Optimization of ballast plan in launch jacket load-out

Struct Multidisc Optim (2009) 38:267–288DOI 10.1007/s00158-008-0287-7

INDUSTRIAL APPLICATION

Optimization of ballast plan in launch jacket load-out

Adi Kurniawan · Guowei Ma

Received: 21 January 2008 / Revised: 22 April 2008 / Accepted: 19 May 2008 / Published online: 19 July 2008© Springer-Verlag 2008

Abstract The calculation of ballast plan in load-out op-erations is generally performed assuming a rigid barge.This assumption may not be reliable since in realitythe barge is flexible. Having incorrect ballast plan maylead to overstressing of the loaded-out structure. Wepresent a method of finding a more accurate ballastplan, taking into account the flexibility of the barge.This method makes use of a multi-objective evolution-ary algorithm to find the optimum ballast arrangementat every load-out stage. We model the load-out config-uration as a beam on elastic foundation, loaded withdistributed trapezoidal loading representing the loadfrom the structure. Minimizing deflection and curvatureof the beam, as well as maximizing the ballast transferefficiency between the load-out stages are chosen asthe objectives of the algorithm. It is shown that theproposed method is better than the conventional rigidbarge method in terms of minimizing the deflectionand curvature as well as maximizing the ballast transferefficiency.

A. Kurniawan (B)Maritime Research Centre,School of Civil and Environmental Engineering,Nanyang Technological University, Nanyang Avenue,Singapore 639798, Singaporee-mail: [email protected]

G. MaSchool of Civil and Environmental Engineering,Nanyang Technological University, Nanyang Avenue,Singapore 639798, Singaporee-mail: [email protected]

Keywords Load-out · Ballast · Barge ·Beam on elastic foundations ·Multi-objective evolutionary algorithms

1 Introduction

Load-out is the process of transferring a completedoffshore structure, usually a jacket, from the yard ontothe deck of a barge, to be transported to the installationsite. A load-out operation is divided into stages, ateach of which the jacket progressively travels a certaindistance on the barge, until it reaches the required finalposition. Throughout this process the arrangement ofballast in the barge is critical. Having incorrect ballastplan may cause the loaded-out structure (hereafteralso called the jacket) to be overstressed (Ultramarine2007).

Therefore it is a common practice in load-out analy-ses (Curtis and Gilmore 1992; Ferguson et al. 1983;Hofferber et al. 1991; Kenney et al. 1994; Piter et al.1989) to construct an allowable displacement enve-lope. This envelope was constructed by first assumingand enforcing some displacements to the structure.Maximum displacements that could be taken by thestructure without overstressing it were plotted as theallowable displacement envelope. The predicted andthe measured actual displacements must be ensured tobe within the allowable envelope. Furthermore, ballastand barge levels are usually monitored during load-outand shown to be within the structural limitations ofthe barge and the structure (Noble Denton Report No.0013/NDI 2004).

268 A. Kurniawan, G. Ma

Generally, the calculation of ballast plan is per-formed assuming a rigid barge. The objective is to main-tain the deck of the barge horizontally levelled with thequay wall, while the load-out progresses. However, inreality, the barge is flexible. There is also foundationsettlement at the yard.

It is true that sometimes the flexibility of the barge,as well as the yard settlement, was taken into account,but only in the final step of the analysis. Initial stepsdid not take into account the flexibility of the barge.For example, in the method of Ferguson et al. (1983),the land skidways were modelled as elastic but thebarge was assumed to be a rigid body that was alwayshorizontal. The barge was modelled as an elastic beamonly in the validity check of the ballast plan, whichhad previously been calculated assuming a rigid barge.A similar method was adopted by Piter et al. (1989).In their preliminary step, the yard and barge wereassumed to be rigid and all lying in a horizontal plane.The next steps were performed taking into account thestiffness of the yard, but the barge was still assumed tobe rigid. Only in the final step was the flexibility of thebarge considered.

An attempt to develop a simple yet accurate bal-last calculation, which considers the flexibility of thebarge, was made by Liu and Gho (2005). They de-veloped an iterative method to find the ballast. Thenovel component was the use of an empirical value toquantify the relationship between the ballast in the tankcalculated assuming a rigid barge and the readjustedballast calculated assuming a flexible barge. The tankgeometry (width, length, and thickness) and the pointof application of the load were taken as parameters.Nevertheless, some shortcomings could be identified.They considered the flexibility of the barge, but noyard settlement was taken into account. Moreover, theballast arrangements in their case studies were simple,in that either the number of active tanks was limitedto two, or, for cases with more than two active tanks,the ballast in each tank could only be adjusted suchthat all differences between the ballasts in any twoadjacent tanks were equal. These rendered the problemstatically determinate. However, when a considerablenumber of active tanks are used without fixing theballasts in any of these tanks, the problem becomesnontrivial; static equilibrium equations alone will notbe adequate to solve for the ballast in each tank.

Motivated by the need for accuracy and robustness,we hereby propose a method of ballast calculationwhich takes into account the flexibility of the barge andthe settlement of the yard, and puts no restrictions tothe amount of ballast each tank is to have in order

to obtain the most efficient ballast transfer, hence themost efficient load-out operation. The kernel of ourmethod is the use of multi-objective evolutionary algo-rithm (MOEA) to find the optimum ballast plan.

The present paper is concerned with the launchjacket load-out, whereby the jacket slides on horizontalskidways on the barge. The load-out configuration ismodelled as a beam on elastic foundation; the wholejacket-barge system is modelled as a single beam restingon both yard and water, and the jacket load is assumedto be trapezoidal. An analytical expression for thebeam deflection is derived using the method of super-position (Hetényi 1974). We relate the ballast in tanki to a dimensionless variable αi, with values rangingfrom 0 to 1, where 0 denotes an empty tank and 1denotes a fully ballasted tank. A ballast arrangement isrepresented by a vector α = (α1, . . . , αn), where n is thetotal number of ballast tanks. The task of the MOEAis to find a vector α that minimizes the deflection andcurvature of the beam and the change of α from theprevious load-out stage to the current one. A set of α

vectors obtained from all the load-out stages constitutesan optimum ballast plan for the load-out operation.

In the next section we discuss the modelling of theload-out configuration as a beam on elastic foundation.The analytical expression for the beam deflection isderived. In Section 3 we describe the conventional rigidbarge method of ballast calculation. In Section 4 wepresent our proposed method. The MOEA and pro-cedures to find the optimum ballast plan are describedin this section. Results from case studies are then pre-sented in Section 5. Comparisons are made betweenresults from our proposed method and those from theconventional rigid barge method in terms of deflection,curvature, and ballast transfer efficiency. It is shownthat the proposed method is better than the conven-tional one. Finally, we present our main conclusionsin Section 6 and suggest some recommendations forfuture works.

2 Load-out configuration as beam on elasticfoundation

The load-out configuration can be represented as a two-dimensional beam on elastic foundation (Fig. 1) by as-suming that the geometry and loadings are symmetricalabout the centre line dividing the port and starboardsides of the barge, and that there is only longitudinalvariation of ballasts in the barge. Some part of the beamrests on the yard with foundation stiffness k1, and the

Optimization of ballast plan in launch jacket load-out 269

water

ballast load

jacket

barge

jacket load

yard

Fig. 1 Modelling of load-out configuration as beam on elasticfoundation (shown for n = 2)

remaining part rests on water with hydrostatic stiffnessk2. Further assumptions are made:

(1) the jacket is oriented with its base toward thewater;

(2) the jacket weight is trapezoidal, which can beseparated into a uniformly distributed loading anda triangular loading;

(3) the barge is of rectangular box shape; it has uni-form height;

(4) all ballast tanks have equal lengths;1

(5) the analysis is static (Chakrabarti 2005); wind,wave, current, and impact forces are neglected(API RP 2A–WSD 2000);

(6) the barge is restrained horizontally; and(7) initial deformations of the barge are neglected.

Initial ballast is already applied such that it balancesthe excess buoyancy force from the water for a givendraft. The initial ballast is calculated as follows. Theminimum draft dmin of the barge is the draft requiredto balance the barge weight:

dmin = wbarge

k2, (1)

where wbarge is the barge weight, and k2 is the hydro-static stiffness. Given a draft d, which has to be greaterthan dmin, the buoyancy Fb can be calculated as

Fb = k2d . (2)

1Derivations for ballast tanks having different lengths, however,can be made with little modifications.

It follows that the initial ballast needed to balance theexcess buoyancy force is

wint = Fb − wbarge , (3)

and the initial ballast height is

hb int = wint

k2. (4)

We note that the barge weight and buoyancy areconstant for a given draft. Thus, we can exclude them inour calculation by setting the initial ballast as the zeroreference for the ballast load wi in tank i. In any tank,a reduction of ballast from the initial ballast results inan upward ballast force (negative wi), while an additionresults in a downward ballast force (positive wi).

According to Winkler theory, which assumes thatthe reaction forces of the foundation are proportionalat every point to the deflection of the beam at thatpoint, the differential equation for the deflection y ofthe beam on elastic foundation is

EId4 ydx4

= −ky + q , (5)

where k is the foundation stiffness, q is the distributedload acting upon the beam, and x is the distance alongthe beam. It should be noted that since y denotes thedeflection of the beam, y-axis is pointing downward.

We employ the method of superposition (Hetényi1974) to obtain the analytical expression for the beamdeflection. Depending on the loading configuration, thesolution of (5) will involve some combination of theseexpressions:

Aλx = e−λx(cos λx + sin λx) (6a)

Bλx = e−λx sin λx (6b)

Cλx = e−λx(cos λx − sin λx) (6c)

Dλx = e−λx cos λx (6d)

EIλl = 1

2

eλl

sinh λl + sin λl(6e)

EIIλl = 1

2

eλl

sinh λl − sin λl, (6f)

where

λ = 4

√k

4EI. (7)

Standard expressions for deflection, slope, moment,and shearing force for several general loading

270 A. Kurniawan, G. Ma

configurations can be found in Hetényi (1974). Wewill use some of these in the ensuing derivation.

The beam is first cut into three segments, each havingdifferent moments of inertia (Fig. 2). Segment I isdefined as the part of the jacket that rests on the yard,segment II as the part of the jacket on the barge, whilesegment III is the remaining part of the barge. Thelength of segment I is l1, the length of segment II isl2, while the length of each ballast tank is lb . Thus, thelength l3 of segment III for any n is equal to nlb − l2.The total length of the beam is l = l1 + l2 + l3. Thenotations for the trapezoidal loading due to the jacketweight are defined as in Fig. 3.

For each segment we first find the moments andshearing forces produced at the end points of the seg-ment, assuming an infinitely long beam. We have, forsegment I,

Mq1a = q1

4λ21

Bλ1l1 (8a)

Qq1a = q1

4λ1

(1 − Cλ1l1

)(8b)

Mq1

b = Mq1a (8c)

Qq1

b = −Qq1a (8d)

due to the uniformly distributed loading, and

Mq2a = − q2

8λ31

1

l1

(1 − Aλ1l1 − 2λ1l1 Bλ1l1

)(9a)

Qq2a = q2

4λ21

1

l1

(Bλ1l1 − λ1l1Cλ1l1

)(9b)

Mq2

b = − q2

8λ31

1

l1

(Aλ1l1 − 1

)(9c)

Qq2

b = q2

4λ21

1

l1

(Bλ1l1 − λ1l1

)(9d)

a b c d

I II III

I k22 ,I k11 , I k23 ,

Fig. 2 Beam model (shown for n = 2)

a cb

q1

q2 q3

q4

Fig. 3 Trapezoidal loading due to the jacket weight

due to the triangular loading. The sum of momentsand shearing forces in the infinitely long beam dueto the above loadings are denoted as MaI , QaI , Mb I ,and Qb I .

For a beam of finite length, we need to apply end-conditioning forces to satisfy the required conditionsat both ends of the beam. For a beam with free ends,we can simplify the problem by resolving the originalloading into two parts: a symmetrical part and an anti-symmetrical one. The consecutive steps will be:

(a) resolving the original loading into a symmetricaland an antisymmetrical part,

(b) determining the end-conditioning forces in each ofthese parts, and

(c) adding them together afterwards at each endpoint.

By this means we obtain the total value of the end-conditioning forces for the original given loading.

Resolving the original loading into a symmetricaland an antisymmetrical part, we have:

M′I = MaI + MbI

2(10a)

Q′I = QaI − QbI

2(10b)

M′′I = MaI − MbI

2(10c)

Q′′I = QaI + QbI

2, (10d)

Optimization of ballast plan in launch jacket load-out 271

where the prime (′) denotes the symmetrical part,while the double primes (′′) denote the antisymmet-rical part. The end-conditioning forces in each of theparts are

P′0I = 4EIλ1l1 [Q′

I(1 + Dλ1l1) + λ1 M′I(1 − Aλ1l1)]

(11a)

M′0I = − 2

λ1EIλ1l1 [Q′

I(1 + Cλ1l1) + 2λ1 M′I(1 − Dλ1l1)]

(11b)

P′′0I = 4EIIλ1l1[Q′′

I(1 − Dλ1l1) + λ1 M′′I (1 + Aλ1l1)]

(11c)

M′′0I = − 2

λ1EIIλ1l1[Q′′

I(1 − Cλ1l1) + 2λ1 M′′I (1 + Dλ1l1)] .

(11d)

They are combined in the following manner to obtainthe total value of the end-conditioning forces for theoriginal given loading:

P0aI = P′0I + P′′

0I (12a)

P0b I = P′0I − P′′

0I , (12b)

likewise for M′0I and M′′

0I to obtain M0aI and M0b I .By superposition of the original given loading and

the end-conditioning forces at segment I, we can obtainthe displacement and slope at b as

yIb = q1

2k1

(1 − Dλ1l1

)+ q2

4λ1k1

1

l1

(Cλ1l1 − 1 + 2λ1l1

)

+ P0aIλ1

2k1Aλ1l1 + M0aIλ

21

k1Bλ1l1 + P0bIλ1

2k1(13a)

θ Ib = q1λ1

2k1

(Aλ1l1 − 1

)− q2

2k1

1

l1

(Dλ1l1 − 1 + λ1l1

)

− P0aIλ21

k1Bλ1l1 + M0aIλ

31

k1Cλ1l1 − M0b Iλ

31

k1. (13b)

We follow the same procedure for the rest of thesegments. For segment II, the loads are made up of thejacket load and ballast. We have, due to the uniformlydistributed loading,

Mq3

b = q3

4λ22

Bλ2l2 (14a)

Qq3

b = q3

4λ2

(1 − Cλ2l2

)(14b)

Mq3c = Mq3

b (14c)

Qq3c = −Qq3

b , (14d)

and due to the triangular loading,

Mq4

b = − q4

8λ32

1

l2

(1 − Aλ2l2 − 2λ2l2 Bλ2l2

)(15a)

Qq4

b = q4

4λ22

1

l2

(Bλ2l2 − λ2l2Cλ2l2

)(15b)

Mq4c = − q4

8λ32

1

l2

(Aλ2l2 − 1

)(15c)

Qq4c = q4

4λ22

1

l2

(Bλ2l2 − λ2l2

). (15d)

It should be noted that, generally, some part of thetank on which the base of the jacket rests belongs tosegment II, and the remaining part belongs to segmentIII. Denoting the length of the part of the tank whichbelongs to segment II by l′b II , for every ballast load wIIi

in segment II we obtain the distances from the left andright ends of the load to points b and c as follows:

llbIIi = (i − 1)lb (16)

lrbIIi =

{(i − 1)lb + l′b II, if i = m1

ilb , if 1 ≤ i < m1(17)

llcIIi = (m1 − i)lb + l′b II (18)

lrcIIi =

{0, if i = m1

(m1 − i − 1)lb + l′b II, if 1 ≤ i < m1, (19)

where m1 is the number of tanks in segment II, includ-ing the tank on which the base of the jacket rests.

Moments and shearing forces produced at the endpoints of segment II due to all wIIi in segment II,assuming an infinitely long beam, are

MwIIb = −

m1∑i=1

wIIi

4λ22

(Bλ2llb

IIi− Bλ2lrb

IIi

)(20a)

QwIIb =

m1∑i=1

wIIi

4λ2

(Cλ2llb

IIi− Cλ2lrb

IIi

)(20b)

MwIIc =

m1∑i=1

wIIi

4λ22

(Bλ2llc

IIi− Bλ2lrc

IIi

)(20c)

QwIIc =

m1∑i=1

wIIi

4λ2

(Cλ2llc

IIi− Cλ2lrc

IIi

). (20d)

272 A. Kurniawan, G. Ma

The sum of moments and shearing forces in theinfinitely long beam due to the jacket and ballast loadsare denoted as MbII , QbII , McII , and QcII .

We resolve the original loading into a symmetricaland an antisymmetrical part to obtain M′

II , Q′II , M′′

II ,and Q′′

II by replacing MaI , QaI , MbI , and QbI in (10)with MbII , QbII , McII , and QcII , respectively. The end-conditioning forces in each of the parts, P′

0II , M′0II , P′′

0II ,and M′′

0II , can be obtained as in (11) by replacing λ1,l1, M′

I , Q′I , M′′

I , and Q′′I with λ2, l2, M′

II , Q′II , M′′

II , andQ′′

II , respectively. They are combined in the mannerof (12) to obtain the total end-conditioning forces P0bII ,P0cII , M0bII , and M0cII .

The displacement and slope at b and c due to all wIIi

in segment II are as follows:

ywIIb =

m1∑i=1

wIIi

2k2

(Dλ2llb

IIi− Dλ2lrb

IIi

)(21a)

θwIIb =

m1∑i=1

wIIiλ2

2k2

(Aλ2llb

IIi− Aλ2lrb

IIi

)(21b)

ywIIc = −

m1∑i=1

wIIi

2k2

(Dλ2llc

IIi− Dλ2lrc

IIi

)(21c)

θwIIc =

m1∑i=1

wIIiλ2

2k2

(Aλ2llc

IIi− Aλ2lrc

IIi

). (21d)

The displacements and slopes at b and c due to thesuperposition of the original given loading and the end-conditioning forces at segment II are

yIIb = q3

2k2

(1−Dλ2l2

)+ q4

4λ2k2

1

l2

(−Cλ2l2 +1−2λ2l2 Dλ2l2

)

+ywIIb + P0b IIλ2

2k2+ P0cIIλ2

2k2Aλ2l2 + M0cIIλ

22

k2Bλ2l2

(22a)

θ IIb = q3λ2

2k2

(−Aλ2l2 +1)− q4

2k2

1

l2

(Dλ2l2 −1+λ2l2 Aλ2l2

)

+θwIIb + M0bIIλ

32

k2+ P0cIIλ

22

k2Bλ2l2 − M0cIIλ

32

k2Cλ2l2

(22b)

yIIc = q3

2k2

(1 − Dλ2l2

)+ q4

4λ2k2

1

l2

(Cλ2l2 − 1 + 2λ2l2

)

+ywIIc + P0b IIλ2

2k2Aλ2l2 + M0b IIλ

22

k2Bλ2l2 + P0cIIλ2

2k2

(22c)

θ IIc = −q3λ2

2k2

(1 − Aλ2l2

)− q4

2k2

1

l2

(Dλ2l2 − 1 + λ2l2

)

+θwIIc − P0b IIλ

22

k2Bλ2l2 + M0b IIλ

32

k2Cλ2l2 − M0cIIλ

32

k2.

(22d)

For segment III, the load is made up of only theballast loads. As before, denoting the length of the partof the tank which belongs to segment III by l′bIII , forevery ballast load wIIIi in segment III we obtain thedistances from the left and right ends of the load topoints c and d as follows:

llcIIIi =

{0, if i = 1

(i − 2)lb + l′b III, if 1 < i ≤ m2(23)

lrcIIIi = (i − 1)lb + l′b III (24)

lldIIIi =

{(m2 − i)lb + l′b III, if i = 1

(m2 − i + 1)lb , if 1 < i ≤ m2(25)

lrdIIIi = (m2 − i)lb , (26)

where m2 is the number of tanks in segment III, includ-ing the tank on which the base of the jacket rests.

Moments and shearing forces produced at the endpoints of segment III due to the loading in segment III,assuming an infinitely long beam, are

McIII = MwIIIc = −

m2∑i=1

wIIIi

4λ23

(Bλ3llc

IIIi− Bλ3lrc

IIIi

)(27a)

QcIII = QwIIIc =

m2∑i=1

wIIIi

4λ3

(Cλ3llc

IIIi− Cλ3lrc

IIIi

)(27b)

MdIII = MwIIId =

m2∑i=1

wIIIi

4λ23

(Bλ3lld

IIIi− Bλ3lrd

IIIi

)(27c)

QdIII = QwIIId =

m2∑i=1

wIIIi

4λ3

(Cλ3lld

IIIi− Cλ3lrd

IIIi

). (27d)

We resolve the original loading into a symmetricaland an antisymmetrical part to obtain M′

III , Q′III , M′′

III ,and Q′′

III by replacing MaI , QaI , MbI , and QbI in (10)with McIII , QcIII , MdIII , and QdIII , respectively. Theend-conditioning forces in each of the parts, P′

0III , M′0III ,

P′′0III , and M′′

0III , can be obtained as in (11) by replacingλ1, l1, M′

I , Q′I , M′′

I , and Q′′I with λ3, l3, M′

III , Q′III ,

M′′III , and Q′′

III , respectively. They are combined in themanner of (12) to obtain the total end-conditioningforces P0cIII , P0dIII , M0cIII , and M0dIII .

Optimization of ballast plan in launch jacket load-out 273

The displacement and slope at c due to all wIIIi insegment III are

ywIIIc =

m2∑i=1

wIIIi

2k2

(Dλ3llc

IIIi− Dλ3lrc

IIIi

)(28a)

θwIIIc =

m2∑i=1

wIIIiλ3

2k2

(Aλ3llc

IIIi− Aλ3lrc

IIIi

). (28b)

The displacement and slope at c due to the super-position of the original given loading and the end-conditioning forces at segment III are

yIIIc = ywIII

c + P0cIIIλ3

2k2+ P0dIIIλ3

2k2Aλ3l3 + M0dIIIλ

23

k2Bλ3l3

(29a)

θ IIIc = θwIII

c + M0cIIIλ33

k2+ P0dIIIλ

23

k2Bλ3l3 − M0dIIIλ

33

k2Cλ3l3 .

(29b)

At every cutting point there will be a relative dis-placement and slope difference between the ends ofthe neighbouring segments. The next step is to find theshearing force X and moment Y at each of the cuttingpoints, i.e. points b and c, to bring about and maintainthe continuity of the beam. Denoting the deflection andangular displacement due to unit values of each of theseforces by δ

Fki, j and φ

Fki, j , in which i denotes the point

where the force F is acting, j denotes the point wherethe deflection or angular displacement is produced, andk denotes the segment that is considered, we can writethe equations of continuity at b and c as

yIIb − yI

b − δXIb ,b Xb + δ

XIIb ,b Xb + δ

XIIc,b Xc

− δYIb ,b Yb + δ

YIIb ,b Yb + δ

YIIc,b Yc = 0 (30a)

θ IIb − θ I

b − φXIb ,b Xb + φ

XIIb ,b Xb + φ

XIIc,b Xc

− φYIb ,b Yb + φ

YIIb ,b Yb + φ

YIIc,b Yc = 0 (30b)

yIIIc − yII

c − δXIIc,c Xc − δ

XIIb ,c Xb + δXIII

c,c Xc

− δYIIc,c Yc − δ

YIIb ,cYb + δYIII

c,c Yc = 0 (30c)

θ IIIc − θ II

c − φXIIc,c Xc − φ

XIIb ,c Xb + φXIII

c,c Xc

− φYIIc,c Yc − φ

YIIb ,cYb + φYIII

c,c Yc = 0 . (30d)

It can be shown that some of the δFki, j s and φ

Fki, j s are

identical:

δXIIb,b = −δXII

c,c (31a)

δXIIc,b = −δ

XIIb,c (31b)

φXIIb,b = φXII

c,c (31c)

φXIIb,c = φ

XIIc,b (31d)

δYIIb,b = δYII

c,c = −φXIIb,b = −φXII

c,c (31e)

δYIIc,b = δ

YIIb,c = φ

XIIc,b = φ

XIIb,c (31f)

δYIb,b = −φ

XIb,b (31g)

δYIIIc,c = −φXIII

c,c (31h)

φYIIb,b = −φYII

c,c (31i)

φYIIc,b = −φ

YIIb,c . (31j)

Hence, putting (30) in matrix form and making use ofthe above identities, we have⎛⎜⎜⎜⎜⎝

−δXIb,b + δ

XIIb,b −δ

YIb,b + δ

YIIb,b δ

XIIc,b δ

YIIc,b

δYIb,b − δ

YIIb,b −φ

YIb,b + φ

YIIb,b δ

YIIc,b φ

YIIc,b

δXIIc,b −δ

YIIc,b δ

XIIb,b + δXIII

c,c −δYIIb,b + δYIII

c,c

−δYIIc,b φ

YIIc,b δ

YIIb,b − δYIII

c,c φYIIb,b + φYIII

c,c

⎞⎟⎟⎟⎟⎠

×

⎛⎜⎜⎝

Xb

Yb

Xc

Yc

⎞⎟⎟⎠ =

⎛⎜⎜⎝

yIb − yII

bθ I

b − θ IIb

yIIc − yIII

cθ II

c − θ IIIc

⎞⎟⎟⎠ . (32)

The δFki, j s and φ

Fki, j s are

δXIb,b = 2λ1

k1

sinh λ1l1 cosh λ1l1 − sin λ1l1 cos λ1l1

sinh2 λ1l1 − sin2 λ1l1(33)

δXIIb,b = −2λ2

k2

sinh λ2l2 cosh λ2l2 − sin λ2l2 cos λ2l2

sinh2 λ2l2 − sin2 λ2l2(34)

δXIIc,b = 2λ2

k2

sinh λ2l2 cos λ2l2 − sin λ2l2 cosh λ2l2

sinh2 λ2l2 − sin2 λ2l2(35)

δXIIIc,c = −2λ3

k2

sinh λ3l3 cosh λ3l3 − sin λ3l3 cos λ3l3

sinh2 λ3l3 − sin2 λ3l3(36)

274 A. Kurniawan, G. Ma

δYIb,b = −2λ2

1

k1

sinh2 λ1l1 + sin2 λ1l1

sinh2 λ1l1 − sin2 λ1l1(37)

δYIIb,b = −2λ2

2

k2

sinh2 λ2l2 + sin2 λ2l2

sinh2 λ2l2 − sin2 λ2l2(38)

δYIIc,b = 4λ2

2

k2

sinh λ2l2 sin λ2l2

sinh2 λ2l2 − sin2 λ2l2(39)

δYIIIc,c = −2λ2

3

k2

sinh2 λ3l3 + sin2 λ3l3

sinh2 λ3l3 − sin2 λ3l3(40)

φYIb,b = −4λ3

1

k1

sinh λ1l1 cosh λ1l1 + sin λ1l1 cos λ1l1

sinh2 λ1l1 − sin2 λ1l1(41)

φYIIb,b = 4λ3

2

k2

sinh λ2l2 cosh λ2l2 + sin λ2l2 cos λ2l2

sinh2 λ2l2 − sin2 λ2l2(42)

φYIIc,b = −4λ3

2

k2

sinh λ2l2 cos λ2l2 + sin λ2l2 cosh λ2l2

sinh2 λ2l2 − sin2 λ2l2(43)

φYIIIc,c = 4λ3

3

k2

sinh λ3l3 cosh λ3l3 + sin λ3l3 cos λ3l3

sinh2 λ3l3 − sin2 λ3l3(44)

Solving (32), we can obtain the shearing forces andmoments Xb, Yb, Xc, and Yc.

Having obtained Xb, Yb, Xc, and Yc, we can obtainthe beam deflection y(x) by superposition of all theloadings. For 0 ≤ x < l1 (segment I),

y(x) = q1

2k1

(2 − Dλ1x − Dλ1(l1−x)

)

+ q2

4λ1k1

1

l1

(Cλ1x−Cλ1(l1−x)−2λ1l1 Dλ1(l1−x)+4λ1x

)

+ P0aIλ1

2k1Aλ1x + M0aIλ

21

k1Bλ1x + P0b Iλ1

2k1Aλ1(l1−x)

+M0b Iλ21

k1Bλ1(l1−x) + 2Xbλ1

k1

× sinh λ1l1 cos λ1(l1 − x) cosh λ1x

sinh2 λ1l1 − sin2 λ1l1− 2Xbλ1

k1

×sin λ1l1 cosh λ1(l1 − x) cos λ1x

sinh2 λ1l1 − sin2 λ1l1+ 2Ybλ2

1

k1

× 1

sinh2 λ1l1 − sin2 λ1l1

×[ sinh λ1l1(

cosh λ1x sin λ1(l1−x)−sinh λ1x

×cos λ1(l1−x))+sin λ1l1

(sinh λ1(l1−x)

×cos λ1x−cosh λ1(l1−x) sin λ1x)]

. (45)

For l1 ≤ x < l1 + l2 (segment II), the contributionof the ballast loads is calculated as follows. First, thedistances from x to the left and right ends of the ballastload wIIi can be obtained as

llII(x) = |x − l1 − llb

IIi| (46a)

lrII(x) = |x − l1 − lrb

IIi| . (46b)

The deflection at x due to ballast load wIIi at segmentII is

ywIIi =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

wIIi

2k2

(Dλ2ll

II(x) − Dλ2lrII(x)

), if i > j

wIIi

2k2

(2 − Dλ2ll

II(x) − Dλ2lrII(x)

), if i = j

−wIIi

2k2

(Dλ2ll

II(x) − Dλ2lrII(x)

), if i < j

,

(47)

where j is the tank index where x is. The completeexpression for segment II is therefore

y(x)= q3

2k2

(2 − Dλ2(x−l1) − Dλ2(l1+l2−x)

)+ q4

4λ2k2

× 1

l2

(Cλ2(x−l1) − Cλ2(l1+l2−x) − 2λ2l2 Dλ2(l1+l2−x)

+4λ2(x − l1))+

m1∑i=1

ywIIi + P0b IIλ2

2k2Aλ2(x−l1)

+ M0b IIλ22

k2Bλ2(x−l1) + P0cIIλ2

2k2Aλ2(l1+l2−x)

+ M0cIIλ22

k2Bλ2(l1+l2−x) − 2Xbλ2

k2

×(

sinh λ2l2 cos λ2(x − l1) cosh λ2(l1 + l2 − x)

sinh2 λ2l2 − sin2 λ2l2

− sin λ2l2 cosh λ2(x − l1) cos λ2(l1 + l2 − x)

sinh2 λ2l2 − sin2 λ2l2

)

+2Ybλ22

k2

1

sinh2 λ2l2 − sin2 λ2l2

×[ sinh λ2l2(

cosh λ2(l1 + l2 − x) sin λ2(x − l1)

− sinh λ2(l1 + l2 − x) cos λ2(x − l1))

+ sin λ2l2(

sinh λ2(x − l1) cos λ2(l1 + l2 − x)

− cosh λ2(x − l1) sin λ2(l1 + l2 − x))]+ 2Xcλ2

k2

Optimization of ballast plan in launch jacket load-out 275

×(

sinh λ2l2 cos λ2(l1+l2−x) cosh λ2(x−l1)

sinh2 λ2l2−sin2 λ2l2

− sin λ2l2 cosh λ2(l1+l2−x) cos λ2(x−l1)

sinh2 λ2l2−sin2 λ2l2

)

+ 2Ycλ22

k2

1

sinh2 λ2l2 − sin2 λ2l2

× [sinh λ2l2

(cosh λ2(x − l1) sin λ2(l1 + l2 − x)

− sinh λ2(x − l1) cos λ2(l1 + l2 − x))

+ sin λ2l2(

sinh λ2(l1 + l2 − x) cos λ2(x − l1)

− cosh λ2(l1 + l2 − x) sin λ2(x − l1))]

. (48)

Likewise, for l1 + l2 ≤ x ≤ l (segment III), the dis-tances from x to the left and right ends of the ballastload wIIIi are calculated as

llIII(x) = |x − l1 − l2 − llc

IIIi| (49a)

lrIII(x) = |x − l1 − l2 − lrc

IIIi| . (49b)

The deflection at x due to ballast load wIIIi at segmentIII, ywIII

i , can be obtained as in (47) by replacing wIIi,λ2, ll

II(x), and lrII(x) with wIIIi, λ3, ll

III(x), and lrIII(x),

respectively. The complete expression for segmentIII is

y(x)=m2∑i=1

ywIIIi + P0cIIIλ3

2k2Aλ3(x−l1−l2)+

M0cIIIλ23

k2Bλ3(x−l1−l2)

+ P0dIIIλ3

2k2Aλ3(l−x)+ M0dIIIλ

23

k2Bλ3(l−x)− 2Xcλ3

k2

×(

sinh λ3l3 cos λ3(x − l1 − l2) cosh λ3(l − x)

sinh2 λ3l3 − sin2 λ3l3

− sin λ3l3 cosh λ3(x − l1 − l2) cos λ3(l − x)

sinh2 λ3l3 − sin2 λ3l3

)

+2Ycλ23

k2

1

sinh2 λ3l3 − sin2 λ3l3

×[ sinh λ3l3(

cosh λ3(l − x) sin λ3(x − l1 − l2)

− sinh λ3(l − x) cos λ3(x − l1 − l2))

+ sin λ3l3(

sinh λ3(x − l1 − l2) cos λ3(l − x)

− cosh λ3(x − l1 − l2) sin λ3(l − x))]

. (50)

3 Conventional ballast calculation method

In the conventional ballast calculation method, thebarge is assumed to be rigid. The objective of the ballastplan is to maintain the barge to be horizontally levelledwith the quay wall. By fixing the ballasts in severaltanks and selecting any two tanks in which the ballasts

are to be determined, the ballasts in these two tankscan be calculated using the equations of static equilib-rium alone. We have devised an algorithm to illustratethis method.

We use a dimensionless value, αi, i = 1, 2, . . . , n torepresent the ballast load wi in tank i. αi has a valuefrom 0 to 1, and is related to wi as follows:

wi = (αihb − hb int)k2 , (51)

where hb is the height of the ballast tank. Initially, allballasts in the tanks are set to the initial value, wint.Thus, the ballast load wi at this stage is zero and theinitial αi in each tank is

αpi = hb int

hb. (52)

The applied jacket load is assumed to be made upof the weight of the jacket part on the barge. Usingthe notations in Fig. 4, we can calculate the point ofapplication of the equivalent jacket point load as

xjacket = 3q1 + 2q2

6q1 + 3q2ll , (53)

where ll is the length of the loaded part of the barge.The next step is to determine the initial number of

tanks needed to counter the jacket load, ntank, which iscalculated as follows:

ntank = Fjacket

wintlb, (54)

where Fjacket is the jacket load. If ntank > n, then thereis not enough barge capacity. The ntank value is roundedto its two nearest integers, with the smaller integerdenoted as S and the larger as L:

S = �ntank� (55a)

L = �ntank� , (55b)

where � � denotes the floor function, while � � de-notes the ceiling function. If ntank is an integer, then

L = ntank (56a)

S = ntank − 1 . (56b)

Fig. 4 Trapezoidal loadingdue to the jacket weight, forthe conventional rigid bargemethod q1

q2

276 A. Kurniawan, G. Ma

Three cases are considered depending on the posi-tion of xjacket:

Case 1 : xjacket < lb2

Case 2 : lb2 ≤ xjacket < (n − 0.5)lb

Case 3 : xjacket ≥ (n − 0.5)lb

For Case 1, we first set i = L and j = n, where i andj are tank indices. The point of application of ballastforce in tank t, where t is the index of the tank, can becalculated as

xt = (t − 0.5)lb . (57)

The ballast forces in tanks i and j are obtained as

Fi = −Fjacket(x j − xjacket) − Swintlb

(x j − lb S

2

)x j − xi

(58a)

F j = −(Fjacket + Fi − Swintlb ) . (58b)

The change of α in tank i is

�αi = Fi

lb hb k2, (59)

and the new αi is

αi = αpi + �αi . (60)

Likewise for tank j. The values of α1 to αi−1 are thenset to 0.

Next, we set two counters, ci and c j, as 0. The follow-ing calculations are performed while αi < 0 or α j > 1. Ifαi < 0 and α j ≤ 1, then

αi = 0 (61a)

ci = ci + 1 (61b)

i = i + 1 . (61c)

Else, if α j > 1 and αi ≥ 0, then

α j = 1 (62a)

c j = c j + 1 (62b)

j = j − 1 . (62c)

Else, if αi < 0 and α j > 1, then

αi = 0 (63a)

α j = 1 (63b)

ci = ci + 1 (63c)

c j = c j + 1 (63d)

i = i + 1 (63e)

j = j − 1 . (63f)

If i ≥ j, then there is not enough barge capacity. Else,the ballast forces in tank i and j are calculated as

Fi =−Fjacket

(x j − xjacket

)− Swintlb

(x j − lb S

2

)x j − xi

+ciwintlb(x j−xi+lb

ci+12

)+c j (hb −hb int) lb k2lbc j+1

2

x j − xi

(64a)

F j =−[Fjacket+Fi−Swintlb −ciwintlb +cj(hb −hb int)lb k2

],

(64b)

and the values of �α and α for tanks i and j areobtained from (59) and (60).

For Case 2, we first set i and j as follows:

i =

⎧⎪⎨⎪⎩

[xjacket

lb

]− S

2, if S is even⌈xjacket

lb

⌉− S + 1

2, if S is odd

(65a)

j = i + S + 1 , (65b)

where [ ] is the notation for the nearest integerfunction.

If i < 1 and j > n, then there is not enough bargecapacity.

Else, if i ≥ 1 and j ≤ n, then the ballast force in tanki can be calculated as

Fi = − Fjacket(x j − xjacket) − Swintlblb L

2

x j − xi(66)

and the ballast force in tank j as in (58b). The �α andα for tanks i and j are obtained from (59) and (60). Thevalues of αi+1 to α j−1 are set to 0.

Else, if i < 1, then we set i = L and j = n, and thesame calculations as in (58a) to (64) are performed.The �α and α for tanks i and j are obtained from (59)and (60).

Else, if j > n, then we set i = 1 and j = n − S. Theballast force in tank i is calculated as

Fi = − Fjacket(x j − xjacket) + Swintlblb L

2

x j − xi(67)

and the ballast force in tank j as in (58b). The �α andα for tanks i and j are obtained from (59) and (60).The values of α j+1 to αn are set to 0. Next, we set twocounters, ci and c j, as 0, and the following calculationsare performed while αi > 1 or α j < 0. If αi > 1 andα j ≥ 0, then

αi = 1 (68a)

ci = ci + 1 (68b)

i = i + 1 . (68c)

Optimization of ballast plan in launch jacket load-out 277

Table 1 Parameters for theload-out cases

Parameter Case 1 Case 2a Case 2b

Jacket length (m) 100 150 150Jacket weight at the top (kg/m) 3 × 104 7 × 104 7 × 104

Jacket weight at the base (kg/m) 8 × 104 2.2 × 105 2.2 × 105

Barge length (m) 100 150 150Barge weight (kg/m) 8 × 104 2 × 105 2 × 105

Stages 0 to 17 0 to 25 0 to 25Number of tanks 10 10 15Tank length (m) 10 15 10Tank height (m) 7 10 10Draft (m) 6 8.5 8.5I1 (m4) 200 200 200I2 (m4) 500 500 500I3 (m4) 300 300 300E (kg/m2) 2 × 1010 2 × 1010 2 × 1010

Yard stiffness (kg/m2) 1.4 × 108 2 × 108 2 × 108

Hydrostatic stiffness (kg/m2) 3.5 × 104 5 × 104 5 × 104

Number of generations 200 200 200Population size 200 200 200

Else, if α j < 0 and αi ≤ 1, then

α j = 0 (69a)

c j = c j + 1 (69b)

j = j − 1 . (69c)

Else, if αi > 1 and α j < 0, then

αi = 1 (70a)

α j = 0 (70b)

ci = ci + 1 (70c)

c j = c j + 1 (70d)

i = i + 1 (70e)

j = j − 1 . (70f)

If i ≥ j, then there is not enough barge capacity. Else,the ballast forces in tank i and j are calculated as

Fi =−Fjacket

(x j − xjacket

)+ Swintlb

(l − x j − lb S

2

)x j − xi

−ci(hb −hb int)k2lb(x j−xi+lb

ci+12

)− c jwintlb lbc j+1

2

x j − xi

(71a)

F j =−[Fjacket+Fi−Swintlb +ci(hb −hb int)k2lb −c jwintlb

],

(71b)

where l is the total length of the barge. The �α and α

for tanks i and j are obtained from (59) and (60).For Case 3, we first set i = 1 and j = n − S. The same

calculations as in (67) to (71a) are then performed, and

the �α and α for tanks i and j are obtained from (59)and (60).

4 Proposed ballast calculation method

In our method, we utilize multi-objective evolutionaryalgorithm (MOEA) (for an overview of MOEA, see,for example, Deb 2001) to find the optimum ballastplan. The ability of MOEA to find multiple solutionsat a single run and the fact that it can incorporate anynumber of objectives makes it well-suited to tackle theballast calculation problem at hand. In our algorithm,we use three objectives:

(1) minimizing the deflection of the beam model,(2) minimizing the curvature of the beam model,

and(3) maximizing the ballast transfer efficiency between

the load-out stages, that is, minimizing the totalballast removed or shifted from one load-out stageto the next.

We will first give some fundamentals of multi-objective optimization.

4.1 Multi-objective optimization fundamentals2

Multi-objective optimization deals with multiple con-flicting objectives. Normally, the optimal solution ofone objective is not necessarily the optimum for theother objectives. Multi-objective optimization is not

2We are indebted to Nebro et al. (2006) and Deb (2007) for thematerials in this section.

278 A. Kurniawan, G. Ma

restricted to find a unique single solution, but a set ofsolutions called non-dominated solutions. Each solutionin this set is said to be Pareto optimal, and when thesesolutions are plotted in the objective space they arecollectively known as the Pareto front. Without losinggenerality, we assume the minimization of all the ob-jective function values. A multi-objective optimizationproblem (MOOP) is defined as follows:

Definition 1 (MOOP) Find a vector of n decisionvariables x∗ = (x∗

1, x∗2, . . . , x∗

n) which satisfies the k in-equality constraints gi(x) ≥ 0, i = 1, 2, . . . , k, the pequality constraints hi(x) = 0, i = 1, 2, . . . , p, the vari-able bounds xL

i ≤ xi ≤ xUi , i = 1, 2, . . . , n, and mini-

mizes the vector of m objective function values f(x) =( f1(x), f2(x), . . . , fm(x)).

The set of all values satisfying the constraints andvariable bounds defines the feasible decision variablespace �. Any point x ∈ � is a feasible solution. Foreach x ∈ �, there exists a point in the objective space,denoted by f(x) = ( f1(x), f2(x), . . . , fm(x)). A mappingexists between an n-dimensional decision variable vec-tor and an m-dimensional objective vector through theobjective function, constraints, and variable bounds.

As mentioned previously, we seek for the Paretooptima.

Definition 2 (Pareto Optimality) A point x∗ ∈ � isPareto optimal if for every x ∈ � and I = {1, 2, . . . , m}either (∀i ∈ I) fi(x) = fi(x∗) or (∃i ∈ I) fi(x) > fi(x∗).

The above definition states that x∗ is Pareto optimalif there is no feasible vector x which would decreasesome function values without simultaneously increas-ing at least one other function value. Other importantdefinitions associated with Pareto optimality are thefollowing:

Definition 3 (Pareto Dominance) A vector u = (u1,

. . . , um) is said to dominate v = (v1, . . . , vm) (denotedby u � v) if and only if u is partially less than v, that is,(∀i ∈ {1, . . . , m})ui ≤ vi ∧ (∃i ∈ {1, . . . , m})ui < vi.

Definition 4 (Pareto Optimal Set) For a given MOOP,the Pareto optimal set is defined as P∗ = {x ∈ � |(¬∃x′ ∈ �) f(x′) � f(x)}.

Definition 5 (Pareto Front) For a given MOOP and itsPareto optimal set P∗, the Pareto front is defined asPF∗ = {f(x) | x ∈ P∗}.

Obtaining the Pareto front and the Pareto optimalset is the main goal of multi-objective optimization.However, since a Pareto front can contain a large num-ber of points, a good solution must contain a limitednumber of them. These should be as close as possibleto the true Pareto front and well distributed over theentire Pareto front. Otherwise, they would not be veryuseful to the decision maker.

4.2 Multi-objective optimization problem of findingoptimum ballast arrangement

Representing the ballast load wi in tank i by a dimen-sionless value αi as defined in (51), we can formulatethe MOOP of finding optimum ballast arrangementas follows:

For 0 ≤ αi ≤ 1, minimize

f1(α) = yrms,

f2(α) =(

d2 ydx2

)max

,

f3(α)=

⎧⎪⎪⎨⎪⎪⎩

∑i(α′

i −αi)+ ∑k | αk≥α′

k

(αk−α′k), if

∑i

αi ≤∑i

α′i

∑i(αi−α′

i)+∑

k | αk<α′k

(α′k−αk), if

∑i

αi >∑

iα′

i

.

(72)

The first function is the root-mean-square deflection ofthe beam model:

yrms =√

〈y(xi)2〉 , (73)

where 〈 〉 denotes the arithmetic mean. The expres-sion for the beam deflection y(x) is what we have de-rived in Section 2. The second function is the maximumcurvature of the beam. The curvature is obtained bytwice differentiating the deflection numerically. Thethird function is the function to calculate the totalballast removed or shifted from the previous load-outstage to the current one. The prime (′) denotes thevalue at the previous stage. The formulation of thefunction assumes that the ballast in one tank can betransferred to every other tank with equal speed.

4.3 Multi-objective evolutionary algorithm for findingoptimum ballast plan

The first step of the algorithm is to create an N-sizedset of solutions, P. These are generated randomly. A

Optimization of ballast plan in launch jacket load-out 279

solution j is defined by a vector (αj1, . . . , α

jn) in the

decision variable space and its map ( f1(αj), . . . , fm(α j))

in the objective space. n is the total number of deci-sion variables, which, in our case, is equal to the totalnumber of ballast tanks, and m is the number of objec-tive functions, as given in Section 4.2. The mapping isfrom an n-dimensional vector α j to an m-dimensionalvector f(α j).

The next step is selection. In our algorithm, theselection operation is made up of two tasks: (1) theselection of non-dominated solutions, and (2) the dis-card of crowded solutions from those obtained from(1). By (1) we want to keep the solutions closest tothe true Pareto optimal solutions, and by (2) we wantto maximize the spread of solutions. In addition, with(2) the number of solutions are kept within a specifiedlimit, since a large number of solutions slows down theiteration process.

The first task is described as follows:

1. Create two empty sets of temporary non-dominated solutions, Pt and Pt′ .

2. Mark the solution in P having the least f1(α) andput its copy into Pt.

3. Put into Pt′ the solutions in P with f2(α) smallerthan that of the solution we have previouslymarked. Keep the remaining solutions in P.

4. Repeat step 3 for the rest of the objective functionsup to fm, or until there is only one solution in P.

5. Update P = Pt′ and then empty Pt′ .6. Repeat steps 2 to 5 until there are no more solu-

tions in Pt′ .

At the end of this task, we have temporary non-dominated solutions in Pt.

The second task is performed only if the number ofsolutions in Pt is larger than N, which can only happenafter the first iteration/generation. To reduce the sizeof Pt to N, we apply the clustering algorithm. The basicidea is to group solutions close to each other into onecluster, keep one solution in the cluster, and discardthe remaining solutions in the cluster. This is achievedas follows:

1. Calculate the distance in the objective space, Di, j,between two clusters Ci and C j for all pairs ofclusters in Pt. A cluster is defined as a set of points.Thus, initially, each point belongs to a distinctcluster.

Di, j = 1

|Ci||C j|∑

i∈Ci, j∈C j

√√√√ m∑k=1

(fk(αi) − fk(α j)

f maxk (α) − f min

k (α)

)2

,

(74)

0.531 0.531 0.531 0.531 0.531 0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531 0.531 0.531 0.531 0.531

0.531

0.532

0.531 0.531 0.531 0.531 0.531 0.531 0.531 0.531 0.534

0.531 0.531 0.531 0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531 0.531 0.531 0.531

0.531

0.531 0.531 0.531 0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531

0.531 0.531 0.531 0.531

0.531 0.531 0.531

0.531 0.531 0.531

0.531 0.531 0.531

0.531 0.531

0.496

0.303

0.188

0.466

0.000

0.505 0.503

0.339

0.000

0.500

0.387

0.000

0.000

0.000 0.000

0.000

0.000

0.000 0.000 0.000

0.000 0.000 0.000

0.000 0.000 0.000

0.000 0.000 0.000

0.000 0.000 0.000

0.000 0.000 0.000 0.000

0.000 0.000 0.000 0.000

0.000

0.000

0.000

0.291

0.502 0.552

0.5590.348

0

1

2

3

4

5

6

7

8

0.531

0.202

0.565

0.531

0.288 0.347

0.5310.436

0.605

0.531

0.531 0.531

0.531 0.531

0.531 0.531

0.531 0.531

0.531 0.531 0.531

0.5250.399

0.181

0.632

0.2980.413

0.441

0.174

0.5940.464

0.291

0.6850.531

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

17

9

10

Fig. 5 Ballast plan produced by conventional method (case 1)

280 A. Kurniawan, G. Ma

The superscripts max and min denote the maximumand minimum function values, while | | denotesthe size of the cluster. The expression with thesquare root is the normalized distance between anytwo clusters i and j in Pt.

2. Find two clusters, among all the clusters in Pt,which have the minimum Di, j. These clusters arecombined to form one bigger cluster. The numberof clusters in Pt is now reduced by one.

3. Repeat steps 1 to 2 until the number of clusters inPt is reduced to N.

4. Calculate the centroid of each cluster. The centroidci of a cluster Ci is the average coordinate in theobjective space of all the points in the cluster. It iscalculated as

ci = 1

|Ci|∑j∈Ci

(f1(α

j), . . . , fm(α j))

. (75)

5. In each cluster, keep the point with the minimumdistance to the centroid and discard all the otherpoints.

At the end of this task we have N temporary non-dominated solutions in Pt.

After performing the selection operation we haveq ≤ N solutions. These solutions are called the parentsolutions and are labelled 1 to q. The next step is togenerate (2N − q) offspring solutions labelled q + 1 to2N via a variation operator. This is described as follows:

1. Set j = 1 and k = q + 1.2. Calculate the coordinate in the decision variable

space of the offspring solution k as

αki = α

ji + 1

G0.8R , for i = 1, . . . , n , (76)

where R is a random value drawn from a normaldistribution with mean 0 and standard deviation 1,G is the generation number, while 1 and 0.8 in 1

G0.8

are pre-selected parameters.3. If αk

i < 0, then αki = 0. If αk

i > 1, then αki = 1.

4. Update k = k + 1.5. If j = q, then j = 1. Else, j = j + 1.6. Repeat steps 2 to 5 while k ≤ 2N.7. Compute ( f1(α

k), . . . , fm(αk)), for k = q + 1,

. . . , 2N.

At the end of the variation operation we have a popula-tion of 2N solutions made up of the q parent solutionsand the (2N − q) offspring solutions. This becomes thenew population for the next generation.

0.531 0.531 0.531 0.531 0.531 0.531 0.531 0.531 0.531 0.531

0.539 0.577 0.505 0.527 0.498 0.503 0.505 0.564 0.560 0.518

0.352

0.5750.461 0.515 0.526 0.492 0.511 0.583 0.534 0.532

0.193

0.4350.541 0.517 0.500 0.487 0.530 0.616

0.504 0.526

0.074

0.3580.481 0.554 0.480 0.551 0.486

0.606 0.520 0.519

0.000

0.2510.439

0.549 0.469 0.552 0.5000.606 0.518 0.513

0.0190.242

0.413 0.503 0.4240.538 0.506 0.550 0.524 0.547

0.024

0.2490.347 0.425 0.420

0.541 0.497 0.527 0.522 0.557

0.0680.220 0.222

0.400 0.313

0.575 0.509 0.552 0.510 0.545

0

1

2

3

4

5

6

7

8

0.1620.336

0.2130.377

0.140

0.526 0.518 0.545 0.570 0.501

0.1560.322

0.2050.390

0.113

0.459 0.465 0.538 0.576 0.501

0.2020.305

0.1790.399

0.106

0.444 0.3570.497 0.571 0.530

0.240 0.288 0.2070.387

0.094

0.4070.295

0.4490.566 0.534

0.2530.344

0.2400.373

0.094

0.403

0.172

0.438 0.503 0.560

0.300 0.3410.245

0.393

0.107

0.398

0.1680.387 0.417

0.545

0.297 0.3810.263

0.376

0.116

0.4200.203

0.295 0.308

0.545

0.323 0.3950.295

0.3870.186

0.3910.198 0.249 0.222

0.496

0.3430.435

0.2790.382

0.2250.408

0.215 0.210 0.2170.352

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

17

9

10

Fig. 6 Ballast plan produced by proposed method (case 1)

Optimization of ballast plan in launch jacket load-out 281

The above steps are carried out iteratively up to aspecified number of generations, T. At G = T we stopafter the selection operation; we do not carry out thevariation operation. The final solutions contained in Pt

become the optimum solutions for the particular load-out stage. We can then pick one solution from this setof solutions and move on to the next load-out stage.

The steps of our algorithm for a particular load-outstage can be summarized as follows:

1. Create a random population of solutions of size N.Denote this set of solutions by P. Set G = 1.

2. Perform selection of non-dominated solutions fromP. Keep these non-dominated solutions in Pt.If |Pt| > N, then reduce it to N by discardingcrowded solutions from Pt using the clustering al-gorithm. Otherwise, proceed to step 3. At the endof step 2 we have q ≤ N solutions.

3. Generate (2N − q) offspring solutions by variation.At the end of this step we have a total of 2Nsolutions.

4. Denote this set of solutions by P. Set G = G + 1.5. Repeat steps 2 to 4 while G ≤ T. When G = T,

stop at step 2.6. The final solutions in Pt are the optimum solutions.

Any solution picked from Pt is an optimum ballastarrangement for the particular load-out stage.

The algorithm is applied at every load-out stage. Atany particular stage, the algorithm produces a set ofoptimum α vectors. An optimum ballast arrangementis represented by a vector α chosen from this set. Asfor α′, for the first stage α′ is equal to α p, the initial α.

For the second and remaining stages, α′ is the optimumα from the previous stage. A set of selected optimumα vectors for all the load-out stages constitutes anoptimum ballast plan for the load-out operation.

5 Case studies

We compare results obtained from the convention-al rigid barge method with those from the proposedmethod. For the proposed method, selection of opti-mum solutions is done by prioritizing solutions withminimum curvature and deflection than those with min-imum ballast shifted. We first look for solutions withminimum curvature and deflection, and then, amongthese solutions, select that which has the minimumballast shifted.

Three cases are studied, the parameters for whichare shown in Table 1. The dimensions of the jacketand barge in Cases 2a and 2b are larger than those inCase 1. The difference between Case 2a and Case 2b isin the number of ballast tanks. For Case 2b, the numberof ballast tanks is increased to 15. This means that thereare more decision variables for Case 2b than Case 2a,which increases the complexity of the problem. Weassume that there is no tide variation. It follows thatthe draft is constant for all stages.

We define Stage 0 as the stage before the jacket isloaded onto the barge. The initial ballast is applied atthis stage. The incremental progress of the jacket on thebarge from one stage to the next is 6 m, except for Stage

Fig. 7 Root-mean-squaredeflections, maximumcurvatures, ballast removedor shifted, and average ballastobtained from conventionalmethod (solid) and proposedmethod (dashed) for case 1

0

0.0005

0.001

0.0015

0.002

0.0025

0 2 4 6 8 10 12 14 16

0 2 4 6 8 10 12 14 16

rms

defle

ctio

n (m

)

stage

05e − 0111e − 010

1.5e − 0102e − 010

2.5e − 0103e − 010

3.5e − 0104e − 010

0 2 4 6 8 10 12 14 16max

cur

vatu

re (

m–1

)

stage

00.10.20.30.40.50.6

stage0 2 4 6 8 10 12 14 16

stage

0.3

0.35

0.4

0.45

0.5

0.55

aver

age

balla

st (

in

)

balla

st s

hift

ed (

in

)α α

282 A. Kurniawan, G. Ma

0 to 1, which is 1 m. At Stage 1, the jacket has covered adistance of 1 m on the barge; at Stage 2, it has covereda distance of 7 m, and so on up to the final Stage. Sincethe jacket and barge in Cases 2a and 2b are longer thanthose in Case 1, it requires more stages to complete theload-out operation in cases 2a and 2b as compared tocase 1. For case 1, the stages are from 0 to 17; at Stage17, the jacket has covered a distance of 97 m. For Cases2a and 2b, the Stages are from 0 to 25; at Stage 25, thejacket has covered a distance of 145 m.

For the proposed method, both the population sizeand the number of generations used in the MOEA areset to 200.

5.1 Case 1

For case 1, the schematic diagram of the barge withballast plan obtained from the conventional method isshown in Fig. 5. The shaded area represents the ballastor the mass of water occupying the tanks. The directionof the progression of the jacket is from left to right.Thus, tank 1 denotes the tank closest to the yard. Shownin each tank is the α value, whose initial value is foundto be 0.531.

From the figure we can see that Tank 1 is graduallyemptied as the load-out commences (Stage 0 to 4).From Stage 5 onwards, as the jacket progresses to

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.419 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.452

0.2420.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.456

0.077

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.453

0.000

0.366 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.451

0.0000.208

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.066 0.000

0.439 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.000 0.000

0.374 0.450 0.450 0.450 0.450 0.450 0.450 0.432

0.000 0.000

0.2400.450 0.450 0.450 0.450 0.450 0.450 0.422

0.1340.000 0.000

0.388 0.450 0.450 0.450 0.450 0.450 0.450

0.000 0.000 0.000

0.467 0.450 0.450 0.450 0.450 0.450 0.372

0.2170.000 0.000 0.000

0.492 0.450 0.450 0.450 0.450 0.450

0.237

0.000 0.000 0.000

0.3470.450 0.450 0.450 0.450 0.450

0

1

2

3

4

5

6

7

8

11

12

9

10

1 2 3 4 5 6 7 8 9 10

0.274

0.000 0.000 0.0000.190

0.450 0.450 0.450 0.450 0.450

0.333

0.000 0.000 0.000 0.000

0.466 0.450 0.450 0.450 0.450

0.368

0.000 0.000 0.000 0.000

0.3200.450 0.450 0.450 0.450

0.417

0.000 0.000 0.000 0.0000.166

0.450 0.450 0.450 0.450

0.450

0.036 0.000 0.000 0.000 0.000

0.446 0.450 0.450 0.450

0.527

0.000 0.000 0.000 0.000 0.000

0.3090.450 0.450 0.450

0.450

0.1850.000 0.000 0.000 0.000 0.000

0.5600.450 0.450

0.4500.240

0.000 0.000 0.000 0.000 0.000

0.418 0.450 0.450

0.4500.307

0.000 0.000 0.000 0.000 0.000

0.2700.450 0.450

0.450 0.385

0.000 0.000 0.000 0.000 0.0000.115

0.450 0.450

0.450 0.468

0.000 0.000 0.000 0.000 0.000 0.000

0.410 0.450

0.4500.543

0.000 0.000 0.000 0.000 0.000 0.000

0.2670.450

0.450 0.450

0.2210.000 0.000 0.000 0.000 0.000 0.000

0.527

1 2 3 4 5 6 7 8 9 10

13

14

15

16

17

19

20

21

22

23

24

25

18

Fig. 8 Ballast plan produced by conventional method (case 2a)

Optimization of ballast plan in launch jacket load-out 283

the right, the part of the barge with empty tanks alsomoves to the right to counter the jacket load. Thenumber of empty tanks is proportional to the mag-nitude of the jacket load applied onto the barge. Thegreater the magnitude, the larger the number of tanksto be emptied, since greater upward force is needed tocounter the jacket load. As the jacket progresses alongthe barge, more tanks are emptied due to the increasingjacket load.

The ballast plan obtained from the proposed methodis shown in Fig. 6. The ballasts arrange themselves in asinklike pattern which moves to the right as the jacketprogresses. Compared with that from the conventional

method, the ballast pattern obtained from the proposedmethod is less ‘abrupt.’

The variations of root-mean-square deflections,maximum curvatures, ballast removed or shifted, andaverage ballast for all load-out stages are plotted inFig. 7. From the figure we observe that the deflectionobtained from the proposed method is lower than thatfrom the conventional method for all stages.

In terms of maximum curvature, the proposedmethod also gives the better result. The maximum cur-vature for the conventional method increases monoton-ically from Stage 5 onwards and eventually becomes thelargest at the final stage. On the other hand, the maxi-

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.438 0.444 0.463 0.439 0.436 0.447 0.462 0.429 0.478 0.441

0.222

0.468 0.490 0.444 0.440 0.440 0.449 0.3930.544

0.414

0.106

0.3930.498 0.438 0.450 0.435 0.502

0.3510.535

0.424

0.045

0.2990.501 0.439 0.467 0.401

0.5070.359

0.5300.429

0.0310.192

0.473 0.415 0.483 0.4000.515

0.3620.526

0.426

0.0170.121

0.398 0.425 0.468 0.3960.523

0.3690.522

0.424

0.000 0.069

0.331 0.3580.477 0.388

0.5340.385

0.5150.424

0.082 0.0130.229 0.310

0.5210.369

0.5200.377

0.5360.415

0.023 0.0180.154 0.226

0.5490.351

0.5140.388

0.5210.424

0.022 0.026 0.0610.194

0.5130.380

0.476 0.406 0.486 0.448

0.127 0.072 0.006 0.042

0.462 0.430 0.4870.369

0.512 0.434

0.130 0.091 0.008 0.006

0.373 0.3720.503

0.3660.514 0.441

0

1

2

3

4

5

6

7

8

11

12

9

10

1 2 3 4 5 6 7 8 9 10

0.1830.066 0.015 0.000

0.268 0.3430.497

0.3660.518 0.436

0.1760.077 0.053 0.028

0.1700.270

0.448 0.3760.571

0.404

0.1730.084 0.076 0.048

0.141 0.132

0.4760.352

0.5640.409

0.186 0.103 0.133 0.047 0.093 0.054

0.415 0.3660.531

0.431

0.188 0.141 0.1430.035 0.092 0.041

0.320 0.3430.518 0.439

0.228 0.156 0.1480.007

0.118 0.0480.225 0.302

0.484 0.457

0.199 0.243 0.1850.001

0.1210.000

0.178 0.234

0.490 0.446

0.217 0.235 0.2110.015

0.1600.000

0.1030.204

0.3740.491

0.211 0.217 0.2040.092 0.146

0.0000.138 0.064

0.374 0.456

0.238 0.254 0.1960.093

0.2000.004

0.1280.016

0.2560.462

0.234 0.230 0.186 0.1490.266

0.000 0.087 0.0200.165

0.415

0.204 0.218 0.218 0.1520.287

0.046 0.076 0.032 0.072

0.338

0.2450.355

0.1980.089

0.281

0.0550.158

0.035 0.0320.196

13

14

15

16

17

19

20

21

22

23

24

25

18

1 2 3 4 5 6 7 8 9 10

Fig. 9 Ballast plan produced by proposed method (case 2a)

284 A. Kurniawan, G. Ma

mum curvature for the proposed method is relativelymore constant. It should be noted that the values forStage 1 may not be correct due to numerical instability.This will be observed if we plot the curvature of thebeam for Stage 1. The values of curvature fluctuates atsegment II, and the maximum occurs at this segment.

In terms of ballast transfer efficiency, the total bal-last removed or shifted for the proposed method isgenerally lower than that for the conventional method.Hence, for this particular run, the proposed method isalso shown to result in a more efficient ballast transfer.As for the average ballast, the variations of average bal-last obtained from the conventional and the proposedmethods are comparable.

5.2 Case 2a

The schematic diagrams of the barge with ballastplan obtained from the conventional and the proposedmethods are shown in Figs. 8 and 9. As mentionedabove, the number of load-out stages in Cases 2a and2b is higher than that in Case 1. For the conventionalmethod, the ballast plan is similar to that in Case 1,only that the total number of empty tanks is higher thanthat in Case 1, as the jacket is heavier. As in Case 1,the ballast plan obtained from the proposed methodexhibits a sinklike pattern which moves to the right asthe load-out progresses.

From Fig. 10 we observe that the proposed methodgives much better result than the conventional methodin terms of root-mean-square deflection and maximumcurvature. The root-mean-square deflection from the

proposed method is smaller than that from the con-ventional method for all load-out stages. The root-mean-square deflection from the conventional methodgenerally increases as the load-out progresses, and ismuch larger than that from the proposed method es-pecially for the second half of the load-out stages. Onthe contrary, the root-mean-square deflection from theproposed method is relatively constant for all stages.In terms of maximum curvature, the proposed methodalso gives the better result. The maximum curvatureobtained from the conventional method increases mo-notonically from Stage 11 onwards and at the final stageit becomes much larger than that from the proposedmethod. In contrast, the maximum curvature from theproposed method is relatively constant for all stages. Asin Case 1, it should be noted that the values at Stage 1may not be correct due to numerical instability.

The total ballasts removed or shifted for the conven-tional and proposed methods are comparable. For themajority of the stages, the values from the proposedmethod are slightly higher than those from the conven-tional method. However, the values from the proposedmethod are relatively more constant. As for the aver-age ballast, the variations of average ballast obtainedfrom the conventional and the proposed methods arecomparable.

5.3 Case 2b

The schematic diagrams of the barge with ballastplan obtained from the conventional and the proposed

Fig. 10 Root-mean-squaredeflections, maximumcurvatures, ballast removedor shifted, and average ballastobtained from conventionalmethod (solid) and proposedmethod (dashed) for case 2a

00.0020.0040.0060.0080.01

0.0120.014

0 5 10 15 20 25stage

0 5 10 15 20 25stage

02e − 0104e − 0106e − 0098e − 0091e − 009

1.2e − 009

0

0.1

0.2

0.3

0.4

0.5

0 5 10 15 20 25stage

0.150.2

0.250.3

0.350.4

0.45

0 5 10 15 20 25stage

rms

defle

ctio

n (m

)ba

llast

shi

fted

(in

)α

max

cur

vatu

re (

m–1

)av

erag

e ba

llast

(in

)α

Optimization of ballast plan in launch jacket load-out 285

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.405 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.451

0.144

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.453

0.000

0.3440.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.451

0.0000.102

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.448

0.000 0.000

0.3190.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.443

0.0920.000 0.000

0.440 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.000 0.000 0.000

0.3380.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.421

0.1550.000 0.000 0.000

0.388 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0

1

2

3

4

5

6

7

8

0.2210.000 0.000 0.000 0.000

0.5640.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.241

0.000 0.000 0.000 0.000

0.3420.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.309

0.000 0.000 0.000 0.000 0.000

0.529 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.347

0.000 0.000 0.000 0.000 0.000

0.3040.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

11

12

9

10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0.410

0.000 0.000 0.000 0.000 0.000 0.060

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.471

0.000 0.000 0.000 0.000 0.000 0.000

0.2770.450 0.450 0.450 0.450 0.450 0.450 0.450

0.450

0.1140.000 0.000 0.000 0.000 0.000 0.000

0.468 0.450 0.450 0.450 0.450 0.450 0.450

0.614

0.000 0.000 0.000 0.000 0.000 0.000 0.000

0.2600.450 0.450 0.450 0.450 0.450 0.450

0.4500.281

0.000 0.000 0.000 0.000 0.000 0.000 0.000

0.442 0.450 0.450 0.450 0.450 0.450

13

14

15

16

17

0.450 0.365

0.000 0.000 0.000 0.000 0.000 0.000 0.0000.214

0.450 0.450 0.450 0.450 0.450

0.450 0.465

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

0.427 0.450 0.450 0.450 0.450

0.4500.559

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.0000.203

0.450 0.450 0.450 0.450

0.450 0.4500.241

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

0.399 0.450 0.450 0.450

0.450 0.4500.355

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.0000.170

0.450 0.450 0.450

0.450 0.450 0.478

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

0.388 0.450 0.450

0.450 0.450 0.450

0.1820.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

0.5840.450

0.450 0.450 0.4500.310

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

0.363 0.450

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

19

20

21

22

23

24

25

18

Fig. 11 Ballast plan produced by conventional method (case 2b)

methods are shown in Figs. 11 and 12. The barge hasmore ballast tanks than Case 2a, but the proposedmethod produces a ballast plan (Fig. 12) with a verysimilar pattern to that of Case 2a (Fig. 9).

From Fig. 13 we again observe that the proposedmethod gives much better result than the conventionalmethod in terms of root-mean-square deflection andmaximum curvature. The root-mean-square deflectionfrom the proposed method is smaller than that from theconventional method for all load-out stages. The root-

mean-square deflection from the conventional methodgenerally increases as the load-out progresses, and ismuch larger than that from the proposed method forthe second half of the load-out stages. On the contrary,the root-mean-square deflection from the proposedmethod is relatively constant for all stages. In termsof maximum curvature, from Stage 11 onwards theproposed method gives the better result. The maximumcurvature obtained from the conventional method in-creases monotonically from Stage 10 onwards and at

286 A. Kurniawan, G. Ma

0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450 0.450

0.442 0.446 0.464 0.409 0.419 0.403 0.458 0.449 0.487 0.474 0.471 0.429 0.451 0.424 0.463

0.333 0.3550.473 0.386 0.408 0.419 0.501 0.446 0.466 0.439 0.522

0.416 0.439 0.447 0.447

0.278 0.3070.431 0.394 0.440 0.407 0.486 0.450 0.401 0.467 0.494 0.433 0.435 0.451 0.463

0.188 0.2500.425 0.420 0.375 0.375 0.450 0.462 0.439 0.461 0.508 0.484 0.427 0.410 0.460

0.157 0.185 0.2520.408 0.355 0.425 0.472 0.418 0.425 0.463 0.516 0.474 0.407 0.430 0.470

0.093 0.098 0.144

0.378 0.350 0.395 0.463 0.424 0.3820.508 0.533 0.456 0.469 0.432 0.420

0.1050.000

0.1070.310 0.310

0.401 0.462 0.435 0.428 0.458 0.5200.393

0.486 0.459 0.428

0.1220.000 0.003

0.179 0.2600.384 0.458 0.433 0.425 0.464

0.582

0.3190.476 0.534

0.386

0

1

2

3

4

5

6

7

8

0.1550.000 0.029 0.067

0.1940.299

0.475 0.4790.381 0.464 0.539

0.3350.474 0.528

0.405

0.1540.000 0.019 0.000

0.191 0.186

0.461 0.429 0.413 0.473 0.5160.324

0.4560.565

0.399

0.1320.013 0.000 0.000

0.143 0.1250.331

0.446 0.418 0.4140.560

0.387 0.4320.558

0.379

0.1350.031 0.000 0.000

0.114 0.086

0.312 0.362 0.395 0.3960.544

0.3650.482 0.528

0.394

11

12

9

10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0.1730.005 0.000 0.000

0.133 0.0640.249 0.292 0.343 0.374

0.5620.351

0.481 0.5180.408

0.2690.062 0.000 0.008

0.1370.000

0.209 0.1520.335 0.404 0.494

0.3640.485 0.524

0.407

0.2690.131

0.035 0.0230.121

0.0000.181

0.0700.227

0.3930.494

0.3510.489 0.507 0.421

0.2720.125 0.082 0.029 0.110

0.0130.158

0.059 0.1390.349

0.4770.317

0.472 0.5210.421

0.2370.129 0.168

0.0450.137

0.0000.137 0.050 0.013

0.238

0.5180.340 0.417

0.511 0.434

13

14

15

16

17

0.2820.117 0.196

0.0430.155

0.0350.167

0.014 0.0000.210

0.344 0.349 0.3470.546

0.448

0.3010.133

0.2350.079 0.118 0.103 0.146

0.022 0.0000.145

0.267 0.2460.377

0.526 0.447

0.2990.120

0.2340.090

0.2000.090 0.180

0.023 0.0290.147

0.0270.244

0.3330.519 0.460

0.2780.167 0.228

0.0710.225

0.1170.234

0.017 0.0200.137

0.033 0.0990.307

0.465 0.462

0.3050.150

0.2810.062

0.200 0.152 0.199 0.110 0.045 0.1210.024 0.035

0.140

0.469 0.449

0.3390.158 0.230 0.154 0.214 0.144

0.2690.118

0.0260.144

0.000 0.000 0.033

0.385 0.444

0.3330.181 0.256 0.190 0.192 0.173

0.2670.085 0.054

0.1800.026 0.015 0.037

0.2050.365

0.3450.133

0.3070.208 0.214 0.197 0.264

0.110 0.0350.230

0.011 0.016 0.0000.214 0.166

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

19

20

21

22

23

24

25

18

Fig. 12 Ballast plan produced by proposed method (case 2b)

the final stage it becomes much larger than that fromthe proposed method. In contrast, the maximum cur-vature from the proposed method is relatively constantfor all load-out stages. As before, it should be notedthat the values at Stage 1 may not be correct due tonumerical instability.

The total ballast removed or shifted for the conven-tional method is comparable to that obtained from theproposed method although in general the values fromthe proposed method are higher than those from the

conventional method. The variations of average bal-last obtained from the conventional and the proposedmethods are comparable.

6 Concluding remarks

This study deals with the calculation of ballast planin load-out operations. As a proposal to improve theconventional method, which assumes the barge to be

Optimization of ballast plan in launch jacket load-out 287

Fig. 13 Root-mean-squaredeflections, maximumcurvatures, ballast removedor shifted, and average ballastobtained from conventionalmethod (solid) and proposedmethod (dashed) for case 2b

00.0020.0040.0060.008

0.010.0120.014

0 5 10 15 20 25stage

02e − 0104e − 0106e − 0108e − 0101e − 009

1.2e − 009

0 5 10 15 20 25stage

0

0.1

0.2

0.3

0.4

0.5

0 5 10 15 20 25stage

0.150.2

0.250.3

0.350.4

0.45

0 5 10 15 20 25stage

rms

defle

ctio

n (m

)ba

llast

shi

fted

(in

)α

max

cur

vatu

re (

m–1

)av

erag

e ba

llast

(in

)α

rigid, we have developed a more realistic method ofballast calculation, which takes into account the flexi-bility of the barge. The kernel of the method is the useof a multi-objective evolutionary algorithm (MOEA)to find the optimum ballast arrangement at every load-out stage. We modelled the load-out configuration as abeam on elastic foundation and derived the expressionfor the beam deflection.

We have shown that the ballast plan calculated usingthe conventional method is prone to produce excessivedeflection and curvature, especially at later stages ofthe load-out, where the deflection and curvature wereobserved to increase monotonically. On the other hand,the proposed method, which combines three objectives,namely minimizing root-mean-square deflection, mini-mizing maximum curvature, and minimizing the totalballast removed or shifted, was shown to give betteroverall results than those obtained from the conven-tional rigid barge method.

Using MOEA, there is no need of fixing the ballastsin some of the tanks as in the conventional method.Every tank is allowed to have any ballast betweenempty and full, and the algorithm automatically seeksfor the optimum ballasts to be allocated in the tanks.

A more refined beam model used to represent theload-out configuration will surely be useful in improv-ing the accuracy of the deflection and curvature andhence the optimum ballast plan. In particular, correctmodelling of the stiffness at the bow and stern of thebarge will improve the accuracy of the method. Inthis regard, model tests will provide a useful check. Asystematic study using different MOEA parameters canalso be performed to find out which parameters give the

best results. In addition, studies similar to our presentstudy can be carried out, such as those incorporatingtide variations or initial deformations of the barge. Tidevariations can be included by varying the draft at eachload-out stage according to a given tide data. The effectof initial deformations of the barge can be includedby adding constant displacement to the displacementobtained at each load-out stage due to the given load-ing. The values of deflection and curvature which thealgorithm seeks to minimize will then be obtained fromthis total deflection.

Acknowledgements We would like to thank Dr. Gho Wie Min,who brought to our attention the problem discussed in this pa-per, and the anonymous reviewers for their helpful suggestions.The support from Maritime and Port Authority of Singapore isacknowledged.

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