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 =>

TIPSHEET FOR ITM107: MANAGERIAL DECISION MAKING

ITM107 FINAL EXAM REVIEW TEXTBOOK: Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition

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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:

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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

• 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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