OPTIMIZATION ENGR 351 Numerical Methods Instructor: Dr. L.R. Chevalier.

OPTIMIZATION

ENGR 351 Numerical MethodsInstructor: Dr. L.R. Chevalier

x

f(x)

Recall, when determining the root, we were seeking x where f(x) = 0

f(x) = 0

x

f(x)

With optimization, however we are seeking f '(x) = 0

f '(x) = 0

f '(x) = 0

x

f(x)

The maximum occurs when f "(x)<0

f '(x) = 0f "(x) < 0

x

f(x)

The minimum occurs when f "(x)>0

f '(x) = 0f "(x)> 0

Optimization

In some techniques, we determine the optima by solving the root problem f '(x) =0

If f ’(x) is not available analytically, we may use a finite difference approximation to estimate the derivative

Examples of Optimization Problems

Design structures for minimum cost Design water resource project to

mitigate flood damage while yielding maximum hydropower

Design pump and heat transfer equipment for maximum efficiency

Inventory control Optimize planning and scheduling

Methods Presented

One-dimensional Unconstrained Optimization Golden Search Method

Constrained Optimization Graphically Using Excel

Specific Study Objectives

Understand why and where optimization occurs in engineering problem solving

Understand the major elements of the general optimization problem: objective function, decision variables, and constraints

Be able to distinguish between linear and nonlinear optimization, and between constrained and unconstrained problems

Be able to define the golden ratio and understand how it makes 1-D optimization efficient

Be able to solve a 2-D linear programming problem graphically

Specific Study Objectives

Mathematical Background

An optimization or mathematical programming problem is generally stated as:

find x which minimizes or maximizes f(x) subject to

di(x) ai i = 1,2,……mei(x) = bi i=1,2,……p



di(x) ai i = 1,2,……mei(x) = bi i=1,2,……p

x is the design vector (n-dimensions)f(x) is the objective function



di(x) ai i = 1,2,……mei(x) = bi i=1,2,……p

di(x) are inequality constraintsei(x) are equality constraints

If f(x) and the constraints are linear, we have linear programming

If f(x) is quadratic, and the constraints are linear, we have quadratic programming

If f(x) in not linear or quadratic, and/or the constraints are nonlinear, we have nonlinear programming




di(x) ai i = 1,2,……mei(x) = bi i=1,2,……p

Without these, we have unconstrained optimization



di(x) ai i = 1,2,……mei(x) = bi i=1,2,……p

With them, we have constrained optimization

1-D Unconstrained Optimization

f(x)

x

global maximum

global minimumlocal minimum

local maximum

Here we see a multimodal case…however we want the global max or min!

1-D Unconstrained Optimization

We will consider the Golden Section Search method which is based on the Golden Ratio

......61803.02

15

Golden Ratio and Fibonacci Numbers

O.61803

1

This proportion was considered aesthetically pleasing by the Greeks

The Parthenon5th century BC


The Golden Ratio is related to an important mathematical series known as the Fibonacci numbers

0,1,1,2,3,5,8,13,21,34…..

Each number after the first two represents the sum of the preceding two. Note the ratio of consecutive numbers


0,1,1,2,3,5,8,13,21,34…..

0/1=01/1=11/2=0.52/3=0.6673/5==0.65/8=0.6258/13=0.615

Continue and the ratio approaches the golden ratio!

01 0.000001 1.000002 0.500003 0.666675 0.600008 0.6250013 0.6153821 0.6190534 0.6176555 0.6181889 0.61798144 0.61806233 0.61803

......61803.02

15


1-D Unconstrained Optimization: The Golden-Section Search

f(x)

x

Pick two points,xu and xl

xl xu

lo = xu-xl


f(x)

xxl xu

lo = xu-xl

We will now need a two new points based on the constraints l0 = l1 + l2 l1/l0 = l2/l1


f(x)

xxl xu

lo = xu-xl

Substitutingl1/(l1+l2) = l2/l1

If the reciprocal is taken, and R = l2/l11+R = 1/RR2 + R - 1 = 0


f(x)

xxl xu

lo = xu-xl

R2 + R - 1 = 0

This can be solved for the positive root

61803.02

15

2

1411


f(x)

xxl xu

lo = xu-xl

dxx

dxx

xxd

u

l

lu

2

1

2

15

Evaluate the function at these points. Two results can occur.

f(x)

x

xl xu


x1d

dx2

f(x)

x

xl xu


x1d

dx2

Here, f(x1) > f(x2)

f(x)

x

xl xu


x1d

dx2

Eliminate the domain to the left of x2

f(x)

x

xl xu


x1d

dx2

x2 becomes xl for the next round

f(x)

x

xl xu


x1d

dx2

If f(x1)<f(x2), eliminate points to the right of x1

f(x)

x

xl xu


x1d

dx2

Back to the first case, here the new xl is x2

f(x)

x

xl xu


old x1=x2

xl

Because of the Golden Ratio, the previous x1 becomes the current x2

d

f(x)

x

xu


x2

xld

lul xxxx

2

151

f(x)

x

xu


x2

xld

Repeat this algorithm until f(x) stabilizes


Let’s review the spreadsheet fileopt-a.xls

xl f(xl) x2 f(x2) x1 f(x1) xu f(xu) d

0.0000 0.0000 1.5279 1.7647 2.4721 0.6300 4.0000 -3.1136 2.47210.0000 0.0000 0.9443 1.5310 1.5279 1.7647 2.4721 0.6300 1.52790.9443 1.5310 1.5279 1.7647 1.8885 1.5432 2.4721 0.6300 0.94430.9443 1.5310 1.3050 1.7595 1.5279 1.7647 1.8885 1.5432 0.58361.3050 1.7595 1.5279 1.7647 1.6656 1.7136 1.8885 1.5432 0.36071.3050 1.7595 1.4427 1.7755 1.5279 1.7647 1.6656 1.7136 0.2229

10sin2

2xx Find the maximum of

ExamplePerform three iterations of the golden section search to maximizef(x) = -1.5x6 - 2x4 +12x

using the initial guessesxl=0 and xu =2

0

2

4

6

8

10

0.0 0.5 1.0 1.5

x

f(x)

Solutionxl f(xl) x2 f(x2) x1 f(x1) xu f(xu) d

0.0000 0.0000 0.7639 8.1879 1.2361 4.8142 2.0000 -104.0000 1.23610.0000 0.0000 0.4721 5.5496 0.7639 8.1879 1.2361 4.8142 0.76390.4721 5.5496 0.7639 8.1879 0.9443 8.6778 1.2361 4.8142 0.4721

Reference opt-a.xls

Use of Solver

If SOLVER is not under Tools, you’ll have to add it Use <TOOLS - ADD INS> command Choose SOLVER ADD-IN If not available as an option, you will

need to install it from the original MS Office CD

Reference opt-a.xls

Constrained Optimization

Linear programming (LP) is an optimization approach that deals with meeting a desired objective

- maximizing profit- minimizing cost

Both the objective function and the constraints are linear in this case


Objective function

Maximize Z = c1x1 +c2x2 +…..cnxn

orMinimize Z = c1x1 +c2x2 +…..cnxn

where ci = payoff of each unit of the jth activity xi = magnitude of the jth activity

Constrained OptimizationObjective function

Maximize Z = c1x1 +c2x2 +…..cnxn

orMinimize Z = c1x1 +c2x2 +…..cnxn

where ci = payoff of each unit of the jth activity xi = magnitude of the jth activity

Hence, Z is the total payoff due to the total number of activities, n


The constraints can be represented by:

ai1x1 +bi2x2+…..ainxn bi

where aij = amount of the ith resource that is consumed for each unit of the jth activity

bi = amount of the ith resource available


Finally, we add the constraint that all activities must have a positive value

xi 0

Setting up the general problem

Similar to Problem 15.1 p. 377

Gas processing plant that receives a fixed amount of raw gas each week

Capable of processing two grades of heating gas (regular and premium)

High demand for the product (I.e. guaranteed to sell)

Each grade yields a different profit

Each grade has different production time and on-site storage constraints

Facility is only open 120hrs/weekUsing the factors in the table on

the next page, develop a linear programming formulation to maximize profits for this operation.

Setting up the general problem

Parameters

Note: a metric ton, or tonne, is equal to 1000 kg)

Resource Regular Premium Resource Availability

120

Storage 9 6Profit (/tonne) 150 175

77

10 8

Product

Raw Gas

(m3/tonne)Production Time (hr/tonne)

7 11

Parameters

Let x1 = amount of regular and x2 = amount of premium


120


77

10 8

Product

Raw Gas


7 11

Objective Function

Total Profit = 150 x1 + 175 x2

Maximize Z = 150 x1 + 175 x2


120


77

10 8

Product

Raw Gas


7 11

Objective Function

Total Profit = 150 x1 + 175 x2

Maximize Z = 150 x1 + 175 x2Objectivefunction


120


77

10 8

Product

Raw Gas


7 11

Constraints

7x1 + 11x2 77 (material constraint)10x1 + 8x2 120 (time constraint)x1 9 (storage constraint)x2 6 (storage constraint)x1,x2 0 (positivity constraint)


120


77

10 8

Product

Raw Gas


7 11

0

2

4

6

8

10

12

14

0 5 10 15

x1

x 2

Graphical Solution

Graphical Solution

Now we need to add the objective function to the plot. Start with

Z = 0 (0=150x1 + 175x2)

and

Z = 500 (500=150x1 + 175x2)

-4

-2

0

2

4

6

8

10

12

14

-5 0 5 10 15

x1

x 2

Z=1550

Still in feasible regionx1 8x2 2

Graphical Solution

Excel Solution: Using Solver

Solver Parameters

Note: See Example 15.3 p. 388

Solver Solution

Recall graphical solutionx1 8x2 2

Example

Develop the equations (objective function and constraints) needed to optimize the problem on the next slide.

ExampleA construction site requires a minimum of 10,000 yd3 of sand and gravel mixture. The mixture must contain no less than 5000 yd3 of sand and no more than 6000 yd3 of gravel. The material may be obtained from two sites

1 5 30 702 7 60 40

Delivery Cost

($/yd3)

% Sand % GravelSite

Excel Solution

C 0.00 Objective Functionx1 0.00x2 0.00

Constraintsx1 + x2 0.00 >= 10,0000.3x1 +0.6x2 0.00 >= 50000.7x1+0.4x2 0.00 <= 6000x1 0.00 >= 0x2 0.00 >= 0

Excel SolutionC 63333.33 Objective Functionx1 3333.33x2 6666.67

Constraintsx1 + x2 10000.00 >= 10,0000.3x1 +0.6x2 5000.00 >= 50000.7x1+0.4x2 5000.00 <= 6000x1 3333.33 >= 0x2 6666.67 >= 0

Date post:	15-Jan-2016
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OPTIMIZATION ENGR 351 Numerical Methods Instructor: Dr. L.R. Chevalier.

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