A Lasserre-based (1 + ) -approximation for Makespan Scheduling - - PowerPoint PPT Presentation

A Lasserre-based (1 + ε)-approximation for Makespan Scheduling with Precedence Constraints

Elaine Levey and Thomas Rothvoss

Makespan Scheduling w. Precedence Constraints

Input:

◮ Unit-size jobs J with precedence constraints ◮ Number m of identical machines

1 2 3 4 5 6 7 8 9 10 11 12 13 machine 1 machine 2 machine m

Makespan Scheduling w. Precedence Constraints

Input:

◮ Unit-size jobs J with precedence constraints ◮ Number m of identical machines

Goal: Schedule jobs as to minimize makespan 1 2 3 4 5 6 7 8 9 10 11 12 13 machine 1 machine 2 machine m 1 2 3 4 5 6 7 8 9 10 11 12 13 makespan

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 9

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 9

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 9

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 9

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 9

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 9

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 9

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9 10

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9 10 11

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9 10 11 12

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9 10 11 12 13

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9 10 11 12 13 Analysis:

◮ There is a chain active at all non-busy times!

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9 10 11 12 13 ≤ OPT Analysis:

◮ There is a chain active at all non-busy times! ◮ Length of chain ≤ OPT

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9 10 11 12 13 ≤ OPT ≤ OPT Analysis:

◮ There is a chain active at all non-busy times! ◮ Length of chain ≤ OPT ◮ Length of busy times ≤ OPT

List Scheduling (Graham ’66)

Algorithm: (1) FOR slots at time t = 1 TO ∞ DO: (2) Select any job that has all predecessors completed 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 m 1 2 3 4 5 6 7 8 9 10 11 12 13 ≤ OPT ≤ OPT Analysis:

◮ There is a chain active at all non-busy times! ◮ Length of chain ≤ OPT ◮ Length of busy times ≤ OPT

⇒ 2-apx ((2 − 2

m) for pj = 1; 2 − 1 m for arbitrary pj)

A linear programming formulation

T

• t=1

xjt = 1 ∀j ∈ J

• j∈J

xjt ≤ m ∀t ∈ [T]

• t′≤t

xit′ ≤

• t′≤t+1

xjt′ ∀i ≺ j ∀t ∈ [T] 0 ≤ xjt ≤ 1 ∀j ∈ J ∀t ∈ [T] Interpretation: xjt =

• 1

if run j at time t

• therwise

A linear programming formulation

T

• t=1

xjt = 1 ∀j ∈ J

• j∈J

xjt ≤ m ∀t ∈ [T]

• t′≤t

xit′ ≤

• t′≤t+1

xjt′ ∀i ≺ j ∀t ∈ [T] 0 ≤ xjt ≤ 1 ∀j ∈ J ∀t ∈ [T] Interpretation: xjt =

• 1

if run j at time t

• therwise

1 2 3 4 5 6 time 1 2 3 4 5 6 1 2 3 4 5 6 M1 M2

◮ Integrality gap: 2 − Θ( 1 m)

Main Theorem

Unit size job makespan minimization:

◮ (2 − Θ( 1 m))-apx in poly-time [Graham ’66]

Main Theorem

Unit size job makespan minimization:

◮ (2 − Θ( 1 m))-apx in poly-time [Graham ’66] ◮ No (2 − ε)-apx under variant of UGC [Svensson ’10]

Main Theorem

Unit size job makespan minimization:

◮ (2 − Θ( 1 m))-apx in poly-time [Graham ’66] ◮ No (2 − ε)-apx under variant of UGC [Svensson ’10]

Open problems:

◮ Is P3 | pj = 1, prec | Cmax NP-hard? [Garey Johnson ’79]

Main Theorem

Unit size job makespan minimization:

◮ (2 − Θ( 1 m))-apx in poly-time [Graham ’66] ◮ No (2 − ε)-apx under variant of UGC [Svensson ’10]

Open problems:

◮ Is P3 | pj = 1, prec | Cmax NP-hard? [Garey Johnson ’79] ◮ Is there a PTAS for P3 | pj = 1, prec | Cmax?

[Part of # 1 of “Open problems in Scheduling”; Schuurman, Woeginger ’99]

Main Theorem

Unit size job makespan minimization:

◮ (2 − Θ( 1 m))-apx in poly-time [Graham ’66] ◮ No (2 − ε)-apx under variant of UGC [Svensson ’10]

Open problems:

◮ Is P3 | pj = 1, prec | Cmax NP-hard? [Garey Johnson ’79] ◮ Is there a PTAS for P3 | pj = 1, prec | Cmax?

[Part of # 1 of “Open problems in Scheduling”; Schuurman, Woeginger ’99]

Theorem (Levey, R. 15)

The integrality gap of LP is 1 + ε after (log n)O(log log n)-Lasserre rounds (for constants m and ε > 0).

◮ Running time n(log n)O(log log n) ◮ Sherali-Adams suffices

Lift-and-project methods K

• int. hull
• f K

b b b

Lift-and-project methods K

• int. hull
• f K

b b b b

Las1(K)

b b b

Lift-and-project methods K

• int. hull
• f K

b b b b

Las2(K)

b b b

Lift-and-project methods K

• int. hull
• f K

b b b b Las3(K)

b b b

Lift-and-project methods K

• int. hull
• f K

b b b b

Lasn(K)

b b b

Round-t Lasserre relaxation

◮ Given: K = {x ∈ Rn | Ax ≥ b}.

Round-t Lasserre relaxation

◮ Given: K = {x ∈ Rn | Ax ≥ b}. ◮ Introduce variables for I ⊆ [n], |I| ≤ t

yI ≡

• i∈I

(xi = 1)

Round-t Lasserre relaxation

◮ Given: K = {x ∈ Rn | Ax ≥ b}. ◮ Introduce variables for I ⊆ [n], |I| ≤ t

yI ≡

• i∈I

(xi = 1)

◮ y{i} = xi

Round-t Lasserre relaxation

◮ Given: K = {x ∈ Rn | Ax ≥ b}. ◮ Introduce variables for I ⊆ [n], |I| ≤ t

yI ≡

• i∈I

(xi = 1)

◮ y{i} = xi ◮ y∅ = 1

Round-t Lasserre relaxation

◮ Given: K = {x ∈ Rn | Ax ≥ b}. ◮ Introduce variables for I ⊆ [n], |I| ≤ t

yI ≡

• i∈I

(xi = 1)

◮ y{i} = xi ◮ y∅ = 1

Round-t Lasserre relaxation

(yI∪J)|I|,|J|≤t

• i∈[n] AℓiyI∪J∪{i} − bℓyI∪J
• |I|,|J|≤t
• ∀ℓ ∈ [m]

y∅ = 1

Round-t Lasserre relaxation

◮ Given: K = {x ∈ Rn | Ax ≥ b}. ◮ Introduce variables for I ⊆ [n], |I| ≤ t

yI ≡

• i∈I

(xi = 1)

◮ y{i} = xi ◮ y∅ = 1

Round-t Lasserre relaxation

(yI∪J)|I|,|J|≤t

• i∈[n] AℓiyI∪J∪{i} − bℓyI∪J
• |I|,|J|≤t
• ∀ℓ ∈ [m]

y∅ = 1 moment matrix

Round-t Lasserre relaxation

◮ Given: K = {x ∈ Rn | Ax ≥ b}. ◮ Introduce variables for I ⊆ [n], |I| ≤ t

yI ≡

• i∈I

(xi = 1)

◮ y{i} = xi ◮ y∅ = 1

Round-t Lasserre relaxation

(yI∪J)|I|,|J|≤t

• i∈[n] AℓiyI∪J∪{i} − bℓyI∪J
• |I|,|J|≤t
• ∀ℓ ∈ [m]

y∅ = 1 moment matrix slack moment matrix

Round-t Lasserre relaxation

◮ Given: K = {x ∈ Rn | Ax ≥ b}. ◮ Introduce variables for I ⊆ [n], |I| ≤ t

yI ≡

• i∈I

(xi = 1)

◮ y{i} = xi ◮ y∅ = 1

Round-t Lasserre relaxation

(yI∪J)|I|,|J|≤t

• i∈[n] AℓiyI∪J∪{i} − bℓyI∪J
• |I|,|J|≤t
• ∀ℓ ∈ [m]

y∅ = 1 moment matrix slack moment matrix

◮ Solvable in time nO(t)mO(1)

Inducing on one variable

Lemma

For y ∈ Last(K), t ≥ 1, i ∈ [n], y ∈ conv{z ∈ Last−1(K) | zi ∈ {0, 1}}. y Last 1 y{i} R2n−1

Inducing on one variable

Lemma

For y ∈ Last(K), t ≥ 1, i ∈ [n], y ∈ conv{z ∈ Last−1(K) | zi ∈ {0, 1}}. y Last

b b

z(0) z(0)

{i} = 0

z(1) z(1)

{i} = 1

Last−1 1 y{i} R2n−1

Inducing on one variable

Lemma

For y ∈ Last(K), t ≥ 1, i ∈ [n], y ∈ conv{z ∈ Last−1(K) | zi ∈ {0, 1}}. y Last

b b

z(0) z(0)

{i} = 0

z(1) z(1)

{i} = 1

Last−1 1 y{i} R2n−1 Operation: “Induce on xi = 1” y Last

b z

zi = 1 Last−1 1 y{i} R2n−1

Inducing on one variable

Lemma

For y ∈ Last(K), t ≥ 1, i ∈ [n], y ∈ conv{z ∈ Last−1(K) | zi ∈ {0, 1}}. y Last

b b

z(0) z(0)

{i} = 0

z(1) z(1)

{i} = 1

Last−1 1 y{i} R2n−1 Operation: “Induce on xi = 1” y Last

b z

zi = 1 Last−1 1 y{i} R2n−1

◮ Observation: Conditioning only shrinks support!

Partition

T

Partition

T

◮ Partition time horizon into binary family

Partition

T

◮ Partition time horizon into binary family

Partition

T

◮ Partition time horizon into binary family

Partition

T . . . log(T)

◮ Partition time horizon into binary family

Partition

T . . . log(T) supp(j) supp(j′) supp(j′′)

◮ Partition time horizon into binary family

Partition

T . . . log(T)

◮ Partition time horizon into binary family ◮ Assign jobs to min interval containing support

The algorithm

. . . . . . time

The algorithm

. . . . . . O(log log n)2 time (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T

The algorithm

. . . . . . O(log log n)2 time (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T

The algorithm

. . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs

The algorithm

I . . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs (3) For each interval I directly below middle levels: Recursively call algorithm I with {j | supp(j) ⊆ I} → valid schedule for bottoms jobs

The algorithm

I . . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs (3) For each interval I directly below middle levels: Recursively call algorithm I with {j | supp(j) ⊆ I} → valid schedule for bottoms jobs

The algorithm

I . . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs (3) For each interval I directly below middle levels: Recursively call algorithm I with {j | supp(j) ⊆ I} → valid schedule for bottoms jobs

The algorithm

I . . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs (3) For each interval I directly below middle levels: Recursively call algorithm I with {j | supp(j) ⊆ I} → valid schedule for bottoms jobs

The algorithm

I . . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs (3) For each interval I directly below middle levels: Recursively call algorithm I with {j | supp(j) ⊆ I} → valid schedule for bottoms jobs

The algorithm

I . . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs (3) For each interval I directly below middle levels: Recursively call algorithm I with {j | supp(j) ⊆ I} → valid schedule for bottoms jobs

The algorithm

. . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs (3) For each interval I directly below middle levels: Recursively call algorithm I with {j | supp(j) ⊆ I} → valid schedule for bottoms jobs (4) Schedule top jobs

The algorithm

. . . . . . O(log log n) time bottom middle top (1) Apply conditioning until max chain length of jobs in first O(log log n)2 levels down to ε′T (2) Among first O(log log n)2 levels pick O(log log n) of consecutive middle levels and discard jobs (3) For each interval I directly below middle levels: Recursively call algorithm I with {j | supp(j) ⊆ I} → valid schedule for bottoms jobs (4) Schedule top jobs

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T.

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T.

I I1 I2

◮ Pick an interval I

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T.

I I1 I2

◮ Pick an interval I and consider jobs assigned to it.

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T. j ≥ δ|I|

I I1 I2

◮ Pick an interval I and consider jobs assigned to it. ◮ Consider j with δ|I| successors

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T. j ≥ δ|I|

I I1 I2

◮ Pick an interval I and consider jobs assigned to it. ◮ Consider j with δ|I| successors ⇒ condition on xj,I2 = 1

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T. j ≥ δ|I|

I I1 I2

◮ Pick an interval I and consider jobs assigned to it. ◮ Consider j with δ|I| successors ⇒ condition on xj,I2 = 1 ◮ Needed ≤ 1 δ times per interval

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T. j ≥ δ|I|

I I1 I2

◮ Pick an interval I and consider jobs assigned to it. ◮ Consider j with δ|I| successors ⇒ condition on xj,I2 = 1 ◮ Needed ≤ 1 δ times per interval; 2O(log log n)2 intervals

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T. j ≥ δ|I|

I I1 I2

◮ Pick an interval I and consider jobs assigned to it. ◮ Consider j with δ|I| successors ⇒ condition on xj,I2 = 1 ◮ Needed ≤ 1 δ times per interval; 2O(log log n)2 intervals ◮ Loose ≤ 1 δ · 2O(log log n)2 Lasserre-rounds

Reducing chain length

Lemma

After conditioning 2O(log log n)2 · 1

ε′ times, maximum length chain

among top jobs is ≤ ε′ · T. j ≥ δ|I|

I I1 I2

◮ Pick an interval I and consider jobs assigned to it. ◮ Consider j with δ|I| successors ⇒ condition on xj,I2 = 1 ◮ Needed ≤ 1 δ times per interval; 2O(log log n)2 intervals ◮ Loose ≤ 1 δ · 2O(log log n)2 Lasserre-rounds ◮ Longest chain in top jobs: δT · O(log log n)2

Scheduling top jobs (I) . . .

j top: bottom:

Scheduling top jobs (I) . . .

j top: bottom:

Scheduling top jobs (I)

◮ For top job j set

[release time(j), deadline(j)] := fully contained bottom intervals

. . .

j top: bottom: release time(j) deadline(j)

Scheduling top jobs (I)

◮ For top job j set

[release time(j), deadline(j)] := fully contained bottom intervals

◮ Set Q := polylog(n) as top/bottom gap

. . .

j top: bottom: release time(j) deadline(j) Q

Scheduling top jobs (I)

Lemma

Can assign top jobs respecting release times/deadlines, ignoring

• prec. constraints & discarding 2Tm

Q

jobs.

◮ For top job j set

[release time(j), deadline(j)] := fully contained bottom intervals

◮ Set Q := polylog(n) as top/bottom gap

. . .

j top: bottom: release time(j) deadline(j) Q

Scheduling top jobs (I)

Lemma

Can assign top jobs respecting release times/deadlines, ignoring

• prec. constraints & discarding 2Tm

Q

jobs.

◮ For top job j set

[release time(j), deadline(j)] := fully contained bottom intervals

◮ Set Q := polylog(n) as top/bottom gap ◮ Hall’s Theorem: Loss for jobs J′ is 2 Q-fraction of slots

. . .

top: bottom: Q

EDF (1)

Setting: time

EDF (1)

Setting:

◮ Jobs with precedence constraints

time

EDF (1)

Setting:

◮ Jobs with precedence constraints ◮ Maximum chain length C

length C time

EDF (1)

Setting:

◮ Jobs with precedence constraints ◮ Maximum chain length C ◮ Release times, deadlines only at beginning/end of p blocks

length C time 1 2 p

EDF (1)

Setting:

◮ Jobs with precedence constraints ◮ Maximum chain length C ◮ Release times, deadlines only at beginning/end of p blocks ◮ Capacity cap(t) ∈ {0, . . . , m}

length C time 1 2 p

EDF (1)

Setting:

◮ Jobs with precedence constraints ◮ Maximum chain length C ◮ Release times, deadlines only at beginning/end of p blocks ◮ Capacity cap(t) ∈ {0, . . . , m} ◮ There exists a schedule ignoring precedence constraints

length C time 1 2 p

EDF (1)

Setting:

◮ Jobs with precedence constraints ◮ Maximum chain length C ◮ Release times, deadlines only at beginning/end of p blocks ◮ Capacity cap(t) ∈ {0, . . . , m} ◮ There exists a schedule ignoring precedence constraints

length C time 1 2 p Earliest Deadline First: (1) FOR t = 1 TO T DO

(2) Process available job with minimum deadline (3) If any job not done by deadline discard it

EDF (2)

Analysis: 1 2 p possible release times / deadlines time

EDF (2)

Analysis:

◮ Suppose in some interval I we discard K jobs

> K discarded

EDF (2)

Analysis:

◮ Suppose in some interval I we discard K jobs ◮ Assume w.l.o.g. last discarded job had highest deadline

> K discarded

EDF (2)

Analysis:

◮ Suppose in some interval I we discard K jobs ◮ Assume w.l.o.g. last discarded job had highest deadline ◮ Take maximum interval [t1, t2] ⊇ I so that in each block, there

are ≤ C idle times > K discarded ≤ C idle times each > C idle times t1 t2

EDF (2)

Analysis:

◮ Suppose in some interval I we discard K jobs ◮ Assume w.l.o.g. last discarded job had highest deadline ◮ Take maximum interval [t1, t2] ⊇ I so that in each block, there

are ≤ C idle times

◮ All jobs scheduled or discarded in [t1, t2] have both release +

deadline in [t1, t2]. > K discarded ≤ C idle times each > C idle times t1 t2

EDF (2)

Analysis:

◮ Suppose in some interval I we discard K jobs ◮ Assume w.l.o.g. last discarded job had highest deadline ◮ Take maximum interval [t1, t2] ⊇ I so that in each block, there

are ≤ C idle times

◮ All jobs scheduled or discarded in [t1, t2] have both release +

deadline in [t1, t2].

◮ K ≤ idle slots in [t1, t2] ≤ pmC (feasibility w/o prec.-constr.)

> K discarded ≤ C idle times each > C idle times t1 t2

The end

Open problems:

◮ Are c(ε, m)-Lasserre rounds enough? ◮ .. or at least (log(n))c(ε,m) ◮ What about Pm | prec | Cmax (i.e. arbitrary running times)?

The end

Open problems:

◮ Are c(ε, m)-Lasserre rounds enough? ◮ .. or at least (log(n))c(ε,m) ◮ What about Pm | prec | Cmax (i.e. arbitrary running times)?

Thanks for your attention