153 Linear Programming Problem. 154 which can be written in the form:- Assuming a feasible solution exists it will occur at a corner of the feasible region.

1

Linear Programming Problem

1 1 2 2

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

min ( )

s.t.

bounds , 1, , ( )

n n

n n

n n

m m mn n m

i i i

f c x c x c x

a x a x a x b

a x a x a x b

a x a x a x b

l x u i n

xx

2

which can be written in the form:- min ( )

xx c x

Ax b

l x u

s.t.

f T

Assuming a feasible solution exists it will occur at a corner of the feasible region. That is at an intersection of equality constraints. The simplex linear programming algorithm systematically searches the corners of the feasible region in order to locate the optimum.

3

Example: Consider the

problem:

max ( ) . .

. ,

. ,

. ,

,

xx

s.t. .8 (a)

.05 (b)

.1 (c)

f x x

x x

x x

x x

x x

81 108

0 0 44 24 000

0 01 2 000

0 0 36 6 000

0 0

1 2

1 2

1 2

1 2

1 2

4

Plotting contours of f(x) and the constraints produces:

0 0.5 1 1.5 2 2.5 3 3.5 4

x 104

0

0.5

1

1.5

2

2.5

3x 10

4

x1

x2

solution

(b)

(c)

(a)

f increasing

5

The maximum occurs at the intersection of (a) and (b):- .8

.05

giving

0 0 44 24 000

0 01 2 000 26 207 6 897

81 108 286 760

1 2

1 2 1 2

1 2

x x

x x x x

f x x

. ,

. , , , ,

. . ,At the other intersections (corners) of the feasible region:

intersection x f

x1=0, (c) (0, 16,667) 180,000

(b), (c) (15,000, 12,500) 256,500

(a), (b) (26,207, 6,897) 286,760

(a), x2=0 (30,000, 0) 243,000

max

6

Solution using MATLAB Optimisation Toolbox Routine LPf=[-8.1,-10.8];A=[0.8 0.44;0.05 0.1;0.1 0.36];b=[24000;2000;6000];vlb=[0;0];vub=[40000;30000];[x,lambda,how]=lp(f,A,b,vlb,vub);disp('solution x ='),disp(x)disp('f ='),disp(-f*x)disp('constraint values ='),disp(A*x-b)disp('Lagragian multipliers'),disp(lambda)disp(how)

solution x = 1.0e+004 * 2.6207 0.6897f = 2.8676e+005constraint values = 0.0000 0.0000 -896.5517Lagragian multipliers 4.6552 87.5172 0 0 0 0 0ok

DEMO

7

GENETIC ALGORITHMS

Refs: - Goldberg, D.E.: ‘ Genetic Algorithms in Search, Optimization and Machine Learning’ (Addison Wesley,1989)

Michalewicz, Z.: ‘Genetic Algorithms + Data Structures = Evolution Programs’ (Springer Verlag, 1992)

8

Genetic Algorithms are search algorithms based on the mechanics of natural selection and natural genetics. They start with a group of knowledge structures which are usually coded into binary strings (chromosomes). These structures are evaluated within some environment and the strength (fitness) of a structure is defined. The fitness of each chromosome is calculated and a new set of chromosomes is then formulated by random selection and reproduction. Each chromosome is selected with a probability determined by it’s fitness and, hence, chromosomes with the higher fitness values will tend to survive and those with lower fitness values will tend to become extinct.

9

The selected chromosomes then undergo certain genetic operations such as crossover, where chromosomes are paired and randomly exchange information, and mutation, where individual chromosomes are altered. The resulting chromosomes are re-evaluated and the process is repeated until no further improvement in overall fitness is achieved. In addition, there is often a mechanism to preserve the current best chromosome (elitism).

“Survival of the fittest”

10

Genetic Algorithm Flow DiagramInitial

Populationand Coding

Selection“survival of the

fittest”

Elitism

Crossover

Mutation

qMating

11

Components of a Genetic Algorithm (GA)

• a genetic representation

• a way to create an initial population of potential solutions

• an evaluation function rating solutions in terms of their “fitness”

• genetic operators that alter the composition of children during reproduction

• values of various parameters (population size, probabilities of applying genetic operators, etc)

12

Differences from Conventional Optimisation

•GAs work with a coding of the parameter set, not the parameters themselves

•GAs search from a population of points, not a single point

•GAs use probabilistic transition rules, not deterministic rules

•GAs have the capability of finding a global optimum within a set of local optima

13

Initial Population and CodingConsider the problem:where, without loss of generality, we assume that f is always +ve (achieved by adding a +ve constant if necessary). Also assume:

max ( ),x

x xf n

a x b i ni i i , , , , . 1 2

Suppose we wish to represent xi to d decimal places. That is each range needs to be cut into (bi-ai).10d equal sizes. Let mi be the smallest integer such that

Then xi can be coded as a binary string of length mi.

a bi i

( ). .b ai id mi 10 2 1

x a decimalb a

i ii imi

( ).'binary string'2 1

Also, to interpret the string, we use:

14

Each chromosome (population member) is represented by a binary string of length:m mi

i

n

1

where the first m1 bits map x1 into a value from the range [a1,b1], the next group of m2 bits map x2 into a value from the range [a2,b2] etc; the last mn bits map xn into a value from the range [an,bn].To initialise a population, we need to decide upon the number of chromosomes (pop_size). We then initialise the bit patterns, often randomly, to provide an initial set of potential solutions.

15

Selection (roulette wheel principle)

We mathematically construct a ‘roulette wheel’ with slots sized according to fitness values. Spinning this wheel will then select a new population according to these fitness values with the chromosomes with the highest fitness having the greatest chance of selection. The procedure is:

16

1)Calculate the fitness value eval(vi) for each chromosome vi (i = 1,...,pop_size)

2)Find the total fitness of the population

F eval vii

pop size

( )

_

1

3)Calculate the probability of a selection, pi, for each chromosome vi (i = 1,...,pop_size)p

eval v

Fii

( )

4)Calculate a cumulative probability qi for each chromosome vi (i = 1,...,pop_size)

q pi ij

i

1

17

The selection process is based on spinning the roulette wheel pop_size times; each time we select a single chromosome for a new population as follows:1) Generate a random number r in the range [0,1]

2) If r < q1, select the first chromosome v1; otherwise select the ith chromosome vi such that: q r q i pop sizei i 1 2 , ( _ )

Note that some chromosomes would be selected more than once: the best chromosomes get more copies and worst die off.

“survival of the fittest”All the chromosomes selected then replace the previous set to obtain a new population.

18

p1

p2p3

p4

p5

p6

p7

p8

p9

p10 p11

p12

segment area proportional to pi, i=1,...,12

example:

19

CrossoverWe choose a parameter value pc as the probability of crossover. Then the expected number of chromosomes to undergo the crossover operation will be pc.pop_size. We proceed as follows:- (for each chromosome in the new population)1)Generate a random number r from the range

[0,1].

2)If r < pc, then select the given chromosome for crossover.

ensuring that an even number is selected. Now we mate the selected chromosomes randomly:-

20

For each pair of chromosomes we generate a random number pos from the range [1,m-1], where m is the number of bits in each chromosome. The number pos indicates the position of the crossing point. Two chromosomes:

1 2 1

1 2 1pos p

pos pos

os m

mc

b b b b b

c c c c

are replaced by a pair of their offspring (children)

11

2

2

11

popos

p

s m

po o ms s

c c

c

b b

c

b

b bc

21

MutationWe choose a parameter value pm as the probability of mutation. Mutation is performed on a bit-by-bit basis giving the expected number of mutated bits aspm.m.pop_size. Every bit, in all chromosomes in the whole population, has an equal chance to undergo mutation, that is change from a 0 to 1 or vice versa. The procedure is:

For each chromosome in the current population, and for each bit within the chromosome:-1) Generate a random number r from the range

[0,1].

2) If r < pm, mutate the bit.

22

Elitism

It is usual to have a means for ensuring that the best value in a population is not lost in the selection process. One way is to store the best value before selection and, after selection, replace the poorest value with this stored best value.

23

Example max ( ) sin( ) . ,x

f x x x x 10 10 1 2

-1 -0.5 0 0.5 1 1.5 2-1

-0.5

0

0.5

1

1.5

2

2.5

3

x

f(x)

global max

24

Let us work to a precision of two decimal places. then the chromosome length m must satisfy:

2 1 10 2 1 2 301 92 ( )b g m m m

Also let pop_size = 10, pc = 0.25, pm = 0.04

To ensure that a positive fitness value is always achieved we will work on val = f(x) + 2

25

Consider that the initial population has been randomly selected as follows (giving also the corresponding values of x, val, probabilities and accumulated probabilities)

population x val p q v1 0 1 1 1 0 1 1 0 1 0.39 2.89 0.09 0.09 v2 0 0 0 1 1 1 0 1 0 -0.66 3.63 0.11 0.20 v3* 1 1 0 1 0 0 0 1 0 1.45 4.44 0.14 0.34 v4 1 0 1 0 1 1 1 1 0 1.05 4.04 0.13 0.47 v5 0 0 1 0 0 1 1 0 1 -0.55 2.45 0.08 0.55 v6 0 0 0 1 0 0 1 1 0 -0.78 2.48 0.08 0.63 v7 0 0 0 0 0 1 1 1 0 -0.92 2.51 0.08 0.71 v8 0 0 0 1 1 1 0 0 0 -0.67 3.53 0.11 0.82 v9 0 1 1 1 1 1 1 1 1 0.50 3.05 0.09 0.91 v10 1 0 0 1 1 0 0 1 1 0.80 3.06 0.09 1.00

* fittest member of the population

Note for v1:

dec v

x

F val p

( )

.

..

..

12 3 5 6 7

9

1 2 2 2 2 2 237

1237 3

2 10 39

32 082 89

32 080 09

26

Selection

Assume 10 random numbers, range [0,1], have been obtained as follows:-0.47 0.61 0.72 0.03 0.18 0.69 0.83 0.68 0.54 0.83These will select:

0.47 0.61 0.72 0.03 0.18 0.69 0.83 0.68 0.54 0.83 v4 v6 v8 v1 v2 v7 v9 v7 v5 v9giving the new population:

27

Note that the best chromosome v3 in the original population has not been selected and would be destroyed unless elitism is applied.

Population before selection

selection

Population after selection

v1 0 1 1 1 0 1 1 0 1 4 1 0 1 0 1 1 1 1 0 v2 0 0 0 1 1 1 0 1 0 5 0 0 0 1 0 0 1 1 0 v3 1 1 0 1 0 0 0 1 0 0 0 0 1 1 1 0 0 0 v4 1 0 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 1 v5 0 0 1 0 0 1 1 0 1 9 0 0 0 1 1 1 0 1 0 v6 0 0 0 1 0 0 1 1 0 2 0 0 0 0 0 1 1 1 0 v7 0 0 0 0 0 1 1 1 0 6,8 0 1 1 1 1 1 1 1 1 v8 0 0 0 1 1 1 0 0 0 3 0 0 0 0 0 1 1 1 0 v9 0 1 1 1 1 1 1 1 1 7,10 0 0 1 0 0 1 1 0 1 v10 1 0 0 1 1 0 0 1 1 0 1 1 1 1 1 1 1 1

28

Assume the 10 random numbers:- 1 2 3 4 5 6 7 8 9 100.07 0.94 0.57 0.36 0.31 0.14 0.60 0.07 0.07 1.00These will select v1, v6, v8, v9 for crossover.

Now assume 2 more random numbers in the range [1,8] are obtained:-

Crossover (pc = 0.25)

7 20 335. . bits 8 and 4.

29

Mating v1 and v6 crossing over at bit 8:-

v

v1

6

1 0 1 0 1 1 1 1 0

0 0 0 0 0 1 1 1 0no change

Mating v8 and v9 crossing over at bit 4:-

v

v8

9

0 0 0 0 0 1 1 1 0

0 0 1 0 0 1 1 0 1

produces

v

v8

9

0 0 0 0 0 1 1 0 1

0 0 1 0 0 1 1 1 0

giving the new population:-

30

population beforecrossover

population aftercrossover

v1 1 0 1 0 1 1 1 1 0 1 0 1 0 1 1 1 1 0v2 0 0 0 1 0 0 1 1 0 0 0 0 1 0 0 1 1 0v3 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0v4 0 1 1 1 0 1 1 0 1 0 1 1 1 0 1 1 0 1v5 0 0 0 1 1 1 0 1 0 0 0 0 1 1 1 0 1 0v6 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0v7 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1v8 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 0 1v9 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1 1 1 0v10 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1

bit for mutation

31

mutation (pm = 0.04)

Suppose a random number generator selects bit 2 of v2 and bit 8 of v9 to mutate, resulting in:-

population aftermutation

x val

v1 1 0 1 0 1 1 1 1 0 1.05 4.04v2 0 1 0 1 0 0 1 1 0 -0.02 3.02v3 0 0 0 1 1 1 0 0 0 -0.67 3.53v4 0 1 1 1 0 1 1 0 1 0.39 2.89v5 0 0 0 1 1 1 0 1 0 -0.66 3.63v6 0 0 0 0 0 1 1 1 0 -0.92 2.51v7 0 1 1 1 1 1 1 1 1 0.50 3.05

v8** 0 0 0 0 0 1 1 0 1 -0.92 2.37v9 0 0 1 0 0 1 1 0 0 -0.55 2.45v10 0 1 1 1 1 1 1 1 1 0.50 3.05

Total fitness F = 30.54

** weakest

32

ElitismSo far the iteration has resulted in a decrease in overall fitness (from 32.08 to 30.54). However, if we now apply elitism we replace v8 in the current population by v3 from the original population, to produce:

population after mutation

x val

v1 1 0 1 0 1 1 1 1 0 1.05 4.04 v2 0 1 0 1 0 0 1 1 0 -0.02 3.02 v3 0 0 0 1 1 1 0 0 0 -0.67 3.53 v4 0 1 1 1 0 1 1 0 1 0.39 2.89 v5 0 0 0 1 1 1 0 1 0 -0.66 3.63 v6 0 0 0 0 0 1 1 1 0 -0.92 2.51 v7 0 1 1 1 1 1 1 1 1 0.50 3.05 v8 1 1 0 1 0 0 0 1 0 1.45 4.44 v9 0 0 1 0 0 1 1 0 0 -0.55 2.45 v10 0 1 1 1 1 1 1 1 1 0.50 3.05

Total fitness F = 32.61

33

resulting now in an increase of overall fitness (from 32.08 to 32.61) at the end of the iteration.

The GA would now start again by computing a new roulette wheel and repeating selection, crossover, mutation and elitism; repeating this procedure for a pre-selected number of iterations.

34

-1 0 1 2-1

0

1

2

3iteration 40

0 20 403

3.5

4

4.5

5best and average values

-1 0 1 20

10

20

30x distribution

0 10 20

0

10

20

30

chromosomes

Final results from a MATLAB GA program using parameters:

pop_size = 30, m = 22, pc = 0.25, pm=0.01

35

1.8500 4.8500 1.8503 4.8502 1.8500 4.8500 1.8496 4.8495 1.8500 4.8500 1.8500 4.8500 0.3503 2.6497 1.8504 4.8502 1.8269 4.3663 1.8504 4.8502 1.8503 4.8502 1.8500 4.8500 1.8265 4.3520 1.8503 4.8502 1.8386 4.7222 1.8500 4.8500 1.8496 4.8495 1.8500 4.8500 1.8503 4.8502 1.8504 4.8502 1.8500 4.8500 1.8500 4.8500 1.8503 4.8502 1.8500 4.8500 1.8496 4.8495 1.8496 4.8495 1.8503 4.8502 1.8500 4.8500 1.8500 4.8500 1.8968 3.1880

x val x val

x val

Tabulated results:

The optimum val = 4.8502 at x = 1.8504

Hence:

remembering that val(x) = f(x) + 2

max ( ) sin( ) . ,

. .x

f x x x x

x

10 10 1 2

2 8502 18504

at

36

DEMO

37

ON-LINE OPTIMISATION - INTEGRATED SYSTEM

OPTIMISATION AND PARAMETER ESTIMATION (ISOPE)

An important application of numerical optimisation is the determination and maintenance of optimal steady-state operation of industrial processes, achieved through selection of regulatory controller set-point values. Often, the optimisation criterion is chosen in terms of maximising profit, minimising costs, achieving a desired quality of product, minimising energy usage etc. The scheme is of a two-layer hierarchical structure:-

38

OPTIMISATION(based on steady-state model)

set points

outputs

Note that the steady-state values of the outputs are determined by the controller set-points assuming, of course, that the regulatory controllers maintain stability.

REGULATORY CONTROLe.g. PID Controllers

controlsignalsmeasurements

INDUSTRIAL PROCESSinputs

39

The set points are calculated by solving an optimisation problem, usually based on the optimisation of a performance criterion (index) subject to a steady-state mathematical model of the industrial process. Note that it is not practical to adjust the set points directly using a ‘trial and error’ technique because of process uncertainty and non-repeatability of measurements of the outputs. Inevitably, the steady-state model will be an approximation of the real industrial process, the approximation being both in structure and parameters. We call this the model-reality difference problem.

40

ISOPE PrincipleROP - Real Optimisation Problem• Complex• Intractable

MOP - Model Based Optimisation Problem• Simplified (e.g... Linear - Quadratic)• Tractable

???

Can We Find the Correct Solution of ROP ByIterating on MOP in an Appropriate Way

YES

By Applying Integrated System Optimisation AndParameter Estimation - ISOPE

41

Iterative Optimisation and Parameter Estimation

In order to cope with model-reality differences, parameter estimation can be used giving the following standard two-step approach:-

42

1. Apply current set points values and, once transients have died away, take measurements of the real process outputs. use these measurements to estimate the steady-state model parameters corresponding to these set point values. This is the parameter estimation step.

2. Solve the optimisation problem of determining the extremum of the performance index subject to the steady-state model with current parameter values. This is the optimisation step and the solution will provide new values of the controller set points.

The method is iterative applied through repeated application of steps 1 and 2 until convergence is achieved.

43

Standard Two-Step Approach

MODEL BASED OPTIMISATION

min ( )

s.t. ( , )

J

c

c,y

y f c

c

REGULATORYCONTROL

REAL PROCESS

y f c* *( )

PARAMETER ESTIMATION

*( , ) ( )

y c y c

y

y*

y*

44

Example min ( )

*

cJ y c

y c

y c

where (model)

(reality)

2

2 1

2 2

real solutionJ y c c c c c

dJ

dcc c c

y J

* *

*

* *

( ) ( ) ( )

( ) .

. .

2 2 1 2 2 1

0 4 2 1 2 0 0 4

18 0 2

2 2 2 2 2 2

At a min,

giving and

Now consider the two-step approach:-

parameter estimation

y c y c

c c c

( , ) ( )*

i.e. 2 1 1

45

optimisationmin ( )

cJ y c

y c

s.t. with given

2 2 2

J c cdJ

dcc c

c c c

( ) , ( )

( ) .

2 2 2 2

0 2 2 2 0 1 0 5

2 2

for a min when

Hence, at iteration k:

k k

k k

c

c

1

1 0 51 .

i.e. c c ck k k 1 1 0 5 05 0 5 0 5. . . .This first-order difference equation will converge (i.e. stable) since 05 1 05 05 0 3333. . . . giving c c c

and y c J 2 1 16667 0 2222 13333. , . , .

46

HENCE, THE STANDARD TWO STEP APPROACH DOES NOT CONVERGE TO THE CORRECT

SOLUTION!!!

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0

0.2

0.4

0.6

0.8

1

c

J*

Standard Two-Step Final Solution c = 0.3333 J* = 0.2222

1

2 3

4 5

final solution

47

Integrated Approach

The standard two-step approach fails, in general, to converge to the correct solution because it does not properly take account of the interaction between the parameter estimation problem and the system optimisation problem.

Initially, we use an equality v = c to decouple the set points used in the estimation problem from those in the optimisation problem. We then consider an equivalent integrated problem:-

48

* *

min ( , )

s.t. ( , ) ( , )

( ) ( )

model based optimisation

decoupling

parameter

( , ) ( ,

)

J

cc y

y f c y c

v c

y v f v f v y v

estimation

This is clearly equivalent to the real optimisation problem ROP

min ( , )

( )

*

* *

cc y

y f c

s. t.

J

49

If we also write the model based optimisation problem as: (by eliminating y in J(c,y))

min ( , )

where ( , ) , ( , )

F

F Jc

c

c c f c

giving the equivalent problem:-

*

min ( , )

s.t.

( , ) ( )

F

cc

v = c

y v y v

50

Form the Lagrangian:

*( , , ) ( , ) ( ) ( , ) ( )T TL F c v c v c y v y v

with associated optimality conditions:

*

(1)

(2)

(3)

TT

T

L F

L

L F

c c

v

0

y y0

v v

y0

together with:-

*( , ) ( )

v c

y v y v

51

Condition (1) gives rise to the modified optimisation problem:

min ( , ) TF c

c c

which is the same as:-

min ( , )

s.t. ( , )

TJ

c

c y c

y f c

and the modifier is given from (2) and (3) :-

* TT T

F

y y y

v v

52

Modified Two-Step ApproachMODIFIER

*

where ( , ) , )

TT T

F

F J

y y y

c c

c c,f(c

c,y

PARAMETER ESTIMATION

*( , ) ( )

y c y c

y

c

REGULATORYCONTROL

REAL PROCESS

y f c* *( )

y*

y*

min ( )

s.t. ( , )

TJ

c

c,y c

y f c

MODEL BASED OPTIMISATION

53

Modified two-step algorithm

The starting point of iteration k is an estimated set point vector vk.

Step 1 : parameter estimationApply the current set points vk and measure the corresponding real process outputs y*k. Also, compute the model outputs yk = f(vk,) and determine the model parameters k such that yk = y*k.

Step 2 : modified optimisation(i) compute the modifier vector

*

( , )k

k

T Tk kk k k

k k kF

y y yv

v v

54

(ii) solve the modified optimisation problem

min ( ) ( )

s.t. ( , )

k T

k

J

c

c,y c

y f c

to produce a new estimated set point vector ck.

(iii) update the set points (relaxation)

v v K c vk k k k 1 ( )

where the matrix K, usually diagonal, is chosen to regulate stability. (Note: if K = I, then vk+1 = ck).

We then repeat from Step 1 until convergence is achieved.

55

Acquisition of derivatives

The computation of modifier requires the derivatives:-

*

[ ]

model based - not

usually a probl

em

on the real measurements

- quite

a

proble

m

TFF

y

vy

y

v

56

Some methods for obtaining y

v

*

(i) applying perturbations to the set points and computing the derivatives by finite differences.

(ii) using a dynamic model to estimate the derivatives (recommended method).

(iii) estimate the derivative matrix using Broyden’s method (recommended method).

57

Note:- Digital filtering can be used with effect to smooth the computational values of .

Note:- When = 0 we arrive back to the standard two-step approach. From the expression for we see that the standard two step approach will only achieve the correct solution when the model structure is chosen such that:

y

c

y

cLNM

OQP

LNM

OQP

*

a condition rarely achieved in practice.

58

Example: Consider the same example as used previously:

min ( )

*

cJ y c

y c

y c

where (model)

(reality)

2

2 1

2 2

where the real solution is :- c = 0.4, y* = 1.8, J* = 0.2

parameter estimation This is unchanged. Hence: v 1

where

**

2 2

1, 1; 2 1 2

, ( , ) ( 2) 2( 2)

y y yy v y v

v v

F J v y v v v F v

Hence: 1 2 1 2 2 2 2

1b gbgb g( ) ( )v v

modifier* TT T

F

y y y

v v

59

modified optimisation

2 2

2 2

min ( 2)

s.t. , , given

i.e. ( 2)

2( 2) 2 0

1 0.5 0.25

m

m

cy c c

y c

F c c c

Fc c

cc

60

At iteration k (with K = 1)

k k

k k k

k k k

k k

v

v

c

v c

1

2 2

1 05 0 251

( )

. .

i.e. v v v v

v v

k k k k

k k

1

1

1 0 5 1 0 25 2 1 2

15 1

. ( ) . ( )( ( ) )

.

IF this difference equation converges the result will be:-

v v v Note c v 15 1 0 4. . : ) (

61

which is the correct solution

However, it does not converge since

(the eigenvalue is outside the unit circle)

Hence, it is necessary to apply relaxation in the algorithm to produce the iterative scheme:-

15 1.

k k

k k k

k k k

k k k k k k

v

v

c

v v g c v gc g v

1

2 2

1 0 5 0 25

11

( )

. .

( ) ( )

where g is a gain parameter (g > 0).

62

Then:-

v g v g v g v gk k k k 1 15 1 1 1 2 5( . ) ( ) ( . )

This will converge provided

1 2 5 1 0 0 8 . .g g, i.e.

(Hence, typically use g=0.4)

Then: v g v g v ( . ) .1 2 5 0 4

Then: 14 0 4 0 4. , . , . c

and: y y J * *. , .18 0 2

i.e. THE CORRECT REAL SOLUTION IS OBTAINED

63

ISOPE (THE MODIFIED TWO-STEP APPROACH) ACHIEVES THE CORRECT STEADY STATE REAL PROCESS OPTIMUM IN SPITE OF MODEL-REALITY DIFFERENCES

64-2 -1.5 -1 -0.5 0 0.5 1 1.5 20

5

10

15

20

25

Modified Two-Step g = 1 (divergence)

v

J*

12

3

4

5

650 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0

0.2

0.4

0.6

0.8

1

Modified Two-Step Final Solution v = 0.3999 J* = 0.2

v g = 0.2

J*

1

2

34 5

final solution

66

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0

0.2

0.4

0.6

0.8

1

Modified Two-Step Final Solution v = 0.4 J* = 0.2

v g = 0.4

J*

1

2

When g = 0.4 convergence is achieved in a single iteration. This is because the eigenvalue = 0. (|1 - 2.5g| = 0)

67

Example:

1 2

2 2 2 21 2 1 2

*1 1

*2 1 2

1 1 1

2 1 2 2

,min ( 2) ( 3)

2 1s.t.

3 2reality

0.1 mod

5.5el

c cJ y y c c

y c

y c c

y c

y c c

680 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

v1

v2

standard J = 0.6563

modified J = 0.2353

69

DEMO