Example 4.7
We can think about the link between two points as a pipe.
The cross section of the pipe represents the bandwidth,
and the length of the pipe represents the delay. We can say
the volume of the pipe defines the bandwidth-delay
product, as shown in Figure 4.7.
The bandwidth-delay product defines
the number of bits that can fill the link.
4.17
Figure 4.7 Filling the link with bits in case 2
4.18
Example 4.8
A network link is transferring a data file in two minutes period. The figure shown
below shows the throughput vs. time. Use this figure to calculate the following:
1) Average data transfer rate for this whole time.
2) Amount of data transferred between second 70 and second 120 ?
P Kbps
500
30
4.19
70
120
t sec
Solution
Size of data=0.5*30*500+40*500+0.5*50*500= 40000 KB
1)
Average data transfer =S/T=40000/120= 334.3333 Kbps
2)
Amount of data transferred =0.5*50*500= 12500 bit
4.20
Example 4.9
We are sending a 30 Mbit MP3 file from a source host to a destination host.
All links in the path between source and destination have8 a transmission rate
of 10 Mbps. Assume that the propagation speed is 2 * 10 meters/sec, and the
distance of the each link is 10,000 km. the processing time of the router is
0.01 sec
a) Suppose there is only one link between source and destination. Find the
latency?
b)Suppose there is three links between source and destination. The router is
found between each two links Find the latency?
Sol:
a)Latency = Transmission delay +Propagation delay
6 7
7
8
=30* 10 /10 +10 /(2 * 10 ) = 4.05 sec
b)Latency = 3*Transmission delay +3*Propagation delay + 2* Processing
time(waiting time)
6 7
7
8
=3*( 30* 10 /10 )+ 3*(10 /(2 * 10 ))+2*0.01=3*4.05+2*0.01=9.17
4.21
4.22
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Computer Networks
Al-Mustansiryah University
Elec. Eng. Department College of Engineering
Fourth Year Class
Chapter 5
Transmission Media
5.1
Figure 5.1 Transmission medium and physical layer
5.2