Improved approximation algorithms for a bilevel knapsack problem

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Contents lists available atScienceDirect

Theoretical

Computer

Science

www.elsevier.com/locate/tcs

Improved

approximation

algorithms

for

a

bilevel

knapsack

problem

Xian Qiu

a

,

∗

,

1

,

Walter Kern

b

a_College_of_Computer_Science,_Zhejiang_University,_China

b_Department_of_Applied_Mathematics,_University_of_Twente,_Netherlands

a

r

t

i

c

l

e

i

n

f

o

a

b

s

t

r

a

c

t

Articlehistory:

Received20September2014 Receivedinrevisedform13May2015 Accepted13June2015

Availableonline19June2015 CommunicatedbyX.Deng Keywords: Bilevel Knapsack Approximationalgorithm Stackelberg

Westudythe Stackelberg/bilevelknapsackproblemasproposedbyChenand Zhang[1]: Consider two agents, a leader and a follower. Each has his own knapsack. (Knapsack capacitiesare possibly different.) Asusual, there is aset of itemsi=1, . . . , n ofgiven weights wiandproﬁtspi.Itisallowedtopackitemi intobothknapsacks,butinthiscase

thecorrespondingproﬁtforeachplayerbecomespi+ai,whereai isagiven(positiveor

negative)number.Theobjectiveistofindapackingfortheleadersuchthatthetotalprofit ofthetwoknapsacksismaximized,assumingthatthefolloweractsselfishly.Wepresent tightapproximationalgorithmsforallsettingsconsideredin[1].

1. Introduction

The standardknapsackproblemisoneofthemostfundamentalandwell-studiedproblemsincombinatorial optimiza-tion: There isa knapsackofprescribed capacityW and n items withgivensize wi andprofit pi.The taskisto selecta set ofitemsoftotalsize atmostW and maximumtotalprofit. Afirst bilevelvariant(in theformofa

Stackelberg game)

was introducedbyDempeandRichter

[2]

:Therearetwodecisionmakers(players)–a

leader and

a

follower –

aswellasa (universal)knapsackwithflexiblecapacityandasetofitemswithgivensizesasabove,yetitemprofitsmayvaryw.r.t. the leader andthefollower,respectively.Theleaderfirstdeterminesthecapacityoftheknapsack,andafterwardsthefollower, assumedtobeselfish,packsitemstotheknapsack,maximizinghisownprofit.The(leader’sbilevel)problemistocompute the knapsackcapacitysuch that the leader’s profit–definedby a linearfunction oftheknapsack capacityplus histotal profitofpackeditems–ismaximized.

Severalotherbilevelvariantsofknapsackhavebeenproposedaswell. Forexample,Mansi etal.

[12]

studyasettingin whichboth theleaderandthefollowerpackitemsintoaknapsack(offixedcapacity).DeNegre [17]investigatesabilevel version whereboth playersown aprivate knapsackeachandpackitemsfroma commonitemset.Again,theleader acts first,selectingasetofitemsforhisownknapsack,thenthefollowerpacksitemsfromtheremainingitemsetintohisown knapsack,seekingtomaximize histotalprofit.The objectiveofthe(hostile)leaderisto choosehissetofitemssuchthat thefollower’sprofitisminimized.

*

Correspondingauthor.

E-mailaddresses:xianqiu@zju.edu.cn(X. Qiu),w.kern@utwente.nl(W. Kern).

1 _Supported_by_the_Natural_Science_Foundation_of_Zhejiang_Province,_No._LQ15A010001_and_the_Fundamental_Research_Funds_for_the_Central_Universities ofChina.

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Cases Approx. ratios[1] Lower bounds ai≤0 2+ 1.5 ai≥0 W1>W2 W1<W2 1+√2+ 2+ 1.5 2 Fig. 1. Known lower bounds.

Inthispaperweconsideryetanothervariantofthebilevelknapsackproblem,duetoChenandZhang

[1]

.Inthissetting, again,eachplayerhashisownknapsackofﬁxedcapacities W1and

W

2,respectively.Items1

,

. . . ,

n have ﬁxedweights wi andproﬁts

p

i.Thecharacteristicfeatureofthemodelin

[1]

isthatitemsmaybe

double-packed, i.e. packed

bybothplayers. Incaseitem

i is

packedonlybyoneplayer,itaccountsforaproﬁtof

p

i,asusual, however,if

i is

packedbybothplayers, itsproﬁt(forbothplayers)ismodiﬁed to pi

+

ai forgiven

proﬁt modiﬁer a

i

∈ R

.Again,thesettingisthatofaStackelberg game,andtheobjectiveistoexhibit an optimalpackingfortheleader,

i.e.,

one thatmaximizesthetotal proﬁtassuming that the second player (the follower) actsselﬁshly (disregarding the impact anydouble packing mayhave on the items packedby theleader).Asamotivatingexample,ChenandZhangmentionthecaseoftwoinvestors,say,the government andacompany withbudgets W1 and W2,respectively.Items correspondtopotential projects ofcost wi andreward pi, resp. pi

+

ai with

a

i

>

0 ifbothplayersinvestinprojecti. Dependingontheapplication, thenumbers

a

i maybepositive ornegative (“double booking”).Incaseall

a

i are positive,ChenandZhang

[1]

call itthe

beneﬁcial model and

ifall

a

i are negative,itisreferredtoasthe

competitive model.

Bileveloptimizationis oftencomputationally diﬃcultandlikely toextend beyondNP.In thelast decades, bilevel and multilevel optimizationhavereceived much attentionin theliterature (cf. books by Migdalas, Pardalos andVärbrand [4]

andDempe

[3]

,asurveybyColsonetal.

[5]

).DempeandRichter

[2]

introducedamixedintegerbilevelprogramfortheir problemvariant andproposed an algorithm basedon branch andbound. Afterwards,a dynamic programmingalgorithm forthisproblemwasgivenby Brotcorneetal.

[6]

.Recently, Capraraetal.

[7]

provedthat theﬁrstthreeproblemvariants mentionedaboveare

₂P-hard(probablythefourthoneisaswell),

i.e.,

thereisnowayofformulatingthemassingle-level integer programs of polynomial size unless the polynomial hierarchy collapses (cf. [7]for more details).In particular, they showedthattheﬁrsttwovariants(cf. DempeandRichter

[2]

,Mansietal.

[12]

)donotpossessapolynomialapproximation algorithmwithﬁniteworst caseguaranteeunless P

=

NP and proposed apolynomial timeapproximation schemeforthe thirdvariant(cf. DeNegre

[17]

),whichisknownastheﬁrstapproximationschemefora

₂P-hardproblem.Forothervariants andrelatedproblems,

cf.

[8–12].

Regarding the problem to be considered in this paper, Chen and Zhang [1] proposed a

(

2

+

)

-approximation algo-rithmforthecompetitivemodel(ai

≤

0),and,forthebeneﬁcialmodel(ai

≥

0),a

(

1

+

√

2

+

)

-approximationforthecase

W1

>

W2 anda

(

2

+

)

-approximationforthecase

W

1

≤

W2.

In thispaper, we presentbetter approximation algorithms forthe beneﬁcial model aswell asthe competitive model andshowthat theapproximationratiosaretight ineachcase, i.e., theapproximationratios canbemadearbitrarily close to theknownlower bounds(cf. Fig. 1). The mainingredients ofour approachare: An

-approximation ofthemaximum proﬁtproblemincasebothplayerscooperate–whichmaybeofindependentinterest,

cf.

(P3)inSection2–andafactor revealingLPforestimatingthequalityofourapproximationalgorithms(cf. Jainetal.

[18]

).

Therestofthepaperisorganizedasfollows:Inthesectionbelow,we formallyintroducethebilevel problem(cf.(P1)

inSection2)andits“cooperative”counterpart(cf.(P3)).InSection3,wedescribeapolynomialtimeapproximationscheme (PTAS)forthecooperativeproblemversion

(P3)

.InSection4,wepresentnewapproximationalgorithmsandanalyzetheir approximationratios.Finally,inSection5,wementionsomeopenproblems.

2. Bilevelknapsackwithindependentknapsacks

Let W1, W2 be capacities ofthe knapsacks ownedby player 1 (leader)and player2 (follower), respectively. Let A

=

{

1

,

2

, . . . ,

n

}

be a set of itemsof weight wi, proﬁt pi and “double packing modiﬁer” ai for all i

∈

A. Let xi

,

y

i

∈ {

0

,

1

}

indicatewhetheritem

i is

packedbyplayer1andplayer2,respectively.

Recallthatthe proﬁtofitem i is modiﬁed to pi

+

ai ifi is packedby bothplayers. Thusthe leader’s problemcan be formulatedasabilevelintegerprogramasfollows:

max x n

i=1 pi

(

xi

+

yi

)

+

2 n

i=1 aixiyi (P1) s.t. n

i=1 wixi

≤

W1

,

xi

∈ {

0

,

1

} ,

i

=

1

,

2

, . . . ,

n

,

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max y n

i=1 piyi

+

n

i=1 aixiyi (P2) s.t. n

i=1 wiyi

≤

W2

,

yi

∈ {

0

,

1

} ,

i

=

1

,

2

, . . . ,

n

.

Notethatforﬁxed

x there

mayexistmultiplecorrespondingoptimalsolutions y for player2.Inouranalysis,wealways assumeaworstcasescenario,

i.e. we

focusona“pessimistic”versionoftheabovebilevelproblem,whereplayer2chooses an optimalsolution y of (P2)minimizing the objectiveof

(P1)

. Wecall theoutcome ofthispessimistic versionthe com-petitive optimum and –asinthepaperbyChen andZhang

[1]

–compareittotheso-calledcooperative optimum, i.e., the maximumtotalproﬁtthetwoplayerscouldachieve.Thelattercanbeexpressedbya(singlelevel)integerprogram

max n

i=1 pi

(

xi

+

yi

)

+

2 n

i=1 aixiyi s.t. n

i=1 wixi

≤

W1

,

n

i=1 wiyi

≤

W2

,

xi

,

yi

∈ {

0

,

1

} .

(P3)

Example.Considertwoknapsacksofcapacity

W

1

=

1 (fortheleader)andcapacityW2

=

2 (forthefollower).Theitemset

containstwoitemsofweights

w

1

=

1,

w

2

=

2,proﬁts

p

1

=

2, p2

=

1 andmodiﬁers

a

1

= −

1 (a2 isarbitrary).

Observethatplayer1onlyhastwooptions: Packingitem1orpackingnothing.Ifplayer1packsitem1,then player2 gets amaximalprofit1 byeitherpackingitem1oritem2,resultinginatotalprofit2 or3 respectively.Ifplayer1packs nothing,thenplayer2packsitem1,resultinginatotalprofit2.Thus,thecooperativeoptimummaybeaslargeas1

.

5 times thecompetitive optimum.Weaimatshowingthatthisexampleisaworst caseexampleinthesense thatifallmodiﬁers

aiarenegative,thentheratiobetweenthecooperativeandcompetitiveoptimumisboundedby1.5andthatsolutions

x for

theleader’sproblemensuringaratioof1

.

5

+

canbefoundinpolynomialtime.Similarly,wepresenttightboundsforthe caseofnon-negative

a

i and,eventually,themixedcasewithbothpositiveandnegativedoublepackingmodiﬁers.

2.1. Dynamic programming for (P3)

We derivea straightforwardpseudo-polynomialalgorithmfor

(P3)

basedondynamic programming.Let fi

(

a

,

b

)

be the maximumtotalproﬁtwithknapsackcapacities

a

,

b w.r.t. itemset

{

1

, . . . ,

i

}

,for

a

,

b

,

i

∈ Z

+_,₁

_≤

_i

_≤

_{n, a}

_≤

_W₁_,

_b

_≤

_W₂_._The recursiveformulaisdeﬁnedby

f_i+1

(

a

,

b

)

=

max

{

fi

(

a

,

b

),

fi

(

a

−

wi+1

,

b

)

+

pi+1

,

fi

(

a

,

b

−

wi+1

)

+

pi+1

,

fi

(

a

−

wi+1

,

b

−

wi+1

)

+

2pi+1

+

2ai+1

}.

Theinitialconditionis f1

(

a

,

b

)

=

_max

_{

_p

1

,

2p1

+

2a1

} ,

a

=

w1

,

b

=

w1

,

p1

,

a

=

w1

,

b

<

w1or a

<

w1

,

b

=

w1

,

0

,

a

<

w1

,

b

<

w1

.

It isstraightforwardtocheckthat fn

(

W1

,

W2

)

returnsthemaximumproﬁtfor

(P3)

andthealgorithmhasarunningtime O

(

nW1W2

)

.

3. Polynomialtimeapproximationschemefor(P3)

As aﬁrststep,wecompute approximatelyoptimalcooperativesolutions.Itiswell knownthatknapsackcanbesolved byafullypolynomialtimeapproximationscheme(FPTAS)(cf.[13,14]).As

(P3)

with

a

i

→ −∞

becomesamultipleknapsack problemwithtwo knapsacks,wecannot expectanFPTASfor

(P3)

unless P

=

NP (cf.[15]).Inthefollowing,we seekfora polynomialtimeapproximationscheme(PTAS)for

(P3)

.

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We startwithsome notations.Foranyset S of items, let w

(

S

)

and p

(

S

)

denotethe totalweight andthetotal proﬁt of itemsin S, respectively, i.e., w

(

S

)

=

i∈Swi and p

(

S

)

=

i∈Spi. Since theproﬁts of itemsmaybe modiﬁed dueto double-packing, p

(

S

)

mayalsodenotethemodiﬁedtotalproﬁtofitemsin S if nomisunderstandingispossible.

As it turnsout,

negative items,

i.e., those with

a

i

<

0,and

non-negative items (those

withai

≥

0) can be treated inde-pendently. Therefore,we simplifymattersby ﬁrst assuming thatall items arenegative (non-negative itemswill be dealt afterwards).Fortechnicalreasons(tobeexplainedintheproof)we slightlygeneralizeourproblem,assuming thatcertain items,say,itemsintheset

N

1

⊆

N, arenotallowedtobedouble-packed.Welet

n

1

= |

N1

|

denotethenumberofitemsthat

areprescribedtobesingle,and

n

2isthenumberofitemsin

N

2

=

N

\

N1thatmaybedouble-packed.Thus

n

1

+

n2

=

n. Our

prooffortheapproximationratiowillbebyinductionon

n

1

+

2n2.

Wedescribe an O

(

1

−

1

/

k

)

-approximationalgorithm–againdenotedbyALG –proceedingina similarwaytothatof Sahni

[16]

.Onedifferenceisthatforadouble-packableitem

i we

distinguishbetweenits

primary copy i with

anassociated proﬁt pi andits

secondary copy i with

proﬁt pi

+

2ai

<

pi.Let S∗

=

S∗₁

∪

S∗₂

=

supp x∗

∪

supp y∗ be anoptimalsolution of(P3).Observethat–aswedistinguishprimaryandsecondarycopies–theset S∗maybeunderstoodasa

set rather

than amultiset.InphaseI,ALGseeksto“guess”the

k most

proﬁtableitemsfromtheoptimumsolution

S

∗₁

∪

S∗₂ toincludethem intheinitialpacking.Incaseanitem

i is

double-packedin

S

∗,and(theprimarycopyof)

i belongs

tothe

k most

proﬁtable items, wewantALGto includealsothesecondary copyintotheinitialpacking.Forthisreason, welet ALGstartfromall initialpackings withupto

k primary

itemsplussome oftheir secondarycopiesandlet S

=

S1

∪

S2 be thissetofitems,

with

S

1

,

S

2packedonknapsacks1and2,respectively.

Thesecondphase,again,considerstheremainingitemsinorderofnon-increasingproﬁtrates.Asinthesingleknapsack case,itproceedsinatrueonlinemannerasexplainedbelow.Inparticular,wheneverALGchecksanitem

i,

itimmediately decidesuponpackingornotpacking

i,

butdoesnotyetdecidewhether

i should

bedouble-packed.Notethat,aswestickto thecaseofnegativeitemshere,wehave

p

i

+

2ai

<

pi,sothatprimaryitemscomewithhigherproﬁtratesandarechecked forinclusionbeforetheircorrespondingsecondarycopiesarrive.

Summarizing,inphaseII,ALGconsidersthe(copiesof)itemsin

{

1

, . . . ,

n

} \

S in orderofnon-increasingproﬁtrates ri1

≥

ri2

≥ . . . ≥

rim

(

m

=

n1

+

2n2

− |

S

|)

deﬁnedasexplainedabove.

WheneverALG checksa primaryitem

i

=

it,theitem ispackedwherever itﬁts. Incaseitdoesnot ﬁtanywhere,the itemisskipped.WheneverALGchecksasecondaryitem i

=

it,itisperfectlyclearonwhichknapsack

i should

bepacked andALGseekstoaccommodateitem

i there,

say,onknapsack1,by“switching”singleitemsfromknapsack 1toknapsack 2 ifnecessary.Moreprecisely,ALG considersallsingle (primary)itemsfrom

{

1

,

. . . ,

n

}\

S currently packedonknapsack1in someorderandswitchesthemontoknapsack2whenevertheyﬁtthere,untileitheritem

i can

eventuallybeaccommodated onknapsack1ornofurthersingleitemcanbeswitchedtoknapsack2whileitem

i still

cannotbeaddedtoknapsack1.In thelattercase,item

i

=

it isskipped.Theorderinwhichitemsareconsideredforswitchingfromknapsack1toknapsack2 isnotrelevant,butitisconvenienttousea“lastinﬁrstout”orderasswitchingrule.

Notethatanypackeditemremainspacked,onlytheassignmenttoaparticularknapsackmayberevised(possiblyeven severaltimes),duetoswitching.

Lemma1.

ALG as described above yields a

(

1

−

2

/

k

)

-approximation for (P3)(assuming that all items are negative).

Proof. Theproof isbyinductionon

n

1

+

2n2.The claimisobviouslytruefor

n

1

+

2n2

≤

k. Indeed,whenevertheoptimal

solution

S

∗

=

S∗₁

∪

S∗₂contains2k orlessitems,thenALGwillexhibit

S

∗ inphaseI.

Hence, let us assume that n1

+

2n2

>

k and follow the computation of ALG, assuming that in phase I, S

=

S1

∪

S2

containingthe

k largest

proﬁtitemsfrom

S

∗arepackedinthesamewayandwiththesamemultiplicitiesasintheoptimal solution.Let

i

=

it denotetheﬁrstitemthatisskippedbyALG.If

i

∈

N1,theproofisalmostidenticaltothesingleknapsack

case:As wi exceedstheremainingcapacity,say,

c

1

,

c2 respectively,forbothknapsacks,thetotalremainingcapacity

c

1

+

c2

satisﬁes

c

1

+

c2

<

2wi.Ifwecompare thecurrenttotalproﬁt Pt−1 ofALG, justafteritem

i

t−1 hasbeenplaced,thetotal

usedcapacityequals

W

1

+

W2

− (

c1

+

c2

)

and–sinceALGpacksinorderofnon-increasingproﬁtrates–nootherpacking

canachieveahigherproﬁtonthispart.Consequently,nootherpackingalgorithmcanachieveatotalproﬁtlargerthanor equalto Pt−1

+

2

(

c1

+

c2

)

ri(theamountthatwouldbeachievedbypackinganitemwithproﬁtrateequalto

r

i,butsmaller size(c1

,

c2 respectively), therebyexhausting exactlybothknapsack capacities).Hence, theﬁnal totalproﬁt P achieved by

ALGcomparestotheoptimumproﬁt P∗asfollows: P∗

≤

P

+ (

c1

+

c2

)

ri

<

P

+

2wiri

=

P

+

2pi

.

Thus, ifi

∈

S∗₁

∪

S∗₂,then pi

≤

P∗

/

k and we get P

≥

P∗

(

1

−

2

/

k

)

asrequired. Else,ifi

∈

/

S∗1

∪

S∗2,then P and P∗ remain

unchangedifweremoveitem i,therebydecreasing

n

1

+

2n2 byone,sotheclaimfollowsbyinduction.

The same argumentworks without anychange in casei

=

it

∈

N2 is a primary item, exceptthat by removing i, we

decrease

n

1

+

2n2 by two. Thus assume nowthat i

=

it

∈

N2 isalready packed,say, onknapsack 1,but thesecondary i

doesnot ﬁtonknapsack2,evenafterswitchingamaximal setofitemsfromknapsack2toknapsack1.Thesituationcan be described asfollows: Besides the initial packing S1

,

S2, thereis a certain set D of itemsthat aredouble-packed and

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Fig. 2. Packing negative items.

sets J1

,

J

2 ofsingle/primary itemson knapsack1and2,respectively. Byassumption,the remaining capacities

c

1 and

c

2

on knapsack 1 and2 satisfy: c2

<

wi (item i does not ﬁt) and c1

≤

wj for every j

∈

J2 (no j

∈

J2 can be switched to

knapsack 1).Let D∗ denotetheset ofitemsthat aredouble-packedinthe optimumsolution.

Fig. 2

belowillustrates the currentsituation.

First note that, again, we mayassume that i

∈

J1

∩

D∗.Otherwise, ifi

∈

J1

\

D∗,prescribing i as single wouldreduce n1

+

2n2 butotherwise affectneither ALG nor the optimum solution, and the claim would follow by induction. Hence i

=

it

∈

J1

∩

D∗ indeed.Hence pi

≤

1_kp

(

S1

∪

S2

)

≤

1_kP∗ follows.

Asimilarargumentshowsthatwemayassume j

∈

/

S∗₁

∪

S∗₂ forany j

∈

J2:If j

∈

S∗1

∪

S∗2,then pj

≤

1_kp

(

S∗1

∪

S∗2

)

≤

1kP∗. Thus if c1

,

c2 are the remaining capacities on knapsack 1 and 2 respectively, then (due to the fact that ALG proceeds

accordingtonon-increasingproﬁtrates)wehave

r

j

≥

riandhence P∗

≤

P

+ (

c1

+

c2

)

ri

≤

P

+

c1rj

+

c2ri

<

P

+

wjrj

+

wiri

=

P

+

pj

+

pi

≤

P

+

2 kP

∗

andtheclaimwouldfollow.Hencewemayassumeindeedthat J2

∩ (

S1∗

∪

S∗2

)

= ∅

.

Aswehaveseenabove,wemayassumethat

p

i

≤

P∗

/

k, sowecanaffordskippingthesecondaryitem

i

=

it.Theproblem isthat

i

t+1

,

it+2 etc.mightbesimilarsecondaryitemsthatwehavetoskip.Thus howmuchskippingcanweafford?The

answerisgivenby thesizeof J2 andD

\

D∗:Astheitemsin J2 arenot packedintheoptimumsolutionandtheitemsin D

\

D∗ areatleastnotdouble-packedintheoptimumsolution.ALGspendsatotalof

w

=

w

(

J2

)

+

w

(

D

\

D∗

)

packingitems

atrate

r

≥

ri which arenot packedby an optimalalgorithm, say,OPT. Instead,OPTpacksthe secondaryitems

i

=

it and subsequentsecondaryitemsatrate

r

≤

ri.Thuswecanaffordskippingsecondaryitemsoftotalsize

w without

fallingback behind OPT(theonethatbuilds S∗

=

S∗₁

∪

S∗₂).Thegoodnewsisthat,roughly,nomorethanthisisabouttocome:IfOPT buildstheoptimumsolution

S

∗₁

,

S

∗₂,thenitpacks

S

2 and

D

∩

D∗onknapsack2,sotheremainingspaceforsecondaryitems i

∈

S∗₁

∪

S∗₂ isboundedby

W2

−

w

(

S2

)

−

w

(

D

∩

D∗

)

≤

w

(

D

\

D∗

)

+

w

(

J2

)

+

wi

=

w

+

wi

.

Thusthetotalsizeofsecondaryitemsthatweskipcanbecompensatedbytheproﬁtwegainwhilepacking J2and

D

\

D∗,

exceptforanamountofatmost

w

iri

=

pi.

Thus atleast after skipping that many secondary itemssomething else must happen: Either a single item i that we skip or a secondary item j

∈

J2 that we cannot accommodate on knapsack 1. But then, again, even without anyeffort

of reassigning items, we conclude that j

∈

S∗₁

∪

S∗₂ (otherwise prescribe j as single and apply induction), implying that

pj

≤

P∗

/

k and, hence–disregardingatotalofw for theskippedsecondary itemscompensatedby J2 and D

\

D∗ –allwe

looseisboundedby

p

i (onknapsack2)andpj(onknapsack1).Summarizing,weget P

≥ (

1

−

2

/

k

)

P∗alsointhiscase. WeliketostressthattheaboveargumentholdsalsowhenthesecondaryitemsthatALGskipsdonotcomeinarow.For example,itmightwellhappenthatALGskips

i

=

it,but

i

t+1isa(primaryorsecondary)itemthatﬁtswellonknapsack 2,

thereby enlargingD or J2.Itmayalsohappenthat–possiblyafterreassigningsome itemsin J1toknapsack2–anitem j

=

it+1 canbeaccommodatedonknapsack1.Inanycase,wheneverthenextsecondaryitem j to beputonknapsack2is

skippedbyALG,thenthisisbecauseALGhastriedinvaintoaccommodate j by switchingsingleitemsfromknapsack2to knapsack1.Accordingtotheswitchingrule,ALGreassignsitemsina“lastinﬁrstout”order,ensuringthattheset J2that

wasblockingthesecondaryitem

i

=

it willstaypackedonknapsack2.(Thus,atalaterstage,wemightevenhavealarger set J ₂

⊃

J2, J2

∩ (

S∗1

∪

S∗2

)

= ∅

,tocompensateforskipping.)

2

It is an easy exercise tobound the runningtime:There are O

(

nk+1

₎

_sets _{S of}_size

_|

_S

_{| ≤}

_{k to}_choose _for_the_primary

items. Given S, weare lefttoﬁxforeach

i

∈

S whether todouble-pack i or not,and, inthelattercase,wheretopackit (knapsack1or2).Thusintotalthenumberof“guesses”isboundedby

O

(

nk+13k

)

.

Nextletusturntothebeneﬁcialmodel,whichturnsouttobeeasier:Assumethatallitemsin

N are

non-negative,

i.e.,

ai

≥

0.Ourapproximationalgorithm–denotedby ALG–inthiscase,will againhavea“guessing” phaseI, followedbya phase II, whereitemsare processedinorder ofnon-increasingprofitrates.Notethat thistime, however,double-packing yields higher profit rates than single packing, so ALG will first decide about double-packingitem i (at profit rate ri

=

(

pi

+

a1

)/

wi) and then – in case of non-acceptance consider single-packing i at a later stage. Correspondingly, in the beneﬁcial model,it doesnot makesense to distinguishbetweena primary andsecondary copyofitem i (as both arrive

(6)

atthesametime).Thus, inwhat followswewill interpretan optimalsolution of(P3)asamultiset S∗

=

S∗₁

∪

S∗₂, where

D∗

=

S∗₁

∩

S∗₂ isthesetofdouble-packeditems.

Inphase I,ALG guesses aset D

⊆ {

1

,

. . . ,

n

}

ofsize

|

D

|

≤

k (as acandidatefora setofmostproﬁtabledouble-packed itemsinan optimalsolution) aswellasaset S

⊆ {

1

,

. . . ,

n

}\

D,

|

S

|

≤

k of mostproﬁtablesingle-packeditems. Inaddition, itguessesanumber

l

≤

n indicating thetotalnumberofitemsthatshouldbedouble-packedbyALG(ontopofthealready chosenset

D).

InphaseII,ALGprocessestheremainingitemsinorderofnon-increasingproﬁtratesasusual.Moreprecisely, let ri1

=

pi1

+

ai1 wi1

≥ . . . ≥

ril

=

pil

+

ail wil

bethe

l highest

double-packingproﬁtratesin

N

2

\

D. ThenALGdouble-packsasmuchaspossibleof

i

1

,

. . . ,

il (ontopof D) andthencontinues(afterpacking S1 andS2)withsinglepackingoftheremainingitems(i.e.,itemsin

(

N1

∪

N2

)

\(

D

∪

S

∪

{

i1

,

. . . ,

il

})

)inorderofnon-increasingproﬁtrates.

Thereare O

(

n2(k+1)

₎

_possible_choices_for

_{D and S,}

_and_O

₍

₂k

₎

_bipartitions_of

_S.

_Together_with_the_different_choices_for

_l,

thereare

O

(

n2k+32k

)

differentchoicesofparameters.

Lemma2.

For suitable choice of parameters D

,

S1

,

S

2and l, ALG yields an approximation ratio

(

1

−

4

/

k

)

.

Proof. Theproofisbyinductionon

n

1

+

2n2.Let

S

∗1

,

S

∗2beanoptimalsolutionandlet

W

∗

:=

w

(

S1∗

∩

S∗2

)

bethetotalsizeof

double-packeditems.If

S∗₁

∩

S∗₂

≤

k, thenALGwillguess

D

=

S∗₁

∩

S∗₂ exactly,otherwiselet D denote thesetof

k highest

proﬁt

(

=

pi

+

ai

)

itemsin

S

∗1

∩

S∗2 andassume

l is

chosen suchthat w

(

D

)

+

wi1

+ . . . +

wil

≤

W∗ but w

(

D

)

+

wi1

+ . . . +

wil+1

>

W∗.

ThusALG,withthischoiceofparameters

D and l will

double-pack

i

1

,

. . . ,

il(ontopof

D),

butallfurtheritemswillgetat mostsingle-packed.Weclaimthatthetotaldouble-packingproﬁtisatleast

(

p

+

a

)(

S∗₁

∩

S₂∗

)(

1

−

2_k

)

,where

(

p

+

a

)(

S∗₁

∩

S∗₂

)

:=

i∈S∗₁∩S∗₂

(

pi

+

ai

)

.Indeed,ifi

=

il+1

∈

S∗1

∩

S∗2,thisfollowsin the–by now–usual mannerfrom pi

+

ai

≤

1

k

(

p

+

a

)(

D

)

. Else,if

i

∈

/

S∗₁

∩

S∗₂,wemightprescribe

i as

single,thusdecreasing

n

1

+

2n2 andtheclaimfollowsbyinduction.

ThusALGperformsverywellw.r.t.doublepacking,achievingalmostoptimalproﬁt,byusinglessspaceingeneral.Sowe arelefttoanalyzethesinglepackingpart.Ifatmost

k items

aresingle-packedintheoptimumsolution S∗₁

∪

S∗₂,ALGwill exhibitthese(alongwiththecorrectassignmenttoknapsack1andknapsack2).Else,let S

=

S1

∪

S2 denotethe

k highest

proﬁtsingle-packeditemsintheoptimumsolution(with Si packedonknapsack

i).

AsALG,whenitstarts single-packing items, hasatleastasmuchcapacityleft asOPToneachknapsack, S1 and S2 canbe single-packedwithoutanyproblem.

Theremaining itemsare thenprocessedinorderofnon-increasingproﬁtratesandtheremaining partoftheproof isby nowroutine:Let

i

=

it betheﬁrstitemthatcannotbeaccommodated,neitheronknapsack1noronknapsack2.(Herewe donot needto trytoreassign anyitems.)Then thetotalproﬁt, say P , includingsingle-packeditemsobtainedby ALGat thistime,isatleasta

(

1

−

4

/

k

)

-fractionof P∗thetotalproﬁtintheoptimumsolution:

Incase

i

∈

/

S∗₁

∩

S∗₂,theclaimfollowsbyinductionon

n

1

+

2n2 ifweprescribe

i as

single item.Else,if

i

∈

S∗₁

∩

S∗₂,then pi

≤

pi

+

ai

≤

_k1

(

p

+

a

)(

S∗₁

∩

S∗₂

)

≤

1_kP∗.Hence,addingacopyofitem

i to

bothknapsacks(therebyviolatingtheir capacity constraints)wouldyieldaproﬁtlargerthan

P

∗

(

1

−

2_k

)

.(Thefactor1

−

2_k takescareofthepossiblelossinthedouble-packing case.)Thus,intotal,whenskippingitem

i,

wecanensureatotalproﬁtofatleastP∗

(

1

−

2_k

)

−

2pi

≥

P∗

(

1

−

4_k

)

.

2

Remark1.Weliketopointoutthatnegativeitemscannotbetreatedinsuchaneasyway:Consider,forexamplealistof2n itemswithpi

=

10,

a

i

= −

5 plus2n items withpi

=

5

,

ai

= −

andwi

=

1 forall 4n items.Anoptimumpackingfortwo knapsacksofcapacity3n eachwouldsingle-packallitemswith

p

i

=

10 anddouble-packallremainingitems.Yet,a simple greedyheuristiclikethe“positiveitemvariant”ofALGwouldalwaysstartdouble-packingthe“highproﬁt”items–atleast wecannotpreventitfromdoingsobyprescribingthenumber

l of

itemstobedouble-packed.

We are left to combinethe two algorithms for negative and non-negative items in order to deal with “mixed” sets ofitems. The onlyway wefound worksin asense by “brute force”:We guesstheamount W₁+ andW₂+ ofcapacityan optimum solutionuses on knapsack 1and2 fornon-negative itemsrespectively andthen splitthe problemaccordingly intoonewithnegativeandone withnon-negativeitems.Inourcase,itissuﬃcienttoapproximate

W

₁+ and

W

₂+ uptoa factor1

/

k, soweactuallyguess

m

i

:=

log1+1/kWi

andrunthealgorithmwithW

˜

_i+

= (

1

+

1

/

k

)

mi insteadof

W

_i+.Thetotal numberofguessesweneedis O

(

klogWi

)

for

i

=

1

,

2,thus

O

(

k2

(

logW1

)(

logW2

))

intotal.

Thisintroduces another possibleloss oforder1

/

k on thetotal proﬁtgainedwithpositive items, so thatafter all,the combinedalgorithmALG willyieldatotal proﬁtP with P

≥ (

1

−

1

/

k

)(

1

−

4

/

k

)

P∗,where P∗ istheoptimumproﬁt. Thus wehaveshown

Theorem1.

For ﬁxed k, there exists a polynomial time algorithm ALG that approximates

(P3)up to a factor

(

1

−

5

/

k

)

in time O

(

n2k+52klog W1log W2

)

.

(7)

4. Approximationalgorithmsfortheleader

Let S1 and S2 be optimalsolutions forthestandard single knapsackinstances withknapsack capacities W1 and W2,

resp., and proﬁts pi for all items. In other words, S1 and S2 denote the support of optimal solutions of the following

problemwith

W

=

W1 and

W

=

W2,resp.:

max n

i=1 pixi s.t. n

i=1 wixi

≤

W

,

xi

∈ {

0

,

1

} ,

i

=

1

,

2

, . . . ,

n

.

(P4)

Let

(

S∗₁

,

S

∗2

)

beanoptimalsolutionofthecooperativerelaxation

(P3)

.Weﬁrstnotethatthevalue

OPT of (P3)

satisﬁes

OPT

=

p

(

S∗₁

)

+

p

(

S∗₂

)

+

2h∗

,

where h∗

:=

i∈S∗₁∩S∗₂

ai

.

(1)

Furthermore,wetriviallyhave

p

(

S∗₁

)

≤

p

(

S1

)

and p

(

S∗2

)

≤

p

(

S2

).

(2)

As

(

S∗₁

,

S

∗₂

)

canbefoundbydynamicprogramming(cf. Section2.1)andcanbeapproximatedarbitrarilycloselybyaPTAS (cf. Section3),weﬁrstpresentanalgorithmbyassumingthat S∗₁

,

S

∗₂

,

S

1and

S

2 canbefoundexactly.Afterwards,weshow

thatweloseafactorof

ifthesesolutionsareapproximatedcorrespondingly(cf. Section4.3).We(ﬁrst)treatthebeneﬁcial andthecompetitivemodelseparately.

4.1. The beneﬁcial model: ai

≥

0

We presenta very simple algorithm and show that the approximation ratios are tight in both cases (W1

≥

W2 and W1

<

W2).

Algorithm1.Packoneof

S

∗₁ andS1 (whicheverresultsinamaximumtotalproﬁt1).

Assume ﬁrstthat player1packs S∗₁ andplayer 2packssome set S.

ˆ

Leth

ˆ

=

_i_∈_S∗

1∩ˆS2ai.Then p

( ˆ

S2

)

+ ˆ

h

≥

p

(

S ∗

2

)

+

h∗.

Denoteby

ALG the

valueobtainedbythealgorithm.Thus ALG

=

p

(

S∗₁

)

+

p

( ˆ

S2

)

+

2h

ˆ

≥

p

(

S∗₁

)

+

p

(

S∗₂

)

+

h∗

+ ˆ

h

.

Since

p

( ˆ

S2

)

≤

p

(

S2

)

,wehave

ˆ

h

≥

p

(

S∗₂

)

−

p

( ˆ

S2

)

+

h∗

≥

p

(

S∗2

)

−

p

(

S2

)

+

h∗

,

implying ALG

≥

p

(

S₁∗

)

+

2p

(

S₂∗

)

−

p

(

S2

)

+

2h∗

.

(3)

Nowassumethatplayer1packs S1.Thenplayer2willgetatleast

p

(

S2

)

bypacking S2,implying ALG

≥

p

(

S1

)

+

p

(

S2

).

Besides,weobservethefollowing“knapsackconstraints”:

p

(

S1

)

≥

p

(

S2

)

if W1

≥

W2

,

(4)

p

(

S1

)

≤

p

(

S2

)

if W1

≤

W2

.

(5)

Let

α

=

ALG

/

OPT. Wederivethefollowinglinearprogramforestimatingtheapproximationratio:

(8)

minimize

α

subject to p

(

S∗₁

)

+

p

(

S∗₂

)

+

2h∗

=

1

,

α

≥

p

(

S∗₁

)

+

2p

(

S∗₂

)

−

p

(

S2

)

+

2h∗

,

α

≥

p

(

S1

)

+

p

(

S2

),

p

(

S1

)

≥

p

(

S∗1

),

p

(

S2

)

≥

p

(

S∗2

),

p

(

S1

),

p

(

S2

),

p

(

S∗1

),

p

(

S∗2

),

h∗

≥

0

and the knapsack constraints hold

.

(6)

Theminimumvalue equals2

/

3 if

W

1

≥

W2 and1

/

2 if

W

1

<

W2,proving that

OPT

/

ALG is atmost3

/

2 resp. 2,matching

thelowerbounds(cf.[1]).

4.2. The competitive model: ai

<

0

Inprinciple,wecouldalsoapply

Algorithm 1

tothiscase:Ifplayer1packs

S

∗1,similartotheaboveargument,wehave ALG

≥

p

(

S∗₁

)

+

2p

(

S∗₂

)

−

p

(

S2

)

+

2h∗

.

In case of packing S1, we want to show that ALG

≥

p

(

S1

)

+

p

(

S2

)

+

2h, where h

=

i∈S1∩S2ai. This is clearly true ifplayer 2 packs S2. Otherwise,player 2packs a set,say, S

ˆ

2,with h

ˆ

=

_i_∈_S₁_∩ˆ_S₂ai,such that p

( ˆ

S2

)

+ ˆ

h

≥

p

(

S2

)

+

h. As p

( ˆ

S2

)

≤

p

(

S2

)

,thisimpliesh

ˆ

≥

h. Thus,

ALG

≥

p

(

S1

)

+

p

( ˆ

S2

)

+

2h

ˆ

≥

p

(

S1

)

+

p

(

S2

)

+

h

+ ˆ

h

≥

p

(

S1

)

+

p

(

S2

)

+

2h

,

asclaimed.

Nowweobtainthefollowinglinearprogramboundingtheapproximationratioof

Algorithm 1

.

min

α

s.t. p

(

S∗₁

)

+

p

(

S∗₂

)

+

2h∗

=

1

,

α

≥

p

(

S∗₁

)

+

2p

(

S∗₂

)

−

p

(

S2

)

+

2h∗

,

α

≥

p

(

S1

)

+

p

(

S2

)

+

2h

,

p

(

S1

)

≥

p

(

S∗1

),

p

(

S2

)

≥

p

(

S∗2

),

p

(

S1

),

p

(

S2

),

p

(

S∗1

),

p

(

S2∗

)

≥

0

,

h

,

h∗

≤

0

.

(7)

Letting p

(

S∗₁

)

=

p

(

S1

)

=

p

(

S2

)

=

1, h

= −

1 and

h

∗

=

p

(

S₂∗

)

=

0,we can easily seethat theoptimal objectivevalue is0.

We observe that in thisworst-case instance the penalty h is too large: A better choice for player 1 isto pack nothing. Then player2mustpackS2,implying

ALG

≥

p

(

S2

)

.Addingthissimpleconstrainttotheaboveprogram yieldsan optimal

objectivevalue0

.

5.

Thus,excludingitemswithlargepenaltyfrom

S

1canimprovetheperformanceofALG,however,itdoesnotyieldatight

approximation yet.The idea forfurtherimprovement isnot only toavoid packingitemswithlarge penalties butalsoto packitemswith

small penalties.

Wedeﬁnethefollowingsets:

S+₂

=

i

∈

S2

| |

ai

| >

pi 2

,

S−₂

=

S2

\

S+₂

,

(8)

where

S

+₂

,

S

−₂ aresetsofitemshavinglargepenaltiesandsmallpenaltiesrespectively.Ouralgorithmcannowbedescribed asfollows.

Algorithm2.Packoneof

S

∗1

,

S1

\

S+2 and

∅

(whicheverresultsinamaximumtotalproﬁt).

Toanalyzeitsperformance,wedistinguishthreecases:

(9)

Case2.Player1packs S1

\

S+2.Weprovethat

ALG

≥

p

(

S1

)

+

p

(

S−2

)

+

2h−,where

h

−

=

i∈S−₂ai.Clearly,thisistruewhen player 2packs S2.Ifplayer 2packssome otherset,say

ˆ

S2

=

S2,withh

ˆ

=

_i_∈(_S₁_\_S+

2)∩ˆS2ai,then p

( ˆ

S2

)

+ ˆ

h

≥

p

(

S2

)

+

h −_, implyingh

ˆ

≥

h− (recallthat

p

( ˆ

S2

)

≤

p

(

S2

)

).Hence,

ALG

≥

p

(

S1

)

−

p

(

S2+

)

+

p

( ˆ

S2

)

+

2h

ˆ

≥

p

(

S1

)

−

p

(

S+₂

)

+

p

(

S2

)

+

h−

+ ˆ

h

≥

p

(

S1

)

−

p

(

S₂+

)

+

p

(

S2

)

+

2h−

=

p

(

S1

)

+

p

(

S−₂

)

+

2h−

.

Case3.Player1packsnothing.Thenplayer2canguaranteeatotalproﬁt p

(

S2

)

bypacking

S

2.Thus,

ALG

≥

p

(

S2

)

.

Furthermore,inadditiontotheknapsackconstraints

(4)

,

(5)

wehavethefollowingconstraints:

p

(

S2

)

=

p

(

S+₂

)

+

p

(

S−₂

)

and h−

≥ −

p

(

S−₂

)

2

.

Thisﬁnallyyieldsthefollowinglinearprogramwithanoptimalvalueof2

/

3 (forboth

W

1

≥

W2 and

W

1

<

W2).

minimize

α

subject to p

(

S∗₁

)

+

p

(

S∗₂

)

+

2h∗

=

1

,

α

≥

p

(

S∗₁

)

+

2p

(

S₂∗

)

−

p

(

S2

)

+

2h∗

,

α

≥

p

(

S1

)

+

p

(

S−2

)

+

2h−

,

α

≥

p

(

S2

),

p

(

S2

)

=

p

(

S+2

)

+

p

(

S−2

),

h−

≥ −

p

(

S − 2

)

2

,

p

(

S1

)

≥

p

(

S∗1

),

p

(

S2

)

≥

p

(

S∗2

),

p

(

S1

),

p

(

S2

),

p

(

S+2

),

p

(

S−2

),

p

(

S∗1

),

p

(

S∗2

)

≥

0

,

h∗

,

h−

≤

0

.

(9)

Hence

OPT

/

ALG

≤

1

.

5,whichistight(cf. theexampleinSection2).

Remark2. Since player2 actsafterplayer 1,thereadermaywonder how toﬁnd themaximal totalproﬁtover all cases (in

Algorithms 1 and

2).Thiscanbedonebycheckingtheinequalities“

α

≥ . . .

”in

(6)

and

(9)

respectively:Ifanyofthese inequalitiesgetstight,thealgorithmmaypicktheassociatedset(forplayer1).

4.3. Approximating S∗₁and S∗₂

Algorithms 1 and 2 mayequally well be applied, ifwe replace S1

,

S

2

,

S

∗1 and S∗2 by corresponding

-approximations

(to be computed as explained in Section 2.1, say, S

˜

1

,

˜

S2

,

˜

S∗1 and S

˜

∗2 with corresponding S

˜

+2

= {

i

∈ ˜

S2

| |

ai

|

>

pi

/

2

}

and

˜

S−₂

= ˜

S2

\˜

S₂+). A close lookat the analysisof Algorithms 1 and 2 revealsthat optimality of

(

S∗₁

,

S

∗₂

)

is only used in (1)

and(3).Itiseasytoseethat

(1)

becomes

(

1

−

)

OPT

≤

p

( ˜

S₁∗

)

+

p

( ˜

S₂∗

)

+

2h

˜

∗

,

whereh

˜

∗

=

i∈˜S∗₁∩ ˜S∗₂ ai

.

Notethatthe“

˜

-version”of

(2)

maybeassumedtobesatisﬁedwithoutanychanges.(Replacetheapproximation S

˜

1 by

˜

S∗₁ incasep

( ˜

S1

)

<

p

( ˜

S∗1

)

.)Thesameis trueforthe knapsackconstraints.Thusall thatchanges inthelinearprograms in

theprevioustwosectionsisthattherighthandsideoftheﬁrstconstraintchangesfrom1 to

(

1

−

)

.Thus,asimplescaling argumentshowsthattheresultingoptimumwilldifferbyatmostan O

(

)

-fractionfromtheoriginalvalue.

Remark3.To be precise, theabove argument is validonly ifwe assume that player 2 is ableto solve his (lowerlevel) problemexactly. Ifalsoplayer 2applies

-approximation algorithmsinstead, weget anotherfactorof

(

1

−

)

introduced in(3).

(10)

4.4. The mixed case

Finally,letusturntothecasewheretheitemsetcontains bothpositiveandnegativeitems.Thiscaseiseasilyreduced tothe twocases(beneﬁcialandcompetitive modelrespectively)considered above: Allwehaveto doistosplit theitem set A into theset A+and A−ofnon-negativeandnegativeitems,respectivelyandtoguess–again,uptoacertainfactor, say, 1

/

k – the associated parts of knapsacks 1 and2 that are ﬁlled with non-negative andnegative items, resp., in an optimalsolution

S

∗1

∪

S∗2of

(P3)

.Solvingthe“pure”(beneﬁcialresp.competitive)casesfortheapproximatelycorrectchoice

ofcorrespondingknapsackcapacitieswillthengivesolutionsfortheleader’sproblemthatdifferfromthecooperativevalue byatmostafactorof

(

2

+

)

attheexpenseofanadditionalO

((

log W1

)(

log W2

))

factorintherunningtime.

5. Remarks

Wehavepresentedauniﬁedapproachforcomputingsolutionstotheleader’sproblemwithapproximatelyoptimalratio, ifcomparedtotheoutcomeofthecooperativeversionoftheproblem.Withoutfurthergoingintodetails,wementionthat inthelowerboundexamples(cf. theexampleinSection2andtheexamplesin

[1]

),themaximumcooperativevalueequals themaximumvalue of(the optimistic versionof) thebilevel problem.Thus ourresultsalso providetightboundsforthe ratiobetweentheoptimisticandpessimisticversionofthebilevelproblemitself.

A natural question to ask is aboutside payments: Player 1 can certainly enforce the optimistic value of the bilevel problemwitharbitrarily small side payments.Thus, it seemsnatural toalso investigateapproximation algorithms inthe optimisticsetting.

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