Computer Aided Ship Design Part I. Optimization Method

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Computer Aided Ship Design, I-1. Overview of Optimal Design, Fall 2013, Myung-Il Roh 1

N a v a l A r c h it e c tu r e & O c e a n E n g in e e r in g

Computer Aided Ship Design, I-1. Overview of Optimal Design, Fall 2013, Myung-Il Roh

Computer Aided Ship Design Part I. Optimization Method

Term Project

September, 2013 Prof. Myung-Il Roh

Department of Naval Architecture and Ocean Engineering, Seoul National University of College of Engineering

Computer Aided Ship Design Lecture Note

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N a v a l A r c h it e c tu r e & O c e a n E n g in e e r in g

Computer Aided Ship Design, I-1. Overview of Optimal Design, Fall 2013, Myung-Il Roh

Term Project

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Overview (1/2)

þ Objective

n To develop a program for the determination of optimal principal dimensions of a ship by using a constrained optimization method

þ Optimization Method

n Exterior Penalty Function Method

n Hooke & Jeeves Method or Nelder & Mead Simplex Method for minimizing the above penalty function

Æ Select one of them!

{ }

1

( , ) ( ) max ( ), 0

m

k k j

j

r f r g

=

F x = x + å x { } 2

1

( , ) ( ) max ( ), 0

m

k k j

j

r f r g

=

é ù

F x = x + å ë x û

or

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Overview (2/2)

þ Implementation

n Any program language (C++[Recommended], FORTRAN) or tool (Matlab, MS Excel) can be used.

n However, the grading is different according to the language or tool what you select.

n Evaluate the validity of your program by applying it to all test examples and discuss its results in your report.

n You can refer materials on the internet, but do not copy!

þ Due date: 23:59 on 24 th November, 2013

þ Submissions

n Report for the term project (MS word file) n Source files including an executable file.

n After compressing all files in one file (e.g., YourStudentNumber.zip) and

upload to our eTL homepage.

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Test Examples #1

1 2 3 4

x 1 x 2

A

B C

g 2 = 0

g 3 = 0

g 1 = x 1 2 + x 2 2 - 6.0 = 0 )

3 , 3

* (

= x

x (0) = (1, 1)

f = -25 f = -20

f = -10 f = -3

0 )

(

0 )

(

0 0

. 6 1

1 6

) 1 (

1 3

1 2

2 2 2

1 1

£ -

=

£ -

=

£ -

+

=

x g

x x

g

x x x Minimize

Subject to

2 1 2

2 2

1 3

)

( x x x x

f x = + -

Optimal Solution:

3 )

( ), 3 , 3

( *

* = x = -

x f

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2 2

1 2 1 2

( , ) 25 ( 5) ( 5)

f x x = - é ë - x - - x - ù û Subject to

10 )

, (

0 )

, (

10 )

, (

0 )

, (

0 4

32 )

, (

2 2

1 5

2 2

1 4

1 2

1 3

1 2

2 2 1

2 1 1

£

=

£ -

=

£

=

£ -

=

£ +

+ -

=

x x

x g

x x

x g

x x

x g

x x

x g

x x

x x g

Solution

815 .

4 )

, ( , 808 .

3 ,

374 .

4 2 * 1 * 2 *

*

1 = x = f x x = -

x Minimize

Test Examples #2

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-2

-1

0

1

2 ^-2

-1 0

1 2

0 50000 100000 150000 200000

-2

-1

0

1

2

x

₁

f(x

₁

, x

₂

)

x

₂

)}

27 36

48 12

32 18 ( ) 3 2 ( 30 {

)}

3 6

14 3

14 19 ( ) 1 (

1 { ) , (

2 2 2

1 2

2 1 1

2 2 1

2 2 2 1 2 2

1 1 2

2 1 2

1

x x

x x x x x

x x

x f

+ -

+ +

-

× -

+

×

+ +

- +

-

× + + +

=

0 2 )

, ( , 0 2 )

, (

, 0 2

) , ( , 0 2

) , (

2 2

1 4 1

2 1 3

2 2

1 2 1

2 1 1

£ -

=

£ -

=

£ - -

=

£ - -

=

x x

x g x

x x g

x x

x g x

x x g

Subject to Minimize

Goldstein-Price Function

A : Global Minimum

B : Local Minimum

C : Local Minimum

D : Local Minimum

x

₁^*

= 0.0, x

₂^*

= -1.0, f(x

₁^*

, x

₂^*

) = 3.0

x

₁^*

= -0.6, x

₂^*

= -0.4, f(x

₁^*

, x

₂^*

) = 30.0

x

₁^*

= 1.2, x

₂^*

= 0.8, f(x

₁^*

, x

₂^*

) = 840.0

x

₁^*

= 1.8, x

₂^*

= 0.2, f(x

₁^*

, x

₂^*

) = 84.0

-2 -1 0 1 2

x

₁

x

₂

A B

C

D

Test Examples #3

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Test Examples #4

- Determination of the Optimal Principal Dimensions of a Ship (1/4)

§ Find: L, B, C B

l Hydrostatic equilibrium(Weight equation)

1.6 2/3 3

( , , , )

( ) ( )

s B sw given B

given s o power d B

L B T C C DWT LWT L B D C

DWT C L B D C L B C L B T C V r a

× × × × × = +

= + × × + + × × + × × × × ×

Æ Indeterminate Equation: 3 variables(L, B, C B ), 2 equality constraints((a), (b))

l Recommended range of obesity coefficient considering maneuverability of a ship

( )

.

...

H req H

V = C × × × L B D b

l Required cargo hold capacity(Volume equation)

( ) 0.15 ... ( )

/ C B

L B < c

( )

... a

§ Given: DWT, V H.req , D, T s , T d

2.0

( )

C s ¢ L B D

® × × + ® C power ¢ × (2 × × B T d + × × 2 L T d + × L B V ) ×

³

is (Volume)

^2/3

and means the submerged area of the ship.

So, we assume that the submerged area of the ship is equal to the submerged area of the rectangular box.

( L B T C × ×

_d

×

_B

)

2/3

B T

L D

Simplify ① Simplify ②

It can be formulated as an optimization problem to minimize an objective function.

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2.0 3

( , , B ) PS s ( ) PO o PM power (2 d 2 d )

f L B C = C × C ¢ × L × B + D + C × C × × L B C + × C ¢ × × × B T + × × L T + × L B V ×

§ Minimize: Building Cost

§ Subject to

l Hydrostatic equilibrium(Simplified weight equation)

2.0 3

( , , , )

( ) (2 2 )

s B sw given B

given s o power d d

L B T C C DWT LWT L B D C

DWT C L B D C L B C B T L T L B V r a

× × × × × = +

¢ ¢

= + × × + + × × + × × × + × × + × ×

.

... ( )

H req H

V = C × × L B D × b

( ) 0.15 ... ( )

/ C B

L B < c

( )

... d

( )

... a '

§ Find: L, B, C B

§ Given: DWT, V H.req , D, T s , T d

Test Examples #4

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Test Examples #4

Item Basis ship(150,000ton Bulk Carrier) Design ship(160,000ton Bulk Carrier) Notes

Principal Dimensions

L

_OA

abt. 274.00 m max. 284.00 m

L

_BP

264.00 m ?

B

_mld

45.00 m ?

D

_mld

23.20 m 23.20m

T

_mld

16.90 m 17.20 m

T

_scant

16.90 m 17.20 m

C

_B

0.8214 ?

Deadweight 150,960 ton 160,000 ton at 17.20 m

Speed 13.5 kts 13.5 kts 90 % MCR (with 20 % SM)

M / E

TYPE B&W 5S70MC

NMCR 17,450 HP×88.0 RPM Derating Ratio

= 0.9

DMCR 15,450 HP×77.9 RPM E.M = 0.9

NCR 13,910 HP×75.2 RPM

F O C

SFOC 126.0 g/HP․H

Standardize to NCR

TON/DAY 41.6

Cruising Range 28,000 N/M 26,000 N/M

Midship Section

Single Hull Double Bottom/Hopper

/Top Side Wing Tank

Single Hull Double Bottom/Hopper

/Top Side Wing Tank

Capacity

Cargo abt. 169,380 m

³

abt. 179,000 m

³

Including Hatch Coaming

Fuel Oil abt. 3,960 m

³

Total

Fuel Oil abt. 3,850 m

³

Bunker Tank Only

Given data for basis ship and owner’s requirements for design ship

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Test Examples #4

Given data for basis ship(150,000 ton Bulk Carrier)

Item Value

Lightweight 18,269 ton

Hull structural weight (Ws) 15,289 ton

Outfit weight (Wo) 1,694 ton

Machinery weight (Wm) 1,281 ton

C

_s’

, C

_o

, C

_power

Calculate!

Coefficient for hull structural cost (C

_PS

) 972.80 Coefficient for outfitting cost (C

_PO

) 20,256 Coefficient for machinery cost (C

_PM

) 7,760

Ca Calculate!

r

_sw

1.025 ton/m

³

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[Reference] Transformation of an Equality Constraint into Two Inequality Constraints

þ For convenience, one equality constraint into two inequality constraints, as follows.

( ) 0

h x = 0 £ h ( ) x £ 0

1 ( ) ( ) 0

g x = - - e h x £

( )

e h e

- £ x £

Æ Æ

Æ

2 ( ) ( ) 0

g x = h x - £ e

e : positive small value