Chapter 6 Solved Problems
1. Evaluate the following expressions without using MATLAB. Check the answers with MATLAB. (a)(b) (c)(d)
Solution
>> % Part (a) >> 6*4>32-3 ans = 0 >> % Part (b) >> y=4*3-7<15/3>-1 y = 1 >> % Part (c) >> y=2*(3<8/4+2)^2<(-2) y = 0 >> % Part (d) >> (5+~0)/3==3-~(10/5-2) ans = 1 >>
1 2 Chapter 6: Solved Problems
2. Given: , , . Evaluate the following expressions without using MATLAB. Check the answers with MATLAB. (a)(b) (c)(d)
Solution
>> d=6 d = 6 >> e=4 e = 4 >> f=-2 f = -2 >> % Part (a) >> y=d+f>=e>d-e y = 0 >> % Part (b) >> y=e>d>f y = 1 >> % Part (c) >> y=e-d<=d-e==f/f y = 1 >> % Part (d) >> y=(d/e*f
3. Given: v = [–2 4 1 0 2 1 2 ] and w = [2 5 0 1 2 –1 3 ]. Evaluate the follow- ing expressions without using MATLAB. Check the answers with MAT- LAB. (a)~v ==~w (b) w > = v (c) v > ~ –1*w (d) v > –1*w
Solution
>> v=[-2 4 1 0 2 1 2] v = -2 4 1 0 2 1 2 >> w=[2 5 0 1 2 -1 3] w = 2 5 0 1 2 -1 3 >> % Part (a) >> ~v==~w ans = 1 1 0 0 1 1 1 >> % Part (b) >> w>=v ans = 1 1 0 1 1 0 1 >> % Part (c) >> v>~1*w ans = 0 1 1 0 1 1 1 >> % Part (d) >> v>-1*w ans = 0 1 1 1 1 0 1 >> 4 Chapter 6: Solved Problems
4. Use the vectors v and w from Problem 3. Use relational operators to create a vector u that is made up of the elements of v that are smaller than or equal to the elements of w.
Solution
Command Window:
>> v=[-2 4 1 0 2 1 2]; >> w=[2 5 0 1 2 -1 3]; >> u=v(v<=w) u = -2 4 0 2 2
5. Evaluate the following expressions without using MATLAB. Check the answers with MATLAB. (a)0|7&9&–3 (b)7>6&~0<=2 (c) ~4<5|0>=12/6 (d) – 7<–5<–2&2+3<=15/3
Solution
>> % Part (a) >> 0|7&9&-3 ans = 1 >> % Part (b) >> 7>6&~0<=2 ans = 1 >> % Part (c) >> ~4<5|0>=12/6 ans = 1 >> % Part (d) >> -7<-5<-2&2+3<=15/3 ans = 0 >> Chapter 6: Solved Problems 5
6. Use loops to create a matrix in which the value of each element is two times its row number minus three times its column number. For example, the value of element (2,5) is .
Solution
Script File
for i=1:4 for j=1:6 A(i,j)=2*i-3*j; end end A
Command Window:
A = -1 -4 -7 -10 -13 -16 1 -2 -5 -8 -11 -14 3 0 -3 -6 -9 -12 5 2 -1 -4 -7 -10
7. Write a program that generates a vector with 30 random integers between –20 and 20 and then finds the sum of the all the elements that are divisible by 3.
Solution
Script File:
v=randi([-20 20],1,30) n=length(v); S=0; for i=1:n if rem(v(i),3)==0 S=S+v(i); end end S 6 Chapter 6: Solved Problems
Command Window:
v = Columns 1 through 12 10 -10 0 8 16 19 2 -15 -14 - 10 14 -10 Columns 13 through 24 13 -11 18 -6 -12 -10 5 -1 -6 14 3 2 Columns 25 through 30 17 -9 11 10 -5 3 S = -24
8. Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for loop the program examine each element of the vec- tor. If the element is positive its value is doubled. If the element is negative its value is tripled. The program displays the vector that was entered and the modified vector. Execute the program and when the program ask the user to input a vector type randi([-10 20],1,19). This creates a 19-element vector with random integers between –10 and 20.
Solution
Script File: v=randi([-10 20],1, 19) n=length(v); for i=1:n if v(i) > 0 v(i)=2*v(i); end if v(i) < 0 v(i)= 3*v(i); end end vNew=v
Command Window:
v = Columns 1 through 15 Chapter 6: Solved Problems 7
-8 -9 6 14 18 -6 7 4 -10 0 -5 14 -1 6 -5 Columns 16 through 19 8 -2 10 11 vNew = Columns 1 through 15 -24 -27 12 28 36 -18 14 8 -30 0 -15 28 -3 12 -15 Columns 16 through 19 16 -6 20 22
9. Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for loop the program eliminates all the negative ele- ments. The program displays the vector that was entered and the modified vector, and a message that says how many elements were eliminated. Exe- cute the program and when the program ask the user to input a vector type randi([-15 20],1,25). This creates a 25-element vector with random integers between –15 and 20.
Solution
Script File:
v=randi([-15 20],1, 25) n=length(v); j=0; for i=1:n if v(i) >= 0 j=j+1; vNew(j)=v(i); end end elim=n-j; vNew fprintf('%.0f elements were eliminated.\n',elim)
Command Window:
v = Columns 1 through 13 8 Chapter 6: Solved Problems
17 6 7 15 13 5 -9 -7 16 - 14 2 -9 20 Columns 14 through 25 10 3 1 -13 9 -14 -13 3 -12 14 14 11 vNew = Columns 1 through 13 17 6 7 15 13 5 16 2 20 10 3 1 9 Columns 14 through 17 3 14 14 11 8 elements were eliminated.
10. The daily high temperature (°F) in New York City and Denver, Colorado during the month of January 2014 is given in the vectors below (data from the U.S. National Oceanic and Atmospheric Administration). NYC = [33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39] DEN = [39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28] where the elements in the vectors are in the order of the days in the month. Write a program in a script file that determines and displays the following information: (a) The average temperature for the month in each city (rounded to the nearest degree). (b) The number of days that the temperature was above the average in each city. (c) The number of days that the temperature in Denver was higher than the temperature in New York.
Solution
Script File:
clear, clc NYC = [33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 ... 39 36 45 33 18 19 19 28 34 44 21 23 30 39]; DEN = [39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 ... 52 62 45 62 40 25 57 60 57 20 32 50 48 Chapter 6: Solved Problems 9
28]; %Part (a) NYCav=round(mean(NYC)); DENav=round(mean(DEN)); fprintf('The averager temperature in New York City is: % g F.\n',NYCav) fprintf('The averager temperature in Denver is: % g F.\n',DENav) %Part (b) nNYC=sum(NYC>NYCav); nDEN=sum(DEN>DENav); fprintf('During % g days the temperature in New York City was above the average.\n',nNYC) fprintf('During % g days the temperature in Denver was above the average.\n',nDEN) %Part (c) DENhNYC=sum(DEN>NYC); fprintf('During % g days the temperature in Denver was higher than in New York City.\n',DENhNYC)
Command Window:
The averager temperature in New York City is: 35 F. The averager temperature in Denver is: 44 F. During 15 days the temperature in New York City was above the average. During 18 days the temperature in Denver was above the average. During 22 days the temperature in Denver was higher than in New York City. 10 Chapter 6: Solved Problems
11. The Pascal’s triangle can be displayed as elements in a 10 0 0 00 lower-triangular matrix as shown on the right. Write a 11 0 0 00 MATLAB program that creates a matrix that dis- plays n rows of Pascal’s triangle. Use the program to create 12 1 0 00 4 and 7 rows Pascal’s triangles. (One way to calculate the 13 3 1 00 value of the elements in the lower portion of the matrix is 14 6 4 10 15101051 .)
Solution
Script File:
n=6; pt=zeros(n); for i=1:n for j=1:i pt(i,j)=factorial(i-1)/(factorial(j- 1)*factorial(i-j)); end end pt
Command Window:
pt = 1 0 0 0 0 0 1 1 0 0 0 0 1 2 1 0 0 0 1 3 3 1 0 0 1 4 6 4 1 0 1 5 10 10 5 1 Chapter 6: Solved Problems 11
12. Tribonacci numbers are the numbers in a sequence in which the first three elements are 0, 1, and 1, and the value of each subsequent element is the sum of the previous three elements: 0, 1, 1, 2, 4, 7, 13, 24, ... Write a MATLAB program in a script file that determines and displays the first 25 Fribonacci numbers.
Solution
Script File:
F(1)=0; F(2)=1; F(3)=1; for i=4:25 F(i)=sum(F([i-3:i-1])); end F
Command Window:
F = Columns 1 through 6 0 1 1 2 4 7 Columns 7 through 12 13 24 44 81 149 274 Columns 13 through 18 504 927 1705 3136 5768 10609 Columns 19 through 24 19513 35890 66012 121415 223317 410744 Column 25 755476 12 Chapter 6: Solved Problems
13. The reciprocal Fibonacci constant ψ is defined by the infinite sum: ∞ ψ = n = 1 where Fn are the Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, ... . Each element in this sequence of numbers is the sum of the previous two. Start by setting the first two elements equal to 1, then . Write a MATLAB pro- gram in a script file that calculates ψ for a given n. Execute the program for n = 10,, 50 and 100.
Solution
Script File: n=50; F(1)=1; F(2)=1; for i=3:n F(i)=F(i-2)+F(i-1); end xi=sum(1./F)
Command Window:
xi = 3.359885666243178
14. The value of π can be estimated from: ∞ n = 0 Write a program (using a loop) that determines π for a given n. Run the pro- gram with n = 10, n = 100, and n = 1000. Compare the result with pi. (Use format long.)
Solution
Script File: format long n=input('Enter a value for n '); S=0; for i=0:n S=S+(-1)^i/(2*i+1)^3; end Chapter 6: Solved Problems 13
ESpi=(32*S)^(1/3)
Command Window:
Enter a value for n 10 ESpi = 3.141642788603785 >> ed6_HW6_14 Enter a value for n 100 ESpi = 3.141592719141050 >> ed6_HW6_14 Enter a value for n 1000 ESpi = 3.141592653657139
15. The value of π can be estimated from the expression:
Write a MATLAB program in a script file that determine π for any number of terms. The program asks the user to enter the number of terms, and then calculates the corresponding value of π . Execute the program with 5, 10, and 40 terms. Compare the result with pi. (Use format long.)
Solution
Script File:
format long n=5; S2=0; S=1; for i=1:n S2=sqrt(2+S2); S=S*S2/2; end piEst=2/S
Command Window: 14 Chapter 6: Solved Problems
piEst = 3.140331156954753
With n=10:
piEst = 3.141591421511200
With n=40:
piEst = 3.141592653589794
16. Write a program that (a) generates a vector with 20 random integer elements with integers between 10 and 30, (b) replaces all the elements that are not even integers with random integers between 10 and 30, (c) repeats (b) until all the elements are even integers. The program should also count how many times (b) is repeated before all the elements are even integers. When done, the program displays the vector and a statement that states how many itera- tions were needed for generating the vector.
Solution
Script File: vInitial=randi([10 30],1, 20) for i=1:30 c=0; for j=1:20 if rem(vInitial(j),2) ~=0 vInitial(j)=randi([10 30]); c=1; end end if c ==0 break end end vFinal=vInitial fprintf('%.0f iterations were done.\n',i-1)
Command Window: Chapter 6: Solved Problems 15
vInitial = 27 12 22 17 26 20 20 28 17 19 30 10 30 13 24 22 24 17 23 27 vFinal = 10 12 22 28 26 20 20 28 30 16 30 10 30 14 24 22 24 18 12 12 4 iterations were done.
17. A vector is given by x = [9 –1.5 13.4 13.3 –2.1 4.6 1.1 5 –6.1 10 0.2]. Using conditional statements and loops, write a program that rearranges the elements of x in order from the smallest to the largest. Do not use MAT- LAB’s built-in function sort.
Solution
Script file:
clear, clc x=[9 -1.5 13.4 13.3 -2.1 4.6 1.1 5 -6.1 10 0.2] n=length(x); for i=1:n-1 for j=i+1:n if x(j) Command Window: x = 9.0000 -1.5000 13.4000 13.3000 -2.1000 4.6000 1.1000 5.0000 -6.1000 10.0000 0.2000 x = -6.1000 -2.1000 -1.5000 0.2000 1.1000 16 Chapter 6: Solved Problems 4.6000 5.0000 9.0000 10.0000 13.3000 13.4000 >> 18. The Pythagorean theorem states that a2 + b2 = c2 . Write a MATLAB pro- gram in a script file that finds all the combinations of triples a, b, and c that are positive integers all smaller or equal to 50 that satisfy the Pythagorean theorem. Display the results in a three-column table in which every row cor- responds to one triple. The first three rows of the table are: 3 4 5 5 12 13 6 8 10 Solution Script file: clear, clc id=1; for k=1:50 for j=k+1:50 for i=j+1:50 if i^2==k^2+j^2 a(id)=k; b(id)=j; c(id)=i; id=id+1; end end end end table=[a' b' c'] Command Window: table = 3 4 5 5 12 13 6 8 10 7 24 25 8 15 17 9 12 15 9 40 41 Chapter 6: Solved Problems 17 10 24 26 12 16 20 12 35 37 14 48 50 15 20 25 15 36 39 16 30 34 18 24 30 20 21 29 21 28 35 24 32 40 27 36 45 30 40 50 >> 19. Write a MATLAB program in a script file that finds and displays all the numbers between 100 and 999 whose product of digits is 6 times the sum of the digits. (e.g. 347 since ). Use a for loop in the pro- gram. The loop should start from 100 and end at 999. Solution Script File: for n=100:999 h=fix(n/100); d=fix((n-100*h)/10); s=n-h*100-d*10; mul=h*d*s; add=6*(h+d+s); if mul == add n end end Command Window: n = 18 Chapter 6: Solved Problems 268 n = 286 n = 347 n = 374 n = 437 n = 473 n = 628 n = 682 n = 734 n = 743 n = 826 n = 862 20. A safe prime is a prime number that can be written in the form where p is also a prime number. For example, 47 is a safe prime since and 23 is also a prime number. Write a computer program that finds and displays all the safe primes between 1 and 1000. Do not use MATLAB’s built-in function isprime. Solution Script file: clear, clc k=1; for n=1:500; nispr=1; nsafeisPrime=0; for i=2:fix(n/2) if rem(n,i)==0 nispr=0; Chapter 6: Solved Problems 19 break end end if nispr==1 nsafe=2*n+1; nsafeisPrime=1; for j=2:fix(nsafe/2) if rem(nsafe,j)==0 nsafeisPrime=0; break end end end if nsafeisPrime==1 SafePrimeNum(k)=nsafe; k=k+1; end end SafePrimeNum Command Window: SafePrimeNum = Columns 1 through 7 3 5 7 11 23 47 59 Columns 8 through 14 83 107 167 179 227 263 347 Columns 15 through 21 359 383 467 479 503 563 587 Columns 22 through 26 719 839 863 887 983 20 Chapter 6: Solved Problems 21. Sexy primes are two prime numbers that the difference between them is 6. For example, 23 and 29 are sexy primes since . Write a computer pro- gram that finds all the sexy primes between 1 and 300. The numbers should be displayed in a two-column matrix where each row display one pair. Do not use MATLAB’s built-in function isprime. Solution Script file: clear, clc k=1; for n=1:300; nispr=1; nSexyeisPrime=0; for i=2:fix(n/2) if rem(n,i)==0 nispr=0; break end end if nispr==1 nsexy=n+6; nSexyeisPrime=1; for j=2:fix(nsexy/2) if rem(nsexy,j)==0 nSexyeisPrime=0; break end end end if nSexyeisPrime==1 SexyPrimeNum(k,1)=n; SexyPrimeNum(k,2)=nsexy; k=k+1; end end SexyPrimeNum Command Window: Chapter 6: Solved Problems 21 SexyPrimeNum = 1 7 5 11 7 13 11 17 13 19 17 23 23 29 31 37 37 43 41 47 47 53 53 59 61 67 67 73 73 79 83 89 97 103 101 107 103 109 107 113 131 137 151 157 157 163 167 173 173 179 191 197 193 199 223 229 227 233 233 239 251 257 257 263 263 269 271 277 277 283 >> 22 Chapter 6: Solved Problems 22. A Mersenne prime is a prime number that is equal to , where n is an inte- ger. For example, 31 is a Mersenne prime since . Write a computer program that finds all the Mersenne primes between 1 and 10,000. Do not use MATLAB’s built-in function isprime. Solution Script file: clear, clc k=1; for N=3:10000; nispr=1; for i=2:fix(N/2) if rem(N,i)==0 nispr=0; break end end if nispr==1 n=ceil(log(N)/log(2)); if N == 2^n-1 MersennePrimes(k)=N; k=k+1; end end end MersennePrimes Command Window: MersennePrimes = 3 7 31 127 8191 Chapter 6: Solved Problems 23 23. A perfect number is a positive integer that is equal to the sum of its positive divi- sors except the number itself. The first two perfect numbers are 6 and 28 since and . Write a computer program that finds the first four perfect numbers. Solution Script file: clear, clc j=1; for N=2:10000 clear div div(1)=1; k=2; for i=2:fix(N/2) if rem(N,i)==0 div(k)=i; k=k+1; end end if sum(div)==N PerfectN(j)=N; j=j+1; end end PerfectN Command Window: PerfectN = 6 28 496 8128 >> 24 Chapter 6: Solved Problems 24. A list of exam scores (S) (in percent out of 100%) is given: 72, 81, 44, 68, 90, 53, 80, 75, 74, 65, 50, 92, 85, 69, 41, 73, 70, 86, 61, 65, 79, 94, 69. Write a computer program that calculates the average (Av) and standard deviation (Sd) of the scores, which are rounded to the nearest integer. Then, the program determines the letter grade of each of the scores according to the following scheme: Score (%) Letter grade AB Score (%) Letter grade CD Score (%) Letter grade F The program displays the values of Av and Sd followed by a list that shows the scores and the corresponding letter grade (e.g. 72% Letter grade C). Solution Script file: clear,clc G=[72 81 44 68 90 53 80 75 74 65 50 92 85 69 41 73 70 86 61 65 79 94 69]; Av=round(mean(G)); S=round(std(G)); fprintf('Average grade is %3.0f%%. Standard deviation is %3.0f%%.\n',Av,S) disp(' ') n=length(G); for i=1:n if G(i) > Av+1.3*S fprintf('%3i%% Letter grade A\n',G(i)) elseif G(i) > Av+0.5*S & G(i) < Av+1.3*S fprintf('%3i%% Letter grade B\n',G(i)) elseif G(i) > Av-0.5*S & G(i) < Av+0.5*S fprintf('%3i%% Letter grade C\n',G(i)) elseif G(i) > Av-1.3*S & G(i) < Av-0.5*S fprintf('%3i%% Letter grade D\n',G(i)) elseif G(i) < Av-1.3*S fprintf('%3i%% Letter grade F\n',G(i)) end Chapter 6: Solved Problems 25 end Command Window: Average grade is 71%. Standard deviation is 14%. 72% Letter grade C 81% Letter grade B 44% Letter grade F 68% Letter grade C 90% Letter grade A 53% Letter grade D 80% Letter grade B 75% Letter grade C 74% Letter grade C 65% Letter grade C 50% Letter grade F 92% Letter grade A 85% Letter grade B 69% Letter grade C 41% Letter grade F 73% Letter grade C 70% Letter grade C 86% Letter grade B 61% Letter grade D 65% Letter grade C 79% Letter grade B 94% Letter grade A 69% Letter grade C >> 26 Chapter 6: Solved Problems 25. The Taylor series expansion for ax is: ∞ n = 0 Write a MATLAB program that determines ax using the Taylor series expansion. The program asks the user to type a value for x. Use a loop for adding the terms of the Taylor series. If cn is the nth term in the series, then the sum Sn of the n terms is . In each pass calculate the esti- mated error E given by . Stop adding terms when . The program displays the value of ax. Use the program to calculate: (a) 23.5 (b) 6.31.7 Compare the values with those obtained by using a calculator. Solution Script File: a=input('Enter a value for a '); x=input('Enter a value for x '); S=1; for n=1:30 cn=log(a)^n/factorial(n)*x^n; Sn=S+cn; if abs((Sn-S)/S)<0.000001 S=Sn; break end S=Sn; end S Command Window: Enter a value for a 2 Enter a value for x 3.5 S = 11.313707964994668 Calculator value: 11.3137085 Enter a value for a 6.3 Enter a value for x 1.7 Chapter 6: Solved Problems 27 S = 22.849612550742957 Calculator value: 22.84961748 26. Write a MATLAB program in a script file that finds a positive integer n such that the sum of all the integers 123+++… +n is a number between 100 and 1000 whose three digits are identical. As output, the program displays the integer n and the corresponding sum. Solution Script File: for n=1:100 Sm=sum(1:n); if Sm >= 100 & Sm <= 1000 h=fix(Sm/100); d=fix((Sm-100*h)/10); s=Sm-h*100-d*10; if h == d & h == s n Sm break end end end Command Window: n = 36 Sm = 666 28 Chapter 6: Solved Problems 27. The following are formulas for calculating the Training Heart Rate (THR): where MHR is the maximum heart rate given by (https://en.wikipedia.org/ wiki/Heart_rate): For male: , for female: , RHR is the resting heart rate, and INTEN the fitness level (0.55 for low, 0.65 for medium, and 0.8 for high fitness). Write a program in a script file that determines the THR. The program asks users to enter their gender (male or female), age (number), resting heart rate (number), and fitness level (low, medium, or high). The program then displays the training heart rate (rounded to the nearest integer). Use the program for determining the train- ing heart rate for the following two individuals: (a) A 19-year-old male, resting heart rate of 64, and medium fitness level. (b) A 20-year-old female, resting heart rate of 63, and high fitness level. Solution Script File: G=input('Please enter gender (Male or Female): ','s'); Age=input('Enter age: '); Rhr=input('Enter resting heart rate: '); Int=input('Enter fitness level (low, medium, or high): ','s'); switch Int case 'low' Intn=0.55; case 'medium' Intn=0.65; case 'high' Intn=0.8; end switch G case 'Male' Mhr=203.7/(1+exp(0.033*(Age-104.3))); case 'Female' Mhr=190.2/(1+exp(0.0453*(Age-107.5))); end THR=round((Mhr-Rhr)*Intn+Rhr) Command Window: (a) Please enter gender (Male or Female): Male Chapter 6: Solved Problems 29 Enter age: 19 Enter resting heart rate: 64 Enter fitness level (low, medium, or high): medium THR = 147 (b) Please enter gender (Male or Female): Female Enter age: 20 Enter resting heart rate: 63 Enter fitness level (low, medium, or high): high THR = 162 28. Body Mass Index (BMI) is a measure of obesity. In standard units, it is cal- culated by the formula where W is weight in pounds, and H is height in inches. The obesity classifi- cation is: BMI Classification Below 18.5 Underweight 18.5 to 24.9 Normal 25 to 29.9 Overweight 30 and above Obese Write a program in a script file that calculates the BMI of a person. The pro- gram asks the person to enter his or her weight (lb) and height (in.). The program displays the result in a sentence that reads: “Your BMI value is XXX, which classifies you as SSSS,” where XXX is the BMI value rounded to the nearest tenth, and SSSS is the corresponding classification. Use the program for determining the obesity of the following two individuals: (a) A person 6 ft 2 in. tall with a weight of 180 lb. (b) A person 5 ft 1 in. tall with a weight of 150 lb. Solution Script file: 30 Chapter 6: Solved Problems clear, clc W=input('Please input your weight in lb: '); h=input('Please input your height in in.: '); BMI=703*W/h^2; if BMI<18.5 fprintf('\nYour BMI value is %.1f, which classifies you as underweight\n\n',BMI) elseif BMI<25 fprintf('\nYour BMI value is %.1f, which classifies you as normal\n\n',BMI) elseif BMI<30 fprintf('\nYour BMI value is %.1f, which classifies you as overweight\n\n',BMI) else fprintf('\nYour BMI value is %.1f, which classifies you as obese\n\n',BMI) end Command Window: Please input your weight in lb: 180 Please input your height in in.: 74 Your BMI value is 23.1, which classifies you as normal Please input your weight in lb: 150 Please input your height in in.: 61 Your BMI value is 28.3, which classifies you as overweight >> Chapter 6: Solved Problems 31 29. Write a program in a script file that calculates the cost of renting a car according to the following price schedule: Sedan SUV Duration of rent Daily Free Cost of Daily Free Cost of rate miles Additional rate miles Additional (per day) mile (per day) mile 1-6 days $79 80 $0.69 $84 80 $0.74 7-29 days $69 100 $0.59 $74 100 $0.64 30 or more days $59 120 $0.49 $64 120 $0.54 The program asks the user to enter the type of car (Sedan or SUV), the number of days, and the number of miles driven. The program then displays the cost (rounded to cents) for the rent. Run the program three times for the following cases: (a) Sedan, 10 days, 769 miles. (b) SUV, 32 days, 4,056 miles. (c) Sedan, 3 days, 511 miles. Solution Script file: clear, clc T=input('Enter the type of car (Sedan, or SUV) ','s'); D=input('Enter the number of days '); M=input('Enter the number of miles '); switch T case 'Sedan' if D <= 6 if M <= D*80 cost=79*D; else cost=79*D+(M-D*80)*0.69; end elseif D <= 29 if M <= D*100 cost=69*D; else cost=69*D+(M-D*100)*0.59; end 32 Chapter 6: Solved Problems else if M <= D*120 cost=59*D; else cost=59*D+(M-D*120)*0.49; end end case 'SUV' if D <= 6 if M <= D*80 cost=84*D; else cost=84*D+(M-D*80)*0.74 end elseif D <= 29 if M <= D*100 cost=74*D; else cost=74*D+(M-D*100)*0.64; end else if M <= D*120 cost=64*D; else cost=64*D+(M-D*120)*0.54 end end end fprintf('The cost of the rent is $%5.2f\n',cost) Command Window: Enter the type of car (Sedan, or SUV) Sedan Enter the number of days 10 Enter the number of miles 769 The cost of the rent is $690.00 >> Enter the type of car (Sedan, or SUV) SUV Enter the number of days 32 Chapter 6: Solved Problems 33 Enter the number of miles 4056 The cost of the rent is $2164.64 >> Enter the type of car (Sedan, or SUV) Sedan Enter the number of days 3 Enter the number of miles 511 The cost of the rent is $423.99 >> 34 Chapter 6: Solved Problems 30. Write a program that determines the change given back to a customer in a self-service checkout machine of a supermarket for purchases of up to $50. The program generates a random number between 0.01 and 50.00 and dis- plays the number as the amount to be paid. The program then asks the user to enter payment, which can be one $1 bill, one $5 bill, one $10 bill, one $20 bill, or one $50 bill. If the payment is less than the amount to be paid, an error message is displayed. If the payment is sufficient, the program calcu- lates the change and lists the bills and/or the coins that make up the change, which has to be composed of the least number each of bills and coins. For example, if the amount to be paid is $2.33 and a $10 bill is entered as pay- ment, then the change is one $5 bill, two $1 bills, two quarters, one dime, one nickel, and two pennies. Execute the program three times. Solution Script file: clear, clc AmToPy=randi(5000)/100; fprintf('The amount to be paid is: $%5.2f\n',AmToPy) Cash=input('Please enter payment in dollars (1, 5, 10, 20 or 50) '); if Cash fprintf('%2.0f nickels\n',R(7)) fprintf('%2.0f cents\n',R(8)) end Command Window: The amount to be paid is: $13.93 Please enter payment in dollars (1, 5, 10, 20 or 50) 20 The change is: 0 $20 bills 0 $10 bills 1 $5 bills 1 $1 bills 0 quarters 0 dimes 1 nickels 2 cents The amount to be paid is: $27.35 Please enter payment in dollars (1, 5, 10, 20 or 50) 50 The change is: 1 $20 bills 0 $10 bills 0 $5 bills 2 $1 bills 2 quarters 1 dimes 1 nickels 0 cents The amount to be paid is: $40.02 Please enter payment in dollars (1, 5, 10, 20 or 50) 50 The change is: 0 $20 bills 0 $10 bills 1 $5 bills 4 $1 bills 3 quarters 2 dimes 0 nickels 3 cents >> 36 Chapter 6: Solved Problems 31. The concentration of a drug in the body CP can be modeled by the equation: where DG is the dosage administered (mg), Vd is the volume of distribution –1 (L), ka is the absorption rate constant (h ), ke is the elimination rate con- stant (h–1), and t is the time (h) since the drug was administered. For a cer- tain drug, the following quantities are given: DG = 150 mg, Vd = 50 L, –1 –1 ka = 1.6 h , and ke = 0.4 h . (a) A single dose is administered at t = 0 . Calculate and plot CP versus t for 10 hours. (b) A first dose is administered at t = 0 , and subsequently four more doses are administered at intervals of 4 hours (i.e., at t = 4,, 8 12 , 16 ). Calcu- late and plot CP versus t for 24 hours. Solution Part (a) Script file: clear, clc t=0:0.1:10; n=length(t); DG=150; Vd=50; C=DG/Vd; ke=0.4; ka=1.6; K=ka/(ka-ke); Con=C*K; Cp=Con*(exp(-ke*t)-exp(-ka*t)); plot(t,Cp) xlabel('Time (hr)') ylabel('Drug Concentration (mg/L)') Chapter 6: Solved Problems 37 Figure: Part (b) Script file: clear, clc t=0:0.2:24; n=length(t); DG=150; Vd=50; %C=1.5*100/50; C=DG/Vd; ke=0.4; ka=1.6; K=ka/(ka-ke); Con=C*K; for i=1:n Cp(i)=Con*(exp(-ke*t(i))-exp(-ka*t(i))); if t(i)>4 Cp(i)=Cp(i)+Con*(exp(-ke*(t(i)-4))-exp(- ka*(t(i)-4))); end if t(i)>8 Cp(i)=Cp(i)+Con*(exp(-ke*(t(i)-8))-exp(- ka*(t(i)-8))); end if t(i)>12 Cp(i)=Cp(i)+Con*(exp(-ke*(t(i)-12))-exp(- 38 Chapter 6: Solved Problems ka*(t(i)-12))); end if t(i)>16 Cp(i)=Cp(i)+Con*(exp(-ke*(t(i)-16))-exp(- ka*(t(i)-16))); end end plot(t,Cp) xlabel('Time (hr)') ylabel('Drug Concentration (mg/L)') Figure: Chapter 6: Solved Problems 39 32. One numerical method for calculating the cubic root of a number, 3 P is Hal- ley’s method. The solution process starts by choosing a value x1 as a first esti- mate of the solution. Using this value, a second, more accurate value x2 is calculated with, which is then used for calculating a third, still more accurate value x3 , and so on. The general equation for calculat- ing the value of xi + 1 from the value of xi is . Write a MATLAB program that calculates the cubic root of a number. In the program use x1 = P for the first estimate of the solution. Then, by using the general equation in a loop, calculate new, more accurate values. Stop the looping when the estimated relative error E defined by is smaller than 0.00001. Use the program to calculate: (a)(3 800 b)(3 59071 c) Solution Script File: P=input('Enter a number for calculating its cubic root: '); xi=P*(P^3+2*P)/(2*P^3+P); for n=1:30 xip1=xi*(xi^3+2*P)/(2*xi^3+P); if abs((xip1-xi)/xi)<0.00001 QRP=xip1; break end xi=xip1; end fprintf ('the qubic root of %g is %g\n',P,QRP) Command Window: Enter a number for calculating its cubic root: 800 the qubic root of 800 is 9.28318 Enter a number for calculating its cubic root: 59071 the qubic root of 59071 is 38.9456 Enter a number for calculating its cubic root: -31.55 the qubic root of -31.55 is -3.15985 40 Chapter 6: Solved Problems 33. Write a program in a script file that converts a measure of area given in units of either m2, cm2, in2, ft2, yd2, or acre to the equivalent quantity in different units specified by the user. The program asks the user to enter a numerical value for the size of an area, its current units, and the desired new units. The output is the size of the area in the new units. Use the program to: (a) Convert 55 in2 to cm2. (b) Convert 2400 ft2 to m2. (c) Convert 300 cm2 to yd2. Solution Script file: clear, clc Ain=input('Enter the value of the area to be converted: '); AinUnits=input('Enter the current units (sqm, sqcm, sqin, sqft or sqyd): ','s'); AoutUnits=input('Enter the new units (sqm, sqcm, sqin, sqft or sqyd): ','s'); error=0; switch AinUnits case 'sqm' Asqcm=Ain*1E4; case 'sqcm' Asqcm=Ain; case 'sqin' Asqcm=Ain*2.54^2; case 'sqft' Asqcm=Ain*(2.54*12)^2; case 'sqyd' Asqcm=Ain*(3*2.54*12)^2; otherwise error=1; end switch AoutUnits case 'sqm' Aout=Asqcm*1E-4; case 'sqcm' Aout=Asqcm; case 'sqin' Aout=Asqcm/2.54^2; Chapter 6: Solved Problems 41 case 'sqft' Aout=Asqcm/(2.54*12)^2; case 'sqyd' Aout=Asqcm/(3*2.54*12)^2; otherwise error=1; end if error disp('ERROR current or new units are typed incorrectly.') else fprintf('A= %g %s\n',Aout,AoutUnits) end Command Window: Enter the value of the area to be converted: 55 Enter the current units (sqm, sqcm, sqin, sqft or sqyd): sqin Enter the new units (sqm, sqcm, sqin, sqft or sqyd): sqcm A= 354.838 sqcm >> Enter the value of the area to be converted: 2400 Enter the current units (sqm, sqcm, sqin, sqft or sqyd): sqft Enter the new units (sqm, sqcm, sqin, sqft or sqyd): sqm A= 222.967 sqm >> Enter the value of the area to be converted: 300 Enter the current units (sqm, sqcm, sqin, sqft or sqyd): sqcm Enter the new units (sqm, sqcm, sqin, sqft or sqyd): sqyd A= 0.0358797 sqyd >> 42 Chapter 6: Solved Problems 34. In a one-dimensional random walk, the position x of a walker is computed by: xj = xj + s where s is a random number. Write a program that calculates the number of steps required for the walker to reach a boundary . Use MATLAB’s built-in function randn(1,1) to calculate s. Run the program 100 times (by using a loop) and calculate the average number of steps when B = 10 . Solution Script File: clear, clc for j=1:100 x=0; for i=1:1000 x=x+randn(1,1); if abs(x)>10 break end end if i==1000 disp('ERROR: Boundary was not reached in 1000 steps') end steps(j)=i; end AveNumSteps=mean(steps) Command Window: AveNumSteps = 123.7700 >> Chapter 6: Solved Problems 43 35. The Sierpinski triangle can be implemented in MATLAB by plotting points iteratively according to one of the following three rules that are selected ran- domly with equal probability. Rule 1: xn + 1 = 0.5xn , yn + 1 = 0.5yn Rule 2: xn + 1 = 0.5xn + 0.25 , Rule 3: xn + 1 = 0.5xn + 0.5 , yn + 1 = 0.5yn Write a program in a script file that calculates the x and y vectors and then plots y versus x as individual points (use plot(x,y,‘^’)). Start with x1 = 0 and y1 = 0 . Run the program four times with 10, 100, 1,000, and 10,000 iterations. Solution Script file: clear, clc x(1)=0; y(1)=0; for i=2:10000 n=randi(3); if n==1 x(i)=0.5*x(i-1); y(i)=0.5*y(i-1); elseif n==2 x(i)=0.5*x(i-1)+0.25; y(i)=0.5*y(i-1)+sqrt(3)/4; elseif n==3 x(i)=0.5*x(i-1)+0.5; y(i)=0.5*y(i-1); end end plot(x,y,'^') axis equal 44 Chapter 6: Solved Problems Figure 10 iterations: Figure 100 iterations: Figure 1000 iterations: Chapter 6: Solved Problems 45 Figure 10,000 iterations: 46 Chapter 6: Solved Problems 36. The roots of a cubic equation can be calculated using the following procedure: Set: , , and . Calculate: , where and . If the equation has complex roots. If all roots are real and at least two are equal. The roots are given by: , , and . If all roots are real and are given by: , , and , where . Write a MATLAB program that determines the real roots of a cubic equa- tion. As input the program asks the user to enter the values of a3, a2, a1, and a0 as a vector. The program then calculates the value of D. If the equations has complex roots the message “The equation has complex roots” is dis- played. Otherwise the real roots are calculated and displayed. Use the pro- gram to solve the following equations: (a)(b) (c) Solution Script File: v=input('Enter the values of a3, a2, a1, and a0 as a vector: '); a=v(2)/v(1); b=v(3)/v(1); c=v(4)/v(1); Q=(3*b-a^2)/9; R=(9*a*b-27*c-2*a^3)/54; D=Q^3+R^2; if abs(D)<0.01 D=0; end if D > 0 disp('The equation has complex roots.') elseif D ==0 S=nthroot(R,3); x1=2*S-a/3; x2=-S-a/3; Chapter 6: Solved Problems 47 x3=x2; fprintf('The roots are: %g , %g , and %g.\n',x1,x2,x3) elseif D < 0 Qs=sqrt(-Q); th=acosd(R/sqrt(-Q^3)); x1=2*Qs*cosd(th/3)-a/3; x2=2*Qs*cosd(th/3+120)-a/3; x3=2*Qs*cosd(th/3+240)-a/3; fprintf('The roots are: %g , %g , and %g.\n',x1,x2,x3) end Command Window: (a) Enter the values of a3, a2, a1, and a0 as a vector: [5 - 34.5 36.9 8.8] The roots are: 5.5 , -0.2 , and 1.6. (b) Enter the values of a3, a2, a1, and a0 as a vector: [2 - 10 24 -15] The equation has complex roots. (c) Enter the values of a3, a2, a1, and a0 as a vector: [2 - 1.4 -20.58 30.87] The roots are: -3.5 , 2.1 , and 2.1. 48 Chapter 6: Solved Problems 37. The overall grade in a course is determined from the grades of 10 homework assignments, 2 midterms, and a final exam, using the following scheme: Homeworks: Homework assignments are graded on a scale from 0 to 80. The grade of the two lowest assignments is dropped and the average of the 8 assignments with the higher grades constitutes 20% of the course grade. Midterms and final exam: Midterms and final exams are graded on a scale from 0 to 100. If the average of the midterm scores is higher than, or the same as, the score on the final exam, the average of the midterms constitutes 40% of the course grade and the grade of the final exam constitutes 40% of the course grade. If the final exam grade is higher than the average of the midterms, the average of the midterms constitutes 30% of the course grade and the grade of the final exam constitutes 50% of the course grade. Write a computer program in a script file that determines the course grade for a student. The program first asks the user to enter the 10 home- work assignment grades (in a vector), two midterm grades (in a vector), and the grade of the final. Then the program calculates a numerical course grade (a number between 0 and 100). Execute the program for the following cases: (a) Homework assignment grades: 65, 79, 80, 50, 71, 73, 61, 70, 69, 74. Mid- term grades: 83, 91. Final exam: 84. (b) Homework assignment grades: 70, 69, 83, 45, 90, 89, 52, 78, 100, 87. Midterm grades: 87, 72. Final exam: 90. Solution Script file: clear, clc HW=input('Enter ten homework grades (0-80) in a vector '); MT=input('Enter two midterm grades (0-100) in a vector '); FE=input('Enter the final exam grade (0-100) '); HWsorted=sort(HW); HW8=HWsorted(3:10); HWtoFinalGrade=mean(HW8)/100*20; MTave=mean(MT); if MTave >= FE Grade=round(0.4*MTave+0.4*FE+HWtoFinalGrade); else Grade=round(0.3*MTave+0.5*FE+HWtoFinalGrade); end fprintf('Course grade: % i\n',Grade) Chapter 6: Solved Problems 49 Command Window: Enter ten homework grades (0-80) in a vector [65 79 80 50 71 73 61 70 69 74]; Enter two midterm grades (0-100) in a vector [83 91] Enter the final exam grade (0-100) 84 Course grade: 83 >> Enter ten homework grades (0-80) in a vector [70 69 83 45 90 89 52 78 100 87] Enter two midterm grades (0-100) in a vector [87 72] Enter the final exam grade (0-100) 90 Course grade: 86 >> 50 Chapter 6: Solved Problems 38. Keith number is a number (integer) that appears in a Fibonacci-like sequence that is based on its own decimal digits. For two-decimal digit num- bers (10 through 99) a Fibonacci-like sequence is created in which the first element is the tens digit and the second element is the units digit. The value of each subsequent element is the sum of the previous two elements. If the number is a Keith number, then it appears in the sequence. For example, the first two-decimal digit Keith number is 14, since the corresponding Fibo- nacci-like sequence is 1, 4, 5, 9, 14. Write a MATLAB program that deter- mines and displays all the Keith numbers between 10 and 99. Solution Script file: clear, clc p=1; for i=1:9 for j=0:9 N=10*i+j; a=[i, j]; for k=3:20 a(k)=a(k-2)+a(k-1); if a(k)==N KN(p)=N; p=p+1; break end if a(k)>N break end end end end KN Command Window: KN = 14 19 28 47 61 75 >> Chapter 6: Solved Problems 51 39. The following MATLAB commands creates a sine-shaped signal y(t) that contains random noise: t = 0:.05:10; y = sin(t)-0.1+0.2*rand(1,length(t)); Write a MATLAB program that use these commands to creates a noisy sine-shaped signal. Then the program smooths the signal by using the three- points moving average method. In this method the value of every point i, except the first and last, is replaced by the average of the value of three adja- cent points (i–1, i, and i+1). Make a plot that display the noisy and smoothed signals. Solution Script file: clear, clc t = 0:.05:10; y = sin(t)-0.1+0.2*rand(1,length(t)); n=length(t); ys=y; for k=1:10 for j=2:n-1 ys(j)=(ys(j-1)+ys(j)+ys(j+1))/3; end end plot(t,y,t,ys,'linewidth',2) % Plot both signals. legend('Original signal','Signal with AWGN'); 52 Chapter 6: Solved Problems Figure: