Distance formula - Part II

Continuing from my last post, where I told you that the distance between two points can be determined by finding the length of the hypotenuse of the right angle triangle formed from the two points.

We already know that to find the length of the hypotenuse, we apply the Theorem of Pythagoras, which says that c^2 = a^2 + b^2. (The square of the hypotenuse is equal to the sum of the squares of the remaining two sides). So from this, then, we see that we need to determine the lengths of the two sides that we have created by extending lines through our points to join at a right angle. And to find these lengths, all we need to know are the coordinates of our ! 2 points!

For a general triangle, then, we have something like this:

To find the horizontal length, it is just the difference between the two x-coordinates (ie. x2-x1). Think of it as taking a stick that is x2 units long, and chopping off a length of that stick that is x1 units long. The stick that you are left with, x2-x1, is the length of our horizontal side.

The same reasoning applies to find the vertical length, which is the difference between the two y-coordinates (ie. y2-y1). One comment I will make here, is that since we are talking ab! out a length of a side, the length has to be the absolute diff! erence b etween the two points (ie. you can't have a negative side length).

So then, for our general triangle, we have our 2 lengths, and let's call the hypotenuse "d" (as in, the distance between the two points).

So then, if we apply the Theorem of Pythagoras to this triangle we have created, we can come up with the distance formula very easily!

c^2 = a^2 + b^2...... which we can change to read:

d^2 = (x2-x1)^2 + (y2-y1)^2

And so we have:

d = sqrt [(x2-x1)^2 + (y2-y1)^2]

Let's quickly try with 2 points. You can draw the triangle out as I have above to follow along more closely. I will just do the quick calculation for you though.

Find the distance between the points (1,2) and (3,5).

d = sqrt [(3-1)^2 + (5-2)^2]

d = sqrt [(2)^2 + (3)^2]

d = sqrt [4 + 9]

d = sqrt [13]

And that's all there is to it. I hope that I've been able to clearly explain how to derive the distance formula. Once you know where a lot of these formulas come from, you'll never have to worry about memorizing them again! :)

trigonometric identities solver

Square Roots and Great Comebacks

From the time I learned about them, I've been fascinated (probably to an unseemly amount) by the square root. I remember reading about a method for calculating square roots by long hand. There's no point, really; we have calculators to do that for us. (If you have some spare time and you enjoy this sort of thing, see if you can figure out the algorithm from the example at left.)

What use are square roots, anyway, aside from solving math problems about ! the diameters of circular lawns? (Have you ever seen any of those? They must encircle those conical swimming pools we dealt with in calculus class.) Here's one use: They can tell you when how big a lead your favorite basketball team needs to be secure in a win.

A few years ago, I derived a rule for determining when a lead was safe in a basketball game--specifically, an NBA game. (It matters, because the shot clock is different between an NBA game and a WNBA game and a NCAA men's game and a NCAA women's game.) You take the square root of the number of seconds left, and add three. For instance, if there's 3:45 left in the game, that's 225 seconds. Square root of 225 is 15, and you add 3, so an 18-point lead is pretty darned safe with 3:45 left. The "add 3" is for a trey at the buzzer. Go ask the Miami Heat about that 'round about now.



Pretty keen, huh? Although--not to put too fine a point on it--well-known sports statistician Bill James also came up with this very same rule. We'll call it independent discovery, at least on my part. I have no idea whether James stole it from me. Give him the benefit of the doubt, though.

But why? Why should this rule work? Why isn't it just the time remaining divided by some rate at which the team that's behind catches up? If a team can make up a 15 points in 225 seconds and then cap that with a trey to make up the 18, why can't it make up 33 points in 7:30? Or 63 points in 15:! 00?

And the sort-of answer to that is, it can. It's just terribly unlikely. Of course, it's already unlikely that a team can make up 15 points in 3:45, but it's still in the realm of possibility. Asking a team to do that twice in a row is just too much. If it was 100 to 1 against doing it once, doing it twice in a row would be 10,000 to 1 against. On the other hand, making up the same 15 points in twice the time is obviously easier. So in twice the time (7:30, natch), you should be able to make up some deficit in between. According to both me and Bill James--and honestly, are you going to go against both of us?--that deficit is 15 times the square root of 2. That's about 21, and if you add the 3 at the end it makes it 24.

Where on earth does this come from? One place is the drunkard's walk, otherwise known as the random walk (but I think "drunkard's walk" is more evocative). In this mathematical scenario, the eponymous drunkard starts off at some placemark--a lamppost, say. Each moment in time, he takes a step, but in a completely random direction. Might be in the same direction as the last step, might be in the opposite direction, might be anything. So after a bunch of steps, he might end up back at the lamppost where he started...or he might be home.

Odds are, though, he'll be at some intermediate distance. How far from the lamppost? Well, the first step is g! oing to take him one step away for sure. We'll represent this by saying that d(1) = 1, where d(t) is the distance of the drunkard from the lamppost at time t. OK, now what about d(2)? Before that second step, he's one step away from the lamppost. His second step might take him two steps away, if he walks in the same direction, or zero steps away, if he walks in the opposite direction (back toward the lamppost). On average, though, he'll walk in some intermediate direction: let's say, perpendicular to his current progress from the lamppost. The Pythagorean theorem says then that

[d(2)]² = [d(1)]² + 1² = 1² + 1² = 1 + 1 = 2

or, in other words! , d(2) = √2. We can go further. We've already got two examples where d(t) = √t and we'd like to get more. To do that, we'll use a process called induction. Suppose that you have a value of t for which d(t) = √t ; we'll now try to show that d(t) = √(t + 1) . Using the same argument as before--that the drunkard walks in some intermediate direction, we get

[d(t +! 1)]² = [! d(t)]² + 1² = [√t]² + 1² = t + 1

and then we directly get d(t + 1) = √(t + 1) . So as long as we can find a t wh! ere d(t) = √t , we're set; it's true for all greater values of t. But we already have such a value: t = 1! (And t = 2, for that matter.) It turns out, then, that the drunkard's walk, after time t, takes him a distance √t away from the lamppost.



Now, a couple of things. First, this isn't anything like a rigorous demonstration of the square root property of the drunkard's walk. You can look that up if you like. But if you work at it a little, it gives you an inkling of the intuition behind it. Secondly, though, and here we're back on track a bit: What has all this got to do with basketball games?

A basketball game is an alternating sequence of possessions. In each possession, the team with the ball is of course trying to score, and the other team is of course trying to prevent it from scoring. When the ball changes hands, the roles are reversed. In each individual possession, the effect on the score is biased: Only the team with the ball can score, usually. But in each pair of possessions, that bias cancels out, since both teams get a chance with the ball. The margin in the game can move in any direction--just like the drunkard's walk.

If the drunkard starts off 50 steps from home, he could conceivably get home in just 50 steps. But it's ridiculously unlikely: Each of those 50 steps would have to be in exactly the right direction. The square root property ! tells us he'll probably be just a bit over 7 steps from the la! mppost; it would take 2500 steps to get him, on average, 50 steps from his starting point. After those 2500 steps, is he guaranteed to be home? Nope. He still has to be walking in the right direction. But it's at least plausible now.

In the same way, a basketball team that's down 18 points could conceivably make that up by scoring six three-pointers in a row while holding their opponents scoreless. If they did that by fouling and their opponents obliged by missing all of their free throws, the whole deficit could be made up in half a minute or so. But that's as unlikely as the drunkard walking 50 steps in exactly the right direction. Instead, a team will make up its deficit in halting fashion, sometimes making up three points, but other times giving up a point, or staying even, in any particular pair of possessions. The drunkard's walk, in other words, and that's why the square root rules great combacks.

I was going to follow this up with a discussion of sociology and mobilizing people, but this post is getting long (see, I do notice it!) and I'll defer that till next time.

square root rules

Solving Equations Using Cramer's Rule

Here is the video on how to solve equations using Cramer's Rule:





The slides for the! above video are as follows:

Slide 1 of 6:



Slide 2 of 6:



Slide 3 of 6:



Slide 4 of 6:



Slide 5 of 6:



Slide 6 of 6:

solving linear equation

Algebra Tutor will help you solve your problem

When I am at college, I really hate algebra. My Professor who taught is old. He was intelligent but unfortunately not able to transfer knowledge. Sometimes when he is explaining the lessons, I'm sleepy and do not understand.

Until even now, I still do not like everything about algebra. A friend who attended the math major had offered to pursuit of algebra, and I agreed. But the result is i still do not understand.
But one thing is then disturbing me. Algebra is just like the backbone to any field because algebra has applications in all the different fields. So, it is very important to study alg! ebra for our career. Until then I decided to learn algebra online. I start looking for information algebra tutor and I found online algebra tutor.

I'm in the right place, it's the best place for Free online Algebra help that will help to solve my problem about Algebra. Algebra tutoring really help me to improve my algebra. I do not need to move from my room, just sitting in front of the computer and I can learn about algebra. This is very helpful in saving transportation costs and save my time. Live tutors are available around the clock, 24 / 7 365 for Algebra help. If you have problems with Algebra 2, there is Algebra 2 Help.

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Knowledge of Math

Introduction to algebra fraction calculator:

The term algebra states the constants and variables in the equation. Algebra is one of the familiar topics in mathematics. Algebraic fraction is similar to the usual fraction but a difference is that the numerator and the denominator are algebraic expressions. Algebraic fraction simplifying is similar to that of algebraic fractions. Simplifying algebraic expressions is very easy and simple. In online, students can learn about algebraic expressions. In online, algebraic calculator is helpful to solve the given problem. In online calculator, when the denominator and numerator value is entered, calculator automatically simplifies the problem.

Example : Simplifying Algebraic Fractions

Simplify 3b/9b².

Solution:

The given problem is an algebraic expression.,

3b = 3*b

9b² 9*b*b

Like terms gets cancelled,

On simplifying this,

Answer = 1/3b

Need More help with Algebra 1

simplifying algebraic expressions calculator

Example 7.28: Bubble plots

A bubble plot is a means of displaying 3 variables in a scatterplot. The z dimension is presented in the size of the plot symbol, typically a circle. The area or radius of the circle plotted is proportional to the value of the third variable. This can be a very effective data presentation method. For example, consider Andrew Gelman's recent re-presentation of health expenditure/survival data/annual number of doctor visits per person. On the other hand, Edward Tufte suggests that such representations are ambiguous, in that it is often unclear whether the area, radius, or height reflects the third variable. In addition, he reports that humans tend not to be good judges of relative area.

However, other means of presentin! g three dimensions on a flat screen or piece of paper often rely on visual cues regarding perspective, which some find difficult to judge.

Here we demonstrate SAS and R bubble plots using the HELP data set used in our book. We show a plot of depression by age, with bubble size proportional to the average number of drinks per day. To make the plot a little easier to read, we show this only for female alcohol abusers.

SAS

In SAS, we can use the bubble statement in proc gplot. We demonstrate here the use of the where data set option (section 1.5.1) for subsetting, which allows us to avoid using any data steps. SAS allows the circle area or radius to be proportional to the third variable; we choose the radiu! s for compatibility with R. We alter the size of the circles ! for the same reason. We also demonstrate options for coloring in the filled circles.

libname k "c:\book";

proc gplot data = k.help (where=((female eq 1) 
                    and (substance eq "alcohol")));
bubble cesd*age=i1 / bscale = radius bsize=60 
                           bcolor=blue bfill=solid;
run;

R

In R, we can use the symbols() function for the plot. Here we also demonstrate reading in data previously saved in native R format (section 1.1.1), as well as the subset() function and the with() function (the latter appears in section 1.3.1). The inches option is an arbitrary scale factor. We note that the symbols() function has a great deal of additional capability-- it can substitute squares for circles for plotting the third variable, and add additional dimensions with rectangles or stars. ! Proportions can be displayed with thermometers, and boxplots can also be displayed.

load(url("http://www.math.smith.edu/sasr/datasets/savedfile"))
femalealc = subset(ds, female==1 & substance=="alcohol")
with(femalealc, symbols(age, cesd, circles=i1, 
                                 inches=1/5, bg="blue"))

The results are shown below. It appears that younger women with more depressive symptoms tend to report more drinking.

scale factor examples

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A "Thank You" note from my student

I received this email from my student a couple of days ago.

"Dear Daniel,

Your math tutoring was perfect... the test was a challenge but I felt so well prepared. I called today to get my mark. I got 94%!!!!!!

Thank you again!"

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Snappy Ed App Saturday (#1, Number Line)

Ever consider the educational possibilities for an iTouch or iPhone? Using educational apps?

A good friend is set to begin teaching 3rd grade this fall. The district received a grant, enabling every third grade student to have an iTouch. She (and I) are looking for apps to promote skills practice and concept development in math. She'll be using them in her classroom. I'll be giving them to my student when we're sitting at an appointment or standing in line at the grocery store. Over time I will post apps that appeal to a variety of ages and ability levels. I'm less interested in drill/flashcard/traditional "story problem" material than I am in games and activities that present skill practice and/or concept development in a unique (fun!) way.

As long as I continue to find quality apps, I'll post a couple new links each Saturday, usually by math topic.

I'd love your help! Please comment with names of your favorites; I'm looking for FREE or very inexpensive apps. Many free apps seem to be slowly adding advertisements, although I haven't found it to be a major problem. Yet.

Let's begin with two number line apps (at posting they are both FREE):

Number Line
Player drags circles containing a fraction, decimal or percent onto the number line in order from lowest to highest. Here's a demo.






Pearl Diver

I listed this earlier this week, but want to mention it again.  I've played it several times since then and marvel at it's design.

In this game, available on the iPad/iPhone/iPod Touch (or play online) the "pearl diver" must swim to the bottom of the sea, while avoiding electronic eels. The "sea" is a number line with increasing levels of difficulty. I played with numberlines that included: 1-10, negative to positive numbers, fractions, etc. One time I made it to the sixth level or so (my excuse is that I was interrupted by a little helper!) and could barely figure out the mixed fractions fast enough to survive. My 7yo enjoys the first several levels; he isn't ready for the more difficult problems. He loves the "cutting Sushi" activity between levels. Players cut Sushi into fractional parts and are rewarded points based on how close they get to the target. Very fun, interesting game for numberline skills practice and development of number sense. [Note: you can play online -OR- download for iPod]

number line template

Help on math homework

Introduction about mathematics homework help tutorial
In this article you will get help on maths homework with the help of the tutor through online. Online help will be easier for the students to their homework problems in mathematics. Students can get instant help from any part of the world. This tutorial also include word problems. Below are some of the mathematics homework problems.

Help on Homework Problem in Mathematics:

1. Add 324 + 261

Tutor Solution
3 2 4 Add the ones digit 4 + 1 = 5
2 7 1 + Add tens digit 2 + 7 = 9
---------- Add hundred’s digit 3+ 2 = 5
5 9 5
---------- This could also help us on formula! for circumference

Keep reading may be in the next lesson let me help you with Algebra 2 Calculator

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Telegraph TOI : IIT JEE 2010 Cut-off Raises Brows

Telegraph :: 08 June 2010

IIT cut-off raises brows

CHARU SUDAN KASTURI

New Delhi, June 7: A steep hike in subject cut-offs in this year’s IIT Joint Entrance Examination (JEE) has raised concerns within the IIT fraternity over whether corrective measures employed to compensate question paper errors possibly helped undeserving candidates.

The concerns were dismissed by a top IIT-JEE official as “tilting at windmills”. But the fact that the aggregate cut-off did not change much in 2010, suggesting that the difficulty level of questions was similar to previous years’, offers legitimacy to these concerns.

The IIT-JEE this year was ravaged by errors in instructions given in question papers. One of the major errors involved instructions, which erroneously switched the labelling on physics and math answer scripts, which are read by an optical mark reader (OMR) machine.

Instructors at all test centres announced the error once it was recognised, but many students had already started marking answer scripts by that time — and these answer scripts were not replaced with fresh ones. Amid fears that many students might have marked physics answers where they were to mark math answers and math instead of physics for no fault of theirs, the IITs announced a “corrective measure”.

The OMR machine would mark answer sheets meant for physics (or math) for both subjects and award students the higher ! score. This way, the institutes argued, a student who followed! the ins tructions and marked correct physics (or math) answers on the math (or physics) answer sheet would not lose out.

But in an affidavit before Delhi High Court, an IIT Kharagpur computer science professor, Rajeev Kumar, argued that this corrective measure could inadvertently benefit undeserving students. Below par students whose incorrect physics answers matched with correct math answers — or the other way round — could receive scores much higher than they deserve, Kumar argued.

A handful of such questions — where an incorrect answer matched the correct answer of the other subject — would suffice to clear the average cut-off in that subject, he argued.

Kumar’s concerns have now triggered a debate within the IITs and among IIT-JEE observers over whether his arguments need a relook, following ! the hike in subject cut-offs announced by the IITs on Saturday! .

The IITs first shortlist students who clear subject cut-offs in each of the three subjects tested — physics, chemistry and math. An aggregate cut-off of total scores across the three subjects is set next, based on the total number of seats available to finalise selected candidates.

The subject cut-offs in physics, chemistry and math this year are 19, 19 and 17 respectively as opposed to 8, 11 and 11 in these subjects last year, according to data released on Saturday by the IITs. The sam! e formula — the average marks of all students in a subject is set as the subject cut-off — was followed both in 2009 and 2010.

The hike could normally be explained by easier questions — or better average performance — this year. But such a scenario would also mean much higher aggregate scores of students on average — 55 as opposed to 30 last year. This, in turn, would translate into a correspondingly higher aggregate cut-off.

The aggregate cut-off in the 2010 JEE, according to the IITs, was 19! 0 as opposed to 178 last year — a difference statistical! ly much smaller than the variation between aggregate averages in 2009 and this year. In other words, while the average scores of students improved, the toppers performed almost similar to their counterparts last year — suggesting the question papers were overall not much easier in 2010 compared with 2009.

Some candidates scoring higher than they deserved, because of the corrective measure, would offer a possible explanation for the hike in subject marks without a corresponding jump in the performance of qualified candidates.

http://www.telegraphindia.com/1100608/jsp/nation/story_12541547.jsp  

Times of India :: 08 June 2010

IIT JEE average mark almost doubles but question remains

Akshay Mukul, TNN, June 08, 2010

New Delhi: For the first time in four years, the IIT-JEE average marks announced on Saturday has shown double digit scores in all three subjects — physics, chemistry and mathematics.  The aggregate average almost doubled — from 29 in 2009 to 55 in 2010 out of a total 489 — while the cut-off for successful candidates showed a marginal improvement from 178 last year to 190 this year. 

Is the increase in cut-off — calculated on the basis of average performance of candidates — a case of improved standards or is it due to the corrective measures applied by IITs to neutralise errors in the JEE question paper as pointed out by TOI? 

In order to paper over its mistakes, IIT-JEE, for the first time, evaluated each answer sheet in two ways — sequential question numberwise and subject headingwise. The higher score of the two systems was taken as a candidate’s score. This strategy wa! s adopted to offset the interchange of subject headings of mathematics and physics in question paper and Optical Response Sheet. IIT is not disclosing the marks scored by both ways of evaluation. Now, IIT insiders are pointing out faults with this method. 

Normally, IITs evaluate the JEE question paper by the sequential method.  A senior faculty member of IIT-Delhi said, “Sequencewise evaluation is how evaluation should be done. But subject ! heading-wise evaluation leads to going through mathematics ans! wers on the basis of instructions which was for physics. A similar thing happened in case of physics answers. The second way of evaluation may give higher score to candidates who performed poorly.” JEE is a traditionally low-scoring examination. 

The cut-off for general category candidates this year is 190, a slight improvement from last year’s 178. However, this year’s average marks for mathematics, physics and chemistry is 17, 19 and 19 respectively, making the average marks 55 out of 489. The same procedure was applied for deciding cut-off in 2009 and 2010. If we compare the average marks of 2009 and 2010, it is clear that cut-off of physics this year is almost 2.4 times last year’s cut-off. Mathematics cut-off is 1.5 times more than last year. 

This is much higher than the subject average of 2009 that was 11 (mathematics), 8 (physics) and 11 (chemistry). The average marks last year was 29 out of 489. Thus, the overall performance in 2010 is 1.9 times better than last year. The marginal increa! se in cut-off — 178 to 190 — is indicative that th! e perfor mance of the selected candidates has marginally improved. However, the overall performance is almost double. “A clearer picture,” a source said, “will emerge only when IITs reveal both ways of evaluation.”

http://timesofindia.indiatimes.com/india/IIT-JEE-average-scores-double-cutoff-goes-up-to-190-from-178/articleshow/6021912.cms

math answers

Word Problems and Unit Conversions

For some people, working with variables that have different units is very easy. They can conceptualize the relationship between the different variables in a way that seems natural to them, yet remains very difficult to explain to others. There are even more people who cannot do this type of math so automatically, and often struggle to solve problems... myself included. However, there is a trick to solving these questions, and the solution lies, quite simply, in keeping track of the units. I will try to demonstrate.

I will start simply, with multiplication and division. You know that if one number is on the top and on the bottom of a division line, you can cancel them out to reduce things:

You also know that you can do this when you are working with variables, like x and y:

So then, it's only natural that you can do this with units as well!

Now, here is the trick. If you are struggling with the concept of a word or other problem that deals with units, change the way that you are looking at it, to focus on the units involved. They will guide you towards the final solution.

Let's try a basic physics problem: How far will an object travel in 10 seconds if it is moving at a constan! t speed of 25 meters/second? This one might not be too di! fficult to conceptualize, but it demonstrates my point. Watch what I do with the units at the start, if you are not familiar with this part.

As you can see, the units cancel out to leave just the 'meters.'

Let's try one that is a bit more complicated: If a force of 10 N pushes a 10 kg block along a flat, frictionless surface for 5 seconds, what is the block's average velocity? This one is undoubtedly more difficult to do in your head, but if you keep track of the units (what you have, and what you need), then you should be able to do it. For those who aren't in physics, it is easier to know that the uni! t Newton, N, is equal to (kg) x (m) / (s)^2... blogger is still terrible with superscripts and equations. :(

As you can see, we wanted velocity (m/s) in the end, and so I just worked with the units until I got what I wanted. From a physics point of view, I used Force = Mass x Acceleration, and Velocity = Acceleration x Time.

Currency conversions are a common everyday example of where you can use this trick. If I say that the exchange rate is 0.985 $US / $CAN, and I want to know how much $US I can exchange for $232 CAN:

Naturally, it's best to try to understand the concepts of the questions, and the more complex questions won't be quite so easy to muddle through just by tracking the units. However, hopefully you can see the benefit of looking at these types of problems from this angle, and it will help you to solve them. I learned this technique in high school, and I still benefit from it. As always, feel free to drop me a line if you would like me to clear anything up. :)

help with math word problems

Fall Problem Solving

The following open-ended assessments require students to apply mathematical concepts and skills to solve problems and explain their thinking using words, pictures and numbers.

• Candy Corn presents a triangular numbers problem using a candy corn pattern. Younger students might use candy corn to model the problem. A sample solution shows how older students might use an input-output table to model the pattern and find the solution without the use of manipulatives.
• Younger students will enjoy analyzing and completing Fall Patterns.  They should then classify the pattern and explain their reasoning as part of the class discussion. It is possible that students will see different patterns in some of the items so their explanation and justification are very important mathematical discourse.
• Annual Fall Parade challenges students to use the triangular pattern to figure out how many students are in the fourth grade. Given the number of full rows, students must apply the pattern and use effective recording (picture, table, etc.) to explain their reasoning.
• See more Problem Solving Resources from the Mathwire collections.  All problems are classroom-ready in PDF format.  Many contain a sample solution for teacher use.
  
  
  
  
  
  
  
  

graphing coordinates to make a picture

Online Statistics Tutorial

We had our 3rd Parent Teacher Conference last Sunday. The conference helps analyze the students performance in the class. All parents were given at least ten to fifteen minutes time to talk with respective teacher/adviser regarding the performance of students in the class. Honestly, even without seeing my daughter's report card, I already knew how she responds in the class. My daughter got a superior grade in IQ Test and English and I'm just so proud of her but not with this subject, statistic. I didn't bother to blog about her superior grade in other subjects because I'd focussed on the subject where she got an average grade. I'm looking for Statistics help for her. I'll be happy and contented with above av! erage grade if ever she did not achieve superior grade in Statistic and so we're now actually looking for Statistics tutor.

My two girls have a hard time understanding Statistic. As early as today I already know that they will encounter harder Statistics problems later. My older daughter has been asking me what course in college she will take without any Statistics questions. But I told her that everything now can be learned online. I told her there's free Statistics help to run to! . I en couraged her to take a course that she loves even with Math or Statistic. Everything is available online and Statistics answers can be found at Tutor Vista. Over four million are now depending on online tutorial and everything is right there at the comfort of your home with unlimited tutoring session. Free Demo Session is given to first time user and so everything is just made easy with online tutorial.

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Comprehensive Online Math Tutoring By TutorVista.com

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Combining Graph Types

I got a call from my good friend Sandee Cohen (aka vectorbabe) late Friday and our conversation revolved around the topic of graphs in Illustrator. At one point, Sandee had mentioned that she was working with a client who often compared two sets of data using two different graph types -- the result looked as if you had overlayed one graph directly upon another. This can be done in Excel, but Sandee wondered if that could be done in Illustrator as well. My response was "Yes! Illustrator can do that..." and rather than describe to Sandee how it's done, I promised her I'd post it up here on my blog so that everyone else can benefit from it as well...

It's that time of year, and I'm deciding who to draft for my baseball fantasy team. So of course, I've got baseball stats on my mind. And I wondered if it might be interesting to compare the number of home runs a player hits over the course of many seasons to the number of strikeouts that players has had over the same pe! riod of time. It might be interesting to see if there's a trend (for example, as home runs go up, do strikeouts also go up, etc.).

So here are the steps I took to create a chart that compared this data -- using Illustrator's Graph tool.

1. I needed some data on a player, and I picked an easy target -- Barry Bonds. After all, he's been playing for 20 years, has plenty of home runs, and isn't known to strike out very often. It's amazing what you can find on the internet. With the help of Google, not ten seconds passed before I accessed the career stats for Barry -- and I quickly entered the information into an Excel spreadsheet. I should note that using Excel isn't necessary -- I could input data directly into Illustrator, but I added it to Excel for a variety of reasons. First, I'd like to use this data later and it would be nice to have the file handy. Second, most people get data sent to them in Excel format, so this tutorial would be more "complete" ! this way. And third, I wanted to point out something important! about h ow you format data that will be used in Illustrator.

I wanted to list all the years Barry has been playing, which runs from 1986 through 2006. But Illustrator looks at the numbers 1986, 1987, etc. as actual data numbers, not years. If you want a number to be perceived as a value in Illustrator (and not a data point), you must enclose that number in quotation marks, as you can see below.



2. I copied the data from Excel and then switched to Illustrator, and created a new document.

3. I selected the Column Graph tool and dragged out a wide rectangle. This action defined the area for the graph and brought up the Graph Data window. I then chose Edit > Paste to bring the data from Excel into my graph's data window, which looked like this:



Again, it's important to emphasize that I had to place quotation mar! ks around the years, or Illustrator would have charted those numbers as data points.

4. I applied the data to my graph by clicking on the checkmark icon in the Graph Data window (upper right corner). I then closed the Graph Data window. My resulting graph appeared as so:



I know, the graph doesn't look very charming, and because there are so many years, the labels across the bottom of the graph are all overlapping each other. Don't worry -- we'll make everything look fabulous soon enough. Which by the way, is one of the reasons why it's NEVER a good idea to have a client sit with you while you work (I can just hear them saying "but I don't WANT the chart to be colored gray...").

5. OK, so the first order of business is to use a different type of chart to show the different sets of data. I like the column format for the number of strikeouts, but it would be totally cool if I could di! splay the home runs as a line graph, so it would be easier to ! track th em. Using the Group Selection tool, I clicked TWICE on the rectangle next to the word Home Runs in the LEGEND of the chart.

6. At this point, all of the values for the Home Runs in my chart are selected. I then double-clicked on the Graph tool in the Tools palette (toolbar, toolbox, whatever you call it). This action brings up the Graph Type dialog box.

7. Since I wanted to display the Home Runs as a line, I selected the Line option in the Graph Type dialog, and clicked OK. The chart now updated to display the Home Runs as a line graph, while keeping the Strikouts as a column graph. The result looked like this:



8. Now that the data and the graph types were complete, I was able to make the graph look "pretty". I used colors to differentiate the years that Barry was on the Pirates (yellow bars) and on the Giants (orange bars), I adjusted the type to look nice, changed the color of the! line, and added a drop shadow to the line as well. The final graph looked like this:



Looking at the final chart, one can quickly see that as of 1999, Barry has had significantly fewer strikeouts compared to his hitting more home runs. I would argue that the number of walks have also increased during those years, which would have an effect on that, and maybe adding walks to the graph would help visualize that as well -- consider that homework for my dedicated readers...

So there you have it -- a single chart with different graph types displaying different data points. If you need to edit the data, you can simply edit that one chart, and everything will update nicely. Even the formatting will remain intact, as the final graph was never ungrouped.

Thanks for the request Sandee!

Types of Graph

Transportation Problem

Transportation Method

A transportation tableau is given below. Each cell represents a shipping route (which is an arc on the network and a decision variable in the LP formulation), and the unit shipping costs are given in an upper right hand box in the cell.

To solve the transportation problem by its special purpose algorithm, the sum of the supplies at the origins must equal the sum of the demands at the destinations (Balanced transportation problem).

• If the total supply! is greater than the total demand, a dummy destination is added with demand equal to the excess supply, and shipping costs from all origins are zero.

• Similarly, if total supply is less than total demand, a dummy origin is added. The supply at the dummy origin is equal to the difference of the total supply and the total demand. The costs associated with the dummy origin are equal to zero.

When solving a transportation problem by its special purpose algorithm, unacceptable shipping routes are given a cost of +M (a large number).

Develop an Initial Solution

Two methods to get the initial solution:

• Northwest Corner Rule

• Minimum Cell-Cost Method

Northwest Corner Rule

1. Starting from the northwest corner of the transportation tableau, allocate as much quantity as possible to cell (1,1) from Origin 1 to Destination 1, within the supply constraint of source 1 and the demand! constraint of destination 1.

2. The first allocation ! will sat isfy either the supply capacity of Source 1 or the destination requirement of Destination 1.

Æ' If the demand requirement for destination 1 is satisfied but the supply capacity for Source 1 is not exhausted, move on to cell (1,2) for next allocation.

Æ' If the demand requirement for destination 1 is not satisfied but the supply capacity for Source 1 is exhausted, move to cell (2,1)

Æ' If the demand requirement for Destination 1 is satisfied and the supply capacity for Source 1 is also exhausted, move on to cell (2,2).

3. Continue the allocation in the same manner toward the southeast corner of the transportation tableau until the supply capacities of all sources are exhausted and the demands of all destinations are satisfied.

Initial tableau developed using Northwest Corner Method

Total Cost = 12(400)+13(100)+4(700)+9(100)+12(200)+4(500)= 142,000

Minimum Cell-Cost Method

Although the North-west Corner Rule is the easiest, it is not the most attractive because our objective is not included in the process.

Steps of Minimum Cell-Cost Method

1. Select the cell with the minimum cell cost in the tableau and allocate as much to this cell as possible, but within the supply and demand constraints.

2. Select the cell with the next minimum cell-cost and allocate as much to this cell as possible within the demand a! nd supply constraints.

3. Continue the procedure until! all of the supply and demand requirements are satisfied. In a case of tied minimum cell-costs between two or more cells, the tie can be broken by selecting the cell that can accommodate the greater quantity.

Initial tableau developed using Minimum Cell-Cost Method

Total Cost = 12(300)+4(200)+4(700)+10(100)+9(200)+4(500)= 120,000

MODI Method (for obtaining reduced costs)

Associate a number, ui, with each row and vj with each column.

• Step 1: Set u1 = 0.

•! ; Step 2: Calculate the remaining ui's and vj's by solving the relationship cij

= ui + vj for occupied cells.

• Step 3: For unoccupied cells (i,j), the reduced cost = cij - ui - vj.

Step 1: For each unoccupied cell, calculate the reduced cost by the MODI method. Select the unoccupied cell with the most negative reduced cost. (For maximization problems select the unoccupied cell with the largest reduced cost.) If none, STOP.

Step 2: For this unoccupied cell, gen! erate a stepping stone path by forming a closed loop with this! cell an d occupied cells by drawing connecting alternating horizontal and vertical lines between them. Determine the minimum allocation where a subtraction is to be made along this path.

Step 3: Add this allocation to all cells where additions are to be made, and subtract this allocation to all cells where subtractions are to be made along the stepping stone path. (Note: An occupied cell on the stepping stone path now

becomes 0 (unoccupied). If more than one cell becomes 0, make only one unoccupied; make the others occupied with 0's.)

GO TO STEP 1.

Example: Acme Block Co. (ABC)

Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, and Eastwood -- 10 tons. Acme has two plants, each of which can produce 50 tons per week. Delivery cost per ton from each plant to each suburban location is shown below.

How should end of week shipments be made to fill the above orders?

Since total supply = 100 and total demand = 80, a dummy destination is created with demand of 20 and 0 unit costs.

Iteration 1: Tie for least cost (0), arbitrarily select x14. Allocate 20. Reduce s1 by 20 to 30 and delete the Dummy column.

Iteration 2: Of the remaining cells the least cost is 24 for x11. Allocate 25. Reduce s1 by 25 to 5 and eliminate the Northwood column.

Iteration 3: Of the remaining cells the least cost is 30 for x12. Allocate 5. Reduce the Westwood column to 40 and eliminate the Plant 1 row. Iteration 4: Since there is only one row with two cells left, make the final allocations of 40 and 10 to x22 and x23, respectively.

1. Set u1 = 0

2. Since u1 + vj = c1j for occupied cells in row 1, then v1 = 24, v2 = 30, v4 = 0.

3. Since ui + v2 = ci2 for occupied cells in column 2, then u2 + 30 = 40, hence u2 = 10.

4. Since u2 + vj = c2j for occupied cells in row 2, then 10 + v3 = 42, hence v3 = 32.

Calculate the reduced costs (circled numbers on the p! revious slide) by cij - ui - vj.

Unoccupied Cell Reduced Cost (1,3) 40 - 0 - 32 = 8 (2,1) 30 - 24 -10 = -4 (2,4) 0 - 10 - 0 = -10

Iteration 1:

The stepping stone path for cell (2,4) is (2,4), (1,4), (1,2), (2,2). The allocations in the subtraction cells are 20 and 40, respectively. The minimum is 20, and hence reallocate 20 along this path. Thus for the next tableau:

x24 = 0 + 20 = 20 (0 is its current allocation) x14 = 20 - 20 = 0 (blank for the next tableau) x12 = 5 + 20 = 25

x22 = 40 - 20 = 20

The other occupied cells remain the same.

1. Set u1 = 0.

2. Since u1 + vj = cij for occupied cells in row 1, then v1 = 24, v2 = 30.

3. Since ui + v2 = ci2 for occupied cells in column 2, then u2 + 30 = 40, or u2 =

10.

4. Since u2 + vj = c2j for occupied cells in row 2, then 10 + v3 = 42 or v3 = 32;

and, 10 + v4 = 0 or v4 = -10.

Iteration 2

Calculate the reduced costs (circled numbers on the previous slide) by cij - ui - vj.

Unoccupied Cell Reduced Cost (1,3) 40 - 0 - 32 = 8 (1,4) 0 - 0 - (-10) = 10 (2,1) 30 - 10 - 24 = -4

The most negative reduced cost is = -4 determined by x21. The stepping stone path for this cell is (2,1),(1,1),(1,2),(2,2). The allocations in the subtraction cells are 25 and 20 respectively. Thus the new solution is obtained by reallocating 20 on the stepping stone path. Thus for the next tableau:

x21 = 0 + 20 = 20 (0 is its current allocation)

x11 = 25 - 20 = 5

x12 = 25 + 20 = 45

!

x22 = 20 - 20 = 0 (blank for the next tableau)

!

The o ther occupied cells remain the same.

1. Set u1 = 0

2. Since u1 + vj = c1j for occupied cells in row 1, then v1 = 24 and v2 = 30.

3. Since ui + v1 = ci1 for occupied cells in column 2, then u2 + 24 = 30 or u2 =6.

4. Since u2 + vj = c2j for occupied cells in row 2, then 6 + v3 = 42 or v3 = 36, and 6 + v4 = 0 or v4 = -6.

Iteration 3

Calculate the reduced costs (circled numbers on the previous slide) by cij - ui - vj.

Unoccupied Cell Reduced Cost

(1,3)	40 - 0 - 36 =	4
(1,4)	0 - 0 - (-6) =	6
(2,2)	40 - 6 - 30 =	4

Since all the reduced costs are non-negative, this is the optimal tableau.

linear programming problems(LPP)