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Ethernet

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Review Media Access Control – Broadcast media shared by all stations – MAC is used to determine who gets the right to send Developed protocol (All contention based) – P0. Send at will. – P1. ALOHA. – P2. Slotted ALOHA – P3. Carrier Sense. (Ask which curve is for which) 1-persistent Non-persistent P-persistent – P4. CSMA+CD

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Contention Free? The contention based protocols allows contention, so works well under light load. How to design contention-free protocols? The key is: let other people know that I am going to send

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Contention Free Protocols – Bitmap bit-map method. – control frame contain N bits, each station send 1 bits to indicate whether it has a frame to send – at the end of the control frame, every station knows all stations that want to send, the station can send in order. – example: – Performance: d/(d+1) channel utilization rate for high load. N bits delay for low load. (d is the frame size).

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Contention Free Protocols: Binary Count Down each station sends the address bits in some order The bits in each position from different stations are ORed. As soon as a station sees that a high-order bit position that is 0 is overwrite by 1, it gives up. Eventual, only one station (with largest station number among all the competitors) gets the channel. Performance: – channel utilization rate: d/(d+log(N)) for high load – log(N) bits delay for low load. – Contention field can serve as the address field. This protocol assumes that delays are negligible!

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Contention Free Protocols – Token Ring Passes tokens among stations and only stations got the token is allowed to send

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Limited Contention Protocols Want to be as contention-based protocols (ALOHA) under light load and as contention- free protocols under heavy load (bitmap) Can we achieve that? The trick is to dynamically control the size of the group that can contend for a slot – under light load, only one group including everyone and everyone can try for each slot – this is aloha – under heavy load, the group size is small and each group can only try for his slot – when the size of the group is 1, this is bitmap

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Limited Contention Protocols So, how to determine the right size of the group? Each station monitors the load (because it is broadcast media). If there are N stations and n have frames to send, divide into N/n groups

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Limited Contention Protocols What happens if there is a contention inside a group? Many ways to deals with it. One way we will talk about is to send all frames of this group, then move to the next group. Suggestions?

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Limited Contention Protocols 0 1 Like a binary search. Example: 2 3 45 6 ABC* D E*F*GH* Slot 0: C*, E*, F*, H* (all nodes under node 0 can try), conflict slot 1: C* (all nodes under node 1 can try), C sends slot 2: E*, F*, H*(all nodes under node 2 can try), conflict slot 3: E*, F* (all nodes under node 5 can try), conflict slot 4: E* (all nodes under E can try), E sends slot 5: F* (all nodes under F can try), F sends slot 6: H* (all nodes under node 6 can try), H sends.

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Ethernet (802.3) 1-persistent CSMA/CD + binary exponential backoff Carrier sense: station listens to channel first 1-persistent: If idle, station may initiate transmission Collision detection: continuously monitor channel and if collision, abort transmission immediately, and wait for a random time binary exponential backoff (new, how to pick the random time): each time slot is 51.2 us first collision, retransmission interval = random number between [0,1] second collision, interval = random number between [0,1,2,3] kth collision, interval = random number between [0, 2^k-1] upper bound 1023 slots.

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Why binary exponential backoff Why not pick a random number from a fixed interval? Why a fixed small interval not good? Why a fixed large interval not good?

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Ethernet Frame Format (a) DIX Ethernet, (b) IEEE 802.3

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Minimum Frame Size Why a minimum frame size is needed? How long does it take for a station to notice a collision?

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Worst case

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Minimum Frame Size So, if maximum delay is t, the minimum frame size is 2t*bit rate. t is about 50us. So the minimum frame size of 10M Ethernet is 512 bits. What if the speed goes up?

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Ethernet Performance Suppose there are k stations. Let p be the probability that a station has a frame to send when the channel is idle. Assume it is independent across stations, and is independent for one station at different times. Find the average number of collisions before a frame is sent. First, the probability that one station got the chance to send is A=kp(1-p)^{k-1}. Second, maximized when p=1/k. So A is bounded by (1- 1/k)^{k-1}. Third, each contention is indepedent, so average number of collision is 1/A, which is e when k is large. Each contention is 2t, so channel efficiency is P/P+2et.

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Switched Ethernet Stations connect to a switch using dedicated lines. Input frames are buffered. So no collision!

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Ethernet Physical medium – thin cable/thick cable/twisted pair/fiber 10Base5 500 meters thick (cable) Ethernet 100 nodes/seg 10Base2 200 meters thin (cable) Ethernet 30 nodes/seg 10BaseT 100 meters twist pair 1024 nodes/seg 10BaseF 2000 meters fiber optics 1024 nodes/seg 10Base5/10Base2, cable connected to each machine 10BaseT -- connecting to a hub 10BaseF -- between building Connecting

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Ethernet Fast Ethernet Keep everything in Ethernet, make the clock faster 100Mbps. Cable – 100Base-T4 100m category 3 UTP, 4 lines. – 100Base-Tx 100m category 5 twisted pair – 100Base-Fx 2000m Fiber optic

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Review: Medium Access Control Sublayer –What is the problem to be addressed in this sublayer? –Protocols that allow collision Pure ALOHA Slotted ALOHA.

Review: Medium Access Control Sublayer –What is the problem to be addressed in this sublayer? –Protocols that allow collision Pure ALOHA Slotted ALOHA.

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