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Consider  1 Mbps error-free line , The maximum frame size is 1000 bits .New packets are generated about 1 second apart . Timeout interval is 10m/sec if the special acknowledgement timer were eliminated unnecessary timout would occur . How any times average msg would have been transimitted ?

a) Only once

b) Twice

C) thrice

d) None

5 Answers

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  • Transmission delay (Tt) = L / B = 1000 bits / 10^6 bits per sec = 1 msec.
  • After the packet is put on the link, the time out timer is started which is 10 msec long.
  • The next packet is transmitted after 1 sec = 1000 msec.
  • If no acknowledgment is received within 10 msec, the packet will be retransmitted.
  • We have been asked how many times the average message be transmitted i.e. how many retransmissions are possible.
  • Retransmission occurs or not depends on the propagation delay (Tp).
  • If Tp is more, time out will occur and retransmission will take place but if Tp is less, then there will be no time out.
  • Since propagation delay (Tp) is not given in the question, therefore we can not say anything.
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I am still not sure what you meant by "time interval is 10m/sec". If I am correct, it is a typing error. That should be 10msec (milli second).

10^6 bits in 1 sec

therefore bits that can be transferred before the timeout is = 10^6 * 10/1000 = 10000 bits

and avg  msg size is (0+1000)/2=500 bits 

number of frames (or msgs) that can be transferred are 10000/500=20

D) NONE

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Given BW=1Mbps   and frame size =1000 bits

in 1 second -----> 10^6 bits  We can transfer that much bits

and a timeout occurs  10msec( typo in question) so in 10 mesc------> 10^4 bits we can transfer that much bits

(10^4)/10^3 = 10  frame can be transmitted

Option :D" NONE
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I think to make sure transmissions occur RTT>Timeout.RTT=2*Tp.As no information is given regarding Tp,so we cannot say how many transmissions will occur in avg.

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