Improved Approximation Algorithms for Budgeted Allocations Lecture - PowerPoint PPT Presentation

Improved Approximation Algorithms for Budgeted Allocations

Lecture outlines 1. Presenting the budgeted allocation problem 2. Defining some basic terms we are going to use 3. The 3/2-approximation algorithm 4. Analyzing its correctness and running time 5. Handling a private case of the problem 2

Part 1 The budgeted allocation problem 3

The budgeted allocation problem • Motivation: online sponsored search advertising 4

The budgeted allocation problem • Our model: o 𝑊 - a set of keywords o 𝑉 - a set of bidders o Each bidder 𝑗 ∈ 𝑉 has: o A known daily budget 𝐶 𝑗 o A nonnegative bid 𝑐 𝑗𝑘 for every keyword 𝑘 ∈ 𝑊 o The profit extracted from a bidder is the minimum between: o His budget, 𝐶 𝑗 o The sum of the 𝑐 𝑗𝑘 ’s for keywords allocated to it 5

The budgeted allocation problem • Our goal: to find an allocation of keywords to bidders that maximizes the total profit o The sum of the profits extracted from all bidders 6

The budgeted allocation problem • The online version: o The keywords arrive one-by-one in an online fashion o The bids for keyword 𝑘 ∈ 𝑊 are revealed only when 𝑘 arrives o Then, we allocate the keyword to one of the bidders 7

The budgeted allocation problem • The offline version: o We can see all of the bids before allocating keywords o We assume that 𝑛𝑏𝑦 𝑗,𝑘 𝑐 𝑗𝑘 ≤ 𝑛𝑗𝑜 𝑗 𝐶 𝑗 o Two versions: o Uniform: o each keyword 𝑘 ∈ 𝑊 has a single price 𝑐 𝑘 o 𝑐 𝑗𝑘 ∈ 0, 𝑐 𝑘 o Non-uniform: o the 𝑐 𝑗𝑘 values can be arbitrary for each 𝑗, 𝑘 8

The budgeted allocation problem • The non-uniform version representation: o 𝑊 - the group of keywords o 𝑉 - the group of bidders o 𝐶 𝑗 - the budget of bidder 𝑗 o 𝑐 𝑗𝑘 - the bid of bidder 𝑗 for keyword 𝑘 o 𝑦 𝑗𝑘 - represent whether keyword 𝑘 is allocated to bidder 𝑗 o Our goal: 𝑦 𝑗𝑘 ≤ 1 ∀𝑘 ∈ 𝑊 𝑛𝑏𝑦 𝑛𝑗𝑜 𝐶 𝑗 , 𝑐 𝑗𝑘 𝑦 𝑗𝑘 𝑡. 𝑢. 𝑗∈𝑉 𝑦 𝑗𝑘 ∈ 0,1 ∀𝑗 ∈ 𝑉, 𝑘 ∈ 𝑊 𝑗∈𝑉 𝑘∈𝑊 9

The fractional allocation • We are going to use o A linear relaxation of our integer program 𝑦 𝑗𝑘 ≤ 1 ∀𝑘 ∈ 𝑊 𝑛𝑏𝑦 𝑛𝑗𝑜 𝐶 𝑗 , 𝑐 𝑗𝑘 𝑦 𝑗𝑘 𝑡. 𝑢. 𝑗∈𝑉 0 ≤ 𝑦 𝑗𝑘 ≤ 1 ∀𝑗 ∈ 𝑉, 𝑘 ∈ 𝑊 𝑗∈𝑉 𝑘∈𝑊 𝒋 𝑪 𝒋 𝒄 𝒋𝟐 𝒄 𝒋𝟑 1 10 10 8 2 20 10 0 3 16 0 16 10

The fractional allocation • We are going to use o A linear relaxation of our integer program 𝑦 𝑗𝑘 ≤ 1 ∀𝑘 ∈ 𝑊 𝑛𝑏𝑦 𝑛𝑗𝑜 𝐶 𝑗 , 𝑐 𝑗𝑘 𝑦 𝑗𝑘 𝑡. 𝑢. 𝑗∈𝑉 0 ≤ 𝑦 𝑗𝑘 ≤ 1 ∀𝑗 ∈ 𝑉, 𝑘 ∈ 𝑊 𝑗∈𝑉 𝑘∈𝑊 o An algorithm for rounding the optimal fractional solution o Our solution is feasible o The total profit loss is no larger than 1/3 of the profit of the optimal solution 11

Part 2 Some basic terms we are going to use 12

Graph representation • The graph induced by the fractional allocation: o A bipartite graph over 𝑉 ∪ 𝑊 o An edge 𝑗, 𝑘 for every 𝑦 𝑗𝑘 > 0 o With weight 𝑥 𝑗𝑘 = 𝑐 𝑗𝑘 𝑦 𝑗𝑘 13

Active bidders • A bidder is active if it has at least one neighbor of degree 2 or more in 𝐻 . • Otherwise, it is inactive 14

Saturated bidders • A bidder is saturated if 𝑛𝑗𝑜 𝐶 𝑗 , 𝑐 𝑗𝑘 𝑦 𝑗𝑘 = 𝐶 𝑗 𝑘∈𝑊 • Otherwise, it is unsaturated 15

Saturated bidders • Given a feasible allocation for the fractional version, it’s possible to create a graph 𝐻 s.t. o 𝐻 represents a solution with no loss of profit regarding the given allocation o 𝐻 is a forest o There is at most 1 unsaturated bidder per tree o If there is such bidder, it’s placed at the root of the tree o If there isn’t, an arbitrary bidder is placed at the root • It is possible to construct 𝐻 in polynomial time 16

k-chains • Consider a path 𝑗 1 , 𝑘 1 , 𝑗 2 , 𝑘 2 , … , 𝑗 𝑙−1 , 𝑘 𝑙−1 , 𝑗 𝑙 – Consisting of 2𝑙 − 1 nodes – Starting and ending with a bidder – Such that • The keywords all have degree exactly 2 • Bidder 𝑗 1 is the highest node on the path • For all other bidders 𝑗 2 , 𝑗 3 , … , 𝑗 𝑙 – Any keyword not on the path that is adjacent to bidder 𝑗 𝑚 has degree exactly 1 • The graph formed by this path is a 𝑙 − 𝑑ℎ𝑏𝑗𝑜 17

Interesting nodes • We round the fractional allocation using interesting nodes • There are 4 types of interesting nodes 18

Interesting nodes • We round the fractional allocation using interesting nodes • There are 4 types of interesting nodes: 1. The root of the whole tree, if the tree consists of a 2-chain 19

Interesting nodes • We round the fractional allocation using interesting nodes • There are 4 types of interesting nodes: A bidder 𝑤 who’s subtree contains at least one 3-chain rooted at 𝑤 2. 20

Interesting nodes • We round the fractional allocation using interesting nodes • There are 4 types of interesting nodes: A bidder 𝑤 who’s subtree consists of more than one 2-chain rooted 3. at 𝑤 21

Interesting nodes • We round the fractional allocation using interesting nodes • There are 4 types of interesting nodes: 4. A keyword with at least 2 children, such that each child is a root of a 1-chain or a 2-chain 22

Interesting nodes • Lemma: a tree that has a keyword with degree more than 1 must have at least one interesting node 23

Interesting nodes • Partial proof: – Given a tree, we can ignore keywords with degree 1 • Since they don’t affect k -chains – Therefore, we can assume all leaves are bidders – There are several possible cases: 1. There is only one keyword in the tree – the first type 2. All keywords have no grandsons – the third or fourth type Let 𝑘 be a keyword with distance ≤ 3 from all of its descendants 3. 1. If it has more than one son – the third or the fourth type 2. If its only son has two or more sons – the third or the fourth type 3. If its only son has exactly one son – the second or the fourth type 4. If its only son has no sons – find another keyword 24

Part 3 The 3/2-approximation algorithm 25

Constructing the feasible solution • We are going to turn the fractional allocation into a feasible one • We will do it in iterations • at each iteration: o The degree of at least one keyword becomes 1 o At least one active bidder becomes inactive • The result – a feasible allocation 26

Constructing the feasible solution • Rounding subgraphs: • The act of reducing the degree of at least one keyword in the subgraph • There are 4 types of roundings – Each one is related to a different kind of an interesting node • Reminder: a tree that has a keyword with degree more than 1 must have at least one interesting node 27

Constructing the feasible solution • Type 1 rounding: – 𝑗 is the interesting node • Can be unsaturated, with unused budget – 𝑘 is its child of degree 2 – 𝑙 is its grandchild • It’s saturated – There are two ways to round the subgraph • We choose the one that incurs the smaller loss 28

Constructing the feasible solution • Type 1 rounding: 𝒄 𝒌 𝒚 𝒌 𝒄 𝒌 𝒚 𝒌 B – Example: i 10 8 0.5 4 k 20 10 0.5 5 29

Constructing the feasible solution • Type 1 rounding: 𝒄 𝒌 𝒚 𝒌 𝒄 𝒌 𝒚 𝒌 B – Example: i 7 5 0.6 3 k 25 25 0.4 10 30

Constructing the feasible solution • Type 2 rounding: – There are 4 possible ways to round the subgraph 31

Constructing the feasible solution • Type 3 rounding: – We take a partial subtree rooted at the interesting node – And proceed as in type 2 rounding 32

Constructing the feasible solution • Type 4 rounding: – There are 3 possible cases: 1. There are no 2-chains attached to the interesting node 2. There are no 1-chains attached to the interesting node 3. There are at least one 1-chain and one 2-chain attached to the interesting node – Each one is treated differently 33

Constructing the feasible solution • Type 4 rounding: first case – There are no 2-chains attached to the interesting node • We retain the edge with the largest weight and delete all others 34

Constructing the feasible solution • Type 4 rounding: second case – There are no 1-chains attached to the interesting node • While it’s possible, we choose 3-chains and treat them like in type 2 rounding 35

Constructing the feasible solution • Type 4 rounding: second case – There are no 1-chains attached to the interesting node • While it’s possible, we choose 3-chains and treat them like in type 2 rounding 36

Constructing the feasible solution • Type 4 rounding: third case – There are at least one 1-chain and one 2-chain attached to the interesting node • We choose an 1-chain and a 2-chain, and treat them as in type 2 rounding 37