optimizing architectural layout design via mixed integer...

Optimizing Architectural Layout Design via

Mixed Integer Programming

Authored byKamol Keatruangkamala and Krung Sinapiromsaran, Ph.D.

Department of Mathematic, Faculty of Science, Chulalongkorn University, Thailand

CAADFutures 2005, Vienna, AustriaJune 22, 05

Contents

IntroductionOptimization of Geometry

- Design variables and parameters- Problem objective- Layout design constraints

Experiments- Computational time- Comparison between MIP and Nonlinear programming- Practical case study

Conclusion

1

Optimizing Architectural Layout Design via Mixed Integer Programming

Introduction1. Architectural Layout Design involves both

quantifiable and qualifiable goals, preferenceand constraints.

1

2. The previous research- Nonlinear programming (Imam and Mir, M., 1989)

- Evolutionary approach (Michalek and Papalambros, 2002)

- Only guarantee thelocally optimal solution

- Consume a largeamount of time

- Only guarantee the convergenceto the locally optimal solution

- Consume a large amount of time

Nonlinear Programming Evolutionary Approach

The disadvantage of Nonlinear programming and Evolutionary approaches


Introduction1. This research proposes the novel approach with

the Mixed Integer Programming model (MIP).

2

1. Guarantee the global optimal

solution.

2. Reduce the search space

via additional reduction

constraints.

3. The weigthed sum of the

multiobjective optimization

1. Only guarantee the locally

optimal solution or convergence

of the solution

2. Deal with a complex

mathematical expressions

3. Can not exhaustively search

throughout the problem space

(Evolutionary Approaches)

This research (MIP)

2. The comparison from the previous researches.

Previous researches

Kamol Keatruangkamala and Krung Sinapiromsaran, Ph.D.

Opimization of GeometryDesign variables and Parameters

1

xi = X coordinate of the top left corner of room i

yi = Y coordinate of the top left corner of room i

wi = the horizontal width of room i

hi = the vertical height of room i


W = The width of the boundary area

H = The height of the boundary area

wmin,i , wmax,i = the minimal and maximal

width of room ihmin,i ,,hmax,i = the minimal and maximal vertical

height of room iTij = minimal contact length between

room i and j

Opimization of GeometryObjective of Problem

2

The multiobjective was represented by the weighted summation of the three objectives from architectural views for fixed room i.

1) First room positioning (ui,1)2) Minimal room distance (ui,2)3) Approximated maximal area (ui,3)

Minimize ( ui,1(xi + yi) + ui,2 Σ (absolute distance)

- ui,3 Σ (approximate area) )


Opimization of GeometryLayout Design Constraints

3

ARCHITECTURAL CONSTRAINTS

FUNCTIONALCONSTRAINTS

DIMENSIONALCONSTRAINTS

- Connectivity constraints

- Fixed position constraints*

- Unused grid cell constraints*

- Boundary constraints

- Fixed Border constraint*

- Non-intersecting constraints

- Overlapping constraints

- Length constraints

- Ratio constraints

Note : * feasible space reductionKamol Keatruangkamala and Krung Sinapiromsaran, Ph.D.

Software Development 4

a b


c d

ExperimentsComputation time

1

Types of experiments4,5,6,7,8 and 10 rooms are used for experiments.

1. Testing the speed and the connectivity patternsEach run was performed using 12 configurations of the six room sizes.

(more than 70 examples)

2. Testing the speed and the room proportionsThis research divides the room proportion in 3 casescase 1 : speed equality of room proportioncase 2 : speed inequality of room proportioncase 3 : speed different of room ratio


Experiments 2Each run composes of 12 configurations in 4 patterns.

A B C D

1 2 3

451

2

3

4

51 3 5

42

1 3 5

42

1 2 3

451

2

3

4

5

1 3 5

42

1 3 5

42

1 2 3

45

1 2 3

45

1 2 3

45

1 2 3

45

PATTERN

CASE AEquality of room proportion

CASE BInquality of room proportion

CASE CDifferent of roomratio

ratio = 1 ratio = 0.7 ratio = 0.5 ratio = 0.3


ExperimentsComputation time

3

All experiments were carried on a PC computer using Pentium 1 GHz and 256 Mb of memories.


ExperimentsComparison between MIP and MINLP

4

The GLPK solver for MIP, CPLEX solver for MIP and DICOPT solver for MINLP are compared

Time (sec.) Variables (non-zero) Objective valueMIP MIP MINLP MIP MIP MINLP MIP MIP MINLP

GLPK CPLEX DICOPT GLPK CPLEX DICOPT GLPK CPLEX DICOPT

4 rm. 1.0 1.0 1.0 594 593 517 32 32 32

5 rm. 69.3 17.4 56.2 856 855 725 80 80 126

6 rm. 430 135 957 1152 1151 935 125 125 183

RoomSized

Comparison between MIP and Nonlinear programming


ExperimentsLayout Design with Multiobjectives

1

We allow the architect to select his/her alternatives by using the weight values (ui,1, ui,2 and ui,3) in the objective function.

Show the five alternative global solutions of 5 room configurationsbased on the first room position for 1, 2, 3, 4 and 5.


Experiments 2


ExperimentsThe Practical Case Study

1

A simple practical floor plan was solved by the architect and by our software with the time limit of 1000 seconds. The problem composes of eight rooms in 20x20 sq.m.

The detailed parameters are shown in the table (scale in meter)

Room Name min width

Max width min height

max height Ratio Connect Fixed wall

1. Garage x2 5 6 6 6 0.5 2,3 South

2. Living 4 6 5 6 0.5 1,3 None

3. Hall 3 6 3 6 0.5 1,2,4,5,6,7 None

4. Ms Bedroom 5 6 5 6 0.5 3,6 None

5. Bedroom 4 5 4 5 0.5 3 None

6. Bath 2 3 2 3 0.5 3,4 None

7. Dining 5 6 5 6 0.5 3,8 None

8. Kitchen 4 6 4 6 0.5 7 None


ExperimentsThe Practical Case Study

2

A single floor plan of regular house

The comparison shows the similarity between the MIP solution and the architect’s layout.


ConclusionConclusions

1

1. The MIP model under the reduced feasible space can be solved using GLPK and CPLEX to find the global optimal solution.

2. For a larger sized-problem, the fixed position constraints, unused grid cell and fixed boundary help reducing the computational time, considerably.

3. The multiobjectives offer the optimal architectural layout design alternatives according to architect’s preferences.

Suggestions

1. For a large architectural layout design problem, the parallel MIP solver are required.

2. The machine learning methodology should be adopted to speed up the computational time.


The end

Question and AnswerThank your for your attention


Sponsor by M.S.(CAAD)Program in Computer-Aided Architecture Design

Functional Constraint 1


xi + wi > xj – W*(pij + qij) i to the left of j , pij = 0, qij = 0

yj + hj > yi – H*(1 + pij - qij) i above j, pij = 0, qij = 1

xj + wj > xi – W*(1 - pij + qij) i to the right of j, pij = 1, qij = 0

yi + hi > yj – H*(2 - pij - qij) i below j, pij = 1, qij = 1

Connectivity constraintexplains the relationship between different rooms and two binary variables (pij, qij) control the room connections.

W

H room i

room j

xi,yi

xj,yjhi

hjwi

wj

i j

ij

ij

ij

Control variablesFormulae connection

BACK

Functional Constraints 2


Unused grid cell constraintdetermines the unusable area and two binary variables (sik, tik) to identify the connection of unused grid cell, kth.

xi > xu,k + 1 – W*(sik + tik) unused space to left of i sik = 0, tik = 0

xu,k > xi + wi – W*(1+ sik - tik) unused space to right of i sik = 0, tik = 1

yi > yu,k+ 1 – H*(1-sik + tik) unused space to top of i sik = 1, tik = 0

yu,k > yi + hi – H*(2 – sik – tik) unused space to bottom of i sik = 1, tik = 1

W

H room i room j

xi,yi

xj,yjhi

hjwi

wjunused grid cell

u i

ui

iu

ui


BACK

Functional Constraints 3


xi + wi < W the width location boundary

yi + hi < H the height location boundary

Boundary constraintforces a room location to be inside a boundary.

Fixed Border constraint : addresses the absolute placement of the room. This constraint is divided into four types: north, south, east and west.

W

H room i hi

wi

Fixed Position constraintFixed the room positioning in a space.

xi = fixed x coordinate

yi = fixed y coordinate

(fixed xi, fixed yi)

yi = 0

Fixed north

iFixed east

i

yi + hi = H

Fixed south

ixi + wi = W

Fixed west

i

xi = 0

BACK

Dimensional Constraints 4


Non-intersection constraintprevents two rooms from occupying the same using and two previous binary variables pij and qij.

W

H room i room j

xi,yi

xj,yjhi

hj

wj


xi + wi < xj + W*(pij + qij) i to the left of j pij = 0, qij = 0

yj + hj < yi + H*(1 + pij - qij) i above j pij = 0, qij = 1

xj + wj < xi + W*(1 - pij + qij) i to the right of j pij = 1, qij = 0

yi + hi < yj + H*(2 - pij - qij) i below j pij = 1, qij = 1

i j

ij

ij

ij

BACK

wi



W

H

room i

room j

xi,yi

xj,yjhi

hj

wj

Formulae Control variables

i j

ij

ij

ij

Overlapping constraintTwo rooms are touching with each other with the minimal contact length defined by the value (Tij) between room i and room j.

0.5*(wi + wj) – (xj – xi) > Tij – W*(pij + qij) i to the left j pij = 0, qij = 0

0.5*(hi + hj) – (yj – yi) > Tij – H*(2– pij – qij) i to the top of j pij = 1, qij = 1

0.5*(wi + wj) – (xi – xj) > Tij – W*(1– pij + qij) i to the right of j pij = 1, qij = 0

0.5*(hi + hj) – (yi – yj) > Tij – H*(1+ pij – qij) i to the bottom of j pij = 0, qij = 1

Tij

BACK



W

room i

xi,yi

hi

wi

H

Length constraintminimal or maximal length of the bounded size of each room.

wmin <

wmin wmax

hmin

hmax

wi < wmax range of width of room i

hmin < hi < hmax range of height of room i

Ratio constraintrestricts the length between horizontal and vertical dimension. A binary variable (ri) uses to select the constraint satisfying horizontal and vertical ratio.

wi + ri*(W + H) > R*hi horizontal ratio, r = 0

hi + (1 – ri)*(W + H) > R*wi vertical ratio, r = 1

Formulae

BACK

Control variables

xi + wi >

xj – W*(pij + qij)

yj + hj >

yi – H*(1 + pij - qij)

xj + wj >

xi – W*(1 - pij + qij)

yi + hi >

yj – H*(2 - pij - qij)

i j ij ij i

j

BACK

pij= 0, qij = 0 pij= 0, qij = 1 pij= 1, qij = 0 pij= 1, qij = 1

xi + wi > xj

yj + hj > yi – H

xj + wj > xi – W

yi + hi > yj – 2H

xi + wi > xj – W

yj + hj > yi

xj + wj > xi – 2W

yi + hi > yj – H

xi + wi > xj - W

yj + hj > yi – 2H

xj + wj > xi

yi + hi > yj – H

xi + wi > xj – 2W

yj + hj > yi – H

xj + wj > xi – W

yi + hi > yj

Connectivity constraint - control variables


BACK

sik= 0, tik = 0 sik= 0, tik = 1 sik= 1, tik = 0 sik= 1, tik = 1

Unused grid cell - control variables


xi >

xu,k+ 1 – W*(sik + tik)

xu,k >

xi + wi – W*(1+ sik - tik)

yi >

yu,k+ 1 – H*(1-sik + tik)

yu,k >

yi + hi – H*(2 – sik – tik)

xi > xu,k+ 1

xu,k > xi+ wi– W

yi > yu,k+1– H

yu,k > yi+ hi– 2H

xi > xu,k+ 1 - W

xu,k > xi+ wi

yi > yu,k+1– 2H

yu,k > yi+ hi– H

xi > xu,k+ 1 - W

xu,k > xi+ wi – 2W

yi > yu,k+1

yu,k > yi+ hi– H

xi > xu,k+ 1 - 2W

xu,k > xi+ wi-W

yi > yu,k+1– H

yu,k > yi+ hi

u i uiiu

ui

BACK


xi + wi < xj

yj + hj < yi – H

xj + wj < xi – W

yi + hi < yj – 2H

xi + wi < xj – W

yj + hj < yi

xj + wj < xi – 2W

yi + hi < yj – H

xi + wi < xj - W

yj + hj < yi – 2H

xj + wj < xi

yi + hi < yj – H

xi + wi < xj – 2W

yj + hj < yi – H

xj + wj < xi – W

yi + hi < yj

Non-intersection - control variables


xi + wi <

xj + W*(pij + qij)

yj + hj <

yi + H*(1 + pij - qij)

xj + wj <

xi + W*(1 - pij + qij)

yi + hi <

yj + H*(2 - pij - qij)

i j ij ij i

j

BACK

Overlapping - control variables



0.5*(wi+wj)–(xj–xi)

> Tij–W*(pij+qij)

0.5*(hi+ hj)–(yj–yi)

> Tij–H*(2– pij – qij)

0.5*(wi+wj)–(xi – xj)

> Tij–W*(1– pij+qij)

0.5*(hi+hj)–(yi– yj)

> Tij–H*(1+pij–qij)


>Tij

0.5*(hi+hj)–(yj – yi)

>Tij – 2H

0.5*(wi+wj)–(xi–xj)

>Tij – W

0.5*(hi+hj)–(yi – yj)

>Tij – H


>Tij - 2W

0.5*(hi+hj)–(yj – yi)

>Tij


>Tij – W

0.5*(hi+hj)–(yi – yj)

>Tij – H


>Tij - W

0.5*(hi+hj)–(yj – yi)

>Tij - H


>Tij

0.5*(hi+hj)–(yi – yj)

>Tij – 2H


>Tij - W

0.5*(hi+hj)–(yj – yi)

>Tij - H


>Tij – 2W

0.5*(hi+hj)–(yi – yj)

>Tij – H

i ji

ji

jij

BACK

Ratio constraint - control variable


r = 0 r = 1

wi + ri*(W + H) > R*hi

hi + (1 – ri)*(W + H) > R*wi

wi + ri*(W + H) > 0

hi + (1 – ri)*(W + H) > 0

wi + ri*(W + H) > hi

hi + (1 – ri)*(W + H) > wi

Optimization of Geometry

Approximated areaApproximated area is calculated by

maximizing the minimal width and height

of each room.

wi

hi

h

w

approximated area1/x = max{min(hi,wi)}

area : f(x) = 1/x

area : max{ min(hi ,wi) }

hi

wi

Ai

area Ai = hiwi

hi = area Aiwi

f(x) = 1/x

::

max : zist. : zi < wi

zi < hi

Rectangular area

Optimization model


Optimization of Geometry

Absolute room distanceMinimizing the distance of room i and j

with absolute value on x and y axis.

xi

yi

y

x

(xi,yi)

min : Zx(i,j) + Zy(i,j)st. : xi - xi < Zx(i,j)

xj - xi < Zx(i,j)yi - yi < Zy(i,j)yj - yi < Zy(i,j)

Optimization model

||(xi,…xn)-(yi,…,yn)|| = Σi=1n |xi - yi|

||(xi,yi) - (xj,yj) || = |xi - yi| + |xj - yj|2 dimensions

One norm distance

n dimensions

xj

yj (xj,yj)

(xi-yi)-(xj-yj)


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