Using Fractional Primal-Dual to Schedule Split Intervals with - - PowerPoint PPT Presentation

Using Fractional Primal-Dual to Schedule Split Intervals with Demands

Reuven Bar-Yehuda (CS, Technion) Dror Rawitz (CRI, University of Haifa)

– p.1

Scheduling t-intervals with Demands

The problem: Set of jobs that require the utilization of a resource Amount of resource is normalized to 1

– p.2

Scheduling t-intervals with Demands

The problem: Set of jobs that require the utilization of a resource Amount of resource is normalized to 1 Job j:

t-interval

demand, dj ∈ [0, 1] weight, wj

– p.2

Scheduling t-intervals with Demands

The problem: Set of jobs that require the utilization of a resource Amount of resource is normalized to 1 Job j:

t-interval

demand, dj ∈ [0, 1] weight, wj A schedule S:

∀p ∈ R,

• j : p∈j∈S

dj ≤ 1

– p.2

Scheduling t-intervals with Demands

The problem: Set of jobs that require the utilization of a resource Amount of resource is normalized to 1 Job j:

t-interval

demand, dj ∈ [0, 1] weight, wj A schedule S:

∀p ∈ R,

• j : p∈j∈S

dj ≤ 1

Goal:

max

S

• j∈S

wj

– p.2

2-intervals Example

j5 d5 = 0.5 j4 d4 = 0.8 j3 d3 = 0.4 j2 d2 = 0.2 j1 d1 = 0.7

S = {j1, j4} is a feasible solution S = {j1, j2, j4} is a maximal solution

(with respect to set inclusion)

– p.3

Scheduling t-intervals with Demands

LP relaxation:

max

• j

wjxj

s.t.

• j : p∈j

djxj ≤ 1 ∀p ∈ R xj ∈ [0, 1] ∀j

b b b b b b b b b

– p.4

Scheduling t-intervals with Demands

LP relaxation:

max

• j

wjxj

s.t.

• j : p∈j

djxj ≤ 1 ∀p ∈ R Right-end-points xj ∈ [0, 1] ∀j

Observation: It is enough to consider right end-points of segments

b b b b b b b b b

– p.4

Scheduling t-intervals with Unit Demands

Observation: Scheduling t-intervals with unit demands reduces to IS on t-intervals graphs Poly-time solvable for t = 1 (IS on interval graphs)

– p.5

Scheduling t-intervals with Unit Demands

Observation: Scheduling t-intervals with unit demands reduces to IS on t-intervals graphs Poly-time solvable for t = 1 (IS on interval graphs) Positive result [BHNSS02]:

2t-approximation algorithm using a novel extension of

the local ratio technique, called fractional local ratio

– p.5

Scheduling t-intervals with Unit Demands

Observation: Scheduling t-intervals with unit demands reduces to IS on t-intervals graphs Poly-time solvable for t = 1 (IS on interval graphs) Positive result [BHNSS02]:

2t-approximation algorithm using a novel extension of

the local ratio technique, called fractional local ratio Negative results [BHNSS02]: APX-hard (even on t-union graphs) Cannot be approximated within O(t/ log t) unless P=NP

– p.5

Scheduling t-intervals with Demands

Observation: NP-hard even when t = 1, since knapsack is the special case where all intervals intersect

– p.6

Scheduling t-intervals with Demands

Observation: NP-hard even when t = 1, since knapsack is the special case where all intervals intersect Our results:

6t-approximation algorithm that uses a new extension of

primal-dual schema, called fractional primal-dual Fractional Primal-Dual ≡ Fractional Local Ratio Contiguous allocation of a non-fungible resource: Bi-criteria approx alg. that computes 4t-approx solutions that may need up to 4 times the given amount

• f resource

– p.6

Scheduling t-intervals with Demands

Observation: NP-hard even when t = 1, since knapsack is the special case where all intervals intersect Our results:

6t-approximation algorithm that uses a new extension of

primal-dual schema, called fractional primal-dual Fractional Primal-Dual ≡ Fractional Local Ratio Contiguous allocation of a non-fungible resource: Bi-criteria approx alg. that computes 4t-approx solutions that may need up to 4 times the given amount

• f resource

In this talk: we focus on fractional primal-dual

– p.6

Primal-Dual Algorithms — Maximization

(P) max n

j=1 wjxj

s.t. n

j=1 aijxj ≤ bi

∀i xj ≥ 0 ∀j (D) min m

i=1 biyi

s.t. m

i=1 aijyi ≥ wj

∀j yi ≥ 0 ∀i wT x∗ bT y∗ wT x∗

I

– p.7

Primal-Dual Algorithms — Maximization

(P) max n

j=1 wjxj

s.t. n

j=1 aijxj ≤ bi

∀i xj ≥ 0 ∀j (D) min m

i=1 biyi

s.t. m

i=1 aijyi ≥ wj

∀j yi ≥ 0 ∀i wT x∗ bT y∗ wT x∗

I

bT y wT x bT y/r

Idea: Find an integral primal solution x and a dual solution y s.t. wTx ≥ bTy/r

= ⇒ wTx ≥ bTy/r ≥ OPT(P)/r ≥ OPT/r

– p.7

Primal-Dual Algorithms — Maximization

(P) max n

j=1 wjxj

s.t. n

j=1 aijxj ≤ bi

∀i xj ≥ 0 ∀j (D) min m

i=1 biyi

s.t. m

i=1 aijyi ≥ wj

∀j yi ≥ 0 ∀i wT x∗ bT y∗ wT x∗

I

bT y wT x bT y/r

Idea: Find an integral primal solution x and a dual solution y s.t. wTx ≥ bTy/r

= ⇒ wTx ≥ bTy/r ≥ OPT(P)/r ≥ OPT/r

Question: How do we find such solutions?

– p.7

Primal-Dual Schema

Idea: Find an integral primal x and dual y that satisfy: Primal:

∀j, xj > 0 ⇒ m

i=1 aijyi = wj

Relaxed Dual:

∀i, yi > 0 ⇒ bi/r ≤ n

j=1 aijxj ≤ bi

– p.8

Primal-Dual Schema

Idea: Find an integral primal x and dual y that satisfy: Primal:

∀j, xj > 0 ⇒ m

i=1 aijyi = wj

Relaxed Dual:

∀i, yi > 0 ⇒ bi/r ≤ n

j=1 aijxj ≤ bi

n

• j=1

wjxj

P

=

n

• j=1

m

• i=1

aijyi

• xj =

m

• i=1

 

n

• j=1

aijxj   yi

RD

≥ 1 r

m

• i=1

biyi

– p.8

Primal-Dual Schema

Idea: Find an integral primal x and dual y that satisfy: Primal:

∀j, xj > 0 ⇒ m

i=1 aijyi = wj

Relaxed Dual:

∀i, yi > 0 ⇒ bi/r ≤ n

j=1 aijxj ≤ bi

n

• j=1

wjxj

P

=

n

• j=1

m

• i=1

aijyi

• xj =

m

• i=1

 

n

• j=1

aijxj   yi

RD

≥ 1 r

m

• i=1

biyi

Question: How do we find such solutions?

– p.8

Primal-Dual Schema

Idea: Find an integral primal x and dual y that satisfy: Primal:

∀j, xj > 0 ⇒ m

i=1 aijyi = wj

Relaxed Dual:

∀i, yi > 0 ⇒ bi/r ≤ n

j=1 aijxj ≤ bi

n

• j=1

wjxj

P

=

n

• j=1

m

• i=1

aijyi

• xj =

m

• i=1

 

n

• j=1

aijxj   yi

RD

≥ 1 r

m

• i=1

biyi

Question: How do we find such solutions? In each step we use only good constraints:

yi > 0 = ⇒ bi/r ≤ n

j=1 aijxj ≤ bi (∀ maximal solution)

x contains elements that are not over paid: xj = 1 = ⇒ m

i=1 aijyi = wj

– p.8

Primal-Dual vs. Fractional Primal-Dual

Standard Primal-Dual: In each iteration we consider a single valid constraint [BT95,BR01] Solution is compared to an optimal solution with respect to this constraint The superposition of these “local optima” may be significantly worse than the “global optimum”

– p.9

Primal-Dual vs. Fractional Primal-Dual

Standard Primal-Dual: In each iteration we consider a single valid constraint [BT95,BR01] Solution is compared to an optimal solution with respect to this constraint The superposition of these “local optima” may be significantly worse than the “global optimum” Fractional Primal-Dual: An optimal fractional solution x∗ is used as a yardstick In each iteration we consider a single constraint that is satisfied by x∗ but may be invalid

– p.9

Fractional Primal-Dual Algorithms

A new LP: Let P be the original LP relaxation Algorithm constructs constraints that are satisfied by x∗ Let P′ be the LP that consists of the objective function

• f P

, and the constraints used

⇒ x∗ is a feasible solution of P′

– p.10

Fractional Primal-Dual Algorithms

A new LP: Let P be the original LP relaxation Algorithm constructs constraints that are satisfied by x∗ Let P′ be the LP that consists of the objective function

• f P

, and the constraints used

⇒ x∗ is a feasible solution of P′ ⇒ OPT(P) ≤ OPT(P′)

– p.10

Fractional Primal-Dual Algorithms

A new LP: Let P be the original LP relaxation Algorithm constructs constraints that are satisfied by x∗ Let P′ be the LP that consists of the objective function

• f P

, and the constraints used

⇒ x∗ is a feasible solution of P′ ⇒ OPT(P) ≤ OPT(P′)

The dual: Algorithm constructs a solution y′ of D′ (dual of P′) It finds an integral solution x s.t. wTx ≥ bT y′/r

⇒ wTx ≥ bTy′/r ≥ OPT(P′)/r ≥ OPT(P)/r ≥ OPT/r

– p.10

Fractional Complementary Slackness Conds.

Find x and y′ such that Primal:

∀j, xj > 0 ⇒ m

i=1 aijy′ i = wj

Relaxed Dual:

∀i, y′

i > 0

⇒ ci/r ≤ n

j=1 aijxj and

n

j=1 aijx∗ j ≤ ci

– p.11

Fractional Complementary Slackness Conds.

Find x and y′ such that Primal:

∀j, xj > 0 ⇒ m

i=1 aijy′ i = wj

Relaxed Dual:

∀i, y′

i > 0

⇒ ci/r ≤ n

j=1 aijxj and

n

j=1 aijx∗ j ≤ ci

In this case:

n

• j=1

wjxj =

n

• j=1

m

• i=1

aijy′

i

• xj =

m

• i=1

n

• j=1

aijxj

• y′

i ≥ 1

r

m

• i=1

ciy′

i

– p.11

Fractional Complementary Slackness Conds.

Find x and y′ such that Primal:

∀j, xj > 0 ⇒ m

i=1 aijy′ i = wj

Relaxed Dual:

∀i, y′

i > 0

⇒ ci/r ≤ n

j=1 aijxj and

n

j=1 aijx∗ j ≤ ci

In this case:

n

• j=1

wjxj =

n

• j=1

m

• i=1

aijy′

i

• xj =

m

• i=1

n

• j=1

aijxj

• y′

i ≥ 1

r

m

• i=1

ciy′

i

x is r-approximate, since cT y′ ≥ OPT(P′) ≥ wTx∗ = OPT(P) ≥ OPT

– p.11

Scheduling t-intervals with Demands

Two special cases:

• 1. wide jobs: dj > 1

2 for all j

• 2. narrow jobs :dj ≤ 1

2 for all j

– p.12

Scheduling t-intervals with Demands

Two special cases:

• 1. wide jobs: dj > 1

2 for all j

• 2. narrow jobs :dj ≤ 1

2 for all j

Wide jobs ≡ unit demands

⇒ 2t-approximation algorithm [BHNSS02]

– p.12

Scheduling t-intervals with Demands

Two special cases:

• 1. wide jobs: dj > 1

2 for all j

• 2. narrow jobs :dj ≤ 1

2 for all j

Wide jobs ≡ unit demands

⇒ 2t-approximation algorithm [BHNSS02]

Narrow jobs: 4t-approximation algorithm using fractional primal-dual

– p.12

Scheduling t-intervals with Demands

Two special cases:

• 1. wide jobs: dj > 1

2 for all j

• 2. narrow jobs :dj ≤ 1

2 for all j

Wide jobs ≡ unit demands

⇒ 2t-approximation algorithm [BHNSS02]

Narrow jobs: 4t-approximation algorithm using fractional primal-dual General case: Solve separately and return solution of greater weight

– p.12

Scheduling t-intervals with Demands

Two special cases:

• 1. wide jobs: dj > 1

2 for all j

• 2. narrow jobs :dj ≤ 1

2 for all j

Wide jobs ≡ unit demands

⇒ 2t-approximation algorithm [BHNSS02]

Narrow jobs: 4t-approximation algorithm using fractional primal-dual General case: Solve separately and return solution of greater weight The schedule returned is 6t-approximate since either

OPT(narrow) ≥ 2

3OPT, or OPT(wide) ≥ 1 3 OPT

– p.12

Scheduling t-intervals - Narrow Jobs

N(j)

△

= set of jobs that intersect j N[j]

△

= N(j) ∪ {j} j

– p.13

Scheduling t-intervals - Narrow Jobs

N(j)

△

= set of jobs that intersect j N[j]

△

= N(j) ∪ {j} j

Lemma: ∃j such that x∗ satisfies the constraint:

(1 − dj)zj +

k∈N(j) dkzk ≤ 1 − 2dj + 2t

– p.13

Scheduling t-intervals - Narrow Jobs

N(j)

△

= set of jobs that intersect j N[j]

△

= N(j) ∪ {j} j

Lemma: ∃j such that x∗ satisfies the constraint:

(1 − dj)zj +

k∈N(j) dkzk ≤ 1 − 2dj + 2t

A j-maximal solution x satisfies:

(1 − dj)xj +

k∈N(j) dkxk ≥ 1 − dj

– p.13

Scheduling t-intervals - Narrow Jobs

N(j)

△

= set of jobs that intersect j N[j]

△

= N(j) ∪ {j} j

Lemma: ∃j such that x∗ satisfies the constraint:

(1 − dj)zj +

k∈N(j) dkzk ≤ 1 − 2dj + 2t

A j-maximal solution x satisfies:

(1 − dj)xj +

k∈N(j) dkxk ≥ 1 − dj

For narrow jobs: (1 − 2dj + 2t)/(1 − dj) ≤ 4t

– p.13

Open Problems

Find t-approximation algorithm in the unit demands case (even only for t-union graphs) Improve approximation ratio in the general case Find applications of fractional primal-dual/local ratio (rectangle packing [LNO02,R05])

– p.14