On Revenue Optimal Combinatorial Auctions - IFP Group at the ...

On Revenue Optimal Combinatorial Auctions:
Structure and Efficiency
Vineet Abhishek
Department of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
CSL Communication Seminar
August 30, 2010
Joint work with Bruce Hajek
Thanks to Steven Williams for many helpful suggestions

Example - Wireless Spectrum Auctions
FCC uses auctions to sell rights to use wireless spectrum.
Companies such as AT&T, Verizon, etc, bid for spectrum licenses.
Each bundle of licenses has a value for AT&T, Verizon, etc.
A bundle of licenses might be more valuable than the individual
ones.
Objective - maximize FCC’s revenue from spectrum sale.

What Makes It Hard?
Private values:
The exact value of a bundle of licenses to AT&T is known only to
AT&T.
FCC can only have a rough estimate of it.
Strategic behavior:
AT&T wants to maximize its own profit. Different from FCC’s
objective.
AT&T can misreport its value of a bundle of licenses.

Combinatorial Auctions (CAs)
Provides a common framework for such problems:
Seller - FCC.
Buyers - AT&T, Verizon, etc.
Items - spectrum licenses.
Buyers can compete for any bundle of items.
Allocation and payments based on the competition.
Design a CA to maximize revenue.

Our Contributions
Most of the literature on CAs is on efficient auctions.
Maximizing the total value generated through the allocation.
Theoretical results on revenue optimal auctions apply only under
simple settings.
Our contributions:
Quantifying how different a revenue optimal auction is from an
efficient auction.
Characterizing revenue optimal auction for a special class of CAs
problems.

Outline
1 Review of revenue optimal auctions
2 Efficiency loss in a revenue optimal auctions
3 Revenue maximization for a special class of CAs

Outline
1 Review of revenue optimal auctions
2 Efficiency loss in a revenue optimal auctions
3 Revenue maximization for a special class of CAs

Model
N buyers, multiple items.
The bundles desired by the buyers are publicly known.
Each desired bundle has the same value for a buyer.
A buyer is a winner if he gets any one of his desired bundles.

Model
A Bayesian framework
The value of a buyer n:
A realization of a discrete random variable X n .
X n ∈ {xn 1 , xn 2 , . . . , xn
Kn
}, where 0 ≤ xn 1 < xn 2 < . . . < xn Kn
.
p i n P(X n = x i n), assume p i n > 0.
One-dimensional private information:
Beliefs:
The exact realization of X n is known only to buyer n.
X n ’s are independent across the buyers.
The probability distributions of X n ’s are common knowledge.

Model
Feasible allocations
Combinatorial constraints restrict the set of possible winners, e.g.,
Single item auction - at most one buyer can win.
Auction of S identical items - subsets of buyers of size at most S.
Each buyer wants a specific bundle of items - subsets of buyers
with disjoint bundles.
A collection of all possible sets of winners.
Assume that A is downward closed:
If A ∈ A and B ⊆ A, then B ∈ A.

The Components of an Auction
.Bid
vector
v
v 1
. .
Allocation rule
p(v)
Payment rule
M(v)
Winners
Buyer 1’s
payment py
. . .
v N
Auction
Buyer N’s
payment
π(v) = a probability distribution over A, given v.
π A (v) = P(the set of buyers A ∈ A win simultaneously, given v).
The payoff of a buyer = value of the allocation - payment made.

The High Level Idea
Design a game for revenue maximization:
An auction induces a game of incomplete information among the
buyers.
Maximize the expected revenue at its Bayes-Nash equilibrium
(BNE).
Reduce it to a constrained optimization problem:
Truth-telling as a BNE - incentives to be truthful.
This is with no loss of optimality by the revelation principle.
Voluntary participation - nonnegative payoff from participation.

The Constraints
First, focus on one buyer:
q(x i ) conditional probability of winning for the buyer for bid x i .
m(x i ) expected payment made by the buyer for bid x i .
Incentives to be truthful:
q(x i )x i − m(x i ) ≥ q(x k )x i − m(x k ).
Nonnegative payoff from participation:
q(x i )x i − m(x i ) ≥ 0.

Geometric Interpretation
Let A 0 (0, 0) and A i (q(x i ), m(x i )).
A k must lie above or on the line through A i with slope x i .
m(.)
slope = x 3
A 2
slope = x2
A 3 slope = x 1
A 1
A 0 q(x 1 ) q(x 3 ) q(x 2 ) q(.)
x 1 < x 2 < x 3
Non-monotone q(.) : no m(.) can meet the constraints!

Geometric Interpretation
Let A 0 (0, 0) and A i (q(x i ), m(x i )).
A k must lie above or on the line through A i with slope x i .
m(.)
slope = x 3
A 2
slope = x2
A 3 slope = x 1
A 1
A 0 q(x 1 ) q(x 3 ) q(x 2 ) q(.)
x 1 < x 2 < x 3
Non-monotone q(.) : no m(.) can meet the constraints!

Geometric Interpretation
Let A 0 (0, 0) and A i (q(x i ), m(x i )).
A k must lie above or on the line through A i with slope x i .
m(.)
A
3
slope = x 3
slope = x 2
A 2 slope = x 1
A 1
A 0 q(x 1 ) q(x 2 ) q(x 3 ) q(.)
x 1 < x 2 < x 3
Monotone q(.) : a suitable m(.) can meet the constraints.

Simplifying the Constraints
A payment rule satisfying the constraints exists if and only if
q(x 1 ) ≤ q(x 2 ) ≤ . . . ≤ q(x K ).
The maximum m(.) for a given q(.) is given by:
m(x i ) =
i∑
k=1
[
]
q(x k ) − q(x k−1 ) x k .
For the above m(.), E [m(X)] = E [q(X)w(X)].
w(.) is the virtual-valuation function of the buyer.
It suffices to find only an optimal allocation rule.

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
−x 2
−x 3
−x 4

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
−x 2
−x 3
−x 4

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
−x 2
−x 3
−x 4

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
−x 2
−x 3
−x 4

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
−x 2
−x 3
−x 4

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
w( x
1 )
−x 2
−x 3
slopes are virtual valuations, w
−x 4

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
w( x
1 )
−x 2
−x 3
slopes are virtual valuations, w
−x 4

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
w( x
1 )
−x 2 w( x
2 )
−x 3
slopes are virtual valuations, w
−x 4

Graphical Construction of Virtual Valuations
Let X ∈ {x 1 , x 2 , x 3 , x 4 } w.p. {p 1 , p 2 , p 3 , p 4 }.
(0,0) 0)
p 1 p 2 p 3 p 4 (1,0)
−x 1
w( x
1 )
−x 2 w( x
2 )
w( x
3 )
w(x( 4 )
−x 3
slopes are virtual valuations, w
−x 4

The Optimal Auction Problem
Extending the analysis to N buyers:
[ ∑N
]
Under truth-telling, the expected revenue = E
n=1 M n(X) ,
where the random vector X {X 1 , X 2 , . . . , X N }.
The Optimal Auction Problem (OAP)
maximize
π,M
[ N
]
∑
E M n (X) ,
n=1
subject to: truth-telling and voluntary participation.

Simplifying OAP
Revenue(π) maximum revenue to the seller for a given π,
[ N
]
∑
= E q n (X n )w n (X n ) (from single buyer analysis),
n=1
[ ∑
= E π A (X) ( ∑
w n (X n ) )] (by relating q n ’s to π).
A∈A
n∈A

Simplifying OAP
Revenue(π) maximum revenue to the seller for a given π,
[ N
]
∑
= E q n (X n )w n (X n ) (from single buyer analysis),
n=1
[ ∑
= E π A (X) ( ∑
w n (X n ) )] (by relating q n ’s to π).
A∈A
n∈A
Suggests allocating from argmax
A∈A
[ ∑
w n (v n ) ] , for a bid vector v.
n∈A
Works if w n (x 1 n ) ≤ w n (x 2 n ) ≤ . . . ≤ w n (x Kn
n ).
Otherwise, the resulting q n ’s need not be monotone.

Simplifying OAP
Replace w n ’s by monotone virtual valuations (MVVs), w n .
p 1 p 2 p 3 p 4 (1,0)
(0,0)
w( x
1 )
−x 1
w( x
2 )
−x 2 _
w( x
1 )
_
4
4
_
w ( x ) = w(
x )
w( x
2 )
_
3
3
w ( x ) = w(
x )
−x 3
slopes are virtual valuations, w _
slopes are monotone virtual valuations, w
−x 4

The Maximum Weight Algorithm (MWA)
Given a bid vector v:
1 Compute w n (v n ) for each n.
2 π(v) = a probability distribution on argmax
A∈A
[ ∑
w n (v n ) ] .
n∈A
3 Collect payments given by:
M n (v) =
∑ [
Qn (xn, i v −n ) − Q n (xn i−1 , v −n ) ] xn,
i
i:xn≤v i n
where Q n (v) = P(buyer n wins, given the bid vector v).

Reserve Prices
No buyer n with w n (v n ) < 0 is a winner under MWA.
Follows since A is downward-closed and ∅ ∈ A.
Equivalently, the seller sets a reserve price for each buyer n.
Follows since MVVs are nondecresing functions.
The reserve price for buyer n = minimum v n such that w n (v n ) > 0.

Outline
1 Review of revenue optimal auctions
2 Efficiency loss in a revenue optimal auctions
3 Revenue maximization for a special class of CAs

Efficient Auctions
An allocation of items among buyers generates value for the
items.
The realized social welfare (RSW) = total value generated
through the allocation of items.
An efficient auction maximizes the realized social welfare.
For a bid vector v, an efficient allocation is a probability
( ∑ )
distribution on argmax v n .
A∈A
n∈A

Some Questions
How different is a revenue optimal auction from an efficient
auction?
What causes these differences?
How to quantify this difference?
What are the underlying parameters?
We answer these.

Revenue vs Efficiency - Example 1
The role of reserve prices:
Buyer 1 Buyer 2
1/3 2/3 1/3 2/3
$1 $2 $1 $2

Revenue vs Efficiency - Example 1
The role of reserve prices:
Buyer 1 Buyer 2
1/3 2/3 1/3 2/3
$1 $2 $1 $2
Efficient auction
Sell to the highest bidder at the second
highest price.
Revenue = 2*(2/3) 2 + 1 ‐ (2/3) 2 @ $1.45.
MSW = 2*( 1 ‐ (1/3) 2 ) + (1/3) 2 @ $1.89.

Revenue vs Efficiency - Example 1
The role of reserve prices:
Buyer 1 Buyer 2
1/3 2/3 1/3 2/3
$1 $2 $1 $2
Efficient auction
Revenue optimal auction
Sell to the highest bidder at the second
Set reserve price = $2, sell to anyone
highest price.
who bids $2.
Revenue = 2*(2/3) 2 + 1 ‐ (2/3) 2 @ $1.45. Revenue = 2*( 1 ‐ (1/3) 2 ) @ $1.77.
MSW = 2*( 1 ‐ (1/3) 2 ) + (1/3) 2 @ $1.89. RSW = 2*( 1 ‐ (1/3) 2 ) @ $1.77.

Revenue vs Efficiency - Example 2
The buyer with the highest valuation need not win:
Buyer 1 Buyer 2
1/2 1/2 1/2 1/2
$5 $10 $1 $2

Revenue vs Efficiency - Example 2
The buyer with the highest valuation need not win:
Buyer 1 Buyer 2
1/2 1/2 1/2 1/2
$5 $10 $1 $2
Efficient auction
Always sell to buyer 1.
Price charged can be at most $5.
MSW = (10 + 5)*0.55 = $7.5.

Revenue vs Efficiency - Example 2
The buyer with the highest valuation need not win:
Buyer 1 Buyer 2
1/2 1/2 1/2 1/2
$5 $10 $1 $2
Efficient auction
Revenue optimal auction
Always sell to buyer 1.
Sell to buyer 1 if v 1 = $10 at $10, else sell
to buyer 2 at $1.
Price charged can be at most $5. Revenue = (10 + 1)*0.5 = $5.5.
MSW = (10 + 5)*0.55 = $7.5. RSW = (10 + 0.5*(2 05*(2+ 1))*0.5 = $5.75.

Revenue vs Efficiency - Example 3
Efficiency loss in multiple items case:
A Buyer 1
B Buyer 2 C Buyer 3
D
1/2 1/2
1/2 1/2 1/2 1/2
$1 $1.6 $1 $1.6 $1 $1.6
Buyers 1 & 2, or Buyer 2 & 3 cannot win simultaneously.

Revenue vs Efficiency - Example 3
Efficiency loss in multiple items case:
A Buyer 1
B Buyer 2 C Buyer 3
D
1/2 1/2
1/2 1/2 1/2 1/2
$1 $1.6 $1 $1.6 $1 $1.6
Buyers 1 & 2, or Buyer 2 & 3 cannot win simultaneously.
w n ($1) = $0.4, w n ($1.6) = $1.6.
For the bid vector ($1, $1.6, $1):
Efficient auction allocates to buyers 1 and 3.
Revenue optimal auction allocates to buyer 2.

Quantifying Efficiency Loss
RSW by an allocation rule π:
[ ∑
RSW(π, X; A) = E π A (X) ( ∑ ) ]
X n .
A∈A n∈A
Maximum social welfare (MSW) (by an efficient allocation):
[
( ∑ ) ]
MSW(X; A) = E X n .
max
A∈A
n∈A
The RSW by an optimal allocation ≤ MSW.

Quantifying Efficiency Loss
For an optimal allocation rule π o , define the efficiency loss ratio
(ELR) as:
ELR(π o , X; A) MSW(X; A) − RSW(πo , X; A)
MSW(X; A)
The Worst Case ELR Problem
maximize ELR(π o , X; A),
X
subject to: K n ≤ K for all n, and (max n xn Kn )/(min n xn 1 ) ≤ r.
Denote the worst case ELR by η(r, K ; A) for r ≥ 1 and K ≥ 1.

Worst Case ELR Computation - Example
Buyer 1
1‐p p 0 < L < H
r = H/L
L H

Worst Case ELR Computation - Example
Buyer 1
1‐p p 0 < L < H
r = H/L
L H
If pr < 1, then w(L) > 0. The optimal allocations is efficient.
If pr ≥ 1, then w(L) ≤ 0. The buyer wins only if he bids H.

Worst Case ELR Computation - Example
Buyer 1
1‐p p 0 < L < H
r = H/L
L H
If pr < 1, then w(L) > 0. The optimal allocations is efficient.
If pr ≥ 1, then w(L) ≤ 0. The buyer wins only if he bids H.
For pr ≥ 1, the RSW = pH, while the MSW = pH + (1 − p)L.
ELR = (1 − p)/(pr + 1 − p).

Worst Case ELR Computation - Example
Buyer 1
1‐p p 0 < L < H
r = H/L
L H
If pr < 1, then w(L) > 0. The optimal allocations is efficient.
If pr ≥ 1, then w(L) ≤ 0. The buyer wins only if he bids H.
For pr ≥ 1, the RSW = pH, while the MSW = pH + (1 − p)L.
ELR = (1 − p)/(pr + 1 − p).
ELR is maximized by setting p = 1/r.
Hence, η = (r − 1)/(2r − 1).

ELR Bound - Binary Valued Buyers
Proposition (Abhishek and Hajek 2010)
Suppose A is downward closed. Given any r > 1, the worst case ELR
for binary valued buyers, denoted by η(r, 2; A), satisfies:
η(r, 2; A) ≤ (r − 1)/(2r − 1).
The worst case ELR for N buyers (not necessarily identically
distributed) is no worse than it is for single buyer!
Holds for arbitrary downward closed A.
The worst case ELR ≤ 1/2 uniformly over all r.

ELR Bound - Single Item Auctions with i.i.d. Buyers
Moving beyond binary valuations but restricting to single item.
Reduction to a relatively simpler optimization problem involving
only the common probability vector of the buyers.
Lower and upper bounds (asymptotically tight as K → ∞) on the
worst case ELR.
The worst case ELR → 0 as N → ∞ at the rate O ( (1 − 1/r) N) .
The worst case ELR → 1 as K → ∞ and r → ∞.

Outline
1 Review of revenue optimal auctions
2 Efficiency loss in a revenue optimal auctions
3 Revenue maximization for a special class of CAs

Limitations of General CAs
Implementation complexity:
Allocation problem is NP-hard in general.
Large communication overhead.
Current state of the art:
General results on revenue maximization known only for
one-dimensional problems.
Joint problem of tractability and characterization.

Single-Minded Buyers
A relatively more tractable class of CAs.
Each buyer is interested only in a specific bundle.
Both the bundle and its value are known only to the buyer.
Any bundle of items not containing his desired bundle has zero
value for him.

Why Study Single-Minded Buyers Model?
Key is to understand and handle complementarity.
A bundle as a whole can be more valuable than the sum of values
of its parts.
Single-minded buyers - an extreme case of complementarity.
Problem is still nontrivial (two dimensional):
A buyer can misreport both his bundle and its value.
No general result on revenue maximization is known even for
this extreme case.
Some real examples where buyers are single minded.

Extending the Model to Single-Minded Buyers
The type of a buyer n = (bundle, value).
Bundles of a buyer n - a realization of a random set B n .
Value of a buyer n - a realization of a discrete random variable X n .
Two-dimensional private information:
The exact realization of (B n , X n ) is known only to buyer n.
Beliefs:
(B n , X n )’s are independent across the buyers.
The joint probability distribution of (B n , X n ) is common knowledge.

Extending the Model to Single-Minded Buyers
The bid vector (b, v) = reported vector of (bundles, values).
The collection of all possible sets of winners, A(b), depends on b.
A ∈ A(b) if the reported bundles of buyers in set A are disjoint.
The allocation rule π and payments M depend jointly on (b, v).
The virtual-valuation function, w n (b n , v n ), is now constructed from
the condition random variable (X n |B n = b n ).

A Candidate for the Revenue Optimal Auction
Given a bid vector (b, v):
Construct a conflict graph G b :
A node n for each buyer n.
An edge e n,m if b n ∩ b m ≠ ∅.
Assign a weight w n (b n , v n ) to node n.
Set of winners = maximum weight independent set of G b .
Winner’s payment = the minimum value he needs to report to win.

Does It Work?
Buyer 1 Buyer 2
1
1/2 1/2
Bundle {A}
Bundle {A} Bundle {A,B}
1
1/2 1/2 9/10 1/10
$1 $2 $4 $2 $4
__ __ __ __ __
W = $1 W = $0 W = $4 W = $16/9 W = $4

Does It Work?
Buyer 1 Buyer 2
1
1/2 1/2
Bundle {A}
Bundle {A} Bundle {A,B}
1
1/2 1/2 9/10 1/10
$1 $2 $4 $2 $4
__ __ __ __ __
W = $1 W = $0 W = $4 W = $16/9 W = $4
For the bid ({A}, $4), buyer 2 gets the bundle {A} for $4.
For the bid ({A, B}, $4), buyer 2 gets the bundle {A, B} for $2.

Does It Work?
Buyer 1 Buyer 2
1
1/2 1/2
Bundle {A}
Bundle {A} Bundle {A,B}
1
1/2 1/2 9/10 1/10
$1 $2 $4 $2 $4
__ __ __ __ __
W = $1 W = $0 W = $4 W = $16/9 W = $4
For the bid ({A}, $4), buyer 2 gets the bundle {A} for $4.
For the bid ({A, B}, $4), buyer 2 gets the bundle {A, B} for $2.
Thus, buyer 2 will misreport type ({A}, $4) as ({A, B}, $4).
A counter example!

What Went Wrong?
An arbitrary joint distribution of (bundles, values) causes
problems.
If w n (b n , v n ) is increasing in b n , buyer n will bid for a larger bundle.
A possible fix - assume B n and X n to be independent.
Works well mathematically, but an unreasonable assumption.
A reasonable assumption - the value of the larger of two bundles
is likely to be higher.
What is the precise condition for optimality?

The Sufficient Condition
Hazard rate order
Let Z 1 , Z 2 be nonnegative random variables. Z 1 ≤ h Z 2 , if
Prob(Z 1 > z|Z 1 > ẑ) ≤ Prob(Z 2 > z|Z 2 > ẑ) for all z ≥ ẑ.
Stronger than the first order stochastic dominance.
The sufficient condition
Let s ⊆ t be two bundles. Then (X n |B n = s) ≤ h (X n |B n = t).
The larger bundle is likely to have higher value.
The condition is intuitive for single-minded buyers.

Main Result
Proposition (Abhishek and Hajek 2010)
If the conditional distribution of X n given B n = s is nondecreasing in s,
in the hazard rate ordering on probability distributions, then
1 w n (s, v n ) ≥ w n (t, v n ) if s ⊆ t,
2 the candidate auction satisfies truth-telling and voluntary
participation constraints,
3 the candidate auction is revenue optimal.

Concluding Remarks
A revenue optimal auction can be very different from an
efficient auction.
Characterizing revenue optimal CA in full generality - an open
problem.
A possible direction - reducing strategy space and imposing some
structure on private information.
The single-minded buyers model - an initial step towards solving
the general CA problems.