Math 15 Lecture 10 University of California, Merced Scilab Programming – No. 1

Math 15 Lecture 10Math 15 Lecture 10University of California, MercedUniversity of California, Merced

ScilabProgramming – No. 1

Course Lecture ScheduleWeek Date Concepts Project Due

1

2 January 28 Introduction to the data analysis

3 February 4 Excel #1 – General Techniques

4 February 11 Excel #2 – Plotting Graphs/Charts Quiz #1

5 February 18 Holiday

6 February 25 Excel #3 – Statistical Analysis Quiz #2

7 March 3 Excel #4 – Regression Analysis

8 March 10 Excel #5 – Interactive Programming Quiz #3

9 March 17 Introduction to Computer Programming - Part - I

March 24 Spring Recesses

10 March 31 Introduction to Computer Programming - Part - II (4/4) Project #1

11 April 7 Programming – #1 Quiz #4

12 April 14 Programming – #2

13 April 21 Programming – #3 Quiz #5

14 April 28 Programming – #4

15 May 5 Programming - #5 Quiz #6

16 May 12 Movies / Evaluations Project #2

Final May 19 Final Examination (3-6pm COB 116)

How do you like Scilab so far?

Great

Tool!

Outline

1. Introduction to Computer

Programming

2. More programming

4

What is a computer programming? A computer programming is a sequence of

instructions that specifies how to perform a computation. The computation might be something mathematical,

such as solving a system of equations or finding the roots of a polynomial,

but it can also be a symbolic computation, such as searching and replacing text in a document or (strangely enough) compiling a program.

Basically, all computers need someone to tell them what to do!

The way of programming The single most important skill for a scientist is

problem solving. Problem solving means the ability to

formulate problems, think creatively about solutions, and express a solution clearly and accurately.

As it turns out, the process of computer programming is an excellent opportunity to practice problem-solving skills.

October, 2007

Bis 180

When does computer programming become useful?

Let’s calculate populations of A and B species at t = 156 if the population growths of A and B are following this system of equations:

given that A = 10 and B = 5 at t = 0.

ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

8

We can find A and B for next generation readily.


ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

9

001

001

9.01.0

1.09.0

BAB

BAA

59.0101.0

51.0109.0

1

1

B

A

5.55.41

5.95.09

1

1

B

A

One obvious, but not best, way to do this calculation.

To find A and B at t = 156:

10

9.59.01.0

1.91.09.0

112

112

BAB

BAA

5.55.41

5.95.09

1

1

B

A

22.69.01.0

78.81.09.0

223

223

BAB

BAA

155155156

155155156

9.01.0

1.09.0

BAB

BAA

Even though this is an obvious way to do this calculation, nobody wants to do this way.

Is there any other way to do this?

ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

Another way to do this.Let’s use the matrix to solve this system of

equations:

To find A and B for t = 1,

For t = 2,

For t = 3,

ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

11

t

t

t

t

B

A

B

A

9.01.0

1.09.0

1

1

0

0

1

1

9.01.0

1.09.0

B

A

B

A

0

0

2

0

0

1

1

2

2

9.01.0

1.09.0

9.01.0

1.09.0

9.01.0

1.09.0

9.01.0

1.09.0

B

A

B

A

B

A

B

A

0

0

3

0

0

2

2

2

3

3

9.01.0

1.09.0

9.01.0

1.09.0

9.01.0

1.09.0

9.01.0

1.09.0

B

A

B

A

B

A

B

A

0

0

2

0

0

1

1

2

2

9.01.0

1.09.0

9.01.0

1.09.0

9.01.0

1.09.0

9.01.0

1.09.0

B

A

B

A

B

A

B

A

Another way to do this.Let’s use the matrix to solve this system of

equations:

For t = 156,

For t = n,

ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

12

t

t

t

t

B

A

B

A

9.01.0

1.09.0

1

1

0

0

156

156

156

9.01.0

1.09.0

B

A

B

A

0

0

9.01.0

1.09.0

B

A

B

An

n

n

0

0

156

156

156

9.01.0

1.09.0

B

A

B

A

Transition Matrix

Initial Conditions

UC Merced

Any Questions?

Well…The matrix approach is certainly a easy way to

solve the equations.

But I need to plot (A vs. t ) and (B vs. t) between

What is the better way to do this operation?

ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

14

0

0

9.01.0

1.09.0

B

A

B

An

n

n

1560 t

Scilab programming will make your life easy!!

Here is the example:If you know how to program in Scilab, only

9 lines to do the job.

15

T =[0.9 0.1; 0.1 0.9]; // Transition matrixP0=[10;5]; // Initial Conditions

Pt = P0;t = [0];for n=1:156 Pt = [Pt,T^n*P0]; t = [t,n];end

plot(t,Pt(1,:),'-o',t,Pt(2,:),'-s')

Before going further, Some new matrix operations.Simple Matrix Generation

Some basic matrices can be generated with a single command:Command Generated matrix

zeros All zeros

ones All ones

eye Identity matrix

rand Random elements matrix

-->zeros(3,2) ans = 0. 0. 0. 0. 0. 0.

# of rows # of columns

-->ones(1,3) ans = 1. 1. 1.

-->eye(2,2) ans = 1. 0. 0. 1

Some new matrix operations. – cont.rand()

random matrix generator. rand(m1,m2) is a random matrix of dimension m1 by

m2 . Default: Random numbers are uniformly distributed in

the interval (0,1).

# of rows # of columns

-->A = rand(5,3) A = 0.2113249 0.6283918 0.5608486 0.7560439 0.8497452 0.6623569 0.0002211 0.6857310 0.7263507 0.3303271 0.8782165 0.1985144 0.6653811 0.0683740 0.5442573

Some new matrix operations. – Cont.Concatenation

Concatenation is the process of joining smaller size matrices to form bigger ones.

This is done by putting matrices as elements in the bigger matrix.

A=[1 2 3]; B = [4 5 6]; C = [7;8;9]

-->[A B] // to expand columns ans = 1. 2. 3. 4. 5. 6.

-->A=[A B] A = 1. 2. 3. 4. 5. 6.

-->A=[A B] A = 1. 2. 3. 4. 5. 6. 4. 5. 6.

This doesn’t mean A is equal to [A B]. In the programming, this means [A B] is assigned to A.

Some new matrix operations. – Cont.Concatenation

Numbers of rows for two matrices are different. Concatenation must be row/column consistent.

A=[1 2 3]; B = [4 5 6]; C = [7;8;9]

-->[A C] !--error 5inconsistent column/row dimensions

-->[A;B] // to expand rows ans = 1. 2. 3. 4. 5. 6.

-->A=[A;B] // Expand A A = 1. 2. 3. 4. 5. 6.-->A=[A;B] A = 1. 2. 3. 4. 5. 6. 4. 5. 6.

-->A=[A C] A = 1. 2. 3. 7. 4. 5. 6. 8. 4. 5. 6. 9.

UC Merced

Any Questions?

Programming – Use Scilab Editor The "Editor" menu : launches the SCILAB default editor

SciPad. (Use help scipad for additional information). The SciPad editor can be used to create SCILAB script and function files. As well as to create input data files, or to edit output data files.

21

Editor – cont.To Execute the program from SciPad.

Click “Load into Scilab” under Execute menu.

Don’t forget to save your program before closing the editor.The file extensions used by scilab are sce and sci.To save a file, click for the menu File and choose Save.

SCILAB Programming: Simple population Model

Simplest model of a population growth:

Fecundity and death rates; f and d.P – Changes in population

where Pt = P(t) = the site of the population measured on day t.

Pt+1 – Pt is the difference or change in population between two consecutive days.

23

PdfPdPfP

ttt PdfPPP 1

SCILAB Programming: Simple population Model – cont.

where = 1+fd. Population ecologists often refer to the constant as the finite growth

rate.Generally, this type of equations is referred to as a

difference equation.Formula expressing values of some quantity in terms

of previous value:24

ttt PdfPPP 1

t

tttt

P

PdfPdfPP

)1(1

tt QfQ 1

SCILAB Programming: Malthusian Model – cont.

For the values f = .1, d = 0.3 ( =1+f d), and P0 = 500, our entire model is now

First equation is a difference equation.Second equation is called its initial condition.

25

tt PP 1

50007.1 01 PPP tt

SCILAB Programming: Malthusian Model – cont.

For the values =1.07, and P0 = 500, our entire model is now

26

50007.1 01 PPP tt

Day Population

0 500

1

2

3

4

53550007.107.1 01 PP

57250007.107.1 212 PP

61350007.107.1 323 PP

65550007.107.1 434 PP

How to program this sequence of events in Scilab.

SCILAB Programming:

To find a population of Day 5, Scilab Commands are:

To find a table of population vs. Day

27

50007.1 01 PPP tt

007.1 PP tt

P0=500; // Initial Conditiont=5 // Day 5 Pt=(500)^t*P0 // Difference Equation

P0=500; // Initial Conditiont=1 // Day 1 Pt=(500)^t*P0t=2 // Day 2 Pt=(500)^t*P0… Is this best we can do?

SCILAB Programming:

28

Day Population

0

1

2

3

4

n

01

1 07.1 PP

02

2 07.1 PP

03

3 07.1 PP

04

4 07.1 PP

Sequence iteration: Repeating identical or similar tasks

007.1 PP nn

00

0 07.1 PP

Repeating identical or similar tasks without making errors is something that computers do well and people do poorly.

for loop statementThe Basic Structure

for variable = starting_value: increment: ending_value

commands

end

The loop will be executed a fixed number of times specified by (starting_value: increment:

ending_value) or the number of elements in the array variable.

Slightly modified version:for variable = array

commands

end

for loop statement– cont.1st Example:

30

for i = 1:1:5 P = i * 10 end

--> P = 10. P = 20. P = 30. P = 40. P = 50.

loop i P

1 1 110

2 2 210

3 3 310

4 4 410

5 5 510

Numbers between 1 and 5 with increment of 1loop i P

1 1 110

2 2 210

3 3 3*10

4 4 410

5 5 510

for i = 1:5 P = i * 10 end

Default increment is 1, so

for loop statement– cont.2nd Example:

31

for i = 1:3:10 P = i * 5 end

--> P = 5. P = 20. P = 35. P = 50.

loop i P

1 1 15

2 4 45

3 7 75

4 10 105

Numbers between 1 and 10 with increment of 3

for loop statement– cont.3rd Example:

32

t = [0, 0.5, 1, 5, 10, 100];for i = t P = i * 5 end

--> P = 0. P = 2.5 P = 5. P = 25. P = 50. P = 500. loop i P

1 0 15

2 0.5 0.55

3 1 15

4 5 55

5 10 105

6 100 1005

The variable, i, will be assigned from the array t.

Let’s use “for loop” to program Simple Population Model

33

50007.1 01 PPP tt

007.1 PP tt

P0 =500;for t=0:10 Pt=(1.07)^t*P0 end

--> Pt = 500. Pt = 535. Pt = 572.45 Pt = 612.5215 Pt = 655.39801 Pt = 701.27587 Pt = 750.36518 Pt = 802.89074 Pt = 859.09309 Pt = 919.22961 Pt = 983.57568

Let’s calculate populations for Day 1 and Day 10

Now let’s program the first example – Populations of A and B species.

Let’s use the matrix to solve this system of equations:


ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

34

0

0

9.01.0

1.09.0

B

A

B

At

t

t

T=[0.9 0.1;0.1 0.9];P0=[10;5];for t=0:9 Pt=T^t*P0 end 0PTP t

t

Populations of A and B species. – cont.

35

--> Pt = 10. 5. Pt = 9.5 5.5 Pt = 9.1 5.9 Pt = 8.78 6.22 Pt = 8.524 6.476 Pt = 8.3192 6.6808 Pt = 8.15536 6.84464 Pt = 8.024288 6.975712 Pt = 7.9194304 7.0805696 Pt = 7.8355443 7.1644557

T=[0.9 0.1;0.1 0.9];P0=[10;5];for t=0:9 Pt=T^t*P0 end

UC Merced

Any Questions?

Well, the original questions were …

The matrix approach is certainly a easy way to solve the equations.

But I need to plot (A vs. t ) and (B vs. t) between

What is the better way to do this operation?

ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

37

0

0

9.01.0

1.09.0

B

A

B

An

n

n

1560 t

OK. We can use ‘for loop’ to do recursive operation. Now, I would like to know how to plot .



plot(t,Pt(1,:),'-o',t,Pt(2,:),'-s')

Now we understand parts of the earlier program:

0

0

9.01.0

1.09.0

B

A

B

At

t

t

38

We don’t know what operations in these green circles yet.


Pt = P0;t = [0]for n=1:156 Pt = [Pt,T^n*P0]; t = [t,n]end

plot(t,Pt(1,:),'-o',t,Pt(2,:),'-s')

To plot, we need data in the array formats.Example:

So we need to create arrays to plot

theta = linspace(0,2*%pi,100);x = sin(theta);y = cos(theta);plot(theta,x,theta,y,'--')


Pt = P0;

Making arrays and Expanding arrays

40

• Again, this doesn’t means Pt is equal to P0. But, rather, P0 is assigned to a new matrix Pt.

• At this point, Matrices P0 and Pt have exactly the same elements.

Let’s read the previous program line by line:

P0 = 10. 5.

Pt = 10. 5.

Making arrays and Expanding arrays – cont.



Make a new array, t, with one element.

•Expanding elements in the array, t.•For each loop, add an element, n, to the array t.

•Expanding elements in the matrix, Pt.•For each loop, add an element, TnP0, to the matrix Pt.

n is assigned from 1 and 156 with increment of 1

What is happening?



42

t = 0.

Pt = 10. 5.

loop n t Pt

1 1 0, 1 10 9.55 5.5

2 2 0, 1, 2 10 9.5 9.15 5.5 5.9

3 3 0, 1, 2, 3 10 9.5 9.1 8.85 5.5 5.9 6.2

4 4 0, 1, 2, 3, 4 10 9.5 9.1 8.8 8.55 5.5 5.9 6.2 6.5

Pop. A

Pop. B

loop n t Pt

1 1 0, 1 10 9.55 5.5

2 2 0, 1, 2 10 9.5 9.15 5.5 5.9

3 3 0, 1, 2, 3 10 9.5 9.1 8.85 5.5 5.9 6.2

4 4 0, 1, 2, 3, 4 10 9.5 9.1 8.8 8.55 5.5 5.9 6.2 6.5



plot(t,Pt(1,:),'-o',t,Pt(2,:),'-s')

One more line to understand this program:

43

To understand this, let’s review some matrix expressions.

Matrix operations in Scilab For example, the following is a matrix:

In Scilab,

X=[5 8 1;4 0 2]X(2,1) // element of 2nd row and 1st column

First subscript in a matrix refers to the row and the second subscript refers to the column.

X(:,2) //All elements in 2nd Column

X(1,:) // All elements in 1st row.

204

185X

ans = 5. 8.

1.

ans = 8 0

ans = 4



plot(t,Pt(1,:),'-o',t,Pt(2,:),'-s')

Now we understand this program:

0

0

9.01.0

1.09.0

B

A

B

At

t

t

45

This means that plotting all elements of 1st row against t with the option, ‘-O’.

UC Merced46

Any Questions?

Next LectureMore Programming in Scialb

Logical expressionsif statement

Solving some more difference equations, such as

47

tttt

ttttt

QPQQ

QPPPP

6.13.0

5.013.11

1

1

ttt

ttt

QPQ

QPP

9.01.0

1.09.0

1

1

Well. Can I make Scilab outputs nicer?Answer is Yes!

48

--> Pt = 10. 5. Pt = 9.5 5.5 Pt = 9.1 5.9 Pt = 8.78 6.22 Pt = 8.524 6.476 Pt = 8.3192 6.6808 Pt = 8.15536 6.84464 Pt = 8.024288 6.975712 Pt = 7.9194304 7.0805696 Pt = 7.8355443 7.1644557

Day = 0 Pop A= 10.00 Pop B= 5.00

Day = 1 Pop A= 9.50 Pop B= 5.50

Day = 2 Pop A= 9.10 Pop B= 5.90

Day = 3 Pop A= 8.78 Pop B= 6.22

Day = 4 Pop A= 8.52 Pop B= 6.48

Day = 5 Pop A= 8.32 Pop B= 6.68

Day = 6 Pop A= 8.16 Pop B= 6.84

Day = 7 Pop A= 8.02 Pop B= 6.98

Day = 8 Pop A= 7.92 Pop B= 7.08

Day = 9 Pop A= 7.84 Pop B= 7.16

Documents

Math 15 Lecture 10 University of California, Merced Scilab Programming – No. 1