Strong LP Formulations & Primal-Dual Approximation Algorithms David Shmoys (joint work Tim...

Strong LP Formulations & Primal-Dual Approximation Algorithms

David Shmoys(joint work Tim Carnes & Maurice Cheung)

June 23, 2011

Introduction The standard approach to solve

combinatorial integer programs in practice – start with a “simple” formulation & add valid inequalities

Our agenda: show that same approach can be theoretically justified by approximation algorithms

An ®-approximation algorithm produces a solution of cost within a factor of ® of the optimal solution in polynomial time

Introduction Primal-dual method

a leading approach in the design of approximation algorithms for NP-hard

problems Consider several capacitated covering

problems - covering knapsack problem

- single-machine scheduling problems Give constant approximation algorithms

based on strong linear programming (LP) formulations

Approximation Algorithms and LP Use LP to design approximation algorithms

Optimal value for LP gives bound on optimal integer programming (IP) value

Want to find feasible IP solution of value within a factor of ® of optimal LP solution

Key is to start with “right” LP relaxation LP-based approximation algorithm produces

additional performance guarantee on each problem instance

Empirical success of IP cutting plane methods suggests stronger formulations - needs theory!

Primal-Dual Approximation Algorithms

Do not even need to solve LP!! Each min LP has a dual max LP

of equal optimal value Goal: Construct feasible integer solution

S along with feasible solution D to dual of LP relaxation such thatcost(S) · ®¢cost(D) · ®¢LP-OPT · ®¢OPT

) ®-approximation algorithm

Adding Valid Inequalities to LP LP formulation can be too “weak” if there

is “big” integrality gap - OPT/LP-OPT is often unbounded

Fixed by adding additional inequalities to formulation Restricts set of feasible LP solutions Satisfied by all integer solutions, hence

“valid” Key technique in practical codes to

solve integer programs

Knapsack-Cover Inequalities

Carr, Fleischer, Leung and Phillips (2000) developed valid knapsack-cover inequalities and LP-rounding algorithms for several capacitated covering problems Requires solving LP with ellipsoid

method Further complicated since inequalities

are not known to be separable GOAL: develop a primal-dual analog!

Highlights of Our Results For each of the following problems, we have a primal-

dual algorithm that achieves a performance guarantee of 2

Min-Cost (Covering) Knapsack Single-Demand Capacitated

Facility Location

Single-Item Capacitated Lot-Sizing We extend the knapsack-cover inequalities to

handle this more general setting Single-Machine Minimum-Weight Late Jobs 1|| wj Uj Single-Machine General Minimum-Sum Scheduling 1||

fj

Used valid knapsack-cover inequalities developed by Carr, Fleischer, Leung and Phillips as LP formulation

Highlights of Our Results For each of the following problems, we have a primal-

dual algorithm that achieves a performance guarantee of 2

Min-Cost (Covering) Knapsack Single-Demand Capacitated

Facility Location

Single-Item Capacitated Lot-Sizing We extend the knapsack-cover inequalities to

handle this more general setting Single-Machine Minimum-Weight Late Jobs 1|| wj Uj Single-Machine General Minimum-Sum Scheduling 1||

fj

Used valid knapsack-cover inequalities developed by Carr, Fleischer, Leung and Phillips as LP formulation

Min-Sum 1-Machine Scheduling 1|| fj

Each job j has a cost function fj(Cj) that is non-negative non-decreasing function of its completion time Cj

Goal: minimize j fj(Cj) What is known? Bansal & Pruhs (FOCS ’10) gave

first constant-factor algorithm Main result of Bansal-Pruhs adds release dates,

and permits preemption – result is O(loglog(nP))-approximation algorithm

OPEN QUESTIONS – Is a constant-factor doable?

10 Open Problems Better constant factors? Any constant factor? Primal-dual when

rounding is known? But – nothing of the type –

good constant factor is known, but is a factor of 1+possible

for any >0?

Min-Sum 1-Machine Scheduling 1|| fj

Each job j has a cost function fj(Cj) that is non-negative non-decreasing function of its completion time Cj

Goal: minimize j fj(Cj) What is known? Bansal & Pruhs (FOCS ’10) gave

first constant-factor algorithm Main result of Bansal-Pruhs adds release dates,

and permits preemption – result is O(loglog(nP))-approximation algorithm

OPEN QUESTIONS – Is a constant-factor doable?- Can 1+² be achieved w/o release dates?

Primal-Dual for Covering Problems Early primal-dual algorithms

Bar-Yehuda and Even (1981) – weighted vertex cover

Chvátal (1979) – weighted set cover Agrawal, Klein and Ravi (1995)

Goemans and Williamson (1995) generalized Steiner (cut covering)

problems Bertismas & Teo (1998) Jain & Vazirani (1999)

uncapacitated facility location problem Inventory problems

Levi, Roundy and Shmoys (2006)

Minimum (Covering) Knapsack Problem Given a set of items F each with a

cost ci and a value ui

Want to find a subset of items with minimum cost such that the total value is at least D

testing

minimize i F ci xi

subject to i F ui xi D

xi {0,1} for each i F

Bad Integrality Gap

Consider the min knapsack problem with the following two items

Integer solution must take item 1 and incurs a cost of 1

LP solution can take all of item 2 and just 1/D fraction of item 1, incurring a cost of 1/D

c1 = 1u1 = D

c2 = 0u2 = D-1

Integer solutionLP solution

=1

1=D= D

Knapsack-Cover Inequalities

Proposed by Carr, Fleischer, Leung and Phillips (2000)

Consider a subset A of items in F

If we were to take all items in A, then we still need to take enough items to meet leftover demand

D

1 2 3 4 5 6 7

A = {1,2,3}

D –

u

(A)

u(A) = i 2 A ui

i 2 F n A ui xi ¸ D-u(A)

Knapsack-Cover Inequalities

This inequality adds nothing new, but we can now restrict the values of the items

where

since these inequalities only need to be valid for integer solutions

D

1

2

3

4 5 6 7

A = {1,2,3}

D –

u(A

)i 2 FnA ui xi ¸ D-u(A)

i 2 FnA ui(A) xi ¸ D-u(A)

ui(A) = min{ ui , D-u(A) }

Strengthened Min Knapsack LP When A = ; the knapsack-cover inequality becomes

which is the original min knapsack inequality New strengthened LP is

i 2 FnA ui(A) xi ¸ D-u(A)

) i 2 F ui xi ¸ D

Minimize i 2 F ci xi

subject to i 2 FnA ui(A) xi ¸ D- u(A), for each subset A

xi ¸ 0, for each i 2 F

Dual Linear Program

Dual of LP formed by knapsack-cover inequalities

optDual := max A µ F (D-u(A))v(A)

subject to µ F : i ui(A) v(A) · ci , for each i 2 F

v(A) ¸ 0, for each A µ F

Primal-Dual

D - u(A) = 5

Increase v(A)

D = 5

1.25 2 0.25 2 0.75

Dual Variables:

Initially all zeroA = ;A = {3}Dual Variables:

v(;) = 0.25

D - u(A) = 3

Increase v(A)

A = {3}

Primal-Dual

D - u(A) = 3

Increase v(A)

D = 5

1.25 1 0.25 2 0.75

Dual Variables:

v(;) = 0.25

A = {3}

Primal-Dual

D - u(A) = 3

Increase v(A)

D = 5

1.25 1 0.25 2 0.75

Dual Variables:

v(;) = 0.25

A = {3}

Primal-Dual

D - u(A) = 3

Increase v(A)

D = 5

1 1.75 0 2.92 0.5

Dual Variables:

v(;) = 0.25

Dual Variables:

v(;) = 0.25

v({3}) = 0.5

A = {3,5}

D - u(A) = 1

Increase v(A)

A = {3}

Primal-Dual

D - u(A) = 1

Increase v(A)

D = 5

1 1.75 0 2.92 0.5

A = {3,5}Dual Variables:

v(;) = 0.25

v({3}) = 0.5

A = {3}

Primal-Dual

D - u(A) = 1

Increase v(A)

D = 5

1 1.25 0 7.25 0

A = {3,5}Dual Variables:

v(;) = 0.25

v({3}) = 0.5

Dual Variables:

v(;) = 0.25

v({3}) = 0.5

v({3,5}) = 1

A = {3,5,1}

D - u(A) = -1

Increase v(A)Stop!

Primal-Dualc1 = 2.5u1 = 2

c2 = 2u2 = 1

c3 = 0.5u3 = 2

c4 = 10u4 = 5

c5 = 1.5u5 = 2

Primal-Dual

Cost = 4.5

c1 = 2.5u1 = 2

c2 = 2u2 = 1

c3 = 0.5u3 = 2

c4 = 10u4 = 5

c5 = 1.5u5 = 2

Opt. Integer

Cost = 4

Primal-Dual Summary

Start with all variables set to zero and solution A as the empty set

Increase variable v(A) until a dual constraint becomes tight for some item i

Add item i to solution A and repeat Stop once solution A has large enough

value to meet demand D Call final solution S and set xi = 1 for all

i 2 S

Analysis

Let l be last item added to solution S If we increased dual variable v(A)

then l was not in A Thus if v(A) > 0 then A µ ( S \ l ) Since u( S\ l ) < D

then u((S\ l )\A) < D – u(A)

Analysis (continued)

We have u((S\ l )\A) < D – u(A) if v(A) > 0

Cost of solution is

Dual

LP

Primal-Dual Theorem

For the min-cost covering knapsack problem, the LP relaxation with knapsack-cover inequalities can be used to derive a (simple) primal-dual 2-approximation algorithm.

Knapsack-Cover Inequalities Everywhere

Bansal, Buchbinder, Naor (2008) Randomized competitive algorithms for generalized caching (and weighted paging)

Bansal, Gupta, & Krishnaswamy (2010) 485-approximation algorithm for min-sum set cover

Bansal & Pruhs (2010)O(log log nP)-approximation algorithm

for general 1-machine preemptive scheduling + O(1) with identical deadlines

Minimum-Weight Late Jobs on 1 Machine

Each job j has processing time pj , deadline dj, weight wj

Choose a subset L of jobs of minimum-weight to be late - not scheduled to complete by deadline

This problem is (weakly) NP-hard can be solved in O( n j pj ) time [Lawler & Moore], (1+²)-approximation in O(n3/²) time [Sahni]

If there also are release dates that constrain when a job may start, no approximation result is possible - focus on max-weight set of jobs scheduled on time [Bar-Noy, Bar-Yehuda, Freund, Naor, & Schieber] - allow preemption [Bansal & Pruhs]

What if all deadlines are the same? Total processing time is j pj ! P WLOG assume schedule runs through

[0,P] Deadline D ) at least P-D units of

processing are done after D So just select set of total processing at

least P-D of minimum total weight , minimum-cost covering knapsack

problem

Same Idea for General Deadlines

Total processing time is j pj P WLOG assume schedule runs through [0,P] Assume d1 · d2 · … · dn

Deadline di ) among all jobs with deadlines · di , P(i)-di units of processing are done after di where S(i) = { j : dj di } and P(i) = j S(i) pjMinimize wj yj

subject to j S(i) pj yj ¸ P(i)-di, i=1,…,n yj ¸ 0, j=1,…,n

Strengthened LP – Knapsack CoversMinimize wj yj

subject to j S(L,i) pj (L,i) yj D(L,i), for each L,i

where S(L,i) = { j: dj di , j L}

D(L,i) = max{ j S(L,i) pj - di , 0}

pj (L,i) = min{ pj , D(L,i) }

Dual: Maximize D(L,i) v(L,i)subject to (L,i): j 2 S(L,i) pj(L,i) v(L,i) · wj for each j

v(L,i) ¸ 0 for each L,i

Primal-Dual Summary Start with all dual variables set to 0 and

solution A as the empty set Increase variable v(A,i) with largest D(A,i)

until a dual constraint becomes tight for some item i

Add item i to solution A and repeat Stop once solution A is sufficient so

remaining jobs N-A can be scheduled on time Examine each item j in A in reverse order

and delete j if reduced late set is still feasible Call final solution L* and set yj = 1 for all j 2 L*

Highlights of the Analysis

Lemma. Suppose current iteration increases v(L,i), and let L(i) be jobs put in final late set L* afterwards. Then 9 job k L(i) so that L*-{k} is not feasible.

Note: in previous case, since all deadlines were equal, the last job l added satisfies this property.Here, the reverse delete process is set exactly to ensure that the Lemma holds.

Highlights of the Analysis

Lemma. Suppose current iteration increases v(L,i), and let L(i) be jobs put in final late set L* afterwards. Then 9 job k L(i) so that L*-{k} is not feasible.

Lemma. j: j i, j L(i) – {k} pj(L,i) < D(L,i) if v(L,i)>0.

Highlights of the Analysis

Lemma. Suppose current iteration increases v(L,i), and let L(i) be jobs put in final late set L* afterwards. Then 9 job k L(i) so that L*-{k} is not feasible.

Lemma. j: j i, j L(i) – {k} pj(L,i) < D(L,i) if v(L,i)>0.

Fact. pk(L,i) D(L,i) (by definition of pk(L,i) )

Highlights of the Analysis

Lemma. Suppose current iteration increases v(L,i), and let L(i) be jobs put in final late set L* afterwards. Then 9 job k L(i) so that L*-{k} is not feasible.

Lemma. j: j i, j L(i) – {k} pj(L,i) < D(L,i) if v(L,i)>0.

Fact. pk(L,i) D(L,i) (by definition of pk(L,i) )

Corollary. j: j i, j L(i) pj(L,i) < 2D(L,i) if v(L,i)>0.

Previous Analysis (flashback)

We have u((S\ l )\A) < D – u(A) if v(A) > 0

Cost of solution is

Dual

LP

Highlights of the Analysis

Lemma. Suppose current iteration increases v(L,i), and let L(i) be jobs put in final late set L* afterwards. Then 9 job k L(i) so that L*-{k} is not feasible.

Lemma. j: j i, j L(i) – {k} pj(L,i) < D(L,i) if v(L,i)>0.

Fact. pk(L,i) D(L,i) (by definition of pk(L,i) )

Corollary. j: j i, j L(i) pj(L,i) < 2D(L,i) if v(L,i)>0.

Highlights of the Analysis

Corollary j: j i, j L(i) pj(L,i) < 2D(L,i) if v(L,i)>0.

Same trick here:j L* wj = j L* (L,i): j 2 S(L,i) pj(L,i) v(L,i)

= (L,i) v(L,i) j: j i, j L(i) pj(L,i)

· (L,i) 2D(L,i) v(L,i)

· 2 OPT

Primal-Dual Theorem

For the 1-machine min-weight late jobs scheduling problem with a common deadline, the LP relaxation with knapsack-cover inequalities can be used to derive a (simple) primal-dual 2-approximation algorithm.

General 1-Machine Min-Cost SchedulingEach job j has its own nondecreasing cost function fj (Cj ) – where Cj denotes completion time of job j

Assume that all processing times are integerGoal: construct schedule to minimize total cost

incurredLP variables – xjt =1 means job j has Cj =t

Knapsack cover constraint: for each t and L,require that total processing time of jobs finishing at time t or later is sufficiently large

Primal-Dual Theorem(s)

For 1-machine min-cost scheduling, LP relaxation with knapsack-cover inequalities can be used to derive a (simple) primal-dual pseudo-polynomial 2-approximation algorithm.

For 1-machine min-cost scheduling, LP relaxation with knapsack-cover inequalities can be used to derive a (simple) primal-dual (2+²)-approximation algorithm.

“Weak” LP Relaxation Total processing time is j pj P WLOG assume schedule runs

through [0,P] xjt = 1 means job j completes at time

tMinimize fj(t) xjt

subject to t 1,…,P xjt = 1, j=1,…,nj {1,…n} s {t,…,P}

pj xjs D(t)

t=1,…,P xjt 0 j=1,…,n; t=1,…,P

where D(t) = P-t+1.

Strong LP Relaxation Total processing time is j pj P WLOG assume schedule runs through

[0,P] xjt = 1 means job j completes at time tMinimize fj(t) xjt

subject to t {1,…,P} xjt = 1, for all jjL

s {t,…,P} pj (L,t) xjs ¸ D(L,t) for

all L,t xjt ¸ 0, for all j,t

where D(t) = P-t+1, D(L,t) = max{0, D(t)-jL pj }, and

pj(L,t) = min{pj, D(L,t)}.

Primal and Dual LPMinimize fj(t) xjt

subject to t {1,…,P} xjt = 1, for all j

j L s=t,…,P pj (L,t) xjs ¸ D(L,t) for all L,t xjt ¸ 0, for all j,tD(t) = P-t+1 D(L,t) = max{0, j

L pj –t+1}

pj(L,t) = min{pj, D(L,t)}

Maximize L t D(L,t) v(L,t)

subject to L: j Lt=1,…,s pj(L,t) v(L,t) fj(s) for all j,s

v(L,t) 0 for all L,t

Primal-Dual Summary Start w/ all dual variables set to 0 and each At = ; Increase variable v(At,t) with largest D(At,t) until

a dual constraint becomes tight for some item i (break ties by selecting latest time)

Add item i to solution As for all s · t and repeat Stop once solution A is sufficient so remaining

jobs N-A satisfy all demand constraints Focus on pairs (j,t) where t is latest job j is in At

and perform a reverse delete Set dj =t for job j by remaining pairs (j,t) Schedule in Earliest Due Date order

Primal-Dual Theorem

For 1-machine min-cost scheduling, LP relaxation with knapsack-cover inequalities can be used to derive a (simple) primal-dual pseudo-polynomial 2-approximation algorithm.

Removing the “Pseudo” with a (1+²) Loss This requires only standard

techniques For each job j, partition the potential

job completion times {1,…,P} into blocks so that within block the cost for j increases by · 1+²

Consider finest partition based on all n jobs

Now consider variables xjt that assign job j to finish in block t of this partition.

All other details remain basically the same.

Fringe Benefit: more general models, such as possible periods of machine non-availability

Primal-Dual Theorem

For 1-machine min-cost scheduling, LP relaxation with knapsack-cover inequalities can be used to derive a (simple) primal-dual (2+²)-approximation algorithm.

Some Open ProblemsGive a constant approximation algorithm for 1-machine min-sum scheduling with release dates allowing preemption Give a (1+²)-approximation algorithm for 1-machine min-sum scheduling, for arbitrarily small ² > 0Give an LP-based constant approximation algorithm for capacitated facility locationUse “configuration LP” to find an approximation algorithm for bin-packing problem that uses at most ONE bin more than optimal

Thank you!

Any questions?