Approximation Algorithms for Stochastic Optimization by gjjur4356

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```									Approximation Algorithms for
Stochastic Optimization

Chaitanya Swamy
Caltech and U. Waterloo

Joint work with David Shmoys
Cornell University
Stochastic Optimization
• Way of modeling uncertainty.
• Exact data is unavailable or expensive – data is
uncertain, specified by a probability distribution.
Want to make the best decisions given this
uncertainty in the data.
• Applications in logistics, transportation models,
financial instruments, network design, production
planning, …
• Dates back to 1950’s and the work of Dantzig.
Stochastic Recourse Models
Given   : Probability distribution over inputs.
or hedge against uncertainty.
Uncertainty evolves through various stages.
Learn new information in each stage.
Can take recourse actions in each stage – can augment
earlier solution paying a recourse cost.

Choose initial (stage I) decisions to minimize
(stage I cost) + (expected recourse cost).
2-stage problem           k-stage problem
 2 decision points        k decision points
stage I
0.2
stage I          0.2             0.4
0.3             stage II
0.3 0.1      0.02
0.5

stage II scenarios

scenarios in stage k
2-stage problem                 k-stage problem
 2 decision points              k decision points
stage I
0.2
stage I                 0.2             0.4
0.3             stage II
0.3 0.1      0.02
0.5

stage II scenarios

scenarios in stage k

Choose stage I decisions to minimize
expected total cost =
(stage I cost) + Eall scenarios [cost of stages 2 … k].
2-Stage Stochastic Facility Location
Distribution over clients gives
the set of clients to serve.

Stage I: Open some facilities in
advance; pay cost fi for facility i.
facility   stage I facility   Stage I cost = ∑(i opened) fi .
client set D
2-Stage Stochastic Facility Location
Distribution over clients gives
the set of clients to serve.

Stage I: Open some facilities in
advance; pay cost fi for facility i.
facility   stage I facility      Stage I cost = ∑(i opened) fi .

Actual scenario A = {       clients to serve}, materializes.
Stage II: Can open more facilities to serve clients in A; pay cost
fiA to open facility i. Assign clients in A to facilities.
Stage II cost = ∑ i opened in fiA + (cost of serving clients in A).
scenario A
Want to decide which facilities to open in stage I.
Goal: Minimize Total Cost =
(stage I cost) + EA D [stage II cost for A].

How is the probability distribution specified?
• A short (polynomial) list of possible scenarios
• Independent probabilities that each client exists
• A black box that can be sampled.
Approximation Algorithm
Hard to solve the problem exactly.
Even special cases are #P-hard.
Settle for approximate solutions. Give polytime
algorithm that always finds near-optimal solutions.
A is a a-approximation algorithm if,
•A runs in polynomial time.
•A(I) ≤ a.OPT(I) on all instances I,
a is called the approximation ratio of A.
Overview of Previous Work
• polynomial scenario model: Dye, Stougie & Tomasgard;
Ravi & Sinha; Immorlica, Karger, Minkoff & Mirrokni.
• Immorlica et al.: also consider independent activation model
proportional costs: (stage II cost) = l(stage I cost),
e.g., fiA = l.fi for each facility i, in each scenario A.
• Gupta, Pál, Ravi & Sinha (GPRS04): black-box model but also
with proportional costs.
• Shmoys, S (SS04): black-box model with arbitrary costs.
approximation scheme for 2-stage LPs + rounding procedure
“reduces” stochastic problems to their deterministic versions.
for some problems improve upon previous results.
Boosted Sampling (GPRS04)
Proportional costs: (stage II cost) = l(stage I cost)
Note: l is same as s in previous talk.
– Sample l times from distribution
– Use “suitable” algorithm to solve deterministic
instance consisting of sampled scenarios (e.g., all
sampled clients) – determines stage I decisions
Analysis relies on the existence of cost-shares that
can be used to share the stage I cost among
sampled scenarios.
Shmoys, S ’04 vs. Boosted sampling
Both work in the black-box model: arbitrary distributions.

– Can handle arbitrary costs Need proportional costs:
in the two stages.         (stage II cost) = l(stage I cost)
l can depend on scenario.
– LP rounding: give an        Primal-dual approach: cost-
algorithm to solve the      shares obtained by exploiting
stochastic LP.              structure via primal-dual schema.

– Need many more samples Need only l samples.
to solve stochastic LP.
Stochastic Set Cover (SSC)
Universe U = {e1, …, en }, subsets S1, S2, …, Sm  U, set S has
weight wS.
Deterministic problem: Pick a minimum weight collection of
sets that covers each element.
Stochastic version: Set of elements to be covered is given by
a probability distribution.
– choose some sets initially paying wS for set S
– subset A  U to be covered is revealed
– can pick additional sets paying wSA for set S.
Minimize (w-cost of sets picked in stage I) +
EA U [wA-cost of new sets picked for scenario A].
A Linear Program for SSC
For simplicity, consider wSA = WS for every scenario A.
wS : stage I weight of set S
pA : probability of scenario A  U.
xS : indicates if set S is picked in stage I.
yA,S : indicates if set S is picked in scenario A.
Minimize ∑S wSxS + ∑AU pA ∑S WSyA,S
subject to,
∑S:eS xS + ∑S:eS yA,S ≥ 1   for each A  U, eA
xS, yA,S ≥ 0   for each S, A.
Exponential number of variables and exponential number
of constraints.
A Rounding Theorem
Assume LP can be solved in polynomial time.
Suppose for the deterministic problem, we have an
a-approximation algorithm wrt. the LP relaxation, i.e.,
A such that A(I) ≤ a.(optimal LP solution for I)
for every instance I.
e.g., “the greedy algorithm” for set cover is a
log n-approximation algorithm wrt. LP relaxation.
Theorem: Can use such an a-approx. algorithm to get a
2a-approximation algorithm for stochastic set cover.
Rounding the LP
Assume LP can be solved in polynomial time.
Suppose we have an a-approximation algorithm wrt. the LP
relaxation for the deterministic problem.
Let (x,y) : optimal solution with cost OPT.
∑S:eS xS + ∑S:eS yA,S ≥ 1       for each A  U, eA
 for every element e, either
∑S:eS xS ≥ ½    OR       in each scenario A : eA, ∑S:eS yA,S ≥ ½.
Let E = {e : ∑S:eS xS ≥ ½}.
So (2x) is a fractional set cover for the set E can round to get
an integer set cover S for E of cost ∑SS wS ≤ a(∑S 2wSxS) .
S is the first stage decision.
Rounding (contd.)
Sets       Set in S

Elements      Element in E
A

Consider any scenario A. Elements in A E are covered.
For every e A\E, it must be that ∑S:eS yA,S ≥ ½.
So (2yA) is a fractional set cover for A\E  can round to
get a set cover of W-cost ≤ a(∑S 2WSyA,S) .
Using this to augment S in scenario A, expected cost
≤ ∑SS wS + 2a.∑ AU pA (∑S WSyA,S)   ≤ 2a.OPT.
Rounding (contd.)
An a-approx. algorithm for deterministic problem gives a
2a-approximation guarantee for stochastic problem.
In the polynomial-scenario model, gives simple polytime
approximation algorithms for covering problems.
• 2log n-approximation for SSC.
• 4-approximation for stochastic vertex cover.
• 4-approximation for stochastic multicut on trees.
Ravi & Sinha gave a log n-approximation algorithm for
SSC, 2-approximation algorithm for stochastic vertex
cover in the polynomial-scenario model.
Rounding the LP
Assume LP can be solved in polynomial time.
Suppose we have an a-approximation algorithm wrt. the LP
relaxation for the deterministic problem.
Let (x,y) : optimal solution with cost OPT.
∑S:eS xS + ∑S:eS yA,S ≥ 1       for each A  U, eA
 for every element e, either
∑S:eS xS ≥ ½      OR     in each scenario A : eA, ∑S:eS yA,S ≥ ½.
Let E = {e : ∑S:eS xS ≥ ½}.
So (2x) is a fractional set cover for the set E can round to get
an integer set cover S of cost ∑SS wS ≤ a(∑S 2wSxS) .
S is the first stage decision.
A Compact Convex Program
pA : probability of scenario A  U.
xS : indicates if set S is picked in stage I.
Minimize h(x) = ∑S wSxS + ∑AU pAfA(x) s.t. xS ≥ 0 for each S
(SSC-P)
where     fA(x) = min. ∑S WSyA,S
s.t.    ∑S:eS yA,S ≥ 1 – ∑S:eS xS for each eA
yA,S ≥ 0               for each S.

Equivalent to earlier LP.
Each fA(x) is convex, so h(x) is a convex function.
The General Strategy
1. Get a (1+e)-optimal fractional first-stage solution
(x) by solving the convex program.
2. Convert fractional solution (x) to integer solution
– decouple the two stages near-optimally
– use a-approx. algorithm for the deterministic problem
to solve subproblems.

Obtain a c.a-approximation algorithm for the
stochastic integer problem.
Many applications: set cover, vertex cover, facility
location, multicut on trees, …
Solving the Convex Program
Minimize h(x) subject to xP.               h(.) : convex
Ellipsoid method                                            P
y
• Need a procedure that at any point y,
if yP, returns a violated inequality
which shows that yP
Solving the Convex Program
Minimize h(x) subject to xP.               h(.) : convex
Ellipsoid method                                                     P
• Need a procedure that at any point y,
if yP, returns a violated inequality               y h(x) ≤ h(y)
which shows that yP
d
of h(.) at y
d is a subgradient of h(.) at u, if "v, h(v)-h(u) ≥ d.(v-u).
m

• Given such a procedure, ellipsoid runs in polytime and returns
points x1, x2, …, xkP such that mini=1…k h(xi) is close to OPT.

Computing subgradients is hard.        Evaluating h(.) is hard.
Solving the Convex Program
Minimize h(x) subject to xP.             h(.) : convex
Ellipsoid method                                               P
• Need a procedure that at any point y,
if yP, returns a violated inequality            y h(x) ≤ h(y)
which shows that yP
if yP, computes an approximate               d'
m
d' is an e-subgradient at u,
if "vP, h(v)-h(u) ≥ d'.(v-u) – e.h(u).

• Given such a procedure, can compute point xP such that
h(x) ≤ OPT/(1-e) + r without ever evaluating h(.)!
Putting it all together
Get solution x with h(x) close to OPT.

Sample initially to detect if OPT is large – this
allows one to get a (1+e).OPT guarantee.

Theorem: (SSC-P) can be solved to within a factor
of (1+e) in polynomial time, with high probability.
Gives a (2log n+e)-approximation algorithm for
the stochastic set cover problem.
A Solvable Class of Stochastic LPs
Minimize h(x) = w.x + ∑AU pAfA(x) s.t. x P  m 0
≥

where    fA(x) = min. wA.yA + cA.rA
s.t.   DA rA + TA yA ≥ jA – TA x
yA m, rA n, yA, rA ≥ 0.

Theorem: Can get a (1+e)-optimal solution for this class of
stochastic programs in polynomial time.
Includes covering problems (e.g., set cover, network design,
multicut), facility location problems, multicommodity flow.
Moral of the Story
• Even though the stochastic LP relaxation has
exponentially many variables and constraints, we
can still obtain near-optimal fractional first-stage
decisions
• Fractional first-stage decisions are sufficient to
decouple the two stages near-optimally
• Many applications: set cover, vertex cover, facility
locn., multicommodity flow, multicut on trees, …
• But we have to solve convex program with many
samples (not just l)!
Sample Average Approximation
Sample Average Approximation (SAA) method:
– Sample initially N times from scenario distribution
– Solve 2-stage problem estimating pA with frequency of occurrence
of scenario A
How large should N be to ensure that an optimal solution to
sampled problem is a (1+e)-optimal solution to original problem?
Kleywegt, Shapiro & Homem De-Mello (KSH01):
– bound N by variance of a certain quantity – need not be polynomially
bounded even for our class of programs.
S, Shmoys ’05 :
– show using e-subgradients that for our class, N can be poly-bounded.
Charikar, Chekuri & Pál ’05:
– give another proof that for a class of 2-stage problems, N can be
poly-bounded.
Multi-stage Problems
Given   : Distribution over inputs.          k-stage problem
 k decision points
Stage I : Make some advance decisions
– hedge against uncertainty.                                    stage I
0.2            0.4
Uncertainty evolves in various stages.                    0.3             stage II
0.5
Learn new information in each stage.
Can take recourse actions in each
stage – can augment earlier solution
paying a recourse cost.
scenarios in stage k
Choose stage I decisions to minimize
expected total cost =
(stage I cost) + Eall scenarios [cost of stages 2 … k].
Multi-stage Problems
Fix k = number of stages.
LP-rounding: S, Shmoys ’05
– Ellipsoid-based algorithm extends       Computing e-subgradients
– SAA method also works                   is significantly harder,
black-box model, arbitrary costs          need several new ideas
Rounding procedure of SS04 can be easily adapted: lose an
O(k)-factor over the deterministic guarantee
– O(k)-approx. for k-stage vertex cover, facility location, multicut on
trees; k.log n-approx. for k-stage set cover
Gupta, Pál, Ravi & Sinha ’05: boosted sampling extends but with
outcome-dependent proportional costs
– 2k-approx. for k-stage Steiner tree (also Hayrapetyan, S & Tardos)
– factors exponential in k for k-stage vertex cover, facility location
Open Questions
• Combinatorial algorithms in the black box model
and with general costs. What about strongly
polynomial algorithms?
• Incorporating “risk” into stochastic models.
• Obtaining approximation factors independent of k
for k-stage problems.
Integrality gap for covering problems does not increase.
Munagala has obtained a 2-approx. for k-stage VC.
• Is there a larger class of doubly exponential LPs
that one can solve with (more general) techniques?
Thank You.

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