Primal-dual schema algorithm for metric uncapacitated facility location. Mainly referenced Chapter 24 of "Approximation Algorithms" by Vazirani.

1. Approximation Algorithms
Chapter 24. Facility Location
2017.06.02.
Sangwoo Mo
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2. Table of Contents
• LP Duality
• Facility Location
• Algorithm
• Analysis
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3. Table of Contents
• LP Duality
• Facility Location
• Algorithm
• Analysis
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4. LP Duality
• Primal (𝓟):
minimize ∑ 𝑐& 𝑥&
(
&)*
subject to ∑ 𝑎,& 𝑥& ≥ 𝑏,
(
&)* 𝑖 = 1, … , 𝑚
𝑥& ≥ 0 𝑗 = 1, … , 𝑛
• Dual (𝓓):
maximize ∑ 𝑏, 𝑦,
:
,)*
subject to ∑ 𝑎,& 𝑦,
:
,)* ≤ 𝑐& 𝑗 = 1, … , 𝑛
𝑦, ≥ 0 𝑖 = 1, … , 𝑚
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5. LP Duality (Example)
• Let’s consider a simple example
minimize 7𝑥* + 𝑥> + 5𝑥@
subject to 𝑥* − 𝑥> + 3𝑥@ ≥ 10
5𝑥* + 2𝑥> − 𝑥@ ≥ 6
𝑥*, 𝑥>, 𝑥@ ≥ 0
• Let 𝑓 be objective, 𝑓* and 𝑓> be L.H.S. of constraints respectively
• The goal is finding the lower bound of the objective 𝑓
• Since 𝑓, and 𝑥, are bounded below, if we can represent 𝑓 with
positive combination of 𝑓, and 𝑥,, be can bound 𝑓 below
• ex) 7𝑥* + 𝑥> + 5𝑥@ ≥ 𝑥* − 𝑥> + 3𝑥@ + 5𝑥* + 2𝑥> − 𝑥@ ≥ 16
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6. LP Duality (Example)
• Let 𝑓 be objective, 𝑓* and 𝑓> be L.H.S. of constraints respectively
• The goal is finding the lower bound of the objective 𝑓
• Let 𝑦, be positive weight for 𝑓, such that 𝑓 ≥ ∑ 𝑦, 𝑓,
• For each coefficient of ∑ 𝑦, 𝑓,, it should be upper bounded by 𝑓
• It leads the dual program of 𝑦,, which also forms LP
• For previous example, the dual program is
maximize 10𝑦* + 6𝑦>
subject to 𝑦* + 5𝑦> ≤ 7
−𝑦* + 2𝑦> ≤ 1
3𝑦* − 𝑦> ≤ 5
𝑦*, 𝑦> ≥ 0
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7. Duality Theorem
• LP duality theorem: Let 𝑥∗, 𝑦∗ are optimal solutions of primal and
dual programs respectively. Then ∑ 𝑐& 𝑥&
∗(
&)* = ∑ 𝑏, 𝑦,
∗:
,)* .
• Weak duality theorem: Let 𝑥, 𝑦 are feasible solutions of primal and
dual programs respectively. Then ∑ 𝑐& 𝑥&
(
&)* ≥ ∑ 𝑏, 𝑦,
:
,)* .
• Proof) ∑ 𝑐& 𝑥&& ≥ ∑ ∑ 𝑎,& 𝑦,, 𝑥&& = ∑ ∑ 𝑎,& 𝑥&& 𝑦,, ≥ ∑ 𝑏, 𝑦,,
• Complementary slackness: From equality condition, we obtain
• Primal C.S.: For each 𝑗, either 𝑥& = 0 or ∑ 𝑎,& 𝑦,, = 𝑐&
• Dual C.S.: For each 𝑖, either 𝑦, = 0 or ∑ 𝑎,& 𝑥&, = 𝑏,
𝑝∗
= 𝑑∗
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8. Table of Contents
• LP Duality
• Facility Location
• Algorithm
• Analysis
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9. Facility Location
• (Metric Uncapacitated) Facility Location:
• Let 𝐺 be a bipartite graph with bipartition (𝐹, 𝐶)
• Let 𝑓, be the cost of opening facility 𝑖
• Let 𝑐,& be the cost of connecting city 𝑗 to facility 𝑖
• Find the optimal way to connect cities to open facilities
Source: http://www.or.uni-bonn.de/~vygen/files/tokyo.pdf 9/34

10. Facility Location (Formal Ver.)
• (Metric Uncapacitated) Facility Location:
• Let 𝐺 be a bipartite graph with bipartition (𝐹, 𝐶)
• Let 𝑓, be the cost of opening facility 𝑖
• Let 𝑐,& be the cost of connecting city 𝑗 to facility 𝑖
• Then
minimize ∑ 𝑐,& 𝑥,&,∈N,&∈O + ∑ 𝑓, 𝑦,,∈N
subject to ∑ 𝑥,&,∈N ≥ 1 𝑗 ∈ 𝐶
𝑦, − 𝑥,& ≥ 0 𝑖 ∈ 𝐹, 𝑗 ∈ 𝐶
𝑥,& ∈ {0,1} 𝑖 ∈ 𝐹, 𝑗 ∈ 𝐶
𝑦, ∈ {0,1} 𝑖 ∈ 𝐹
• Let 𝑖 ∈ 𝐼 iff 𝑦, = 1, and 𝜙 𝑗 = 𝑖 iff 𝑥,& = 1
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11. Facility Location (Primal-Dual schema)
• Primal 𝓟 (LP Relaxation):
minimize ∑ 𝑐,& 𝑥,&,∈N,&∈O + ∑ 𝑓, 𝑦,,∈N
subject to ∑ 𝑥,&,∈N ≥ 1 𝑗 ∈ 𝐶
𝑦, − 𝑥,& ≥ 0 𝑖 ∈ 𝐹, 𝑗 ∈ 𝐶
𝑥,& ≥ 0 𝑖 ∈ 𝐹, 𝑗 ∈ 𝐶
𝑦, ≥ 0 𝑖 ∈ 𝐹
• Dual (𝓓):
minimize ∑ 𝛼&&∈O
subject to 𝛼& − 𝛽,& ≤ 𝑐,& 𝑖 ∈ 𝐹, 𝑗 ∈ 𝐶
∑ 𝛽,&&∈O ≤ 𝑓, 𝑖 ∈ 𝐹
𝛼& ≥ 0 𝑗 ∈ 𝐶
𝛽,& ≥ 0 𝑖 ∈ 𝐹, 𝑗 ∈ 𝐶
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12. Facility Location (C.S. conditions)
• Complementary Slackness conditions:
① ∀𝑖 ∈ 𝐹, 𝑗 ∈ 𝐶: 𝑥,& > 0 ⇒ 𝛼& − 𝛽,& = 𝑐,&
② ∀𝑖 ∈ 𝐹: 𝑦, > 0 ⇒ ∑ 𝛽,&&∈O = 𝑓,
③ ∀𝑗 ∈ 𝐶: 𝛼& > 0 ⇒ ∑ 𝑥,&,∈N = 1
④ ∀𝑖 ∈ 𝐹, 𝑗 ∈ 𝐶: 𝛽,& > 0 ⇒ 𝑦, = 𝑥,&
• Suppose optimal solution is LP is integral, then
• By condition ④, if 𝑖 ∈ 𝐼 but 𝜙 𝑗 ≠ 𝑖 then 𝑦, ≠ 𝑥,&, thus 𝛽,& = 0
• By condition ②, ∀𝑖 ∈ 𝐼, ∑ 𝛽,&&:[ & ), = 𝑓,
• By condition ①, if 𝜙 𝑗 = 𝑖 then 𝛼& − 𝛽,& = 𝑐,&
• It implies that total price of city 𝜶𝒋 is sum of 𝒄𝒊𝒋: edge price and
𝜷𝒊𝒋: contribution of city 𝑗 opening facility 𝑖
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13. Facility Location (C.S. conditions)
• It implies that total price of city 𝜶𝒋 is sum of 𝒄𝒊𝒋: edge price and
𝜷𝒊𝒋: contribution of city 𝑗 opening facility 𝑖
Source: http://www.or.uni-bonn.de/~vygen/files/tokyo.pdf
𝜶 𝟏 = 𝒄 𝟏𝟏 + 𝜷 𝟏𝟏
𝜶 𝟐 = 𝒄 𝟏𝟐 + 𝜷 𝟏𝟐
𝒇 𝟏 = 𝜷 𝟏𝟏 + 𝜷 𝟏𝟐
𝒄 𝟏𝟏
𝒄 𝟏𝟐
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14. Table of Contents
• LP Duality
• Facility Location
• Algorithm
• Analysis
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15. Algorithm
• Phase 1: Collect all candidate facilities
• Let (𝑖, 𝑗) be tight if 𝛼& ≥ 𝑐,&, and (𝑖, 𝑗) be special if 𝛽,& > 0
• Initialize every city not connected, 𝛼&, 𝛽,& = 0
• While not every city connected:
• Increase 𝛼& of every unconnected cities by 1
• If (𝑖, 𝑗) is tight, increase 𝛽,& by 1 to satisfy 𝛼& = 𝑐,& + 𝛽,&
• If 𝑓, = ∑ 𝛽,&& , let facility 𝑖 be temporarily open
• Every city 𝑗 tight to 𝑖 is now connected to 𝑖
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16. Algorithm
• Phase 2: Prune unnecessary facilities
• Let 𝑖 and 𝑖d are conflicting if there is 𝑗 such that 𝛽,&, 𝛽,e& > 0
• Consider graph 𝐹f such that nodes are temporarily open, and
edge implies two nodes are conflicting
• Choose maximal independent set 𝐼 ⊂ 𝐹f and let 𝑖 ∈ 𝐼 be open
• If 𝜙(𝑗) ∈ 𝐼, keep connection and let 𝑗 be directly connected to 𝜙(𝑗)
• Else, let 𝑖 be any neighbor of 𝜙(𝑗), connect 𝑗 to 𝑖, and let 𝑗 be
indirectly connected to 𝑖
• Note that 𝛽,& = 0 if 𝑗 is indirectly connected to 𝑖
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17. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
Source: http://www.showme.com/sh?h=6W3DJpI&jw_version=6 17/34

18. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 1 11 11
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19. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 2 22 22
(1) (1) (1)𝜷𝒊𝒋
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20. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 3 33 33
(2) (2) (2)𝜷𝒊𝒋 (1) (1) (1)
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21. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 3 33 33
(2) (2) (2)𝜷𝒊𝒋 (1) (1) (1)
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22. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 3 44 43
(2) (2) (3)𝜷𝒊𝒋 (1) (2) (2)
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23. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 3 44 43
(2) (2) (3)𝜷𝒊𝒋 (1) (2) (2)
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24. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 3 54 53
(2) (2) (3)𝜷𝒊𝒋 (1) (2) (3)
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25. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 3 54 53
(2) (2) (3)𝜷𝒊𝒋 (1) (2) (3)
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26. Algorithm (Example)
3 4 6
1 2 1 2 1 2 10
𝜶𝒋 3 104 53
(2) (2) (3)𝜷𝒊𝒋 (1) (2) (3)
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27. Algorithm (Example)
3 4 6
𝜶𝒋 3 104 53
1 2 1 2 1 2 10(2) (2) (3)𝜷𝒊𝒋 (1) (2) (3)
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28. Algorithm (Example)
3 4 6
𝜶𝒋 3 104 53
1 2 1 2 10(2) (3)𝜷𝒊𝒋 (1) (3)
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29. Algorithm (Example)
3 4 6
𝜶𝒋 3 104 53
1 2 2 10(2)𝜷𝒊𝒋 (2) (3)
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1(2)

30. Table of Contents
• LP Duality
• Facility Location
• Algorithm
• Analysis
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31. Theorem
• Theorem The solution of algorithm satisfy
h 𝑐,& 𝑥,&
,∈N,&∈O
+ 3 h 𝑓, 𝑦,
,∈N
≤ 3 h 𝛼&
&∈O
• Corollary The dual solution is 3-approximation algorithm
• Proof) Let 𝛼& = 𝛼&
i
+ 𝛼&
j
, each contributes to facility/edge
• If 𝑗 is directly connected, let 𝛼&
i
= 𝛽,& and 𝛼&
j
= 𝑐,& where 𝑖 = 𝜙(𝑗)
• If 𝑗 is indirectly connected, let 𝛼&
i
= 0 and 𝛼&
j
= 𝛼&
• Goal: prove that
① ∑ 𝑓, 𝑦,,∈N = ∑ 𝛼&&∈O
i
② ∑ 𝑐,& 𝑥,&,∈N,&∈O ≤ 3 ∑ 𝛼&&∈O
j
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32. Theorem
• Lemma 1 ∑ 𝑓, 𝑦,,∈N = ∑ 𝛼&&∈O
i
• Proof)
h 𝑓, 𝑦,
,∈N
= h 𝑓,
,∈k
= h h 𝛼&
i
&:[ & ),,∈k
= h 𝛼&
i
&∈O
• Lemma 2 ∑ 𝑐,& 𝑥,&,∈N,&∈O ≤ 3 ∑ 𝛼&
j
&∈O
• Proof) For directly connected city, 𝑐[(&)& = 𝛼&
j
≤ 3𝛼&
j
• Goal: prove that 𝑐[(&)& ≤ 3𝛼&
j
for indirectly connected cities too
• Then
h 𝑐,& 𝑥,&
,∈N,&∈O
= h 𝑐[ & &
&∈O
≤ 3 h 𝛼&
j
&∈O
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33. Theorem
• Lemma 2 ∑ 𝑐,& 𝑥,&,∈N,&∈O ≤ 3 ∑ 𝛼&
j
&∈O
• Goal: prove that 𝑐[(&)& ≤ 3𝛼&
j
for indirectly
connected city 𝑗
• Let 𝑗 is temporarily connected to 𝑖′, and indirectly connected to 𝑖
• Then there is 𝑗′ such that specially connected to both 𝑖 and 𝑖′
• Let 𝑡* and 𝑡> be time that 𝑖 and 𝑖′ be opened
• Since both (𝑖, 𝑗d) and (𝑖d, 𝑗d) are special, they must be tight before
either 𝑖 or 𝑖′ be opened i.e. 𝛼&d ≤ min(𝑡*, 𝑡>)
• Since (𝑖d, 𝑗) is tight, 𝛼& ≥ 𝑡>, and it leads 𝛼& ≥ 𝛼&e
• By tightness, 𝛼& ≥ 𝑐,e&, 𝛼&e ≥ 𝑐,&e and 𝛼&e ≥ 𝑐,e&e
• Thus, by triangle inequality, 𝑐,& ≤ 𝑐,e& + 𝑐,&e + 𝑐,e&e ≤ 3𝛼&
= 3𝛼&
j
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34. Tight Example
• Consider example below
• After Phase 1, both facility 1 and 2 are temporality open
• If chosen independent set is {1}, the total cost is 𝜖 + 1 + 3(𝑛 − 1),
where the optimal cost is 𝑛 + 1 𝜖 + 𝑛
𝝐 𝝐(𝒏 + 𝟏)𝝐 (𝒏 + 𝟏)𝝐
(𝟏) 𝟐
𝟐
𝟐
(𝟏)
(𝟏)
(𝟏)
(𝟏)
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35. FYI: Better Approximation Ratios
Source: http://www.or.uni-bonn.de/~vygen/files/tokyo.pdf 35/34