At Time T= Dtrans Where Is the Last Bit of the Packet?

This basic problem begins to explore multiplication delay and transmission system delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rateRbits per second. Conjecture that the deuce hosts are separated bym meter, and suppose the propagation speed on the connectedness iss meters/sec. Emcee A is to send a package of sizeL bits to Host B.

a.Express the extension delay, dprop, in price of m and s.

Hither, there are 2 hosts: A and B, both connected by a single inter-group communication of rateR bps. The two hosts A, B are abutting by mmeter. The propagation speed along the link up is s meters/sec. In the demonstration above(provided away Dr. Roseine), we have the symbols and their units defined. R= bits/sec, distance= m, s= m/second, and L= bits

Based on the definition provided on page 37 of Estimator Networking: A Top-Down Approach, the multiplication delay is dprop= d/s with d representing the distance in meters between two hosts and s representing the time in second.  We infer from this formula to find the formula for dprop here. It's important to note, however, that while the formulas may at times be similar, they are not always going to be the homophonic exact formula. Therefore, we should only use this formula as a reference from which to figure out a new formula from the information.

Deriving from the original defined dprop formula, the formula we use of goods and services here is d/s where d represents space in meters between the deuce hosts and s represents the meters/SEC.

b.Determine the TRM of the packet, dtrans, in terms of L and R.

We want to express the transmission time dtrans in terms of L and R. The transmission delay formula given on page 37 is dtrans= L/R, with L denoting the length of the packet by L bits and R being the transmitting grade of R bits/sec.
The formula for the transmission delay of Host A and B is:
dtrans= L/R

c.Ignoring processing and queuing delays, obtain an expression for the end-to-end delay.

We want to find the expression for the end-to-end delay patc ignoring processing and queuing delays.
The expression of throughout delay is defined on page 42 as:

dend-to-end= dshor + dtrans

Seeing that the end-to-end delay is the summation of the delay processes, and so the end-to-ending delay expression ignoring queuing and processing delays can be observed As:

dend-to-end= dtrans + dprop

We only have one link up here, then N=1. We don't motivation N here however as the result is the same with 1*a number versus just having the number there(for example, 1*5 is 5, and 5 is 5), sol we testament exclude it from the formula.

d.Say Host A begins to transmit the bundle at prison term t=0. At metre d=dtrans, where is the last bit of the packet?

To determine where the last bit of the packet is, we protract the connecter of Hosts A and B arsenic shown to a higher place in the representative done by Dr. Roseine during class. Host A is sending a packet to Host B. Therefore t=0 is located at Legion A, since Horde A is the source sending the packet to its finish Host B, and at t=0, no time has passed. Packets can't be transmitted to their destination without all of the bits organism routed first as shown in the diagrams above. A queue of bits of packet starts to grow.

e.Hypothesize dprop is greater than dtrans.At time t=dtrans, where the first bit of the packet?

At the very last bit as shown above in the demo given by Dr. Roseine, it is all but back to where the sending host was. This is destination boniface. These bits start to stack forming a queue from the goal punt towards the original source Host A.When dprop>dtrans and t=dtrans, t=dtrans<dprop. As shown in the picture above, the first bit of the bundle goes into the wires.

g.Suppose s= 2.5*10^8, L=120 bits, and R= 56 kbps. Observe the distance m so that dprop equals dtrans.

s= 2.5*10^8
L= 120 bits
R= 56 kbps

We want to find the distance in meters so that dprop=dtrans

As shown above in the word picture taken during class of the example done by Dr. Roseine, we lucubrate the expressions for dprop and dtrans. So dendtoend= dprop+drans is evaluated to: d/s + L/R. Because dprop=dtrans, d/s=L/R. Then we cross-multiply, divide by R to some sides, and get d= L/R * s for our distance formula. We past substitute L, R, and s with their values. So we get d= (120 bits/56 kbps) * (2.5*10^8 m/sec). The answer is d= 535,714 meters. In other words, the distance of the two hosts A and B when propagation delay and transmittal delay are equal is 535,714 meters.

At Time T= Dtrans Where Is the Last Bit of the Packet?

Source: https://evanmeshberg.wordpress.com/2013/01/28/p6/

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