Consider a packet of length L which begins at end system A and travels over thre

Question

Consider a packet of length L which begins at end system A and travels over three links to a destination end system. These three linksare connected by two packet switches. Let di, si, and Ri denote the length, propagation speed, and the transmission rate of link i, for i=1,2,3. The packet switch delays each packet by dproc.

a. Assuming no queuing delays, in terms of di, si, and Ri (i=1,2,3), and L, what is the total end-–-to-–-end delay for the packet?

b. Suppose now the package is 1,500 bytes, the propagation speed on all three links is 2.5 108 m/s, the transmission rates of all three links are 2Mbos, the packet switch processing Delay is 3 msec., the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km. For these values, what is the endtoend delay? m/s, the transmission rates of all three links are 2Mbos, the packet switch processing Delay is 3 msec., the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km. For these values, what is the endtoend delay?

C. Suppose you would like to urgently deliver 40 terabytes from Boston to Los Angeles. You have available 100 Mbps dedicate link for data transfer. Would you prefer to transmit the data via this link or instead use FedEx overnight delivery? Explain.

Explanation / Answer

Let L denote the length of a packet. This packet begins at end system A and travels over 3 links to a destination end system. N= the number of the links, so here N=3. The three links are connected by two packet switches.Let di denote the length of link i, si the propagation speed of link i, and Ri= transmission rate of link i. For i= 1,2,3. The packet switch delays each packet by dproc and there is no dqueue. We want to find the total dend-to-end in terms of di, si, Ri, and L. So the formula used for dend-to-end(page 42, Computer Networking: A Top-Down Approach) is:
dend-to-end= N(dproc + dtrans + dprop)
= N(dproc+(L/Ri)+ (di meters/si sec))

Given the following:
N=3
L= 1,500 bytes
si= 2.5*10^8 m/s on all links
Ri= 2 Mbps
dproc= 3 msec
d1= 5,000 km
d2= 4,000 km
d3= 1,000 km
We want to find the end-to-end delay dend-to-end.
We must plug in the values for each case of i.
For i=1, N(dproc+(L/R1)+ (d1 meters/s1 m/sec))
=3(0.003 sec+(1,500*8 bits/2,000,000 bits/sec)+ (5,000,000 m/2.5*10^8 m/sec))
=0.087 seconds

i=2; dend-to-end= N(dproc+(L/R2)+(d2 meters/s2 m/sec))
dend-to-end= 3(0.003 sec+(1,500*8 bits/2,000,000 bits/sec)+ (4,000,000 m/2.5*10^8 m/sec))
=0.075 seconds
i=3; dend-to-end= N(dproc+(L/R3)+(d3 meters/s3 m/sec))
dend-to-end= 3(0.003 sec+(1,500*8 bits/2,000,000 bits/sec)+ (1,000,000 m/2.5*10^8 m/sec))
=0.039 seconds

We now add each case together:
0.087 sec+0.075 sec+0.039 sec= 0.201 sec

The total end-to-end delay of the three links is 0.201 seconds.

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