259 Lecture 10 Spring 2017
Advanced Excel Topics – More User
Defined Functions; Conditional
Statements
Outline
 More User Defined Functions!




Perp_Bisector_Slope()
Perp_Bisector_Intercept()
Quadratic_1()
Quadratic_2()
 Pseudo Code
 IF…THEN…ELSE Statements
 Modified Quadratic Function
 A Piecewise Defined Function
2
Perp_Bisector_Slope()
 Example 1: Create a user defined
function Perp_Bisector_Slope() that
will return the slope of the line that is
the “perpendicular bisector” of the
line segment determined by the
points (x1,y1) and (x2,y2).
3
Perp_Bisector_Slope()
 Given:
 Points (x1,y1) and (x2,y2).
 Line L is the perpendicular bisector of the
line segment determined by the points
(x1,y1) and (x2,y2).
 Find:
 The slope of line L.
4
Perp_Bisector_Slope()
 Pseudo Code:
(1) Find the slope of the line segment
determined by the points (x1,y1) and
(x2,y2), and then store the result in the
variable m1.
(2) Find the negative reciprocal of m1
and then store the result in the variable
m2 .
(3) Return the value stored in the
variable m2 as the result of the user
defined function.
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Perp_Bisector_Slope()
Function Perp_Bisector_Slope(x1 As Double, y1 As
Double, x2 As Double, y2 As Double) As Double
'Returns the slope of the line that is the
'"perpendicular bisector" of the line segment
'determined by points (x1,y1) and (x2,y2).
'The function does NOT work for vertical lines
Dim m1 As Double, m2 As Double
m1 = (y2 - y1) / (x2 - x1)
'negative reciprocal of m1
m2 = -1 / m1
Perp_Bisector_Slope = m2
End Function
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Perp_Bisector_Intercept()
 Example 2: Create a user defined
function Perp_Bisector_Intercept()
that will return the y-intercept of the
line that is the “perpendicular
bisector” of the line determined by
the points (x1,y1) and (x2,y2).
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Perp_Bisector_Intercept()
 Given:
 Points (x1,y1) and (x2,y2).
 Line L is the perpendicular bisector of the
line segment determined by the points
(x1,y1) and (x2,y2).
 Find:
 The y-intercept of line L.
8
Perp_Bisector_Intercept()
 Pseudo Code:
(1) Find the slope of the line by using the
Perp_Bisector_Slope(x1, y1, x2, y2) function previously
defined, and then store the result in the variable m.
(2) Find the x-component of the midpoint of the line
segment, and then store the result in the variable
Midpoint_X.
(3) Find the y-component of the midpoint of the line
segment, and then store the result in the variable
Midpoint_Y.
(4) Return the y-intercept of line L, by substituting the
coordinates of the midpoint into the expression y -m*x.
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Perp_Bisector_Intercept()
Function Perp_Bisector_Intercept(x1 As Double, y1 As Double, x2 As
Double, y2 As Double) As Double
'Returns the y-intercept of the line that is the
'perpendicular bisector" of the line segment
'determined by points (x1,y1) and (x2,y2).
'The function does NOT work for vertical lines
Dim m As Double
Dim Midpoint_X As Double, Midpoint_Y As Double
m = Perp_Bisector_Slope(x1, y1, x2, y2)
Midpoint_X = (x1 + x2) / 2
Midpoint_Y = (y1 + y2) / 2
'b = y - m*x
'put in a known point (Midpoint_X, Midpoint_Y) on the line
Perp_Bisector_Intercept = Midpoint_Y - m * Midpoint_X
End Function
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Quadratic_1()
 Example 3: Create a user defined
function Quadratic_1() that will return
the first real root of a quadratic
equation Ax2 + Bx + C = 0.
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Quadratic_1()
 Given:
 A, B, and C
2
2

B

B
 4 AC
 Roots of Ax + Bx + C = 0 are
2
2A

B

B

4
AC
and
 Find:
2A
 The first real root of a quadratic equation
Ax2 + Bx + C = 0.
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Quadratic_1()
 Pseudo Code:
(1) Substitute the A, B, and C values
into the formula
 B  B 2  4 AC
2A
and return the result of this
expression.
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Quadratic_1()
Function Quadratic_1(A As Double, B As Double, C As
Double) As Double
'Returns the first real root of a quadratic equation
'Ax^2+Bx+C=0
'Does NOT return a complex root
Quadratic_1 = (-B + Sqr(B ^ 2 - 4 * A * C)) / (2 * A)
End Function
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Quadratic_2()
 Example 4: Create a user defined
function Quadratic_2() that will return
the second real root of a quadratic
equation Ax2 + Bx + C = 0.
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Quadratic_2()
 Given:
 A, B, and C
2
2

B

B
 4 AC
 Roots of Ax + Bx + C = 0 are
2
2A

B

B

4
AC
and
 Find:
2A
 The second real root of a quadratic
equation Ax2 + Bx + C = 0.
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Quadratic_2()
 Pseudo Code:
(1) Substitute the A, B, and C values
into the formula
 B  B 2  4 AC
2A
and return the result of this
expression.
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Quadratic_2()
Function Quadratic_2(A As Double, B As Double, C As
Double) As Double
'Returns the second real root of a quadratic equation
'Ax^2+Bx+C=0
'Does NOT return a complex root
Quadratic_2 = (-B - Sqr(B ^ 2 - 4 * A * C)) / (2 * A)
End Function
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IF…THEN…ELSE Statements
 Just like in Excel, we can create
conditional statements within VBA!
 One of the most useful conditional
statements in VBA is the
IF…THEN…ELSE statement!
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IF…THEN…ELSE Syntax

Syntax:
If condition_1 Then
result_1
ElseIf condition_2 Then
result_2
...
ElseIf condition_n Then
result_n
Else
result_else
End If






How the command works:
condition_1 to condition_n are
evaluated in the order listed.
Once a condition is found to be
true, the IF…THEN…ELSE
statement will execute the
corresponding code and not
evaluate the conditions any
further.
result_1 to result_n is the code
that is executed once a condition
is found to be true.
If no condition is met, then the
Else portion of the
IF…THEN…ELSE statement will
be executed.
Note that the ElseIf and Else
portions are optional.
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Quadratic_1_Improved()
 Example 5: Modify the Quadratic_1()
user defined function to return the
first root (real or complex).
 For example x2+4x+8=0 should
return (2+2i) and x2+5x+6=0 should
return -2.
 Then modify the VBA code for this
new function to make a user defined
function to find the other root!
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Quadratic_1_Improved()
 Given: A, B, and C
Roots of Ax2 + Bx + C = 0 are:
 B  B 2  4 AC
2A
B


2A
and
B 2  4 AC
2A
i and
 B  B 2  4 AC
2A
B


2A
if
B 2  4 AC
2A
i if
B 2  4 AC  0
B 2  4 AC  0
 Find: The first root of a quadratic
equation Ax2 + Bx + C = 0 .
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Quadratic_1_Improved()

Pseudo Code:
(1) If B^2 – 4AC
the formula
0, then substitute the A, B, and C values into
 B  B 2  4 AC
2A
and return the value of this expression.
(2) Otherwise, put the A, B, and C values into the formula
and store the result in the variable Real_Part.
(3) Substitute the A, B, and C values into the formula

B
2A
B 2  4 AC
2A
and store the result in the variable Imaginary_Part.
(4) Return the string value “{Real_Part} + {Imaginary_Part} i “
by substituting the actual numeric values for Real_Part and
Imaginary_Part into {Real_Part} and {Imaginary_Part} .
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Quadratic_1_Improved()
Function Quadratic_1_Improved(A As Double, B As Double, C As Double) As Variant
'Returns the first root of a quadratic equation Ax^2+Bx+C=0
'Returns both real and complex roots
Dim Real_Part As Double, Imaginary_Part As Double
If B ^ 2 - 4 * A * C >= 0 Then
Quadratic_1_Improved = (-B + Sqr(B ^ 2 - 4 * A * C)) / (2 * A)
Else
Real_Part = -B / (2 * A)
Imaginary_Part = Sqr(Abs(B ^ 2 - 4 * A * C)) / (2 * A)
Quadratic_1_Improved = Str(Real_Part) + "+ " + Str(Imaginary_Part) + "i"
End If
End Function
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A piecewise “user defined
function” f(x)
 Example 6: Use VBA IF statement(s)
to create the following piecewise
“user defined function” f(x):
 x3
 2
f ( x)   x
 x 3

x0 

if 0  x  3
if x  3 
if
 Plot a graph of this function!
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A piecewise “user defined
function” f(x)
Function f(x As Double) As Double
'Returns x^3, if x<0
'Returns x^2, if 0<=x<=3
'Returns Sqr(x-3), if x>3
If x < 0 Then
f=x^3
End If
If (x >= 0) And (x <= 3) Then
f=x^2
End If
If x > 3 Then
f = Sqr(x-3)
End If
End Function
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References
 User Defined Functions Notes – John
Albers
 IF…THEN…ELSE Statements (p. 20 of
this lecture) http://www.techonthenet.com/excel/f
ormulas/if_then.php
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