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TOPIC 2 OPTIMISATION

1. Multi Variables Optimisation 2. Constrained Optimisation I 3. Constrained Optimisation II

Reading: Jacques Ian (2006), Mathematics for Economics and Business, 5th Edition Prentice Hall - Chap 5.1, 5.4, 5.5, 5.6 & Appendix 3 Chiang, A and Wainwright, K (2005), Fundamental methods of mathematical economics, Mc Graw Hill - Chap 12.1, 12.2, 12.3

1. Multi Variables Optimisation

Refresher: Optimisation of a function

Suppose have a function like this: f(x) = 2x + 10 If want to optimise the function, follow a few steps: Step 1: Differentiate the function: f'(x) (also denoted fx or dy/dx) Step 2: Set the differential equal to zero: f '(x) = 0. Step 3: Solve for value of x To establish what type of turning point(s) you have found, steps 4 & 5 Step 4: Find the second differential: f''(x) (also denoted fxx or d2y/dx2) Step 5: Evaluate the second differential at the turning point(s).

• If f ''(x) > 0, you have a minimum point. • If f ''(x) < 0, you have a maximum point. • If f ''(x) = 0, you may have a point of inflection.

1. Multi Variables Optimisation

1.0 Differentiation

Many relationships involve more than two variables. E.g. Demand for a good is a function of its price, advertising, price of other goods, etc E.g. y = f(x1, x2, …, xn) Interested in impact of change in one/all of the variables on y Since there are more than just 1 variable, we distinguish between partial derivative and total derivative Partial: change in y when only 1 of the x’s is changing. The partial derivative of y with respect to xi is denoted as δy/δxi or yxi or δf/δxi or fxi and is found by differentiating y with respect to xi holding all the other variables constant. Total: change in y when all x’s are changing at the same time. It is denoted as dz/dxi. Note difference between δ (partial derivative) & d (total derivative)

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1. Multi Variables Optimisation

1.1 Partial Differentiation

Examples of partial differentiation Means g is a function of x1,x2,x3. Could • g(x1, x2, x3) = x13 + x22 - 3x3 δg/δx1 = 3x12 δg/δx2 = 2x2 • z = f(x,y) = 5x + δz/δx = 5 + y2 xy2 - 10 δz/δy = 2xy also have written g=f(x1, x2,x3)=…

δg/δx3 = -3

Application: Production Functions Q = f(K,L) = 60 KL - 3K3 - 2L2 What are the marginal products of labour (MPL) and capital (MPK)? MPK = δQ/δK = 60L - 9K2 MPL = δQ/δL = 60K -4L

1. Multi Variables Optimisation

1.2 Total Differentiation

The total differential of a function of more than one variable measures the change in the dependent variable (y) brought about by a small change in each of the independent variables (x). If z = f (x1, x2,……xn) dz = zx1dx1 + zx2dx2 +...+ zxndxn where dx1……dxn are small changes in the independent variables. Example: Find the total differential of z = x4 + 6xyw + 3y3 - 2w There are three variables here: x, y, w and the total differential will involve partially differentiating with respect to each of these variables: dz = (4x3 + 6yw) dx + (6xw + 9y2) dy + (6xy - 2) dw differential w.r.t.x differential w.r.t.y differential w.r.t.w

1. Multi Variables Optimisation

1.3 Second Order Partial Derivatives

A second order partial derivative means that a function has been differentiated twice with respect to one of the independent variable while the other variables are held constant. Consider z = f(x, y) • First order partial derivatives: fx = δz/δx fy = δz/δy • Second order: Function already been differentiated once w.r.t. x fxx = (fx)x =

z x x

And differentiated a second time w.r.t. x again Function already been differentiated once w.r.t. y

fyy = (fy)y =

z y y

And differentiated a second time w.r.t. y again

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1. Multi Variables Optimisation

1.4 Second order Cross Partial Derivatives

There are also cross partial second order derivatives, i.e., the function is differentiated a second time, but with respect to another independent variable now, while the other variables are held constant. Function already been differentiated once w.r.t. x fxy = (fx)y =

z y x

And differentiated a second time w.r.t. y Function already been differentiated once w.r.t. y

fyx = (fy)x =

z x y

And differentiated a second time w.r.t. x

1. Multi Variables Optimisation

1.5 Second Order Partial and Cross Derivatives: Examples

Consider Z = 3x2y3 + 4x3 - 5y2 zx = 6xy3 + 12x2 zy = 9x2y2 – 10y zxx = 6y3 + 24x zyy = 18x2y – 10 zxy = 18xy2 zyx = 18xy2

Function differentiated once w.r.t. x Function differentiated once w.r.t. y Function differentiated a second time w.r.t. x Function differentiated a second time w.r.t. y

Function differentiated a second time w.r.t. y Function differentiated a second time w.r.t. x

1. Multi Variables Optimisation

1.6 Optimisation of More than One Variable

Optimisation of function with more than one variable is similar to that for a single variable. The First order conditions determine the location of the optimal point(s). They are called the critical or turning points. All the first order differentials must equal zero simultaneously, that is, fx1, fx2, ..., fxn = 0 The second order conditions determine the nature of that point. An optimal point may be a maximum, a minimum, a point of inflection or a saddle point. At the critical points, second order cross partial derivatives are equal, i.e. fxy = fyx (Young’s Theorem)

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1. Multi Variables Optimisation

1.6 Optimisation of More than One Variable..

Consider z = f(x, y). Following rules determine the exact nature of a turning point Maximum fx = 0 fy = 0 fxx.fyy > (fxy)2 fxx, fyy < 0 Minimum fx = 0 fy = 0 fxx.fyy > (fxy)2 fxx, fyy > 0 Saddle fx = 0 fy = 0 fxx.fyy < (fxy)2 fxx & fyy have same sign Inflection fx = 0 fy = 0 fxx.fyy < (fxy)2 fxx & fyy have different sign

Note: if fxx.fyy = (fxy)2 test is inconclusive

1. Multi Variables Optimisation

1.6 Optimisation of More than One Variable

How these turning points look like in a three dimensional space:

Maximum point Minimum point

Inflection point Saddle point

1. Multi Variables Optimisation

1.6 Optimisation of More than One Variable

Example f(x, y) = x3 - 3x + xy2 First Order Differentials fx = 3x2 - 3 + y2 = 0 …………… (1) fy = 2xy = 0 …………… (2) •From (2), when x = 0, y = 0 •Replace these values in (1): when x = 0 y2 = 3 and y = +3 or -3 when y = 0 x2 = 1 and x = +1 or -1 critical/turning points are: (0, 3); (0, -3); (1, 0); (-1, 0)

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1. Multi Variables Optimisation

1.6 Optimisation of More than One Variable..

Second order differentials: fxx = 6x fyy = 2x fxy = 2y fyx = 2y We need to evaluate fxx, fyy, fxy and fyx at each turning point. At (0, 3) Replace x = 0 and y =3 fxx = 0 fxx.fyy < (fxy)2 0 < 12 fyy = 0 fxx, fyy (same sign) fxy = 23 Saddle Point fyx = 23

1. Multi Variables Optimisation

1.6 Optimisation of More than One Variable

At (0, -3) fxx = 0 fyy = 0 fxy = fyx = -23 At (1, 0) fxx = 6 fyy = 2 fxy = fyx = 0 At (-1, 0) fxx = -6 fyy = -2 fxy = fyx = 0 fxx.fyy < (fxy)2 0 < 12 fxx, fyy (same sign) Saddle Point fxx.fyy > (fxy)2 12 > 0 fxx, fyy > 0 (6, 2 > 0) Minimum Point

fxx.fyy > (fxy)2 12 > 0 fxx, fyy < 0 (-6, -2 < 0) Maximum Point

1. Multi Variables Optimisation

1.7 Optimisation of More than One Variable - Application

A firm in a perfectly competitive market sells two goods, QA and QB at a price of £10 and £8 respectively. If total costs are TC = 2QA2 + 2QAQB + QB2, what is the maximum level of profits for the firm? Use the second order conditions to check that it is a maximum. (profit) = TR - TC TR = p.q = 10QA + 8QB = 10QA + 8QB - 2QA2 - 2QAQB - QB2 First Order Condition: These are simultaneous equations which A = 10 - 4QA - 2QB = 0 we can solve to obtain values for QA & QB B = 8 - 2QA - 2QB = 0 QA = 1 and QB = 3 Revise your understanding of how to solve simultaneous equations = 17 Are QA = 1 & QB = 3 turning points?

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1. Multi Variables Optimisation

1.7 Optimisation of More than One Variable

We need to check whether this is a maximum point (at this point, profits are supposed to be maximised with these output levels) Second Order Condition: AA = -4 BB = -2 AB = -2 BA = -2 fxx.fyy = AA.BB = 8 (fxy)2 = (AB)2 = (BA)2 = 4 fxx.fyy > (fxy)2 , i.e. 8>4 AA and BB are negative So we have a maximum point. That is, output levels QA = 1 & QB = 3 give a maximum profit level of £17.

2. Constrained Optimisation I

2.1. Constraints

Maximising or minimising some variable is often subject to some constraint Consider maximising a utility function

• U = 2X + 3Y • The amount of good X and Y that will maximise utility is close to infinity • But we cannot buy an infinite amount of X and Y (why?)

Consider minimising a total cost of production function

• • • • TC = 4X1 + 6 X22 + 7X3 The minimum value will be zero with zero unit of output of each good But this is not feasible because we need to produce an output! So the output level becomes a constraint

Therefore, most optimisation (minimisation or maximisation) problems are subject to some constraints (budget, output)

2. Constrained Optimisation I

2.1. Constraints

Consider maximising the following utility function U = f(X, Y) subject to a given budget constraint. Price of X and Y is given as PX and PY and the consumer has an income of M. Therefore the budget constraint is given as PXX + PYY = M Graphically, we have:

Y

X

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2. Constrained Optimisation I

2.1. Constraints

At equilibrium, the slope of the budget constraint is equal to the slope of the utility function (MU means marginal utility) So

PX Y PY X

U U

X Y

MU X MU Y

(sign?!)

So Price ratio = MRS (Marginal Rate of Substitution) or

MU X MU Y PX PY

Ratio of MU to Price is the same for all goods

2. Constrained Optimisation I

2.1. Constraints

Consider minimising the following Total Cost function PXX + PYY Subject to producing w amount of output Q = f(X, Y) = w

Y

Q=w X

2. Constrained Optimisation I

2.2. The Substitution Method

How do we optimise subject to a constraint? Use the substitution method We substitute the constraint in the objective function and then do the maximisation procedure. Example: Maximise Z=2x2-3xy+2y+10 subject to y –x = 0 Step 1: Rearrange constraint as y in terms of x y=x Step 2: Substitute y = x in Z Z = 2x2-3xx+2x+10 Z = 2x2-3x2+2x+10 Z = -x2+2x+10 This is what we seek to optimise now

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2. Constrained Optimisation I

2.2. The Substitution Method

Step 3: Optimise Z in the conventional way Z = -x2+2x+10

dz 2 x 2 0 dx

Thus, x = 1

When x = 1, y = 1

d 2z 2 so turning point (1,1) is a maximum point dx 2

2. Constrained Optimisation I

2.3. The Substitution Method: Application 1

A firm faces a production function Q= 4LK + L2 and buys the inputs K and L at prices per unit of $1 and $2, respectively. If it has a budget of $105, what combination of K and L should it use in order to produce the maximum possible output? Also, verify that the ratio of Marginal product to price is the same for both factors. Step 1: rewrite the constraint in a form for substitution 1K + 2L = 105 K = 105-2L Step 2: Substitute the constraint into the production function Q= 4L(105-2L) + L2 Q= 420L – 7L2

2. Constrained Optimisation I

2.3. The Substitution Method: Application 1

• Step 3: Optimisation of Q= 420L – 7L2

dQ 42014L 0 dL

• Therefore L = 30, and replacing L = 30 in the constraint, we can solve for K K = 105-2(30) = 45 • (30, 45) is a turning point • We check for second order condition

d 2Q 14 0 dL2

• Thus point (30, 45) is a maximum point • Replace L = 30 and K = 45 in objective function to get maximum output. • Maximum Output is hence Q= 4(30)(45) + (30)2 = 6300

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2. Constrained Optimisation I

2.3. The Substitution Method: Application 1

MPL = δQ/δL = 4K + 2L • MPK = δQ/δK = 4L • At equilibrium:

MPL MPK PL PK MPL 4 K 2 L 4(45) 2(30) 120 PL 2 2 MPK 4 L 4(30) 120 PK 1 1

• Thus, the ratio of Marginal Product to price is the same for both factors

2. Constrained Optimisation I

2.4. The Substitution Method: Application 2

A firm faces a Cobb Douglas production function Q= 2K1/2L1/2 and can buy the inputs K and L at prices per unit of $4 and $3, respectively. What is the cheapest way of producing 160 units of output? • Here the aim is to minimise total cost subject to producing an output of 160. The objective function is therefore the total cost function and the constraint is the production function. So we have: Minimise TC = 4K + 3L Subject to 2K1/2L1/2 =160 • Step 1: Rewrite the constraint 2K1/2L1/2 =160 L = 6400/K • Step 2: Substitution TC = 4K + 3(6400/K) TC = 4K + 19200/K

2. Constrained Optimisation I

2.4. The Substitution Method: Application 2

• Step 3: Optimisation of TC = 4K + 19200/K

dTC 19200 4 0 dK K2

• Solving for K and L, we get K = 69.28 and L = 92.38 • Therefore (69.28, 92.38) is a turning point • Check for second order conditions:

d 2TC 38400 0 dK 2 K3

Note: we replace these values in the original cost function, i.e. TC=4K+3L

• So it is a minimum point. • Replacing these values in the TC function, the minimum cost of producing 160 units of output is given as $554.26

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3. Constrained Optimisation II

3.1. The Lagrangian Multiplier (LM) Approach

Substitution Method has certain disadvantages

• Becomes more complicated when deal with more than 2 variables • Does not provide additional information from maximisation/minimisation process

Lagrange Multiplier Approach overcomes these problems, and hence:

• Makes it easier to deal with more than 2 variables • Enables us to obtain additional information from the maximisation/minimisation process, namely the Langrage multiplier ()

3. Constrained Optimisation II

3.1. The Lagrangian Multiplier (LM) Approach

If we want to optimise an objective function, f(x1...xn) subject to a constraint, g(x1...xn) = M, the LM approach proceeds as follows: Step 1: Define the Lagrangian function • L(x1..xn, ) = f(x1....xn) + [M - g(x1...xn)] where is known as the Lagrange multiplier. Step 2: Find the partial derivatives, Lx1,...Lxn, L • Set partial derivatives equal to zero and solve for x1...xn and . Step 3: Check for second order conditions The Lagrange Multiplier () can be thought of as the effect on the objective function of a unit change in the constraint (marginal effect)

3. Constrained Optimisation II

3.2. The Lagrange Multiplier Approach: Example

Optimise f(x, y) = x2 - 3xy + 12x subject to 2x + 3y = 6 Step 1: Set up the Lagrangian function L = x2 - 3xy + 12x + (6 - 2x - 3y) Step 2: Find the partial derivatives Lx = 2x - 3y + 12 - 2= 0 ……. (1) Ly = -3x -3 = 0 ……. (2) L = 6 - 2x -3y = 0 ……. (3) From (2): = -x Substituting (4) into (1): 2x -3y + 12 + 2x = 0 3y = 4x + 12 ……. (4)

……. (5)

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3. Constrained Optimisation II

3.2. The Lagrange Multiplier Approach: Example

Substituting (5) into (3): 6 - 2x - 4x -12 = 0 -6x = 6 x = -1 ……. (6) Substituting (6) into (5): y = 8/3 Substituting (6) into (4): =1 = 1 implies that a unit change in the constraint will increase the objective function by 1.

3. Constrained Optimisation II

3.3. The Lagrange Multiplier Approach: Application

Suppose a utility function is given by: U = 40x0.25y0.5 with prices of x and y as px = 4, py = 10 and income M = 60 What level of x and y will maximise utility? What is the meaning of the Lagrange multiplier in this case? Budget contraint: 4x + 10Y = 60 Lagrange Function: L = 40x0.25y0.5 + (60 - 4x - 10y) First Order Conditions Lx = 10x-0.75 y0.5 - 4 = 0 2.5x-0.75 y0.5 = Ly = 20x0.25y-0.5 - 10 = 0 2x0.25y-0.5 = L = 60 - 4x - 10y = 0 ….

… (1) … (2) (3)

3. Constrained Optimisation II

3.3. The Lagrangian Multiplier Approach: Application

Equating (1) and (2) 2.5x-0.75 y0.5 = 2x0.25y-0.5 4x = 5y …… (4) Substituting (4) into (3): 60 - 5y - 10y = 0 60 = 15y y=4 From (4) x=5 From (1) or (2) = 1.495 1.5

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3. Constrained Optimisation II

3.3. The Lagrangian Multiplier Approach: Application

What is the meaning of ? If we relax the budget constraint by a small amount, utility will increase by times that amount. So if we give the consumer an extra £1 income, utility will go up by about 1.5 units. It is also called the marginal utility of income. Note that we could have set up the Lagrangean function as follows: L = 40x0.25y0.5 + (4x + 10y - 60) We would still get same answer for x and y, but = -1.5 The interpretation of is not appropriate (the consumer’s total satisfaction should increase, and not decrease if you give him more income!) Therefore, should be careful when defining the Lagrangian function!

4. Exercises

Exercise 1

The total cost (TC) & total revenue (TR) functions for a company are as follows: TC = 2q2 +2a.q + 2a2 TR = 18q + 12a + a.q where q is the quantity of output and a is the expenditure on advertising. (i) Write down the profit function for this firm. (ii) Find the profit maximising levels of output and advertising. (iii) Find the maximum level of profits

4. Exercises

Exercise 2

The output function of a firm is given by Q = 120L + 200K – L2 – 2K2 where L = quantity of labour and K = quantity of capital. Unit labour cost is £5 and unit capital cost is £8. (i) Use the Lagrangian method to find the maximum level of output the firm can produce with a budget of £130. (ii) Explain the meaning of the Lagrangian multiplier in this example.

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4. Exercises

Exercise 3

A consumer has the following utility function over two goods, X and Y: U = 20 X0.5Y0.5 (i) If the price of X is 20, the price of Y is 5 and the consumer’s income is 300, find the maximum value of utility that the consumer can achieve. (ii) Calculate the value of the Lagrangian multiplier and explain its meaning.

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...Research Method Hypothesis and Theory Hypotheses can be developed and tested to recognize the relationships between categories. Silverman (1991:1) defined hypothesis as a ‘testable proposition’. The appearance of an apparent relationship or connection between categories will need to be tested in order to find out whether there is an actual relationship (Saunders, 1997:344). The importance of hypothesis is that it will bring a specific direction and focus to a research study. The theory on the other hand, is usually drawn from the hypothesis. Theories are usually generated from attempts at explaining observations and thus prediction or expectations can be made (Gill, 1991:25). Deduction and Induction There are two methods of establishing what is true or false and of drawing conclusion. These two methods are deduction and induction. Induction is made by empirical evidence based, while deduction is logic based. Through induction, a general conclusion can be made from empirical observation. It goes by the process of assumption to conclusion (Ghauri, 1995:8). From deduction, conclusions are draw through logical reasoning and it is not necessary to be reality. When an observation is made to generate a theory with consistent facts, it is called induction, on the contrary deduction involves the gathering of facts to confirm or disprove hypothesized relationships among variables that have been deduced from proposition or earlier theories (Ghauri, 1995:9). Research method and......

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...Research Methods Jessica February 2012 What is the difference between direct and indirect observational methods of research? Direct observation is when researchers observe the behavior while it is occurring. Indirect (unobtrusive) observation is when researchers examine physical traces and archival records. (Zechmeister, Zechmeister, & Shaughnessy 2001) Direct observation of behavior can be seen in simply psychology: question and answer, as well as simple observation of a person’s daily activities. The researcher can choose to change the atmosphere, or change the study to intervene and observe the changes, while indirect observations main goal is to be unseen and non-influential on the behavior that is being observed as to take down all natural information. My friend attempted to use direct observation in this study. He went out and observed people’s behavior while intervening with his own behavior in hopes to prove his hypothesis. There is so much wrong with his way of thinking in this study. The first problem is his observational bias as he is not only observing others and the conversation but also himself. There is no scientific way to observe yourself without having some bias in opinion. Beyond that, he uses behavioral sampling, and cannot come to a precise conclusion based soley on his words and the reactions of the people he is in conversation with. There are many factors to consider. Where are the people coming from, where are they going? Are they having a......

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...Measurement and Survey Research Measurement and Survey Research Syed S. Hossain Institute of Statistical Research and Training University of Dhaka, Bangladesh Syed S. Hossain Institute of Statistical Research and Training University of Dhaka, Bangladesh Measurement and Survey Research Measurement and Survey Research Fundamental ideas construct validity (the degree inference can made from study to theory) Reliability (the quality of measurement) random and systematic error, Reliability and validity related Scales of measurements Nominal Ordinal Interval Ratio Syed S. Hossain Institute of Statistical Research and Training University of Dhaka, Bangladesh Measurement and Survey Research Arithmetic strength of scales of measurements Levels Nominal Ordinal Interval Arithmetic Counting Counting Ranking Counting Ranking Addition/Subtraction Counting Ranking Addition/Subtraction Multiplication/Division Features Categories Categories Ranks Categories Ranks Has equal units Categories Ranks Has equal units Has absolute zero Examples Religion Economic class IQ score Ratio Weight Syed S. Hossain Institute of Statistical Research and Training University of Dhaka, Bangladesh Measurement and Survey Research Information strength of Scales of measurements Syed S. Hossain Institute of Statistical Research and Training University of Dhaka, Bangladesh Measurement and Survey Research Types of Survey research Types Questionnaire mail......

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...Submission Form |Name: |Vigneshwaran Palanisamy | |Email: |1118562@rgu.ac.uk | |Course: |Msc Purchasing & Supply Chain Management | |Module: |BSM577 - Research Methods | |Assignment and Title: |Implementing E-procurement in Indian organisation : surveys of SMEs | |Date: |7/1/13 | |For the attention of: |Dr Elizabeth Tait | WORD COUNT: ABSTRACT: This report was undertaken for the commitment of future research in dissertation. E-commerce is gaining a lot more attention in current global market. E-commerce models such as B2B, B2C, B2E and B2G has been successful due to many implementation and adoption of standardized process tools like EDI, shipping, tracking, payment and delivery among the suppliers around the globe through a strong supply network. The most vital element of B2B model is the E-procurement. E-procurement is the process for acquiring......

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...PORTFOLIO 1. What is research and what is a research carried out for? Find a research report in an applied linguistics journal (such as TESOL Quarterly, Language Learning) and point out the objectives and how these objectives are achieved. a. What is research? Research has been defined in a number of different ways. A broad definition of research is given by Martyn Shuttleworth - "In the broadest sense of the word, the definition of research includes any gathering of data, information and facts for the advancement of knowledge." Another definition of research is given by Creswell who states that - "Research is a process of steps used to collect and analyze information to increase our understanding of a topic or issue". It consists of three steps: Pose a question, collect data to answer the question, and present an answer to the question. The Merriam-Webster Online Dictionary defines research in more detail as "a studious inquiry or examination; especially : investigation or experimentation aimed at the discovery and interpretation of facts, revision of accepted theories or laws in the light of new facts, or practical application of such new or revised theories or laws" Scientific research is a systematic way of gathering data, a harnessing of curiosity. This research provides scientific information and theories for the explanation of the natureand the properties of the world. It makes practical applications possible. Scientific research is......

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...METHODS OF RESEARCH 1. Historical Method Purpose: To reconstruct the past objectively and accurately, often in relation to the tenability of a hypothesis. Example: A study reconstructing practices in the study of social studies in the Philippines during the past fifty years; tracing the history of agrarian reform in the Philippines; Philippine education since 1946; testing the hypotheses that Francis Bacon is the real author of the “Works of William Shakespeare”. 2. Descriptive Method Purpose: To describe systematically a situation or area of interest factually and accurately. Examples: Population census studies, public opinion surveys, fact-finding surveys, status studies, task analysis studies, questionnaire and interview studies, observation studies, job descriptions, surveys of the literature, documentary analyses, anecdotal records, critical incident reports, test score analyses, and normative data. 3. Developmental method Purpose: To investigate patterns and sequences of growth and/or change as a function of time. Examples: A longitudinal growth study following an initial sample of 200 children from six months of age to adulthood; a cross-sectional growth study investigating changing patterns of intelligence by sampling groups of children at ten different age levels; a trend study projecting the future growth and educational needs of a community from past trends and recent building estimates. 4. Case and field......

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