Switching Units

Types of switching elements Telephone switches switch samples Datagram routers switch datagrams ATM switches switch ATM cells INPUTS OUTPUTS Lecture 5

Repeaters, bridges, routers, and gateways Repeaters/Hubs: at physical level (L1) Bridges: at datalink level (L2) based on MAC addresses discover attached stations by listening Routers: at network level (L3) participate in routing protocols Application level gateways: at application level (L7) treat entire network as a single hop Gain functionality at the expense of forwarding speed for best performance, push functionality as low as possible Lecture 5

Types of services Packet vs. circuit switches packets have headers and samples don’t Connectionless vs. connection oriented connection oriented switches need a call setup setup is handled in control plane by switch controller connectionless switches deal with self-contained datagrams Lecture 5

Other switching unit functions Participate in routing algorithms to build routing tables Next Lecture! Resolve contention for output trunks buffer scheduling Previous Lecture! Admission control to guarantee resources to certain streams Lecture 5

Requirements Capacity of switch is the maximum rate at which it can move information, assuming all data paths are simultaneously active Primary goal: maximize capacity subject to cost and reliability constraints Circuit switch must reject call if can’t find a path for samples from input to output goal: minimize call blocking Packet switch must reject a packet if it can’t find a buffer to store it awaiting access to output trunk goal: minimize packet loss Subgoal: Don’t reorder packets Lecture 5

Internal switching In a circuit switch, path of a sample is determined at time of connection establishment No need for a sample header--position in frame is enough In a packet switch, packets carry a destination field Need to look up destination port on-the-fly Datagram lookup based on entire destination address Cell lookup based on VCI – used as an index to a table Other than that, switching units are very similar Lecture 5

Blocking in packet switches Can have both internal and output blocking Internal no path to output Example: head of line blocking. Output output link busy If packet is blocked, must either buffer or drop it Lecture 5

Dealing with blocking Overprovisioning Buffers Backpressure internal links much faster than inputs Buffers at input or output Backpressure if switch fabric doesn’t have buffers, prevent packet from entering until path is available Parallel switch fabrics increases effective switching capacity Lecture 5

Three generations of packet switches Different trade-offs between cost and performance Represent evolution in switching capacity, rather than in technology With same technology, a later generation switch achieves greater capacity, but at greater cost All three generations are represented in current products Lecture 5

First generation switch computer CPU linecard linecard queues in memory linecard Most Ethernet switches and cheap packet routers Bottleneck can be CPU, host-adaptor or I/O bus, depending Lecture 5

Second generation switch computer bus front end processors or line cards Port mapping intelligence in line cards Bottleneck is the bus (or ring) Lecture 5

Third generation switches Third generation switch provides parallel paths (fabric) OLC ILC NxN packet switch fabric OUT OLC IN ILC OLC ILC control Lecture 5

Third generation (contd.) Features self-routing fabric output buffer is a point of contention unless we arbitrate access to fabric potential for unlimited scaling, as long as we can resolve contention for output buffer Lecture 5

Switching - Fabric

Switching: abstract model Number of connections: from few (4 or 8) to huge (100K) Lecture 5

Multiplexors and demultiplexors Multiplexor: aggregates sessions N input lines Output runs N times as fast as input Demultiplexor: distributes sessions one input line and N outputs that run N times slower Can cascade multiplexors De-Mux 1 2 N 1 2 N MUX Lecture 5

Time division switching Key idea: when demultiplexing, position in frame determines output link Time division switching interchanges sample position within a frame: Time slot interchange (TSI) M U X D E TSI Lecture 5

Time Slot Interchange (TSI) : example sessions: (1,3) (2,1) (3,4) (4,2) 1 2 3 4 1 2 2 4 4 3 2 1 3 1 4 2 3 1 4 3 Read and write to shared memory in different order Lecture 5

TSI Simple to build. Multicast: easy (why?) Limit is the time taken to read and write to memory For 120,000 telephone circuits Each circuit reads and writes memory once every 125 ms. Number of operations per second : 120,000 x 8000 x2 each operation takes around 0.5 ns => impossible with current technology Need to look to other techniques Lecture 5

Space division switching Each sample takes a different path through the switch, depending on its destination Crossbar: Simplest possible space-division switch Crosspoints can be turned on or off i n p u t s outputs Lecture 5

Crossbar - example inputs output sessions: (1,2) (2,4) (3,1) (4,3) 1 2 Lecture 5

Crossbar Advantages: Drawbacks simple to implement simple control strict sense non-blocking Multicast Single source multiple destination ports Drawbacks number of crosspoints, N2 large VLSI space vulnerable to single faults Lecture 5

Time-space switching Precede each input trunk in a crossbar with a TSI Delay samples so that they arrive at the right time for the space division switch’s schedule Crosspoint: 4 (not 16) memory speed : x2 (not x4) 2 1 4 3 MUX 1 2 3 4 TSI 1 2 4 3 DeMux Lecture 5

Finding the schedule Build a routing graph Feasible schedule nodes - input links session connects an input and output nodes. Feasible schedule Computing a schedule compute perfect matching. 1 2 3 4 Lecture 5

Time-Space: Example time 1 time 2 2 1 2 1 TSI 3 4 4 3 3 2 1 4 2 1 3 4 3 1 2 4 2 1 4 3 TSI TSI Internal speed = double link speed Lecture 5

Time-space-time (TST) switching Allowed to TSI both on input and output Gives more flexibility => lowers call blocking probability TSI Lecture 5

Internal Non-Blocking Types Re-arrangeable Can route any permutation from inputs to outputs. Strict sense non-blocking Given any current connections through the switch. Any unused input can be routed to any unused output. Wide sense non-blocking. There exists a specific routing algorithm, s.t., for any sequence of connections and releases, Any unused input can be routed to any unused output, assuming all the sequence was served by the routing algorithm. Lecture 5

Circuit switching - Space division graph representation transmitter nodes receiver nodes internal nodes Feasible schedule edge disjoint paths. cost function number of crosspoints (complexity of AxB is AB) Lecture 5

Crossbar - example 1 2 3 4 4 1 2 3 Lecture 5

Another Example inputs outputs Lecture 5

Another Example outputs inputs sessions: (1,3) (2,6) (3,1) (4,4) (5,2) (6,5) inputs outputs Lecture 5

Clos Network Clos(N, n , k) : N - inputs/outputs; cross-points: 2 (N/n)nk + k(N/n)2 nxk (N/n)x(N/n) kxn 2x2 3x3 2x2 N=6 n=2 k=2 N 2x2 3x3 2x2 k N/n N/n Lecture 5

Clos Network - strict sense non-blocking Holds for k  2n-1 Proof Methodology: Recall: IF [A,B  S and |A|+|B| > |S|] then A∩ B≠Ø S= The k middle switches A = middle switches reachable from the inputs B = middle switches reachable from the outputs Our case: |S|=k |A| ≥ k-(n-1) |B| ≥ k-(n-1) Lecture 5

Clos Network - strict sense non-blocking Holds for k  2n-1 Proof: Consider an idle input and output Input box connected to at most n-1 middle layer switches output box connected to at most n-1 middle layer switches There exists an ”unused" middle switch good for both. n x k k x n n-1 Lecture 5

Example Clos(8,2,3) Need to route a new call N=8 n=2 k=3 4x4 3x2 2x3 Lecture 5

Clos Network Why is k=n internally blocking? nxk (N/n)x(N/n) kxn N=6 Lecture 5

Clos Network - re-arrangable Holds for k  n Proof: Consider the routing graph. find a perfect matching. route the perfect matching through a single middle switch! remaining network is Clos(N-N/n,n-1,k-1) summary: smaller circuit weaker guarantee Multicast ? 1 2 3 4 Lecture 5

Recursive Construction: basis The basic element: The dimension: r=0 The two states: Lecture 5

Recursive Construction: Benes Network r-1 dimension N/2 size r-1 dimension N/2 size Lecture 5

Example 16x16 Lecture 5

Benes Networks Symmetry Size: Rearrangable Proof I: F(N) = 2(N/2)*4 + 2F(N/2) = O(N log N) Rearrangable Clos network with k=2 n=2 Proof I: Build routing graph. Find 2 matchings route one in the upper Benes and the other in the lower. Lecture 5

Greedy permutation routing Start with an arbitrary node i1 set i1 to upper. At the output, o1 , a new constraint, set o2 to lower. Continue until no new constraint. Completing a cycle. Continue until done. Solve for the upper and lower Benes recursively. Lecture 5

Example: Benes Network for r=2 I1 1 2 3 4 5 6 7 8 I2 level 0 switches level 2r switches Lecture 5

Example ) 1 2 3 4 5 6 7 8 1 5 6 8 4 2 3 7 ( I1 I2 level 0 switches 1 2 3 4 5 6 7 8 1 5 6 8 4 2 3 7 ) I1 1 2 3 4 5 6 7 8 I2 level 0 switches level 2r switches Lecture 5

Example ) 1 2 3 4 5 6 7 8 1 5 6 8 4 2 3 7 ( I1 I2 level 0 switches 1 2 3 4 5 6 7 8 1 5 6 8 4 2 3 7 ) I1 1 2 3 4 5 6 7 8 I2 level 0 switches level 2r switches Lecture 5

Example ) 1 2 3 4 5 6 7 8 1 5 6 8 4 2 3 7 ( I1 I2 level 0 switches 1 2 3 4 5 6 7 8 1 5 6 8 4 2 3 7 ) I1 1 2 3 4 5 6 7 8 I2 level 0 switches level 2r switches Lecture 5

Example ) 1 2 3 4 5 6 7 8 1 5 6 8 4 2 3 7 ( I1 I2 level 0 switches 1 2 3 4 5 6 7 8 1 5 6 8 4 2 3 7 ) I1 1 2 3 4 5 6 7 8 I2 level 0 switches level 2r switches Lecture 5

Strict Sense non-Blocking N/2 x N/2 . . N/2 x N/2 N/2 x N/2 Lecture 5

Properties Size: Better parameters: F(N) = 2N*6 + 3F(N/2) = O( N1.58 ) strict sense non-blocking Clos network with k=3 n=2 Better parameters: n=sqrt{N}, k=2sqrt{N}-1 recursive size sqrt{N} x sqrt{N} Circuit size N log2.58 N Lecture 5

Cantor Networks m copies of Benes network. For m = log N its strict sense non-blocking Network size N log2 N Example Lecture 5

Cantor Network m=4 Lecture 5

Proof Sketch: Benes network: 2 log N -1 layers, N/2 nodes in layer. Middle layer= layer log N -1 Consider the middle layer of the Benes Networks. There are Nm/2 nodes in in all of them combined. Bound (from below) the number of nodes reachable from an input and output. If the sum is more than Nm/2: There is an intersection there has to be a route. Lecture 5

Proof Sketch: Let A(k) = number of nodes reachable at level k. A(0)=m A(1)= 2A(0)-1 A(2)=2A(1)-2 A(k)=2A(k-1) - 2k-1 = 2k A(0) - k 2k-1 A(log N -1) = Nm/2 - (log N -1) N/4 Need that: 2A(log N -1) > Nm/2. 2[Nm/2 - (log N -1) N/4] > Nm/2. Hold for m> log N-1. Lecture 5

Advanced constructions There are networks of size O(N log N). the constants are huge! Basic paradigm also applies to large packet switches. Lecture 5

Proof Sketch: Let A(k) = number of nodes reachable at level k. A(0)=m A(1)= 2A(0)-1 A(2)=2A(1)-2 A(k)=2A(k-1) - 2k-2 = 2k-1 A(1) - (k-1) 2k-2 A(log N) = Nm/2 - (log N -1) N/4 Need that: 2A(log N) > Nm/2. Hold for m> log N-1. Lecture 5