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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) | 72 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 (3.9k 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 (3.9k 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


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

Total time for source to transmit is same whether we use pipelining or not.

But Total delay to reach receiver will change if pipeling used .right ??

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