Assortment Optmzaton under MNL Haotan Song Aprl 30, 2017 1 Introducton The assortment optmzaton problem ams to fnd the revenue-maxmzng assortment of products to offer when the prces of products are fxed. It gans tremendous popularty and seems to be well-studed already. Ths report focuses on the assortment optmzaton problem under the MNL model wthout any capacty constrants and s organzed as follows. We frst gve the defnton and two examples, one of them showng that the greedy scheme cannot work (orgnal); then we state a crucal theorem and gve four proofs from dfferent aspects Proof 1 and 3 are orgnal, Proof 2 belongs to T arek. 2 Assortment Optmzaton under MNL Let N denote the unverse of n tems and 0 denote the non-purchase opton. The assortment optmzaton problem s to fgure out a set of products S to offer so as to maxmze the potental revenue Rev(S). Under the MNL model, the probablty of choosng tem from S after normalzaton s gven by P( S) = 1 + j S v, j..d. where we have the normal settng that U = V + ɛ, ɛ Gumbel(0, µ)and = exp(µv ). Let r denote the revenue of tem, then the expected revenue of assortment S s gven by Rev(S) = S r P( S) = r 1 + S j S v. j Wthout loss of generalty we may assume the tem revenues are n the decreasng order, that s r 1 r 2 r n. Snce t s a decson problem that s to decde for each tem whether to select or not, we can formulate t nto a bnary optmzaton problem. Problem 2.1 [Bnary Formulaton]. The assortment optmzaton problem under the MNL model can be wrtten as the followng unconstraned bnary optmzaton problem: n max Rev(S) = max =1 r x S {1,2,,n} x {0,1} n 1 + n =1 v. x 1
Before we gong to propertes of ths problem, let us take a look at the example below. Eample 2.2. Root tem 0 tem 1 tem 2 tem 3 tem 0 1 2 3 r 0 10 6 3 1 0.4 1.2 0.6 Startng wth S (0) =, we now consder to add tems nto S n order to ncrease the average revenue. Note that r 2 v 2 = 36/11 > r 1v 1 = 20/7 > r 3v 3 = 9/8. 1 + v 2 1 + v 1 1 + v 3 We may add tem 2 to S (0) and take S (1) = S (0) {2} for the next step wth Rev(S (1) ) = r 2v 2 1 + r 2 = 36/11. What should we add next or should we stop? Thnk of Rev(S) as the average revenue of the assortment S. Snce r 3 = 3 < 36/11 < r 1 = 10, addng tem 3 to S (1) wll brng down the revenue, whle addng tem 1 wll ncrease t. Takng S (2) = S (1) {1}, we obtan Rev(S (2) ) = Rev(S(1) ) (1 + v 2 ) + r 1 v 1 1 + v 2 + v 1 = 56/13. Snce addng tem 3 wll only decrease the revenue, we stop. Indeed, by exhaustng all the possbtes, we can confrm that S = S (2). Example 2.2 gves us an mpresson that t seems we can compute the optmal assortment n a greedy manner. However, t s not always the case. Eample 2.3 [Greedy does not work]. It s easy to check that tem 0 1 2 3 4 r 0 12 10 7 6 1 0.5 0.9 2 5 max S r 1 + = r 4v 4 1 + v 4 = 5. However, one can also easly check that S = {1, 2, 3} wth Rev(S ) = and consequently, tem 4 / S. 12 0.5 + 10 0.9 + 7 2 1 + 0.5 + 0.9 + 2 = 145 22, Although the greedy algorthm does not work, both two examples gve us an ntuton that the optmal assortment S seems to consst of contnuous products from tem 1 to tem for some 1 n. Indeed, we have the followng theorem. Theorem 2.4. Then the optmal soluton to Problem 2.1 must be of the form x = [1,, 1, 0,, 0] whch means the optmal assortment S = {1, 2,, } for some. 2
For example, f n Problem 2.1 we have n = 3 and r 1 > r 2 > r 3, we can mmedately know that none of {[1, 0, 1], [0, 1, 1], [0, 1, 0], [0, 0, 1]} can be the optmal. There are varous proofs for ths theorem and here we demonstrate four of them. The four proofs gve dfferent formulatons and nsghts some are neat tself whle some s complcated but compatble wth addtonal constrants and thus I want to cover them all here. The frst one s qute straghtforward. Proof 1. If S, then {j N r j > r } S. Otherwse, we can add such j s to S to ncrease the revenue. In fact, there s a underlyng structure there that Rev(S {k}) s a convex combnaton of Rev(S) and r k for any k / S: Rev(S {k}) = r kv k + S r ( 1 + v k + v k S v = Then for any S, we have 1 + v k + S }{{} α Rev(S ) = α r + (1 α )Rev(S \ {}). Snce Rev(S ) Rev(S \ {}), we have Rev(S \ {}) r. Therefore, S = { r Rev(S )} = { r > Rev(S )}. ) ( 1 + S r k + v ) 1 + v k + S r S 1 + S }{{} v. }{{} 1 α Rev(S) Based on Theorem 2.4 and Proof 1, the algorthm to compute S can be very smple. Input: r 1 r 2 r n,{ } n =1 Output: S Intalzaton: S =, k = 1; whle Rev(S) < r k & k n do S = S {k}; k = k + 1; end The dea of the second proof s to look at the frst-order condton, whch s a typcal strategy for an unconstraned dfferentable optmzaton problem, of the relaxed objectve functon defned as n =1 R(x) = r x 1 + n =1 v, x [0, 1] n. x Proof 2. Take the partal dervatve w.r.t. any x and we obtan R(x) = r (1 + n =1 v x ) v n =1 r x x = (1 + n =1 x ) 2 1 + n =1 x (r R(x)). If at the current x we have r < R(x), then R(x) < 0 whch gves a descent drecton for x. By x notng ths, we clam that the optmal assortment must have the followng form S = { r > R(x )} where x = argmax R(x). x [0,1] n 3
Otherwse, assume such that r > R(x ) but x < 1, then ncreasng x by a suffcently small ɛ leads to a hgher revenuen; the other hand, assume such that r < R(x ) but x > 0, then decreasng x by a suffcently small ɛ leads to a hgher revenue. Both contradct the optmalty. Note that we dd not say that max R(x) = max R(x) n Proof 2, but the analyss tells us x [0,1] n x {0,1} n f r > R(x ) then x = 1 f r < R(x ) then x = 0 whch mples that the optmal revenue wll be acheved at a vertex. The next proof uses no dervatve and has an emphass on the objectve functon value at corner ponts, but essentally t s the same as Proof 2. Proof 3. When R(x) < r j for some j, then we have j R(x) r x + r j v n j 1 + =1 j v = r x n x + v j 1 + =1 n =1 v r x + r j v j (1 x j ) x 1 + n =1 x + v j (1 x j ) = ( n =1 r x )[1 + n =1 x + v j (1 x j )] [ n =1 r x + r j v j (1 x j )](1 + n =1 x ) (1 + n =1 x )[1 + n =1 x + v j (1 x j )] = ( n =1 r x )v j (1 x j ) r j v j (1 x j )(1 + n = (1 + n =1 x )[1 + n v j (1 x j ) =1 x ) =1 x + v j (1 x j )] 1 + n =1 x + v j (1 x j ) (R(x) r j) < 0, whch mples that ncreasng x j to 1 gves a hgher revenue. Smlarly, f R(x) > r j, then j R(x) r x 1 + v j x j j v = x 1 + n =1 v (R(x) r j ) < 0, x v j x j whch mples that decreasng x j to 0 gves a hgher revenue. Then we have the same argument as the last part of Proof 2 and conclude that S = { r > R(x )}. The next proof seems more complcated wth the prevous but actually s very mportant where we wll present a LP formulaton here. We defne the followng LP max r y =1 s.t. y 0 + y = 1 =1 y y 0, 1 n y 0 Denote the optmal soluton to the above LP by Y and then we want to show that Rev(S ) = Y. Proof 4. In partcular, let 1 ŷ 0 = 1 + j S v j ŷ = 1 { S } 1 +, 1 n j S v j 4 = r ŷ = Rev(S ). =1
It s easy to check that such ŷ s a feasble soluton to the LP, and consequently Rev(S ) Y. In fact, we can nterpret such ŷ as the market share of product. To see Rev(S ) Y, we frst wrte down the dualty of the LP mn s.t. θ θ α 0 = θ + α r, 1 n α 0 By the dualty theorem below, the optmal soluton to the dual problem s also Y. Notaton p s the prmal optmal value; d s the dual optmal value; p = f the prmal problem s nfeasble; d = f the dual problem s nfeasble; p = f the prmal problem s unbounded; d = f the dual problem s unbounded; Dualty Theorem If the prmal or dual problem s feasble, then p = d. Moreover, f p = d <, then both optma are attaned. By a trval transformaton α α, we have an equvalent form of the dual problem, whch s neater. mn s.t. θ θ α = α r θ, 1 n α 0 Note that r 1 θ r 2 θ r n θ. Consequently, for any θ, k such that r θ 0 f k and r θ < 0 otherwse. The constrants α r θ wll not be tght for α 0 > r θ. For k, by reducng α to r θ, we only relax the last constrant and everythng stll mantans feasble. Snce that does not change θ, we have the same objectve functon value, and consequently, there exsts an optmal soluton to the dual problem where the frst l constrants of α r θ are tght and the rest are not tght. By the complmentary slackness, we have { y = y 0, 1 j y = 0, j + 1 j n whch mples y0 = y = 1 1 + l j=1 v j 1 + l j=1 v j 5, 1 l.
Ths shows that the assortment S = {1, 2,, l} yelds the optmal value for the LP. Therefore Rev(S ) Rev(S) = Y. The nce thng for ths LP formulaton s more compatble n the sense that t allows the possblty to add more constrants, such as the capacty constrants whch I wll not cover n ths report. References J. Davs, G. Gallego, H. Topaloglu, Assortment plannng under the multnomal logt model wth totally unmodular constrant structures. Workng paper. 6