P10

P10. Consider a packet of length L which begins at end system A and travels over three links to a destination end system. These three links are connected by two packet switches. Let di, si, and R denote the length, propagation speed, and transmission rate of link i, for i= 1, 2, 3. The packet switch delays each packet by dproc. 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 packet? Suppose now the packet is 1,500 bytes, the propagation speed on all three links is 2.5*10^8 m/s, the transmission rates of all three links are 2 Mbps, 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 end-to-end delay?

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 d_i denote the length of link i, s_i  the propagation speed of link i, and R_i= 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.