**CHAPTER ONE : INTRODUCTION TO MANAGERIAL ECONOMICS**

**A. Definition of Managerial Economics:**

It is application of economic theory and the tools of analysis of decision science to examine how a firm can achieve its aims or objectives most efficiently.

**B. Firm in macroeconomic analysis**

Circular flow of income describes the way in which a country’s economy flows backward and forward between the sectors in the economy. It shows how the household and firm interact in the resource and product markets. A simple circular flow assumes that the economy is divided into only into two sectors.

#### a) 2 Sector Economy - The Simple Circular Flow

#### We assumed that:

#### i) households receive their income from the firm by providing factors of production they own

ii) firms sell their entire output(supply of goods & services) to households

iii) households spend their entire income on goods and services. So all goods produced are sold.

Purchase of goods and services

supply of goods & services

supply of factors of production

wages, interest, rent, profit

(i) The household is the owner of factors of production and they are suppliers and sellers of factors of production (resources) to the firms whilst firms are the buyers.

(ii) The bottom half of the diagram shows the flow of the factors of production owned by households to the firms.

(iii) The firms in return pay wages, rent, interest and profits to household as income.

(iv) The top half of the flow shows the flow of goods and services produced from households to firms and the corresponding flow of money payments for goods and services from households to firms as households’ consumption.

(v) The circular flow of income illustrates the basic principle of national income accounting where the value of total outputs equals the value of total income.

(vi) The above diagram shows how two sectors interact in the

__product market__and__resource market.__(vii) The illustration assumed that household spends all their income to buy goods and services.

(viii) But, in the real life, household does not spend all their income for consumption. They also save part of their income.

(ix) So, what happens if the household does not consume all their income to buy goods and services but save part of it in the Financial Institutions?

**C. The Rationale of the firm**

It would be very costly for individual household to enter into each production and distribution process. Thus, a firm should exist to purchase these goods and transform them into goods and services for sale. Resource owners (households) then purchase these goods and services with the income generated form the sale of their services or resources.

(The economies generated in production and distribution would lower the cost of production and provide higher returns to resource owners.

**I. Objectives of the firm**

The primary goal of a firm is to maximize profit (minimize cost).

What is profit?

Profit is a reward for:

Bearing risks

An entrepreneur will bear all the risk associated with production. The reward for the risk is profit.

Imperfect market mechanism

In imperfect market, firms can generate profit since there is less competition

Monopoly status

If a firm is a monopoly, it is able to curtail/prevent other firms form entering the market. Thus, it is able to enjoy profits for long period of time.

Innovations

Development of new products, new production techniques, and new modes of marketing will provide higher return.

**Function of profit**

Profits act as a signal for reallocation of resources to reflect changing demand and taste.

**How to calculate profit?**

Profit = TR – TC

Accounting profit = TR – explicit costs

Economic profit = TR – explicit costs – implicit costs

Other goals:

Sales maximization

Revenue Maximization

Market share maximization

Employment

Working environment for workers

Provide good product and services to customer

Act as a good citizen

**II. Decision Problems**

Firms usually face many constraints such as:

1. Legal constraint

- it includes the array of federal, state, and local laws that must be obeyed by all citizens, both individual and corporate. Areas where managers seem to have some legal difficulty include environmental law, especially those relating to pollution and the disposal of hazardous wastes, and employment law, including wrongful termination and sexual harassment matters.

2. Moral constraint

- it applies to actions that are not illegal but are sufficiently inconsistent with generally accepted standards of behavior to be considered improper.

3. Contractual constraint

- it binds the firm because of some prior agreement such as a long-term lease on a building, or a contract with a labor union that represents the firm’s employees.

4. Financial constraint

- it occurs when a department of a firm is assigned a budget for the next year and managers are given orders to maximize production subject to this budgeted amount.

5. Technological constraint

- it sets physical limits on the amount of output per unit of time that can be generated by particular machines or workers.

**LESSON TWO : ECONOMIC OPTIMIZATION/BASIC TRAINING**

**A. Methods of Differentiation**

The slope of a function y = f(x) is the change in y divided by the corresponding change in x. In order to find the value of slope, there is a need to use derivatives.

For a function y=f(x), the derivative, written as dy/dx is the slope of the function at a particular point on the function.

**Rules of differentiation:**

1.

**The Derivative of a constant**. The derivative of any constant is always zero.

Example:

y = 10

dy/dx = 0

2.

**The derivative of a power function**. If y = ax

^{n}then, dy/dx = n*ax^{(n – 1)} Example:

y = 10x

^{2} dy/dx = (2)10x

^{(2-1)}= 20x3.

**The derivative of a constant times a function**. If y = a*f(x), where x is constant, dy/dx is a.

Example:

y = 10 x

^{(1-1)} dy/dx = (1)10X

^{(0) }= 104.

**The derivative of a sum or difference**. If y = f(x) + g(x), then dy/dx = f’(x) + g’(x).

Example:

y = 20x + 10

dy/dx = (1)20x

^{(1-1}) + 0 = 205.

**The derivative of a product function.** If y = f(x)g(x), then dy/dx = f’(x)g(x) + f(x)g’(x)

Where f’(x) = df/dx and g’(x) =dg/dx

Example:

y = (x

^{2}– 3)(x^{3}+ 4x + 2) Assume that f(x) = (x

^{2}– 3) and g(x) = (x^{3}+ 4x + 2) f’(x) = df/dx = 2x and g’(x) =dg/dx =(3x

^{2 }+ 4) so to find dy/dx , we use the following formula:

dy/dx = f’(x)g(x) + f(x)g’(x)

Thus, dy/dx = (2x)(x

^{3}+ 4x + 2) + (x^{2}– 3)(3x^{2}+ 4) = [2x

^{4}+ 8x^{2}+ 4x] + [3x^{4}+ 4x^{2 }– 9x^{2}^{ }+ 12] dy/dx = 5x

^{4}+ 3x^{2}+ 4x + 12 If the value of x is given (x=1) then

dy/dx = 5(1)

^{4}+ 3(1)^{2}+ 4(1) + 12 = 24

6.

**The derivative of a quotient function**. If y = f(x)/g(x), then

dy/dx = g(x)f’(x) – f(x)g’(x)

[g(x)]

^{2} Example:

y = x

^{2}-4x x

^{2} dy/dx = x

^{2}(2x-4) – (x^{2}– 4x)(2x) (x

^{2})^{2} = (2x

^{3 }– 4x^{2}) – (2x^{3}– 8x^{2}) (x

^{4}) = 2x

^{3 }– 4x^{2}– 2x^{3}+ 8x^{2} (x

^{4}) = 4x

^{2} x

^{4} dy/dx = 4/x

^{2}**Exercise 1A:**

Determine the derivatives of each of the following functions.

a) Y = 20

b) Y = 6X

^{2}+ 8X + 50c) Y = (X

^{2}- 2)(X^{2}+ 4X +10)d) Y = X

^{2}+ 3X + 4 X

^{2}– 4e) Q = 120 – 0.2P

^{2}f) C = 2,500 – 100X

^{2}+ 4X^{3}**B. First and Second Derivatives**

i) The maximum or minimum point of a function, y = f(x) can be found by setting the

**first derivative**of the function to**zero**and solving the equation for the value(s) of x. Example 1B:

If y = 200x – 0.4x

^{2}, find the optimum value of x. Find dy/dx = 0 allows you to find the optimum value of x.

dy/dx = 200 – 0.8x = 0

0.8x = 200

X = 200/0.8

x = 25

ii) When the first derivative of a function is zero, the function is at a

**maximum**if the second derivative is**negative**, or at a**minimum**if the second derivative is**positive**. From example 1B, the first derivative of y = 200x – 0.4x

^{2 }^{ }is dy/dx = 200 – 0.8x.

Setting it to zero, the optimum value of x is 25. To determine whether this value of x is minimum or maximum, you have to find the second derivative of the function.

Since, dy/dx = 200 – 0.8x

d

^{2}y/dx^{2}= -0.8 since d

^{2}y/dx^{2}is less than zero, then x = 25 is the maximum value. Example 2B:

If total revenue, TR = 200Q – 2Q

^{2}, what is the output level that can maximize TR? Maximize TR =

__dTR__= MR = 0 dQ

= 200 – 4Q = 0

4Q = 200

Q = 50

Thus, in order to maximize TR, output should be equal to 50 units.

To determine whether Q = 50 is indeed the maximum value, you have to find the second derivative of the TR function.

dTR/dQ = 200 – 4Q

d

^{2}TR/dQ^{2}= -4 (less than zero)This shows that Q = 50 is indeed a maximum value.

iii. Sometimes, setting the first derivatives to zero will give two extremum values. In order to identify which one is the optimum value, one has to substitute the two values into the second derivative.

Example 3B:

TR = 7Q – 0.1Q

^{2} TC = 10 + 8Q – 0.3Q

^{2}+ 0.01Q^{3} Find the value of Q that will maximize profit.

Profit = TR – TC

= -10 – Q + 0.2Q

^{2}– 0.01Q^{3}__d(Profit)__= -1 + 0.4Q – 0.03Q

^{2 }= 0

dQ

= (3Q – 10) (Q – 10) = 0

Thus, 3Q – 10 = 0 OR Q – 10 = 0

Q = 10/3 Q = 10

There should be only one optimum value. Therefore, substitute these values into the second derivative.

__d__= 0.4 – 0.06Q

^{2}(Profit) dQ

^{2} substitute Q = 10/3 into the second derivative

__d__= 0.4 – 0.06(10/3)

^{2}(Profit) dQ

^{2} = 0.2 (positive value = minimum)

__d__= 0.4 – 0.06(10)

^{2}(Profit) dQ

^{2} = -0.2 (negative value = maximum)

The value of Q that will maximize profit is Q = 10.

iv. To be an optimal value, the value must fulfill both first order condition (setting first derivative equals zero) and second order condition (finding the second derivative.)

**Exercise1B:**

1. Optimize the following functions and find the value of x at the optimum point. Then determine whether x is a minimum or maximum point.

a) y = 3x

^{2}– 4x + 25 b) y = (3x – 2 )

^{2}**2.**

**Given the following total revenue and total cost functions of a firm:**

TR = 22Q – 0.5Q

^{2} TC = 1/3Q

^{3}– 8.5Q^{2}+ 50Q + 90 determine,

a) the level of output at which the firm maximizes its total profit

b) the maximum profit that the firm could earn.

**C. Partial Derivative**

In the previous section, the exercises given involve differentiation of simple function i.e. one dependent and one independent variables. However, functions are normally in the form of multivariate functions i.e. one dependent but many independent variables. In order to find the derivative of the multivariate function, we have to find the partial derivative of the function.

Example:

y = 10 - 8x

^{2}- 130z + xz + 2z^{2} i) Find the partial derivatives of the function with respect to each independent variable x and z and set these derivatives equal to zero.

dy/dx = - 16x + z = 0 ---------- (1)

dy/dz = -130 + x + 4z = 0 --------- (2)

ii)

**Solve the two equations simultaneously for X and Z.** From (1), Substitute z = 16x in (2) gives

-130 + x + 4(16x) = 0

-130 + x + 64x = 0

-130 + 65x = 0

65x = 130

x = 130/65

x = 2

Substitute x = 2 in z = 16x

z = 16x

= 16 (2)

z = 32

**Exercise 1C:**

1. Find the partial derivative for the following functions.

a) y = 4 – x

^{2}– 2z + xz + 2z^{2} b) AC = x

^{2}+ 2y^{2}– 2xy – 2x -6y + 20 2. For the following total-profit function of a firm:

= 144X – 3X

^{2}– XY – 2Y^{2}+ 120Y - 35 determine;

i) the level of output of each commodity at which the firm maximizes its total profit,

ii) the value of the maximum amount of the total profit of the firm

3. For the following function,

Y = 50 + 18X + 10Z – 5XZ – 2X

^{2} find the value of X and Z that will maximize the function. What is the maximum value of Y?

**D. Lagrangian : Solving a constrained optimization problem**

In many decision problems, there are constraints imposed that limit the options of choices available to the decision makers.

Example:

TC = 3x

^{2}+ 6y^{2}– xy x = output produced by line 1

y = output produced by line 2

Objective function : TC = 3x

^{2}+ 6y^{2}– xyConstraint function : x + y = 20

Methods:

1. normal/substitution method

2. lagrangian method

__Method 1: Normal/substitution method__ TC = 3x

^{2}+ 6y^{2}– xy ------ (1) x + y = 20 ------ (2)

X = 20 – y

Substitute (2) into (1):

TC = 3(20 – y)

^{2}+ 6y^{2}– (20 – y)y TC = 1200 – 140y + 10y

^{2}---------- (3) d(TC)/dy = -140 + 20y = 0

y = 7 -------- (4)

Substitute (4) into (2) will give x = 13 -------- (5)

Substitute (4) and (5) into (1) will give TC = 710

Thus, at x = 13 and y = 7, total cost is minimized at 710.

__Method 2 : lagrangian__Lagrangian is a process to combine the objective function and the constraint function with a multiplier (lambda).

Steps:

(i) Objective function : TC = 3x

^{2}+ 6y^{2}– xy(ii) Constraint function : x + y = 20

(iii) Rewrite the constraint function as follows:

x + y – 20 = 0

**OR**

20 – x – y = 0

(iv) Lagrangian function :

L

_{TC}= 3x^{2}+ 6y^{2}– xy - λ(x + y – 20)**OR**

L

_{TC}= 3x^{2}+ 6y^{2}– xy + λ(20 – x - y)(v) Take all partial derivatives and set equal to zero

(vi) Solve simultaneous equation

(v) Interpret lambda

Solution:

L

_{TC}= 3x^{2}+ 6y^{2}– xy – λ(x + y – 20)L

_{TC}= 3x^{2}+ 6y^{2}– xy – xλ – yλ + 20λdL

_{TC}/dx = 6x – y – λ = 0 ------ (1)dL

_{TC}/dy = 12y – x – λ = 0 ------ (2)dL

_{TC}/dλ = – x – y + 20 = 0 ------ (3)(1) – (2) : 7x – 13y = 0 ------ (4)

(4) : x = 13y/7 ------ (5)

Sub.(5) into (3) : –(13y/7) – y + 20 = 0

[–13y – 7y + 140] = 0

-------------------------

7

[–13y – 7y + 140] = 0 ×7

– 20y – 140 = 0

Y = 7 ------ (6)

Substitute (6) into (4):

7x – 13y = 0

7x – 13(7) = 0

7x – 91 = 0

X = 91/7

x = 13 ------ (7)

Substitute (6) and (7) into (1):

6(13) – 7 – λ = 0

78 – 7 – λ = 0

71 – λ = 0

λ = 71

Substitute (6) and (7) into TC function; TC = RM710

λ = 71 means if output (constraint) were to increase by one unit from 20 to 21, then the TC (obj. function) will increase by 71.

**Exercise 1D:**

1. Firm AMC total profit function is given as

= 80x – 2x

^{2}– xy – 3y^{2}+ 100y It faces the constraint that the output of commodity x plus output of commodity y must be 12.

a. Use the Lagrangian Multiplier method to find profit maximization.

b. Interpret the value of lambda.

2. Oct 2003 Part A Question 2

3. April 2009 Part A Question 2

4. Oct 2009 Part A Question 6

**Chapter 4**

**Demand Estimation by Using Regression Analysis**

Regression Analysis a statistical method used to establish a relationship between a variable (Dependent Variable) and other factors that will affect it (Independent Variables).

This relationship can be expressed as a functional form:

**Q = a**

_{0}+ a_{1}A + a_{2}B + a_{3}CDemand Estimation for a product or service using regression analysis is important in the business world especially to the corporate executives and managers because it will enable them to make reasonable forecast for their goods and services in the near future. The manager can narrow down those factors that are important in influencing their sales and thereby formulate appropriate strategies or policies to achieve their management objectives.

The actual process of Regression Analysis can be very complex but it can be summarized into

**FOUR**important steps:- Model Specification: Set the objective and identify the important variables which have influence on the dependent variable.
- Data collected for all the variables specified.
- Choice of a function form

e.g. Linear or non-linear form

- Estimation and interpretation of results.

**1. Model Specification**

If we want to study the factors affecting the demand for automobiles (Qx) in the country, we must identify the most important variables that are believed to affect the demand for automobiles

e.g. a) Price of the automobile (Px)

b) Per capita income (Yc)

c) No. of working population (L)

d) Rate of interest, etc (I)

Qx = f(Px, Yc, L, I,…..)

**2.**

**Data collection on the variables.**

2 types of data :

**a) Time Series Data**

Data is collected for each variable over time (yearly, quarterly, monthly or daily, etc)

**b) Cross-Sectional Data**

Data are collected for same time period but from different section or geographical area of the society.

Types of data to be used depend on the availability of data.

a)

**Primary data**– Data collected from the field through market survey, sampling, & etc.b)

**Secondary data**– These are published data by relevant authority such as Statistical Department, Economic Reports, etc.**3.**

**Specifying the form of Equation.**

i) The simplest model to deal with and the one which is often also the most realistic is the linear model.

**e.g. Qx = a**

_{0}+_{ }a_{1}Px + a_{2}Y + a_{3 }L + a_{4}I + ……..+ ea

_{0},a_{1},….,a_{4}are parameters (coefficients) to be estimatede = disturbance term or error term

ii) Non- Linear model

Sometimes a non-linear form may be the data better than a linear equation.

Qx = a

_{0}Px^{α1}.Yc^{ }^{α2}. L^{ }^{α3}. I^{ }^{α4}(Power Function)**4.**

**Testing the (Econometric) Result**

To evaluate the regression results several statistics are examined.

a) The sign of each estimated coefficient must be checked to see if it conforms to what is expected on the theoretical grounds.

b) Coefficient of Determination, R

^{2}c) t – tests (coefficient)

d) Durbin-Watson statistics, etc.

e) The F-statistics (F-stats)

Note : The statistical procedure in solving Multiple Regression Problems can be very complicated. Fortunately there are many computer software’s available to achieve our objective.

i.e TSP (Time-Series Processor) or SPSS can be used to solve our problems.

**REGRESSION ANALYSIS**It describes the way in which one variable is related to another. Regression analysis derives an equation that can be used to estimate the unknown values of one variable on the basis of known values of another variable.

(a) Simple Regression Analysis

Y = a + bX where Y is sales volume & X is advertising expenditure

# Example 1

(Taken from ECO556 Manual Table 4.1, page 136 )

Year | Sales (Y) (million dollars) | Advertising Expenditure (X) (million dollars) |

1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 | 44 58 48 46 42 60 52 54 56 40 | 10 13 11 12 11 15 12 13 14 9 |

The result from computer print out :

LS// Dependent variable is SAL SMPL range 1986 - 1995 Number of observation 10 | ||||

Variable | Coefficient | Std. Error | T-Stat | 2-Tail Sig. |

| | | | |

C ADV | 7.6000000 3.5333333 | 6.332345 0.5222813 | 1.2001912 6.751919 | 0.264 0.000 |

| ||||

R-squared 0.851212 Mean of dependent var 50.00000 Adjusted R-squared 0.832614 S.D of dependent var 6.992059 S.E. of regression 2.860653 Sum of squared resid 65.46667 Durbin-Watson stat 1.224915 F-statistic 45.76782 Log likelihood -23.58417 |

^ ^ ^

Y = a + bX

^ ^ ^

=> Y = 7.6 + 3.53X

(b) Multiple Regression Analysis

Y = a

_{1}+ b_{1}X_{1}+ b_{ 2}X_{2}where Y is sales volume , a

_{1}is the interceptX

_{1}is advertising expenditure , b_{1}is the Y/X_{1}, marginal effect of adv on salesX

_{2}is price of the product , b_{2}is the Y/X_{2}, marginal effect of price on sales# Example 2

(Taken from ECO556 Manual Table 4.3, page 141 )

Year | Sales (Y) (million dollars) | Advertising Expenditure (X1) (million dollars) | Price (X2) (million dollars) |

1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 | 44 58 48 46 42 60 52 54 56 40 | 10 13 11 12 11 15 12 13 14 9 | 1 1.2 2 1.8 2.1 0.8 1.4 2.0 1.5 1.0 |

The result from computer print out :

LS// Dependent variable is SAL SMPL range 1986 - 1995 Number of observation 10 | ||||

Variable | Coefficient | Std. Error | T-Stat | 2-Tail Sig. |

| | | | |

C ADV P | 11.60403 3.4936051 -2.3836921 | 6.9633945 0.5078770 1.9495316 | 1.6665152 6.8788413 -1.2226999 | 0.140 0.000 0.261 |

| ||||

R-squared 0.877397 Mean of dependent var 50.00000 Adjusted R-squared 0.842367 S.D of dependent var 6.992059 S.E. of regression 2.776058 Sum of squared resid 53.94549 Durbin-Watson stat 1.41 F-statistic 25.04734 |

^ ^ ^ ^

Y = a

_{1}+ b_{1}X_{1}+ b_{2}X_{2}^ ^ ^ ^

=> Y = 11.60 + 3.49X

_{1}- 2.38X_{2 }**Evaluation of Results (Computer Printouts)**

These are the importance statistical results should be interpreted:

- The sign of each estimated coefficient
- Coefficient of determination (R
^{2}) - Standard error of estimate (Se)
- The t-statistics (t-stats)
- The F-statistics (F-stats)

**Interpretation :**

a. The sign of each estimated coefficient must be checked to see if it conforms to what is expected on the theoretical grounds.

^ ^ ^

From Example 1: Y = 7.6 + 3.53X

The estimated function show positive value (+ 3.53) , so it conforms to the expected economic theory. If we spend $1 on Advertisement (X) then the Sales(Y) will increase by 3.53 units.

b. Coefficient of determination (R

^{2}) The value of R

^{2}ranges from ‘0’ to ‘1’R

^{2 }= ‘0’ (it shows that none of the independent variables explain the changes in the dependent variable)R

^{2 }= ‘1’ (it shows that all the changes in the dependent variable is explained by the variation in the independent variables)R

^{2 }= ‘0.85’ (it shows that 85% of the changes in the dependent variables is explained by the variation in the independent variables, advertising expenditure. The other 15% cannot be explaine by the regression analysis. This may be due to the omission of some important independent variables.)c. Standard error of estimate (Se)

It is a measure of dispersion of data points from the line of best fit (regression line). Actual points do not lie on the regression line but are dispersed above and below the line. Thus, the value predicted by regression line will be subjected to error. Therefore, the Se measures the probable error in the predicted value.

For example, data from table 4.1, when the advertising expenditure is $9 the sales is $40. If we use the regression results, the sales is $39.37. Therefore the value predicted will have an error.

The std. error of estimation can be calculated by using the following formula:

n ٨

Se = Σ (Y t – Y)

^{2} t=1

n - k

Se is useful to estimate the range within which the dependent variable will lie at a specified probability. At 95% probability the dependent variable will lie in the predicted interval of :

٨

Y

__+__t_{n – k}* Se٨

Where Y is the predicted value of dependent value based on the regression,

n – k is the degree of freedom (df), it is used to get the critical value for students’ distribution, n is the number of observation and k is the number of coefficient estimated.

Example :

Se = 2.8 At 95% confidence interval of sales when Adv. Exp. (X) = 9 and ٨ ٨

Y = 39.37 then Y

__+__t_{n – k}* Se => 39.37

__+__(2.306)(2.8)ð 39.37

__+__6.457Thus, at 95% C.I. when adv. Exp. Is $9 million, the range of Sales from $32.913 to $45.827 million

d. T-Statistics

The t-statistics is used in t – test to determine if there is a significant relationship between the dependent and each of the independent variable. To do this test, we

need the std. error of coefficient (Sb) and calculate the ‘t’ value. Then we compare the calculated ‘t’ value and the critical ‘t’ value from the student ‘t’ distribution table.

The ‘t’ value is calculated by dividing the value of coefficient (b) by Sb :

٨

Calculated t =

__b__ ٨

Sb

i.e : Calculated t =

__3.53__= 6.79 0.52

To calculate the critical value from student ‘t’ distribution table:

n – k = 10 – 2 = 8 df at 95% C.I and the ‘t critical ‘ = 2.306

Since t computed ( 6.79) > t critical (2.306) then adv.exp. is statistically significant in explaining the variations in sales at 95% C.I.

Note: if there is more than one independent variable then you have to test significance for all the independent vars.

e. Durbin-Watson Statistics

It indicates that whether the presence or absence of auto correlation means the problem that can arise in regression analysis with time series data.

There are 3 possibilities where autocorrelation or multi-co linearity problem can arise:

· When independent variables are interrelated or duplicated

· Where independent variables have been miss- specified

· Where important independent variables are found missing.

f. F-statistics

It is another test of overall explanatory power of regression analysis. (Refer pg 147 manual)

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