1. Approximation algorithms for combinatorial allocation problems
Uriel Feige (Microsoft Research)
Jan Vondrák (Princeton University)

2. Maximizing social welfare
m items
n players
Distribute items among players, to maximize:
where
utility of set S to player i.

3. Classes of utility functions
We always assume: monotonicity
normalization
Additional conditions, weaker to stronger:
Subadditivity: …”no complementarity”
XOS: … maximum of linear functions
Submodularity:
… decreasing marginal values
These conditions prevent “complementarity”,
i.e. items increasing each other’s utility.

4. Models of accessing utility functions
Value queries: For a given set S, we can ask what is wi(S) = ?
Demand queries: For given prices pj, we can ask what is the set S
maximizing wi(S) – p(S) ?
Example:
w(S) = min{2|S|,4} $2 $1 $5 $3 $1 $1
w(S)-p(S) = 2
Note: It might be NP-hard to answer demand queries, even for wi(S) submodular.
When demand queries are considered, we assume that wi(S) are such that
this is not the case, or the players have sufficient power to answer such queries.

5. Related problem The Generalized Assignment Problem (GAP):
Utility functions are linear:
Each item j has value vij and size sij for player i.
The set allocated to player i must satisfy a capacity constraint
Goal: maximize

6. Previous work ½-approximation, optimal unless P=NP Max SubAdd Welfare
[Feige ’06]
Max SubAdd Welfare
All results with the assumption
of a demand oracle
except for GAP.
(1-1/e)-approximation,
optimal unless P=NP
[Dobzinski, Nisan, Shapira ’05]
[Feige ’06]
Max XOS Welfare
Max SubMod Welfare
GAP
(1-1/e)-approximation
[Dobzinski, Shapira ’06]
(1-1/e)-approximation
[Fleischer, Goemans, Mirrokni, Sviridenko ’06]

7. The New Results
There is a (1-1/e+ε)-approximation algorithm for Maximum
Submodular Welfare (with a demand oracle).
There is a (13/17)-approximation algorithm for Maximum
Submodular Welfare with 2 players.
There exists ε>0 such that it is NP-hard to approximate
Maximum Submodular Welfare within 1- ε, even with a demand
oracle (and even for two players).
There is a (1-1/e+ε)-approximation algorithm for GAP.

8. How is 1-1/e achieved?
Exponential size; however, can be solved in many cases
(bounded-size utility functions, GAP [FGMS])
in general, using the demand oracle.
There is an optimal solution of polynomial size.

9. (1-1/e)-approximation
in a nutshell
Each player: probability distribution xi,S => sample a random set Si
E[wi(Si)] = share of player i in the LP S6 S9 S3 S4 S1 S8 S5 S2 S7
S10
Issues: (1) some items in multiple sets => resolve conflicts.
(2) some items are not allocated at all.
Pr[item not allocated] = (1-p1) (1-p2) (1-p3)…(1-pn) < 1/e.

10. Fair Contention Resolution
Problem: Players compete for an item, independently
with probabilities p1,p2,p3,…,pn.
Random subset A of them show up.
How do you decide who gets it, so that
players are rewarded proportionally to pi ?
Fair Contention Resolution technique:
If A is empty, nothing.
If A = {k}, player k gets the item.
If |A|>1, then give it to each competing player k with this probability.
Lemma: Each player, conditioned on competing for the item,
receives it with the same probability.

11. Improvement for fixed n
Every player receives the item with the same conditional probability
=> this probability must be equal to
Since , we have
this technique achieves a factor of 1–(1-1/n)n > 1–1/e
(e.g., 3/4 for two players).
But: we want to achieve an improvement independent of n.

12. How to improve 1–1/e ?? First idea: it should be trivial.
The reason why we get 1-1/e … 1/e-fraction of items unused.
Distribute them among players.
How much profit does it yield?
Sometimes none.

13. Example n players, nn items arranged in a hypercube
Utility of player i: projection onto dim. i
Fractional solution: xi,H(i,j)=1/n
H(i,j) = {x ∈[n]n: xi=j}
LP = nn
But:
Sampling one set for each player covers nn – (n-1)n items.
This is 1 – (1-1/n)n of the optimum.
We cannot improve this by distributing the leftovers, in any way.

14. Same example for 2 players
Player 1 wants one from each row.
Player 2 wants one from each column. π1
w1(S) = |π1(S)|, w2(S) = |π2(S)|
If anybody blocks a row/column,
We can get only 3 out of 4. π2
Solution
Player 1 values each column at $30.
Player 2 values each row at $30.

15. Improvement over ¾ for two players?
Let player 1 sample S, player 2 sample T.
Define pj=Pr[j ∈ S], qj=Pr[j∈T].
2. We can assume pj=qj=1/2, otherwise
Fair Contention Resolution gives ρ>3/4. S T
3. Let player 1 go first and take S, player 2 takes the rest:
=> player 2 gets at least 1/2 of her desired value.
4. Similarly let player 2 go first, etc.
=> on the average, each player gets 3/4 of their share.
Is it possible that the complement any good set for player 1
is worth only 1/2 of player 2’s good set ??

16. Improvement over ¾ for two players
We can assume:
Define new random sets: let player 1 sample two sets S,S’
let player 2 sample two sets T,T’ S T S’ T’ S’
~S’ T ~T
=> Y is a good set for player 1:

17. Improvement over ¾ for two players, cont.
We can assume: E[w1(Y)] = E[w1(S)], E[w2(Z)] = E[w2(T)].
Trick of construction: Y and Z are not independent.
By resolving conflicts between Y and Z, we lose only 3/16
rather than 1/4. We retrieve 13/16 LP in this case.

18. Improvement for n players, general case
If the fractional solution is unbalanced (there are dominant players
for many items) then use the Fair Allocation Procedure.
If the fractional solution is balanced, divide players into two groups;
use the ideas for 2 players to treat the 2 groups.
Sample 2 sets Si , Si’ for each player i∈A; Ti , Ti’ for i∈B.
Let
Resolve contention in each group separately.
As in the case of two players, define:

19. Choose randomly one of four allocation schemes:
probability
players in A
players in B
Lemma: For a balanced solution, this achieves at least LP.

20. Improvement for GAP
Consider a fractional solution xi,S and the standard randomized
rounding algorithm. 1/e-fraction of items still available.
Remaining items always bring value here, but maybe we cannot
pack them due to capacity constraints.
We can assume that
(1) the fractional solution is balanced
(2) item values are uniform
(3) all sets S allocated to player i have roughly similar values
(Otherwise previous algorithms gain compared to 1-1/e.)
In this case, we prove that there is enough value in migrant items.
Migrant items have the property that with some probability, each
player can pack migrant items from two sets instead of one.

21. Summary of results Allocation Problem Approximation factor Integrality
gap
Hardness of
Maximum
XOS
Welfare
1 – 1/e
Max 2-player
3/4
Submodular
13/17
5/6
1 – ε1
1 – 1/e
0.782
1 – ε2
Generalized
Assignment
1 – 1/e
4/5
1 – ε3