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8.10E: Exercises - Mathematics


Practice Makes Perfect

Recognize the Relationship Between the Solutions of an Equation and its Graph

In the following exercises, for each ordered pair, decide:

  1. Is the ordered pair a solution to the equation?
  2. Is the point on the line?

Exercise (PageIndex{1})

y=x+2

  1. (0,2)
  2. (1,2)
  3. (−1,1)
  4. (−3,−1)

Answer
  1. yes; no
  2. no; no
  3. yes; yes
  4. yes; yes

Exercise (PageIndex{2})

y=x−4

  1. (0,−4)
  2. (3,−1)
  3. (2,2)
  4. (1,−5)

Exercise (PageIndex{3})

(y=frac{1}{2} x-3)

  1. (0,−3)
  2. (2,−2)
  3. (−2,−4)
  4. (4,1)

Answer
  1. yes; yes
  2. yes; yes
  3. yes; yes
  4. no; no

Exercise (PageIndex{4})

(y=frac{1}{3} x+2)

  1. (0,2)
  2. (3,3)
  3. (−3,2)
  4. (−6,0)

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

Exercise (PageIndex{5})

(y=3 x-1)

Answer

Exercise (PageIndex{6})

(y=2 x+3)

Exercise (PageIndex{7})

(y=-2 x+2)

Answer

Exercise (PageIndex{8})

(y=-3 x+1)

Exercise (PageIndex{9})

(y=x+2)

Answer

Exercise (PageIndex{10})

(y=x-3)

Exercise (PageIndex{11})

(y=-x-3)

Answer

Exercise (PageIndex{12})

(y=-x-2)

Exercise (PageIndex{13})

(y=2 x)

Answer

Exercise (PageIndex{14})

(y=3 x)

Exercise (PageIndex{15})

(y=-4 x)

Answer

Exercise (PageIndex{16})

(y=-2 x)

Exercise (PageIndex{17})

(y=frac{1}{2} x+2)

Answer

Exercise (PageIndex{18})

(y=frac{1}{3} x-1)

Exercise (PageIndex{19})

(y=frac{4}{3} x-5)

Answer

Exercise (PageIndex{20})

(y=frac{3}{2} x-3)

Exercise (PageIndex{21})

(y=-frac{2}{5} x+1)

Answer

Exercise (PageIndex{22})

(y=-frac{4}{5} x-1)

Exercise (PageIndex{23})

(y=-frac{3}{2} x+2)

Answer

Exercise (PageIndex{24})

(y=-frac{5}{3} x+4)

Exercise (PageIndex{25})

(x+y=6)

Answer

Exercise (PageIndex{26})

(x+y=4)

Exercise (PageIndex{27})

(x+y=-3)

Answer

Exercise (PageIndex{28})

(x+y=-2)

Exercise (PageIndex{29})

(x-y=2)

Answer

Exercise (PageIndex{30})

(x-y=1)

Exercise (PageIndex{31})

(x-y=-1)

Answer

Exercise (PageIndex{32})

(x-y=-3)

Exercise (PageIndex{33})

(3 x+y=7)

Answer

Exercise (PageIndex{34})

(5x+y=6)

Exercise (PageIndex{35})

2x+y=−3

Answer

Exercise (PageIndex{36})

(4x+y=−5)

Exercise (PageIndex{37})

(frac{1}{3} x+y=2)

Answer

Exercise (PageIndex{38})

(frac{1}{2} x+y=3)

Exercise (PageIndex{39})

(frac{2}{5} x-y=4)

Answer

Exercise (PageIndex{40})

(frac{3}{4} x-y=6)

Exercise (PageIndex{41})

(2 x+3 y=12)

Answer

Exercise (PageIndex{42})

4x+2y=12

Exercise (PageIndex{43})

3x−4y=12

Answer

Exercise (PageIndex{44})

2x−5y=10

Exercise (PageIndex{45})

x−6y=3

Answer

Exercise (PageIndex{46})

x−4y=2

Exercise (PageIndex{47})

5x+2y=4

Answer

Exercise (PageIndex{48})

3x+5y=5

Graph Vertical and Horizontal Lines

In the following exercises, graph each equation.

Exercise (PageIndex{49})

x=4

Answer

Exercise (PageIndex{50})

x=3

Exercise (PageIndex{51})

x=−2

Answer

Exercise (PageIndex{52})

x=−5

Exercise (PageIndex{53})

y=3

Answer

Exercise (PageIndex{54})

y=1

Exercise (PageIndex{55})

y=−5

Answer

Exercise (PageIndex{56})

y=−2

Exercise (PageIndex{57})

(x=frac{7}{3})

Answer

Exercise (PageIndex{58})

(x=frac{5}{4})

Exercise (PageIndex{59})

(y=-frac{15}{4})

Answer

Exercise (PageIndex{60})

(y=-frac{5}{3})

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

Exercise (PageIndex{61})

y=2x and y=2

Answer

Exercise (PageIndex{62})

y=5x and y=5

Exercise (PageIndex{63})

(y=-frac{1}{2} x) and (y=-frac{1}{2})

Answer

Exercise (PageIndex{64})

(y=-frac{1}{3} x) and (y=-frac{1}{3})

Mixed Practice

In the following exercises, graph each equation.

Exercise (PageIndex{65})

y=4x

Answer

Exercise (PageIndex{66})

y=2x

Exercise (PageIndex{67})

(y=-frac{1}{2} x+3)

Answer

Exercise (PageIndex{68})

(y=frac{1}{4} x-2)

Exercise (PageIndex{69})

y=−x

Answer

Exercise (PageIndex{70})

y=x

Exercise (PageIndex{71})

x−y=3

Answer

Exercise (PageIndex{72})

x+y=−5

Exercise (PageIndex{73})

4x+y=2

Answer

Exercise (PageIndex{74})

2x+y=6

Exercise (PageIndex{75})

y=−1

Answer

Exercise (PageIndex{76})

y=5

Exercise (PageIndex{77})

2x+6y=12

Answer

Exercise (PageIndex{78})

5x+2y=10

Exercise (PageIndex{79})

x=3

Answer

Exercise (PageIndex{80})

x=−4

Everyday Math

Exercise (PageIndex{81})

Motor home cost. The Robinsons rented a motor home for one week to go on vacation. It cost them $594 plus $0.32 per mile to rent the motor home, so the linear equation y=594+0.32x gives the cost, yy, for driving xx miles. Calculate the rental cost for driving 400, 800, and 1200 miles, and then graph the line.

Answer

$722, $850, $978

Exercise (PageIndex{82})

Weekly earnings. At the art gallery where he works, Salvador gets paid $200 per week plus 15% of the sales he makes, so the equation y=200+0.15x gives the amount, yy, he earns for selling x dollars of artwork. Calculate the amount Salvador earns for selling $900, $1600, and $2000, and then graph the line.

Writing Exercises

Exercise (PageIndex{83})

Explain how you would choose three (x) - values to make a table to graph the line (y=frac{1}{5} x-2)

Answer

Answers will vary.

Exercise (PageIndex{84})

What is the difference between the equations of a vertical and a horizontal line?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all goals?


Works for any number of numbers taken from standard input:

Disclaimer:
Technicaly, this isn't required to work by C++ standard. The minimum iterator category required for minmax_element is ForwardIterator which stream iterators are not. Once an input iterator is dereferenced or incremented, its copies are no longer guaranteed to be dereferenceable or comparable to other iterators. It Works On My Machine TM . :)

You can do something like this:

This reads numbers from standard input until eof (it does not care how many you have - 5 or 1,000,000).

This is not an efficient answer but it still works

you can use similar logic to fid the smallest value

Let max will hold the maximum of 5 numbers. Assign the first number to max. Take the 2nd number and compare it with max if the the 2nd number is greater than max then assign it to max else do nothing. Next take the 3rd number and compare it with max , if the 3rd number is greater than max assign it to max else do nothing. Do the same for 4th and 5th number. Finally max will hold the maximum of 5 number.


How to create Generalized Liner Model (GLM)

  • age: age of the individual. Numeric
  • education: Educational level of the individual. Factor.
  • marital.status: Marital status of the individual. Factor i.e. Never-married, Married-civ-spouse, .
  • gender: Gender of the individual. Factor, i.e. Male or Female
  • income: Target variable. Income above or below 50K. Factor i.e. >50K, <=50K
  • Step 1: Check continuous variables
  • Step 2: Check factor variables
  • Step 3: Feature engineering
  • Step 4: Summary statistic
  • Step 5: Train/test set
  • Step 6: Build the model
  • Step 7: Assess the performance of the model
  • step 8: Improve the model

Your task is to predict which individual will have a revenue higher than 50K.

In this tutorial, each step will be detailed to perform an analysis on a real dataset.

Step 1) Check continuous variables

In the first step, you can see the distribution of the continuous variables.

  • continuous <- select_if(data_adult, is.numeric): Use the function select_if() from the dplyr library to select only the numerical columns
  • summary(continuous): Print the summary statistic

From the above table, you can see that the data have totally different scales and hours.per.weeks has large outliers (.i.e. look at the last quartile and maximum value).

  • 1: Plot the distribution of hours.per.week
  • 2: Standardize the continuous variables
  1. Plot the distribution

Let's look closer at the distribution of hours.per.week

The variable has lots of outliers and not well-defined distribution. You can partially tackle this problem by deleting the top 0.01 percent of the hours per week.

We compute the top 2 percent percentile

98 percent of the population works under 80 hours per week.

You can drop the observations above this threshold. You use the filter from the dplyr library.

You can standardize each column to improve the performance because your data do not have the same scale. You can use the function mutate_if from the dplyr library. The basic syntax is:

You can standardize the numeric columns as follow:

Step 2) Check factor variables

  • Select the categorical columns
  • Store the bar chart of each column in a list
  • Print the graphs

We can select the factor columns with the code below:

  • data.frame(select_if(data_adult, is.factor)): We store the factor columns in factor in a data frame type. The library ggplot2 requires a data frame object.

The dataset contains 6 categorical variables

The second step is more skilled. You want to plot a bar chart for each column in the data frame factor. It is more convenient to automatize the process, especially in situation there are lots of columns.

  • lapply(): Use the function lapply() to pass a function in all the columns of the dataset. You store the output in a list
  • function(x): The function will be processed for each x. Here x is the columns
  • ggplot(factor, aes(get(x))) + geom_bar()+ theme(axis.text.x = element_text(angle = 90)): Create a bar char chart for each x element. Note, to return x as a column, you need to include it inside the get()

The last step is relatively easy. You want to print the 6 graphs.

Note: Use the next button to navigate to the next graph

Step 3) Feature engineering

Recast education

From the graph above, you can see that the variable education has 16 levels. This is substantial, and some levels have a relatively low number of observations. If you want to improve the amount of information you can get from this variable, you can recast it into higher level. Namely, you create larger groups with similar level of education. For instance, low level of education will be converted in dropout. Higher levels of education will be changed to master.

  • We use the verb mutate from dplyr library. We change the values of education with the statement ifelse

In the table below, you create a summary statistic to see, on average, how many years of education (z-value) it takes to reach the Bachelor, Master or PhD.

Recast Marital-status

Step 4) Summary Statistic

It is time to check some statistics about our target variables. In the graph below, you count the percentage of individuals earning more than 50k given their gender.

Next, check if the origin of the individual affects their earning.

The number of hours work by gender.

The box plot confirms that the distribution of working time fits different groups. In the box plot, both genders do not have homogeneous observations.

You can check the density of the weekly working time by type of education. The distributions have many distinct picks. It can probably be explained by the type of contract in the US.

  • ggplot(recast_data, aes( x= hours.per.week)): A density plot only requires one variable
  • geom_density(aes(color = education), alpha =0.5): The geometric object to control the density

To confirm your thoughts, you can perform a one-way ANOVA test:

The ANOVA test confirms the difference in average between groups.

Non-linearity

Before you run the model, you can see if the number of hours worked is related to age.

  • ggplot(recast_data, aes(x = age, y = hours.per.week)): Set the aesthetic of the graph
  • geom_point(aes(color= income), size =0.5): Construct the dot plot
  • stat_smooth(): Add the trend line with the following arguments:
    • method='lm': Plot the fitted value if the linear regression
    • formula = y

    In a nutshell, you can test interaction terms in the model to pick up the non-linearity effect between the weekly working time and other features. It is important to detect under which condition the working time differs.

    Correlation

    The next check is to visualize the correlation between the variables. You convert the factor level type to numeric so that you can plot a heat map containing the coefficient of correlation computed with the Spearman method.

    • data.frame(lapply(recast_data,as.integer)): Convert data to numeric
    • ggcorr() plot the heat map with the following arguments:
      • method: Method to compute the correlation
      • nbreaks = 6: Number of break
      • hjust = 0.8: Control position of the variable name in the plot
      • label = TRUE: Add labels in the center of the windows
      • label_size = 3: Size labels
      • color = "grey50"): Color of the label

      Step 5) Train/test set

      Any supervised machine learning task require to split the data between a train set and a test set. You can use the "function" you created in the other supervised learning tutorials to create a train/test set.

      Step 6) Build the model

      To see how the algorithm performs, you use the glm() package. The Generalized Linear Model is a collection of models. The basic syntax is:

      You are ready to estimate the logistic model to split the income level between a set of features.

      • AIC (Akaike Information Criteria): This is the equivalent of R2 in logistic regression. It measures the fit when a penalty is applied to the number of parameters. Smaller AIC values indicate the model is closer to the truth.
      • Null deviance: Fits the model only with the intercept. The degree of freedom is n-1. We can interpret it as a Chi-square value (fitted value different from the actual value hypothesis testing).
      • Residual Deviance: Model with all the variables. It is also interpreted as a Chi-square hypothesis testing.
      • Number of Fisher Scoring iterations: Number of iterations before converging.

      The output of the glm() function is stored in a list. The code below shows all the items available in the logit variable we constructed to evaluate the logistic regression.

      # The list is very long, print only the first three elements

      Each value can be extracted with the $ sign follow by the name of the metrics. For instance, you stored the model as logit. To extract the AIC criteria, you use:

      Step 7) Assess the performance of the model

      Confusion Matrix

      The confusion matrix is a better choice to evaluate the classification performance compared with the different metrics you saw before. The general idea is to count the number of times True instances are classified are False.

      To compute the confusion matrix, you first need to have a set of predictions so that they can be compared to the actual targets.

      • predict(logit,data_test, type = 'response'): Compute the prediction on the test set. Set type = 'response' to compute the response probability.
      • table(data_test$income, predict > 0.5): Compute the confusion matrix. predict > 0.5 means it returns 1 if the predicted probabilities are above 0.5, else 0.

      Each row in a confusion matrix represents an actual target, while each column represents a predicted target. The first row of this matrix considers the income lower than 50k (the False class): 6241 were correctly classified as individuals with income lower than 50k (True negative), while the remaining one was wrongly classified as above 50k (False positive). The second row considers the income above 50k, the positive class were 1229 (True positive), while the True negative was 1074.

      You can calculate the model accuracy by summing the true positive + true negative over the total observation

      The model appears to suffer from one problem, it overestimates the number of false negatives. This is called the accuracy test paradox. We stated that the accuracy is the ratio of correct predictions to the total number of cases. We can have relatively high accuracy but a useless model. It happens when there is a dominant class. If you look back at the confusion matrix, you can see most of the cases are classified as true negative. Imagine now, the model classified all the classes as negative (i.e. lower than 50k). You would have an accuracy of 75 percent (6718/6718+2257). Your model performs better but struggles to distinguish the true positive with the true negative.

      Precision vs Recall

      Precision looks at the accuracy of the positive prediction. Recall is the ratio of positive instances that are correctly detected by the classifier

      • mat[1,1]: Return the first cell of the first column of the data frame, i.e. the true positive
      • mat[1,2] Return the first cell of the second column of the data frame, i.e. the false positive
      • mat[1,1]: Return the first cell of the first column of the data frame, i.e. the true positive
      • mat[2,1] Return the second cell of the first column of the data frame, i.e. the false negative

      You can test your functions

      When the model says it is an individual above 50k, it is correct in only 54 percent of the case, and can claim individuals above 50k in 72 percent of the case.

      You can create the /> score based on the precision and recall. The /> is a harmonic mean of these two metrics, meaning it gives more weight to the lower values.

      Precision vs Recall tradeoff

      It is impossible to have both a high precision and high recall.

      • Imagine, you need to predict if a patient has a disease. You want to be as precise as possible.
      • If you need to detect potential fraudulent people in the street through facial recognition, it would be better to catch many people labeled as fraudulent even though the precision is low. The police will be able to release the non-fraudulent individual.

      The ROC curve

      The Receiver Operating Characteristic curve is another common tool used with binary classification. It is very similar to the precision/recall curve, but instead of plotting precision versus recall, the ROC curve shows the true positive rate (i.e., recall) against the false positive rate. The false positive rate is the ratio of negative instances that are incorrectly classified as positive. It is equal to one minus the true negative rate. The true negative rate is also called specificity. Hence the ROC curve plots sensitivity (recall) versus 1-specificity

      To plot the ROC curve, we need to install a library called RORC. We can find in the conda library. You can type the code:

      conda install -c r r-rocr --yes

      We can plot the ROC with the prediction() and performance() functions.

      • prediction(predict, data_test$income): The ROCR library needs to create a prediction object to transform the input data
      • performance(ROCRpred, 'tpr','fpr'): Return the two combinations to produce in the graph. Here, tpr and fpr are constructed. Tot plot precision and recall together, use "prec", "rec".

      Step 8) Improve the model

      You need to use the score test to compare both model

      The score is slightly higher than the previous one. You can keep working on the data a try to beat the score.


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      Solutions for Chapter 10.8: Discrete Mathematics and Its Applications 7th Edition

      Solutions for Chapter 10.8

      • 10.8.1E: In Exercises 1–4 construct the dual graph for the map shown. Then f.
      • 10.8.3E: In Exercises 1–4 construct the dual graph for the map shown. Then f.
      • 10.8.2E: In Exercises 1–4 construct the dual graph for the map shown. Then f.
      • 10.8.4E: In Exercises 1–4 construct the dual graph for the map shown. Then f.
      • 10.8.5E: In ExeEcises 5–11 find the chromatic number of the given graph.
      • 10.8.6E: In Exercises 5–11 find the chromatic number of the given graph.
      • 10.8.7E: In Exercises 5–11 find the chromatic number of the given graph.
      • 10.8.8E: In Exercises 5–11 find the chromatic number of the given graph.
      • 10.8.10E: In Exercises 5–11 find the chromatic number of the given graph.
      • 10.8.9E: In Exercises 5–11 find the chromatic number of the given graph.
      • 10.8.11E: In Exercises 5–11 find the chromatic number of the given graph.
      • 10.8.12E: For the graphs in Exercises 5–11, decide whether it is possible to .
      • 10.8.13E: Which graphs have a chromatic number of 1 ?
      • 10.8.17E: Schedule the final exams for Math 115. Math 116. Math 185. Math 195.
      • 10.8.14E: What is the least number of colors needed to color a map of the Uni.
      • 10.8.15E: What is the chromatic number of Wn?
      • 10.8.16E: Show that a simple graph that has a circuit with an odd number of v.
      • 10.8.18E: How many different channels are needed for six stations located at .
      • 10.8.19E: The mathematics department has six committees, each meeting once a .
      • 10.8.20E: A zoo wants to set up natural habitats in which to exhibit its anim.
      • 10.8.21E: An edge coloring of a graph is an assignment of colors to edges so .
      • 10.8.24E: Show that the edge chromatic number of a graph must be at least as .
      • 10.8.25E: Show that if G is a graph with n vertices, then no more than n/2 ed.
      • 10.8.22E: Suppose that n devices are on a circuit board and that these device.
      • 10.8.23E: Find the edge chromatic numbers of
      • 10.8.26E: Find the edge chromatic number of Kn when n is a positive integer.
      • 10.8.27E: Seven variables occur in a loop of a computer program. The variable.
      • 10.8.29E: This algorithm can be used to color a simple graph: First, list the.
      • 10.8.28E: What can be said about the chromatic number of a graph that has Kn .
      • 10.8.30E: Use pseudocode to describe this coloring algorithm.
      • 10.8.31E: Show that the coloring produced by this algorithm may use more colo.
      • 10.8.32E: Show that Cn is chromatically 3-critical whenever n is an odd posit.
      • 10.8.33E: Show that Wn is chromatically 4-critical whenever n is an odd integ.
      • 10.8.34E: Show that W4 is not chromatically 3-critical.
      • 10.8.35E: Show that if G is a chromatically k-critical graph, then the degree.
      • 10.8.36E: Find these values:
      • 10.8.37E: Let G and H be the graphs displayed in Figure 3. Find
      • 10.8.38E: What is ?k(G) if G is a bipartite graph and k is a positive integer?
      • 10.8.39E: Frequencies for mobile radio (or cellular) telephones are assigned .
      • 10.8.40E: Show that every planar graph G can be colored using six or fewer co.
      • 10.8.43E: Show that g(5) = 1. That is, show that all pentagons can be guarded.
      • 10.8.42E: Show that g(3) = 1 and g(4) = 1 by showing that all triangles and q.
      • 10.8.41E: Show that every planar graph G can be colored using five or fewer c.
      • 10.8.44E: Show that g(6) = 2 by first using Exercises 42 and 43 as well as Le.
      • 10.8.45E: Show that . [Hint: Consider the polygon with 3k vertices that resem.
      • 10.8.46E: Solve the art gallery problem by proving the art gallery theorem, w.
      Textbook: Discrete Mathematics and Its Applications
      Edition: 7
      Author: Kenneth Rosen
      ISBN: 9780073383095

      Chapter 10.8 includes 46 full step-by-step solutions. Since 46 problems in chapter 10.8 have been answered, more than 362704 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. Discrete Mathematics and Its Applications was written by and is associated to the ISBN: 9780073383095. This textbook survival guide was created for the textbook: Discrete Mathematics and Its Applications, edition: 7.

      The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + . + cnvn = dl wI + . + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)

      Remove row i and column j multiply the determinant by (-I)i + j •

      S. Permutation with S21 = 1, S32 = 1, . , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1 eigenvectors are columns of the Fourier matrix F.

      Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and

      dim(V) = number of vectors in any basis for V.

      A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.

      If A has full column rank n, then A+ = (AT A)-I AT has A+ A = In.

      A combination other than all Ci = 0 gives L Ci Vi = O.

      Ln = 2,J, 3, 4, . satisfy Ln = L n- l +Ln- 2 = A1 +A

      , with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U

      The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A - AI) if no eigenvalues are repeated always meA) divides peA).

      Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.

      If N NT = NT N, then N has orthonormal (complex) eigenvectors.

      Symmetric matrix with positive eigenvalues and positive pivots. Definition: x T Ax > 0 unless x = O. Then A = LDLT with diag(D» O.

      Those extremes are reached at the eigenvectors x for Amin(A) and Amax(A).

      Each equation gives a plane in Rn the planes intersect at x.

      Combinations of VI, . ,Vm fill the space. The columns of A span C (A)!

      Columns of n by n identity matrix (written i ,j ,k in R3).

      If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.


      Abstract

      This work pertains to the implicit high order compact discretization of the Navier-Stokes (N-S) equations on nonuniform grid. Subsequently, the discretization is used to approximate the Boussinesq equation as well. Contrary to earlier works on nonuniform grids this newly developed scheme is based on a comparatively smaller five-point stencil and leads to an algebraic system of equations with constant coefficients. The scheme carries the flow variable and its gradients as unknown and is seen to report back truncation accuracy of order four for linear flow problems even in a nonuniform mesh. Temporally the scheme is second-order accurate. Both primitive and vorticity-streamfunction formulations have been successfully tackled using the proposed formulation. Verification and validation studies were carried out to establish the efficiency of the formulation in conjunction with both Dirichlet and Neumann boundary conditions. Simulation of interior and exterior flow problems near-critical Hopf bifurcation points using a comparatively lesser number of grid points helps establish the robustness of the scheme. The numerical solution obtained by solving the Boussinesq equation for the problem of natural convection reveals the wider applicability of the scheme involving heat transfer.


      Assessment Tasks

      Weekly quiz

      Due: Weekly
      Weighting: 10%

      • Understanding of the tools and methods that are used in the geosciences these are organised in three modules: o Tools of the geoscientist o Hot rocks o Rocks under stress
      • Competence in applying geo-scientific principles to understanding the world around you
      • Capacity to employ appropriate geo-scientific tools to solve problems and to interpret the results
      • Understanding scientific methodology
      • Competence in accessing, using and synthesising appropriate information
      • Application of knowledge to solving problems and evaluating ideas and information
      • Capacity to present ideas clearly with supporting evidence

      Case studies

      Due: TBA
      Weighting: 45%

      Case studies (includes Hartley quiz + field notes 5% Mt. Todd, 15% Volcanoes, 10%, Hartley 15%)

      • Understanding of the tools and methods that are used in the geosciences these are organised in three modules: o Tools of the geoscientist o Hot rocks o Rocks under stress
      • Competence in applying geo-scientific principles to understanding the world around you
      • Capacity to employ appropriate geo-scientific tools to solve problems and to interpret the results
      • Understanding scientific methodology
      • Competence in accessing, using and synthesising appropriate information
      • Application of knowledge to solving problems and evaluating ideas and information
      • Capacity to present ideas clearly with supporting evidence

      Final examination

      Due: University Examination Period
      Weighting: 45%

      • Understanding of the tools and methods that are used in the geosciences these are organised in three modules: o Tools of the geoscientist o Hot rocks o Rocks under stress
      • Competence in applying geo-scientific principles to understanding the world around you
      • Capacity to employ appropriate geo-scientific tools to solve problems and to interpret the results
      • Understanding scientific methodology
      • Competence in accessing, using and synthesising appropriate information
      • Application of knowledge to solving problems and evaluating ideas and information
      • Capacity to present ideas clearly with supporting evidence

      In this paper, a novel population-based, nature-inspired optimization paradigm is proposed, which is called Harris Hawks Optimizer (HHO). The main inspiration of HHO is the cooperative behavior and chasing style of Harris’ hawks in nature called surprise pounce. In this intelligent strategy, several hawks cooperatively pounce a prey from different directions in an attempt to surprise it. Harris hawks can reveal a variety of chasing patterns based on the dynamic nature of scenarios and escaping patterns of the prey. This work mathematically mimics such dynamic patterns and behaviors to develop an optimization algorithm. The effectiveness of the proposed HHO optimizer is checked, through a comparison with other nature-inspired techniques, on 29 benchmark problems and several real-world engineering problems. The statistical results and comparisons show that the HHO algorithm provides very promising and occasionally competitive results compared to well-established metaheuristic techniques. Source codes of HHO are publicly available at http://www.alimirjalili.com/HHO.html and http://www.evo-ml.com/2019/03/02/hho.

      Ali Asghar Heidari is now a Ph.D. research intern at Department of Computer Science, School of Computing, National University of Singapore (NUS), Singapore. Currently, he is also an exceptionally talented Ph.D. candidate at the University of Tehran and he is awarded and funded by Iran’s National Elites Foundation (INEF). His main research interests are advanced machine learning, evolutionary computation, meta-heuristics, prediction, information systems, and spatial modeling. He has published more than 14 papers in international journals such as Information Fusion, Energy Conversion and Management, Applied Soft Computing, and Knowledge-Based Systems.

      Seyedali Mirjalili is a lecturer at Griffith University and internationally recognized for his advances in Swarm Intelligence (SI) and optimization, including the first set of SI techniques from a synthetic intelligence standpoint – a radical departure from how natural systems are typically understood – and a systematic design framework to reliably benchmark, evaluate, and propose computationally cheap robust optimization algorithms. Dr Mirjalili has published over 80 journal articles, many in high-impact journals with over 7000 citations in total with an H-index of 29 and G-index of 84. From Google Scholar metrics, he is globally the 3rd most cited researcher in Engineering Optimization and Robust Optimization. He is serving an associate editor of Advances in Engineering Software and the journal of Algorithms.


      1 Introduction

      The NLopt (Non-Linear Optimization) library (v2.4.2) johnson2010nlopt is a rich collection of optimization routines and algorithms, which provides a platform-independent interface for their use for global and local optimization. The library has been widely used for practical implementations of optimization algorithms as well as for benchmarking new algorithms.

      The work in this paper is based on the I A C O R -LS algorithm proposed by Liao, Dorigo et al. iacor:algo This algorithm introduced the local search procedure in the original I A C O R technique, specifically Mtsls1 by Tseng et al. mtsls for local search. The I A C O R was an extension of the A C O R algorithm for continuous optimization with the added advantage of a variable size solution archive. The premise of our work lies in improving the local search strategy adopted by I A C O R -LS, by allowing algorithms other than Mtsls1 to be used for local search.

      We present a comparison of using various algorithms from the NLopt library for local search procedure in the I A C O R -LS algorithm. In order to introduce a hybrid approach for local search, we use a parameter that probabilistically determines whether to use the Mtsls1 algorithm or the NLopt library algorithm. In case of stagnation, we switch between Mtsls1 or the NLopt algorithm based on the algorithm being used in the previous iteration. The objective is to rigorously analyze the effect of using various optimization algorithms in the local search procedure for I A C O R -LS, as well as provide results on benchmark functions to enable a naive researcher to choose an algorithm easily. To the best of our knowledge, available works in literature have not provided exhaustive comparisons using optimization algorithm libraries on ant colony based approaches, other than rios2013derivative

      . However, surveys on state-of-art in multi-objective evolutionary algorithms

      zhou2011multiobjective , differential evolution das2011differential and real-parameter evolutionary multimodal optimization das2011real have appeared in literature.

      The rest of the paper is organized as follows. Section 2 discusses our hybrid approach which allows using Mtsls1 alongwith an NLopt library algorithm for local search phase of I A C O R -LS. This is followed by a discussion on the NLopt library in Section 3 . We present our results and a discussion in Section 4 , followed by the conclusions in Section 5 .


      Solutions for Chapter 9.8: Discrete Mathematics with Applications 4th Edition

      • 9.8.1E: In any sample space S, what is P(?)?
      • 9.8.2E: Suppose A, B, and C are mutually exclusive events in a sample space.
      • 9.8.3E: Suppose A and B are mutually exclusive events in a sample space S,C.
      • 9.8.4E: Suppose A and B are events in a sample space S with probabilities 0.
      • 9.8.5E: Suppose A and B are events in a sample space S and suppose that P(A.
      • 9.8.6E: Suppose U and V are events in a sample space S and suppose that P(U.
      • 9.8.7E: Suppose a sample space S consists of three outcomes: 0, 1, and 2. L.
      • 9.8.8E: Redo exercise 7 assuming that P(A) = 0.5 and P(B) = 0.4.Reference:S.
      • 9.8.9E: Let A and B be events in a sample space S, and let C = S ? (A ? B).
      • 9.8.10E: Redo exercise 9 assuming that P(A) = 0.7, P(B) = 0.3, and P(A n B) .
      • 9.8.11E: Prove that if S is any sample space and U and V are events in S wit.
      • 9.8.12E: Prove that if S is any sample space and U and V are any events in S.
      • 9.8.23E: A gambler repeatedly bets that a die will come up 6 when rolled. Ea.
      Textbook: Discrete Mathematics with Applications
      Edition: 4
      Author: Susanna S. Epp
      ISBN: 9780495391326

      Since 13 problems in chapter 9.8 have been answered, more than 92484 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Discrete Mathematics with Applications , edition: 4. Discrete Mathematics with Applications was written by and is associated to the ISBN: 9780495391326. Chapter 9.8 includes 13 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions.

      Ax = b is solvable when b is in the column space of A then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.

      Ax = AX with x#-O so det(A - AI) = o.

      A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.

      A must be positive definite the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA-1 yll2 = Y T(AAT)-1 Y = 1 displayed by eigshow axis lengths ad

      Constant along each antidiagonal hij depends on i + j.

      Diagonal entries = 1, off-diagonal entries = 0.

      The m by n edge-node incidence matrix has a row for each edge (node i to node j), with entries -1 and 1 in columns i and j .

      A symmetric matrix with eigenvalues of both signs (+ and - ).

      Nullspace of AT = "left nullspace" of A because y T A = OT.

      A combination other than all Ci = 0 gives L Ci Vi = O.

      The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).

      The columns of N are the n - r special solutions to As = O.

      The diagonal entry (first nonzero) at the time when a row is used in elimination.

      Column and row spaces = lines cu and cv.

      = number of pivots = dimension of column space = dimension of row space.

      Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.