tipsheet for itm107 final exam review · operation: interchange row 1 and row 3 2) add multiples of...
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Chapter3.3(Gauss-JordanEliminationMethod)
• TheprocessusedtosolveasystemofequationsiscalledtheeliminationmethodorGauss-JordanEliminationMethod
• Thegoalistoreducethecoefficientmatrixtoanidentitymatrix,sowecanreadtheanswerfromtherightcolumn.
• Therearethreedifferentoperationsthatcanbeusedtoreducethematrixwhicharecalledelementaryrowoperations
1) Interchangetworows2) Addamultipleofonerowtoanotherrow3) Multiplyarowbyanonzeroconstant
Example
Solvethefollowinglinearsystem:
Thesystemcanberepresentedbytheaugmentedmatrix:
Steps:
1) Get a 1 in row 1, column 1:Operation: Interchange row 1 and row 3
2) Add multiples of row 1 to the other two rows (row 2 & row 3)to get zeros in the other entry of column 1:
Operation: (2 x R1) + R2 à R2 =>
(-1 x R1) + R3 à R3 =>
3) Use rows below row 1 to get a 1 in row 2, column 2:Operation: (-1/3 x R2) à R2 =>
4) Add multiples of row 2 to the other two rows (row 1 and row 3)to get zeros in the other entry of column 2:
Operation: (-4 x R2) + R1 à R1 =>
(7 x R2) + R3 à R3 =>
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5) Use rows below row 2 to get a 1 in row 3, column 3:Operation: (-3 x R3) à R3 =>
6) Add multiples of row 3 to the other two rows (row 1 and row 2)to get zeros in the other entry of column 3:
Operation: (-1/3 x R3) + R1 à R1 =>
(-2/3 x R3) + R2 à R2 =>
7) Repeat the process until it cannot be continued:Operation: Since all rows have been used, the matrix is in its reduced form
Initial: 2x + 5y + 4z = 4 Final: x + 0y + 0z = 3 x = 3
x + 4y + 3z = 1 0x + y + 0z = -2 => y = -2
x – 3y -2z = 5 0x + 0y + z = 2 z = 2
Chapter 3.4 (Inverse of a square matrix)
Equation:
If A = , then A-1 =
Example:
Find the inverse of A =
Solution:
A-1 =
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Chapter4.1(LinearInequalitiesintwoVariables)
Example:
• Graphthefollowingsolution• Beginbygraphingtheequations3x–2y=4andx+y=3(fromx+y–3=0)
bytheinterceptmethod:Findywhenx=0andfindxwheny=0.
• (1)x=0ày=-2,y=0àx=¾;(2)x=0ày=3,y=0àx=3• Wegraph3x–2y=4asasolidlineandx+y=3asadashedline• Thepointsthatsatisfybothoftheseinequalitiesliein
theintersectionofthetwoindividualsolutionregions
• Whenthetwolinesforma“corner”,thepointssatisfythetwoinequalities
Example:
Findthemaximumandminimumvalues(iftheyexist)ofC=x+ysubjecttotheconstraints:
3x+2y>=12; x+3y>=11; x>=0,y>=0
• Notethatthefeasibleregionisnotclosedandbounded,so,wemustcheckwhetheroptimalvaluesexist.
ThischeckisdonebygraphingC=x+yforselected
valuesofCandnotingthetrend.
• Thecorners(0,6)and(11,0)canbeidentifiedfromthegraph.Thethirdcorner,(2,3),canbefoundbysolvingtheequations
3x+2y=12andx+3y=11 simultaneously.
3x+2y=12
(x-3) -3x–9y=-33
-7y=-21=>y=3;x=12–2(3)/3=>6/3=>x=3
• ExaminingthevalueofCateachcornerpoint,wehaveAt(0,6)C=x+y=>0+6=>6;At(11,0)C=x+y=>11+0=>11;At(2,3)C=x+y=>2+3=>5
MinimumValueofC=x+yis5at(2,3);MaximumValueofC=x+ydoesnotexist
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SensitivityAnalysis
• SensitivityAnalysisistheprocedureofdeterminingthemarginaleffectofinputchangesontheoptimalsolutionofamodel
• Therearetwodifferentsensitivityanalysischangescanbeobserved1) ObjectiveFunctionCoefficient(OFC)2)Constraints’right-handside(RHS)
Example(ImpactofOFC)
OriginalObjectiveFunction:7x+5y(Profit)
SubjectTo:3x+4c<=2400(CarpentryHours)
2x+y<=1000(PaintingHours)
y<=450(Max#ofChairs)
x<=100(Min#oftables)
x>=0,y>=0(Nonnegativity)
Whatifprofitcontributionfortableschangedfrom$7(7x)to$8(8x).Howwouldthisaffecttheoptimalsolution:
NewObjectiveFunction:8x+5y
OriginalOptimalPoint(320,360)
OriginalValue->7(320)+5(360)=4040
RevisedOptimalPoint(320,360)
RevisedValue->8(320)+5(360)=4360
Thereisnoeffectonthefeasibleregion
Howevertheslopeoftheisoprofitlinechanges
ImpactofRHSChanges
• ImpactofRHSchangesdependsoniftheconstraintisbindingornonbinding.• Bindingconstraintspassthroughtheoptimalcornerandhaveazeroslack• Nonbindingconstraintshaveanonzeroslackorsurplusvalueattheoptimalsolution.• Slackisthedifferentbetweentheright-handsideandleft-handsideofconstraint.• Surplusisthedifferentbetweentheright-handsideandlefthandsideofaconstraint.
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ShadowPrice
• Theshadowpriceofaconstraintisthechangeintheoptimalvalueoftheobjectivefunctionforaone-unitincreaseintheRHSofthatconstraint.(Valueofanextraunitofresource)
Example
Paintinghoursisincreasedto1300hours:
NewProfit=$4820
OldProfit=$4040
ProfitIncrease=$780
The300additionalpaintinghoursbrings$780inprofits.Eachadditionalpaintinghourwillincreasetheprofitby:$780/300=>$2.60àShadowprice=>$2.60
Chapter2.1(QuadraticFunction)
• ThegeneralequationofaQuadraticFunctionisY=f(x)=ax2+bx+cWherea,bandcarerealnumbersanda=0
Twomethodsofsolvingquadraticequations
1. Factoring2. TheQuadraticFormula
Example
1) Solve:(y–4)x(y+3)=8
Step1:y2–y–12=8
Step2:y2–y–20=0
Step3:(y–5)(y+4)=0=>Y=5;Y=-4
Note:
• Whensolvingaquadraticequation,wecanusethesignoftheradicandinthequadraticformula• Referb2–4acasthequadraticdiscriminant• Ifb2–4ac>0àEquationhastwodistinctrealsolutions• Ifb2–4ac=0àEquationhasexactlyonerealsolution• Ifb2–4ac<0àEquationhasnorealsolutions
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Chapter2.2(QuadraticFunctions;Parabola)
Example
Forthefunctiony=4x–x2,determinewhetheritsvertexismaximumpointoraminimumpointandfindthecoordinatesofthispoint
TheProperform:y=-x2+4x=0àa=-1àparabolaopensdownward(maximum)
Thevertexoccursatàx=-b/2a=-4/2(-1)=2
They-coordinatesofthevertexisf(2)=-(2)2+4(2)=4=>Coordinatesare(2,4)
• Wecantranslatetheparabolaverticallytoproduceanewparabolathatissimilartothebasicparabola.• Thefunctiony=x2+bhasagraphwhichsimplylookslikethestandardparabolawiththevertexshiftedbunits
alongthey-axis.Thevertexwillbelocatedat(0,b)• Ifbispositive,thentheparabolamovesupward• Ifbisnegative,thentheparabolamovesdownward
Chapter5.1(ExponentialFunctions)
• Exponentialfunctionsareafunctionoftheformwherebisapositiverealnumber,andinwhichtheargumentxoccursasanexponent.
Example:
Graphy=10x
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GraphofExponentialGrowthFunction GraphsofExponentialDecayFunctions
Function:y=f(x)=C(ax)(C>0,a>1) Function:y=f(x)=C(a-x)(C>0,a>1)or
y-intercept:(0,C) y=f(x)=C(bx)(C>0,0<b<1)
GraphShape: y-intercept:(0,C)
GraphShape:
Domain:Allrealnumbers;Range:y>0 Domain:Allrealnumbers;Range:y>0
Asymptote:Thex-axis(negativehalf) Asymptote:Thex-axis(positivehalf)
Chapter5.2(LogarithmicFunctionsandtheirProperties)
• LogarithmicFunctionsaretheinverseofanexponentialfunctionLogarithmicandExponentialFormTable
PropertiesofLogarithms
PropertyI:Ifa>0,a=/1,thenlogaax=x,foranyrealnumberx
• Toprovethisresult,weusetheexponentialformofy=logaaxisay=ax,soy=x.Thisis,logaax=x.
Ex:(a)log443=3
PropertyII:Ifa>0,a=/1,thenalogax=x,foranypositiverealnumberx
UsePropertyIItosimplifyeachofthefollowing.
Ex:2log24=4
PropertyIII:Ifa>0,a=/1,andMandNarepositiverealnumbers,then
Loga(MN)=logaM+logaN
Ex:log2(4x16)=log24+log216=2+4=6
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PropertyIV:Ifa>0,a=/1,andMandNarepositiverealnumbers,then
Loga(M/N)=logaM-logaN
Ex:Log3(9/27)=log39–log327=2-3=-1
PropertyV:Ifa>0,a=/1,MisapositiverealnumberandNisanyrealnumber,then
Loga(MN)=NlogaM
Ex:Log3(92)=2log39=2x2=4
Chapter7.1(Probability;Odds)
IfaneventEcanoccurinn(E)=kwaysoutofn(S)=nequallylikelyways,then:Pr(E)=n(E)/n(S)=k/n
Example
Ifanumberistobeselectedatrandomfromtheintegers1through12,whatistheprobabilitythatitisdivisibleby4?
Answer:ThesetS={1,2,3,4,5,6,7,8,9,10,11,12}isanequiprobablesamplespaceforthisexperimentThesetE={4,8,12}containsthenumbersthataredivisibleby4.Thus
Pr(divisibleby3)=n(E)=3=1
n(S)124
IfaneventEiscertaintooccur,Econtainsalloftheelementsofthesamplespace,S.
HencethesumoftheprobabilityweightsofEisthesameasthatofS,so
Pr(E)=1 ifEiscertaintooccur
IfeventEisimpossible, Pr(E)=0, ifEisimpossible
Example:Supposeacoinistossed3times.
(a) Constructanequiprobablesamplespacefortheexperiment.
(b) Findtheprobabilityofobtaining0heads.
(c) Findtheprobabilityofobtaining2heads
(a) {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT},whereHHTindicatesthatthefirsttwotosseswereheadsandthethirdwasatail.
(b) Becausethereare8equallylikelypossibleoutcomes,n(S)=8.Onlyoneoftheeightpossibleoutcomes,E={TTT},gives0heads,son(E)=1.ThusPr(0heads)=n(E)/n(S)=1/8
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(c) Theevent“twoheads”isF={HHT,HTH,THH},soPr(2heads)=n(F)=3
OddsandProbability
• Wesometimesuseoddstodescribethelikelihoodthataneventwilloccur.TheoddsinfavorofaneventEoccurringandtheoddsagainstEoccurringarefoundasfollows.
IftheprobabilitythateventEoccursinPr(E)=/1,thentheoddsthatEwilloccurare
OddsinfavourofE=Pr(E)/1-Pr(E)
TheoddsthatEwillnotoccur,ifPr(E)=/0,areOddsagainstE=1-Pr(E)/Pr(E)
Iftheprobabilityofdrawingaqueenfromadeckofplayingcardsis1/13,whataretheodds(a) infavorofdrawingaqueen?(b)againstdrawingaqueen?
(a) 1/13=1/13=1
1–1/1312/13=12Theoddsinfavorofdrawingaqueenare1to12,whichwewriteas1:12.
(b) 12/13=12
1/131Theoddsagainstdrawingaqueenare12to1,denoted12:1.
Chapter7.2(UnionsandIntersectionofEvents:One-TrialExperiments)
IfEandFaretwoeventsinasamplespaceS,thenthe
TheintersectionofEandFisEÇF={a:aÎEandaÎF}
TheunionofEandFisEÈF={a:aÎEoraÎF}
ThecomplementofEisE¢={a:aÎSandaÏE}
Example
Acardisdrawnfromaboxcontaining15cardsnumbered1to15.Whatistheprobabilitythatthecardis(a)evenanddivisibleby3?
a) IfweletErepresent“even-numbered”andDrepresent“numberdivisibleby3,”wehave
E={2,4,6,8,10,12,14)andD={3,6,9,12,15}
MutuallyExclusiveEvents
WesaythateventsEandFaremutuallyexclusiveifandonlyifEÇF=f.
ThusPr(EÈF)=Pr(E)+Pr(F)–0=Pr(E)+Pr(F)
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Example
Afirmisconsideringthreepossiblelocationsforanewfactory.TheprobabilitythatsiteAwillbeselectedis1/3andtheprobabilitythatsiteBwillbeselectedis1/5.
Onlyonelocationwillbechosen.
(a) WhatistheprobabilitythatsiteAorsiteBwillbechosen?
(a) Thetwoeventsaremutuallyexclusive,so;
Pr(SiteAorSiteB)=>Pr(SiteAÈSiteB)
=>Pr(SiteA)+Site(B)
=>1/3+1/5=>8/15
Chapter7.3(ConditionalProbability:TheProductRule)
TheConditionalProbabilitythatAoccurs,giventhatBoccursisdenotedasPr(A|B).Isgivenby:Pr(A|B)=n(AÇB)/n(B)
TheProductRule
IfAandBareprobabilityevents,thentheprobabilityofevent“AandB”isPr(AandB),anditcanbefoundbyoneortheotherofthesetwoformulas
1) Pr(AandB)=Pr(AÇB)=Pr(A)xPr(A|B)
2) Pr(AandB)=Pr(AÇB)=Pr(B)xPr(A|B)
IndependentEvents
TheeventsAandBareindependentifandonlyif:
Pr(A|B)=Pr(A)andPr(B|A)=Pr(B)
Example
Abagcontains4redmarbles,5whitemarblesand3blackmarbles.Findtheprobabilityofgettingaredmarbleonthefirstdraw,ablackmarbleontheseconddraw,andawhitemarbleonthethirddraw.
a) Ifthemarblesaredrawnwithreplacement
b) Ifthemarblesaredrawnwithoutreplacement
a) Themarblesarereplacedaftereachdraw,sothecontentsarethesameoneachdraw.Thustheeventsare
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independent.(LetE1be“redon1st”,E2be“blackon2nd”,E3be“whiteon3rd”
Pr(E1ÇE2ÇE3)=Pr(E1)xPr(E2)xPr(E3)=>4/12x5/12x3/12=60/1000=>15/250
b) Themarblesarenotreplaced,sotheeventsareindependent
Pr(E1ÇE2ÇE3)=Pr(E1)xPr(E2|E1)xPr(E3|E1andE2)=>4/12x5/11x3/9=>15/297
Chapter7.4(ProbabilityTreesandBayesFormula)
• Probabilitytreesprovideasystematicwaytoanalyzeprobabilityexperimentsthathavetwoormoretrialsorthatusemultiplepathswithinthetree.
Example
Abagcontains5redballs,4blueballsand3whiteballs.Twoballsaredrawnoneaftertheotherwithoutreplacement.Drawatreerepresentingtheexperimentandfind:
a) Pr(blueonfirstdrawandwhiteonseconddraw) b) Pr(whiteonbothdraws)
c) Pr(drawingablueballandwhiteball) d) Pr(secondballisred)
a) Pr(blueon1standwhiteon2nd)=4/12x3/11=>1/11
b) Pr(whiteon1stand2nd)=3/12x2/11=>1/22
c) Pr(firstBlue,thenWhite)+Pr(firstWhite,thenBlue)
=>(4/12x3/11)+(2/12x4/11)=>2/11
d) Pr(R,thenR)+Pr(B,thenR)+Pr(W,thenR)=>(5/12x4/11)+(4/12x5/11)+ (3/12x5/11)=>5/12
InBayesproblems,weknowtheresultofthesecondstageofatwo-stageexperimentandwanttofindtheprobabilityofaspecifiedresultinthefirststage.
TheprobabilitythatE1occursinthefirststage,giventhatF1hasoccurredinthesecondstage,
isPr(E1|F1)=Pr(E1ÇF1)/Pr(F1)
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BayesFormulaandTrees=ProductofbranchprobabilitiesonpathleadingtoF1throughE1
SumofallbranchproductsonpathsleadingtoF1
SummaryofProbabilityandFormulas
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