Computer Networks

Question 1

a)

The consultants proposed scheme indeed produces a viable link between the two locations. The explanation is given below.

The distance between the locations are 40 meters.

The transmission power is Pt = 80mW

The sensitivity of the receiver is Pr = -80dBm If we convert dBm into mW we have:

Pr = -80dBm = 10^(-80/10) = 10^-8 mW

The operating frequency of IEEE 802.11g protocol is 2.4 GHz and in Hz is 2.4*10^9. So frequency

f = 2.4*10^9 Hz

The transmission range of a wireless node with the about characteristics is in a free space model without any obstacles in the line of sight is:

The transmission range d is given from the following formula d = (Pt/Pr)*c2 4πf

c is the speed of light and it equals to 3*10^8 m/s

If we replace the values of the above formula we get that transmission range is d = 889.70 meters

That means that the transmitters of the proposed solution can ideally transmit a packet up to a distance of 889.70 meters and the packet will be received by the receiver successfully. In the case scenario, given that the distance between the transmitter and the receiver is only 40 meters, we can assume that the proposed solution provides a very strong link and data can travel between the two locations.

b)

IEEE 802.11g protocol claims a data rate of 54 Mb/s. We assume that there is a Bit Error Rate (BER) of 1*10^-3 at the receiver which means that for every 1000 bits of data received 1 bit has an error.

Within the period of 1 second 802.11 protocol can receive 54Mbits or 54.000.000 bits. When the BER is applied to that link, the number of error bits is 54.000.000*10^3 = 54.000 error bits. So from the 54.000.000 bits that are sent only (54.000.000 - 54.000) = 53.946.000 bits are successfully received giving us a rate of 53,946 Mb/s.

Time to transmit without the BER applied:

20GB of data = 160Gbits of data = 160.000 Mbits The rate is 54Mbit/sec which means the transmission time is t = Amount of data/ data rate = 160.000 / 54 = 2963...