fault tolerant facility location

Fault Tolerant Facility Location

Chaitanya Swamy David ShmoysCornell University

Theory Seminar 042002

Metric Facility Location

F set of facilitiesD set of clients

Facility i has facility cost fi

cij distance between any i and j in D F

Client j wants to be connected to rj distinct facilities

3

2

2

client

facility


We want to

1) Pick a set S of facilities to open

2) Assign each client j to rj open facilities

Goal Minimize total facility cost of S + sum of distances(service cost)

3

2

2client

facility

openfacility


rj=1

bullLP rounding Shmoys Tardos amp Aardal Chudak amp Shmoys Sviridenko

bullPrimal-dual algorithms Jain amp Vazirani Markakis Mahdian Saberi amp Vazirani(MMSV01) Jain Mahdian amp Saberi(JMS02)

bullLocal search Koropulu Plaxton amp Rajaraman Guha amp Khuller Charikar amp Guha

Best approx - Mahdian Ye amp Zhang (MYZ02) 152

Previous Work


Previous Work (contd)

Uniform requirements rj=r

bullMarkakis et al (MMSV01) 1861

Non-uniform requirements rj

bullJain amp Vazirani O(log rmax)

bullGuha Meyerson amp Munagala 247

Our Results

bullNon-uniform rj get a 2076-approx

bullrj=r can extend JMS02 MYZ02 to

get a 152-approx


LP Formulation

Primal

Min i fiyi + ji cijxij

st

i xij ge rj j

xij le yi i j

yi le 1 i

xij ge 0 yi ge 0 i j


LP Formulation (contd)

Max j rjvj - i zi

st

vj le wij + cij i j

j wij le fi + zi i

vj ge 0 wij ge 0 zi ge 0 i j

Dual


Complementary Slackness

Primal Slackness Conditions

bullxij gt 0 vj = wij + cij

bullyi gt 0 j wij = fi + zi

Dual Slackness Conditions

bullvj gt 0 j xij = rj

bullwij gt 0 xij = yi

bullzi gt 0 yi = 1


4-approximation outline

Basic Idea vj lsquopaysrsquo for each cij stxij gt 0

Bound service cost for each copy of j by ρvj total service cost leρj

rjvj

Problem Have ndashzis in the dual

But zi gt 0 yi = 1 So can open these facilities and charge all of this cost to the LP

2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm

Phase 1 Clustering Ensures that each copy j(c) has a nearby open facilityIterative algorithm

S = j|rj gt 0 Fj = i|xij gt 0 in fi order

Start of iteration

1 Pick j with smallest vj

2 Cluster is M Fj with iM yi = rj

2

51

2

j

client in Sfacility in some Fj

Cluster M


0

X XX

30

2

j

3 Open rj cheapest facilities in M

4 For k st Fk M connect rj copies to opened facilities Decrease rk set Fk=Fk-M

End of iteration

client in S

facility in some Fj

client not in S

X facility removed from

Fj

Cluster M

facility opened from M


Analysis Phase 1

Solution is feasible each j is connected to rj distinct facilities

Lemma Facility cost lei fiyi

Proof Cost of rj cheapest facilities in M lerj (avg cost) = iM fiyi These facilities donrsquot get used again


Analysis (contd)

Lemma For any j and c service cost of copy j(c) le3vj

Proof

vk le vj since k was chosen as cluster center

Service cost le vj + 2vk le 3vj

Cluster M

j(c)le vj

le vk

le vk k


The Algorithm (contd)

Phase 2 Taking care of ndashzis

1 Open all (unopened) i st yi = 1

2 For any j if xij = yi = 1 disconnect a copy of j and connect it to i

j

rj = 3

X

i with yi = 1i with yi lt

1 and open


Analysis Phase 2

Lemma Cost of phase 2 = fi + cij = j ljvj ndash i zi

Proof Each i with zi gt 0 is opened For iL1 all j st wij gt 0 are connected to it So

vj = (service cost) + (fi +

zi)

j ljvj = fi + cij + i zi

Let L1 = i | yi = 1

Lj = i | xij = 1 L1 and lj = |Lj|

iL1 jiLj

j|iLj

jiLjiL1


Finally hellip

Theorem Total cost le 4 times the optimal cost

Proof Total cost le

i fiyi + 3j (rj ndash lj)vj + fi + cij

facility cost of phase 1 cost for

copies connected by

phase 1

cost of phase 2

lei fiyi + 3j (rj ndash lj)vj + (j ljvj ndash i

zi )

lei fiyi + 3(j rjvj ndash i zi )

le4OPT

iL1 jiLj


A Randomized Algorithm

Idea Open i with probability ρyi

Expected facility cost le ρi fiyi

Hope that each copy j(c) has a nearby facility open and service cost decreases

Not quitehellip no facility may be open

Cluster facilities open ge 1 facility in each cluster


Phase 1 Pruning out ndashzis

Open all i st yi = 1

For each j if xij = yi = 1 connect j to i

Let Lj = i | xij = 1 and lj = |Lj|

Cost = j ljvj ndash i zi

Lj

Fj10

rrsquoj = residual reqmt = 6

Lrsquoj

Phase 2

Open all i st frac12 le yi lt 1

For each j let Lrsquoj = i | frac12 le xij lt 1

Connect copies of j to i Lrsquoj

Lose a factor of 2

facilities opened in

phases 1 2yi = 1 frac12 le yi lt 1yi lt frac12

Set L1

Set L2


Phase 3

Notation facwt(S j) = iS xij

1 Form clusters Each cluster has facwt ge frac12

2 Open facilities Open ge 1 facility in a cluster ndash used as a backup facility Open facility i with prob 2yi

3 Assign facilities to copies Each copy j(c) gets a preferred set of facilities ndash P(j(c)) with facwt ge frac12 For c d P(j(c)) P(j(d)) =

4 Connect clients Connect j(c) to the nearest i open in P(j(c)) or to a backup facility


ClusteringAfter phases 1 and 2 Fj = i | xij lt frac12 Sort these by cij and distribute among the rrsquoj copies

Cj(c) = avg service cost of the cth copy denote c Cj(c) = cijxij by Cj

Initial Fj before any iterations

Cj(1)

Cj(2)

Cj(3)3

i Fj

client j

Want the following properties

Clusters to be disjoint

Each cluster have facwt ge frac12

Each j be connected to rrsquoj clusters

iFj


Iterative algorithm

S = j | rrsquoj gt 0

aj = lsquoactiversquo copy of j initially = 1

Ĉj(aj) = avg distance to the first k

facilities in Fj gathering facwt ge frac12

say these facilities lsquoserversquo j

Will maintain Ĉj(aj) le Cj(aj)

X

X

X1

Fj after some iterations

X i removed from Fj

i Fjserving jĈj(3)

facilities serving j

aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration

1 Choose j in S with minimum vj + Ĉj(aj)

2 Form cluster M = facilities serving j Note facilities are not split

3 For k st Fk M decrease rrsquok advance ak set Fk = Fk ndash M

2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj

X facility removed from Fj

(aj)


Opening Facilities

Central facilities opened in 2 steps

1 Open exactly 1 facility in M i opened with prob qyi Acts as backup

denoted b(k ) for each k st Fk M

2 Open each i in M indep with prob (2-q)yi and independent of step 1

Non-central facilities

Cluster M

k

open with prob 2yi independent of other choices

j

(ak

)


Let Sj(c) = avg dist from j to P(j(c))

= ( cijxij)facwt(P(j(c))

j)

Then c Sj(c) le 2Cj

Distributing Facilities

iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))

Copy c gets a preferred set P(j(c))Preferred sets are disjoint

Ensure facwt(P(j(c)) j) ge frac12 for all c

Possible to do so since each xij lt frac12

facility in Fj


Analysis

Feasibility follows from

1 Facilities in phases 1 2 not reused

2 After clustering j is connected to rrsquoj disjoint clusters backups are distinct

3 Preferred sets are disjoint

So j connected to rj distinct facilities


Facility cost

Recall L1 = i | yi = 1

Phase 2 incur a factor of 2

Phase 3 each i is opened with probability 2yi

Expected facility cost le 2 fiyifor phases 2 3

iL1


Bounding backup cost denoted by B rv

D event that no i in P(j(c)) is open

Lemma E[B|D] le 2vj + Cj(c)

Proof 2 cases

Service cost I

iM Fj st cik le Ĉj(d)

Also vk + Ĉj(d) le vj + Ĉj(c) le vj + Cj(c)

k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)

le Ĉj(d) in expectatio

n

1)

2)

backup = b(j(c))


Service Cost II

Fix j c Let X(c) = service cost of j(c)

Let di = cij pi = prob i is opened = 2yi

B(c) = backup costD(c) = event that no iP(j(c)) is

openp = Pr[D(c)] = (1-p1)hellip(1-pm) le e-1

davg = weighted avg of the dis

= (i pidi)(i pi) = Sj(c)

d1

d2 dm

P(j(c)) sorted by increasing cij

j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip

+ (1-p1)hellip(1-pm-1)pmdm]

+ pE[B(c)|D(c)]

le (1-p)davg + p[2vj + Cj(c)]

le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]

Let X = c X(c) = service cost of j

c Sj(c) le 2Cj and c Cj(c) le 2Cj

Summing over all c = 1helliprrsquoj

E[X] le (1-e-1)2Cj + e-1(2rrsquojvj + Cj)

le 2Cj + 2e-1rrsquojvj


Putting it all together

Phase 1 pay the optimal LP cost

Phases 2 3

bull Facility cost twice LP facility cost

bull Service costLose a factor of 2 for phase 2Phase 3 cost is 2(LP service cost)+2e-1(dual value)

Overall cost for le (2+2e)(LP cost) phases 2 3

Total cost le (2+2e)OPT


How to improve this

bull Distribute facilities more equitably (in an expected sense) among copies - decreases prob of lsquobadrsquo event

bull Better analysis ndash maximum distance within a cluster can be bounded by 2Cj(c)

bull Balance phases 2 and 3


Summary of Results

bullGive a 2076-approx algorithm for non-uniform rjs Based on LP rounding using complem slackness

bullFor rj = r extend the primal-dual algorithm of (JMS02) to get a 152-approximation

bullFault tolerant k medians with rj = r

a Primal-dual algorithm (JMS02) gives a 4-approx using Lagrangean relaxation

b LP rounding gives a factor of 8


Open Questions

1 Reduce gap between rj = r non-uniform rj

2 Combinatorial algorithms for non-uniform rj primal-dual local-search

3 Constant-factor approx for fault tolerant k medians with non-uniform rjs



PowerPoint Presentation

Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions



F set of facilitiesD set of clients

Facility i has facility cost fi

cij distance between any i and j in D F

Client j wants to be connected to rj distinct facilities

3

2

2

client

facility


We want to




3

2

2client

facility

openfacility


rj=1





Previous Work








Our Results



get a 152-approx


LP Formulation

Primal


st

i xij ge rj j

xij le yi i j

yi le 1 i




Max j rjvj - i zi

st

vj le wij + cij i j

j wij le fi + zi i


Dual









bullzi gt 0 yi = 1





rjvj



2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


We want to




3

2

2client

facility

openfacility


rj=1





Previous Work








Our Results



get a 152-approx


LP Formulation

Primal


st

i xij ge rj j

xij le yi i j

yi le 1 i




Max j rjvj - i zi

st

vj le wij + cij i j

j wij le fi + zi i


Dual









bullzi gt 0 yi = 1





rjvj



2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


rj=1





Previous Work








Our Results



get a 152-approx


LP Formulation

Primal


st

i xij ge rj j

xij le yi i j

yi le 1 i




Max j rjvj - i zi

st

vj le wij + cij i j

j wij le fi + zi i


Dual









bullzi gt 0 yi = 1





rjvj



2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions








Our Results



get a 152-approx


LP Formulation

Primal


st

i xij ge rj j

xij le yi i j

yi le 1 i




Max j rjvj - i zi

st

vj le wij + cij i j

j wij le fi + zi i


Dual









bullzi gt 0 yi = 1





rjvj



2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


LP Formulation

Primal


st

i xij ge rj j

xij le yi i j

yi le 1 i




Max j rjvj - i zi

st

vj le wij + cij i j

j wij le fi + zi i


Dual









bullzi gt 0 yi = 1





rjvj



2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions



Max j rjvj - i zi

st

vj le wij + cij i j

j wij le fi + zi i


Dual









bullzi gt 0 yi = 1





rjvj



2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions









bullzi gt 0 yi = 1





rjvj



2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions





rjvj



2

j(1) j(2)

le vj

le vj

view as rj

copiesj(c) cth copy


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


The Algorithm



Start of iteration



2

51

2

j


Cluster M


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


0

X XX

30

2

j



End of iteration

client in S

facility in some Fj

client not in S


Fj

Cluster M



Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Analysis Phase 1





Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Analysis (contd)


Proof



Cluster M

j(c)le vj

le vk

le vk k






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions






j

rj = 3

X


1 and open


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Analysis Phase 2




zi)


Let L1 = i | yi = 1


iL1 jiLj

j|iLj

jiLjiL1


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Finally hellip


Proof Total cost le



copies connected by

phase 1

cost of phase 2


zi )


le4OPT

iL1 jiLj














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions














Lj

Fj10


Lrsquoj

Phase 2




Lose a factor of 2



Set L1

Set L2


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Phase 3










Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions





Cj(1)

Cj(2)

Cj(3)3

i Fj

client j





iFj


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Iterative algorithm







X

X

X1


X i removed from Fj

i Fjserving jĈj(3)


aj = 3

4X

(aj)

(aj)

(aj)


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Start of iteration




2j(3) Cluster M

aj = 1

4

1

XX X

aj = 4 Cluster M

aj = 2

3

client in S

facility in some Fj


(aj)


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Opening Facilities






Cluster M

k


j

(ak

)




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions




j)

Then c Sj(c) le 2Cj


iP(j(c))

j

rrsquoj = 3

P(j(1))P(j(2))

P(j(3))




facility in Fj


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Analysis







Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Facility cost





iL1





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions





Proof 2 cases

Service cost I



k(d)

j(c)

le vj

le Ĉj(d)le vk

B

k(d)

j(c)

le vj

le vk

iM Fj cik gt Ĉj(d)


n

1)

2)

backup = b(j(c))


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Service Cost II







d1

d2 dm


j(c)

i P(j(c))


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Then

E[X(c)] = [p1d1 + (1-p1)p2d2 + hellip


+ pE[B(c)|D(c)]


le (1-e-1)Sj(c) + e-1[2vj +

Cj(c)]









Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions




Phases 2 3






How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


How to improve this





Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Summary of Results







Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions


Open Questions







Previous Work


LP Formulation




The Algorithm

Slide 11

Analysis Phase 1

Analysis (contd)


Analysis Phase 2

Finally hellip


Slide 18

Slide 19

Clustering

Slide 21

Slide 22

Opening Facilities


Analysis

Facility cost

Service cost I

Service Cost II

Slide 29


How to improve this

Summary of Results

Open Questions

fault tolerant facility location

Documents

facility i

i fiyi j

service cost vj

rj open facilities

cost of rj cheapest

copy of j

vj wij cij i

rj distinct facilities