Circuit switching

Question

If two host are connected by a link and in middle of the link there is a switch between them, then transmission and propagation time both doubled.

But if for the same link, there is no switch between that two hosts, then propagation time and transmission time will be once

Why that happens? Though both are operating on same link, why only a switch doubles propagation delay and transmission delay time?

Comments

In circuit switching resources are allocated (buffer ,bandwidth) early ,before packet is sent .So packet will reach to the destination definately.

As we can see time will be doubled when switch is there due to circuit switching.

Let me know if I m wrong.

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this is the concept of packet switching not circuit switching.

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I am not getting why switching doubles transmission and propagation times?

@Prateek I know, that happens. But my question is why that happens?

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As transmission time is depend on L/B(L=length,B=bandwith) by using switching this time will be increases.

but we also know that propagation time =d/v(d=distance,v=velocity) by using switching i think the distance between source and destination remain sakme but how it become double when we put a switch between them???

Answer 1

IN QUESTION THEY SPECIFIED ABOUT PROPAGATION TIME FOR WHOLE LINK OR INDIVIDUAL LINK. IF THEY GIVE PROPAGATION TIME OF INDIVIDUAL LINK THEN PROPAGATION TIME DOUBLE IN CASE OF SWITCH BETWEEN THEM. OTHERWISE IT WILL BE PT REMAIN.

BUT TRANSMISSION TIME IS ALWAYS DOUBLED BECAUSE TRANSMISSION TIME IS THE TIME TO PUT DATA ON CABLE.