intelligent control problems with solutions

7/24/2019 Intelligent Control Problems with Solutions

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EE 5322: Assignment #1

Due on Tuesday, Sept 8, 2015

Ajay Pratap Yadav

1001248915

September 7, 2015

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Ajay Pratap Yadav1001248915 EE 5322 (F. Lewis): Assignment #1

Question #1

a. Compute the 20 day MA. Plot on the same figure as the stock closing price.

b. Plot the stock minus the 20 day MA.

c. Compute and plot the 20 day moving sample variance.

d. On the same figure, plot the stock closing price, the 20 day MA, and the MA plus three

times the 20 day standard deviation the MA minus three times the 20 day standard deviation.

The last two lines are known as the Bollinger Bands, after John Bollinger.

Answer

Note: I have first plotted the figures followed by code.

(a)

0 50 100 150 200 250 3005

10

15

20

25

30

Days

Price

Closing Price

20Moving Avg

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(b)

0 50 100 150 200 250 3006

4

2

0

2

4

6

8

Days

Pricedifferenceb/wStockand20MA

(c)

0 50 100 150 200 250 3000

1

2

3

4

5

6

7

8

9

Days

20daymo

vingsamplevariance

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(d)

0 50 100 150 200 250 3000

5

10

15

20

25

30

35

Days

Prices

Closing Price

Mavg

Bol band

MATLAB Code

1 % NOTE: FUNCTIONS ARE WRITTEN FIRST FOLLOWED BY MAIN CODE

2 function out = MA(x,t) %Calculating Moving Avg

3 n = size(x);

4 out = zeros(n(1),1);

5 for i=1:t

6 out(i) = sum(x(1:i))/i;

7 end

8 for i=t+1:n(1)

9 out(i) = sum(x(i(t1):i))/t;

10 end

11 end

12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

13 function out = MV(x,MA,t) %Calculating Moving Variance

14 n = size(x);

15 out = zeros(n(1),1);

16

17 for i = t+1:n(1)

18 a=0;

19 for j=1:t

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20 a=a+(x(ij+1)MA(i))2;

21 end

22 out(i) = a/t;

23 end

24

25 for i = 1:t

26 a=0;

27 for j=1:i

28 a=a+(x(ij+1)MA(i))2;

29 end

30 out(i) = a/i;

31 end

32 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

33 % Main Code for question #1

34 %EE5322: Home Work 1

35 %Ajay Yadav 1001248915

36 clc

37 clear

38 load ('HW1 data');

39 data = Data;

40 t = 20;%size of window

41 N = size(data);

42 x = data(:,2);

43 M avg = MA(x,t);

44 diff = data(:,2)M avg;

45 plot(data(:,1),data(:,2));

46 xlabel('Days');ylabel('Price');

47 hold on

48 plot(M avg,'k')

49 legend('Closing Price','20Moving Avg');grid on

50 figure

51 plot(data(:,1),diff)

52 xlabel('Days');ylabel('Price difference b/w Stock and 20 MA ');

53 grid on

54 M var = MV(x,M avg,t);

55 figure

56 plot(data(:,1),M var )

57 xlabel('Days');ylabel('20 day moving sample variance');

58 grid on59

60 Bol1 = M avg+3*sqrt(M var);

61 Bol2 = M avg3*sqrt(M var);

62 figure

63 plot(data(:,1),x)

64 hold on

65 plot(M avg,'g')

66 hold on

67 plot(Bol1,'r')

68 hold on

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69 plot(Bol2,'r')

70 legend('Closing Price', 'M avg', 'Bol band')

71 xlabel('Days');ylabel('Prices')

Question #2

a. Compute and plot the 20 day moving skew.

b. Compute and plot the 20 day moving kurtosis.

c. Can you use these statistics to find a leading indicator for movements in the stock?

i.e. how can we predict using statistics when the stock is about to break its trend (change its

pattern)?

Answer

(a)

0 50 100 150 200 250 3002.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

Days

20daymovingSKE

W

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(b)

0 50 100 150 200 250 3003

2

1

0

1

2

3

4

5

6

7

Days

20daymovingKURTOSIS

(c)

0 50 100 150 200 250 3000

5

10

15

20

25

30

35

Days

Prices

Closing Price

Mavg

Bol band

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0 50 100 150 200 250 3003

2

1

0

1

2

3

4

5

6

7

Days

20daymovingKURTOSIS

0 50 100 150 200 250 3002.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

Days

20

daymovingSKEW

Figure 1: 20-day Skew and 20-day Kurtosis8 Page 8 of 16


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We can predict the change in movement of the stocks from its skew and kurtosis. To

demonstrate this, lets observe the 20-day skew and 20-day kurtosis along with the plots

of closing price. From figure 1, we can observe that sudden rise in skew and kurtosis

(shown by black arrows) coincides with points on stock prices graph where the stocks

touch the upper Bollinger Band. These are the points when its best to sell stocks.

Matlab Code

1 % NOTE: FUNCTIONS ARE WRITTEN FIRST FOLLOWED BY MAIN CODE

2 function out = MA(x,t) %Calculating Moving Avg

3 n = size(x);

4 out = zeros(n(1),1);

5 for i=1:t

6 out(i) = sum(x(1:i))/i;

7 end

8 for i=t+1:n(1)

9 out(i) = sum(x(i(t1):i))/t;

10 end

11 end

12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

13 function out = MV(x,MA,t) %Calculating Moving Variance

14 n = size(x);

15 out = zeros(n(1),1);

16

17 for i = t+1:n(1)

18 a=0;

19 for j=1:t

20 a=a+(x(ij+1)MA(i))2;

21 end

22 out(i) = a/t;

23 end

24

25 for i = 1:t

26 a=0;

27 for j=1:i

28 a=a+(x(i

j+1)

MA(i))2;29 end

30 out(i) = a/i;

31 end

32 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

33 % Main Code for question #2

34 clc

35 clear


37 data = Data;

38 t = 2 0;%size of window

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39 N = size(data);

40 x = data(:,2);

41 M avg = MA(x,t);

42 %Calculate Skew

43 M var = MV(x,M avg,t); %sigma square44 skew = zeros(N(1),1);

45 kurt = zeros(N(1),1);

46 for i = t+1:N(1)

47 a = 0;b = 0;

48 for j = 1:t

49 a = a + (x(ij+1) M avg(i))3;

50 b = b + (x(ij+1) M avg(i))4;

51 end

52 skew(i) = a/(t*(M var(i))1.5); % factor 1.5 balances the square

53 kurt(i) = (b/(t*(M var(i))2))3;

54 end

55 for i = 1:t

56 a = 0;b = 0;

57 for j=1:i

58 a = a + (x(ij+1)M avg(i))3;

59 b = b + (x(ij+1)M avg(i))4;

60 end

61 skew(i) = a/(i*(M v ar(i))+0.00001)1.5; %Small constant added to avoid ...

0 in denominator

62 kurt(i) = (b/(i*(M va r(i))+0.00001)2)3;%Small constant added to ...

avoid 0 in denominator

63 end

64

65 plot(data(:,1),skew)

66 xlabel('Days');ylabel('20 day moving SKEW');grid on;

67 figure

68 plot(data(:,1),kurt)

69 xlabel('Days');ylabel('20 day moving KURTOSIS');grid on

Question #3

a. Compute and plot the overall autocorrelation.

b. Compute and plot the overall autocovariance.

Answer

Note: I have calculated the autocorrelation and autocovariance for different lags or delays

starting from 1 to 253.

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(a)

0 50 100 150 200 250 3000

50

100

150

200

250

300

lag (n)

Autocorrelation

Matlab Code

1 %Question # 3

2 %Overall Auto correlation

3 clc

4 clear


6 data = Data;

7 N = size(data);

8 x = data(:,2);

9 autocor = zeros(N(1),1);

10 for i = 1:N(1)

11 a = 0;

12 for j = 1:N(1)

13 if i+j N(1)

14 a = a + x(j)*x(j+i);

15 end

16 end

17 autocor(i) = a/N(1);

18 end

19 plot(autocor,'o')

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20 xlabel('lag (n)');ylabel('Autocorrelation');grid on;

(b)

0 50 100 150 200 250 30010

5

0

5

10

15

20

25

lags (n)

Au

tocovariance

Matlab Code

1 % H W 1 3 b

2 % Overall auto covariance

3 clc

4 clear


6 data = Data;

7 N = size(data);8 x = data(:,2);

9 x mean = mean(x);

10 autovar = zeros(N(1),1);

11 for i = 1:N(1)

12 a = 0;

13 for j = 1:N(1)

14 if i+j N(1)

15 a = a + (x(j)x mean)*(x(j+i)x mean);

16 end

17 end

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18 autovar(i) = a/N(1);

19 end

20 plot(autovar,'o')

21 xlabel('lags (n)');ylabel('Auto covariance');grid on;

Question #4

Any news about predicting movements in this stock? Is it time to buy this stock now?

Answer

For this problem, I have used a neural network to perform 1-day ahead prediction of stock

closing prices. I am assuming that the next day prices are affected by todays price, 20-day

moving average, 20-day moving variance, 20-day skew and kurtosis. Therefore, the output

of the neural networks is the next day stock price while todays price, its 20-day moving

average, 20-day moving variance, 20-day skew and 20-day kurtosis are inputs.

The neural network model used is a simple feed forward network with 5 inputs, 10 hidden

neurons and 1 output. Training is done using back propagation algorithm. The first 220

days are used for training and the prediction is made for the next 34 days. From figure 2,

we can observe that the neural network is able to forecast the movements of closing stock

prices.

Note: I believe this network will fail if the stock market crashes due to factors such as war.

Also, the recent market crash due to Chinese currency devaluation are also difficult to predictusing this particular model.

My neural network model predicts that the next day (255th day) closing price is 25.75 while

the price of 254th day is 19.03. Therefore, we can buy the stocks.

Matlab Code

1 function [y]=weightconv(Wo,Wi)

2

insize=size(Wi);outsize=size(Wo);3 z=Wo';

4 for i=1:insize(1)

5 z=[z;Wi(i,:)'];

6 end

7 y=z;

8 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

9 %Forecasting Stock Prices Using Stocks

10 clc

11 clear all

12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5

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0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

X: 220Y: 0.6486

Days

NormalizedStockPrices

Actual Prices

Predicted Prices

Predicted Values

Training Data

Figure 2: Training and forecasting data


14 data = Data;

15 t = 2 0;%size of window

16 N = size(data);

17 x = data(:,2);

18 M avg = MA(x,t);

19 M var = MV(x,M avg,t);

20 load('skew.mat')

21 load('kurt.mat') %loading the presaved 20 day skew and Kurt

22 %%%%creating a Pattern Matrix%%%%%%%%

23 pattern = zeros(N(1),6); % 5 inp and 1 outputs

24 pattern(:,1:5) = [x M avg M var skew kurt];

25 pattern(1:253,6) = x(2:254);

26 pattern = pattern(1:253, :);

27

28 %%%%%Normalizing the Data points%%%%%%%%

29 pmax = zeros(1,6); pmin = zeros(1,6);

30 for i = 1:6

31 pmax(i) = max(pattern(:,i));

32 pmin(i) = min(pattern(:,i));

33 end

34 for i = 1:253

35 pattern(i,:) = (pattern(i,:) pmin)./(pmax pmin);

36 end

37

38 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

39 N=253; %number of pattern

40 %Q=0.8*N;

41 %Q=N;

42 Q = 220; % Number of patterns used for NN training.

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43 Ni=5;Nj=10;

44 Runs=1;

45 output=pattern(:,Ni+1);cc=0;accu=zeros(Runs,1);alpha=0.9;itercount=zeros(Runs,1);

46 for epo=1:Runs

47 tic48 Wi=2*rand(Nj,Ni+1)1; Wi ori=Wi; %hidden weight

49 Wo=2*rand(1,Nj+1)1; Wo ori=Wo;

50 % load('input weight5.mat')

51 % load('output weight5.mat')

52 % Wi=Wi ori;Wo=Wo ori;

53 Wi=zeros(Nj,Ni+1);

54 Wo=zeros(1,Nj+1);

55

56 m=Nj+1+(Ni+1)*Nj;

57 W=zeros(m,1);

58 Si=[ones(N,1) pattern(:,1:Ni)]; %input signals

59 V=[1; zeros(Nj,1)];

60 Jo=zeros(1,Nj+1); %jacobian wrt out

61 Ji=zeros(Nj,Ni+1);% jacobian wrt inp

62 mu=0.45;

63 % mu=1;

64 % mu=0.7;

65 tol=.01;

66 iter=2000;W=weightconv(Wo,Wi);Wmin=2*rand(m,1)1;r=1;

67

68 for kk=1:iter

69 error=0;

70 for i=1:Q

71

72 in=Si(i,:)';out=output(i);

73 h1=Wi*in;

74 v=1./(1+exp(h1));

75 V=[1;v]; %neuron signal

76 h2=Wo*V;

77 y=1/(1+exp(h2));

78 error=(outy)2+error;

79 Jo=y*(1y)*V';

80 for j=1:Nj

81 Ji(j,:)=y*(1

y)*Wo(1+j)*V(1+j)*(1

V(1+j))*in';82 end

83 % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Back Propagation+ momentum

84 J=weightconv(Jo,Ji);

85 e=(outy);

86 deno=norm(J'*e)2;

87 rate=1;

88 Wi=Wi+mu*rate*Ji*e+alpha*Wi;

89 Wo=Wo+mu*rate*Jo*e+alpha*Wo;

90 Wi=Ji*e;

91 Wo=Jo*e;

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92 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5 BP

93 % J=weightconv(Jo,Ji);

94 % e=(outy);

95 % deno=norm(J'*e)2;

96 % rate=1;97 % Wi=Wi+mu*rate*Ji*e;

98 % Wo=Wo+mu*rate*Jo*e;

99

100 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%55

101

102 end

103 epo

104 kk

105 Err(epo,kk)=sqrt((error)/Q);

106 Rate(epo,kk)=rate;

107 if Err(epo,kk)

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