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Quantitative Techniques Deepthy Sai Manikandan. Topics: Linear Programming Linear Programming Transportation Problem Transportation Problem Assignment.

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Presentation on theme: "Quantitative Techniques Deepthy Sai Manikandan. Topics: Linear Programming Linear Programming Transportation Problem Transportation Problem Assignment."— Presentation transcript:

1 Quantitative Techniques Deepthy Sai Manikandan

2 Topics: Linear Programming Linear Programming Transportation Problem Transportation Problem Assignment problem Assignment problem Queuing Theory Queuing Theory Decision Theory Decision Theory Inventory Management Inventory Management Simulation Simulation Network Analysis Network Analysis

3 LINEAR PROGRAMMING

4 Linear Programming It is a mathematical technique for optimum allocation of scarce or limited resources to several competing activities on the basis of given criterion of optimality, which can be either performance, ROI, cost, utility, time, distance etc. It is a mathematical technique for optimum allocation of scarce or limited resources to several competing activities on the basis of given criterion of optimality, which can be either performance, ROI, cost, utility, time, distance etc.

5 Steps Define decision variables Define decision variables Formulate the objective function Formulate the objective function Formulate the constraints Formulate the constraints Mention the non-negativity criteria Mention the non-negativity criteria

6 Components & Assumptions Objective Objective Decision Variable Decision Variable Constraint Constraint Parameters Parameters Non-negativity Non-negativity Proportionality Proportionality Addivity Addivity Divisibility Divisibility Certainity Certainity

7 Problem: An animal feed company must produce at least 200 kgs of a mixture consisting of ingredients x1 and x2 daily. x1 costs Rs.3 per kg. and x2 Rs.8 per kg. No more than 80 kg. of x1 can be used and at least 60 kg. of x2 must be used. Formulate a mathematical model to the problem. An animal feed company must produce at least 200 kgs of a mixture consisting of ingredients x1 and x2 daily. x1 costs Rs.3 per kg. and x2 Rs.8 per kg. No more than 80 kg. of x1 can be used and at least 60 kg. of x2 must be used. Formulate a mathematical model to the problem.

8 Solution: Minimize Z = 3x1 + 8x2 Minimize Z = 3x1 + 8x2 Subject to x1 + x2 >= 200 x1 <= 80 x1 <= 80 x2 >= 60 x2 >= 60 X1 >= 0, x2 >= 0 X1 >= 0, x2 >= 0

9 Graphical Solution Formulate the problem Formulate the problem Convert all inequalities to equations Convert all inequalities to equations Plot the graph of all inequalities Plot the graph of all inequalities Find out the feasilble region Find out the feasilble region Find out the corner points Find out the corner points Substitute the objective function Substitute the objective function Arrive at the solution Arrive at the solution

10 Problem: Maximize Z = 60x1+50x2 Maximize Z = 60x1+50x2 subject to 4x1+10x2 <= 100 subject to 4x1+10x2 <= 100 2x1+1x2 <= 22 2x1+1x2 <= 22 3x1+3x2 <= 39 3x1+3x2 <= 39 x1,x2 >= 0

11 Solution : 4x1+10x2=100(0,10)(25,0) 4x1+10x2=100(0,10)(25,0) 2x1+x2=22 (0,22)(11,0) 2x1+x2=22 (0,22)(11,0) 3x1+3x2=39 (0,13)(13,0) 3x1+3x2=39 (0,13)(13,0) 0 x2 x1 10 13 22 111325 E C B A D

12 A (0,0) = 60*0+50*0 = 0 B (11,0) = 60*11+50*0 = 660 C (9,4) = 60*9+50*4 = 740 D (5,8) = 60*5+50*8 = 700 E (0,10) = 60*0+50*10 = 500 Max Z is at C (9,4) and Z = 740 Z = 60x1 + 50x2

13 TRANSPORTATION PROBLEM

14 Transportation Problem A special kind of optimisation problem in which goods are transported from a set of sources to a set of destinations subject to the supply and demand constraints. The main objective is to minimize the total cost of transportation. A special kind of optimisation problem in which goods are transported from a set of sources to a set of destinations subject to the supply and demand constraints. The main objective is to minimize the total cost of transportation.

15 Initial Basic Feasible Solution North West Corner Method North West Corner Method Least Cost Method Least Cost Method Vogel’s Approximation Method Vogel’s Approximation Method The solution is said to be feasible when one gets (m+n-1) allotments.

16 Assignment Problem It is a problem of assigning various people, machines and so on in such a way that the total cost involved is minimized or the total value is maximized. It is a problem of assigning various people, machines and so on in such a way that the total cost involved is minimized or the total value is maximized.

17 QUEUING THEORY

18 Queuing Theory A flow of customers from finite/infinite population towards the service facility forms a queue due to lack of capacity to serve them all at a time. A flow of customers from finite/infinite population towards the service facility forms a queue due to lack of capacity to serve them all at a time. Input Output Server

19 Measures Traffic intensity Traffic intensity Average system length Average system length Average queue length Average queue length Average waiting time in queue Average waiting time in queue Average waiting time in system Average waiting time in system Probability of queue length Probability of queue length

20 Queuing & cost behavior Cost of service Cost of waiting Total cost

21 DECISION THEORY

22 Decision Theory The decision making environment Under certainity Under certainity Under uncertainity Under uncertainity Under risk Under risk

23 Decision making under uncertainity Laplace Criterion Laplace Criterion Maxmin Criterion Maxmin Criterion Minmax Criterion Minmax Criterion Maxmax Criterion Maxmax Criterion Minmin Criterion Minmin Criterion Salvage Criterion Salvage Criterion Hurwicz Criterion Hurwicz Criterion

24 Inventory management Inventory is vital to the sucessful functioning of manufacturing and retailing organisations. They may be raw materials, work-in-progress, spare parts/consumables and finished goods. Inventory is vital to the sucessful functioning of manufacturing and retailing organisations. They may be raw materials, work-in-progress, spare parts/consumables and finished goods.

25 Models Deterministic Inventory Model Deterministic Inventory Model Inventory Model with Price breaks Inventory Model with Price breaks Probabilistic Inventory Model Probabilistic Inventory Model

26 Basic EOQ Model Slope=0 Total cost Carrying cost Ordering cost Minimu m total cost Optimal order qty

27 SIMULATION

28 Simulation It involves developing a model of some real phenomenon and then performing experiments on the model evolved. It is descriptive in nature and not an optimizing model. It involves developing a model of some real phenomenon and then performing experiments on the model evolved. It is descriptive in nature and not an optimizing model.

29 Process Definition of the problem Definition of the problem Construction of an appropriate model Construction of an appropriate model Experimentation with the model Experimentation with the model Evaluation of the results of simulation Evaluation of the results of simulation

30 NETWORK ANALYSIS PERTCPM

31 A project is a series of activities directed to the accomplishment of a desired objective. A project is a series of activities directed to the accomplishment of a desired objective. PERT PERT CPM CPM Network Analysis / Project Management

32 CPM-Critical Path Method Activities are shown as a network of precedence relationship using Activity- On-Arrow (A-O-A) network construction. Activities are shown as a network of precedence relationship using Activity- On-Arrow (A-O-A) network construction. There is single stimate of activity time There is single stimate of activity time Deterministic activity time Deterministic activity time

33 Project Evaluation & Review Technique Activities are shown as a network of precedence relationships using A-O-A network construction. Activities are shown as a network of precedence relationships using A-O-A network construction. Multiple time estimates Multiple time estimates Probabilistic activity time Probabilistic activity time

34 Crashing Crashing is shortening the activity duration by employing more resources. Crashing is shortening the activity duration by employing more resources. cost slope = Cc – Cn/ Tn - Tc cost slope = Cc – Cn/ Tn - Tc


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