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Assignment on Transmission

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Added on  2020-05-04

Assignment on Transmission

   Added on 2020-05-04

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TutorialsProblem 1.Minimum framesize=Bandwith *delay(transmission)But delay(transmission)=2*propagation delay.Propagation delay therefore=(length wire/speed of signal) =0.000005sec=5μsDelay(transmission)=2*2μs=10usFrame size=20μs**5mbs=100bitsProblem 2Assuming that the probability that the frame is sent successfully after transmission is equal to the probability that frames are damages Q-1 Pq=P^(q-1-(q-1))=p^(q-1)(1-p)The mean transmission number therefore isThe mean transmission therefore is:¿∑q−1∞αpq=11−p=E(x)=∑q−1∞qpq=∑q−i0rPa∞❑ (i-p)∑α−1∞αpq−1 = (1-p)[11−p+p1−p]E(x)=(11−p)
Assignment on Transmission_1
Problem 3Using the poisons distribution for random packet dataP(x=x)=e^-yxx! Where x=0Where e=2.718282Problem 4Stations=100Distribution=Distance =4kmsRate=5MpsDate packet=1000 bitsPropagation time=5mbs/kmDistance between stations=4000100 40m(i)Maximum number of packets present=Bandwith *Delay Delay=2*propagation =2(length of the wire/speed of the signal) =2(4*5)/2*0.8 =1.6μs Throughout =5mbps*1.6μs =2bits/s=8bits per second
Assignment on Transmission_2
=8*1000=8000 bits/sec(ii)if bus length=1km Delay=2*(0.2μseckm0.2μs)=2*0.2μs =0.4μsThroughout=5MbPs*0.4μs=2bits/sec 2*1000=2000bits/s (iii).if the transmission capacity is doubled to 10mBps Throughout = 10Mbs*1.64s=16 bits/sec 16*1000=16000bits/sec(iv) if the packets are 10,000 bits long Throughout=15Mbps *1.6=8 bits = 8*1000 =8000 bits/sec.
Assignment on Transmission_3

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