Optimal Targeting of Customers for a Last-Minute Sale R. Cominetti, - PowerPoint PPT Presentation

Optimal Targeting of Customers for a Last-Minute Sale R. Cominetti, J. Correa, J. San Mart´ ın Universidad de Chile Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008

Want to sell business class upgrades Who should get the offer? Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 1

The Process — (1) Set of customers N Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 2

(2) Address the offer to S ⊆ N Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 3

(3) Customers accept/reject A ⊆ S ⊆ N Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 4

(4) Winner is chosen at random Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 5

The Single Item case Setting Implication set of clients i ∈ N select S ⊆ N to offer different revenue for each client v i want high revenue clients different acceptance probability p i want high pbb clients until sold out first respondent wins last minute no time for regret revenue = discount price − normal price × prob buys anyway Goal balance probabilities and revenues so that the selected S ⊆ N maximizes expected revenue Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 6

The Problem — Discrete Version • Expected revenue for S ⊆ N is: revenue if prob that prob that A accepts A accepts S \ A rejects � �� � � �� � � �� � v ( A ) � � � V S = · p i · (1 − p i ) | A | A ⊆ S i ∈ A i ∈ S \ A Problem: find V ∗ = max S ⊆ N V S • V S can be computed in O ( n 3 ) using convolutions Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 7

The Problem — Continuous Version • Offer is made to each client with probability x i P [ i accepts ] = P [ Y i = 1] = x i · p i = y i revenue if prob that prob that A accepts A accepts N \ A rejects � �� � � �� � � �� � v ( A ) � � � V ( y ) = · y i · (1 − y i ) | A | A ⊆ N i ∈ A i ∈ N \ A Problem: find V ∗ = 0 ≤ y i ≤ p i V ( y ) max • Both problems are equivalent since V ( y ) is linear in each variable y i Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 8

Threshold Strategies • If offer is made to a customer reporting v i , shouldn’t we also consider those customers with higher values? • Find a threshold value V and offer to all clients such that v i ≥ V We assume v 1 ≥ v 2 ≥ · · · ≥ v n • Optimal threshold found in O ( n 4 ) : max 1 ≤ i ≤ n V { 1 ,...,i } • Typical in revenue management ... but... Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 9

Threshold Strategies are not optimal! V (1) = 1 2 · 2 = 1 p i v i V (1 , 2) = 1 4 · 2 + 1 4 · 1 + 1 4 · 3 2 + 1 4 · 0 = 1 . 125 0.5 2 0.5 1 V (1 , 2 , 3) = 1 4 · 2 . 9 2 + 1 4 · 1 . 9 2 + 1 4 · 3 . 9 3 + 1 4 · 0 . 9 = 1 . 15 1 0.9 V (1 , 3) = 1 2 · 2 . 9 2 + 1 2 · 0 . 9 = 1 . 175 • Every subset can be optimal • Sorting by probability or expected value is also sub-optimal Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 10

Heuristics: 10 customers, 200 instances Algorithm % opt Min ratio Avg ratio Time 100.0 1.0000 1.0000 1611.2 optimal 93.0 0.9916 0.9998 23.5 threshold 47.5 0.8168 0.9771 0.4 lp-relax 74.0 0.9775 0.9988 21.5 lp2-relax 99.0 0.9918 0.9999 15.4 in-out Problem complexity is open Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 11

Threshold is 1 2 -optimal: LP relaxation Rewrite objective function as V ( y ) = � i ∈ N v i π i 1 π i = P [ i accepts and wins ] = y i E [ 1+ S i ] S i = � j � = i Y j (number of competitors) � = ⇒ 0 ≤ π i ≤ p i and i ∈ N π i ≤ 1 Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 12

Threshold is 1 2 -optimal: LP relaxation Rewrite objective function as V ( y ) = � i ∈ N v i π i 1 π i = P [ i accepts and wins ] = y i E [ 1+ S i ] S i = � j � = i Y j (number of competitors) � = ⇒ 0 ≤ π i ≤ p i and i ∈ N π i ≤ 1 Consider the relaxation (upper bound) V ∗ ≤ V LP = �� � i ∈ N v i y i : � max i ∈ N y i ≤ 1 0 ≤ y i ≤ p i and use it to get a 1 2 -optimal threshold strategy Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 13

Polynomial approximation algorithm alg LP • LP solution in O ( n ) : find largest k with � k i =1 p i ≤ 1 and set  p i if i ≤ k   1 − � k y i = LP i =1 p i if i = k + 1   0 otherwise Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 14

Polynomial approximation algorithm alg LP • LP solution in O ( n ) : find largest k with � k i =1 p i ≤ 1 and set  p i if i ≤ k   1 − � k y i = LP i =1 p i if i = k + 1   0 otherwise • y LP is a randomized strategy equivalent to � select { 1 , . . . , k } with probability [ � k +1 i =1 p i − 1] /p k +1 select { 1 , . . . , k + 1 } with probability [1 − � k i =1 p i ] /p k +1 Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 15

Polynomial approximation algorithm alg LP • LP solution in O ( n ) : find largest k with � k i =1 p i ≤ 1 and set  p i if i ≤ k   1 − � k y i = LP i =1 p i if i = k + 1   0 otherwise • y LP is a randomized strategy equivalent to � select { 1 , . . . , k } with probability [ � k +1 i =1 p i − 1] /p k +1 select { 1 , . . . , k + 1 } with probability [1 − � k i =1 p i ] /p k +1 • De-randomize in O ( n 3 ) : max between V { 1 ,...,k } and V { 1 ,...,k +1 } Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 16

In the example: = 1 y LP 1 2 p i v i = 1 y LP 2 2 0.5 2 y LP = 0 3 0.5 1 = 1 2 · 2 + 1 V LP 2 · 1 = 1 . 5 1 0.9 = 1 4 · 2 + 1 4 · 1 + 1 4 · 3 2 + 1 V ( y LP ) 4 · 0 = 1 . 125 Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 17

LP is a 1 Theorem: alg 2 − approximation algorithm Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 18

LP is a 1 Theorem: alg 2 − approximation algorithm Proof: By Jensen’s inequality 1 1 1 1+1 = 1 1 E [ 1+ S i ] ≥ 1+ E ( S i ) = j ≥ j � = i y LP 2 1+ � hence V ( y LP ) = � � 1+ S i ] ≥ 1 1 i ∈ N v i y LP i E [ i ∈ N v i y LP i 2 so that V ∗ ≥ V ( y LP ≥ 1 2 V ∗ . LP ) ≥ 1 2 V � Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 19

Alternative: Hyperbolic relaxation y i y i π i ≥ j � = i y j ≥ 1+ � 1+ � j ∈ N y j V ∗ ≥ � i ∈ N v i y i max 1+ � i ∈ N y i 0 ≤ y i ≤ p i • Common-lines problem in transit equilibrium (Chriqui&Robillard’75) • Optimum is a threshold strategy • Linear-time algorithm: max k [ v 1 p 1 + · · · + v k p k ] / [1 + p 1 + · · · + p k ] • Also a 1 2 -approximation algorithm Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 20

Improved 2 3 -approximation Let x = P [ S = 0] so that x + � i ∈ N π i = 1 1 Moreover π i = y i E [ 1+ S i ] with 1 1 · P [ S i =0] + 1 E [ 1+ S i ] ≤ 2 · P [ S i > 0] 1 = 2 (1 + P [ S i =0]) 1 x = 2 (1 + 1 − y i ) and then y i ≤ p i implies π i ≤ p i x 2 (1 + 1 − p i ) Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 21

Hence we get the alternative LP relaxation V ∗ ≤ V max � = i ∈ N v i z i LP 2 z i ≤ p i x 2 (1 + 1 − p i ) x + � i ∈ N z i = 1 x, z i ≥ 0 Algorithm alg LP 2 • Find a basic optimal solution ( z ∗ , x ∗ ) for LP 2 2 z ∗ LP 2 • Set y = i ...either 0 or p i except for one value! 1+ x ∗ i 1 − pi • De-randomize y LP 2 to get a set of the form { 1 , . . . , k } Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 22

LP 2 is a 2 Theorem: alg 3 -approximation Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 23

LP 2 is a 2 Theorem: alg 3 -approximation V ( y ) = � 1+ S i ] ≥ � v i y i 1 Proof: i ∈ N v i y i E [ i ∈ N 1+ � j � = i y j we get V ( y LP 2 ) ≥ � i ∈ N v i z ∗ LP 2 Replacing y i γ i with i 2(1+ x ∗ ) γ i = 1 − pi ) . i )(1+ x ∗ (3 − x ∗ − 2 z ∗ Since V ∗ ≤ � 3 . This is obvious if x ∗ = 0 . i ∈ N v i z ∗ i we need γ i ≥ 2 Else, since ( z ∗ , x ∗ ) is a basic solution exactly one of the two inequalities x ∗ i = p i involving z ∗ i is tight: if z ∗ i > 0 then z ∗ 2 (1 + 1 − p i ) so that 2(1+ x ∗ ) 1 − pi ) ≥ 2 γ i = 3 . (3 − p i − x ∗ 1 − pi )(1+ x ∗ � Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 24