HOW TO PASS ENGR 391 - Numerical Methods in Engineering Past finals-merged solution new update versi

HOW TO PASS ENGR 391 - Numerical Methods in Engineering Past finals-merged solution new update versions Concordia University

Home ► My courses ► ENGR-391-2201-AA ► Topic 4 ► Final exam

Started on State Completed on Time taken Grade

Tuesday, 23 June, 7:00 PM Finished Tuesday, 23 June, 9:59 PM 2 hours 59 mins 25.33 out of 35.00 (72%)

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Question 1 Complete Not graded

By checking the "yes" answer below I confirm 1. that I have neither given nor received unauthorized aid to answer the questions of this assignment.

2. I agree to follow the rules in regard of online assignments as they are posted in the announcement forum of this class and sent to me by email via moodle Select one: a. Yes I agree b. No I do not agree

Your answer is correct.

The correct answer is: Yes I agree

Question 2

A first order initial value problem is solved using Euler’s method.

Correct

If we reduce the step size h = xi+1 − xi to half of it’s previous value, how do you expect the local and global truncation errors of the solution to change?

Mark 1.00 out of 1.00

Select one:

a. local and global errors will be about a quarter of their previous values b. local and global errors will be about a half of their previous values c. local and global error will be a quarter and a half of their previous values, respectively d. local and global errors will be a half and a quarter of their previous values, respectively

Your answer is correct.

The correct answer is: local and global error will be a quarter and a half of their previous values, respectively

Question 3

We want to approximate the defined integral I = ∫ 5 f(x)dx.

Incorrect

When using the trapezoidal rule in its single version we get I ≃ 153.7.

Mark 0.00 out of

Knowing that f(4) = 24.2, how much will be the approximation of I if we use the composite trapezoidal rule with m = 2 sub-intervals ?

1.00

3

Select one: a. Not enough information to answer the question b. None of them c. 101.05 d. 104.2 e. 94.9 f. 145.6

Your answer is incorrect.

We have that

f(3)+f(5)

(5 − 3) = f(3) + f(5) =

Using the composite trapezoidal rule with m = 2 we have (h = 5−3 = 1):

I ≃ 1 [f(3) + f(5) + 2 ⋅ f(4)] = The correct answer is: 101.05

Question 4

The numerical stability of a method for solving initial value problems depends on

Correct Mark 1.00 out of 1.00

Select one: a. only the numerical method b. only on the number of significant digits used c. both the numerical method and the differential equation d. only the differential equation

Your answer is correct.

It depends on the differential equation and as well on the chosen method (different methods can lead to stable or unstable solutions for a same differential equation). It depends as well on how the algorithm is used (for example which step size is chosen.

The correct answer is: both the numerical method and the differential equation

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Question 5

The Lagrange polynomial that passes through the 3 data points is given by

Incorrect Mark 0.00 out of

yi | 9.2 | 9.9 | 2.3

1.00

How much is the value of L1 (x) in x = 7.6 ?

Give at least 4 significant figures Answer: 1.067

L1 (x) is given by

Reference: lecture on Lagrange interpolation of topic 4 "Regression and interpolation"

The correct answer is: 3.977

Question 6

Consider the linear equation ax = b with solution r = ba .

Correct

How much is the maximal relative error magnification factor for this equation for the approximation xr of the solution r?

Mark 1.00 out of 1.00

Select one: a. b. |axr| c. 1 d.

Your answer is correct.

Written in matrix form the system writes: [a]x = [b]. Here the coefficient matrix A is a 1x1 matrix.

The maximal error magnification factor is given by the conditioning number of the coefficient matrix A. In our case the coefficient matrix A is [a].

The correct answer is: 1

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Question 7

Choose the correct statement

Correct Mark 1.00 out of 1.00

Select one: a. The order of convergence for the bisection method is 1 2

b. For a given equation, every algorithm will converge eventually but with different orders of convergence c. The asymptotic error constant

λ for the newton method is − 1

d. The fixed-point method converges linearly

Your answer is correct.

The correct answer is: The fixed-point method converges linearly

Question 8 Correct Mark 1.00 out of

Consider the following data set: x | 1.0| 2.0| 5.0| 7.0 y | 2.1| 2.9| 6.1| 8.3

1.00

The linear least square fitting will give the following model: Select one:

Your answer is correct.

Reference: lectures on least square regression of topic 4 The correct answer is:

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Question 9

Consider the following initial value problem

Correct Mark 1.00 out of 1.00

with What is

when using Euler's method with a step size

Select one: a. none of them b.

c. d. e.

Your answer is correct.

The correct answer is:

Question 10

Which statement is correct regarding curve fitting and interpolation

Incorrect Mark 0.00 out of 1.00

Select one: a. more data points ensure better interpolation b. interpolation is more accurate than least-squares regression, because interpolation passes exactly by all data points

c. given

data points, when calculating a Lagrange interpolating polynomial, we

will find a

th order polynomial

d. Gauss Quadrature is always the best interpolation method

Your answer is incorrect.

The correct answer is: given polynomial, we will find a

data points, when calculating a Lagrange interpolating th order polynomial

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Question 11 Correct Mark 1.00 out of 1.00

The error that arises when only the first few terms of a series expansion are considered for calculations is

Select one: a. a relative error b. an absolute error c. a round-off error d. a truncation error

Your answer is correct.

Reference: lecture on truncation errors of topic 1 The correct answer is: a truncation error

Question 12

Which statement is true for solving a nonlinear equation by numerical methods

Correct Mark 1.00 out of 1.00

Select one: a. Fixed-point iteration method always converges b. Newton’s method needs the value of the 1 -order derivative c. Bisection is an open method d. False-position method is the same as the Fixed-Point method

Your answer is correct.

The correct answer is: Newton’s method needs the value of the 1 -order derivative

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Question 13

Select the correct statement:

Correct Mark 1.00 out of 1.00

Select one: a. Round-off errors can be avoided when using exact algorithms (algorithms involving no mathematical approximations) b. All algorithms will contain truncation errors c. Using more significant digits in our calculations will reduce the truncation errors d. Truncation errors can be reduced using algorithms based on higher order Taylor series expansions and round-off errors can be reduced by caring forward more digits during the calculations

Your answer is correct.

The correct answer is: Truncation errors can be reduced using algorithms based on higher order Taylor series expansions and round-off errors can be reduced by caring forward more digits during the calculations

Question 14 Incorrect Mark 0.00 out of 1.00

In composite Newton-Cotes quadrature methods, which type of error increases with increasing number of sub-divisions?

Select one or more:

a. Round-off errors b. Depends on the method c. None. Errors only decrease d. Truncation errors e. Programming errors

Your answer is incorrect.

Reference: lecture on order of convergence of topic 7 The correct answer is: Round-off errors

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Question 15 Correct Mark 1.00 out of 1.00

Four estimates of an integral are calculated using trapezoidal method with 2, 4, 8 and 16 segments.

What is the best order of accuracy we can reach by applying Romberg method to these estimates?

Select one: a.

b.

c.

d.

Your answer is correct.

Reference: Lecture on Romberg integration from topic 7 The correct answer is:

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Question 16 Correct

You have to solve a system of equations Ax=b. An answer with a relative error smaller than 10 is required.

Mark 1.00 out of

You use the PA=LU decomposition algorithm.

1.00

Knowing that cond(A)

, how many significant digits at least you need to carry

forward in your calculations?

Select one: a. Doesn't matter, as PA=LU is an exact algorithm (no approximations). As long we use all digits from the system we solve we always get an exact answer. b. 6 c. 11

d. 16 e. 5

Your answer is correct.

As cond(A)

, it implies that we will lose about 5 digits when solving the system

due to round-off errors. To reach the set precision at least 5+6=11 digits have to be carried forward in calculations.

Reference: Lecture on applications of the matrix condition number of topic 3

The correct answer is: 11

Question 17 Correct

Find an approximation of a root with a relative error below

of the following

equation

Mark 1.00 out of 1.00

Use one of the methods you learned in the course Answer: -4.37602432

To verify that your approximation

is indeed below the asked tolerance you need to

estimate the relative error of your approximation using

below

and check if it is

or not.

The correct answer is: -0.360817777

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Question 18

We want to find an approximation

of a root

of the following functions

with

Correct Mark 2.00 out of 2.00

a relative error

less than

.

Choose the ones for which the proposed approximation tolerance

satisfies the above specified

Select one or more: a.

and and

b.

c.

and

d.

and

e.

and

Your answer is correct.

One has to check if the estimated relative error

is below

or

not.

The correct answers are:

and , and ,

and

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Question 19

Consider the following system

Correct Mark 0.50 out of 0.50

Solve this system using LU decomposition where

The component

of the vector

has the form

obtained by solving

is equal to

obtained by solving

is equal to

obtained by solving

is equal to

Answer: 47

The correct answer is: 47

Question 20

The component

of the vector

Correct Mark 0.50 out of

Answer: 37

0.50

The correct answer is: 37

Question 21

The component

of the vector

Correct Mark 0.50 out of

Answer: 42

0.50

The correct answer is: 42

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Question 22

The component

of the vector

obtained by solving

is equal to

obtained by solving

is equal to

obtained by solving

is equal to

Correct Mark 0.50 out of

Answer: 3

0.50

The correct answer is: 3

Question 23

The component

of the vector

Correct Mark 0.50 out of

Answer: -7

0.50

The correct answer is: -7

Question 24

The component

of the vector

Correct Mark 0.50 out of

Answer: 6

0.50

The correct answer is: 6

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Question 25 Incorrect Mark 0.00 out of

The attached table (download it as excel sheet) lists the cumulative cases increases of COVID19 infections in Quebec until end of May 2020.

and daily

Use least square regression to adjust the model

.

1.00

Use at least 12 significant figures in your calculations and report the results for the coefficients and with at least 5 significant figures Coefficient Answer: 8257.05

Define the column vector vector

, the vector of the daily increases and the column

, the vector of the cumulative cases.

We apply the technique of normal equations. The coefficient matrix in octave:

can be built like this

>> A = [x x.^2] And the model parameters are found by solving the normal equations

>> A'*A\A'*y 0.0720312853 -0.0000012734

The correct answer is: 0.072031

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Question 26

Coefficient

Incorrect Mark 0.00 out of

Answer: 21.837855

1.00

The correct answer is: -1.2734E-6

Question 27

Compute the root mean square error (RMSE) of the adjusted model

Incorrect Mark 0.00 out of

Use at least 12 signification figures in your calculations and report the result with at least 3 significant figures

1.00

RMSE Answer: 361759855

The RMSE can be computed from the residuals. In octave that can be done by (if a is the column vector with the model parameters and x and y the vectors as defined previously) >> norm(A*a-y)/sqrt(length(x)) ans = 223.70

The correct answer is: 223

Question 28 Incorrect

According this model, how many total cases (i.e. daily increase drops to zero) are expected to happen in Quebec?

Mark 0.00 out of

Give your answer with 5 significant figures (you can round the result to the nearest integer)

1.00

Answer: 8278.887

For this we compute for which

Two roots:

we will get no increase. So we look for solving

(which is when wave started) and

(which is the end of the wave, we look for) The correct answer is: 56567

/

Question 29

We want to develop a quadrature formula using three interpolating points

Correct

and

,

for the following defined integral of a function

:

Mark 1.00 out of 1.00

.

Determine the coefficients

,

of the quadrature formula. Give your answers

with at minimum two significant figures.

Coefficient Answer: 0.291

The quadrature formula with the three interpolating points

,

and

is:

.

Using Lagrange interpolation the coefficient

is

The correct answer is: 0.29

Question 30

Coefficient

Correct Mark 1.00 out of

Answer: 1.04

1.00

The correct answer is: 1.0

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Question 31

Coefficient

Correct Mark 1.00 out of

Answer: -0.333

1.00

The correct answer is: -0.33

Question 32

Approximate the following integral using Gauss quadrature with three interpolating points

Correct Mark 1.00 out of 1.00

Use at least 12 significant digits in calculations and provide at least 6 significant digits in your answer.

Answer: 6.998678

First step is to make a change of variable such to have the integration interval equal to -1 to +1:

In a second step we apply the Gauss quadrature formula using the integral after the change of variable.

The correct answer is: 6.9986778

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Question 33 Correct

Use Romberg integration approximation

to approximate the following defined

integral

Mark 1.00 out of 1.00

Start Romberg with the trapezoidal rule such that

is the approximation of

with

one sub-interval Use 16 digits in your calculations and provide the answer with at least 7 significant figures.

Answer: 1.3169647

The Romberg table becomes

1.299038106 1.304172724 1.313581341 1.316112542

1.305884263 1.316717547 1.316956276

1.317439765 1.316972191

1.316964769

The first column is computed using the composite trapezoid rule with and 8 subdivisions.

Reference: lecture on Romberg integration of topic 7 The correct answer is: 1.3169647

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Question 34

We want to approximate the following integral using the composite Simpson 1/3 rule:

Partially correct Mark 0.33 out of 2.00

Which step sizes are in the range where the error of the composite Simpson 1/3 rule is essentially only due to truncation error?

Select one or more: a. b. c. d. e.

Your answer is partially correct.

You have correctly selected 2.

We compute the approximation of

for various number

of sub-intervals.

Using Richardson's error formula we can estimate the absolute error and plot it in function of

.

/

From this figure one can read for which step sizes the total error is mainly due to the truncation error The correct answers are:

,

,

/

Question 35 Correct

Compute an approximation of

, the solution in

of the following ordinary

differential equation

Mark 1.00 out of 1.00

with

using the Trapezoid method with

Use 16 significant figures in your calculations and give your answer with at least 6 significant figures

Answer: 3.384015504815365

Applying the trapezoid method leads to (note the "2" in front of the left-hand side of the equation) 0.00000 0.10000 0.20000 0.30000 0.40000 0.50000 0.60000 0.70000 0.80000 0.90000 1.00000

1.00000 1.10525 1.22278 1.35555 1.50755 1.68419 1.89308 2.14528 2.45740 2.85567 3.38402

The correct answer is: 3.38402

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Question 36

Consider the following initial value problem

Correct Mark 2.00 out of 2.00

with We will use the trapezoid method to compute approximations of the solution of this initial value problem.

In all your calculations use 16 significant figures For which values of the step size

can you obtain an approximation of

solution of the initial value problem in

, the

, with an absolute error less than

?

Select one or more: a. b. c. d. e.

Your answer is correct.

We solve the initial value problem wit the trapezoid method suing different step sizes and estimate the error with Richardson's method Plotting this error estimation versus

results in

/

From this plot one can answer the question The correct answers are:

,

/

Home ► My courses ► ENGR-391-2201-AA ► Topic 4 ► Final exam

Started on Tuesday, 23 June 2020, 7:00 PM State Finished Completed on Tuesday, 23 June 2020, 9:59 PM Time taken 2 hours 59 mins Grade 20.00 out of 35.00 (57%) Question 1 Complete Not graded

Your answer is correct.

The correct answer is: Yes I agree

Question 2 Correct

Mark 1.00 out of 1.00

Your answer is correct.

The correct answer is: it has a smaller truncation error

Question 3 Correct Mark 1.00 out of 1.00

Your answer is correct.

Reference: First lectures on algorithms in topic 1 The correct answer is: Numerical algorithms provide you a number. Further tools are needed to validate this number /

Question 4 Incorrect Mark 0.00 out of

1.00

Your answer is incorrect.

It depends on the differential equation and as well on the chosen method (different methods can lead to stable or unstable solutions for a same differential equation). It depends as well on how the algorithm is used (for example which step size is chosen.

The correct answer is: both the numerical method and the differential equation

Question 5 Incorrect Mark 0.00 out of

Your answer is incorrect.

r = 0 is indeed a root of f(x) as f(0) = 0.

We have f ′ (x) = 3x2 cos(x) − 1 and f ′ (0) = −1.

Consequently r = 0 is a single root.

The correct answer is: Of multiplicity 1

/

Question 6 Correct Mark 1.00 out of

1.00

P2(x) = 8.7Lo(x) + 5.4L1 (x) + 5.5L2 (x)

1

1

xo x1

2

Comment: Question has no answer with the generated numbers

Question 7

Let T be the matrix which represents the row operations which adds 3 times row 1 to row 2 (i.e.

Incorrect

r2 ← R2 + 3R1 ).

Mark 0.00 out of

What does the matrix T2 represent as row operations?

1.00

Select one: a. r2 ← 2R2 + 9R1 b. r2 ← R2 + 9R1 c. r2 ← 2R2 + 6R1 d. r2 ← R2 + 6R1

Your answer is incorrect.

T2 will apply twice the row operations "add 3 times row 1 to row 2".

This results in the operations "adds 2x3=6 times row 1 to row 2" Reference : lecture on LU decomposition in topic 3 The correct answer is: r2 ← R2 + 6R1

/

Question 8

∫ 510 f(x)dx is exaclty

Correct Mark 1.00 out of

1.00

−1 1 b. 2.5 ∫ −1 f(2.5x + 7.5)dx 1 c. ∫ −1 f(2.5x + 7.5)dx 1 d. 5 ∫ −1 f(2.5x + 7.5)dx

1

The correct answer is: 2.5 ∫ −1 f(2.5x + 7.5)dx

Question 9 Incorrect Mark 0.00 out of 1.00

dy dt

=t

with y(to) = yo

yo + 3hto + 3h2 yo + 2hto + 2h2

yo − 3hto + 4h2 yo + 3hto + 3h3

Your answer is incorrect.

The correct answer is: yo + 3hto + 3h2

Question 10 Correct Mark 1.00 out of 1.00

Your answer is correct.

The correct answer is: Round off errors can be reduced by increasing the number of significant digits used in calculation

/

Question 11 Incorrect Mark 0.00 out of

1.00

Your answer is incorrect.

The point x = 1.5 is in the middle of the interpolating points x = 1 and x = 2. Consequently, using linear interpolation between these two points, we find the corresponding y value to be 3.7 Reference: lecture on spline interpolation of topic 4 The correct answer is: 3.7

Question 12 Incorrect Mark 0.00 out of 1.00

Your answer is incorrect.

The correct answer is: given n data points, when calculating a Lagrange interpolating polynomial, we will find a (n − 1)th order polynomial

Question 13 Correct Mark 1.00 out of 1.00

Your answer is correct.

In fact the only conclusion we can make about A is that the naive Gauss elimination algorithm will fail. But even if this algorithm will fail, it doesn't imply that a system Ax = b has no solution. It can very well have a solution and still the algorithm will fail.

The correct answer is: none of them /

Question 14 Correct Mark 1.00 out of

Your answer is correct.

Reference: lecture on order of convergence from topic 7 The correct answer is: The result becomes always more accurate than when using a low number of subdivisions

Question 15 Incorrect Mark 0.00 out of 1.00

Your answer is incorrect.

Reference: lecture on challenging problems in topic 2 The correct answer is: results usually in large error magnification factors

/

Question 16

3

Incorrect Mark 0.00 out of

1.00

Knowing that f(4) = 24.2, how much will be the approximation of I if we use the composite trapezoidal rule with

m = 2 sub-intervals ?

We have that

f(3)+f(5)

(5 − 3) = f(3) + f(5) =

Using the composite trapezoidal rule with m = 2 we have (h =

5−3

= 1):

Question 17 Incorrect Mark 0.00 out of 1.00

your approximation using

∣ f(xr) ∣ and check if it is below 10−7 or not. ∣ f ′(xr)⋅xr ∣

/

Question 18 Correct

than 10−5.

Mark 2.00 out of

2.00

f(x) = ex + 2−x + 2 cos x − 6 and xr = 1.82938543 f(x) = (x − 2)2 − ln x and xr = 3.0571034 f(x) = x − 0.5(sin x + cos x) and xr = 0.72287 f(x) = 3x2 − ex and xr = 0.91091849 f(x) = ln(x − 1) + cos(x − 1) and xr = 1.3977486157

∣ f(xr) 1 ∣ is below 10−5 or not. ∣ f ′(xr) xr ∣ x x − The correct answers are: f(x) = e + 2 + 2 cos x − 6 and xr = One has to check if the estimated relative error

Question 19 Correct Mark 0.50 out of 0.50

A

l The component d1 of the vector d obtained by solving Ld = b is equal to

The correct answer is: -16

Question 20

The component d2 of the vector d obtained by solving Ld = b is equal to

Correct Mark 0.50 out of

Answer: -2

0.50

The correct answer is: -2

/

Question 21 Correct Mark 0.50 out of

0.50

The correct answer is: 0

Question 22 Correct Mark 0.50 out of 0.50

The correct answer is: -4

Question 23 Correct Mark 0.50 out of 0.50

The correct answer is: 2

Question 24 Correct Mark 0.50 out of 0.50

The correct answer is: 0

/

Question 25 Incorrect Mark 0.00 out of

1.00

Coefficient ao

-0.0000012734

Question 26

Coefficient a1

Incorrect Mark 0.00 out of

-.0192516702

1.00

The correct answer is: -1.2734E-6

/

Question 27 Incorrect Mark 0.00 out of

1.00

Question 28 Incorrect Mark 0.00 out of 1.00

For this we compute for which N we will get no increase. So we look for solving

0 = aoN + a1N 2 Two roots: N = 0 (which is when wave started) and N = −ao/a1 (which is the end of the wave, we look for) The correct answer is: 56567

Question 29

We want to develop a quadrature formula using three interpolating points x0 = 0, x1 = 0.2 and x2 = 1 for the

Correct Mark 1.00 out of 1.00

Determine the coefficients c0 , c1 c2 of the quadrature formula. Give your answers with at minimum two significant figures.

Coefficient c0

2).

Using Lagrange interpolation the coefficient c0 is

/

Question 30

Coefficient c1

Correct Mark 1.00 out of

1.00

The correct answer is: 1.0

Question 31

Coefficient c2

Correct Mark 1.00 out of 1.00

The correct answer is: 0.29

Question 32

3

Correct Mark 1.00 out of

I=

1.00

∫ x ln x dx 2

1

First step is to make a change of variable such to have the integration interval equal to -1 to +1: 3

I=

1

∫ x ln x dx = ∫ (t + 2) ln(t + 2) dt 2

1

2

−1

In a second step we apply the Gauss quadrature formula using the integral after the change of variable.

The correct answer is: 6.9986778

/

Question 33

2√3

Correct Mark 1.00 out of

1.00

I=

∫

dx

0

1.299038106 1.304172724 1.313581341 1.316112542

1.305884263 1.316717547 1.316956276

1.317439765 1.316972191

1.316964769

The first column is computed using the composite trapezoid rule with m = 1, 2, 4 and 8 subdivisions.

/

Question 34 Incorrect

2 ex−1 sin x

dx

Mark 0.00 out of

2.00

10−2 10−3 10−4 10−5 10−6

b−a 2m

−2

The correct answers are: 10

−3

, 10

1 2m

−4

, 10

/

Question 35 Correct Mark 1.00 out of

dy

2 dt = 1 + y2

1.00

0.00000 0.10000 0.20000 0.30000 0.40000 0.50000 0.60000 0.70000 0.80000 0.90000 1.00000

1.00000 1.10525 1.22278 1.35555 1.50755 1.68419 1.89308 2.14528 2.45740 2.85567 3.38402

/

Question 36 Correct Mark 2.00 out of

2.00

dy = 2(t + 1)y dt

with y(0) = 1

10−4 10−3 10−2

10−1 100

/