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Suppose that x bits of user data are to be transmitted over K-hop path in a packet-switched network as a series of packets each containing p data bits and h header bits with x>>(p+h). The bit rate of lines is b bps and propagation delay is negligible. What is the time taken by the source to transmit total bits?

  1. (p+h) x/b bits
  2. (p+h) x/pb bits
  3. p x/b bits
  4. hx/pb bits

Please tell the answer with explanation.

asked in Computer Networks by (61 points) | 51 views

2 Answers

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<p>2nd option..source will complete transferring when it has send all packets.</p>

<p>Total packets that need to be send is x/p and each packet has data of (p+h) so transmission time of 1 packet is datasize/bandwidth=(p+h)/b....but you have to send total of x/p so total time that would be required for transferring woulf be x/p*(p+h)/b..</p>
answered by Active (3.8k points)
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packet switching is used so each packet will follow its previous packet in pipelining manner except first packet

total number of bits to send =x bits ,Number of hops=k ,bandwidth = b bps ,each packet contains p data bits and h header bits

number of packets =x/p packets

transmission time (TT)=L/B =(p+h)/b

time taken by first packet =1*k*TT seconds

time taken by all other packets =((x/p)-1)*1*TT seconds

so total time taken=1*k*TT+((x/p)-1)*1*TT =(k+((x/p)-1))*(p+h)/b seconds

time taken by source to transmit =(x/p)*(p+h)/b=x(p+h)/pb
answered by Active (3.4k points)
edited by
@anil ji

question has asked for time taken to transfer by source not the time when its recieved at other end..!
edited ,thnks for noticing

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