Basic Optimization Theory

Analysis, Modelling and Simulation of Energy Systems, SEE-T9Mads Pagh Nielsen

Basic Optimization Theory

• LP (Linear Programming)

• NLP (Non-Linear Programming)

• IP (Integer Programming)

• MIP (Mixed Integer Programming)

• MINLP (Mixed Integer Non-Linear Programming)


Types Of Optimization

• Parameter Optimization

• Configuration Optimization

• Operational Optimization

• Topology Optimization


Topology Optimization


The General Non-Linear Problem

b

bb

subject to:min. max.

m21m

2212

1211

21

)(

)(

)(

)( or

xxxg

xxxgxxxg

xxxf

n

n

n

n

,..., ,

,..., ,

,..., ,

,..., ,

Objective Function

Constraints

Design Space


Extrema: Ordering Situation

0 10 20 30 40 500

200

400

600

800

1000

$

Order Quantity

Total Cost

Transport Cost

Ordering Cost

OPT.


Extrema: Heat Exchanger

0 100 200 300 400 5000

200

400

600

800

1000

$

Heat Exchanger Area

Total Cost

Material Cost

Energy Cost

OPT.


Local vs. Global Extrema

-1,2

-1

-0,8

-0,6

-0,4

-0,2

0

0,2

0,4

0 2 4 6 8 10 12 14

f(x)=sin(30x)/x

f(x)

x

localmin

globalmax stationary

point

localmin

localmax


Convexityf(x) f(x)

x x

f(x) is a convex function if and only if for any given two points x1 and x2 in the function domain and for any constant 0 1

f(x1 +(1- )x2) f(x1)+(1- )f(x2)

x1 x2

f(x1)

f(x2)

x1+(1-)x2

f(x1)+(1- )f(x2)

f(x1 +(1- )x2)

Convex Function Concave Function


The Hessian • The gradient vector of f at x

n

2

1

...

)(

x

f

x

f

x

f

xf

nn

2

n1

2

1n

2

11

2

2 ...

...

...

...

...)(

xx

f

xx

f

xx

f

xx

f

xf

• The Hessian Matrix of f at x


Conditions for convexity

• H(x) is positive semi-definite if and only if xTHx ≥ 0 for all x and there exists and x 0 such that xTHx = 0. => Convexity

• H(x) is positive definite if and only if xTHx > 0 for all x 0.

• H(x) is indefinite if and only if xTHx > 0 for some x, and xTHx < 0 for some other x.

How can we use Hessian to determine whether or not f(x) is convex?


Possible Solutions To Convex Problemsobjective function

level curve

optimal solution

Feasible Region

linear objective,nonlinear constraints

objective function level curve

optimal solution

Feasible Region

nonlinear objective,nonlinear constraints

objective function level curve

optimal solution

Feasible Region

nonlinear objective,linear constraints

objective function level curves

optimal solution

Feasible Region

nonlinear objective,linear constraints


Line SearchLine search techniques are in essence optimization algorithms for one

dimensional minimization problems.

They are often regarded as the backbones of nonlinear optimization algorithms.

Typically, these techniques search a bracketed interval.

Often, unimodality is assumed.

Line search techniques are in essence optimization algorithms for one dimensional minimization problems.

They are often regarded as the backbones of nonlinear optimization algorithms.

Typically, these techniques search a bracketed interval.

Often, unimodality is assumed.

Exhaustive search requires N = (b-a)/ + 1 calculations to search the above interval, where is the resolution.

a bx*


Bracketing Algorithm

Two point search (dichotomous search) for finding the solution to minimizing ƒ(x):

0) assume an interval [a,b]

1) Find x1 = a + (b-a)/2 - /2 and x2 = a+(b-a)/2 + /2 where is the resolution.

2) Compare ƒ(x1) and ƒ(x2)

3) If ƒ(x1) < ƒ(x2) then eliminate x > x2 and set b = x2

If ƒ(x1) > ƒ(x2) then eliminate x < x1 and set a = x1

If ƒ(x1) = ƒ(x2) then pick another pair of points

4) Continue placing point pairs until interval < 2

a bx1 x2


Golden Section Search

a

a

b

b

a - b

In Golden Section, you try to have b/(a-b) = a/b

which implies b*b = a*a - ab

Solving this gives a = (b ± b* sqrt(5)) / 2

a/b = -0.618 or 1.618 (Golden Section ratio)

Note that 1/1.618 = 0.618

Discard


Golden Section Search Algorithm

a bx1 x2

{Initialize}

x1 = a + (b-a)*0.382

x2 = a + (b-a)*0.618

f1 = ƒ(x1)

f2 = ƒ(x2)

{Loop}

if f1 > f2 then

a = x1; x1 = x2; f1 = f2

x2 = a + (b-a)*0.618

f2 = ƒ(x2)

else

b = x2; x2 = x1; f2 = f1

x1 = a + (b-a)*0.382

f1 = ƒ(x1)

end


The 2D case

A

C

B

Local optimal solution

Feasible Region

D

EF

G

Local and global optimal solution

X1

X2


The 2D case

(From John Rasmussen, 1999)


Example: Heat ExchangerProblem: Find the radius of tubes in a heat exchanger tomaximize the total surface area.

(The magnitude of pressure drops are not considered)


Setting up the problem:

What is the best value of r ?What if we added a maximum allowable pressure drop?


Ideal Rankine Cycle Balance• Assumptions: steady flow process, no generation, neglect KE and PE changes for all four devices,• First Law: 0 = (net heat transfer in) - (net work out) + (net energy flow in)

0 = (qin - qout) - (Wout - Win) + (hin - hout)

T

s

1

2

3

4

• 1-2: Pump (q=0) Wpump = h2 - h1 = v(P2-P1)

• 2-3: Boiler (W=0) qin = h3 - h2

• 3-4: Turbine (q=0) Wout = h3 - h4

• 4-1: Condenser (W=0) qout = h4 - h1Thermal efficiency = Wnet/qin = 1 - qout/qin = 1 - (h4-h1)/(h3-h2)

Wnet = Wout - Win = (h3-h4) - (h2-h1)


T

s

12

3

4

(a) lower pressure(temp)

s

T

1

2

(b) Superheating (c) increase pressure

s

T

1

2

3

4

Low qualityhigh moisture content

1

2

4

Red area = increase in W netBlue area = decrease in W net

Thermal efficiency can be improved by manipulating the temperatures and/or pressures in various components(a) Lowering the condensing pressure (lowers TL, but decreases quality, x4 )(b) Superheating the steam to a higher temperature (increases TH but requires higher temp materials)(c) Increasing the boiler pressure (increases TH but requires higher temp/press materials)

Possibility for improvement of thermal efficiency:


Reheating• The optimal way of increasing the boiler pressure without increasing the moisture

content in the exiting vapor is to reheat the vapor after it exits from a first-stage turbine and redirect this reheated vapor into a second turbine.

boiler

high-Pturbine

Low-Pturbine

pump

condenser1

2

3

4

56

T

s

1

2

3 5

6

4

high-Pturbine

low-Pturbine

4


Reheating

• Reheating allows one to increase the boiler pressure without increasing the moisture content in the vapor exiting from the turbine.

• By reheating, the average temperature of the vapor entering the turbine is increased, thus, it increases the thermal efficiency of the cycle.

• Multistage reheating is possible but not practical. One major reason is because the vapor exiting will be superheated vapor at higher temperature, thus, decrease the thermal efficiency.

• Energy analysis: Heat transfer and work output both change

•

qin = qprimary + qreheat = (h3-h2) + (h5-h4)

Wout = Wturbine1 + Wturbine2 = (h3-h4) + (h5-h6)


Regeneration• From 2-2’, the average temperature is very low, therefore, the heat addition process is at

a lower temperature and therefore, the thermal efficiency is lower.• Use a regenerator to heat the liquid (feedwater) leaving the pump before sending it to

the boiler. This increases the average temperature during heat addition in the boiler, hence it increases efficiency.

T

s1

2

2’

3

4

Lower tempheat addition

T

s1

23

4

5

6

7

Use regenerator to heat up the feedwater

higher tempheat addition


Regeneration Cycle• Improve efficiency by increasing feedwater temperature before it enters the boiler.• Two Options:

– Open feedwater : Mix steam with the feedwater in a mixing chamber.– Closed feedwater: No mixing.

Pump 24

Pump 1

OpenFWHboiler

condenser1

23

5

67

1

23

4

5

6

7

T

s

Open FWH

(y) (1-y)

(y)

(1-y)


Analysis Of Regenerative Cycle

• Assume y percent of steam is extracted from the turbine and is directed into open feedwater heater.

• Energy analysis:

qin = h5-h4, qout = (1-y)(h7-h1),

Wturbine, out = (h5-h6) + (1-y)(h6-h7)

Wpump, in = (1-y)Wpump1 + Wpump2

= (1-y)(h2-h1) + (h4-h3)

= (1-y)v1(P2-P1) + v3(P4-P3)

• In general, more feedwater heaters result in higher cycle efficiencies.• However this does not mean that it is necessary a practical optimal solution!

(Note: Ideal pumps)

Basic Optimization Theory

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