Dr.Shoeb_ME212_Lec-2-NUMERICAL METHODS_g

vERY GOOD

1. Introduction to
Numerical
Methods
Numerical Methods (ME 212)
Author:Dr.ShoebAhmed Syed
Papua New Guinea University of Technology
Department of Mechanical Engineering
1

2. SourcesofError

3. Twosourcesofnumericalerror
1)
2)
Roundoff error
Truncationerror
3

4. Round-offError
4

5. RoundoffError
•Causedby representing
approximately
anumber
1
3
0.333333
1.4142...2
5

6. GET THESE RULES FOR ROUNDING OFF
NUMBERS
CASE1:
Inroundingoffnumbers,thelastfigurekeptshouldbeunchangedifthe
firstfiguredroppedislessthan5.Forexample,ifonlyonedecimalis
tobekept,then6.422becomes6.4.
CASE2:
Inroundingoffnumbers,thelastfigurekeptshouldbeincreasedby1if
thefirstfiguredroppedisgreaterthan5.Forexample,ifonlytwo
decimalsaretobekept,then6.4872becomes6.49.Similarly,6.997
becomes7.00.
CASE3:
Inroundingoffnumbers,ifthefirstfiguredroppedis5,andallthefigures
followingthefivearezerooriftherearenofiguresafterthe5,thenthe
lastfigurekeptshouldbeunchangedifthatlastfigureiseven.For
example,ifonlyonedecimalistobekept,then6.6500becomes6.6.
Forexample,ifonlytwodecimalsaretobekept,then7.485becomes
7.48.

7. GET THESE RULES FOR ROUNDING OFF
NUMBERS
CASE4:
Inroundingoffnumbers,ifthefirstfiguredroppedis5,andallthefigures
followingthefivearezerooriftherearenofiguresafterthe5,thenthe
lastfigurekeptshouldbeincreasedby1ifthatlastfigureisodd.For
example,ifonlytwodecimalsaretobekept,then6.755000becomes
6.76.
Forexample,ifonlytwodecimalsaretobekept,8.995becomes9.00.
CASE5:
Inroundingoffnumbers,ifthefirstfiguredroppedis5,andthereareany
figuresfollowingthefivethatarenotzero,thenthelastfigurekept
shouldbeincreasedby1.Forexample,ifonlyonedecimalistobe
kept,then6.6501becomes6.7.
Forexample,ifonlytwodecimalsaretobekept,then7.4852007
becomes7.49.

8. Problemscreatedby roundoff error
•28Americanswerekilled on February25,
1991by an IraqiScud missilein Dhahran,
SaudiArabia.
•The
and
patriot defensesystemfailedtotrack
intercepttheScud.Why?
6

9. Problemwith Patriotmissile
•Clockcycleof1/10secondswas
representedin24-bitfixedpoint
registercreatedanerrorof9.5x
10
-8
seconds.
•Thebatterywasonfor100
consecutivehours,thuscausing
aninaccuracyof
s 3600s
1hr
=9.510
−8
=0.342s
100hr
0.1s
7

10. Problem(cont.)
Theshift calculatedin the rangingsystem
ofthe missilewas 687 meters.
•
•Thetargetwas consideredto beout of
137rangeat
meters.
adistancegreaterthan
8

11. Some disasters caused by
numerical errors?
➢Explosion of the Ariane 5
➢EURO page: Conversion Arithmetic's
➢The Vancouver Stock Exchange
➢Rounding error changes Parliament makeup

12. Explosion of the Ariane 5

13. Explosion of the Ariane 5

14. EURO page: Conversion
Arithmetic's

15. EURO page: Conversion
Arithmetic's

16. The Vancouver Stock Exchange

17. Rounding error changes
Parliament makeup

18. Rounding error changes
Parliament makeup

19. Effectof CarryingSignificant
DigitsinCalculations
9

20. RULESFORSIGNIFICANTFIGURES
1.Allnon-zeronumbersAREsignificant.Thenumber33.2hasTHREEsignificantfigures
becauseallofthedigitspresentarenon-zero.
2.Zerosbetweentwonon-zerodigitsAREsignificant.2051hasFOURsignificantfigures.
Thezeroisbetweena2anda5.
3.LeadingzerosareNOTsignificant.They'renothingmorethan"placeholders."The
number0.54hasonlyTWOsignificantfigures.0.0032alsohasTWOsignificantfigures.
Allofthezerosareleading.
4.TrailingzerostotherightofthedecimalAREsignificant.ThereareFOURsignificant
figuresin92.00.
92.00isdifferentfrom92:ascientistwhomeasures92.00millilitersknowshisvalueto
thenearest1/100thmilliliter;meanwhilehiscolleaguewhomeasured92millilitersonly
knowshisvaluetothenearest1milliliter.It'simportanttounderstandthat"zero"does
notmean"nothing."Zerodenotesactualinformation,justlikeanyothernumber.You
cannottagonzerosthataren'tcertaintobelongthere.

21. RULESFORSIGNIFICANTFIGURES
5.TrailingzerosinawholenumberwiththedecimalshownAREsignificant.Placinga
decimalattheendofanumberisusuallynotdone.Byconvention,however,this
decimalindicatesasignificantzero.Forexample,"540."indicatesthatthetrailingzeroIS
significant;thereareTHREEsignificantfiguresinthisvalue.
6.TrailingzerosinawholenumberwithnodecimalshownareNOTsignificant.Writing
just"540"indicatesthatthezeroisNOTsignificant,andthereareonlyTWOsignificant
figuresinthisvalue.
7.ExactnumbershaveanINFINITEnumberofsignificantfigures.Thisruleappliesto
numbersthataredefinitions.Forexample,1meter=1.00meters=1.0000meters=
1.0000000000000000000meters,etc.
8.Foranumberinscientificnotation:Nx10
x
,alldigitscomprisingNAREsignificantby
thefirst6rules;"10"and"x"areNOTsignificant.5.02x10
4
hasTHREEsignificant
figures:"5.02.""10and"4"arenotsignificant.
Rule8providestheopportunitytochangethenumberofsignificantfiguresinavalueby
manipulatingitsform.Forexample,let'strywriting1100withTHREEsignificantfigures.
Byrule6,1100hasTWOsignificantfigures;itstwotrailingzerosarenotsignificant.Ifwe
addadecimaltotheend,wehave1100.,withFOURsignificantfigures(byrule5.)But
bywritingitinscientificnotation:1.10x10
3
,wecreateaTHREE-significant-figurevalue.

23. Truncationerror
•Errorcausedbytruncatingor
mathematicalapproximating
procedure.
a

24. ExampleofTruncationError
Takingonly a fewtermsofaMaclaurinseriesto
e
x
approximate
2 3
x x
e
x
=1+x+++....................
2!3!
Ifonly 3 termsareused,
x
2
 
x
Error=e −1+x+



Truncation
2!

25. AnotherExampleof Truncation
Error
f(x)xUsingafinite toapproximate
f(x+x)−f(x)
f(x)
x
P
ine
Figure1.Approximatederivativeusing finiteΔx
secant line
tangent l
Q

26. AnotherExampleof Truncation
Error
Usingfinite
integral.
rectanglestoapproximatean
y
90
y =x
2
60
30
0 x
0 1.5 3 4.5 6 7.5 9 10.5 12

27. Example1—Maclaurinseries
e
1.2
Calculatethevalueofwithanabsolute
relativeapproximateerroroflessthan1%.
2 3
=1+1.2+
1.2
+
1.2
e
1.2
+...................
2!3!
6termsarerequired.Howmanyarerequiredtoget
atleast1 significantdigitcorrectinyouranswer?
n
e
1.2
E
a

a%
1 1
2 2.2 1.2 54.545
3 2.92 0.72 24.658
4 3.208 0.288 8.9776
5 3.2944 0.0864 2.6226
6 3.3151 0.020736 0.62550

28. Example2—Differentiation
f(x+x)−f(x)
x
2
f(x)=f(3)Find
and
for usingf(x)
x
x=0.2
f(3+0.2)−f(3)
f
'
(3)=
0.2
f(3.2)−f(3)
2 2
3.2 −3 10.24−9 1.24
0.2
== 6.2= ==
0.2 0.2 0.2
Theactualvalueis
f
'
(x)=2x,f
'
(3)=23=6
Truncationerroristhen,6−6.2=−0.2
Canyou find thetruncationerrorwithx=0.1

29. Example3—Integration
Usetworectanglesofequalwidthto
approximatethearea underthecurvefor
overtheinterval[3,9]x
2
(x)=f
9

3
x
2
dx
y
90
y=x
2
60
30
0 x
0 3 6 9 12

30. Integrationexample(cont.)
Choosinga widthof3, wehave
9
x
2
dx=(x
2
)(6−3)+(x
2
)

3
(9−6)
x=3 x=6
=(3
2
)3+(6
2
)3
=27+108=135
Actualvalueisgivenby
9
x
3

9

9
3
−3
3

x
2
dx= = =234 

3 33

3
Truncationerroristhen
234−135=99
Canyou find thetruncationerrorwith4 rectangles
26
?

31. PropagationofErrors

32. PropagationofErrors
Innumerical methods,the calculations are not
madewithexactnumbers.How dothese
inaccuraciespropagatethroughthecalculations?

33. Example1:
Findthebounds for thepropagationin addingtwo
if one is calculating X+Ywhere
X= 1.5 0.05
Y= 3.4 0.04
Solution
numbers.Forexample
Maximumpossiblevalue of X1.55andY= 3.44=
Maximumpossiblevalue of XY=1.55+ 3.44= 4.99+
possiblevalue of X=1.45and Y= 3.36.Minimum
possiblevalue of X+Y= 1.45+ 3.36= 4.81Minimum
Hence
4.81≤ X + Y≤4.99.

34. Propagationof ErrorsInFormulas
Iff is a functionofseveralvariables
possible value of
X
1,X
2,X
3,.......,X
n−1,
is
X
n
thenthemaximum the errorinf
∂f ∂f ∂f ∂f
∆f≈ ∆X
1+ ∆X
2+.......+ ∆X
n−1+ ∆X
n
∂X
1 ∂X
2 ∂X
n−1 ∂X
n

35. Example2:
squareThestrain inanaxial member
sectionis given by
ofa cross-
F
∈=
Given
h
2
E
F=72±0.9N
h=4±0.1mm
E=70±1.5GPa
Findthe maximumpossible
strain.
error in themeasured

36. Example2:
Solution
72
∈=
(4×10
−3
)
2
(70×10
9
)
64.286×10
−6
64.286µ
=
=
∂∈ ∂∈ ∂∈
∆∈= ∆F+∆h+∆E
∂F ∂h ∂E

37. Example2:
∂∈ 1 ∂∈ ∂∈
−
F
−
2F
= ==
h
2
E∂F
Thus
∆E
h
2
E
2
h
3
E ∂E∂h
1
∆F
2F
∆h
F
∆E= + +
h
2
E h
3
E h
2
E
2
2×72
×0.0001
1
×0.9= +
(4×10
−3
)
2
(70×10
9
) (4×10
−3
)
3
(70×10
9
)
72
×1.5×10
9
+
(4×10
−3
)
2
(70×10
9
)
2
=5.3955µ
Hence
∈=(64.286µ±5.3955µ)

38. Example3:
Subtractionof numbersthat are nearly equalcan create unwanted
inaccuracies.Usingthe formulaforerrorpropagation,showthatthisistrue.
Solution
Let
z=x−y
Then
∂z
∆z=∆x+∆y
∂x
=
=
(1)∆x+
∆y
(−1)∆y
∆x+
Sotherelativechangeis
∆x+∆y∆z
z
=
x−y
∂z
∂y
(−

39. Example3:
Forexample if
x=2±0.001
y=2.003±0.001
0.001+0.001∆z
z
=
|2−2.003|
0.6667
66.67%
=
=

40. MeasuringErrors

41. Whymeasureerrors?
1)Todetermine the accuracyof
numericalresults.
2)Todevelop stoppingcriteriafor
iterativealgorithms.

42. TrueError
◼Definedasthedifferencebetweenthetrue
valuein acalculationandtheapproximate
valuefoundusinganumericalmethodetc.
TrueError=TrueValue–ApproximateValue

43. Example—TrueError
can bef′(x)Thederivative, ofa functionf(x)
approximatedby theequation,
f(x+h)−f(x)
f'(x)≈
h
h=0.3and7e
0.5x
f(x)=
a)
b)
c)
If
Findthe approximatevalueoff'(2)
True
True
value of
error for
f'(2)
part (a)

44. Example(cont.)
Solution:
a)For andx=2 h=0.3
f(2+0.3)−f(2)
f'(2)≈
0.3
f(2.3)−f(2)
=
0.3
7e
0.5(2.3)
−7e
0.5(2)
=
0.3
22.107−19.028
=10.263=
0.3

45. Example(cont.)
Solution:
b)The exactvalue of can be foundbyusingf'(2)
ourknowledge ofdifferentialcalculus.
7e
0.5x
f(x)=
f'(x)=7×0.5×e
0.5x
3.5e
0.5x
=
So thetrue value of is
f'(2)
3.5e
0.5(2)
f'(2)=
=9.5140
Trueerror is calculatedas
E
t =TrueValue –ApproximateValue
=9.5140−10.263=−0.722

46. RelativeTrueError
◼Defined astheratio betweenthetrue
error,and thetrue value.
TrueError
∈
t)=RelativeTrueError(
TrueValue

47. Example—RelativeTrueError
Followingfromthe previousexample fortrue error,
findthe relative true error for atf'(2)7e
0.5x
f(x)=
withh=0.3
Fromtheprevious example,
E
t =−0.722
Relative True Error is definedas
TrueError
∈=
t
TrueValue
−0.722
= =−0.075888
9.5140
as a percentage,
∈
t=−0.075888×100%=−7.5888%

48. ApproximateError
◼Whatcan be done
knownor are very
◼Approximate error
iftrue values are not
difficulttoobtain?
isdefined asthe
difference between the present
approximationand theprevious
approximation.
ApproximateError(E
a)=PresentApproximation–PreviousApproximation

49. Example—ApproximateError
For at findthe following,x=27e
0.5x
f(x)=
f′(2)
f′(2)
a)
b)
c)
using
using
h=0.3
h=0.15
f′(2)approximateerror forthevalueof forpartb)
Solution:
a)For andx=2 h=0.3
f(x+h)−f(x)
f'(x)≈
h
f(2+0.3)−f(2)
f'(2)≈
0.3

50. Example
Solution:(cont.)
f(2.3)−f(2)
(cont.)
=
0.3
7e
0.5(2.3)
−7e
0.5(2)
=
0.3
22.107−19.028
= =10.263
0.3
andb)Forx=2 h=0.15
f(2+0.15)−f(2)
f
'
(2)≈
0.15
f (2.15)−f (2)
=
0.15

51. Example
Solution:(cont.)
(cont.)
7e
0.5(2.15)
−7e
0.5(2)
=
0.15
20.50−19.028
=9.8800=
0.15
c)So
E
a
the approximateerror,isE
a
=PresentApproximation–PreviousApproximation
=9.8800−10.263
=−0.38300

52. RelativeApproximateError
◼Defined astheratio between the
approximate error
approximation.
(∈
a)
and the present
ApproximateError
RelativeApproximateError =
PresentApproximation

53. Example—Relative Approximate Error
7e
0.5x
f(x)=
For at ,findtherelative approximatex=2
errorusingvalues from
Solution:
andh=0.3 h=0.15
f′(2)=10.263From
using
E
a
Example 3,the approximatevalue of
and usingh=0.3 h=0.15f′(2)=9.8800
=PresentApproximation–PreviousApproximation
=9.8800−10.263
=−0.38300

54. Example
Solution:(cont.)
ApproximateError
(cont.)
∈
a=
PresentApproximation
−0.38300
=−0.038765=
9.8800
as a percentage,
∈
a =−0.038765×100%=−3.8765%
Absoluterelative approximateerrors
becalculated,
may alsoneed to
∈
a=|−0.038765|=0.038765or3.8765
%

55. HowisAbsoluteRelative Error usedas
stopping criterion?
a
If whereis a pre-specified tolerance,then|∈
a|≤∈
s ∈
s
nofurtheriterationsare necessaryand the process
stopped.
is
Ifat leastmsignificant digitsarerequiredtobe
correct in the finalanswer,then
0.5×10
2−m
%|∈ |≤
a

56. TableofValues
7e
0.5x
=f(x)
For at withvaryingstepsize,x=2 h
h
f′(2) ∈
a m
0.3 10.263 N/A 0
0.15 9.8800 3.877% 1
0.10 9.7558 1.273% 1
0.01 9.5378 2.285% 1
0.001 9.5164 0.2249% 2

57. 1. http://numericalmethods.eng.usf.edu
2.
http://numericalmethods.eng.usf.edu/topics/sources_of_err
or.html
3. http://numericalmethods.eng.usf.edu/topics/propagatio
n_of_errors.html
4. http://numericalmethods.eng.usf.edu/topics/measuring
_errors.html
References

58. THEEND