Topology, Capacity and Flow Assignment Computer Communication Networks: Analysis and Design (Klei. Vol. 2, Chap. 5)

Transcription

1 Topology, apacty and Flow Assgnment omputer ommuncaton Networks: Analyss and Desgn (Kle. Vol. 2, hap. 5) Orgnal ateral Prepared by: Professor James S. edtch Lecturer:Prof. assmo Tornatore Typesetter: Dr. Anpeng Huang

2 The problem The analyss of stochastc flows n packet networks s extremely complex We study some the more mportant problems arsng n the desgn process Desgn varables: Routng procedure (flow assgnment) hannel capacty Topologcal confguraton Queung dscplne, packet numberng and sequencng, error control, etc.. 2

3 Packet vs. rcut Swtchng A X Y Z f A f X f Y f Z B Packet Swtchng Tempo Tme L/f 0 L h /f 0 L p /f 0 Space Spazo End-to-end Rtardo end-to-end delay Propagazone Propagaton Trasmssone Transmsson Elaborazone Processng Attesa Queung

4 Packet vs. rcut Swtchng A Space Spazo X Y Z B Propagazone Propagaton Trasmssone Transmsson Elaborazone Processng Setup Instaurazone phase rcut Swtchng Rtardo End-to-end end-to-end delay Data Dat Release Rlasco phase Tempo Tme

5 A. Evoluton of network structures A.a. Prvate/most expensve (star network) 5

6 A.b. Least cost/slowest (mnmal spannng tree) ultdrop lnes pollng or contenton 6

7 A.c. ompromse/multplexng or concentratng (statstcal multplexer) 7

8 A.d. Large network combnatons of (c) 8

9 B. Network modelng channels, N nodes, message (packet) swtchng Assumptons: () Lnks and nodes perfectly relable (combnatoral n nature) () Nodal processng tme neglgble (for readng, error checkng) () Nodal storage nfnte (v) Fxed (or random) routng (v) Offered traffc Posson, packet length negatve exponental 9

10 B.. Arrvng traffc Posson wth avg. rate [msg/sec] N N j k j k jk jk total average flow enterng n the system B.2. essages essage lengths exponentally dstrbuted wth mean bts 0

11 B.3. hannel model // queue hannel I, channel capacty n bts/sec (bps) avg. channel flow n msg/sec Note(): Note(2): propagaton tme P Total average flow nsde the network (!)

12 B.4. hannel cost (n $) D d ( ) d s a generc cost functon 2

13 B.5. Average message delay T E [message delay] E [message delay for message from j to k] Z jk T N N j k jk Z jk N N j k jk Z jk fracton of traffc over j k node par 3

14 . Network desgn problems.. Gven: node locatons,, traffc matrx { jk }.2. Objectve functon T.3. Parameters:,, topology τ.4 onstrant 4 optmzaton problems can be defned that dffer only n the set of permssble desgn varables -> d ( ) D 4

15 .3. The 4 problems are: Problem Gven nmze w.r.t. s.t. A, τ T D Increasng complexty FA, τ T 0 < FA τ T, I D TFA - T, I, τ D.4. Dual problems Swap D wth T No dual for FA 5

16 D. Delay analyss D.. odel for T Lttle s result appled: N T T T T : one-way or two-way message flow Frst take-away: global T can be expressed as sum of local T 6

17 D.2. Independence assumpton (a) and channel model (b) D.2.a. essage lengths at nodes are ndependent at each node, and have exponentally dstrbuted length b wth p.d.f. b avg. message length p( b) e b 0 bts In fact, message lengths are not ndependent at each node snce a message enters the network wth a gven length and retans that length from source to destnaton. Ths ndependence assumpton s based on () large numbers of messages passng through each node and () the moderate connectvty of the network, whch support the assumpton that packet lengths can be approxmated as ndependent, exponentally-dstrbuted message lengths. 7

18 D.2.b. // channel model and T T Posson arrvals: Exponental servce: T x ρ x messages/sec. x x sec/message [ ] nb : message sec 8

19 It follows that Other results (easly obtanable): T T N N q ρ ρ 2 W 9 ρ ont d:

20 D.2.c. Other effects Note: so far we neglected:. control traffc, 2.propagaton and nodal processng delay D.2.c.. Average data message delay when control traffc s present ' ' T x W avg. flow of data messages avg. length of data message avg. flow of all traffc avg. length of all messages avg. data message delay avg. servce tme for data messages avg. watng tme for all messages 20 nb: all traffc data + control

21 x W T + ' ' ' ' W x T ' ' ' ' + + T ' ' ' ' 2 The avg. system tme (wth control traffc) can be now easly calculated as

22 D.2.c.2. Propagaton delay and nodal processng delay P propagaton delay [sec/msg] Depends upon the medum and the length of the lnk, e.g., ground staton to geosynchronous satellte s on the order of 20 to 35 msec. K avg. nodal processng tme [sec/msg] K P K T ' ' ' ' 22

23 A smple example 23

24 Queston: How should order to mnmze T? Routng: all of va 2 of va and of va 3 all of 23 va T msg/sec msg/sec be splt n T 225 msec 24

25 D.2.d. Delay T vs Offered load Fxed routng D.2.d.. [Low load] Decrease such that So, no queung n the network, all delay s due to servce tme (.e., transmsson delay) T << T T 0 25

26 D.2.d.. [Hgh load] Increase untl some channel saturates,.e.,. For that channel T, and, therefore so does T T. Let * be the correspondng value of. 26

27 n D.2.e. Average path length avg. path length no. of lnks through whch a message passes n proceedng from ts source to ts destnaton averaged over all source-destnaton node pars 27

28 + + n But Hence, ( ) + ( 3 + 4) n n where 3 28

29 In general, π jk n n jk N path from j to k no. of lnks n π jk N j k j k jkn jk 29

30 But, jk Snce msg/sec wll traverse n jk lnks n passng through the network Hence, n where Usng ths result, we get N N j k T 0 jk n jk n See formula at pp. 22 for no-load delay wth fxed routng. 30

31 Example: Same network as n delay analyss example above 0 ; 4, 2, 3 5 So, Also, usng ether expresson for n 2 0. T 0, we have T 0 22,2 msec Ths s the avg. msg. delay when the network s very lghtly loaded. 3

32 E. apacty Assgnment (A) Problem jk Gven: node locatons,,,, ; T { } mn w.r.t. s.t. τ { } d ( ) D N.B.: the actual problem requres selecton of from a fnte set, e.g., {00bps, Gbps, etc.}, but, to fnd closed form soluton, we have to relax ths requrement 32

33 E.. Lnear costs and t can solve usng the Lagrangan:,2,..., ) ( d d D d T s.t. mn + D d T J β + D d β 0,2,... j J j 0 ) ( 2 + j j j j j d J β 33 Let us assume: Then the optmzaton problem can be wrtten as

34 ( ) j j j d β 2 j j j j d β + + d d d D β e d D d d D β 34 Solvng the Lagrange s equatons: We can then fnd β by «formng the constrant»: «Excess dollars»

35 j, 2,, (*) {mn. requred capacty to satsfy flow reqmnt.} + {sq. root allocaton of normalzed excess capacty} If we substtute j n the o.f. Note: from (*) + j j j e j j d d d D j T j j j e d d d D 35 ont d:

36 j j j e d d d D T mn j j j e d d D 36 ont d: 2 2 e e d D d D 2 mn e d D n T Note: D e must be greater than 0!

37 ont d: Specal case: Hence, D d for all s the constrant Usng the defnton of D e at pp. 3 D e j j + De j T mn D e 2 37

38 Example: FLOWS BUDGET (d ) D34 () 6 step: calculate APAITIES 6 3 D e j j 000 j + 0 bps j j + j (n kbps) 2kbps 2kbps 6kbps T step: calculate AVG DELAY mn 3 ( 0 ) 6 D e ( ) sec 34kbps 687.5msec 38

39 E.2. Other cost functons Approxmaton of actual cost structures a. Logarthmc cost (quantty dscount) A s proportonal D d logα + b a b. Power law (ARPANET study - 970) D g d α 0 < ( α ) 2 < α α g 0 αβ β Lagrange multpler (teratve soluton requred) d 2 39

40 E.3. Other forms of capacty assgnment Heurstc approaches based on the noton of the network beng balanced n some physcally meanngful way E.3.a. All lnks n the network have the same avg. utlzaton a ρ 0 < a < Fx a by specfyng the desred avg. utlzaton, e.g., T a Proportonal A ρ 0.6 a ( a) T 40

41 E.3.b. All lnks n the network have the same lnk delay T τ Smply, let us fx τ by specfyng the desred avg. delay for each lnk, (e.g., to T msec), then: + τ T nτ 4

42 Example: bts [msg/sec] [msg/sec] Fnd capactes so that all T are equal and T0 msec n. T τ 00 msec n 2000 τ 0.sec 2800bps 2. 8kbps 2400bps 2. 4kbps bps 3. 0kbps 3 42

43 F. Flow Assgnment (FA) problem F.. Gven τ and { }, mnmze T wth respect to { } subject to capacty constrants, 0, external traffc requrements { jk }, and flow conservaton T 43

44 F.2. FA propertes. ultcommodty-flow non-lnear optmzaton problem 2. losed-form soluton not possble 3. Use teratve computatonal algorthm 4. Soluton shall gve both the amount of flows { } and the correspondng optmal (mnmum delay) routng for all commodtes 5. TT(, 2. ) s a convex functon of the s 6. apacty constrants, 0, are convex 7. Fundamental theorem. Any mnmum of a convex functon over a convex set s the global mnmum 8. Best known and most effectve algorthm for solvng ths problem s the Flow Devaton (FD) algorthm 9. FD algorthm based on two basc notons: a. Shortest path flows b. Flow devaton whch reduces mnmzaton of TT(, 2. ) over all the smultaneously to mnmzaton of T(a) where a s a scalar, 0 a 44

45 F.3. Shortest path flow part of FD F.3.. Lnk length, l l T ( ) 2 > 0 l F.3.2. Problem statement mn φ l φ subject to capacty, total flow, and commodty flow constrants (a) Lnear programmng problem (b) any effcent algorthms avalable 45

46 F.4. FD Algorthm F.4.. FD algorthm nteraton ndex f (n) ( n, n 2. n ) flow vector on n th teraton f(0)ntal feasble flow (see Klenrock vol. II, pp ) a. Set n0 b. Solve shortest-path flow problem wth wegths l and let ( ) [ ( ) ( ) ( )] n n n n φ φ, φ2,..., φ ( n+ ) ( n) ( n) c. Set f ( a) f + aφ and fnd 0 a such that T s mnmzed c. Any search method s ok, e.g, Fbonacc search ( ) ε ( ) ( ) ( ) ( ( n+) f ) n n+ d. If T f T f where ε > 0 s the accuracy threshold, stop. Otherwse, set n n + and return to Step 2. 46

47 F.4.2. FD algorthm flow chart ε Test ( accuracy threshold) (A) T ( ( n) ) ( n+ f T f ) > ε (B) T ( ( n) ) ( n+ f T f ) ε 47

48 F.5. FA Problem F.5.. Plausblty of FD algorthm ( n) n+ Assume gven and let f f ( ) ( ) n n a f + aφ Then, and T T ( n+ ) ( n) ( f ) T f ( ( n+ ) ) ( n) f T f T ( n+ ) ( n) ( ) + ( ) ( f f ) n f ( ) ( n) ( ( n) ( n) + a l φ f ) ( ( n+) f ) Therefore, to mnmze to wthn frst n order n a, t s necessary to choose the φ such that ( n) ( n) s mnmzed. l φ T ( ) 48

49 F.5.2. Problem set-up example T ( n) 4 ( n) ( n) onservaton of total avg. flow: If there were lnearly ndependent, then the problem would be over

50 a. Shortest path flow subproblem 4 where mn l φ l φ ( ) 2 φ subject to ( refers to ) φ + φ φ φ3 + φ4 2 and 0 φ β where β 0.99 for example b. FD algorthm gves the and Tmn,but not the routng 50

51 c. Extenson to nclude commodty flow 2 Lnk :, 2 Lnk 2: u 3, Lnk 3: Lnk 4: u 2 3 u 23 2 u 32 3 u 2 3 u 3 onservaton of commodty flow: 2 3 u 2 + u u 3 + u3 3 3 u u u2 + u3 2 u2 + u32 2 u3 + u23 3 u u3 3 u u23 u u3 6 eqns. of whch 4 are lnearly ndependent 5

52 c. Extenson to nclude commodty flow (cont.) Shortest path flow subproblem Subject to and mn φ u 2 + u2 φ2 u 3 + u3 3 2 φ3 u 23 φ4 u 32 l φ where φ ( ) u2 + u3 u u3 u u32 2 u3 + u φ β l FD algorthm gves, Tmn, jk u ab 52

53 G. FA Problem G.. Gven τ, mnmze T wth respect to { } and { } subject to d D and 0 where T Note: When we combne A and FA, we are no longer able to gve globally optmal solutons, but rather descrbe procedures that fnd local mnma for T 53

54 G.2. Approach a. ombne A and FA algorthms b. Snce { } s have to be assgned, t can be shown that fxed routng s optmal. Ths mples settng a n the FD algorthm c. It can also be shown that shortest path routng gves only local mnma for T. Ths means that several ntal feasble flows f (0) must be tred 54

55 G.3.Flow chart of FA algorthm 55

56 G. 4. Algorthm propertes a. onverges for each f (0) snce the number of possble shortest-path flows s fnte b. Snce l 0, there are no negatve loops n the shortestpath flow algorthm l T n ( / ) j d d + ( / ) De De c. The form of l that results from the A part of algorthm s such that lm l 0 See Klenrock s Vol. 2, Eq / 2 It mples that once and become zero, they reman there d. Snce only local mnma are possble, one has to be clever n the f (0) that are used j d / 2 > 0 56

57 H. TFA Problem (Dual Form) H.. Gven node locatons and { jk }, mnmze D ( ) wth respect to τ, { } and { } subject to T and connectvty constrants. d T AX 0 <,2,... 0 < N(N -) Note: no. of lnks varable! Note(2): n the sldes s not reported (..normalzed) 57

58 H.2. Approach a. Real desgn requres selecton of dscrete capactes b. Heurstc soluton procedure va teratve use of FA and choces of τ c. Only local mnma possble d. d ( ) termnaton cost + (lne cost) x (length of lne) expressed n $/month ( ) α a d d + 58

59 H.3. TFA 2 Problem BE (oncave Branch Elmnaton) Algorthm a. Select τ (0), e.g., fully-connected b. For each channel, assume a power-law approxmaton and choose d and α n α D d 0 c. Execute the FA 2 (mnmze D subject to constrant TT AX ). At each teraton n FA 2, use a lnearzed value for capacty about the current value of flow. If at any step n FA 2, the connectvty constrant s volated, proceed to Step (d) wth the prevous step FA 2 results; otherwse, let FA 2 run to completon d. Dscretze the { } to the nearest actual lne capactes 0 avalable such τ that < and T T AX 59

60 BE Algorthm (contnued) e. Refne the flow assgnment usng the dual form of the FD algorthm where and adjust the { } accordngly so that < and T T AX f. Repeat Steps 3-5 for varous feasble f (0) g. Repeat Steps -6 for a number of dfferent τ (0) + AX j j j d T d d D l 60

61 ARPANET BE Study a. jk kbps over all possble jk pars gves 650kbps b. T AX 200 msec c. onnectvty 2 d. D c + d α 6

62 e. hannel capactes and costs 62

63 f. Results ) 30 ntal topologes studed 2) BE algorthm requred to 2 seconds PU tme on 360/9 per topology 3) 0.8 α.0 provded good ft to capacty costs and BE worked well for these values 4) Two best solutons on next sldes 63

64 Best Soluton of ARPANET BE 64

65 Best Soluton of ARPANET BE a. Dstrbuton of capactes vs. lnk lengths b. τ (0) was a fully-connected topology c. Soluton has 6 channels d. ost $89,580/month 65

66 Second Best Soluton of ARPANET BE 66

67 Second Best Soluton of ARPANET BE (cont.) a. τ (0) was a low-connected topology b. Both ntal and fnal topologes had 29 channels c. Uses hgher speed lnes (50 and 00kbps) for medum and long dstances; 230kbps for short dstances d. ost $ 94,288/month 67

68 onvex Sets A set S s sad to be a convex set f, for any two ponts n the set, the lne jonng those two ponts s also n the set athematcally, S s a convex set f for any two vectors x () and x (2) n S, the vector ( ) ( 2) x x + ( ) x s also n S for any number between 0 and Examples: Fgures A. and A.2 represent convex sets, whle Fgure A.3 s not a convex set. 68

69 Theorems Theorem The set of all feasble solutons to a lnear programmng problem s a convex set Theorem 2 The ntersecton of convex sets s a convex set (Fgure A.4) Theorem 3 The unon of convex sets s not necessarly a convex set (Fgure A.4) Theorem 4 A hyperplane s a convex set Theorem 5 A half-space s a convex set 69

70 A convex combnaton of vectors x (), x (2),, x (k) s a vector x such that ( ) ( 2) ( k ) x x + 2x + + k x k 0 for,2,, k Vertex An extreme pont or vertex of a convex set s a pont n the set that cannot be expressed as the mdpont of any two ponts n the set. For example, consder the convex set S ( x, x ) 2 0 x. 2, 0 x2 Ths set has four extreme ponts gven by (0,0), (0,2), (2,0), and (2,2) { 2} 70

71 Hyperplane & Half Space A hyperplane s the set of all ponts x satsfyng cxz for a gven vector c 0 and a scalar z The vector c s called the normal to the hyperplane. H {( x x, x ) 2x 3x + 5} x, 3 For example, s a hyperplane. A half-space s the set of all ponts x satsfyng cx z or cx z for a gven vector c 0 and a scalar z 7

72 onvex Functon The fgure llustrates the defnton of a convex functon of a sngle varable. A functon of n varables f(x) defned on a convex set D s sad to be a convex functon f and only f for any two ponts x () and x (2) D and 0, f [ ( ) ( ) ( ) ] 2 ( ) x + x f x ( ) ( ) ( ( 2) + f ) x 72

73 Propertes of onvex Functons a. The chord jonng any two ponts on the curve always falls entrely on or above the curve between those two ponts b. The slope or frst dervatve of f(x) s ncreasng or at least non-decreasng as x ncreases c. The second dervatve of f(x) s always non-negatve for all x n the nterval d. The lnear approxmaton of f(x) at any pont n the nterval always underestmates the true functon value e. For a convex functon, a local mnmum s always a global mnmum 73

74 Fgure A.6 llustrates property 4 74

75 Fgure A.6 llustrates property 4 The lnear approxmaton of f(x) at the pont x 0, denoted by, s obtaned by gnorng the second and other hgher order terms n the Taylor s seres expanson For a convex functon, property 4 mples that for all x The gradent of a functon f(x, x n ) s gven by The Hessan matrx of a functon f(x, x n ) s an nxn symmetrc matrx gven by ( ) 0 ; ~ x x f ( ) ( ) ( )( ) ; ~ x x x f x f x x f + ( ) ( ) ( )( ) x x x f x f x f + ( ) n n x f x f x f x x f δ δ δ δ δ δ,...,,,..., 2 ( ) f x x f x x H j n f 2 2,..., δ δ δ 75

76 Backup sldes 76

77 Example: [msg/sec] 200bts [msg/sec] Requre all T to be the same and that n. T 0msec 2800bps 2. 8kbps T τ 00 msec n 2000 τ 0.sec 2400bps 2. 4kbps bps 3. 0kbps 3 77

78 FA Problem Example: Ground network and satellte system T.5 Determnaton of T ncludes 0.5 sec. Assumed propagaton delay for each of the three satellte lnks. 78