connected by a directlink of R=10Mbps. Suppose the propagation speed over the li

Question

connected by a directlink of R=10Mbps. Suppose the propagation speed over the

link is 2.5*10^8meters/sec. Considering sending a file of 100,000 bits from A toB

as one big message.Ignore the processing and the queuing delays, i.e. weonly

examine transmissionand propagation delays.

a. How long doesit take to send the file? Note, we count the time fromthe

moment the first bitof the file leaves A until the moment the last bit arrivesB.

b. What is themaximum number of bits that will be in the link at any

given time?

c. For how long time,the number of bits in the link is equal to the

maximum number of bits (thequantity you derived in b.)?

Explanation / Answer

Dear, A and B, are separated by 20kilometers    R=10Mbps    the propagation speed over the link is2.5*10^8 meters/sec    sending a file of 100,000 bits from A to B The Propagation Delay dprop = m/s = (20km *1000m/km) / (2.5*10^8)m/s = 1/(8*10^5) sec    Transmission Delay dtrans = L/R =100,000/(10*1000000) = 1/ 100 = 0.01sec Ignore the processing and the queuing delays, i.e. weonlyexamine transmission and propagationdelays The last bit get pushed out of A's interface in L / Rsec, the bit takes m/s sec to reach B,          so the totalend- to- end delay is = L/R + m/s = (1/100)sec + (1/(8*10^5) sec) =0.01000125sec (b) A first bit takes 1/(8*10^5) sec to reach B once itleaves A. During this time, how many bits have been injected intothe wire by A?         1/(8*10^5)sec×(10*10^6 bits/sec) = 305 bits. So the maximum number of bits on the link at any given time is305. Thus Bandwidth-delay product is the maximum number of bits onthe link at any given time. I hope this will Helps you!!!!!!!!!!!!!!!!!!!!!!

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