King M.R., Mody N.A. Numerical and Statistical Methods for Bioengineering: Applications in MATLAB

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A MATLAB program is written to find the sum of 0.2 + 0.2 + 0.2 + ... 250 000 times and initially

converts all variables to the single data type since MATLAB stores all numbers as double by default. The

true final sum of the addition process is 0.2 × 250 000 = 50 000. We then check the computed sum with

the exact solution. In the code below, the variables are converted to the single data type by using the

single conversion function.

MATLAB program 1.1

% This program calculates the sum of a fraction added to itself n times

% Variable declaration and initialization

fraction = 0.2; % fraction to be added

n = 250000; % is the number of additions

summation = 0; % sum of the input fraction added n times

% Converting double precision numbers to single precision

fraction = single (fraction);

summation = single(summation);

% Performing repeated additions

for l = 1:n % l is the looping variable

summation = summation + fraction;

end

summation

The result obtained after running the code is

sum =

4.9879925e+004

Thepercentagerelativeerrorwhenusingthe single data type is 0.24%. The magnitude

of this error is not insigniﬁcant considering that the digital ﬂoating-point representation

of single numbers is accurate to the eighth decimal point. The percentage relative error

when using double precision for representing ﬂoating-point numbers is 3.33 × 10

–12

Double precision allows for much more accurate representation of floating-point numbers and should be

preferred whenever possible.

In the real world, when performing experiments, procuring data, or solving

numerical problems, you usually will not know the true or exact solution; otherwise

you would not be performing these tasks in the ﬁrst place! Hence, calculating the

relative errors inherent in your numerical solution or laboratory results will not be

as straightforward. For situations concerning numerical methods that are solved

iteratively, the common method to assess the relative error in your solution is to

compare the approximations obtained from two successive iterations. The relative

error is then calculated as follows:

iþ1

 m

iþ1

(1:4)

where i is the iteration count and m is the approxim ation of the solution that is being

sought. With every successive iteration, the relative error is expected to decrease,

1.3 Methods used to measure error

thereby promising an incremental improvement in the accuracy of the calculated

solution.

The importance of having different deﬁnitions for error is that you can choose

which one to use for measuring and subsequently controlling or limiting the amount

of error generated by your computer program or your calculations. As the magni-

tude of the measured value |m| moves away from 1, the relative error will be

increasingly different from the absolute error, and for this reason relative error

measurements are often preferred for gauging the accuracy of the result.

1.4 Significant digits

The “signiﬁcance” or “qu ality” of a numeric approximation is dependent on the

number of digits that are signiﬁcant. The number of signiﬁcant digits provides a

measure of how accurate the approximation is compared to its true value. For

example, suppose that you use a cell counting device to calculate the number of

bacterial cells in a suspended culture medium and arrive at the number 1 097 456.

Box 1.4 Measurement of cells in a cell-counting device

Suppose that you are working in a clinical laboratory and you need to measure the number of stem cells

present in two test tube solutions. From each test tube you make ten samples and run each of the ten

samples obtained from the same test tube through a cell-counter device. The results are shown in

Table 1.3.

Test tube 1 Test tube 2

Mean count of ten samples 20 890 4 526 750

Maximum count 21 090 4 572 007

Minimum count 20 700 4 481 490

Difference (max. – min.) 390 90 517

At first glance, you might question the results obtained for test tube 2. You interpret the absolute

difference in the minimum and maximum counts as the range of error possible, and that difference for

test tube 2 is approximately 206 times the count difference calculated for test tube 1! You assume that

the mean count is the true value of, or best approximation to, the actual stem cell count in the test tube

solution, and you continue to do some number crunching.

You find that the maximum counts for both test tubes are not more than 1% larger than the values of

the mean. For example, for the case of test tube 1: 1% of the mean count is 20 890 × 0.01 = 208.9, and

the maximum count (21 090) – mean count (20 890) = 200.

For test tube 2, you calculate

ðmaximum count  mean countÞ

mean Count

4 572 007  4 526 750

4 526 750

¼ 0:01:

The minimum counts for both test tubes are within 1% of their respective mean values. Thus, the

accuracy of an individual cell count appears to be within ±1% of the true cell count. You also look up

the equipment manual to find a mention of the same accuracy attainable with this device. Thus, despite

the stark differences in the absolute deviations of cell count values from their mean, the accuracy to

which the cell count is determined is the same for both cases.

Types and sources of numerical error

However, if the actual cell count is 1 030 104, then the number of signiﬁcant digits in

the measured value is not seven. The measurement is not accurate due to various

experimental and other errors. At least ﬁve of the seven digits in the num ber

1 097 456 are in error. Clearly, this measured value of bacterial count onl y gives

you an idea of the order of magnitude of the quantity of cells in your sample. The

number of signiﬁcant ﬁgures in an approximation is not determined by the countable

number of digits, but depends on the relative error in the approximation. The digit 0

is not counted as a signiﬁcant ﬁgure in cases where it is used for the purpose of

positioning the decimal point or when used for ﬁlli ng in one or more blank positions

of unknown digits.

The method to determine the number of signiﬁcant digits, i.e. the number of correct

digits, is related to our earlier discussion on the method of “rounding” ﬂoating-point

numbers. A number with s signiﬁcant digits is said to be correct to s ﬁgures, and

therefore, in absolute terms, has an error of within ±0.5 units of the position of the sth

ﬁgure in the number. In order to calculate the number of signiﬁcant digits in a number

, ﬁrst we determine the relative error between the approximation m

and its true

value m. Then, we determine the value of s,wheres is the largest non-negative integer

that fulﬁls the following condition (proof given in Scarborough, 1966):

 mj

jmj

 0:5  10

s

; (1:5)

where s is the number of signiﬁcant digits, known with absolute certainty, with which

approximates m .

Example 1.5 Number of significant digits in rounded numbers

You are given a number m = 0.014 682 49. Round this number to

(a) four decimal digit places after the decimal point,

(b) six decimal digits after the decimal point.

Determine the number of significant digits present in your rounded number for both cases.

The rounded number m

for (a) is 0.0147 and for (b) is 0.014 682. The number of significant digits

in (a) is

j0:0147  0:01468249j

j0:01468249j

¼ 0:0012 ¼ 0:12  10

2

0:5  10

2

The number of significant digits is therefore two. As you may notice, the preceding zero(s) to the left of the

digits “147” are not considered significant since they do not establish any additional precision in the

approximation. They only provide information on the order of magnitude of the number or the size of the

exponent multiplier.

Suppose that we did not round the number in (a) and instead chopped off the remaining digits to arrive at

the number 0.0146. The relative error is then 0.562 × 10

–2

< 0.5 × 10

–1

, and the number of significant

digits is one, one less than in the rounded case above. Thus, rounding improves the accuracy of the

approximation.

For (b) we have

j0:014682  0:01468249j=j0:01468249j¼0:334  10

4

0:5  10

4

The number of significant digits in the second number is four.

When performing arithmetic operations on two or more quantities that have differing

numbers of signiﬁcant digits in their numeric representation, you will need to determine

1.4 Significant digits

the acceptable number of signiﬁcant digits that can be retained in your ﬁnal result.

Speciﬁc rules govern the number of signiﬁcant digits that should be retained in the ﬁnal

computed result. The following discussion considers the cases of (i) addition and

subtraction of several numbers and (ii) multiplication and division of two numbers.

Addition and subtraction

When adding or subtracting two or more numbers, all numbers being added or

subtracted need to be adjusted so that they are multiplied by the same power of 10.

Next, choose a number (operand) being added or subtracted whose last signiﬁcant

digit lies in the leftmost position as compared to the position of the last signiﬁcant digit

in all the other numbers (operands). The position of the last or rightmost signiﬁcant

digit of the ﬁnal computed result corresponds to the position of the last or rightmost

signiﬁcant digit of the operand as chosen from the above step. For example:

(a) 10.343+4.56743+

0.62 =15.53043. We can retain only two digits after the decimal

point since the last number being added, 0.62 (shaded), has the last signiﬁcant digit in

the second position after the decimal point. Hence the ﬁnal result is 15.53.

(b) 0.0000345+ 23.56 × 10

−4

+8.12 ×10

−5

=0.345×10

−4

+ 23.56  10

−4

+ 0.812 ×

−4

=24.717×10

−4

. The shaded number dictates the number of digits that can be

retained in the result. After rounding, the ﬁnal result is 24.72× 10

−4

(c)

0.23 − 0.1235=0.1065. The shaded number determines the number of signi ﬁcant

digits in the ﬁnal and correct computed value, which is 0.11, after rounding.

Multiplication and division

When multiplying or dividing two or more numbers, choose the operand that has the

least number of signiﬁcant digits. The number of signiﬁcant digits in the computed

result corresponds to the number of signiﬁcant digits in the chosen ope rand from the

above step. For example:

(a)

1.23 × 0.3045 = 0.374 535. The shaded number has the smallest number of signiﬁ-

cant digits and therefore governs the number of signiﬁcant digits present in the

result. The ﬁnal result based on the correct number of signi ﬁcant digits is 0.375, after

rounding appropriately.

(b) 0.00 234/

1.2 = 0.001 95. The correct result is 0.0020. Note that if the denominator is

written as 1.20, then this would imply that the denominat or is precisely known to

three signiﬁcant ﬁgures.

(c)

301/0.045 45 = 6622.662 266. The correct result is 6620.

1.5 Round-off errors generated by floating-point operations

So far we have learned that round-off errors are produced when real numbers are

represented as binary ﬂoating-point numbers by computers, and that the ﬁnite

available memory resources of digital machines limit the precision and range of

the numeric representation. Round-off errors resulting from (1) lack of perfec t

precision in the ﬂoating-point representation of numbers and (2) arithmetic manip-

ulation of ﬂoating-point numbers, are usually insigniﬁcant and ﬁt well within the

tolerance limits applicable to a majority of engineering calculations. However, due

to precision limitations, certain arithmetic operations performed by a computer can

Types and sources of numerical error

produce large round-off errors that may either create problems during the evalua-

tion of a mathematical expression, or generate erroneous results. Round-off errors

are generated during a mathematical operation when there is a loss of signiﬁcant

digits, or the absolute error in the ﬁnal result is large. For example, if two numbers of

considerably disparate magnitudes are added, or two nearly equal numbers are

subtracted from each other, there may be signiﬁcant loss of information and loss

of precision incurred in the process.

Example 1.6 Floating-point operations produce round-off errors

The following arithmetic calculations reveal the cause of spontaneous generation of round-off errors.

First we type the following command in the Command Window:

format long

This command instructs MATLAB to display all floating-point numbers output to the Command Window

with 15 decimal digits after the decimal point. The command format short instructs MATLAB to

display four decimal digits after the decimal point for every floating-point number and is the default

output display in MATLAB.

(a) Inexact representation of a floating-point number

If the following is typed into the MATLAB Command Window

a = 32.8

MATLAB outputs the stored value of the variable a as

32.799999999999997

Note that the echoing of the stored value of a variable in the MATLAB Command Window can be

suppressed by adding a semi-colon to the end of the statement.

(b) Inexact division of one floating-po int number by another

Typing the following statements into the Command Window:

a = 33.0

b = 1.1

a/b

produces the following output:

1.100000000000000

ans =

29.999999999999996

a = 1e10

b =1e-10

c = a+b

1.5 Round-off errors due to floating-point arithmetic

typed into the Command Window produces the following output:

1.000000000000000e+010

1.000000000000000e-010

1.000000000000000e+010

Evidently, there is no difference in the value of a and c despite the addition of b to a.

(d) Loss of a significant digit due to subtraction of one number from another

Typing the following series of statements

a = 10/3

b=a– 3.33

c = b*1000

produces this output:

3.333333333333334

0.003333333333333

3.333333333333410

If the value of c were exact, it would have the digit 3 in all decimal places. The appearance of a 0 in the

last decimal position indicates the loss of a significant digit.

Example 1.7 Calculation of round-off errors in arithmetic operations

A particular computer has memory constraints that allow every floating-point number to be represented by

a maximum of only four significant digits. In the following, the absolute and relative errors stemming from

four-digit floating-point arithmetic are calculated.

(a) Arithmetic operations involving two numbers of disparate values

(i) 4000 + 3/2 = 4002. Absolute error = 0.5; relative error = 1.25 × 10

–4

. The absolute error is O(1), but the

relative error is small.

(ii) 4000 – 3/2 = 3998. Absolute error = 0.5; relative error = 1.25 × 10

–4

(b) Arithmetic operations involving two numbers close in value

(i) 5.0/7 + 4.999/7 = 1.428. Absolute error = 4.29 × 10

–4

; relative error = 3 × 10

–4

(ii) 5.0/7 – 4.999/7 = 0.7143 – 0.7141 = 0.0002. Absolute error = 5.714 × 10

–5

; relative error = 0.4 (or

40%)! There is a loss in the number of significant digits during the subtraction process and hence a

considerable loss in the accuracy of the result. This round-off error manifests from subtractive cancel-

lation of significant digits. This occurrence is termed as loss of significance.

Example 1.7 demonstrates that even a single arithmetic operation can result in

either a large absolute error or a large relative error due to round-off. Addition of

two numbers of largely disparate values or subtraction of two numbers close in

value can cause arithmetic inaccuracy. If the result from subtraction of two nearly

equal numbers is multiplied by a large number, great inaccuracies c an result in

Types and sources of numerical error

subsequent computations. With knowledge of the types of ﬂoating-point opera-

tions prone to round-off errors, we c an re write mathematical e quations in a

reliable and robust manner to prevent round-off errors from corrupting our

computed results.

Example 1.8 Startling round-off errors found when solving a quadratic equation

Let’s solve the quadratic equation x

þ 49:99x  0:5 ¼ 0.

The analytical solution to the quadratic equation ax

þ bx þ c ¼ 0 is given by

b þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

; (1:6)

b 

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

: (1:7)

The exact answer is obtained by solving these equations (and also from algebraic manipulation of the

quadratic equation), and is x

= 0.01; x

= –50.

Now we solve this equation using floating-point numbers containing no more than four significant digits.

The frugal retainment of significant digits in these calculations is employed to emphasize the effects of

round-off on the accuracy of the result.

Now,

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

= 50.01 exactly. There is no approximation involved here. Thus,

49:99 þ 50:01

¼ 0:01;

49:99  50:01

¼50 :

We just obtained the exact solution.

Now let’s see what happens if the equation is changed slightly. Let’s solve x

þ 50x  0:05 ¼ 0.We

have

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

= 50.00, using rounding of the last significant digit, yielding

50 þ 50: 00

¼ 0:000;

50  50: 00

¼50:00:

The true solution of this quadratic equation to eight decimal digits is x

= 0.000 999 98 and x

–50.000 999 98. The percentage relative error in determining x

is given by

0:000 999 98  0

0:000 999 8

 100 ¼ 100%ð!Þ;

and the error in determining x

ð50:00099998  50:00Þ

50:00099998

 100 ¼ 0:001999  0:002%:

The relative error in the result for x

is obviously unacceptable, even though the absolute error is still small.

Is there another way to solve quadratic equations such that we can confidently obtain a small relative error

in the result? Look at the calculation for x

closely and you will notice that two numbers of exceedingly

close values are being subtracted from each other. As discussed earlier in Section 1.5, this naturally has

disastrous consequences on the result. If we prevent this subtraction step from occurring, we can obviate

the problem of subtractive cancellation.

When b is a positive number, Equation (1.6) will always result in a subtraction of two numbers in the

numerator. We can rewrite Equation (1.6)as

1.5 Round-off errors due to floating-point arithmetic

b þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac



b þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

b þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

: (1:8)

Using the algebraic identity a þ b

ðÞ

 a  b

ðÞ

¼ a

 b

to simplify the numerator, we evaluate

Equation (1.8) as

2c

b þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

: (1:9)

The denominator of Equation (1.9) involves the addition of two positive numbers. We have therefore

successfully replaced the subtraction step in Equation (1.6) with an addition step. Using Equation (1.9) to

obtain x

, we obtain

2 0:05

50 þ 50:00

¼ 0:001:

The percentage relative error for the value of x

is 0.002%.

Example 1.8 illustrates one method of algebraic manipulation of equations to

circumvent round-off errors spawning from subtractive cancellation. Problem 1.5

at the end of this chapter uses another technique of algebraic manipulation for

reducing round-off errors when working with polynomials.

Box 1.5 Problem encountered when calculating compound interest

You sold some stock of Biotektronics (see Box 1.3) and had a windfall. You purchased the stocks when

they had been newly issued and you decided to part with them after the company enjoyed a huge

success after promoting their artificial knee joint. Now you want to invest the proceeds of the sale in a

safe place. There is a local bank that has an offer for ten-year cer tificates of deposit (CD). The scheme is

as follows:

(a) Deposit x dollars greater than $1000.00 for ten years and enjoy a return of 10% simple interest for

the period of investment

(b) Deposit x dollars greater than $1000.00 for ten years and enjoy a return of 10% (over the period of

investment of ten years) compounded annually, with a $100.00 fee for compounding

investment of ten years) compounded semi-annually or a higher frequency with a $500.00 fee for

compounding.

You realize that to determine the best choice for your $250 000 investment you will have to do some

serious math. You will only do business with the bank once you have done your due diligence. The final

value of the deposited amount at maturity is calculated as follows:

P 1 þ

100n



;

where P is the initial principal or deposit made, r is the rate of interest per annum in percent, and n is

based on the frequency of compounding over ten years.

Choice (c) looks interesting. Since you are familiar with the power of compounding, you know that a

greater frequency in compounding leads to a higher yield. Now it is time to outsmart the bank. You

decide that, since the frequency of compounding does not have a defined lower limit, you have the

authority to make this choice. What options do you have? Daily? Every minute? Every second? Table 1.4

displays the compounding frequency for various compounding options.

Types and sources of numerical error

A MATLAB program is written to calculate the amount that will accrue over the next ten years.

Table 1.4.

Compounding frequency n

Semi-annually 20

Monthly 120

Daily 3650

Per hour 87 600

Per minute 5.256 × 10

Per second 3.1536 × 10

Per millisecond 3.1536 × 10

Per microsecond 3.1536 × 10

Table 1.5.

Compounding frequency n Maturity amount ($)

Semi-annually 20 276 223.89

Monthly 120 276 281.22

Daily 3650 276 292.35

Per hour 87 600 276 292.71

Per minute 5.256 × 10

276 292.73

Per second 3.1536 × 10

276 292.73

Per millisecond 3.1536 × 10

276 291.14

Per microsecond 3.1536 × 10

268 133.48

MATLAB program 1.2

% This program calculates the maturity value of principal P deposited in a

% CD for 10 years compounded n times

% Variables

p = 250000; % Principal

r = 10; % Interest rate

n = [20 120 3650 87600 5.256e6 3.1536e8 3.1536e11 3.1536e14 3.1536e17];

% Frequency of compounding

% Calculation of ﬁnal accrued amount at maturity

maturityamount = p*(1+r/100./n).^n;

fprintf(‘%7.2f\n’,maturityamount)

In the MATLAB code above, the ./ and .^ operators (dot preceding the arithmetic operator) instruct

MATLAB to perform element-wise operations on the vector n instead of matrix operations such as matrix

division and matrix exponentiation.

The results of MATLAB program 1.2 are presented in Table 1.5. You go down the column to observe

the effect of compounding frequency and notice that the maturity amount is the same for compounding

1.5 Round-off errors due to floating-point arithmetic

How do round-off errors propagate in arithmetic computations? The error in the

individual quantities is found to be additive for the arithmetic operations of addi-

tion, subtraction, and multiplication. For example, if a and b are two numbers to be

added having errors of Δa and Δb, respectively, the sum will be (a + b)+(Δa + Δb),

with the error in the sum being (Δa + Δb). The result of a calculation involving

a number that has error E , raised to the power of a, will have approximately an

error of aE.

If a number that has a small error well within tolerance limits is fed into a series

of arithmetic operations that produce small deviations in the ﬁnal calculated value of

the same order of magnitude as the initial error or less, the numerical algorithm or

method is said to be stable. In such algorithms, the initial errors or variations in

numerical values are kept in check during the calculations and the growth of error is

either linear or diminished. However, if a slight change in the initial values produces

a large change in the result, this indicates that the initial error is magniﬁed through

the course of successive computations. This can yield results that may not be of much

use. Such algorithms or numerical systems are unstable.InChapter 2, we discuss the

source of ill-conditioning (instability) in systems of linear equations.

1.6 Taylor series and truncation error

You will be well aware that functions such as sin x, cos x, e

, and log x can be

represented as a series of inﬁnite terms. For example,

¼ 1 þ

þ;  ∞

∞: (1:10)

The series shown above is known as the Taylor expan sion of the function e

. The

Taylor series is an inﬁnite series representation of a differentiable function f (x). The

series expansion is made about a point x

at which the value of f is known. The terms

in the series are progressively higher-order derivatives of fx

ðÞ. The Taylor expan-

sion in one variable is shown below:

fxðÞ¼fx

ðÞþf

ðÞxx

ðÞþ

ðÞx  x

ðÞ

þþ

ðÞx  x

ðÞ

þ;

(1:11)

every minute versus compounding every second.

Strangely, the maturity amount drops with even larger

compounding frequencies. This cannot be correct.

This error occurs due to the large value of n and therefore the smallness of r=100n. For the last two

scenarios, r=100n = 3.171 × 10

–13

and 3.171 × 10

–16

, respectively. Since floating-point numbers

have at most 16 significant digits, adding two numbers that have a magnitude difference of 10

results

in the loss of significant digits or information as described in this section. To illustrate this point fur ther,

if the calculation were performed for a compounding frequency of every nanosecond, the maturity

amount is calcuated by MATLAB to be $250 000 – the principal amount! This is the result of 1+

ðr=100nÞ = 1, as per floating-point addition. We have encountered a limit imposed by the finite

precision in floating-point calculations.

The compound-interest problem spurred the discovery of the constant e. Jacob Bernoulli (1654–1705)

noticed that compound interest approaches a limit as n → ∞; lim

n!∞

1 þ



, when expanded using the

binomial theorem, produces the Taylor series expansion for e

(see Section 1.6 for an explanation on the

Taylor series). For starting principal P, continuous compounding (maximum frequency of compound-

ing) at rate r per annum will yield $ Pe

at the end of one year.

Types and sources of numerical error