The Auction Algorithm
Shahar Paz
Assignment Problem
• Given:
– 𝑛 Persons, 𝑛 Objects
– 𝐴(𝑖) – Objects that may be assigned with person 𝑖
– 𝑎𝑖𝑗 – Benefit of assigning object 𝑗 with person 𝑖
• Target:
– Assign each person 𝑖 with unique object 𝑗𝑖 ∈ 𝐴(𝑖)
– Maximize total benefit 𝑛𝑖=1 𝑎𝑖𝑗𝑖
Assignment Problem
4
2
1
1
1
-2
3
2
Assignment Problem
4
2
Assignment benefit:
2+1–2+3=4
1
1
1
-2
3
2
Assignment vs. Min-Cost-Flow
• Special case of Min-Cost-Flow
• Nodes:
– Persons have 𝑏(𝑖) = 1 (1 supply unit)
– Objects have 𝑏(𝑗) = −1 (1 demand unit)
• Arcs:
– (𝑖, 𝑗) 𝑗 ∈ 𝐴(𝑖)
– 𝑢(𝑖, 𝑗) = 1
– 𝑐 𝑖, 𝑗 = −𝑎𝑖𝑗
(1 capacity unit)
(cost is -benefit)
Assignment vs. Min-Cost-Flow
• Min-Cost-Flow can turn into Assignment
– Arcs => new nodes
b(i) i
c=0
c(i,j)
u(i,j)
b(j) j
i b(i)-u(i,j)
b=u(i,j) ij
c=c(i,j)
j b(j)
Assignment vs. Min-Cost-Flow
• Min-Cost-Flow can turn into Assignment
– Arcs => new nodes
– New nodes split by supply (old arc capacity)
c=0
i b(i)-u(i,k)
b=u(i,j) ij
b=1 ij1
i b(i)-u(i,k)
b=1 ij2
c=c(i,j)
j b(j)-u(j,k)
b=1 ij3
j b(j)-u(j,k)
Assignment vs. Min-Cost-Flow
• Min-Cost-Flow can turn into Assignment
– Arcs => new nodes
– New nodes split by supply (old arc capacity)
– Old nodes split by demand
b=1 ij1
i b(i)-u(i,k)
b=1 ij2
b=1 ij3
j b(j)-u(j,k)
b=1 ij1
i1 b=-1
b=1 ij2
i2 b=-1
b=1 ij3
j b(j)-u(j,k)
Assignment vs. Min-Cost-Flow
• Min-Cost-Flow can turn into Assignment
– Arcs => new nodes
– New nodes split by supply (old arc capacity)
– Old nodes split by demand
– #new nodes
•
𝑖,𝑗 𝑢(𝑖, 𝑗)
– #old nodes
•
𝑖(
𝑗𝑢
𝑖, 𝑗 − 𝑏 𝑖 ) =
= 𝑖,𝑗 𝑢 𝑖, 𝑗
𝑖,𝑗 𝑢
𝑖, 𝑗 −
𝑖 𝑏(𝑖)
Auction Algorithm
• Init
– Empty assignment 𝑆, zero prices 𝑝𝑗
• Iterate until 𝑆 = 𝑛:
– Bidding phase
• Some unassigned persons bid for most desired objects
• Bids must be greater then current objects’ prices
– Assignment phase
• Prices of objects increased to max bid
• Assignment 𝑆 change accordingly
Auction Algorithm
• Bidding phase:
– Select subset 𝐼 of unassigned persons
– For each person 𝑖 ∈ 𝐼:
• 𝑗𝑖 = arg max {𝑎𝑖𝑗 −𝑝𝑗 }
(most desired object)
𝑗∈𝐴(𝑖)
• 𝑣𝑖 = max {𝑎𝑖𝑗 −𝑝𝑗 } = 𝑎𝑖𝑗𝑖 − 𝑝𝑗𝑖
𝑗∈𝐴(𝑖)
• 𝑤𝑖 =
max {𝑎𝑖𝑗 −𝑝𝑗 }
𝑗𝑖 ≠𝑗∈𝐴(𝑖)
(object’s benefit)
(second-desired benefit)
• Set bid of person 𝑖 for object 𝑗𝑖 :
• 𝑏𝑖𝑗𝑖 = 𝑝𝑗𝑖 + 𝑣𝑖 − 𝑤𝑖 + 𝜖 = 𝑎𝑖𝑗𝑖 − 𝑤𝑖 + 𝜖
Auction Termination
• Observations
– Objects receiving bids get assigned
– Assigned objects remain assigned
– Assigned persons might get unassigned
– Only on termination no unassigned persons and
objects
Auction Termination
• Observation
– Each time object 𝑗 receives a bid,
𝑝𝑗 increase by at least 𝜖.
• Conclusion
– Object receives ∞ bids ⇔ get price ∞
• Proof
– 𝑣𝑖 ≥ 𝑤𝑖
– 𝑝𝑗′ = 𝑝𝑗 + 𝑣𝑖 − 𝑤𝑖 + 𝜖 ≥ 𝑝𝑗 + 𝜖
Auction Termination
• Observation
– If person 𝑖 bid ∞ times,
all objects in 𝐴 𝑖 get price ∞
• Proof
– 𝐴∞
𝑗 𝑟𝑒𝑐𝑒𝑖𝑣𝑒 ∞ 𝑏𝑖𝑑𝑠 }
𝑖 = 𝑗 ∈𝐴 𝑖
– As of some iteration:
• Person 𝑖 only bid for objects in 𝐴∞
𝑖
• All objects in 𝐴∞
𝑖 has price ∞
– But 𝑗 ∈ 𝐴 𝑖 − 𝐴∞
𝑖 has higher benefit (const)
Auction Termination
• Proposition:
– Feasible solution ⟺ Auction algorithm terminates
• Proof: ⟹
𝐼 ∞ = 𝑖 ∈ 𝐼 𝑖 𝑏𝑖𝑑 ∞ 𝑡𝑖𝑚𝑒𝑠 }
𝐽∞ = 𝑗 ∈ 𝐽 𝑗 𝑔𝑒𝑡 ∞ 𝑏𝑖𝑑𝑠 }
– As of some iteration:
– All objects in 𝐽∞ are assigned
• Must be assigned with persons from 𝐼 ∞ (price ∞)
– Only persons from 𝐼∞ bid
• Not all persons in 𝐼 ∞ are assigned
– 𝐼∞ > 𝐽∞ , but 𝐴 𝐼∞ ⊆ 𝐽∞
Duality & CS
• Dual problem:
– Minimize
𝑖
max 𝑎𝑖𝑗 − 𝑝𝑗 +
𝑗∈𝐴 𝑖
– Over all price vectors 𝑝𝑗
𝑗 𝑝𝑗
𝑛
𝑗=1
• Complementary Slackness:
– Given assignment
𝑗𝑖 𝑛𝑖=1
and prices 𝑝𝑗
– 𝑎𝑖𝑗𝑖 − 𝑝𝑗𝑖 = max 𝑎𝑖𝑗 − 𝑝𝑗
𝑗∈𝐴 𝑖
𝑛
𝑗=1
Duality & CS
• Observation:
– Any Primal  Any Dual
• Proof:
– 𝑎𝑖𝑗𝑖 − 𝑝𝑗𝑖 ≤ max 𝑎𝑖𝑗 − 𝑝𝑗
𝑗∈𝐴 𝑖
–
𝑖
–
𝑖 𝑎𝑖𝑗𝑖
−
𝑗 𝑝𝑗
–
𝑖 𝑎𝑖𝑗𝑖
≤
𝑖
𝑎𝑖𝑗𝑖 − 𝑝𝑗𝑖 ≤
≤
𝑖
max { 𝑎𝑖𝑗 − 𝑝𝑗 }
𝑗∈𝐴 𝑖
𝑖
max { 𝑎𝑖𝑗 − 𝑝𝑗 }
𝑗∈𝐴 𝑖
max { 𝑎𝑖𝑗 − 𝑝𝑗 } +
𝑗∈𝐴 𝑖
𝑗 𝑝𝑗
Duality & CS
• Observation:
– Any Primal  Any Dual
• Conclusion:
– Primal  Optimal Primal  Optimal Dual  Dual
– Primal = Dual  optimal Primal & Dual
– Primal = Dual  CS
Duality & CS
• Observation:
– CS  Primal = Dual
• Proof:
– 𝑎𝑖𝑗𝑖 − 𝑝𝑗𝑖 = max 𝑎𝑖𝑗 − 𝑝𝑗
𝑗∈𝐴 𝑖
–
𝑖
–
𝑖 𝑎𝑖𝑗𝑖
−
𝑗 𝑝𝑗
–
𝑖 𝑎𝑖𝑗𝑖
=
𝑖
𝑎𝑖𝑗𝑖 − 𝑝𝑗𝑖 =
=
𝑖
max { 𝑎𝑖𝑗 − 𝑝𝑗 }
𝑗∈𝐴 𝑖
𝑖
max { 𝑎𝑖𝑗 − 𝑝𝑗 }
𝑗∈𝐴 𝑖
max { 𝑎𝑖𝑗 − 𝑝𝑗 } +
𝑗∈𝐴 𝑖
𝑗 𝑝𝑗
Auction Optimality
• -CS:
– 𝑎𝑖𝑗𝑖 − 𝑝𝑗𝑖 ≥ max 𝑎𝑖𝑗 − 𝑝𝑗 − 𝜖
𝑗∈𝐴 𝑖
• Observation:
– Auction preserves -CS
• Proof:
– 𝑝𝑗′𝑖 = 𝑎𝑖𝑗𝑖 − 𝑤𝑖 + 𝜖
– 𝑎𝑖𝑗𝑖 − 𝑝𝑗′𝑖 = 𝑤𝑖 − 𝜖 =
≥
max
𝑗𝑖 ≠𝑗∈𝐴 𝑖
max
𝑗𝑖 ≠𝑗∈𝐴 𝑖
𝑎𝑖𝑗 − 𝑝𝑗′ − 𝜖
𝑎𝑖𝑗 − 𝑝𝑗 − 𝜖
Auction Optimality
• Proposition:
– Auction algorithm is 𝑛𝜖-optimal
• Proof:
– 𝑎𝑖𝑗𝑖 − 𝑝𝑗𝑖 ≥ max 𝑎𝑖𝑗 − 𝑝𝑗 − 𝜖
𝑗∈𝐴 𝑖
–
𝑖 𝑎𝑖𝑗𝑖
−
𝑖 𝑝𝑗𝑖
≥
–
𝑖 𝑎𝑖𝑗𝑖
+ 𝑛𝜖 ≥
𝑖
𝑖
max 𝑎𝑖𝑗 − 𝑝𝑗 −
𝑗∈𝐴 𝑖
max 𝑎𝑖𝑗 − 𝑝𝑗 +
𝑗∈𝐴 𝑖
𝑖𝜖
𝑗 𝑝𝑗
– 𝑃𝑟𝑖𝑚𝑎𝑙 + 𝑛𝜖 ≥ 𝐷𝑢𝑎𝑙 ≥ 𝑂𝑝𝑡𝑖𝑚𝑎𝑙 ≥ 𝑃𝑟𝑖𝑚𝑎𝑙
-Scaling
• Choosing 
1
𝑛
– If data is integer, 𝜖 < guarantees optimality
– Small  causes small changes, bad performance
• -Scaling
– Run Auction with large 𝜖 0 and prices 𝑝𝑗0 = 0
• Compute 𝜖 0 -CS prices 𝑝𝑗′0
– Run Auction with
𝜖𝑘
=
𝜖𝑘−1
𝜃
• Compute 𝜖 𝑘 -CS prices 𝑝𝑗′𝑘
and prices 𝑝𝑗𝑘 = 𝑝𝑗′𝑘−1
-Scaling
• Adjustments for -Scaling:
– On each iteration only 1 person bid
– First scaling phase starts with initial prices 0
– Other scaling phases start with previous prices
– On scaling phase 𝑘, use
𝑘
𝑎𝑖𝑗
=
𝑎𝑖𝑗
𝜖𝑘
𝜖 𝑘 as benefits
𝑘
• So 𝑎𝑖𝑗
is integer multiple of 𝜖 𝑘
– Final prices of scaling phase k satisfy
𝑘
𝑘+1
𝜖 𝑘 -CS with 𝑎𝑖𝑗
, and 𝑟𝜖 𝑘+1 -CS with 𝑎𝑖𝑗
.
-Scaling
• Candidate-list of person 𝑖:
– Holds pairs 𝑗 ∈ 𝐴 𝑖 , 𝑝𝑗′ of potential next-bids
• Modified bidding phase:
– search 𝑐𝑎𝑛𝑑 𝑖 for 2 pairs with 𝑝𝑗′ = 𝑝𝑗
– If not found:
• Recalculate 𝑐𝑎𝑛𝑑 𝑖 and bid normally
– Else:
• Remove all encountered pairs except second match
• Select first match 𝑗 and set bid of 𝑝𝑗 + 𝜖
-Scaling Complexity
• Candidate-list complexity:
– Calculation of 𝑐𝑎𝑛𝑑 𝑖 takes 𝑂 𝐴 𝑖
• But only happen when 𝑣𝑖 decrease
– Using 𝑐𝑎𝑛𝑑 𝑖 takes 𝑂 𝐴 𝑖
until recalculation
• Auction complexity
– 𝑖 𝑂 𝐴 𝑖 ∗ #𝑣𝑖 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠
– Each decrease of 𝑣𝑖 is of at least 𝜖
–
𝑖𝑂
𝐴 𝑖
∗
∆𝑣𝑖
𝜖
-Scaling Complexity
• Proposition:
– Running Auction with prices satisfying 𝑟𝜖-CS
gives Δ𝑣𝑖 = 𝑂 𝑟𝑛𝜖
• Proof:
– 𝑝0 – initial prices, 𝑆 0 – corresponding assignment
– 𝑆, 𝑝 – partial assignment & prices,
generated by Auction algorithm
just before last bid of some person 𝑖1
-Scaling Complexity
• Proof cont:
– Find path (𝑖1 , 𝑗1 , … , 𝑖𝑚 , 𝑗𝑚 )
• 𝑖𝑘 , 𝑗𝑘 ∈ 𝑆 0 , 𝑖𝑘+1 , 𝑗𝑘 ∈ 𝑆
• 𝑖1 , 𝑗𝑚 unassigned under 𝑆
– Observe:
• 𝑎𝑖1 𝑗1 − 𝑝𝑗01 + 𝑟𝜖 ≥ max 𝑎𝑖1 𝑗 − 𝑝𝑗0 = 𝑣𝑖01
𝑗∈𝐴 𝑖1
• 𝑎𝑖𝑘 𝑗𝑘 − 𝑝𝑗0𝑘 ≥ 𝑎𝑖𝑘𝑗𝑘−1 − 𝑝𝑗0𝑘−1 − 𝑟𝜖
∀𝑘 = 2 … 𝑚
•
𝑎𝑖𝑘𝑗𝑘 − 𝑝𝑗0𝑘 ≥ 𝑎𝑖𝑘𝑗𝑘−1 − 𝑝𝑗0𝑘−1 + 𝑣𝑖01 − 𝑚𝑟𝜖
•
𝑎𝑖𝑘𝑗𝑘 − 𝑝𝑗0𝑚 ≥ 𝑎𝑖𝑘 𝑗𝑘−1 + 𝑣𝑖01 − 𝑚𝑟𝜖
-Scaling Complexity
• Proof cont:
– Find path (𝑖1 , 𝑗1 , … , 𝑖𝑚 , 𝑗𝑚 )
• 𝑖𝑘 , 𝑗𝑘 ∈ 𝑆 0 , 𝑖𝑘+1 , 𝑗𝑘 ∈ 𝑆
• 𝑖1 , 𝑗𝑚 unassigned under 𝑆
– Observe:
• 𝑣𝑖1 = max 𝑎𝑖1𝑗 − 𝑝𝑗 ≥ 𝑎𝑖1 𝑗1 − 𝑝𝑗1
𝑗∈𝐴 𝑖1
• 𝑎𝑖𝑘 𝑗𝑘−1 − 𝑝𝑗𝑘−1 ≥ 𝑎𝑖𝑘𝑗𝑘 − 𝑝𝑗𝑘 − 𝜖
∀𝑘 = 2 … 𝑚
•
𝑎𝑖𝑘𝑗𝑘−1 − 𝑝𝑗𝑘−1 + 𝑣𝑖1 ≥ 𝑎𝑖𝑘 𝑗𝑘 − 𝑝𝑗𝑘 − 𝑚𝜖
•
𝑎𝑖𝑘𝑗𝑘−1 + 𝑣𝑖1 ≥ 𝑎𝑖𝑘𝑗𝑘 − 𝑝𝑗𝑚 − 𝑚𝜖
-Scaling Complexity
• Proof cont:
– We got:
•
𝑎𝑖𝑘𝑗𝑘 − 𝑝𝑗0𝑚 ≥ 𝑎𝑖𝑘 𝑗𝑘−1 + 𝑣𝑖01 − 𝑚𝑟𝜖
•
𝑎𝑖𝑘𝑗𝑘−1 + 𝑣𝑖1 ≥ 𝑎𝑖𝑘𝑗𝑘 − 𝑝𝑗𝑚 − 𝑚𝜖
– Summing:
• 𝑣𝑖01 − 𝑣𝑖1 ≤ 𝑝𝑗𝑚 − 𝑝𝑗0𝑚 + 𝑟 + 1 𝑚𝜖
– 𝑗𝑚 isn’t assigned under 𝑆
• 𝑗𝑚 didn’t receive any bid, has initial price 𝑝𝑗𝑚 = 𝑝𝑗0𝑚
– 𝑣𝑖01 − 𝑣𝑖1 = 𝑂 𝑟𝑛𝜖
-Scaling Complexity
• What is r?
• 𝜖 𝑘 -CS:
𝑘
𝑘
𝑘
– 𝑎𝑖𝑗
−
𝑝
≥
max
𝑎
−
𝑝
−
𝜖
𝑗
𝑗
𝑖𝑗
𝑖
𝑖
𝑘
– 𝛿 = 𝑎𝑖𝑗
−
𝑗∈𝐴 𝑖
𝑘+1
𝑎𝑖𝑗
𝑘
𝑘+1
• 𝛿 ≤ 𝑎𝑖𝑗
− 𝑎𝑖𝑗 − 𝑎𝑖𝑗
− 𝑎𝑖𝑗 ≤ 𝜖 𝑘 + 𝜖 𝑘+1 = (𝜃
-Scaling Complexity
• Conclusion:
– Scaling phase Complexity:
•
𝑖𝑂 𝐴 𝑖
∗
∆𝑣𝑖
𝜖
=
𝑖𝑂
𝐴 𝑖
– Number of phases:
• 𝜖 0 ,𝜖 – first & last values
• 𝑂(log 𝜖 0 /𝜖 )
• Total -Scaling Complexity:
– 𝑂(𝑛𝐴 log 𝜖 0 /𝜖 )
∗ 𝑂 𝑛 = 𝑂 𝑛𝐴
Infeasibility
• Auction never terminates for infeasible input
– One cannot tell if the algorithm won’t terminate,
or just haven’t yet
• Ways to detect infeasibility:
– Adding arcs with highly negative benefit
• − 2𝑛 − 1 𝐶
– Checking 𝑣𝑖 against lower bound
• − 2𝑛 − 1 𝐶 − 𝑛 − 1 𝜖 − max{𝑝𝑗0 }
𝑗
– Checking for augmenting path
Questions?
Thank you