Topology, Capacity and Flow Assignment Computer Communication - PowerPoint PPT Presentation

Topology, Capacity and Flow Assignment Computer Communication Networks: Analysis and Design (Klei. Vol. 2, Chap. 5) Original Material Prepared by: Professor James S. Meditch Lecturer:Prof. Massimo Tornatore Typesetter: Dr. Anpeng Huang 1

The problem • The analysis of stochastic flows in packet networks is extremely complex • We study some the more important problems arising in the design process • Design variables: – Routing procedure (flow assignment) – Channel capacity – Topological configuration – Queuing discipline, packet numbering and sequencing, error control, etc.. 2

Packet vs. Circuit Switching X Y Z f X f Y f A f Z B A Space Spazio Propagation Propagazione Transmission Trasmissione Elaborazione Processing Attesa Queuing Packet Ritardo end-to-end L/f 0 End-to-end delay Switching L h /f 0 L p /f 0 Tempo Time

Packet vs. Circuit Switching X Y Z A B Space Spazio Propagation Propagazione Transmission Trasmissione Elaborazione Processing Setup phase Instaurazione Circuit End-to-end Ritardo Switching end-to-end delay Data Dati Release phase Time Rilascio Tempo

A. Evolution of network structures A.a. Private/most expensive (star network) 5

A.b. Least cost/slowest (minimal spanning tree) Multidrop lines – polling or contention 6

A.c. Compromise/multiplexing or concentrating (statistical multiplexer) 7

A.d. Large network – combinations of (c) 8

B. Network modeling M channels, N nodes, message (packet) switching Assumptions: (i) Links and nodes perfectly reliable (combinatorial in nature) (ii) Nodal processing time negligible (for reading, error checking) (iii) Nodal storage infinite (iv) Fixed (or random) routing (v) Offered traffic Poisson, packet length negative exponential 9

B.1. Arriving traffic • Poisson with avg. rate [msg/sec] N N ∑∑ γ γ = γ • = jk jk = = total average flow entering j 1 k 1 j ≠ in the system k B.2. Messages • Message lengths exponentially distributed with 1 mean bits µ 10

B.3. Channel model M/M/1 queue Channel I λ C i , i C • = channel capacity in bits/sec (bps) i λ • = avg. channel flow in msg/sec i M ∑ λ = λ • Note(1): Total average flow inside i the network ( ≠ γ !) = 1 i • Note(2): propagation time P i 11

B.4. Channel cost (in $) M ∑ = ( ) D d i C i = i 1 • d is a generic cost function 12

B.5. Average message delay • T = E [message delay] ≅ • E [message delay for message from j to k ] Z jk γ N N N N 1 ∑∑ ∑∑ = = γ jk T Z Z γ γ jk jk jk = = = = j 1 k 1 j 1 k 1 fraction of traffic over j–k node pair 13

C. Network design problems 1 C.1. Given: node locations, , traffic matrix { γ jk } µ C.2. Objective function T C.3. Parameters: C i , λ i , topology τ C.4 Constraint M ∑ = d ( C ) D i i = 1 i 4 optimization problems can be defined that differ only in the set of permissible design variables -> 14

C.3. The 4 problems are: Problem Given Minimize w.r.t. s.t. λ i , τ CA T C i D C i , τ ≤ λ < µ λ i 0 C FA T Increasing i i complexity τ C i, λ I CFA T D C i, λ I, τ TCFA - T D C.4. Dual problems Swap D with T No dual for FA 15

D. Delay analysis D.1. Model for T Little’s result applied: M ∑ = γ = λ N T i T i = i 1 M 1 ∑ = λ T i T γ i = i 1 λ : one-way or two-way message flow i First take-away: global T can be expressed as sum of local T i 16

D.2. Independence assumption (a) and channel model (b) D.2.a. Message lengths at nodes are independent at each node, and have exponentially distributed length b with p.d.f. − µ = µ ≥ b p ( b ) e b 0 1 avg. message length = bits µ • In fact, message lengths are not independent at each node since a message enters the network with a given length and retains that length from source to destination. This independence assumption is based on (i) large numbers of messages passing through each node and (ii) the moderate connectivity of the network, which support the assumption that packet lengths can be approximated as independent, exponentially-distributed message lengths. 17

D.2.b. M/M/1 channel model and T C i T i λ Poisson arrivals: messages/sec. i 1 Exponential service: sec/message = x µ i C [ ] message i µ i = nb : C sec 1 µ x x C 1 = = = = i i i T λ − ρ − λ µ − λ i 1 1 x C − i 1 i i i i i µ C 18 i

Cont’d: λ M 1 ∑ = It follows that i T γ µ − λ C = i 1 i i Other results (easily obtainable): λ λ = λ = ρ = i N T i µ − λ i i i µ i C C i i i λ i ρ µ 2 C = 1 = i N i W − ρ q i µ − λ i C i i i 19

D.2.c. Other effects Note: so far we neglected: 1. control traffic, 2.propagation and nodal processing delay D.2.c.1. Average data message delay when control traffic is present λ = avg. flow of data messages i 1 = avg. length of data message µ λ = avg. flow of all traffic ' i 1 = avg. length of all messages µ ' = avg. data message delay T i = avg. service time for data messages x i = avg. waiting time for all messages W i 20 nb: all traffic = data + control

The avg. system time (with control traffic) can be now easily calculated as λ ' 1 i = + = µ ' x T W x C = µ i i W C i i i µ − λ i ' ' C i i i λ ' i µ ' C 1 = + i T µ − λ µ i ' ' C C i i i   λ ' i   λ µ = ∑ ' M C 1 +   i i T γ µ − λ µ ' '   C C = 1 i i i i     21

D.2.c.2. Propagation delay and nodal processing delay P i = propagation delay [sec/msg] Depends upon the medium and the length of the link, e.g., ground station to geosynchronous satellite is on the order of 120 to 135 msec. K = avg. nodal processing time [sec/msg]   λ ' i   λ µ ' M C 1 ∑ = + + + +   i i T K P K γ µ − λ µ i ' '   C C = i 1 i i i     22

A simple example 23

γ Question: How should be split in 13 order to minimize T? γ C Routing: all of via 12 γ 1 C C 1 of via and 13 γ 1 2 5 of via C 13 γ 3 C all of via 23 2 λ 1 = λ 2 = λ 3 = msg/sec 4 2 5 γ = γ + γ + γ = msg/sec 10 12 13 23 λ   = ∑ 3 1 1 4 2 5 = + + i T   γ µ − λ − − −   C 10 12 4 6 2 9 5 = 1 i i i = T 225 msec 24

D.2.d. Delay T vs Offered load γ λ M 1 Fixed ∑ = i T γ µ − λ C routing = i 1 i i λ << µ D.2.d.1. [Low load] Decrease γ such that ∀ i C i i λ M 1 ∑ → i T γ µ C = i 1 i So, no queuing in the network, all delay is due to service time (i.e., transmission delay) λ M 1 ∑ ≅ i T γ µ 0 C = i 1 i 25

D.2.d.1. [High load] Increase γ until some channel λ → µ saturates, i.e., . C i i → ∞ For that channel , T i and, therefore so does T ≅ T i . Let γ * be the corresponding value of γ . 26

D.2.e. Average path length n = avg. path length no. of links through which a message passes in proceeding from its source to its destination averaged over all source-destination node pairs 27

γ = γ + γ + γ 12 13 14 γ + γ + γ γ + γ + γ + γ + γ + γ . 1 . 2 . 3 ( ) ( ) = = 12 13 14 12 13 14 13 14 14 n γ γ λ = γ + γ + γ But 1 12 13 14 λ = γ + γ 2 13 14 λ = γ 3 14 λ + λ + λ 3 1 ∑ = = λ Hence, 1 2 3 n γ γ i = i 1 λ 3 ∑ = λ ≅ λ n where γ i 28 = i 1

In general, π = “path” from j to k π jk = n no. of links in jk jk γ jk n N N ∑∑ = jk n γ = = j 1 k 1 j ≠ k 29

M N N ∑ ∑∑ λ ≅ λ = γ But, n i jk jk = = = i 1 j 1 k 1 γ Since msg/sec will traverse n jk links in passing jk through the network Hence, λ M ∑ = λ = λ n where γ i = i 1 λ See formula Using this result, we get i at pp. 22 M λ ∑ = T n µ 0 C = 1 i i for no-load delay with fixed routing. 30

Example: Same network as in delay analysis example above γ = λ 1 = λ 3 = λ 2 = 10 4 5 2 ; , , 11 = = So, n 1 . 1 10 Also, using either expression for , we have T 0 0 ≈ T 122 , 2 msec This is the avg. msg. delay when the network is very lightly loaded. 31

E. Capacity Assignment (CA) Problem 1 τ γ { } λ Given: node locations, , , , ; jk i µ { } M ( ) ∑ = T C d C D min w.r.t. s.t. i i i = 1 i N.B.: the actual problem requires selection of C i from a finite set, e.g., C i ∈ {100Mbps, 1Gbps, etc.}, but, to find closed form solution, we have to relax this requirement 32