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1 HOW MUCH WILL I SPEND ON GAS? Outcome (lesson objective) The students will use the current and future price of gasoline to construct T-charts, write algebraic equations, and plot the equations on a graph. Student/Class Goal Students will determine their gasoline cost for a month s time. They will use this information to calculate yearly gas cost. Time Frame Two 1 ½ hour classes One 3 hour class Standard Use Math to Solve Problems and Communicate NRS EFL 3-6 COPS Understand, interpret, and work with pictures, numbers, and symbolic information. Apply knowledge of mathematical concepts and procedures to figure out how to answer a question, solve a problem, make a prediction, or carry out a task that has a mathematical dimension. Define and select data to be used in solving the problem. Determine the degree of precision required by the situation. Solve problem using appropriate quantitative procedures and verify that the results are reasonable. Communicate results using a variety of mathematical representations, including graphs, charts, tables, and algebraic models. Materials Graph paper 40x30 paper is included at the end of the lesson (Worksheet 4) Colored pencils (if available) Calculators How Do I Find My Gas Mileage? Handout T-Chart Illustration How to Graph a Linear Equation in 5 Quick Steps Student Resource Activity Addresses Components of Performance Students will work with basic operations and patterns to complete a T-chart. Students will construct a graph and plot points on it. Students will use problem solving to determine how to figure out their monthly gas cost. Students will determine appropriate intervals on the x and y axes of the graph. Students will round the number when necessary to plot the points on the graph. Students will recognize if a data set is not reasonable by observing the other data sets and observing the points on the graph (is it on the line?) Students will communicate the results of the data by constructing a graph, T-chart, algebraic formula and in writing. Learner Prior Knowledge Basic understanding of gas prices and what is meant by miles per gallon (mpg), found at lesson, Pumped Up Gas Prices. Instructional Activities Step 1 - Discuss with students the current gasoline price and what they expect to happen to gas prices in the future. Ask students about how much they are spending on gas each month and how many miles the car they are driving gets per gallon? Students may need to research this information by actually calculating their mileage on the handout How Do I Find My Gas Mileage? or by visiting Cars and mpg This web site lists the in-city and highway mileage for 1985 or newer vehicles. If using this site, students will need to decide what type of driving they do during a month (city or highway) and estimate their car s mileage. Step 2 - Using \$4 per gallon as the price of gas, construct, with the class, a T-chart showing the relationship between the number of gallons purchased(x) and the total cost of the gas(y). See T-Chart Illustration for an example. As a class, look at the T-chart that was constructed. Give the students time to discover the relationship between the number of gallons purchased(x) and the total price of the gas(y). Express this relationship verbally (The number of gallons purchased times \$4 will equal the total cost.). Discuss how this relationship can be written as an equation as a function of x? (4x=y) Step 3 Next, with the students, graph the relationship/equation the class discovered during Step 2. The How to Graph a Linear Equation in 5 Quick Steps, a student resource from their math journals, provides basic guidelines for graphing equations. Help the students determine appropriate intervals and labels for the x and y axes and a title for the graph. Plot several points together and then let the students plot the remainder of the points. Draw a line passing through the points and label the line with the equation written in Step 2. Step 4 - Discuss with the students how many miles they drive each month. Share with the class that many people drive about 1,000
4 Gallons purchased (x) T-CHART ILLUSTRATION Total gas cost (y) 1 \$ \$ x30 Grid
5 How to Graph a Linear Equation in 5 Quick Steps Step 1 Construct a T-chart of Values Using your equation, construct a T-chart of values if one has not been done already. Substitute some simple numbers into the equation for x or y. If x=1, what is y? If x=10, what is y? If y=0, what is x? Each pair of values in your T-chart will become a point on the graph. (See illustration 1 for an example of a T-chart) Step 2 Decide on the interval for each axis Before starting the graph, look at the T-chart to determine the highest value for y found on the chart. Look at the values needed for x. Using graph paper, count the number of lines on the x and y axes. Use these numbers to determine the intervals on each axis. (If you use the graph paper at the end of this lesson there are 30 spaces on the x axis and 40 spaces on the y axis.) If the largest total cost/y value that needs to be graphed is \$80 and there are 40 lines on the y axis, let each line on the y axis represent \$2. The number of gallons of gas/x value that goes with \$80 is 20. There are 30 lines, so to make it simple one line will equal one gallon. Be sure the students realize they do not need to put a number next to every line. For example, the x might be labeled on every 5 th line (five gallons) and the y axis might also be labeled on every 5 th line (or \$10). This is a good step to do in pencil. That way if the interval you selected did not work out, the numbers can be erased any you can start over. Step 3 Label each Axis Decide what labels need to be added to the x and y axis. What do the numbers on the x-axis represent? What do the numbers on the y-axis represent? Usually the labels will match the descriptions/labels of x and y on the T-chart. (Note: When graphing equations involving elapsed time, time is traditionally represented by x) Step 4 Plot the points Using each pair of points from the T-chart, plot the points on the graph. Every point does not need to be plotted. Just be sure you have at least 3. Using a ruler, draw a line through the points you have plotted. Write your equation next to the line. Step 5 Give the graph a title Decide on a title for the graph. Make sure it accurately represents what is being shown on the graph. Does it explain the relationship between x and y? How to Graph a Linear Equation in 5 Quick Steps Student Resource
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# 11.2: Arithmetic Sequences
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###### Learning Objectives
• Find the common difference for an arithmetic sequence.
• Write terms of an arithmetic sequence.
• Use a recursive formula for an arithmetic sequence.
• Use an explicit formula for an arithmetic sequence.
Companies often make large purchases, such as computers and vehicles, for business use. The book-value of these supplies decreases each year for tax purposes. This decrease in value is called depreciation. One method of calculating depreciation is straight-line depreciation, in which the value of the asset decreases by the same amount each year.
As an example, consider a woman who starts a small contracting business. She purchases a new truck for $$25,000$$. After five years, she estimates that she will be able to sell the truck for $$8,000$$. The loss in value of the truck will therefore be $$17,000$$, which is $$3,400$$ per year for five years. The truck will be worth $$21,600$$ after the first year; $$18,200$$ after two years; $$14,800$$ after three years; $$11,400$$ after four years; and $$8,000$$ at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value.
## Finding Common Differences
The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is $$-3,400$$.
The sequence below is another example of an arithmetic sequence. In this case, the constant difference is $$3$$. You can choose any term of the sequence, and add $$3$$ to find the subsequent term.
###### ARITHMETIC SEQUENCE
An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference. If $$a_1$$ is the first term of an arithmetic sequence and $$d$$ is the common difference, the sequence will be:
$\{a_n\}=\{a_1,a_1+d,a_1+2d,a_1+3d,...\}$
###### Example $$\PageIndex{1}$$: Finding Common Differences
Is each sequence arithmetic? If so, find the common difference.
1. $$\{1,2,4,8,16,...\}$$
2. $$\{−3,1,5,9,13,...\}$$
Solution
Subtract each term from the subsequent term to determine whether a common difference exists.
1. The sequence is not arithmetic because there is no common difference.
$$2-1={\color{red}1} \qquad 4-2={\color{red}2} \qquad 8-4={\color{red}4} \qquad 16-8={\color{red}8}$$
1. The sequence is arithmetic because there is a common difference. The common difference is $$4$$.
$$1-(-3)={\color{red}4} \qquad 5-1={\color{red}4} \qquad 9-5={\color{red}4} \qquad 13-9={\color{red}4}$$
Analysis
The graph of each of these sequences is shown in Figure $$\PageIndex{1}$$. We can see from the graphs that, although both sequences show growth, (a) is not linear whereas (b) is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line.
Figure $$\PageIndex{1}$$
###### Q&A
If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference?
No. If we know that the sequence is arithmetic, we can choose any one term in the sequence, and subtract it from the subsequent term to find the common difference.
###### Exercise $$\PageIndex{1A}$$
Is the given sequence arithmetic? If so, find the common difference.
$$\{18, 16, 14, 12, 10,…\}$$
Answer
The sequence is arithmetic. The common difference is $$–2$$.
###### Exercise $$\PageIndex{1B}$$
Is the given sequence arithmetic? If so, find the common difference.
$$\{1, 3, 6, 10, 15,…\}$$
Answer
The sequence is not arithmetic because $$3−1≠6−3$$.
## Writing Terms of Arithmetic Sequences
Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of $$n$$ and $$d$$ into formula below.
$a_n=a_1+(n−1)d$
###### How to: Given the first term and the common difference of an arithmetic sequence, find the first several terms.
1. Add the common difference to the first term to find the second term.
2. Add the common difference to the second term to find the third term.
3. Continue until all of the desired terms are identified.
4. Write the terms separated by commas within brackets.
###### Example $$\PageIndex{2}$$: Writing Terms of Arithmetic Sequences
Write the first five terms of the arithmetic sequence with $$a_1=17$$ and $$d=−3$$.
Solution
Adding $$−3$$ is the same as subtracting $$3$$. Beginning with the first term, subtract $$3$$ from each term to find the next term.
The first five terms are $$\{17,14,11,8,5\}$$
Analysis
As expected, the graph of the sequence consists of points on a line as shown in Figure $$\PageIndex{2}$$.
Figure $$\PageIndex{2}$$
###### Exercise $$\PageIndex{2}$$
List the first five terms of the arithmetic sequence with $$a_1=1$$ and $$d=5$$.
Answer
$$\{1, 6, 11, 16, 21\}$$
###### How to: Given any the first term and any other term in an arithmetic sequence, find a given term.
1. Substitute the values given for $$a_1$$, $$a_n$$, $$n$$ into the formula $$a_n=a_1+(n−1)d$$ to solve for $$d$$.
2. Find a given term by substituting the appropriate values for $$a_1$$, $$n$$, and $$d$$ into the formula $$a_n=a_1+(n−1)d$$.
###### Example $$\PageIndex{3}$$: Writing Terms of Arithmetic Sequences
Given $$a_1=8$$ and $$a_4=14$$, find $$a_5$$.
Solution
The sequence can be written in terms of the initial term $$8$$ and the common difference $$d$$.
$$\{8,8+d,8+2d,8+3d\}$$
We know the fourth term equals $$14$$; we know the fourth term has the form $$a_1+3d=8+3d$$.
We can find the common difference $$d$$.
\begin{align*} a_n&= a_1+(n-1)d \\ a_4&= a_1+3d \\ a_4&=8+3d\qquad \text{Write the fourth term of the sequence in terms of }a_1 \text{ and } d. \\ 14&=8+3d\qquad \text{Substitute }14 \text{ for } a_4. \\ d&=2\qquad \text{Solve for the common difference.} \end{align*}
Find the fifth term by adding the common difference to the fourth term.
$$a_5=a_4+2=16$$
Analysis
Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation $$a_n=a_1+(n−1)d$$.
###### Exercise $$\PageIndex{3}$$
Given $$a_3=7$$ and $$a_5=17$$, find $$a_2$$.
Answer
$$a_2=2$$
## Using Recursive Formulas for Arithmetic Sequences
Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is $$5$$, then each term is the previous term plus $$5$$. As with any recursive formula, the first term must be given.
$$a_n=a_n−1+d$$
for $$n≥2$$
###### Note: RECURSIVE FORMULA FOR AN ARITHMETIC SEQUENCE
The recursive formula for an arithmetic sequence with common difference $$d$$ is:
$a_n=a_n−1+d$
for $$n≥2$$
###### How to: Given an arithmetic sequence, write its recursive formula.
1. Subtract any term from the subsequent term to find the common difference.
2. State the initial term and substitute the common difference into the recursive formula for arithmetic sequences.
###### Example $$\PageIndex{4}$$: Writing a Recursive Formula for an Arithmetic Sequence
Write a recursive formula for the arithmetic sequence.
$$\{−18, −7, 4, 15, 26, …\}$$
Solution
The first term is given as $$−18$$. The common difference can be found by subtracting the first term from the second term.
$$d=−7−(−18)=11$$
Substitute the initial term and the common difference into the recursive formula for arithmetic sequences.
$$a_1=−18$$
$$a_n=a_{n−1}+11$$
for $$n≥2$$
Analysis
We see that the common difference is the slope of the line formed when we graph the terms of the sequence, as shown in Figure $$\PageIndex{3}$$. The growth pattern of the sequence shows the constant difference of 11 units.
Figure $$\PageIndex{3}$$
###### Q&A
Do we have to subtract the first term from the second term to find the common difference?
No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference.
###### Exercise $$\PageIndex{4}$$
Write a recursive formula for the arithmetic sequence.
$$\{25, 37, 49, 61, …\}$$
Answer
\begin{align*}a_1 &= 25 \\ a_n &= a_{n−1}+12 , \text{ for }n≥2 \end{align*}
## Using Explicit Formulas for Arithmetic Sequences
We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept.
$$a_n=a_1+d(n−1)$$
To find the $$y$$-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence.
The common difference is $$−50$$, so the sequence represents a linear function with a slope of $$−50$$. To find the $$y$$-intercept, we subtract $$−50$$ from $$200$$: $$200−(−50)=200+50=250$$. You can also find the $$y$$-intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure $$\PageIndex{4}$$.
Figure $$\PageIndex{4}$$
Recall the slope-intercept form of a line is $$y=mx+b$$. When dealing with sequences, we use $$a_n$$ in place of $$y$$ and $$n$$ in place of $$x$$. If we know the slope and vertical intercept of the function, we can substitute them for $$m$$ and $$b$$ in the slope-intercept form of a line. Substituting $$−50$$ for the slope and $$250$$ for the vertical intercept, we get the following equation:
$$a_n=−50n+250$$
We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is $$a_n=200−50(n−1)$$, which simplifies to $$a_n=−50n+250$$.
###### Note: EXPLICIT FORMULA FOR AN ARITHMETIC SEQUENCE
An explicit formula for the $$n^{th}$$ term of an arithmetic sequence is given by
$a_n=a_1+d(n−1)$
###### How to: Given the first several terms for an arithmetic sequence, write an explicit formula.
1. Find the common difference, $$a_2−a_1$$.
2. Substitute the common difference and the first term into $$a_n=a_1+d(n−1)$$.
###### Example $$\PageIndex{5}$$: Writing the nth Term Explicit Formula for an Arithmetic Sequence
Write an explicit formula for the arithmetic sequence.
$$\{2, 12, 22, 32, 42, …\}$$
Solution
The common difference can be found by subtracting the first term from the second term.
\begin{align*} d &= a_2−a_1 \\ &= 12−2 \\ &= 10 \end{align*}
The common difference is $$10$$. Substitute the common difference and the first term of the sequence into the formula and simplify.
\begin{align*}a_n &= 2+10(n−1) \\ a_n &= 10n−8 \end{align*}
Analysis
The graph of this sequence, represented in Figure $$\PageIndex{5}$$, shows a slope of $$10$$ and a vertical intercept of $$−8$$.
Figure $$\PageIndex{5}$$
###### Exercise $$\PageIndex{5}$$
Write an explicit formula for the following arithmetic sequence.
$$\{50,47,44,41,…\}$$
Answer
$$a_n=53−3n$$
### Finding the Number of Terms in a Finite Arithmetic Sequence
Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence.
###### How to: Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms.
1. Find the common difference $$d$$.
2. Substitute the common difference and the first term into $$a_n=a_1+d(n–1)$$.
3. Substitute the last term for $$a_n$$ and solve for $$n$$.
###### Example $$\PageIndex{6}$$: Finding the Number of Terms in a Finite Arithmetic Sequence
Find the number of terms in the finite arithmetic sequence.
$$\{8, 1, –6, ..., –41\}$$
Solution
The common difference can be found by subtracting the first term from the second term.
$$1−8=−7$$
The common difference is $$−7$$. Substitute the common difference and the initial term of the sequence into the nth term formula and simplify.
\begin{align*} a_n &= a_1+d(n−1) \\ a_n &= 8+−7(n−1) \\ a_n &= 15−7n \end{align*}
Substitute $$−41$$ for $$a_n$$ and solve for $$n$$
\begin{align*} -41&=15-7n\\ 8&=n \end{align*}
There are eight terms in the sequence.
###### Exercise $$\PageIndex{6}$$
Find the number of terms in the finite arithmetic sequence.
$$\{6, 11, 16, ..., 56\}$$
Answer
There are $$11$$ terms in the sequence.
### Solving Application Problems with Arithmetic Sequences
In many application problems, it often makes sense to use an initial term of $$a_0$$ instead of $$a_1$$. In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula:
$a_n=a_0+dn$
###### Example $$\PageIndex{7}$$: Solving Application Problems with Arithmetic Sequences
A five-year old child receives an allowance of $$1$$ each week. His parents promise him an annual increase of $$2$$ per week.
1. Write a formula for the child’s weekly allowance in a given year.
2. What will the child’s allowance be when he is $$16$$ years old?
Solution
1. The situation can be modeled by an arithmetic sequence with an initial term of $$1$$ and a common difference of $$2$$.
Let $$A$$ be the amount of the allowance and $$n$$ be the number of years after age $$5$$. Using the altered explicit formula for an arithmetic sequence we get:
$$A_n=1+2n$$
2. We can find the number of years since age $$5$$ by subtracting.
$$16−5=11$$
We are looking for the child’s allowance after $$11$$ years. Substitute $$11$$ into the formula to find the child’s allowance at age $$16$$.
$$A_{11}=1+2(11)=23$$
The child’s allowance at age $$16$$ will be $$23$$ per week.
###### Exercise $$\PageIndex{7}$$
A woman decides to go for a $$10$$-minute run every day this week and plans to increase the time of her daily run by $$4$$ minutes each week. Write a formula for the time of her run after $$n$$ weeks. How long will her daily run be $$8$$ weeks from today?
Answer
The formula is $$T_n=10+4n$$, and it will take her $$42$$ minutes.
###### Media
Access this online resource for additional instruction and practice with arithmetic sequences.
## Key Equations
recursive formula for nth term of an arithmetic sequence $$a_n=a_{n−1}+d$$ $$n≥2$$ explicit formula for nth term of an arithmetic sequence $$a_n=a_1+d(n−1)$$
## Key Concepts
• An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant.
• The constant between two consecutive terms is called the common difference.
• The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example $$\PageIndex{1}$$.
• The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example $$\PageIndex{2}$$ and Example $$\PageIndex{3}$$.
• A recursive formula for an arithmetic sequence with common difference dd is given by $$a_n=a_{n−1}+d$$, $$n≥2$$. See Example $$\PageIndex{4}$$.
• As with any recursive formula, the initial term of the sequence must be given.
• An explicit formula for an arithmetic sequence with common difference $$d$$ is given by $$a_n=a_1+d(n−1)$$. See Example $$\PageIndex{5}$$.
• An explicit formula can be used to find the number of terms in a sequence. See Example $$\PageIndex{6}$$.
• In application problems, we sometimes alter the explicit formula slightly to $$a_n=a_0+dn$$. See Example $$\PageIndex{7}$$.
### Contributors and Attributions
This page titled 11.2: Arithmetic Sequences is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
• Was this article helpful? | 6,153 | 21,128 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.200616 |
https://www.aimsuccess.in/2017/12/questions-asked-in-ibps-so-prelims-exam_31.html | 1,725,862,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00789.warc.gz | 608,847,563 | 82,913 | # Questions Asked in IBPS SO Prelims Exam 31 December 2017
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
Here is the detailed Questions Asked in IBPS SO Prelims Exam 31 December 2017. Read today’s Exam Analysis along with this article to know what kind of questions are asked in today’s exam. This will be very beneficial for you if you haven’t appeared for your exam yet. Even if you did appear for the exam already, you can analyze your performance by having a look at the analysis published.
## Quant Questions Asked in IBPS SO Pre Exam 31 Dec 2017
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
Directions (1 – 5): Answer the questions on the basis of the information given below.
Eight friends A, B, C, D, E, F, G and H are seating around a circular table. Some of them are facing inside & others are facing outside. Opposite direction means if one is facing inside the centre, second one face outside the centre, and vice versa.
Three people are seating between F and D and both face opposite direction to each other. E is seating second to right of both D and F, and face opposite direction as F faces. C is seating third to right of E, who is not opposite to B. G is neighbour of both E and D, and seating second to right of B, who is not a neighbour of E. G is third to left of A, who is third to right of both H and G. B face towards the centre and is seating second to left of C.
1. If all the friends are seating according to alphabetical order in anti clock wise direction, starting from A, how many friends remain at same position (excluding A)?
A) One
B) Two
C) Three
D) Four
E) None
2. If D interchanged his position with H and, F interchanged his position with G, who sits immediately right of H in new arrangement?
A) F
B) G
C) E
D) A
E) None
3. Who is seating third to left of B?
A) H
B) C
C) A
D) F
E) None
4. How many persons are facing away from the centre?
A) Three
B) Four
C) Five
D) Two
E) None
5. Who among the following pairs are facing same direction and seating opposite to each other ?
A) H, B
B) C, G
C) F, D
D) E, A
E) None
Directions (6 – 10): Answer the questions on the basis of the information given below.
Ten friends are sitting on twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In Row 1: A, B, C, D and E are seated and all of them are facing south, and in Row 2: P, Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row.
All of them like different colors – Red, Green, Black, Yellow, White, Blue, Brown, Purple, Pink and Grey, but not necessarily in the same order.
There are two seats between Q and the vacant seat. Q does not like White, Red and Purple. E is not an immediate neighbor of C. B likes Grey. Vacant seat of row 1 is not opposite to S and is also not at any of the extreme ends of Row-1.The one who likes Black sits opposite to the one, who sits third to the right of the seat, which is opposite to S. C is not an immediate neighbor of D. T, who likes neither White nor Blue, does not face vacant seat. D faces R. The vacant seats are not opposite to each other. Two seats are there between C and B, who sits third right of the seat, on which the person who likes Brown is sitting. S sits third to the right of seat on which R sits and likes Yellow. The one who likes Pink faces the one who likes Yellow. The persons who like Red and Purple are adjacent to each other. The vacant seat in row 1 is not adjacent to D.Q sits at one of the extreme ends. E neither likes Pink nor faces the seat which is adjacent to the one who likes Blue. The one who likes White is not to the immediate right of the one who likes Yellow. The person who likes Green doesn’t face the person who likes Purple.
6. How many persons are sitting between T and the one who likes yellow color?
A) None
B) One
C) Two
D) Three
E) None of these
7. Which of the following faces the vacant seat of Row – 2?
A) The one who like white color
B) A
C) D
D) The one who likes grey color
E) Cannot be determined
8. Who is sitting at the immediate left of person who likes purple color?
A) E
B) D
C) The one who likes black color
D) The one who likes green color
E) The one who likes grey color
9. Who amongst the following sits at the extreme end of the row?
A) R, Q
B) E, S
C) T, C
D) C, D
E) None of these
10. If Q is made to sit on vacant seat of his row, then how many persons are there between the persons who sit opposite to Q now and who sat opposite to Q previously?
A) Two
B) Three
C) Four
D) None
E) One
Directions (Q. 11–15): Study the following arrangement of series carefully and answer the questions given below:
C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S
11. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both?
A) None
B) Three
C) One
D) More than three
E) Two
12. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end?
A) 8
B) 2
C) A
D) Other than given options
E) O
13. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter?
A) None
B) Two
C) One
D) Three
E) More than three
14. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group?
A) 6 &*
B) 5 X O
C) F L 4
D) G O 8
E) 9 K %
15. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement?
A) I
B) Other than given options
C) 8
D) J
E) *
1. B
2. A
3. D
4. B
5. A
6. C
7. D
8. E
9. C
10. E
11. B
12. E
13. C
14. A
15. D
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
## General Awareness Asked in IBPS SO Pre Exam 31 Dec 2017
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
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## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript:
Market Power: Monopoly and Monopsony
Chapter 9
Review of Perfect Competition
P = LMC = LRAC Normal profits or zero economic profits in the long run Large number of buyers and sellers Homogenous product Perfect information Firm is a price taker
Review of Perfect Competition
Q P Market D S Q0 P0 Q P Individual Firm P0 D = MR = P q0 LRAC LMC
Monopoly Monopoly One seller - many buyers
One product (no good substitutes) Barriers to entry Price Maker The monopolist is the supply-side of the market and has complete control over the amount offered for sale. Monopolist controls price but must consider consumer demand Profits will be maximized at the level of output where marginal revenue equals marginal cost.
Average & Marginal Revenue
The monopolist’s average revenue, price received per unit sold, is the market demand curve. Monopolist also needs to find marginal revenue, change in revenue resulting from a unit change in output. Finding Marginal Revenue As the sole producer, the monopolist works with the market demand to determine output and price. An example can be used to show the relationship between average and marginal revenue Assume a monopolist with demand: P = 6 - Q
Average and Marginal Revenue
1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7
We can also see algebraically that Q. maximizes profit
We can also see algebraically that Q* maximizes profit. Profit π is the difference between revenue and cost, both of which depend on Q: As Q is increased from zero, profit will increase until it reaches a maximum and then begin to decrease. Thus the profit-maximizing Q is such that the incremental profit resulting from a small increase in Q is just zero (i.e., Δπ /ΔQ = 0). Then But ΔR/ΔQ is marginal revenue and ΔC/ΔQ is marginal cost. Thus the profit-maximizing condition is that , or
Monopolist’s Output Decision
\$ per unit of output D = AR MR MC AC P1 Q1 P* Q* P2 Q2 Lost profit Lost profit Quantity
Monopoly: An Example
Monopoly: An Example
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
(Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost:
Example of Profit Maximization
\$/Q 10 20 40 Profit = (P - AC) x Q = (\$30 - \$15)(10) = \$150 MC AC P=30 Profit AR AC=15 MR 5 10 15 20 Quantity
A Rule of Thumb for Pricing
Produce one more unit brings in revenue (1)(P) = P With downward sloping demand, producing and selling one more unit results in small drop in price P/Q. Reduces revenue from all units sold, change in revenue: Q(P/Q)
A Rule of Thumb for Pricing
A Rule of Thumb for Pricing
A Rule of Thumb for Pricing
A Rule of Thumb for Pricing
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity
Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market
The Effect of a Tax In competitive market, a per-unit tax causes price to rise by less than tax: burden shared by producers and consumers Under monopoly, price can sometimes rise by more than the amount of the tax. To determine the impact of a tax: t = specific tax MC = MC + t
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
The Multi-plant Firm For some firms, production takes place in more than one plant each with different costs Firm must determine how to distribute production between both plants Production should be split so that the MC in the plants is the same Output is chosen where MR=MC. Profits is therefore maximized when MR=MC at each plant We can show this algebraically: Q1 and C1 is output and cost of production for Plant 1 Q2 and C2 is output and cost of production for Plant 2 QT = Q1 + Q2 is total output Profit is then: = PQT – C1(Q1) – C2(Q2)
The Multi-plant Firm Firm should increase output from each plant until the additional profit from last unit produced at Plant 1 equals 0
Deadweight Loss from Monopoly Power
\$/Q Because of the higher price, consumers lose A+B and producer gains A-C. Lost Consumer Surplus MC Deadweight Loss B A QC PC C C Qm Quantity
Monopsony A monopsony is a market in which there is a single buyer.
An oligopsony is a market with only a few buyers. Monopsony power is the ability of the buyer to affect the price of the good and pay less than the price that would exist in a competitive market. Typically choose to buy until the benefit from last unit equals that unit’s cost Marginal value is the additional benefit derived from purchasing one more unit of a good Demand curve – downward sloping Marginal expenditure is the additional cost of buying one more unit of a good Depends on buying power
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
Monopsonist Buyer ME S = AE PC P*m D = MV Q*m QC \$/Q Monopsony
ME above S Quantity where ME = MV: Qm Price from Supply curve: Pm ME D = MV S = AE Q*m P*m Competitive P = PC Q = Q+C PC QC Quantity
Monopoly and Monopsony
Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher
Monopoly and Monopsony
\$/Q MV ME S = AE Q* P* PC QC Monopsony Note: ME = MV; ME > AE; MV > P
Monopoly and Monopsony
MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
My Classes & Students
- - - - - - | 9,383 | 34,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2013-20 | latest | en | 0.864713 |
https://tutorialspoint.dev/algorithm/dynamic-programming-algorithms/longest-increasing-path-matrix | 1,638,582,624,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00618.warc.gz | 615,987,150 | 10,677 | # Longest Increasing Path in Matrix
Given a matrix of N rows and M columns. From m[i][j], we can move to m[i+1][j], if m[i+1][j] > m[i][j], or can move to m[i][j+1] if m[i][j+1] > m[i][j]. The task is print longest path length if we start from (0, 0).
Examples:
```Input : N = 4, M = 4
m[][] = { { 1, 2, 3, 4 },
{ 2, 2, 3, 4 },
{ 3, 2, 3, 4 },
{ 4, 5, 6, 7 } };
Output : 7
Longest path is 1 2 3 4 5 6 7.
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
Below is the implementation of this approach:
## C++
`// CPP program to find longest increasing ` `// path in a matrix. ` `#include ` `#define MAX 10 ` `using` `namespace` `std; ` ` ` `// Return the length of LIP in 2D matrix ` `int` `LIP(``int` `dp[][MAX], ``int` `mat[][MAX], ``int` `n, ``int` `m, ``int` `x, ``int` `y) ` `{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-1 && y == m-1) ` ` ``return` `dp[x][y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-1 || y == m-1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(mat[x][y] < mat[x+1][y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x+1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(mat[x][y] < mat[x][y+1]) ` ` ``result = max(result, 1 + LIP(dp, mat, n, m, x, y+1)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` `} ` ` ` `// Wrapper function ` `int` `wrapper(``int` `mat[][MAX], ``int` `n, ``int` `m) ` `{ ` ` ``int` `dp[MAX][MAX]; ` ` ``memset``(dp, -1, ``sizeof` `dp); ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ``int` `mat[][MAX] = { ` ` ``{ 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, ` ` ``}; ` ` ``int` `n = 4, m = 4; ` ` ``cout << wrapper(mat, n, m) << endl; ` ` ` ` ``return` `0; ` `} `
/div>
## Java
`// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `|| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/* Driver program to test above function */` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. `
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
## C#
`// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 || y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/* Driver code */` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */`
Output:
```7
```
Time Complexity: O(N*M). | 3,679 | 9,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-49 | latest | en | 0.647698 |
https://www.physicsforums.com/threads/a-question-about-e-mc-2.816217/ | 1,531,801,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00244.warc.gz | 948,265,113 | 20,084 | 1. May 28, 2015
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
Staff Emeritus
No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t2, that we could say that E is inversely proportional to time squared. But what you found is E= md2/t2, where m is a constant and d depends on t. So md2 isn't a constant.
3. May 28, 2015
### PeroK
No, but it means that Energy has a "dimension" of $ML^2T^{-2}$.
Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above.
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation?
Thanks.
5. May 28, 2015
### Deepak K Kapur
One more question (silly one).
Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence.
Then, how on earth can we multiply a concrete entity with an abstract one??
6. May 28, 2015
### Fredrik
Staff Emeritus
1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
7. May 28, 2015
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
### Deepak K Kapur
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
### Deepak K Kapur
Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
I would say that aspects of the real world are represented by abstract mathematical things in the theory.
12. May 28, 2015
### Staff: Mentor
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties??
14. May 28, 2015
### Deepak K Kapur
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
Why not, this seems to be a sensible interpretation...
15. May 28, 2015
### ShayanJ
t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$
So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$.
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
### Staff: Mentor
Well, two reasons:
1. Because the units are still there and you are pretending they aren't. The "1" has units kg-m2. The equation should read E=1 kg-m2/t2.
2. The equation isn't describing the motion of a kilogram of mass, the piece you broke apart is describing the speed of light ("c" is the speed of light). So the 1m is the distance traveled by light in "t" time: time is a constant here, not a variable and you can't change that. So when you plug in m=1 kg, and d= 1m, t=1/300,000,000s and it all simplifies to E=9x10^16 Joules.
Science includes the implied assumption that the observed properties are real and there is nothing else to "concern with" or not "concern with". Beyond that, you're getting into philosophy, not science.
Last edited: May 29, 2015
17. May 30, 2015
### Deepak K Kapur
I dont get your point fully...
1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
### ArmanCham
Time constant means this ;
c=x/t If we assume x is one then you cannot change t cause c is constant.So you are telling us E=1/t2 here you assumed x equal one.Then you assumed t =2 We cannot change time If you choose x =1.If you change time you can see easily that you are breaking fundemental physics law "Speed of light is constant"
Here the math c=x/t you said x =1 then t=1/c.Now we found t.Its a number but you are telling it can be three or four.Thats nonsense.
19. May 30, 2015
### Staff: Mentor
You are mistaken. C is just a universal constant, that happens to be the speed of light.
You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem.
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
### Deepak K Kapur
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
Actually i was expecting the meaning of this equation as follows..
Matter will change into energy when...................(some relation to the speed of light)
Cant you elaborate this equation in this way?
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
If they are same, then do they look different to the 'observer' only or are they 'really' different?
25. May 30, 2015
### Staff: Mentor
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy. | 2,517 | 8,849 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-30 | latest | en | 0.912886 |
https://www.o.vg/unit/speed/yard-second-to-picometer-second.php | 1,718,342,225,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00128.warc.gz | 830,006,804 | 10,495 | Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
From yd/s to pm/s: 1 yd/s = 914400000003.66 pm/s;
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s).
To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66.
Example : convert 25 yd/s to pm/s:
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
## How to Convert Picometer/second to Yard/second?
As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s).
To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12.
Example : convert 45 pm/s to yd/s:
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
Example : convert 45 yd/s to pm/s:
45 yd/s = 45 ÷ 1.0936132983333E-12 pm/s = pm/s
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s.
### How many Picometer/second are in a Yard/second?
There are 914400000003.66 pm/s in one yd/s. In turn, one pm/s is equal to 1.0936132983333E-12 yd/s.
### How many yd/s is equal to 1 pm/s?
1 pm/s is approximately equal to 1.0936132983333E-12 yd/s.
### What is the yd/s value of 8 pm/s?
The Yard/second value of 8 pm/s is yd/s. (i.e.,) 8 x 1.0936132983333E-12 = yd/s.
### yd/s to pm/s converter in batch
Cite this Converter, Content or Page as:
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters | 1,819 | 4,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-26 | latest | en | 0.706668 |
https://takechargeofyourmoney.blog/category/calculate-savings/ | 1,498,392,808,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320491.13/warc/CC-MAIN-20170625115717-20170625135717-00612.warc.gz | 816,117,911 | 19,665 | ## How to calculate the interest you can earn
Here’s a quick explanation on how to calculate the interest that you can earn on your savings (or pay on your debt).
Let’s use an example of R1,500 (don’t worry about the currency – the maths is the same) and you deposit it in your bank account that earns 3.5% interest.
First thing to note is that the interest rate is always given as the annual rate (unless very specifically detailed otherwise). If you use a calculator (or a spreadsheet such as Excel) and do the following : 1500 x 3.5% the answer you get is 52.50.
Thus R1,500 invested for 1 year at 3.5% interest will earn R52.50. That’s easy!
To calculate what the monthly interest is, simply divide 52.50 by 12 and you will get R4.38 (rounded up) interest for 1 month. You could be more accurate and divide 52.5 by 365.4 (number of days in the year) and then multiply by the exact number of days in the month (e.g. 30).
E.g. 52.5 divided 365.4 multiplied by 30 = R4.31 (a very slight difference to the first answer we received)
So, if you invest R1,500 for 1 year at an interest rate of 3.5%, the interest at the end of the year would be R52.50. However, banks and all financial institutions calculate interest on a daily basis and pay it out monthly. If you withdrew each months interest (and leave the original amount of money in the account) then you would effectively earn “simple” interest – just as we calculated above and you would have R52.50 (assuming no fees or charges)
However, if you leave the interest that you earn each month in your account, then the bank would calculate the next months interest on your original amount plus whatever extra interest is already in your account. So each month you would earn slightly more. This is what is called “compound” interest.
In this case the difference is not great, but the larger the starting amount is, and the longer time you keep reinvesting the interest, the larger the difference is! In fact, after a few years, compound interest can be really large. Remember of course that we are assuming that no fees or charges apply.
This example may seem simple, but there is no need to complicate things!
## Find Extra Money
We’d all love some extra money I’m sure, so here are some steps to literally find it.
Step 1 : Look at your bank statement
Look at your bank statement as well as your credit card statement for the past month and make a list of all automatic payments that are deducted (eg internet bill, mobile phone, insurance, Netflix, etc)
Step 2 : Analyze all payments
One-by-one look at each automatic payment and ask:
• Do I need this service / subscription or whatever it is?
• Do I need all the features I pay for? Can I downgrade at all or pay for less?
• If it is a store card, do you need the extra’s such as insurance, magazine, etc?
• If it is a debt repayment, how much interest am I paying? What must I do to pay it off?
Step 3 : Insurance
Pay special attention to your insurance payment. Look at your latest insurance policy or contact the company for it. Are the values correct? Now, get new quotes and see if you can find a cheaper option.
Look at optional features you may have on any insurance, store card or debt policies. Perhaps you have the same benefits offered by different policies; in that case you can cancel some.
Step 4 : Look for wasted money
Look at your monthly expenses and take special note of the following:
• Eating Out / Take-Aways
• Groceries
• Clothing
• Entertaining at home
• Alcohol bought
• Luxuries
• Telephone
• Electricity
• Internet
• Make-up
Can you spend less on any of the above? Try for just 1 month to spend less than the previous month and you will see that you really can survive quite easily!
Important note is that what you save today should not be spent tomorrow, don’t feel tempted to spend your savings on other things. See when saving is not really saving.
Step 5 : Be strict with yourself
Make a list of every area where you feel you can save a little and be strict with yourself over the next month to actually do this!
## When saving ins’t saving
I am constantly looking at how and where I spend my money and deciding whether what I am doing is worth it or not. There needs to be balance between enjoying live but at the same time living well within ones means and achieving ones financial goals. All it takes to be conscious about what you spend. I have come across times though when I think I am saving money when in fact I am not.
### Hosting people vs Going out
The first instance is the fallacy that entertaining at home is cheaper than going out. This can be true, but not necessarily all the time. It is great to host friends and family and one should enjoy every moment of it; life is too short not too. I am definitely not implying that one should not host people in your home, but the point of this post is about saving money. If you specifically decide to host people at your home instead of going out (in order to save) then you need to ensure that you do actually spend less money.
Think of the average restaurant bill that you would be paying if you went out, and now look at what you are about to spend in order to host your friends. Flowers, candles, snacks, wine, groceries, dessert, etc… It is easy to go overboard and spend more than you would have. (Especially if you’re married to a chef, which is the case for me.) The point is not to be stingy though. If you wish to save money then calculate the costs of going versus what you would like to spend when entertaining and chose the cheaper option.
### A picnic vs lunch out
As with the first example, if you decide to go on a picnic in order to save money (rather than an expensive restaurant), then be sure to do some calculations. It’s funny how in our minds we convince ourselves that because we’re taking the cheaper option, we can now spend more. A bottle of bubbly, expensive cheeses, some tapas, etc… It’s really easy to spend more on the picnic than in a restaurant.
###### The point of these examples is that if you choose to do something in order to save some money, keep that in mind and be sure to actually save! Don’t let yourself be tempted to buy the more expensive items to compensate for the cheaper option. You will actually end up spending more.
Now, take the money you calculated you would save and immediately transfer it into your savings account! Do it now before you find something else to spend it on!
## Calculate the Future Value of an investment
In this article we looked at the effect of extra payments into ones home loan (mortgage bond). That’s all good and well, but what if you don’t have a loan? Or perhaps you would rather invest in a Unit Trust or Interest Bearing bank account.
Microsoft Excel has a simple formula called “FV” which will help you to calculate the future value of your investment. This does not account for changing interest rates or complicated scenarios, but it will give you a simple tool to give you an idea of how your money can grow.
To start with a simple example, let’s save 600 each month for 10 years (120 months) at an interest of 6%. The answer as you see is 98,327.61.
That’s pretty easy. Let’s quickly look at the input parameters:
Rate: This is the interest rate per period. Generally speaking, financial institutions will quote an annual interest rate. In this example it is 6%. To get the monthly interest rate we simply use 6%/12. If we change the example an decide to make quarterly payments we would need to use 6%/4.
Nper: This is the number of periods (in total) for which we will make a payment. If we wish to pay monthly, this is the number of months.
Pmt: This is the amount we wish to pay each period. This figure cannot change over the lifetime of the investment. But, see further down how we can circumvent this potential problem. Excel expects the payment to be a negative figure as payments out of your account would “minus” from your account. You will see that if you use a positive figure, your answer will show negative. The figure will be the same though. (If that is confusing, try it yourself)
Pv: The present value. If you are adding money to an existing bank account, use Pv as the present value in the account. If this to calculate a new investment leave it blank or type a zero in the box.
Type: This is to specify whether the payment is being made at the beginning of each period, or at the end. This will affect the interest that you earn. By default Excel will assume you are paying at the end of each period.
So, in the example above we assumed a constant 600 payment each month for 10 years. But what if you decide to pay 10% more each year? For that we will need to spice things up a bit. See how I have calculated the Future Value for 12 months at a time. I then use the answer as my Present Value for the following year. The monthly payment is simply increased by 10% in each row.
## How much is your coffee costing you?
I love coffee! The very thought of drinking less makes me quiver with fear. But, sometimes one must take the emotion out of decisions and just look at cold, hard facts.
In a previous article on being more conscious with cash, I discussed how I buy 2 coffees each day, along with a breakfast and lunch. I’m not good at planning meals to bring to work but I have decided to spend slightly less. The savings I calculated was only R150 per week, but I’ve decided to look at how this could affect my home loan (mortgage bond) if I pay R600 extra per month.
These figures may not make sense to you if you use a different currency and your countries interest rates may be significantly different. I’ll post something soon about how you can do these calculations yourself.
The calculations can get messy so take note of the following assumptions:
Home Loan Value: R1,500,000
Interest Rate: 11% (annual)
Total Loan period: 20 years
For this example I will assume 2 years of the loan are already complete, and that up to now no additional payments have been made. Also, the R600 p.m. will remain constant (with other words I won’t save more in years to come)
This may all sound complicated, but the end result is simple to see. If I pay R600 p.m. extra into my bond until my bond is paid off I will:
• Save a total of R236,000 (rounded to nearest thousand)
• Pay the loan off 22 months earlier
That’s pretty amazing don’t you think? Considering that I am making this extra saving by cutting down on my weekly coffee/food expenses at my office. In fact, it was quite easy to find the extra R150 per week. Imagine if I actually analyze my expenses properly and relook at my insurance policies, health care, mobile phone contract, bank charges, etc. Imagine the savings I can make then!
What if you don’t have a home loan?
That’s really not a problem. If you invest this money in either a unit trust or interest bearing account for the next 15 – 20 years you will have quite a large lump sum. We’ll look at how to use the Future Value formula in Excel to calculate this. | 2,574 | 11,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-26 | longest | en | 0.956167 |
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### Ching chi tu journal 5
1. 1. Ching Chi Tu Journal#5 m1 geometry
2. 2. Perpendicular Bisector A line perpendicular to a segment at the segment’s midpoint. Perpendicular Bisector Theorem – If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. Converse of the Perpendicular bisector Theorem – If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
3. 3. Converse P.B. Thm H A Ex 1. f = g Ex 2. d = e Ex 3. b = c T T f = g; g = c; i = h ce ab; bg cb Ex 1. a = h Ex 2. i = j Ex 3. cg = bg T
4. 4. Angle Bisector Angle Bisector - A ray that divides an angle into two congruent angles. Angle Bisector Theorem – If a point is on the bisector of an angle, then it is equidistant from the sides of the angle. Converse of Angle Bisector Theorem – If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle
5. 5. Ex 1. c bisects <abc; point L is equidistant from both sides. Ex 2. i bisects <ghi; point J is equidistant from both sides. Ex 3. e bisects <def; point K is equidistant from both sides. H A.B. Thm Converse Ex 1. L is equidistant from both sides; c must be an a.b. Ex 2. J is equidistant from both sides; i must be an a.b. Ex 3. K is equidistant from both sides; f must be an a.b.
6. 6. Concurrency To be concurrent means for three or more lines to intersect at one point. The point of concurrency of perpendicular bisectors form a circumcenter. Circumcenter - where three perpendicular bisectors of a triangle are concurrent. It is equidistant from the vertices of the triangle
7. 7. Circumcenter D is the circumcenter of triangle abc, it is equidistant from points a, b, and c. Right = circumcenter midpoint of hypotenuse
8. 8. Circumcenter L is the circumcenter of triangle jkl, it is equidistant from points j, k, and l. Acute = circumcenter inside
9. 9. Circumcenter H is the circumcenter of triangle efg, it is equidistant from points e, f, and g. Obtuse = circumcenter outside
10. 10. Concurrency The point of concurrency of angle bisectors forms an incenter. Incenter – The point of concurrency of three angle bisectors of a triangle. It is equidistant from the sides of the triangle. Concurrency Concurrency
11. 11. Incenter Incenter always inside triangle D is the incenter of triangle abc, it is equidistant to all sides.
12. 12. Incenter Incenter always inside triangle A is the incenter of triangle xyz, it is equidistant to all sides.
13. 13. Incenter Incenter always inside triangle D is the incenter of triangle abc, it is equidistant to all sides.
14. 14. Concurrency Median – segments whose endpoints are a vertex of the triangle and the midpoint of the opposite side. The point of concurrency of medians is called a centroid. Centroid – Point of concurrency of the medians of a triangle. It is located 2/3 of the distance from each vertex to the midpoint of the opposite side
15. 15. Centroid Ex 1. g is the centroid of triangle abc Ex 2. o is the centroid of triangle egc Ex 3. k is the centroid of triangle adg Centroid = Point of balance
16. 16. Concurrency Altitude – a perpendicular segment from a vertex to the line containing the opposite side. The point of concurrency of three altitudes of a triangle creates an orthocenter. Orthocenter – point where three altitudes of a triangle are concurrent. Nothing special.
17. 17. Orthocenter D is the orthocenter of triangle abc with lines e, d, and f as altitudes.
18. 18. Orthocenter D is the orthocenter of triangle abc with lines e, d, and f as altitudes.
19. 19. Orthocenter D is the orthocenter of triangle abc with lines e, d, and f as altitudes.
20. 20. Midsegment Midsegment of a triangle – a segment that joins the midpoints of two sides of the triangle. Every triangle has three midsegments, forming a midsegment triangle. Triangle Midsegment Theorem – A midsegment of a triangle is parallel to a side of the triangle, and its length is half the length of that side.
21. 21. Midsegment Ex 1. f BC Ex 2. d AC Ex 3. e BA They are all midsegments and are parallel to the lines written beside them.
22. 22. Relationships In any triangle, the longest side is always opposite to the largest angle. Small side = small angle.
23. 23. <ul><li>Ex 3. </li></ul><ul><li><a is the smalles angle, so its opposite angle is also the smallest. </li></ul><ul><li>Side b is the largest side so its opposite angle is the largest. </li></ul><ul><li>Ex 2. </li></ul><ul><li><c is the largest angle so its opposite side is also the largest. </li></ul><ul><li>Side a is the smallest side so its opposite angle is also the smallest. </li></ul><ul><li>Ex. 1 </li></ul><ul><li><a is the largest angle so therefore side a, its opposite side, must be the largest side. </li></ul><ul><li><c has the smallest angle so this means that side c is the smallest side in the triangle. </li></ul>
24. 24. Inequality Exterior Angle Inequality – The exterior angle of a triangle is larger than either of the non-adjacent angles.
25. 25. Ex 1. m<e is larger than all non adjacent angles such as b and y. Ex 2. m<c is greater than all non adjacent angles such as a and y. Ex 3. m<o is greater than all non adjacent angles such as a and b.
26. 26. Inequality Triangle Inequality Theorem – The two smaller sides of a triangle must add up to MORE than the length of the 3 rd side.
27. 27. Ex 1. 2.08 + 1.71 is less than 5.86 so it doesn’t form a triangle. Ex 2. 3.02 + 2.98 = 6 But it doesn’t form a triangle, it forms a line/segment. Ex 2. 1.41 + 1.41 is greater than 2 so therefore, it does form a triangle.
28. 28. Indirect Proofs <ul><li>Steps for indirect proofs: </li></ul><ul><li>Assume the opposite of the conclusion is true </li></ul><ul><li>Get the contradiction </li></ul><ul><li>Prove it is true </li></ul>
29. 29. Ex 1. Ex 2. Prove: a supplement of an acute angle cannot be another acute angle Contradiction Prove: a scalene triangle cannot have two congruent midsegments. Ex 3. Prove: an isosceles triangle cannot have a base angle that is a right angle Angle addition postulate 89 + 89 does not equal 180 Definition of acute angles Acute angles are less than 90 degrees Definition of supplementary angles Two angles add up to form 180 degrees Given Assuming that the supplement of an acute angle is another acute angle. Definition of a scalene triangle. a scalene triangle doesn’t have congruent sides definition of a midsegment. A midsegment can be congruent if there are congruent sides. Definition of a midsegment A midsegment is half the length of the line it is parallel to Given Assuming that a scalene triangle has two congruent midsegments. Substitution 90 + m<b + 90 have to equal 180. m<b must equal 0 Triangle sum theorem M<a + m<b + m<c have to equal 180 Definition of right angles M<a and m<c are both 90 Isosceles triangle theorem. Angle c is congruent to angle a, so angle c is also a right angle. Given Assuming that triangle abc has a base angle that is a right angle. let angle a be the right angle. Triangle abc is an isosceles triangle.
30. 30. Prove: triangle abc cannot have a base angle that is a right triangle Given: triangle abc is an isosceles triangle with base ac. Assuming that triangle abc has a base angle that is a right angle. let angle a be the right angle By the isosceles triangle theorem, angle c is congruent to angle a, so angle c is also a right angle. By the difinitio of right angle m<a = 90 and m<r = 90. By the triangle sum theorem m<a + m<b + m<c =180. By substitution, 90 + m<b + 90 = 180 so m<b = 0.
31. 31. HINGE The Hinge Theorem – if two triangles have 2 sides congruent, but the third side is not congruent, then the triangle with the larger included angle has the longer 3 rd side. Converse – if two triangles have 2 congruent sides, but the third side is not congruent, then the triangle with the larger 3 rd side has the longer 3 rd angle.
32. 32. Ex 1. As <a is larger than <b, side d is larger than side c Ex 2. As <a is smaller than <b, side e is smaller than side b Ex 3. As <b is greater than <a side f is larger than side b Hinge Converse Ex 1. As side d is larger than side c, <a must be larger than <b Ex 3. As side b is smaller than side f, <a must be smaller than <b. Ex 2. As side b is larger than side e, <b must also be larger than <a.
33. 33. Special Relationships 45-45-90 triangle - in all 45-45-90 triangles, both leg are congruent, and the length of the hypotenuse is the length of a leg radical 2. 30-60-90 – In all 30-60-90 triangles, the length of the hypotenuse is twice the length of the short leg; the longer side is the length of the smaller leg radical 3.
34. 34. 45-45-90 Ex 1. b = 4 √2 Ex 3. b = 6.92 √2 Ex 2. b = 4 √2
35. 35. 30-60-90 Ex 1. b = 4.62 c = 4 Ex 2. h = 4.62 i = 4 Ex 3. e = 8 f = 6.93 | 2,613 | 9,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-34 | latest | en | 0.809972 |
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© Scott Evans, Ph.D. 1 Introduction to Biostatistics, Harvard Extension School Confidence Intervals
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### Transcript of Introduction to Biostatistics, Harvard Extension School © Scott Evans, Ph.D.1 Confidence Intervals.
© Scott Evans, Ph.D. 1
Introduction to Biostatistics, Harvard Extension School
Confidence Intervals
© Scott Evans, Ph.D. 2
Introduction to Biostatistics, Harvard Extension School
Statistical Inference
Statistical Inference Inferences regarding a population are
made based on a sample Inferences about population parameters (e.g., μ) are made by examining sample statistics (e.g., sample mean)
© Scott Evans, Ph.D. 3
Introduction to Biostatistics, Harvard Extension School
Statistical Inference
Statistical Inference 2 primary approaches
Hypothesis Testing (of a population parameter) Estimation (of a population parameter)
Point Estimation Sample mean is an estimate of the population mean
Interval Estimation Confidence Intervals
Hypothesis Testing vs. Estimation Closely related
© Scott Evans, Ph.D. 4
Introduction to Biostatistics, Harvard Extension School
Confidence Interval
A range of values associated with a parameter of interest (such as a population mean or a treatment effect) that is calculated using the data, and will cover the TRUE parameter with a specified probability (if the study was repeated a large number of times)
© Scott Evans, Ph.D. 5
Introduction to Biostatistics, Harvard Extension School
Confidence Interval (CI)
Based on sample statistics (estimates of population parameters)
The width of the CI provides some information regarding the precision of the estimate
Provides both a range of plausible values and a test for the parameter of interest
© Scott Evans, Ph.D. 6
Introduction to Biostatistics, Harvard Extension School
Confidence Interval
Roughly speaking: an interval within which we expect the true parameter to be contained.
Based on the notion of repeated sampling (similar to hypothesis testing)
© Scott Evans, Ph.D. 7
Introduction to Biostatistics, Harvard Extension School
Confidence Interval
The “percent” indicates the probability (based on repeated sampling) that the CI covers the TRUE parameter Not the probability that the parameter falls in the
interval The CI is the random entity (that depends on the random
sample) The parameter is fixed A different sample would produce a different interval,
however the parameter of interest remains unchanged.
© Scott Evans, Ph.D. 8
Introduction to Biostatistics, Harvard Extension School
Illustration
We are analyzing a study to determine if a new drug decreases LDL cholesterol. We measure the LDL of 100 people before
administering the drug and then again after 12 weeks of treatment.
We then calculate the mean change (post-pre) and examine it to see if improvement is observed.
© Scott Evans, Ph.D. 9
Introduction to Biostatistics, Harvard Extension School
Illustration
Assume now that the drug has no effect (i.e., the true mean change is 0). Note that in reality, we never know what the true mean is.
We perform the study and note the mean change and calculate a 95% CI
Hypothetically, we repeat the study an infinite number of times (each time re-sampling 100 new people)
© Scott Evans, Ph.D. 10
Introduction to Biostatistics, Harvard Extension School
Illustration
Study 1, Mean=–7, 95% CI (-12, 2) Study 2, Mean=–2, 95% CI (-9, 5) Study 3, Mean=4, 95% CI (-3, 11) Study 4, Mean=0, 95% CI (-7, 7) Study 5, Mean=-5, 95% CI (-12, 2) . . .
© Scott Evans, Ph.D. 11
Introduction to Biostatistics, Harvard Extension School
Illustration
Remember the true change is zero (i.e., the drug is worthless)
95% of these intervals will cover the true change (0). 5% of the intervals will not cover 0
In practice, we only perform the study once. We have no way of knowing if the interval
that we calculated is one of the 95% (that covers the true parameter) or one of the 5% that does not.
© Scott Evans, Ph.D. 12
Introduction to Biostatistics, Harvard Extension School
Example
Two treatments are being compared with respect to “clinical response” for the treatment of nosocomial (hospital-acquired) pneumonia
The 95% CI for the difference in response rates for the two treatment groups is (-0.116, 0.151)
What does this mean?
© Scott Evans, Ph.D. 13
Introduction to Biostatistics, Harvard Extension School
Example (continued)
Based upon the notion of repeated sampling, 95% of the CIs calculated in this manner would cover the true difference in response rates. We do not know if we have one of the 95% that
covers the true difference or one of the 5% that does not.
Thus we are 95% confident that the true between-group difference in the proportion of subjects with clinical response is between -0.116 and 0.151
© Scott Evans, Ph.D. 14
Introduction to Biostatistics, Harvard Extension School
Caution
Thus, every time that we calculate a 95% CI, then there is a 5% chance that the CI does not cover the quantity that you are estimating. If you perform a large analyses, calculating
many CIs for many parameters, then you can expect that 5% of the will not cover the the parameters of interest.
© Scott Evans, Ph.D. 15
Introduction to Biostatistics, Harvard Extension School
CIs and Hypothesis Testing
To use CIs for hypothesis testing: values between the limits are values for which the null hypothesis would not be rejected At the α=0.05 level:
We would fail to reject H0: treatment change=0 since 0 is contained in (-0.116, 0.151)
We would reject H0: treatment change=-20 since -20 is not contained in (-0.116, 0.151)
© Scott Evans, Ph.D. 16
Introduction to Biostatistics, Harvard Extension School
CIs and Hypothesis Testing
What would be the conclusion of these hypothesis tests if we wanted to test at α=0.01 or α=0.10?
© Scott Evans, Ph.D. 17
Introduction to Biostatistics, Harvard Extension School
Confidence Interval Width
It is desirable to have narrow CIs Implies more precision in your estimate Wide CIs have little meaning
In general, with all other things being equal: Smaller sample size wider CIs Higher confidence wider CIs Larger variability wider CIs
© Scott Evans, Ph.D. 18
Introduction to Biostatistics, Harvard Extension School
In Practice
It is a good idea to provide confidence intervals in an analyses They provide both a test and an estimate
of the magnitude of the effect (which p-values do not provide)
© Scott Evans, Ph.D. 19
Introduction to Biostatistics, Harvard Extension School
For Illustration
Consider the neuropathy example (insert neuropathy_example.pdf)
Sensitivity 48.5 w/ 95% CI (36.9, 60.3)
© Scott Evans, Ph.D. 20
Introduction to Biostatistics, Harvard Extension School
CIs
Similar to hypothesis testing: We may choose the confidence level (e.g.,
95% α=0.05) CIs may be:
1-sided: (-∞, value) or (value, ∞), or 2-sided (value, value)
© Scott Evans, Ph.D. 21
Introduction to Biostatistics, Harvard Extension School
CIs
Some things to think about Scale of the variable(s): continuous vs.
binary (next class) 1-sample vs. 2-sample
CI for a population mean CI for the difference between 2 population
means
σ known or unknown
© Scott Evans, Ph.D. 22
Introduction to Biostatistics, Harvard Extension School
CIs
Insert CI1.pdf | 1,783 | 7,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-23 | latest | en | 0.766616 |
https://networklessons.com/cisco/ccie-routing-switching-written/bgp-4-byte-number | 1,726,599,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00293.warc.gz | 389,925,065 | 33,336 | # BGP 4-Byte AS Number
Lesson Contents
Before January 2009, we only had 2 byte AS numbers in the range of 1-65535. 1024 of those (64512-65534) are reserved for private AS numbers.
Similar to IPv4, we started running out of AS numbers so IANA increased the AS numbers by introducing 4-byte AS numbers in the range of 65536 to 4294967295.
There are three ways to write down these new 4-byte AS numbers:
• Asplain
• Asdot
• Asdot+
Asplain is the most simple to understand, these are just regular decimal numbers. For example, AS number 545435, 4294937295, 4254967294, 2294967295, etc. These numbers are simple to understand but prone to errors. It’s easy to make a configuration mistake or misread a number in the BGP table.
Asdot represents AS numbers less than 65536 using the asplain notation and AS numbers above 65536 with the asdot+ notation.
Asdot+ breaks the AS number in two 16-bit parts, a high-order value, and a low-order value, separated by a dot. All older AS numbers can fit in the second part where the first part is set to 0. For example:
• AS 6541 becomes 0.6541
• AS 54233 becomes 0.54233
• AS 544 becomes 0.544
For AS numbers above 65535, we use the next high order bit value and start counting again at 0. For example:
• AS 65536 becomes 1.0
• AS 65537 becomes 1.1
• AS 65538 becomes 1.2
These numbers are easier to read but harder to calculate than the asplain numbers, it’s also a bit trickier to create regular expressions.
If you want to convert an asplain AS number to an asdot+ AS number, you take the asplain number and see how many times you can divide it by 65536. This is the integer that we use for the high order bit value.
Then, you take the asplain number and deduct (65536 * the integer) to get your low order bit value. In other words, this is the formula:
``````integer (high order bit value) = asplain / 65536
remainder (low order bit value) = asplain - (integer * 65536)
asdot value = integer.remainder``````
Here are two examples:
``````#AS 5434995
5434995 / 65536 = 82
5434995 - (82 * 65536) = 61043
asdot = 82.61043``````
``````#AS 1499547
1499547 / 65536 = 22
1499547 - (22 * 65536) = 57755
asdot = 22.57755``````
Once you understand how the conversion is done, you can use the APNIC asplain to asdot calculator to convert this automatically and make your life a bit easier.
BGP speakers that support 4-byte AS numbers advertise this via BGP capability negotiation and there is backward compatibility. When a “new” router talks to an “old” router (one that only supports 2-byte AS numbers), it can use a reserved AS number (23456) called AS_TRANS instead of its 4-byte AS number. I’ll show you how this works in the configuration.
## Configuration
Cisco routers support the asplain and asdot representations. The configuration is pretty straightforward and I’ll show you two scenarios:
• Two routers that both have 4-byte AS support.
• Two routers where one router only has 2-byte AS support.
### 4-byte AS support
We have two routers:
Both routers support 4-byte AS numbers. You can see this when you configure the AS number:
``````R1(config)#router bgp ?
<1-4294967295> Autonomous system number
<1.0-XX.YY> Autonomous system number``````
As you can see, this IOS router supports asplain and asdot numbers. Let’s pick asplain and establish a BGP neighbor adjacency:
``````R1(config)#router bgp 12000012
R1(config-router)#neighbor 192.168.12.2 remote-as 12000012``````
``````R2(config)#router bgp 12000012
R2(config-router)#neighbor 192.168.12.1 remote-as 12000012``````
You can see the asplain AS numbers in all bgp show commands:
``````R1#show ip bgp summary
BGP router identifier 192.168.12.1, local AS number 12000012
BGP table version is 1, main routing table version 1
Neighbor V AS MsgRcvd MsgSent TblVer InQ OutQ Up/Down State/PfxRcd
192.168.12.2 4 12000012 5 5 1 0 0 00:01:02 0``````
If you want, you can change the representation to the asdot format:
``````R1(config-router)#bgp asnotation ?
dot asdot notation``````
Let’s change it:
``R1(config-router)#bgp asnotation dot``
You will now see the asdot format in all show commands:
``````R1#show ip bgp summary
BGP router identifier 192.168.12.1, local AS number 183.6924
BGP table version is 1, main routing table version 1
Neighbor V AS MsgRcvd MsgSent TblVer InQ OutQ Up/Down State/PfxRcd
192.168.12.2 4 183.6924 6 6 1 0 0 00:02:38 0``````
AS 12000012 now shows up as AS 183.6924.
Configurations
Want to take a look for yourself? Here you will find the startup configuration of each device.
R1
``````hostname R1
!
ip cef
!
interface GigabitEthernet0/1
!
router bgp 183.6924
bgp asnotation dot
neighbor 192.168.12.2 remote-as 183.6924
!
end``````
R2
``````hostname R2
!
ip cef
!
interface GigabitEthernet0/1
!
router bgp 12000012
neighbor 192.168.12.1 remote-as 12000012
!
end``````
### 2-byte AS support
Let’s use two routers. R1 only supports 2-byte AS numbers, R2 supports 4-byte AS numbers:
R1 has no clue what an AS number above 65535 is:
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## Forum Replies
1. Hi Rene,
Very Good Stuff.A quick questions for you …
How R2 know R1 only 2 byte supported before sending any open message ??
br//zaman
2. Hi Zaman,
It doesn’t. When R2 receives a reply from R1 without the “support for 4-octet AS number capability” in its OPEN message, it knows that R1 doesn’t support it.
You can see it in the wireshark capture:
https://www.cloudshark.org/captures/d8e5e9240959
Take a look at the 1st packet from R2 and the 2nd packet from R1. R1 is missing the capability.
Hope this helps!
Rene
3. Hi Rene,
I have question in 2-Byte & 4-Byte AS compatibility situation.
In case of we adding R3 (AS3) and connect it to R2. R1 advertises one prefix (for example: 1.1.1.1/32) to R2 and R2 will forward to R3.
What will be in the AS path on R3? (22222222 1 or 23456 1)?
Thanks,
Minh
4. Hello Minh
This is an excellent question. Essentially what you are asking (allow me to put it more generally) is how is the AS path displayed when there is a 2-byte AS compatible router in the mix?
Well, if you display the AS path on the 2-byte AS compatible router, then you will see the 23456 AS in place of the incompatible 4-byte AS number. So essentially the AS_TRANS attribute replaces the 4-byte AS number. However, the AS4 PATH number is an attribute that is received by the 2-byte AS compatible device, and is transitive. It may not understand it, but it
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5. Thanks Laz for the explanation. | 2,024 | 7,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.835501 |
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# Microneconomics Essay
1707 words - 7 pages
1) a. If we do not have scarce resources, will we have a law of demand? Will we observe price rationing for goods?
The law of demand states the relationship between quantity demanded and price, showing that the lower the price, the higher the demand and vice versa. If we do not have scarce resources, there will still be a law of demand, because all humans are greedy. This means that we will always want more of what is there and demand always initially exceeds supply, but supply will then catch up, and over time will fall behind again, although this bottlenecking' is always temporary. This can be seen in fibre optic cables, as they catapulted the amount of information able to be ...view middle of the document...
2) a. Explain the difference between point elasticity of demand and arc elasticity of demand.
Point elasticity is the measure of responsiveness or sensitivity of quantity demanded to changes in price. However, arc elasticity is the measure of elasticity between two points on the demand curve. At the two points move closer together on the demand curve, it approaches point elasticity.
Point elasticity is calculated by the percentage change in quantity demanded divided by the percentage change in price, which is shown in formula by: ∆Q% P1
∆P% x Q1
Arc elasticity is calculated the same as point elasticity, although instead of the percentage change, it is calculated by the actual difference between the points. This is shown in formula by: Q2-Q1 P1+P2
P2-P1 x Q2+Q1
This formula is used when there is the change in Price and the Quantity, which allows arc elasticity to be more accurate than point elasticity.
b. Construct on the same graph two straight-line demand curves with the same intercept on the vertical axis, with one curve flatter than the other. Can you make a general statement about which one is the more elastic curve? If so, what is it?
This demand curve shows two straight lines. Line A is steeper and represents a less elastic curve while Line B is flatter and represents a more elastic curve. Point elasticity is the measure of responsiveness or sensitivity of quantity demanded to changes in price. If the elasticity is higher (more than 1), it means that there will be a larger response in demand with price, which is illustrated by curve B. One reason for lower elasticity is brand loyalty, with consumers sticking to one brand all the time. A reason for higher elasticity is the number of substitute products, because as soon as one product reduces its price, it will gain more sales than its substitutes.
3) It has been argued that in order to fight drug abuse sensibly, policy makers must understand the demand for drugs
a. Would you expect that the price elasticity of demand for illegal drugs to be greater than, or less than one?
The price elasticity of illegal drugs is less than one, which is inelastic. This is because drug addicts are will to pay anything to get their hit that they crave. If the person is craving the drug very badly, they will be willing to pay a lot of money, as they don't care f or substitutes and need the drug as soon as possible. Even if the price increases, the addicts still need the drug.
b. Given your answer to a, what happens to total expenditure on illegal drugs when their price increases? What do you think would happen to the amount of drug related crime?
Because drug addicts need the drugs, they are going to pay for the drugs regardless of the price, which means that total expenditure of the users will increase. This could result in the users having to leave more essential expenses to pay for drugs, as the drug opportunity cost far outweighs other...
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