Payoffs in Location Games

Payoffs in Location Games

Shuchi Chawla1/22/2003

joint work with Amitabh Sinha, Uday Rajan & R.

Ravi

Shuchi Chawla, Carnegie Mellon Univ2

Caffeine wars in Manhattan

Sam owns Starcups; Trudy owns Tazzo

Every month both chains open a new outlet – Sam chooses a location first, Trudy follows

Indifferent customers go to the nearest coffee shop

At the end of n months, how much market share can Sam have?

Trudy knows n, Sam doesnt


An artist’s rendering of Manhattan

$$

T$$T

$$

Sam’s Starcups T Trudy’s Tazzo


Why bother?

Product Placement many features to choose from – high

dimension high cost of recall – cannot modify earlier

decisions

Service Location cannot move service once located


Some history…

The problem was first introduced by Harold Hotelling in 1929

Acquired the name “Hotelling Game”

Originally studied on the line with n players moving simultaneously

Extensions to price selection


Formally…

Given: (M,L,F) – Metric space, Location set, Distribution of demands

At step i, S first picks si2 L. Then T picks ti2 L

si = si(s1,…,si-1,t1,…,ti-1) ; ti = ti(s1,…,si,t1,…,ti-1) S is an online player: does not know n

Payoff for S at the end of n moves is:

(M,L,F)(T) = 1 - (M,L,F)(S)

(M,L,F)(S) = s(v,S)<(v,T)dF(v) + ½s(v,S)=(v,T)dF(v)


The second mover advantage

Note that if 8 i, ti = si

(M,L,F)(S) = (M,L,F)(T) = ½

T can always guarantee a payoff of ½ Can S do the same?

We will show that S cannot guarantee ½ but at least some constant fraction depending on M


Some more notation

M(S)= minL,F minn minT (M,L,F)(S)

M(S) is the worst case performance of strategy S on any metric space in M

M = maxS M(S)

M(1) – defined analogously when n=1


Our Results

R d(1) = 1/(d+1)

½ 1/(d+1) · R d · 1/(d+1)


The 1-D case: Beaches & Icecream

Assume a uniform demand distribution for simplicity

S moves at ½ , no move of T can get more than ½

) R (1) = ½



No subsequent move of T can get > ½ Recall T’s strategy to obtain ½ : repicate S’s

moves

S can use the same strategy for moves si>1

s1 = ½ ; si = ti-1



(tn) · ½

(t1,…,tn-1) = (s2,…,sn)

) (S) ¸ (s2,…,sn) ¸ ¼


Median and Replicate

Given a 1-move strategy with payoff obtain an n-move strategy with payoff /2

Use 1-move strategy for the first move,

Replicate all other moves of player 2

Last move of player 2 gets at most 1-, the rest get at most half of the remaining : /2

“MEDIAN”

“REPLICATE”


Locn Game in the Euclidean plane

Thm 1: R 2(1) = 1/3

Thm 2: 1/6 · R 2· 1/3

Proof of Thm 2: Use Median and Replicate with Thm

1


R 2(1) · 1/3

Condorcét voting paradoxL1

L2L3

D3

D2

D1

D1 : L1 > L3 > L2

D2 : L2 > L1 > L3

D3 : L3 > L2 > L1

S gets only 1/3 of the demand

The vote is inconclusive

$$

T


R 2(1) ¸ 1/3

Our goal: 9 a location s such that 8 t, (s,t) ¸ 1/3

Outline:

Construct a digraph on locationsG contains edge u!v iff (u,v) 1/3Show that G contains no cycles) G has a sink s


> 2/3 demand

R 2(1) ¸ 1/3

Each edge defines a half-space containing at least 2/3 of the demand

A cycle defines an intersection of half-spaces


If not:

R 2(1) ¸ 1/3

All triplets of half spaces must intersect!

Contradiction!!

< 1/3 < 1/3

< 1/3


R 2(1) ¸ 1/3

Helly’s Theorem

Given a collection {C1,C2,…,Cn} of convex sets in Euclidean space :

If every triplet of the sets has a non empty intersection, then Å1·i·n Ci ;

) all half-spaces defined by the graph G contain a common point P


R 2(1) ¸ 1/3

Let u1,…,uk be a cycle in G

Then,(P,ui+1) · (P,ui)

because P is in the half-space defined by the edge ui!ui+1

) (P,ui) = (P,uj) 8 i,j

P

P is the intersection of hyperplanes bisecting the edges


R 2(1) ¸ 1/3

P

Let demand at P be Then each half-space has a total demand of at least 2/3 + /2

>1/3

> 2/3 - Contradiction!!


The d-dimensional case

Results on R 2 extend nicely to R d:

Thm 3 : R d(1) = 1/(d+1)

Thm 4 : 1/2(d+1) · R d · 1/(d+1)


Condorcét instance in d-dimensions

As before we should have Di: Li > Li+1 > Li+2 > … > Li-1

Embedding in R d+1 : Li ´ ( 0 , … , 0 , 1 , 0 , … , 0) Di ´ (d-i,d-i+1,…,1-,2, 3 , … )

Project all points down to the d dimensional plane containing {L1,…,Ld+1} – relative distances between Li and Dj are preserved


Lower bound in d-dimensions

As before, define a directed graph on locations with each half-space containing d/(d+1) demand

Every set of d half-spaces must intersect

By Helly’s Theorem all half-spaces must have a non empty intersection. Assume WLOG that the origin lies in this intersection.

(Skip)



Assume for the sake of contradiction that a cycle exists.

Each point in the intersection is equi-distant from all vertices in the cycle

We want this to hold for at most some d+1 half-spaces

Arrive at a contradiction just as before.



Let i be a vector representing the i-th edge in the cycle; Let p represent some point in the intersection

Then, p¢i¸ 0 8 i; i i = 0

9 a collection of · d+1 vectors i such that ii = 0 with i¸ 0

Then, p¢ii = 0.

But p¢i¸ 0 8 i, so, p¢i = 0 for the d+1 vectors.

Thus every point in the intersection of these half-spaces is equi-distant from all vertices in the cycle.


Concluding Remarks

Results hold even when demands lie in some high dimensional space

We can obtain tighter results in the line when n is bounded.


Open Problems

Closing the factor-of-2 gap for

Convergence with n If S knows a lower/upper bound on n, is there a better

strategy? Can he do better as n gets larger – we believe so

Brand loyalty What about demand in the intermediate steps? fraction of demand at every time step becomes

loyal to the already opened locations. The rest carries on to the next step.


Questions?

Payoffs in Location Games

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