Eee 241 Lecture 06

5/29/2011

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Lecture 6

Summaryand Solved Problemsfor the first mid-term exam

EEE 241 Computer Programminghttp://eee241.gantep.edu.tr/

summary and solved problems for MT1 2

This document summarizes the topics that youneed to know for the first mid-term exam.

They include:

• C++ Structure• Basic I/O• Data types• Operators• Intrinsic functions• Strings• Control structures:

Selection and Loops

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C++ Structure


The main function.Input – Calculate – Output

#include <iostream>

#include <cmath>

using namespace std;

int main() {

double circleArea;

cin >> circleArea;

double r = sqrt(circleArea/M_PI);

cout << "radius = " << r << endl;

return 0;

}

Include the headers forthe services you need.

So far in this course we have used:

iostream

and cmath

optional

Basic I/O


Example run:Input two reals and two integers:

a = 34.5, b = 6.234, c = 78, d = 36

#include <iostream>


int main() {

cout << "Input two reals and two integers: ";

double a, b;

int c, d;

cin >> a >> b >> c >> d;

cout << " a = " << a << ", b = " << b

<< ", c = " << c << ", d = " << d << endl;

return 0;

}

4. Output the results

3. Input the values and assign

2. Declare the variables

1. Request input from the user

34.5 6.234 78 36

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The are many data types and qualifiers in C++. You can solve most problems using only two: double and int.

int

Use this data type for representing integer numbers

Remember the range is limited to:

-2,147,483,648 (232) to 2,147,483,647 (232-1)

Examples:

int k, i=174;

k = -456;

Data Types


Note that variables can beinitialised at declaration.

double

Use this data type for representing floating point numbers.

Remember that the precision is limited to about 15 significant figures,and the range is limited as follows:

negative tiny positivehuge huge

-10+308 -10-324 10-324 +10+308

Examples:

double Weight;

double electronMass = 9.10938188e-31;


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There are many operators in C++ but all you need are:

Important!be careful with integer division and assignments:

13/5 results in 2 (the division of two integers results in an integer)but 13.0/5 results in 2.6 because integer 5 is automatically cast to 5.0but int n=2.6 results in 2 being assigned to n because n is type integer!

Operators


Operator Description Example Result

+ Addition 13 + 5 18

- Subtraction 13 – 5 8

* Multiplication 13 * 5 65

/ Division 13.0/5

13 / 5

2.6

2

%

(integers only)Modulo a % b

13 % 5

(remainder of a/b)3

++ Increment k++ add 1 to k

Sometimes you need to cast:

double y;

int k=8, n=5;

y=k/double(n);

There are many intrinsic functions in C++, a common group being math functions provided by the cmath header:

#include <cmath>

With this inclusion you can now use the following functions:

cos(), sin(), exp(), sqrt(),log(), log10(), pow(), .....

full list at

http://www.cplusplus.com/reference/clibrary/cmath/

For other C++ library groups see:

http://www.cplusplus.com/reference/clibrary/

Intrinsic functions


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Example:

Intrinsic functions


// Velocity of a body after falling s metres

#include <iostream>

#include <cmath>


int main() {

double g = 9.81; // m/s^2

double s = 100.; // m

double v = sqrt(2*g*s);

cout << v << " m/s" << endl;

return 0;

}

44.2945 m/sOutput

Example

Strings


#include <iostream>


int main () {

string s1 = "one_", s2, s3;

s2 = "two";

s3 = s1 + s2;

cout << s3 << endl;

return 0;

}

one_twoOutput

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Three basic selection control structures:

Control: Selection


if (condition) {

statements

.

.

}

if (condition) {

statements1

.

.

} else {

statements2

.

.

}

if (condition1) {

statements1

.

.

} else if (condition2) {

statements2

.

.

} else {

statements3

.

.

}

If a condition is truethen the block, defined by the braces {...}, is executed.

Remember that variables defined inside the block are local to that block. i.e they are not visible outside of the defining block.


if ( x < y ) { ... } if x is less than yif ( x <= y ) { ... } if x is less than or equal to yif ( x > y ) { ... } if x is greater than yif ( x >= y ) { ... } if x is greater than or equal to yif ( x == y ) { ... } if x is equal to yif ( x != y ) { ... } if x is not equal to y

if ( x > 2 && y == 3 ) { ... } if x is greater than 2and y is equal to 3

if ( x > 2 || y == 3 ) { ... } if x is greater than 2 or y is equal to 3if ( !(x < y) ) { ... } if x is not less than yif ( x ) { ... } this is false if x==0, otherwise it is true

Relational and logical operators

Not that the operator == makes a comparison while the operator = makes an assignment. Common mistake:

if ( x = y ) { ... DON’T DO THIS!

The value of y is assigned to x, and the relation is always true!

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Example (also exemplifies a “flow chart”)

Note that variable c needs to be declaredoutside the if block so that it is visible atthe output statement (see scoping).

start

Input a

b = a + 2

a > 2 c = ln(a+b)

Output a, c

TF

end

c = b5

#include <iostream>

#include <cmath>


int main() {

double a, b, c;

cin >> a;

b = a + 2.0;

if ( a > 2.0 ) {

c = log(a+b);

} else {

c = pow(b,5.0);

}

cout << a << " "

<< c << endl;

return 0;

}

Control: Loops


Three basic loop (repetitive) structures:

while (condition) {

statements

.

.

}

do {

statements

.

.

} while (condition);

for (initialisation; condition; increment) {

statements

.

.

}

Also, the break statement allows you to break out of a loop.The continue statement executes the next turn in the loop.

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Example (also exemplifies “pseudo code”)These are all equivalent counted loops

#include <iostream>


int main() {

for (int i=1; i<7; i++) {

double x = 0.1*i;

cout <<i<<" "<<x<< endl;

}

}

Pseudo code:

loop i = 1 to 6

x = 0.1i

output i, x

end loop#include <iostream>


int main() {

int i=1;

while (i<7) {

double x = 0.1*i;

cout <<i<<" "<<x<<endl;

i++;

}

}

#include <iostream>


int main() {

int i=1;

do {

double x = 0.1*i;


i++;

} while (i<7);

}

#include <iostream>


int main() {

int i=1;

while (1) {

double x = 0.1*i;


if (i==6) break;

i++;

}

}

1 0.1

2 0.2

3 0.3

4 0.4

5 0.5

6 0.6

That’s all the C++ syntax you needfor the first mid-term exam!

After the break we willlook at some solved problems ......

+

16summary and solved problems for MT1

Don’t forget to bring your

calculator to the exam!

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Solved Problems


This section presents problems and their solutions.

Problems:

1. Exponential growth of bacteria (revisited)

2. Heat flow in foods


Exponential growth of bacteria

Recall that in Lecture 2 we wrote a program to calculate the number nof bacteria after filtration and growth periods t1 and t2.

1. Exponential growthof bacteria (revisited)

input , Q, t1, t2

m = Q t1 n = m e t2

outputm , n

begin

end

= 0.468

Given a 24-hour period (i.e. t1+t2 = 24) wewant to optimize t1 and t2 to maximise the final number of bacteria n.

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#include <iostream>

#include <cmath>


int main () {

double b = 0.468; // growth constant [1/hr]

double a, Q, t1, t2; // inputs

cout << "Input the volume number of bacteria [1/L]: ";

cin >> a;

cout << " Input the volume flow rate [L/hr]: ";

cin >> Q;

cout << " Input is the filtration duration [hr]: ";

cin >> t1;

cout << " Input the growth duration [hr]: ";

cin >> t2;

double m = a*Q*t1; // Number of filtered bacteria

double n = m*exp(b*t2); // Number of bacteria after growth

cout << " Number filtered bacteria = " << m << endl;

cout << "Number of bacteria after growth = " << n << endl;

}

Exponential growth of bacteriaThe program developed in Lecture 2


Input the volume number of bacteria [1/L]: 0.25

Input the volume flow rate [L/hr]: 36

Input is the filtration duration [hr]: 5

Input the growth duration [hr]: 19

Number filtered bacteria = 45

Number of bacteria after growth = 327310

Example outputs of the program developed in Lecture 2

Input the volume number of bacteria [1/L]: 0.25

Input the volume flow rate [L/hr]: 36

Input is the filtration duration [hr]: 4

Input the growth duration [hr]: 20

Number filtered bacteria = 36

Number of bacteria after growth = 418118

24 hrs

24 hrs

Let’s rewrite the program to use a loop to search values of t1 to find the maximum value of n. We can make the values of and Q constant so

that the only variable is t1 with t2 = 24 - t1.



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#include <iostream>

#include <cmath>


int main () {

double a = 0.25; // volume number of bacteria [1/L]

double Q = 36.0; // volume flow rate [L/hr]

double b = 0.468; // growth constant [1/hr]

for (int i=1; i<=20; i++) { // loop over twenty periods

double t1 = i*0.2; // Filtration period (0.2-hour steps)

double t2 = 24-t1; // Growth period

double m = a*Q*t1; // Number of filtered bacteria

double n = m*exp(b*t2); // Number of bacteria after growth

cout << fixed << "t1=" << t1 << " n=" << n << endl;

} // end loop

return 0;

}


The modified program - to search for a maximum.


t1=0.200000 n=123770.789824

t1=0.400000 n=225422.984054

t1=0.600000 n=307921.128718

t1=0.800000 n=373876.581528

t1=1.000000 n=425586.959955

t1=1.200000 n=465071.359521

t1=1.400000 n=494101.783271

t1=1.600000 n=514231.178315

t1=1.800000 n=526818.434839

t1=2.000000 n=533050.666589

t1=2.200000 n=533963.059029

t1=2.400000 n=530456.541976

t1=2.600000 n=523313.517015

t1=2.800000 n=513211.846226

t1=3.000000 n=500737.287359

t1=3.200000 n=486394.541410

t1=3.400000 n=470617.061292

t1=3.600000 n=453775.754781

t1=3.800000 n=436186.701040

t1=4.000000 n=418117.987514

Exponential growth of bacteriaSearch in steps of 0.2 hours

t1=1.800000 n=526818.434839

t1=1.850000 n=528929.386168

t1=1.900000 n=530660.886035

t1=1.950000 n=532029.357415

t1=2.000000 n=533050.666589

t1=2.050000 n=533740.140107

t1=2.100000 n=534112.581268

t1=2.150000 n=534182.286124

t1=2.200000 n=533963.059029

t1=2.250000 n=533468.227729

t1=2.300000 n=532710.658031

t1=2.350000 n=531702.768037

t1=2.400000 n=530456.541976

t1=2.450000 n=528983.543623

t1=2.500000 n=527294.929345

We see the maximum occurs at

t1 = 2.15 hours (t2 = to 21.85 hours)

Any value 1.9 < t1 < 2.4 is good.

Refined the search (0.05 hours):


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Graphical view of the search for a maximum

Exponential growth of bacteriaDiscussion

Our numerical solution is t1 = 2.15 0.05 hours

and any value in the range 1.9 < t1 < 2.4 hours is good.

The analytical solution looks like this:

m = Q t1 n = m e t2 = Q t1 e t2 = Q t1 e (24- t1)

For a maximum dn/dt1 = 0So d/dt1 [ Q t1 e (24- t1) ] = 0 So d/dt1 [ Q t1 e 24 e -t1 ] = 0 Q e 24 e - t1 - Q t1 e 24 e - t1 = 0 Q e 24 e - t1 (1- t1) = 0(1- t1) = 0 => t1 = 1/ = 1/ 0.468 = 2.14 hours

That was not too difficult; but often an analytical solution is difficult

or impossible to obtain and so we use a numerical method

requires Computer Programming!

Also our numerical method shows us that a wide range of values are

close to optimal, i.e. we gained some extra information by using a

numerical method. 24summary and solved problems for MT1

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2. Heat flow in foods


The Problem

A solid food has a temperature

difference T between two sides.Heat flows via conduction at arate Q through a surface area Aaccording to the equation:

Q = k A T / L

where k is the thermal conductivity given by:

k = 0.25 mc + 0.155 mp + 0.160 mf + 0.135 ma + 0.580 mm

where m are the food mass fractions (mc + mp + mf + ma + mm = 1) c = “carbohydrate”, p = “protein”, f = “fat”, a = “ash”, and m = “moisture”.

See “Fourier's Law of Conduction”

A [m2]

L [m]T [oC]

Q [W]

Heat flow in foods


The Task

Write a program that inputs the required data andoutputs the rate of heat flow through the food.

Analysis

We want Q = k A T / L

[ 0.25 mc + 0.155 mp + 0.160 mf + 0.135 ma + 0.580 mm ]

condition that food mass fractions mc + mp + mf + ma + mm = 1

input: A, T, L

input: mc , mp , mf , ma , mm

Heat flow in foods

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Flow chart

input:A, T, L

input:

mc , mp , mf ,

ma , mm

output: Q

mc+mp+mf +ma+mm = 1 ?

start

false

k = 0.25 mc + 0.155 mp + 0.160 mf

+ 0.135 ma + 0.580 mm

true

Q = k A T / L

end

Heat flow in foods


Algorithm

(Follow the flow chart, but add some instructions to the user)

1. output “enter the mass fractions mc , mp , mf , ma , mm”

2. input mc , mp , mf , ma , mm

3. if ( mc+mp+mf + ma+mm ≠ 1 )output “Mass fraction sum should equal 1!”go back to step 1

4. k = 0.25 mc + 0.155 mp + 0.160 mf + 0.135 ma + 0.580 mm

5. input A, T, L

6. Q = k A T / L

7. output Q

Heat flow in foods

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Implementation in C++ (Follow the algorithm but adding more instructions)

#include <iostream>


int main() {

double Mc, Mp, Mf, Ma, Mm;

bool badSum;

do {

cout << "Input the Carbohydrate mass fraction: "; cin >> Mc;

cout << "Input the Protein mass fraction: "; cin >> Mp;

cout << "Input the Fat mass fraction: "; cin >> Mf;

cout << "Input the Ash mass fraction: "; cin >> Ma;

cout << "Input the Moisture mass fraction: "; cin >> Mm;

double sum = Mc + Mp + Mf + Ma + Mm;

badSum = ( sum < 0.999 || sum > 1.001 ); // allow 0.1% error

if ( badSum ) cout << "Mass fraction sum (" << sum

<< ") should equal 1!" << endl;

} while ( badSum );

Heat flow in foods


.... continued from previous page

we have at this point values for Mc, Mp, Mf, Ma, Mm ....

// compute the thermal conductivity

double k = 0.250*Mc + 0.155*Mp + 0.160*Mf + 0.135*Ma + 0.580*Mm;

// Input the area, temperature difference and length

double A, dT, L;

cout << "Input the conduction surface area [m^2]: "; cin >> A;

cout << "Input the temperature difference [C]: "; cin >> dT;

cout << "Input the conduction length [m]: "; cin >> L;

// calculate and output the rate of heat flow

double Q = k * A * dT / L;

// Output the result

cout << "Rate of heat flow = " << Q << " Watts" << endl;

return 0;

}

Heat flow in foods

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Input the Carbohydrate mass fraction: 0.15

Input the Protein mass fraction: 0.18

Input the Fat mass fraction: 0.23

Input the Ash mass fraction: 0.02

Input the Moisture mass fraction: 0.52

Mass fraction sum (1.1) should equal 1!

Input the Carbohydrate mass fraction: 0.15

Input the Protein mass fraction: 0.08

Input the Fat mass fraction: 0.23

Input the Ash mass fraction: 0.02

Input the Moisture mass fraction: 0.52

Input the conduction surface area [m^2]: 0.2

Input the temperature difference [C]: 25.

Input the conduction length [m]: 0.05

Rate of heat flow = 39.1 Watts

Example run:

Heat flow in foods

In this lecture we summarised the C++ syntax and structure you need for the first mid-term exam.

You can find the lecture slides in the course website:

http://eee241.gantep.edu.tr/

For further reading, search for the relevant sections at:

http://cpp.gantep.edu.tr/

andhttp://www.learncpp.com/

Finally


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http://www1.gantep.edu.tr/~eee241/labs.php

This week’s lab exercise sheet contains some

example first mid-term exam questions.

First try solving the questions by hand,

then program your solutions in the lab hour.

THE FIRST MID-TERM EXAM IS NEAR

DON’T FORGET TO BRING A CALCULATOR

TO THE EXAM!

Exercises for Lab 6

#include <iostream>

#include <cmath>

#include <cstdlib>


int main() {

double Area;

cin >> Area;

double r = sqrt(Area/M_PI);

int n, m;

cin >> n >> m;

int k = pow(n,m);

string name;

cin >> name;

string s = "Hello " + name;

bool p=true;

bool q = (n>m);

cout << r << " " << k

<< " " << s << " " << p

<< " " << q << endl;

return 0;

}

if (condition) {

statements

}

if (condition) {

statements1

} else {

statements2

}

if (condition1) {

statements1


statements2


statements3

} else {

statements4

}

for (int i=1; i<7; i++) {

statements

}

int i=1;

while (i<7) {

statements

i++;

}

int i=1;

do {

statements

i++;

} while (i<7);

int i=1;

while (true) {

statements

if (i==6) break;

i++;

}

Loop structures [i = 1 to 6]

if ( x < y ) ...

if ( x <= y ) ...

if ( x > y ) ...

if ( x >= y ) ...

if ( x == y ) ...

if ( x != y ) ...

if ( !(x < y) ) ...

if ( x ) ...

if ( x > 2 && y == 3 ) ...

if ( x > 2 || y == 3 ) ...

sin() cos() tan()

asin() acos() atan()

fabs() sqrt() pow()

exp() log() log10()

int() double() rand()

Basic structure, headers, data types, I/O

Selection structures

logical operators

Intrinsic functions

Also the continue statement

EEE241 MT1 reference sheet 27/02/11

This information will be provided in the exam.

Return only your answer sheet to the invigilator.

Cheating is a serious offence, do not pass /

receive any information to/from another student.

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Eee 241 Lecture 06

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