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

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(2) Address the offer to S ⊆ N

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(3) Customers accept/reject A ⊆ S ⊆ N

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(4) Winner is chosen at random

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The Single Item case

Setting Implication set of clients i ∈ N select S ⊆ N to offer different revenue for each client vi want high revenue clients different acceptance probability pi 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

A accepts

prob that

A accepts

prob that

S\A rejects

VS =

• A⊆S

v(A) |A| ·

• i∈A

pi ·

• i∈S\A

(1 − pi) Problem: find V ∗ = max

S⊆N VS

• VS can be computed in O(n3) using convolutions

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The Problem — Continuous Version

• Offer is made to each client with probability xi

P[i accepts] = P[Yi = 1] = xi · pi = yi

revenue if

A accepts

prob that

A accepts

prob that

N\A rejects

V (y) =

• A⊆N

v(A) |A| ·

• i∈A

yi ·

• i∈N\A

(1 − yi) Problem: find V ∗ = max

0≤yi≤pi V (y)

• Both problems are equivalent since V (y) is linear in each variable yi

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Threshold Strategies

• If offer is made to a customer reporting vi, shouldn’t we also consider

those customers with higher values?

• Find a threshold value V and offer to all clients such that vi ≥ V

We assume v1 ≥ v2 ≥ · · · ≥ vn

• Optimal threshold found in O(n4): 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 V (1, 2) = 1

4 · 2 + 1 4 · 1 + 1 4 · 3 2 + 1 4 · 0

= 1.125 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 V (1, 3) = 1

2 · 2.9 2 + 1 2 · 0.9

= 1.175 pi vi 0.5 2 0.5 1 1 0.9

• Every subset can be optimal
• Sorting by probability or expected value is also sub-optimal

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Heuristics: 10 customers, 200 instances

Algorithm % opt Min ratio Avg ratio Time

• ptimal

100.0 1.0000 1.0000 1611.2 threshold 93.0 0.9916 0.9998 23.5 lp-relax 47.5 0.8168 0.9771 0.4 lp2-relax 74.0 0.9775 0.9988 21.5 in-out 99.0 0.9918 0.9999 15.4 Problem complexity is open

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Threshold is 1

2-optimal: LP relaxation

Rewrite objective function as V (y) =

i∈N viπi

πi = P[i accepts and wins] = yi E[

1 1+Si]

Si =

j=i Yj (number of competitors)

= ⇒ 0 ≤ πi ≤ pi and

• i∈N πi ≤ 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 viπi

πi = P[i accepts and wins] = yi E[

1 1+Si]

Si =

j=i Yj (number of competitors)

= ⇒ 0 ≤ πi ≤ pi and

• i∈N πi ≤ 1

Consider the relaxation (upper bound) V ∗ ≤ V LP = max

0≤yi≤pi

• i∈N viyi :

i∈N yi ≤ 1

• and use it to get a 1

2-optimal threshold strategy

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Polynomial approximation algorithm alg

LP

• LP solution in O(n): find largest k with k

i=1 pi ≤ 1 and set

y

LP

i =

     pi if i ≤ k 1 − k

i=1 pi if i = k + 1

• therwise

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Polynomial approximation algorithm alg

LP

• LP solution in O(n): find largest k with k

i=1 pi ≤ 1 and set

y

LP

i =

     pi if i ≤ k 1 − k

i=1 pi if i = k + 1

• therwise
• yLP is a randomized strategy equivalent to
• select {1, . . . , k} with probability [k+1

i=1 pi − 1]/pk+1

select {1, . . . , k + 1} with probability [1 − k

i=1 pi]/pk+1

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Polynomial approximation algorithm alg

LP

• LP solution in O(n): find largest k with k

i=1 pi ≤ 1 and set

y

LP

i =

     pi if i ≤ k 1 − k

i=1 pi if i = k + 1

• therwise
• yLP is a randomized strategy equivalent to
• select {1, . . . , k} with probability [k+1

i=1 pi − 1]/pk+1

select {1, . . . , k + 1} with probability [1 − k

i=1 pi]/pk+1

• De-randomize in O(n3): max between V{1,...,k} and V{1,...,k+1}

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In the example: pi vi 0.5 2 0.5 1 1 0.9 yLP

1

= 1

2

yLP

2

= 1

2

yLP

3

= 0 V LP = 1

2 · 2 + 1 2 · 1

= 1.5 V (yLP) = 1

4 · 2 + 1 4 · 1 + 1 4 · 3 2 + 1 4 · 0

= 1.125

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Theorem: alg

LP is a 1

2−approximation algorithm

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Theorem: alg

LP is a 1

2−approximation algorithm

Proof: By Jensen’s inequality E[

1 1+Si] ≥ 1 1+E(Si) = 1 1+

j=i yLP j ≥

1 1+1 = 1 2

hence V (yLP) =

i∈N vi yLP i E[ 1 1+Si] ≥ 1 2

• i∈N vi yLP

i

so that V ∗ ≥ V (y

LP) ≥ 1

2V

LP ≥ 1

2V ∗.

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Alternative: Hyperbolic relaxation

πi ≥

yi 1+

j=i yj ≥

yi 1+

j∈N yj

V ∗ ≥ max

0≤yi≤pi

• i∈N viyi

1+

i∈N yi

• Common-lines problem in transit equilibrium

(Chriqui&Robillard’75)

• Optimum is a threshold strategy
• Linear-time algorithm: maxk [v1p1 + · · · + vkpk]/[1 + p1 + · · · + pk]
• 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 πi = 1

Moreover πi = yi E[

1 1+Si] with

E[

1 1+Si]

≤ 1 · P[Si=0] + 1

2 · P[Si>0]

=

1 2(1 + P[Si=0])

=

1 2(1 + x 1−yi)

and then yi ≤ pi implies πi ≤ pi

2 (1 + x 1−pi)

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Hence we get the alternative LP relaxation V ∗ ≤ V

LP2

= max

i∈N vizi

zi ≤ pi

2 (1 + x 1−pi)

x +

i∈N zi = 1

x, zi ≥ 0 Algorithm algLP2

• Find a basic optimal solution (z∗, x∗) for LP2
• Set y

LP2

i

=

2z∗

i

1+ x∗

1−pi

...either 0 or pi except for one value!

• De-randomize yLP2 to get a set of the form {1, . . . , k}

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Theorem: alg

LP2 is a 2

3-approximation

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Theorem: alg

LP2 is a 2

3-approximation

Proof: V (y) =

i∈N viyiE[ 1 1+Si] ≥ i∈N viyi 1+

j=i yj

Replacing y

LP2

i

we get V (yLP2) ≥

i∈N viz∗ i γi with

γi =

2(1+x∗) (3−x∗−2z∗

i )(1+ x∗ 1−pi).

Since V ∗ ≤

i∈N viz∗ i we need γi ≥ 2

• 3. This is obvious if x∗ = 0.

Else, since (z∗, x∗) is a basic solution exactly one of the two inequalities involving z∗

i is tight: if z∗ i > 0 then z∗ i = pi 2 (1 + x∗ 1−pi) so that

γi =

2(1+x∗) (3−pi− x∗

1−pi)(1+ x∗ 1−pi) ≥ 2

3.

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Multiple items

When there are m available items the value of a strategy y is

revenue if

A accepts

prob that

A accepts

prob that

S\A rejects

V (y) =

• A⊆N
• min
• 1, m

|A|

• v(A)

·

• i∈A

yi ·

• i∈S\A

(1 − yi)

Journ´ ees Franco-Chiliennes d’Optimisation — Universit´ e de Toulon — Mai 2008 25

Multiple items

When there are m available items the value of a strategy y is

revenue if

A accepts

prob that

A accepts

prob that

S\A rejects

V (y) =

• A⊆N
• min
• 1, m

|A|

• v(A)

·

• i∈A

yi ·

• i∈S\A

(1 − yi) =

• i∈N

viyiE[min{1,

m 1+Si}]

Problem: find V ∗ = max

0≤yi≤pi V (y)

Complexity is open though V (y) is computed in O(n3) by convolution

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The trivial extension of LP gives an upper bound V LP = max

• i∈N viyi :

i∈N yi ≤ m and 0 ≤ yi ≤ pi

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The trivial extension of LP gives an upper bound V LP = max

• i∈N viyi :

i∈N yi ≤ m and 0 ≤ yi ≤ pi

• which guarantees a constant factor of the optimum:
• V (y) =

i∈N viyiE[min{1, m 1+Si}] ≥ ρ(y) i∈N viyi

• ρ(y) = E[min{1,

m 1+S}] ≥ 1 − 1 m+1

• 1 + Var(S) P(S ≥ m)

find lower bound for ρ(y)...

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Asymptotically optimal bound since Var(S) =

i yi(1 − yi) ≤ m and P(S ≥ m) ≤ 1, we get

ρ(y) ≥ 1 −

1 √m+1

so that V (yLP) is within

1 √m+1 from optimal ...but not better!

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Sharper bound ρ(y) ≥ 1 − √

1+σ2 m+1 min{1, 1 2 + M σ }

M = max

u≥0

√ 2u e−2u

∞

• k=0

(uk

k!)2 ∼ 0.46882235549939533

“if the mean number of successes in n independent heterogeneous Bernoulli trials is an integer m then the median is also m”

(Jogdeo-Samuels’62, Siegel’01)

P[S ≥ m] ≤ 1

2 + P[S = m] ≤ 1 2 + M σ

(latter estimate from C+Vaisman’08)

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Extensions — Buy by Bulk

Dean, Goemans, Vondrak 04

• The difference is that customer i ∈ N demands si ≥ 1 units
• Problem is now NP-hard: for pi = 1 we get knapsack
• Computing V (y) is actually #P-complete

(e.g. count the number of perfect matchings in a bipartite graph)

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Extensions — Multistage

• Medium-term: address offers in k successive rounds.
• Find an adaptive online algorithm that computes an optimal subset
• f clients to address the offer in the next round.

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Extensions — Pricing

• Personalized contact (email): price can be different for each client
• Pbb of accepting is a decreasing function of price pi : R+ → [0, 1]
• Select a price for each customer to maximize expected revenue

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