Question

Note: b denotes bits and B denotes Bytes (1 Byte = 8 bits). Consider a packet...

Note: b denotes bits and B denotes Bytes (1 Byte = 8 bits).

Consider a packet of length L, which begins at source and travels over seven links to a destination. These links are connected through six routers. Let di, si, and Ri denote the length, propagation speed, and the transmission rate of link i, for i = 1 to 7. The processing delay at each router is d-proc. The queuing delay at each router is d-que. What is the total end-to-end delay for the packet in terms of di, si, Ri (i = 1 to 7), and L?

Now suppose, the packet is 2,500 Bytes, the propagation speed on all links is 2.5x108m/s (meters per second), the transmission rates of all links are 1.6 Mbps. The length of first and second links are 5,000 Km (kilometer), the length of third, fourth, fifth and sixth links are 4,000 Km, and the length of seventh links is 1,000 Km. Processing delay in each router is 2 msec (mili-second) and queuing delay in each router is 7 msec. For these values, what is the end-to-end delay? Write down your calculations.

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Answer #1

Variables:

L = size of packet

di = array of lengths of link

si = array of propagation speeds of link

Ri = array of transmission speeds of link

dProc = processing delay at each router

dQue = queuing delay at each router

number of links = 7

number of routers = number of links - 1 = 6

=7 Transmission Delay = (L/R[i]) 21

i=7 Propagation Delay = {(d[i]/s[i]

Processing Delay = 6* dProc

Queuing Delay = 6 *dQue

Total EndToEnd Delay = Transmission Delay + Propagation Delay + Processing Delay + Queuing Delay

i=7 TotalEndToEndDelay = (L/R[i] + d[i]/s[i]) + 6* (dProc+dQue) i- 1

answer:

L = 2500 Bytes = 2500*8 bits = 2 * 10^4 bits

Propagation speed = si = 2.5 * 10^8 m/s for i=1 to i=7

Transmission rate = Ri = 1.6 Mb/s = 1.6 * 10^6 b/s for i=1 to i=7

d1 = d2 = 5000 km = 5 * 10^6 m

d3 = d4 = d5 = d6 = 4000 km = 4 * 10^6 m

d7 = 1000 km = 1 * 10^6 m

dProc = 2 ms = 2 * 10^-3 s

dQue = 7 ms = 7 * 10^-3 s

Total EndTo End Delay = 7/(2*104)/(1.6*106) +2((5*10%)/(2.5*10%)) + 4(4 * 106)/(2.5* 108)) + 10/(2.5 * 109) + 6(2* 10-3+7* 10

TotalEndToEndDelay = 249.5 * 10-3s = 249.5ms

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