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Truss Design and Convex Optimization
RobertM.Freund
April 13, 2004
c2004MassachusettsInstituteof Technology.
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1 Outline
• PhysicalLawsof aTrussSystem• TheTrussDesignProblem• Second-OrderConeOptimization• TrussDesignandSecond-OrderConeOptimization• SomeComputationalResults• Semi-DefiniteOptimization
• TrussDesignandSemi-DefiniteOptimization• TrussDesignandLinearOptimization
• Extensionsof theTrussDesignProblem
2 Physical Laws of a Truss System
Atruss isastructureind= 2 or d=3dimensions,formedbynnodesandmbars joiningthesenodes. Figure1showsanexampleof atruss.
2
3
4
5
6
Figure 1: Example of a truss in d = 2 dimensions, with n = 6 nodes andm= 13 bars.
Examplesof trusses includebridges,cranes,andtheEiffelTower.
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Thedatausedtodescribeatruss is:
• asetof nodes(given inphysicalspace)• asetof bars joiningpairsof nodeswithassociateddataforeachbar:
– the lengthLk of bark
– theYoung’smodulusE k of bark
– thevolumetk of bark
• anexternalforcevectorF onthenodes
Nodes
can
be
static
(fixed
in
place)
or
free
(movable
when
the
truss
is stressed). The allowable movements of the nodes defines the degrees of freedom(dof)of thetruss. WesaythatthetrusshasN degreesof freedom.Of course,N ≤ nd.
Movements of nodes in the truss will be represented by a vector u of N displacements ,whereu∈ .
N Theexternalforceonthetruss isgivenbyavectorF ∈ .Displacementsof thenodesinthetrusscauseinternalforcesof compres-
sionand/orexpansiontoappearinthebarsinthetruss. Letf k denotetheinternalforceof bark. Thevectorf
∈ m isthevectorof forcesof thebars.
Figure
2
shows
an
example
of
a
truss
problem.
In
the
figure,
there
is
a
singleexternal forceappliedtonode3 inthedirection indicated. Thiswillresult in internal forcesalongthebars inthetrussandwillsimultaneouslycausesmalldisplacements inallof thenodes.
In Figure 2, nodes 1 and 5 are fixed, and the other nodes are free. Tosimplify notation we will label the 13 different bars by the nodes that heylink to: in generalthebark will joinnodes iand j. Thebarswillbe: 12,13, . . .,56.
The internalforcef k of barkcanbepositiveornegative:
•
If
f k ≥ 0,bark hasbeenexpandedand its internal forcecounteractstheexpansionwithcompression,seeFigure3.
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2
3
4
5
6
1
f
f
F13
12
3
Figure2: Atrussproblem.
∆ > 0 L k
Figure
3:
A
bar
under
expansion.
L k
∆ < 0
Figure4: Abarundercompression.
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• If f
k
≤0,barkhasbeencompressedanditsinternalforcecounteracts
the
compression
with
expansion,
see
Figure
4.
2.1 Physical Laws
2.1.1 Force Balance Equations
Inastatictrusstheinternalforceswillbalancetheexternalforcesineverydegreeof freedom. That is,theforcesaroundeach freenodemustbalance.This is a law of conservation of forces. (This is akin to material balanceequations innetworks,forexample.)
For
example,
consider
the
balance
of
forces
on
node
3:
xcoordinate: −f 13 −f 23 cos(π/4) +f 35 +f 36 cos(π/4) =−F 3xy coordinate: +f 23 sin(π/4) +f 34 +f 36 sin(π/4) =−F 3y
FortheentiretrusswewriteN linearequationsthatrepresentthebal-anceof forcesineachdegreeof freedom.
Inmatrixnotationthiscanbewrittenas
Af =−F
whereA isanN
×mmatrix,andwhereeachcolumnof A,denotedasak,
is
the
projection
of
the
bar
onto
the
degrees
of
freedom
of
the
nodes
that
barkmeets.
Inourexample,wehave:
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12 13 14 16 23 24 25 34 35 36 45 46 56
1√ 1 2√
0 0 0 0
0 0 0 0 0 0 2x52
−1 0 0 0 − 1√ 0 − 1√ 52
0 0 0 0 0 0
2y
0 −1 0 0 − 1√ 0 0 0 1 1√
22
0 0 0 3x 1√ 0 0 1 0 1√ 0 0 0 0 0 0 0 3yA=
22
1√ 0 0 −1 0 0 0 0 1√ 0 0 1 0 4x − 22
1√ 0 0 0 0 −1 0 0 − 1√ 0 0 0 0 4y− 22
2√ 0 0 0 0 0 − 1√ 0 0 0
6x0 −1 0−25
1√ 0
0
0
0 0
−1
√ 0 0
0
−
0 0
−1 6y25
2.1.2 Constitutive Relationship between Forces and Distortion
of a Bar
If bar k has length Lk and cross-sectional area Ack, Young’s modulus E k,
and its length ischangedby∆k,thenthe internalforcef k isgivenby:
∆kf k =E kA
c .kLk
However,
for
our
purposes
it
will
be
easier
to
work
instead
with
the
volume tk of bark,which is:
kLk .tk =Ac
Wethereforecanwritetheconstitutiverelationshipas:
E kf k =
L2tk∆k .
k
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2
2.1.3 Distortion and Displacements
Displacements in the nodes will cause the lengths of the bars to change.SupposethatL12 =L13 =L35 = 1.0,withallotherbarsmeasuredpropor-tionally. InFigure5,weshowadisplacementof:
u= (−,−,0,0,0,0,0,0) .
F3
2
3
4
5
6
1
Figure5: Exampleof asmalldisplacement inanode.
Table
1
shows
the
new
lengths
of
the
bars
under
the
displacement
u
=
(−,−,0,0,0,0,0,0). Thethirdcolumnof thetableshowsthelinearization(first-order Taylor approximation) of the lengths of the bars, and the lastcolumnshowsthecomputationof theinnerproduct:
(ak)T u .
Notethatif issmall,thenthedistortionof barkisnicelyapproximatedbythe linearexpression:
∆k =−(ak)T u .
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Bark OldLength NewLength LinearizedLength LinearizedChange∆k
(ak
)T u
12 1
1 + 2( −1) 1− − 13 1 1 1 0 014 1√ 1√ 1√ 0 016
23
5√ 2
5√ 2 + 22
5√ 2
0
0
0
0 24
25
1√ 5
1 + 2( + 1)
5 + 2( + 1)
1 + √ 5 + 1√
5
1√
5
−1−√ 5
34 1 1 1 0 035 1√ 1√ 1√ 0 036 2
√
2
√
2
√
0 0
45
2
2
2
0
0
46 1 1 1 0 056 1 1 1 0 0
Table1: Changes inthe lengthsof thebarsundernodaldisplacement
Wethereforewritetheinternalforceonbark duetoafeasibledisplace-mentu as:
E kf k =− 2 tk(ak)T uLk
LetthematrixB bethefollowingdiagonalmatrix:
L21
E 1t10
E 1t1
L21
0
B = ... , B−1 = ... .
0 L2m
E mtm0 E mtm
L2m
Thenwecanwritetheaboverelationshipas:
f =−B−1AT u ,
which
is:
Bf +AT u = 0 .
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2.1.4 Equilibrium Conditions and Compliance
Nowwecanwritethephysical lawsof thetrusssystemas:
Af = −F (conservationof forces)
Bf +AT u = 0 (forces,distortions,anddisplacements)
The compliance of the truss is the work (or energy) performed by thetruss,which isthesumof theforcestimesdisplacements:
1F T u .
2
“1
nience.
Notice that the larger the compliance, the more work or displacementthatthetrussundergoesunderthe loadF . Ideally,wewould likethecom-pliancetobeasmallnumber.
Combiningtheequilibriumconditions,wewrite:
We really do not need the fraction2
”, but we will keep it for conve-
F
=
−Af
= AB−1AT u
= Ku
wherethematrixK isdefinedtobe:
K =AB−1AT .
ThematrixK iscalledthestiffness matrix of thetruss. NotethatK isanSPSDmatrix.
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Usingthis,wecanwritethecomplianceas:
1 1F T u = uT Ku .
2 2
Notefromthisthatthecompliancewillalwaysbeanonnegativequantity,asintuitionsuggests. Moregenerally,of course,wewouldexpectthestiffnessmatrix K to be SPD, and so the compliance will generally be a positivequantity.
2.1.5 Solving the Equations and Computing the Compliance
Wesolvetheequationsystem:
Ku =F
toobtainthenodaldisplacementsu,andthenobtaintheforcesonthebarsbycomputing:
f =−B−1AT u . Foreachbark,this lastexpression issimply:
E kf k =
−
2 tk(ak)T
u .
Lk
Thecompliance isthencomputedas:
1 1F T u = uT Ku .
2 2
2.2
Viewing the Equilibrium Conditions as the Solution to
an Optimization Problem
We can also view the truss equilibrium conditions as the solution to anoptimizationproblem. Considerthefollowingoptimizationproblem:
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m1 1 L2
OP: minimizef ̃
k f ̃22 tk E k
k
k=1
m
s.t. akf ̃k =−F . k=1
ProblemOPseeksaforcevectorf ̃ thatsatisfiestheforcebalanceequa-tions:
m
−F = akf ̃k =Af̃ , k=1
andthatminimizestheweightedsumof squaresof theforces:
m 1 L21 k f ̃22 tk E k
k .k=1
ProblemOP isaconvexquadraticproblem. TheKKToptimalitycon-ditionsforthisproblemare:
m
akf k =−F k=1
L2 T k f k +aku= 0 for k= 1, . . . , m . tkE k
Noticethatwecanrewritetheseoptimalityconditionsas:
Af =−F
Bf +AT u= 0 ,
which
are
precisely
the
equilibrium
conditions
for
the
truss
system.
Thisshowsthatthetrusssystemisnature’ssolutiontoanoptimizationproblem.
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Theoptimalobjectivefunctionvalueof theoptimizationproblemOPis:
m1 1 L2 1 1 1 1k f 2 = f T Bf =− f T AT u=− uT Af = F T u ,2 tk E k
k 2 2 2 2k=1
which is the compliance of the truss. Therefore we see that the optimalobjectivevalueof OP isthecomplianceof thetrusssystem.
3 The Truss Design Problem
Let us now consider the volumes tk of the bars k to be design parametersthatwewishtodetermine. Wecanwritethesystemof equations:
F =Ku
as:
F = Ku
= AB−1AT u
m
B−1AT = (ak) uk
k=1
m
= −(ak)f kk=1
m E k
= tkL2
(ak)(ak)T u
k=1 k
= K (t)u
wherethestiffnessmatrix isnowwrittenas:
m E k
K (t) = tkL2
(ak)(ak)T .
k=1 k
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NoteinthisexpressionthatK =K (t)isaweightedsumof rank-onematri-ces
(weighted
by
the
volumes
tk). Putanotherway,K =K (t) isaweightedsumof outer-productmatrices.
Indesigningthetruss,wemustchoosethevolumestk of thebarssubjectto linearconstraintsonthevolumesof thebars:
M t ≤ d
t ≥ 0 ,
wheretypicallytheseconstraintsincludeupperandlowerboundsonthe
volumes
of
certain
bars,
as
well
as
an
overall
cost
or
volume
constraint:
m
tk ≤ V . k=1
The criterion intruss design is to choose the volumes tk of the bars soastominimizethecomplianceof thetruss,namely:
1minimizet,u F
T u .2
Such
a
truss
will
be
the
most
resistant
to
the
external
force
F .
Thesingle-loadtrussdesignproblemcanbestatedas:
1(TDP
1): minimizet,u F
T u2
s.t. K (t)u=F
M t ≤ d
t≥ 0mu∈ N , t ∈ .
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whichisthesameas:
1(TDP
1): minimizet,u F
T u2
m
s.t. tkE k (ak)(ak)
T u=F L2
k=1 k
M t ≤ d
t≥ 0m
u∈
N
, t∈ .
Thedecisionvariableshereareuandt. However,oneshouldreallythinkof thedecisionvariablesastonly,sinceoncetischosen,uwillbedeterminedbythesolutiontothesystemof equations:
m
tkE k
(ak)(ak)T u=F .
L2k=1 k
Note that as written, the truss design problem TDP is not a convex
problem.
4 A Convex Version of the Truss Design Problem
Recallthatforgivenvolumestk onthebarsof thetruss,thatthecomplianceof thetruss istheoptimalobjectivevalueof thequadraticproblem:
m1 1 L2
OP: minimizef ̃
k f ̃22 tk E k
k
k=1
m
s.t.
akf ̃k =
−F .
k=1
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The truss design problem is then to choose the values of the volumest
= (t1, . . . , tm) on the bars so that the optimal solution of OP (which wehave shown is the compliance of the truss) is minimized, subject to theconstraintsont:
M t ≤ d
t ≥ 0 .
Wewritethisallas:
1 1 L2kf 2
(TDP2)
:
minimizef,t 2
tk E k ktk>0
s.t. akf k =−F tk>0
Mt ≤ d
t≥ 0mf ∈ m, t ∈ .
Notice here that we have made two modifications to the optimizationproblem. First,wehavemodifiedthesummation intheobjectivefunction,sothatbarswithzerovolume(tk =0)arenolongercounted,sincetheywillnotexist. Second,wechangedthenotation,usingf insteadof f ̃. Becausethe summations with the tk > 0 is rather awkward mathematically, wefurtherre-writeTDP2 asfollows:
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m1(TDP
2): minimizef,t,s
2sk
k=1
m
s.t. akf k =−F k=1
M t ≤ d
L2k f 2 ≤ tksk fork= 1, . . . mE k k
t≥ 0, s ≥ 0mf ,t ,s∈ .
Inthisproblem,wehavereplacedthequantity
1 L2 k f 2tk E k
k
withthenewvariablesk subjecttothecondition:
L2k f 2E k
k ≤ tksk .
Wecanfurtherre-writethisnowas:
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m1(TDP
2): minimizef,t,s
2sk
k=1
s.t. Af =−F
M t ≤ d
L2k f 2 ≤ tksk fork= 1, . . . mE k k
t
≥0, s
≥0
mf ,t ,s∈ .
Thislastproblemhasalinearobjectivefunction,andlinearconstraintsexceptfortheconstraints:
L2k f 2E k
k ≤ tksk .However, it isprettyeasytoshowthattheconstraints:
L2k f 2 ≤ tksk, tk ≥ 0, sk ≥ 0E k
k
describe
a
convex
region,
and
so
this
formulation
is
actually
a
convex
opti-mizationproblem.
5 Second-Order Cone Optimization
Asecond-orderconeoptimizationproblem(SOCP)isanoptimizationprob-lemof theform:
T SOCP: minx c x
s.t.
Ax
=
b
Qix +di ≤ giT x +hi , i = 1, . . . , k .
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Inthisproblem,thenormvisthestandardEuclideannorm:√
v := vT v .
The norm constraints in SOCP are called “second-order cone” con-straints.
NoticefirstthatSOCP isaconvexproblem,sincethefunction:
Qix +di − giT
x +
hi
isaconvexfunction.
Noticealsothat linearoptimization isaspecialcaseof SOCP, if weset
Qi = 0, di = 0, hi = 0, and gi tobetheith unitvectorfori= 1, . . . , n.
Alsonoticethatanyconvexquadraticconstraintcanbeconverted intoasecond-orderconstraint. Toseethis,supposewehaveaconstraint:
1xT Qx +q T x +r≤ 0 ,
2
whereQisSPSD.Wecanfactor
Q=M T M
forsomematrixM ,andthenwriteourconstraintas:
1√ 2
M x , q tx +r + 1
2
≤
−q tx − r+ 1 2
.
If yousquarebothsidesandcollectterms,youwillseethatthisisindeedequivalenttotheoriginalconvexquadraticconstraint.
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Finally,notethatwecanalways formulateaconvexquadraticproblemas
an
SOCP,
since
we
can
create
a
new
variable
xn+1 andwriteourquadraticproblemas:
minx,xn+1 xn+1
s.t....
T 1xT Qx +q x≤ xn+1 ,2
which
can
be
further
re-written
as
an
SOCP.
6 Truss Design and Second-Order Cone Optimiza
tion
Wenowpresentasecond-orderconeoptimizationproblemthatisequivalentto the truss design problem TDP2. Recall this version of the truss designproblem:
m1(TDP
2): minimizef,t,s
2sk
k=1
s.t. Af =−F
M t ≤ d
L2k f 2 ≤ tksk fork= 1, . . . mE k
k
t≥ 0, s ≥ 0mf ,t ,s∈ .
Letusnowmakethesimplechangeof variables:
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1wk = 2 tk +1sk2
yk = −1 tk +1sk2 2
Thensk =wk +yk andtk =wk− yk, and so:
2 2tksk = (wk− yk)(wk +yk) = wk− yk,
andsotheconstraintL2k
f 2
k ≤ tkskE k
becomes:
L2kf 2 2E k
k ≤ wk− y2 .kWecanrearrangethisandtakesquarerootstoobtain:
L2k f 2 +y2 ≤ wk ,E k
k k
whichwecan inturnwriteas:
Lk
yk,√ f k ≤ wk .
E k
Thereforewecanre-writeTDP2 as:
1(CTDP): minimizef,w,y e
T (w +y)2
s.t. Af =−F
yk,√ Lk f k
≤ wk , k = 1, . . . , mE k
M (w
− y)
≤ d
mw ,y ,f ∈ .
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NoticethatCTDP isasecond-orderconeproblem. Theobjectivefunc-tion and the first and third set of constraints are linear equations and in-equalities. Thesecondsetof constraintsaresecond-order-coneconstraints,sincetheLHS isthenormof a2-dimensionalvector:
(v1, v2):=
yk ,
Lk√ E k
f k
,
andtheRHSisthe linearexpression:
wk.
Proposition 6.1 Suppose that (f ,w ,y) is a feasible or optimal solution of CTDP.
Let:
t=w −y and s=w+y.Then
(f ,t ,s) is the corresponding feasible or optimal solution of TDP 2.
6.1
Other Formulations and Formats for the Truss Design
Problem
We
have
just
seen
how
to
formulate
the
truss
design
problem
as
the
convex
optimization problem TDP2 and more specially as the second-order coneoptimization problem CTDP. In addition to these two formulations of theproblem,thereisawaytoformulatethetrussdesignproblemasaninstanceof amoregeneraltypeof convexproblemcalleda“semi-definiteoptimizationproblem” (SDO for short). The topic of SDO and its application to trussdesign isdiscussedhereininSections8and9.
Finally,whentheonlyconstraintsonthevolumevariablest1, . . . , tm arenonnegativityconditionsandatotal-volumeconstraintof theform:
m
tk ≤V , k=1
then the truss design problem can actually be converted to a linear opti-mizationproblem. This isshownherein inSection10.
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7 Some Computational Results
Inthissectionwereportsomecomputationalresultsonsolvingtrussdesignproblems. Wesolvedthreedifferenttypesof trussdesignproblems. Thefirsttype is a basic bridge design, and is illustrated in Figure 6. In the figure,thearrowscollectivelycomprisetheforcevectorF appliedtothestructure,andthecircles(intheleftlowerandrightlowercorners)arethefixednodesof the structure. The stars in the figures are the other nodes in the trussstructure. Figure7showsthesetof barsthatcanbeusedtoconstructthebridge. These bars were generated by considering bars between any twonodeswhose“distance”fromoneanotherwasthreenodesor less.
7
6
5
4
3
2
1
0
0 2 4 6 8 10 12 14 16
Figure6: Theforcesandnodesforthebasicbridgedesignmodel.
Thesecondtypeof problemthatwesolvedistheproblemof designingahangingsupportforaloadsuchasatrafficsign,andisillustratedinFigure
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0 2 4 6 8 10 12 14 16
0
1
2
3
4
5
6
7
Figure 7: The set of possible bars for the basic bridge design problem.
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8. The circles on the left part of the figure correspond to the fixed nodes, andthe arrow on the right side is a force that will be applied (the gravitationalforce of the traffic sign).
0 5 10 15 20 25 30
0
2
4
6
8
10
12
14
16
18
20
Figure 8: The forces and nodes for the hanging sign design model.
The third type of problem that we solved is the problem of designinga crane to handle a gravitational load, and is illustrated in Figure 9. Thecircles on the bottom of the figure correspond to the fixed nodes, and thearrow on the right side is a force that will be applied (the gravitational forceof the traffic sign).
For the problems shown in Figures 6, 8, and 9, the truss design modelsolutions are shown in Figures 10, 11, and 12. In these figures, the thickness
of the image of the bars is drawn proportional to the volumes of the bars inthe optimal solutions.
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40
35
30
25
20
15
10
5
0
0 2 4 6 8 10 12 14 16 18 20
Figure9: Theforcesandnodesforthecranedesignmodel.
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7
6
5
4
3
2
1
0
0 2 4 6 8 10 12 14 16
Figure10: Optimalsolutiontothebasicbridgedesignproblem.
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20
18
16
14
12
10
8
6
4
2
0
0 5 10 15 20 25 30
Figure11: Optimalsolutiontothehangingsigndesignproblem.
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40
35
30
25
20
15
10
5
0
0 2 4 6 8 10 12 14 16 18 20
Figure12: Optimalsolutiontothecranedesignproblem.
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Notice inFigure10thatthebasicbridgedesigndoesnotautomaticallyinclude
horizontal
bars
for
a
“road
surface”
on
the
base
of
the
structure.
In
order to force there to be such a road surface, we must add lower boundsonthevolumesof thosebarsonthebaseof structure. If weaddsuchlowerbounds,theoptimalbridgedesign isasshown inFigure13.
7
6
5
4
3
2
1
0
0 2 4 6 8 10 12 14 16
Figure13: Optimalsolutionof thebridgedesignproblem,withlowerboundsonthe“roadsurface”barvolumes.
7.1 Details of Problems Solved
Forthebridgedesignproblem,we included lower-boundconstraintsonthevolumesof theroad-surfacebarsaswellasaconstraintonthetotalvolume
of
the
structure.
With
the
lower-bound
constraints
in
the
model,
the
model
mustbesolvedeitherasasecond-orderconeproblem(SOCP)orasasemi-definiteoptimizationproblem(SDO).Forthehangingsigndesignproblem
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Name Sizeof grid Nodes Arcs Degreesof freedom MaximumVolumen
m
N
Bridge-16×7 16×7 136 1,547 268 1,600Bridge-20×10 20×10 231 2,878 458 2,000Bridge-30×10 30×10 341 4,368 678 4,000Sign-10×20 10×20 231 2,878 444 1,000Sign-20×30 20×30 651 9,058 1,280 2,000Sign-30×40 30×40 1,271 18,438 2,512 3,000Crane-10×20 10×20 96 818 188 2,000Crane-20×40 20×40 291 3,188 578 5,000Crane-30×40 30×40 401 4,678 798 10,000
Table
2:
Dimensions
of
the
truss
design
problems.
aswellasforthecranedesignproblem,theonlyconstraintonthevolumeswas a constraint on the total volume of the truss structure. This made itpossibletosolvetheseproblemsusing linearoptimization.
We solved three different instances of each of the bridge, hanging sign,and cranedesignproblems, with differentnumbers of nodes correspondingtodifferentmeshes. Table2showsdetailsof themodelsthatwesolved. Theright-mostcolumninthetableshowsthevalueof thevolumeconstraintonthevolumeof thetrussstructure.
Table3showsthenumberof variablesandthenumberof inequalitiesintheseproblems,andTable4showsthenumberof variablesandthenumberof equations/constraints/matrixdimensionof theseproblems.
Table 5 shows the compliance of the optimal truss design for the nineproblemsthatweresolved.
Table 6 shows the iterations of the interior-point algorithms that wereused to solve the truss design problems. For the linear problems and thesecond-order cone problems, we used LOQO to solve the model. For the
semi-definite
optimization
models,
we
used
an
SDO
algorithm
called
SDPa.Therunningtimesof thesemethods isshown inTable7.
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ProblemLP
Model
SOCP
Model
Variables Inequalities Variables Inequalities2m 2m 3m m+ 1 + LBs
Bridge-16×7 † † 4,641 1,564Bridge-20×10 † † 8,634 2,899Bridge-30×10 † † 13,104 4,399Sign-10×20 5,756 5,756 8,634 2,879Sign-20×30 18,116 18,116 27,174 9,059Sign-30×40 36,876 36,876 55,314 18,439Crane-10×20 1,636 1,636 2,454 819Crane-20
×40 6,376 6,376 9,564 3,189
Crane-30×40
9,356
9,356
14,034
4,679
Table3: Numberof variablesand inequalitiesinthetrussdesignproblems.†The linearoptimizationmodelcannotbeused forthemodifiedbridgede-signmodel.
ProblemLPModel SOCPModel SDOModel
Variables Equations Variables Constraints MatrixDimension2m N 3m N +m+1+LBs N +m+ 2
Bridge-16×7
†
†
4,641
1,832
1,685Bridge-20×10 † † 8,634 3,357 3,111
Bridge-30×10 † † 13,104 5,077 4,711Sign-10×20 5,756 444 8,634 3,323 3,111Sign-20×30 18,116 1,280 27,174 10,339 9,711Sign-30×40 36,876 2,512 55,314 20,951 19,711Crane-10×20 1,636 188 2,454 1,007 916Crane-20×40 6,376 578 9,564 3,767 3,481Crane-30×40 9,356 798 14,034 5,477 5,081
Table 4: Number of variables and equations in the truss design problems.
†The linearoptimizationmodelcannotbeused forthemodifiedbridgede-
sign
model.
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Problem LP SOCP SDP
Bridge-16×7 † 27.43365 27.15510Bridge-20×10 † 52.58314 52.10850Bridge-30×10 † 132.46038 ‡Sign-10×20 0.77279 0.77293 0.77279Sign-20×30 2.52003 2.52056 ‡Sign-30×40 4.08573 4.08681 ‡Crane-10×20 28.67827 28.67840 28.67753Crane-20×40 286.24854 286.24954 286.19340Crane-30×40 596.73177 596.73324 596.63740
Table
5:
Compliance
of
the
optimized
truss
design
for
the
truss
design
prob-lems. †Thelinearoptimizationmodelcannotbeusedforthemodifiedbridgedesignmodel. ‡ThedatatoruntheSDOmodelcouldnotbeprepared forthisinstanceduetomemoryrestrictions.
Problem LP SOCP SDO
Bridge-16×7 † 38 36Bridge-20
×10
† 55 43
Bridge-30×10
†
47
‡
Sign-10×20 18 61 37Sign-20×30 21 47 ‡Sign-30×40 24 53 ‡Crane-10×20 17 52 34Crane-20×40 31 158 58Crane-30×40 31 144 60
Table 6: Number of iterations of the interior-point algorithm to solve thetruss design problems. †The linear optimization model cannot be used forthemodifiedbridgedesignmodel. ‡ThedatatoruntheSDOmodelcouldnot
be
prepared
for
this
instance
due
to
memory
restrictions.
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Problem LP SOCP SDO
Bridge-16×7 † 93.90 257.01Bridge-20×10 † 460.96 1,088.09Bridge-30×10 † 904.86 ‡Sign-10×20 3.33 513.37 1,081.54Sign-20×30 32.08 4,254.08 ‡Sign-30×40 111.03 26,552.74 ‡Crane-10×20 0.52 41.12 446.98Crane-20×40 6.46 1,299.12 33,926.20Crane-30×40 10.67 1,599.85 95,131.49
Table7: Runningtime(inseconds)of the interior-pointalgorithmtosolve
the
truss
design
problems.
†The
linear
optimization
model
cannot
be
used
for the modified bridge design model. ‡The data to run the SDO modelcouldnotbepreparedforthis instanceduetomemoryrestrictions.
8 Semi-Definite Optimization
If S isak × kmatrix,thenS isasymmetricpositivesemi-definite(SPSD)matrix if S issymmetric(S ij =S ji foranyi, j= 1, . . . , k) and
v
T
Sv
≥ 0
for
any
v
∈ k
.
If S isak × kmatrix,thenS isasymmetricpositivedefinite(SPD)matrixif S issymmetricand
vT Sv >0 forany v∈ k, v = 0.
LetS
k
denote
the
set
of
symmetric
k× k matrices, and letS
k
denote
the
+setof symmetricpositivesemi-definite(SPSD)k × kmatrices. SimilarlyletS k denotethesetof symmetricpositivedefinite(SPD)k× kmatrices.++
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LetS and X beanysymmetricmatrices. Wewrite “S
0”todenotethat
S
is
symmetric
and
positive
semi-definite,
and
we
write
“S
X ” to
denote that S − X 0. We write “S 0” to denote that S is symmetricandpositivedefinite,etc.
Remark 1 S k ={S ∈ S k | S 0} is a convex set in k2.+Proof: Suppose that S, X ∈ S k+. Pick any scalars α, β ≥ 0 for whichα +β =1. Foranyv∈ k, we have:
vT (αS +βX )v=αvT Sv +βv T Xv ≥ 0,
whereby
αS
+
βX
∈ S k
+.
This
shows
that
S k
is
a
convex
set.
+
q.e.d.
Asemi-definiteoptimizationproblem isanoptimizationproblemof thefollowingtype:
SDO: minimizey bT y
m
s.t. C + yiAi 0i=1
My
≥ g .
wherethematricesC, A1, . . . , Am aresymmetricmatrices. Oneconvenientway of thinking about this problem is as follows. Given values of the mscalarvariablesy1, . . . , ym,theobjective istominimizethelinearfunction:
m
biyi .i=1
Theconstraintsof SDOstatethatthevariablesy= (y1, . . . , ym)mustsatisfy
the
linear
inequalities:
M y ≥ g
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aswellastheconditionthatthematrixS ,definedby:
m
S :=C + yiAi ,i=1
mustbepositivesemi-definite. That is,m
S :=C + yiAi0.i=1
We illustratethisconstructionwiththefollowingexample:
SDO
:
minimizey1,y2 11y1 + 19y2
1 2 3 1 0 1 0 2 8
s.t. 2 9 0 +y10 3 7 +y22 6 0 =S 03 0 7 1 7 5 8 0 4
3y1 + 7y2 ≤12
2y1 +y2 ≤6 .
whichwecanrewrite inthefollowingform:
SDO: minimize 11y1 + 19y2
s.t.
1 + 1y1 + 0y2 2 + 0y1 + 2y2 3 + 1y1 + 8y2
2 + 0y1 + 2y2 9 + 3y1 + 6y2 0 + 7y1 + 0y2 03 + 1y1 + 8y2 0 + 7y1 + 0y2 7 + 5y1 + 4y2
3y1 + 7y2 ≤
12
2y1 +y2 ≤6.
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Remark 2 SDO is a convex minimization problem.
Proof: Theobjectivefunctionof SDOisalinearfunction,whichisconvex.yand˜Supposethat¯ yaretwofeasiblesolutionsof SDO,andlety=αȳ+βỹ ,
whereα,β ≥0 and α+β =1. Then:m
S :=C + ȳiAi 0i=1
and
m
X :=C + ỹiAi 0.i=1
Therefore,
m m
C +
yiAi = C +
(αȳ+β ̃y)Aii=1 i=1
m m
= α C + ȳiAi +β C + ỹiAii=1 i=1
=
αS
+
βX
0
.
Thisshowsthatthefeasibleregionof SDO isaconvexset.q.e.d.
Semi-definite optimization is a unifying model in optimization. Linearoptimization, quadratic optimization, second-order cone optimization, aswell as certain other convex optimization problems, can all be shown tobe special cases of semi-definite optimization. Furthermore, semi-definiteoptimizationhasapplicationsthatspanconvexoptimization,discreteopti-
mization,
and
control
theory.
In
fact,
semi-definite
optimization
may
very
wellbecomethecanonicalwaythatoptimizerswillthinkaboutoptimizationinthenextdecade.
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9 Truss Design and Semi-Definite Optimization
We now present an SDO problem that is equivalent to the truss designproblemTDP.Thisproblem is:
(STDP): minimizet,θ θ
s.t.θ
F T
m
F tkE k (ak)(ak)
T 0L2
k=1 k
M t
≤d
t≥ 0mθ∈ , t ∈ .
NoticethatSTDP isasemi-definiteoptimizationproblem. Theequiva-lenceof TDPandSTDPisaconsequenceof thefollowingtwopropositions:
Proposition 9.1 Suppose that (t, u) is a feasible solution of TDP. Let:
θ
=
F T u .
Then (t, θ) is a feasible solution of STDP, and θ=F T u.
Proposition 9.2 Suppose that (t, θ) is a feasible solution of STDP. Then there exists a vector u for which (t, u) is feasible for TDP, and in fact:
F T u≤ θ .
Together, Propositions 9.1 and 9.2 demonstrate that any solution of STDPtranslates intoasolutiontoTDP.Therefore, inordertosolveTDP,
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∗ ∗wecansolveSTDP for(t , θ )andthensolvethe following linearequation
∗system
for
u
:
mt∗kE k T ∗
L2aka u =F .k
k=1 k
AsameanstoprovePropositions9.1and9.2,wefirstshowthefollowing:
Remark 3 Given a vector v, a square matrix M , and a scalar θ, then:
θ vT
Q
:=
0
v M
if and only if:
T M 0 , there exists u satisfying Mu =v , and θ≥v u .
Proof of Remark 3: Toseewhythisistrue,letusfirstsupposethatQ0.ThenclearlyM 0,sinceM isformedbyasubsetof thecomponentsthatform Q. Now, let us suppose that there is no u such that Mu = v. ThisimpliesthatthereexistsλsuchthatλT M =0andλT v > 0. Then:
−(−, λT )
θ vT
=
θ2 −
2v
T λ +
λT M λ
=
θ2 −
2v
T λ
<
0v M λ
for small enough, which is a contradiction. Therefore there exists usuchthatM u =v. Furthermore,forsuchau, we have:
−1 T 0≤(−1, u T ) θ vT
=θ−2vT u +uT M u =θ−v uv M u
T whichshowsthatθ≥v u.Now
let
us
prove
the
reverse
of
these
implications.
Suppose
that
M
0,
thereexistsusatisfyingMu =v, and θ≥vT u. Since M 0 we only have toprovethat:
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−1 T δ (x) := (−1, x T ) θ vT
=θ−2v x +xT Mx ≥0 ∀x . v M x
Butnoticethatδ (x)isaconvexquadraticfunctionof x(sinceinpartic-ularM 0),andtherefore
∇δ (x) = −2v+ 2M x = 0
is
a
necessary
and
sufficient
condition
for
the
unconstrained
minimization
of δ (x). Sinceusatisfiesthiscondition, it istruethat
T T δ (x)≥δ (u) = θ−2v u +uT M u =θ−v u≥0 ∀x .
q.e.d.
Nowletusprovethepropositions.
Proof of Proposition 9.1: Suppose that (t, u) is a feasible solution of TDP.Let:
θ=F T u ,
and let
m E kM = tk
L2(ak)(ak)
T ,k=1 k
and letv=F . Then
m E k
Mu = tkL2
(ak)(ak)T u=F ,
k=1 k
andnoticeaswellthat
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=
M
0
because M is the nonnegative sum of rank-one SPSD matrices. FromRemark3,
θ F T
θ vT
m
F tkE k (ak)(ak)
T v M 0 .
L2 k=1 k
Therefore (t, θ) is feasible for STDP with objective function value θ =F T u.
q.e.d.
Proof of Proposition 9.2: Suppose that (t, θ) is a feasible solution of STDP.ThenfromRemark3,thereexistsavectoruforwhich:
m
tkE k
(ak)(ak)T u=F ,
L2k=1 k
and
θ≥F
T
u.
Therefore
(t, u)
is
feasible
for
TDP
and
θ≥F
T
u.
q.e.d.
10 Truss Design and Linear Optimization
Considerthefollowingspecial,butnotunusual,instanceof thetrussdesignproblem:
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(TDP): minimizet,u F T u
m
s.t. tkE k (ak)(ak)
T u=F L2
k=1 k
m
tk ≤ V k=1
t≥ 0mu
∈ N , t
∈ .
In this problem, the only constraint on the volumes of the bars tk is avolume constraint limiting the total volume of the bars to not exceed thegivenvalueV . Substitutingthenotation:
m E kK (t) := tk
L2(ak)(ak)
T ,k=1 k
wecanalsoconvenientlywriteourproblemas:
(TDP):
minimizet,u F T u
s.t. K (t)u=F
m
tk ≤ V k=1
t≥ 0mu∈ N , t ∈ .
In
this
section,
we
show
that
when
TDP
has
this
particularly
simple
form,thenitcanbesolvedvialinearoptimization. Beforedoingso,wefirstwritedownadualproblemassociatedwithTDP inthisform:
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(DTDP): maximizev,z −2F T v− V z
s.t. 2 L2T kakv ≤ E k z , k = 1, . . . , m y∈ N .
InorderforDTDPtobeanhonestdualof TDP,wenowpresentaweakdualityresult:
Proposition 10.1 Suppose that (t, u)is feasible for TDP and that (v, z)is feasible for DTDP.Then:
F T u≥ −2F T v− V z .
Proof: For k= 1, . . . , m, we have
E k T tk ≥ 0 and L2
(akv)2 ≤ z.
k
Multiplyingandsummingterms,weobtain:
m
T T vT K (t)v=
tkE k(v ak) (akv)≤
m
tk z≤ V z. L2
k=1 k k=1
Also,
0≤ (u +v)T K (t)(u +v) = uT K (t)u + 2uT K (t)v+vT K (t)v= F T u + 2vT F +vT K (t)v
≤ F T u + 2vT F +V z. ThereforeF T u≥ −2F T v− V z.q.e.d.
Considerthefollowingpairof primalandduallinearoptimizationmod-els:
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m √ Lk(LP): minimizef +,f − E k
f k+ +f k
−k=1
s.t. A(f + − f −) = −F
f + ≥ 0 , f − ≥ 0mf +, f − ∈ .
(LD): maximizey−
F T y
s.t.T −√ Lk ≤ aky≤ √ Lk , k= 1, . . . , mE k E k
y∈ N .
We will prove the following important result that shows how to usesolutionsof LPandLDtoconstructsolutionstoTDP:
¯Proposition 10.2 Suppose that (f ̄+, f −) is an optimal solution of LP and that ȳ is an optimal solution of LD, and consider the following assignment
of
variables:
m
¯R =
√ Lk f ̄+ +f k−E k kk=1t̄ k =
V √
Lk ¯f ̄+ +f k−
k= 1, . . . , m , R E k
k
Rū = − ȳ
V R
v̄ = ȳV R2
z̄ = V 2
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V
Then (¯u) solves TDP and (¯ z) solves DTDP.t, ¯ v, ¯
Proof: Notethatbythecomplementary slackness propertyof linear opti-mization,wehave:
T + Lk +akyf = √ f k, k = 1, . . . , m , k E kT − −Lk −akyf = √ f k, k = 1, . . . , m .k E k
m
We first show that (t, u) is feasible forTDP. Note that t≥0 and tk =k=1
R=V . Also
R
RK (t)u = − K (t)y
V tkE kR m T = −V L2
akakyk=1 k
Lk − T = −R V m
√ E k (f + +f k)akakykV R E k L2k=1 k√ m E k + T − T = − ak (f kaky+f kaky)Lkk=1 m −=
−
ak (f
+
−
f
−
k) =
−A(f
+
−
f
) =
F.kk=1
Wenextshowthat(v, z)isfeasibleforDTDP.Toseethis,observethatfork= 1, . . . , m, we have:
R2 R2 L2T T (akv)2 =
V 2(aky)
2 ≤ √ Lk2
=z k,V 2 E k E k
andso(v, z)isfeasible.Finally,weshowthatthesesolutionsexhibitstrongduality:
−2R 2R2 R2 R2 R−2F T v−V z= F T y−V R2
= − = = R R= (−F T y) = F T u.V V 2 V V V V V
q.e.d.
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11 Extensions of the Truss Design Problem
• Multiple Loads. We consider a finite set of external loads on thetruss:
F ={F 1, F 2, . . . , F J } ⊂ IRN .
In this case we might consider solving two types of design problems.Thefirstistodesignthetrussstructureconservatively,soastomini-mizethemaximumcompliance:
minimizet max j=1,...,J F T j u j
Mt ≤ d,t≥ 0 s.t. K (t)u j =F j.
Alternatively, we might consider the “average-case” design problem:letλ j denotetherelativefrequencyorimportanceassociatedwiththe
J
truss structure undergoing the external force F j, where λ j = 1.0. j=1
The following optimization model minimizes the average complianceof thetrussstructure:
J
minimizet λ jF jT u j
j=1
Mt ≤ d,t≥ 0 s.t. K (t)u j =F j.
• Self-weights. Thetrussdesignproblemthatwehaveformulatedpre-sumesthatthetrussstructureitself isnotaffectedbyitsownweight.Tocorrectforthis,wecansimplyaddanexternalforcecorresponding
to
the
gravitational
force
associated
with
bar
k,
and
linearly
propor-tionaltotk fork= 1, . . . , m. Letusdenotebygk ∈ IRN thevectorthatprojects the gravitational force of bar k onto the appropriate nodes.
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Then we can write the TDP of a truss under a single external loadvector
F
as:
T m(TDP): minimizet,u F + tkgk u
k=1
m
s.t. K (t)u= F + tkgkk=1
M t ≤ d
t≥
0
mu∈ N , t ∈ .
It is straightforward to convert this problem into a convex problem, justasforthebasecaseconsideredearlier.
• Reinforcement. Inthisproblem,wearegivenanexistingtrussstruc-ture,andwemustdeterminehowtostrengthenit. Tosolvethisprob-lem,wesimplyadd lowerboundsonthevolumesof theexistingbarsequaltothecurrentvolumesof thebars:
(TDP): minimizet,u F T u
s.t. K (t)u=F
M t ≤ d
t≥ 0
tk ≥ tk , k = 1, . . . , m mu
∈ N , t
∈ .
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• Robustness of Solutions. Theoptimalsolutionof theTDPmight
yield
a
truss
design
that
is
not
rigid
for
a
different
external
force
than
theoneused,andcouldchangeconsiderablyevenunderasmallchangein the external force. To obtain a robust solution, we must considermultiple external loads that will contain the possible loads that thetrusswillbesubjectto. Thiscanbemodeledbyconvexoptimizationaswell.
• Buckling Constraints. Thetrussdesignproblemthatwehavepre-sentedignoresthefactthatif abarisundergreatcompressionitmightactuallycollapse insteadof counteractingthatexternal forcewith itsinternalforce. Forthiswehavetoadd lowerboundsontheallowable
internal
forces
in
terms
of
the
design
variables
tk and
the
geometryof thebars. Theseconstraintscausetheresultingproblemto lose its
convexstructure.
• Other Considerations . . . .
47