Question

In: Computer Science

Subject: Computer Networks Chapter:packets 2.0. Suppose the path from A to B has a single switch...

Subject: Computer Networks

Chapter:packets

2.0. Suppose the path from A to B has a single switch S in between: A───S───B. Each link has a propagation delay of 60 µsec and a bandwidth of 2 bytes/µsec.

(a). How long would it take to send a single 600-byte packet from A to B?

(b). How long would it take to send two back-to-back 300-byte packets from A to B?

(c). How long would it take to send three back-to-back 200-byte packets from A to B?

If there is a diagram please draw the diagram

Solutions

Expert Solution

Answer:-----------
a). For a single 600 byte packet from A to B we divide into two steps.
From A to S
Bandwidth = 2 bytes/µsec
Propagation delay = 60 µsec
Bandwidth delay = 600 bytes / 2 bytes/µsec = 300 µsec
So from A to S for a single 600 byte packet will take = 300 µsec + 60 µsec = 360 µsec.

From S to B
Bandwidth = 2 bytes/µsec
Propagation delay = 60 µsec
Bandwidth delay = 600 bytes / 2 bytes/µsec = 300 µsec
So from S to B for a single 600 byte packet will take = 300 µsec + 60 µsec = 360 µsec.

Total from A to B = 360 µsec + 360 µsec = 720 µsec


b).
From A to S
Bandwidth = 2 bytes/µsec
Propagation delay = 60 µsec
Packets = 2
Bandwidth delay = 300 bytes / 2 bytes/µsec= 150 µsec

From S to B
Bandwidth = 2 bytes/µsec
Propagation delay = 60 µsec
Packets = 2
Bandwidth delay = 300 bytes/ 2 bytes/µsec= 150 µsec

As there will be same delay from S to B, in fact the first packet will have been even fully transmitted from S to B while both packets will have been received by the Switch S.
Therefore here:Time taken to send two back-to-back 300-byte packets from A to B
= (60 µsec +150µsec) + ( 60 µsec +150 µsec) = 210 µsec + 210 µsec = 420 µsec

c).
From A to S

Bandwidth = 2 bytes/µsec
Propagation delay = 60 µsec
Packets = 3
Bandwidth delay = 200 bytes / 2 bytes/µsec= 100 µsec

From S to B
Bandwidth = 2 bytes/µsec
Propagation delay = 60 µsec
Packets = 3
Bandwidth delay = 300 bytes/ 2 bytes/µsec= 100 µsec

As there will be same delay from S to B, in fact the first packet will have been even fully transmitted from S to B while all packets will have been received by the Switch S.
Therefore here:Time taken to send three back-to-back 200-byte packets from A to B
= (60 µsec +100µsec) + ( 60 µsec +100 µsec) + ( 60 µsec +100 µsec) = 160 µsec + 160 µsec + 160 µsec = 480 µsec


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