approximation algorithms for stochastic combinatorial optimization part i: multistage problems...

Approximation Algorithms for Stochastic Combinatorial Optimization

Part I: Multistage problems

Anupam GuptaCarnegie Mellon University

stochastic optimization

Question: How to model uncertainty in the inputs? data may not yet be available obtaining exact data is difficult/expensive/time-consuming

Goal: make (near)-optimal decisions given some predictions (probability distribution on potential inputs).

Studied since the 1950s, and for good reason: many practical applications…

Approximation Algorithms

Recent development of approximation algorithms for NP-hard stochastic optimization problems.

Several different models, several different techniques.

I will give an overview of some of the models/results/ideas in the two talks (today and on Thursday).

two representative formulations for stochastic Steiner tree

the Steiner tree problem

Input: a metric spacea root vertex ra subset R of terminals

Output: a tree T connecting R to rof minimum length/cost.

Facts: NP-hard and APX-hard

MST is a 2-approximation cost(MST(R [ r)) ≤ 2 OPT(R)

[Byrka et al. STOC ’10] give a 1.39-approximation

model I: “two-stage” Steiner tree

The Model:Instead of one set R, we are given probability distribution ¼over subsets of nodes.

E.g., each node v independently belongs to R with probability pv

Or, may be explicitly defined over a small set of “scenarios”

pA = 0.6 pB = 0.25 pC = 0.15

model I: “two-stage” Steiner tree

Stage I (“Monday”)

Pick some set of edges EM

at costM(e) for each edge e

Stage II (“Tuesday”)

Random set R is drawn from ¼Pick some edges ET,R so that EM [ ET,R connects R to root

but cost changes to costT,R (e)

Objective Function: costM (EM) + E¼ [ costT,R (ET,R) ]

pA = 0.6 pB = 0.25 pC = 0.15

approximation algorithm

Objective Function:costM (EM) + E¼ [ costT,R (ET,R) ]

Optimum: Sets EM* and ET,R* which

achieve expected cost Z*

A c-approximation:Find sets EM and ET,R that

achieve expected cost c.Z*

for some small factor c.pA = 0.6 pB = 0.25 pC = 0.15

model II: average-case online

The Model:Instead of being given set R up-front, the nodes in R arrive one-by-one.

Node must be connected to the root as soon as it arrives.

However, the requests not drawn by an adversary, but randomly fromsome known stochastic process.

E.g., each node v independently chosen with probability pv


Measure of Goodness:Usual measure is competitive ratio

So we consider

or

E¾ [ cost of algorithm A on ¾ ]

E¾ [ OPT(set ¾) ]

E [ cost of algorithm A on ¾ ]

OPT(set ¾) max¾

cost of algorithm A on ¾

OPT(set ¾)E¾

a tale of two models…

(At least) two ways of modeling Steiner tree

problems become harder in the two-stage model(if no inflation, we can just solve the problem on Tuesday.)

problems become easier in the online stochastic model(we know that the “adversary” is just a stochastic process.)

Let us consider these two models in some detail now…

roadmap

the two-stage model eg: stochastic vertex cover (using LPs) how to solve two-stage stochastic LPs the sample average approximation method

the stochastic online model eg: stochastic Steiner tree (using sampling)

models with recourse

The problem instance is revealed in “stages” initially we perform some anticipatory actions at each stage, more information released we may take some more recourse actions at this point

Initially, given “guesses” about final problem instance(i.e., given probability distribution ¼ over problem instances)

Want to minimize:Cost(Initial actions) + E¼ [ cost of recourse actions ]

representations of ¼

“Explicit scenarios” model Complete listing of the sample space

“Black box” access to probability distribution generates an independent random sample from ¼

Also, independent decisions Each vertex v appears with probability pv indep. of others.

example I: vertex cover

vertex cover = set of vertices that hit all edges.

Finding a minimum cost vertex cover is NP-hard. 2-approx: several algorithms

easy one: solve the linear program relaxation and round

Boolean variable x(v) = 1 iff vertex v chosen in the vertex cover

minimize v c(v) x(v)

subject tox(v) + x(w) ≥ 1 for each edge (v,w) in edge set E

andx’s are in {0,1}

integer-program formulation

stochastic vertex cover

Explicit scenario model: M scenarios explicitly listed.Edge set Ek appears with prob. pk

Vertex costs c(v) on Monday, ck(v) on Tuesday if scenario k appears.

Pick V0 on Monday, Vk on Tuesday

such that (V0 [ Vk) covers Ek.

Minimize c(V0) + Ek [ ck(Vk) ] p1 = 0.1 p2 = 0.6 p3 = 0.3

Boolean variable x(v) = 1 iff v chosen on Monday, yk(v) = 1 iff v chosen on Tuesday if scenario k realized

minimize v c(v) x(v) + k pk [ v ck(v) yk(v) ]

subject to[ x(v) + yk(v) ] + [ x(w) + yk(w) ] ≥ 1 for each k, edge (v,w) in Ek

andx’s, y’s are Boolean

integer-program formulation

minimize v c(v) x(v) + k pk [ v ck(v) yk(v) ]

subject to[ x(v) + yk(v) ] + [ x(w) + yk(w) ] ≥ 1 for each k, edge (v,w) in Ek

Now choose V0 = { v | x(v) ≥ ¼ }, and Vk = { v | yk(v) ≥ ¼ }

Note: if we have explicit multi-stage solution with k stages, gives 2k approximation

We are increasing variables by factor of 4 we get a 4-approximation

linear-program relaxation

solving the LP and rounding

This idea useful for many stochastic problems Set cover, Facility location, some cut problems

Tricky when the sample space is exponentially large exponential number of variables and constraints natural (non-trivial) approaches have run-times depending on the variance

of the problem…

Shmoys and Swamy approach: consider this doubly-exponential vertex cover LP in black-box model can approximate it arbitrarily well, smaller run-times. solution has exponential size,

but we need only polynomially-large parts of it at a time.

roadmap

the two-stage model eg: stochastic vertex cover (using LPs) how to solve two-stage stochastic LPs the sample average approximation method

the stochastic online model eg: stochastic Steiner tree (using sampling)

Approximation Algorithms for Stochastic Problems

Part II: Online (Stochastic) Problems

Anupam GuptaCarnegie Mellon University

the Steiner tree problem

Input: a metric spacea root vertex ra subset R of terminals

Output: a tree T connecting R to rof minimum length/cost.

Facts: NP-hard and APX-hard

MST is a 2-approximation cost(MST(R [ r)) ≤ 2 OPT(R)

[Byrka et al. STOC ’10] give a 1.39-approximation


Requests:generated by a given stochastic process, have to be connected to the root.

Measure of Goodness:Want to minimize

E¾ [ cost of algorithm A on ¾ ]

E¾ [ OPT(set ¾) ]

recap: the standard online model

The Greedy Algorithm

Given: metric space, root vertex r, requests come online

measure of goodness

Competitive ratio of algorithm A:

max Cost(algorithm A on sequence ¾) ¾ Cost(optimum Steiner tree on ¾)


[Imase Waxman ’91] The greedy algorithm is O(log k) competitive for sequences of length k.


[Imase Waxman ’91] The greedy algorithm is O(log k) competitive for sequences of length k.And this is tight.

stochastic online model

random arrivals

Suppose demands are nodes in V drawn uniformly at random, independently of previous demands.

uniformity: not important could have (given) probability ¼: V [0,1]independence: important, lower bounds otherwise

Measure of goodness:

E¾ [ cost of algorithm A on ¾ ]E¾ [ OPT(set ¾) ]

Assume: know the length k of the sequence

greed is (still) bad

[Imase Waxman ’91] The greedy algorithm is £(log k) competitive for sequences of length k.

Tight example holds also for (uniformly) random demands.

however, we can prove…

For stochastic online Steiner tree, “augmented greedy” with known length k achieves

E¾ [ cost of algorithm A on ¾ ]E¾ [ OPT(set ¾) ]

≤ 4

Augmented greedy

1. Sample k vertices A = {a1, a2, …, ak} independently.

2. Build an MST T0 on these vertices A [ root r.

3. When actual demand points xt (for 1 · t · k) arrives,greedily connect xt to the tree Tt-1

Augmented greedy

Proof for augmented greedy

Let S = {x1, x2, …, xk}

Claim 1: E[ cost(T0) ] ≤ 2 £ E[ OPT(S) ]

Claim 2: E[ cost of k augmentations in Step 3 ] ≤ E[ cost(T0) ]




Ratio of expectations ≤ 4


Let S = {x1, x2, …, xk}

Claim 2: ES,A[ augmentation cost ] ≤ EA[ MST(A [ r) ]

Claim 2a: ES,A[ x2S d(x, A [ r) ] ≤ EA[ MST(A [ r) ]

Claim 2b: Ex,A[ d(x, A [ r) ] ≤ (1/k) EA[ MST(A [ r) ]





Claim 2b: Ex,A[ d(x, A [ r) ] ≤ (1/k) EA[ MST(A [ r) ]

= E[ distance from one random point to (k random points [ r) ]

≥ (1/k) * k * Ey, A-y[ distance(y, (A-y) [ r) ]

≥ E[ distance from one random point to (k-1 random points [ r) ]

≥ E[ distance from one random point to (k random points [ r) ]


Let S = {x1, x2, …, xk}

Claim 1: E[ cost(T0) ] ≤ 2 £ E[ OPT(S) ]

Claim 2: E[ cost of k augmentations in Step 3 ] ≤ E[ cost(T0) ]




Ratio of expectations ≤ 4

approximation algorithms for stochastic combinatorial optimization part i: multistage problems...

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