Chapter 4
Medium Access Control Sublayer

The main theme of the chapter is how to allocate a single broadcast channel (versus point to point) among competing users. Various approaches using static and dynamic channel allocation will be examined. Special attention will be given Ethernet.

4.1.1 Static Channel Allocation in LANs and MANs - Important to remember bursty nature of data communications. WANs not considered because normally connected via a point to point channel.

• Frequency Division Multiplexing  - Frequency of one channel divided (usually evenly) among n users. Each user appears to have full channel of full frequency/n. Wastes bandwidth when user has nothing to send or receive, which is often the case in data communications. Other users cannot take advantage of unused bandwidth.
• Time Division Multiplexing - Time of one channel divided (usually evenly) among n users. Each user appears to have full channel for time/n. Same problems as FDM.
• Analysis of Static Channel Allocation - Static allocation is intuitively a bad idea when considering that for an n divisions of a channel, any one user is limited to only 1/n channel bandwidth whether other users where accessing the channel or not. By limiting a user to only a fraction of the available channel, the delay to the user is increased over that if the entire channel were available.

Intuitively, static allocation results in restricting one user to one channel even when other channels are available. Consider the following two diagrams, each with one user wanting to transmit 400 bits over a 1 bit per second channel. The left diagram would have a delay 4 times greater than the diagram on the right, delaying 400 seconds using 1 channel versus 100 second delay using the 4 channels.

Static Allocation 1 channel per user

Dynamic Allocation up to 4 channels per user

One user with entire channel

Generally we want delay to be small. More formally the mean time delay for one channel, T is:

For each of n subchannels, the mean delay time using FDM (T_FDM) is:

C/n = Channel capacity bps (constant) l/n = input rate, frames/sec. m  = average bits per frame TFDM  = 1/ (m(C/n)-(l/n)) = n/(mC-l) = nT

Generally, dividing a channel statically into n channels increases the average delay by a factor of n agreeing with the intuitive result from the diagrams.

4.1.2 Dynamic Channel Allocation in LANs and MANs - Obviously static allocation of a multi-access channel is not generally desirable when overall channel usage is low. Note that channel use is bursty with long periods of inactivity punctuated by short bursts of activity.

4.2 Multiaccess Protocols - Dynamic channel allocation

4.2.1 ALOHA - Radio broadcast on common frequency to transmit. The figure at right illustrates 4 independent stations each attempting to transmit at the same time. Because of the use of a common frequency, when more than one station broadcasts, the signal is effectively destroyed and must be retransmitted.

Pure ALOHA - The first three assumptions are valid along with the following two:

Important points

• Station transmits whenever ready.
• Collision detected through radio feedback. Sending station cannot receive own transmission on collision with other transmitter.
• On collision, sender waits random time (to prevent a certain collision when colliding stations retransmitted) before trying again. When a transmission collides the sender continues sending the full frame to completion.
• Contention period - The time interval in which frames can overlap (i.e. collision can occur).
• Assuming fixed size frames, for Pure ALOHA the contention period begins when one station starts transmitting.
• Because of no carrier sense (i.e. a sender does not check whether channel in use) the contention period lasts for two frame times, since a new station first frame bit transmission could overlap the last bit of the transmitting station's frame.
• Efficiency
• S = channel throughput; the mean new frames generated (per frame time), if S>1 exceeds channel capacity so 0<S<1 assumed. Successfully sending one new frame every two frame times would yield S=0.5 throughput.
• G=channel load (i.e. offered traffic); retransmissions (per frame time) + new frames S, G>= S.
• when no collisions,  retransmissions = 0, G = S.
• when high loads, collisions likely then G>S.
• P₀=e^-2G, probability transmission does not collide for contention interval of 2 frames.
• S=GP₀=Ge^-2G (assuming all new frames get sent eventually). Channel throughput is (retransmitted frames + new frames)*probability of a collision
• Maximum throughput  - At G=0.5, meaning that a frame transmission is attempted by all stations combined about 1/2 the available time. Note that 5 stations each transmitting 100% of the possible frame time would produce G=5.0 offered traffic, far exceeding the channel capacity. Makes some intuitive sense due to the contention period of 2 frames; transmitting more than 1/2 the available frame times would guarantee that most transmissions would result in a collision. 

Then maximum throughput S = Ge^-2G = 0.5e^-2*0.5 = 0.5e^-1 = 1/2e = 1/2*2.71 = 0.184 new frames per frame time. Channel utilization is about 18%.

Pure ALOHA Contention Period is 2 Frame Times

Slotted ALOHA - The assumption of continuous time has been changed from Pure Aloha:

Important points different from Pure ALOHA

• Slot beginning defined by a central synchronization signal received by all stations simultaneously.
• Station transmits whenever ready at the beginning of a slot.
• Contention period - The time interval in which frames can overlap (i.e. collision can occur).
• Assuming fixed size frames, for Slotted ALOHA the contention period begins when slot starts.
• Because of a common reference at the slot start the contention period lasts for one frame time. Two stations that begin transmission simultaneously will overlap.
• Efficiency
• P₀=e^-G, greater probability transmission does not collide since contention interval is now only 1 frame time.
• S=GP₀=Ge^-G  throughput (assuming all new frames get sent eventually).
• Maximum throughput S - At G=1.0 meaning that a frame transmission is attempted by all stations combined each possible time slot which makes intuitive sense because G>1 means all frames cannot be sent while G<1 means that opportunity to send frames is missed. 

Then S = Ge^-G = 1e^-1 = 1/e = 1/2.71 = 0.368 new frames per frame time. Channel utilization is about 36% or double that of Pure Aloha due to halving the cost of a collision.

Slotted ALOHA Contention Period is 1 Frame Time

• Expected number of transmissions per new frame actually transmitted - Recall G=channel load (total number of frames attempted, new and retransmissions, per frame time). The number of transmissions required to send a new frame is E = e^G, growing exponentially with G. Small changes in the load G can then drastically impact a channel's throughput performance S, as illustrated by the following graph.

4.2.2 Carrier Sense Multiple Access Protocols - Persistent and Non-persistent. The first three assumptions are valid along with the following 4 and 5:

• 2t (Minimum time to detect a collision) where t = propagation time between the farthest points of a channel. Suppose Station 1 below transmits a frame that takes t time to arrive at Station 2. Just before frame arrival, Station 2 detects an idle channel and starts transmitting, a collision occurs and is detected by Station 2 almost immediately. However, another t time must pass before Station 1 detects the collision. Stations must transmit for 2t to guarantee collision detection.
• Slot size - Minimum time to detect a collision, 2t.
• Seized channel - After a station has broadcast for 2t time it has exclusive use of the channel since all other stations have had time to sense the channel is in use and will not transmit until the channel is sensed to be idle.

CSMA/CD (Carrier Sense Multiple Access with Collision Detection) - CSMA improved over ALOHA by only transmitting when the channel was idle, but still transmitted the full frame because did not detect a collision occurred to stop. A station can transmit a frame that was much longer than the contention slot 2t time, for example, Ethernet has a contention slot size of 64 bytes but a maximum frame size of 1526 bytes.

Collision detection serves to stop transmitting as soon as a collision occurs, reducing wasted time due to the collision to a maximum of 2t.

1. Problem - What is the minimum frame size (slot size) in bits for:
• L = channel length = 5 km
• p = propagation time per meter = 5*10^-10 sec/m
• b = bit rate = 4 Mb/sec = 4*10⁶ bit/sec
• Bit time = 1/b = 1 bit/4*10⁶ bit/sec = .25*10^-6 sec/bit
• 2t = contention time = 2*L*p = 2*5000m*5*10^-10sec/m = 5*10^-6 sec.
• S = contention slot size = 2*L*p/(1/b) = 2t/(1/b) =  5*10^-6 sec/.25*10^-6 sec/bit = 20 bits
2. Problem - Suppose bit rate was increased from 4 Mb/sec to 10 Mb/sec?
• Bit time = 1/10⁷ sec/bit = 1*10^-7 sec/bit
• S = contention slot size = 2t/(1/b) =  5*10^-6 sec/1*10^-7 sec/bit = 50 bits
3. Problem - What is S for 802.3 standard (Ethernet)
• L = 2.5 km.
• p = 1024*10^-11 sec/m
• b = 10 Mb/sec = 1*10⁷ bits/sec.
• Bit time = 1*10^-7 sec/bit
• 2t = contention time = 2*2.5 km*1024*10^-11 sec/m =5120000m*10^-11sec./m.= 512*10^-7sec.
• S = contention slot size = 2t/(1/b) = 512*10^-7sec/10^-7sec/bit = 512 bit slot
4. Problem - Long channel length (L) is desirable as is higher bit rates (b) however large slot sizes increase the cost due to collisions.
  

What happens to S, the slot size, as L or b increases?       S = 2t/(1/b) = 2*L*p/(1/b)

4.2.3 Collision-Free Protocols - Each of N stations assumed to have an address 0 to N-1. Examine which station gets channel after successful transmission.
 

Basic bit map - Contention slot used to make reservation for transmitting by placing a 1 if station wants to transmit. Stations 1, 3, and 7 make reservation during contention slot, transmit in numerical order during appropriate data slot. All stations can see reservations broadcast by other stations. Utilization high under heavy load since can approach 100% minus the contention slots. Constant delay under light load because a single transmitter must wait through the contention slot time.

Binary countdown - Reduces delay caused by contention slot period. Station wanting to send broadcasts address as bit string starting with high order bit. Addresses from all stations are ORed as arrive (i.e. a system with 256 stations would require an 8-bit address with bits broadcast 7-0; all bit 7's of four station addresses broadcast at same time, ORing 0+0+0+1 = 1). Any station seeing a 1 in a position where their address is 0 drops out and must wait for next contention slot. After all address bits are broadcast the last remaining station transmits. Utilization can be 100% under heavy load. Delay is function of address, station 1111 would enjoy short delay compared to station 0000.

4.2.6 Wireless LAN Protocols

Key problems using CSMA:

• Single frequency creates potential of collision.
• hidden station (a) - Station B is in range of station A and C but A and C are out of each other's range and cannot sense each other's transmission. C transmitting would collide at B but not sensed by A.
• exposed station (b) - With station A and D out of each others range, station B can transmit to A; station C can transmit to D without interference. However, C falsely concludes it cannot transmit because it senses B transmission.
• CSMA works on wire because all stations receive transmission, generally not on wireless because only a subset may receive transmission.

MACA

Multiple Access with Collision Avoidance - Sender stimulates receiver to output short frame that is detected by adjacent stations, preventing them from transmitting.

1. A sends RTS (Request To Send) to B, a 30 byte frame with the length of data frame to follow.
2. B responds with a CTS (Clear To Send) with the copied length of data frame as an acknowledgement.
   • Any station other than B receiving RTS must remain silent long enough for CTS to be transmitted by B. Station C and B receives RTS from A, D out of range.
   • Any station other than A receiving CTS must remain silent long enough for data frame to be immediately transmitted by A. Station D and A receives CTS from B with data frame length, C out of range.
3. A receives CTS and starts sending data frame. C can transmit also after waiting for but not receiving CTS since out of range of B.
4. Collisions can occur if A receives RTS from C and B simultaneously (RTS collision). A does not send CTS causing both B and C use binary exponential backup algorithm to delay a random time before resending RTS.

4.3 IEEE 802 Standards

4.3.1 Ethernet 802.3

Collision detection at sender A can take as long as 2t

• Manchester encoding - To define bit boundaries. Using straight binary encoding, 0000000 could be a long series of low voltages, difficult to determine between bit boundaries since sender and receiver clocks may drift apart. Each bit using Manchester encoding has a voltage change, a 1 might be a high to low voltage and a 0 a low to high voltage. Differential Manchester encoding has better noise immunity by relating the start voltage of a bit to the end voltage of the previous, a 1 is represented by a voltage transition at the start, a 0 by the lack of a transition. Requires double the bandwidth of straight binary encoding.

• Frame format - The frame must be a minimum of 64 bytes due to collision detection. It contains:
• Preamble - 7 bytes for synchronization (10101010101...)
• Start - Framing byte of 10101011
• Destination address - 2 or 6 (unique NIC card address) for receiver to recognize its own frames.
• Source address - 2 or 6 (unique NIC card address) for receiver to recognize the sender.
• Data Length - 2 bytes allows data up 65,535 bytes but limit is 1500.
• Data - 0 to 1500 bytes from higher layer.
• Padding - 0 to 46 bytes in case frame needs to be filled out to 64 bytes.
• Checksum - 4 bytes based on a cyclic redundancy check (CRC) algorithm.
• Example Ethernet frame  Station 1 sending an 'a' to Station 2

• Binary Exponential Backoff - Dynamically adapts to number of stations trying to transmit. As a station experiences repeated collisions the probability that it will attempt to transmit when the channel is idle decreases.
• On a collision each station waits a random number of slots (a slot is minimum 64 bytes long).
• To prevent all stations randomly picking the same number of  slots, on repeated collisions each station randomly picks from a progressively larger number of slots to wait.
• On a collision the range of slots to wait grows to 2ⁿ for the nth collision.
• On the 1st collision a station waits a random number of slots between 0 and 1 (i.e. 2¹ slots), the probability of two colliding stations will attempt to transmit is then 1/2*1/2=1/4, as is the probability that neither will attempt. The probability that only one will attempt is 1/4+1/4=1/2.
• On the 6th collision, a station waits a random number of slots between 0 and 63 (i.e. 2⁶ slots), the probability of two stations that have collided 6 times will both attempt to transmit is then 1/64*1/64=1/4096.
• The average wait can become very long with the maximum wait interval reached after 10 collisions at 0 to 1023 slots (i.e. 2¹⁰ slots with a minimum contention interval of 51.2 msec (one slot time) yields a maximum delay of 1024*51.2 msec =0.524288 sec. or on average about 0.261244 sec. for this attempt in addition to previous attempts).
• Acknowledgments - CSMD/CD offers no acknowledgements, another frame must be sent as a receipt creating additional network traffic.
• Performance - Normally each station has a different transmit probability, the assumption here is that all stations are the same.
• contention interval - The interval during which frames collide. For a slot 64 bytes long, a series of 5 collisions before transmitting a frame results in a contention period of 5 slots * 64 bytes or 320 bytes. For a data rate of 10 Mbps and 51.2 msec. slot interval the contention interval would be 5*51.2 msec.=256 msec.
• k - number of stations always ready to transmit
• p - probability a station is ready to transmit
• A - probability of no collision, some station acquires slot.
• A = kp(1-p)^k-1 when  k=1, A=1*p(1-p)^1-1 =p (1 station to transmit)
•  

  A is maximum when p=1/k For k=4 stations: p=1/4=0.25 A = kp(1-p)k-1=4*0.25(1-0.25)4-1=1*(1-0.25)3 = 0.753 =  0.421 For k=100 stations:         p=1/100=0.01         A = kp(1-p)k-1=100*0.01(1-0.01)100-1=1*(1-0.01)99 = 0.9999 =  0.369 A -> 1/e = 1/2.71 = 0.367 as k -> infinity

•  A(1-A)^j-1 is probability contention interval has exactly j stations attempting to transmit. A, the probability of no collision, (1-A) the probability of one collision, (1-A)^j-1 the probability of j-1 collisions.

Delay based on average number of slots per contention interval. For j=0 to infinity SjA(1-A)j-1 = 1/A the average number of stations attempting to transmit during one contention interval. The smaller the probability A of no collision, the greater the delay.

• w mean contention interval - For a slot of duration 2t, the mean contention interval is w =2t/A. For optimal p the average contention slots is never greater than e, so w <= 2te = 5.4t.

Channel efficiency = P/(P+2t/A).  P is mean time to transmit a frame. As t (slot size) increases the efficiency goes down. Recall what affects t, the channel length and the bit rate, both of which are desirable to increase. Channel efficiency = 1/(1+ 2BLe/cF). As frame length increases efficiency increases but delay would also increase (e.g. if the delay for a given frame size were 1, doubling the frame size would increase delay to 2). From the equation, as bandwidth B or channel length L increase, efficiency decreases due to larger contention slots and corresponding cost of a collision. Increasing F  or c increases channel efficiency. F - frame length B - bandwidth L - channel length c - propagation rate e - optimal case of e contention slots per frame

Efficiency of 802.3 with 64 byte contention slots and different size frames.  Larger frames are more efficient; smaller frames have a higher percentage of overhead and require more transmissions, increasing the probability of a collision and the overall cost of transmitting the frame.

• Switched 802.3 LANs - Provides collision free transmission between switched elements.
• Common collision domain - Multiple ports connected to common switch point using normal collision handling. Switch points are connected collision free using high speed back plane (1 Gbps).
• Individual collision domain - Each port individual collision domain so no collisions can occur. Port transmission is buffered and sent directly to port on same board or across back plane collision free.
• Connected switches - Switches connected over Ethernet would share a collision domain. For example, A1-F1 can communicate collision free across the switch back plane but A1-F1 communicates with A2-F2 over the usual Ethernet collision domain.

4.3.7 Fast Ethernet 802.3u

100Base-TX

• Category 5 UTP wiring
• Full duplex using two twisted pairs

Category 5 UTP - Can handle clock rater of 125 MHz., maximum distance for 100Mbps is 100m.

4.3.8 Gigabit Ethernet 802.3z

• Backward compatible with original 802.3 (10 Mbps)
• Uses switch or hub, switched only discussed here.
• Point-to-point between two stations or more often a switch (hub not discussed here)
• Switch creates point-to-point connection with individual collision domains, no contention so CSMA/CD not required.

Token Ring 802.5

• Topology - Point-to-point though all stations can see traffic as passes through each computer interface, ring is unidirectional.

• Bit time - 1 Mbps channel has bit time of 1 bit /1,000,000 bits/second = 1 msecond/bit.
• Bit length - With a propagation rate of 200 meters/msecond, a 1 Mbps channel has bit length of 200 meters (i.e. a bit time of 1 msecond on a medium with propagation at 200 meters/1 msecond). A 1000 meter ring can then hold 5 bits on wire. Each active station adds a one bit delay due to buffering one bit by interface, a 1000 meter ring with 10 active stations would have a 15 bit delay.
• Token - Token is 24 bits of a special pattern used to claim channel. Ring must have sufficient delay to contain token, since ring changes as machines turned ON and OFF, an artificial delay may need to be added by RING MONITOR. Note in the diagram below that the first 2 bytes of the token are the first 3 bytes of the data frame.

• Ring Monitor - Central control for ring management.  Responsible for:
• adding delays when necessary for ring to hold token,
• removing orphan (sets bits in frame that its been seen first time),
• removes checksum error frames,
• regenerates token if none seen after n*token holding time, etc.
• Occasionally transmits an active monitor present frame.
• Monitor selection
• First machine turned on normally becomes monitor after transmitting CLAIM TOKEN frame and receiving it back..
• Should monitor crash other stations can transmit CLAIM TOKEN frame, if it returns before any others seen, that monitor becomes monitor.
• Transmission - Station claims token by setting a single bit in the 24-bit token (recall each station has a 1-bit buffer through which all ring transmissions pass). Can then transmit as many frames as possible until token holding time expires.
• Acknowledge - Receiving station passes message on, bit-by-bit, marking acknowledge bit of frame as it passes. How would Ethernet acknowledge?
• Draining - Since each bit passed completely around ring back to transmitter, transmitter responsible for removing or draining bits off ring and regenerating token. Token and frame header differ by only 1-bit (note the first three bytes of token and data).
• Token holding time - Normally 10 millisecond or 10^-2 sec, the time limit that one station can transmit after claiming token. Any number of frames can be sent in that time. On a 1 Mbps ring or 10⁶ bps for 10^-2 sec, 10⁴ bits or 10,000 bits could be transmitted. Most token-ring systems run at 16 Mbps so the 160,000 bits could be sent.
• Delay - To transmit, a station must claim the token. The token is passed from station to station, each may claim the token and transmit for token holding time.
• The minimum delay for a station that has just finished transmitting to transmit again is the time to reclaim the token, or at least 24-bit time.

What is the minimum delay on Ethernet?

• The maximum delay to a station that just finished transmitting to transmit again is a factor of the number of stations on the ring and the token holding time. Assuming all N stations transmit for the token holding time of 10 milliseconds, the delay is approximately N*10 milliseconds.

 

What is the maximum delay on Ethernet for N stations trying to transmit. Would acknowledgments affect either calculation?

• Monitor Selection - Usually first station turned on. If monitor dies, any other station can send a CLAIM TOKEN control frame after no ACTIVE MONITOR PRESENT frame has been seen in the frame control field. If the CLAIM TOKEN returns that station is the new monitor. Danger is that monitor make not die but only fail to do parts of its job.
• Breaks - Wiring center, Multi-Access Unit, closes relays when stations connect to ring, opens to maintain the ring when disconnect through power off, resets, or channel breaks.
• Efficiency - Can approach 100% under heavy load when every station transmitting. How does Ethernet perform under heavy load?

 

Ethernet No priorities Non deterministic (not absolutely predictable), as speed increases (bandwidth) efficiency drops because larger contention frame size produces larger minimum packets, greater loss on collisions. Collisions increase with number of transmitting stations. No delay when only one transmitter. Delay not deterministic under high loads. Distributed control via CMSA/CD.

Token Ring Priorities possible Deterministic, performance easily calculated by knowing number of stations and token holding time. No collisions occur. Efficiency always high, increasing with number of transmitting stations. Delay fixed when one transmitter. Delay deterministic under any load when no priorities. Centralized control requiring monitor to control ring management. Possibility monitor could go berserk and no way to remove monitor.

4.3.9 802.2 Logical Link Control - The top half of the data link layer that interfaces to the network layer. Hides the details of whether 802.3 or 802.5 at the medium access layer from the network layer to provide a consistent interface to Ethernet, Token-Ring, etc.

• LLC header add
ed on sender:
1. Sequence number
2. Acknowledgement number
• LLC header copied into MAC frame
• Receiver removes LLC header
• Three service options
1. unreliable datagram service
2. acknowledged datagram service
3. connection oriented service

4.4 WIRELESS LANS

4.4.1 802.11 Protocol Stack

802.11 Protocol Stack

4.4.2 Physical Layer

802.11b

• Operates in 2.4-GHz ISM short-range radio band, same as garage door openers and cordless telephones.

• HR-DSSS (High Rate Direct Sequence Spread Spectrum) communicates at maximum of 11 Mbps. using CDMA. Slower than 802.11a but range about 7 times greater.

• CDMA uses entire 2.4-GHz spectrum by:

  • Each station assigned unique m-bit code called chip-sequence that corresponds to a binary 1.

  • When multiple stations transmit at same time, the signals add rather than garble.

  • To send binary 0 station sends one's complement of chip-sequence.

    • Station with 00110011 chip-sequence sends binary 101 as: 00110011 11001100 00110011

  • Example: Convenient to think in bipolar form of 0=-1 and 1=+1.

    1. Chip sequence assigned to 4 stations.

    2. Chip sequence in bipolar form. C chip-sequence = 01011100 = -1+1-1+1+1+1-1-1 in bipolar form.

    3. Six samples.

       • S₂ = B+C is the addition of chip sequence for B and C sending a 1.

    4. Recovery of C transmission of binary 1 from S₂.

       • Receiver computes inner product of C and S₂, multiplies then adds, and divides by 8 (the number of chips).

         •     -2       0     0      0    +2    +2     0    -2    Received

         • *  -1    +1    -1    +1    +1    +1    -1    -1   C chip sequence in bipolar form

         •     +2      0     0      0     +2   +2     0     -2 = 8/8 = 1

       • Chip sequence chosen so all other chip sequences are orthogonal (i.e. their inner product is 0).

• Recover A from S₄

4.4.3 802.11 MAC Sublayer Protocol

4.7 Data Link Layer Switching - Bridges

Bridges connect two or more LANs by forwarding frames received. Operate at data link layer so is best effort, if frame cannot be delivered, too bad.

• Reasons for a bridge
• Connect 802.x to 802.y, the bridge would require two NICs each connected to the respective LAN.
• Load balancing among several LANs. A heavily congested LAN can be split into two or more LANs.
• Distance, 802.3 has 2.5 km limit
• Reliability, isolate LAN segments from each other's faults. Suppose an Ethernet machine goes transmit wild.
• Security, can filter addresses allowing only those specified across the bridge.
• Logical Link Control
• Provides consistent interface to network layer regardless of 802.x protocol.
• Placed in data load of 802.x frame, consists of header, sequence and acknowledge numbers. Supports unreliable and acknowledged datagram and connection-oriented services.
• Allows connection of 802.x to 802.y LANs. Figure 4-35.
• 802.x to 802.y Connections
• 802.3 to 802.3, only problem occurs due to collisions on destination LAN preventing delivery of frame.
• 802.11 to 802.3, use different frame formats requiring reformatting and calculation of checksum, takes time.
• 802.3z to 802.3, 1000Mbps to 10 Mbps
• 802.3 to 803.5, must generate priority bit.
• 802.5 to 802.3
• Must lie about Acknowledge and Copy bit, since bridge must pass along received bits immediately before sending 802.3 frame.
• Arbitrarily large 802.5 frame may not fit on 802.3 and would be discarded.
• 802.5 to 802.5, again must lie about Acknowledge and Copy bit.

Transparent Bridges

• Administration - none, just connect to LAN.
• Frame - standard with no special forwarding information.
• Forwarding
• if destination and source LAN same discard since no need to bridge.
• if destination and source LAN different, forward on to a known destination LAN segment.
• if destination LAN unknown, flood on all but arrival LAN.
• Backward learning - when frame arrives the source address and LAN is marked in table so future frames to that destination can be forwarded. Note that does not learn silent host addresses, non-transmitting destinations are always flooded.
• Table updates - to accommodate topology changes, addresses are updated as new source frame addresses arrive.
• Table purges - periodically table entries are purged so that machines can be moved, otherwise the learned address of machine that has been moved is wrong until it transmits.

Spanning Tree Bridges

• Cycles - when multiple paths through network due to redundant bridges, possible that flooding can cause frames to be passed endlessly between bridges.
• Spanning tree (text p.369) - a tree where every LAN is reachable from a root. It is desirable that this be a minimal spanning tree based upon a metric such as channel speed, cost, etc. Recall that a tree has no cycles.
• Construction - the lowest numbered NIC is selected as tree root for which the minimum path to all LANs is determined. Minimum path is unique hence redundant paths are eliminated.
• Changes - bridges continue to run spanning tree algorithm since topology may change as bridges or channels added or removed.

The following illustrates a network with multiple transparent bridges and a spanning tree following the minimum path to all LANs for the network at right. The numbers on the connecting LANs represent some metric such as channel cost or bandwidth that could be used to determine the minimal spanning tree. The spanning tree must still connect all network segments but without cycles a necessary condition to prevent flooded frames from generating redundant frames.

When multiple bridges connect the same LAN, such as LAN O connected by B₂₄ and B₁₁ and LAN G connected by B₁₂ and B₁₁, one of the bridges will not transmit to that LAN, in this case B₁₁ is effectively disconnected from both LAN G and O.

Backward learning - Each bridge builds a table as frames arrive containing the source host and arrival LAN. For example, host H₂ on LAN Y sending a frame to host H₁ on LAN X would cause the frame to be flooded across the spanning tree and an entry to be added in each of the bridge routing tables along the spanning tree:

When H₁ responds by transmitting a frame onto LAN X the bridge B₂₄ forwards the frame toward H₂ on LAN N. Each bridge along the path from H₁ to H₂ would record the source host H₁ on arrival LAN in its table. The table below reflects routing tables after the transmissions by hosts H₁ and H₂.

a) Interconnected LANs b) A spanning tree covering the LANs.

Transparent Bridges - Advantages/Disadvantages

• No administrative setup or maintenance when topology changes due to reorganization or failures.

• Frame unchanged, contains only data payload.

• Not optimal use of redundant bridges under heavy traffic since some are removed to build spanning tree.

• Bridge software relatively complex to maintain table, backward learning, updating, spanning tree algorithm. Host software simple.

Source Routing Bridges

• Administration, must number each bridge uniquely for a LAN and each LAN uniquely for network.
• Data frame from host contains complete path from source to destination as bridge and LAN number.
• Forwarding, bridge searches frame path for its own LAN and address, forwards to next LAN in path.
• Route learning
1. Source host sends special discovery frame which is flooded by bridges on all but arrival LAN, each bridge adding their address allowing frame can be discarded if seen again (i.e. if bridge sees a frame with its own address means that the frame is in a cycle).
2. Destination host sends back the received frame, which now includes source to destination path, back to source, giving it source to destination path. Possible only first frame responded to, since arriving first it would have presumably followed best path.


The following figure shows one possible discovery frame created for host H₂ to transmit to H₁. The discovery frame contains the source and destination hosts accumulating the bridges crossed as the frame is flooded. Each bridge would broadcast the discovery frame on all but the originating LAN. Presumably the discovery frame reaching host H₁ first would be returned as the best route between H₁and H₂. Note that 27 discovery frames would converge on H₁.

Source Routing Bridges - Advantages/Disadvantages

• Requires administration on setup and whenever topology changes for any reason. Difficult to detect duplicate server or LAN numbers since problems evident only on certain paths.
• Server simple, hosts complex. Many more hosts than servers. Each host must be capable of source routing.
• Flooded frames grows exponentially. If network contained 9 bridges in a heirarchy as below, a discovery frame from the H host  is copied by each of the three level 3 bridges, flooding nine frames to level 2 bridges, which flood 27 frames to level 1 bridges:



• On bridge failure, host does not know whether destination or bridge failed, must timeout and send new discovery frame. Since discovery frame is flooded, early timeouts could incur serious overhead. Major bridge failure forces many simultaneous discovery frames from affected hosts.
• Presumably allows multiple paths with multiple bridges.
• Splitting an existing LAN requires making changes to all host software to enable source routing.

Remote Bridges

Connect two or more distant LANs using bridges at connection of LAN to a point-to-point line. Common for bridge to place complete MAC frame in payload field of PPP and transmit on point-to-point line. Receiving bridge copies payload field onto destination LAN. Only problem is getting frames to correct LAN, requires translation of MAC address to an IP address when using PPP.

FDDI - Fiber Distributed Data Interface

• 100 Mbps token ring LAN up to 200 km and 1000 stations.
• Commonly used as backbone to join copper LANs via bridge with 802.x and FDDI NIC.
• Uses cheaper multimode fiber since relatively low speed. Multimode fiber uses inexpensive and reliable LEDs rather than laser as source and less precise fiber. Cannot transmit as far or in as high a bandwidth as a laser.
• Consists of two rings, one in each direction. If break occurs in one the other is used as backup. If both break, each station can connect the two rings, forming a new ring with about twice the length.
• For fault tolerance, class A stations connect to both rings and can heal breaks, class B stations connect to one ring and cannot heal breaks.
• Does not use Manchester encoding for synchronization on every bit, which is self-clocking but requires double the bandwidth. Instead, uses a long synchronizing preamble and requires clock with not more than 0.005 per cent drift, ensuring that 4500 byte frames can be sent.
• Follows 802.5 closely but does not have to wait for frame to return before regenerating token. Since 200 km and 1000 stations introduce considerable delay waiting for frame to return before regenerating token, this allows multiple frames to exist on ring simultaneously.
• Also supports synchronous frames for circuit-switched PCM and ISDN traffic generated every 125 microseconds by a master station. Each frame has 16 bytes of non-circuit switched data and 96 circuit-switched data bytes (for 4 PCM 24 channel frames). Stations reserve a time slot within a synchronous frame, holding it till released.

Fast Ethernet - 802.u

• Backward compatible with 802.3 10Base-T wiring but supports 100Mbps

Three wiring standards

100Base-T4, 100m, half duplex?, using all 8 category 3 UTP wires (internal phone wire) to hub that forms one collision domain.

100Base-TX, 100m, full duplex, only 4 of the category 5 wires to hub used, hub forms one collision domain.

100Base-FX, 2000m, full duplex, using fiber optics to switched hub. Since frames are too long for collision algorithm, with switched hub each connection is it own collision domain, allowing all stations to receive and transmit simultaneously.

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