Question

Question#1

1. What is the difference between a BSS and an ESS?

2. Discuss the three types of mobility in a wireless LAN.

Question#2

A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (Including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.611US, what is the minimum size of the frame?

Question#3

Assume that, in a Stop-and-Wait ARQ system, the bandwidth of the line is 1 Mbps, and 1 bit takes 20 ms to make a round trip. What is the bandwidth- delay product? If the system data frames are 1000 bits in length, what is the utilization percentage of the link?

Answer 1

Question#1

1. A Basic Service Set (BSS) consists of a group of computers and one Access Point (AP) linked to a wired Local Area Network (LAN), whereas there are more than one Access Points in an Extended Service Set (ESS).

2. The three types of mobility in a wireless LAN are:

i) No-transition mobility - A station with no-transition mobility is either stationary (not moving) or moving only inside a BSS.

ii) BSS-transition mobility - A station with BSS-transition mobility can move from one BSS to another, but the movement is confined inside one ESS.

iii) ESS-transition mobility - A station with ESS-transition mobility can move from one ESS to another.

Question#2

The minimum frame transmission time is T?ᵣ = 2 × Tp

= 2 x 25.611  μs.

This means, in the worst case, a station needs to transmit for a period of 51.2 μs to detect the collision.
The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet.