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I read in Bus topology that "the network is occupied until the destination computer accepts the data". Is this statement correct?

If yes, then what happens if the destination computer is not present? How does the network handle this problem? Of course the bus topology shouldn't be occupied this whole time.

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3 Answers 3

up vote 4 down vote accepted

If you're talking about Token Ring (which no one uses anymore), the packet will flow around the ring until it comes back to you. At which point, you don't pass it on.

[update]

Think of your bus as a single wire. Along that wire speakers are attached to it (and earth ground.) Then add key-switches along the line to signal Morse code. Any switch can make all the speakers click, however, while a message is being sent, only one operator can be keying the message. The bus is thus "occupied" as long as the message is being sent. When done, anyone else can key their message. It doesn't matter if the message is intended for an operator who isn't listening, the message is still "put on the wire".

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my question is related to bus topology.in bus topology data cannot flow around the ring as we have terminators at both the end to prevent bouncing of packets.correct me if i am wrong?? –  rock321987 Aug 4 at 19:41
    
if you're the one with a terminator, then you know you're at the end. Think "hub", it's job is done when it's copied the frame to all the other ports. –  Ricky Beam Aug 4 at 19:52
    
it means that it is like broadcasting of signal to all the computer and the job is done?? –  rock321987 Aug 4 at 19:56

I would say the network is occupied (bus topology) as long as someone is sending data. Once the medium is free (no other node is transmitting), any node can start transmitting. One way this process is governed in Ethernet is via Carrier Sense Multiple Access/Collision Detection.

If the destination node is unreachable, this is not handled at layer 2 (again talking about Ethernet, but as far as I know, there's no other protocol that would handle this at OSI layer 2). The sent frame is lost and this must be detected by an upper layer protocol, such as TCP. TCP works based on acknowledgement, if a certain piece of the sent data is not acknowledged as received by the destination within a certain time, the sending device will try again.

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this implies that the above statement is incorrect as the medium becomes free as soon as the data is sent?? –  rock321987 Aug 4 at 19:37
2  
Yes, as there is no concept of "accepting data", a node can only listen for data (and transmit data). Acceptance of data would indicate some signal being transmitted saying so, how else would the other nodes know the medium is free? This is not the case at layer 2. Line sensing is implemented, which tells the transmitter logic whether or not someone else is transmitting. Line sensing cannot tell if another node is actually "accepting" (reading) data from the bus. –  RedShift Aug 4 at 19:58
    
that's beautiful.. +1 for this –  rock321987 Aug 4 at 20:02

In a Bus topology, all transceivers connected to the same bus can "hear" each other. This means that they are all a part of the same collision domain. When transceivers transmits something on the bus, the others have to be quiet to avoid a collision.

An example of this, is the old Half - Dulpex Ethernet, that used coaxial cables to form a bus topology, with CSMA/CD to avoid and cure collisions.

With that in mind, I think that the correct phrasing is that the bus is occupied while there is still data on the line, regardless of the receiver. Note that I say "data on the line" instead of "while data is being transmitted" since the data remains on the line for a while after the initial transmission due to propagation delay.

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