Please include working too! Thank you! :) magine that you have conducted an expe
ID: 3712785 • Letter: P
Question
Please include working too! Thank you! :)
magine that you have conducted an experiment to send some data using a full duplex l protocol (as discussed in lectures). The link that you used has 10 Mb/s capacity, you were sending the data in frames of 1400 bytes, and receiving an AC ink from location A to location B, 1540 kilometres away, using the Simple K/NAK frame of bytes. Please assume the speed of propagation in the medium is 2x108 m/s. The following TWO questions relate to this experiment 11. Assume that you sent 30 frames back-to-back, and that no NAK frames were received in response, so the A-B link's loss probability was 0. How long should it take to send all 30 frames? Express the time in milliseconds. A. 11.91 B. 41.35 C. 33.65 D. 49.05 E. 19.61 12. Now you realise that the A-B link's loss probability has changed. You measure the link's loss probability, and find that it is now 0.1. Assume that you will send 30 frames and that for each frame, if you receive a NAK you will resend the frame, and continue to do that until you receive an ACK. At every stage, consider only whole frames for re- sending. What would have been the effective bit rate for this transfer? Express the rate in Mb/s. A. 7.51 B. 52.41 C. 9.08 D. 6.41 E. 72.63Explanation / Answer
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Only 2 questions solved.
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Some Calculations based on the data given:
Link bandwidth (capacity) = 10 Mb/s = 10 x 106 bits/s
Distance = 1540 km = 1540 x 1000 meters
Frame size = 1400 bytes = 1400 x 8 bits
ACK / NAK size = 64 bytes = 64 x 8 bits
Speed of Propagation in the medium = 2 x 108 m/s
Propagation Delay , tprop = distance / speed of progation of medium
= 1540 x 1000 / 2 x 108
= 7.7 x 10-3 sec
= 7.7 ms
Time taken for 1 frame to transmit: ttrans = Total bits for 1 frame/ bandwidth
= 1400 x 8 / 10 x 106 (both are converted to bits)
= 1.12 ms
Time taken for 1 ACK/NAK msg to transmit: tntrans = tatrans = Total bits for 1 msg / bandwidth
= 64 x 8 / 10 x 106
= 0.0512 ms
Q11 Anwer:)
Assumed, 30 frames are send back to back. i.e., the packets are transmitted without waiting for ACK or NAK packets.
No NAK frames were received. So, no need to send the frames again. (i.e. no errors in transmission).
A-B link loss probability = 0
1 ACK will be recieved for the transmitted 30 frames to indicate total data is recieved.
So, Effective transmission time for all 30 packets = 30 * transmission time for frame + Propagation delay for frame + transmission time for ACK + propagation delay for ACK
= 30 * 1.12 + 7.7 + 0.0512 + 7.7
= 33.6 + 7.7 + 7.7512
= 41.3 ms + 7.7512 ms = 49.05 ms
Q12 Answer:
Link’s loss probability is = 0.1
That means for every 10 frames 1 frame will be lost (i.e. errors)
For 30 frames, the total number of loss frames = 3
So, 3 NAK frames will be received, 3 frames have to be resend and 3 ACK will be received.
1 ACK packet at the end of transmission.
Therefore Total transmission time = transmission time and propgation time for (30 frames + 3 NAK + 3 frames + 3 ACK + 1 ACK)
= (30 * 1.12 + 7.7 )+(3 * 0.0512 + 7.7) + (3 * 1.12 + 7.7) + (3 *0.0512 + 7.7) + (0.0512 + 7.7)
= 75.812 ms
For 10 Mbps speed ------------------ transferred time is 49.05 ms
For ??? Mbps speed ------------------ transferred time is 75.812 ms
i.e., less/more * 10 = 49.05 / 75.812 * 100 = 6.47 Mbps
But,the effective rate will be very less than this as we are not taking loss probability for the 3 re-transmitted frames. We are only taking into account for 30 frames send at the start.
So, from the options it will be near to 6.47 ... which is 6.41 Mbps
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