A Series of ILP Models for the Optimization of Water Distribution Networks

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1 A Seres of ILP Models for he Opzaon of Waer Dsrbuon Neworks NIKHIL HOODA 1,*, OM DAMANI 1 and ASHUTOSH MAHAJAN 2 1 Deparen of Copuer Scence and Engneerng, Indan Insue of Technology, Bobay 2 Deparen of Indusral Engneerng and Operaons Research, Indan Insue of Technology, Bobay 1 Inroducon Absrac. The desgn of rural drnkng waer schees consss of opzaon of several nework coponens lke ppes, anks, pups and valves. The szng and confguraon of hese nework confguraons needs o be such ha he waer requreens are e whle a he sae e beng cos effcen so as o be whn governen nors. We developed he JalTanra syse o desgn such waer dsrbuon neworks. The Ineger Lnear Progra (ILP) odel used n JalTanra and descrbed n our prevous work solved he proble opally, bu ook a sgnfcan aoun of e for larger neworks, an hour for a nework wh 100 nodes. In hs curren work we descrbe a seres of hree proveens of he odel. We prove ha hese proveens resul n gher odels,.e. he se of pons of lnear relaxaon s srcly saller han he lnear relaxaon for he nal odel. We es he seres of hree proved odels along wh he nal odel over egh neworks of varous szes and show a dsnc proveen n perforance. The 100 node nework now akes only 49 seconds o solve. These changes have been pleened n JalTanra, resulng n a syse ha can solve he opzaon of real world rural drnkng waer neworks n a aer of seconds. The JalTanra syse s free o use, and s avalable a hps:// MS receved ; revsed ; acceped Keywords. Waer Dsrbuon, Opzaon, Ineger Lnear Progra, Ppe Daeer Selecon, Tank Confguraon Selecon Pped waer dsrbuon neworks are used o ranspor drnkng waer fro coon waer sources o several deand areas. Therefore, he desgn of such neworks s an poran proble and has been suded n varous fors over several decades. A ypcal pped waer dsrbuon nework, as shown n fgure 1, consss of several nfrasrucure coponens lke ppes, anks, pups and valves. The locaon and szng of hese coponens are deerned as par of he nework desgn. The nework consss of one or ore sources of waer and several deand nodes. Each of hese deand nodes are descrbed by her elevaons, deand and nu pressure requreens. These nodes are conneced by several lnks along whch ppes have o be lad ou o ranspor he waer fro he source o each of he nodes. The nework layou can be looped/cyclc (ypcally urban) or branched/acyclc (ypcally rural). As he waer flows hrough he ppes, he waer pressure head reduces due o frconal losses. Ths loss, coonly referred o as headloss, depends on varous facors lke he daeer, roughness, flow and lengh of he ppe. The ppe daeer selecon proble consss of assgnng ppe daeers o each lnk n he nework. Ths selecon s o be ade fro a dscree se of coercally avalable ppe daeers. Each lnk can be broken up no ulple segens, each conssng of ppes of dfferen da- *For correspondence eers or each lnk can be resrced o jus one ppe daeer. In he os basc proble forulaon oher coponens lke anks, pups and valves are no consdered. Several approaches have been consdered over he years, rangng fro radonal opzaon echnques lke lnear prograng (LP) [7] [4], non lnear prograng (NLP) [9], neger lnear prograng (ILP) [8] o ea-heursc echnques lke genec prograng [10], abu search [1], shuffled frogs [2] ec. Neworks, however, do no conss of ppes and nodes exclusvely. Oher coponens lke anks, pups and valves are also par of any nework desgn. Typcally, rural waer neworks are gravy fed,.e. waer fro he source flows downsrea o he varous deand nodes n he nework. The waer head along he lnks decreases gradually due o headloss. For ceran nodes, here gh no be enough waer head n he syse o ensure her deands are sasfed. In such cases, pups are used o provde addonal head o he syse. Dependng on he nework confguraon pups can be nsalled a he source or a varous pons n he nework, as per requreen. Though he use of pups gh decrease he cos of ppes (snce ppes wh saller daeers would be requred), and n soe cases her use ay be unavodable, hey cause a sgnfcan burden on he nework operaon snce pupng requres elecrc supply. Therefore, apar fro a one-e capal cos of nfrasrucure, here s now an addonal operaonal cos of runnng he waer nework. Conversely, n ceran neworks he source gh be a 1

2 2 Fgure 1. Coponens of a ypcal Rural Pped Waer Schee. Waer s puped fro he Source o he Waer Treaen Plan (WTP) and hen o he Mass Balancng Reservor (MBR). The Prary Nework hen ranspors waer fro he MBR o he Tanks/Elevaed Sorage Reservors (ESRs), and hen fnally he Secondary Nework connecs he Tanks/ESRs o ndvdual vllages. (couresy: CTARA, IITB) a sgnfcanly hgher elevaon han he res of he nework. Ths would resul n excess pressure hroughou he nework whch ay cause ppes o burs. Therefore ppes wh hgher pressure rang would need o be nsalled, causng a sgnfcan ncrease n capal cos. In such cases, pressure reducng valves aybe eployed o arfcally reduce he excess pressure n he syse. Valves ay also be nsalled o resrc flow hrough ceran ppes for anenance and operaonal purposes. Tanks help provde buffer capacy o he nework. Snce deand vares wh e, anks can be flled durng low deand perods and provde waer n es of hgher deands. They can also be used o anage dsrbuon of waer and ac as neredary sources. Ths s parcularly relevan n he case of areas where waer s scarce. Tanks are flled fro he source and hey n urn ac as secondary sources o he fnal deand nodes. The nflow and ouflow of he anks s anaged o ensure equable and ely dsrbuon of waer. In he absence of such a syse, upsrea nodes wh hgh pressures wll draw ajory of he waer fro he source leadng o nsuffcen supply o downsrea nodes. Pups were he earles coponen o be consdered durng nework opzaon, n addon o he selecon of ppe daeers, alhough hey were resrced o a sngle pup a he source [11]. Tanks and valves were ncorporaed whn ea heursc fraeworks [1] [2] [10]. The neworks consdered n hese sudes are urban, where he role of anks s o ac as buffers o be used durng perods of hgh deand. The choce o be ade s he locaon, sze and hegh of he ank. The nuber of anks o be nsalled s fxed. Bu as enoned earler, waer scarce areas use anks as secondary sources raher han buffers. The deand nodes are paroned and allocaed o ndvdual anks. The source supples waer o he anks n a prary nework. The anks n urn supply waer o her allocaed nodes n secondary neworks. The cos of he schee can vary sgnfcanly dependng upon he nuber of anks and he paron of nodes o hese anks. Indan governen bodes use sofware lke WaerGes [14] and BRANCH [6] o desgn waer dsrbuon neworks ([3], [12], [13], [16]). These only conss of ppe daeer opzaon. Oher coponens.e pups, anks and valves are consdered anually by he desgn engneer, relyng on hs/her experence and nuon [12]. We have pleened a nework degn sofware, JalTanra, whch ncludes anks, pups and valves n addon o he ppe daeer selecon proble. JalTanra s a web based applcaon ha s free o use and avalable a hps:// JalTanra has been offcally adoped by he governen of Maharashra as a ool n he desgn process of her drnkng waer schees. In [4] we presened he frs verson of JalTanra ha ncluded jus he ppe daeer opzaon for branched neworks (ypcal n he case of rural areas). I used a LP odel and hus solved he proble quckly and opally. Ths allowed even neworks of a housand nodes o be solved n a couple of seconds. In [5] we exended he odel o nclude anks. The added coplexy of consderng boh prary and secondary neworks sulaneously, requred an ILP odel. Alhough sll opal n ers of cos, he e aken was sgnfcanly worse. In he presen work we descrbe hree sgnfcan proveens ha were ade o he odel. These proveens reduced he e aken o opze he larger neworks by orders of agnude. The e aken o opze a 150 node nework has gone fro over 40 nues o 5 seconds, and a 200 node nework whch could no be solved whn 24 hours now akes jus 70 seconds. The proveens conss of ghenng he se of consrans used o descrbe he ILP odel. Consder he exaple shown n fgure 2. The pons represen he neger pons over whch we are ryng o opze. The lnes represen

3 3 a d S b c e S Proble Forulaon Inpu: General: prary/secondary supply hours, nu/ axu headloss per k, axu waer speed Source node: head Node: elevaon, waer deand, nu pressure requreen Fgure 2. Consrans a, b, c and d descrbe he area S whch represens he lnear relaxaon of he se of 7 neger pons n wo densons. Inroducng he consran e cus off he area S 2 fro he lnear relaxaon whle sll ananng he sae se of neger pons. he consrans ha encopass hose neger pons. When solvng he lnear pon (LP) relaxaon, he enre se S s consdered. By nroducng he consran e, we can sll capure he sae neger pons whle cung off a par (S 2 ) of he lnear relaxaon. Snce a saller soluon space s now consdered whle solvng he LP relaxaon, hs speeds up he opzaon. For each of he hree proveens presened, we prove ha he newer se of consrans have a lnear relaxaon ha s a src subse of he lnear relaxaon of he older se, whle ananng he sae se of neger pons. In parcular, for he ank confguraon proveen we show ha he newer subse of consrans s as gh as possble,.e. he lnear relaxaon has no fraconal pons. Snce he overall odel s coplex, whle dscussng each proveen, we only consder a sall subse of relevan consrans a a e. The res of he paper s srucured as follows: In secon 2 we descrbe he opzaon proble forulaon and he nal odel used o solve he proble. In secons 3, 4 and 5 we descrbe he hree proveens. For each proveen, we frs repea he relevan subse of consrans fro he nal odel, hen provde he new se of consrans of he proved odel and hen fnally prove ha he proved odel s srcly beer han he nal odel. In secon 6 we descrbe an nal aep a an alernave edge based approach o odellng he proble. The nal odel and he hree pos proveen odels were esed on egh neworks of varyng szes. Secon 7 provdes he perforance deals of hese ess. Fnally we provde our concludng houghs n secon 8. 2 Inal Model (Model 1) As dscussed above, drnkng waer dsrbuon neworks consss of varous coponens. To opze he cos of such neworks, several npus us be consdered and for each coponen several paraeers us be deerned. We frs explcly forulae he proble ha we are aepng o solve. Then n secon 2.2 we provde deals of he nal ILP odel used o solve he proble. Lnk: sar/end node, lengh Exsng Ppes: sar/end node, lengh, daeer, parallel allowed, roughness Coercal ppe daeer: roughness, cos per un lengh Tanks: axu ank heghs, ank capacy facor, nodes ha us/us no have anks, capal cos able Pups: nu pup sze, effcency, desgn lfee, capal/energy cos, dscoun/neres rae, ppes ha canno have pups Valves: locaon, pressure rang Oupu: Lengh and daeer of ppe segens for each lnk Paronng he se of lnks no prary and secondary nework Locaon, hegh and sze of Tanks Se of nodes beng served by each Tank Locaon and power of Pups Objecve: Mnze oal capal cos (ppe + ank + pup) and oal energy cos (pup) Consrans: Pressure a each node us exceed nu pressure specfed Waer deand us be e a each node 2.2 Model Deals The ppe daeer selecon n he odel s represened by he connuous varable l j whch represens he lengh of he j h ppe daeer coponen of he h lnk n he nework. Ths deernes he capal cos of he ppes. The ank allocaon s represened by he bnary varable s n whch s rue f he ank a he n h node n he nework provdes waer o he h node n he nework. The choce of ank allocaon varables, fxes he oal deand ha each ank serves.e. he varable d n. Ths n urn deernes he capal cos of he

4 4 anks. Apar fro he cos consderaons, each node n us also have s nu pressure consran sasfed. The head a each node, h n s dependen on he headloss hl n he lnks of he nework. Ths headloss depends on he ppe varables l j and he ank varables s n enoned earler. In addon, he nroducon of pups/valves ncreases/decreases he headloss respecvely. The deals of he paraeers, varables, objecve funcon and consrans of he odel are as follows: Paraeers: NL: Nuber of lnks n he nework NP: Nuber of coercal ppe daeers D j : Daeer of j h coercal ppe daeer C j : Cos per un lengh of j h coercal ppe daeer NN: Nuber of nodes n he nework NE: Nuber of rows n he ank cos able B k : Base cos of he k h row of he ank cos able UN k : Un cos of he k h row of he ank cos able UP k : Upper l capacy for he k h row of he ank cos able LO k : Lower l capacy for he k h row of he ank cos able CP: Capal cos of pups per un kw EP: Energy cos of pups per un kwh DF: Dscoun facor for he energy cos over he enre schee lfee PH: Nuber of hours of waer supply n he prary nework S H: Nuber of hours of waer supply n he secondary nework Y: Lfee of schee n years INFR: Inflaon rae INT R: Ineres rae L : Lengh of he h lnk P n : Mnu pressure requred a node n E n : Elevaon of he n h node DE n : Waer deand of he n h node DE: The oal waer deand of he nework VH : Head reducon by valve n h lnk HL p j : Headloss for he jh daeer of he h lnk, f s par of he prary nework HL s j : Headloss for he jh daeer of he h lnk, f s par of he secondary nework FL p : Flow n h lnk f s par of he prary nework FL s : Flow n h lnk f s par of he secondary nework R j : Roughness of j h coercal ppe daeer T n : Mnu ank hegh allowed T ax : Maxu ank hegh allowed ρ: Densy of waer g: Acceleraon due o gravy η: Effcency of pup PP n : Mnu pup power allowed PP ax : Maxu pup power allowed A n : Se of nodes ha are ancesors of node n D n : Se of nodes ha are descendans of node n C n : Se of nodes ha are chldren of node n P n : Paren node of node n I n : Incong lnk for node n O n : Se of ougong lnks fro node n Connuous Varables: l j : Lengh of he j h ppe coponen of he h lnk l p j : Lengh of he jh ppe coponen of he h lnk, f lnk s par of he prary nework hl : Toal headloss across lnk d n : Toal deand served by ank a node n z nk : Toal deand served by ank a node n, f cosed by he k h row of he ank cos able p : Power of pup nsalled a lnk p p : Power of pup nsalled a lnk, f lnk s par of prary nework p s : Power of pup nsalled a lnk, f lnk s par of secondary nework ph : Head provded by pup a lnk h n : Waer head a node n n : Hegh of ank a node n h n: Effecve head provded o lnk by s sarng node n

5 5 Bnary Varables: e nk : 1 f ank a n h node s cosed by he k h row of ank cos able f : 1 f lnk s par of he prary nework, 0 f par of he secondary nework es n : 1 f source for lnk s s sarng node n s n : 1 f ank a node n s source for node pe : 1 f a pup s nsalled a lnk Objecve Funcon: The objecve funcon s sply he su of capal cos of he ppes, anks, pups and valves used n he nework. In addon, we also have he operaonal cos of he pups. Ths operaonal cos s copued as he presen value of he oal cos over he schee lfee. O(.) = + Consrans: NL C j (D j )l j + =1 NL CP p =1 NN e nk (B k + UN k (d n LO k )) n=1 NL NL + EP DF PH p p + S H p s where DF = =1 Y n=1 =1 ( ) n INFR 1 + INTR (1) The oal lengh of he ppe daeer segens us equal o he oal lnk lengh: L = l j, = 1... NL (2) The pressure a each node us exceed he nu pressure requred: P n h n (E n + n ), n = 1... NN (3) Across every lnk here s headloss hl. Ths headloss depends on he flow, lengh and daeer of he ppe daeer coponen. We use he Hazen-Wllas equaon [15] o calculae he headloss. The headloss across a lnk also depends on he pup and valve nsalled across, f any. The valves are sply npu paraeers o he odel snce hey are anually fxed. The consrans relaed o he pup head ph are descrbed furher below. The flow hrough he lnk depends on wheher he lnk s par of he prary or secondary nework: hl = (HL p j lp j + HLs j (l j l p j )) ph + VH, HL p j = HL s j = FL s = ( FL p D 4.87 j ( FL s D 4.87 j R j ) 1.852, = 1... NL (4) = 1... NL, j = 1... NP (5) R j ) 1.852, = 1... NL, j = 1... NP (6) FL p PH, = 1... NL (7) S H The head h n a each node n s calculaed by he effecve head h provded by s paren node and he headloss hl across he lnk connecng wo nodes. The effecve head n urn depends on wheher he lnk has he ank a he sarng node as s source. Ths s represened by he Boolean varable es : h n = h hl, n = 1... NN, = P, = I n (8) h = ( + E ) es + h (1 es ), es = s (1 f ), = 1... NN, O (9) = 1... NN, O (10) Nex, we look a he consrans relang o he ank allocaon. The frs ank consran s o ensure ha every ank hegh s beween paraeers T n and T ax. T n n T ax (11) We hen look a he consrans ha deal wh allocaon of deand nodes o anks. s n s 1 f ank a node n serves he deand of node. If a node n does no serve s own deand.e. s par of a secondary nework, hen all s downsrea nodes wll also be par of a secondary nework. s s nn, n = 1... NN, D n (12) If a node n does no serve s own deand, hen canno serve he deand of s downsrea nodes. s n s nn, n = 1... NN, D n (13) For every node n, only one upsrea node can serve s deand. s n = 1, n = 1... NN, A n {n} (14) The oal deand d n served by node n s he su of he deands of he downsrea nodes ha serves.e. all such ha s n = 1. d n = s n DE, n = 1... NN, D n {n} (15)

6 6 For a node n, s ncong ppe wll have prary flow only f he node serves self. f = s nn, n = 1... NN, = I n (16) If a node n serves node.e. s j = 1, each node o n he pah fro n o belongs o a secondary nework and herefore canno serve self. s n 1 s oo, n = 1... NN, D n, o D n A (17) Nex, we have he consrans ha relae he deand ha a ank serves o s cos varables e nk. Noe ha we requre z nk n our objecve funcon: z nk = e nk d n, n = 1... NN, k = 1... NE (18) Snce every ank can be cosed usng exacly one row, he su of e nk for a gven n us be 1: e nk = 1, n = 1... NN (19) Nex we have consrans ha ake sure ha he ank capacy d n les beween he nu and axu capacy of he seleced row of he cos able: For n = 1... NN, k = 1... NE : LO k e nk d n (20) DE e nk + d n UP k + DE (21) Nex, we look a consrans relaed o pups. The pup power p relaes o he pup head ph n he followng way: p = p p + p s, = 1... NL, (22) p p = (ρ g FLp ph ) f, = 1... NL, (23) η p s = (ρ g FLs ph ) (1 f ), = 1... NL (24) η Fnally, he pup power for each pup us le beween nu and axu allowed pup power. Ths s pleened usng he bnary varable pe. PP n pe p PP ax pe, = 1... NL (25) Ths coplees he descrpon of he nal odel. Alhough hs odel provdes opal resuls n ers of capal cos, he e aken o solve neworks rses rapdly wh ncreased nework sze. In he nex hree secons we go over hree proveens ade eravely o hs nal odel. For each proveen, we frs descrbe he subse of varables and consrans fro he nal odel ha are beng consdered. We nex provde he proved se of consrans. Fnally, we prove how he lnear relaxaon of he proved se s a src subse of he lnear relaxaon of he nal se. 3 Ppe Headloss Iproveen 3.1 Inal Model We descrbe a par of he odel whose purpose s o deerne he ppe daeers chosen for each lnk n he nework. Each lnk can conss of ulple ppe daeers. Also, each lnk can be par of he prary nework or he secondary nework. The headloss across he lnk depends on hese choce of ppe daeers and wheher belongs o he prary or secondary nework. The se of varables and daa used for hs purpose are defned as follows. Consder a nework of NL lnks. Le NP be he nuber of ppe daeers avalable. Varables: Daa: l j = lengh of he j h ppe daeer coponen of lnk, = 1... NL, j = 1... NP, l p j = lengh of he jh ppe daeer coponen of lnk, f lnk s par of he prary nework, = 1... NL, j = 1... NP, hl = headloss across lnk, = 1... NL, f = 1 of lnk s par of he prary nework, 0 f s par of he secondary nework, = 1... NL. L = Lengh of lnk, = 1... NL, HL p j = Un headloss for he j h ppe daeer coponen of lnk, f s par of prary nework, = 1... NL, j = 1... NP, HL s j = Un headloss for he jh ppe daeer coponen of lnk, f s par of secondary nework, = 1... NL, j = 1... NP. Consrans: The frs consran capures l p j as a produc of l j and f : l p j = l j f, = 1... NL, j = 1... NP, (26) Equaon (26) consss of a produc of wo varables, and s herefore a non-lnear equaon. Forunaely snce f s a bnary varable, we can lnearze he equaon usng he followng nequales: 0 l p j, = 1... NL, j = 1... NP, (27) l p j L f, = 1... NL, j = 1... NP, (28) l j L (1 f ) l p j, = 1... NL, j = 1... NP, (29) l p j l j, = 1... NL, j = 1... NP, (30) The su of all ppe daeer coponens us equal he lnk lengh: l j = L, = 1... NL, (31)

7 7 Nex we have he equaon for hl, whch s he su all headloss coponens conrbued by he dfferen ppe daeer coponens of lnk : hl = P j l p j + S j (l j l p j ), = 1... NL. (32) Fnally we have consrans ha relae o he bounds for he varables: l j 0, = 1... NL, j = 1... NP, (33) f { 0, 1 }, = 1... NL, (34) Snce here exss a l p j for each lnk and ppe daeer cobnaon n he nework, a large nuber of lnear decoposons of equaon (26) need o be done. In he nex secon we show an proved odel ha has he sae feasble 0-1 se of values bu wh a gher LP bound resulng n beer perforance. 3.2 Iproved Model (Model 2) In order o decopose he produc of varables n (26), a large nuber of consrans needed o be added. Ths s avoded n he new odel by no explcly defnng l p j. Insead s relaon o l j and f s plc. In he nex secon we show ha new odel s beer, n ha has a gher LP bound han he old odel. Varables: We nroduce one new varable, whch s slar o l p j bu for he secondary nework: l s j = lengh of he jh ppe daeer coponen of lnk, f lnk s par of he secondary nework, and 0 f lnk s par of he prary nework = 1... NL, j = 1... NP. Consrans: The frs consran sply saes ha l j s he su of he prary and secondary coponens,.e l p j and l s j respecvely: l j = l p j + ls j, = 1... NL, j = 1... NP, (35) For a gven lnk, eher all l p j are 0 or all ls j are 0, dependng on he value of f. And he su of he non-zero coponens us equal he lengh of he lnk L. The frs wo nequales of he new odel capure hs: l p j = L f, = 1... NL, (36) l s j = L (1 f ), = 1... NL. (37) Nex we have he equaon for hl, whch s he su all headloss coponens conrbued by he dfferen ppe daeer coponens of lnk. For he new odel we equvalenly use l s j nsead of l j l p j due o equaon (35): hl = P j l p j + S j l s j, = 1... NL. (38) Fnally as before we have he bounds for he varables: l j 0, = 1... NL, j = 1... NP, (33) l p j l s j 0, = 1... NL, j = 1... NP, (27) 0, = 1... NL, j = 1... NP, (39) f { 0, 1 }, = 1... NL. (34) We now prove ha he proved odel s gher han he nal odel, ha s he lnear relaxaon of he proved odel s a src subse of he lnear relaxaon of he nal odel. Le S 1 be he se of pons belongng o he nal odel and S 2 be he se of pons belongng o he proved odel. Le R 1 and R 2 be he se of pons correspondng o he LP relaxaons of S 1 and S 2 respecvely. Boh R 1 and R 2 are defned by he sae se of consrans ha descrbe he nal ses S 1 and S 2, excep for he consran (34) whch refers o he bnary naure of f. Insead, he connuous bounds for f s defned as follows: 0 f 1, = 1... NL. (40) Proposon 1. R 2 s a src subse of R 1.e. R 2 R 1. We prove R 2 s a src subse of R 1 n wo seps. Frs we show ha R 2 s a subse of R 1 and hen we show ha R 2 s no equal o R 1. Cla 1.1. R 2 R 1,.e. for every pon P, P R 2 P R 1. Proof. Consder a pon P R 2. I sasfes he consrans (27), (33) and (35) - (40). We prove ha also les n R 1 by showng ha sasfes he consrans (27)-(33) and (40). Consrans (27), (33) and (40) are rvally sasfed snce hey are coon for boh ses. For = 1... NL, j = 1... NP Provng (28) : l p j L f l p j = L f (36) {usng l p j 0 (27)} l p j L f Hence sasfed. Provng (29) : l j L (1 f ) l p j l s j = L (1 f ) (37) {usng l s j 0 (39)}

8 8 l s j L (1 f ) {usng l j = l p j + ls j (35)} l j l p j L (1 f ) {rearrangng} l j L (1 f ) l p j Hence sasfed. Provng (30) : l p j l j 0 l s j (39) {usng l j = l p j + ls j (35)} Proof. Take pon Q(l, l p, l s, hl, f ) = ([L/2, L/2], [L/2, L/2], [0, 0], L, 1/2). Here (n, ) = (1, 2) and (L, P, S ) = (L, [1, 1], [1, 1]) where L 0. We show Q R 1 snce sasfes all he consrans. 27 : 28 : 0 l p 11, replacng values we ge 0 L/2 0 l p 12, replacng values we ge 0 L/2 l p 11 L 1 f 1, replacng values we ge L/2 L/2 l p 12 L 1 f 1, replacng values we ge L/2 L/2 0 l j l p j {rearrangng} l p j l j Hence sasfed. Provng (31) : l j = L l p j = L f (36) l s j = L (1 f ) (37) {addng equaons} (l p j + ls j ) = L f + L (1 f ) {usng l j = l p j + ls j (35) and splfyng} l j = L Hence sasfed. 29 : 30 : 31 : 32 : 33 : l 11 L 1 (1 f 1 ) l p 11, replacng values we ge L/2 L(1 1/2) L/2 l 12 L 1 (1 f 1 ) l p 12, replacng values we ge L/2 L(1 1/2) L/2 l p 11 l 11, replacng values we ge L/2 L/2 l p 12 l 12, replacng values we ge L/2 L/2 l 11 + l 12 = L 1, replacng values we ge L/2 + L/2 = L hl 1 = P 11 l p 11 + P 12l p 12 + S 11(l 11 l p 11 ) + S 12(l 12 l p 12 ), replacng values we ge L = L/2 + L/ l 11 0, replacng values we ge L/2 0 l 12 0, replacng values we ge L/2 0 Provng (32) : hl = P j l p j + S j (l j l s j ) hl = P j l p j + S j l s j (38) {usng l j = l p j + ls j (35)} hl = P j l p j + S j (l j l s j ) Hence sasfed. Therefore pon P R 1, snce sasfes he consrans (27)-(33) and (40). Therefore R 2 R 1. Cla 1.2. There exss a pon Q such ha Q R 1 and Q R : 0 f 1 1, replacng values we ge 0 1/2 1/2 Therefore pon Q R 1. To show ha Q R 2 consder equaon 37: l s 11 + ls 12 = L 1(1 f 1 ), replacng values we ge = L/2 Therefore Q R 2. Cla 1.1 shows ha R 2 R 1. Cla 1.2 shows ha R 2 R 1. These wo ogeher ply R 2 R 1. Proposon 1 herefore shows ha he LP relaxaon of he new odel (R 2 ) has a gher bound han he LP relaxaon of he old odel (R 1 ).

9 9 4 Tank Cos Iproveen 4.1 Inal Model We descrbe a par of he odel whose purpose s o deerne he capal cos of each ank n he nework. Snce he ank cos s a pece-wse lnear funcon, we need o deerne whch row n he ank cos able does he ank capacy fall n. Each row n he ank cos able has nu and axu capacy values. If he ank capacy s whn hese values, hen ha row s used o copue he ank s cos. Bnary varables are used o capure for each ank, he row n he cos able ha s chosen o copue he cos. The se of varables and daa used for hs purpose are defned as follows. Consder a nework of n locaons. Le be he nuber of lnear coponens of he pecewse lnear cos of consrucon of a ank. Varables: Daa: e nk = 1 f he ank a locaon n s cosed usng he k h row of he ank cos able, n = 1... NN, k = 1... NE, z nk = capacy of he ank a locaon n f s cosed usng he k h row of he ank cos able, 0 oherwse, n = 1... NN, k = 1... NE, d n = capacy of he ank a locaon n, n = 1... NN. LO k = nu capacy ha he k h row of he ank cos able can sasfy, k = 1... NE, UP k = axu capacy ha he k h row of he ank cos able can sasfy, k = 1... NE, DE = value of he oal waer deand n he nework. Consrans: The frs consran relaes he ank capacy correspondng o he k h row as a produc of he ank capacy and he bnary choce varable e nk : z nk = e nk d n, n = 1... NN, k = 1... NE, (41) Snce equaon (41) consss of a produc of wo varables, s a non-lnear equaon. Forunaely snce e nk s a bnary varable, we can lnearze he equaon usng he followng nequales: 0 z nk, n = 1... NN, k = 1... NE, (41.a) z nk DEe nk, n = 1... NN, k = 1... NE, (41.b) d n DE(1 e nk ) z nk, n = 1... NN, k = 1... NE, (41.c) z nk d n, n = 1... NN, k = 1... NE, (41.d) Snce every ank can be cosed usng exacly one row, he su of e nk for a gven n us be 1: e nk = 1, n = 1... NN, (42) Nex we have consrans ha ake sure ha he ank capacy d n les beween he nu and axu capacy of he seleced row of he cos able: LO k e nk d n, n = 1... NN, k = 1... NE, (43) DEe nk + d n UP n + D, n = 1... NN, k = 1... NE, (44) Fnally we have consrans ha relae o he bounds for he varables: DE d n, n = 1... NN, (45) DE UP k, k = 1... NE. (46) e nk { 0, 1 }, n = 1... NN, k = 1... NE, (47) d n 0, n = 1... NN. (48) Snce here exss a z nk for each ank and row of cos able cobnaon, a large nuber of lnear decoposons of equaon (41) need o be done. Ths resuls n poor perforance of he odel. In he nex secon we show an proved odel ha has he sae feasble 0-1 se of values bu wh a gher LP bound resulng n beer perforance. We hen show ha he proved odel has a gher bound. 4.2 Iproved Model (Model 3) As dscussed n he prevous secon, he prncpal ssue wh he old odel was equaon (41), where z n k s expressed as a produc of wo varables. In order o decopose he varables, a large nuber of consrans needed o be added. Ths s avoded n he new odel by no explcly defnng z nk. Insead s relaon o e nk and d n s plc. In he nex secon we frs show ha new odel s beer n ha has a gher LP bound han he old odel and hen we go on o show ha he LP for he new odel has gh soluons. The varables rean sae for he new odel. The frs wo nequales of he odel provde he bounds for z nk n ers of e nk and he nu(lo k ) and axu(up k ) capaces for each row of he cos able: LO k e nk z nk, n = 1... NN, k = 1... NE, (49) z nk UP k e nk, n = 1... NN, k = 1... NE, (50) The nex equaon for he odel reans unchanged, represens he fac ha each row of he cos able s chosen exacly once for each ank: e nk = 1, n = 1... NN, (42)

10 10 Nex, we have a slar equaon bu hs e relaed o he varable z nk. The su of all z nk values for a gven ank us equal d n : z nk = d n, n = 1... NN, (51) In fac along wh he prevous hree equaons of he odel, one can ply ha exacly one of he z nk values wll be non zero for a specfc ank and herefore wll be equal o d n. Ths herefore capures he non-lnear consran ha equaon 41 of he old odel capured. The reanng consrans relae o he bounds for he varables: DE d n, n = 1... NN, (45) DE UP k, k = 1... NE. (46) e nk { 0, 1 }, n = 1... NN, k = 1... NE, (47) d n 0, n = 1... NN, (48) z nk 0, n = 1... NN, k = 1... NE. (41.a) Le S 1 be he se of pons belongng o he old odel and S 2 be he se of pons belongng o he new odel. Le R 1 and R 2 be he se of pons correspondng o he LP relaxaons of S 1 and S 2 respecvely. Boh R 1 and R 2 are defned by he sae se of consrans ha descrbe he nal ses S 1 and S 2, excep for he consrans (47) whch refers o he bnary naure of e nk. Insead, he connuous bounds for e nk s defned as follows: Hence sasfed. Provng (41.c) : d n DE(1 e nk ) z nk z nk = d n (51) k =1 z nk + {splng su} k =1,k k z nk {rearrangng} k =1,k k = d n z nk = d n z nk (53) e nk = 1, (42) k =1 {splng su} e nk + k =1,k k {rearrangng} k =1,k k e nk = 1 e nk = 1 e nk, (54) 0 e nk 1, n = 1... NN, k = 1... NE. (52) Slar o secon 3, we prove ha he LP relaxaon of he new odel s gher han he LP relaxaon of he old odel. We do so by showng ha R 2 s a src subse of R 1. We hen go on o show ha R 2 has no fraconal corner pons and hus canno be ghened furher. Proposon 2. R 2 s a src subse of R 1.e. R 2 R 1 As n secon 3, we prove R 2 s a src subse of R 1 n wo seps. Frs we show ha R 2 s a subse of R 1 and hen we show ha R 2 s no equal o R 1. z nk DEe nk {su over k } k =1,k k z nk DE {usng (53), (54)} k =1,k k d n z nk DE(1 e nk ) {rearrangng} d n DE(1 e nk ) z nk Hence sasfed. e nk (41.b) Cla 2.1. R 2 R 1,.e. for every pon P, P R 2 P R 1. Proof. Consder a pon P R 2. I sasfes he consrans (41.a), (42) and (45)-(52). We prove ha also les n R 1 by showng ha sasfes he consrans (41.a)-(48) and (52). Consrans (41.a), (42), (45)-(48) and (52) are rvally sasfed snce hey are coon for boh ses. For n = 1... NN, k = 1... NE Provng (41.b) : z nk DEe nk z nk UP k e nk (50) {usng DE UP k (46)} z nk DEe nk Provng (41.d) : z nk d n z nk = d n, (51) {usng 0 z nk (41.a)} z nk d n Hence sasfed. Provng (43) : LO k e nk d n LO k e nk z nk (49) {usng z nk d n (41.d)} LO k e nk d n

11 11 Hence sasfed. Provng (44) : d n DE(1 e nk ) z nk {usng z nk UP k e nk (50)} d n DE + DEe nk UP k e nk {usng 0 e nk 1 (52)} d n DE + DEe nk UP k {rearrangng} d n + DEe nk UP k + DE Hence sasfed. d n + DEe nk UP k + DE (41.c) DEe 11 + d 1 UP 1 + DE, replacng values we ge 2d 1/2 + d d + 2d DEe 12 + d 1 UP 2 + DE, replacng values we ge 2d 1/2 + d 2d + 2d 46 : DE UP 1, replacng values we ge 2d d DE UP 2, replacng values we ge 2d 2d 45 : DE d 1, replacng values we ge 2d d Therefore pon P R 1, snce sasfes all he consrans. Therefore R 2 R 1. Cla 2.2. There exss a pon Q such ha Q R 1 and Q R 2. Proof. Take a pon Q(z, e, d) = ([d, d], [1/2, 1/2], d). Here (n, ) = (1, 2), (LO k, UP k, DE) = ([0, d], [d, 2d], 2d) where d 0. We show Q R 1 snce sasfes all he consrans. 41.a : 41.b : 41.c : 41.d : 0 z 11, replacng values we ge 0 d 0 z 12, replacng values we ge 0 d z 11 DEe 11, replacng values we ge d 2d/2 z 12 DEe 12, replacng values we ge d 2d/2 d 1 DE(1 e 11 ) z 11, replacng values we ge d 2d(1 1/2) d d 1 DE(1 e 12 ) z 12, replacng values we ge d 2d(1 1/2) d z 11 d 1, replacng values we ge d d z 12 d 1, replacng values we ge d d 52 : 0 e 11 1, replacng values we ge 0 1/2 1 0 e 12 1, replacng values we ge 0 1/2 1 Therefore pon Q R 1. To show ha Q R 2 consder equaon 51: z 11 + z 12 = d, replacng values we ge d + d = 2d Therefore Q R 2. Proposon 2.1 shows ha R 2 R 1. Proposon 2.2 shows ha R 2 R 1. These wo ogeher ply R 2 R 1. Proposon 2 herefore shows ha he LP relaxaon of he new odel (R 2 ) has a gher bound han he LP relaxaon of he old odel (R 1 ). We nex show ha n fac R 2 has he ghes bound possble by showng ha a pon wh fracon value for e nk wll never be a corner pon. Proposon 3. If pon P R 2 has a fraconal value for e nk, P canno be a corner pon of R 2. Proof. Consder a pon P(z, e, d) R 2 wh a leas one fraconal value for e nk.e. 0 < e n k < 1 for soe n, k. Le e n k =. Consruc anoher pon P 1 ha has he sae coponens of P for n n. For n = n ake (z, e, d) as follows: For k = 1... NE z n k = 0 42 : e 11 + e 12 = 1, replacng values we ge 1/2 + 1/2 = 1 43 : LO 1 e 11 d 1, replacng values we ge 0 1/2 d LO 2 e 12 d 1, replacng values we ge d 1/2 d 44 : P. z n k = zp n k, f or k k 1 e n k = 0 e n k = ep n k, f or k k 1 d k = dp n zp n k 1 Here zn P k, e P n k, d P n are he correspondng values of pon

12 12 We show ha P 1 R 2 snce sasfes all he consrans: 49 : LO k e n k z n k LO k 0 0 LO k e n k z n k, k k (defnon) en P LO k k 1 zp n k 1, k k (defnon) LO k en P k zp n k, k k (0 < < 1) 50 : Sasfed snce P R 2. z n k UP k e n k 0 UP k 0 z n k UP k e n k, k k zp n k e P n k (defnon) 1 UP k 1, k k (defnon) zn P k UP ken P k, k k (0 < < 1) 42 : 51 : Sasfed snce P R 2. NE e n k = 0 + =,k k = 1 ep n k 1,k k e n k ep n k 1 (splng su) (defnon) en P k = 1) NE ( = 1 1 (defnon) = 1 (0 < < 1) NE z n k = 0 + =,k k,k k z n k zp n k 1 = dp n ep n k 1 (splng su) NE ( (defnon) zn P k = dp n ) 52 : = d n (defnon) en P k = 1 (P R 2) e P n k + +,k k e P n k = 1,k k e P n k = 1,k k e P n k = 1 (splng su) (defnon) (rearrangng) 0 e P n k 1, k k (e P n k 0) 0 ep n k 1 1, k k (1 > 0) 0 e n k 1, k k (defnon) 48 : zn P k = dp n (P R 2) z P n k + 41.a :,k k z P n k = dp n,k k z P n k = dp n zp n k (splng su) (rearrangng) d P n zp n k 0 (zp n k 0) dp n zp n k 0 1 (1 > 0) d n 0 (defnon) z P n k 0, k k (P R 2 ) zp n k 1 0, k k (1 > 0) z n k 0, k k (defnon) Therefore P 1 R 2. Slar o P 1 consruc pon P 2 havng he sae coponens as P for n n. For n = n ake (z, e, d) as follows: For k = 1... NE z n k = zp n k z n k = 0 f or k k e n k = 1 e n k = 0 f or k k

13 13 d n = zp n k As before zn P k, e P n k, d P n are he correspondng values of pon P. We show ha P 2 R 2 snce sasfes all he consrans: 49 : LO k e P n k zp n k (P R 2) LO k z P n k (defnon) LO k zp n k (0 < ) LO k z n k (defnon) LO k 1 z n k LO k e n k z n k (defnon) LO k e n k z n k, k k LO k 0 0, k k (de f non) 50 : z P n k UP k ep n k (P R 2) z P n k UP k (defnon) UP k (0 < ) z n k UP k 1 (defnon) zp n k z n k UP k e n k (defnon) z n k UP k e n k, k k 0 UP k 0, k k (defnon) 42 : 51 : NE e n k = e n k +,k k e n k (splng su) = (defnon) NE z n k = z n k + = zp n k,k k z n k (splng su) + 0 (defnon) = d n (defnon) 52 : 48 : 41.a : e n k = 1 (defnon) e n k = 0, k k (defnon) z P n k 0 (P R 2) zp n k 0 (0 < ) d n 0 (defnon) z P n k 0 (P R 2) 0 (0 < ) z n k 0 (defnon) zp n k z n k 0, k k 0 0, k k (defnon) Therefore P 2 R 2. P 1 = (( zp n k 1, 0), ( ep n k d P, 0), zn P k 1 1 P 2 = ((0, zp n k ), (0, 1), zp n k ) P 1 (1 ) + P 2 = ((zn P k, zp n k ), (ep n k, ), d P ) = P Snce P s a pon wh fraconal e k ha can be represened as a lnear cobnaon of wo oher pons belongng o R 2, P canno be a corner pon of R 2. Ths ples ha LP relaxaon (R 2 ) of he new odel wll provde only neger soluons. Therefore he new odel has he ghes bound possble. 5 Tank Confguraon Iproveen For a gven nework of nodes and lnks, one aspec of he proble s o deerne he locaon of anks and he se of downsrea nodes ha are o be served by each ank. We need a se of consrans o odel a vald nework confguraon. For a gven branched nework layou wh a sngle source, a vald nework confguraon s one n whch: 1. Each node needs o be provded waer, by exacly one of s ancesors (ncludng self). 2. If a node n provdes waer o self.e. has a ank, only hen can provde waer o s descendans. 3. If a node n ges waer fro anoher ank, hen all s descendans canno ge waer fro heselves. )

14 14 4. If node n provdes waer o one of s descendans k, hen he nodes along he pah connecng he canno serve heselves. In he followng secon we repea he se of consrans ha odel such a nework as lad ou n secon 2. We hen show ha he odel s no gh,.e. s lnear relaxaon s no guaraneed o have negral corner pons. In secon 5.2 we hen descrbe an proved odel and prove s ghness. In secon 6 we descrbe an alernae edge based approach o odellng he nework. 5.1 Inal Model Daa: Consder a ree nework of n nodes. A n = Nodes ha are ancesors of node n, n = 1... NN. D n = Nodes ha are descendans of node n, n = 1... NN. C n = Nodes ha are chldren of node n, n = 1... NN. P n = Paren node of node n, n = 1... NN. Varables: s n = 1 f ank a n h node serves he deand of h node, n = 1... NN, D n {n}. Consrans: Then we can use he followng se of consrans o descrbe he se of vald nework confguraons as descrbed earler: s s nn, n = 1... NN, D n, (55) s n s nn, n = 1... NN, D n, (56) s n = 1, n = 1... NN, A n {n}, (57) s n 1 s oo, n = 1... NN, D n, o D n A, (58) s n { 0, 1 }, n = 1... NN, D n {n}. (59) s 23 s 22, s 11 = 1, s 12 + s 22 = 1, s 13 + s 23 + s 33 = 1, s 13 1 s 22, (56.c) (57.a) (57.b) (57.c) (58.a) 0 s 11, s 12, s 13, s 22, s 23, s 33 1, (60) Snce s 11 = 1, we replace s value n he consrans and replace repeang consrans o ge he followng se: s 33 s 22, s 23 s 22, s 11 = 1, s 12 + s 22 = 1, s 13 + s 23 + s 33 = 1, s 13 1 s 22, (55.c) (56.c) (57.a) (57.b) (57.c) (58.a) 0 s 12, s 13, s 22, s 23, s 33 1, (60) Consder a pon P defned as: P{s 11, s 12, s 13, s 22, s 23, s 33 } = {1, 1 2, 0, 1 2, 1 2, 1 2 }. Snce sasfes all he consrans, P S. We now show ha P canno be descrbed as a lnear cobnaon of wo dsnc pons ha belong o S. Consder wo pons Q, R S such ha: s Q 11 = sr 11 = 1 s P 13 = 0 P = Q + (1 )R, 0 < < 1 {57.a} {defnon} s Q 13 = sr 13 = 0 {60.c} (61) s 33 + s 23 2s 22 1 s Q 13 2sQ 22 {addng 55.c and 56.c} {57.c} 1 2 sq 22 {61} Proposon 4. The lnear relaxaon of S s no gh. Proof. Le he lnear relaxaon of se S be S. Insead of consran 59 we wll have he followng consran: 0 s n 1, n = 1... NN, D n {n}. (60) Consder a sall nework of 3 nodes wh node 1 as roo and node 2 chld of node 1, and node 3 chld of node 2. For a pon o belong o S, he followng consrans us be e: s 22 s 11, s 33 s 11, s 33 s 22, s 12 s 11, s 13 s 11, (55.a) (55.b) (55.c) (56.a) (56.b) 1 2 sr 22 s Q 22 = sr 22 = 1 2 s 12 + s 22 = 1 s Q 12 = sr 12 = 1 2 s 33 s 22 s Q s 23 s 22 s Q {Slarly} {s22 P = 1 2 } (62) {57.b} {62} {55.c} {62} {56.c} {62}

15 s Q 23 = sq 33 = 1 2 s R 23 = sr 33 = 1 2 {57.c} {Slarly} Therefore P = Q = R. Snce P canno be expressed as a lnear cobnaon of wo dsnc pons, P s a corner pon of S. And snce P conans non-negral values for s n, se S s no gh. 5.2 Iproved Model (Model 4) A new odel s proposed whch alhough anans he sae srucure as he nal odel, does so usng gher consrans. The prary nsgh abou he srucure s expressed n he second consran enoned below. A node serves s chld j f and only f serves all he nodes downsrea of j. Consder se S 2 defned by he followng se of consrans: s n = 1, n = 1... NN, A n {n}, (57) s n = s nk, n = 1... NN, C n, k D, (63) s n { 0, 1 }, n = 1... NN, D n {n}. (59) Proposon 5. The lnear relaxaon of S 2 s gh. Proof. Le he lnear relaxaon of se S 2 be S 2. Insead of consran 59 we wll have he followng consran: 0 s n 1, n = 1... NN, D n {n}. (64) We wll show ha S 2 s gh by showng any pon P, wh a non-neger coponen can be expressed as a lnear cobnaon of wo dsnc pons fro S 2. Consder a pon P S 2 wh 0 < s n n = < 1 for soe n. Le n be he frs such node n he pah fro roo. Cla 5.1. s nn = 1, n A n Proof. s nn canno be fraconal snce n s he frs such node fro roo by defnon. Assue s nn = 0 for soe n A n. Le E nn = (D n {n}) (A n {n }). s nn = 0 {usng s n = 1 (57)} s n = 1 A n s n = 1 A n {n } {splng su} s n + s kn = 1 A n, k E nn k {usng s n = s nk (63)} s n + s kn = 1 A n, k E nn k {usng s n = 1 fro above} 1 + s kn = 1 k E nn k {splfyng} s kn = 0 k E nn k {usng 0 s n 1 (64)} s kn = 0 k E nn {snce n E nn } s n n = 0 15 Bu hs s a conradcon snce we know s n n s fraconal. Therefore s nn canno be fraconal and canno be 0. s nn = 1, n A n (65) Cla 5.2. s n = 0, n A p, D n, p = P n Proof. s = 1, A n {usng s n = 1 (57)} s n = 0, n A p, A n, j D n, p = P n {usng s n = s nk (63)} s n = 0, n A p, D, p = P n (66) Consder a pon Q wh s n n = 0: s n = s P n, (A n D n {n }) (67) s n = sp n 1, n A n, j D n (68) s n = 0, n (D n {n }), D n (69) Cla 5.3. Pon Q S 2 Proof. We prove ha pon Q belongs o S2 by showng sasfes he consrans (57), (63) and (64). For nodes ha are no downsrea or upsrea of n, s n values are sae as ha of pon P. Therefore hey sasfy he consrans snce P belongs o S 2. For he res of he nodes: For n A n : Provng (57) : s n = 1 {usng s nn = 1 (65)} s nn = 1, n A n {usng s n = 0 (66)}

16 16 s n = 0, n A n, A n {sung over } s n = 1, n A n Hence sasfed. Provng (63) : s n = s nk {usng s n = s nk (63)} s P n = s P nk, {dvdng by (1 ) snce 1 } n A n, C n, k D s P n 1 = sp nk 1, n A n, C n, k D {usng s n = sp n 1 (68)} s n = s nk, n A n, C n, k D Hence sasfed. Provng (64) : 0 s n 1 {usng s n = 1 (57)} sn P = 1, A n {n } {splng su} sn P + sp n n = 1, A n {usng sn P n = } sn P = 1, A n {usng s P n 0 (64)} 0 s P n 1, A n {dvdng by (1 ) snce 1} 0 sp n 1 1, A n {usng s n = sp n 1 (68)} 0 s n 1, A n Hence sasfed. For n D n {n }: Provng (57) : s n = 1 {usng s n = 1 (57)} sn P = 1, A n { } {usng s n = 0 (66)} s P kn + sp n n = 1, k = P n {usng s P n n = } s P kn = 1, k = P n {usng s n = sp n 1 (68)} s kn = 1, k = P n {usng s n = 0 (69)} s n = 1, Hence sasfed. Provng (63) : s n = s nk {usng s n = 0 (69)} s n = 0 s n = s nk Hence sasfed. Provng (64) : 0 s n 1 {usng s n = 0 (69)} s n = 0 Hence sasfed. Therefore pon Q S 2. n D n {n }, A n n D n {n }, D n n D n {n }, D n, k C n Slarly consder pon R wh s n n = 1: n D n {n }, D n s n = s P n, (A n D n {n }) (70) s n = 0, n A n, D n (71) s n = sp n, n (D n {n }), D n {n} (72) Cla 5.4. Pon R S 2 Proof. We prove ha pon R belongs o S2 by showng sasfes he consrans (57), (63) and (64) For nodes ha are no downsrea or upsrea of n, s n values are sae as ha of pon P. Therefore hey sasfy he consrans snce P belongs o S 2. For he res of he nodes: For n A n : Provng (57) : s n = 1 {usng s nn = 1 (65)} s nn = 1, n A n {usng s n = 0 (66)} s n = 0, n A n, A n {sung over } s n = 1, n A n Hence sasfed.

17 Provng (63) : s n = s nk {usng s n = 0 (71)} s n = 0 s n = s nk Hence sasfed. Provng (64) : 0 s n 1 {usng s n = 0 (71)} s n = 0 Hence sasfed. For n D n {n }: Provng (57) : s n = 1 {usng s n = 1 (57)} sn P = 1, n A n, D n n A n, D n, k C n n A n, D n A n {n} {splng su} sn P + skn P = 1, D n {n }, k A n k {usng skn P = 1 } k sn P =, D n {n } {dvdng by snce 0 } sn P = 1, D n {n } {usng s n = sp n (72)} s n = 1, D n {n } Hence sasfed. Provng (63) : s n = s nk {usng s n = s nk (63)} s P n = s P nk, {dvdng by snce 0 } s P n C n, k D = sp nk, C n, k D {usng s n = sp n (72)} s n = s nk, Hence sasfed. Provng (64) : 0 s n 1 C n, k D {usng s n = 1 (57)} sn P = 1, A n {n} {splng su} sn P + skn P = 1, D n {n }, k A n k {usng skn P = 1 } k sn P =, D n {n } {usng 0 s P n (64)} 0 s P n, D n {n } {dvdng by snce 0 } 0 sp n 1, D n {n } {usng s n = sp n (72)} 0 s n 1, D n {n } Hence sasfed. Therefore pon R S 2. Cla 5.5. P s a lnear cobnaon of pons Q and R.e. P = (1-)Q + R Proof. For (A n D n {n }): {usng s P n = s Q n (67) and s P n = s R n (70)} s P n = s Q n = s R n s P n = (1 )s Q n + s R n For n A n, D n : sn Q = sp n 1 {68} s R n = 0 {71} s P n = (1 )s Q n + s R n For n D n, D n : s Q n = 0 {69} s R n = sp n s P n = (1 )s Q n + s R n {72} Therefore P s a lnear cobnaon of pons Q and R 17 Snce any general pon P wh a fraconal coponen can be expressed as lnear cobnaon of wo oher pons n he se S2, ples ha such a pon P canno be a corner pon and herefore se S2 s gh.

18 18 Ths concludes he dscusson on he hree proveens ade o he nal ILP odel. Experenal resuls of he perforance of he odel afer each proveen s presened n secon 7. Alhough we have shown he ghness of varous subses of he proved odel, he overall se of consrans of he odel s sll no gh. As such, here reans roo for furher proveens o he odel. In he nex secon we descrbe an nal aep a an alernave approach o he proble. Insead of usng node varables s j o paron he prary and secondary nework, purely edge based varables are used. 6 Edge Based Model An alernave approach o he node based represenaon of he nework s o have an edge based represenaon. Here nsead of he focus beng whch ank serves whch node, he focus s on whch ppes n he nework are par of he prary nework and whch ppes are par of he secondary nework. Consder a ree nework of NE edges: Daa: C = Se of ppes ha are edaely downsrea of ppe, = 1... NE. U = Se of ppes ha are upsrea of ppe, = 1... NE. D = Se of ppes ha are downsrea of ppe, = 1... NE. S = Se of ppes ha are edaely downsrea of he source. Varables: f = 1 f h edge belongs o he prary nework and = 0 f h edge belongs o he secondary nework, = 1... e The prary nework connecs he source o he anks, and he secondary nework connecs he anks o downsrea nodes. Therefore ppes sarng fro he source us belong o he prary nework. Also, secondary ppes us be downsrea of he prary ppes. And once a ppe s secondary, hen any ppes downsrea can no longer be prary. We can use he followng se of consrans o descrbe he se S 3 of vald nework confguraons: f = 1, S, (73) f j f, = 1... NE, j C, (74) f { 0, 1 }, = 1... NE. (75) Proposon 6. The lnear relaxaon of S 3 s gh. Proof. Le he lnear relaxaon of se S 3 be S 3. Insead of consran 75 we wll have he followng consran: 0 f 1, = 1... NE. (76) We wll show ha S 3 s gh by showng any pon P, wh a non-neger coponen can be expressed as a lnear cobnaon of wo dsnc pons fro S 3. Consder a pon P S 3 wh 0 < f = < 1 for soe. Le be he frs such edge n he pah fro source. Cla 6.1. f = 1, U Proof. f canno be fraconal snce s he frs such edge fro source by defnon. If f = 0, hen by 74 for all s downsrea edges j, f = 0. Bu s downsrea of and f = 0. Therefore f canno be fraconal and canno be 0. f = 1, U (77) Consder a pon Q wh f = 0: Cla 6.2. Pon Q S 3 f = f P, (D { }) (78) f = 0, (D { }) (79) Proof. For all he edges no downsrea of, consrans 73, 74 and 76 are sasfed snce he values are sae as pon P and P S 3. Seng f = 0 for all downsrea also anans he consrans rvally. Therefore pon Q S 3. Slarly consder pon R wh f = 1: Cla 6.3. Pon R S 3 f = f P, (D { }) (80) f = f P, (D { }) (81) Proof. We prove ha pon R belongs o S3 by showng sasfes he consrans (73), (74) and (76). For edges ha are no downsrea of, f values are sae as ha of pon P. Therefore hey sasfy he consrans snce P S 3. For he res of he edges: For (D { }): (73) s rvally rue snce (and s downsrea edges) canno be conneced o he source snce for pon P, f 0. Provng (74) : {usng f P j f j f f P (74)} f j P f P, (D { }), j C {dvdng by snce 0 } f P j f P, {usng f = f P f j f, Hence sasfed. (81)} Provng (76) : 0 f 1 (D { }), j C (D { }), j C

19 19 {usng f P f P P (74) and 0 f (76)} 0 f P f P (D { }) {usng f P = } 0 f P (D { }) {dvdng by snce 0 } 0 f P 1 (D { }) {usng f = f P (81)} 0 f 1 (D { }) Hence sasfed. Therefore pon R S 3. Cla 6.4. P s a lnear cobnaon of pons Q and R.e. P = (1-)Q + R Proof. For (D { }): f P f P = f Q = (1 ) f Q For (D { }): = f R {78, 80} + f R f Q = 0 {79} f R f P = f P = (1 ) f Q + f R {81} Therefore P s a lnear cobnaon of pons Q and R Snce any general pon P wh a fraconal coponen can be expressed as lnear cobnaon of wo oher pons n he se S3, ples ha such a pon P canno be a corner pon and herefore se S3 s gh. The perforance of he edge based odel s worse han odel 4. Alhough we prove ha he LP relaxaon of he se of consrans descrbed by S3 s gh, he LP relaxaon objecve for he overall odel s worse. Ths s due o changes n oher consrans of he odel, ha are requred snce n hs odel only edge based varables are consdered. 7 Resuls The hree ppe cos/ank cos/ank allocaon proveens were appled sequenally o he nal odel (odel 1) o gve odel 2/odel 3/odel 4 respecvely. These 4 odels were esed over egh dfferen neworks of varyng szes n order o es her perforance and scalably: Real World Neworks: Three of he neworks, Khard, Shahpur and Mokhada are real lfe neworks fro Maharashra sae n Inda. These regons conss of rbal vllages ha regularly face exree waer sress durng suer onhs and as a resul have o be provded waer usng ankers. Arfcal Neworks: The oher fve neworks are arfcally creaed o es he perforance of he odels across dfferen nework szes (10 o 200). Each of he s a randoly generaed branched nework. Ranges for he node and lnk properes are as follows: Nuber of chldren nodes: 1 o 5, Elevaon (n eres): 100 o 300, Deand (n lres per second): 0.01 o 5, Lengh of lnks (n eres): 500 o For all four odels, he proble saeen reans he sae, o opze he oal ppe and ank cos of he nework. The nuber of bnary and connuous varables scale wh he sze of he nework. Snce all four odels solve he proble opally, he fnal capal cos of he ppes and anks s he sae. The perforance of each odel s easured n ers of hree ercs: he oal e aken n seconds, he sze of he branch and bound ree and he objecve value of he LP relaxaon. For each of he egh neworks, he e aken proves wh each odel, resulng n odel 4 provdng he bes perforance. Typcally he e aken scales wh he sze of he nework. However, hs need no always be he case. For exaple, alhough gen50 has ore nodes (50 nodes) han Mokhada (37 nodes), s solved n lesser aoun of e. Ths s because apar fro he nuber of nodes beng a facor, he nework confguraon also aers whle solvng he odel. 8 Concluson In he presen work we looked a he cos opzaon of rural drnkng waer schees. These schees conss of several nework coponens lke ppes, anks, pups and valves. We frs descrbe an nal ILP odel ha was used o solve he opzaon. Alhough opal, he odel ook a sgnfcan aoun of e for larger neworks, an hour for a nework wh 100 nodes. We hen descrbe a seres of hree proveens of he odel. For each proveen we prove ha he proved odel s gher han he nal odel. We hen fnally presen he perforance resuls of he hree proved odels along wh he nal odel over egh neworks of varous szes. The 100 node nework now akes only 49 seconds o solve. Thus we show ha ghenng he ILP odel can resul n sgnfcan proveens n ers of perforance. Ths enables praconers o consder greaer nuber of eraons of he desgn for large neworks, snce each eraon can be opzed n a aer of seconds. And alhough proveens were ade n several consrans, he overall odel s