lec 1 intro to OT

8/6/2019 lec 1 intro to OT

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Introduction to Optimization

Dr. Nasir M Mirza

Optimization TechniquesOptimization Techniques

Email: [email protected]


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nasir m mirza Slide 2

BooksRecommended Books:

1. Kalyanmoy Deb, optimization for engineering design,

algorithms and examples, Prentice Hall (2005).

2. M. Asghar BhattiPractical Optimization Methods With

Mathematics Applications, Springer-Verlag N.Y, Inc., 2000.


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Optimization is derived from the Latin wordoptimus, the best and characterizes the

activities involved to find the best. People have been optimizing for ages, but the

roots for modern day optimization can be

traced to the Second World War.

Optimization


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Operational Research was defined by the Operational

Research Society of Great Britain as follows:

"Operational research is the application of the methods of science

to complex problems arising in the direction and management of

large systems of men, machines, materials and money inindustry, business, government, and defense. The distinctive

approach is to develop a scientific model of the system,

incorporating measurements of factors such as chance and risk,

with which to predict and compare the outcomes of alternative

decisions, strategies or controls. The purpose is to help

management determine its policy and actions scientifically."

Operational Research was defined by the Operational

Research Society of Great Britain as follows:

"Operational research is the application of the methods of science

to complex problems arising in the direction and management of

large systems of men, machines, materials and money in

industry, business, government, and defense. The distinctive

approach is to develop a scientific model of the system,

incorporating measurements of factors such as chance and risk,

with which to predict and compare the outcomes of alternative

decisions, strategies or controls. The purpose is to helpmanagement determine its policy and actions scientifically."

Operational Research


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Without computers, Operation Research and optimizationwould not be what they are today.

Earlier mathematical models (such as calculus, Lagrangemultipliers) relied on sophistication of technique to solvethe problem.

Methods of mathematical optimization (e.g., LinearProgramming) rely far less on mathematicalsophistication than they do on an unusual adaptabilityto the mode of solution inherent in the modern digital

computer.

The simplicity of these methods of mathematics coupled withtheir iterative processes makes them very useful.

Use of Computers


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Consider the case of Linear Programming

The first large scale computer became available in

1946 at the University of Pennsylvania, USA. This was just one year before the development of

simplex method.

The simplex method for linear programming consists

only of a few steps and these steps require only themost basic mathematical operations which a computeris well suited to handle.

However, these steps must be repeated over and overbefore one finally obtains an answer.

The first successful computer solution of a LP problemwas in January 1952 on the National Bureau ofStandards SEAC computer in USA.

Computers and Linear Programming


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Need for Optimization Optimization problems arise naturally in many different

disciplines. In engineering design problems often goal is

to maximize or minimize certain parameter or a variable.

For example optimization algorithms are often used in

aerospace design activity to minimize overall weight.

For chemical engineers design and operation of a

process plant should have optimal rate of production.

Amechanical engineer designs a component to

minimize the cost or to maximize the component life.

An electrical engineer designs a network to achieve

minimum time for communications.


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Need for Optimization A structural engineer designing a multi-story building

must choose materials in the building with a safe

structure and an economical as well.

A portfolio manager for a large mutual fund company

must choose investments that generate the largest

possible rate of return for its investors A plant manager in a manufacturing facility must

schedule the plant operations such that the plant

produces products that maximize company's revenues.

A scientist in a research laboratory may be interested in

finding a mathematical function that best describes an

observed physical phenomenon.


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Need for Optimization All these situations have the following three

things in common.

There is an overall goal, or objective, for the activity.

For the structural engineer, the goal may be tominimize the cost of the building,

for the portfolio manager it is to maximize the rateof return,

for the plant manager it is to maximize the

revenue, and for the scientist, the goal is to minimize the

difference between the prediction from themathematical model and the physical observation.


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Need for Optimization In addition to the overall goal, there are

constraints, that must be met.

The structural engineer must meet safetyrequirements dictated by applicable buildingstandards.

The portfolio manager must keep the risk ofmajor

losses below levels determined by the company'smanagement.

The plant manager must meet customer demandsand work within available work force and rawmaterial limitations.

For the laboratory scientist, there are no othersignificant requirements.


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Need for Optimization In all situations, choices available to meet the goals and

requirements.

The choices are known as optimization variables. For example, from safety point of view, it does not matter

whether a building is painted purple or pink, and therefore thecolor of a building would not represent a good optimization

variable. On the other hand, the height of one story could be a possible

design variable because it will determine overall height of thebuilding, which is an important factor in structural safety.

The variables that do not affect the goals are clearly notimportant.


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Optimization All engineering tasks involve either minimization or

maximization of an objective function.

It will be very difficult to discuss formulation of each

type of engineering optimization problem in one course.

However a designer can learn different types of

optimization techniques and latter choose optimal

algorithm for his or her problem.

First we shall learn optimal problem formulation.


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Optimization Problem Some define the formulation of problem as

taking statements,

defining general goals and requirements of a givenactivity, and

converting them into a series of well-defined

mathematical statements. Others say the formulation of Opt. problem is:

1. Selecting one or more optimization variables,

2. Choosing an objective function, and3. Identifying a set of constraints.


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The objective functionand the constraintsmustall be functions of one or more optimization

variables.

Let us have examples on this.


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Maximizing AreaEXAMPLE 1: Find the dimensions of the rectangulargarden of greatest area that can be fenced off (all four sides)

with 300 meters of fencing.SOLUTION

The garden will be in the shape of a rectangle. The perimeter of it

is to be 300 meters. Lets make a picture of the garden, labeling

the sides.

x x

y

y

Since we know the perimeter is 300 meters, we can now constructan equation based on the variables contained within the picture.

x +x +y +y = 2x + 2y = 300 (Constraint Equation)


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Maximizing AreaNow, we wish to maximize is area. Therefore, we will needan equation that contains a variable representing area.

(Objective Equation)A = xy

- CONTINUED

Now we will rewrite the objective equation in terms of A(the variable we wish to optimize) and either xor y. Since itdoesnt make a difference which one we select, we willselect x.

2x + 2y= 300; This is the constraint equation.

2y= 300 2 ; y= 150 x

Now we substitute 150 x foryin the objective equation sothat the objective equation will have only one independent

variable.


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Maximizing Area-CONTINUED

A =xyThe objective equation is given as

Now we will graph the resultant function;

A =x(150 x)Replacey with 150 x.A = 150x x2Finally we get

0

1000

2000

3000

4000

5000

6000

0 50 100 150

x

Area(A)


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Maximizing Area-CONTINUED

Since the graph of the function is obviously a parabola, then themaximum value of A (along the vertical axis) would be found at the onlyvalue of xfor which the first derivative is equal to zero.

A = 150x x2The area function is:

A = 150 2xDifferentiating we get:

150 2x = 0;Let the derivative equal to zero. x = 75

Therefore, the slope of the function equals zero whenx = 75.

This is thex-value for where the function is maximized. Then,

2x + 2y = 300; 2(75) + 2y = 300; y = 75

So, the dimensions of the garden will be 75 m x75 m.


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Minimizing CostEXAMPLE 2:

SOLUTION

A rectangular garden of area 75 square feet is to be surrounded on three sides

by a brick wall costing $10 per foot and on one side by a fence costing $5 per

foot. Find the dimensions of the garden such that the cost of materials isminimized.

Below is a picture of the garden. The red side represents the side that is fenced.

x x

y

y

The quantity that we will be minimizing is cost. Therefore, our objective

equation will contain a variable representing cost, C.


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Minimizing CostEXAMPLE 2: -CONTINUEDONTINUED

(Objective Equation)

C = (2x +y)(10) +y(5)

C = 20x + 10y + 5y

C = 20x + 15y

Now we will determine the constraint equation. The only piece of information

we have not yet used in some way is that the area is 75 square feet. Using this,we create a constraint equation as follows.

75 =xy (Constraint Equation)

Now we rewrite the constraint equation, isolating one of the variables therein.

75 =xy

75/y =x


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0

200

400600

800

1000

1200

1400

1600

1800

2000

0 50 100 150

y

Cost(C)

Minimizing CostEXAMPLE 2: -CONTINUEDONTINUED

The objective equation is: C = 20x + 15y

Now we rewrite the objective equation using the substitution we just acquired

from the constraint equation.

Replacex with 75/y: C = 20(75/y) + 15y;

Simplify,

C = 1500/y + 15y

Now we use this

equation tosketch a graph of

the function.


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Minimizing CostEXAMPLE 2: -CONTINUED

It appears from the graph that there is exactly one relativeextremum, a relative minimum around x= 10 or x= 15.

To know exactly where this relative minimum is, we needto set the first derivative equal to zero and solve

The object function is: C = 1500/y + 15y

Differentiating we get: C = -1500/y2 + 15

Set the function equal to 0.

-1500/y2 + 15 = 0;

15y2 = 1500

y2 = 100; y = 10.


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Minimizing CostEXAMPLE 2: -CONTINUED

Therefore, we know that cost will be minimized when y=10. Now we will use the constraint equation to determine

the corresponding value for x.

75 =xyThe constraint equation is:

75 =x(10)Replacey with 10:

7.5 =xLet us solve forx:

So the dimensions that will minimize cost,

arex = 7.5 ft andy = 10 ft.


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Minimizing Surface AreaEXAMPLE 3:

SOLUTION

(Volume) A canvas wind shelter for the beach has a back, two

square sides, and a top. Find the dimensions for which the

volume will be 250 cubic feet and that requires the least possibleamount of canvas.

Below is a picture of the wind shelter.

The quantity that we will be maximizing is surface area.Therefore, our objective equation will contain a variablerepresenting surface area, A.

x

y

x


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Minimizing Surface Area

EXAMPLE 3: -CONTINUED

A = xx +xx +xy +xy

A = 2x2 + 2xy (Objective Equation)

Now we will determine the constraint equation. The only piece of

information we have not yet used in some way is that the volumeis 250 ft3. Using this, we create a constraint equation as follows.

250 =x2y (Constraint Equation)

Now we rewrite the constraint equation, isolating one of the

variables therein.

250 =x2y; 250/x2 =y

Sum of the areas of the sides:


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0

500

1000

1500

2000

2500

3000

3500

4000

-5 5 15 25 35 45

x

Area(A)


EXAMPLE 3: -CONTINUED

The objective equation is:

Now we rewrite the objective equation using the substitution we just acquired

from the constraint equation.

Replacey with 250/x2 :

Simplify.

Now we use this

equation to sketch agraph of the function.

A = 2x2 + 2xy

A = 2x2 + 2x(250/x2)

A = 2x2 + 500/x


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Minimizing Surface AreaEXAMPLE 3: -CONTINUEDONTINUEDIt appears from the graph that there is exactly one relative extremum, a

relative minimum aroundx = 5. To know exactly where this relative

minimum is, we need to set the first derivative equal to zero and solve (sinceat this point, the function will have a slope of zero).

Differentiating we get

Then set the function equal to 0. 4x - 500/x2 = 0

4x = 500/x2

4x3

= 500x3 = 125

x = 5

A = 2x2 + 500/x

A = 4x 500/x2


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EXAMPLE 3: -CONTINUEDTherefore, we know that surface area will be minimized when

x = 5. Now we will use the constraint equation to determinethe corresponding value fory.

250 =x2yThe constraint equation:

250 = (5)2 y; y = 10

Replacex with 5.

So the dimensions that will minimize surface area,arex = 5 ft andy = 10 ft.

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lec 1 intro to OT