Global Transportation for Sustainability- myassignmenthelp.com

Questions: 1.Why did the writer of the current week's perusing think that its important to look at the European and North American frameworks; would they say they are actually that unique? 2.How do expenses of transportation sway worldwide exchange? Answers: 1. As I would see it, the creators decision to look at the transportation frameworks in the European and the North America as crucial. In the first place, the transportation framework in the two topographical locales is firmly related. They are both very much kept up and set up. Much the same as in the United States, the European nations have a closeness to one another (Clausen Voll, 2013). Moreover, both have built up rules to upgrade the development of merchandise and items inside and outside their separate fringes. Both the North America and EU nations utilize comparable circulation centers or focuses, for example, trains and pipelines. Additionally, the two of them have a comparability in the transportation of merchandise framework. Finally, both the cargo terminals and gear are exclusive in the two locales. In the midst of the likenesses, there are additionally existing contrasts between the frameworks. The distinctions are realized by issues, for example, duties, charges, hardware, association, showcase center, proprietorship, and separation. In Europe, there is a division of activities and foundation association for the reasons for bookkeeping while in North America, tasks and framework are isolated by district. Second, the EU advertise is traveler situated while the North Americas is Freight arranged (Clausen Voll, 2013). 2. Clearly, the transportation cost greatly affects the worldwide exchange in light of the fact that an association will likewise pick a most efficient methods for transport. The financial development in the EU has incited an expansion of trucks occupied with the cross-fringe exchanges (Leinbach Capineri, 2007). The European nations can without much of a stretch vehicle products over their fringe utilizing trucks. The pattern can without much of a stretch jeopardize the utilization of different methods of transport like air and water. Notwithstanding, North America can just utilize air and boat goes to move products universally. Trucks can't be utilized on the grounds that the district is encircled by the Oceans (Leinbach Capineri, 2007). References Clausen, U., Voll, R. (2013, Feb 14). examination of North American and European railroad frameworks. Recovered from https://download.springer.com/static/pdf/596/art%3A10.1007%2Fs12544-013-0090-4.pdf? originUrl=https://link.springer.com/article/10.1007/s12544-013-0090-4token2=exp=1449775695~acl=/static/pdf/596/art%253A10.1007%252Fs12544-013-009 Leinbach, T. R., Capineri, C. (2007). Globalized Freight Transport : Intermodality, E-business, Logistics and Sustainability. Cheltenham, UK: Edward Elgar.

European States in 18th century, The French Revolution Assignment

European States in eighteenth century, The French Revolution - Assignment Example There was an endeavor to legitimize the monarchial framework and its laws. In France, there was a weakening of the government, trailed by the French Revolution, whose Jacobin program saw the conclusion to chapel force, and changes in the measurement, money related and lawful frameworks, in light of discerning idea and the perfect of social equity. Illuminated Politics, generally, was showed in ‘Enlightened Despots’ in Europe, whose force laid not on divine right, however on the need of efficient government for the government assistance of the individuals. Frederick the Great of Prussia presented more noteworthy strict opportunity, prodded financial enhancements and classified the legitimate framework. Joseph II of Austria likewise set out on state-supported enhancements and shortened the forces of the Catholic Church. In the eighteenth century, the Industrial Revolution saw Britain develop as a worldwide force, with particular favorable circumstances over the mainland states. A central point was British accomplishment in working up huge settlements, especially in the East and West Indies, and North America, which gave merchandise for exchange mainland Europe and furthermore a business opportunity for local products. The solidness of abroad exchange was ensured by Britain’s maritime force, which forestalled war-time interruptions, in contrast to France. Populace development and urbanization, and rising ways of life prompted expanded mass utilization and empowered large scale manufacturing. Rather than the ancien system and primitive structure predominant in mainland Europe, Britain’s prosperous working class took an interest in the political and financial framework. A developing vote based system and the shortening of outright monarchial impact over property advanced monetary developm ent. Britain’s transport arrange was further developed than that of the mainland. This encouraged the modest vehicle of the results of the Industrial Revolution. Britain’s Agricultural Revolution changed over little family land property, or

Ironies of Emancipation Essay

The article â€Å"Ironies of Emancipation: Changing Configurations of Women’s Work in the ‘Mission of Sisterhood’ to Indian Women† by Jane Haggis is an article that was composed from a women's activist point of view to pose explicit inquiries about how the impact of the crucial sisterhood either aided or hurt Indian ladies in the 1800’s. The creator takes the position that Indian ladies were bound to relationships all the more intensely in view of the nearness of female teachers in India. The creator brings up that the female ministers turned out to be right around a â€Å"symbol of liberation as far as British royal feminism† and hence alone, the Indian populace was discontent with the manner in which the females introduced themselves and didn't change (Haggis123). The evangelist work, in of itself, was â€Å"women’s work† and was a manner by which the British ladies had the option to apply their own autonomy while doing the desire of God. This article concentrates a lot on how the religion of the time influenced the manner by which British and Indian ladies conveyed and connected with one another and was an explanation behind the result, either the disappointment or the achievements, of the Mission of Sisterhood. The creator portrays how these ladies were relied upon to do their â€Å"women’s work† regardless of the household obligations that were anticipated from a Victorian spouse and mother, for the most part on the grounds that in Travencore ladies had an alternate job as far as parenthood in light of the fact that their kids were regularly sent away to class to get legitimate British instruction. This left them ready to seek after different obligations, for example, their strategic the Indian ladies (Haggis 119). This article makes a ton of valid statements about how British ladies were engaged with preacher work in India in Victorian occasions and clarifies their job in their general public, both at home and abroad. It’s fascinating to take note of that the Indian ladies were maybe held back in light of the impact of these ladies on their way of life, rather than being changed over. Work Cited Haggis, Jane. â€Å"Ironies of Emancipation: Changing Configurations of ‘Women’s Work’ in the ‘Mission of Sisterhood’ to Indian Women. † Feminist Review 65 (2000): 108-126. JSTOR. 5 Feb. 2007.

Changing the Scope of Practice for All Nurses Free Essays

Changing The Scope Of Practice For All Nurses October 21, 2012 Changing The Scope Of Practice For All Nurses The Institute of Medicine and the Robert Wood Johnson Foundation has proposed numerous progressions happen in the field of training for attendants to aid the Affordable Care Act to be a full accomplishment in changing human services. They have distributed the report â€Å"The Future of Nursing Leading Change Advancing Health† suggesting that the extent of training for all medical attendants which incorporate the enrolled nurture, advance practice nurture, guaranteed nurture anesthetist, and the ensured nurture birthing specialist to have the option to rehearse at their instruction and aptitude level through proceeding with abilities. Presently it is dependent upon all attendants to choose if changing the extent of training for medical caretakers is a positive or negative thought. We will compose a custom exposition test on Changing the Scope of Practice for All Nurses or then again any comparable theme just for you Request Now This paper will show how changing the extent of training will empower medical caretakers to think about their patient all the more viably. Attendants will need to take care of business, and show the entire world how significant nursing is to giving medicinal services to all individuals. What's more, for the Affordable Care Act to really change the medicinal services framework to all the more likely assistance the individuals and networks of the United States, the extent of training for attendants should change to more readily address the issues of people in general. Training The Institute of Medicine report empowers attendants who graduate with a partner degree to proceed with their instruction to at any rate the baccalaureate level. Medical caretakers at the baccalaureate level are then urged to proceed to the bosses or doctorate certificate. The various establishments and the central government acquainting subsidizing openings and grants with understudies attempting to discover ways for schools and government to encourage nurse’s training to more significant level to be increasingly moderate (â€Å"The Future of Nursing† 2010. p. 177). It is imperative to stretch the need to pick up medical caretakers with the information to show our future attendants. Medical caretakers will likewise should be engaged with proceeding with training and proceeded with abilities. Where attendants keep awake to date with current practice and can exhibit information and aptitude in the training (â€Å"The Future of Nursing† 2010. . 202). Likewise, outfitting attendants with the abilities and information to deal with patients of today, who have more comorbidities than any other time in recent memory. Furnishing medical attendants with the information and aptitude to preform the competency as well as comprehend the how and why it functions, which will be a key factor in changing medicinal services. One proposition is that â€Å"Interprofessional instruction of doctors, medical attendants, and other wellbeing experts, just as new strategies for improving and showing competency all through one’s career†¦Ã¢â‚¬  (â€Å"The Future of Nursing† 2010. p. 165). This will permit better correspondence, coordinated effort, and regard between everybody engaged with tolerant consideration. The Institute of Medicine likewise talks about various approaches to make attendants proceeding with their training increasingly reasonable. Essential Care The manner in which essential consideration attendants convey care to patients will change if the report by the Institute of Medicine is received. The greater part of the progressions will influence the propelled practice medical attendants, by permitting them to practice to their maximum capacity of information and ability level. They will have the option to deal with patients without needing a doctor on the premises or administering their work consistently. The Institute of Medicine gives a model where in Philadelphia there is an attendant overseen human services community. This inside is controlled by Nurses, Nurse Practitioners, and Social Workers. Other human services experts help when required like nutritionists, doctors, and others when required. Spots like this would have the option to serve regions where there is almost no doctors accessible to meet the social insurance needs of the individuals who live there. Authority By perusing the report, â€Å"The Future of Nursing Leading Change Advancing Health†, it portrays how significant medical attendants are turning out to be more grounded pioneers for the future progression of nursing. With all the progressions accompanying the Affordable Care Act, this report prescribes that medical caretakers should play an enormous contributing move in the improvement of approaches to keep up quiet focused medicinal services, ensuring that human services is available to all, and is reasonable. More grounded administration abilities can be created through proceeded with training. By medical caretakers proceeding onward to higher degree training, they are in this manner building up the administration abilities they should work all the more strongly with others both in the social insurance field and out. Medical caretakers should assume a job in all parts of strategy making. From working in the emergency clinics and networks as promoters for patients which is characterized as â€Å"The nurture is exhibiting the estimation of other-centeredness to propel the soundness of an individual† (Creasia Friberg, 2011, p. 80). Medical caretakers should sit on sheets both in the emergency clinics and in the open division, administering to political authorities, and even potentially holding a political office themselves. It will take all medical attendants to roll out an improvement from nursing understudies to propel practice medical attendants, instructors, and nursing associations authority (â€Å"The Future of Nursing† 2010. p. 221-251). Objectives for Practice For the nursing extent of training to change successfully, it should begin in nursing school. Nursing understudies should begin pointing toward an objective of advanced education. Realizing they should be more ready with the picked up information on proceeding to a baccalaureate, experts, or doctorate qualification in nursing to have the option to deal with the consistently expanding human services needs. Medical caretakers should keep up their insight and ability level for forward-thinking practice through competency based learning all through their profession. They should be the future heads all through the nursing calling. Each medical attendant realizing that nursing comprises of a wide range of angles, trying to help in the progression of every single future attendant. A few attendants decide to go into the clinical field, network nursing, instruction, propelled practice, explore, enactment, strength affiliations, and a lot more to additionally nursing calling and social insurance. Taking everything into account, by changing the extent of training for medical caretakers, they will have the option to convey tolerant consideration that coordinates their insight and aptitude level. This will permit more medicinal services experts predominantly nurture, to rehearse in territories that have needed adequate human services suppliers to address the issues of the general population. The open will profit by having medical attendant experts accessible in country, rural, and urban territories where doctors are rare. In this manner the open will approach moderate, top notch social insurance to all who are out of luck. References Creasia, J. , Friberg, E. (2011). Reasonable Foundations the Bridge to Professional Nursing Practice. (fifth ed. , p. 80). Recovered from http://pageburstls. elsevier. com/books/978-0-323-06869-7 The Future of Nursing Leading Change, Advancing Health (2010). Recovered October 13, 2012, from http://www. iom. edu/Reports/2010/The-Future-of-Nursing-Leading-Change-Advancing-Health. aspx. Step by step instructions to refer to Changing the Scope of Practice for All Nurses, Papers

The New SAT Math What’s Changing

The New SAT Math What’s Changing SAT/ACT Prep Online Guides and Tips Beginning in March 2016, there will be a recently overhauled SAT. The new SAT just has two areas: Evidence-Based Reading and Writing and Math. While the vast majority are centered around the progressions to the Reading and Writing segment, there have been a couple of changes to the SAT Math area that are essential to know. What are these changes? By what method will your SAT investigation technique need to change? I’ll dig into that and more in this guide. Math: The Major Changes in the 2016 New SAT How about we experience every one of the significant adjustments to the math segment of the test. Two Sections: One With Calculator, One With No Calculator On the old SAT, the entire math segment permitted you to utilize an adding machine. On the new SAT, the math area is separated into two bits: one which permits adding machine and one which doesn't. The non-mini-computer part will consistently be the third segment of the test. The number cruncher segment will consistently be the fourth segment of the test. Try not to fear the no-number cruncher area. The explanation you’re not permitted an adding machine is you ought to have the option to unravel these inquiries without one. A portion of the abilities required to respond to these no number cruncher questions include: Basic math (expansion, deduction, increase, division) Streamlining single conditions or expressions (utilizing the FOIL strategy) Illuminating an arrangement of two conditions Knowing square roots (or having the option to locate a square root by increasing) Being acquainted with forces (and how to reconfigure powers). These inquiries can get fairly testing. Here is an example no adding machine question (from an official practice SAT) that expects you to utilize your insight into powers: On the off chance that $3x-y=12$, what is the estimation of ${8^x}/{2^y}$? A) $2^12$B) $4^4$C) $8^2$D) The worth can't be resolved from the data given. Answer Explanation: One methodology is to communicate $${8^x}/{2^y}$$ so the numerator and denominator are communicated with a similar base. Since 2 and 8 are the two forces of 2, subbing $2^3$ for 8 in the numerator of ${8^x}/{2^y}$ gives $${(2^3)^x}/{2^y}$$ which can be revamped $${2^(3x)}/{2^y}$$ Since the numerator and denominator of have a typical base, this articulation can be revamped as $2^(3xâˆ'y)$. In the inquiry, it expresses that $3x âˆ' y = 12$, so one can substitute 12 for the type, $3x âˆ' y$, giving that the $${8^x}/{2^y}= 2^12$$ The last answer is A. Here is an example no adding machine question that expects you to rearrange: On the off chance that $x3$, which of coming up next is comparable to $1/{1/{x+2}+1/{x+3}}$? A) ${2x+5}/{x^2+5x+6}$ B) ${x^2+5x+6}/{2x+5}$ C) $2x+5$ D) $x^2+5x+6$ Answer Explanation: So as to discover the appropriate response, you have to revise the first expression and to do that you have to increase it by ${(x+2)(x+3)}/{(x+2)(x+3)}$. At the point when you duplicate through, you ought to get ${(x+2)(x+3)}/{(x+2)+(x+3)}$. Keep improving by duplicating $(x+2)(x+3)$ in the numerator and streamlining the denominator by finishing the expansion of $(x+2)+(x+3)$. You should then get: $${x^2+5x+6}/{2x+5}$$ That matches answer decision B, so that is the last answer! Less Emphasis on Geometry Geometry took up around 25-35% of inquiries on the old SAT, however it will presently represent under 10% of inquiries on the new SAT. The inquiries will remain generally the equivalent, yet there will basically be less of them. Here is an example Geometry question from another SAT practice test: Answer Explanation: The volume of the grain storehouse can be found by including the volumes of the considerable number of solids of which it is made (a chamber and two cones). The storehouse is comprised of a chamber (with stature 10 feet and base sweep 5 feet) and two cones (each with tallness 5 ft and base range 5 ft). The recipes given toward the start of the SAT Math area (Volume of a Cone $V={1}/{3}ï€r^2h$ and Volume of a Cylinder $V=Ï€r^2h$) can be utilized to decide the absolute volume of the storehouse. Since the two cones have indistinguishable measurements, the all out volume, in cubic feet, of the storehouse is given by $$V_(silo)=Ï€(5)^2(10)+(2)({1}/{3})ï€(5)^2(5)=({4}/{3})(250)ï€$$ which is roughly equivalent to 1,047.2 cubic feet. The last answer is D. Additionally, fairly incidentally, despite the fact that the quantity of Geometry questions is diminishing, the College Board chose to give you more Geometry recipes in the reference segment, which is toward the start of the SAT Math segments. The reference area records a few equations and laws for you to utilize when addressing questions. Here is the old reference area: Here is the new reference area: Notwithstanding the recipes remembered for the old reference area, the College Board has incorporated the volume equations for a circle, cone, and pyramid. Likewise, the College Board gives you an extra law of Geometry: the quantity of radians of circular segment around is 2ï€. For a full rundown of gave equations and recipes you ought to retain, read our manual for recipes you should know. Need to become familiar with the SAT yet worn out on perusing blog articles? At that point you'll adore our free, SAT prep livestreams. Planned and driven by PrepScholar SAT specialists, these live video occasions are an extraordinary asset for understudies and guardians hoping to study the SAT and SAT prep. Snap on the catch underneath to enlist for one of our livestreams today! Expanded Focus on Algebra Variable based math will currently represent the greater part of the inquiries in the SAT math segment. While variable based math was constantly a piece of the math segment, it’s now being stressed significantly more. These inquiries can be precarious on the grounds that they request that you apply variable based math in special manners. A portion of the polynomial math aptitudes required to prevail on the SAT math segment include: Comprehending direct conditions Comprehending an arrangement of conditions Making straight conditions or arrangement of conditions to tackle issues (utilized in the model beneath). Making, dissecting, settling and diagramming exponential, quadratic, and other non-straight conditions. The accompanying model variable based math question is from a genuine new SAT practice question: Answer Explanation: To tackle this issue, you ought to make two conditions utilizing two factors ($x$ and $y$) and the data you’re given. Let $x$ be the quantity of left-gave female understudies and let $y$ be the quantity of left-gave male understudies. Utilizing the data given in the issue, the quantity of right-gave female understudies will be $5x$, and the quantity of right-gave male understudies will be $9y$. Since the complete number of left-gave understudies is 18 and the complete number of right-gave understudies is 122, the arrangement of conditions beneath must be valid: $$x + y = 18$$ $$5x + 9y = 122$$ At the point when you understand this arrangement of conditions, you get $x = 10$ and $y = 8$. Accordingly, 50 of the 122 right-gave understudies are female. Accordingly, the likelihood that a right-gave understudy chose indiscriminately is female is ${50}/{122}$, which to the closest thousandth is 0.410. The last answer is A. Expanded Focus on Modeling The new SAT math segment has another kind of inquiry which pose to you to consider what conditions or models mean. You will be given a model or condition and be approached to clarify what certain parts mean or speak to. These inquiries are strange in light of the fact that they're posing to you to accomplish something you once in a while do: they solicit you to examine the importance from the number or variable in setting instead of explain the condition. Here is an example displaying question from another SAT practice test: Kathy is a fix expert for a telephone organization. Every week, she gets a cluster of telephones that need fixes. The quantity of telephones that she has left to fix toward the finish of every day can be assessed with the condition $P=108-23d$, where $P$ is the quantity of telephones left and $d$ is the quantity of days she has worked that week. What is the importance of the worth 108 in this condition? A) Kathy will finish the fixes inside 108 days.B) Kathy begins every week with 108 telephones to fix.C) Kathy fixes telephones at a pace of 108 for each hour.D) Kathy fixes telephones at a pace of 108 every day. Answer Explanation: In the given condition, $108$ is the estimation of $P$ in $P = 108 âˆ' 23d$ when $d = 0$. When $d = 0$, Kathy has worked $0$ days that week. As it were, $108$ is the quantity of telephones left before Kathy has begun work for the week. In this way, the significance of $108$ in the given condition is that Kathy begins every week with $108$ telephones to fix since she has worked $0$ days and has $108$ telephones left to fix. The last answer is B. Further developed Topics Expansion of Trigonometry Trigonometry had never been asked on the SAT Math section†¦ as of recently! Trigonometry will currently represent the same number of as 5% of math questions. You'll be tried on your insight into sine and cosine. Here is an example trigonometry question from a genuine new SAT practice test: In triangle $ABC$, the proportion of edge $∠B$ is 90â °, $BC=16$, and $AC=20$. Triangle $DEF$ is like triangle $ABC$, where vertices $D$, $E$, and $F$ compare to vertices $A$, $B$, and $C$, separately, and each side of triangle $DEF$ is $1/3$ the length of the relating side of triangle $ABC$. What is the estimation of sin$F$? (This is a framework being referred to, not different decision, so there are no answer decisions recorded with the inquiry.) Answer Explanation: Triangle ABC is a correct triangle with its correct point at B. Subsequently, $ov {AC}$ is the hypotenuse of right triangle ABC, and $ov {AB}$ and $ov {BC}$ are the legs of right triangle ABC. As indicated by the Pythagorean hypothesis, $$AB =√{20^2-16^2}=√{400-256}=√{144}=12$$ Since triangle DEF is like triangle ABC, with vertex F comparing to vertex C, the proportion of $angle ∠{F}$ rises to the proportion of $angle ∠{C}$. In this manner, $sin F = sin C$. From the side lengths of triangle ABC, $$sinF ={opposite side}/{hypotenuse}={AB}/{AC}={12}/{20

Numerical differential equation analysis package - Free Essay Example

The Numerical Differential Equation Analysis package combines functionality for analyzing differential equations using Butcher trees, Gaussian quadrature, and Newton-Cotes quadrature. Butcher Runge-Kutta methods are useful for numerically solving certain types of ordinary differential equations. Deriving high-order Runge-Kutta methods is no easy task, however. There are several reasons for this. The first difficulty is in finding the so-called order conditions. These are nonlinear equations in the coefficients for the method that must be satisfied to make the error in the method of order O (hn) for some integer n where h is the step size. The second difficulty is in solving these equations. Besides being nonlinear, there is generally no unique solution, and many heuristics and simplifying assumptions are usually made. Finally, there is the problem of combinatorial explosion. For a twelfth-order method there are 7813 order conditions! This package performs the first task: finding the order conditions that must be satisfied. The result is expressed in terms of unknown coefficients aij, bj, and ci. The s-stage Runge-Kutta method to advance from x to x+h is then where Sums of the elements in the rows of the matrix [aij] occur repeatedly in the conditions imposed on aij and bj. In recognition of this and as a notational convenience it is usual to introduce the coefficients ci and the definition This definition is referred to as the row-sum condition and is the first in a sequence of row-simplifying conditions. If aij=0 for all ij the method is explicit; that is, each of the Yi (x+h) is defined in terms of previously computed values. If the matrix [aij] is not strictly lower triangular, the method is implicit and requires the solution of a (generally nonlinear) system of equations for each timestep. A diagonally implicit method has aij=0 for all ij. There are several ways to express the order conditions. If the number of stages s is specified as a positive integer, the order conditions are expressed in terms of sums of explicit terms. If the number of stages is specified as a symbol, the order conditions will involve symbolic sums. If the number of stages is not specified at all, the order conditions will be expressed in stage-independent tensor notation. In addition to the matrix a and the vectors b and c, this notation involves the vector e, which is composed of all ones. This notation has two distinct advantages: it is independent of the number of stages s and it is independent of the particular Runge-Kutta method. For further details of the theory see the references. ai,j the coefficient of f(Yj(x)) in the formula for Yi(x) of the method bj the coefficient of f(Yj(x)) in the formula for Y(x) of the method ci a notational convenience for aij e a notational convenience for the vector (1, 1, 1, ) Notation used by functions for Butcher. RungeKuttaOrderConditions[p,s] give a list of the order conditions that any s-stage Runge-Kutta method of order p must satisfy ButcherPrincipalError[p,s] give a list of the order p+1 terms appearing in the Taylor series expansion of the error for an order-p, s-stage Runge-Kutta method RungeKuttaOrderConditions[p], ButcherPrincipalError[p] give the result in stage-independent tensor notation Functions associated with the order conditions of Runge-Kutta methods. ButcherRowSum specify whether the row-sum conditions for the ci should be explicitly included in the list of order conditions ButcherSimplify specify whether to apply Butchers row and column simplifying assumptions Some options for RungeKuttaOrderConditions. This gives the number of order conditions for each order up through order 10. Notice the combinatorial explosion. In[2]:= Out[2]= This gives the order conditions that must be satisfied by any first-order, 3-stage Runge-Kutta method, explicitly including the row-sum conditions. In[3]:= Out[3]= These are the order conditions that must be satisfied by any second-order, 3-stage Runge-Kutta method. Here the row-sum conditions are not included. In[4]:= Out[4]= It should be noted that the sums involved on the left-hand sides of the order conditions will be left in symbolic form and not expanded if the number of stages is left as a symbolic argument. This will greatly simplify the results for high-order, many-stage methods. An even more compact form results if you do not specify the number of stages at all and the answer is given in tensor form. These are the order conditions that must be satisfied by any second-order, s-stage method. In[5]:= Out[5]= Replacing s by 3 gives the same result asRungeKuttaOrderConditions. In[6]:= Out[6]= These are the order conditions that must be satisfied by any second-order method. This uses tensor notation. The vector e is a vector of ones whose length is the number of stages. In[7]:= Out[7]= The tensor notation can likewise be expanded to give the conditions in full. In[8]:= Out[8]= These are the principal error coefficients for any third-order method. In[9]:= Out[9]= This is a bound on the local error of any third-order method in the limit as h approaches 0, normalized to eliminate the effects of the ODE. In[10]:= Out[10]= Here are the order conditions that must be satisfied by any fourth-order, 1-stage Runge-Kutta method. Note that there is no possible way for these order conditions to be satisfied; there need to be more stages (the second argument must be larger) for there to be sufficiently many unknowns to satisfy all of the conditions. In[11]:= Out[11]= RungeKuttaMethod specify the type of Runge-Kutta method for which order conditions are being sought Explicit a setting for the option RungeKuttaMethod specifying that the order conditions are to be for an explicit Runge-Kutta method DiagonallyImplicit a setting for the option RungeKuttaMethod specifying that the order conditions are to be for a diagonally implicit Runge-Kutta method Implicit a setting for the option RungeKuttaMethod specifying that the order conditions are to be for an implicit Runge-Kutta method $RungeKuttaMethod a global variable whose value can be set to Explicit, DiagonallyImplicit, or Implicit Controlling the type of Runge-Kutta method in RungeKuttaOrderConditions and related functions. RungeKuttaOrderConditions and certain related functions have the option RungeKuttaMethod with default setting $RungeKuttaMethod. Normally you will want to determine the Runge-Kutta method being considered by setting $RungeKuttaMethod to one of Implicit, DiagonallyImplicit, and Explicit, but you can specify an option setting or even change the default for an individual function. These are the order conditions that must be satisfied by any second-order, 3-stage diagonally implicit Runge-Kutta method. In[12]:= Out[12]= An alternative (but less efficient) way to get a diagonally implicit method is to force a to be lower triangular by replacing upper-triangular elements with 0. In[13]:= Out[13]= These are the order conditions that must be satisfied by any third-order, 2-stage explicit Runge-Kutta method. The contradiction in the order conditions indicates that no such method is possible, a result which holds for any explicit Runge-Kutta method when the number of stages is less than the order. In[14]:= Out[14]= ButcherColumnConditions[p,s] give the column simplifying conditions up to and including order p for s stages ButcherRowConditions[p,s] give the row simplifying conditions up to and including order p for s stages ButcherQuadratureConditions[p,s] give the quadrature conditions up to and including order p for s stages ButcherColumnConditions[p], ButcherRowConditions[p], etc. give the result in stage-independent tensor notation More functions associated with the order conditions of Runge-Kutta methods. Butcher showed that the number and complexity of the order conditions can be reduced considerably at high orders by the adoption of so-called simplifying assumptions. For example, this reduction can be accomplished by adopting sufficient row and column simplifying assumptions and quadrature-type order conditions. The option ButcherSimplify in RungeKuttaOrderConditions can be used to determine these automatically. These are the column simplifying conditions up to order 4. In[15]:= Out[15]= These are the row simplifying conditions up to order 4. In[16]:= Out[16]= These are the quadrature conditions up to order 4. In[17]:= Out[17]= Trees are fundamental objects in Butchers formalism. They yield both the derivative in a power series expansion of a Runge-Kutta method and the related order constraint on the coefficients. This package provides a number of functions related to Butcher trees. f the elementary symbol used in the representation of Butcher trees ButcherTrees[p] give a list, partitioned by order, of the trees for any Runge-Kutta method of order p ButcherTreeSimplify[p,,] give the set of trees through order p that are not reduced by Butchers simplifying assumptions, assuming that the quadrature conditions through order p, the row simplifying conditions through order , and the column simplifying conditions through order all hold. The result is grouped by order, starting with the first nonvanishing trees ButcherTreeCount[p] give a list of the number of trees through order p ButcherTreeQ[tree] give True if the tree or list of trees tree is valid functional syntax, and False otherwise Constructing and enumerating Butcher trees. This gives the trees that are needed for any third-order method. The trees are represented in a functional form in terms of the elementary symbol f. In[18]:= Out[18]= This tests the validity of the syntax of two trees. Butcher trees must be constructed using multiplication, exponentiation or application of the function f. In[19]:= Out[19]= This evaluates the number of trees at each order through order 10. The result is equivalent to Out[2] but the calculation is much more efficient since it does not actually involve constructing order conditions or trees. In[20]:= Out[20]= The previous result can be used to calculate the total number of trees required at each order through order10. In[21]:= Out[21]= The number of constraints for a method using row and column simplifying assumptions depends upon the number of stages. ButcherTreeSimplify gives the Butcher trees that are not reduced assuming that these assumptions hold. This gives the additional trees that are necessary for a fourth-order method assuming that the quadrature conditions through order 4 and the row and column simplifying assumptions of order 1 hold. The result is a single tree of order 4 (which corresponds to a single fourth-order condition). In[22]:= Out[22]= It is often useful to be able to visualize a tree or forest of trees graphically. For example, depicting trees yields insight, which can in turn be used to aid in the construction of Runge-Kutta methods. ButcherPlot[tree] give a plot of the tree tree ButcherPlot[{tree1,tree2,}] give an array of plots of the trees in the forest {tree1, tree2,} Drawing Butcher trees. ButcherPlotColumns specify the number of columns in the GraphicsGrid plot of a list of trees ButcherPlotLabel specify a list of plot labels to be used to label the nodes of the plot ButcherPlotNodeSize specify a scaling factor for the nodes of the trees in the plot ButcherPlotRootSize specify a scaling factor for the highlighting of the root of each tree in the plot; a zero value does not highlight roots Options to ButcherPlot. This plots and labels the trees through order 4. In[23]:= Out[23]= In addition to generating and drawing Butcher trees, many functions are provided for measuring and manipulating them. For a complete description of the importance of these functions, see Butcher. ButcherHeight[tree] give the height of the tree tree ButcherWidth[tree] give the width of the tree tree ButcherOrder[tree] give the order, or number of vertices, of the tree tree ButcherAlpha[tree] give the number of ways of labeling the vertices of the tree tree with a totally ordered set of labels such that if (m, n) is an edge, then mn ButcherBeta[tree] give the number of ways of labeling the tree tree with ButcherOrder[tree]-1 distinct labels such that the root is not labeled, but every other vertex is labeled ButcherBeta[n,tree] give the number of ways of labeling n of the vertices of the tree with n distinct labels such that every leaf is labeled and the root is not labeled ButcherBetaBar[tree] give the number of ways of labeling the tree tree with ButcherOrder[tree] distinct labels such that every node, including the root, is labeled ButcherBetaBar[n,tree] give the number of ways of labeling n of the vertices of the tree with n distinct labels such that every leaf is labeled ButcherGamma[tree] give the density of the tree tree; the reciprocal of the density is the right-hand side of the order condition imposed by tree ButcherPhi[tree,s] give the weight of the tree tree; the weight (tree) is the left-hand side of the order condition imposed by tree ButcherPhi[tree] give (tree) using tensor notation ButcherSigma[tree] give the order of the symmetry group of isomorphisms of the tree tree with itself Other functions associated with Butcher trees. This gives the order of the tree f[f[f[f] f^2]]. In[24]:= Out[24]= This gives the density of the tree f[f[f[f] f^2]]. In[25]:= Out[25]= This gives the elementary weight function imposed by f[f[f[f] f^2]] for an s-stage method. In[26]:= Out[26]= The subscript notation is a formatting device and the subscripts are really just the indexed variable NumericalDifferentialEquationAnalysis`Private`$i. In[27]:= Out[27]//FullForm= It is also possible to obtain solutions to the order conditions using Solve and related functions. Many issues related to the construction Runge-Kutta methods using this package can be found in Sofroniou. The article also contains details concerning algorithms used in Butcher.m and discusses applications. Gaussian Quadrature As one of its methods, the Mathematica function NIntegrate uses a fairly sophisticated Gauss-Kronrod-based algorithm. The Gaussian quadrature functionality provided in Numerical Differential Equation Analysis allows you to easily study some of the theory behind ordinary Gaussian quadrature which is a little less sophisticated. The basic idea behind Gaussian quadrature is to approximate the value if an integral as a linear combination of values of the integrand evaluated at specific points: Since there are 2n free parameters to be chosen (both the abscissas xi and the weights wi) and since both integration and the sum are linear operations, you can expect to be able to make the formula correct for all polynomials of degree less than about 2n. In addition to knowing what the optimal abscissas and weights are, it is often desirable to know how large the error in the approximation will be. This package allows you to answer both of these questions. GaussianQuadratureWeights[n,a,b] give a list of the pairs (xi, wi) to machine precision for quadrature on the interval a to b GaussianQuadratureError[n,f,a,b] give the error to machine precision GaussianQuadratureWeights[n,a,b,prec] give a list of the pairs (xi, wi) to precision prec GaussianQuadratureError[n,f,a,b,prec] give the error to precision prec Finding formulas for Gaussian quadrature. This gives the abscissas and weights for the five-point Gaussian quadrature formula on the interval (-3, 7). In[2]:= Out[2]= Here is the error in that formula. Unfortunately it involves the tenth derivative of f at an unknown point so you dont really know what the error itself is. In[3]:= Out[3]= You can see that the error decreases rapidly with the length of the interval. In[4]:= Out[4]= Newton-Cotes As one of its methods, the Mathematica function NIntegrate uses a fairly sophisticated Gauss-Kronrod based algorithm. Other types of quadrature formulas exist, each with their own advantages. For example, Gaussian quadrature uses values of the integrand at oddly spaced abscissas. If you want to integrate a function presented in tabular form at equally spaced abscissas, it wont work very well. An alternative is to use Newton-Cotes quadrature. The basic idea behind Newton-Cotes quadrature is to approximate the value of an integral as a linear combination of values of the integrand evaluated at equally spaced points: In addition, there is the question of whether or not to include the end points in the sum. If they are included, the quadrature formula is referred to as a closed formula. If not, it is an open formula. If the formula is open there is some ambiguity as to where the first abscissa is to be placed. The open formulas given in this package have the first abscissa one half step from the lower end point. Since there are n free parameters to be chosen (the weights) and since both integration and the sum are linear operations, you can expect to be able to make the formula correct for all polynomials of degree less than about n. In addition to knowing what the weights are, it is often desirable to know how large the error in the approximation will be. This package allows you to answer both of these questions. NewtonCotesWeights[n,a,b] give a list of the n pairs (xi, wi) for quadrature on the interval a to b NewtonCotesError[n,f,a,b] give the error in the formula Finding formulas for Newton-Cotes quadrature. option name default value QuadratureType Closed the type of quadrature, Open or Closed Option for NewtonCotesWeights and NewtonCotesError. Here are the abscissas and weights for the five-point closed Newton-Cotes quadrature formula on the interval (-3, 7). In[2]:= Out[2]= Here is the error in that formula. Unfortunately it involves the sixth derivative of f at an unknown point so you dont really know what the error itself is. In[3]:= Out[3]= You can see that the error decreases rapidly with the length of the interval. In[4]:= Out[4]= This gives the abscissas and weights for the five-point open Newton-Cotes quadrature formula on the interval (-3, 7). In[5]:= Out[5]= Here is the error in that formula. In[6]:= Out[6]= Runge-Kutta Methods From Wikipedia, The Free Encyclopedia Jump to: navigation, search In numerical analysis, the Runge-Kutta methods (German pronunciation:[kta]) are an important family of implicit and explicit iterative methods for the approximation of solutions of ordinary differential equations. These techniques were developed around 1900 by the German mathematicians C. Runge and M.W. Kutta. See the article on numerical ordinary differential equations for more background and other methods. See also List of Runge-Kutta methods. Contents 1 The common fourth-order Runge-Kutta method 2 Explicit Runge-Kutta methods o 2.1 Examples 3 Usage 4 Adaptive Runge-Kutta methods 5 Implicit Runge-Kutta methods 6 References 7 External links The Common Fourth-Order Runge-Kutta Method One member of the family of Runge-Kutta methods is so commonly used that it is often referred to as RK4, classical Runge-Kutta method or simply as the Runge-Kutta method. Let an initial value problem be specified as follows. Then, the RK4 method for this problem is given by the following equations: where yn + 1 is the RK4 approximation of y(tn + 1), and Thus, the next value (yn + 1) is determined by the present value (yn) plus the product of the size of the interval (h) and an estimated slope. The slope is a weighted average of slopes: k1 is the slope at the beginning of the interval; k2 is the slope at the midpoint of the interval, using slope k1 to determine the value of y at the point tn + h / 2 using Eulers method; k3 is again the slope at the midpoint, but now using the slope k2 to determine the y-value; k4 is the slope at the end of the interval, with its y-value determined using k3. In averaging the four slopes, greater weight is given to the slopes at the midpoint: The RK4 method is a fourth-order method[needs reference], meaning that the error per step is on the order of h5, while the total accumulated error has order h4. Note that the above formulae are valid for both scalar- and vector-valued functions (i.e., y can be a vector and f an operator). For example one can integrate Schrdingers equation using the Hamiltonian operator as function f. Explicit Runge-Kutta Methods The family of explicit Runge-Kutta methods is a generalization of the RK4 method mentioned above. It is given by where (Note: the above equations have different but equivalent definitions in different texts). To specify a particular method, one needs to provide the integer s (the number of stages), and the coefficients aij (for 1 j i s), bi (for i = 1, 2, , s) and ci (for i = 2, 3, , s). These data are usually arranged in a mnemonic device, known as a Butcher tableau (after John C. Butcher): 0 c2 a21 c3 a31 a32 cs as1 as2 as,s 1 b1 b2 bs 1 bs The Runge-Kutta method is consistent if There are also accompanying requirements if we require the method to have a certain order p, meaning that the truncation error is O(hp+1). These can be derived from the definition of the truncation error itself. For example, a 2-stage method has order 2 if b1 + b2 = 1, b2c2 = 1/2, and b2a21 = 1/2. Examples The RK4 method falls in this framework. Its tableau is: 0 1/2 1/2 1/2 0 1/2 1 0 0 1 1/6 1/3 1/3 1/6 However, the simplest Runge-Kutta method is the (forward) Euler method, given by the formula yn + 1 = yn + hf(tn,yn). This is the only consistent explicit Runge-Kutta method with one stage. The corresponding tableau is: 0 1 An example of a second-order method with two stages is provided by the midpoint method The corresponding tableau is: 0 1/2 1/2 0 1 Note that this midpoint method is not the optimal RK2 method. An alternative is provided by Heuns method, where the 1/2s in the tableau above are replaced by 1s and the bs row is [1/2, 1/2]. If one wants to minimize the truncation error, the method below should be used (Atkinson p.423). Other important methods are Fehlberg, Cash-Karp and Dormand-Prince. Also, read the article on Adaptive Stepsize. Usage The following is an example usage of a two-stage explicit Runge-Kutta method: 0 2/3 2/3 1/4 3/4 to solve the initial-value problem with step size h=0.025. The tableau above yields the equivalent corresponding equations below defining the method: k1 = yn t0 = 1 y0 = 1 t1 = 1.025 k1 = y0 = 1 f(t0,k1) = 2.557407725 k2 = y0 + 2 / 3hf(t0,k1) = 1.042623462 y1 = y0 + h(1 / 4 f(t0,k1) + 3 / 4 f(t0 + 2 / 3h,k2)) = 1.066869388 t2 = 1.05 k1 = y1 = 1.066869388 f(t1,k1) = 2.813524695 k2 = y1 + 2 / 3hf(t1,k1) = 1.113761467 y2 = y1 + h(1 / 4 f(t1,k1) + 3 / 4 f(t1 + 2 / 3h,k2)) = 1.141332181 t3 = 1.075 k1 = y2 = 1.141332181 f(t2,k1) = 3.183536647 k2 = y2 + 2 / 3hf(t2,k1) = 1.194391125 y3 = y2 + h(1 / 4 f(t2,k1) + 3 / 4 f(t2 + 2 / 3h,k2)) = 1.227417567 t4 = 1.1 k1 = y3 = 1.227417567 f(t3,k1) = 3.796866512 k2 = y3 + 2 / 3hf(t3,k1) = 1.290698676 y4 = y3 + h(1 / 4 f(t3,k1) + 3 / 4 f(t3 + 2 / 3h,k2)) = 1.335079087 The numerical solutions correspond to the underlined values. Note that f(ti,k1) has been calculated to avoid recalculation in the yis. Adaptive Runge-Kutta Methods The adaptive methods are designed to produce an estimate of the local truncation error of a single Runge-Kutta step. This is done by having two methods in the tableau, one with order p and one with order p 1. The lower-order step is given by where the ki are the same as for the higher order method. Then the error is which is O(hp). The Butcher Tableau for this kind of method is extended to give the values of : 0 c2 a21 c3 a31 a32 cs as1 as2 as,s 1 b1 b2 bs 1 bs The Runge-Kutta-Fehlberg method has two methods of orders 5 and 4. Its extended Butcher Tableau is: 0 1/4 1/4 3/8 3/32 9/32 12/13 1932/2197 7200/2197 7296/2197 1 439/216 8 3680/513 -845/4104 1/2 8/27 2 3544/2565 1859/4104 11/40 16/135 0 6656/12825 28561/56430 9/50 2/55 25/216 0 1408/2565 2197/4104 1/5 0 However, the simplest adaptive Runge-Kutta method involves combining the Heun method, which is order 2, with the Euler method, which is order 1. Its extended Butcher Tableau is: 0 1 1 1/2 1/2 1 0 The error estimate is used to control the stepsize. Other adaptive Runge-Kutta methods are the Bogacki-Shampine method (orders 3 and 2), the Cash-Karp method and the Dormand-Prince method (both with orders 5 and 4). Implicit Runge-Kutta Methods The implicit methods are more general than the explicit ones. The distinction shows up in the Butcher Tableau: for an implicit method, the coefficient matrix aij is not necessarily lower triangular: The approximate solution to the initial value problem reflects the greater number of coefficients: Due to the fullness of the matrix aij, the evaluation of each ki is now considerably involved and dependent on the specific function f(t,y). Despite the difficulties, implicit methods are of great importance due to their high (possibly unconditional) stability, which is especially important in the solution of partial differential equations. The simplest example of an implicit Runge-Kutta method is the backward Euler method: The Butcher Tableau for this is simply: It can be difficult to make sense of even this simple implicit method, as seen from the expression for k1: In this case, the awkward expression above can be simplified by noting that so that from which follows. Though simpler then the raw representation before manipulation, this is an implicit relation so that the actual solution is problem dependent. Multistep implicit methods have been used with success by some researchers. The combination of stability, higher order accuracy with fewer steps, and stepping that depends only on the previous value makes them attractive; however the complicated problem-specific implementation and the fact that ki must often be approximated iteratively means that they are not common. References J. C. Butcher, Numerical methods for ordinary differential equations, ISBN 0471967580 George E. Forsythe, Michael A. Malcolm, and Cleve B. Moler. Computer Methods for Mathematical Computations. Englewood Cliffs, NJ: Prentice-Hall, 1977. (See Chapter 6.) Ernst Hairer, Syvert Paul Nrsett, and Gerhard Wanner. Solving ordinary differential equations I: Nonstiff problems, second edition. Berlin: Springer Verlag, 1993. ISBN 3-540-56670-8. William H. Press, Brian P. Flannery, Saul A. Teukolsky, William T. Vetterling. Numerical Recipes in C. Cambridge, UK: Cambridge University Press, 1988. (See Sections 16.1 and 16.2.) Kaw, Autar; Kalu, Egwu (2008), Numerical Methods with Applications (1st ed.), www.autarkaw.com. Kendall E. Atkinson. An Introduction to Numerical Analysis. John Wiley Sons 1989 F. Cellier, E. Kofman. Continuous System Simulation. Springer Verlag, 2006. ISBN 0-387-26102-8. External links Runge-Kutta Runge-Kutta 4th Order Method Runge Kutta Method for O.D.E.s Numerical integration First order methods Euler method Backward Euler Semi-implicit Euler Exponential Euler Second order methods Verlet integration Velocity Verlet Crank-Nicolson method Beemans algorithm Midpoint method Heuns method Newmark-beta method Leapfrog integration Higher order methods Runge-Kutta methods List of Runge-Kutta methods Linear multistep method Retrieved from https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods Categories: Numerical differential equations | Runge-Kutta methods This page was last modified on 28 November 2009 at 11:21. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. See Terms of Use for details. Wikipedia is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Contact us Privacy policy About Wikipedia Disclaimers Higher Order Taylor Methods Marcelo Julio Alvisio Lisa Marie Danz May 16, 2007 Introduction Differential equations are one of the building blocks in science or engineering. Scientists aim to obtain numerical solutions to differential equations whenever explicit solutions do not exist or when they are too hard to find. These numerical solutions are approximated though a variety of methods, some of which we set out to explore in this project. We require two conditions when computing differential equations numerically. First, we require that the solution is continuous with initial value. Otherwise, numerical error introduced in the representation of the number in computer systems would produce results very far from the actual solution. Second, we require that the solution changes continuously with respect to the differential equation itself. Otherwise, we cannot expect the method that approximates the differential equation to give accurate results. The most common methods for computing differential equations numerically include Eulers method, Higher Order Taylor method and Runge-Kutta methods. In this project, we concentrate on the Higher Order Taylor Method. This method employs the Taylor polynomial of the solution to the equation. It approximates the zeroth order term by using the previous steps value (which is the initial condition for the first step), and the subsequent terms of the Taylor expansion by using the differential equation. We call it Higher Order Taylor Method, the lower order method being Eulers Method. Under certain conditions, the Higher Order Taylor Method limits the error to O(hn), where n is the order used. We will present several examples to test this idea. We will look into two main parameters as a measure of the effectiveness of the method, namely accuracy and efficiency. Theory of the Higher Order Taylor Method Definition 2.1 Consider the differential equation given by y0(t)= f(t,y), y(a)= c. Then for ba, the nth order Taylor approximation to y(b) with K steps is given by yK, where {yi} is defined recursively as: t0 = a y0 = y(a)= c ti+1 = ti + h h2 f hn n1f yi+1 = yi + hf(ti,yi)+ (ti,yi)+ +(ti,yi) 2 t n! tn1 with h =(b a)/K. It makes sense to formulate such a definition in view of the Taylor series expansion that is used when y(t) is known explicitly. All we have done is use f(t,y) for y0(t), ft(t,y) for y00(t), and so forth. The next task is to estimate the error that this approximation introduces. We know by Taylors Theorem that, for any solution that admits a Taylor expansion at the point ti, we have h2 hn h(n+1) y(ti+1)= y(ti)+ hy0(ti)+ y00(ti)+ + y(n)(ti)+ y(n+1)() 2 n!(n + 1)! where is between ti and ti+1 Using y0 = f(t,y), this translates to h2 f hn (n1)fh(n+1) (n)f y(ti+1)= y(ti)+hf(ti,yi)+ (ti,yi)++(ti,yi)+ (,y()) 2 t n! t(n1) (n + 1)! t(n) Therefore, the local error, that is to say, the error introduced at each step if the values calculated previously were exact, is given by: 1 (n)f Ei =(hn+1)(,y()) (n + 1)! tn which means that 1 (n)f max (hn+1)(,y()) Ei [a,b] (n + 1)! tn 23 We can say Ei = O(hn+1). Now, since the number of steps from a to b is proportional to 1/h, we multiply the error per step by the number of steps to find a total error E = O(hn). In Practice: Examples We will consider differential equations that we can solve explicitly to obtain an equation for y(t) such that y0(t)= f(t,y). This way, we can calculate the actual error by subtracting the exact value for y(b) from the value that the Higher Order Taylor method predicts for it. To approximate values in the following examples, the derivatives of f(t,y) were computed by hand. MATLAB then performed the iteration and arrived at the approximation. Notice that the definitions given in the previous section could also have been adapted for varying step size h. However, for ease of computation we have kept the step size constant. In our computations, we have chosen step size of (b a)/2k, which resulted in K =2k evenly spaced points in the interval. Example 3.1 We consider the differential equation 1+ t y0(t)= f(t,y)= 1+ y with initial condition y(1) = 2. It is clear that y(t)= t2 +2t +6 1 solves this equation. Thus we calculate the error for y(2) by subtracting the approximation of y(2) from y(2), which is the exact value. Recall that we are using h =2k because (b a)=1. The following table displays the errors calculated. k = 1 k = 2 k = 3 k = 4 order = 1 .0333 .0158 .0077 .0038 order = 2 .0038 .0009 .0002 .0001 order = 3 .0003269 .0000383 .0000046 .0000006 Runge-Kutta Methods The Taylor methods in the preceding section have the desirable feature that the F.G.E. is of order O(hN ), and N can be chosen large so that this error is small. However, the shortcomings of the Taylor methods are the a priori determination of N and the computation of the higher derivatives, which can be very complicated. Each Runge-Kutta method is derived from an appropriate Taylor method in such a way that the F.G.E. is of order O(hN ). A trade-off is made to perform several function evaluations at each step and eliminate the necessity to compute the higher derivatives. These methods can be constructed for any order N. The Runge-Kutta method of order N = 4 is most popular. It is a good choice for common purposes because it is quite accurate, stable, and easy to program. Most authorities proclaim that it is not necessary to go to a higher-order method because the increased accuracy is offset by additional computational effort. If more accuracy is required, then either a smaller step size or an adaptive method should be used. The fourth-order Runge-Kutta method (RK4) simulates the accuracy of the Taylor series method of order N = 4. The method is based on computing yk+1 as follows: (1) yk+1 = yk + w1k1 + w2k2 + w3k3 + w4k4, where k1, k2, k3, and k4 have the form (2) k1 = h f (tk , yk ), k2 = h f (tk + a1h, yk + b1k1), k3 = h f (tk + a2h, yk + b2k1 + b3k2), k4 = h f (tk + a3h, yk + b4k1 + b5k2 + b6k3). By matching coefficients with those of the Taylor series method of order N = 4 so that the local truncation error is of order O(h5), Runge and Kutta were able to obtain the 490 CHAP. 9 SOLUTION OF DIFFERENTIAL EQUATIONS following system of equations: (3) b1 = a1, b2 + b3 = a2, b4 + b5 + b6 = a3, w1 + w2 + w3 + w4 = 1, w2a1 + w3a2 + w4a3 = 1 2, w2a2 1 + w3a2 2 + w4a2 3 = 1 3 , w2a3 1 + w3a3 2 + w4a3 3 = 1 4 , w3a1b3 + w4(a1b5 + a2b6) = 1 6 , w3a1a2b3 + w4a3(a1b5 + a2b6) = 1 8 , w3a2 1b3 + w4(a2 1b5 + a2 2b6) = 1 12 , w4a1b3b6 = 1 24 The system involves 11 equations in 13 unknowns. Two additional conditions must be supplied to solve the system. The most useful choice is (4) a1 = 1 2 and b2 = 0. Then the solution for the remaining variables is (5) a2 = 1 2 , a3 = 1, b1 = 1 2 , b3 = 1 2 , b4 = 0, b5 = 0, b6 = 1, w1 = 1 6 , w2 = 1 3 , w3 = 1 3 , w4 = 1 6 The values in (4) and (5) are substituted into (2) and (1) to obtain the formula for the standard Runge-Kutta method of order N = 4, which is stated as follows. Start with the initial point (t0, y0) and generate the sequence of approximations using (6) yk+1 = yk + h( f1 + 2 f2 + 2 f3 + f4) 6 , SEC. 9.5 RUNGE-KUTTA METHODS 491 where (7) f1 = f (tk , yk ), f2 = f tk + h 2 , yk + h 2 f1 , f3 = f tk + h 2 , yk + h 2 f2 , f4 = f (tk + h, yk + h f3). Discussion about the Method The complete development of the equations in (7) is beyond the scope of this book and can be found in advanced texts, but we can get some insights. Consider the graph of the solution curve y = y(t) over the first subinterval [t0, t1]. The function values in (7) are approximations for slopes to this curve. Here f1 is the slope at the left, f2 and f3 are two estimates for the slope in the middle, and f4 is the slope at the right (a)). The next point (t1, y1) is obtained by integrating the slope function (8) y(t1) y(t0) = _ t1 t0 f (t, y(t)) dt. If Simpsons rule is applied with step size h/2, the approximation to the integral in (8) is (9) _ t1 t0 f (t, y(t)) dt h 6 ( f (t0, y(t0)) + 4 f (t1/2, y(t1/2)) + f (t1, y(t1))), where t1/2 is the midpoint of the interval. Three function values are needed; hence we make the obvious choice f (t0, y (t0)) = f1 and f (t1, y(t1)) f4. For the value in the middle we chose the average of f2 and f3: f (t1/2, y(t1/2)) f2 + f3 2 . These values are substituted into (9), which is used in equation (8) to get y1: (10) y1 = y0 + h 6 f1 + 4( f2 + f3) 2 + f4 . When this formula is simplified, it is seen to be equation (6) with k = 0. The graph for the integral in (9) is shown in Figure 9.9(b). 492 CHAP. 9 SOLUTION OF DIFFERENTIAL EQUATIONS y t m1 = f1 m2 = f3 m3 = f4 m4 = f4 (t0, y0) y = y(t) (t1, y(t1)) t0 t1/2 t1 (a) Predicted slopes mj to the solution curve y = y(t) z t (t0, f1) (t1/2, f2) (t1/2, f3) (t1, f4) t0 t1/2 t1 (b) Integral approximation: h 6 y(t1) y0 = ( f1 + 2f2 + 2f3 + f4) Figure 9.9 The graphs y = y(t) and z = f (t, y(t)) in the discussion of the Runge-Kutta method of order N = 4. Step Size versus Error The error term for Simpsons rule with step size h/2 is (11) y(4)(c1) h5 2880 . If the only error at each step is that given in (11), after M steps the accumulated error for the RK4 method would be (12) _M k=1 y(4)(ck) h5 2880 b a 5760 y(4)(c)h4 O(h4). The next theorem states the relationship between F.G.E. and step size. It is used to give us an idea of how much computing effort must be done when using the RK4 method. Theorem 9.7 (Precision of the Runge-Kutta Method). Assume that y(t) is the solution to the I.V.P. If y(t) C5[t0, b] and {(tk , yk)}M k=0 is the sequence of approximations generated by the Runge-Kutta method of order 4, then (13) |ek| = |y(tk ) yk| = O(h4), |_k+1| = |y(tk+1) yk hTN (tk , yk)| = O(h5). SEC. 9.5 RUNGE-KUTTA METHODS 493 In particular, the F.G.E. at the end of the interval will satisfy (14) E(y(b), h) = |y(b) yM| = O(h4). Examples 9.10 and 9.11 illustrate Theorem 9.7. If approximations are computed using the step sizes h and h/2, we should have (15) E(y(b), h) Ch4 for the larger step size, and (16) E y(b), h 2 C h4 16 = 1 16 Ch4 1 16 E(y(b), h). Hence the idea in Theorem 9.7 is that if the step size in the RK4 method is reduced by a factor of 12 we can expect that the overall F.G.E. will be reduced by a factor of 1. Example 9.10. Use the RK4 method to solve the I.V.P. y_ = (t y)/2 on [0, 3] with y(0) = 1. Compare solutions for h = 1, 12 , 14 , and 18 . Table 9.8 gives the solution values at selected abscissas. For the step size h = 0.25, a sample calculation is f1 = 0.0 1.0 2 = 0.5, f2 = 0.125 (1 + 0.25(0.5)(0.5)) 2 = 0.40625, f3 = 0.125 (1 + 0.25(0.5)(0.40625)) 2 = 0.4121094, f4 = 0.25 (1 + 0.25(0.4121094)) 2 = 0.3234863, y1 = 1.0 + 0.25 0.5 + 2(0.40625) + 2(0.4121094) 0.3234863 6 = 0.8974915. _ Example 9.11. Compare the F.G.E. when the RK4 method is used to solve y_ = (ty)/2 over [0, 3] with y(0) = 1 using step sizes 1, 12 , 14 , and 18 Table 9.9 gives the F.G.E. for the various step sizes and shows that the error in the approximation to y(3) decreases by about 1 16 when the step size is reduced by a factor of 1/2. E(y(3), h) = y(3) yM = O(h4) Ch4 where C = 0.000614. _ A comparison of Examples 9.10 and 9.11 and Examples 9.8 and 9.9 shows what is meant by the statement The RK4 method simulates the Taylor series method of order N = 4. For these examples, the two methods generate identical solution sets {(tk , yk)} 494 CHAP. 9 SOLUTION OF DIFFERENTIAL EQUATIONS Table 9.8 Comparison of the RK4 Solutions with Different Step Sizes for y_ = (t y)/2 over [0, 3] with y(0) = 1 yk tk h = 1 h = 12 h = 14 h = 18 y(tk ) Exact 0 1.0 1.0 1.0 1.0 1.0 0.125 0.9432392 0.9432392 0.25 0.8974915 0.8974908 0.8974917 0.375 0.8620874 0.8620874 0.50 0.8364258 0.8364037 0.8364024 0.8364023 0.75 0.8118696 0.8118679 0.8118678 1.00 0.8203125 0.8196285 0.8195940 0.8195921 0.8195920 1.50 0.9171423 0.9171021 0.9170998 0.9170997 2.00 1.1045125 1.1036826 1.1036408 1.1036385 1.1036383 2.50 1.3595575 1.3595168 1.3595145 1.3595144 3.00 1.6701860 1.6694308 1.6693928 1.6693906 1.6693905 Table 9.9 Relation between Step Size and F.G.E. for the RK4 Solutions to y_ = (t y)/2 over [0, 3] with y(0) = 1 Step size, h Number of steps, M Approximation to y(3), yM F.G.E. Error at t = 3, y(3) yM O(h4) Ch4 where C = 0.000614 1 3 1.6701860 0.0007955 0.0006140 12 6 1.6694308 0.0000403 0.0000384 14 12 1.6693928 0.0000023 0.0000024 18 24 1.6693906 0.0000001 0.0000001 over the given interval. The advantage of the RK4 method is obvious; no formulas for the higher derivatives need to be computed nor do they have to be in the program. It is not easy to determine the accuracy to which a Runge-Kutta solution has been computed. We could estimate the size of y(4)(c) and use formula (12). Another way is to repeat the algorithm using a smaller step size and compare results. A third way is to adaptively determine the step size, which is done in Program 9.5. In Section 9.6 we will see how to change the step size for a multistep method. SEC. 9.5 RUNGE-KUTTA METHODS 495 Runge-Kutta Methods of Order N = 2 The second-order Runge-Kutta method (denoted RK2) simulates the accuracy of the Taylor series method of order 2. Although this method is not as good to use as the RK4 method, its proof is easier to understand and illustrates the principles involved. To start, we write down the Taylor series formula for y(t + h): (17) y(t + h) = y(t) + hy_ (t) + 1 2 h2 y__ (t) + CT h3 + , where CT is a constant involving the third derivative of y(t) and the other terms in the series involve powers of h j for j 3. The derivatives y_ (t) and y__ (t) in equation (17) must be expressed in terms of f (t, y) and its partial derivatives. Recall that (18) y_ (t) = f (t, y). The chain rule for differentiating a function of two variables can be used to differentiate (18) with respect to t, and the result is y__ (t) = ft (t, y) + fy(t, y)y_ (t). Using (18), this can be written (19) y__ (t) = ft (t, y) + fy(t, y) f (t, y). The derivatives (18) and (19) are substituted in (17) to give the Taylor expression for y(t + h): y(t + h) = y(t) + h f (t, y) + 1 2 h2 ft (t, y) + 1 2 h2 fy(t, y) f (t, y) + CT h3 + . (20) Now consider the Runge-Kutta method of order N = 2, which uses a linear combination of two function values to express y(t + h): (21) y(t + h) = y(t) + Ah f0 + Bhf1, where (22) f0 = f (t, y), f1 = f (t + Ph, y + Qhf0). Next the Taylor polynomial approximation for a function of two independent variables is used to expand f (t, y) (see the Exercises). This gives the following representation for f1: (23) f1 = f (t, y) + Phft (t, y) + Qhfy(t, y) f (t, y) + CPh2 + , 496 CHAP. 9 SOLUTION OF DIFFERENTIAL EQUATIONS where CP involves the second-order partial derivatives of f (t, y). Then (23) is used in (21) to get the RK2 expression for y(t + h): y(t + h) = y(t) + (A + B)h f (t, y) + BPh2 ft (t, y) + BQh2 fy(t, y) f (t, y) + BCPh3 + . (24) A comparison of similar terms in equations (20) and (24) will produce the following conclusions: h f (t, y) = (A + B)h f (t, y) implies that 1 = A + B, 1 2 h2 ft (t, y) = BPh2 ft (t, y) implies that 1 2 = BP, 1 2 h2 fy(t, y) f (t, y) = BQh2 fy(t, y) f (t, y) implies that 1 2 = BQ. Hence, if we require that A, B, P, and Q satisfy the relations (25) A + B = 1 BP = 1 2 BQ = 1 2 , then the RK2 method in (24) will have the same order of accuracy as the Taylors method in (20). Since there are only three equations in four unknowns, the system of equations (25) is underdetermined, and we are permitted to choose one of the coefficients. There are several special choices that have been studied in the literature; we mention two of them. Case (i): Choose A = 12 . This choice leads to B = 12 , P = 1, and Q = 1. If equation (21) is written with these parameters, the formula is (26) y(t + h) = y(t) + h 2 ( f (t, y) + f (t + h, y + h f (t, y))). When this scheme is used to generate {(tk , yk)}, the result is Heuns method. Case (ii): Choose A = 0. This choice leads to B = 1, P = 12 , and Q = 12 . If equation (21) is written with these parameters, the formula is (27) y(t + h) = y(t) + h f t + h 2 , y + h 2 f (t, y) . When this scheme is used to generate {(tk , yk)}, it is called the modified Euler-Cauchy method. Numerical Methods Using Matlab, 4th Edition, 2004 John H. Mathews and Kurtis K. Fink ISBN: 0-13-065248-2 Prentice-Hall Inc. Upper Saddle River, New Jersey, USA https://vig.prenhall.com/ Deriving the Runge-Kutta Method Deriving the midpoint method The Taylor method is the gold standard for generating better numerical solutions to first order differential equations. A serious weakness in the Taylor method, however, is the need to compute a large number of partial derivatives and do other symbolic manipulation tasks. For example, the second order Taylor method for the equation y( t) = f(t,y(t)) is yi+1 = yi + h f(ti ,yi ) + h2 2 f t ( ti ,yi ) + f ( ti ,yi ) f y ( ti ,yi ) Higher order formulas get even uglier. The Midpoint method arises from an attempt to replace the second order Taylor method with a simpler Euler-like formula yi+1 = yi + h f(ti + ,yi + ) We can solve for the best values for and by applying a first order Taylor expansion to the term f(ti + ,yi + ): yi+1 = yi + h f ( ti ,yi ) + f t ( ti ,yi ) + f y ( ti ,yi ) + 2f t y ( ti ,yi ) The choices of and that make this look as close as possible to the second order Taylor formula above are = h2 = h2 f(ti ,yi ) leading to the so-called midpoint rule: 1 yi+1 = yi + h f(ti + h2 ,yi + h2 f(ti ,yi)) This formula has a simple interpretation. Essentially what we are doing here is driving an Euler estimate half way across the interval [ti , ti+1] and computing the slope f(ti + h2 ,yi + h2 f(ti ,yi)) at that midpoint. We then rewind back to the point ( ti ,yi ) and drive an Euler estimate all the way across the interval to ti+1 using this new midpoint slope in place of the old Euler slope. The Runge-Kutta Method The textbook points out that it is possible to derive similar methods by starting with more complex Euler-like formulas with more free parameters and then trying to match those Euler-like methods to higher order Taylor formulas. The Runge-Kutta method is essentially an attempt to match a more complex Euler-like formula to a fourth order Taylor method. The problem with this is that the Euler-like formula needed to match all the complexity of the fourth order Taylor method formula is quite complex. The textbook states in exercise 31 at the end of section 5.4 that the formula required is yi+1 = yi + h6 f(ti ,yi ) + h3 f(ti + 1 h,yi + 1 h f(ti ,yi)) + h3 f (ti + 2 h,yi + 2 h f (ti + 2 h, yi + 3 h f ( ti ,yi ))) + h6 f (ti + 3 h, yi + 3 h f (ti + 4 h, yi + 5 h f ( ti + 6 h,yi+ 7 h f ( ti ,yi)))) It is very messy to do so, but this form can expanded out and matched against the Taylor formula of order four. This allows us to solve for all the unknown coefficients. A somewhat cleaner alternative derivation is based on the following argument. Another way to solve for yi+1 is to compute this integral !t i+1 t i y( t) dt = y(ti+1) y(ti ) = yi+1 yi We can imagine beginning to compute the integral by noting that y( t) = f(t,y(t)) !t i+1 t i y( t) dt = !t i+1 t i f ( t,y( t)) dt 2 Unfortunately, we can not do the integral on the right hand side exactly, because we dont know what y(t) is. That is, after all, the unknown we are trying to solve for. Even though we cant compute the integral on the right exactly, we can estimate it. For example, applying Simpsons rule to the integral produces the estimate !t i+1 t i f ( t,y( t)) dt h3 f ( ti ,y( ti))+4 f ti+ti+1 2 ,y ti+ti+1 2 + f ( ti+1,y( ti+1)) The Runge-Kutta method takes this estimate as a starting point. The thing we need to do to make this estimate work is to find a way to estimate the unknown terms y((ti + ti+1) /2) and y(ti+1) . The first step is to rewrite the estimate as h3 f ( ti ,y( ti))+2 f ti+ti+1 2 ,y ti+ti+1 2 +2 f ti+ti+1 2 ,y ti+ti+1 2 + f ( ti+1,y( ti+1)) We write the middle term twice because we are going to develop two different estimates for y((ti + ti+1) /2). The thinking is that the mistakes we make in developing those two interior estimates may partly cancel each other out. Here is how we will develop our estimates. 1. y(ti ) is just yi . We estimate the first y((ti + ti+1) /2) by driving the original Euler slope k1 = f(ti ,yi ) half-way across the interval: 2. k1 = f(ti ,y(ti)) y ti + ti+1 2 yi + h/2 k1 As in the midpoint rule, we compute a second slope at that midpoint we just estimated. We then rewind to the start and drive that slope half-way across the interval again. 3. k2 = f(ti + h/2,yi + h/2 k1 ) y ti + ti+1 2 yi + h/2 k2 We use the second estimated midpoint to compute another slope and then drive that slope all the way across the interval. 4. 3 We use the second estimated midpoint to compute another slope and then drive that slope all the way across the interval. 4. k3 = f(ti + h/2,yi + h/2 k2 ) y(ti+1) = yi + h k3 k4 = f(ti + h,yi + h k3 ) Substituting all of these estimates into the Simpsons rule formula above gives yi+1 yi = !t i+1 t i f ( t,y( t)) dt h3 f ( ti ,y( ti))+ 2 f ti+ti+1 2 ,y ti+ti+1 2 +2 f ti+ti+1 2 ,y ti+ti+1 2 + f ( ti+1,y( ti+1)) or yi+1 = yi + h3 (k1 + 2 k2 + 2 k3 + k4 ) Summary Of The Method k1 = f(ti ,yi ) k2 = f(ti + h/2,yi + h/2 k1 ) k3 = f(ti + h/2,yi + h/2 k2 ) k4 = f(ti + h,yi + h k3 ) yi+1 = yi + h3(k1 + 2 k2 + 2 k3 + k4 ) 4 Taylor Series Methods: To derive these methods we start with a Taylor Expansion: y(t+_t) _ y(t) + _ty0(t) + 1 2 _t2y00(t) + + 1 r! y(r)(t)_tr. Lets say we want to truncate this at the second derivative and base a method on that. The scheme is, then: yn+1 = yn + fn_t + f0 tn 2 _t2. The Taylor series method can be written as yn+1 = yn +_tF (tn, yn,_t) where F = f + 1 2_tf0. If we take the LTE for this scheme, we get (as expected) LTE(t) = y(tn +_t) y(tn) _t f(tn, y(tn)) 1 2 _tf0(tn, y(tn)) = O(_t2). Of course, we designed this method to give us this order, so it shouldnt be a surprise! So the LTE is reasonable, but what about the global error? Just as in the Euler Forward case, we can show that the global error is of the same order as the LTE. How do we do this? We have two facts, y(tn+1) = y(tn) + _tF (tn, y(tn),_t), and yn+1 = yn +_tF (tn, yn,_t) where F = f + 1 2_tf0. Now we subtract these two |y(tn+1) yn+1| = |y(tn) yn +_t (F(tn, y(tn)) F(tn, yn)) + _tLTE| _ |y(tn) yn|+_t |F(tn, y(tn)) F(tn, yn)|+_t|LTE| . Now, if F is Lipschitz continuous, we can say en+1 _ (1 + _tL)en+_t|LTE|. Of course, this is the same proof as for Eulers method, except that now we are looking at F, not f, and the LTE is of higher order. We can do this no matter which Taylor series method we use, how many terms we go forward before we truncate. Advantages And Disadvantages Of The Taylor Series Method: advantages a) One step, explicit b) can be high order c) easy to show that global error is the same order as LTE disadvantages Needs the explicit form of derivatives of f. 4 Runge-Kutta Methods To avoid the disadvantage of the Taylor series method, we can use Runge-Kutta methods. These are still one step methods, but they depend on estimates of the solution at different points. They are written out so that they dont look messy: Second Order Runge-Kutta Methods: k1 = _tf(ti, yi) k2 = _tf(ti + __t, yi + _k1) yi+1 = yi + ak1 + bk2 lets see how we can chose the parameters a,b, _, _ so that this method has the highest order LTE possible. Take the Taylor expansions to express the LTE: k1(t) = _tf(t, y(t)) k2(t) = _tf(t + __t, y + _k1(t) = _t _ f(t, y(t) + ft(t, y(t))__t+ fy(t, y(t))_k1(t) + O(_t2) _ LTE(t) = y(t+_t) y(t) _t a _t f(t, y(t))_t b _t (ft(t, y(t))__t+ fy(t, y(t)_k1(t) + f(t, y(t))_t + O(_t2) = y(t+_t) y(t) _t af(t, y(t)) bf(t, y(t)) bft(t, y(t))_ bfy(t, y(t)_f(t, y(t))+ O(_t2) = y0(t) + 1 2 _ty00(t) (a + b)f(t, y(t)) _t(b_ft(t, y(t))+ b_f(t, y(t))fy(t, y(t)) + O(_t2) = (1 a b)f + ( 1 2 b_)_tft + ( 1 2 b_)_tfyf + O(_t2) So we want a = 1 b, _ = _ = 1 2b . Fourth Order Runge-Kutta Methods: k1 = _tf(ti, yi) (1.3) k2 = _tf(ti + 1 2 _t, yi + 1 2 k1) (1.4) k3 = _tf(ti + 1 2 _t, yi + 1 2 k2) (1.5) k4 = _tf(ti+_t, yi + k3) (1.6) yi+1 = yi + 1 6 (k1 + k2 + k3 + k4) (1.7) The second order method requires 2 evaluations of f at every timestep, the fourth order method requires 4 evaluations of f at every timestep. In general: For an rth order Runge- Kutta method we need S(r) evaluations of f for each timestep, where S(r) = 8 : r for r _ 4 r + 1 for r = 5 and r = 6 _ r + 2 for r _ 7 5 Practically speaking, people stop at r = 5. Advantages of Runge-Kutta Methods 1. One step method global error is of the same order as local error. 2. Dont need to know derivatives of f. 3. Easy for Automatic Error Control. Automatic Error Control Uniform grid spacing in this case, time steps are good for some cases but not always. Sometimes we deal with problems where varying the gridsize makes sense. How do you know when to change the stepsize? If we have an rth order scheme and and r + 1th order scheme, we can take the difference between these two to be the error in the scheme, and make the stepsize smaller if we prefer a smaller error, or larger if we can tolerate a larger error. For Automatic error control yo are computing a useless (r+1)th order shceme . . . what a waste! But with Runge Kutta we can take a fifth order method and a fourth order method, using the same ks. only a little extra work at each step.

Macroeconomic Implications Education As Foundation Of Nations Economy - 825 Words

Macroeconomic Implications: Education As A Foundation Of Nations Economy (Essay Sample) Content: Macroeconomics Name Instructor’s name Institution Education as A Foundation of Nations Economy According to me, education is simply a continuous learning process that equips one with the knowledge to face all aspects of life challenges. Education illustrates its significance to the growth of the economy as a promoter of production quality, technology advancement and human development. Education helps individuals to attain success in various fields and also helps in setting higher objectives which helps them surpass the ordinary persons, especially in the job market. In this article, am going to discuss the educational system in both Ireland and Taiwan and its macroeconomic implications. Irish Education System In Ireland, education is mandatory for all children aged 6-16 until completion of the three-year second level education. The education system comprises of primary, secondary, vocational and tertiary education. Under primary or first-level education entirely covers basic education including science education and environment, social, mathematics, language, health education, physical integration, and education arts (drama and music). Secondary education offers several kinds of post-primary schools; comprehensive, community, secondary and vocational schools. After primary education, the majority of pupils go to vocational/secondary schools till they attain a leaving certificate after the 12th grade. Second level education constitutes a three-year junior series accompanied by a two or three-year senior series. In vocational education, 30% of secondary school students are offered with practical skills that can help them earn a living. Third-level education. It comprises of a technological sector, university sector and colleges of education. It also includes independent private institutions. There are universities which govern themselves and the offer degree programs at bachelor, masters and doctorate level. The whole population of Ireland as per 2018 statistics is approximated to be 4,803,748 people. The population that attend primary, secondary and higher education institution is approximated to be 1,091,632 people. This implies that at least 50% of this statistical population have attained primary education, 34% have acquired secondary education and at 16% have acquired post-secondary education. Taiwan Education System In Taiwan, the ministry of education has the responsibility of setting and maintaining policies governing education and also managing all public education. The education system is made up of primary education -elementary schools. Secondary education-junior, senior and vocational schools. Higher education includes institutions such as universities, colleges and technical institutes ADDIN ZOTERO_ITEM CSL_CITATION {"citationID":"OI1kfnQG","properties":{"formattedCitation":"(Woo, 1991)","plainCitation":"(Woo, 1991)","noteIndex":0},"citationItems":[{"id":498,"uris":["http://zotero.org/users/local/MqS91Xa3/items/855BMJQK"],"uri":["http://zotero.org/users/local/MqS91Xa3/items/855BMJQK"],"itemData":{"id":498,"type":"article-journal","title":"Education and economic growth in Taiwan: A case of successful planning","container-title":"World Development","page":"1029–1044","volume":"19","issue":"8","source":"Google Scholar","shortTitle":"Education and economic growth in Taiwan","author":[ {"family":"Woo","given":"Jennie Hay"}],"issued":{"date-parts":[["1991"]]}}}],"schema":"https://github.com/citation-style-language/schema/raw/master/csl-citation.json"} (Woo, 1991). Taiwan as a province roughly homes a population of 23 million people out of which 5,384,926 people are formally literate at various levels. Total of 40% of this population attends primary education, 31% secondary school level and 29% post-secondary education. Macroeconomic Implications of Education In as much as education seems to benefit an individual, it is one of the foundations of economic growth in many countries and in this case, Ireland and Taiwan. some of the positive macroeconomic implications of education include employment. From statistics employment figures in both countries have illustrated a considerable increase, because the educated young graduates have occupied several net professional job opportunities offered by the government. Employed people are capable of meeting their basic needs and thus contributing to the overall income of the nation. One’s income is another one’s expenditure. Education is an investment. In the long run edu...