6 Programming with MATLAB.pdf

8/11/2019 6 Programming with MATLAB.pdf

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Programming with

MATLAB

Cheng-Liang ChenPSELABORATORY

Department of Chemical EngineeringNational TAIWAN University


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Chen CL 1

Relational Operators

Six relational operators

Relational operator Meaning

< Less than.

Greater than.

>= Greater than or equal to.

== Equal to.

= Not equal to.

Note: single = is assignmentor replacement


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Chen CL 2

Result of comparison: (can be used as a variable)

1 if the comparison is true

0 if the comparison is false

x = 2 ;

y = 5 ;

z = x < y,...

w = x > y

z =

1

w =

0

Compare arrays

x = [ 6, 3, 9];

y = [14, 2, 9];

z = (x < y),...

p = (x ~= y),...

q = (x > 8)

z =

1 0 0p =

1 1 0

q =

0 0 1

z = x(x


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Chen CL 3

Logical Operators and Functions

Three logical operators

Operator Name Definition

NOT A returns an array the same dimension asA; the new array hasones whereA is zeroandzeros where A is nonzero.

& ANDA&B returns an array the same dimension asA and B; the new array has ones where bothA and B have nonzero elements and zeros

where either A orB is zero.| OR

A|B returns an array the same dimension as Aand B; the new array hasones where at leastone element in A or B is nonzero and zeros

where A andB are both zero.


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Chen CL 4

Element-by-element operations (exceptNOT)

x = [ 6, 3, 9];

y = [14, 2, 9];

z = ~x,...p = x > y

q = x y)

% = ( x < = y )

z =

0 0 0

p =

0 1 0

q =

1 0 1

r =

1 0 1z = 0&3

w = 2&3

z =0

w =

1

z = [5,-3,0,0]&[2,4,0,5]

z =

1 1 0 0

C C


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Chen CL 5

Order of Precedence for Operator Types

First Parentheses; evaluated starting with the innermost pair.

Second Arithmeticoperators and NOT(

); evaluated fromleft to right.

Third Relationaloperators; evaluated fromleft to right.Fourth Logical AND.

Fifth Logical OR.

z = 5 < 6 & 1 % ( 5 < 6 ) & 1

z =

1

x = [ 6, 3, 9];

y = [14, 2, 9];

a = [ 4, 3, 12];

z = x > y & a ,...

% = (x>y) & aw = x > y & x > a

% = (x>y) & (x>a)

z =

0 1 0

w =

0 0 0

~(4 > 5) is equivalent to 5 >= 4

5 < x < 10 is equivalent to (5


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Chen CL 6

z = ~3==7|4==6 %((~3)==7)|(4==6)

z =

0

x = xor([3,0,6],[5,0,0])

y = [3,0,6]&[5,0,0]z = [3,0,6]|[5,0,0]

x =

0 0 1

y =

1 0 0z =

1 0 1

x = [1, 1, 0, 0];

y = [1, 0, 1, 0];

Truth_table = [x, y, ~x, x|y, x&y, xor(x,y)]Truth_table =

1 1 0 1 1 0

1 0 0 1 0 1

0 1 1 1 0 1

0 0 1 0 0 0

x y x x|y x&y xor(x, y)true true false true true falsetrue false false true false true

false true true true false true

false false true false false false

Ch CL 7


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Chen CL 7

Logical Operators andfindFunction

Logical function Definition

any(X) Returns a scalar, which is 1 if any of the elements in thevectorx is nonzeroand0 otherwise.

any(A) Returns a row vector having the same number of column asA and containing ones and zeros, depending on whether or

not the corresponding column of the matrix Acontains anynonzero elements.

all(X) Returns a scalar, which is1 if all the elements in the vectorx are nonzeroand0 otherwise.

all(A) Returns a row vector having the same number of column

as the matrix A and containing ones and zeros, dependingon whether or not the corresponding column of A has allnonzero elements.

find(A) Computes an array containing the indices of the nonzeroelementsof the array A.

Ch CL 8


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Chen CL 8

Logical function Definition

[u,v,w] = find(A) Computes the arraysuandv containing the row and columnindices of the nonzero elements of the arrayAand computes

the array w containing the values of the nonzero elements.The array w may be omitted.

finite(A) Returns an array of the same dimension as A with oneswhere the elements ofA are finite and zeros elsewhere.

isnan(A) Returns an array of the same dimension as A with ones

where A has NaN and zeros elsewhere. (NaN stands forNota Number, which means an undefined result.)

isinf(A) Returns an array of the same dimension as A, with oneswhere A has inf and zeros elsewhere.

isempty(A) Returns a 1 ifA is an empty matrix and 0 otherwise.

isreal(A) Returns a 1 ifA has no elements with imaginary parts and0 otherwise.

xor(A,B) Returns an array the same dimension as A andB; the newarray has ones where either A or B is nonzero, but notboth, and zeros where A and B are either both nonzero orboth zero.

Ch CL 9


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Chen CL 9

Logical Operators andfindFunctionExample

x = [5,-3,0,0,8];y = [2, 4,0,5,7];

L_nonzero = find(x&y) ,...

values = y(x&y) ,...

how_many = length(values)

L_nonzero =1 2 5

values =

2 4 7

how_many =

3

Chen CL 10


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Chen CL 10

Test Your Understanding

T4.2-1Ifx= [5,3, 18, 4] and y = [9, 13, 7, 4], what will be the resultof the following operations ? UseMATLABto check your answer.

1. z= y > x2. z=x&y3. z=x|y4. z=xor(x, y)

Chen CL 11


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Chen CL 11

Example: Height and Speed of A Projectile

The height and speed of a projectile (such as a thrown ball) launched with a speed

ofv0 at an angle A to the horizontal are given by h(t) =v0tsinA 0.5gt2

,v(t) =

v20 2v0gtsinA+g2t2, where g is the acceleration due to gravity.

The projectile will strike the ground when h(t) = 0, which gives the time to hit,thit= 2(v0/g) sin(A). Suppose that A= 40

o, v0= 20 meters/second, andg= 9.81 meters/second2. Use the MATLAB relational and logical operators to

find the times when the height is no less than 6 meters and the speed issimultaneously no greater than 16 meters/second. In addition, discuss anotherapproach to obtaining a solution.

Solution:

The key to solving this problem with relational and logical operators is to use thefind command to determine the times at which the logical expression(h>=6)&(v


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Chen CL 12


% Set the values for initial speed, gravity, and angle.

v0 = 20; g = 9.81; A=40*pi/180;% Compute the time to hit.

t_hit = 2*v0*sin(A)/g;

% Comput the arrays containing time, height, and speed.

t = [0 : t_hit/100 : t_hit];

h = v0*t*sin(A) - 0.5*g*t.^2;

v = sqrt(v0^2 - 2*v0*g*sin(A)*t + g^2*t.^2);

h6=6*t.^0; v16=16*t.^0;

subplot(3,1,1)

plot(t,h,t,h6,Linewidth,2)

set(gca,LineWidth,2,FontSize,12)

xlabel(\bf Time,FontSize,12)ylabel(\bf Height,FontSize,12)

subplot(3,1,2)

plot(t,v,t,v16,Linewidth,2)

set(gca,LineWidth,2,FontSize,12)

xlabel(\bf Time,FontSize,12)

ylabel(\bf Velocity,FontSize,12)

Chen CL 13


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Chen CL 13


% Determine when the height is no less than 6,

% and the speed is no greater than 16.

y = h>=6 & v=6 & v


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Chen CL 14


t_1 =

0.8649

t_2 =

1.7560

Chen CL 15


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Chen CL 15


T4.2-2Consider the problem given in the previous Example. Use relationaland logical operators to find the times for which either theprojectiles height is less than 4 meters or the speed is greater than17 meters per second. Plot h(t) and v(t) to confirm your answer.

Chen CL 16


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Chen CL 16

Conditional Statements

% Decision making (I)

If I get a raise,

I will buy a new car

. (period)

% Decision making (II)

If I get at least a $100 per week raise,I will buy a new car;

else,

I will put the raise into savings

. (period)

% Decision making (III)If I get at least a $100 per week raise,

I will buy a new car;

else, if the raise is greater than $50,

I will buy a new stereo;

otherwise,

I will put the raise into savings

. (period)

Chen CL 17


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Chen CL 17

TheifStatement

if logical expression

statements

end

i f x > = 0y = sqrt(x)

end

if x >= 0, y = sqrt(x), end

Chen CL 18


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Chen CL 18

NestedifStatements

if logical expression 1


statementsend

end

if (logical expression 1) & (logical expression 2)

statements

end

Chen CL 19


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Chen CL 19

z =

x+

y

w = log x 3log yz = 0 ; w = 0 ;

if (x >= 0)

if (y >= 0)

z = sqrt(x) + sqrt(y)

w = log(x) - 3*log(y)

end

end

z = 0 ; w = 0 ;

if (x >= 0)&(y >= 0)z = sqrt(x) + sqrt(y)

w = log(x) - 3*log(y)

end

%%% Nested if statements


statement group 1


statement group 2

end

end

Chen CL 20


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Chen CL 20

TheelseStatement

if logical expression

statement group 1

else

statement group 2end

y=

x for x 0

ex 1 for x = 0y = sqrt(x)

else

y = exp(x) - 1

end

Chen CL 21


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Note: For logical expression of an array, the test returns a value of

trueonly ifALL the elements of the logical expression are nonzero !

x = [4, -9, 25];i f x < 0

disp(All elements of)

disp(x are negative.)

else

y = sqrt(x)

end

y =

2 0+3.000i 5

x = [4, -9, 25];i f x > = 0

y = sqrt(x)

else

disp(Some of the elements)

disp( of x are negative.)

end

Some of the elements

of x are negative.

Chen CL 22


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TheelseifStatement


statement group 1

elseif logical expression 2

statement group 2

elsestatement group 3

end

Chen CL 23


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y=

ln x if x 5x if 0 x = 5

y = log(x)

else

i f x > = 0y = sqrt(x)

end

end

% Note: no action for x=-2 ?

i f x > = 5

y = log(x)

elseif x >= 0y = sqrt(x)

end

y=

ln x if x >10

x if 0 x 10ex 1 if x 1 0

y = log(x)

elseif x >= 0

y = sqrt(x)

else

y = exp(x) - 1

end

Chen CL 24


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i f x > 1 0

y = log(x)

i f y > = 3

z = 4*yelseif y >= 2.5

z = 2*y

else

z = 0

end

else

y = 5*x

z = 7*x

end

Chen CL 25


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T4.3-1Enter the script file shown above, whose flowchart is show in Figure

4.3-6. Run the file for the following values ofx. Check the program

results by hand: x= 11, 25, 2, 13.

T4.3-2

Given a number x and the quadrant q(q = 1, 2, 3, 4), write a

program to compute sin1(x) in degrees, taking into account the

quadrant. The program should display an error message if

|x

|>1.

Chen CL 26


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Strings and Conditional Statements

name = Leslie Student

name =

Leslie Student

number = 123

number =123

size(name)

ans =

1 14

name(8)

ans =

S

first_name = name(1:6)

first_name =Leslie

full_name = [name(1:6), C. ,name(7:14)]

full_name =

Leslie C. Studentfull_name(8) = F

full_name =

Leslie F. Student

findstr(full_name,e)

ans =

2 6 15

% letter e occurs in 2nd,

% 6th, and 15th columns

Chen CL 27


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Input Prompts and Output Messages

One of the most important applications for stringsis to create

Input PromptsandOutput Messages

x = input(prompt,string)

response = input(Do you want to continue? Y/N [Y]: ,s);

if ( isempty(response) | (response == Y) | (response == y) )response = Y

else

response = N

end

Chen CL 28


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forLoops

for loop variable = m:s:n

statements

end

for k = 5:10:35x = k^2

end

for x = 0:2:10, y = sqrt(x), end

Chen CL 29


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Nested Loops

function A = specmat(n)A = ones(n);

for r = 1:n

for c = 1:n

if (r>1)&(c>1)

s = A(r-1,c) + A(r,c-1);

if s


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T4.4-1

Write a program to produce the following matrix:

A=

4 8 12

10 14 18

16 20 24

22 26 30

Chen CL 31


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Rules for UsingforLoops m:s:n

The step value smay be negative.For example, k= 10 : 2 : 4 produces k= 10, 8, 6, 4.

Ifs is omitted, the step value defaults to one.

Ifs is positive, the loop will not be executed ifm > n.

Ifs is negative, the loop will not be executed ifm is less than n.

Ifm equals n, the loop will be executed only once.

If the step value s is not an integer, round-off errors can cause the

loop to execute a different number of passes than intended.

Chen CL 32


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breakandcontinueCommands

for k = 1:10

x = 50 - k^2;i f x < 0

break

end

y = sqrt(x)end

% The program execution

% jumps to

% ------HERE------

% if the break command

% is executed

x = [10,1000,-10,100];

y = NaN*x;

for k = 1:length(x)

if x(k) < 0

continue

end

y(k) = log10(x(k));

end

y

y =1 3 NaN 2

Chen CL 33


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Vector as Loop Variables

A = [1,2,3,4,5,6];

f o r v = Adisp(v)

end

12

3

4

5

6

A = [1,2,3,4,5,6];

n = 3 ;for k = 1:n

v = A(:,k)

end

v =1

v =

2

v =

3

Chen CL 34


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Implied Loops

x = [0:5:100];

y = cos(x);

for k = 1:21

x = (k-1)*5;

y(k) = cos(x);end

y = find(x>0)

j=0;

for i=1:length(x)

if x(i)>0j = j + 1 ;

y(j)=i;

end

end

Chen CL 35


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T4.4-1

Write a for loop that is equivalent to the command sum(A), where

A is a matrix.

Chen CL 36


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Example: Data Sorting

A vector x has been obtained from measurements. Suppose we want to consider

any data value in the range0.1< x


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%%%%% Method 1: use a for loop %%%%%

x = [1.92,0.05,-2.43,-0.02,0.09,0.85,-0.06];

y = [];z = [];

for k = 1:length(x)if abs(x(k)) >= 0.1

y = [y,x(k)];

else

z = [z,x(k)];

end

end

xnew = [y, zeros(size(z))]

xnew =

1.9200 -2.4300 0.8500 0 0 0 0

%%%%% Method 2: use find function %%%%%

x = [1.92,0.05,-2.43,-0.02,0.09,0.85,-0.06];

y = x(find(abs(x) >= 0.1));

z = zeros(size(find(abs(x)


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EX: Flight of An Instrumented Rocket

All rockets lose weight as burn fuel; thus the mass of the system is variable. Thefollowing equations describe the speed v and height h of a rocket launched

vertically, neglecting air resistance. They can be derived from Newtons law [Beerand Johnston 1997].

v(t) = u ln

m0m0qt

gt

h(t) = u

q

(m0

qt) l n (m0

qt)

+u ln (m0+ 1) t gt2

2 m0u

q ln(m0)

where m0 is the rockets initialmass, qis the rate as which the rocket burns fuelmass (assumed constant), u is the exhaust velocity of the burned fuel relative to

the rocket (also assumed constant), and g is the acceleration due to gravity. Let bbe the burn time, after which all the fuel is consumed. Thus the rockets masswithoutfuel is me= m0 qb. At the burn time, the rocket height is

h(b) = u

q(me) l n (me) +u ln (m0+ 1) b

gb2

2 m0u

q ln(m0)

Chen CL 39

F b h k i l d h d h d d h i h


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For t > b the rocket engine no longer produces thrust, and the speed and heightare given by

v(t) = v(b) g(t b) (fort > b)h(t) = h(b) +v(b)(t b) g(t b)

2

2 ()

The time tp to reach the peak height is found by setting v(t) = 0. The result is

tp= b+v(b)/g. Substituting this expression into the expression (*) for h(t) gives

the following expression for the peak height: hp=h(b) +v2

(b)/(2g). The time atwhich the rocket hits the ground is thit= tp+

2hp/g.

Suppose the rocket is carrying instruments to study the upper atmosphere, and we

need to determine the amount of time spent above 50, 000 feet as a function of

the burn time b (and thus as a function of the fuel mass qb). Assume that we are

given the following values: me= 100 slugs, q= 1 slug per second, u= 8000 feet

per second, and g= 32.2 feet per second2. If the rockets maximum fuel load is

100 slugs, the maximum value ofb is 100/q= 100 seconds.

Write a MATLAB program to solve this problem.

Chen CL 40

S l i


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Solution

Pseudocode for developing the program appears in the following. A for loop is alogical choice to solve this problem because we know the burn timebandthit, thetime it takes to hit the ground. A MATLAB program to solve this problemappears in the following. It has two nested for loops. The inner loop is over timeand evaluates the equations of motion at times spaced1/10 of a second apart.This loop calculates the duration above 50, 000 feet for a specific value of theburn timeb. We can obtain more accuracy by using a smaller value of the timeincrement dt. The outer loop varies the burn time in integer values from b= 1 to

b= 100. The final result is the vector of durations for the various burn time. Thfollowing Figure gives the resulting plot.

Chen CL 41


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Enter data.

Increment burn time from 0 to 100. For each burn-time value:

Computem0, vb, hb, hp.

Ifhp hdesired,Computetp, thit.

Increment time from 0 to thit.

Compute height as a function of time, using

the appropriate equation, depending on whether

burnout has occurred.

Compute the duration above desired height.

End of the time loop.

Ifhp< hdesired, set duration equal to zero.

End of the burn-time loop.Plot the result.

Chen CL 42% Script file rocket1.m:% Computes flight duration as a function of burn time


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% Computes flight duration as a function of burn time.

% Basic data values.

m_e = 100; q = 1; u = 8000; g = 32.2;

dt = 0.1; h_desired = 50000;

for b = 1:100 % Loop over burn time.

burn_time(b) = b;

% % Following lines implement the formulas in text.m_0 = m_e + q*b;

v_b = u*log(m_0/m_e) - g*b;

% h_b = ((u*m_e)/q)*log(m_e/(m_e+q*b))+u*b-0.5*g*b^2;

h_b = ((u*m_e)/q)*log(m_e) + u*(log(m_0)+1)*b ...

- 0.5*g*b^2 - ((u*m_0)/q)*log(m_0);

h_p = h_b + v_b^2/(2*g);

if h_p >= h_desired;

% % Calculate only if peak height > desire height.t_p = b + v_b/g; %Compute peak time.

t_hit = t_p + sqrt(2*h_p/g); %Compute time to hit.

% for p = 0:t_hit/dt:t_hit

for p = 0:t_hit/dt

% % Use a loop to compute the height vector.

k = p+1; t = p*dt; time(k) = t;

% k = p + 1; t = p*dt; time(k) = t;

i f t < = b% % Burnout has not yet occurred.

h(k) = (u/q)*(m_0 - q*t)*log(m_0 - q*t) ...

+ u*(log(m_0) + 1)*t - 0.5*g*t^2 ...

- (m_0*u/q)*log(m_0);

else

% % Burnout has occurred.

h(k) = h_b + v_b*(t - b) - 0.5*g*(t - b)^2;

end

end

% % Compute the duration.

duration(b) = ...

length(find(h>=h_desired))*dt;else

% % Rocket did not reach desired heig

duration(b) = 0;

end

end % Plot the results.

plot(burn_time,duration) ,...

xlabel(Burn Time (sec)),...

ylabel( Duration (sec)),...title(Duration Above 50,000 Feet)

0 10 20 30 40 50 60 70 80 90 1000

20

40

60

80

100

120

140

160

180

Burn Time (sec)

Duration(sec)

Duration Above 50,000 Feet

Chen CL 43

hil L


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whileLoops

The while loop is used when

the looping process terminates

because a specified condition

is satisfied, and the number of

passes in not known in advance.

while logical expression

statements

end

A practical application ofwhileloop is when we want the loop to

continue as long as a certain statement in true

Chen CL 44

x = 5 ; x=1;


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x = 5 ;

while x < 25

disp(x)

x = 2*x-1;

end

5

9

17

x=1;

while x ~= 5

disp(x)

x = x+1;

end

1

2

3

4

Chen CL 45

E l Ti t R h S ifi d H i ht


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Example: Time to Reach a Specified Height

Consider the variable-mass rocket treated in previous Example.Write a MATLAB program to determinehow long it takes for the

rocket to reach 40, 000feet if the burn time is 50seconds.

Solution: The pseudocode appears in the following Table. Because we do not

know the time required, a while loop is convenient to use. The program performs

the task and is a modification of the previous program. Note that the new

program allows for the possibility that the rocket might not reach 40, 000 feet. It

is important to write your program to handle all such foreseeable circumstances.

The answer given by the program is 53 seconds.

Chen CL 46


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Enter data.

Computem0, vb, hb, hp.

Ifhp hdesired,Use a while loop to increment time and compute height until desired heightis reached.

Compute height as a function of time, using the appropriate equation,

depending on whether burnout has occurred.

End of the results.

Display the results.

Ifhp< hdesired, rocket cannot reach desired height.

Chen CL 47

% Script file rocket2 m % Compute time to reach desired height


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% Script file rocket2.m % Compute time to reach desired height.

% Set the data values.

h_desired = 40000; m_e = 100; q = 1;

u = 8000; g = 32.2; dt = 0.1; b = 50;

% Compute values at burnout, peak time, and height.

m_0 = m_e + q*b; v_b = u*log(m_0/m_e) - g*b;

h_b = ((u*m_e)/q)*log(m_e/(m_e+q*b))+u*b- .5*g*b^2;t_p = b + v_b/g; h_p = h_b + v_b^2/(2*g);

% If h_p > h_desired, compute time to reached h_desired.

if h_p > h_desired

h = 0 ; k = 0 ;

while h < h_desired % Compute h until h = h_desired.

t = k*dt; k = k + 1;

i f t < = b

h = (u/q)*(m_0 - q*t)*log(m_0 - q*t) ... % Burnout has not yet occurred.+ u*(log(m_0) + 1)*t - 0.5*g*t^2 ...

- (m_0*u/q)*log(m_0);

else

h = h_b - 0.5*g*(t - b)^2 + v_b*(t - b); % Burnout has occurred.

end

end

% Display the results.

disp(The time to reach the desired height is:)

disp(t)

else

disp(Rocket cannot achieve the desired height.)

end

The time to reach the desired height is:

53

Chen CL 48

Test Yo U de sta di


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T4.4-3

Rewrite the following code using a whileloop to avoid using thebreakcommand.

for k = 1:10

x = 50 - k^2;

i f x < 0

break

end

y = sqrt(x)

end

T4.4-4

Find to two decimal places the largest value ofx before the error in

the series approximation ex

1 +x+x2/2 +x3/6 exceeds 1

percent. (answer: x= 0.83)

Chen CL 49

The switch Structure


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TheswitchStructureTheswitchstructure provides an alternative to using theif,elseif,

andelsecommands.

Anything programmed usingswitchcan also be programmed usingifstructures. However, theswitchcan provide more readable codes.

switch input expression

(scalar or string)

case value1statement group 1

case value2

statement group 2

...

...

otherwise

statement group n

end

switch angle

case 45

disp(Northeast)

case 135

disp(Southeast)

case 225

disp(Southwest)

case 315disp(Northwest)

otherwise

disp(Direction Unknow

end

Chen CL 50


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Use a string variable for inputexpression to result in veryreadable programs

t = [0:100]; x = exp(-t).*sin(t);

response = ...

input(Type min, max, or sum.,s)

response = lower(response);switch response

case min

minimum = min(x)

case max

maximum = max(x)

case sumtotal = sum(x)

otherwise

disp(You have not entered a proper choice.)

end

Theswitchstatement can handlemultiple conditions in a singlecase statement by enclosing thecasevalue in a cell array

switch angle

case {0,360}

disp(North)

case {-180,180}disp(South)

case {-270,90}

disp(East)

case {-90,270}

disp(West)

otherwisedisp(Direction Unknown)

end

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T4.5-1

Write a program using the switch structure to input one angle,whose value may be 45,45, 135degrees, and display the quadrant(1, 2, 3, or4) containing the angle.

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Debugging MATLAB Programs


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DebuggingMATLABPrograms

function y = fun1(x)

avg = sum(x)/length(x);

y = fun2(avg,x);

function above = fun2(x,avg)

above = length(find(x>avg));

above = fun1([1,2,3,4,10])

above =

3

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Debugging A Loop


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Debugging A Loop

function z = invest(x,r)

z = 0 ;

y = 1+0.01*r;

for k = 1:length(y)

z = z*y+x(k);

end

>>total = invest([1000,1500,2000],10)

total =

1000

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Example: A College Enrollment Model (I)


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Example: A College Enrollment Model (I)

As a example of how simulation can be used for operations research, consider thefollowing college enrollment model. A certain college wants to analyze the effect

of admissions and freshman retention rate on the colleges enrollment so that it

can predict the future need for instructors and other resources. Assume that the

college has estimates of the percentages of students repeating a grade or leaving

school before graduating. Develop a matrix equation on which base a simulation

model that can help in this analysis.

Solution:Suppose that the current freshman enrollment is 500 students and the collegedecides to admit 1000 freshman per year from now on. The college estimates that10 percent of the freshman class will repeat the year. The number of freshman inthe following year will be 0.1(500) + 1000 = 1050, then it will be0.1(1050) + 1000 = 1105, and so on. Let x1(k) be the number of freshman in

year k, where k= 1, 2, 3, 4, 5, 6, . . .. Then in year k+ 1, the number of freshman

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is given by


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is given by

x1(k+ 1) = 10 percent of previous freshman class repeating freshman year +1000new freshman

= 0.1x1(k) + 1000

Because we know the number of freshman in the first year of our analysis (whichis 500), we can solve this equation step to predict the number of freshman in thefuture.Let x2(k) be the number of sophomores in year k. Suppose that 15 percent of the

freshman do not return and that 10 percent repeat freshman year. Thus 75percent of the freshman class returns as sophomore. Suppose also 5 percent of thesophomores repeat the sophomore year and that200 sophomores each year transferfrom other schools. Then in year k+ 1, the number of sophomores is given by

x2(k+ 1) = 0.75x1(k) + 0.05x2(k) + 200

To solve this equation we need to solve the freshman equation at the sametime, which is easy to do with MATLAB. Before we solve these equations, let usdevelop the rest of the model.Let x3(k) andx4(k) be the number of juniors and seniors in year k. Suppose that5 percent of the sophomores and junior leave school and that 5 percent of the

sophomores, juniors, and seniors repeat the grade. Thus 90 percent of the

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sophomores and juniors return and advance in grade. The models for juniors and


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sophomores and juniors return and advance in grade. The models for juniors andseniors are

x3(k+ 1) = 0.9x2(k) + 0.05x3(k)

x4(k+ 1) = 0.9x3(k) + 0.05x4(k)

These four equations can be written in the following matrix form:

x1(k+ 1)

x2(k+ 1)

x3(k+ 1)

x4(k+ 1)

=

0.1 0 0 0

0.75 0.05 0 0

0 0.9 0.05 0

0 0 0.9 0.05

x1(k)

x2(k)

x3(k)

x4(k)

+

1000

200

0

0

In the following example, we will see how to use MATLAB to solve such

equations.

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T4.7-1

Suppose that 70 percent of the freshman, instead of75 percent,return for the sophomore year. How does the previous equation

change ?

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Example: A College Enrollment Model (II)


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Example: A College Enrollment Model (II)

To study the effects of admissions and transfer policies, generalize the enrollmentmodel in previous example to allow for varying admissions and transfers.Solution: Let a(k) be the number of new freshman admitted in the spring ofyeark for the following year k+ 1and let d(k) be the number of transfers into thefollowing years sophomore class. Then the model becomes

x1(k+ 1) = c11x1(k) +a(k)

x2(k+ 1) = c21x1(k) +c22x2(k) +d(k)

x3(k+ 1) = c32x2(k) +c33x3(k)

x4(k+ 1) = c43x3(k) +c44x4(k)

where we have written the coefficients c21, c22, and so on in symbolic, rather thannumerical, form so that we can change their values if desired.This model can be represented graphically by a state transition diagram, like the

one shown in the following Figure.

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Such diagrams are widely used to represent time-dependent and probabilisticprocesses. The arrows indicate how the models calculations are updated for eachvalues ofx1(k), x2(k), x3(k), andx4(k); that is, by the vector x(k), which is

called the state vector. The elements of the state vector are the state variablesdepend on both the previous values and the input a(k) andd(k).The four equations can be written in the following matrix form:

x1(k+ 1)

x2(k+ 1)x3(k+ 1)

x4(k+ 1)

=

c11 0 0 0

c21 c22 0 00 c32 c33 0

0 0 c43 c44

x1(k)

x2(k)x3(k)

x4(k)

+

a(k)

d(k)0

0

or more compactly as

x(k+ 1) =Cx(k) +b(k)

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where


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x(k) =

x1(k+ 1)

x2(k+ 1)

x3(k+ 1)

x4(k+ 1)

b(k) =

a(k)

d(k)

0

0

and

C=

c11 0 0 0

c21 c22 0 0

0 c32 c33 0

0 0 c43 c44

Suppose that the initial total enrollment of1480 consists of500 freshman, 400sophomores,300 juniors, and 280 seniors. The college wants to study, over a10-year period, the effects of increasing admissions by 100 each year and transfersby50 each year until the total enrollment reaches 4000; then admissions and

transfers will be held constant. Thus the admissions and transfers for the next 10years are given by

a(k) = 900 + 100k

d(k) = 150 + 50k

fork= 1, 2, 3, . . . until the colleges total enrollment reaches 4000; then

admissions and transfers are held constant at the previous years levels. We cannot

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determine when this event will occur without doing a simulation. The following


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g gTable gives the pseudocode for solving this problem.

Enter the coefficient matrix C and the initial enrollment vector x.

Enter the initial admissions and transfers, a(1) andd(1).

Set the first column of the enrollment matrix E equal tox.

Loop over years 2 to10.

If the total enrollment is 4000, increase admissions by 100 and transfersby50 each year.If the total enrollment is >4000, hold admissions and transfers constant.

Update the vector x, usingx = Cx + b.

Update the enrollment matrix E by adding another column composed ofx.

End of the loop over years 2 to10.

Plot the results.

Because we know the length of the study (10 years), a for loop is a natural choice.

We use an if statement to determine when to switch from the increasing

admissions and transfer schedule to the constant schedule. A MATLAB script file

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to predict the enrollment for the next 10 years appears in the following. The


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Figure shows the resulting plot. Note that after year 4 there are more sophomores

than freshmen. The reason is that increasing transfer rate eventually overcomes

the effect of the increasing admission rate.

% Script file enroll1.m Computes college enrollment.

% Models coefficients.

C = [.1, 0, 0, 0; .75, .05, 0, 0; 0, .9, .05, 0; 0, 0, .9, .05];

% Initial enrollment vector.

x = [500; 400; 300; 280];

% Initial admissions and transfers.

a(1) = 1000; d(1) = 200;

% E is the 4 x 10 enrollment matrix.

E(:,1) = x;

% Loop over years 2 to 10.

for k = 2:10

% The following describes the admissions

% and transfer policies

if sum(x)


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x = C*x+b;

E(:,k) = x;

end

% Plot the results. (Demo in the class)

plot(E), hold,...

plot(E(1,:),o), plot(E(2,:),+), ...

plot(E(3,:),*), plot(E(4,:),x), ...

xlabel(Year), ylabel(Number of Students),...

gtext(Frosh),gtext(Soph),gtext(Jr),gtext(Sr),...

title(Enrollments as a Function of Time)

In actual practice this program would be run many times to analyze the effects of

different admissions and transfer policies and to examine what happens if different

values are used for the coefficients in the matrix C (indicating different dropout

and repeat rates).

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T4.7-1

In the previous program, line 16 and 17 compute the values ofa(k)and d(k). These lines are repeated here:

a(k) = 900 + 100 kd(k) = 150 + 50 k;

Why does the program contain the line a(1) = 1000; d(1) = 200 ?

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Thank You for Your AttentionQuestions Are Welcome

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