Combinatorial Optimization in Container Scheduling Michael Pinedo Stern School of Business, New York...

Combinatorial Optimization Combinatorial Optimization in Container Scheduling in Container Scheduling

Michael PinedoStern School of Business, New York University

with Byung-Cheon Choi, Kangbok Lee (NYU), Joseph Leung (NJIT), Dirk Briskorn (Cologne).

CMMI, Boston, July, 2012

Contents

Preliminary

Scenarios, Examples, and

Equivalent Scheduling Problems

Assumptions / Parameters / Constraints / Objectives

Settings

Motivation

Summary

Future Study

Model

Problems

Conclusion

Results Result 1 / Result 2 / Result 3 / Result 4

Other Objective functions

Motivation

A shipping company has

a number of containers to be delivered from source ports to destination ports

ships that have fixed routing schedules.

with

Which container should be delivered by which ship?

Motivation

Qingdao

Ningbo

Yangshan

New York

Norfolk

Savannah

Busan

Qingdao

201

0-0

3-1

5

201

0-0

3-2

2

201

0-0

3-2

9

201

0-0

4-0

5

201

0-0

4-1

2

201

0-0

4-1

9

201

0-0

4-2

6

201

0-0

5-0

3

201

0-0

5-1

0

http://www.hapag-lloyd.com/en/products_and_services/services_between_asia_and_north_america.html#TPD_east

Service : North/Central China Eastcoast Express (NCE)

Model

Assumptions


Unidirectional route

Ship 1

Ship 2

Ship 3

• There is a sequence of ports• A ship may skip some ports

Port A B C D E F

visitingnon-visiting

1. Unidirectional Route


2. Non-Overtaking

Model

Assumptions


2. Non-Overtaking

Non-overtaking

Ports

• Overtaking case Ports

time

• Non-Overtaking example

ship 1

ship 2

time

ship 1ship 2

ship 3


2. Non-Overtaking

3. No Transfer

Model

Assumptions

3. No Transfer

Example transfer

• Transfer case

Ship 1

Ship 2

Container

Source port

Destination port

loading

loading unloading

unloading

Transfer


Model

Containers

Assumptions / Parameters / Constraints / ObjectivesContainers Ships

• Release date / Due date rj, dj

• Importance (weight) factor

wj

• Two types of Size (1 or 2)

Twenty-foot Equivalent Unit = 1 TEU

Forty-foot Equivalent Unit (FEU) = 2 TEU

20-ft8 ft

8.6 ft

40-ft8 ft

8.6 ft

sj

• Source port / Destination port

uj, vj

Model

• Number of ships m

• Capacity : ci

Ships

• Predetermined Routing Schedule

Ship i

ti,k

ti,1 ti,2 ti,3 ti,5

: time when ship i visits port k.

port 1 2 3 4 5 6

Assumptions / Parameters / Constraints / ObjectivesContainers Ships

Model

Constraints


•A container can be delivered only by a ship that visits both the source port and destination port of the container.

•A container can be delivered only by a ship that visits the source port of the container at a time after the release date of the container.

•Ship Eligibility Constraints

By source and destination ports

By release dates

• At any time, the total size of containers being loaded at a ship cannot exceed the capacity of the ship.

•Capacity Constraints

Model

• Delivery Time (Dj)

Terminology

• Tardiness (Tj)

The arrival time of the ship at the destination port of the container.

Maximum { 0, Delivery Time – Due date }

•The maximum tardiness

Objectives

•The weighted Delivery Time

•The weighted number of tardy containers•The weighted tardiness

Tmax

wjDj

wjUj

wjTj

Main topic


• Tardy Container (Uj)Uj = 0 if Tj = 0, Uj = 1 if Tj > 0

With a target objective function value T ,

Consider a feasibility problem to find a feasible schedule not violating the modified due date dj + T.

If the answer is Yes, T*max ≤ T .

If the answer is No, T*max > T .

ModelAssumptions / Parameters / Constraints / Objectives

A Feasibility Problem for Tmax

T*max { max{0, Dj – dj} | j = 1, …, n }

= { max{0, ti, – dj} | j = 1, …, n , i = 1, …, m } =

vj

The optimal objective function value

A simple model

•All ships visit all ports following the order of the ports

•All containers have a common source port

•All containers are released at time 0.

Problems An Example / Scenarios

•Each container has a last available ship.

size 1

size 2

ship

capacity

•Containers of each size can be ordered in increasing order of the last available ship

A connectivity problem in a layered graph


• n1 and n2 represent the number of containers of size 1 and size 2

• h1a and h2

b represent the last available ship for a-th container of size 1 and b-th container of size 2, respectively.

(ai, bi) (ai+1, bi+1)• and are connected to one another if ai ≤ ai+1 and bi ≤ bi+1

(ai, bi)•A node at layer i represents a partial schedule with

ai and bi containers of size 1 and size 2 each are assigned to ships 1, …, i.

if h1a≥ i , h2

b ≥ ii i

and ai + 2bi = ckk=1

i

A connectivity problem in a layered graph


(0,0)

(c1, 0) (c1-2, 1) (c1-2n2, n2)

(c1+c2, 0) (c1+c2-2, 1) (c1+c2-2n2, n2)

(n1,n2)

(ck-2,1)k=1

m-1 ( ck-2n2, n2)k=1

m-1

layer 0

:

…layer 1

layer 2

layer m-1

layer m

…

…

: ::

(c1+c2-4, 2)

(ck, 0)k=1

m-1


Category Options Symbols

Arbitrary routes (special case: nested routes)

A (N)

Ship routings Identical routes (all ships visit all ports)

Rou

tin

g I

# of Ports Fixed number of ports (constant)Arbitrary number of ports

Np

Release Date

Source Port

Container Size

Con

tain

er

Zero release dates (all containers are ready)Arbitrary release dates

Common source ports (distribution center)Arbitrary source ports

Unit container size only

Container size of 1 or 2

rj = 0rj

uj = 1uj

sj = 1sj = 1 or 2

Problems

Example:

I , Np | rj=0, uj, sj = 1 or 2 | Tmax

Container Characteristics

Objective FunctionRouting Environment

An Example / Scenarios

rj uj sj

Fixed # of ports Arbitrary # of ports

Identical Arbitrary Identical Arbitrary

0 1 1

0 1 1 or 2

rj1 1

rj1 1 or 2

0 uj1

0 uj1 or 2

rj uj1

rj uj1 or 2

The Computational Complexities

P: polynomial time solvable, NP: NP-hard

Problems

Routing

Con

tain

er P

An Example / Scenarios

ports

Ship 1

0

Ship 2

Ship 3

c1=3

c2=4

c3=2

uj

vj

5 10 15 20

sourceport

destinationport

Container j : (uj, vj) = (8,16)

I , | rj = 0, uj , sj = 1 | Tmax

due daterestriction

Since rj= 0, the first ship is available.

Results Result 1 / Result 2 / Result 3 / Result 4.

Results

Equivalent Problem :

Hierarchical Interval Scheduling with T machine Types: HIS(T)

Job j : fixed starting time (bj) and completion time (ej)Job j of type i : job j can be scheduled on machine with type k for 1 k i.

job j of type 2 with (bj, ej) = (8,16)

time

m/c type 1

0

m/c type 2m/c type 3

m1=3m2=4m3=2

bj ej

5 10 15 20

Example:

Result 1 / Result 2 / Result 3 / Result 4.

Strongly NP-hard for T ≥ 3Kolen et al. 2007. NRL

Results

I , | rj , uj = 1, sj = 1 or 2 | Tmax

1 2 3 m

Capacity

c

12

… Ship

size = 1 or 2

the firstavailable ship

the lastavailable ship


Resource constrained scheduling

Pc | rj , pj = 1, sizej = 1 or 2 | Tmax

1 2 3 m time

Machinec

12

…

Results

Related (and in some way equivalent) Problem:

OPEN Complexity Problem by P. Brucker and A. Krämer. EJOR (1996)P | rj , pj = 1, |{ sizej }| = K | Tmax for constant K

Partial Results on P | rj , pj = 1, sizej = 1 or m | Tmax

O(n4) Philippe Baptiste and Baruch Schieber. J. Scheduling (2003)O(n3) Christoph Dürr and Mathilde Hurand, Algorithmica (2009)


Data Structure

I , | rj , uj = 1, sj = 1 or 2 | Tmax

1 2 3 4

11:4J

11:3J

11:2J

11:1J

12:4J

12:3J

12:2J

13:4J

14:4J

13:3J

5

15:1J

15:2J

15:3J

15:4J

15:5J

1:khJ

The set of containers

of size 1

that are eligible to ships h to k.

ships


• n1(n2) : the number of containers of size 1 (size 2).

• m ships with capacity ci

N(i , ai , bi) :

N(0, 0, 0) :

indicates that ai containers of size 1 and

bi containers of size 2

have been assigned to ships 1, 2, …, i.

Solution Structure

single node for the case of i = 0.

N(m, n1, n2) :

single node for the case of i = m.

211

2nncm

ii

Parameters and assumptions


Node Feasibility conditions

1 2 3 4 m…

…

…

…

…

11:4J

11:3J

11:2J

11:1J

12:4J

12:3J

12:2J

13:4J

14:4J

13:3J

1:3mmJ

1:2 mmJ

1:1mmJ

1:mmJ

1:1mJ

1:

11

1:

1ih

mihkkih

kih

JnaJ

2:

11

2:

1ih

mihkkih

kih

JnbJ

k

iikk cba

1

2


Arc Feasibility conditions

1 2 3 4 m…

…

…

…

…

11:4J

11:3J

11:2J

11:1J

12:4J

12:3J

12:2J

13:4J

14:4J

13:3J

1:3mmJ

1:2 mmJ

1:1mmJ

1:mmJ

1:1mJ

1:

1ih

lihkkl Jaa

N(k , ak , bk) and N(l , al , bl) are connected to each other for k < l,

if

2:

1ih

lihkkl Jbb

k lk+1

Results Result 1 / Result 2 / Result 3 / Result 4

is reduced toI , | rj , uj = 1, sj = 1 or 2 | Tmax

N(0, 0, 0)

N(m, n1, n2)

……

……

……

Layer 0

Layer 1

Layer 2

Layer m1

Layer m

N(m1, am-12, bm-1+1) N(m1, am-1, bm-1) N(m1, am-1+2, bm-11)

N(1, a12, b1+1) N(1, a1, b1) N(1, a1+2, b11)

N(2, a22, b2+1) N(2, a2, b2) N(2, a2+2, b21)…

the problem of finding a (m + 1)-clique in an (m + 1)-layered graph.

N(0, 0, 0)

N(m, n1, n2)

……

……

……

Layer 0

Layer 1

Layer 2

Layer m1

Layer m

N(m1, am-12, bm-1+1) N(m1, am-1, bm-1) N(m1, am-1+2, bm-11)

N(1, a12, b1+1) N(1, a1, b1) N(1, a1+2, b11)

N(2, a22, b2+1) N(2, a2, b2) N(2, a2+2, b21)…


I , | rj , uj = 1, sj = 1 or 2 | Tmax can be solved in O( (mn2 + m3n) log mn ) time.


Additionally, we proved that

• There is a clique with size (m+1) there exists a feasible schedule.

There exist a polynomial time algorithm to construct a feasible schedule with a given (m+1) clique in (n+m2) time.

Trivial.

• There exists an algorithm to find a (m+1)-clique in polynomial time if

it exists.

In general, to find a k-clique in a k-layered graph is NP-complete.

However, based on a special property of the graph, we can find a (m+1)-clique in O(mn2 + m3n)time.

For, the binary search on the target objective function value takes time of O( log mn ).

Additional assumptions• m is enough large to load all containers by MS rule.

• All ships have the same capacity c, i.e. ci = c for i = 1, …, m.

• The inter-departure times between consecutive ships at port l = , i.e. ti,l = t1,l + (i-1) for i = 2, …, m and l = 1, …, Np

Minimum Slack (MS) first priority rule for

I , | rj , uj

= 1, sj = 1 or 2 | Tmax


The error bound of MS rule

c

nnm 21 2where min

1

1)()( maxmax

c

mOPTTMST min

capacity

c

time

The optimal schedule

mmin 2

: : : : : : :

(mmin-1)

The schedule by MS rule

capacityc

time

1

1minmin c

mm

: : : : : : :

2

1minm

:

The worst case example of MS rule

size 1

size 2


mmin(mmin-1)

rj uj sj



0 1 1 P P P P0 1 1 or 2 P P P NP

rj1 1 P P P P

rj1 1 or 2 P OPEN P NP

0 uj1 P P NP NP

0 uj1 or 2 P P NP NP

rj uj1 OPEN OPEN NP NP

rj uj1 or 2 OPEN OPEN NP NP

The Time complexities of the problems (Tmax)

P: polynomial time solvable, NP: NP-hard, OPEN: unknown

Summary for the Maximum Tardiness objective


I , | rj = 0, uj = 1, sj = 1 or 2 | wj Dj

with identical inter-departure times

• All inter-departure times are identical.• In an optimal schedule, containers of each size are ordered in non-increasing order of weight. • We can add dummy containers of size 1 with zero weight.

Additional Assumptions & Observation


I , | rj = 0, uj = 1, sj = 1 or 2 | wj Dj with identical inter-

departure times(ai, bi)•A node at layer i represents a partial schedule with

ai and bi containers of size 1 and size 2 each being assigned to ships 1, …, i.

k=1

if ai + 2bi = cki

(ai, bi) (ai+1, bi+1)• are connected if ai ≤ ai+1 and bi ≤ bi+1

and its weight is

where wj and vj are weight and destination of j-th container of

size

1

2

1

1

1,1

2

1,1

1i

i

j

i

i

j

b

bjvij

a

ajvij

twtw


A Shortest Path Problem : O(mn2)

(0,0)

(c1, 0) (c1-2, 1) (c1-2n2, n2)

(c1+c2, 0) (c1+c2-2, 1) (c1+c2-2n2, n2)

(n1,n2)

(ck-2,1)k=1

m-1 ( ck-2n2, n2)k=1

m-1

layer 0

:

…layer 1

layer 2

layer m-1

layer m

…

…

: ::

(c1+c2-4, 2)

(ck, 0)k=1

m-1

1 2 3 4 5 61 2 3 4 5 6 7 8


I , | rj , uj = 1, sj = 1 or 2 | wj Dj with identical inter-departure times

•Order containers in non-increasing order of wj/sj.WSS (Weighted Smallest Size first) rule

1

1

2

2

3

3

4

4

5

5

6

61 2 3 4 5 6 7 8

1

1

2

2

3

4

4

5

5

6

61

2

3

4

5

6

7

8

3 ship

capacity

c

2

3

3

4

4 5

5

6

6

2

3

4

5

6 7

8

1

1

21 ship

capacity

cc-1

Obj. value = A Obj. value = B

OPT WSS

A B

c c+1

= 1 + c 1

≤ ≤

rj uj sj



0 1 1 P P P P0 1 1 or 2 P P OPEN, NP

rj1 1 P P P P

rj1 1 or 2 OPEN, OPEN, OPEN, NP

0 uj1 P P NP NP

0 uj1 or 2 P P NP NP

rj uj1 OPEN, OPEN, NP NP

rj uj1 or 2 OPEN, OPEN, NP NP

The Time complexities of the problems ( wjDj)

P : polynomial time solvable, NP: NP-hard, OPEN: unknown : polynomial time solvable for identical weight (wj = 1 ) case : polynomial time solvable for identical inter-departure times case : polynomial time solvable for identical weight and identical inter-departure times case

Summary for Weighted Delivery dates objective


I , | rj =0, uj = 1, sj = 1 or 2 | wj Uj with identical inter-departure times

N(i , a1, a2, b1, b2) :

N(1, 0, 0, 0, 0) :

indicates that a1 containers of size 1 and a2 containers of size 2 have been determined to be assigned to ships 1, 2, …, i. or not.

b1, and b2 are current numbers of

containers of size 1 and 2 on ship i, respectively.

Solution Structure

only node for the case of i = 1.

N(m+1, n1, n2, 0, 0) : only node for the case of i = m+1.

•In an optimal schedule, on-time (non-tardy) containers of each size are ordered in non-decreasing order of due date.

Observation


I , | rj =0, uj = 1, sj = 1 or 2 | wj Uj with identical inter-departure times

N(i , a1, a2, b1, b2)

N(i , a1+1, a2, b1, b2 )

N(i , a1+1, a2, b1+1, b2

)

N(i , a1, a2+1, b1 , b2 )

N(i , a1, a2+1, b1 , b2+1)

N(i+1, a1, a2, 0, 0 )

Assign (a1+1)th container of size 1

Not assign (a1+1)th container of size 1

Assign (a2+1)th container of size 2

Not assign (a2+1)th container of size 2

Close ship i and open ship i+1

A Shortest Path Problem : O(mn4)

At most

five outgoing arcs

Future Research

• Some open problems of minimizing ( Tmax / wjDj / wjUj / wjTj )• Other objectives: Minimizing shipping cost by selecting ships.

Open Problems

• Worst case analysis or experimental analysis of a simple priority rule • For example : assign containers in the decreasing order of

A simple priority rule

wj

sj T(uj, vj)

Larger weight

Shorter distance

Smaller size

wj

sj T(uj, vj)

T(uj, vj)

uj vj

Ports

Distance from the source port to the destination port

Future Research

• A refrigerator container requires electricity • A ship has a limited space for such containers.

Additional restrictions

• Nested routes case (between identical and arbitrary routes cases)

Other routings

Ship 1Ship 2Ship 3

visiting only important ports

visiting all ports

• Analysis of the performance of algorithms with real data.• Development of efficient algorithms for practical use.

Empirical work

Combinatorial Optimization in Container Scheduling Michael Pinedo Stern School of Business, New York...

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