or presentation

8/2/2019 Or Presentation

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Operations Research 7Operations Research 7thth

EditionEdition

HAMDY A.TAHAHAMDY A.TAHA

Presentation Prepared by:Presentation Prepared by:

Nasser Mohammed AliNasser Mohammed Ali

Submitted to :Submitted to :

DR. Farouk ShaabanDR. Farouk Shaaban


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IntroductionIntroduction

Operations Research (OR)

It is a scientific approach to determine

the optimum (best) solution to a decisionproblem under the restriction of limited

resources. Using the mathematical

techniques to model, analyze, and solve theproblem


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Phases of OR

1. Definition of the problem

2. Model Construction

3. Solution of the model

4. Model validity5. Implementation of the solution


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Definition of the problem includesDefinition of the problem includes

The description of the decision variables

(alternatives)

The determination of the objective of thestudy

The specification of the limitations under

which the modeled system operates


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Model ConstructionModel Construction

Translating the real world problem into

mathematical relationships (the most

suitable model to represent the system, LP,dynamic, integer,)


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Solution of the modelSolution of the model

Using wellUsing well--defined optimization techniques.defined optimization techniques.

An important aspect of model solution isAn important aspect of model solution is

sensitivity analysis.sensitivity analysis.


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Model validityModel validity

Testing and evaluation of the model. ATesting and evaluation of the model. A

common method for testing a validity of acommon method for testing a validity of a

model is to compare its performance withmodel is to compare its performance withsome past data available for the actualsome past data available for the actual

system.system.


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Implementation of the solutionImplementation of the solution

Implementation of the solution of validatedImplementation of the solution of validated

model involves the translation of themodel involves the translation of the

model's results into instructions issued inmodel's results into instructions issued inunderstandable form to the individuals.understandable form to the individuals.


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Basic components of the modelBasic components of the model

1.1. Decision VariablesDecision Variables

2.2. Objective FunctionObjective Function

3.3. ConstraintsConstraints


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Decision VariablesDecision Variables

It is the unknown to determined from theIt is the unknown to determined from the

solution of a model (what does the modelsolution of a model (what does the model

seek to determine). It is one of the spe

cificseek to determine). It is one of t

he spe

cificdecisions made by a decision maker (DM).decisions made by a decision maker (DM).


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Objective FunctionObjective Function

It is the end result (goal) desired to beIt is the end result (goal) desired to be

achieved by the system. A commonachieved by the system. A common

objective is to maximize profit orobjective is to maximize profit orminimize cost. It is expressed as aminimize cost. It is expressed as a

mathematical function of the systemmathematical function of the system

decision variables.decision variables.


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ConstraintsConstraints

These are the limitations imposed on theThese are the limitations imposed on the

variables to satisfy the restriction of thevariables to satisfy the restriction of the

modeled system. They m

ust be expressed asmodeled system. T

hey m

ust be expressed asmathematical functions of the systemmathematical functions of the system

decision variables (D.V).decision variables (D.V).


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Example 1:Example 1:

A company manufacturesA company manufactures two products A&B.two products A&B.

withwith4 & 3 units4 & 3 units. A&B take. A&B take 3&2 minutes3&2 minutes

respectively to be machined. The total timerespectively to be machined. The total time

available at machining departmentavailable at machining department is 800 hoursis 800 hours

(100 days or 20 weeks). A market research(100 days or 20 weeks). A market research

showed thatshowed that at leastat least 10000 units of A10000 units of A and notand not

more thanmore than 6000 units of B6000 units of B are needed. It isare needed. It isrequired to determine the number ofunits ofrequired to determine the number ofunits of

A&B to be produced to maximize profit.A&B to be produced to maximize profit.


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Example 2:Example 2: Feed mix problemFeed mix problem

A farmer is interested in feeding his cattle atA farmer is interested in feeding his cattle at

minimum cost.minimum cost. Two feeds are used A&BTwo feeds are used A&B. Eachcow. Eachcow

must getmust get at leastat least 400 grams/day of protein400 grams/day of protein,, at leastat least

800 grams/day ofcarbohydrates800 grams/day ofcarbohydrates, and, and not more thannot more than100 grams/day of fat100 grams/day of fat. Given that. Given that A contains 10%A contains 10%

protein, 80% carbohydrates and 10% fat while Bprotein, 80% carbohydrates and 10% fat while B

contains 40% protein, 60% carbohydratescontains 40% protein, 60% carbohydrates andand no fatno fat..

A costs 2 L.E/kgA costs 2 L.E/kg, and, and B costs 5 L.E/kg.B costs 5 L.E/kg. Formulate theFormulate theproblem to determine the optimum amount of eachproblem to determine the optimum amount of each

feed to minimize cost.feed to minimize cost.


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Problem FormulationProblem Formulation

Decision variablesDecision variablesX1= weight of feed A kg/day/animalX1= weight of feed A kg/day/animal

X2= weight of feed B kg/day/animalX2= weight of feed B kg/day/animal

Objective FunctionObjective FunctionMaximize Z= 2XMaximize Z= 2X11 + 5X+ 5X22

ConstraintsConstraintsProteinProtein 0.1 X0.1 X

11+ 0.4 X+ 0.4 X

22 0.4 0.4

CarbohydratesCarbohydrates 0.8 X0.8 X11 + 0.6 X+ 0.6 X22 0.8 0.8

FatsFats 0.1 X0.1 X11 0.1 0.1

XX11, X, X22 0 0


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Example 3:Example 3: Blending ProblemBlending Problem

An iron ore fromAn iron ore from 4 mines4 mines will be blended.will be blended.

The analysis has shown that, in order toThe analysis has shown that, in order to

obtain suitable tensile properties, minim

umobtain s

uitable tensile properties, minim

umrequirements must be met forrequirements must be met for3 basic3 basic

elements A, B, and Celements A, B, and C. Each of the. Each of the 4 mines4 mines

contains different amounts of thecontains different amounts of the 3 elements3 elements

((see the tablesee the table). Formulate to find the least). Formulate to find the least

cost blend for one ton of iron ore.cost blend for one ton of iron ore.


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Decision variablesDecision variablesX1= Fraction of ton to be selected from mine number1X1= Fraction of ton to be selected from mine number1

X2= Fraction of ton to be selected from mine number 2X2= Fraction of ton to be selected from mine number 2




Minimize Z= 800 X1 + 400 X2 + 600 X3 + 500 X4Minimize Z= 800 X1 + 400 X2 + 600 X3 + 500 X4


10 X10 X11+ 3X+ 3X22 + 8X+ 8X33 + 2X+ 2X44 55

90 X90 X11 + 150 X+ 150 X22 + 75 X+ 75 X33+ 175 X+ 175 X44 1010

45 X45 X11 + 25 X+ 25 X22 + 20 X+ 20 X33+ 37 X+ 37 X44 3030

XX11 + X+ X22 + X+ X33 + X+ X44 11

XX11, X, X22, X, X33, X, X44 00

u

uu

u

!


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Example 4:Example 4: Inspection ProblemInspection Problem

A company hasA company has 2 grades2 grades of inspectors 1&2. It isof inspectors 1&2. It isrequired thatrequired that at leastat least 1800 pieces1800 piecesbe inspectedbe inspectedper 8per 8

hour dayhour day.. Grade 1Grade 1 inspectors can check pieces at theinspectors can check pieces at the

rate of 25 perhourrate of 25 perhourwith anwith an accuracy of 98%.accuracy of 98%. Grade 2Grade 2

inspectors can check at theinspectors can check at the rate of15 pieces perhourrate of15 pieces perhourwith anwith an accuracyaccuracy of 95%.of 95%. Grade 1costs 4 L.E/hourGrade 1costs 4 L.E/hour,,

grade 2 costs 3 L.E/hourgrade 2 costs 3 L.E/hour.. Each time an error is madeEach time an error is made

by an inspectorcosts the company 2 L.Eby an inspectorcosts the company 2 L.E. There are. There are 88

grade 1grade 1 andand 10 grade 210 grade 2 inspectors available. Theinspectors available. Thecompany wants to determine the optimal assignmentcompany wants to determine the optimal assignment

of inspectors which will minimize the total cost ofof inspectors which will minimize the total cost of

inspection/day.inspection/day.


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Decision variablesDecision variablesX1= Number of grade 1 inspectors/day.X1= Number of grade 1 inspectors/day.

X2=X2=Number of grade 2 inspectors/day.Number of grade 2 inspectors/day.


Cost of inspection = Cost of error + Inspector salary/dayCost of inspection = Cost of error + Inspector salary/day

Cost of grade 1/hour= 4 + (2 X 25 X 0.02) = 5 L.ECost of grade 1/hour= 4 + (2 X 25 X 0.02) = 5 L.E

Cost of grade 2/hour= 3 + (2 X15 X 0.05) = 4.5 L.ECost of grade 2/hour= 3 + (2 X15 X 0.05) = 4.5 L.E

Minimize Z= 8 (5 X1 + 4.5 X2) = 40 X1 + 36 X2Minimize Z= 8 (5 X1 + 4.5 X2) = 40 X1 + 36 X2


XX11

8 8

XX22 10 10

8(25) X8(25) X11+ 8(15)X+ 8(15)X22 1800 1800

200 X200 X11+ 120 X+ 120 X22 1800 1800

XX11,, XX22 0 0


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Example 5:Example 5: TrimTrim--loss Problem.loss Problem.

A company produces paper rolls with aA company produces paper rolls with astandard width of 20 feet. Each special customerstandard width of 20 feet. Each special customer

orders with different widths are produced byorders with different widths are produced by

slitting the standard rolls. Typical orders areslitting the standard rolls. Typical orders are

summarized in the following tables.summarized in the following tables.


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Possible knife settingsPossible knife settings

Formulate to minimize the trim loss and the numberFormulate to minimize the trim loss and the number

of rolls needed to satisfy the order.of rolls needed to satisfy the order.

Figure

2.9


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Problem FormulationProblem FormulationDecision variablesDecision variables

XXjj ==Number of standard rolls to be cutNumber of standard rolls to be cut accordingaccording toto

setting j j =1, 2, 3, 4, 5, 6setting j j =1, 2, 3, 4, 5, 6

Number of 5 feet rolls producedNumber of 5 feet rolls produced = 2 X= 2 X22 + 2 X+ 2 X33 + 4 X+ 4 X44 + X+ X55

Number of 7 feet rolls producedNumber of 7 feet rolls produced =X=X11 + X+ X22+ 2 X+ 2 X55 Number of 9 feet rolls producedNumber of 9 feet rolls produced =X=X11 + X+ X33+ 2 X+ 2 X66 Let YLet Y11, Y, Y22, Y, Y33be the number of surplus rolls of the 5, 7, 9 feetbe the number of surplus rolls of the 5, 7, 9 feet

rolls thusrolls thus

YY11= 2 X= 2 X22 + 2 X+ 2 X33 + 4 X+ 4 X44 + X+ X55 -- 150150YY22=X=X11 + X+ X22+ 2 X+ 2 X55 -- 200200

YY33=X=X11 + X+ X33+ 2 X+ 2 X66 -- 300300

The total trim losses = L (4XThe total trim losses = L (4X11 +3 X+3 X22+ X+ X33 + X+ X55 + 2 X+ 2 X66 + 5Y+ 5Y11++

7Y7Y22+ 9Y+ 9Y33)) **Where L is the length of the standard roll.Where L is the length of the standard roll.


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MinimizeMinimize Z= L(4XZ= L(4X11 +3 X+3 X22+ X+ X33 + X+ X55 + 2 X+ 2 X66 ++5Y5Y11+ 7Y+ 7Y22+ 9Y+ 9Y33))


2 X2 X22+ 2 X+ 2 X33+ 4 X+ 4 X44+ X+ X55 -- YY11 == 150150XX11+ X+ X22 + 2 X+ 2 X55 -- YY22 = 200= 200

XX11+ X+ X33 + 2 X+ 2 X66 -- YY33 == 300300

X1, X2, X3, X4, X5, X6X1, X2, X3, X4, X5, X6 0 0

Y1, Y2, Y3Y1, Y2, Y3 0 0


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General form of a LP problem with mGeneral form of a LP problem with m

constraints and n decision variables is:constraints and n decision variables is:

Maximize Z = CMaximize Z = C11XX11+ C+ C22XX22+ + C+ + CnnXXnn ConstraintsConstraints

AA1111XX11 + A+ A1212XX22++ A++ A1n1nXXnn B B11

AA2121XX11 + A+ A2222XX22++ A++ A2n2nXXnn B B22

..

..

..

..

AAm1m1XX11+ A+ Am2m2XX22+...+ A+...+ AmnmnXXnn B Bmm

XX11

, X, X22,, X,, X

nn 0 0


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OROR

MaximizeMaximize


!

!

n

jjjxcz

1

!

e

n

jijij bxa

1

mi ,...,2,1, !

xj 0u nj ,...,2,1, !


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WhereWhere

nn==Number of activities.Number of activities.

XXjj= Level of activity= Level of activity jj

CCjj= Contribution of the objective function/unit of= Contribution of the objective function/unit ofactivity jactivity j

mm= Number of resources= Number of resources

BBii = Amount of resource i available.= Amount of resource i available.AAijij= Amount of resource i consumed by one unit of= Amount of resource i consumed by one unit of

activity jactivity j


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Terminology of solutions for a LP model:Terminology of solutions for a LP model:

A SolutionA Solution

Any specifications of values of XAny specifications of values of X11, X, X22, , X, , Xnn is calledis calleda solution.a solution.

A Feasible SolutionA Feasible SolutionIs a solution for which all the constraints areIs a solution for which all the constraints are

satisfied.satisfied.

An Optimal SolutionAn Optimal Solution

Is a feasible solution that has the most favorable valueIs a feasible solution that has the most favorable valueof the objective function (largest of maximize orof the objective function (largest of maximize orsmallest for minimize)smallest for minimize)

Notes


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If there is exactly one optimal solution it mustIf there is exactly one optimal solution it must

be acorner point feasible sol

ution.be a

corner point feasible sol

ution.

If there are multiple optimal solutions, then atIf there are multiple optimal solutions, then atleast two of them must be adjacent cornerleast two of them must be adjacent corner--

point feasible solutions.point feasible solutions.

Two cornerTwo corner--point feasible solutions are said topoint feasible solutions are said to

be adjacent if the line segment connectingbe adjacent if the line segment connectingthem lies on the boundary of the feasiblethem lies on the boundary of the feasibleregion (one of the constraints).region (one of the constraints).


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Graphical SolutionGraphical Solution

Construction of the LP modelConstruction of the LP model

Example 1: The Reddy Mikks CompanyExample 1: The Reddy Mikks Company

Reddy Mikks produces both interior and exterior paintsReddy Mikks produces both interior and exterior paintsfrom two raw materials, M1&M2. The following tablefrom two raw materials, M1&M2. The following table

provides the basic data of the problem.provides the basic data of the problem.


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A market survey indicates that the dailyA market survey indicates that the daily

demand for interior paint cannot exceed that ofdemand for interior paint cannot exceed that ofexterior paint by more than 1 ton. Also, theexterior paint by more than 1 ton. Also, the

maximum daily demand of interior paint is 2maximum daily demand of interior paint is 2

ton.ton. Reddy Mikks wants to determine the optimumReddy Mikks wants to determine the optimum

(best) product mix of interior and exterior(best) product mix of interior and exterior

paints that maximizes the total daily profitpaints that maximizes the total daily profit


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Decision variablesDecision variables

XX11=Tons produced daily of exterior paint.=Tons produced daily of exterior paint.

XX22=Tons produced daily of interior paint.=Tons produced daily of interior paint.

Objective FunctionObjective FunctionMaximize Z= 5 XMaximize Z= 5 X11 + 4 X+ 4 X22


6 X6 X11+4 X+4 X22 2424

XX11+2 X+2 X22 66-- XX11+ X+ X22 11

XX22 22

XX11, X, X22 00

e

ee

e

u


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Any solution that satisfies all the constraints of theAny solution that satisfies all the constraints of themodel is a feasible solution. For example, X1=3 tonsmodel is a feasible solution. For example, X1=3 tons

and X2=1 ton is a feasible solution. We have anand X2=1 ton is a feasible solution. We have an

infinite number of feasible solutions, but we areinfinite number of feasible solutions, but we are

interested in the optimum feasible solution that yieldsinterested in the optimum feasible solution that yieldsthe maximum total profit.the maximum total profit.


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Graphical SolutionGraphical Solution

The graphical solution is valid only for twoThe graphical solution is valid only for two--

variable problem which is rarely occurred.variable problem which is rarely occurred.

The grap

hical sol

ution in

cludes two basi

c

The grap

hical sol

ution in

cludes two basi

csteps:steps:

1.1. The determination of the solution space thatThe determination of the solution space that

defines the feasible solutions that satisfyalldefines the feasible solutions that satisfyall

the constraints.the constraints.2.2. The determination of the optimum solutionThe determination of the optimum solution

from among all the points in the feasiblefrom among all the points in the feasible

solution space.solution space.


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ABCDEF consists of an infinite number of points; weABCDEF consists of an infinite number of points; we

need a systematic procedure that identifies theneed a systematic procedure that identifies the

optimum solutions. The optimum solution isoptimum solutions. The optimum solution isassociated with a corner point of the solution space.associated with a corner point of the solution space.

To determine the direction in which the profitTo determine the direction in which the profit

function increases we assign arbitrary increasingfunction increases we assign arbitrary increasing

values of10 and 15values of10 and 15

5 X5 X11 + 4 X+ 4 X22=10=10

AndAnd 5 X5 X11 + 4 X+ 4 X22=15=15

The optimum solution is mixture of 3 tons of exteriorThe optimum solution is mixture of 3 tons of exteriorand 1.5 tons of interior paints will yield a daily profitand 1.5 tons of interior paints will yield a daily profit

of 21000$.of 21000$.


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Example 2:Example 2: The Diet ProblemThe Diet Problem

Farm uses at least 800lb of special feed daily.Farm uses at least 800lb of special feed daily.The special feed is a mixture ofcorn andThe special feed is a mixture ofcorn and

soybean with the following composition.soybean with the following composition.

The food mixture must contain at least 30%The food mixture must contain at least 30%protein and at most 5% fiber. They want toprotein and at most 5% fiber. They want to

determine the daily minimum cost feed mix.determine the daily minimum cost feed mix.


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Decision variablesDecision variables

X1= lb ofcorn in the daily mix.X1= lb ofcorn in the daily mix.

X2= lb of soybean in the daily mix.X2= lb of soybean in the daily mix.


Minimize Z= 0.3 XMinimize Z= 0.3 X11+ 0.9 X+ 0.9 X22 ConstraintsConstraints

X1+X1+ X2 800X2 800

0.09X1+ 0.6 X2 0.3 (X1+X2) 0.21X10.09X1+ 0.6 X2 0.3 (X1+X2) 0.21X1--0.3 X2 00.3 X2 0

0.02X1+0.06 X2 0.05(X1+X2) 0.03X10.02X1+0.06 X2 0.05(X1+X2) 0.03X1--0.01X2 00.01X2 0X1,X1, X2X2 00

u

uue

u e


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Slack, Surplus, and Unrestricted

variables

Slack Variable: Forconstraints of the type ( ) the

R.H.S normally represents the limit on the availability

of a resources and the L.H.S represents the usage of

this limited resource by the different activities(variables) of the model. A slack represents the

amount by which the available amount of the

resource exceeds its usage by the activities.

For example (6 X1+4 X2 24) is equivalent to(6 X1+4 X2 + S1= 24) provided that S1 0. The slack

variable S1=24-6 X1-4 X2 represents the unused

amount of raw material M1.


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Surplus VariableSurplus Variable:: It is used in the constraints of typeIt is used in the constraints of type(( ) normally set minimum specification) normally set minimum specificationrequirements. Surplus represents the excess of therequirements. Surplus represents the excess of theL.H.S over the minimum requirement. For exampleL.H.S over the minimum requirement. For example(X(X11+X+X22 800) is equivalent to (X800) is equivalent to (X11+X+X22 -- SS11= 800)= 800)provided that Sprovided that S11 0, this signifies that a surplus0, this signifies that a surplus

amount of feed over the minimum requirement willamount of feed over the minimum requirement willbe producedbe produced

Unrestricted Variable:Unrestricted Variable: The variable whichcan beThe variable whichcan bepositive or negativepositive or negative


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The Simplex MethodThe Simplex Method

IntroductionIntroduction

It is a general algebraic method to solve a set ofIt is a general algebraic method to solve a set of

linear equations. We use simplex method to getlinear equations. We use simplex method to get

extreme (orcorner) point solution. We must firstextreme (orcorner) point solution. We must firstconvert the model into the standard LP form by usingconvert the model into the standard LP form by using

slack or surplus variables to convert the inequalityslack or surplus variables to convert the inequality

constraints into equations. Our interest in the standardconstraints into equations. Our interest in the standard

LP form lies in the basic solutions of theLP form lies in the basic solutions of thesimultaneous linear equations.simultaneous linear equations.


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Standard LP Form and its BasicStandard LP Form and its Basic

SolutionSolution

Example 1:Example 1: Express the following LP modelExpress the following LP modelin standard form.in standard form.

Maximize Z= 2XMaximize Z= 2X11+ 3X+ 3X22 +5 X+5 X33S.TS.T XX11+ X+ X22 -- XX33 --55

--6X6X11+ 7X+ 7X22 -- 9 X9 X33 4 4

XX11 + X+ X22 + 4 X+ 4 X33 =10=10

XX11, X, X22 00

XX33 unrestrictedunrestricted


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Standard LP formStandard LP form

Maximize Z= 2XMaximize Z= 2X11 + 3 X+ 3 X22+5 X+5 X++

33--5 X5 X--33

S.TS.T --XX11-- XX22+3 X+3 X++

33--3 X3 X--33+ X+ X44 = 5= 5

--6X6X11+7 X+7 X22--9 X9 X++

33+9 X+9 X--33+ X+ X55 = 4= 4

XX11+ X+ X22+4 X+4 X++

33--4 X4 X--33 =10=10

X1, X2, X+3, XX1, X2, X+3, X--3, X4, X53, X4, X5 0 0


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Determination of basic solutionsDetermination of basic solutions

The standard LP form includes mThe standard LP form includes m

simultaneous linear equations in n unknownssimultaneous linear equations in n unknowns

or variables (m


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The resulting solution is basic solution if allThe resulting solution is basic solution if all

values are satisfying nonvalues are satisfying non--negativity. The nonnegativity. The non--

basic solution is feasible basic solutionbasic solution is feasible basic solution

otherwise, it is infeasible. The maximumotherwise, it is infeasible. The maximum

number of possible basic solution for mnumber of possible basic solution for m

equations in n unknowns isequations in n unknowns is

)!(!

!

mnm

n


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Maximize Z= 5XMaximize Z= 5X11 + 6X+ 6X22

S.TS.T 2X2X11 + 3X+ 3X22 18 18

2X2X11 + X+ X22 12123X3X11 + 3X+ 3X22 = 24= 24

XX11, X, X22 0 0


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Maximize Z=5XMaximize Z=5X11 + 6X+ 6X22

S.TS.T 2X2X11 + 3X+ 3X22 + X+ X33 =18=18

2X2X11 + X+ X22 + X+ X44 =12=123X3X11 + 3X+ 3X22 + X+ X55 = 24= 24

XX11, X, X22, X, X33, X, X44, X, X55 0 0


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Maximum number of possible basicMaximum number of possible basic

solutions is:solutions is:

There are 3 constraints so, m=3. And we haveThere are 3 constraints so, m=3. And we have

5 variables so, n=55 variables so, n=5

10!2!3!3*4*5

!2!3!5

)!35(!3!5 !!!


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The basic feasible solutions are the corner pointsThe basic feasible solutions are the corner points


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The Simplex AlgorithmThe Simplex Algorithm

We solve the Reddy Mikks model, whereWe solve the Reddy Mikks model, whereX1=Tons produced daily of exterior paint.X1=Tons produced daily of exterior paint.

X2=Tons produced daily of interior paint.X2=Tons produced daily of interior paint.

Exterior and interior paints are produced from twoExterior and interior paints are produced from twotypes of raw materials M1 and M2types of raw materials M1 and M2

MaximizeMaximize ZZ --5 X5 X11-- 4 X4 X22+0 X+0 X33+0 X+0 X44+0 X+0 X55+0X+0X66=0=0

6 X6 X11+4 X+4 X22+ X+ X33 24 24

XX11+2 X+2 X22+ X+ X44 6 6

-- XX11+ X+ X22 + X+ X55 11

XX22 + X+ X66 2 2

XX11, X, X22, X, X33, X, X44, X, X55, X, X66 0 0


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The variables X3, X4, X5, X6 are the slacksThe variables X3, X4, X5, X6 are the slacks

associated with the four (associated with the four () constraints.) constraints.


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The starting basic feasible solution isThe starting basic feasible solution is

Is the starting solution optimal?

No, since the coefficients ofX1 and X2 are still negative so they

can increase the profit of Z.


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We choose XWe choose X11with more negative (with more negative (--5) i.e.; X5) i.e.; X11 is theis the

entering variableentering variable..

Since none of the Z-row coefficients associated with the non-basic

variables X3, X4 is negative the last table is optimal.


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The rules for selecting the entering and leavingThe rules for selecting the entering and leaving

variables are referred to as the optimality andvariables are referred to as the optimality and

feasibility conditionsfeasibility conditions

1)1) Optimality conditionsOptimality conditions

2)2) Feasibility ConditionFeasibility Condition


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1.1. Optimality conditions:Optimality conditions:

The entering variable inThe entering variable in

maximizationmaximization(minimization) problem is(minimization) problem is

the nonthe non--basic variablebasic variable

having the most negativehaving the most negative

(positive) coefficient in(positive) coefficient in

the Zthe Z--row. Ties arerow. Ties arebroken arbitrarily. Thebroken arbitrarily. The

optimum is reached at theoptimum is reached at the

iteration where all the Ziteration where all the Z--

row coefficients of therow coefficients of thenonnon--basic variables arebasic variables are

nonnon--negative (nonnegative (non--

positive)positive)

2.2. Feasibility Condition:Feasibility Condition:

for both the maximum andfor both the maximum and

minimu

m problems th

eminimu

m problems th

eleaving variables is theleaving variables is the

basic variables associatedbasic variables associated

with the smallest nonwith the smallest non--

negative ratio. Ties arenegative ratio. Ties are

broken arbitrarily.broken arbitrarily.


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The steps of the simplex method are:The steps of the simplex method are:

1.1. Determine a starting basic feasible solutionDetermine a starting basic feasible solution

2.2. Select an entering variable using theSelect an entering variable using theoptimality condition. Stop if there is nooptimality condition. Stop if there is no

entering variable.entering variable.3.3. Select a leaving variable using the feasibilitySelect a leaving variable using the feasibility

condition.condition.

4.4. Determine the new basic solution by usingDetermine the new basic solution by usingthe appropriate Gaussthe appropriate Gauss--Jordan computation.Jordan computation.

5.5. Go to step 1Go to step 1


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Special Cases in Simplex MethodSpecial Cases in Simplex Method

1.1. DegeneracyDegeneracy

2.2. Alternative optima (Infinity ofSolution)Alternative optima (Infinity ofSolution)

3.3. Unbou

ndedS

olu

tionUnbou

ndedS

olu

tion4.4. NonNon--existing orInfeasible Solutionexisting orInfeasible Solution


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Degeneracy:Degeneracy: one or more basicone or more basic

variable(s) has zero valuevariable(s) has zero value..

If you find more than one leaving variableIf you find more than one leaving variable

(i.e.; we have two or more variables having the(i.e.; we have two or more variables having the

same ratio in the R.H.S) so one or more of thesame ratio in the R.H.S) so one or more of the

basic variable(s) will be equal zero in the nextbasic variable(s) will be equal zero in the next

iteration. This condition indicates that theiteration. This condition indicates that the

model has at least one redundant constraint.model has at least one redundant constraint.


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Maximize Z= 3 XMaximize Z= 3 X11+ 9X+ 9X22

XX11+4 X+4 X22 88

XX11+2 X+2 X22 44

XX11, X, X22 00



S.TS.T XX11+4X+4X22+ X+ X33 = 8= 8XX11+ 2X+ 2X22 + X+ X44= 4= 4

XX11, X, X22, X, X33, X, X44 00


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X1=0 , X2 =2 , X3 =0 , X4=0 , Z=18Is it possible to stop at the second iteration (when degeneracyfirst appears) even though it is not optimum? The answer is NO,

because the solution may be temporarily degenerate.


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Alternative optimaAlternative optima (Infinity of(Infinity of

Solution):Solution):

When the objective function is parallel to aWhen the objective function is parallel to a

binding constraint. So the objective functionbinding constraint. So the objective function

(Z) will have the same optimal value at more(Z) will have the same optimal value at more

than one solution pointthan one solution point..


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Maximize Z= 2 XMaximize Z= 2 X11+ 4 X+ 4 X22

XX11+2 X+2 X22 5 5

XX11+ X+ X2222 4 4

XX11, X, X22 0 0



S.TS.T XX11+2X+2X22+ X+ X33 = 8= 8XX11+ X+ X22+ X+ X44 = 4= 4

XX11, X, X22 , X, X33 , X, X44 0 0


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First Optimum Solution:

X1*=0 X2*=5/2 X3*=0 X4*=3/2 Z*=10

Second Optimum Solution:

X1*=3 X2*=1 X3*=0 X4*=0 Z*=10

In the second table:

X1=0

X2= 5/2

In the third table:

X1= 3

X2=1

The general form:

X1 = (0) + (1-) (3) = 3-3

X2 = (5/2) + (1-) (1) =1+3/2

= 0 1


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Unbounded Solution:Unbounded Solution:

In which the objective function (Z) can beIn which the objective function (Z) can be

increased indefinitely without violating and ofincreased indefinitely without violating and of

the constraints i.e.; the solution space isthe constraints i.e.; the solution space is

unbounded in at least one directionunbounded in at least one direction..


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Maximize Z= 2 X1 + X2Maximize Z= 2 X1 + X2

S.T X1S.T X1 -- X2X2 1010

2X12X1 4040

X1 , X2X1 , X2 00 Standard LP formStandard LP form

Maximize Z=2XMaximize Z=2X11 +X+X22S.TS.T XX11--XX22+ X+ X33 =10=10

2X2X11 + X+ X44 = 40= 40

XX11, X, X22 , X, X33 , X, X44 0 0


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BothXBothX11, X, X22 are candidates for entering the solution. But all theare candidates for entering the solution. But all the

constraint coefficient underXconstraint coefficient underX22 are negative or zero, meaning thatare negative or zero, meaning that

XX22 can be increased indefinitely without violating any of thecan be increased indefinitely without violating any of the

constraints. Any increase in Xconstraints. Any increase in X22 will increase Z so it can bewill increase Z so it can be

increased as Xincreased as X22 . Thus, the problem has no bounded . Thus, the problem has no bounded

solution.solution.


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NonNon--existing or Infeasible Solution:existing or Infeasible Solution:

If the constraints are not satisfiedIf the constraints are not satisfied

simultaneously, the model has no feasiblesimultaneously, the model has no feasible

solution. This situation can never occur if allsolution. This situation can never occur if all

the constraints are of the type (assuming nonthe constraints are of the type (assuming non--negative R.H.Sconstraints) because the slacksnegative R.H.Sconstraints) because the slacks

provide a feasible solution.provide a feasible solution.


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Maximize Z= 3 XMaximize Z= 3 X11+2 X+2 X22

S.TS.T 2 X2 X11 + X+ X22 2 2

3 X3 X11 +4X+4X22 12 12

XX11 , X, X22 0 0

T i M d l d IT i M d l d I


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Transportation Model and ItsTransportation Model and Its

VariantsVariants

Definition of the transportation modelDefinition of the transportation model

The transportation model is a special class ofThe transportation model is a special class of

the LP problem. It deals with the situation inthe LP problem. It deals with the situation in

which a commodity is shipped from sourcewhich a commodity is shipped from source

(e.g. factories) to destinations (e.g.(e.g. factories) to destinations (e.g.

warehouses). The objective is to determine thewarehouses). The objective is to determine the

amou

nts sh

ipped from each

sou

rce to ea

chamo

unts s

hipped from ea

chsou

rce to ea

chdestination that minimize the total shippingdestination that minimize the total shipping

cost while satisfying both the supply limits andcost while satisfying both the supply limits and

the demand requirements.the demand requirements.


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WhereWhere

aaii: represents the amount available at source i where i=1,2,3 m: represents the amount available at source i where i=1,2,3 m

bbjj: represents the demand by destination j where j=1,2,3 n: represents the demand by destination j where j=1,2,3 n

ij: arc from source i to destination j (road)ij: arc from source i to destination j (road)CCijij: cost forunit shipped from source i to destination j: cost forunit shipped from source i to destination j

XXijij: units supplied from source i to destination j: units supplied from source i to destination j

The objective is to determine Xij that will minimize the total

transportation cost while satisfying all the supply and demand restriction.


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There are three plants A, B, C and twoThere are three plants A, B, C and two

destinations 1and 2, the distance between themdestinations 1and 2, the distance between them

is as followsis as follows

The mile costs 8 cents find the cost/unit roundThe mile costs 8 cents find the cost/unit round

to $to $


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We define

XX1111 as the amount shipped from A to 1as the amount shipped from A to 1

XX1212 as the amount shipped from A to 2as the amount shipped from A to 2

XX2121 as the amount shipped from B to 1as the amount shipped from B to 1

XX2222 as the amount shipped from B to 2as the amount shipped from B to 2

XX3131 as the amount shipped from C to 1as the amount shipped from C to 1

XX3232 as the amount shipped from C to 2as the amount shipped from C to 2

The plants A, B, and C produces 1000, 1500,The plants A, B, and C produces 1000, 1500,and1200 cars respectively. The destination 1&2and1200 cars respectively. The destination 1&2

demands are 2300 and 1400 cars.demands are 2300 and 1400 cars...


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The LP model isThe LP model is

Minimize Z=80 XMinimize Z=80 X1111

+215 X+215 X1212 +100 X+100 X2121 +108+108

XX2222 +102 X+102 X3131 +68 X+68 X3232S.TS.TXX

1111+ X+ X

1212 =1000 . (A)=1000 . (A)

XX2121 + X+ X2222 =1500 . (B)=1500 . (B)

XX3131 + X+ X3232 =1200 . (C)=1200 . (C)

XX1111

+ X+ X2121+ X+ X3131 =2300 .. (1)=2300 .. (1)

XX1212 + X+ X2222+ X+ X3232 =1400 .. (2)=1400 .. (2)

XXijij 00 i=1, 2, 3i=1, 2, 3 j=1, 2j=1, 2


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These constraints are all equations because theThese constraints are all equations because the

total supply from t

he th

ree plants=

1

000 +total supply from t

he th

ree plants=

1

000 +1500 + 1200 = 3700 cars, equals the total1500 + 1200 = 3700 cars, equals the total

demands at the two destinations = 2300 + 1400demands at the two destinations = 2300 + 1400

= 3700 cars.= 3700 cars.

The LP model can be solved by the simplexThe LP model can be solved by the simplex

method. However, the special structure of themethod. However, the special structure of the

constraints allows us to solve the problemconstraints allows us to solve the problem

more conveniently using the transportationmore conveniently using the transportationtableau.tableau.


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Wh

en th

e total su

pply does not equ

al th

e totalWh

en th

e total su

pply does not equ

al th

e totaldemand, the transportation model is said to bedemand, the transportation model is said to beunbalanced. We must add a dummy source orunbalanced. We must add a dummy source ordestination.destination.

E l 2E l 2


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Example 2:Example 2: In example (1) suppose that plant B capacity isIn example (1) suppose that plant B capacity is

1300 cars (instead of1500). The total supply1300 cars (instead of1500). The total supply=3500cars and the total demand = 3700cars.So=3500cars and the total demand = 3700cars.So

the total supply < the total demand. Therefore, wethe total supply < the total demand. Therefore, we

have to add dummy source withcapacity 200have to add dummy source withcapacity 200

carscars..

E l 3E l 3


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Example 3Example 3:: In example (1) suppose that the demand of theIn example (1) suppose that the demand of the

destination 1 is 1900 cars (instead of 2300). Thedestination 1 is 1900 cars (instead of 2300). Thetotal supply =3700cars and the total demand =total supply =3700cars and the total demand =

3300cars. So the total supply > the total demand.3300cars. So the total supply > the total demand.

Therefore, we have to add dummy destinationTherefore, we have to add dummy destination

with 400 cars.with 400 cars.


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The Transportation AlgorithmThe Transportation Algorithm

Step (1):Step (1): Determine an Initial Basic FeasibleDetermine an Initial Basic FeasibleSolution (IBFS) by using one of the followingSolution (IBFS) by using one of the followingmethods.methods.

1.1. The NorthThe North--West Corner Method OR,West Corner Method OR,

2.2. The LeastThe Least--Cost Method OR,Cost Method OR,

3.3. Vogel Approximation Method.Vogel Approximation Method.

In general, the number of basic variables in a BFSIn general, the number of basic variables in a BFSis given by the number of independent constraints,is given by the number of independent constraints,

thus the number of variables that can take a positivethus the number of variables that can take a positivevalue is limited to (total number of sources (m) +value is limited to (total number of sources (m) +total number of destinations (n)total number of destinations (n)--1).1).

Step (2):Step (2): Improve the IBFSImprove the IBFS

1.1. Determine an entering variable.Determine an entering variable.

2.2. Determine a leaving variable.Determine a leaving variable.

3.3. Obtain a new BFSObtain a new BFS4.4. If the solution is optimal stop. Otherwise; repeat step (2).If the solution is optimal stop. Otherwise; repeat step (2).


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Solve the following transportation problem.Solve the following transportation problem.


S l tiS l ti


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SolutionSolution

Step (1): finding the IBFSStep (1): finding the IBFS

a)a) NorthNorth--West Corner MethodWest Corner Method

Transportation Cost = 5(10) +10(2) +5(7) +15(9) +5(20) +10(18) =520


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Steps ofNorthSteps ofNorth--West MethodWest Method

Allocate as much as possible to a selected cell XAllocate as much as possible to a selected cell X1111

Cross out the row orcolumn with zero supply of demand (ifCross out the row orcolumn with zero supply of demand (if

both are satisfied at the same time, only one should be crossedboth are satisfied at the same time, only one should be crossed

out).out).

If exactly one row orcolumn is left uncrossed out stop.If exactly one row orcolumn is left uncrossed out stop.

Otherwise go to step(1)Otherwise go to step(1)


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b)b) The LeastThe Least--Cost Method:Cost Method: this method gives athis method gives a

better starting solution because it takes the costbetter starting solution because it takes the cost

into considerationinto consideration

Transportation Cost =15(2) +15(9) +0(7) +10(20) +5(4) +5(18) =475Transportation Cost =15(2) +15(9) +0(7) +10(20) +5(4) +5(18) =475


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Steps of LeastSteps of Least--Cost MethodCost Method

Assign as much as possible to the variable with the smallestAssign as much as possible to the variable with the smallest

unit cost cell.unit cost cell.

Cross out the row orcolumn with zero supply of demand (ifCross out the row orcolumn with zero supply of demand (if

both are satisfied at the same time, only one should be crossedboth are satisfied at the same time, only one should be crossed

out).out).

If exactly one row orcolumn is left uncrossed out stop.If exactly one row orcolumn is left uncrossed out stop.

Otherwise go to step(1)Otherwise go to step(1)


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c)c) The Vogel Approximation Method:The Vogel Approximation Method: thisthis

method provides a better starting solutionmethod provides a better starting solution

(optimal orclosed to the optimal solution)(optimal orclosed to the optimal solution)

Steps of Vogel ApproximationSteps of Vogel Approximation


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Steps of Vogel ApproximationSteps of Vogel Approximation

MethodMethod In each row and column, subtract the lowest two costs fromIn each row and column, subtract the lowest two costs from

each other.each other.

Determine the row orcolumn with the largest difference.Determine the row orcolumn with the largest difference.

Allocate as much as possible to the variable with the least costAllocate as much as possible to the variable with the least costin the selected row orcolumn. Any row orcolumn with zeroin the selected row orcolumn. Any row orcolumn with zerosupply or would no be used in computing the following step.supply or would no be used in computing the following step.

If exactly one row orcolumn remains uncrossed out stop.If exactly one row orcolumn remains uncrossed out stop.Otherwise, go to step (1).Otherwise, go to step (1).

If only one row orcolumn with positive supply or demandIf only one row orcolumn with positive supply or demandremains uncrossed out determine the basic variable by theremains uncrossed out determine the basic variable by the

leastleast--cost method.cost method. If all uncrossed out rows and columns have assigned zeroIf all uncrossed out rows and columns have assigned zero

supply and demand, determine the zero basic variable by thesupply and demand, determine the zero basic variable by theleastleast--cost method.cost method.


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X31=5

X12=15

X23

=15

X14=0, X34=5, X24=10

Transportation Cost =15(2) +15(9) +0(4) +10(20) +5(4) +5(18)= 475

It is expected to produce better starting solution than least-

cost method.


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Improving the IBFSImproving the IBFS

Determine an entering variable using methodDetermine an entering variable using method

of multipliers (U, V method), In this methodof multipliers (U, V method), In this method

associate the multipliers with row I andassociate the multipliers with row I and

column j of the transportation tableau.column j of the transportation tableau.

yy For eachcurrent basic variableFor eachcurrent basic variable

UUii + V+ Vjj = C= Cijij for each basicXfor each basicXijijyy For nonFor non--basic variablebasic variable

C'C'ijij = U= Uii + V+ Vjj -- CCijij for each nonfor each non--basicXbasicXijij

Choose the nonChoose the non--basic variable with the largest positivebasic variable with the largest positive


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Choose the nonChoose the non--basic variable with the largest positivebasic variable with the largest positive

coefficient (C'coefficient (C'ijij) to be the entering variable. The solution is) to be the entering variable. The solution is

optimal if and only if all C'optimal if and only if all C'ijij areare nonnon--positive.positive.

ConsiderConsiderthe same example:the same example:


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The entering variable must have the maximum C'The entering variable must have the maximum C'ijijwhich is Xwhich is X3131 = 9= 9

To determine the leaving variable make a closedTo determine the leaving variable make a closedloop of basic variable starting with maximum C'loop of basic variable starting with maximum C'ijij ofofnonnon--basic variable keepingbasic variable keeping::

1.1. Supply and demand requirements remain satisfied.Supply and demand requirements remain satisfied.

2.2. No negative shipments are allowed through any of theNo negative shipments are allowed through any of theroutes.routes.

Therefore, =5 and XTherefore, =5 and X3131will be basic with value 5will be basic with value 5and Xand X2222 is a basic variable but with value 0 and Xis a basic variable but with value 0 and X1111will be nonwill be non--basic (leaving variable)basic (leaving variable)

The cost of the IBFS by Northwest corner methodThe cost of the IBFS by Northwest corner methodwas 520$was 520$

The first iteration cost is 475$ (less which is better)The first iteration cost is 475$ (less which is better)

In the second iterationIn the second iteration


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In the second iterationIn the second iteration

Get the entering variable with maximum C'Get the entering variable with maximum C'ijij = C'= C'1414 = 4= 4

The Leaving variable is XThe Leaving variable is X2424 =10=10

XX1414 =10 and the new solution is:=10 and the new solution is:

All C'All C' i i f lli i f ll b i i bl hb i i bl h


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All C'All C'ijij is negative for all nonis negative for all non--basic variables, so thebasic variables, so the

solution is optimal which is:solution is optimal which is:

XX1212 =5, X=5, X1414 =10, X=10, X2222 =10, X=10, X2323 =15, X=15, X3131 =5, X=5, X3434 =5=5

With optimum costWith optimum cost = 5*2 +10*11 + 10*7 + 15*9= 5*2 +10*11 + 10*7 + 15*9

+5*4 + 5*18+5*4 + 5*18

=1

0+11

0+70+1

35+20+90=

1

0+11

0+70+1

35+20+90= 435= 435


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The Assignment ModelThe Assignment Model

ThereThere areare nn machinesmachines MM11,, MM22,, MM33,, MnMn.. AndAnd nn

differentdifferent jobsjobs JJ11,, JJ22,, ..,, JnJn..

TheThe jobsjobs areare toto bebe assignedassigned toto thesethese machinesmachines.. TheThe

machinemachine costcost forfor eacheachjobjob dependsdepends onon thethe machinemachine totowhichwhich itit isis assignedassigned.. EachEach machinemachine cancan workwork onlyonly onon

oneone jobjob.. TheThe problemproblem isis toto assignassign thethe jobsjobs toto thethe

machinesmachines toto minimizeminimize thethe totaltotal costcost ofof machiningmachining..

NumberNumber ofof possiblepossible assignmentassignment == n!n!

IfIf n=n=55,, thenthen 55!! == 120120 waysways

The Linear ProgrammingThe Linear Programming


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The Linear ProgrammingThe Linear Programming

Formulation of the ProblemFormulation of the Problem Decision variablesDecision variables

XXijij job i is assigned to machine jjob i is assigned to machine j


MinimizeMinimize


i=1, 2 ni=1, 2 n

j=1, 2 nj=1, 2 n

XXijij =1=1 if the job i is assigned to the machine jif the job i is assigned to the machine j

XXijij =0=0 other wiseother wise

XCij

n

j

n

iijZ ! !

!

1 1

!

!

n

jijX

1

1

1

1

!

!

n

iij

X

The Assignment AlgorithmThe Assignment Algorithm


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The Assignment AlgorithmThe Assignment Algorithm

(Hungarian Algorithm)(Hungarian Algorithm)

It is the most efficient way or solving theIt is the most efficient way or solving the

assignment problem. It is based on the ideaassignment problem. It is based on the idea

that if a constant is added or subtracted fromthat if a constant is added or subtracted from

any row orcolumn the optimal assignment isany row orcolumn the optimal assignment isnot affected.not affected.

The Algorithm StepsThe Algorithm Steps


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The Algorithm StepsThe Algorithm Steps

1.1. Subtract the smallest value in eachcolumn from all the valuesSubtract the smallest value in eachcolumn from all the values

in that column.in that column.2.2. Subtract the smallest value in each row from all the values inSubtract the smallest value in each row from all the values in

that row.that row.

3.3. Note: step 1 & 2 will produce a cost matrix containing at leastNote: step 1 & 2 will produce a cost matrix containing at leastone zero element in each row orcolumn.one zero element in each row orcolumn.

4.4. Determine a feasible assignment using the cells with zero cost.Determine a feasible assignment using the cells with zero cost.I this is possible stop, as optimal solution is found. This step isI this is possible stop, as optimal solution is found. This step isdone as follows:done as follows:* Determine the number of lines required to cover all zeros.* Determine the number of lines required to cover all zeros.

* If the number of lines is equal to n, stop (This is the optimal* If the number of lines is equal to n, stop (This is the optimalassignment).assignment).

* If the number of lines is less than n then go to (4).* If the number of lines is less than n then go to (4).

5.5. Subtract the smallest value not covered with lines from allSubtract the smallest value not covered with lines from alluncovered values and add it to the value at the intersection ofuncovered values and add it to the value at the intersection ofthe lines then go to (3).the lines then go to (3).

6.6. Note: Obtain the minimum cost from the original table.Note: Obtain the minimum cost from the original table.


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Solve the following assignment problem.Solve the following assignment problem.

S l iS l i


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SolutionSolution::

The optimal assignment is XThe optimal assignment is X1212, X, X2121, X, X3333

Job1 is assigned to machine2Job1 is assigned to machine2



The cost =10 + 9 + 8 = 27$The cost =10 + 9 + 8 = 27$

E l 2E l 2


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Solve the following assignment problemSolve the following assignment problem..

S l iS l i


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SolutionSolution::

The optimal assignment is XThe optimal assignment is X1515, X, X2222, X, X3333, X, X44,44, XX5151



Job3 is assigned to machine3Job3 is assigned to machine3Job4 is assigned to machine4Job4 is assigned to machine4


The cost = 3 + 4 + 2 + 5 + 3 =17$The cost = 3 + 4 + 2 + 5 + 3 =17$

N t k M d lN t k M d l


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Network ModelsNetwork Models

There is a multitude of operations research situationsThere is a multitude of operations research situationsthat can be modeled and solved as networks (nodesthat can be modeled and solved as networks (nodesconnected with branches)connected with branches)

We have 5 network algorithms such as:We have 5 network algorithms such as:

1.1. Minimal spanning tree.Minimal spanning tree.

2.2. Shortest route algorithm.Shortest route algorithm.

3.3. MinimumMinimum--cost capacitated network algorithm.cost capacitated network algorithm.

4.4. Critical Path Method (CPM).Critical Path Method (CPM).5.5. Program Evaluation and Review TechniqueProgram Evaluation and Review Technique

(PERT).(PERT).

Project planning and control withProject planning and control with


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Project planning and control withProject planning and control with

PERTPERT--CPMCPM

The successful management of largeThe successful management of large--scale projectsscale projects

requires careful planning, scheduling and control ofrequires careful planning, scheduling and control of

numerous interrelated activities especially when thesenumerous interrelated activities especially when these

activities

have to be performed in a spe

cifieda

ctivities

have to be performed in a spe

cifiedtechnological sequence, to aid in these tasks specialtechnological sequence, to aid in these tasks special

procedures based on the use of network techniquesprocedures based on the use of network techniques

were developed. The most important of thesewere developed. The most important of these

procedures are:pro

cedures are:

oo Critical Path Method (CPM).Critical Path Method (CPM).

oo Program Evaluation and Review Technique (PERT).Program Evaluation and Review Technique (PERT).

A li ti f PERT & CPMA li ti f PERT & CPM


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Applications ofPERT & CPMApplications ofPERT & CPM

1.1. Construction projects ( buildings, highways,Construction projects ( buildings, highways,houses, bridges)houses, bridges)

2.2. Preparations of bids and proposals for largePreparations of bids and proposals for largeprojects.projects.

3.3. Maintenance planning of oil refineries, ship repairMaintenance planning of oil refineries, ship repairand other large operations.and other large operations.

4.4. Manufacture and assembly of large items as aeroManufacture and assembly of large items as aeroplans and computers.plans and computers.

5.5. Development of new weapons system.Development of new weapons system.

6.6. Simple projects such as cleaning and painting.Simple projects such as cleaning and painting.

Basic TerminologyBasic Terminology


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Basic TerminologyBasic Terminology

A Project:A Project:

A project defines a combination of interrelatedA project defines a combination of interrelated

activities that must be done in a certain order beforeactivities that must be done in a certain order before

the entire taskcan be completed.the entire taskcan be completed.

An Activity:An Activity: It is an effort required to complete a plan of theIt is an effort required to complete a plan of the

project. Most activities can not start until someproject. Most activities can not start until some

others are completed.others are completed.

Real Activity:Real Activity:

Consumes resources and time.Consumes resources and time.


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Waiting Activity:Waiting Activity:Consumes time.Consumes time.

Dummy Activity:Dummy Activity:

Consumes no time and no reso

urces.

It represents t

heCons

umes no time and no reso

urces.

It represents t

hedependence of one activity upon another.dependence of one activity upon another.

An Event:An Event:

Represents a point in time. It signifies the beginningRepresents a point in time. It signifies the beginning

or ending of some activities.or ending of some activities.

Network RepresentationNetwork Representation


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Network RepresentationNetwork Representation

An activity is represented by a line or an arrow (usually withAn activity is represented by a line or an arrow (usually with

time estimate written along it). The arrow begins and endstime estimate written along it). The arrow begins and ends

with an event which is represented by a node.with an event which is represented by a node.

A dummy activity is represented by dashed line arrow or solidA dummy activity is represented by dashed line arrow or solid

arrow with zero time estimate.arrow with zero time estimate.

The arrow must be oriented from left to right.The arrow must be oriented from left to right.

The network must have one initial node and one ending node.The network must have one initial node and one ending node.

Any two events may not be connected by more than oneAny two events may not be connected by more than one

activity.activity.

Event numbers must not be duplicated in the network.Event numbers must not be duplicated in the network.

Each activity is represented by one and only one arrow. NoEach activity is represented by one and only one arrow. No

single activity can be represented twice in the network.single activity can be represented twice in the network.

Numbering a networkNumbering a network


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Numbering a networkNumbering a network1.1. An initial event is one whichhas arrows coming outAn initial event is one whichhas arrows coming out

of it only. Number it (1).of it only. Number it (1).

2.2. Delete all arrows emerging from event (1). ThisDelete all arrows emerging from event (1). This

will create at least one more initial event, numberwill create at least one more initial event, number

these new events(s) as (2, 3).these new events(s) as (2, 3).

3.3. Delete all emerging arrows from these numberedDelete all emerging arrows from these numbered

events which will create new initial events andevents which will create new initial events and

continue until the last event whichhas no arrowscontinue until the last event whichhas no arrows

emerging from it.emerging from it.

Solution by Linear ProgrammingSolution by Linear Programming


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y g gy g g

FormulationFormulation

Each node represents a variable.Each node represents a variable.

Each arc represents a constraint.Each arc represents a constraint.



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pp

Consider the following Network problem.Consider the following Network problem.

SolutionSolution


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Minimize Z= t5Minimize Z= t5 t1t1ConstraintsConstraints

tt22 tt11 3 3

tt33 tt11 1 1

tt33 tt22 0 0tt44 tt33 22

tt44 tt22 4 4

tt55 tt44 5 5

ttii 0 0 i=1, 2, 3, 4, 5i=1, 2, 3, 4, 5

This problem can be solved by simplex method ZThis problem can be solved by simplex method Z** is theis theminimum time required to complete the project.minimum time required to complete the project.

Solution by Network AnalysisSolution by Network Analysis


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Solution by Network AnalysisSolution by Network Analysis

CPM Calculations:CPM Calculations:

Let DLet Dijij=Duration (completion time) of activity ij=Duration (completion time) of activity ij

ESESii=The earliest start time at which event (i)=The earliest start time at which event (i)occursoccurs

LFLFii=The latest finish time at which event (i)=The latest finish time at which event (i)

occurs without delaying the project beyond itsoccurs without delaying the project beyond its

earliest time.earliest time.


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ESESii . Forward. Forward

ESESii = max [ES= max [ESii+ D+ Dijij] for all (i, j)] for all (i, j)

ESESii = max [(2+8), (1+6), (3+5)]= max [(2+8), (1+6), (3+5)]

= max [10, 7, 8] =10= max [10, 7, 8] =10

Start from the initial node and move to theStart from the initial node and move to the

terminal node for initial node ESterminal node for initial node ES11

= 0= 0


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LFLFii . Backward. BackwardESESii = min [LF= min [LFjj -- DDijij] for all (i, j)] for all (i, j)

LFLFii = min [(10= min [(10--3), (93), (9--5), (135), (13--4)]4)]

= min [7, 4, 9] =10= min [7, 4, 9] =10

Start from the terminal node and move backwardStart from the terminal node and move backward

to the initial node for terminal node LFto the initial node for terminal node LFnn =ES=ESnn



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Determine the critical path for the projectDetermine the critical path for the project

network given below (all the durations are innetwork given below (all the durations are in

days)days)

SolutionSolution


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The critical path is defined by 1The critical path is defined by 1--22--44--55--66

The duration of the project = 25 daysThe duration of the project = 25 days

EndEnd


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EndEnd

Hope the best for every one in his lifeHope the best for every one in his life

Good luckGood luck

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