· Introduction to the concept about Ethernet design limitation (see also Ch7 Min

Question

· Introduction to the concept about Ethernet design limitation (see also Ch7 Minicase I.):

Question 1

One important issue in designing shared Ethernet lies in making sure that if a computer transmits a frame, any other computer that attempts to transmit at the same time will be able to hear the incoming frame before it stops transmitting, or else a collision might go unnoticed.

For example, assume that we are on earth and send an Ethernet frame over a very long distance (maybe Moon). If a computer on the Moon starts transmitting at the same time as we do on Earth and finishes transmitting before our frame arrives at the Moon, there will be a collision, but neither computer will detect it; the frame will be garbled, but no one will know why. So, in designing Ethernet, we must make sure that the length of the cable in the LAN is shorter than the distance a shortest possible frame travels. Otherwise, a collision could go undetected.

Consider this Analogy/Example:

First: It will take 16 balls / 8 balls per minute = 2 minutes time to drop all 16 balls.

Second: In the 2 minutes time, the belt will advance 2 meter per minute x 2 minutes = 4 meters => The trail of 16 balls is 4 meters long.

Assumptions for Question:

The effective speed is 40 million meters per second (While electricity in the cable travels a bit slower than the speed of light, once you include delays in the electrical equipment in transmitting and receiving the signal, the speed is much slower)

The frame size is 64 bytes (the smallest possible Ethernet frame size, including the 33 byte overhead).

The Question:   What distance in meters will this message travel (the length of the message in meters) over the medium when using Ethernet protocol with speed 10 Mbps?

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Hint:   First calculate the number of seconds it would take to transmit the frame.

Second calculate the number of meters the signal would travel in that time, and you have the total length of the frame.

Question 2

The distance a 64 byte frame will travel if using 1GbE is ______ meters.

Explanation / Answer

1 byte breaks down into 8 bits, so 64 bytes break down into 512 bits
10 megabits per second actually breaks down into 10,485,760 bits per second. So in one second you can transmit 10,485,760 bits to some distance.
To determine how long it takes to send 512 bits

1 second / 10,485,760 bits = ? seconds / 512 bits
1 second * 512 bits / 10,485,760 bits = 4.8828125 × 10^-5 seconds = .000048828125 seconds

Now we know that it takes .000048828125 seconds to send 64 bytes of data, can apply that information to the effective speed i.e 40,000,000 meters per second.

If it only takes .000048828125 seconds to send 64 bytes of data , and in a full second the data could travel 40,000,000 meters, then how far will it travel in only .000048828125?

40,000,000 meters per second * .00048828125 seconds = 1,953.125 meters

Meters are traveled per bit are

40,000,000 meters per second / 10,485,760 bits per second = 3.81469727 meters per bit
Knowing there are 512 bits per 64 byte message

512 bits * 3.81469727 meters per bit = 1,953.125 meters

b) 40,000,000 meters per second / 1 * 1024 *1024 *1024
     = 40,000,000 meters per second / 10,73,741,824 bits per seconds = 0.03725290 meters per bit

      512 x 0.03725290 meters per bit = 19.07348 meters

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