ch05solutionssolved editWord文档格式.docx
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⇒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=
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