a UK billion (which is a million, million, and is gramatically correct, I think!). I think one thousand million should really be called a killion, but that doesn't sound too good.

Friday, January 19, 2001 at 20:37:54

Public-key encryption is the best way to secure data. With this method a user generates two electronic keys, typically with hundreds or thousands of bits. These keys are special number and relate to extremely large prime numbers (as it is difficult to factorise large prime numbers. For example, I have two prime numbers (small ones), and when I multiple them together I get the value of:

1,354,657

What was the original prime numbers [answer at the bottom of this page]? With public key encryption these numbers typically have thousands of bits, which gives values from 1 to 1,797,693,134, 862, 315,907,729,305,190,789, ...... (in total, it has 309 digits). Image finding the factors for two number that are this long?

One of the keys is kept private and no-one knows it apart from the user (typically it is kept under password protection), and the other is public, and is broadcast to anyone who wants to send the user an encrypted message. This could be send by e-mail, contained on a site with a collection of public keys, or could be put on a WWW page. When the someone want to send a message to the user they use the receiptants public key to encrypt the data. The only key that can decrypt the encrypted data is the receiptiants private key. Thus it is a totally secure method. It is so secure that governments are extremely worried as they cannot listen to user transmission (and they often do this with telephone messages, in fact it is thought that the UK govenment listens to every single telephone call and they look for specific key phases, such as bomb, IRA, and so on). The UK government has just passed a bill that legislates that someone who uses public-key encryption must hand-over their private key when they demand. So don't loose your private key or you'll end up in jail.

To make it easy to manage keys, the user normally uses a key-ring in which destination keys are added to the key ring (see on the right-hand side), and when they want to send a message they simply take the key from the key-ring. Many organisations, especially military and defence organisations, encrypt most of their data when transmitting over public-access channels, and will also encrypt the data which is written to these computer systems. The weak link in all this, as always, is the password which protects the private key. If someone gets this password they can easily copy it and then decrypt and data which is sent to the user, and can even pretend to the user.

Tuesday, January 23, 2001 at 10:16:29

128 bits. It will also have self configuration of the networking address, just in the same way as IPX, which takes the network address and then adds the MAC address of the network card (in order to make it unique).

Tuesday, January 23, 2001 at 23:06:00

The VCI header is the part that is added to the data. It is 5 bytes long (40 bits). This is made up of a 24-bit (3 bytes) VCI label, an 8-bit control field (1 byte), and a 8-bit checksum (1 byte). Note that the total length of the ATM cell is 53 bytes (with the header of five bytes and data of 48 bytes).

Wednesday, January 24, 2001 at 12:30:00

The cell header has three main fields:

	24-bit VCI label. This is made up of a VCI and a VPI GFC, PT and CLP (8 bits) which make up the Control part, followed by the HEC which is the checksum.
	8-bit Control field. GFC (4 bits) - generic control bits, PT (3 bits - payload) and CLP (1 bit - defines if the cell can be dropt if the network is busy)
	8-bit Checksum. Header error control field (8 bits).

 

Wednesday, January 24, 2001 at 06:52:09

Bridges and switches have similarities:

	They isolate collision domains. Both a bridge and a switch contain collisions within a network segment.
	They transmit broadcasts from one network segment to another (leading to a broadcast storm on the network).
	They learn where nodes are located based on MAC addresses.

The big difference is that switches allow simultaneous communications between any two nodes, at the same time. Also switches are used to create LANs (in the same way that a hub does), whereas bridges are used to segment networks. Thus a switch is optimised to communicate with individual nodes, whereas a bridge interfaces to network segments. Switches can also be used to create vLANs where collisions are completely elliminated, and broadcast domains can be programmed by software. This enhances security, uses the bandwidth more efficiently, and so on. Bridges also are setup so that they can sense when there is multiple route to a destination, and can easily switch between the two, when one become inoperative (this uses the Spanning-tree protocol).

 

Wednesday, May 30, 2001 at 08:00

A hub is basically just the same as have a common cable which connects all the nodes to the common bus. Thus if two nodes transmit at the same time they will cause a collision, and no nodes can transmit. Each of the nodes which caused the collision will wait for a ramdom period before getting access to the network. The key is that the transmit at a rate of 10Mbps, but THEY ONLY TRANSMIT ONE DATA FRAME (which is a maximum of 1500 bytes). Thus after they transmit this frame they must contend to get access to the network. Thus if the network is busy they might not get access for some time. As the traffic increases on the network, there will be more collision, thus there will be more wasted time. Thus causes the maximum throughput to reduce to around 5Mbps.

 

Wednesday, May 30, 2001 at 08:00

In a hub all the transmit lines are connected together, thus if the node detects a collision on its RX line, it stops transmitting and sends a jamming signal to the rest of the network. It is thus important for the node to 'listen' to the network as it is transmitting.