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Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Optimization TechniquesPrepared by Dr. S. AcharyaDepartment of Mathematics School of Applied Sciences KIIT University, Bhubaneswar, Odisha

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Outline

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Introduction Model Formulation Solution Procedure Basic Preliminaries Graphical Methods

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Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Introduction

Optimization is an act of obtaining best results under given restrictions. In several engineering design problems, engineers have to take many technological and managerial decisions at several stages. The objective of such decisions is to either minimize the effort required or to maximize the desired benet. The optimum seeking methods are known as Optimization Techniques. It is a part of Operations Research (OR). OR is a branch of Mathematics concerned with some techniques for nding best solutions.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Some Applications

Some Applications of Optimization Techniques are given as: Optimal Design of Solar Systems, Electrical Network Design, Energy Model and Planning, Optimal Design of Components of a System, Planning and Analysis of Existing Operations Optimal Design of Motors, Generators and Transformers, Design of Aircraft for Minimum Weight, Optimal Design of Bridge and Building. Optimization Techniques are divided into two different types, namely Linear Models and Non-Linear Models. At rst we shall discuss about some of the Linear Models.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Steps involved in mathematical programming

Conversion of stated problem into a mathematical model that abstracts all the essential elements of the problem. Exploration of different solutions of the problem. Finding out the most suitable or optimum solution. Linear programming requires that all the mathematical functions in the model be linear functions.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Linear Programming Models

Linear Models are known as Linear Programming Problem (LPP). Mathematical statement of a linear model is stated as follows: Find x1 , x2 , x3 , . . . , xn so as ton

Optimize(min / max) : Z =j=1

cj xj

(1)

subject ton

aij xj bi , i = 1, 2, 3, . . . , mj=1

(2)

xj 0, j = 1, 2, 3, . . . , n

(3)

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Linear Programming Models (Contd.)

Model (I)n

max : Z =j=1

cj xj

(4)

subject ton

aij xj (, =, )bi , i = 1, 2, 3, . . . , mj=1

(5) (6)

xj 0, j = 1, 2, 3, . . . , n

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Model (II)n

min : Z =j=1

cj xj

(7)

subject ton

aij xj (, =, )bi , i = 1, 2, 3, . . . , mj=1

(8) (9)

xj 0, j = 1, 2, 3, . . . , n

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Standard form of LPP

After introducing slack, surplus and articial variables a LPP can be put in standard form. Add a slack variable xn+i , where xn+i 0 forn

aij xj bi , bi 0j=1 n

j=1

aij xj + xn+i = bi

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Substract a surplus variable xn+i and add an articial variable xn+1+i , where xn+i , xn+1+i 0 forn

aij xj bi , bi 0j=1 n

j=1

aij xj xn+i + xn+1+i = bi

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Add an articial variable xn+i , where xn+i 0 forn

aij xj = bi , bi 0j=1 n

j=1

aij xj + xn+i = bi .

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


After introducing slack, surplus and articial variables a LPP can be put in standard form. Model (I)N

max : Z =j=1

cj xj

(10)

subject toN

aij xj = bi , i = 1, 2, 3, . . . , mj=1

(11)

xj 0, j = 1, 2, 3, . . . , N

(12)

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Model (II)N

min : Z =j=1

cj xj

(13)

subject toN

aij xj = bi , i = 1, 2, 3, . . . , mj=1

(14)

xj 0, j = 1, 2, 3, . . . , N

(15)

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Model Formulation

Steps Involved Identify the unknown variables to be determined (decision variables) and represent them in terms of algebraic symbols. Identify all the restrictions or constraints in the problem and express them as linear equations or inequalities which are linear functions of the unknown variables. Identify the objective or criterion and represent it as a linear function of of the decision variables, which is to be maximized or minimized.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Product Mix Problem

The N. P. Company produces two products: I and II. The raw material requirements, space needed for storage, production rates, and selling prices for these products are given in Table 1.Table 1: Production Data for N.P. Company

Requirements Storage Space (ft 2 /unit) Raw materials (lb/unit) Production Rate (units/hour ) Selling Price ($/unit)

Product I 4 5 60 13

Product II 5 3 30 11

The total amount of raw material available per day for both products is 1575 lb. The total storage space for all products is 1500 ft 2 , and a maximum of 7 hours per day can be used for production. All products manufactured are shipped out of the storage area at the end of the day. Therefore, the two products must share the total raw material, storage space, and production time. The company wants to determine how many units of each product to produce per day to maximize its total income.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Formulation of PMP

The company has decided that it wants to maximize its sale income, which depends on the number of units of product I and II that it produces. Therefore, the decision variables, x1 and x2 can be the number of units of products I and II, respectively, produced per day. The object is to maximize the equation: Z = 13x1 + 11x2 subject to the constraints on storage space, raw materials, and production time. Each unit of product I requires 4ft 2 of storage space and each unit of product II requires 5ft 2 . Thus a total of 4x1 + 5x2 ft 2 of storage space is needed each day. This space must be less than or equal to the available storage space, which is 1500ft 2 . Therefore,4X1 + 5X2 1500 Similarly, each unit of product I and II produced requires 5 and 3 lb, respectively, of raw material. Hence a total of 5xl + 3x2 lb of raw material is used.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Formulation of PMP (Contd.)

This must be less than or equal to the total amount of raw material available, which is 1575 lb. Therefore, 5x1 + 3x2 1575 Prouct I can be produced at the rate of 60 units per hour. Therefore, it must take I minute or 1/60 of an hour to produce I unit. Similarly, it requires 1/30 of an hour to produce 1 unit of product II. Hence a total of x1 /60 + x2 /30 hours is required for the daily production. This quantity must be less than or equal to the total production time available each day. Therefore, x1 /60 + x2 /30 7 or x1 + 2x2 420. Finally, the company cannot produce a negative quantity of any product, therefore x1 and x2 must each be greater than or equal to zero.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Formulation of PMP (Contd.)

The linear programming model for this example can be summarized as: max : Z = 13x1 + 11x2 subject to 4X1 + 5X2 5x1 + 3x2 x1 + 2x2 X1 X2 1500 1575 420 0 0

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Solution Procedure

An LPP can be solved by the following method Graphical Method (Only 2-variable problem), Simplex Method, Big-M Method/ Charnes Penalty Method, Two-Phase Simplex Method, Revised Simplex Method, Dual Simplex Method, Primal-Dual Simplex Method, Interior Point Method.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Basic Solution

Given a system AX = b of m linear equations in n variables (n > m), the system is consistent and the solutions are innite if r (A) = m, m < n i.e. Rank of A is m where m < n. We may select any m variables out of n variables. Set the remaining (n m) variables to zero. The system AX = b becomes BXB = b. If it has a solution then then XB = B 1 b. XB is called basic solution. n n Maximum possible basic solutions= m = nm . If value of all the variables (decision variables) are non-negative, then such basic solution is known as Basic Feasible Solution (B.F.S.).

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Basic Solution (Contd.)

Find Basic Solution: x1 + x2 + x3 x1 + 4x2 = 10

+ x4 = 16

Sl. No. 1 2 3 4 5 6

Non-Basic Variables x1 = 0, x2 = 0 x1 = 0, x3 = 0 x1 = 0, x4 = 0 x2 = 0, x3 = 0 x2 = 0, x4 = 0 x3 = 0, x4 = 0

Basic Variables x3 = 10, x4 = 6 x2 = 10, x4 = 24 x2 = 4, x3 = 6 x1 = 10, x4 = 6 x1 = 16, x3 = 6 x1 = 8, x2 = 2

There are six Basic Solutions. Only four are Basic Feasible Solutions. Sl. No. (2) and (5) are not Basic Feasible Solutions (B.F.S.).

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Point in n-dimensional space: A point X = (x1 , x2 , x3 , . . . , xn )T has n co-ordinates xi , i = 1, 2, 3, . . . , n. Each of them are real numbers. Line Segment in n-dimensions: Let X1 be the coordinates of A and X2 be the coordinates of B. The line segment joining these two points is given by X (), 0 1 i.e. L = {X ()|X () = X1 + (1 )X2 , 0 1}

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Denition Hyperplane: A hyperplane H is dened as: H = {X |C T X = b} c1 x1 + c2 x2 + . . . + cn xn = b A hyperplane has n 1 dimensions in an n-dimensional space. In 2-dimensional space hyperplane is a line. In 3-dimensional space it is a plane. A hyperplane divides the n-dimensional space into two closed half spaces as: (i) c1 x1 + c2 x2 + . . . + cn xn b (ii) c1 x1 + c2 x2 + . . . + cn xn b

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Basic Preliminaries (Contd.)

Denition Convex Sex: A convex set S is a collection of points such that if X1 and X2 are any two points in the set, the line segment joining them is also in the set S. Let X = X1 + (1 )X2 , 0 1 If X1 , X2 S, then X S. Denition Convex Polyhedron and Convex Polytope: A convex polyhedron is a set S (a set of points) which is common to one or more half spaces. A convex polyhedron that is bounded is called a convex polytope. Denition Extreme Point: It is a point in the Convex set S which does not lie on a line segment joining two other points of the set S.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Denition Feasible Solution: In a LPP any solution X which satisfy AX = b and X 0 is called a feasible solution. Denition Basic Solution: This is a solution in which (n m) variables are set equal to zero in AX = b. It has m equations and n unknowns n > m Denition Basis: The collection of variables not in set equal to zero to obtain the basic solution is the basis. Denition Basic Feasible Solution (B.F.S.): The basic solution which satisfy the conditions X 0 is called B.F.S.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Denition Non-degenerate B.F.S.: It is a B.F.S. which has exactly m positive xi out of n. Denition Optimal Solution: A B.F.S. which optimizes the objective function is called an optimal solution. Theorem The intersection of any number of convex sets is also convex. Proof: Let R1 , R2 , . . . , Rk be convex sets and their intersection be R i.e.k

R=i=1

Ri

Let X1 and X2 R. Then X1 + (1 )X2 R, where 0 1. Thus X Ri , i = 1, 2, . . . , k.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Theorem The feasible region of a linear programming problem (LPP) forms a convex set. Proof: The feasible region of LPP is dened as: S = {X |AX = b, X 0} Let the points X1 and X2 be in the feasible set S so that AX1 = b, X1 0; AX2 = b, X2 0. Let 0 1. we have 0, 1 0. Now A[X1 + (1 )X2 ] = b + (1 )b = b AX = b. Thus the point X satises the constraints if 0 1, X 0.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Theorem In general a LPP has either one optimal solution or no optimal solution or innite number of optimal solutions. Any local minimum/maximum solution is a global minimum/maximum solution of a LPP. (LPP) max : Z = C T X subject to : AX = b X 0 X is a maximizing point of the LPP. (LPP) min : Z = C T X subject to : AX = b X 0 X is a minimizing point of the LPP.

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Theorem Every B.F.S. is an extreme point of the convex set of the feasible region. Proof: Let X = (x1 , x2 , x3 , . . . , xm , xm+1 , xm+2 , . . . , xn ) be a BFS of the LPP where x1 , x2 , x3 , . . . , xm are basic variables. Now x1 = b1 , x2 = b2 , x3 = b3 , . . . , xm = bm , x1 , x2 , x3 , . . . , xm 0. This feasible region forms a convex set. To show X is an extreme point, we must show that there do not exist feasible solutions Y and Z such that X = Y + (1 )Z , 0 1 Let Y = (y1 , y2 , y3 , . . . , ym , ym+1 , ym+2 , . . . , yn ) and T Z = (z1 , z2 , z3 , . . . , zm , zm+1 , zm+2 , . . . , zn ) Last (n m) components gives yj + (1 )zj = 0, j = m + 1, m + 2, . . . , n. Since 0, (1 ) 0, yj 0, zj 0, it gives yj = zj = 0, j = m + 1, m + 2, . . . , n. This shows that Y = Z = X . So, X is an extreme point.T T

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods


Theorem Let S be a closed bounded convex polyhedron with p number of extreme points Xi , i = 1, 2, . . . , p. Then any vector X S can be written asp p

X =i=1

i Xi , i 0,i=1

i = 1

Theorem Let S be a closed convex polyhedron. Then the minimum of a linear function over S attained at an extreme point of S. Proof:Suppose X minimizes the objective function Z = C T X over S and minimizes does not occur at an extreme point. From the denition of minimum C T X < C T Xi , i = 1, 2, . . . , p with p number of extreme points. For 0 i 1, i C T X < i C T Xi , i = 1, 2, . . . , p

Introduction

Model Formulation

Solution Procedure

Basic Preliminaries

Graphical Methods

Basic Preliminaries (Contd.)p p

i C X

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