People have been thinking about network problems for a long time Koenigsberg Bridge problem (Euler, 1736)

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1 Networks People have been thnkng about network problems for a long tme Koengsberg Brdge problem (Euler, 1736) Can you cross all 7 brdges exactly once on a walk? Lesson 8-1, p. 1

2 Types of Network Flow Problems (Wnston) Transportaton problem (Htchcock, 1941) Gven a set of sources, destnatons Have source capactes, destnaton demands Know shppng cost/unt for each source-destnaton par Mnmze total cost of meetng demands at destnatons Assgnment Problem Restrcted transportaton problem Each source can supply 1 object Each destnaton demands 1 object Transshpment Problem Transportaton problem wth ntermedate (transshpment) ponts Generalzes both transportaton and assgnment problem Lesson 8-1, p. 2

3 Other Network Models n Wnston Shortest path problem (Djkstra1959) Fnd the shortest route between an orgn and a destnaton Algorthms range smple greedy algorthm to very complex Specal case of the transshpment problem Maxmum flow problem Maxmze flows from a sngle source to a sngle destnaton Important dual result: fnds the mnmum cut n a network Mnmum spannng tree Objectve s to connect all nodes n a network Want to mnmze the total length of the connectons Project management (PERT-CPM) Fnd tme to complete a lnked set of tasks Is actually a longest path problem Lesson 8-1, p. 3

4 Some Network Models Not n Wnston Crculaton problem Transshpment problem wth no source or demand nodes Example: arlne routng schedules Generalzed flow problem Some of the materal flowng on the network s lost n transmsson Example: electrcal power transmsson, water dstrbuton Multcommodty flow problem More than one type of materal s flowng on the network Dfferent materals consume network capacty, but may have dfferent transmsson costs Network nterdcton problem Bad guys are movng on a network; good guys try to stop them Each sde has to choose what paths to use or nterdct Lesson 8-1, p. 4

5 Network Models Not n Wnston (cont d) Network relablty problem Fnd the maxmum relablty set of routes n network In certan cases, can be recast as an MCNFP Network models wth sde constrants Knapsack problem: optmze the goodness of a set of objects that must ft nto a fnte-szed knapsack Travelng Salesman Problem: fnd the mnmum-cost tour among a set of destnatons, but only vst each destnaton once Varous vehcle routng problems Network optmzaton lterature s ggantc Semnal text s Ahuja, Magnant, and Orln (1993); 846 pages! Huge number of applcatons Inspraton for much of the research nto algorthmc effcency Lesson 8-1, p. 5

6 Concentraton n Ths Course I wll emphasze the transshpment problem Otherwse known as the mnmum-cost network flow problem (MCNFP) Reasons: Transportaton, assgnment, max flow, shortest path problems are all cases of MCNFP All commercal software mplements a modfed smplex method based on MCNFP To explot ths capablty, you have to formulate n terms of MCNFP Specalzed transportaton and assgnment problem algorthms are largely unnecessary (e.g., steppng-stone and Hungaran methods) Lesson 8-1, p. 6

7 Network Jargon To model a problem as a network, we need some termnology A graph G conssts of: A set of nodes (N) A set of arcs (A), whch connect the nodes So, G = (N,A) specfes the topology of the network Graph characterstcs Let,j be ndces for the nodes Then the par (, dentfes an arc Ths notaton allows us to defne thngs such as: Unt transportaton costs on an arc (C j ) Supples at each source node (S ) Demands at each demand node (D j ) Lesson 8-1, p. 7

8 From Network to Optmzaton Model Consder the transshpment network below: -4-3 (-6,5) (10,4) (13,5) (-4,2) (13,9) (11,7) What does all ths mean? Numbers n crcles are node labels 2 2 (11,8) Numbers next to nodes are supples (>0) and demands (<0) Numbers on arcs are (cost/unt to shp, max flow on arc) Assume we want to mnmze total cost 5-1 Lesson 8-1, p. 8

9 Optmzaton Model (Wnston format) Let x j be the flow on arc (, Then, the model s: mn z subject to x 12 all x x x j x = 11x x , x x x x + x x 4, x x x x , x x + x 4x x + x 2, x x 8, x 6x = 6 (node1 balance) = 2 (node 2 balance) = - 4 (node 3 balance) = - 3 (node 4 balance) = -1(node 5 balance) , x x 9 54 Lesson 8-1, p. 9

10 Some Key Ponts FLOW MUST BALANCE! For pure supply nodes, flow out must equal supply For pure demand nodes, flow n must equal demand For transshpment nodes, flow out must equal flow n Note the example has nodes that demand and transshp What do we do f supply s unequal to demand? Create dummy node to absorb (shp) excess supply (demand) Create costs that make sense n the model Arc costs can be negatve Can also force flow on arcs by puttng n lower bounds Sgn conventons Flow out s postve Flow n s negatve Lesson 8-1, p. 10

11 Optmzaton Model (Algebrac format) Indces, j = nodes {1,2,3,4,5} Subsets ARCS(, = connectons between nodes Data C j = cost per unt to shp on arc,j L j = lower bound on flow on arc,j U j = upper bound on flow on arc,j SD = supply or demand at node Varables x j = flow on arc,j Lesson 8-1, p. 11

12 Algebrac Format (cont d) So, the general MCNFP s: mn z = subject to C j, j ARCS (, * x j FLOW IN FLOW OUT L x x U j j ARCS (, j j j x j = SD j ARCS ( j, ) for all ARCS(, for all nodes One MPL program can represent any MCNFP! So, are we done yet? Lesson 8-1, p. 12

13 Answer s NO The trck wth networks s to translate the problem nto a network structure Many thngs that don t appear to be networks are See Wnston, pp on the nventory problem Many, many other models can be represented by a network So, the challenge s to: Determne f the problem can be represented by a network If so, come up wth the nodes, arcs, costs, and capactes Why do we want to do ths? Speed: network codes are much faster than normal LP smplex Integralty: f a problem can be represented by an MCNFP, t wll have ntegral answers (f the supples, demands and bounds are nteger) Lesson 8-1, p. 13

14 Common MCNFP Models - Transportaton Transportaton problem Ths s an MCNFP wth no transshpment nodes Nodes are dvded nto two dsjont sets, SOURCES and SINKS So, the formulaton becomes: mn z subject to j j ARCS (, j ARCS (, 0 x j x x = j, j ARCS (, = = U C * x SUPPLY DEMAND j j for all nodes j for all nodes for all ARCS(, SOURCES j SINKS Wnston shows a specal algorthm for ths, but t s unnecessary Lesson 8-2, p. 1

15 Common MCNFP Models - Assgnment A very smple model (too smple to be of much use) We are matchng demands to supples Each demand (supply) node demands (supples) 1 unt Number of supply and demand nodes are equal So, ths model s: mn subject to j j ARCS (, j ARCS (, x j z x x = j, j ARCS (, = 1 for all nodes = 1for all nodes { 0,1} for all ARCS(, Agan, Wnston shows a specal (Hungaran) algorthm for ths - t s not needed C * x j SOURCES j SINKS Lesson 8-2, p. 2

16 Common MCNFP Problems - Max Flow Ths s useful, but s a twst on the MCNFP formulaton We are maxmzng flow from a sngle source (s) to a sngle destnaton (d) Ths flow, however, s a varable, v Ths model s: max subject to 0 x z j ARCS j = v v for = s xj x j = 0 for all nodes (, j ARCS ( j, ) v for = d U for all ARCS(, j Note here that we have to nclude a return arc from d to s to ensure the flow balances s, d Lesson 8-2, p. 3

17 Shortest (Longest) Path Problem Ths can also be represented by an MCNFP Costs are arc lengths Flow 1 unt from the orgn s to the destnaton d Mnmze (maxmze) the sum of the arcs used Formulaton: mn ( or max) z j, j ARCS(, * x subject to 1, = s xj x j = 0, s and j ARCS(, j ARCS( j, ) 1, = d 0 x 1 for all ARCS(, j = C j d NOTE: lots of smple algorthms (e.g., Djkstra) avalable for ths problem Lesson 8-2, p. 4

18 Convertng a Problem to a Network Many hard problems become easy f you can convert them to a network Example: open-pt mnng problem An open pt mne can be represented as a set of blocks Each block has a net proft w f you choose to extract t But, you have to remove the blocks above t to get to t Ths problem s called a maxmum weght closure The graph s a set of nodes wth weghts The arcs show dependences among nodes A closure s a set of nodes wth no outgong arcs The objectve s to fnd the closure wth maxmum weght Lesson 8-2, p. 5

19 Example Closure Problem A(-10) B(-10) C(-20) D(-20) E(30) F(-5) G(5) H(50) I(10) Number n parentheses s payoff for choosng that node Example: A, B, and E s a closure, wth total weght = 10 Lesson 8-2, p. 6

20 Convertng a Closure to a Max Flow Problem Add two new nodes, s and t Connect all nodes wth postve payoffs wth arcs from node s Connect all nodes wth negatve payoffs arcs to node t Make the upper bound on all new arcs the absolute value of the weght of the node Make the upper bound on the orgnal arcs nfnte Solve a maxmum flow problem wth ths network Lesson 8-2, p. 7

21 The Maxmum Weght Closure as a Max Flow T A(-10) B(-10) C(-20) D(-20) E(30) H(50) F(-5) I(10) G(5) Upper bound on flow for all new arcs s the absolute value of the node they connect to (except for T to S) S Lesson 8-2, p. 8

22 Gettng the Answer T A(-10) B(-10) C(-20) D(-20) E(30) F(-5) H(50) I(10) G(5) 15 5 Objectve functon value: 60 Flows shown on dagram But what nodes are n the closure? And what s the weght? 25 S 10 Lesson 8-2, p. 9

23 Asde: the Max-Flow Mn-Cut Theorem In a maxmum flow problem, the optmal flow s equal to the capacty of the mnmum cut A cut s a set of arcs that dvdes the network nto two sets of nodes, one contanng the source (S) and the other the snk (T) Call these sets of nodes N 1 and N 2 Each arc n the cut set has one endpont n N 1 and another n N 2 Consequences: Solvng the max flow problem also gves the mnmal set of arcs that can dsconnect the network The arcs n the cut wll all be at ther upper bounds A large network can have many cutsets May have to resort to a separate algorthm to fnd them all Lesson 8-2, p. 10

24 Fndng the Mn Cut n the Closure Problem T A(-10) B(-10) C(-20) D(-20) E(30) F(-5) G(5) Marked arcs are at ther upper bounds Note that the sum of those bounds s the max flow H(50) I(10) 5 10 S Lesson 8-2, p. 11

25 Delete the Arcs n the Mn Cut T A(-10) B(-10) C(-20) D(-20) E(30) F(-5) G(5) Now we see the partton of the nodes... H(50) I(10) Whch partton s the maxmum weght closure? And what s ts weght? S Lesson 8-2, p. 12

26 Some Fnal Notes We can solve the max weght closure problem drectly: max z W x subject to x x j for all, j ARCS(, x = { 0,1} for all People convert t to a network because: There are specal max flow algorthms avalable that do not requre expensve LP solvers It s relatvely easy to code these algorthms and they run quckly However, you must do added work to fnd the soluton See (the Remote Interactve Optmzaton Testbed) webste for a demo Lesson 8-2, p. 13

27 The Crtcal Path Method (CPM) Recent evoluton of project schedulng Methodology depended on who was n charge After WW I, the Gantt (bar) chart became a popular method But, bar charts had lmted ablty to depct complex relatonshps DuPont and Remngton Rand Unvac developed a new method n the late 1950 s Approach was to depct the project as a network Am was provde a means to nvestgate tradeoffs n project cost and duraton Came to be known as CPM Lesson 8-3, p. 1

28 CPM Formulaton CPM s essentally a longest-path problem (and can be depcted as an MCNFP) Consder the followng example (from Schrage): Actvty Job # Tme Predecessors Dg basement A 3 none Pour foundaton B 4 A Pour basement floor C 2 B Install floor josts D 3 B Install Walls E 5 B Install rafters F 3 C,E Install floorng G 4 D Rough nteror H 6 G Install roof I 7 F Fnsh nteror J 5 I,H Landscape K 2 C,E Lesson 8-3, p. 2

29 Network Representaton (Actvty-on-Arc, or AOA) Nodes represent precedences Arcs represent actvtes and completon tmes Project tme s the longest path from 1 to 9 What s the crtcal path? 6 4 (G) 6 (H) 3 (D) 4 7 (I) 8 5 (J) (A) 4 (B) 3 2 (C) 3 (F) (E) 2 (K) 5 Lesson 8-3, p. 3

30 A Better Representaton: Actvty on Node (AON) Here s the same problem wth the nodes as actvtes 4 C 2 2 F 3 I s A B 4 D 3 G H 4 6 J E 5 K 2 d Ths representaton s far superor to AOA AOA frequently requres dummy arcs to depct precedences Mnmzng the number of dummy arcs s a dffcult problem We wll not use AOA representatons n ths course Lesson 8-3, p. 4

31 CPM as a Longest Path Problem Just maxmze the shortest path formulaton: max subject to j ARCS ( 0 x z j = C j, j ARCS (, * x xj x j = 0,, j ARCS ( j, ) 1 for all ARCS(, j 1, = s s and 1, = d d ( us ) ( u ) ( u ) d However, we wll work (for now) wth the dual of ths problem The ndces, j (wth start s and fnsh d) now represent jobs The varable u s the start tme for each job Let C be the completon tme of job Lesson 8-3, p. 5

32 Dual Formulaton Here s what the dual looks lke: mn u u z = u subject to j u d C u for all, unrestrcted for all s j ARCS(, The dual s not a network! The total tme s the dfference between u d and u s The rest of the constrants enforce precedences, completon tmes The dual s easer to formulate (and extend) than the MCNFP Ths formulaton does let us get at what the orgnal researchers wanted to nvestgate, though... Lesson 8-3, p. 6

33 Project Crashng Addresses trades between expendtures, completon tme Assume that: You know the cost per unt tme to crash a job (CC ) You know the mnmum job completon tme (MIN ) TOT s the total desred project tme Formulaton, where cr s the amount a job s crashed: mn z u u u d j 0 u u cr = subject to s TOT + C C CC unrestrcted for all cr * cr MIN for all, j ARCS(, Lesson 8-3, p. 7

34 Just-In-Tme Schedulng In ths model, some jobs must start wthn a certan amount of tme of other jobs Let S j be the max length of tme between the start of job and the start of job j How do we modfy the formulaton to handle ths? mn z = u subject to u u u j j u u d C + S u j for all, for all, unrestrcted for all s j ARCS(, j wth S j 0 Lesson 8-3, p. 8

35 Another Twst Suppose nstead we penalze the tme dfference between the completon of job and the start of job j Let: P j be the late penalty per unt tme TOT be the total desred project tme The followng formulaton mnmzes these penaltes: mn z = subject to u u u d j u u s j, j ARCS (, TOT C ( u u C ) for all, unrestrcted for all * P j j ARCS(, Lesson 8-3, p. 9

36 How Do You Fnd the Crtcal Path? Suppose you solve the example problem n MPL You get task start tmes, but don t know whch ones are crtcal The key s to look at the dual values of the constrants, whch represent the arcs Any arc wth a nonzero dual value s on the crtcal path j Slack Shadow Prce s A 0 1 A B 0 1 B C -3 0 B D -2 0 B E 0 1 C F 0 0 C K D G 0 0 E F 0 1 E K F I 0 1 G H 0 0 H J 0 0 I J 0 1 J d 0 1 K d 0 0 Lesson 8-3, p. 10

37 MPL Code TITLE CPM; { Schrage CPM example; MPL must be } { n case senstve mode! } INDEX node := (s,a,b,c,d,e,f,g,h,i,j,k,d); := node; j := node; DATA { prec s used to defne precedence arcs } prec[,j] := [s,a,1, A,B,1, B,C,1, B,D,1, B,E,1, C,F,1, C,K,1, D,G,1, E,F,1, E,K,1, F,I,1 G,H,1, H,J,1, I,J,1, J,d,1, K,d,1]; { note 0 (dummy) duraton tmes for s and d } dur[] := (0,3,4,2,3,5,3,4,6,7,5,2,0); VARIABLES u[node]; MODEL mn span = u["d"] - u["s"]; SUBJECT TO precedence[,j] where prec[,j]>0: u[node:=j] - u[node:=] > dur[]; END Lesson 8-3, p. 11