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Two hosts are connected via a packet switch with $10^7$ bits per second links. Each link has a propagation delay of $20$ microseconds. The switch begins forwarding a packet $35$ microseconds after it receives the same. If $10000$ bits of data are to be transmitted between the two hosts using a packet size of $5000$ bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is ______.
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4 votes
4 votes

packet switching  follows pipelining and the main hero is pipelining  in this question. 

2 votes
2 votes
First packet takes all the way:-

$500 \mu s+20\mu s+35 \mu s+500\mu s + 20 \mu s=1075\mu s$

 

Second packet rolls in pipelined fashion. So just consider $T_t$

=> $500 \mu s$

Total: $1575 \mu s$
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1 votes
1 votes
There are 10000/5000 = 2 packets.

Transmission time for 1 packet = 5000 bits / 10^7 bps = 500 micro sec.

Packet 1 reaches in 500 (transmission from source) + 20 ( propagation to switch) +35 ( switch delay) + 500 (transmission from switch) + 20 (propagation to receiver) = 1075 micro sec.

Since packet 2 starts 500 micro sec after packet 1, the last bit of it reaches in 1075 + 500 = 1575 micro sec.

Correct me if I am wrong.
Answer:

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