Math 15 Lecture 12 University of California, Merced Scilab Programming – No. 3.

Math 15 Lecture 12Math 15 Lecture 12University of California, MercedUniversity of California, Merced

ScilabProgramming – No. 3

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 SCILAB - Part - I

March 24 Spring Recesses

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

11 April 7 Introduction to SCILAB - Part - III Quiz #4

12 April 14 Programming – #1

13 April 21 Programming – #2 Quiz #5

14 April 28 Programming – #3

15 May 5 Programming - #4 Quiz #6

16 May 12 Movies / Evaluations Project #2

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

Project #2 – Due May 12th

Projects can be performed individually or in groups of three, with following rules:A team consists of at most 3 people—no copying between

teams! Teams turn in one project report and get the same grade.Team project report must include a title page, where a team

describe each team member’s contribution.10% bonus for labs done individually

Individual projects must not be copied from anyone else

Scilab

Materials will be available on April 30th.

Math 15 Final – May 19 (COB 116)

100 pts. total2 hours (3pm – 5pm)50 questions (2 pts. each)

10% from 1st lecture40% Excel Related50% Scilab Related

Mostly Multiple choices and fill-in-blanksSimilar to quizzes.One problem will ask you to make a small

programmingOpen notes (Max. 5 sheets)

But No computers will be allowed.

12%60Computer Labs

20%100In-Class Quizzes

20%100Assignments

100%500Total

20%100Final Exam

14%70Project #2

14%70Project #1

% Final Grade

PointsActivity

12%60Computer Labs

20%100In-Class Quizzes

20%100Assignments

100%500Total

20%100Final Exam

14%70Project #2

14%70Project #1

% Final Grade

PointsActivity

Grading for Math 15

Activity Points

Assignments 120

In-Class Quizzes 120

Computer Labs 70

Project #1 70

Project #2 70

Final Exam 100

Total 550

Projected points

+~45 extra points (by optional hw.)Over 275D

Over 325C

Over 375B

Over 425A

Total points achieved

Grade

Over 275D

Over 325C

Over 375B

Over 425A

Total points achieved

Grade

Any Questions?

Outline

Today – more programming

1.More for-loop

2.Logical expressions

if statement

7

Review: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.

We can describe this system of difference equations in terms of matrices.

ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

8

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

equations:

For t = n,

ttt

ttt

BAB

BAA

9.01.0

1.09.0

1

1

9

t

t

t

t

B

A

B

A

9.01.0

1.09.0

1

1

0

0

9.01.0

1.09.0

B

A

B

An

n

n

Transition Matrix

Initial Conditions

Now you know how to program in Scilab.

A system of two linear difference equations

10

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')

ttt

ttt

QPQ

QPP

9.01.0

1.09.0

1

1

In addition to the previous example:

Including lab 9 and HW #8

We are only dealing with linear equations. In many cases, a prediction from the linear equation cannot really be accurate for very long. So what’s next?

t

tttt

P

PdfPdfPP

)1(1 Lecture 10

Dolphin training problem Lecture 11

Now, what if a population is described by the following equation:

These is not a linear difference equation, so you cannot define the simple matrix.

How can we solve this equation?

12

K

PrPP t

tt 111

where K and r are positive.You may see this equation in Bis 1 or/and Bis 180.

Logistic model

The parameter K and r in this model have direct biological interpretations:If P<K, the population will increase.If P>K, the population will decrease.

K is called the carrying capacity of the environment, because it represents the maximum number of individuals that can be supported over a long period.

13

K

PrPP t

tt 111

Logistic model – cont.

If P<<K,

14

K

PrPP t

tt 111

0K

Pt rPK

PrPP t

ttt

1111

The model becomes a simple population model.

tt PP 1

UC Merced

Any Questions?

Let’s program this logistic model.

If P00 = 2, let’s program to graph how the population of this model change.

16

K

PrPP t

tt 111

where K = 100 and r = 0.7

Let’s program this discrete logistic model – cont.

17

2100

17.01 01

P

PPP t

tt

P=2; // Initial Condition

Pt = [P]; // Set up a new array, Pt, for populationst = [0]; // Set up a new array, t, for generations

for n=1:15 // Up to 15 generation P = P * (1 + 0.7*(1.0 – P/100)); // discrete logistic model Pt = [Pt P]; // Expanding the array, Pt. t = [t,n]; // Expanding the array, t.end

plot(t,Pt,'-o')

Let’s program this discrete logistic model – cont.

18

2100

17.01 01

P

PPP t

tt


Pt = [P]; // Set up a new array, Pt, for populationst = [0]; // Set up a new array, t, for generations

for n=1:15 // Up to 15 generation P = P * (1 + 0.7 *(1.0 – P/100)); // discrete logistic model Pt = [Pt P]; // Expanding the array, Pt. t = [t,n]; // Expanding the array, t.end

plot(t,Pt,'-o')

0 5 10 150

10

20

30

40

50

60

70

80

90

100

Generation

Po

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Review: for loop statementfor loop is used to repeat a command or a

group of commands for a fixed number of times.

The Basic Structurefor 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.

UC Merced

Any Questions?

Now, what if two populations are described by the following equations:

Again, these are not linear difference equations, so you cannot define the simple matrix.

How can we solve this system of nonlinear difference equations?

Well, we can do the same!

21

tttt

ttttt

QPQQ

QPPPP

6.14.0

5.013.11

1

1

where Initial Conditions, P0 = 1.1 and Q0 = 0.5

Let’s program this nonlinear population model.

22

//Initial ConditionsP_current=1.1;Q_current=0.5;

// Set up a new array, Pt, for the population, PPt = [P_current];// Set up a new array, Qt, for the population, QQt = [Q_current];

t = [0]; // Set up a new array, t, for generations

for n=1:15 // Up to 15 generation P_next = P_current * (1 + 1.3 *(1.0 - P_current))-0.5*P_current*Q_current; Q_next = 0.4*Q_current+1.6*P_current*Q_current; Pt = [Pt P_next]; // Expanding the array, Pt. Qt = [Qt Q_next]; // Expanding the array, Qt t = [t,n]; // Expanding the array, t.end

plot(t,Pt,'-o',t,Qt,'-s')legend('P','Q')

5.0

1.1

6.14.0

5.013.11

0

0

1

1

Q

P

QPQQ

QPPPP

tttt

ttttt

Let’s program this nonlinear population model – cont.






0 5 10 150.5

0.6

0.7

0.8

0.9

1.0

1.1

Generation

Po

pu

latio

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P

Q

This program is not correct at all!

So, what is wrong on this program?

24






5.0

1.1

6.14.0

5.013.11

0

0

1

1

Q

P

QPQQ

QPPPP

tttt

ttttt

So, what is wrong on this program? – cont.

25


5.0

1.1

6.14.0

5.013.11

0

0

1

1

Q

P

QPQQ

QPPPP

tttt

ttttt

Let’s look at for loop statement carefully.

26

loop n P_current Q_current P_next Q_next

initial 1.1 0.5

1 1 1.1 0.5 0.682 1.08

2 2 1.1 0.5 0.682 1.08

3 3 1.1 0.5 0.682 1.08

4 4 1.1 0.5 0.682 1.08

for n=1:15P_next = P_current * (1 + 1.3 *(1.0 - P_current)) - 0.5*P_current*Q_current;Q_next = 0.4*Q_current+1.6*P_current*Q_current;end


initial 1.1 0.5

1 1 1.1 0.5 0.682 1.08

2 2 1.1 0.5 0.682 1.08

3 3 1.1 0.5 0.682 1.08

4 4 1.1 0.5 0.682 1.08

Why does this happen?We have never updated P_current & Q_current

within the for-loop.

What we want to happen:

27


initial 1.1 0.5

1 1 1.1 0.5 0.682 1.08

2 2 0.682 1.08 0.596 1.610

3 3 0.596 1.610 0.429 2.179

4 4 0.429 2.179 0.280 2.368

5.0

1.1

6.14.0

5.013.11

0

0

1

1

Q

P

QPQQ

QPPPP

tttt

ttttt


initial 1.1 0.5

1 1 1.1 0.5 0.682 1.08

2 2 0.682 1.08 0.596 1.610

3 3 0.596 1.610 0.429 2.179

4 4 0.429 2.179 0.280 2.368

OK – How to program this:

28

5.0

1.1

6.14.0

5.013.11

0

0

1

1

Q

P

QPQQ

QPPPP

tttt

ttttt


initial 1.1 0.5

1 1 1.1 0.5 0.682 1.08

2 2 0.682 1.08 0.596 1.610

3 3 0.596 1.610 0.429 2.179

4 4 0.429 2.179 0.280 2.368

for n=1:15P_next = P_current * (1 + 1.3 *(1.0 - P_current)) - 0.5*P_current*Q_current;Q_next = 0.4*Q_current+1.6*P_current*Q_current;

P_current = P_next;Q_current = Q_next;end

Re-assign values to P_current and Q_current!

Correct Program is

29




for n=1:15 // Up to 15 generation P_next = P_current * (1 + 1.3 *(1.0 - P_current))-0.5*P_current*Q_current; Q_next = 0.4*Q_current+1.6*P_current*Q_current; Pt = [Pt P_next]; // Expanding the array, Pt. Qt = [Qt Q_next]; // Expanding the array, Qt t = [t,n]; // Expanding the array, t. P_current = P_next; Q_current = Q_next;end


5.0

1.1

6.14.0

5.013.11

0

0

1

1

Q

P

QPQQ

QPPPP

tttt

ttttt

0 5 10 150.0

0.5

1.0

1.5

2.0

2.5

Generations

Po

pu

latio

ns

P

Q

Predator-Prey model

This is a simple predator-prey model.P – prey populationQ – Predator population

30

tttt

ttttt

QPQQ

QPPPP

6.13.0

5.013.11

1

1

where Initial Conditions, P0 = 1.1 and Q0 = 0.5

0 5 10 150.0

0.5

1.0

1.5

2.0

2.5

Generations

Po

pu

latio

ns

P

Q

More Generations (up to 100)

0 10 20 30 40 50 60 70 80 90 1000.0

0.5

1.0

1.5

2.0

2.5

Geneations

Po

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0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.10.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

2.2

2.4

P

Q

P

Q

Phase plot

subplot(2,1,1)plot(t,Pt,'-o',t,Qt,'-s')legend('P','Q')subplot(2,1,2)plot(Pt,Qt,'-o')

Here are Scilab commands to generate two graphs in one page

UC Merced

Do you have a question?

Now, let’s program this modified discrete logistic model.

Initial population, P00 = 2r = 0.7What if K is not constant.

i.e. K = 100 for first 10 generation, then K = 150 for rest of generations.

Let’s program to graph how the population of this model change.

33

K

PrPP t

tt 111

It means:

34

20 P

10017.011

ttt

PPP

15017.011

ttt

PPP

Initial population

First 10 generations

From 11th generation

if statementThe main statement used for selecting from alternative actions

based on test results.This construction provides a logical branching for computations

if <test1><statement1>

end

Syntax

If the <test1> is non-zero or True, then <statement1> is executed.

if statement – cont.Other syntax


else<statement2>

end


elseif <test2><statement2>

else:<statement3>

end

If-else structure. If <test1> is true, <statement1> is executed; otherwise <statement2> is executed.

If-elif-else structure. If <test1> is true, <statement1> is executed; If <test1> is false, but <test2> is true, <statement2> is executed; otherwise <statement3> is executed.

if statement examples

if x < 0printf (“it must be a negative number”)

end

if x < 0printf (“It must be a negative number”)

elseif x >0printf (“It must be a positive number”)

elseprintf (“It must be zero”)

end

Example 1

Example 2

Comparison or Test Operators

Operator Meaning Example Evaluates To

== equal to or congruent to “A” == “A” True

!= not equal to 8 != 5 True

> greater than 5 > 8 False

< less than 5 < 8 True

>= greater than or equal to 5 >= 8 False

<= less than or equal to 5 <= 5 True

Here is how:

39

10150

100100 17.01

2

1

0

tK

tK

K

PPP

P

ttt


Pt = P; // Set up a new array, Pt, for populationst = [0]; // Set up a new array, t, for generations

for n=1:20 // Up to 20 generation if n <= 10 K = 100 // if n is less than and eqaul to 10, K = 100 else K = 150 // other n’s – K = 150 end P = P * (1 + 0.7 *(1.0 – P/K)); // discrete logistic model Pt = [Pt P]; // Expanding the array, Pt. t = [t,n]; // Expanding the array, t.end

plot(t,Pt,'-o')

Here is the plot:

0 2 4 6 8 10 12 14 16 18 200

50

100

150

Generations

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K = 100 K = 150

UC Merced

Any Questions?

Do you have a question?

Next WeekLecture : Last Programming in Scialb

Another loop – while loop

Quiz #6

42

Math 15 Lecture 12 University of California, Merced Scilab Programming – No. 3.

Documents

Transcript of Math 15 Lecture 12 University of California, Merced Scilab Programming – No. 3.