CN Presentation ALOHA Slotted ALOHA Services of DLL

CN Presentation ALOHA Slotted ALOHA

Services of DLL • • Framing Link access Reliable delivery Error detection and correction (bit level error detection and correction) 6/17/2021 Data Link Layer 2

Data Link Layer is divided into two sublayers: 1) LLC: Logical Link Control layer 2) MAC: Media Access Control layer

Media Access Control layer (MAC Layer) In the Open Systems Interconnection (OSI) model of communication, the Media Access Control layer is one of two sublayers of the Data Link Control layer and is concerned with sharing the physical connection to the network among several computers. Each computer has its own unique MAC address. Ethernet is an example of a protocol that works at the Media Access Control layer level.

RANDOM ACCESS In random access, no node is superior to another node and none is assigned control over another. No node permits, or does not permit, another node to send. At each instance, a node that has data to send uses a procedure defined by the protocol to make a decision on whether or not to send.

Random Access MAC protocols • ALOHA • slotted ALOHA • CSMA, CSMA/CD, CSMA/CA

ALOHAnet, also known as the ALOHA System, or simply ALOHA, was a pioneering computer networking system providing the first public demonstration of a wireless packet data network. It refers to a simple communications scheme in which each source (transmitter) in a network sends data whenever there is a frame to send. If the frame successfully reaches the destination (receiver), the next frame is sent. If the frame fails to be received at the destination, it is sent again.

Pure ALOHA • unslotted Aloha: simpler, synchronization • when frame first arrives: transmit immediately • collision probability increases: Disadvantages of Pure ALOHA: 1) Time is wasted 2) Data is lost 3) Collisions were more no

Slotted ALOHA An improvement to the original ALOHA protocol was "Slotted ALOHA", which introduced discrete timeslots and increased the maximum throughput. A station can send only at the beginning of a timeslot, and thus collisions are reduced.

Pure ALOHA • New arrivals are transmitted immediately (no slots) – No need for synchronization – No need for fixed length packets • In pure ALOHA, the stations transmit frames whenever they have data to send. • When two or more stations transmit simultaneously, there is collision and the frames are destroyed. • In pure ALOHA, whenever any station transmits a frame, it expects the acknowledgement from the receiver. • If acknowledgement is not received within specified time, the station assumes that the frame (or acknowledgement) has been destroyed. • If the frame is destroyed because of collision the station waits for a random amount of time and sends it again. This waiting time must be random otherwise same frames will collide again and again.

• Therefore pure ALOHA dictates that when time out period passes, each station must wait for a random amount of time before resending its frame. This randomness will help avoid more collisions. • If acknowledgement is not received within specified time, the station assumes that the frame (or acknowledgement) has been destroyed. • If the frame is destroyed because of collision the station waits for a random amount of time and sends it again. This waiting time must be random otherwise same frames will collide again and again. • Therefore pure ALOHA dictates that when time out period passes, each station must wait for a random amount of time before resending its frame. This randomness will help avoid more collisions.

Efficiency of Pure Aloha

P(success by given node) = P(node transmits). P(no other node transmits in [t 0 1, t 0]. P(no other node transmits in [t 0, t 0+1] = p. (1 p)N-1 = p. (1 p)2(N-1) Differentiate the above equation to get value of ‘p’. Putting worst case value for ‘n’ = ∞ Efficiency(Throughput) = 1/2 e = 0. 18 That is 18 % efficiency!

A group of N stations share a 56 kbps pure ALOA channel. Each station outputs a 1000 bit frame on an average of once every 100 sec, even if the previous one has not yet been send (e. g. the stations are buffered). What is the maximum value of N ?

Solution. Recall that ALOHA achieves an average throughput of appr 18%, when operating at reasonable load. In an ALOHA network with channel capacity 56 kbps, only 18% of this capacity will be used to deliver meaningful data. With pure ALOHA the usable bandwidth is 0. 184 * 56 kbps = 10. 3 kbps. This 10 kbps must be divided among N hosts, each of which is transmitting an average of 1000 bits every 100 seconds. This corresponds to a transmission rate of 10 bits per second per host. If the channel can support 10 kbps of data, then it can support up to N users, each transmitting at 10 bps Each station requires 10 bps (1000 bit/100 sec), so N = 10300 /10 = 1030 stations.

Slotted ALOHA

• Slotted ALOHA is a refinement over the pure ALOHA. • The Slotted ALOHA requires that time be segmented into slots of a fixed length exactly equal to the packet transmission time. • Every packet transmitted must fit into one of these slots by beginning and ending in precise synchronisation with the slot segments. • A packet arriving to be transmitted at any given station must be delayed until the beginning of

. The Slotted ALOHA requires that time be segmented into slots of a fixed length, T, exactly equal to the packet transmission time. Every packet transmitted must fit into one of these slots by beginning and ending in precise synchronization with the slot segments. A station can send only at the beginning of a timeslot, and thus collisions are reduced Slotted ALOHA requires additional overhead to provide the synchronization required between the different stations in the network. T = Transmission D Vulnerable Time = Time during which collisio

S = Number of frames successfully received for a given unit time G=Data/S lot

Slotted ALOHA Vs Pure ALOHA • Pure ALOHA do not required global time synchronization. • Pure ALOHA station can send data in continuous time manner. • Vulnerable time for the pure ALOHA is 2 X Tfr • Throughput is G * exp( 2 G) [MAX = 0. 18] Slotted ALOHA • It requires global time synchronisation. • It divides the time in slots. • Vulnerable time is Tfr. • Throughput is G * exp( G) [MAX = 0. 36] G=Data/Slot

Problem on Slotted Aloha Ques 1. Let G = 0. 5 [frames/slot] be the total rate at which frames are transmitted in a slotted ALOHA system. What proportion of slots will be collision free? What proportion of slots will be collision free when the system is operating at its maximum throughput?

Solution: Proportion of slots that will be collision free = P[0 frames arriving in the datalink layer in a slot] = e –G = e 0. 5 The value of G maximum throughput condition in slotted ALOHA is given by G = 1. P[0 frames arriving in the datalink layer in a slot] = e −G = e 1 = 0. 368

Ques 2. A large population of ALOHA users manages to generate 50 requests/sec, including both originals and retransmissions. Time is slotted in units of 40 msec. (a) What are the chances of success on the first attempt? (b) What is the probability of exactly k collisions and then a success? (c) What is the expected number of transmission attempts needed?

Solution: a) Slott time(x) = 40 ms load (request/sec)= 50 requests/sec Arrival Rate G(frames/x sec)=50 x (40 x 10 3) = 2 requests/x sec Prob of successful trans in 1 st attempt = e G =e 2=0. 1353 b) Prob of successful trans in (k+1)th attempt = P[collision in k attempts] x P [success in (k+1) attempt] P[collision in k attempts] = (1 e G )k Prob of successful trans in (k+1)th attempt = e G (1 e G )k = 0. 8647 k (0. 1353)

(c)Expected number of Transmissions = ∑ k[Prob of success in k transmissions] = ∑ k[e G (1 e G )k 1 ] =e G ∑ k (1 e G )k 1 = ( e G /(1 e G ) )∑ k (1 e G )k Also, ∑ kxk = (x / (1 x 2)) Expected number of Transmissions = ( e G /(1 e G )) *(1 e G )/1 (1 e G )2