CN Computer Networks Research Group G R Department of Computer - PowerPoint PPT Presentation

Closed-Form Expression for the Collision Probability in the IEEE EPON Registration Scheme Swapnil Bhatia (with Dr. Radim Bartoˇ s) CN Computer Networks Research Group G R Department of Computer Science University of New Hampshire This work was supported in part by the Cisco University Research Program.

✬ ✩ Outline ◮ Introduction to Ethernet Passive Optical Networks (EPONs) ◮ IEEE EPON Registration Scheme ⋄ Description and model ◮ Efficiency of the Registration Scheme ⋄ Two devices ⋄ n devices ◮ Summary and discussion ✫ ✪ IEEE GLOBECOM 2005 2 of 29

✬ ✩ IEEE 802.3ah Ethernet Passive Optical Network ◮ Shared, passive, optical, access network ◮ Directed tree topology ◮ Central arbitrator: OLT ONU m ( Optical Line Terminator ) k 0 2 ONU ONU ONU ◮ Subscriber Device: ONU ONU ONU ( Optical Network Unit ) Backbone OLT ONU ONU ONU ◮ Data rate of 1 Gbps ONU Optical ONU Fiber (bidirectionally) ONU Passive User data Optical Splitter ONU ◮ 20 km reach = ⇒ RTT OLT: Optical Line Terminator ONU: Optical Network Unit ≤ 200 µs ◮ Recently standardized by the IEEE 802.3ah group ✫ ✪ IEEE GLOBECOM 2005 3 of 29

✬ ✩ IEEE EPON Discovery and Registration Scheme ◮ Allow new ONUs to join the network ⋄ Synchronization ◮ Provide new ONUs with transmission parameters ⋄ Logical link identifiers, security (?) and QoS (?) ⋄ Enable RTT calculation by new ONUs ◮ RTT to new ONUs unavailable to the OLT, initially ⋄ = ⇒ a contention-based scheme ✫ ✪ IEEE GLOBECOM 2005 4 of 29

✬ ✩ IEEE EPON Discovery Protocol ◮ OLT reserves a sufficiently OLT ONU large window of time GATE Random delay REGISTER_REQ Discovery ◮ Broadcasts a “Discovery start” window message to all ONUs REGISTER ⋄ Contains length of window GATE ◮ New ONUs follow Random REGISTER_ACK Wait rule: ⋄ Wait for a random period after receiving message ⋄ Transmit request to join network ⋄ Expect a Register reply before next Discovery cycle ⋄ Finish with an Ack , or retry ✫ ✪ IEEE GLOBECOM 2005 5 of 29

✬ ✩ IEEE EPON Discovery Protocol ◮ Discovery window required ONU OLT periodically GATE Random ⋄ Every few minutes delay REGISTER_REQ Discovery window ◮ No subscriber traffic during Discovery window REGISTER ⋄ Including voice, video etc. GATE ◮ = ⇒ Important to choose window size wisely REGISTER_ACK ◮ Questions: ⋄ How do we choose the most efficient window size? ⋄ What do we mean by the efficiency of the Random Wait scheme? ✫ ✪ IEEE GLOBECOM 2005 6 of 29

✬ ✩ Efficiency of the IEEE EPON Discovery Protocol ◮ Observation: Larger the discovery window size (say w ), ⋄ larger the number of successful registrations ⋄ but larger the bandwidth wasted ◮ ∴ High efficiency means: ⋄ Maximal number of successful registrations, with ⋄ Minimal window size ◮ ∴ Efficiency of Discovery Window ⋄ Could be defined as: Number of successful registrations = ρ Size of Discovery window ◮ “Number of successful registrations” is a random variable ✫ ✪ IEEE GLOBECOM 2005 7 of 29

✬ ✩ Efficiency of the IEEE EPON Discovery Protocol ◮ Suppose: ⋄ there are n ONUs ⋄ P s ( n ) is the probability of a successful registration for an ONU, in the presence of n − 1 other ONUs ⋄ T is the duration of the discovery window reserved by the OLT ◮ Efficiency of Discovery Window ρ = n · P s ( n ) T ✫ ✪ IEEE GLOBECOM 2005 8 of 29

✬ ✩ Probability of a Successful Registration: P s ( n ) ◮ For ONU- i , let random variable: � its random RTT, X i � Y i the random wait, and � the time of arrival of the message at the OLT. Z i OLT ONU GATE X ′ i window T = 2 p + w Random Z i Y i delay Reserved X ′ i REGISTER_REQ = ⇒ Z i = X i + Y i 2 X ′ where : X i = i ✫ ✪ IEEE GLOBECOM 2005 9 of 29

✬ ✩ Efficiency of the IEEE EPON Discovery Protocol ◮ Efficiency of Discovery Window ρ = n · P s ( n ) 2 p + w where: is the total number of ONUs in the EPON, n P s ( n ) is the probability of a successful registration for an ONU, in the presence of n − 1 other ONUs, is the maximum propagation delay to any new p ONU, and is the size of the discovery window w ◮ Rule: Choose a window size w that maximizes ρ , given n and p . ◮ Question: How do we find P s ( n )? ✫ ✪ IEEE GLOBECOM 2005 10 of 29

✬ ✩ Probability of a Successful Registration: P s ( n ) ◮ Suppose OLT chooses window size to be w µ s ⋄ = ⇒ Wait Y i is ∈ [0 , w ] ⋄ distributed uniformly (IEEE standard) ◮ Suppose ONUs are located at a maximum propagation delay p ⋄ = ⇒ RTT X i ∈ [0 , 2 p ] ⋄ distributed uniformly (reasonable) ◮ Then, density of Z i = X i + Y i : f Z i ( z ) 1 m M + M m M z M − z m Z i m M + m M m = min(2 p, w ) , M = max(2 p, w ) ✫ ✪ IEEE GLOBECOM 2005 11 of 29

✬ ✩ Probability of a message collision: 1 − P s ( n ) ◮ Consider 1 − P s (2): the probability of a message collision for an ONU, in the presence of 1 other ONU ◮ Consider arrival times Z 1 and Z 2 ( = ⇒ independent) ◮ The joint density of Z 1 , Z 2 : Z 2 M + m f 3 ( z 2 ) 0.01 M k − 0.008 = Z 2 − 0.006 Z 1 f 2 ( z 2 ) 0.004 k = 2 Z 0.002 − k 14 Z 1 12 0 10 14 8 12 z1 6 10 8 f 1 ( z 2 ) 4 Z 1 6 z2 4 2 2 M + m m k M 0 0 f 1 ( z 1 ) f 3 ( z 1 ) f 2 ( z 1 ) ◮ The collision event: | Z 1 − Z 2 | ≤ k where k � message length ✫ ✪ IEEE GLOBECOM 2005 12 of 29

✬ ✩ Probability of a message collision: 1 − P s (2) Z 2 M + m − k M + m R 7 Z 2 M + m − k M + m f 3 ( z 2 ) M + k R 8 M + m − k f 3 ( z 2 ) R 7 R 9 M − k R 12 R 9 M − k M M M + k R 13 k R 8 k − R 4 R 4 = − M + m − k = Z 2 M − k Z 2 − − 1 Z f 2 ( z 2 ) Z 1 M − k R 5 k = f 2 ( z 2 ) R 5 Z 2 m + k m + k − k = R 6 Z 1 Z 2 R 10 m − k − k m Z 1 R 6 R 11 m + k R 1 R 2 m k R 2 f 1 ( z 2 ) m − k m + k R 1 R 3 f 1 ( z 2 ) Z 1 Z 1 R 3 M + m m k M M M + m k m f 1 ( z 1 ) f 3 ( z 1 ) f 2 ( z 1 ) f 1 ( z 1 ) f 2 ( z 1 ) f 3 ( z 1 ) m ≥ k m < k M − m ≥ k ✫ ✪ IEEE GLOBECOM 2005 13 of 29

✬ ✩ Probability of a message collision: 1 − P s (2) Z 2 Z 2 M + m − k M + m − k M + m M + m R 11 R 3 R 2 M + k f 3 ( z 2 ) m + k f 3 ( z 2 ) R 12 R 8 M − k M + m − k M m + k Z 1 − Z 2 = − k R 9 R 1 R 13 M − k 6 M M + k k R 10 f 2 ( z 2 ) R 7 R 7 R 4 f 2 ( z 2 ) m − k M + m − k m m + k k 8 R 1 − = R 6 R 5 M − k 2 Z m k − m + k R 9 1 Z R 2 k f 1 ( z 2 ) Z 1 − Z 2 = k = f 1 ( z 2 ) M − k m − k 2 Z − R 5 1 Z R 4 R 3 Z 1 Z 1 m M + m k m M + m k M M f 3 ( z 1 ) f 1 ( z 1 ) f 3 ( z 1 ) f 1 ( z 1 ) f 2 ( z 1 ) f 2 ( z 1 ) m ≥ k m < k M − m < k ✫ ✪ IEEE GLOBECOM 2005 14 of 29

✬ ✩ Probability of a message collision: 1 − P s (2) Z 2 M + m − k M + m Z 1 − Z 2 = − k Z 1 − Z 2 = − k Z 2 R 1 f 3 ( z 2 ) k k M + m M f 3 ( z 2 ) M R 2 f 2 ( z 2 ) f 2 ( z 2 ) m Z 1 − Z 2 = k Z 1 − Z 2 = k m M + m − k f 1 ( z 2 ) f 1 ( z 2 ) R 3 Z 1 Z 1 M + m M + m M k m M k m f 1 ( z 1 ) f 3 ( z 1 ) f 2 ( z 1 ) f 1 ( z 1 ) f 2 ( z 1 ) f 3 ( z 1 ) M + m > k M + m ≤ k m < M < k ✫ ✪ IEEE GLOBECOM 2005 15 of 29

✬ ✩ Probability of a message collision: 1 − P s (2) Z 2 M + m − k M − k M M + m Z 1 − Z 2 = − k R 1 k R 2 f X ( z 2 ) Z 1 − Z 2 = k M − k M + m − k R 3 m Z 1 k M m M + m f X ( z 1 ) m = 0 , M > k ✫ ✪ IEEE GLOBECOM 2005 16 of 29

✬ ✩ Probability of a message collision: 1 − P s (2) true false k > 0 , M > 0 true false m = 0 true false false true M > k M − m < k false true false true m < k m < k false true M < k true false M + m ≤ k P 5 P 4 P 1 P 6 P 6 P 3 P 2 P 6 P 7 ✫ ✪ IEEE GLOBECOM 2005 17 of 29

✬ ✩ Probability of a message collision: 1 − P s (2) 1 0.8 0.6 0.4 400 0.2 300 200 p 0 0 100 100 200 0 w 300 400 k = 2 . 528 µ s m = min(2 p, w ) , M = max(2 p, w ) ✫ ✪ IEEE GLOBECOM 2005 18 of 29

✬ ✩ How about P s ( n )? ◮ A successful transmission by Z 1 can be expressed as: n � � � | Z 1 − Z i | > k i =2 ◮ ∴ P s ( n ): ∞ � n � n � �� �� � � � � | Z 1 − Z i | > k = | Z 1 − Z i | > k | Z 1 = t · P ( Z 1 = t ) · dt P P i =2 i =2 −∞ ∞ � [1 − F Z 2 ( t + k ) + F Z 2 ( t − k )] ( n − 1) · f Z 2 ( t ) · dt = −∞ ✫ ✪ IEEE GLOBECOM 2005 19 of 29

✬ ✩ How about P s ( n )? f Z i ( z ) 1 m M + M m M z M − z m Z i m M + m t M M + k t − k m − k m + k t + k M − k M + m − k ∞ � n � �� [1 − F Z 2 ( t + k ) + F Z 2 ( t − k )] ( n − 1) · f Z 2 ( t ) · dt � � | Z 1 − Z i | > k = P i =2 −∞ ◮ Must consider relative magnitude of t in addition to m, M, k as before ⋄ Many cases, quite tedious ✫ ✪ IEEE GLOBECOM 2005 20 of 29

✬ ✩ How about P s ( n )? ◮ A simple approximation ⋄ An ONU’s transmission was successful in the presence of n − 1 other ONUs ( P s ( n )) ⋄ = ⇒ it was successful in the presence of each of the n − 1 ONUs, taken one at a time ( P s (2)) P s (2) n − 1 P s ( n ) ≈ ⋄ Done! ✫ ✪ IEEE GLOBECOM 2005 21 of 29