Suppose two hosts, A and B, are separated by 20 kilometers and are connected by

Question

Suppose two hosts, A and B, are separated by 20 kilometers and are connected by a direct link of R=100Mbps. Suppose the propagation speed over the link is 2.5*10^8 meters/sec. Considering sending a file of 100,000 bits from A to B as one big message. Ignore the processing and the queuing delays. For parts a and c, express your final results in the unit of second.

A. How long does it take to send the file? Note, we count the time from the moment the first bit of the file leaves A until the moment the last bit arrives B.

.00108 secs

B. What is the maximum number of bits that will be in the link at any given time? 8000 bits

C. For how long time, the number of bits in the link is equal to the maximum number of bits (the quantity you derived in b.)?

I think I have A and B correct, please check them to make sure though.

What is the formula to answer C?

Explanation / Answer

Given,

Bit Rate = 100 Mbps = 100 x 10^6 Bits per second.

No. of Bits = 100000

Distance = 20km = 20000m

Propogation speed = 2.5 x 10^8 m/s

A. Calculating Transmission Time:

Packet Transmission Time = Size of Packet / Bit rate = 100000 bits / (100 x 10^6 bps) = 0.001 seconds

Propagation delay = Distance / Propagation speed = 20000 m / (2.5 x 10^8 m/s) = 0.00008 seconds

Transmission time = Packet Transmission Time + Propagation delay = 0.001 + 0.00008 = 0.00108 seconds

B. Maximum Number of bits at any given time = Propagation delay x Bit rate = 0.00008 sec x 100 x 10^6 bps = 8000 bits

C. Number of bits = Max number of bits

Bit rate = 100 x 10^6 bps.

So, The time time taken to put 8000 bits on the link is, t = 8000 / (100 x 10^6) = 0.00008 seconds.

From this instant, there will be 8000 bits on the link till 100000 - 8000 = 92000 bits are transmitted completely.

Transmission time for 92000 bits = 92000 / (100 x 10^6) = 0.00092 seconds

Propation delay = 0.00008 seconds

Total transmission time = 0.001 seconds

So, Total time for which number of bits on the link is equal to the maximum number of bits = 0.001 seconds - 0.00008 seconds = 0.000092 seconds

NOTE: In this case, this time and transmission time for 92000 packets are equal because of symmetry between link speed and distance and propagation speed. In other cases, this time will differ.

Hope this helps. Do upvote :)

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