Lecturer: Moni Naor

Algorithmic Game Theory

Uri Feige Robi Krauthgamer Moni Naor

Lecture 11: Combinatorial Auctions

Announcements

• Course resumes to 1400:-16:00

The setting

• Set of alternatives A– Who wins the auction– Which path is chosen– Who is matched to whom

• Each participant: a value function vi:A R• Can pay participants: valuation of choice a with

payment pi is vi(a)+pi

Quasi linear preferences

Mechanism Design

• Mechanisms• Recall: We want to implement a social choice function

– Need to know agents’ preferences – They may not reveal them to us truthfully

• Example:– One item to allocate:

• Want to give it to the participant who values it the most– If we just ask participants to tell us their preferences: may lie

• Can use payments result is also a payment vector p=(p1,p2, … pn)

The Revelation Principle

Theorem: if there exists an arbitrary mechanism implementing a social choice function f in dominant strategies, then there exists an incentive compatible mechanism that implements f

The payments of the players in the incentive compatible mechanism are identical to those obtained at equilibrium in the original mechanism

Proof: by simulation

Direct CharacterizationA mechanism is incentive compatible iff the following hold for all i

and all vi

• The payment pi does not depend on vi but only on the alternative chosen f(vi, v-i) – the payment of alternative a is pa

• The mechanism optimizes for each player: f(vi, v-i) 2 argmaxa (vi(a)-pa)

Expected RevenuesTheorem [Revenue Equivalence]: under very general

conditions, every two Bayesian Nash implementations of the same social choice function if for some player and some type they have the same

expected payment then– All types have the same expected payment to the player– If all player have the same expected payment: the expected

revenues are the same

Not true when there are reservation prices!

Range Voting

• Each voter ranks the candidates in a certain range (say 0-99)

• The votes for all candidates are summed up and the one with highest total score wins

Can be considered as a generalization of approval voting from the range 0-1

No incentive for voter to rate a candidate lower than a candidate they like less.

Vickrey Clarke Grove Mechanism• f(v1, v2, … vn) maximizes i vi(a) over A

– Maximizes welfare

• There are functions h1, h2,… hn where

hi: V1 V2 … Vn R does not depend on vi

we have that:pi(v1, v2, … vn) = hi(v-i) - j i vj(f(v1, v2,… vn))

Clark Pivot rule: Choosing hi(v-i) = maxb 2 A j i vj(b)Payment of i when a=f(v1, v2,…, vn):

pi(v1, v2,… vn) = maxb 2 A j i vj(b) - j i vj(a)

Depends only on chosen alternative

Does not depend on vi

Social welfare (of others) when he participates

Social welfare when he does not participate

Examples of Auctions

• Sponsored Search:– Buying keywords (Google,

Yahoo, MS)• Spectrum Auctions• Ebay• Government procurement • Privatization

FCC Auction 2008

המשך המכרז: אלג'ם העלו מיליארד, וסבן העלה 3.8ל- מיליארד, וניצח.4.11ל-

הפרטת בזק:2005 למאי 29מקור: "הארץ",

שתי קבוצות התמודדו במכרז: קבוצת איייפקס-סבן, וקבוצת אלג'ם.• מיליארד 3.6 מיליארד, סבן: 3.2ההצעות ההתחלתיות - אלג'ם: •

ש"חהתכנון המקורי של משרד האוצר: לחשוף את ההצעות, ולקבל •

הצעות חדשות לערוך מכרז אנגלי עולה. –

נוכח הפער הגדול בין שתי ההצעות: באוצר חששו להמשיך לפי •התכנון המקורי, שמא אלג'ם תתייאש ותפרוש.

היועץ המקצועי, פרופ' מוטי פרי מירושלים, התעקש להמשיך לפי •אין ביטחון מלא, אבל בדרך כלל השקיפות מביאה התכנון: "

לתוצאות הטובות ביותר. יש למסור לכל אחד מהמשתתפים את תוצאות הסבב הראשון... ההצעה העיקרית שלי במכרז הייתה

ליצור מערכת שקופה לחלוטין, שבה כל אחד מהצדדים יודע בדיוק מה קורה ומקבל את כל המידע. ככל שלמתחרים יש יותר ידע,

רמת הסיכון שלהם יורדת, מרכיב ההימור בעסקה יורד, והם ". מוכנים להסתכן בהצעות גבוהות יותר

From Ron Lavi’s slides

VCG: turning mechanisms truthfulUsing exact optimization

VCG prices

truthful mechanism

Strategic players

Exact optimization algorithms for social welfare f

VCG and Computational Complexity

Several optimization must be performed in VCG:• f(v1, v2, … vn) maximizes i vi(a) over A• Clark Pivot rule: hi(v-i) = maxb 2 A j i vj(b)These may be hard optimization problems.

What happens when we replace then with approximations?

Incentive compatibility not guaranteed!

Combinatorial Auctions• Set M of m indivisible items• Set N of n bidders• Preferences are on subsets S µ M – bundles – of items • Valuation function vi: 2M R

– vi(S) - value bidder i receives on bundle S• Must satisfy: monotonicity: v(s) · v(T) for S µ T• Normalized: v() = 0

Allocation: subsets S1, S2, … Sn where Si Å Sj = for i≠jSocial welfare of allocation: i=1

n vi(Si)

Combinatorial Auctions: Example

Allocation: subsets S1, S2, … Sn where Si Å Sj = for i≠j

Social welfare of allocation: i=1n vi(Si)

items {1,…, m}indivisible item players

Interested in any one of red subsets

Issues

• Computational Complexity of finding the allocation maximizing the social welfare

• Representation and communication of the valuation functions vi – May be large since defined for every subset

• How to analyze the strategy of the bidder in light of the obstacles

Today

• Complexity of Approximation– Tight bounds for the single minded bidder

• Connection with communication complexity• Demand queries and the relaxed LP formulation

Single Minded Auctions

A valuation v is single minded if there is a bundle of items S* and value v* 2 R such that – v(S) = v* if S* µ S – v(S) = 0 for all other S

• Very simple to represent: (Si*, vi* )

Allocation problem for single minded bidders:• Given {(Si*, vi* )}i for bidders i=1..n • Find a subset W of winning bids such that Si* Å Sj* =

with maximum social welfare j 2 W vj*

Single Minded Auctions Example

Allocation problem for single minded bidders:• Given {(Si*, vi* )}i for bidders i=1..n • Find a subset W of winning bids such that Si* Å Sj* =

with maximum social welfare j 2 W vj*Example: communication links in a tree • bidders want a path between some two nodes in the tree

– Since there is only one path between any pair of nodes, the bidders are single-minded

Bidder i wants to connect si to ti– Items = edges of tree– Si* = {set of edges connecting (si, ti) – vi* = 1

Complexity of Single Minded Auctions

Allocation problem for single minded bidders:• Given {(Si*, vi* )}I for bidders i=1..n • Find a subset W of winning bids such that Si* Å Sj* = with

maximum social welfare j 2 W vj*Claim: Allocation problem for single minded bidders is NP-hardProof: Reduction from imaximum independent set. For a Graph G=(V,E):

– each node bidder – each edge item – Si* = {e 2 E|I 2 e} and vi*=1.Winning set W must correspond to an independent setWelfare of W is its size |W|Size of max W is exactly size of max independent set!

Size nSize m

Computationally easy casesWhen each Si is of size 2• Allocation problem corresponds to maximum weighted

matching in the corresponding graph– Item node– Bidder edge, vi* is the weightCan be solved in polytime

When the items are on a line. Each Si is a contiguous segment

Finding max weight independent set in in interval graphs possible in poly time

vi*

vi*

item

Approximation of Single Minded Auctions

Allocation problem for single minded bidders:• Given {(Si*, vi* )}I for bidders i=1..n • Find a subset W of winning bids such that Si* Å Sj* =

with maximum social welfare j 2 W vj*Size of max W is exactly size of max independent set

What about approximating the best allocation?W is a c-approximation: if for any other allocation W’

j 2 W’ vj* / j 2 W vj* · cLimited by the approximation ratio of independent set:

Cannot be better than n1- [Hastad]In terms of the number of items: cannot be better than m1/2-

Incentive Compatible Approximation Mechanism

Want to satisfy both incentive compatibility and computational efficiency

Lemma: a mechanism for single-minded bidders where losers pay 0 is incentive compatible iff

• Monotonicity: a bidder who wins with (vi* , Si*) also wins with (vi’, Si’) for any vi’ ¸ vi* and Si’ µ Si*

• Critical payment: a bidder who wins with pays the minimum value needed for winning: infimum of all vi’, where (vi’, Si*) wins.

Incentive Compatible Approximation Mechanism

Want to satisfy both incentive compatibility and computational efficiency

Given {(Si*, vi* )}I for bidders i=1..n • Rank bidders by vi* / √ |Si*|:

– v1* / √ |S1*| ¸ v2* / √ |S2*| ¸ …• Run a greedy algorithm, starting from large to small

– Add to W if not adjacent to any current member of W• Allocation: the set W• Payments: for i 2 W: pi = vj*/ √ (|Sj*| / |Si*|) where j

is smallest index that – Si* Å Sj* ≠ – j wins without i

and 0 if none exists

The value that would have made j appear before i

Approximation Ratio of the MechanismTheorem: Let OPT be allocation maximizing j 2 OPT vj* and W the

output of the greedy algorithm. Thenj 2 OPT vj* · √m j 2 W vj*

Proof:For each i 2 W let OPTi={j 2 OPT, j ¸ i| Si*Å Sj* ≠ }

The set of elements that did not enter W “because” of iWe know that OPT µ [i 2 W OPTi

Will show: j 2 OPTi vj* · √m vi* For all j 2 OPTi we know that vj*· vi*/ √ (|Sj*|/|Si*|)

j 2 OPTi vj* · vi*/ √ |Si*| j 2 OPTi √ |Sj*|

Approximation Ratio of the MechanismWill show: j 2 OPTi vj* · √m vi* For all j 2 OPTi we know that vj*· vi*/ √ (|Sj*|/|Si*|)

j 2 OPTi vj* · vi*/ √ |Si*| j 2 OPTi √ |Sj*| By Caushy-Schwarz:

j 2 OPTi √ |Sj*| · √ |OPTi | √ j 2 OPTi |Sj*| For j 2 OPTi : Si*Å Sj* ≠

Since OPT is an allocation:– these intersections are disjoint and |OPTi | · |Si*| j 2 OPTi |Sj*| · mj 2 OPTi √ |Sj*| · √|Si*| √m

Today

• Complexity of Approximation– Tight bounds for the single minded bidder

• Connection with communication complexity• Demand queries and the relaxed LP formulation

Communication Complexity and Bidding• Two (or more) parties• Each party i can compute valuation function

vi: 2M R by an oracle call – No succinct description

• Want to find optimal allocation by sending as few bits to each other.

• Suppose vi: 2M {0,1}Claim: need to exchange an exponential number of

bits to find optimal allocation

Communication Complexity

Alicex2X

Boby2Y

Let f:X x Y Z

Input is split between two participantsWant to compute outcome: z=f(x,y)

while exchanging as few bits as possible

A protocol is defined by the communication tree

z0 z1 z2 z3 z4 z5 z6 z7 ...

Alice: 0Bob: 1Alice: 0Bob: 1

z5

A ProtocolA protocol P over domain X x Y with range Z is a binary tree

where– Each internal node v is labeled with either

• av:X {0,1} or• bv:Y {0,1}

– Each leaf is labeled with an element z 2 Z• The value of protocol P on input (x,y) is the label of the leaf

reached by starting from the root and walking down the tree. • At each internal node labeled av walk

– left if av(x)=0 – right if av(x)=1

• At each internal node labeled bv walk– left if bv(y)=0 – right if bv(y)=1

– The cost of protocol P on input (x,y) is the length of the path taken on input (x,y)

– The cost of protocol P is the maximum path length

Motivation for studying communication complexity

• Originally for studying VLSI questions• Connection with Turing Machines• Data structures and the cell probe model• Boolean circuit depth• Combinatorial Auctions• …

Communication Complexity of a function

• For a function f:X x Y Z the (deterministic) communication complexity of f (D(f)) is the minimum cost of protocol P over all protocols that compute f

Observation: For any function f:X x Y Z D(f) ≤ log |X| + log |Z|

Example: let x,y µ {1,…,n} and let f(x,y)=max{x [ y} Then D(f) · 2 log n

Combinatorial Rectangles• A combinatorial rectangle in X x Y is a subset R µ X x Y such that R=

A x B for some A µ X and B µ Y Proposition: R µ X x Y is a combinatorial rectangle iff (x1,y1) 2 R and

(x2,y2) 2 R implies that (x1,y2) 2 R

For Protocol P and node v let Rv be the set of inputs (x,y) reaching v

Claim: For any protocol P and node v the set Rv is a combinatorial rectangle

Claim: given the transcript of an exchange between Alice an Bob (but not x and y) possible to determine z=f(x,y)

Fooling Sets• For f:X x Y Z a subset R µ X x Y is f-monochromatic if f is fixed on

R• Observation: any protocol P induces a partition of X x Y into f-

monochromatic rectangles. – The number of rectangles is the number of leaves in P

• A set Sµ X x Y is a fooling set for f if there exists a z 2 Z where– For every (x,y) 2 S, f(x,y)=z – For every distinct (x1,y1), (x2,y2) 2 S either

• f(x1,y2)≠z or• f(x2,y1)≠z

Property: no two elements of a fooling set S can be in the same monochromatic rectangle

Lemma: if f has a fooling set of size t, then D(f) ≥ log2 t

ExamplesEquality: Alice and Bob each hold x,y 2 {0,1}n

– want to decide whether x=y or not.• Fooling set for Equality

S={(w,w)|w 2 {0,1}n } Conclusion: D(Equality) ¸ n

Disjointness: let x,y µ {1,…,n} and let– DISJ(x,y)=1 if |x y|¸ 1 and – DISJ(x,y)=0 otherwise

• Fooling set for Disjointness S={(A, comp(A))|A µ {1,…,n} }

Conclusion: D(DISJ) ¸ n

Bidding as a function

• In bidding the parties are interested in finding a good allocation

• Can be viewed as computing a relation• Can view the

Communication Complexity and BiddingTheorem: every protocol that find the optimal allocation for every

pair of {0,1} valuation v1,v2 must use {m choose m/2} ¼ 2m/√ (/2 m)

bits.Proof: by a fooling set argument• For every valuation v define “dual” v*: v*(S)=1-v(Sc)

– dual” v* is monotoneClaim: Let v ≠ u: then the sequence of bits transmitted on

(v,v*) cannot be the same as that transmitted on (u,u*)Claim Theorem: there are at least {m choose m/2}

different valuations

Number of 1’s

m/2 ‘0’

‘1’

• Claim: Let v ≠ u: then the sequence of bits transmitted on (v,v*) cannot be the same as that transmitted on (u,u*)

Proof: if v ≠ u but same transcript on (v,v*) and (u,u*): then same transcript on (v,u*) and (u,v*).Same allocation, {S, Sc}, is produced in all 4 cases

For some T we have that v(T) ≠ u(T). WLOG: v(T) =1 and u(T)=0, u*(Tc )=1. Hence: v(T) + u*(Tc )=2. Therefore v(S) + u*(Sc ) ¸ 2.However: v(S) + v*(Sc) + u(S) + u*(Sc )=2

We get that u(S) + v*(Sc)· 0 which is suboptimal

Allocation produced•Should be optimal

Approximation?

• Approximating the social welfare with a factor strictly smaller than min{n,m1/2-} requires exponential communication

Today

• Complexity of Approximation– Tight bounds for the single minded bidder

• Connection with communication complexity• Demand queries and the relaxed LP formulation

Demand Queries

Demand query: auctioneer presents a vector of item prices p1, p2, …, pn

the bidder reports a demand bundle for the prices: a set S maximizing

v(S) +j 2 S pi

Value query: auctioneer presents a bundle S, the bidder reports his value v(S) for this bundle.

v(S) never explicitly represented!

Linear Programming Formulation

Linear Programming Relaxation: • Variable xi,S for each bidder i 2 N and subset SMax i,S vi(S) xi,S s.t.

i,S s.t. j 2 S xi,S · 1 for all j 2 MS µ M xi,S · 1 for all i 2 N xi,S ¸ 0 for all S µ M, i 2 N

The integer Program: xi,S 2{0,1}xi,S =1 iff i receives bundle S

Exponential number of variables

For single minded

bidder: just xi,S*

Dual Linear Programming Relaxation

• Variable ui for each bidder i 2 N and pj for each item j 2 M

Minimize i 2 N ui + j 2 M pi s.t.

ui + j 2 S pj ¸ vi(S) for all S µ M, i 2 Nui ¸ 0, pj ¸ 0 for all i 2 N, j2 M

Exponential number of constraints

Interpretation:

pj: price of item jUi: demand of user i

Demand Queries

Demand query: auctioneer presents a vector of item prices p1, p2, …, pn

the bidder reports a demand bundle for the prices: a set S maximizing

v(S) +j 2 S pi

Theorem: Linear Programming Relaxation can be solved in poly time (in n,m and number of bits of precision) using only demand queries with item prices.

v(S) never explicitly represented!

Back to the RLP• Use the dual problem (n+m variables, exp # constraints)

• Ellipsoid Method:– Uses Separation oracle for : when given a candidate solution, either

confirms that it is feasible or respond with a violated constraint • Given (u,p) need to check ui ¸ vi(S) -j 2 S pj

– query user i with demand query p – response Di – See whether ui + j 2 Di pj ¸ vi(D_i)

ui + j 2 S pj ¸ vi(S) for S µ M, i 2 Nui ¸ 0, pj ¸ 0 for i 2 N, j2 M

Value query

From the Dual to the RLP• The Ellipsoid Algorithm makes a polynomial number of calls to the

separation oracle• Each time a constraint is returned: poly # altogether• Remove all other constraints and obtain a “reduced” dual LP

– Ellipsoid Algorithm still returns the same result

• Use the dual of the reduced dual problem to get a solution to the original primal – With a polynomial number of variable

• It is also a solution to the original (with 0’s on all “non”-variables)What to do with the relaxed solution?

Many approximation algorithms use it:Randomized rounding is the obvious choice

ui + j 2 S pj ¸ vi(S) for S µ M, i 2 Nui ¸ 0, pj ¸ 0 for i 2 N, j2 M

Encountered in Separation