CHAPTER 2

2.1 Two possible versions can be developed:

|IF x ( 10 THEN |IF x ( 10 THEN |
|DO |DO |
|x = x – 5 |x = x – 5 |
|IF x < 50 EXIT |IF x < 50 EXIT |
|END DO |END DO |
|ELSE |ELSEIF x < 5 |
|IF x < 5 THEN |x = 5 |
|x = 5 |ELSE |
|ELSE |x = 7.5 |
|x = 7.5 |ENDIF |
|END IF | |
|ENDIF | |

2.2

DO i = i + 1 IF z > 50 EXIT x = x + 5 IF x > 5 THEN y = x ELSE y = 0 ENDIF z = x + y ENDDO

2.3 Note that this algorithm is made simpler by recognizing that concentration cannot by definition be negative. Therefore, the maximum can be initialized as zero at the start of the algorithm.

Step 1: Start Step 2: Initialize sum, count and maximum to zero Step 3: Examine top card. Step 4: If it says “end of data” proceed to step 9; otherwise, proceed to next step. Step 5: Add value from top card to sum. Step 6: Increase count by 1. Step 7: If value is greater than maximum, set maximum to value. Step 7: Discard top card Step 8: Return to Step 3. Step 9: Is the count greater than zero? If yes, proceed to step 10. If no, proceed to step 11. Step 10: Calculate average = sum/count Step 11: End

2.4 Flowchart: [pic]

2.5 Students could implement the subprogram in any number of languages. The following Fortran 90 program is one example. It should be noted that the availability of complex variables in Fortran 90 would allow this subroutine to be made even more concise. However, we did not exploit this feature, in order to make the code more compatible with languages such as Visual BASIC or C.

PROGRAM Rootfind IMPLICIT NONE INTEGER::ier REAL::a, b, c, r1, i1, r2, i2 DATA a,b,c/1.,6.,2./ CALL Roots(a, b, c, ier, r1, i1, r2, i2) IF (ier == 0) THEN PRINT *, r1,i1," i" PRINT *, r2,i2," i" ELSE PRINT *, "No roots" END IF END

SUBROUTINE Roots(a, b, c, ier, r1, i1, r2, i2) IMPLICIT NONE INTEGER::ier REAL::a, b, c, d, r1, i1, r2, i2 r1=0. r2=0. i1=0. i2=0. IF (a == 0.) THEN IF (b /= 0) THEN r1 = -c/b ELSE ier = 1 END IF ELSE d = b**2 - 4.*a*c IF (d >= 0) THEN r1 = (-b + SQRT(d))/(2*a) r2 = (-b - SQRT(d))/(2*a) ELSE r1 = -b/(2*a) r2 = r1 i1 = SQRT(ABS(d))/(2*a) i2 = -i1 END IF END IF END

The answers for the 3 test cases are: (a) (0.3542, (5.646; (b) 0.4; (c) (0.4167 + 1.4696i; (0.4167 ( 1.4696i.

Several features of this subroutine bear mention:
The subroutine does not involve input or output. Rather, information is passed in and out via the arguments. This is often the preferred style, because the I/O is left to the discretion of the programmer within the calling program.
Note that an error code is passed (IER = 1) for the case where no roots are possible.

2.6 The development of the algorithm hinges on recognizing that the series approximation of the cosine can be represented concisely by the summation,

[pic]

where i = the order of the approximation. The following algorithm implements this summation:

Step 1: Start Step 2: Input value to be evaluated x and maximum order n Step 3: Set order (i) equal to one Step 4: Set accumulator for approximation (approx) to zero Step 5: Set accumulator for factorial product (factor) equal to one Step 6: Calculate true value of cos(x) Step 7: If order is greater than n then proceed to step 13 Otherwise, proceed to next step Step 8: Calculate the approximation with the formula [pic] Step 9: Determine the error [pic] Step 10: Increment the order by one Step 11: Determine the factorial for the next iteration [pic] Step 12: Return to step 7 Step 13: End

2.7 (a) Structured flowchart

[pic]

(b) Pseudocode:

SUBROUTINE Coscomp(n,x) i = 1 approx = 0 factor = 1 truth = cos(x) DO IF i > n EXIT approx = approx + (-1)i-1•x2(i-2 / factor error = (true - approx) / true) * 100 DISPLAY i, true, approx, error i = i + 1 factor = factor•(2•i-3)•(2•i-2) END DO END

2.8 Students could implement the subprogram in any number of languages. The following MATLAB M-file is one example. It should be noted that MATLAB allows direct calculation of the factorial through its intrinsic function factorial. However, we did not exploit this feature, in order to make the code more compatible with languages such as Visual BASIC and Fortran.

function coscomp(x,n) i = 1; tru = cos(x); approx = 0; f = 1; fprintf('\n'); fprintf('order true value approximation error\n'); while (1) if i > n, break, end approx = approx + (-1)^(i - 1) * x^(2*i-2) / f; er = (tru - approx) / tru * 100; fprintf('%3d %14.10f %14.10f %12.8f\n',i,tru,approx,er); i = i + 1; f = f*(2*i-3)*(2*i-2); end

Here is a run of the program showing the output that is generated:

>> coscomp(1.25,6)

order true value approximation error 1 0.3153223624 1.0000000000 -217.13576938 2 0.3153223624 0.2187500000 30.62655045 3 0.3153223624 0.3204752604 -1.63416828 4 0.3153223624 0.3151770698 0.04607749 5 0.3153223624 0.3153248988 -0.00080437 6 0.3153223624 0.3153223323 0.00000955

2.9 (a) The following pseudocode provides an algorithm for this problem. Notice that the input of the quizzes and homeworks is done with logical loops that terminate when the user enters a negative grade:

INPUT WQ, WH, WF nq = 0 sumq = 0 DO INPUT quiz (enter negative to signal end of quizzes) IF quiz < 0 EXIT nq = nq + 1 sumq = sumq + quiz END DO AQ = sumq / nq nh = 0 sumh = 0 DO INPUT homework (enter negative to signal end of homeworks) IF homework < 0 EXIT nh = nh + 1 sumh = sumh + homework END DO AH = sumh / nh DISPLAY "Is there a final grade (y or n)" INPUT answer IF answer = "y" THEN INPUT FE AG = (WQ * AQ + WH * AH + WF * FE) / (WQ + WH + WF) ELSE AG = (WQ * AQ + WH * AH) / (WQ + WH) END IF DISPLAY AG END

(b) Students could implement the program in any number of languages. The following VBA code is one example.

Sub Grader() Dim WQ As Double, WH As Double, WF As Double Dim nq As Integer, sumq As Double, AQ As Double Dim nh As Integer, sumh As Double, AH As Double Dim answer As String, FE As Double Dim AG As Double

'enter weights WQ = InputBox("enter quiz weight") WH = InputBox("enter homework weight") WF = InputBox("enter final exam weight") 'enter quiz grades nq = 0 sumq = 0 Do quiz = InputBox("enter negative to signal end of quizzes") If quiz < 0 Then Exit Do nq = nq + 1 sumq = sumq + quiz Loop AQ = sumq / nq 'enter homework grades nh = 0 sumh = 0 Do homework = InputBox("enter negative to signal end of homeworks") If homework < 0 Then Exit Do nh = nh + 1 sumh = sumh + homework Loop AH = sumh / nh 'determine and display the average grade answer = InputBox("Is there a final grade (y or n)") If answer = "y" Then FE = InputBox("final grade:") AG = (WQ * AQ + WH * AH + WF * FE) / (WQ + WH + WF) Else AG = (WQ * AQ + WH * AH) / (WQ + WH) End If MsgBox "Average grade = " & AG End Sub

The results should conform to:

AQ = 437/5 = 87.4 AH = 541/6 = 90.1667

without final [pic]

with final [pic]

2.10 (a) Pseudocode:

IF a > 0 THEN tol = 10–5 x = a/2 DO y = (x + a/x)/2 e = ((y – x)/y( x = y IF e < tol EXIT END DO SquareRoot = x ELSE SquareRoot = 0 END IF

(b) Students could implement the function in any number of languages. The following VBA and MATLAB codes are two possible options.

|VBA Function Procedure |MATLAB M-File |
|Option Explicit |function s = SquareRoot(a) |
|Function SquareRoot(a) |if a > 0 |
|Dim x As Double, y As Double |tol = 0.00001; |
|Dim e As Double, tol As Double |x = a / 2; |
|If a > 0 Then |while(1) |
|tol = 0.00001 |y = (x + a / x) / 2; |
|x = a / 2 |e = abs((y - x) / y); |
|Do |x = y; |
|y = (x + a / x) / 2 |if e < tol, break, end |
|e = Abs((y - x) / y) |end |
|x = y |s = x; |
|If e < tol Then Exit Do |else |
|Loop |s = 0; |
|SquareRoot = x |end |
|Else | |
|SquareRoot = 0 | |
|End If | |
|End Function | |

2.11 A MATLAB M-file can be written to solve this problem as

function futureworth(P, i, n) nn = 0:n; F = P*(1+i).^nn; y = [nn;F]; fprintf('\n year future worth\n'); fprintf('%5d %14.2f\n',y);

This function can be used to evaluate the test case,

>> futureworth(100000,0.06,5)

year future worth 0 100000.00 1 106000.00 2 112360.00 3 119101.60 4 126247.70 5 133822.56

2.12 A MATLAB M-file can be written to solve this problem as

function annualpayment(P, i, n) nn = 1:n; A = P*i*(1+i).^nn./((1+i).^nn-1); y = [nn;A]; fprintf('\n year annual payment\n'); fprintf('%5d %14.2f\n',y);

This function can be used to evaluate the test case,

>> annualpayment(55000,0.066,5)

year annual payment 1 58630.00 2 30251.49 3 20804.86 4 16091.17 5 13270.64

2.13 Students could implement the function in any number of languages. The following VBA and MATLAB codes are two possible options.

|VBA Function Procedure |MATLAB M-File |
|Option Explicit |function Ta = avgtemp(Tm,Tp,ts,te) |
|Function avgtemp(Tm, Tp, ts, te) |w = 2*pi/365; |
|Dim pi As Double, w As Double |t = ts:te; |
|Dim Temp As Double, t As Double |T = Tm + (Tp-Tm)*cos(w*(t-205)); |
|Dim sum As Double, i As Integer |Ta = mean(T); |
|Dim n As Integer | |
|pi = 4 * Atn(1) | |
|w = 2 * pi / 365 | |
|sum = 0 | |
|n = 0 | |
|t = ts | |
|For i = ts To te | |
|Temp = Tm+(Tp-Tm)*Cos(w*(t-205)) | |
|sum = sum + Temp | |
|n = n + 1 | |
|t = t + 1 | |
|Next i | |
|avgtemp = sum / n | |
|End Function | |

The function can be used to evaluate the test cases. The following show the results for MATLAB,

>> avgtemp(22.1,28.3,0,59)

ans = 16.2148

>> avgtemp(10.7,22.9,180,242)

ans = 22.2491

2.14 The programs are student specific and will be similar to the codes developed for VBA, MATLAB and Fortran as outlined in sections 2.4, 2.5 and 2.6. The numerical results for the different time steps are tabulated below along with an estimate of the absolute value of the true relative error at t = 12 s:

|Step |v(12) |((t( (%) |
|2 |49.96 |5.2 |
|1 |48.70 |2.6 |
|0.5 |48.09 |1.3 |

The general conclusion is that the error is halved when the step size is halved.

2.15 Students could implement the subprogram in any number of languages. The following Fortran 90 and VBA/Excel programs are two examples based on the algorithm outlined in Fig. P2.15.

|Fortran 90 |VBA/Excel |
|Subroutine BubbleFor(n, b) |Option Explicit |
| | |
|Implicit None |Sub Bubble(n, b) |
| | |
|!sorts an array in ascending |'sorts an array in ascending |
|!order using the bubble sort |'order using the bubble sort |
| | |
|Integer(4)::m, i, n |Dim m As Integer |
|Logical::switch |Dim i As Integer |
|Real::a(n),b(n),dum |Dim switch As Boolean |
| |Dim dum As Double |
|m = n - 1 | |
|Do |m = n - 1 |
|switch = .False. |Do |
|Do i = 1, m |switch = False |
|If (b(i) > b(i + 1)) Then |For i = 1 To m |
|dum = b(i) |If b(i) > b(i + 1) Then |
|b(i) = b(i + 1) |dum = b(i) |
|b(i + 1) = dum |b(i) = b(i + 1) |
|switch = .True. |b(i + 1) = dum |
|End If |switch = True |
|End Do |End If |
|If (switch == .False.) Exit |Next i |
|m = m - 1 |If switch = False Then Exit Do |
|End Do |m = m - 1 |
| |Loop |
|End | |
| |End Sub |

For MATLAB, the following M-file implements the bubble sort following the algorithm outlined in Fig. P2.15:

function y = Bubble(x) n = length(x); m = n - 1; b = x; while(1) s = 0; for i = 1:m if b(i) > b(i + 1) dum = b(i); b(i) = b(i + 1); b(i + 1) = dum; s = 1; end end if s == 0, break, end m = m - 1; end y = b;

Notice how the length function allows us to omit the length of the vector in the function argument. Here is an example MATLAB session that invokes the function to sort a vector:

>> a=[3 4 2 8 5 7]; >> Bubble(a)

ans = 2 3 4 5 7 8

2.16 Here is a flowchart for the algorithm:

[pic]

Students could implement the function in any number of languages. The following VBA and MATLAB codes are two possible options.

|VBA Function Procedure |MATLAB M-File |
|Option Explicit |function Vol = tankvolume(R, d) |
|Function Vol(R, d) |if d < R |
|Dim V1 As Double, V2 As Double |Vol = pi * d ^ 3 / 3; |
|Dim pi As Double |elseif d 0 th(i) = atan(y(i) / x(i)) + pi; elseif y(i) < 0 th(i) = atan(y(i) / x(i)) - pi; else th(i) = pi; end else if y(i) > 0 th(i) = pi / 2; elseif y(i) < 0 th(i) = -pi / 2; else th(i) = 0; end end th(i) = th(i) * 180 / pi; end ou=[x;y;r;th]; fprintf('\n x y radius angle\n'); fprintf('%8.2f %8.2f %10.4f %10.4f\n',ou);

This function can be used to evaluate the test cases.

>> x=[1 1 1 -1 -1 -1 0 0 0]; >> y=[1 -1 0 1 -1 0 1 -1 0]; >> polar(x,y)

x y radius angle 1.00 1.00 1.4142 45.0000 1.00 -1.00 1.4142 -45.0000 1.00 0.00 1.0000 0.0000 -1.00 1.00 1.4142 135.0000 -1.00 -1.00 1.4142 -135.0000 -1.00 0.00 1.0000 180.0000 0.00 1.00 1.0000 90.0000 0.00 -1.00 1.0000 -90.0000 0.00 0.00 0.0000 0.0000

2.18 Students could implement the function in any number of languages. The following VBA and MATLAB codes are two possible options.

|VBA Function Procedure |MATLAB M-File |
|Function grade(s) |function grade = lettergrade(score) |
|If s >= 90 Then |if score >= 90 |
|grade = "A" |grade = 'A'; |
|ElseIf s >= 80 Then |elseif score >= 80 |
|grade = "B" |grade = 'B'; |
|ElseIf s >= 70 Then |elseif score >= 70 |
|grade = "C" |grade = 'C'; |
|ElseIf s >= 60 Then |elseif score >= 60 |
|grade = "D" |grade = 'D'; |
|Else |else |
|grade = "F" |grade = 'F'; |
|End If |end |
|End Function | |

2.19 Students could implement the functions in any number of languages. The following VBA and MATLAB codes are two possible options.

|VBA Function Procedure |MATLAB M-File |
|(a) Factorial | |
|Function factor(n) |function fout = factor(n) |
|Dim x As Long, i As Integer |x = 1; |
|x = 1 |for i = 1:n |
|For i = 1 To n |x = x * i; |
|x = x * i |end |
|Next i |fout = x; |
|factor = x | |
|End Function | |
| | |
|(b) Minimum | |
|Function min(x, n) |function xm = xmin(x) |
|Dim i As Integer |n = length(x); |
|min = x(1) |xm = x(1); |
|For i = 2 To n |for i = 2:n |
|If x(i) < min Then min = x(i) |if x(i) < xm, xm = x(i); end |
|Next i |end |
|End Function | |
| | |
|(c) Average | |
|Function mean(x, n) |function xm = xmean(x) |
|Dim sum As Double |n = length(x); |
|Dim i As Integer |s = x(1); |
|sum = x(1) |for i = 2:n |
|For i = 2 To n |s = s + x(i); |
|sum = sum + x(i) |end |
|Next i |xm = s / n; |
|mean = sum / n | |
|End Function | |

2.20 Students could implement the functions in any number of languages. The following VBA and MATLAB codes are two possible options.

|VBA Function Procedure |MATLAB M-File |
|(a) Square root sum of squares | |
|Function SSS(x, n, m) |function s = SSS(x) |
|Dim i As Integer, j As Integer |[n,m] = size(x); |
|SSS = 0 |s = 0; |
|For i = 1 To n |for i = 1:n |
|For j = 1 To m |for j = 1:m |
|SSS = SSS + x(i, j) ^ 2 |s = s + x(i, j) ^ 2; |
|Next j |end |
|Next i |end |
|SSS = Sqr(SSS) |s = sqrt(s); |
|End Function | |
| | |
|(b) Normalization | |
|Sub normal(x, n, m, y) |function y = normal(x) |
|Dim i As Integer, j As Integer |[n,m] = size(x); |
|Dim max As Double |for i = 1:n |
|For i = 1 To n |mx = abs(x(i, 1)); |
|max = Abs(x(i, 1)) |for j = 2:m |
|For j = 2 To m |if abs(x(i, j)) > mx |
|If Abs(x(i, j)) > max Then |mx = x(i, j); |
|max = x(i, j) |end |
|End If |end |
|Next j |for j = 1:m |
|For j = 1 To m |y(i, j) = x(i, j) / mx; |
|y(i, j) = x(i, j) / max |end |
|Next j |end |
|Next i | |
|End Sub |Alternate version: |
| | |
| |function y = normal(x) |
| |n = size(x); |
| |for i = 1:n |
| |y(i,:) = x(i,:)/max(x(i,:)); |
| |end |