ie 400 principles of engineering management graphical ... · graphical solution of 2-variable lp...

IE400PrinciplesofEngineeringManagement

GraphicalSolutionof2-variableLPProblems


maxx1 +3x2s.t.

x1+x2 ≤6(1)

- x1+2x2 ≤8(2)

x1,x2 ≥0(3),(4)

Ex1.a)


maxx1+3x2s.t.

x1+x2 ≤6(1)

- x1+2x2 ≤8(2)

x1,x2 ≥0(3),(4)

General Form:max cx

s.t. Ax ≤ b

x ≥ 0

[ ]

úû

ùêë

é=ú

û

ùêë

é=

÷÷ø

öççè

æ==

86

b 2 1-1 1

A

xx

x 3 1c

where,

2

1

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

FeasibleRegion

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0) (6,0)

(0,6)(4/3,14/3)

x1 +3x2 =3

x1 +3x2 =6

x1 +3x2 =0

Recalltheobjectivefunction:

maxx1 +3x2

Alineonwhichallpointshavethesamez-value(z=x1+3x2)iscalled:

§ Isoprofitlineformaximizationproblems,

§ Isocostlineforminimizationproblems.


Considerthecoefficientsofx1 andx2 intheobjectivefunctionandletc=[13].


(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0) (6,0)

(0,6)(4/3,14/3)

x1 +3x2 =3

x1 +3x2 =6

x1 +3x2 =0

objectivefunction:maxx1 +3x2

c=[13]

Notethataswemovetheisoprofit linesinthedirectionofc,thetotalprofitincreases!


Foramaximizationproblem,thevectorc,givenbythecoefficientsofx1 andx2 intheobjectivefunction,iscalledtheimprovingdirection.

Ontheotherhand,-c istheimprovingdirectionforminimizationproblems!


TosolveanLPproblemintwovariables,weshouldtrytopushisoprofit (isocost)linesintheimprovingdirectionasmuchaspossiblewhilestillstayinginthefeasibleregion.


(9,0)(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0) (6,0)

(0,6)(4/3,14/3)

x1 +3x2 =3

x1 +3x2 =6

x1 +3x2 =0

Objective:maxx1 +3x2

c=[13]

x1 +3x2 =9(3,0)

(1,0)

(0,2)

(0,3)

(9,0)(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0) (6,0)

(0,6)(4/3,14/3)


x1 +3x2 =9

(0,3)c=[13]

c=[13]

(12,0)(9,0)(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0) (6,0)

(0,6)(4/3,14/3)


c=[13]

x1 +3x2 =9

(0,3)

c=[13]

x1 +3x2 =12

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0) (6,0)

(0,6)(4/3,14/3)


x1+3x2=46/3

c=[1,3]

(6,0)(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(0,6)(4/3,14/3)


c=[1,3]

Canweimprove(6,0)?

(6,0)(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(0,6)(4/3,14/3)


c=[13]

Canweimprove(0,4)?

(6,0)(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(0,6)(4/3,14/3)


c=[13]Canweimprove(4/3,16/3)?No!No!Westophere

(6,0)(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(0,6)(4/3,14/3)

Objective:maxx1 +3x2c=[13]

(4/3,14/3)istheOPTIMALsolution


Point(4/3,14/3)isthelastpointinthefeasibleregionwhenwemoveintheimprovingdirection.So,ifwemoveinthesamedirectionanyfurther,thereisnofeasiblepoint.

Therefore,pointz*=(4/3,14/3)isthemaximizingpoint.TheoptimalvalueoftheLPproblemis46/3.


Ex1.a)maxx1 +3x2s.t.

x1+x2 ≤6(1)

- x1+2x2 ≤8(2)

x1,x2 ≥0(3),(4)

Ex1.b)maxx1- 2x2s.t.

x1+x2 ≤6(1)

- x1+2x2 ≤8(2)

x1,x2 ≥0(3),(4)

Letusmodifytheobjectivefunctionwhilekeepingtheconstraintsunchanged:


Doesfeasibleregionchange?

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

FeasibleRegion


Whatabouttheimprovingdirection?

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

FeasibleRegion

c=[1-2]

x1 - 2x2 =0

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

FeasibleRegion

c=[1-2]

(6,0)istheoptimalsolution!


Ex1.c)max-x1+x2s.t.

x1+x2 ≤6(1)

- x1+2x2 ≤8(2)

x1,x2 ≥0(3),(4)

Nowletusconsider:

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

c=[-11]


Ex1.c)max-x1+x2s.t.

x1+x2 ≤6(1)

- x1+2x2 ≤8(2)

x1,x2 ≥0(3),(4)

Now,letusconsider:

Question:Isitpossiblefor(5,1)betheoptimal?

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

c=[-11](5,1)

-x1 +x2 =-4

improvingdirection!

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

c=[-11](5,1)

-x1 +x2 =-4

(3,3)

-x1 +x2 =0

improvingdirection!

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

c=[-11](5,1)

-x1 +x2 =-4

(3,3)

-x1 +x2 =0

-x1 +x2 =10/3-x1+x2=4

Point(0,4)istheoptimalsolution!


Ex1.d)max-x1- x2s.t.

x1+x2 ≤6(1)

- x1+2x2 ≤8(2)

x1,x2 ≥0(3),(4)

Now,letusconsider:

Question:Isitpossiblefor(0,4)betheoptimal?

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(3)

(3)

(4)(4)

(6,0)

(0,6)(4/3,14/3)

c=[-1-1]

-x1 - x2 =-4

-x1 - x2 =0

Point(0,0)istheoptimalsolution!


Haveyounoticedanythingabouttheoptimalpoints?

ConvexSet:

AsetofpointsSisaconvexsetifforanytwopointsxЄSandyЄS,theirconvexcombinationλx+(1-λ)yisalsoinSforallλЄ[0,1].

Inotherwords,asetisaconvexsetifthelinesegmentjoininganypairofpointsinSiswhollycontainedinS.

ArethoseConvexSets?

ArethoseConvexSets?

FeasiblesetofanLPx={xЄ Rn:Ax≤b,x≥0}

isaconvexset.

ExtremePoints(CornerPoints):

ApointPofaconvexsetSisanextremepoint(cornerpoint)ifitcannotberepresentedasastrictconvexcombinationoftwodistinctpointsofS.

strictconvexcombination:xisastrictconvexcombinationofx1 andx2if

x=λx1+(1-λ)x2forsomeλЄ(0,1).


Inotherwords,foranyconvexsetS,apointPinSisanextremepoint(cornerpoint)ifeachlinesegmentthatliescompletelyinSandcontainsthepointP,hasPasanendpointofthelinesegment.


Acompanywishestoincreasethedemandforitsproductthroughadvertising.Eachminuteofradioadcosts$1andeachminuteofTVadcosts$2.Eachminuteofradioadincreasesthedailydemandby2unitsandeachminuteofTVadby7units.Thecompanywouldwishtoplaceatleast9minutesofdailyadintotal.Itwishestoincreasedailydemandbyatleast28units.Howcanthecompanymeetitsadvertisingrequirementsat§minimumtotalcost?

Ex2)


minx1 +2x2s.t.

2x1+7x2 ≥28(1)

x1+x2 ≥9(2)

x1,x2 ≥0(3),(4)

x1:#ofminutesofradioadspurchasedx2:#ofminutesofTVadspurchased

(0,4)

x1 +x2 =9

x1

x2

2x1+7x2=28

(0,0)

(1)

(1)

(2)

(2)(3)

(3)

(4)(4)

(9,0) (14,0)

(0,9)

(0,4)

x1 +x2 =9

x1

x2

2x1+7x2=28

(0,0)

(1)

(1)

(2)

(2)(3)

(3)

(4)(4)

(9,0) (14,0)

(0,9)

(7,2) x1+2x2=24

c=[12]

Objective:minx1 +2x2

totalcostx1 +2x2INCREASESinthisdirection!

-c=[-1-2] -c=[-1-2]istheimprovingdirection

(0,4)

x1 +x2 =9

x1

x2

2x1+7x2=28

(0,0)

(1)

(1)

(2)

(2)(3)

(3)

(4)(4)

(9,0) (14,0)

(0,9)

(7,2) x1+2x2=24


-c=[-1-2]

x1+2x2=18

(0,4)

x1 +x2 =9

x1

x2

2x1+7x2=28

(0,0)

(1)

(1)

(2)

(2)(3)

(3)

(4)(4)

(9,0) (14,0)

(0,9)

(7,2)


x1+2x2=18

x1+2x2=14x1+2x2=11

(7,2)istheoptimalsolutionandoptimalvalueis11


Supposethatinthepreviousexample,itcosted $4toplaceaminuteofradioadand$14toplaceaminuteofTVad.

min4x1 +14x2s.t.

2x1+7x2 ≥28(1)

x1+x2 ≥9(2)

x1,x2 ≥0(3),(4)

Ex3)

(0,4)

x1 +x2 =9

x1

x2

2x1+7x2=28

(0,0)

(1)

(1)

(2)

(2)(3)

(3)

(4)(4)

(9,0) (14,0)

(0,9)

(7,2)4x1+14x2=140

Objective:min4x1 +14x2

-c=[-4-14]

4x1+14x2=126

4x1+14x2=56-c=[-4-14]

Inthiscase,everyfeasiblesolutiononthelinesegment[(7,2),(14,0)]isan

optimalsolutionwiththesameoptimalvalue56.

(0,4)

x1 +x2 =9

x1

x2

2x1+7x2=28

(0,0)

(1)

(1)

(2)

(2)(3)

(3)

(4)(4)

(9,0) (14,0)

(0,9)

(7,2)

Objective:min4x1 +14x2

4x1+14x2=56-c=[-4-14]


Thisisalsotrueforanyfeasiblepointontheline

segment[A,B].WesaythatLPhasmultiple oralternateoptimalsolutions.

[ ] ( ) [ ]7 14Line Segment A,B 1 : 0,1

2 0l l lì üé ù é ù

= + - Îí ýê ú ê úë û ë ûî þ

7 7At point A= , the objective function value is: [4 14] 56,

2 2

14 14At point B= , the objective function value is: [4 14] 56.

0 0

é ù é ù=ê ú ê ú

ë û ë û

é ù é ù=ê ú ê ú

ë û ë û


ConsiderthefollowingLPproblem:

maxx1 +3x2s.t.

-x1+2x2 ≤8(1)

x1,x2 ≥0(2),(3)

Ex4)

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

(0,0)

(2)

(2)

(2)

(2)

(3)(3)c=[13]

x1+3x2=0

x1+3x2=12

UnboundedLPProblems:

Theobjectivefunctionforthismaximizationproblem canbeincreasedbymovingintheimprovingdirectioncasmuchaswewantwhilestillstayinginthefeasibleregion.Inthiscase,wesaythattheLPisunbounded.

ForunboundedLPproblems,thereisnooptimalsolutionandtheoptimalvalueisdefinedtobe+∞.

UnboundedLPProblems:

Fortheminimizationproblem,wesaythattheLPisunboundedifwecanDECREASEtheobjectivefunctionvalueasmuchaswewantwhilestillstayinginthefeasibleregion.

Inthiscase,theoptimalvalueisdefinedtobe-∞.


maxx1 +3x2s.t.

x1+x2 ≤6(1)

- x1+2x2 ≤8(2)

x2 ≥6(3)

x1,x2 ≥0(4),(5)

Ex5)

(-8,0)

(0,4)

-x1 +2x2 =8

x1

x2

x1 +x2 =6

(0,0)

(1)

(1)

(2)

(2)

(4)

(4)

(5)(5)

(6,0)

(0,6)(4/3,14/3)

(3)(3)

Nofeasiblesolutionexists!

WesaythatLPproblemisinfeasible.

EveryLPProblemfallsintooneofthe4cases:

Case1:TheLPproblemhasauniqueoptimalsolution(seeEx.1andEx.2)

Case2:TheLPproblemhasalternativeormultipleoptimalsolution(seeEx.3).Inthiscase,thereareinfinitelymanyoptimalsolutions.

Case3:TheLPproblemisunbounded(seeEx.4).Inthiscase,thereisnooptimalsolution.

Case4:TheLPproblemisinfeasible(seeEx.5).Inthiscase,thereisnofeasiblesolution.


Theorem: Ifthefeasibleset{xЄ Rn:Ax≤b,x≥0}isnotempty,thatisifthereisafeasiblesolution,thenthereisanextremepoint.

Theorem: IfanLPhasanoptimalsolution,thenthereisanextreme(orcorner)pointofthefeasibleregionwhichisanoptimalsolutiontotheLP

P.S: Noteveryoptimalsolutionneedstobeanextremepoint!(Rememberthealternateoptimalsolutioncase)

GraphicalSolutionofLPProblems:

WemaygraphicallysolveanLPwithtwodecisionvariablesasfollows:

Step1:Graphthefeasibleregion.

Step2:Drawanisoprofitline(formaxproblem)oran

isocostline(forminproblem).

Step3:Moveparalleltotheisoprofit/isocostlineinthe

improvingdirection.Thelastpointinthefeasibleregionthatcontactsanisoprofit/isocostlineisanoptimalsolutiontotheLP.

ie 400 principles of engineering management graphical ... · graphical solution of 2-variable lp...

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