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Quantitative TechniquesQuantitative Techniques
Asif Bodla
Topics:Topics:
 Linear ProgrammingLinear Programming
 Transportation ProblemTransportation Problem
 ForecastingForecasting
 Assignment problemAssignment problem
 Queuing TheoryQueuing Theory
 Decision TheoryDecision Theory
 Inventory ManagementInventory Management
 SimulationSimulation
 Network AnalysisNetwork Analysis
LINEARLINEAR
PROGRAMMINGPROGRAMMING
Linear ProgrammingLinear Programming
 It is a mathematical technique forIt is a mathematical technique for
optimum allocation of scarce or limitedoptimum allocation of scarce or limited
resources to several competingresources to several competing
activities on the basis of given criterionactivities on the basis of given criterion
of optimality, which can be eitherof optimality, which can be either
performance, ROI, cost, utility, time,performance, ROI, cost, utility, time,
distance etc.distance etc.
StepsSteps
 Define decision variablesDefine decision variables
 Formulate the objective functionFormulate the objective function
 Formulate the constraintsFormulate the constraints
 Mention the non-negativity criteriaMention the non-negativity criteria
Components &Components &
AssumptionsAssumptions
 ObjectiveObjective
 Decision VariableDecision Variable
 ConstraintConstraint
 ParametersParameters
 Non-negativityNon-negativity
 ProportionalityProportionality
 AddivityAddivity
 DivisibilityDivisibility
 CertainityCertainity
Problem:Problem:
An animal feed company mustAn animal feed company must
produce at least 200 kgs of a mixtureproduce at least 200 kgs of a mixture
consisting of ingredients x1 and x2consisting of ingredients x1 and x2
daily. x1 costs Rs.3 per kg. and x2daily. x1 costs Rs.3 per kg. and x2
Rs.8 per kg. No more than 80 kg. of x1Rs.8 per kg. No more than 80 kg. of x1
can be used and at least 60 kg. of x2can be used and at least 60 kg. of x2
must be used. Formulate amust be used. Formulate a
mathematical model to the problem.mathematical model to the problem.
Solution:Solution:
Minimize Z = 3x1 + 8x2Minimize Z = 3x1 + 8x2
Subject to x1 + x2 >= 200Subject to x1 + x2 >= 200
x1 <= 80x1 <= 80
x2 >= 60x2 >= 60
X1 >= 0 , x2 >= 0X1 >= 0 , x2 >= 0
Graphical SolutionGraphical Solution
 Formulate the problemFormulate the problem
 Convert all inequalities to equationsConvert all inequalities to equations
 Plot the graph of all inequalitiesPlot the graph of all inequalities
 Find out the feasilble regionFind out the feasilble region
 Find out the corner pointsFind out the corner points
 Substitute the objective functionSubstitute the objective function
 Arrive at the solutionArrive at the solution
Problem:Problem:
 Maximize Z = 60x1+50x2Maximize Z = 60x1+50x2
subject to 4x1+10x2 <= 100subject to 4x1+10x2 <= 100
2x1+1x2 <= 222x1+1x2 <= 22
3x1+3x2 <= 393x1+3x2 <= 39
x1,x2 >= 0x1,x2 >= 0
Solution :Solution :
4x1+10x2=1004x1+10x2=100 (0,10)(25,0)(0,10)(25,0)
2x1+x2=222x1+x2=22 (0,22)(11,0)(0,22)(11,0)
3x1+3x2=393x1+3x2=39 (0,13)(13,0)(0,13)(13,0)
0
x2
x1
10
13
22
11 13 25
E
C
B
A
D
Z = 60x1 + 50x2Z = 60x1 + 50x2
A (0,0) = 60*0+50*0 = 0A (0,0) = 60*0+50*0 = 0
B (11,0) = 60*11+50*0 = 660B (11,0) = 60*11+50*0 = 660
C (9,4) = 60*9+50*4 = 740C (9,4) = 60*9+50*4 = 740
D (5,8) = 60*5+50*8 = 700D (5,8) = 60*5+50*8 = 700
E (0,10) = 60*0+50*10 = 500E (0,10) = 60*0+50*10 = 500
Max Z is at C (9,4) and Z = 740Max Z is at C (9,4) and Z = 740
TRANSPORTATIONTRANSPORTATION
PROBLEMPROBLEM
Transportation ProblemTransportation Problem
 A special kind of optimisation problemA special kind of optimisation problem
in which goods are transported from ain which goods are transported from a
set of sources to a set of destinationsset of sources to a set of destinations
subject to the supply and demandsubject to the supply and demand
constraints. The main objective is toconstraints. The main objective is to
minimize the total cost ofminimize the total cost of
transportation.transportation.
Initial Basic Feasible SolutionInitial Basic Feasible Solution
 North West Corner MethodNorth West Corner Method
 Least Cost MethodLeast Cost Method
 Vogel’s Approximation MethodVogel’s Approximation Method
The solution is said to be feasible whenThe solution is said to be feasible when
one gets (m+n-1) allotments.one gets (m+n-1) allotments.
Assignment ProblemAssignment Problem
 It is a problem of assigning variousIt is a problem of assigning various
people, machines and so on in such apeople, machines and so on in such a
way that the total cost involved isway that the total cost involved is
minimized or the total value isminimized or the total value is
maximized.maximized.
ForecastingForecasting
Forecasting is the inherentForecasting is the inherent
process for all Organizations.process for all Organizations.
It is extremely important becauseIt is extremely important because
you have to commit resources.you have to commit resources.
Types/techniques ofTypes/techniques of
forecastingforecasting
Three techniques of forecasting:Three techniques of forecasting:
1)1)QualitativeQualitative
2)2)Time SeriesTime Series
3)3)CasualCasual
Queuing TheoryQueuing Theory
Queuing TheoryQueuing Theory
 A flow of customers from finite/infiniteA flow of customers from finite/infinite
population towards the service facilitypopulation towards the service facility
forms a queue due to lack of capacityforms a queue due to lack of capacity
to serve them all at a time.to serve them all at a time.
InputInput OutputOutputServer
MeasuresMeasures
 Traffic intensityTraffic intensity
 Average system lengthAverage system length
 Average queue lengthAverage queue length
 Average waiting time in queueAverage waiting time in queue
 Average waiting time in systemAverage waiting time in system
 Probability of queue lengthProbability of queue length
Queuing & cost behaviorQueuing & cost behavior
Cost of
service
Cost of waiting
Total cost
DECISION THEORYDECISION THEORY
Decision TheoryDecision Theory
The decision making environmentThe decision making environment
 Under certainityUnder certainity
 Under uncertainityUnder uncertainity
 Under riskUnder risk
Decision making underDecision making under
uncertainityuncertainity
 Laplace CriterionLaplace Criterion
 Maxmin CriterionMaxmin Criterion
 Minmax CriterionMinmax Criterion
 Maxmax CriterionMaxmax Criterion
 Minmin CriterionMinmin Criterion
 Salvage CriterionSalvage Criterion
 Hurwicz CriterionHurwicz Criterion
Inventory managementInventory management
 Inventory is vital to the sucessfulInventory is vital to the sucessful
functioning of manufacturing andfunctioning of manufacturing and
retailing organisations. They may beretailing organisations. They may be
raw materials, work-in-progress, spareraw materials, work-in-progress, spare
parts/consumables and finishedparts/consumables and finished
goods.goods.
ModelsModels
 Deterministic Inventory ModelDeterministic Inventory Model
 Inventory Model with Price breaksInventory Model with Price breaks
 Probabilistic Inventory ModelProbabilistic Inventory Model
Basic EOQ ModelBasic EOQ Model
Slope=0 Total cost
Carrying cost
Ordering cost
Minimu
m total
cost
Optimal order qty
SIMULATIONSIMULATION
SimulationSimulation
 It involves developing a model ofIt involves developing a model of
some real phenomenon and thensome real phenomenon and then
performing experiments on the modelperforming experiments on the model
evolved. It is descriptive in nature andevolved. It is descriptive in nature and
not an optimizing model.not an optimizing model.
ProcessProcess
 Definition of the problemDefinition of the problem
 Construction of an appropriate modelConstruction of an appropriate model
 Experimentation with the modelExperimentation with the model
 Evaluation of the results of simulationEvaluation of the results of simulation
NETWORK ANALYSISNETWORK ANALYSIS
PERTPERT
CPMCPM
Network Analysis /Network Analysis /
Project ManagementProject Management
A project is a series of activities directedA project is a series of activities directed
to the accomplishment of a desiredto the accomplishment of a desired
objective.objective.
 PERTPERT
 CPMCPM
CPM-Critical PathCPM-Critical Path
MethodMethod
 Activities are shown as a network ofActivities are shown as a network of
precedence relationship using Activity-precedence relationship using Activity-
On-Arrow (A-O-A) networkOn-Arrow (A-O-A) network
construction.construction.
 There is single stimate of activity timeThere is single stimate of activity time
 Deterministic activity timeDeterministic activity time
Project Evaluation &Project Evaluation &
Review TechniqueReview Technique
 Activities are shown as a network ofActivities are shown as a network of
precedence relationships using A-O-Aprecedence relationships using A-O-A
network construction.network construction.
 Multiple time estimatesMultiple time estimates
 Probabilistic activity timeProbabilistic activity time
CrashingCrashing
 Crashing is shortening the activityCrashing is shortening the activity
duration by employing moreduration by employing more
resources.resources.
cost slope = Cc – Cn/ Tn - Tccost slope = Cc – Cn/ Tn - Tc

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Quantitative techniques

  • 2. Topics:Topics:  Linear ProgrammingLinear Programming  Transportation ProblemTransportation Problem  ForecastingForecasting  Assignment problemAssignment problem  Queuing TheoryQueuing Theory  Decision TheoryDecision Theory  Inventory ManagementInventory Management  SimulationSimulation  Network AnalysisNetwork Analysis
  • 4. Linear ProgrammingLinear Programming  It is a mathematical technique forIt is a mathematical technique for optimum allocation of scarce or limitedoptimum allocation of scarce or limited resources to several competingresources to several competing activities on the basis of given criterionactivities on the basis of given criterion of optimality, which can be eitherof optimality, which can be either performance, ROI, cost, utility, time,performance, ROI, cost, utility, time, distance etc.distance etc.
  • 5. StepsSteps  Define decision variablesDefine decision variables  Formulate the objective functionFormulate the objective function  Formulate the constraintsFormulate the constraints  Mention the non-negativity criteriaMention the non-negativity criteria
  • 6. Components &Components & AssumptionsAssumptions  ObjectiveObjective  Decision VariableDecision Variable  ConstraintConstraint  ParametersParameters  Non-negativityNon-negativity  ProportionalityProportionality  AddivityAddivity  DivisibilityDivisibility  CertainityCertainity
  • 7. Problem:Problem: An animal feed company mustAn animal feed company must produce at least 200 kgs of a mixtureproduce at least 200 kgs of a mixture consisting of ingredients x1 and x2consisting of ingredients x1 and x2 daily. x1 costs Rs.3 per kg. and x2daily. x1 costs Rs.3 per kg. and x2 Rs.8 per kg. No more than 80 kg. of x1Rs.8 per kg. No more than 80 kg. of x1 can be used and at least 60 kg. of x2can be used and at least 60 kg. of x2 must be used. Formulate amust be used. Formulate a mathematical model to the problem.mathematical model to the problem.
  • 8. Solution:Solution: Minimize Z = 3x1 + 8x2Minimize Z = 3x1 + 8x2 Subject to x1 + x2 >= 200Subject to x1 + x2 >= 200 x1 <= 80x1 <= 80 x2 >= 60x2 >= 60 X1 >= 0 , x2 >= 0X1 >= 0 , x2 >= 0
  • 9. Graphical SolutionGraphical Solution  Formulate the problemFormulate the problem  Convert all inequalities to equationsConvert all inequalities to equations  Plot the graph of all inequalitiesPlot the graph of all inequalities  Find out the feasilble regionFind out the feasilble region  Find out the corner pointsFind out the corner points  Substitute the objective functionSubstitute the objective function  Arrive at the solutionArrive at the solution
  • 10. Problem:Problem:  Maximize Z = 60x1+50x2Maximize Z = 60x1+50x2 subject to 4x1+10x2 <= 100subject to 4x1+10x2 <= 100 2x1+1x2 <= 222x1+1x2 <= 22 3x1+3x2 <= 393x1+3x2 <= 39 x1,x2 >= 0x1,x2 >= 0
  • 11. Solution :Solution : 4x1+10x2=1004x1+10x2=100 (0,10)(25,0)(0,10)(25,0) 2x1+x2=222x1+x2=22 (0,22)(11,0)(0,22)(11,0) 3x1+3x2=393x1+3x2=39 (0,13)(13,0)(0,13)(13,0) 0 x2 x1 10 13 22 11 13 25 E C B A D
  • 12. Z = 60x1 + 50x2Z = 60x1 + 50x2 A (0,0) = 60*0+50*0 = 0A (0,0) = 60*0+50*0 = 0 B (11,0) = 60*11+50*0 = 660B (11,0) = 60*11+50*0 = 660 C (9,4) = 60*9+50*4 = 740C (9,4) = 60*9+50*4 = 740 D (5,8) = 60*5+50*8 = 700D (5,8) = 60*5+50*8 = 700 E (0,10) = 60*0+50*10 = 500E (0,10) = 60*0+50*10 = 500 Max Z is at C (9,4) and Z = 740Max Z is at C (9,4) and Z = 740
  • 14. Transportation ProblemTransportation Problem  A special kind of optimisation problemA special kind of optimisation problem in which goods are transported from ain which goods are transported from a set of sources to a set of destinationsset of sources to a set of destinations subject to the supply and demandsubject to the supply and demand constraints. The main objective is toconstraints. The main objective is to minimize the total cost ofminimize the total cost of transportation.transportation.
  • 15. Initial Basic Feasible SolutionInitial Basic Feasible Solution  North West Corner MethodNorth West Corner Method  Least Cost MethodLeast Cost Method  Vogel’s Approximation MethodVogel’s Approximation Method The solution is said to be feasible whenThe solution is said to be feasible when one gets (m+n-1) allotments.one gets (m+n-1) allotments.
  • 16. Assignment ProblemAssignment Problem  It is a problem of assigning variousIt is a problem of assigning various people, machines and so on in such apeople, machines and so on in such a way that the total cost involved isway that the total cost involved is minimized or the total value isminimized or the total value is maximized.maximized.
  • 17. ForecastingForecasting Forecasting is the inherentForecasting is the inherent process for all Organizations.process for all Organizations. It is extremely important becauseIt is extremely important because you have to commit resources.you have to commit resources.
  • 18. Types/techniques ofTypes/techniques of forecastingforecasting Three techniques of forecasting:Three techniques of forecasting: 1)1)QualitativeQualitative 2)2)Time SeriesTime Series 3)3)CasualCasual
  • 20. Queuing TheoryQueuing Theory  A flow of customers from finite/infiniteA flow of customers from finite/infinite population towards the service facilitypopulation towards the service facility forms a queue due to lack of capacityforms a queue due to lack of capacity to serve them all at a time.to serve them all at a time. InputInput OutputOutputServer
  • 21. MeasuresMeasures  Traffic intensityTraffic intensity  Average system lengthAverage system length  Average queue lengthAverage queue length  Average waiting time in queueAverage waiting time in queue  Average waiting time in systemAverage waiting time in system  Probability of queue lengthProbability of queue length
  • 22. Queuing & cost behaviorQueuing & cost behavior Cost of service Cost of waiting Total cost
  • 24. Decision TheoryDecision Theory The decision making environmentThe decision making environment  Under certainityUnder certainity  Under uncertainityUnder uncertainity  Under riskUnder risk
  • 25. Decision making underDecision making under uncertainityuncertainity  Laplace CriterionLaplace Criterion  Maxmin CriterionMaxmin Criterion  Minmax CriterionMinmax Criterion  Maxmax CriterionMaxmax Criterion  Minmin CriterionMinmin Criterion  Salvage CriterionSalvage Criterion  Hurwicz CriterionHurwicz Criterion
  • 26. Inventory managementInventory management  Inventory is vital to the sucessfulInventory is vital to the sucessful functioning of manufacturing andfunctioning of manufacturing and retailing organisations. They may beretailing organisations. They may be raw materials, work-in-progress, spareraw materials, work-in-progress, spare parts/consumables and finishedparts/consumables and finished goods.goods.
  • 27. ModelsModels  Deterministic Inventory ModelDeterministic Inventory Model  Inventory Model with Price breaksInventory Model with Price breaks  Probabilistic Inventory ModelProbabilistic Inventory Model
  • 28. Basic EOQ ModelBasic EOQ Model Slope=0 Total cost Carrying cost Ordering cost Minimu m total cost Optimal order qty
  • 30. SimulationSimulation  It involves developing a model ofIt involves developing a model of some real phenomenon and thensome real phenomenon and then performing experiments on the modelperforming experiments on the model evolved. It is descriptive in nature andevolved. It is descriptive in nature and not an optimizing model.not an optimizing model.
  • 31. ProcessProcess  Definition of the problemDefinition of the problem  Construction of an appropriate modelConstruction of an appropriate model  Experimentation with the modelExperimentation with the model  Evaluation of the results of simulationEvaluation of the results of simulation
  • 33. Network Analysis /Network Analysis / Project ManagementProject Management A project is a series of activities directedA project is a series of activities directed to the accomplishment of a desiredto the accomplishment of a desired objective.objective.  PERTPERT  CPMCPM
  • 34. CPM-Critical PathCPM-Critical Path MethodMethod  Activities are shown as a network ofActivities are shown as a network of precedence relationship using Activity-precedence relationship using Activity- On-Arrow (A-O-A) networkOn-Arrow (A-O-A) network construction.construction.  There is single stimate of activity timeThere is single stimate of activity time  Deterministic activity timeDeterministic activity time
  • 35. Project Evaluation &Project Evaluation & Review TechniqueReview Technique  Activities are shown as a network ofActivities are shown as a network of precedence relationships using A-O-Aprecedence relationships using A-O-A network construction.network construction.  Multiple time estimatesMultiple time estimates  Probabilistic activity timeProbabilistic activity time
  • 36. CrashingCrashing  Crashing is shortening the activityCrashing is shortening the activity duration by employing moreduration by employing more resources.resources. cost slope = Cc – Cn/ Tn - Tccost slope = Cc – Cn/ Tn - Tc