Wireless LAN By B R Chandavarkar CSE Dept

Wireless LAN By, B. R. Chandavarkar, CSE Dept. , NITK, Surathkal Ref: B. A. Forouzan, 5 th Edition

• Wireless communication is one of the fastest-growing technologies. The demand for connecting devices without the use of cables is increasing everywhere. • Two promising wireless technologies for LANs: IEEE 802. 11 wireless LANs, sometimes called wireless Ethernet, and Bluetooth, a technology for small wireless LANs. • IEEE has defined the specifications for a wireless LAN, called IEEE 802. 11, which covers the physical and data link layers. • The standard defines two kinds of services: the basic service set (BSS) and the extended service set (ESS). Basic Service Set • IEEE 802. 11 defines the basic service set (BSS) as the building block of a wireless LAN. • A basic service set is made of stationary or mobile wireless stations and an optional central base station, known as the access point (AP).

Extended Service Set • An extended service set (ESS) is made up of two or more BSSs with APs. • In this case, the BSSs are connected through a distribution system, which is usually a wired LAN. • The distribution system connects the APs in the BSSs. • IEEE 802. 11 does not restrict the distribution system; it can be any IEEE LAN such as an Ethernet. • Note that the extended service set uses two types of stations: mobile and stationary. • The mobile stations are normal stations inside a BSS. • The stationary stations are AP stations that are part of a wired LAN.

Station Types • IEEE 802. 11 defines three types of stations based on their mobility in a wireless LAN: no-transition, BSS-transition, and ESS-transition mobility. • A station with no-transition mobility is either stationary (not moving) or moving only inside a BSS. • A station with BSS-transition mobility can move from one BSS to another, but the movement is confined inside one ESS. • A station with ESS-transition mobility can move from one ESS to another. However, IEEE 802. 11 does not guarantee that communication is continuous during the move. MAC Sublayer • IEEE 802. 11 defines two MAC sublayers: the distributed coordination function (DCF) and point coordination function (PCF).

• Wireless LANs cannot implement CSMA/CD for three reasons: – For collision detection a station must be able to send data and receive collision signals at the same time. This can mean costly stations and increased bandwidth requirements. – Collision may not be detected because of the hidden station problem. – The distance between stations can be great. Signal fading could prevent a station at one end from hearing a collision at the other end.

Hidden station problem 14. 9

Use of handshaking to prevent hidden station problem 14. 10

Exposed station problem 14. 11

Use of handshaking in exposed station problem 14. 12

CSMA/CA and NAV 14. 14

Point Coordination Function (PCF) • The point coordination function (PCF) is an optional access method that can be implemented in an infrastructure network (not in an ad hoc network). • It is implemented on top of the DCF and is used mostly for timesensitive transmission. • PCF has a centralized, contention-free polling access method. • The AP performs polling for stations that are capable of being polled. • The stations are polled one after another, sending any data they have to the AP. • To give priority to PCF over DCF, another set of interframe spaces has been defined: PIFS and SIFS. The SIFS is the same as that in DCF, but the PIFS (PCF IFS) is shorter than the DIFS. • This means that if, at the same time, a station wants to use only DCF and an AP wants to use PCF, the AP has priority.

Example of repetition interval 14. 16

Fragmentation • The wireless environment is very noisy; a corrupt frame has to be retransmitted. • The protocol, therefore, recommends fragmentation-the division of a large frame into smaller ones. It is more efficient to resend a small frame than a large one. Frame Format • The MAC layer frame consists of nine fields – Frame control (FC): The FC field is 2 bytes long and defines the type of frame and some control information. – D: In all frame types except one, this field defines the duration of the transmission that is used to set the value of NAY. In one control frame, this field defines the ID of the frame. – Addresses: There are four address fields, each 6 bytes long. The meaning of each address field depends on the value of the To DS and From DS subfields. – Sequence control: This field defines the sequence number of the frame to be used in flow control. – Frame body: This field, which can be between 0 and 2312 bytes, contains information based on the type and the subtype defined in the FC field. – FCS: The FCS field is 4 bytes long and contains a CRC-32 error detection sequence.

Frame format 14. 18

Subfields in FC field 14. 19

Frame Types • A wireless LAN defined by IEEE 802. 11 has three categories of frames: management frames, control frames, and data frames. • Management Frames: Management frames are used for the initial communication between stations and access points. • Control Frames: Control frames are used for accessing the channel and acknowledging frames. • Data Frames: Data frames are used for carrying data and control information.

Addressing Mechanism • The IEEE 802. 11 addressing mechanism specifies four cases, defined by the value of the two flags in the FC field, To DS and From DS. • Each flag can be either 0 or 1, resulting in four different situations. • The interpretation of the four addresses (address 1 to address 4) in the MAC frame depends on the value of these flags.

• Note that: – Address 1 is always the address of the next device. – Address 2 is always the address of the previous device. – Address 3 is the address of the final destination station if it is not defined by address 1. – Address 4 is the address of the original source station if it is not the same as address 2.

• Example 1: Ten thousand airline reservation stations are competing for the use of a single slotted ALOHA channel. The average station makes 18 requests/hour. A slot is 125 μsec. What is the approximate total channel load? • Average requests for 10, 000 stations = 104 x 18 / (60 x 60) = 50 requests/sec • Average slots number = 1 / (125 x 10 -6) = 8000 slots/sec. • Total channel load: = average requests / average slots number = 50 / 8000 = 0. 0625 • Hence, the total channel load is 0. 0625 request/slot

• Example 2: In a TDM medium access control bus LAN, each station is assigned one time slot per cycle for transmission. Assume that the length of each time slot is the time to transmit 100 bits plus the end-toend propagation delay. Assume a propagation speed of 2 x 108 m/sec. The length of the LAN is 1 km with a bandwidth of 10 Mbps. What is the maximum number of stations that can be allowed in the LAN so that the throughput of each station can be 2/3 Mbps ? • Each slot is = transmission time of 100 bits + propagation delay = 100/10 Mbps + 1 km/2*108 ms = 10μs + 5μs = 15μs • Let there are maximum N number of station, 15μs is the length of cycle and in a whole cycle each user transmit for only 10μs. • Efficiency = 10/15 N • Throughput of each station =(10/15 N)*10 Mbps = 2/3 Mbps • N = (10*10*3) / (15 *2) = 10

• Example 3: On a wireless link, the probability of packet error is 0. 2. A stop and wait protocol is used to transfer data across the link. The channel condition is assumed to be independent from transmission to transmission. What is the average number of transmission attempts requires to transfer 100 packets? • In stop and wait protocol sender will transmits : 100*(1+0. 22 +0. 23 +0. 24 +……………) packets = 100 *(1/(1 -. 02)) =100/0. 8 = 125 (sum of infinite G. P. is a/(a-r))

• Example 4: A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 x 108 m/sec. What is the minimum frame size for this network ? • Tp = (1 km)/( 2 x 108 m/s) = 5 x 10 -6 Seconds • Let S is minimum frame size. • Minimum frame size can be found by formula Tx = 2 x Tp • S/1 Gbps = 2 x 5 x 10 -6 • S = 109 x 10 -5 =104 bits

• Example 5: What is the length of a contention slot in CSMA/CD for (a) a 2 -km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)? , and (b) a 40 -km multimode fiber optic cable (signal propagation speed is 65% of the signal propagation speed in vacuum)? • (a) Signal propagation speed in twin lead is 2. 46 × 108 m/sec. Signal propagation time for 2 km is 8. 13 μsec. So, the length of contention slot is 16. 26 μsec. • (b) Signal propagation speed in multimode fiber is 1. 95 × 108 m/s. Signal propagation time for 40 km is 205. 13 μsec. So, the length of contention slot is 410. 26 μsec.

• Example 6: A group of N stations share a 56 -kbps pure ALOHA channel. Each station outputs a 1000 -bit frame on average once every 100 sec, even if the previous one has not yet been sent (e. g. , the stations can buffer outgoing frames). What is the maximum value of N? • With pure ALOHA, the usable bandwidth = 0. 184 × 56 kbps = 10. 3 kbps. • Each station requires 10 bps, so N = 10300/10 = 1030 stations.

• Example 7: Consider five wireless stations, A, B, C, D, and E. Station A can communicate with all other stations. B can communicate with A, C and E. C can communicate with A, B and D. D can communicate with A, C and E. E can communicate A, D and B. (a) When A is sending to B, what other communications are possible? (b) When B is sending to A, what other communications are possible? (c) When B is sending to C, what other communications are possible?

• (a) Since all stations will see A’s packet, it will interfere with receipt of any other packet by any other station. So, no other communication is possible in this case. • (b) B/s packet will be seen by E, A and C, by not by D. Thus, E can send to D, or A can send to D, or C can send to D at the same time. • (c) This scenario is same as (b).