Technical Note: Capacity Constraints Across Nests in Assortment Optimization Under the Nested Logit Model

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1 Techncal Note: Capacty Constrants Across Nests n Assortment Optmzaton Under the Nested Logt Model Jacob B. Feldman, Huseyn Topaloglu School of Operatons Research and Informaton Engneerng, Cornell Unversty, Ithaca, New York 14853, USA jbf232@cornell.edu, topaloglu@ore.cornell.edu March 16, 2015 Abstract We consder assortment optmzaton problems when customers choose accordng to the nested logt model and there s a capacty constrant lmtng the total capacty consumpton of all products offered n all nests. When each product consumes one unt of capacty, our capacty constrant lmts the cardnalty of the offered assortment. For the cardnalty constraned case, we develop an effcent algorthm to compute the optmal assortment. When the capacty consumpton of each product s arbtrary, we gve an algorthm to obtan a 4-approxmate soluton. We show that we can compute an upper bound on the optmal expected revenue for an ndvdual problem nstance by solvng a lnear program. In our numercal experments, we consder problem nstances nvolvng products wth arbtrary capacty consumptons. Comparng the expected revenues from the assortments obtaned by our 4-approxmaton algorthm wth the upper bounds on the optmal expected revenues, our numercal results ndcate that the 4-approxmaton algorthm performs qute well, yeldng less than 2% optmalty gap on average. 1 Introducton A conventonal approach to modelng demand n revenue management s to assume that each customer arrves nto the system wth the ntenton of purchasng a fxed product. If ths product s avalable for sale, then the customer purchases t. Otherwse, the customer leaves the system wthout makng a purchase. In realty, however, there may be multple products that can potentally serve the needs of a customer, n whch case, customers may make a choce between the products and may substtute a product for another one when ther favorte product s not avalable. Ths knd of a choce process creates nteractons between the demand for the dfferent products, nflatng the demand for an avalable product when some other product s not avalable so that customers satsfy ther needs by substtutng for the avalable product. A common queston that arses n ths settng s what products to make avalable to customers so as to maxmze the expected revenue, gven that customers choose and substtute accordng to a partcular choce model. In ths paper, we consder assortment optmzaton problems when customers choose accordng to the nested logt model and there s lmted capacty for the products n the offered assortment. We consder a settng where we need to decde whch assortment of products to offer. Each arrvng customer chooses among the offered products accordng to the nested logt model. Under the nested logt model, the products are organzed n nests. Each customer, after vewng the offered assortment, decdes ether to make a purchase wthn one of the nests or to leave the system wthout purchasng anythng. If a nest s chosen, then the customer purchases one of the products wthn the chosen nest. There s a capacty constrant lmtng the total capacty consumpton of the 1

2 products n the offered assortment. The goal s to choose an assortment of products to offer so as to maxmze the expected revenue obtaned from each customer. We consder two types of capacty constrants. In the frst type of constrants, each product occupes one unt of space, n whch case, the capacty constrant lmts the total number of products n the offered assortment. We refer to ths type of a capacty constrant as a cardnalty constrant. In the second type of constrants, the capacty consumpton of a product s arbtrary, possbly reflectng the space or captal requrement of a product. We refer to ths type of a capacty constrant as a space constrant. Under a cardnalty constrant, we show that we can obtan an optmal assortment by solvng a lnear program wth O(m 2 n) decson varables and O(m 2 n 4 ) constrants, where m n the number of nests and n s the number of products n each nest. As far as we are aware, the assortment problem was not known to be tractable when customers choose accordng to the nested logt model and there s a cardnalty constrant lmtng the total number of products n the offered assortment. Ths paper gves the frst exact soluton method for ths problem. On the other hand, under a space constrant, we show that we can obtan a 4-approxmate soluton by solvng a lnear program wth O(m) decson varables and O(mn 4 ) constrants. The runnng tme of ths algorthm scales polynomally wth the number of products and the number of nests. To our knowledge, ths paper gves the frst algorthm for the assortment problem that scales polynomally wth the number of nests, when there s a capacty constrant on the space consumpton of all offered products and customers choose accordng to the nested logt model. In addton to gvng algorthms to solve the assortment problem, we gve a tractable lnear program that computes an upper bound on the optmal expected revenue. By comparng the expected revenues of the assortments obtaned by our 4-approxmaton algorthm wth the upper bounds on the optmal expected revenues, we demonstrate that our 4-approxmaton algorthm performs qute well n practce. An attractve approach for modelng the customer choce process s to use the utlty maxmzaton prncple, where a customer assocates a random utlty wth each product and chooses the product wth the largest utlty. Multnomal logt model s one of the popular choce models that are based on the utlty maxmzaton prncple where the utltes of the products are ndependent of each other; see Luce (1959) and McFadden (1974). Due to the ndependence of the utltes of the products, the multnomal logt model mplctly assumes that how hghly a customer evaluates a certan product has nothng to do wth how hghly the same customer evaluates another product. The nested logt model remedes ths shortcomng by organzng the products n nests such that the utltes of the products n the same nest can be dependent on each other; see Tran (2003). Ths feature allows the modeler to capture stuatons where the products n the same nest are alke and how hghly a customer evaluates a certan product can be a good ndcator of how hghly the same customer evaluates another product. Tallur and van Ryzn (2004) and Gallego et al. (2004) consder revenue management problems under customer choce behavor. Durng the course of ther analyses, they gve effcent approaches to solve the assortment problem under the multnomal logt model wthout any 2

3 constrants. Rusmevchentong et al. (2010), Wang (2012) and Wang (2013) consder assortment problems under varants of the multnomal logt model wth a cardnalty constrant on the offered assortment and show that the problem can be solved effcently. Bront et al. (2009), Mendez- Daz et al. (2010) and Rusmevchentong et al. (2013) consder assortment problems where there are multple customer types and customers of dfferent types choose accordng to multnomal logt models wth dfferent parameters. The authors show that the problem s NP-complete, study heurstcs and nvestgate vald cuts for nteger programmng formulatons. Davs et al. (2013) show how to solve the assortment problem under the nested logt model wthout any constrants. L and Rusmevchentong (2014) gve a greedy algorthm for the same problem. Gallego and Topaloglu (2014) study constraned assortment problems under the nested logt model, but they mpose capacty constrants separately on the assortment offered n each nest. Smlar to us, Rusmevchentong et al. (2009) and Desr and Goyal (2013) consder assortment problems under the nested logt model, where there s a constrant on the total capacty consumpton of the products offered n all nests. They gve approxmaton schemes that tradeoff runnng tme wth soluton qualty, but the runnng tme for ther approaches grows exponentally wth the number of nests. For example, to obtan a 4-approxmate soluton, Rusmevchentong et al. (2009) need O(m(m 6 n 6 log(mn)) m ) operatons, whch gets prohbtve when m exceeds two or three but there are practcal applcatons where the number of nests easly exceeds two or three; see Tran et al. (1987), Slade (2009) and Grgolon and Verboren (2013). As mentoned above, Gallego and Topaloglu (2014) study assortment problems under the nested logt model, but they mpose capacty constrants separately on the assortment offered n each nest. To understand ther constrant structure, consder a retaler that s nterested n fndng an assortment of cars to offer to ts customers. We model the demand for cars through a nested logt model, where each car category, such as compact, md-sze and sedan, corresponds to a dfferent nest. The products wthn a nest correspond to cars of dfferent models wthn the car category correspondng to the nest. The constrant structure n Gallego and Topaloglu (2014) assumes that there s a separate fxed amount of space reserved for the cars offered n each car category and the retaler s nterested n fndng a set of cars to offer such that the set of cars offered n each car category does not volate the space reserved for that car category. In contrast, we mpose a capacty constrant on the assortment offered over all nests. Thus, our constrant structure assumes that there s a fxed amount of space avalable for all car categores and the retaler s nterested n fndng a set of cars to offer such that the total amount of space consumed by all offered cars n all car categores does not volate the space avalablty. One lne of attack for assortment optmzaton under the nested logt model has been to dentfy a collecton of good canddate subsets to offer n each nest. Once these collectons are dentfed, t s possble to solve a separate lnear program to pck a subset to offer n each nest so that the combned subsets over all nests provde a good assortment. Ths s the strategy followed by Davs et al. (2013) and Gallego and Topaloglu (2014). We follow an approach smlar to thers n dentfyng 3

4 the collectons of canddate subsets n each nest, but due to the fact that our capacty constrant lmts the total capacty consumpton of the subsets of products offered n all nests, dfferent nests nteract wth each other, makng the assortment problem substantally more dffcult. The lnear programs used by Davs et al. (2013) and Gallego and Topaloglu (2014) become neffectve and we cannot buld on the earler work to fgure out how to pck a subset to offer n each nest so that the combned subsets over all nests provde the hghest possble expected revenue. We get around ths dffculty by usng the followng general methodology. The expected revenue functon under the nested logt model s a fracton. We convert the problem of fndng an assortment that maxmzes the expected revenue nto the problem of fndng the fxed pont of a functon. Computng the value of ths functon at any pont requres solvng an optmzaton problem, whch nvolves fndng a subset of products to offer n each nest so as to maxmze a nonlnear functon. Under a cardnalty constrant, we show that we can consder a small number of canddate subsets n each nest wthout ncurrng any loss. Furthermore, we can maxmze the nonlnear functon by usng dynamc programmng. Under a space constrant, we show that we can consder a small number of canddate subsets n each nest whle ncurrng a constant factor loss. Furthermore, we can maxmze the nonlnear functon wth a constant factor loss by solvng the lnear programmng relaxaton of a multple choce knapsack problem. The dscusson n the prevous paragraph ndcates that our approach s related to maxmzng a fracton. Meggdo (1979) shows how to buld on a tractable algorthm for a combnatoral optmzaton problem wth a lnear objectve functon to develop a tractable algorthm for the same combnatoral optmzaton problem wth a fractonal objectve functon, where the numerator and the denomnator of the fracton are lnear. Hashzume et al. (1987) extend ths result by showng how to buld on an approxmaton algorthm for the former problem to gve an approxmaton algorthm for the latter, but ths extenson requres the exstence of an approxmaton algorthm when the lnear objectve functon has negatve coeffcents. Correa et al. (2010) show how to drop ths requrement. Mttal and Schulz (2013) make generalzatons to sums of fractons. One of the mportant dstngushng features of our assortment problem s that the objectve functon s a fracton where the numerator and the denomnator nvolve nonlneartes. Thus, the general methods n the papers outlned n ths paragraph do not mmedately apply. 2 Problem Formulaton In ths secton, we formulate the assortment optmzaton problem that we want to solve. There are m nests ndexed by M = {1,..., m}. In each nest, there are n products that we can offer to customers and we ndex the products by N = {1,..., n}. Although we assume that each nest has the same number of products, ths assumpton s only for notatonal brevty and t s straghtforward to extend our results to the case where dfferent nests have dfferent numbers of products. Under the nested logt model, a customer decdes ether to make a purchase wthn one of the nests or to leave wthout purchasng anythng. If the customer decdes to make a purchase wthn one of the nests, 4

5 then the customer chooses one of the products offered n ths nest. We let v j be the preference weght assocated wth product j n nest. Gven that we offer the assortment S N of products n nest, we use V (S ) = j S v j to denote the total preference weght of the products n the offered assortment. Under the nested logt model, f we offer the assortment S n nest and a customer has already decded to make a purchase n ths nest, then ths customer chooses product j S n nest wth probablty v j /V (S ). We let r j be the revenue assocated wth product j n nest. In ths case, gven that we offer the assortment S n nest and a customer has already decded to make a purchase n ths nest, the expected revenue that we obtan from ths customer can be wrtten as R (S ) = j S v j V (S ) r j S j = v j r j. V (S ) Assocated wth each nest, there s a parameter γ [0, 1] capturng the degree of dssmlarty between the products n nest. The preference weght of the no purchase opton s v 0. Under the nested logt model, f we offer the assortment (S 1,..., S m ) over all nests wth S N for all M, then a customer chooses nest wth probablty Q (S 1,..., S m ) = V (S ) γ /(v 0 + l M V l(s l ) γ l), whch corresponds to the probablty that a customer s attracted to nest as a functon of the assortment (S 1,..., S m ) offered over all nests. So, f we offer the assortment (S 1,..., S m ) over all nests, then we obtan an expected revenue of Π(S 1,..., S m ) = M Q (S 1,..., S m ) R (S ) = M V (S ) γ R (S ) v 0 + M V (S ) γ from each customer. Our goal s to fnd an assortment of products so as to maxmze the expected revenue from each customer, subject to a capacty constrant on the offered assortment. We consder two types of capacty constrants. In the frst type of constrant, we lmt the total number of products offered over all nests to c. Thus, the set of feasble assortments can be wrtten as {(S 1,..., S m ) : M S c, S N M}. We refer to ths constrant as a cardnalty constrant. In the second type of constrant, we let w j be the space requrement of product j n nest and lmt the total space requrement of the products offered over all nests to c. In ths case, the set of feasble assortments s {(S 1,..., S m ) : M j S w j c, S N M}. We refer to ths constrant as a space constrant. For unformty, we use C (S ) to denote the capacty consumpton of the assortment S offered n nest. We have C (S ) = S under a cardnalty constrant and C (S ) = j S w j under a space constrant. In ths case, we can wrte the set of feasble assortments as {(S 1,..., S m ) : M C (S ) c, S N M} under capacty or space constrants. We want to fnd an assortment that maxmzes the expected revenue from each customer subject to a capacty constrant, yeldng the problem { } z = max Π(S 1,..., S m ), (1) (S 1,..., S m ) : M C (S ) c, S N M 5

6 where C (S ) may correspond to a cardnalty or space constrant. If C (S ) corresponds to a cardnalty constrant, then we can assume wthout loss of generalty that c s an nteger. In ths paper, we show that f we have a cardnalty constrant on the offered assortment, then we can obtan an optmal soluton to problem (1) by solvng a tractable lnear program. On the other hand, f we have a space constrant, then Lemma 2.1 n Rusmevchentong et al. (2009) shows that problem (1) s NP-hard even when there s a sngle nest wth a dssmlarty parameter of one. Therefore, obtanng an optmal soluton to problem (1) under a space constrant s lkely to be ntractable. In ths paper, we show that f we have a space constrant, then we can obtan a 4-approxmate soluton to problem (1) by solvng a tractable lnear program. 3 Fxed Pont Representaton In ths secton, we descrbe the connecton of problem (1) to the problem of computng the fxed pont of a functon. Ths connecton plays an mportant role throughout the paper and t becomes crtcal for constructng an effcent soluton approach for problem (1) under a cardnalty or space constrant. To connect problem (1) to the problem of computng the fxed pont of a functon, we defne the functon f( ) : R + R + as { } f(z) = max (S 1,..., S m) : M C (S ) c, S N M M V (S ) γ (R (S ) z). (2) Offerng the empty assortment over all nests s a feasble soluton to the problem above provdng the objectve value of zero, so that f(z) 0 for all z R +. Consder a value of ẑ that satsfes f(ẑ) = v 0 ẑ, correspondng to the fxed pont of the functon f( )/v 0. Such a value of ẑ 0 always exsts snce f( ) s a decreasng functon and f(0) 0. The next theorem shows that the value of ẑ that satsfes f(ẑ) = v 0 ẑ s useful n dentfyng an optmal soluton to problem (1). Theorem 1 Let ẑ be such that f(ẑ) = v 0 ẑ. If the assortment (Ŝ1,..., Ŝm) satsfes V (Ŝ) γ (R (Ŝ) ẑ) f(ẑ), M then we have Π(Ŝ1,..., Ŝm) z, where z s the optmal objectve value of problem (1). Proof. We clam that z = ẑ. Frst, we show that z ẑ. We let ( S 1,..., S m ) be an optmal soluton to problem (2) when we solve ths problem wth z = ẑ. Thus, we have v 0 ẑ = f(ẑ) = M V ( S ) γ (R ( S ) ẑ). Focusng on the frst and last expressons n ths chan of equaltes and solvng for ẑ yelds ẑ = M V ( S ) γ R ( S )/(v 0 + M V ( S ) γ ) = Π( S 1,..., S m ). Notng that ( S 1,..., S m ) s a feasble soluton to problem (1), we have Π( S 1,..., S m ) z. Usng ths nequalty wth the last chan of equaltes, we obtan z ẑ. Second, we show that z ẑ. Usng (S 1,..., S m) to denote an optmal soluton to problem (1), we have z = Π(S 1,..., S m) = 6

7 M V (S )γ R (S )/(v 0 + M V (S )γ ). Focusng on the frst and last expressons n ths chan of equaltes and solvng for z, we obtan v 0 z = M V (S )γ (R (S ) z ). In ths case, we obtan v 0 ẑ = f(ẑ) M V (S )γ (R (S ) ẑ) M V (S )γ (R (S ) z ) = v 0 z, where the frst nequalty follows by the fact that (S1,..., S m) s a feasble soluton to problem (2) when we solve ths problem wth z = ẑ and the second nequalty uses the fact that z ẑ, whch s shown above. The last chan of nequaltes ndcate that z ẑ, establshng the clam. Snce f(ẑ) = v 0 ẑ, we wrte the nequalty n the theorem as M V (Ŝ) γ (R (Ŝ) ẑ) v 0 ẑ. Solvng for ẑ n ths nequalty yelds ẑ M V (Ŝ) γ R (Ŝ)/(v 0 + M V (Ŝ) γ ) = Π(Ŝ1,..., Ŝm), n whch case, the desred result follows by notng that ẑ = z. Theorem 1 suggests the followng procedure to obtan an optmal soluton to problem (1). We fnd ẑ such that f(ẑ) = v 0 ẑ and solve problem (2) wth z = ẑ to obtan an optmal soluton (Ŝ1,..., Ŝm). In ths case, t s possble to show that (Ŝ1,..., Ŝm) s an optmal soluton to problem (1). To see ths result, we have f(ẑ) = M V (Ŝ) γ (R (Ŝ) ẑ) by the defnton of (Ŝ1,..., Ŝm), n whch case, (Ŝ1,..., Ŝm) satsfes the nequalty n Theorem 1 and we obtan Π(Ŝ1,..., Ŝm) z. Snce the soluton (Ŝ1,..., Ŝm) s feasble to problem (2), t s feasble to problem (1) as well and we have Π(Ŝ1,..., Ŝm) z. Therefore, we have Π(Ŝ1,..., Ŝm) = z, establshng that Π(Ŝ1,..., Ŝm) s an optmal soluton to problem (1), as desred. Later n the paper, we show that we can effcently fnd ẑ that satsfes f(ẑ) = v 0 ẑ when we have a cardnalty constrant. However, fndng such ẑ may be dffcult when we have a space constrant. The next corollary gves an approxmate verson of Theorem 1 that does not requre fndng ẑ such that f(ẑ) = v 0 ẑ. To state ths corollary, we let f R ( ) be an approxmaton to f( ) that satsfes α f R (z) f(z) for all z R + for some α 1. We do not yet specfy how to construct ths approxmaton. We only assume that f R ( ) s a decreasng functon smlar to f( ) and f R (0) 0, n whch case, we can always fnd ẑ 0 satsfyng f R (ẑ) = v 0 ẑ. The next corollary shows that we can use ths value of ẑ to get an approxmaton guarantee for problem (1). Its proof s smlar to that of Theorem 1 and deferred to Onlne Appendx A. Corollary 2 Let f R ( ) be an approxmaton to f( ) that satsfes α f R (z) f(z) for all z R + for some α 1 and ẑ be such that f R (ẑ) = v 0 ẑ. If the assortment (Ŝ1,..., Ŝm) satsfes β M V (Ŝ) γ (R (Ŝ) ẑ) f R (ẑ) for some β 1, then we have α β Π(Ŝ1,..., Ŝm) z, where z s the optmal objectve value of problem (1). 4 Cardnalty Constrant In ths secton, we consder problem (1) under a cardnalty constrant. Thus, we have C (S ) = S throughout ths secton. Frst, we show how to solve problem (2), whch allows us to compute f(z) at a partcular value of z. Second, we show how to fnd a value of ẑ that satsfes f(ẑ) = v 0 ẑ. In 7

8 ths case, notng the dscusson that follows Theorem 1, we can fnd an optmal soluton to problem (1) by fndng a value of ẑ that satsfes f(ẑ) = v 0 ẑ and solvng problem (2) wth z = ẑ. 4.1 Computaton at a Partcular Pont We consder solvng problem (2), whch allows us to compute f(z) at a partcular value of z. The startng pont for our dscusson s a result due to Gallego and Topaloglu (2014), who study the problem of maxmzng the expected revenue obtaned from each customer subject to a separate cardnalty constrant on the assortment offered n each nest. In partcular, for any z R + and b Z +, Gallego and Topaloglu (2014) focus on the problem { } max V (S ) γ (R (S ) z) S : C (S ) b, S N = max S : S b, S N { } V (S ) γ (R (S ) z). (3) The authors construct O(n 2 ) dfferent orderngs of the products such that for any z R + and b Z +, an optmal soluton to problem (3) can be obtaned by sortng the products accordng to one of these orderngs and usng an assortment that ncludes some number of earlest products n ths orderng. In other words, lettng {σ g : g G } wth G = O(n 2 ) be the orderngs constructed by Gallego and Topaloglu (2014) and S (σ g, k) be the assortment that ncludes the frst k products when the products n nest are sorted accordng to the orderng σ g, for any z R + and b Z +, an optmal soluton to problem (3) can always be found n the collecton of assortments {S (σ g, k) : g G, k = 0,..., n}. Snce G = O(n 2 ), there are O(n 3 ) assortments n ths collecton. For notatonal brevty, we use A = {S t : t T } wth T = O(n 3 ) to denote the collecton of assortments {S (σ g, k) : g G, k = 0,..., n} and the next lemma follows. Lemma 3 There exsts a collecton of assortments A = {S t : t T } wth T = O(n 3 ) such that for any z R + and b Z +, an optmal soluton to problem (3) can be found n A. Lemma 3 allows us to focus only on the assortments n the collectons A 1,..., A m n problem (2). In partcular, we can wrte problem (2) equvalently as { } f(z) = max V (S ) γ (R (S ) z). (4) (S 1,..., S m ) : M C M (S ) c, S A M To see that problems (2) and (4) have the same optmal objectve values, we note that problem (2) allows usng assortments of the form (S 1,..., S m ) wth S N for all, whereas problem (4) allows usng assortments of the form (S 1,..., S m ) wth S A for all M. Therefore, the optmal objectve value of problem (2) s at least as large as the optmal objectve value of problem (4). On the other hand, lettng (Ŝ1,..., Ŝm) be an optmal soluton to problem (2) and C (Ŝ) = ˆb, snce (Ŝ1,..., Ŝm) s a feasble soluton to problem (2), we have M ˆb = M C (Ŝ) c. By 8

9 Lemma 3, the collecton of assortments A ncludes an optmal soluton to problem (3) for any z R + and b Z +. Usng ths result wth b = ˆb and notng that Ŝ s a feasble soluton to problem (3) when we solve ths problem wth b = ˆb, t follows that there exsts S A such that V ( S ) γ (R ( S ) z) V (Ŝ) γ (R (Ŝ) z) and C ( S ) ˆb. Addng the last two nequaltes over all M, we have M V ( S ) γ (R ( S ) z) M V (Ŝ) γ (R (Ŝ) z) and M C ( S ) M ˆb c. Snce S A for all M, the last two chans of nequaltes show that ( S 1,..., S m ) s a feasble soluton to problem (4) and the objectve value provded by ths soluton for problem (4) s at least as large as the optmal objectve value of problem (2). Therefore, the optmal objectve value of problem (4) s at least as large as the optmal objectve value of problem (2), establshng that problems (2) and (4) have the same optmal objectve values. The dscusson above ndcates that we can compute f(z) at a partcular value of z by solvng problem (4), nstead of problem (2). However, solvng problem (4) n a brute force fashon s stll dffcult snce there are A 1... A m dfferent combnatons of assortments that we can choose from dfferent nests and the number of such possble combnatons grows exponentally fast wth the number of nests. To solve problem (4) n a tractable fashon, the crtcal observaton s that the objectve functon of ths problem s separable by the nests. If we offer the assortment S n nest, then we obtan a contrbuton of V (S ) γ (R (S ) z). Problem (4) fnds one assortment to offer n each nest to maxmze the total contrbuton subject to the constrant that the total cardnalty of the assortments offered over all nests does not exceed c. Therefore, we can solve problem (4) by usng a dynamc program. In ths dynamc program, the decson epochs correspond to the nests. The state varable n each decson epoch s the remanng capacty left from the earler nests just before choosng the assortment offered n a partcular nest. Fnally, the acton varable n each decson epoch s the assortment offered n a partcular nest. partcular value of z by solvng the dynamc program { J (b z) = max S : C (S ) b S A Thus, we can compute f(z) at a V (S ) γ (R (S ) z) + J +1 (b C (S ) z) }, (5) wth the boundary condton J m+1 ( z) = 0. Under a cardnalty constrant, we can assume that c s an nteger that does not exceed mn, whch s the total number of products n all of the nests. Thus, the state space n the dynamc program above s 0,..., mn. Computng the value functons {J (b z) : b = 0,..., mn, M}, the value of J 1 (c z) corresponds to f(z). The dynamc program n (5) provdes an effcent approach for computng f(z) at a partcular value of z. Snce there are m decson epochs, the state space s 0,..., mn and A = O(n 3 ), ths dynamc program can be solved n O(m 2 n 4 ) operatons. In the next secton, we buld on the dynamc program to fnd ẑ that satsfes f(ẑ) = v 0 ẑ. 9

10 4.2 Fndng the Fxed Pont We consder the problem of fndng ẑ that satsfes f(ẑ) = v 0 ẑ. For ths purpose we use the lnear programmng representaton of the dynamc program n (5). A dynamc program wth fnte states and actons has a lnear programmng representaton. In ths lnear program, there s one decson varable for each state and decson epoch correspondng to the value functon at each state and decson epoch. Inspred by ths lnear program, we propose solvng mn Θ 1 (c) (6) st Θ (b) V (S ) γ (R (S ) z) + Θ +1 (b C (S )) M, b = 0,..., mn, S F (b) (7) Θ 1 (c) = v 0 z, (8) to fnd ẑ satsfyng f(ẑ) = v 0 ẑ. The decson varables are Θ = {Θ (b) : M, b = 0,..., mn} and z n the lnear program above. We use the conventon that Θ m+1 (b) = 0 for all b = 0,..., mn. The set F (b) s gven by F (b) = {S : C (S ) b, S A }, capturng the set of feasble actons at decson epoch and state b. If we drop the second constrant n problem (6)-(8) and mnmze the objectve functon subject to the frst set of constrants for a fxed value of z, then t s well known that the optmal value of the decson varable Θ 1 (c) gves the value functon J 1 (b z) computed through the dynamc program n (5); see Puterman (1994). Interestngly, f we solve problem (6)-(8) as formulated, then the optmal value of the decson varable z gves the value of ẑ satsfyng f(ẑ) = v 0 ẑ. The next theorem shows ths result. Theorem 4 Lettng ( ˆΘ, ẑ) be an optmal soluton to problem (6)-(8), ẑ satsfes f(ẑ) = v 0 ẑ. Proof. We let (Ŝ1,..., Ŝm) be an optmal soluton to problem (2) when we solve ths problem wth z = ẑ. We defne {ˆb : M} as ˆb 1 = c and ˆb +1 = ˆb C (Ŝ) so that ˆb corresponds to the total capacty consumpton of the assortment (Ŝ1,..., Ŝm) n nests 1,..., 1. Snce ( ˆΘ, ẑ) s a feasble soluton to problem (6)-(8), t satsfes the frst set of constrants for state and acton (ˆb, Ŝ) for all M. So, we have ˆΘ (ˆb ) V (Ŝ) γ (R (Ŝ) ẑ) + ˆΘ +1 (ˆb C (Ŝ)) for all M. Notng that ˆb+1 = ˆb C (Ŝ), addng these nequaltes gves v 0 ẑ = ˆΘ 1 (c) M V (Ŝ) γ (R (Ŝ) ẑ) = f(ẑ), where the frst equalty uses the fact that ( ˆΘ, ẑ) satsfes the second constrant n problem (6)-(8) and the second equalty s by the defnton of (Ŝ1,..., Ŝm). So, we have v 0 ẑ f(ẑ). To get a contradcton, assume that v 0 ẑ > f(ẑ) n the rest of the proof and let z be such that f( z) = v 0 z. Compute the value functons J( z) = {J (b z) : M, b = 0,..., mn} through the dynamc program n (5) wth z = z. Notng the way the value functons are computed n (5), we have J (b z) V (S ) γ (R (S ) z) + J +1 (b C (S ) z) for all M, b = 0,..., mn and S A such that C (S ) b, whch ndcates that (J( z), z) satsfes the frst set of constrants n problem (6)-(8). Furthermore, we know that J 1 (c z) provdes the optmal objectve value of problem (4) when ths problem s solved wth z = z, so that J 1 (c z) = f( z) = v 0 z. Thus, the soluton (J( z), z) satsfes the second constrant n problem (6)-(8) as well. The optmal objectve value ˆΘ 1 (c) of 10

11 problem (6)-(8) must be no larger than the objectve value J 1 (c z) at the feasble soluton (J( z), z), mplyng v 0 ẑ = ˆΘ 1 (c) J 1 (c z) = v 0 z. So, we obtan f(ẑ) < v 0 ẑ v 0 z = f( z), but snce f( ) s decreasng, we cannot have v 0 ẑ v 0 z and f(ẑ) < f( z), yeldng a contradcton. To sum up, we can solve the lnear program n (6)-(8) to obtan ẑ satsfyng f(ẑ) = v 0 ẑ. Snce F (b) A = O(n 3 ), there are O(m 2 n) decson varables and M O(mn A ) = O(m 2 n 4 ) constrants n ths lnear program. Once we have ẑ, notng the dscusson that follows Theorem 1, we can solve problem (4) wth z = ẑ to obtan an optmal soluton to problem (1). To solve problem (4) wth z = ẑ, we can use the dynamc program n (5). The dynamc program n (5) can be solved n O(m 2 n 4 ) operatons and the computatonal effort for solvng ths dynamc program s neglgble when compared wth that for solvng the lnear program n (6)-(8). 5 Space Constrant In ths secton, we consder problem (1) under a space constrant. Thus, we have C (S ) = j S w j throughout ths secton. Frst, we show how to construct an approxmaton f R ( ) to f( ) such that 2 f R (z) f(z) for all z R +. Second, we show how to fnd ẑ satsfyng f R (ẑ) = v 0 ẑ. Thrd, we show how to fnd an assortment (Ŝ1,..., Ŝm) such that 2 M V (Ŝ) γ (R (Ŝ) ẑ) f R (ẑ) and M C (Ŝ) c. In ths case, we obtan 4 Π(Ŝ1,..., Ŝm) z by Corollary 2 and (Ŝ1,..., Ŝm) s a feasble soluton to problem (1). Therefore, t follows that (Ŝ1,..., Ŝm) s a 4-approxmate soluton to problem (1) under a space constrant. 5.1 Approxmaton at a Partcular Pont We consder constructng an approxmaton f R ( ) to f( ) that satsfes 2 f R (z) f(z) for all z R +. Smlar to our development under the cardnalty constrant, the constructon of our approxmaton to f( ) bulds on a result that s due to Gallego and Topaloglu (2014). In addton to a separate cardnalty constrant on the assortment offered n each nest, the authors study the problem of maxmzng the expected revenue obtaned from each customer subject to a separate space constrant on the assortment offered n each nest. Wthn ths settng, for any z R + and b R +, Gallego and Topaloglu (2014) focus on the problem { } max V (S ) γ (R (S ) z) S : C (S ) b, S N = max S : j S w j b, S N { } V (S ) γ (R (S ) z). (9) The authors construct O(n 2 ) dfferent orderngs between the products such that for any z R + and b R +, a 2-approxmate soluton to problem (9) can be obtaned by sortng the products accordng to one of these orderngs, droppng the products whose space consumpton exceeds b from consderaton and usng an assortment that ncludes some number of earlest products n ths orderng. In other words, we use {σ g : g G } wth G = O(n 2 ) to denote the orderngs constructed by Gallego and Topaloglu (2014) and S (σ g, k, b ) to denote the assortment that ncludes the frst k products when the products n nest are sorted accordng to the orderng σ g and the products whose 11

12 space consumpton exceeds b are dropped from consderaton. In ths case, Gallego and Topaloglu (2014) show that for any z R + and b R +, a 2-approxmate soluton to problem (9) can always be found n the collecton of assortments {S (σ g, k, b ) : σ g G, k = 0,..., n, b R + }. A crtcal observaton s that we can consder only b {w 1,..., w n }, rather than b R +, wthout changng the collecton of assortments {S (σ g, k, b ) : σ g G, k = 0,..., n, b R + }, snce f b takes a value other than {w 1,..., w n }, then we can decrease the value of b to the closest element n {w 1,..., w n } wthout changng the set of products whose space consumptons exceed b. Therefore, notng that G = O(n 2 ) and we can consder only b {w 1,..., w n }, there are O(n 4 ) assortments n the collecton {S (σ g, k, b ) : σ g G, k = 0,..., n, b R + }. For notatonal brevty, we use A = {S t : t T } wth T = O(n 4 ) to denote the collecton of assortments {S (σ g, k, b ) : g G, k = 0,..., n, b R + } and obtan the next lemma. Ths lemma becomes useful when constructng our approxmaton f R ( ) to f( ). Lemma 5 There exsts a collecton of assortments A = {S t : t T } wth T = O(n 4 ) such that for any z R + and b R +, a 2-approxmate soluton to problem (9) can be found n A. We construct our approxmaton to f( ) by focusng only on the assortments n the collectons A 1,..., A m. Usng the decson varables x = {x (S ) : M, S A }, we defne f R ( ) as f R (z) = max V (S ) γ (R (S ) z) x (S ) (10) M S A st C (S ) x (S ) c (11) M S A x (S ) = 1 M (12) S A x (S ) 0 M, S A, (13) whch corresponds to the optmal objectve value of a lnear program wth M O( A ) = O(mn 4 ) decson varables and O(m) constrants. In problem (2), each assortment S offered n nest provdes a contrbuton of V (S ) γ (R (S ) z). Ths problem fnds one assortment S N to offer n each nest such that the total contrbuton over all nests s maxmzed and the total capacty consumpton over all nests does not exceed c. Smlarly, each assortment S offered n nest provdes a contrbuton of V (S ) γ (R (S ) z) n problem (10)-(13). If we mpose ntegralty constrants on the decson varables n problem (10)-(13), then notng the second set of constrants, ths problem fnds one assortment S A to offer n each nest such that the total contrbuton over all nests n maxmzed and the total capacty consumpton over all nests does not exceed c. We note that f R (z) s decreasng n z. Also, we assume that C (S ) c for all M and S A. If C (S ) > c for some M and S A, then we can drop ths assortment from A snce usng ths assortment n problem (1) would yeld an nfeasble soluton. It s possble to use Lemma 5 to show that our approxmaton f R ( ) to f( ) satsfes 2 f R (z) f(z) for all z R +. To see ths result, we let (Ŝ1,..., Ŝm) be an optmal soluton to problem 12

13 (2) and ˆb = C (Ŝ). Snce (Ŝ1,..., Ŝm) s a feasble soluton to problem (2), we have M ˆb = M C (Ŝ) c. Lemma 5 ndcates that the collecton of assortments A ncludes a 2-approxmate soluton to problem (9) for any z R + and b R +. Usng ths result wth b = ˆb and notng the fact that Ŝ s a feasble soluton to problem (9) when ths problem s solved wth b = ˆb, t follows that there exsts S A such that 2 V ( S ) γ (R ( S ) z) V (Ŝ) γ (R (Ŝ) z) and C ( S ) ˆb. Addng the last two nequaltes over all M, we have 2 M V ( S ) γ (R ( S ) z) M V (Ŝ) γ (R (Ŝ) z) and M C ( S ) M ˆb c. To obtan the desred result, we defne the soluton x to problem (10)-(13) as x ( S ) = 1 for all M and x (S ) = 0 for all M and S A \ { S }. The soluton x s feasble to problem (10)-(13) snce the defnton of x mples that M S A C (S ) x (S ) = M C ( S ) c and S A x (S ) = x ( S ) = 1. Furthermore, the objectve value provded by the soluton x for problem (10)-(13) satsfes M S A V (S ) γ (R (S ) z) x (S ) = M V ( S ) γ (R ( S ) z) M V (Ŝ) γ (R (Ŝ) z)/2 = f(z)/2, where the frst equalty follows from the defnton of x, the nequalty uses the fact that 2 M V ( S ) γ (R ( S ) z) M V (Ŝ) γ (R (Ŝ) z) shown n the prevous paragraph and the second equalty s by the fact that (Ŝ1,..., Ŝm) s an optmal soluton to problem (2). Thus, there exsts a feasble soluton to problem (10)-(13) provdng an objectve value for ths problem that s at least f(z)/2, whch mples that the optmal objectve value f R (z) of problem (10)-(13) satsfes f R (z) f(z)/2, establshng the desred result. 5.2 Fndng the Fxed Pont We consder the problem of fndng the value of ẑ that satsfes f R (ẑ) = v 0 ẑ. Notng that f R (z) s gven by the optmal objectve value of the lnear program n (10)-(13), we use the dual of ths problem to fnd the value of ẑ that satsfes f R (ẑ) = v 0 ẑ. In partcular, assocatng the dual varables and y = {y : M} respectvely wth the two sets of constrants n problem (10)-(13), we propose solvng the lnear program mn c + y (14) M st C (S ) + y V (S ) γ (R (S ) z) M, S A (15) c + y = v 0 z (16) M 0, y s free, z s free M (17) to fnd ẑ satsfyng f R (ẑ) = v 0 ẑ. The decson varables are, y and z n the problem above. If we drop the second constrant n problem (14)-(17) and mnmze the objectve functon subject to the frst set of constrants for a fxed value of z, then ths problem corresponds to the dual of problem (10)-(13). Wth the second constrant added, problem (14)-(17) allows us to fnd ẑ that satsfes f R (ẑ) = v 0 ẑ, as shown n the next theorem. Theorem 6 Lettng ( ˆ, ŷ, ẑ) be an optmal soluton to problem (14)-(17), ẑ satsfes f R (ẑ) = v 0 ẑ. 13

14 Proof. Assocatng the dual varables and y = {y : M} wth the two sets of constrants n problem (10)-(13), the dual of ths problem s f R (z) = mn c + M y (18) st C (S ) + y V (S ) γ (R (S ) z) M, S A (19) 0, y s free M. (20) Therefore, the soluton ( ˆ, ŷ) s feasble to problem (18)-(20) when we solve ths problem wth z = ẑ, whch mples that f R (ẑ) c ˆ + M ŷ = v 0 ẑ, where the equalty follows from the fact that ( ˆ, ŷ, ẑ) s a feasble soluton to problem (14)-(17). To get a contradcton, assume that the last nequalty s strct so that f R (ẑ) < v 0 ẑ. We let z be such that f R ( z) = v 0 z and (, ỹ) be an optmal soluton to problem (18)-(20) when we solve ths problem wth z = z. Thus, we get v 0 z = f R ( z) = c + M ỹ, whch ndcates that (, ỹ, z) s a feasble soluton to problem (14)-(17). In ths case, t follows that v 0 z = f R ( z) = c + M ỹ c ˆ + M ŷ = v 0 ẑ > f R (ẑ), where the frst nequalty s by the fact that (, ỹ, z) s a feasble, but not necessarly an optmal soluton to problem (14)-(17). The last chan of nequaltes yelds v 0 z v 0 ẑ and f R ( z) > f R (ẑ), whch contradct the fact that f R ( ) s a decreasng functon. 5.3 Constructon of an Approxmate Assortment By the earler dscusson n ths secton, our approxmaton f R ( ) to f( ) satsfes 2 f R (z) f(z) for all z R +. Furthermore, we can fnd the value of ẑ that satsfes f R (ẑ) = v 0 ẑ by solvng the lnear program n (14)-(17). In the remander of ths secton, we consder the problem of fndng an assortment (Ŝ1,..., Ŝm) such that 2 M V (Ŝ) γ (R (Ŝ) ẑ) f R (ẑ) and M C (Ŝ) c. In ths case, Corollary 2 mples that we have 4 Π(Ŝ1,..., Ŝm) z and (Ŝ1,..., Ŝm) s a feasble soluton to problem (1). So, (Ŝ1,..., Ŝm) s a 4-approxmate soluton to problem (1). It s a smple exercse n lnear programmng dualty to show that any basc optmal soluton to problem (10)-(13) ncludes at most two fractonal components; see Snha and Zoltners (1979). We let ˆx be a basc optmal soluton to problem (10)-(13) when we solve ths problem wth z = ẑ. We make two observatons. Frst, f ˆx (P ) (0, 1] for some nest M and assortment P A, then notng the second set of constrants n problem (10)-(13), there must be some other assortment Q A such that ˆx (Q ) [0, 1) as well. Second, snce ˆx has as most two fractonal components, there can be no other fractonal component of ˆx. In ths case, notng the second set of constrants n problem (10)-(13) once more, t follows that for each nest M \ { }, there exsts a sngle assortment S such that ˆx ( S ) = 1. Therefore {ˆx ( S ) : M \ { }} {ˆx (P )} {ˆx (Q )} ncludes all components of ˆx takng strctly postve values. In the rest of the dscusson, we assume that the basc optmal soluton ˆx has two fractonal components. In partcular, notng the second set of constrants n problem (10)-(13), ˆx cannot have one fractonal component and the result holds n a straghtforward fashon when ˆx has no fractonal 14

15 components. As descrbed n the prevous paragraph, f the basc optmal soluton ˆx to problem (10)-(13) has two fractonal components, then there exst some nest M and assortments P, Q A such that ˆx (P ), ˆx (Q ) (0, 1) and there s no other fractonal component of ˆx. Wthout loss of generalty, we assume that C (P ) C (Q ). Furthermore, for each nest M \ { }, there exsts a sngle assortment S such that ˆx ( S ) = 1. we construct two assortments (Ŝ1 1,..., Ŝ1 m) and (Ŝ2 1,..., Ŝ2 m) as follows. Usng the soluton ˆx, The frst one of these assortments s constructed as (Ŝ1 1,..., Ŝ1 m) = ( S 1,... S 1, P, S +1,..., S m ). In other words, the assortment (Ŝ1 1,..., Ŝ1 m) uses the components of the soluton ˆx that take value one, along wth the fractonal component of the soluton ˆx wth the smaller capacty consumpton. We construct the second assortment as (Ŝ2 1,..., Ŝ2 m) = (,...,, Q,,..., ), offerng the subset Q n nest, but offerng empty subsets n all of the other nests. A crucal observaton s that the two assortments (Ŝ1 1,..., Ŝ1 m) and (Ŝ2 1,..., Ŝ2 m) as defned above collectvely nclude all components of the soluton ˆx = {ˆx (S ) : M, S A } takng a strctly postve value. In ths case, we get f R (ẑ) = V (S ) γ (R (S ) ẑ) ˆx (S ) V (Ŝ1 ) γ (R (Ŝ1 ) ẑ) + V (Ŝ2 ) γ (R (Ŝ2 ) ẑ) M S A M M { } 2 max V (Ŝ2 ) γ (R (Ŝ2 ) ẑ), M V (Ŝ1 ) γ (R (Ŝ1 ) ẑ), M where the frst nequalty s by the fact that f ˆx (S ) > 0 for some M and S A, then we have S = Ŝ1 or Ŝ = Ŝ2. Therefore, the chan of nequaltes above shows that f we choose (Ŝ1,..., Ŝm) as one of the assortments (Ŝ1 1,..., Ŝ1 m) and (Ŝ2 1,..., Ŝ2 m), then (Ŝ1,..., Ŝm) satsfes 2 M V (Ŝ) γ (R (Ŝ) ẑ) f R (ẑ). Furthermore, we note that both of the solutons (Ŝ1 1,..., Ŝ1 m) and (Ŝ2 1,..., Ŝ2 m) are feasble to problem (1). The soluton (Ŝ2 1,..., Ŝ2 m) s feasble snce ths soluton only offers Q n nest and we have M C(Ŝ2 ) = C (Q ) c, where the last nequalty uses the assumpton that C (S ) c for all M and S A. To see the feasblty of the soluton (Ŝ1 1,..., Ŝ1 m) to problem (1), we observe that M C(Ŝ1 ) = M\{ } C ( S )+C (P ) M\{ } C ( S ) + C (P ) ˆx (P ) + C (Q ) ˆx (Q ) = M S A C (S ) ˆx (S ) c, where the frst nequalty uses the fact that C (P ) C (Q ) and ˆx (P ) + ˆx (Q ) = 1 by the second set of constrants n problem (10)-(13) and the second equalty uses the fact that {ˆx ( S ) : M \ { }} {ˆx (P )} {ˆx (Q )} correspond to all components of ˆx takng strctly postve values. To sum up, we can solve the lnear program n (14)-(17) to fnd ẑ satsfyng f R (ẑ) = v 0 ẑ. Ths lnear program has O(m) decson varables and M O( A ) = O(mn 4 ) constrants. Once we have ẑ, we can solve problem (10)-(13) wth z = ẑ to obtan an optmal soluton ˆx and construct the assortments (Ŝ1 1,..., Ŝ1 m) and (Ŝ2 1,..., Ŝ2 m) as n the prevous paragraph. If we choose (Ŝ1,..., Ŝm) as one of the assortments (Ŝ1 1,..., Ŝ1 m) and (Ŝ2 1,..., Ŝ2 m), then (Ŝ1,..., Ŝm) satsfes 2 M V (Ŝ) γ (R (Ŝ) ẑ) f R (ẑ) and (Ŝ1,..., Ŝm) s a feasble soluton to problem (1). Snce our approxmaton f R ( ) to f( ) satsfes 2 f R (z) f(z) for all z R +, by Corollary 2, we obtan 4 Π(Ŝ1,..., Ŝm) z, so that (Ŝ1,..., Ŝm) s a 4-approxmate soluton to problem (1). Thus, f we check the expected revenue provded by each one of the assortments (Ŝ1 1,..., Ŝ1 m) and (Ŝ2 1,..., Ŝ2 m) for problem (1) and pck the best one, then we obtan a 4-approxmate soluton to problem (1). 15

16 6 Numercal Experments In ths secton, our goal s to numercally test the performance of the 4-approxmaton algorthm descrbed n Secton 5. Snce we can obtan the optmal soluton to problem (1) when we have a cardnalty constrant, we do not provde numercal experments under a cardnalty constrant. 6.1 Numercal Setup In our numercal experments, we randomly generate a large number of problem nstances. We generate each problem nstance by usng the followng procedure. We set the number of nests as m = 3 or m = 5 and the number of products n each nest as n = 15 or n = 30. To come up wth the revenues and preference weghts of the products, we sample C j from the unform dstrbuton over [0, 1]. Smlarly, we sample X j and Y j from the unform dstrbuton over [0.75, 1.25]. We set the revenue and preference weght of product j n nest respectvely as r j = 10 C 2 j X j and v j = 10 (1 C j ) Y j. Through C j, we ensure that the products havng larger revenues generally tend to have smaller preference weghts, ndcatng more expensve products tend to be less attractve. Squarng C j n the expresson for r j skews the dstrbuton of the revenues so that we have a small number of products wth large revenues. Through X j and Y j, we ncorporate dosyncratc nose nto the revenues and preference weghts so that not all products wth large revenues have small preference weghts. We sample the dssmlarty parameter γ for each nest from the unform dstrbuton over [0.25, 0.75]. We set the preference weght v 0 of the no purchase opton such that the probablty of no purchase s 0.4 even when we offer all products n all nests. We sample the space requrement w j of product j n nest from the unform dstrbuton over [1, 10]. We set the capacty avalablty as c = κ M j N w j, correspondng to a κ fracton of the total space consumpton of all products n all nests. We use κ = 0.1, κ = 0.15 or κ = 0.2. We vary (m, n, κ) over {3, 5} {15, 30} {0.1, 0.15, 0.2} to get 12 parameter combnatons. In each parameter combnaton, we generate 5,000 ndvdual problem nstances by usng the approach n the prevous paragraph. For each problem nstance, we use the approach n Secton 5 to obtan a 4-approxmate soluton. In Onlne Appendx B, we also gve a lnear program that provdes an upper bound on the optmal expected revenue for a partcular problem nstance. To assess the qualty of the 4-approxmate soluton, we check the gap between the expected revenue from the 4-approxmate soluton and the upper bound on the optmal expected revenue. 6.2 Numercal Results Our numercal results are gven n Table 1. The frst column n ths table shows the parameter combnatons by usng (m, n, κ). We recall that we generate 5,000 ndvdual problem nstances n each parameter combnaton. For each problem nstance, we use the approach n Secton 5 to obtan a 4-approxmate soluton to problem (1). We use Rev p to denote the expected revenue obtaned by the 4-approxmate soluton for problem nstance p. Usng Bnd p to denote the upper bound on 16

17 the optmal total expected revenue for problem nstance p, the second column n Table 1 gves the average percent gap between Rev p and Bnd p, where average s taken over all 5,000 problem nstances n a parameter combnaton. The thrd and fourth columns respectvely gve the 95th percentle and maxmum of the percent gaps between Rev p and Bnd p over all 5,000 problem nstances n a partcular parameter combnaton. Thus, the second, thrd and fourth columns respectvely gve the average, 95th percentle and maxmum of the data {100 (Bnd p Rev p )/Bnd p : p = 1,..., 5,000}. In ths way, these columns gve an ndcaton of the optmalty gaps of the 4-approxmate solutons. The ffth column gves the number of products n the 4-approxmate soluton, averaged over all 5,000 problem nstances. The sxth column gves the number of problem nstances for whch the percent gap between Rev p and Bnd p s less than 1%. Smlarly, the seventh, eghth and nnth columns gve the number of problem nstances where the percent gap between Rev p and Bnd p s respectvely less than 2.5%, 5% and 10%. The tenth and eleventh columns attempt to gve a feel for the tghtness of the space constrant. The tenth column shows the average number of products n the optmal assortment when there are no space constrants. The eleventh column shows the number of problem nstances for whch ths unconstraned soluton volates the space constrant. For comparson, we also use a greedy algorthm to obtan a heurstc soluton to problem (1). The greedy algorthm starts wth the empty assortment, fnds the product that provdes the largest mprovement n the expected revenue per unt of space consumpton and adds ths product to the assortment, untl there s no avalable space or no mprovement n the expected revenue. The twelfth, thrteenth and fourteenth columns respectvely show the average, 95th percentle and maxmum of the percent gaps between the expected revenue obtaned by the greedy algorthm and the upper bound on the optmal expected revenue. So, usng Gre p to denote the expected revenue obtaned by the greedy algorthm for problem nstance p, these three columns show the average, 95th percentle and maxmum of the data {100 (Bnd p Gre p )/Bnd p : p = 1,..., 5,000}. Our results ndcate that our 4-approxmaton algorthm performs qute well. Over all problem nstances, the average optmalty gap of ths algorthm s no larger than 1.56%. In 58,034 out of all 60,000 problem nstances, the optmalty gaps of the 4-approxmate solutons are less than 5%. As a general trend, the optmalty gaps tend to get smaller as κ gets larger. As κ gets larger, the capacty avalablty gets larger and each product occupes a smaller fracton of the avalable capacty. So, problem nstances where each product occupes a smaller fracton of the avalable capacty appear to be easer to approxmate. Ths observaton s algned wth the ntuton that the lnear programmng relaxaton of a knapsack problem becomes tghter as each tem occupes a smaller fracton of the knapsack capacty. In partcular, t s known that f each tem occupes no larger than a fracton ϵ of the knapsack capacty, then the optmal objectve value of the lnear programmng relaxaton exceeds the optmal objectve value of a knapsack problem by at most a factor of 1/(1 ϵ). The most problematc parameter combnaton n Table 1 s (3, 15, 0.1), correspondng to a small value of κ wth κ = 0.1. Even for ths parameter combnaton, n more than 75% of the problem nstances, the optmalty gap of the 4-approxmate soluton s no larger than 5%. Also, we note that the reported optmalty gaps are pessmstc estmates, snce these optmalty gaps are obtaned by 17