Technical Note: A Simple Greedy Algorithm for Assortment Optimization in the Two-Level Nested Logit Model

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1 Techncal Note: A Smple Greedy Algorthm for Assortment Optmzaton n the Two-Level Nested Logt Model Guang L and Paat Rusmevchentong {guangl, rusmevc}@usc.edu September 12, 2012 Abstract We consder the assortment optmzaton problem under the classcal two-level nested logt model. We establsh a necessary and suffcent condton for the optmal assortment, and use ths optmalty condton to develop a smple greedy algorthm that teratvely removes at most one product from each nest, untl the optmalty condton s satsfed. Our algorthm also explots the beautful lumpy structure of the optmal soluton, where n each nest, a certan set of consecutve products wll always appear together n the optmal assortment. The algorthm s smple, ntutve, and extremely fast. For a problem wth m nests, wth each nest havng n products, the runnng tme s On m log m). Ths s the fastest known runnng tme for ths problem. 1. Introducton Snce Tallur and van Ryzn 2004) have shown the mportance of ncorporatng customer choce behavor n operatons management, the assortment optmzaton problem has receved much attenton from researchers. In ths problem, a frm wshes to determne the revenuemaxmzng assortment of products, assumng that a customer chooses a product from an assortment accordng to a dscrete choce model. As shown n an excellent survey by Kök et al. 2006), ths problem has many applcatons n retal and revenue management. Tallur and van Ryzn 2004) consdered the assortment optmzaton problem under the multnomal logt MNL) model, and gave a beautful algorthm for computng the revenuemaxmzng assortment; see also, Gallego et al. 2004); Kunnumkal and Topaloglu 2008); Faras et al. 2012). Although the MNL model admts a tractable soluton, t suffers from the ndependence of rrelevant alternatves property, whch results n based estmates when alternatves are correlated Tran, 2003). The nested logt model generalzes the MNL model 1

2 by groupng smlar alternatves nto a nest and allowng dfferental substtuton patterns wthn and across nests, thereby partally relaxng the ndependence of rrelatve alternatves restrcton Wllams, 1977; McFadden, 1978). Under the nested logt model, a consumer frst chooses a nest of products, and subsequently chooses a product from the chosen nest. Davs et al. 2011) presented a novel soluton to the assortment optmzaton problem under the two-level nested logt model. For the classcal model 1 wth m nests, wth each nest havng n products, they transform the combnatoral optmzaton problem nto an equvalent lnear program wth Onm) constrants and Om) varables 2, enablng them to compute the optmal soluton n On m 4 ) runnng tme. The lnear programmng framework can also be used to obtan approxmaton algorthms for other NP-hard varants of the classcal model. In ths paper, we present an alternatve soluton to the assortment optmzaton problem under the classcal nested logt model. Our approach s based on a smple greedy algorthm that teratvely removes at most) one product from each nest to ncrease the revenue. The greedy algorthm does not requre any lnear program, and t s extremely fast, wth a runnng tme of On m log m). The algorthm makes use of the necessary and suffcent condton for an optmal assortment Theorem 3), whch to our knowledge, s the frst such optmalty condton for ths problem. In addton, our analyss reveals a beautful lumpy structure of the optmal assortment: surprsngly, wthn each nest, a certan set of consecutve products wll always appear together n the optmal soluton. Moreover, as shown n Theorem 4, by lookng at a certan ndex of each product, we can determne n advance whch products wll appear together! Our greedy algorthm explots ths structure to obtan a fast runnng tme. Problem Descrpton: We brefly revew the setup for the classcal two-level nested logt model, along wth the correspondng optmzaton problem. We have m nests ndexed by {1, 2,..., m}, and each nest has n products ndexed by {1, 2,..., n} 3. We denote the dssmlarty factor of nest by τ 0, 1). A product j n nest s denoted by j, and n settngs where the nest ndex s clear from context, we smply refer to t as product j. Each product j has a revenue r j 0 and a preference weght v j > 0. Wthout loss of generalty, we assume that the products n each nest are sorted n a descendng order of revenues; that s, for each nest = 1, 2,..., m, r 1 r 2 r n 0. We denote the preference weght of the no-purchase opton by v 0 > 0. 1 Under the classcal two-level nested logt model McFadden, 1978; Börch-Supan, 1990), the dssmlarty parameter of each nest s between 0 and 1, ensurng that the model s consstent wth random utlty maxmzaton; see, Tallur and van Ryzn 2004) for an overvew of the random utlty theory. We wll also assume that the no-purchase opton appears as a sngle product n ts own nest. 2 Solvng a lnear program wth h constrants and k varables requres Ohk 3 ) operatons Potra and Wrght, 2000). 3 All of the analyss easly extends to the case where some nests have fewer than n products. 2

3 An assortment S = S 1,..., S m ) conssts of products n each nest, wth S {1, 2,..., n} for all. For any assortment S = S 1,..., S m ), the revenue RevS) s gven by: RevS) = m Q S)Rev S ), =1 where Q S) denotes the probablty that a customer chooses a product from nest, and Rev S ) denotes the expected revenue from assortment S n nest, and Q S) = l S v l ) τ v 0 + m ) k=1 l S v τk and Rev S ) = k lk The correspondng optmzaton problem s gven by: l S r l v l l S v l. Z = max S=S 1,...,S m) RevS) and we denote an optmal assortment by S = S 1,..., S m) and let Z = RevS ) denote the optmal revenue. If there are tes, we can choose the optmal assortment accordng to any pre-determned te-breakng rule Characterzaton of the Optmal Assortments Let N + = {{1}, {1, 2},..., {1, 2,..., n}} denote the collecton of revenue-ordered subsets. As shown n the followng lemma, at the optmal soluton, f the nest s nonempty, then the nest s revenue-ordered. The proof follows from Davs et al. 2011), and we omt the detals. Lemma 1 Optmal Nest Structure). For = 1,..., m, f S, then S N +. Although Lemma 1 shows that S { } N + for all, exhaustve search s not feasble because there are n + 1) m possble confguratons of revenue-ordered subsets among m nests. To develop an effcent algorthm, we wll establsh a necessary and suffcent condton for an optmal assortment. For any ntegers k 1 and k 2, let { {k1, k 1 + 1,..., k 2 } f k 1 k 2, [k 1, k 2 ] = otherwse. 4 For example, we can choose S to be the mnmal optmal assortment; that s, there s no other optmal assortment B = B 1,..., B m ) S such that B S for all. All of our analyss apples to other te-breakng rules. 3

4 Also, for each nest and 1 k 1 k 2 n, let G k 1, k 2 ) be defned by: G k 1, k 2 ) = Rev [k 1, k 2 ]) Rev [1, k 2 ]) l [k f 1 1,k 2 ] v ) + Rev [1, k 2 ]), l l [1,k 2 ] v l where f : [0, 1] [0, 1] s defned by: x 1 τ x f x [0, 1) f x) = 1 x τ f x = 1, and t s easy to verfy that f ) s a contnuous, strctly ncreasng functon, wth f 0) = 0. Also, we defne 0/0 = 0, and thus G 1, k 2 ) = Rev [1, k 2 ]) for all k 2. The followng lemma shows that we can use the ndex functon G, ) to determne f a subset can be removed from S to mprove the revenue. The proof s gven n Appendx A. Lemma 2 When s Removng Subsets Benefcal?). For any assortment S = S 1,..., S m ), a collecton of subsets {A S revenue f and only f : = 1, 2,..., m} can be removed from S to acheve a greater m =1 [ V S ) τ Rev S ) V S \ A ) τ Rev S \ A ) ] m =1 [ V S ) τ V S \ A ) τ ] < RevS), where for any set X, V X) denotes the sum of the preference weghts of the products n X. Moreover, f S = [1, p ] wth p 1, A = [k, p ] wth k p, and A l = for all l, then removng A from S leads to a greater revenue f and only f G k, p ) < RevS). The man result of ths secton s the followng theorem, whch establshes a necessary and suffcent condton for an optmal assortment. Ths theorem forms the bass for our greedy algorthm gven n Secton 3. Theorem 3 Optmalty Condton). Consder any assortment S = [1, p 1 ],..., [1, p m ]) such that S [1, p ] for all. Then, S s optmal f and only f for every nest such that p 1, mn G j, p ) RevS). j= 1,..., p Proof. Suppose that S = S 1,..., S m ) s an optmal soluton, where S = [1, p ] for all. If p 1, then S. Snce S s optmal, removng a subset [k, p ] from S cannot mprove the revenue, and thus, t follows from Lemma 2 that G k, p ) RevS) for all k p, whch s the desred result. To establsh suffcency, consder any assortment S = S 1,..., S m ) such that S S = [1, p ] for all, and S satsfes the condton of the theorem. We wll show that RevS) = Z. For 4

5 each, let E = S \ S = [1, p ] \ S = [k, p ] for some ndex k. By our hypothess, for every nest such that E, G k, p ) RevS), whch means that removng E from S does not ncrease the revenue. It then follows from the frst part of Lemma 2 that RevS) V S ) τ Rev S ) V S \ E ) τ Rev S \ E ) V S ) τ V S \ E ) τ, where for any set X, V X) s the sum of the preference weghts of the products n X. Thus, V S ) τ Rev S ) V S \ E ) τ Rev S \ E ) RevS) mn :E V S ) τ V S \ E ) τ :E [V S ) τ Rev S ) V S \ E ) τ Rev S \ E )] :E [V S ) τ V S \ E ) τ, ] where the second nequalty follows because for any x R k and y R k +, and V S ) τ V S \ E ) τ from the frst part of Lemma 2 that P k =1 x P k =1 y mn =1,...,k x 0 for all. Note that f E =, then S = S. It then follows y, RevS) RevS 1 \ E 1, S 2 \ E 2,..., S m \ E m ) = RevS 1,..., S m) = Z, whch s the desred result. Surprsngly, as shown n the followng theorem, the optmal assortment n each nest s lumpy, wth certan consecutve products always appearng together. Moreover, for each nest, by lookng at the ndex G j, j) of each product j, we can determne n advance whch products wll appear together! Theorem 4 Lumpness of the Optmal Assortments). For every nest, f there exst products j and k such that j < k and G j, j) < G j + 1, j + 1) < < G k, k), then ether {j, j + 1,..., k} S, or {j, j + 1,..., k} S =. Proof. It suffces to prove the result when k = j + 1; that s, we need to show that f G j, j) < G j + 1, j + 1), then ether {j, j + 1} S, or {j, j + 1} S =. There are two cases. Case 1: j + 1 S. In ths case, {j, j + 1} S because S N + by Lemma 1. Case 2: j + 1 / S. In ths case, we wll show that j / S. Suppose on contrary that j S. Then, we have that S = [1, j]. Let Ŝ = S 1,..., [1, j + 1],..., S m). By defnton, RevŜ) RevS ), and snce S s obtaned from Ŝ by removng product j + 1 from nest, t 5

6 follows from Lemma 2 that G j + 1, j + 1) RevŜ). So, we have that G j, j) < G j + 1, j + 1) RevŜ) RevS ), whch mples that Rev S 1,..., S \ {j},..., S m) > RevS ). Ths contradcts the optmalty of S. Therefore, j / S. 3. A Greedy Algorthm Our proposed greedy algorthm generates a sequence of assortments { S t : t = 0, 1,... }, termnatng wth an assortment that satsfes the optmalty condton n Theorem 3. In Stage 1 of the algorthm, we explot the lumpness structure of the optmal assortments shown n Theorem 4, by combnng products that wll always appear together nto a sngle group. A formal descrpton of the algorthm s gven as follows. Greedy Algorthm Stage 1 Lumpng): Compute the ndex G j, j) for every product j n every nest. For each nest, f G j, j) < G j + 1, j + 1), then t follows from Theorem 4 that ether {j, j + 1} S or {j, j + 1} S =. Thus, we can combne the two products together, and replace products j and j + 1 wth a sngle new product wth a revenue Rev [j, j + 1]) = v jr j + v j+1, r j+1, v j + v j+1, and a preference weght v j + v j+1,. Assgn the new product the ndex j, and calculate the new G j, j). Repeat ths process untl we obtan a lst of ndces G j, j) that s non-ncreasng n j. Wthout loss of generalty, assume that at the end of Stage 1, each nest has n products such that G 1, 1) G 2, 2) G n, n). Stage 2 Removal): Let S 1 = [1, n],..., [1, n]). For each teraton t 1, gven S t = [1, J t 1 ], [1, J t 2 ],..., [1, J t m] ), let Mn t = mn : J t 1 G J t, J t ) and Index t = arg mn : J t 1 G J t, J t ). If Mn t RevS t ), the algorthm termnates and outputs S t. Otherwse, f Mn t < RevS t ), then the algorthm generates a new assortment S t+1 = S1 t+1,..., Sm t+1 ) as follows: S t+1 = { S t f Index t S t \ {J t } = [1, J t 1] f = Indext Thus, n each teraton, we remove J t from St f G J t, J t ) = Mnt the optmalty condton gven n Theorem 3, that s, G J t, J t ) < RevSt ). and the product J t volates 6

7 The followng lemma shows that S t always contans the optmal assortment. Lemma 5 Contanment). For all t, S S t. Proof. We wll prove ths result by nducton. It s true for t = 1 by our constructon. Suppose that the lemma s true for t; that s, S S t. We wsh to show that S S t+1 for all. Consder an arbtrary nest. There are two cases to consder: S = St and S St. Case 1: Suppose that S = S t = [1, J t]. If J t = 0, then Index t and S t+1 = S t by our constructon. So, suppose that J t 1. Snce Rev S ) RevS1,..., S \ {J t},..., m) S, t follows from Lemma 2 that G J t, J t ) RevS ) RevS t ), where the last nequalty follows from the optmalty of S. Snce G J t, J t ) RevSt ), S t+1 = S t, whch s the desred result. Case 2: Suppose that S St = [1, J t ]. Then, J t / S and J t 1. Snce S N +, we have that S [1, J t 1]. By our constructon, we wll ether remove J t from S t or do nothng, and n both cases, we have S St+1, whch s the desred result. The man result of ths secton s stated n the followng theorem. Theorem 6 Correctness). The Greedy Algorthm termnates wth an optmal assortment, wth a runnng tme of On m log m). Proof. We wll frst establsh the runnng tme of the algorthm. By defnton, G j, j) = r j Rev [1, j]) ) + Rev [1, j]), f 1 v j P j l=1 v l and note that Rev [1, j]) s a convex combnaton of Rev [1, j 1]) and r j. For smplcty, assume that we can compute the functon f x) for each x n constant tme, and thus, by startng from j = 1, we can compute the ndex G j, j) for each product j n each nest n constant tme. Moreover, for each nest, t s easy to verfy that the lumpng process n Stage 1 requres computatons of On) ndces, and at most On) comparsons. Wth m nests, the total runnng tme for Stage 1 s thus Onm). For a detaled proof, see Appendx B. Durng the removal n Stage 2, the algorthm wll contnue to run as long as a sngle product s removed from any nest. There are nm products; thus, the algorthm wll termnate n Onm) teratons. We wll now show that each teraton takes Olog m) operatons. At the begnnng of Stage 2, we create a self-balancng bnary search tree SB-BST) wth m nodes, where for = 1, 2,..., m, node n the tree corresponds to the ndex G n, n). Ths takes Om log m) operatons; see Chapter 6 n Knuth 1998) for more detals. We use the SB- BST as our data structure because such a tree always mantans a heght of Olog m), allowng for an effcent search operaton. In each teraton t 1, we perform the followng 3 operatons: 7

8 SEARCH: The algorthm searches for the nest wth the mnmum ndex G J t, J t ). Ths can be done n Olog m) operatons under the SB-BST. DELETE: If the mnmum ndex G J t, J t ) s greater than or equal to the revenue of the current assortment, the algorthm termnates. Otherwse, the algorthm removes the product J t from St n nest, and also removes node from the tree. The removal s done n Olog m) operatons because the tree may need to re-balance ts heghts. INSERT: If the product J t n nest s removed and J t > 1, the algorthm adds a new node wth a correspondng ndex G J t 1, J t 1) nto the tree. Agan, the nserton of a new node n SB-BST takes Olog m) operatons. Therefore, each teraton n Stage 2 takes Olog m) operatons. As we have Onm) teratons, the total runnng tme s On m log m). Let S = [1, p 1 ],..., [1, p m ]) denote the assortment at the termnaton of the greedy algorthm. By our constructon, for every nest such that p 1, we have G p, p ) RevS). To complete the proof, we wll show that S satsfes the optmalty condton n Theorem 3; that s, for every nest such that p 1, mn G j, p ) RevS). j=1,2,...,p We wll prove ths by contradcton. Suppose on the contrary that the optmalty condton s volated, and mn j=1,2,...,p G j, p ) < RevS). Let k {1, 2,..., p 1} denote the largest ndex where the optmalty condton s volated; that s, and t follows from Lemma 2 that G k, p ) < RevS) G k + 1, p ), RevS 1,..., S \ [k, p ],..., S m ) > RevS) RevS 1,..., S \ [k + 1, p ],..., S m ) Note that S \ [k, p ] = [1, k 1] and S \ [k + 1, p ] = [1, k ]. Thus, removng product k from [1, k ] n nest ncreases the revenue, and t follows from Lemma 2 that G k, k ) < RevS 1,..., S \ [k + 1, p ],..., S m ) RevS). Snce we lump the products n Stage 1 so that G 1, 1) G p, p ) and k < p, we have G p, p ) G k, k ) < RevS), whch contradcts our hypothess on p! Therefore, mn j=1,2,...,p G j, p ) RevS). Numercal Results: We conduct small numercal experments to compare the runnng tme of our greedy algorthm wth the lnear programmng formulaton of Davs et al. 2011). 8

9 We consder 9 problem classes, where each problem class s characterzed by the number of nests m, and the number of products n n each nest, wth m, n) {10, 50, 100} {10, 50, 100}. For each problem class m, n), we generate 1000 ndependent problem nstances, and for each nstance, r j are sampled from a unform dstrbuton on the nterval [0, 20], v j are unformly dstrbuted on [0, 10], v 0 s unformly dstrbuted on [0, 5], and fnally, the dssmlarty parameter τ for each nest s drawn from a unform dstrbuton on [0, 1]. Table 1 shows the average runnng tme for each problem class. To solve the lnear program, we use the MATLAB LP Solver, and t turns out that both the nteror pont and smplex methods have roughly comparable runnng tmes, so we report only the runnng tme from the nteror pont method. For the greedy algorthm, to smplfy our mplementaton, we use a standard array of sze m to mplement Stage 2, nstead of the self-balancng bnary search tree 5. Even wth ths smple mplementaton, our greedy algorthm stll outperforms the LP n all problem nstances. In columns 3 and 4 of Table 1, we show the average runnng tme n seconds for each problem class. We observe that the greedy algorthm s, on average, faster than the LP by an order of magntude. We also want to compare the runnng tme on every problem nstance. So, for each problem nstance, we compute the rato between the runnng tme of LP and the runnng tme of greedy, and report the mnmum, medan, and maxmum of these ratos n columns 5, 6, and 7. Snce the mnmum s at least 2.2, we see that the greedy s always at least twce as fast as the LP, n every problem nstance. In fact, for large problem nstances, the greedy can be up to 99 tmes faster than the LP. Problem Avg. Runnng Tme Rato Between the LP and Greedy Class n seconds) Runnng Tmes for Each Problem Instance m n LP Greedy Mnmum Medan Maxmum Table 1: Columns 3 and 4 show the average runnng tme n seconds for the LP and greedy algorthms. All numbers are statstcally sgnfcant at 5%. We also consder the rato between the LP and greedy runnng tmes for each problem nstance, and the statstcs of these ratos are reported n columns 5, 6, and 7. 5 By usng the standard array of sze m, the runnng tme of the greedy algorthm ncreases to Onm 2 ) because n each teraton of Stage 2, searchng for the mnmum ndex requres Om) operatons, nstead of Olog m) under the self-balancng bnary search tree. 9

10 4. Concluson We present a smple and fast greedy algorthm for the assortment optmzaton problem under the classcal two-level nested logt model. Our algorthm makes use of the optmalty condton, and explot the lumpy structure of the optmal assortment. As a drecton for future research, t would be nterestng to explore whether the optmalty condton can be establshed for more general choce models, or f a smlar lumpy structure exsts for other choce models. Acknowledgement We would lke to thank Huseyn Topaloglu for ntroducng us to the surprsng connecton between the assortment optmzaton problem under the two-level nested logt model and ts equvalent lnear programmng formulaton. Hs work has nspred us to consder the problem n ths paper. References Börch-Supan, A On the compatblty of nested logt models wth utlty maxmzaton. Journal of Econometrcs 43 3): Davs, J. M., G. Gallego, and H. Topaloglu Assortment optmzaton under varants of the nested logt model. Workng paper, Cornell Unversty. Faras, V. F., S. Jagabathula, and D. Shah Assortment optmzaton under a nonparametrc choce model. Workng Paper, MIT. Gallego, G., G. Iyengar, R. Phllps, and A. Dubey Managng flexble products on a network. Workng Paper, Columba Unversty. Knuth, D. E The Art of Computer Programmng, Volume 3. New York: Addson Wesley Longman Publshng Co., Inc. Kök, A. G., M. Fsher, and R. Vadyanathan Assortment plannng: Revew of lterature and ndustry practce. In Retal Supply Chan Management. New York: Sprnger. Kunnumkal, S., and H. Topaloglu A refned determnstc lnear program for the network revenue management problem wth customer choce behavor. Naval Research Logstcs 55 6): McFadden, D Modelng the choce of resdental locaton. Transportaton Research Record 672): Potra, F. A., and S. J. Wrght Interor-pont methods. Journal of Computatonal and Appled Mathematcs 124 2): Tallur, K., and G. J. van Ryzn Revenue management under a general dscrete choce model of consumer behavor. Management Scence 50 1): Tallur, K., and G. J. van Ryzn The theory and practce of revenue management. New York: Sprnger. Tran, K Dscrete Choce Methods wth Smulaton. New York, NY: Cambrdge Unversty Press. Wllams, H. C. W. L On the formaton of travel demand models and economc evaluaton measures of user beneft. Envronment and Plannng. Envronment and Plannng A 9 3):

11 A. Proof of Lemma 2 Proof. For = 1,..., m, let Ŝ = S \ A. Assume that A for some ; otherwse, the result s trvally true. Snce v j > 0 for all j, m =1 V Ŝ) < m =1 V S ). By defnton, RevS) = = = m =1 V S ) τ Rev S ) v 0 + m =1 V S ) τ v 0 + m =1 V Ŝ) τ ) [ ] m m v 0 + m =1 V S ) τ =1 V Ŝ) τ Rev Ŝ) v 0 + =1 V S ) τ Rev S ) V Ŝ) τ Rev Ŝ) m =1 V Ŝ) + τ v 0 + m =1 V S ) τ v 0 + ) m =1 V Ŝ) τ v 0 + m =1 V S ) τ RevŜ) [ ] m =1 V S ) τ V Ŝ) τ [ ] m + v 0 + =1 V S ) τ Rev S ) V Ŝ) τ Rev Ŝ) m =1 V S ) τ [ ] m =1 V S ) τ V Ŝ ) τ Thus, RevS) s a convex combnaton of RevŜ) and P m =1[ V S ) τ Rev S ) V b S ) τ Rev b S ) ] P m =1[ V S ) τ V b S ) τ ], whch gves the desred result. Suppose that S = [1, p ], A = [k, p ], and A l = for all l. Usng the fact that ) Rev Ŝ) = V S ) V S b Rev S ) + 1 V S ) ) V S b Rev A ), and that ) f 1 V A ) ) ) V Ŝ) = f V S ) V S ) = ) V S b 1 τ ) V S ) V S b ) V S ) 1 V b S ) V S ) = V S ) τ V Ŝ) τ V Ŝ) τ 1 V S ) V Ŝ) τ, we have V S ) τ Rev S ) V Ŝ) τ Rev Ŝ) ) ) = V S ) τ V Ŝ) τ 1 V S ) Rev S ) + V Ŝ) τ 1 V S ) V Ŝ) τ Rev A ) ) = V S ) τ V Ŝ) τ + V Ŝ) τ V Ŝ) τ 1 V S ) Rev S ) + V Ŝ) τ 1 V S ) V Ŝ) τ = V S ) τ V Ŝ) ) [ ) τ V Ŝ) τ 1 V S ) V Ŝ) τ ] Rev A ) Rev S )) + Rev S ) V S ) τ V Ŝ ) τ ) [ ] = V S ) τ V Ŝ) τ Rev A ) Rev S ) ) + Rev S ) = V S ) τ V Ŝ) τ f 1 V A ) V S ) ) G k, p ). ) Rev A ) Therefore V S ) τ Rev S ) V Ŝ) τ Rev Ŝ) V S ) τ V Ŝ ) τ f and only f G k, p ) < RevS). < RevS) 11

12 B. Complexty of the Lumpng Step Stage 1) Proof. For each nest, t suffces to show that the lumpng process n Stage 1 takes On) operatons, n order to create a lst of ndces G j, j) that s non-ncreasng n j. The lumpng process requres two types of operatons: 1) comparson between two ndces, and 2) mergng of two products, whch requres the computaton of the new value of G, ). For smplcty, we assume that computng the ndex G, ) takes constant tme. Whenever we merge two nodes to create a new one, the total number of nodes s reduced by one. Snce we start wth n nodes, the number of mergngs s at most n 1. To complete the proof, we wll show that the number of comparsons s at most On), by creatng a drected graph that keeps track of the total number of comparsons. Create a node j for each product j {1, 2,..., n} n nest. Place these nodes from left to rght n an ncreasng order of ndex, so node 1 wll be the leftmost node, followed by node 2, and node n wll be the rghtmost node. We refer to these n nodes as the orgnal nodes. For each node j, let j l and j r represent the adjacent nodes, on the left and rght of node j, respectvely. Whenever we compare G j, j) wth G k, k), we add a drected edge j, k) from node j to node k. We say that a volaton occurs f G j, j) < G j r, j r ). We start from the left most node node 1). When we are at the orgnal node j, we compare G j, j) wth G j + 1, j + 1), and add an edge j, j + 1). If no volaton occurs, then G j, j) G j + 1, j + 1), and we proceed forward to node j + 1. On the other hand, f a volaton occurs, then we must merge products j and j + 1 together. Ths s done by removng both nodes j and j + 1, and replace them wth a new node j, and we calculate the new ndex G j, j ). Whenever we create a new node j, to ensure the correct orderng of ndces, we need to compare G j, j ) wth the value of ts left adjacent node, denoted by j l. So, when a new node j s created, we add a backward edge j, j l ). If a volaton occurs, we replace both nodes j and j l wth a new node. Otherwse, we have the correct orderng, and we compare node j wth ts rght neghbor j r, and of course, add an edge j, j r) to keep track of ths comparson. The process stops when all exstng nodes are connected and no volaton occurs. Each of the n orgnal nodes except for the rghtmost node) has at most one outgong edge to ts rght neghbor. Moreover, each newly created node has at most two outgong edges one to ts left neghbor and another to ts rght). Snce a new node s created only when a mergng occurs, there are at most n 1 new nodes. Therefore, the total number of edges s bounded by n 1) + 2n 1) = 3n 1), and thus, the total number of comparsons s On). 12