Approximate Revenue Maximization
with Multiple
Items
Sergiu Hart
Noam Nisan
April 2012
Presented by: Nir Shabat
The Problem
• How can a seller maximize its revenue when
selling multiple items to a single buyer?
Single Item
• 𝑥 – Buyer’s item valuation
• 𝑥 is unknown, but we know that 𝑥~𝐹
• Selling at price 𝑝
– Buyer buys at probability 1 − 𝐹 𝑝
– Revenue is 𝑝 ⋅ 1 − 𝐹 𝑝
• Seller chooses price 𝑝∗ that maximize this
expression
Single Item
• Can other mechanisms yield higher revenue?
– Indirect, different prices for different probabilities,
other?
• No… Take-it-or-leave-it at price 𝑝∗ yields
optimal revenue among all mechanisms.
Myerson [1981]
Two Items
•
•
•
•
𝑥, 𝑦 - Buyer’s valuation for item 1 and 2, resp.
𝑥, 𝑦 i.i.d. according to 𝐹
Valuation for both items is 𝑥 + 𝑦 (additive)
Is selling each single item optimally
(separately) is the optimal way for selling both
of them?
• Turns out it isn’t …
Example
• Distribution takes values 1 and 2, each with
prob. ½
• Selling a single item optimally:
– At price 1 (selling always) – revenue is 1
1
2
– At price 2 – revenue is ⋅ 2 = 𝟏
Revenue from both items is 2
Example (Cont.)
• Sell both as a bundle at price 3:
– Buyer buys with probability of
3
4
3
4
Revenue is 3 ⋅ = 𝟐. 𝟐𝟓
More than selling them separately…
Selling Separately vs. Bundling
• Is bundling always better than selling
separately?
• No.
• Example:
– Distribution taking values 0 and 1, each with
probability ½
– Max. revenue with bundle is ¾
– Can sell each item for revenue of ½
In this case selling separately is better…
Other Mechanisms
• Sometimes neither selling separately nor
bundling is optimal.
• Example:
• Distribution taking values 0, 1 and 2 each with
prob. 1/3
– Selling separately or as bundle – revenue is 𝟒/𝟑
– Buyer can buy each single item at price 2, or both
at price 3 – revenue is 𝟏𝟑/𝟗 (better)
Other Mechanisms (cont.)
• Similar case for uniform distribution on
[0,1] (Manelli and Vincent [2006])
• Sometimes not even deterministic (Hart and
Reny [2011]):
– Distribution takes value 1, 2 and 4, with
probabilities 1/6, 1/2, 1/3 resp.
– Buyer chooses between buying one item with
prob. ½ for a price of 1, and buying both items for
a price of 4 - Optimal
Result 1
• For every one-dimensional distributions 𝐹1
and 𝐹2 :
𝑅𝐸𝑉1 𝐹1 + 𝑅𝐸𝑉1 𝐹2
1
≥ ⋅ 𝑅𝐸𝑉2 𝐹1 × 𝐹2
2
Also generalizes for multi-dimensional
distributions.
Result 1 (cont.)
• Generalizes for multiple buyers:
1
𝑛
𝑛
𝑅𝐸𝑉
𝑋 + 𝑅𝐸𝑉
𝑌 ≥ ⋅ 𝑅𝐸𝑉
2
𝑛
𝑋, 𝑌
Where 𝑋 = 𝑋1 , … , 𝑋 𝑛 ∈ 𝑅 𝑛 , Y =
𝑌1 , … , 𝑌 𝑛 ∈ 𝑅 𝑛 are the values of the first item
and the second item to the n buyers, resp.
Result 2
• For equal distributions we get a tighter bound
)𝐹1 = 𝐹2 = 𝐹):
𝑒
𝑅𝐸𝑉1 𝐹 + 𝑅𝐸𝑉1 𝐹 ≥
⋅ 𝑅𝐸𝑉2 𝐹1 × 𝐹2
𝑒+1
•
𝑒
~0.73
𝑒+1
• Conjecture – 0.78 is the tight bound
Result 3
• There exists a constant 𝑐 > 0 s.t. for every
integer 𝑘 ≥ 2, and every one-dimensional
distributions 𝐹1 , … , 𝐹𝑘 :
𝑅𝐸𝑉1 𝐹1 + ⋯ + 𝑅𝐸𝑉1 𝐹𝑘
𝑐
≥
⋅ 𝑅𝐸𝑉𝑘 𝐹1 × ⋯ × 𝐹𝑘
2
log 𝑘
Result 4
• There exists a constant 𝑐 > 0 s.t. for every
integer 𝑘 ≥ 2, and every one-dimensional
distribution 𝐹:
𝑐
𝑅𝐸𝑉1 𝐹 ∗ ⋯ ∗ 𝐹 ≥
⋅ 𝑅𝐸𝑉𝑘 𝐹 × ⋯ × 𝐹
log 𝑘
𝑘
𝑘
Open Problems
• Characterizing optimal auctions:
– When is selling separately optimal?
– When is bundling optimal?
– When are deterministic auctions optimal?
Open Problems
• Upper/Lower bound gaps:
– 2 items, i.i.d. F:
∀𝐹
𝑆𝑅𝐸𝑉 𝐹 × 𝐹
≤ 0.73 …
𝑅𝐸𝑉 𝐹 × 𝐹
∃𝐹
𝑆𝑅𝐸𝑉 𝐹 × 𝐹
≤ 0.78 …
𝑅𝐸𝑉 𝐹 × 𝐹
Open Problems
• Upper/Lower bound gaps:
– 𝑘 items, i.i.d. 𝐹:
∀𝐹
∃𝐹
𝐵𝑅𝐸𝑉 𝐹 ×𝑘
1
≥
×𝑘
𝑆𝑅𝐸𝑉 𝐹
4
𝐵𝑅𝐸𝑉 𝐹 ×𝑘
≤ 0.57
×𝑘
𝑆𝑅𝐸𝑉 𝐹
For all large enough 𝑘
Open Problems
• Upper/Lower bound gaps:
– k items, i.i.d. F:
𝐵𝑅𝐸𝑉 𝐹 × 𝐹
2
𝑒
∀𝐹
≥ ⋅
𝑆𝑅𝐸𝑉 𝐹 × 𝐹
3 𝑒+1
∃𝐹
𝐵𝑅𝐸𝑉 𝐹 × 𝐹
2
≤
𝑆𝑅𝐸𝑉 𝐹 × 𝐹
3
Some more gaps …