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SolutionstoChapter5Exercises

SOLVEDEXERCISES

S1.（a）R’sbest-responseruleisgivenbyy=10√x–x.Lspends$16million,sox=16.ThenR’sbestresponseisy=10√16–16=10（4）–16=40–16=24,or$24million.

（b）R’sbestresponseisy=10√x–x,andL’sbestresponseisx=10√y–y.Solvethesesimultaneously:

x=10（10√x–x）1/2–10√x+x

√x=（10√x–x）1/2

⇒x=10√x–x

⇒2x=10√x

⇒√x=5

⇒x=25

y=10√25–25=25

S2.（a）Xavier’scostshavenotchanged,norhavethedemandfunctions,soXavier’sbest-responseruleisstillthesameasinFigure5.1:

Px=15+0.25Py.Yvonne’snewprofitfunctionisBy=（Py–2）Qy=（Py–2）（44–2Py+Px）=–2（44+Px）+（4+44+Px）Py–2（Py）2.RearrangingordifferentiatingwithrespecttoPyleadstoYvonne’snewbest-responserule:

Py=12+0.25Px.SolvingthetworesponserulessimultaneouslyyieldsPx=19.2andPy=16.8.

（b）Seethegraphbelow.Yvonne’sbest-responsecurvehasshifteddown;ithasthesameslopebutanew,lowerintercept（12ratherthan15）.Yvonneisabletochargelowerpricesduetolowercosts.Thenewintersectionpointoccursat（19.2,16.8）ascalculatedabove.

S3.（a）LaBoulangerie’sprofitis:

Y1=P1Q1–Q1=P1（14–P1–0.5P2）–（14–P1–0.5P2）=–P12+15P1–0.5P1P2+0.5P2–14.

TofindtheoptimalP1withoutusingcalculus,werefertotheresultintheAppendixtoChapter5,rememberingthatP2isaconstantinthissituation.UsingthenotationoftheAppendix,wehaveA=0.5P2–14,B=15–0.5P2,andC=1,sothesolutionis:

P1=B/（2C）=15–0.5P2/

（2）,orP1=7.5–0.25P2.

ThisisLaBoulangerie’sbest-responsefunction.Yougetthesameanswerbysetting

=–2P1+15–0.5P2=0andsolvingforP1.

Similarly,LaFromagerie’sprofitis:

Y2=P2Q2–2Q2=P2（19–0.5P1–P2）–2（19–0.5P1–P2）=–P22+21P2–0.5P1P2+P1–38.

Again,usingthenotationintheAppendix,A=P1–38,B=21–0.5P1,andC=1,whichyields:

P2=B/（2C）=21–0.5P1/

（2）,orP2=10.5–0.25P1.

ThisisLaFromagerie’sbest-responsefunction.Yougetthesameanswerbysetting

=–2P2+21–0.5P1=0andsolvingforP1.

Tofindthesolutionfortheequilibriumpricesanalytically,substituteLaFromagerie’sbest-responsefunctionforP2intoLaBoulangerie’sbest-responsefunction.ThisyieldsP1=7.5–0.25（10.5–0.25P1）,orP1=5.2.GiventhisvalueforP1,youcanfindP2=10.5–0.25（5.2）=9.2.Thebest-responsecurvesareshowninthediagrambelow.

（b）Colludingtosetpricestomaximizethesumofprofitsmeansthatthefirmsmaximizethejoint-profitfunction:

Y=Y1+Y2=16P1+21.5P2–P12–P22–P1P2–52.

Toanswerwithoutcalculus,usetheresultintheAppendix.UsingthatnotationtosolveforP1,A=21.5P2–P22–52,B=16–P2,andC=1,sothesolutionis:

P1=B/（2C）=16–P2/

（2）,orP1=8–0.5P2.

Similarly,solvingforP2,A=16P1–P12–52,B=21.5–P1,andC=1,sothesolutionis:

P2=B/（2C）=21.5–P1/

（2）,orP2=10.75–0.5P2.

SolvingthesetwoequationssimultaneouslyyieldsthesolutionP1=3.5andP2=9.

Youcangetthesameanswerbypartiallydifferentiatingthejoint-profitfunctionwithrespecttoeachprice.ProfitsmustbemaximizedwithrespecttobothP1andP2,soweneed

=16–2P1–P2=0and

=21.5–P1–2P2=0.

（c）Whenfirmschoosetheirpricestomaximizejointprofit,theyactasasinglefirmandignoreanyindividualincentivesthattheymighthavetodeviatefromthejointprofitgoal.However,giventheirpartner’scollusiveprice,eachcompanycanreapmoreprofitindividuallybychargingmore.Forinstance,pluggingthejoint-profit-maximizingvalueofLaBoulangerie’spriceintoLaFromagerie’sindividualbest-responserulewillnotyieldLaFromagerie’sjointprofit-maximizingprice:

P2=10.5–0.25（3.5）=9.6259.

Likewise,pluggingthejoint-profit-maximizingvalueofLaBoulangerie’spriceintoLaFromagerie’sindividualbest-responserulegives:

P1=7.5–0.25（9）=4.753.5.

Thus,thetwojointprofit-maximizingpricesarenotbestresponsestoeachother;thatis,theydonotformaNashequilibrium.

（d）Whenfirmsproducesubstitutes,adropinpriceatonestorehurtsthesalesoftheother.Thus,asyourrivaldropsherprice,youalsowanttodropyourstoattempttomaintainsales（andprofits）.InthebistroexampleinthetextandinExercise1above,thisresultledtobest-responsecurvesthatwerepositivelyslopedandNashequilibriumpricesthatwerelowerthanthejoint-profit-maximizingprices.Here,thefirmsproducecomplements,soadropinpriceatonestoreleadstoanincreaseinsalesattheother.Inthiscase,asonestoredropsitsprice,theothercansafelyincreaseitspricesomewhatandstillmaintainsales（andprofits）.Thus,thebest-responsecurvesarenegativelysloped,andtheNashequilibriumpricesarehigherthanthejoint-profit-maximizingprices.

S4.Torationalizetheninepossibleoutcomes,youneedaseparateargumentforeachone.Weofferjustoneexample,leavingyoutoconstructtherest.Notethatyouneednotconsiderthestrategycombination（A,A）sincethatisaNashequilibriumandthereforerationalizable.Consider（A,C）leadingtothepayoffs（0,2）.CisapossiblebestresponseforColumnifhethinksthatRowisplayingA.WhydoesColumnbelievethis?

BecausehebelievesthatRowbelievesthatColumnisplayingB.ColumnjustifiesthisbeliefbythinkingthatRowbelievesthatColumnbelievesthatRowisplayingC.Thebeliefsinthischainareallperfectlyrationalbecauseeachstrategyofeitherplayerisabestresponse（oramongthebestresponses）tosomestrategyoftherivalplayer.

S5.NomatterwhatbeliefsColumnmightholdaboutwhatRowisplaying,SouthisneverColumn’sbestresponse.ThereforeSouthisnotarationalizablestrategyforColumn.SinceRowrecognizesthis,andsinceEarthisRow’sbestresponseonlyagainstColumn’sSouth,RowdoesnotplayEarth.SinceNorthisColumn’sbestresponseonlyagainstEarth,ColumnwillnotplayNorth.SinceColumnwillneverplayNorthorSouth,WindisneverabestresponseforRow.Theremainingstrategies—WaterandFireforRowandEastandWestforColumn—areusedinthetwopure-strategyNashequilibria,sotheyarecertainlyrationalizable.

S6.Usingthethird-roundrangeofX,wehavethatY=12–X/2mustbeatmost12–9/2=12–4.5=7.5（nothingnewhere）andatleast12–12.75/2=12–6.375=5.625（anarrowingoftherange:

afterthethirdround,thelowerboundwas4.5）.Similarly,usingthethird-roundrangeofY,wefindthatX=15–Y/2mustbeatmost15–4.5/2=15–2.25=12.75（nothingnew）andatleast15–7.5/2=15–3.75=11.25（anarrowingoftherange:

inthethirdround,thelowerboundwas9）.Weseethatinevenroundsthelowerboundsgettighter,andinoddroundstheupperboundsgettighter.

S7.（a）Cart0servesxcustomersandCart1serves（1–x）,wherexisdefinedbytheequationp0 +0.5x2=p1+0.5（1–x）2.Expandingthisequationyieldsp0+0.5x2=p1+0.5–x+0.5x2,andsolvingforxyieldsx=p1–p0+0.5.ThusCart0servesp1–p0+0.5customers,andCart1serves1–x,orp0–p1+0.5,customers.

（b）ProfitsforCart0are（p1–p0+0.5）（p0–0.25）.ProfitsforCart1aresymmetric:

（p0–p1+0.5）（p1–0.25）.ExpandingtheexpressionforCart0profitsyields（p1–p0+0.75）p0–（0.25p1+0.125）.Solvingfortheprofit-maximizingvalueofp0bycompletingthesquareordifferentiatingwithrespecttop0yieldsp0=0.5p1+0.375.Cart1’sbest-responseruleissymmetric:

p1=0.5p0+0.375.

（c）Thegraphisshownbelow.TheNashequilibriumpricesarethevaluesofp0andp1thatsolvesimultaneouslythetwobest-responserulesfoundinpart（b）.SubstitutingCart1’sbest-responseruleintothatforCart0,wefind:

p0=0.5（0.5p0+0.375）+0.375=p0+0.5625.Solvingforp0yieldsp0=0.75（75¢）;Cart1’spriceisp1=0.75（75¢）also.

S8.（a）SouthKorea’sprofitis

YKorea=qKorea*P–cKorea*qKorea=qKorea（180–Q）–30qKorea

=qKorea（180–qKorea–qJapan）–30qKorea=–qKorea2+（180–30）qKorea–qKorea*qJapan.UsingthenotationintheAppendixyieldsA=0,B=150–qJapan,andC=1,soSouthKorea’sbestresponseis:

qKorea=B/（2C）=150–qJapan/

（2）,orqKorea=75–0.5qJapan.

ThisisSouthKorea’sbest-responsefunction.Yougetthesameanswerbysetting

=–2qKorea+150–qJapan=0andsolvingforqKorea.

SinceJapanhasthesamepriceandcostpershipasSouthKorea,Japan’sprofitis

YJapan=–qJapan2+（180–30）qJapan–qKorea*qJapan.

Similarly,Japan’sbest-responsefunctionis:

qJapan=75–0.5qKorea.

（b）Tofindthesolutionfortheequilibriumprices,substituteJapan’sbest-responsefunctionforqJapanintoSouthKorea’sbest-responsefunction.Thisyields:

qKorea=75–0.5qJapan=75–0.5（75–0.5qKorea）=37.5+0.25qKorea

qKorea=50

Therefore:

qJapan=75–0.5（50）=50.

ThepriceofaVLCCisgivenbytheexpressionP=180–QwhereQ=qKorea+qJapan=50+50=100.ThereforeP=180–100=80,or$80million.

SouthKorea’sprofitisYKorea=–qKorea2+（180–cKorea）qKorea–qKorea*qJapan=–（50）2+（180–30）（50）–（50）（50）=–2500+7500–2500=2500.

Likewise,Japan’sprofitisYJapan=–qJapan2+（180–cJapan）qJapan–qKorea*qJapan=2500.Thereforebothcountriesmake$2.5billioninprofits.

（c）SouthKorea’snewbest-responsefunctionis:

qKorea=90–0.5cKorea–0.5qJapan=90–0.5（20）–0.5qJapan=80–0.5qJapan.

Japan’snewbest-responsefunctionis:

qJapan=90–0.5cJapan–0.5qKorea=