Quantitative Analysis Book-revision Kit

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ii Acknowledgement

S TRATHMORE UNIVERSITY ● STUDY PACK

Acknowledgement

This study kit’s success is attributed first and foremost to Saint Jose Maria Escriva

the founder of opus dei and the inspiration to the formation of Strathmore University

and its values.

Much appreciation is also extended to Mr. Randhir Ahluwalia of Strathmore

University for his contribution, Mr. Paul Maloba for his contributon and

compilation of the book.

Acknowledgement is given to the books that were used in research for compilation of

this book; Applied Mathematics for Business Economics and Social Sciences, and

Quantitative Techniques 6th

edition.

We gratefully acknowledge permission to quote from the past examination

papers of the following bodies: Kenya Accountants and Secretaries National

Examination Board (KASNEB); Chartered Institute of Management

Accountants (CIMA); Association of Chartered Certified Accountants

(ACCA).



Instructions for Students iii

QUANTITATIVE TECHNIQUES

Instructions for Students

Quantitative techniques paper in section 4 of the CPA examination is an eight

question paper devided intotwo sections.

Section I has 5 questions and only 3 questions are to be attempted, section II has 3

questions and only 2 are to be attempted. All questions have a total of 20marks.

Section I contains mainly questions on calculus, probability, matrices, test of

hypothesis, regression analysis, sampling and estimation and descriptive statistics.

Section II contains question from operations research and decision theory.

from past papers statitics the topicss frequently feature in exam are network analysis,

matrices, test of hypothesis and calculus. Other topics that are also favorites of theexaminer are regression analysis, time series, decision making techniques, linear

programming, probability and sets theory. It is very important to clearly understand

these topics through and through, emphasize more on working out examples andassignments from this book.

The course demands rigorous practice of of questions to internalize concepts, that’sthe only way to go about it. Work out reinforcing questions at the end of each study

session and compare them with answers given in study session 9. comprehensive

assignments are to be handed in to DLC department for marking.



iv Contents


Contents

Acknowledgement .............................................................................................................. ii

Instructions for Students ................................................................................................... iii Contents ............................................................................................................................. iv

Course Description ............................................................................................................. v

Index .................................................................................................................................. vi

LESSON ONE .................................................................................................................... 1

1. Linear Algebra and Matrices .................................................................................... 1LESSON TWO ................................................................................................................. 45

2. Sets Theory and Calculus ....................................................................................... 45LESSON THREE............................................................................................................. 81

3. Descriptive Statistics and Index Numbers .............................................................. 81LESSON FOUR ............................................................................................................... 118

4. Measures of Relationships and Forecasting ......................................................... 118LESSON FIVE ............................................................................................................... 157

5. Probability ............................................................................................................. 157LESSON SIX .................................................................................................................. 185

6. Sampling and Estimation ...................................................................................... 185LESSON SEVEN ........................................................................................................... 226

7. Decision Theory .................................................................................................... 226LESSON EIGHT ............................................................................................................ 248

8. Operation Research ............................................................................................... 248LESSON NINE .............................................................................................................. 303

9. Revision Aid ......................................................................................................... 303



Course Description v


Course Description

Quantitative techniques is a mathematical paper and fundamental to many proffessionalcoursesit forms the basis of finance and acconting. The course dwells more on practical applicationtypequestions and sometimes even requires one to make inference and decisions based on accurateanalysis of information . This paper also formas a basis for section 5 & 6 papers, management accounting and Financialmanagement. The book has been tailored to make you comfortable in a mathematical environment so as toexcel in the accounting profession.



vi Index


Index



Lesson One 1


LESSON ONE

Linear Algebra and Matrices

Contents- Functions and graphs- Linear equations, higher order equations, inequalities and simultaneous equations- Matrix algebra- Application of matrix algebra to input-output analysis and elementary Markovian

process.



2 Linear Algebra and Matrices


1.1 Functions and graphs A function is a mathematical relationship in which the value of a single dependent variable aredetermined from the values of one or more independent variables. The following is an exampleof a function in which y is said to be a function of x.

y = a + bx

In the above example, both x and y are variables this is because they may assume different valuesthroughout the analysis of the function. On the other hand, a and b are referred to as constantsbecause they assume fixed values. The variable y is a dependant variable in the sense that its values are generated from anindependent variable x. The collection of all the values of the independent variable for which the function is defined isreferred to as the domain of the function corresponding to this we have the range of thefunction, which is the collection of all the values of the dependent variable defined by thefunction The fact that it is a function of x can also be denoted by the following general form

y = f(x)Functions of a single independent variable may either be linear or non linear.Linear functions can be represented by:

y = a + bx Whereas non – linear functions can be represented by functions such as:

i. y = α0 + α 3

1 x + α2x3 ii. y 2 = 3x + 18iii. y = 2x2 + 5x + 7iv. ax2 + bx + cy + d = 0 v. xy = k vi. y = ax

Graph of a function A graph is a visual method of illustrating the behaviour of a particular function. It is easy to seefrom a graph how as x changes, the value of the f( x ) is changing. The graph is thus much easier to understand and interpret than a table of values. For exampleby looking at a graph we can tell whether f( x ) is increasing or decreasing as x increases ordecreases. We can also tell whether the rate of change is slow or fast. Maximum and minimum values ofthe function can be seen at a glance. For particular values of x, it is easy to read the values of f( x )and vice versa i.e. graphs can be used for estimation purposesDifferent functions create different shaped graphs and it is useful knowing the shapes of someof the most commonly encountered functions. Various types of equations such as linear,quadratic, trigonometric, exponential equations can be solved using graphical methods.

Equations

An equation is an expression with an equal sign (=)Equations are classified into two main groups linear equations and non linear equations.Examples of linear equations are

x + 13 = 157x + 6 = 0

Non linear equations in the variable x are equations in which x appears in the second or higherdegrees. They include quadratic and cubic equations amongst others. For example

Where α a b c d k = constants



Lesson One 3


5x2 + 3x + 7 = 0 (quadratic equation)

2x3 + 4x2 + 3x + 8 = 0 (cubic equation)

The solution of equations or the values of the variables for which the equations hold is called theroots of the equation or the solution set.

Solution of Linear EquationSupposing M, N, and P are expressions that may or may not involve variables, then thefollowing constitute some rules which will be useful in the solution of linear equations

Rule 1: Additional ruleIf M = N then M + P = N + P

Rule 2: Subtraction ruleIf M = N, Then M – P = N – P

Rule 3: multiplication ruleIf M = N and P ≠ O then M x P = N x P

Rule 4: Division ruleIf P x M = N and P ≠ O And N/P = Q Q being a raterial number thenM = N/P

Example

i. Solve 3x + 4 = - 8

ii. Solve3

y = - 4

Solutions

i.

3x + 4 = – 83x + 4 – 4 = – 8 – 4 (by subtraction rule)

3x = – 12 (simplifying)

3 12

3 3

x

(by division rule)

x = – 4 (simplifying)

ii. 343

y 3

y = – 12 (simplifying)Solution of quadratic equations

Suppose that we have an equation given as follows

ax2 + bx + c = 0

Where a, b and c are constants, and a≠ 0. such an equation is referred to as the general quadraticequation in x. if b = 0, then we have

ax2 + c = 0





Which is a pure quadratic equation There are 3 general methods for solving quadratic equations; solution by factorization, solutionby completing the square and solution by the quadratic formula.

Solution by Factorization

The following are the general steps commonly used in solving quadratic equations byfactorization

(i) Set the given quadratic equation to zero

(ii) Transform it into the product of two linear factors

(iii) Set each of the two linear factors equal to zero

(iv) Find the roots of the resulting two linear equations

Example

Solve the following equation by factorization

i. 6x2 = 18x

ii. 15x2 + 16x = 15

Solutions

i. 6x2 = 18x

6x2 – 18x = 0 ......................................................... (step 1)

6x(x – 3) = 0 .......................................................... (step 2)

6x = 0 ...................................................................... (step 3)

and x – 3 = 0

∴ x = 0 or x = 3 .................................................. (by step 4)

ii. 15x2 + 16x = 15

15x2 + 16x – 15 = 0 .............................................. (step 1)

(5x – 3) (3x +5) = 0 .............................................. (step 2)

(5x – 3) = 0} Step 3

{3x + 5 = 0}

∴ x = -3

5 or +5

3 ........................................... (step 4)

Solution by Completing the Square

The process of completing the square involves the construction of a perfect square from themembers of the equation which contains the variable of the equation.Consider the equation – 9x2 – bx = 0 The method of completing the square will involve the following steps

i. Make the coefficient of x2 unity

ii. Add the square of ½ the coefficient of x to both sides of the equal sign. The

left hand side is now a perfect square



Lesson One 5


iii. Factorize the perfect square on the left hand side.

iv. Find the square root of both sides

v. Solve for x

Example

Solve by completing the square.

i. 3x2 = 9x

ii. 2x2 + 3x + 1 = 0

Solutions

i. 3x2 = 9x or

(3x2 - 9x = 0)

x2

-

3x = 0 ...................................................................... (Step 1)2 2

2 3 33

2 2 x x

..................................... (Step 2)

23 9

2 4 x

.............................................................. (Step 3)

93

4 x

.............................................................. (Step 4)

∴

2

3

2

3

2

33or

(= 3 or 0)

ii. 2x2 + 3x + 1 = 0 or (2x2 + 3x = -1)

2

3x 1x + = -2 2

…………………...….. (Step 1)

21

4

3

4

3

2

322

2 xX ……… (Step 2)

23 1

x + =4 16

…………………….. (Step 3)

3 3

2 2 x





3 14 16

x + = ±

3 14 4

x = = ±

3 31 14 4 4 4

+ or - -

12 or 1 x x

Solution by Quadratic Formula

Consider the general quadratic equation

2ax + bx + c = 0 where a 0

The roots of the equation are obtained by the following formula:

a

acbb x

2

42

Example

Solve for x by formula

5x2 + 2x – 3 = 0

Solution

a = 5, b = 2, c = - 3

a

acbb x

2

42

)5(2

)3)(5(422 2

x

15

3or x

Inequalities

An inequality or inequation is an expression involving an inequality sign (i.e. >, <, ≤, ≥, i.e.greater than, less than, less or equal to, greater or equal to) The following are some examples of

inequations in variable x.3x + 3 > 5x2 – 2x – 12 < 0

The first is an example of linear inequation and the second is an example of a quadratic inequation.



Lesson One 7


Solutions of inequations

The solutions sets of inequations frequently contain many elements. In a number of cases theycontain infinite elements.

Example

Solve and graph the following inequationx – 2 > 2 ; w x (where x is a subset of w)

Solution

x – 2 > 2 so x – 2 + 2 > 2 + 2

Thus, x>4

The solution set is infinite, being all the elements in w greater than 4

0 1 2 3 4 5 6 7 8 9 10 11

Example

Solve and graph

3x – 7 < - 13;

Solution

3x - 7 < -13

3x - 7 + 7 < -13 + 7

3x < -6

3x -6 <

3 3

x < -2

Rules for solving linear inequations

Suppose M, M1, N, N1 and P are expressions that may or may not involve variables, then thecorresponding rules for solving inequations will be:Rule 1: Addition rule

If M > N and M1> N1

-4 -3 -2 -1 0 2 3 4

….. R Line





Then M + P > N + P and

M1 + P >N1+ P

Rule 2: Subtraction Rule

If M < N and M1 ≥N1

Then M – P < N – P andM1 – P ≥N1 – P

Rule 3: Multiplication rule

If M ≥N and M1 > N1 and P≠ 0

Then MP ≥NP; M1P > N1P

M(-P) ≤ N(-P) and M1(-P) < N1(-P)

Rule 4: Division

If M > N and M1< N1 and P≠ 0

Then M/P > N/P: M1/P < N1/P

M/(-P) < N/(-P) : and M1/(-P) > N1/(-P)

Rule 5: Inversion Rule

If M/P ≤ N/Q where P, Q ≠ 0

M1/P > N1/Q

Then P/M ≥ Q/N and P/M1 < Q/N1

Note: The rules for solving equations are the same as those for solving equations with oneexception; when both sides of an equation is multiplied or divided by a negative number, theinequality symbol must be reversed (see rule 3 & Rule 4 above).

Example

Solve and graph the following:

i. 7 – 2x > - 11 ;

ii. – 5x + 4 ≤ 2x – 10 ;

iii. –3 ≤ 2x + 1 < 7 ;

Solutions

i. 7 - 2x > -11

-2x > -18 (subtraction rule)

-2x -18 < (bydivision rule)

-2 -2

x < 9



Lesson One 9


ii. -5x + 4 2x - 10

-7x + 4 -10 (by subtraction rule)

-7x -14 (by subtraction rule)

x 2 (by division rule)

iii. -3 2x + 1 < 7

-4 2x < 6 (by substraction rule)

-2 x < 3 ( by division rule)

Linear inequation in two variables: relations

An expression of the form

y ≥ 2x – 1

is technically called a relation. It corresponds to a function, but different from it in that,

corresponding to each value of the independent variable x, there is more than one value of thedependent variable yRelations can be successfully presented graphically and are of major importance in linearprogramming.

1.2 Linear simultaneous equations: Two or more equations will form a system of linear simultaneous equations if such equations belinear in the same two or more variables.

Q

-4 -3 -2 -1 0 1 2 3 4 5line

Q

-4 -3 -2 -1 0 1 2 3 4 5line

Q

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11

line





For instance, the following systems of the two equations is simultaneous in the two variables xand y.

2x + 6y = 23

4x + 7y = 10

The solution of a system of linear simultaneous equations is a set of values of the variables

which simultaneously satisfy all the equations of the system.

Solution techniques

a) The graphical technique

The graphical technique of solving a system of linear equations consists of drawing the graphs ofthe equations of the system on the same rectangular coordinate system. The coordinates of thepoint of intersection of the equations of the system would then be the solution.

Example

The above figure illustrates:

Solution by graphical method of two equations

2x + y = 8

x + 2y = 10

The system has a unique solution (2, 4) represented by the point of intersection of the two

equations.

b) The elimination technique

This method requires that each variable be eliminated in turn by making the absolute value of itscoefficients equal in the equations of the system and then adding or subtracting the equations.Making the absolute values of the coefficients equal necessitates the multiplication of eachequation by an appropriate numerical factor.Consider the system of two equations (i) and (ii) below

(2,4)

-1 1 2 3 4 5 6 7 8 9 10 11 12 13

10

9.

8.

7.

6.

5.

4.

3.

2.

1

x + 2 y = 102x + y = 8



Lesson One 11


2x – 3y = 8 …….. ....................................................... (i).

3x + 4y = -5 …….. ...................................................... (ii).

Step 1

Multiply (i) by 3

6x – 9y = 24 …… ...................................................................... (iii).Multiply (ii) By 2

6x + 8y = - 10 …… ................................................................... (iv).

Subtract iii from iv.

17y = -34 ……............................................................................. (v).

y = -2

Step 2

Multiply (i) by 4

8x – 12y = 32 ……. ................................................................... (vi)

Multiply (ii) by 3

9x + 12y = -15 ….. .................................................................... (vii)

Add vi to vii

17x = 17 …….. ............................................................................ (viii)

x = 1

Thus x = 1, y = -2 i.e. {1,-2}

c) The substitution technique

To illustrate this technique, consider the system of two equations (i). and (ii) reproduced below

...... 2x – 3y = 8 …….. (i).

...... 3x + 4y = -5 …… (ii).

The solution of this system can be obtained by

a) Solving one of the equations for one variable in terms of the other variable;b) Substituting this value into the other equation(s) thereby obtaining an equation with one

unknown onlyc) Solving this equation for its single variable finallyd) Substituting this value into any one of the two original equations so as to obtain the

value of the second variable

Step 1

Solve equation (i) for variable x in terms of y

2x – 3y = 8

x= 4 + 3/2 y (iii)

Step 2





Substitute this value of x into equation (ii). And obtain an equation in y only

3x + 4y = -5

3 (4 + 3/2 y) + 4y = -5

8 ½ y = - 17 ……. (iv)

Step 3Solve the equation (iv). For y

8½y = -17

y = -2

Step 4

Substitute this value of y into equation (i) or (iii) and obtain the value of x

2x – 3y = 8

2x – 3(-2) = 8

x = 1

Example

Solve the following by substitution method

2x + y = 8

3x – 2y = -2

Solution

Solve the first equation for y

y = 8 – 2x

Substitute this value of y into the second equation and solve for x3x – 2y = -2

3x – 2 (8-2x) = -2

x = 2

Substitute this value of x into either the first or the second original equation and solve for y

2x + y = 8

(2) (2) + y = 8

y = 4

1.3 MATRICES A matrix is a rectangular array of items or numbers. These items or numbers are arranged inrows and columns to represent some information. The position of an element in one matrix is very important as well be seen later; therefore anelement is located by the number of the row and column which it occupies. The size of a matrix is defined by the number of its rows (m) and column (n).



Lesson One 13


a b ca b

For example = and B = d e f c d

g h i

are (2 x 2) and (3 x 3) matrices since A has 2 rows and 2 columns and B has 3 rows and 3

columns. A matrix A with three rows and four columns is given by one of:

11 12 13 14

21 22 23 24

31 32 33 34

ij

a a a a

A= a a a a

a a a a

or

A = a i = 1, 2, 3

j = 1, 2, 3, 4 where i represents the row number whereas j represents the column number

Properties of matrices

Equal Matrices

Two matrices A and B are said to be equal, that is

ij ijA = B or a = b

If and only if they are identical if they both have the same number of rows and columns and theelements in the corresponding locations in the two matrices should be the same, that is, a ij = bij for all i. And j.

Example

The following matrices are equal

3 4 0 3 4 0

2 2 3 = 2 2 3

5 1 1 5 1 1

Column Matrix or column vector

A column matrix, also referred to as column vector is a matrix consisting of a single column.

For example x =

1

2

n

x

x

.

.

.

x





Row matrix or row vector

It is a matrix with a single row

For example 1 2 3 ny = y , y , y .........y

Transpose of a Matrix

The transpose of an mxn matrix A is the nxm matrix A T obtained by interchanging the rows andcolumns of A.

A = aij

The transpose of A i.e. A T is given by

A T = aij = aji

mxn nxm

ExampleFind the transposes of the following matrices

1 2 3 4

1

2

3

1 5 7

A= 2 1 4

0 9 3

B= b , b , b , b

x

C= x

x

Solution

T

T

1

T 2T

1 2 3 4

3

4

1 5 7 1 2 0

i. A = 2 1 4 = 5 1 9

0 9 3 7 4 3

b

bii. B = b , b , b , b =

b

b



Lesson One 15


T

1

T

2 1 2 3

3

x

iii. C = x x x x

x

Square Matrix

A matrix A is said to be square when it has the same number of rows as columns

e.g.

2 5

A = 3 7 is a square matrix of order 2

B = n × n is a square matrix of the order n

Diagonal matrices

It is a square matrix with zeros everywhere in the matrix except on the principal diagonal

e.g.

3 0 0 9 0 0

A = 0 1 0 , B = 0 0 0

0 0 7 0 0 0

An identity of unity matrix

It is a diagonal matrix in which each of the diagonal elements is a positive one (1)

e.g.

matrixunit33 matrixunit22

100

010

001

3Iand 10

012I

A null or zero matrix

A null or zero matrix is a matrix whose elements are all equal to zero.

Sub matrix The sub matrix of the matrix A is another matrix obtained from A by deleting selected row(s)and/or column(s) of the matrix A.





1 2

7 9 8

e.g, if A = 2 3 6

1 5 0

2 3 6 7 9then A = and A =1 5 0 1 5

are both sub matrices of A

OPERATION ON MATRICES

Matrix addition and subtraction

We can add any number of matrices (or subtract one matrix from another) if they have the samesizes. Addition is carried out by adding together corresponding elements in the matrices.Similarly subtraction is carried out by subtracting the corresponding elements of two matrices as

shown in the following exampleExample: Given A and B, calculate A + B and A – B

6 1 10 5 12 4 7 3

A = 3 4 2 5 B = 0 4 10 4

9 13 6 0 7 3 7 9

6 1 10 5 12 4 7 3 18 3 3 8

A + B = 3 4 2 5 + 0 4 10 4 = 3 0 12 9

9 13 6 0 7 3 7 9 2 16 1 9

6 1 10 5

A - B = 3 4 2 5

9 13

12 4 7 3 6 5 17 2

- 0 4 10 4 = 3 8 8 1

6 0 7 3 7 9 16 10 13 9

If it is assumed that A, B, C are of the same order, the following properties are fulfilled:

a) Commutative law: A + B = B + Ab) Associative law: (A + B) + C = A + (B + C) = A + B + C

Multiplying a matrix by a number

In this case each element of the matrix is multiplied by that number



Lesson One 17


Example

6 1 10 5

If A = 3 4 2 5

9 13 6 0

60 10 100 50

then (10)A = 30 40 20 50

90 130 60 0

Matrix Multiplication

a) Multiplication of two vectors

Let row vector A represent the selling price in shillings of one unit of commodity P, Q, Rrespectively and let column vector B represent the number of units of commodities P, Q, R sold

respectively. Then the vector product A B will be equal to the total sales valuei. e. A B = Total sales value

100

Let A = 4 5 6 and B = 200

300

100

then 4 5 6 200 = 400 + 1,000 + 1,800 = Shs 3,200

300

Rules of multiplication

i. The row vector must have the same number of elements as the column vectorii. The first vector is a row vector and the second is a column vectoriii. The corresponding elements in each vector are multiplied together and the results

obtained are added. This addition is always a single numberGoing back to the example given before

100

A × B = 4 5 6 200 = 4 × 100 + 5 × 200 + 6 × 300=Shs3,200, a single number

300





b) Multiplication of two matrices

Rules

i. Multiplication is only possible if the first matrix has the same number of columns asthe second matrix has rows. That is if A is the order a×b, then B has to be of the

order b×c. If the A×B = D, then D must be of the order a×c.ii. The general method of multiplication is that the elements in row m of the firstmatrix are multiplied by the corresponding elements in columns n of the secondmatrix and the products obtained are then added giving a single number.

We can express this rule as follows

Let A = 11 12 1311 12

21 22 2321 22

b b ba a and b =

b b ba a

Then A B = D = 11 12 13

21 22 23

d d d

d d d

A = 2 x 2 matrix B = 2 x 3 matrix D = 2 x 3 matrix Where

d11 = a11 b11 + a12 b21

d12 = a11 b12 + a12 b22

Example I

6 1 3 0 2 6 3 1 4 6 0 1 5 6 2 1 8 =

2 3 4 5 8 2 3 3 4 2 0 3 5 2 2 3 8

22 5 20 =

18 15 28

Example II

Matrix X gives the details of component parts used in the make up of two products P 1 and P2

matrix Y gives details of products made on each day of the week as follows:

352

243

2P1P

Products

CB A

Parts

X Matrix

11

22

23

32

21

Fri

Thur

Wed

Tues

Mon

PPProducts

Y Matrix

21



Lesson One 19


Use matrix multiplication to find the number of component parts used on each day of the week.Solution:

After careful consideration, it will be easy to decide that the correct order of multiplication is YXX (Note the order of multiplication). This multiplication is compatible and also it gives thedesired answer.

1 2 1×3+2×2 1×4+2×5 1×2+2×5

2 3 2×3+3×2 2×4+3×5 2×2+3×33 4 2

Y × X = × =3 2 3×3+2×2 3×4+2×5 3×2+2×32 5 3

2 2 2×3+2×2 2×4+2×5 2×2+2×3

1 1 1×3+1×2 1×4+1×5 1×2+1×3

5 x 2 matrix 2 x 3 matrix = 5 x 3 matrix

A B C

Mon 7 14 8

Tues 12 23 13

Wed 13 22 12

Thur 10 18 10

Fri 5 9 5

Interpretation

On Monday, number of component parts A used is 7, B is 14 and C is 8. in the same way, thenumber of component parts used for other days can be interpreted.

The determinant of a square matrix

The determinant of a square matrix det (A) or A is a number associated to that matrix. If thedeterminant of a matrix is equal to zero, the matrix is called singular matrix otherwise it is callednon-singular matrix. The determinant of a non square matrix is not defined.

Determination of a 2 x 2 matrix

a bA = = ad - cb

c d

ii. Determinant of a 3 x 3 matrix

a b c e f d f d eA d e f = a b +c

h i g i g hg h i

a ei - fh - b di - gf +c dh - eg

simplify





iii. Determinant of a 4 x 4 matrix

a b c d

e f g hA=

i j k lm n o p

f g h e g h e f h e f g

A = a j k l b i k l +c i j l d i j k

n o p n o p m o p m n o

Simplify 3 x 3 determinants as in ii and then evaluate the 4 x 4 determinants.

Inverse of a matrix

If for an n ( n square matrix A, there is another n ( n square matrix Bsuch that there product is the identity of the order n X n, In, that is A X B

= B X A = I, then B is said to be inverse of A. Inverse if generally

written as A-1

Hence AA-1 = I

Note: Only non singular matrices have an inverse and therefore theinverse of a singular matrix is non defined.

General method for finding inverse of a matrix

In order to introduce the rule to calculate the determinant as well as the

inverse of a matrix, we should introduce the concept of minor andcofactor.

The minor of an element

Given a matrix A = (aij), the minor of an element aij in row i and column j (call it mij), is the value of the determinant formed by deleting row i

and column j in matrix A.

Example

Let A = EMBED Equation.DSMT4

The minors are,



Lesson One 21


11

12

6 1m = = 6×0 3×1 = 3

3 0

5 1m = = 5×0 1×2 = 2

2 0

Similarly

13 21 22 23

31 32 33

5 6 2 3 4 3 4 2m = m = m = m =

2 3 3 0 2 0 2 3

=15 12 = 3 =0 9 = 9 = 0 6 = 6 = 12 4 = 8

2 3 4 3 4 2m = m m

6 1 5 1 5 6 2-18 -16 4-15 -11 24 -10 14

The cofactor of an element

The cofactor of any element aij (known as cij ) is the signed minor associated with that element. The sign is not changed if (i+j) is even and it is changed if (i+j)is odd. Thus the sign alternated whether vertically or horizontally, beginning with a plus in the upper left hand corner.

i.e. 3 x 3 signed matrix will have signs

Hence the cofactor of element a11 is m11 = -3, cofactor of a12 is – m12 = +2 the cofactor ofelement a13 is +m13 = 3 and so on.

Matrix of cofactors of A =

3 2 3

9 6 8

16 11 14

in general for a matrix M =

a b c

d e f g h i

Cofactor of a is written as A, cofactor of b is written as B and so on.Hence matrix of cofactors of M is written as





=

A B C

D E F

G H I

The determinant of a n×n matrix

The determinant of a n×n matrix can be calculated by adding the products of the element in anyrow (or column) multiplied by their cofactors. If we use the symbol ∆ for determinant.

Then ∆ = aA + bB + cC

or

= dD + eE + fF e.t.c

Note: Usually for calculation purposes we take ∆ = aA + bB + cC

Hence in the example under discussion

∆ = (4 – 3) + (2 2) + (3 3) = 1

The ad joint of a matrix

The ad joint of matrix

A B C

D E F

G H I

is written as

A D G

B E H

C F I

i.e. change rows into columns and columns into rows (transpose)

The inverse of the matrix

a b c

d e f

g h i

is written as1

(adjoint of the matrix of cofactors)determinant

-1

A D G1

i.e. A = B E H

C F I

Where ∆ = aA + bB + cC



Lesson One 23


Hence inverse of

4 2 3

5 6 1

2 3 0

is found as follows

∆ = (4 – 3) + (2 2) + (3 ( 3) = 1

A = -3 B = 2 C = 3

D = 9 E = -6 F = -8

G = -16 H = 11 I = 14

EMBED Equation.DSMT4

(note: Check if A ( A-1 = A-1 A = 1)

Solution of simultaneous equationsIn order to determine the solutions of simultaneous equations, we may use either of thefollowing 2 methods

i. The cofactor methodii. Cramers rule

The cofactor method

This method requires that we obtaina) The minors and cofactorsb) The adjoint of the matrix

c) The inverse of the matrixd) Multiply the original by the inverse on both sides of the matrix equation

Example

Solve the following

a) 4x1 + x2 – 5x3 = 8

-2x1 + 3x2 + x3 = 12

3x1 – x2 + 4x 3 = 5

b) 4x1 + 3x3 + 5x3 = 27x1 + 6x2 + 2x3 = 19

3x1 + x2 + 3x3 = 15

c) 4x1 + 2x2 + 6x3 = 28

3x1 + x2 + 2x3 = 20





10x1 + 5x2 + 15x3 = 70

d) 2x1 + 4x2 – 3x3 = 12

3x1 – 5x2 + 2x3 = 13

-x1 + 3x2 + 2x3 = 17

Solution

a) From a, we have

1

2

3

4 1 -5 x 8

-2 3 1 x = 12

3 -1 4 x 5

A X b

We need to determine the minors and the cofactors for the above matrix

Definition

A minor is a determinant of a sub matrix obtained when other elements are detected as shownbelow. A cofactor is the product of (-1) i + j and a minor where

i = Ith row i = 1, 2, 3 …….

j = Jth row j = 1, 2, 3 …….

Cofactor of 4 (a11 ) = (-1) 1+13 1

= 13

1 4

Cofactor of -2 (a21 ) = (-1) 2+1 1 5

= 11 4

Cofactor of 3 (a31 ) = (-1) 3+11 5

= 163 1

Cofactor of 1 (a12 ) = (-1) 1+22 1

= 113 4

Cofactor of 3 (a22 ) = (-1)2+2

4 5 = 31

3 4

Cofactor of -1 (a23 ) = (-1) 2+34 5

= 62 1

Cofactor of -5 (a13 ) = (-1) 1+32 3

= 73 1







1

2

3

1

2

3

13 1 16 4 1 5 x1

11 31 6 2 3 1 x98

-7 7 14 3 1 4 x

13 1 16 81

11 31 6 1298

-7 7 14 5

98 0 0 x 1961 1

0 98 0 x = 49098 98

0 0 98 x 98

1

2

3

1

2

3

1 0 0 x 2= 0 1 0 x = 5

0 0 1 x 1

x 2

x = 5

x 1

X 1 = 2, X 2 = 5, X 3 = 1

c) 4x1 + 2x2 + 6x3 = 28

3x1 + x2 + 2x3 = 20

10x1 + 5x2 + 15x3 = 70

1

2

3

4 2 6 x 28

= 3 1 2 x = 20

10 5 15 x 70

4 2 6 4 2 6 4 2

3 1 2 = 3 1 2 3 1

10 5 15 10 5 15 10 5

= (60 + 40 + 90) – (60 + 40 + 90)

= 0



Lesson One 27


Hence the solutions of x1, x2, and x3 do no exist. The equations are independent

Now work out part (b) on your own.

Cramers Rule in Solving Simultaneous Equations

Consider the following system of two linear simultaneous equations in two variables.

a11 x1 + a12 x2 = b1 ……………(i) a21 x1 + a22 x2 = b2 ……………(ii)

after solving the equations you obtain

1 12

2 221 22 2 121

11 1211 22 12 21

21 22

11 1

21 211 2 21 12

11 1211 22 12 21

21 22

b a

b a b a b ax =

a aa a a a

a a

and

a b

a ba b - a bx =

a aa a - a a

a a

Solutions of x1 and x2 obtained this way are said to have been derived using Cramers rule,practice this method over and over to internalize it. It is advisable for exam situation since it isshorter.Example

Solve the following systems of linear simultaneous equations by Cramers‟ rule:

i) 2x1 – 5x2 = 7

x1 + 6x2 = 9

ii) x1 + 2x2 + 4x3 = 4

2x1 + x3 = 3

3x2 + x3 = 2

Solutions

i. 2x1 – 5x2 = 7

x1 + 6x2 = 9

can be expressed in matrix form as

1

2

x2 5 7 =

x1 6 9

A X b





and applying cramers‟ rule

1

2

7 -5

9 6 87 2x = = = 5

2 -5 17 17

1 6

2 7

1 9 11x = =

2 -5 17

1 6

(ii) can be expressed in matrix form as

1

2

3

1 2 4 x 4

2 0 1 x = 30 3 1 x 2

and by cramers‟ rule

1

4 2 4

3 0 1

2 3 1 22x = =

1 2 4 17

2 0 1

0 3 1

3

1 2 4

2 0 3

0 3 2 7 =

1 2 4 17

2 0 1

0 3 1

x

2

1 4 4

2 3 1

0 2 1 9 =

1 2 4 17

2 0 1

0 3 1

x



Lesson One 29


Solving simultaneous Equations using matrix algebra

i. Solve the equations

2x + 3y = 13

3x + 2y = 12

in matrix format these equations can be written as2 3 x 13

=3 2 y 12

pre multiply both sides by the inverse of the matrix

2 3 = 5

3 2

and inverse of the matrix is

2 32 31 5 5

=3 2 3 25

5 5

Pre multiplication by inverse gives

2 3 2 3

2 3 13 25 5 5 5 = =

3 2 3 2 3 2 12 3

5 5 5 5

Therefore x = 2 y = 3

ii. Solve the equations

4x + 2y + 3z = 4

5x + 6y + 1z = 2

2x + 3y = -1

Solution:

Writing these equations in matrix format, we get

A BX = b4 2 3 x 4

5 6 1 y = 2

2 3 0 z -1

Pre-multiply both sides by the inverse





the inverse of A as found before is A-1 =

3 9 16

2 6 11

3 8 14

3 9 16 4 3 2 x 3 9 16 4 22

2 6 11 5 6 1 y = 2 6 11 2 = -15

3 8 14 2 3 0 z 3 8 14 -1 -18

hence x = 22 y = -15 z = -18

(Note: under examination conditions it may be advisable to check the solution by substitutingthe value of x, y, z into any of the three original equations)

MARKONIKOV CHAIN

Probability Transition Matrices (Brand switching)

These are matrices in which the individual elements are in the form of probabilities. The probabilities represent the probability of one event following another event i.e. theprobability of transition from one event to the next The probabilities of the various changes applied to the initial state by matrix multiplication, givea forecast of the succeeding state.Normally a transition matrix is defined with its columns adding upto one and state vectors ascolumn vectors.In this case the succeeding state is found by pre-multiplying the transition matrix by theproceeding state (column) vectorIf the transition matrices are given with their rows adding up to one, then the succeeding state isfound by post multiplying the transition matrix by the preceding state (row) vector.

Example 1

The probability transition matrix of the switching probabilities, consider that two brands G andX share the market in the ratio of 60% to 40% respectively of customers. If in every week 70%of G‟s customers retain the brand but 30% switch to product x where as 80% of X‟s customersretain brand but 20% percent switch to brand G. Analyse the exchange in share market per week.

G X

State the system next weekG

X

0.7 0.2

0.3 0.8

Column vector for initial market share60G

X 40

Share next week0.7 0.2 60 50

=0.3 0.8 40 50



Lesson One 31


Share week after0.7 0.2 50 45

=0.3 0.8 50 55

and so on

This process can continue till equilibrium is reached.

Let the market share beG

X

∴ 0.7 0.2 G G

=0.3 0.8 X X

0.7G + 0.2 X = G or 0.3G + 0.8X = X0.2X = 0.3G or 0.3X = 0.2X

i.e.G

X =

0.3 3 =

0.2 2

Hence G‟s share is 60% and X‟s share is 40%

Example 2

A marketing division toothpaste manufacturing company has worked out the followingtransition probability matrices concerning the behaviors of customers before and after anadvertising campaign.

Transition probability matrix(before advertising campaign)

TO

FROMOur Brand

(State I) Another Brand

(Sate II)Our brand (State I) 0.8 0.2 Another Brand (sate II) 0.4 0.6

Transition probability matrix(After advertisement)

TO

FROMOur Brand

(State I) Another Brand

(Sate II)Our brand (State I) 0.9 0.1 Another Brand (sate II) 0.5 0.5

If the advertising campaign costs Shs 20,000 per year, would it be worthwhile for the companyto undertake the campaign?





You may suppose there are 60,000 buyers of toothpaste in the market and for each customeraverage annual profit of the company is Shs 2.50

Solution

Let P1 be the fraction share of our brand and P2 be the fraction share of another brand

Before Advertising

1 2 1 2

0.8 0.2

0.4 0.6 P P P P

1 2 1 2 1 20.8 0.4 0.2 0.6 P P P P P P

1 2 1 1 20.8 0.4 and 0.2 0.6 P P P P P P

2 1 1 20.4 0.2 and 0.2 0.4 P P P P

Thus:2 1

3 31 2 and P P

After Advertising

1 2 1 2

0.9 0.1

0.5 0.5 P P P P

1 2 1 2 1 20.9 0.5 0.1 0.5 P P P P P P

2 1 2 10.5 0.1 and 0.5 0.1 P P P P

Thus:5 1

6 61 2 and P P

If there are 60,000 buyers

Before Advertising2

31 P

this implies that,

23 60,000 40,000 customers will buy our

brand40,000 2.5contribution = Sh.100,000

After Advertising5

61 P

this implies that,

56 60,000 50,000 customers will

buy our brand(50,000 2.5) 20,000contribution

= Sh.105,000

the difference between advertising and not advertising is

105,000 – 100,000 = Sh.5,000 in favour of advertising, Thus the advertising campaign is worthwhile.

INPUT – OUTPUT ANALYSIS

The input output analysis is a topic which requires application of matrices The technique analyses the flow of inputs from one sector of the economy to the other sectorsthus the technique is quite useful in studying the interdependence of sectors within a singleeconomy. The input – output analysis was first developed by Prof Leontief hence the Leontief matrix hasbeen developed. See the following example



Lesson One 33


Example (B)INPUT OUTPUT TABLE

TO Final Total

FROM Agric Industry Service Demand Demand (output) Agric 300 360 320 1080 2060

Industry 450 470 410 800 2130Service 610 500 520 270 1900Primary inputs 700 800 650 - - Total inputs 2060 2130 1900 - -

NB: In the above table, one should be able to interpret the table e.g. of the total demand of2060 metric tones from the agriculturalsector; 300 is produced for the agricultural sector, 360 forindustrial sector, 320 for the service sector and 1080 metric tones makes up the final demand. The final demand is the additional demand besides the sectoral demand which is normally madeby other users e.g. government, foreign countries, other manufacturers not included in the othersectors.For production if items besides the inputs from other sectors namely labour capital e.t.c

Technical coefficients : (‘to’ sectors)

Agriculture 300 =300

2060 = 0.14

450 =450

2060 = 0.22

610 =610

2060 = 0.30

Industry 360 =360

2130 = 0.7

470 =470

2130 = 0.22

500 =500

2130 = 0.23

Service 320 =320

1900 = 0.17

410 =410

1900 = 0.22

520 =520

1900 = 0.27

The matrix of technical coefficients is:

TO Final Total

FROM Agric Industry Service Demand Demand (output) Agric 0.14 0.7 0.17 1080 (y 1 ) 2060




S TRATHMORE UNIVERSITY ● STUDY PACK 34

Industry 0.22 0.22 0.22 800 (y 2 ) 2130Service 0.30 0.23 0.27 270(y 3 ) 1900Primary inputs x x x - -

2060(x1 ) 2130(x2 ) 1900(x3 ) - -

From the above table, we may develop the following equations

0.14x1 + 0.7x2 + 0.17x3 + y 1 = x1

0.22x1 + 0.22x2 + 0.22x3 + y 2 = x2

0.30x1 + 0.23x2 + 0.27x3 + y 3 = x3

1 1 1

2 2 2

3 3 3

0.14 0.17 0.17 x y x

0.22 0.22 0.22 x + y = x ...................(*)

0.30 0.23 0.27 x y x

A X Y X

Let the coefficient matrix be represented by

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

a a a y x

a a a y = y x = x

a a a y x

A

Equation (*) may be written as

AX + Y = X

Y = X – AX

Y = X (I-A)

⇒ (I – A) – 1 Y = X

The matrix I – A is known as Leontief Matrix

Technical Coefficients

These show the units required from each sector to make up one complete product in a givensector e.g. in the above matrix of coefficients it may be said that one complete product from theagricultural sector requires 0.14 units from the agricultural sector itself, 0.22 from the industrialsector and 0.30 from the service sectorNB: The primary inputs are sometimes known as “value added”

Example 1Determine the total demand (x) for the industry 1, 2, 3 given the matrix of technical coefficients(A), Capital and the final demand vector B.

1 0.3 0.4 0.1 20

A = 2 0.5 0.2 0.6 B = 10

3 0.1 0.3 0.1 30



Lesson One 35


From the input – output analysis

X = (I – A) – 1 B, Where X =1

2

3

X

X

x

is the demand vector

1

-1

1 0 0 0.3 0.4 0.1

I - A = 0 1 0 - 0.5 0.2 0.6

0 0 1 0.1 0.3 0.1

0.7 -0.4 -0.1

-0.5 0.8 -0.6

-0.1 -0.3 0.9

0.7 -0.4 -0.1

(I - A) = -0.5 0.8 -0.6

-0.1 -0.3 0.9

The matrix of cofactors of I - A is

=

0.54 0.51 0.23

0.39 0.62 0.25

0.32 0.47 0.36

The transpose (adjoint) of the above matrix is

=

0.54 0.39 0.32

0.51 0.62 0.47

0.23 0.25 0.36

Δ of (I - A) = 0.495 – (0.008 + 0.126 + 0.18

= 0.809

-1

0.54 0.39 0.32 0.66 0.48 0.401

(I - A) = 0.51 0.62 0.47 = 0.63 0.77 0.580.809

0.23 0.25 0.36 0.23 0.31 0.44





Therefore X =

1

2

3

0.66 0.48 0.40 20 X

0.63 0.77 0.58 10 = X

0.23 0.31 0.44 30 X

30

X = 37.7

21.9

The total demand from the three industries 1, 2 and 3 is 30 from 1, 37.7 from 2 and 21.9 from 3.

Example 2

Three clients of Disrup, Ltd P, Q and Rare direct competitors in the retail business. In the first week of the year P had 300 customers Q had 250 customers and R had 200 customers. Duringthe second week, 60 of the original customers of P transferred to Q and 30 of the original

customers of P transferred to R. similarly in the second week 50 of the original customers of Qtransferred to P with no transfers to R and 20 of the original customers of R transferred to P with no transfers to Q.

Required

a) Display in a matrix the pattern of retention and transfers of customers from the first to thesecond week (4 marks)

b) Re-expres the matrix that you have obtained in part (a) showing the elements as decimalfractions of the original numbers of customers of P, Q and R (2 marks) Refer to this reexpressed matrix as B

c) Multiply matrix B by itself to determine the proportions of the original customers that havebeen retained or transferred to P, Q and R from the second to the third week. (4 marks)

d) Solve the matrix equation (xyz)B = (xyz) given that x + y + z = 1 (8 marks)e) Interpret the result that you obtain in part (d) in relation to the movement of customers

between P, Q and R (2marks)(Total 20 marks)

Solution

a). Think of each row element as being the point from which the customer originated and eachcolumn element as being the destination e.g. 210 customers move from P to P, 60 movefrom P to Q and 30 move from P to R. The sum of the elements of the first row totalling300, that is the number of customers originally with P.

Hence required matrix is

P Q R ToP 210 60 30 row total 300

FromQ 50 200 0 row total 250

R 20 0 180 row total 200

b). The requirement of this part is to express each element as a decimal fraction of itscorresponding row total. The second row, first element is therefore 50/250, that is 0.2 andthe second element is therefore 200/250 that is 0.8.







e). In proportion terms this solution means that P, Q, and R will in the long term each have onethird of the total customers

Example 4

There are three types of breakfast meal available in supermarkets known as brand BM1, brandBM2 and Brand BM3. In order to assess the market, a survey was carried out by one of themanufacturers. After the first month, the survey revealed that 20% of the customers purchasingbrand BM1 switched to BM2 and 10% of the customers purchasing brand BM1 switched toBM3. similarly, after the first month of the customers purchasing brand BM2, 25% switched toBM1 and 10% switched to BM3 and of the customers purchasing brand BM3 0.05% switched toBM1 and 15% switched to BM2

Required

i. Display in a matrix S, the patterns of retention and transfers of customers from the firstto the second month, expressing percentage in decimal form. (2marks)

ii. Multiply matrix S by itself (that is form S2 ) (5 Marks)iii. Interpret the results you obtain in part ii with regard to customer brand loyalty (3 marks)

Solution

The objective of the first part of the question was to test the candidate‟s ability to formulate andmanipulate a matrix, then interpret the result of such manipulation.a. i. The matrix showing the pattern of retention and transfer from the first to the second

month isBM1 BM2 BM3

0.70 0.20 0.10 BM1

S = 0.25 0.65 0.10 BM2

0.05 0.15 0.80 BM3

(The second element in the first row shows the 20% movement from BM1 to BM2 and so on)

i. The product of matrix S with itself is demonstrated as follows

0.70 0.20 0.10 0.70 0.20 0.10 0.5450 0.2850 0.1700

0.25 0.65 0.10 0.25 0.65 0.10 = 0.3425 0.4875 0.1700

0.05 0.15 0.80 0.05 0.15 0.80 0.1125 0.2275 0.6600

Where for example second element in the first row, that is 0.2850 is the result of multiplying thecorresponding elements of the first row of S by the second column of S and summing theproduct.

0.2850 = 0.70 × 0.20 + 0.20 × 0.65 + 0.10 × 0.15= 0.14 + 0.13 + 0.015 e.t.c.

ii. The resulting matrix may be interpreted in the following wayOf the original customers who buy BM1, 54.5% will remain loyal to the brand inmonth three, 28.5% will have switched to BM2 and 17% will have switched toBM3.



Lesson One 39


Of the original customers who buy BM2, 48.7% will remain loyal to the brand inmonth three 34.25% will have switched to BM1 and 17% will have switched toBM3Of the original customers who buy BM3, 66% will remain loyal to the brand inmonth three 11.25% will have switched to BM1 and 22.75% will have switched toBM2

AlternativelyIn month three 54.5% of the customers buying BM1 are original customers. 34.5%came from BM2 originally and remaining 11.25% have switched from BM3 and soon.

MARKOV CHAINS/PROCESSES

The Markov processes are defined as a set of trials which follow a certain sequence whichdepend on a given set of probabilities known as transition probabilities. These probabilitiesindicate how a particular activity or product moves from one state to another.

Applications of Markov Chains in Business

The Markov processes or chains are frequently applied as follows:-1. Brand Switching

By using the transitional probabilities we can be able to express the manner in which consumersswitch their tastes from one product to another.

2. Insurance industry

Markov analysis may be used to study the claims made by the insured persons and also decidethe level of premiums to be paid in future.

3. Movement of urban population

By formulating a transition matrix for the current population in the urban areas, one can be ableto determine what the population will be in say 5 years.

4. Movement of customers from one bank to another .It is a fact that customers tend to look for efficient banks. Therefore at a certain time when agiven bank installs such machinery as computers it will tend to attract a number of customers who will move from certain banks to efficient ones.

PROPERTIES OF MARKOV CHAINS

1. Each outcome in a markov process belongs to a state space or transition matrix. E.g.

1 2 3

1 11 12 13

2 21 22 23

3 31 32 33

S S S

S P P PS P P P

S P P P

Where S1, S2, S3 are states and P11 P12 e.t.c are probabilities

2. The outcome of each trial depends on the immediate preceding activities but not on theprevious activities

BASIC TERMS IN MARKOV CHAINS





Lesson One 41


Since the elements in A2 and A3 are all positive then A is regular Stochastic matrix.

ABSORBING STATES

A state S i (I = 1, 2, 3 …) of a markov chain is called absorbing if the system remains in the state,S i once it enters there. Thus a state, S i is absorbing if and only if the i th row of the transitionmatrix p has a 1 on the main diagonal and zeroes every where else. See the following example. The following matrix, P is a transition matrix of the markov chain.

1 2 3 4 5

1

2

3

4

5

S S S S S

1 1 1 10S 4 4 4 4

S 0 1 0 0 0

P= S 1 1 10 02 4 4

S0 1 0 0 0

S0 0 0 0 1

The States S2 and S5 are absorbing states since the 2nd and 4th rows have 1 on the main diagonal.





from north

from south

REINFORCING QUESTIONS

Work out these questions and compare with answers given in lesson 9

QUESTION ONE

Determine a) f (1) b) f (-2) for the following functions

1. f(x) = – 5x + 2 2. f(x) = – x 2 + 3x+10

3. f( t ) = 10 – t + t3 4. f(u) = 8

QUESTION TWO

Solve the following simultaneous equations.

a)

b)

c)

QUESTION THREE

Because of inreasing cost increasing cost energy, the population within Maueni district

seem to be shifting from the north to the south the transition matrix S describes themigration behaviour observed between the regions.

0.90 0.10

0.05 0.95S

determine whether the populations will attain an equillibrium condition and if so, the population of the two regions.

QUESTION FOUR

A simple hypothetical economy of three industries A, B and C is represented in the followingtable (data in millions of shillings).

2 5 20

4 4

x y

x y

2 9

3 6

x y

x y

12 4 18

4 6

x y

x y

to north to south



Lesson One 43


Determine the output vector for the economy if the final demand changes to 60 for A, 60 for Band 60 for C

QUESTION FIVE

A tea blender uses two types of tea, T1, and T2, to produce two blends, B1 and B2 for sale. B1 uses 40% of available T1 and 60% of the available T2 whilst B2 uses 50% of the available T1 and25% of the available T2.

Required:a) Given that t1 kilos of T1 and t2 of T2 are made available to produce b1 kilos of B1 and b2

kilos of B2. Express the blending operation in the matrix format.

b) If 400 kilos of T1 and 700 kilos of T2 were made available for blending, what quantities of B1 and B2 would be produced?

c) If 600 kilos of B1 and 450 kilos of B2 were produced, use a matrix method to determine what quantities of T1 and T2 would be used to produce the blends.

QUESTION SIX

Let A =2 2

3 -3

a) Find A2 and A3

b) If F(x) = x3 – 3x2 – 2x + 4IFind F (A)

c) Find the inverse of matrix A

QUESTION SEVEN

A child‟s toy is marketed in three sizes standard size contains 10 squares (S), 15 triangles (T) and6 hexagons (H). The deluxe set contains 15 S, 20 T, 6 H and 4 octagons (O). The super setcontains 24 T, 8 H, 16 H, 16 S and 6 (O). Squares cost 12 pence to produce, triangles cost 8p,hexagons cost 18p and octagons 22p.

The standard set is sold at £6, the deluxe set for £10 and super for £15. The manufacturerproduces 100 standard sets, 80 deluxe sets and 50 super sets per week.

Use matrix form and matrix multiplications to find:

The cost of producing each set. The number of each shape required each week

User

Producer

A B C FinalDemand

TotalOutput

A 80 100 100 40 320B 80 200 60 60 400

C 80 100 100 20 300





Total expenditure on shapes each week

QUESTION EIGHT

Matrix N below shows the number of items of type A, B, and C in warehouses Y and W. Matrixp shows the cost in pence per day of storing (S) and maintaining (M) one item each of A, B andC

A B C S M

A 2 0.5Y 10 12 50

P = B 3 1.5W 60 0 20

C 2 0.5

N

a) Evaluate the matrix (N×P) and say what it represents.

b) Stock movement occurs as follows:

At the start of the day 1:

Withdrawal of 2 type B from warehouse Y, 20 of type A from warehouse W.

At the start of day 2:Delivery of 7 type B and 10 of type C to warehouse Y and 15 of type B to warehouse W.

Evaluate the total cost of storage and maintenance for days 1 and 2.

c) Write down without evaluating a matrix expression which could be used to evaluate thestorage and maintenance cost of items A, B and C for the period from day 1 to 4. Allow for the stock movements on days 1 and 2, as described in part (b). There wereno stock movements on days 3 and 4.



Lesson Two 45


LESSON TWO

Sets Theory and Calculus

- Sets and set theory

- Calculus- Differentiation and integration of polynomial, exponential and logarithmic functions- Application of calculus to economic models

2.1 Sets and set theory A set is a collection of distinct objects. We may consider all the ocean in the world to be a set with the objects being whales, sea plants, sharks, octopus etc, similarly all the fresh water lakes in Africa can form a set. Supposing A to be a set

A = {4, 6, 8, 13}

The objects in the set, that is, the integers 4, 6, 8 and 13 are referred to as the members orelements of the set. The elements of a set can be listed in any order. For example,

A = {4, 6, 8, 13} = {8, 4, 13, 6}Sets are always precisely defined. Each element occurs once and only once in a set.

The notion is used to indicate membership of a set. ∉ represents non membership. However,in order to represent the fact that one set is a subject of another set, we use the notion . A set“S” is a subject of another set “T” if every element in “S” is a member of “T”

Example

If A = {4, 6, 8, 13} then

i) 4 {4, 6, 8, 13} or 4 A; 16 ∉ A

ii) {4, 8} A; {5, 7} A; A A

Methods of set representationCapital letters are normally used to represent sets. However, there are two different methods forrepresenting members of a set:

i. The descriptive method and

ii. The enumerative method

The descriptive method involves the description of members of the set in such a way that onecan determine the elements of the set without difficulty.

The enumerative method requires that one writes out all the members of the set within the curlybrackets.

For example, the set of numbers 0, 1, 2, 3, 4, 5, 6 and 7 can be represented ass followsP = {0, 1, 2, 3, 4, 5, 6, 7} , enumerative method

P = {X/x = 0, 1, 2…7} descriptive method

Or

P = {x/0 ≤ x ≤7} where x is an interger.



46 Sets Theory and Calculus


Finite and infinite sets

A set can be classified as a finite or infinite set, depending upon the number of elements it has. A finite set has a finite number of elements whereas an infinite set has an infinite number ofelements.

For example, set P below has ten elements and is therefore a finite set. Set S, on the other hand,is an infinite set since it has an infinite number of elements.

P = {2, 4, 6…20}

S = {1, 3, 5…}

Universal set The term refers to the set that contains all the elements that an analyst wishes to study. The notation U or ξ is generally used to denote universal sets

The null set or empty set This is a set which contains no elements. It is normally designated by aGreek letter Ø, or { }.

The sets Ø and { Ø } are not the same thing since the former has no elements in it, while thelater has one element in it, namely zero

Equal or equivalent sets Two sets C and D are said to be equal if every member of set C belongs to D and every memberof set D also belongs to C

Complement of a set The complement of set A is written as A΄. This set contains all those elements of universal set which are not in A

Intersection and union

B C denotes the intersection of B and C. it is the set containing all those elements, which

belong to both B and CIf B = {5, 8, 11, 20, 25} and C = {1, 3, 5, 7, 9, 11, 13}

Then B C = {5, 11}

B C = {1, 3, 5, 7, 8, 9, 11, 13, 20 25}

SET OPERATIONS AND SOME LAWS OF SET THEORY

THE VENN DIAGRAM A simple way of representing sets and relations between sets is by means of the Venn diagram. Venn diagram consists of a rectangle that represents the universal set. Subjects of the universalset are represented by circles drawn within the rectangle, or the universe.

Suppose that the universal set is designated by U and the sets A, B and C are subject of U.The Venn diagram below can be used to illustrate the sets as follows



Lesson Two 47


Venn diagram below representing the intersection of set A and B or A B = C is illustrated asfollows

Intersection of sets

U

Example: You are given the universal set

T = {1, 2, 3, 4, 5, 6, 7, 8}

And the following subjects of the universal set:

A = {3, 4, 5, 6,}

B = {1, 3, 4, 7, 8}Determine the intersection of A and B

Solution The intersection of A and B is the subject of T, containing elements that belong to both A and B

A B = {3, 4, 5, 6,} {1, 3, 4, 7, 8}

= {3, 4}

or

U

U

A

B

C

CA B

T

5 3 1 A 6 4 7 B

8





ExampleConsider the following universal set T and its subjects C, D and E

T = {0, 2, 4, 6, 8, 10, 12}C = {4, 8,}D = {10, 2, 0}E = {0}

Findii) D E

iii) C D E

Solutionii) D E = {10, 2, 0} {0} = {0}

D E = Shaded area

ii) C D E = {4, 8} {10, 2, 0} {0} = { } = Ø

Mutually exclusive or disjointed sets

Two sets are said to be disjointed or mutually exclusive if they have no elements in common.Sets P and R below are disjointed

T

DE

T

DC

4; 8 E2; 10

0



Lesson Two 49


Disjointed sets are represented by a null set in this caseP R = Ø

The union of sets Venn diagram representing the union of sets A and B or A B = Shaded area is illustratedbelow;-

A B = Shaded areaExampleConsider the universal set T and its subsets A, B and C below:

T = {a, b, c, d e, f} A = {a, d}B = {b, c, f}C = {a, c, e, f}

Find

ii) A B iii) A C iv) B C v) A B C

Solution

i) A B = {a, d} {b, c, f} = {a, b, c, d, f}

ii) A C = {a, d} {a, c, e, f} = {a, c, d, e, f}

iii) B C = {b, c, f} {a, c, e, f} = {a, b, c, e, f}

iv) A B C = {a, d} {b, c, f } {a, c, e, f} = {a, b, c, d, e, f} = T

Complement of a set Venn diagram representing the complement of a set say A represented by A1 is illustrated below.

T

P R

U

A B





Lesson Two 51


Solving Problems Using Venn DiagramsExample 1Of the 20 girls in a form, 16 play hockey 12 play tennis and 4 play basketball. Every girl plays atleast one game and two play all the three. How many play two and only two games.

SolutionN( ξ ) = 20

H = Hockey T = Tennis B = Basketball

Those who play two and only two games = x + y + z

Using the diagram(14 – x – y) + (10 – y – z) +(2 – x – z ) + 2 + x + y+ z = 2028 – x – y – z = 20

x + y + z = 8

Example 2250 members of a certain society have voted to elect a new chairman. Each member may votefor either one or two candidates. The candidate elected is the one who polls most votes Three candidates x, y z stood for election and when the votes were counted, it was found that

- 59 voted for y only, 37 voted for z only- 12 voted for x and y, 14 voted for x and z- 147 voted for either x or y or both x and y but not for z

-

102 voted for y or z or both but not for xRequired

i) How may voters did not voteii) How many voters voted for x onlyiii) Who won the elections

n(T) = 12n(H) = 16

n(B) = 4

16 – x – y – 2y 12 – y – z – 2

2x z

4 – x – z – 2





Solution

P + 12 + 59 = 147 giving P = 76Q + 59 + 37 = 102 giving Q = 6

i) Those who did not vote= 250 – ( 76 + 12 + 14 + 59 + 6 + 37)= 250 – 204 = 46

ii) x = 76 + 12 + 14 = 102

y = 12 + 59 + 6 = 77z = 37 + 14 + 6 = 57

iii) X won the election

2.2 CALCULUSCalculus is a branch of mathematics which explains how one variable changes in relationship toanother variable. It enables us to find the rate of change of one variable with respect to another variable.

Examplei. The rate at which business revenue is increasing at a particular stage when volume

of sales is increasing.ii. The rate at which costs are changing at a particular stage when volume of sales is

giveniii. The evaluation of „rate of change‟ can help us to identify when the change in one

variable reaches a maximum or minimum.iv. Calculus may be used in production management when the production manager

wants to knowa) How much is to be manufactured in order to maximize the profits, revenues

e.t.c

N( ξ ) = 250

X Y

Z

P

Tail end

14 Q

37



Lesson Two 53


b) How much is to be produced in order to minimize the production costs

Calculus is divided into two sections namely:

Differentiation and integrationDifferentiation deals with the determination of the rates of change of business activities or simply

the process of finding the derivative of a function.Integration deals with the summation or totality of items produced over a given period of time orsimply the reverse of differentiation

The derivative and differentiation The process of obtaining the derivative of a function or slope or gradient is referred to asderivation or differentiation.

The derivative is denoted bydx

dy or f ́ (x) and is given by dividing the change in y variable by the

change in x variable. The derivative or slope or gradient of a line AB connecting points (x,y) and (x+dx, y + dy) is

given by

dx

dy

xdx x

ydy y

x

y

in xChange

yinChange

Where dy is a small change in y and dx is a small change in x variables.

Illustration

x (x + d x)

Rules of Differentiation1. The constant function rule

If given a function y = k where k is a constant thendy

dx = 0

ExampleFind the derivative of (i) y = 5

(x,y) = A

B = (x + d x, y +d y)

dy

y dx

(y + d y)

Line AB





Solution

i. y = 5 d y = 0d x

Illustration

y

5 y = 5

derivative of a constant function x

2. Power function rule

Given a function r y x

1

r dy

Then rxdx Example

Find d y for;d x

(i). y = x7

(ii). y = x2ˠ (iii). y = x-3 (iv). y = x

Solutioni. y = x7

d y = 7x 7-1 = 7x6 d x

ii. y = x2ˠ

d y = 2ˠ x(2ˠ - 1)

dy

5 00

0

dy slope

dx



Lesson Two 55


d x

iii. y = x-3

d y = -3x – 3-1 = -3x-4 d x

iv. y = x

d y = 1x 1-1 = 1.x0= 1 ( since x0=1 )d x

3. Power function multiplied by a constant

If given y = Axr, then d y = rAxr-1 d x

4. The sum rule

The derivative of the sum of two or more functions equals the sum of the derivatives of the

functions.For instanceIf H(x) = h(x) + g(x)

Then d y or H´(x) = h´(x) + g´(x)dx

5. The difference rule

The derivative of the difference of two or more functions equals the difference of the derivativesof the functionsIf H (£) = h(x) – g(x)

Then H´(£) = h´(x) – g´(x)

Examples.

Find the derivatives of

i. y = 3x2 + 5x + 7

ii. y = 4x2 – 2xb

Solution

i. y = 3x2 + 5x + 7

23 5 7

6 5 0

6 5

d x d x d dy

dx dx dx dx

x

x

ii. y = 4x2 – 2xb





2

1

4 2

8 2

b

b

d x d xdy

dx dx dx

x bx

6. The product rule – both factors are functions

The derivative of the product of two functions equals the derivative of the first functionmultiplied by the second function PLUS the derivative of the second function multiplied by thefirst function.

given that . H x h x g x

Then . . H x h x g x h x g x

Example

Find dy fordx

i. y = x2(x)

ii. y = (x2+ 3) (2x3+ x2- 3)

Solution

i. y = x2(x)

2

2

2

2 2

2

. .

.2 .1

2

3

d x d xdy x x

dx dx dx

x x x

x x

x

Note that y = x 2 (x) = x 3. Directly differentiating this we get 3x 2 .

ii. y = (x2+ 3) (2x3+ x2- 3)

2 3 2

3 2 2

3 2 2 2

4 3 2

3 2 3. 2 3 3 .

2 . 2 3 3 . 6 2

10 4 18

d x d x xdy x x x

dx dx dx

x x x x x x

x x x



Lesson Two 57


7. Quotient Rule The derivative of the quotient of two functions equals the derivative of the numerator times thedenominator MINUS the derivative of the denominator times the numerator, all which aredivided by the square of the denominator

If given H (x) =

h x

g x

then2

. .h x g x h x g x H x

g x

For example

Find dy fordx

i.23 x

x

ii. 73 x

x

Solutions

i.23 x

x

2

2

22

3

. 3 .

3

d xd x

x xdy dx dx

dx x

=2

2

3

.23

x

x x x

=2

2

2

22

3

3

3

23

x

x

x

x x

ii. 73x

3x

y

273x

221x

36x

273x

3x373x

23x

dx

dy

Example A farmer of a large farm of poultry announced that egg production per month follows theequation;





w = 3m3 – m2 m2 + 10

Where w – Total no of eggs produced per monthm – amount in kilograms of layers mash feed.

Required

Determine the rate of change of w with respect to m (i.e. the rate at which the number of eggsper month increase or decrease depending on the rate at which the kilos of layers marsh areincreased).

SolutionLet u = 3m3 – m2

∴ du = 9m2 – 2m

dm

Let v = m2 + 10

∴ dv = 2m

dm

∴

2 2 3 2

22

10 9 2 3 2

10

m m m m m mdw

dm m

4 2 3 4 3

22

9 90 2 20 6 2

10

m m m m m m

m

4 2

22

3 90 20

10

m m m

m

8. Chain Rule

This rule is generally applied in the determination of the derivatives of composite functions, which can be defined as a function in which another function can be considered to have takenthe place of the independent variable. The composite function is also referred to as a function ofa function.It is normally of the form y = (2x2 + 3)3. If we let u = (2x2 + 3), then y = u3.In order to differentiate such an equation we use the formula

dx

du

du

dy

dx

dy

Solution

y = (2x2 + 3)3

Let u = 2x2 + 3

∴ du = 2x

dx



Lesson Two 59


Let y = u3

∴ du

dy = 3u2

dy = dy . du = 3u2 x 4x = 12xu2

dx du dx

= 12x(2x2 + 3)2

Example

Consider the function

y = (x2 + 16x + 5)2

which can be decomposed into

y = u2 and u = x2 + 16x + 5. in this case y is a function of (x2 + 16x + 5)

Hence y = f(u) and u = g(x)

dy = dy . dudx du dx

= (2u) (2x +16)

= 2 (x2 + 16x + 5) (2x + 16)

9. The derivative of a function raised to power r; the composite functionrule.

The derivative of a function raised to power r equals to the power r times the function which israised by power (r-1), all of which is multiplied by the derivative of the function

If y = [g(x)]r

Then dy = r[g(x)]r-1 . g´(x)dx

For example

Find5

2 given 3 4dy

y x xdx

Solution4

25 3 4 . 6 4dy x x xdx

Differentiation of an implicit function An Implicit function is one of the y = x2 y + 3x2 + 50. it is a function in which the dependent variable (y) appears also on the right hand side. To differentiate the above equation we use the differentiation method for a product, quotient orfunction of a function.





Lesson Two 61


0dy

dx

dㄫ = 30x2 + 2y + 60dy

Maxima, minima and points of inflexiona) Test for relative maximum

Consider the following function of x whose graph is represented by the figure below

y = f(x)

dy = f´(x)dx

y

Relative maximum point The graph of the function slopes upwards to the right between points A and C and hence has apositive slope between these two points. The function has a negative slope between points Cand E. At point C, the slope of the function is Zero.

Between points X 1 and X 2 0dy

dx Where X 1 ≤ X < X 2

and between X 2 and X 3 0

dy

dx Where X 2 < X ≤ X 3. Thus the first test of the maximum points require that the first derivative of a function equalszero or

0dy

f xdx

The second text of a maximum point requires that the second derivative of a function is negativeor

y f xE

D

C

B

A

x3x2x1x

0dy

dx

0dy

dx





2

20

d y f x

dx

Example

Determine the critical value for the following functions and find out the critical value thatconstitutes a maximum

y = x3 – 12x2 + 36x + 8

Solutiony = x3 – 12x2 + 36x + 8then dy = 3x2 – 24x + 36 +0

dxiii. The critical values for the function are obtained by equating the first derivative of

the function to zero, that is:dy = 0 or 3x2 – 24x + 36 = 0dx

Hence (x-2) (x-6) = 0 And x = 2 or 6 The critical values for x are x = 2 or 6 and critical values for the function are y = 40 or 8ii. To ascertain whether these critical values of x will give rise to a maximum, we apply thesecond text, that is

d2y < 0d2xdy = 3x2 – 24x + 36 anddxd2y = 6x - 24d2x

a) When x = 2 Then d2y = -12 <0

d2x

b) When x = 6

Then d2y = +2 > 0d2x

Hence a maximum occurs when x = 2, since this value of x satisfies the second condition. X = 6does not give rise to a local maximum

b) Tests for relative minimum There are two tests for a relative minimum point

i. The first derivative, that is

dy = f´(x) = 0dx

ii. The second derivative, that isd2y = f´(x) > 0dx2

Example:For the function

h(x) = 1/3 x3 + x2 – 35x + 10



Lesson Two 63


Determine the critical values and find out whether these critical values are maxima or minima.Determine the extreme values of the function

Solution

i. Critical valuesh(x) = 1/3 x3 + x2 – 35x + 10 andh´(x) = x2 + 2x – 35

by first text,then h´(x) = x2 + 2x – 35 = 0or (x-5) (x+7) = 0Hence x = 5 or x = -7

ii. The determinant of the maximum and the minimum points requires that we test the value x = 5 and – 7 by the second textH´´(x) = 2x + 2a) When x = -7 h´´(x) = -12 <0b) When x = 5 h´´(x) = 12>0

There x = -7 gives a maximum point and x = 5 gives a minimum point.

iii. Extreme values of the functionh(x) = 1/3 x3 + x2 – 35x + 10 when x = -7, h(x) = 189 2/3 when x = 5, h(x) = -98 1/3 The extreme values of the function are h(x) = 189 2/3 which is a relative maximum

and h(x) = -98 1/3 , a relative minimum

c) Points of inflexion

Given the following two graphs, points of inflexion can be determined at points P and Q asfollows:

y y=g(x)

P

k1 x





Diagram (i)

y

y =f(x)

Q

k2 x

The points of inflexion will occur at point P wheng ́´(x) = 0 at x = k 1 g ́´(x) < 0 at x < k 1 g ́´ (x) > 0 at x > k 1

and at point Q whenf ́´(x) = 0 at x = k 1 f ́´(x) > 0 at x < k 1 f ́´(x) < 0 at x > k 1

ExampleFind the points of inflexion on the curve of the functiony = x3

Solution The only possible inflexion points will occur where

2

20

d y

dx

From the function given2

2

23 and 6

dy d y x x

dx dx

Equating the second derivative to zero, we have6x = 0 or x = 0 We test whether the point at which x = 0 is an inflexion point as follows

When x is slightly less than 0,2

20

d y

dx which means a downward concavity

When x is slightly larger than 0,2

20

d y

dx which means an upward concavity



Lesson Two 65


Therefore we have a point of inflexion at point x = 0 because the concavity of the curve changesas we pass from the left to the right of x = 0Illustration

y

y=x3

point of

inflexion

0 x

Example1. The weekly revenue Sh. R of a small company is given by

3

14 8112

x R x Where x is the number of units produced.

Required

i. Determine the number of units that maximize the revenueii. Determine the maximum revenueiii. Determine the price per unit that will maximize revenue

Solutioni. To find maximum or minimum value we use differential calculus as follows

3

2

2

2

14 1812

181 .3

12

10 .3.2

12 2

x R x

dR x

dx

d R x x

dx





2

2

2

2

2

1 0 . . 81 0

4

which gives 18 or 18

2

thus when 18; 9 maximum

dR put i e x

dx

x x

d R xdx

d R x which is negative indicating a value

dx

Therefore at x = 18, the value of R is a maximum. Similarly at x = -18, the value of R is a

minimum. Therefore, the number of units that maximize the revenue = 18 units

ii. The maximum revenue is given by

R = 14 + 81 + 18 – (18)3

12= Shs. 986

ii. The price per unit to maximize the revenue is

986 = 54.78 or Shs.54.7818

2.3 INTEGRATION It is the reversal of differentiation An integral can either be indefinite (when it has no numerical value) or definite (have specificnumerical values)

It is represented by the sign ʃ f(x)dx.

Rules of integration

i. The integral of a constant

ʃ adx = ax +c where a = constant

ExampleFind the following

a) ʃ 23dx

b) ʃɤ2dx. (where ɤ is a variable independent of x, thus it is treated as a constant).

Solutioni. ʃ 23dx = 23x + c

ii. ʃɤ2dx. = ɤ2 x + c

ii. The integral of x raised to the power n

11

1

n n x dx x cn



Lesson Two 67


Example

Find the following integrals

a) ʃ x2dx

b) ʃ x-5/2 dx

Solution

5 32 2

2 313

23

)

)

i x dx x c

ii x dx x c

iii). Integral of a constant times a function

af x dx a f x dx

ExampleDetermine the following integrals

i. ʃ ax3dxii. ʃ 20x5dx

Solution3 3

4

4

5 5

610

3

)

) 20 20 20

a

a ax dx a x dx

x c

b x dx x dx

x c

iv). Integral of sum of two or more functions

ʃ {f(x) + g(x)} dx = ʃ f(x)dx + ʃ g(x) dx

ʃ {f(x) + g(x) + h(x)}dx = ʃ f(x)dx + ʃ g(x)dx + ʃ h(x)dx

ExampleFind the following

i. ʃ (4x2 + ½ x-3 ) dx

ii. ʃ (x3/4 + 3/7 x- ½ + x5 )

Solution2 3 2 31 1

2 2

3 24 13 4

) 4 4

=

i x x dx x dx x dx

x x c

3 31 14 2 4 2

7 14 2

5 53 37 7

664 17 7 6

)ii x x x dx x dx x dx x dx

x x x c





5. Integral of a difference

ʃ {f(x) - g(x)} dx = ʃ f(x)dx - ʃ g(x) dx

Definite integrationDefinite integrals involve integration between specified limits, say a and b

The integral

b

a

f x dx Is a definite integral in which the limits of integration are a and b

The integrals is evaluated as follows

1. Compute the indefinite integral ʃ f(x)dx. Supposing it is F(x) + c2. Attach the limits of integration3. Substitute b(the upper limit) and then substitute a (the lower limit) for x.4. Take the difference and the result is the numerical value for the definite

integral.

Applying these steps to the definite integralb

b

aa

f x dx F x c

F b c F a c

F b F a

Example

Evaluate

i. 1

3 (3x 2 + 3)dx

ii. 0

5 (x + 15)dx

Solution

a.

1

3 (3x 2 + 3)dx = [(x 3 + 3x + c)]

= (27 + 9 + c) – (1 + 3 + c)

= 32

b.0

5 (x + 15)dx = [( ½ x2 + 15x + c)] 5

0

= (12 ½ + 75 + c) – (0 + 0 + c)

= 87 ½

The numerical value of the definite integrala

b f(x)dx can be interpreted as the area bounded by

the function f(x), the horizontal axis, and x=a and x=b see figure below



Lesson Two 69


y = f(x)

f(x)

0 a b x

area under curve

Thereforea

b f(x)dx = A or area under the curve

Example1. You are given the following marginal revenue function

1 MR a a q

Find the corresponding total revenue functionSolution

Total revenue 1. MR dq a a q dq

2112

aq a q c

Example 2 A firm has the following marginal cost function

2

1 2 MC a a q a q Find its total cost function.

Solution The total cost C is given by

C = ʃ MC.dq

= ʃ (a + a1q + a2q 2 ).dq

1 22 3

2 3

a aaq q q c

0





Note: Exams focus : Note the difference between marginal function and total function. Youdifferentiate total function to attain marginal function, this is common in exams,total profit = total revenue – total cost.

Example 3. Your company manufacturers large scale units. It has been shown that the marginal (or variable)

cost, which is the gradient of the total cost curve, is (92 – 2x) Shs. thousands, where x is thenumber of units of output per annum. The fixed costs are Shs. 800,000 per annum. It has alsobeen shown that the marginal revenue which is the gradient of the total revenue is (112 – 2x)Shs. thousands.

Requiredi. Establish by integration the equation of the total cost curveii. Establish by integration the equation of the total revenue curveiii. Establish the break even situation for your companyiv. Determine the number of units of output that would

a) Maximize the total revenue andb) Maximize the total costs, together with the maximum total revenue and total

costsSolution

i. First find the indefinite integral limit points of the marginal cost as the first step toobtaining the total cost curve

Thus ʃ (92 – 2x) dx = 92x – x2 + c Where c is constant

Since the total costs are the sum of variable costs and fixed costs, the constant term in theintegral represents the fixed costs, thus if Tc are the total costs then,

Tc = 92x – x2 + 800or Tc = 800 + 92x - x2

ii. As in the above case, the first step in determining the total revenue is to form theindefinite integral of the marginal revenue

Thus ʃ (112 - 2x) dx = 112x – x2 + c Where c is a constant

The total revenue is zero if no items are sold, thus the constant is zero and if Tr represents thetotal revenue, then

Tr = 112x – x2

iii. At break even the total revenue is equal to the total costs

Thus 112x – x2 = 800 + 92x - x2

20x = 800x = 40 units per annum

iv.

a) Tr = 112x – x2

112 2d Tr

x xdx



Lesson Two 71


2

22

d Tr

dx

at the maximum point2

20 that is 112 2 0

d Tr x

dx

x = 56 units per annum

Since2

22

d Tr

dx this confirms the maximum

The maximum total revenue is Shs. (112 x 56 – 56 x 56) x 1000

= Shs. 3,136,000

ii. Tc = 800 + 92 x – x2

92 2d Tc

xdx

2

22

d Tc x

dx

At this maximum point

0d Tc

dx

92 – 2x = 0

92 = 2x

x = 46 units per annum

since

2

22

d Tc x

dx this confirms the maximum

the maximum costs are Shs. (800 + 92 x 46 - 46 x 46) x 1000

= Shs. 2,916,000





1 y

x

REINFORCING QUESTIONS

QUESTION ONE

Find the derivative of

a) y = 6x – x

b)

c)

d)

QUESTION TWO

A cost function isKsh.(c) = Q2 – 30Q + 200

Where Q = quantity of units produced

Find the point of minimum cost.

QUESTION THREE

250 members of a certain society have voted to elect a new chairman. Each member may votefor either one or two candidates. The candidate elected is the one who polls most votes.

Three candidates x, y, z stood for election and when the votes were counted, it was found that,

59 voted for y only, 37 voted for z only12 voted for x and y, 14 voted for x and z147 voted for either x or y or both x and y but not for z102 voted for y or z or both but not for x.

Required:i) How many voters did not vote?ii) How many voters voted for x only?iii) Who won the election?

QUESTION FOUR

The weekly revenue Ksh.R of a small company is given by:R = 14 + 81x – x3 where x is the number of units produced

12

Required:a) Determine the number of units that maximize the revenue.b) Determine the maximum revenue.c) Determine the price per unit that will maximize the revenue

2

1 y

x

1 2 y x



Lesson Two 73


QUESTION FIVE

A furniture firm has two operating departments; Production and sales. The firms‟s operatingcosts are split between these two departments with the resultant period of fixed costs ofShs.20,000 and Shs.6,000 respectively. The production department has a basic variable cost perunit of Shs.6 plus additional variable cost per unit of Shs.0.0002 which relates to all themanufactured items during the period. The sales department has a variable cost per unit ofShs.2. The sales department receives the finished goods from the production department andpay the basic variable cost per unit plus 80% of the same.

NB: Demand Q is given by the following function:

Q = 40,000 – 2,000P, where P is the selling price of the sales department.

Required:a) Calculate the quantity that maximizes the profits of the production department.b) Calculate the selling price that maximizes the profits of the sales department.c) Determine the firm‟s profit as a result of adopting the quantity and selling prices

in i and ii.d) Determine the quantity and selling price that maximize the ship‟s profit. What is the

amount of this profit?

QUESTION SIX

a) Describe how quadratic equations can be used in decision making.

b) The demand for a commodity is given by p = 400 – q . The average total cost of producingthe commodity is given by

where p is the price in shillings and q is the quantity in

kilograms.

Required

i) What does 1000q

in the ATC equation represent economically? (1 mark)

ii) Determine the output that leads to maximum profit and the profit at thelevel of output. (9 marks)

c) Alpha industries sells two products, X and Y, in related markets, with demand functionsgiven by:

Px – 13 + 2X + Y = 0Py – 13 + X + 2Y = 0

The total cost, in shillings, is given by: TC = X + Y

Required:Determine the price and the output for each good which will maximize profits. (7 marks)

(Total: 20 marks)

21000100 5 ATC q q

q





QUESTION SEVEN

a) The following table shows the Fixed Cost (F) and the variable cost (V) of producing 1 unitof X and 1 unit of Y:

ProductX Y

Cost F 5 8 (Shs „000‟) Cost V 4 12

When x units of X and y units of Y are produced, the total fixed cost is Shs.640,000 and total variable cost is Shs.820,000. Express this information as a matrix equation and hence find thequantities of x and y produced using matrix algebra. (10 marks)

The marginal productivity of an industrial operation (the production of electric furnaces) is givenby:

2

60 10 x

f x

Where x is capitalization in millions of shillings. Given that, when the capitalization is Shs.Million they can produce 62 of the furnaces per week.

Required:a) How many furnaces they will be able to produce if their capitalization increased to Shs 10

million.b) What does the term marginal of productivity mean? (10 marks)

(Total: 20 marks)

Compare your solutions with those given in lesson 9



Lesson Two 75


COMPREHESIVE ASSIGNMENT ONEWork out these question for three hours (exam condition) then hand them in to DLC for marking

Instructions: Answer any THREE questions from SECTION I and TWO questions from SECTION II.Marks allocated to each question are shown at the end of the question. Show all your workings

SECTION I

QUESTION ONE

a) Explain the importance of set theory in business. (4 marks)b) By use of matrix algebra, develop the Leontief inverse matrix. (8 marks)c) Digital Ltd. Manufactures and sells floppy disks at Nairobi Industrial Area.

The average total cost (ATC) and Average Revenue (AR) (in thousands of shillings)of producing x floppy disks are given by the following functions:

ATC = 21 5 50050

2 2 x x

x

And

AR = 800 – 2x 2

Where: x is the number of floppy disks produced

Required:i) The profit function (3 marks)ii) The number of floppy disks required to maximize profit (3 marks)iii) The maximum profit (2 marks)

(Total: 20 marks)

QUESTION TWODefine the following terms as used in Markov analysis:

Markov process (2 marks)Equilibrium or steady state (2 marks) Absorbing state (2 marks)Closed state (2 marks)

The manufacturer of Tamu Soft drinks has been facing stiff competition on its main brand Tamu-cola soda. The management is considering an extensive advertising and rebrandingcampaign for Tamu-Cola soda. If the current branding remains, the transition matrix ofconsumer between Tamu-Cola and other brands will be as follows:

The advertising and rebranding campaign is expected to cost Sh.20 million each year.

To Tamu-Cola Others

From Tamu-Cola 0.85 0.15Others 0.25 0.75





There are 40 million consumers of soft drinks in the market and for each consumer the averageprofitability is Sh.5 annually.

Required:i) The equilibrium state proportion of consumers using Tamu-Cola before the advertising

campaign. (4 marks)

ii) The equilibrium state proportion of consumers using Tamu-Cola after the advertisingcampaign. (4 marks)iii) The expected annual profit increase or decrease after the advertising campaign. Would you

recommend the advertising campaign? (4 marks)(Total: 20 marks)

QUESTION THREE

a) A market researcher investigating consumers‟ preference for three brands of beveragesnamely: coffee, tea and cocoa, in Ongata town gathered the following information:

From a sample of 800 consumers, 230 took coffee, 245 took tea and 325 took cocoa, 30took all the three beverages, 70 took coffee and cocoa, 110 took coffee only, 185 took cocoa

only.

Required:i) Present the above information in a Venn Diagram. (4 marks)ii) The number of customers who took tea only. (2 marks)iii) The number of customers who took coffee and tea only. (2 marks)iv) The number of customers who took tea and cocoa only. (2 marks) v) The number of customers who took none of the beverages. (2 marks)

b) i) Explain the importance of the Chi-square significance test (2 marks)ii) The number of books borrowed from Millennium town library during a particular

week was recorded as shown below:Days of the week Monday Tuesday Wednesday Thursday Friday TotalNumber of booksborrowed

132 110 128 105 150 625

Required: Test the hypothesis that the number of books borrowed does not dependon the day of the week at the 1% significance level. (6 marks)

(Total: 20 marks)

QUESTION FOUR

a) The general multiple linear regression equation is expressed as:

1

n

i o i i i

i

Y X

Where Y i is the response variableX i are the explanatory variablesβo is the constant



Lesson Two 77


β1 are the parameters, and

i is the error term

Required:Express the above multiple linear regression equation in a matrix form. Clearly indicate thesize of each vector column and the matrix. (10 marks)

b) Mambo Company Ltd. Manufactures five products V, W, X Y and Z. The company hasdivided its sales team into three regions; A, B and C. The Matrix Q below represent theexpected sales quantities in thousands for each product in each sales region for the comingyear.

Region A B C50 20 35 V40 30 10 W

Q = 25 42.5 5 X products10 15 35 Y

25 17.5 22.5 Z

Each product is manufactured using combinations of four standard components. Thematrix T below indicates the number of units of each component used in producing eachproduct.

Components1 2 3 41 1 2 0 V0 1 1 2 W

T = 3 2 1 1 X products0 0 3 1 Y

1 2 3 1 Z

The manufacture of each component requires the use of certain resources. The matrix Mbelow indicates the quantities of the three standard parts and the number of productionlabour hours and assembly labour hours used to produce one unit of each component.

Part 1 Part 2 Part 3 ProductionsLabour hours

Assemblyhours

2 1 0 2 1 10 3 2 4 3 2

M = 1 2 1 1 6 3 components2 5 4 1 2 4

The costs of the resources in matrix M are Part 1 Sh.20, Part 2 Sh.10, Part 3 Sh.30 whileeach labour hour in the production and assembly departments cost Sh.15 and Sh.5respectively.

Required:i) The total expected demand for each product. (2 marks)ii) The quantities of each component needed in the production process. (3 marks)iii) The quantities of each resource required in the production. (3 marks)





iv) The total cost of producing the required units of each product. (2 marks)(Total: 20 marks)

QUESTION FIVE

The Young Children‟s Fund (YCF) is planning its annual fund-raising campaign for itsDecember school holiday camp for disadvantaged children. Campaign expenditures will beincurred at a rate of Sh.10,000 per day. From past experience, it is known that contributions willbe high during the early stages of the campaign and will tend to fall off as the campaigncontinues. The function describing the rate at which contributions are received is:

C (t) = 100t2 + 200,000 Where t = days of the campaignC (t) = rate at which contributions are received in shillings per day

The fund wants to maximize the net precedes from the campaign.

Required:i) The number of days the campaign should be conducted to maximize the

net proceeds. (3 marks)ii) The total campaign expenditure (2 marks)iii) The total contributions expected to be collected (5 marks)iv) Net proceeds from the campaign (1 mark)

The national office of a car rental company is planning its maintenance for the next year. Thecompany‟s management are interested in determining the company‟s needs for certain repairparts. The company rents saloon cars, station wagons and double cab pick-ups. The matrix Nshown below indicates the number of each type of vehicle available for renting in the fourregions of the country.

Saloons Station

wagons

Double

cabs160 400 500 CoastN = 150 300 200 Central

100 100 150 Western120 400 300 Highlands

Four repair parts of particular interest, because of their cost and frequency of replacement, arefan belts, spark plugs, batteries and tyres. On the basis of studies of maintenance records indifferent parts of the country, the management have determined the average number of repairparts needed per car during a year.

These are summarized in matrix R below:

Saloons Station wagons

Doublecabs

17 16 15 Fan beltsN = 12 8 5 Plugs

9 7 5 Batteries4 7 6 Tyres

Required:i) The total demand for each type of car. (3 marks)



Lesson Two 79


ii) The total number of each repair part required for the fleet. (3 marks)iii) If matrix C below contains the cost per unit in shillings for fan belts, spark plugs, batteries

and tyres, calculate the total cost s for all repair parts. C = (1250,800,6500,8000). (3 marks)(Total: 20 marks)

SECTION II

QUESTION SIX

a) Define the following terms as used in input-output analysis:i) Transactions table. (2 marks)ii) Primary inputs. (2 marks)iii) Technical coefficients. (2 marks)

b) Briefly explain the importance of input-output analysis. (4 marks)

c) A small economy has three main industries which are steel, motor vehicles and construction. The industries are interdependent. Each unit of steel output requires 0.2 units from steel,0.3 units from motor vehicles and 0.4 units from construction. A unit of motor vehiclesoutput requires 0.2 units from steel, 0.4 units from motor vehicles and 0.2 units fromconstruction. A unit of construction output requires 0.3 units from steel, 0.4 units frommotor vehicles and 0.1 units from construction. The final demand is 20 million units fromsteel. 50 million units from motor vehicles and 30 million units from construction.

Required:i) The technical coefficient matrix. (2 marks)ii) Total output of each industry, given that the Lentief‟s inverse matrix is

1__0.192

0.46 0.24 0.260.43 0.60 0.410.30 0.24 0.42

(3 marks)

iii) If the final demand from steel drops by 2 million units, and that from motor vehiclesincreases by 10 million units, but there is no change in the final demand from construction, what would be the change in the total output of constructions? (5 marks)

(Total: 20 marks)

QUESTION SEVEN

a) Explain the purpose of Venn diagram (3 marks)

b) A market study taken at a local sporting goods store, Maua Wahome Stores showed that ofthe 200 people interviewed, 60 owned tents, 100 owned sleeping bags, 80 owned campingstoves, and 40 owned both tents and camping stoves and 40 owned both sleeping bags andcamping stoves.

Required:If 20 people interviewed owned a tent, a sleeping bag and a camping stove, determent howmany people owned only a camping stove. In this case, is it possible for 30 people to ownboth a tent and a sleeping bag, but not a campaign stoves? (6 marks)





“Under One Thousand Shillings” Corner Store is planning to open a new store on thecorner of Main and Crescent Streets. It has asked the „Tomorrow‟s Marketing company‟ todo a market study of randomly selected families within a five kilometers radius of thestore,.the questions it wishes „Tomorrow‟s Marketing Company‟ to ask each home-ownerare:

i) Family income

ii) Family sizeiii) Distance from home to the store siteiv) Whether or not the family owns a car or uses public transport

Required:For each of the four questions, develop a random variable of interest to “Under One ThousandShillings” Corner Store. Denote which of these are discrete and which are continuous random variables. (11 marks)

(Total: 20 marks)

QUESTION EIGHT

Two CPA students were discussing the relationship between average cost and total cost. One

student said that since average cost is obtained by dividing the cost function by the number ofunits Q, it follows that the derivative of the average cost is the same as marginal cost, since thederivative of Q is 1.

Required:Comment on this analysis. (4 marks)

Gatheru and Karibu Certified Public Accountants have recently started to give business adviceto their clients. Acting as consultants, they have estimated the demand curve of a client‟s firm tobe;

AR = 200 – Q

Where AR is average revenue in millions of shillings and Q is the output in units.Investigations of the clients firm‟s cost profile sho ws that marginal cost (MC) is given by:

MC = Q2 – 28Q + 211 (in millions of shillings)

Further investigations have shown that the firm‟s cost when not producing output is Sh.10million.

Required:i) The equation of total cost. (5 marks)ii) The equation of total revenue (2 marks)iii) An expression for profit (2 marks)iv) The level of output that maximizes profit. (5 marks) v) The equation of marginal revenue. (2 marks)

(Total: 20 marks)





82 Descriptive Statistics and Index Numbers


3.1 Descriptive Statisticsa) Statistics

Definition: Statistics viewed as a subject is a process of collecting, tabulating and analyzingnumerical data upon which significant conclusions are drawn.Statistics may also be defined as numerical data, which has been, collected from a given sourceand for a particular purpose e.g. population statistics from the ministry of planning, Agriculturalstatistics from the ministry of AgricultureStatistics may also refer to the values, which have been obtained from statistical calculations e.g.the mean, mode, range e.t.c.

b) Application of statistics

1. Quality Control

Usually there is a quality control departments in every industry which is charged with theresponsibility of ensuring that the products made do meet the customers standards e.g. theKenya bureau of standards (KeBS) is one of the national institutions which on behalf of thegovernment inspects the various products to ensure that they do meet the customers

specification. The KeBS together with other control department have developed quality control charts. Theyuse these charts to check whether the products are up to standards or not.

2. Statistics may be used in making or ordering economic order quantities (EOQ). It is importantfor a business manager to realize that it is an economic cost if one orders a large quantity ofitems which have to be stored for too long before they are sold. This is because the large stockholds a lot of capital which could otherwise be used in buying other items for sale.It is also important to realize that the longer the items are stored in the stores the more will bethe storage costsOn the other hand if one orders a few items for sale he will incur relatively low storage expensesbut may not be able to satisfy all the clients. These may lose their customers if the goods are out

of stock. Therefore it is advisable to work out the EOQ which will be sufficient for the clients ina certain period before delivery. The EOQ will also ensure that minimal costs are incurred in terms of storage3. ForecastingStatistics is very important for business managers when predicting the future of a business forexample if a given business situation involves a dependent and independent variables one candevelop an equation which can be used to predict the output under certain given conditions.4. Human resource managementStatistics may be used in efficient use of human resources for example we may givequestionnaires to workers to find out where the management is weakBy compiling the statistics of those who were signing it may be found useful to analyze such datato establish the causes of resignation thus whether it is due to frustration or by choice.

3.2 Measures of Central Tendency These are statistical values which tend to occur at the centre of any well ordered set of data. Whenever these measures occur they do not indicate the centre of that data. These measures areas follows:

i. The arithmetic meanii. The modeiii. The medianiv. The geometric mean v. The harmonic mean



Lesson Three 83


1. The arithmetic mean

This is commonly known as average or mean it is obtained by first of all summing up the values given and by dividing the total value by the total no. of observations.

i.e. mean =X

n

Where x = no. of values

∑ = summation

n = no of observations

Example

The mean of 60, 80, 90, 120

60+80+90+120

4

350=

4

= 87.5

The arithmetic mean is very useful because it represents the values of most observations in thepopulation. The mean therefore describes the population quite well in terms of the magnitudes attained bymost of the members of the population

Computation of the mean from grouped Data i.e. in classes.

The following data was obtained from the manufacturers of electronic cells. A sample ofelectronic cells was taken and the life spans were recorded as shown in the following table.

Life span hrs No. of cells (f) Class MP(x) X – A = d fd

1600 – 1799 25 1699.5 -600 -15000

1800 – 1999 32 1899.5 -400 -12800

2000-2199 46 2099.5 -200 -9200

2200 – 2399 58 2299.5(A) 0 0

2400 – 2599 40 2499.5 200 8000

2600 – 2799 30 2699.5 400 12000

2800 – 2999 7 2899.5 600 4200

A = Assumed mean, this is an arbitrary number selected from the data, MP = mid point

Arithmetic mean = assumed mean + fd

f

12800= 2299.5 +

238 = 2299.5 +-53.78





= 2245.72 hours

Example 2 – (use of the coded method)

The following data was obtained from students who were registered in a certain college. The table shows the age distribution Age (yrs) No. of Students (f) mid points (x) x-a = d D/c = u fu15 – 19 21 17 -15 -3 -6320 – 24 35 22 -10 -2 -7025 – 29 38 27 -5 -1 -3830 – 34 49 32(A) 0 0 035 – 39 31 37 +5 + 3140 – 44 19 42 +10 +2 38

193 -102

Required calculate the mean age of the students using the coded method

Actual mean = A(assumed mean) + c fu f

102= 32 + 5

193

= 29.36 years

NB. The following statistical terms are commonly used in statistical calculations. They musttherefore be clearly understood.

i) Class limits These are numerical values which limits uq extended of a given class i.e. all the observations in agiven class are expected to fall within the interval which is bounded by the class limits e.g. 15 &19 are class limits as in the table of the example above.

ii) Class boundaries

These are statistical boundaries, which separate one class from the other. They are usuallydetermined by adding the lower class limit to the next upper class limit and dividing by 2 e.g. in

the above table the class boundary between 19 and 20 is 19.5 which is =19+20

2.

iii) Class Mid points

This are very important values which mark the center of a given class. They are obtained byadding together the two limits of a given class and dividing the result by 2.

iv) Class interval/width

This is the difference between an upper class boundary and lower class boundary. The valueusually measures the length of a given class.



Lesson Three 85


2. The mode

- This is one of the measures of central tendency. The mode is defined as a value within afrequency distribution which has the highest frequency. Sometimes a single value may notexist as such in which case we may refer to the class with the highest frequency. Such a classis known as a modal class

- The mode is a very important statistical value in business activities quite oftenbusiness firms tend to stock specific items which are heavily on demand e.g. footwear,clothes, construction materials (beams, wires, iron sheets e.t.c.

- The mode can easily be determined form ungrouped data by arranging the figuresgiven and determining the one with the highest frequency.

- When determining the values of the mode from the grouped data we may use thefollowing methods;-

i. The graphical method which involves use of the histogramii. The computation method which involves use of formula

Example

In a social survey in which the main purpose was to establish the intelligence quotient (IQ) of

resident in a given area, the following results were obtained as tabulated below:

IQ No. of residents Upper class bound CF

1 – 20 6 20 621 – 40 18 40 2441 – 60 32 fo 60 5661 – 80 48 f1 80 10481 – 100 27 f2 100 131101 – 120 13 120 144121 – 140 2 140 146

Required

Calculate the modal value of the IQ‟s tabulated above using

i. The graphical method and

ii. Formular





Graphical method

50

40

30

20

10

20 40 60 80 100 120 140

Value of the mode

Computation method

1 0

1 0 2

f f Mode = L + ×c

2f f f

Where L = Lower class boundary of the class containing the mode

f 0 = Frequency of the class below the modal classf 1 = Frequency of the class containing the modef 2 = frequency of the class above the modal classc = Class interval

48-32 Therefore Mode = 60.50 + ×20

2 48 - 32 - 27

= 69.14

3. The median

- This is a statistical value which is normally located at the center of a given set of data whichhas been organized in the order of magnitude or size e.g. consider the set 14, 17, 9, 8, 20, 32,18, 14.5, 13. When the data is ordered it will be 8, 9, 13, 14, 14.5, 17, 18, 20, 32 The middle number/median is 14.5

- The importance of the median lies in the fact that it divides the data into 2 equal halves. Theno. of observations below and above the median are equal.



Lesson Three 87


- In order to determine the value of the median from grouped data. When data is grouped themedian may be determined by using the following methods

i. Graphical method using the cumulative frequency curve (ogive)ii. The formula

Example

Referring to the table in 105, determine the median using the methods aboveThe graphical method

IQ No of resid UCB Cumulative Frequency0 – 20 6 20 620 – 40 18 40 2440 – 60 32 60 5660 – 80 48 80 10480 – 100 27 100 131100 – 120 13 120 144120 – 140 2 140 146

146

xv

160

140

120

100

80

60

40

20

20 40 60 80 100 120 140 160

Value of the median

n+1 146+1 The position of the median = =

2 2

ii Computation

The formula used is





Lesson Three 89


5. Harmonic mean

This is a measure of central tendency which is used to determine the average growth rates fornatural economies. It is defined as the reciprocal of the average of the reciprocals of all the values given by HM.

1 2 3

1 1 1 1

1

( ... )n x x x HM

Example

The economic growth rates of five countries were given as 20%, 15%, 25%, 18% and 5%Calculate the harmonic mean

1 The HM =

1 1 1 1 1 1( + + + +5 20 15 25 10 5

1=0.2(0.05+ 0.07+0.04 +0.10+0.2)

1=

0.092

10.86%

6. Weighted mean

- This is the mean which uses arbitrarily given weights

- It is a useful measure especially where assessment is being done yet the conditions prevailingare not the same. This is particularly true when assessment of students is being done giventhat the subjects being taken have different levels of difficulties.

Examples

The following table shows that marks scored by a student doing section 3 and 4 of CPA

Subject Scores (x) Weight (w) wxSTAD 65 50 3250BF 63 40 2520FA2 62 45 2340LAW 80 35 2800

QT 69 55 3795FA3 55 60 3300

w = 285 wx = 18005

Weighted mean





Ewx

Ew

18005

285

63.17%

Merits and demerits of the measures of central tendency

The arithmetic mean (a.m)

Merits

i. It utilizes all the observations givenii. It is a very useful statistic in terms of applications. It has several applications in

business management e.g. hypothesis testing, quality control e.t.c.iii. It is the best representative of a given set of data if such data was obtained from a

normal populationiv. The a.m. can be determined accurately using mathematical formulas

Demerits of the a.m.

i. If the data is not drawn from a „normal‟ population, then the a.m. may give a wrongimpression about the population

ii. In some situations, the a.m. may give unrealistic values especially when dealing withdiscrete variables e.g. when working out the average no. of children in a no. offamilies. It may be found that the average is 4.4 which is unrealistic in human beings

The mode

Merits

i. It can be determined from incomplete data provided the observations with thehighest frequency are already known

ii. The mode has several applications in businessiii. The mode can be easily definediv. It can be determined easily from a graph

Demerits

i. If the data is quite large and ungrouped, determination of the mode can be quitecumbersome

ii. Use of the formula to calculate the mode is unfamiliar to most business people

iii.

The mode may sometimes benon existent

or there may be two modes for a givenset of data. In such a case therefore a single mode may not exist

The median

Merits

i. It shows the centre of a given set of dataii. Knowledge of the determination of the median may be extended to determine the

quartiles



Lesson Three 91


iii. The median can easily be defined iv. It can be obtained easily from the cumulative frequency curve v. It can be used in determining the degrees of skew ness (see later)

Demerits

i. In some situations where the no. of observations is even, the value of the medianobtained is usually imaginary

ii. The computation of the median using the formulas is not well understood by mostbusinessmen

iii. In business environment the median has got very few applications

The geometric mean

Merits

i. It makes use of all the values given (except when x = 0 or negative)ii. It is the best measure for industrial growth rates

Demerits

i. The determination of the GM by using logarithms is not familiar process to allthose expected to use it e.g managers

ii. If the data contains zeros or – ve values, the GM ceases to exist

The harmonic mean and weighted mean

Merits – same as the arithmetic meanDemerits – same as the arithmetic mean

3.3 Measures of Dispersion- The measures of dispersion are very useful in statistical work because they indicate

whether the rest of the data are scattered around the mean or away from the mean.

-

If the data is approximately dispersed around the mean then the measure of dispersionobtained will be small therefore indicating that the mean is a good representative of thesample data. But on the other hand, if the figures are not closely located to the meanthen the measures of dispersion obtained will be relatively big indicating that the meandoes not represent the data sufficiently

- The commonly used measures of dispersion area) The rangeb) The absolute mean deviationc) The standard deviationd) The semi – interquartile and quartile deviatione) The 10th and 90th percentile rangef) Variance

a) The range

- The range is defined as the difference between the highest and the smallest values in afrequency distribution. This measure is not very efficient because it utilizes only 2 valuesin a given frequency distribution. However the smaller the value of the range, the lessdispersed the observations are from the arithmetic mean and vice versa

- The range is not commonly used in business management because 2 sets of data mayyield the same range but end up having different interpretations regarding the degree ofdispersion





b) The absolute mean deviation

- This is a useful measure of dispersion because it makes use of all the values given seethe following examples

Example 1

In a given exam the scores for 10 students were as follows

Student Mark (x) x x

A 60 1.8B 45 16.8C 75 13.2D 70 8.2E 65 3.2F 40 21.8G 69 7.2H 64 2.2I 50 11.8 J 80 18.2

Total 618 104.4

Required

Determine the absolute mean deviation

Mean, x =618

10= 61.8

Therefore AMD = X - X 104.4= = 10.44N 10

Example 2

The following data was obtained from a given financial institution. The data refers to the loansgiven out in 1996 to several firms

Firms (f) Amount of loanper firm (x)

fx x x . x x f

3 20000 60000 4157.9 12473.704 60000 240000 35842.1 143368.40

1 15000 15000 9157.9 9157.95 12000 60000 12157.9 60789.506 14000 84000 10157.9 60947.40

Σf = 19 Σfx = 459000 286736.90

Required

Calculate the mean deviation for the amount of items given



Lesson Three 93


459,00024157.9

19

- 286736.90

19

15,091.40

fx X

f

X X AMD

f

Shs

NB if the absolute mean deviation is relatively small it implies that the data is more compact andtherefore the arithmetic mean is a fair sample representative.

c) The standard deviation

- This is one of the most accurate measures of dispersion. It has the followingadvantages;

i. It utilizes all the values given

ii. It makes use of both negative and positive values if they occuriii. The standard deviation reflects an accurate impression of how much the sampledata varies from the mean. This is because its suitability can also be tested usingother statistical methods

Example

A sample comprises of the following observations; 14, 18, 17, 16, 25, 31Determine the standard deviation of this sampleObservation.

x x x

2

x x

14 -6.1 37.2118 -2.1 4.4117 -3.1 9.6116 -4.1 16.8125 4.9 24.0131 10.9 118.81

Total 121 210.56

12120.1

6 X

standard deviation,

2

210.56

6n

x x

= 5.93





Alternative method

x X 2 14 19618 32417 289

16 25625 62531 961

Total 121 2651

2 222651

6

121

6

x

n n

x

= 5.93

Example 2

The following table shows the part-time rate per hour of a given no. of laborers in the month of June 1997.

Rate per hr (x) Shs No. of labourers(f)

fx fx2

230 7 1610 370300400 6 2400 960000350 2 700 245000450 1 450 202500200 8 1600 320000150 11 1650 247500

Total 35 8410 2345300

Calculate the standard deviation from the above table showing how the hourly payment were varying from the respective mean

∴ standard deviation,

22

- fx fx

f f

=2

2345300 8410-

35 35

= 67008.6 577372

= 9271.4

= 96.29

Example 3 – Grouped data



Lesson Three 95


In business statistical work we usually encounter a set of grouped data. In order to determine thestandard deviation from such data, we use any of the three following methods

i. The long methodii. The shorter methodiii. The coded method

The above methods are used in the following examples

Example 3.1

The quality controller in a given firm had an accurate record of all the iron bars produced in may1997. The following data shows those records

i. Using long method

Bar lengths(cm)

No. of bars(f) Class mid point(x)

fx fx2

201 – 250 25 225.5 5637.5 1271256.25251 – 300 36 275.5 9918 2732409301 – 350 49 325.5 15949.5 5191562.25

351 – 400 80 375.5 30040 11280020401 – 450 51 425.5 21700.5 9233562.75451 – 500 42 475.5 19971 9496210.50501 - 550 30 525.5 15765 8284507.50

313 118981.50 47489526

Calculate the standard deviation of the lengths of the bars

∴ standard deviation, σ =

22

- fx fx

f f

=2

47489526 118981.50-

313 313

= 84.99 cm

ii. Using the shorter method

Bar lengths (cm) No. ofbars(f)

mid point (x) x-A = d fd Fd2

201 – 250 25 225.5 -150 -3750 562500

251 – 300 36 275.5 -100 -3600 360000301 – 350 49 325.5 -50 -2450 122500351 – 400 80 375.5 (A) 0 0 0401 – 450 51 425.5 50 2550 127500451 – 500 42 475.5 100 4200 420000501 - 550 30 525.5 150 4500 675000

Total 313 1450 2267500

Calculate the standard deviation using the shorter method quagmire





∴ Standard deviation, σ =

22

fd fd

f f

=

22267500 1450

313 313

= 7244.40 21.50

= 7222.90

= 84.99 cm

iii. Using coded method

Bar lengths(cm) (f) mid point (x) x-A = d d/c = u fu fu2

201 – 250 25 225.5 -150 -3 -75 225251 – 300 36 275.5 -100 -2 -72 144301 – 350 49 325.5 -50 -1 -49 49351 – 400 80 375.5 (A) 0 0 0 0401 – 450 51 425.5 50 1 51 51451 – 500 42 475.5 100 2 84 168501 - 550 30 525.5 150 3 90 270

313 29 907

C = 50 where c is an arbitrary number, try picking a different figure say 45 the answer

should be the same.

Standard deviation using the coded method. This is the most preferable method among the threemethods

22

- fu fu

c f f

2907 29

50

313 313

= 50 × 1.6997

= 84.99

Variance

Square of the standard deviation is called variance.



Lesson Three 97


d) The semi interquartile range

- This is a measure of dispersion which involves the use of quartile. A quartile is amark or a value which lies at the boundary of a division when any given set of datais divided into four equal divisions

- Each of such divisions normally carries 25% of all the observations- The semi interquartile range is a good measure of dispersion because it shows how

the rest of the data are generally spread around the mean- The quartiles normally used are three namely;

i. The lower quartile (first quartile Q1) this usually binds the lower 25% of thedata

ii. The median (second quartile Q2)iii. The upper quartile (third quartile Q3)

The semi-interquartile range,

Q3 - Q1SIR =

2

Example 1

The weights of 15 parcels recorded at the GPO were as follows:16.2, 17, 20, 25(Q1) 29, 32.2, 35.8, 36.8(Q2) 40, 41, 42, 44(Q3) 49, 52, 55 (in kgs)Required

Determine the semi interquartile range for the above data

Q3 Q1 44 - 25 19SIR = = = = 8.5

2 2 2

Example 2 (Grouped Data)

The following table shows the levels of retirement benefits given to a group of workers in agiven establishment.

Retirement benefits £„000

No of retirees (f) UCB cf

20 – 29 50 29.5 5030 – 39 69 39.5 11940 – 49 70 49.5 18950 – 59 90 59.5 27960 – 69 52 69.5 331

70 – 79 40 79.5 37180 – 89 11 89.5 382

Required

i. Determine the semi interquartile range for the above dataii. Determine the minimum value for the top ten per cent.(10%)iii. Determine the maximum value for the lower 40% of the retirees





Solution

The lower quartile (Q1) lies on position

N + 1 382 + 1=

4 4

= 95.75

(95.75 - 50) the value of Q1 = 29.5 + x 10

69

= 29.5 + 6.63

= £36.13

The upper quartile (Q3) lies on position

N + 13

4

382 + 1=3

4

= 287.25

∴ the value of Q3 = 59.5 +287.25 - 279

52

× 10

= 61.08

The semi interquartile range =Q3-Q1

2

61.08 - 36.13=

2 = 12.475

= £12,475

ii. The top 10% is equivalent to the lower 90% of the retirees

The position corresponding to the lower 90%

90= (n + 1) = 0.9 (382 + 1)

100

= 0.9 x 383

= 344.7



Lesson Three 99


∴ the benefits (value) corresponding to the minimum value for top 10%

= 69.5 +344.7 - 331

40 x 10

= 72.925

= £ 72925

iii. The lower 40% corresponds to position

=100

40(382 + 1)

= 153.20

∴ retirement benefits corresponding to its position

= 39.5 +153.2-119

70 x 10

= 39.5 + 4.88

= 44.38

= £ 44380

e. The 10th – 90th percentile range

This is a measure of dispersion which uses percentile. A percentile is a value which separates one

division from the other when a given data is divided into 100 equal divisions. This measure of dispersion is very important when calculating the co-efficient of skewness (seelater)

Example

Using the above data for retirees calculate the 10 th - 90th percentile. The tenth percentile 10th percentile lies on position

10

100(382 + 1) = 0.1 x 383

= 38.3

∴ the value corresponding to the tenth percentile

(38.3 x 10)= 19.5 +

50 = 19.5 + 7.66

= 27.16

The 90th percentile lies on position





Lesson Three 101


iv. Coefficient of variation

=standard deviation

×100mean

Example (see information above)First group of cars: mean = 2000 kms

Standard deviation = 5 kms

∴ C.O.V = 5 x 1002000

= 0.25%

Second group of cars: mean = 3000 kms

Standard deviation = 10kms

∴ C.O.V = 10 x 1003000

= 0.33%

Conclusion

Since the coefficient of variation is greater in the 2nd group, than in the first group we mayconclude that the distances covered in the 1st group are much closer to the mean that in the 2nd group.

Example 2

In a given farm located in the UK the average salary of the employees is £ 3500 with a standarddeviation of £150 The same firm has a local branch in Kenya in which the average salaries are Kshs 8500 with astandard deviation of Kshs.800Determine the coefficient of variation in the 2 firms and briefly comment on the degree ofdispersion of the salaries in the 2 firms.First firm in the UK

C.O.V = 150 x 1003500

= 4.29%

Second firm in Kenya

C.O.V = 800 x 1008500

= 9.4%





Conclusively, since 4.29% < 9.4% then the salaries offered by the firm in UK are much closer tothe mean given them in the case to the local branch in Kenya

COMBINED MEAN AND STANDARD DEVIATIONSometimes we may need to combine 2 or more samples say A and B. It is therefore essential toknow the new mean and the new standard deviation of the combination of the samples.

Combined meanLet m be the combined meanLet x1 be the mean of first sampleLet x2 be the mean of the second sampleLet n1 be the size of the 1st sampleLet n2 be the size of the 2nd sampleLet s1 be the standard deviation of the 1st sampleLet s2 be the standard deviation of the 2nd sample

1 1 2 2

1 2

combined meann x n x

n n

1

2 22 2

1 1 1 2 2 2 2

1 2

combined standard deviationn s n m x n s n m x

n n

Example

A sample of 40 electric batteries gives a mean life span of 600 hrs with a standard deviation of20 hours. Another sample of 50 electric batteries gives a mean lifespan of 520 hours with a standarddeviation of 30 hours.If these two samples were combined and used in a given project simultaneously, determine the

combined new mean for the larger sample and hence determine the combined or pulled standarddeviation.

Size x s40(n1 ) 600 hrs(x1 ) 20hrs (s1 )50 (n1 ) 520 hrs (x2 ) 30 hrs (s2 )

40 600 50 520 50,000555.56

40 50 90Combined mean

Combined standard deviation

2 2 2 240(20 ) 40(555.56- 660) 50(30) 50(555.56-520)

40 50

1600 78996.54 45000 63225.68

90

47.52 hrs



Lesson Three 103


SKEWNESS- This is a concept which is commonly used in statistical decision making. It refers to the

degree in which a given frequency curve is deviating away from the normal distribution- There are 2 types of skew ness namely

i. Positive skew ness

ii. Negative skew ness

1. Positive Skewness- This is the tendency of a given frequency curve leaning towards the left. In a

positively skewed distribution, the long tail extended to the right.In this distribution one should note the following

i. The mean is usually bigger than the mode and medianii. The median always occurs between the mode and meaniii. There are more observations below the mean than above the mean

This frequency distribution as represented in the skewed distribution curve is characteristic ofthe age distributions in the developing countries

2. Negative Skewness This is an asymmetrical curve in which the long tail extends to the left

NB: This frequency curve for the age distribution is characteristic of the age distribution indeveloped countries

- The mode is usually bigger than the mean and median- The median usually occurs in between the mean and mode- The no. of observations above the mean are usually more than those below the

mean (see the shaded region)

MEASURES OF SKEWNESS

- These are numerical values which assist in evaluating the degree of deviation of afrequency distribution from the normal distribution.- Following are the commonly used measures of skew ness.

1. Coefficient Skewness

=mean - median

3Standard deviation

2. Coefficient of skewness

frequency

M o d e

M e d i a n

M e a n

M e a n

M e d i a n

M o d e

Normal distribution

frequency Positively skewedfrequency curve Negatively skewed

frequency curve

Long tail





Lesson Three 105


The standard deviation = c ×

22

- fu fu

f f

=5 ×

2

3086 428-610 610

= 10.68

The Position of the median lies m =n +1

2

=610+1

2 = 305.5

= 60.5 +305.5- 191

120

× 5

= 60.5 +114.4

120 × 5

Median = 65.27

Therefore the Pearsonian coefficient

=66.51-65.27

310.68

= 0.348

Comment The coefficient of skewness obtained suggests that the frequency distribution of the loans given was positively skewed This is because the coefficient itself is positive. But the skewness is not very high implying thedegree of deviation of the frequency distribution from the normal distribution is small

Example 2Using the above data calculate the quartile coefficient of skewness

Quartile coefficient of skewness =Q3+Q1- 2Q2

Q3+Q1

The position of Q1 lies on =

610+1

= 152.754

∴ actual value Q1 =55. 5 +152.75-94

5 58.5397

The position of Q3 lies on =610+1

3 = 458.254





∴ actual value Q3 =70.55 +458.25- 403

835 73.83 × 5

Q2 position: i.e. 2610+1

4 = 305.5

Actual Q2 value305.5-191

12060.5 5 65.27

The required coefficient of skew ness

=73.83 58.53 2 65.27

73.83 58.530.013

ConclusionSame as above when the Pearsonian coefficient was used

KURTOSIS

- This is a concept, which refers to the degree of peaked ness of a given frequencydistribution. The degree is normally measured with reference to normal distribution.

- The concept of kurtosis is very useful in decision making processes i.e. if is afrequency distribution happens to have either a higher peak or a lower peak, then itshould not be used to make statistical inferences.

- Generally there are 3 types of kurtosis namely;-i. Leptokurticii. Mesokurticiii. Platykurtic

Leptokurtica) A frequency distribution which is lepkurtic has generally a higher peak than that

of the normal distribution. The coefficient of kurtosis when determined will be

found to be more than 3. thus frequency distributions with a value of morethan 3 are definitely leptokurtic

b) Some frequency distributions when plotted may produce a curve similar to thatof the normal distribution. Such frequency distributions are referred to asmesokurtic. The degree of kurtosis is usually equal to 3

c) When the frequency curve contacted produces a peak which is lower that thatof a normal distribution when such a curve is said to be platykurtic. Thecoefficient of such is usually less than 3

- It is necessary to calculate the numerical measure of kurtosis. The commonly usedmeasure of kurtosis is the percentile coefficient of kurtosis. This coefficient isnormally determined using the following equation

Percentile measure of kurtosis, K (Kappa) =1

2

Q3-Q1

P90 - P10 ExampleRefer to the table above for loans to small business firms/unitsRequiredCalculate the percentile coefficient of Kurtosis

P90 =90

n +1 = 0.9 610 +1100

= 0.9 (611)



Lesson Three 107


= 549.9 The actual loan for a firm in this position

(549.9) = 80.5 +549.9-538

40 x 5 = 81.99

P10 =10

100 (n + 1) = 0.1 (611) = 61.1

The actual loan value given to the firm on this position is

50.5 +61.1 32

62 x 5 = 52.85

= 0.9 (611)= 549.9

∴ percentile measure of kurtosis

K(Kappa) = ½Q3-Q1

P90- P10

= ½73.83- 58.53

81.99- 52.85

= 0.26Since 0.26 < 3, it can be concluded that the frequency distribution exhibited by the distributionof loans is platykurticKurtosis is also measured by moment statistics, which utilize the exact value of eachobservation.

i. M1 the first moment = M1 =X

n = Mean M1 or M1

M2 =2X

n

M3 =3X

n

M4 =4X

n

3. M2 second moment about the mean M2 or f 2 M2 = M2 – M12

4. M3 third moment about the mean M3 (a measure of the absolute skew ness)M3 = M3 – 3M2M1 + 2M13

5. M4 fourth moment about the mean M4 (a measure of the absolute Kurtosis)M4 = M4 – 4M3M1 + 6M2M12 + 3M14

An alternative formula





M4 =

4 x m f

f Where m is mean

Moment coefficient of KurtosisM4

S4

ExampleFind the moment coefficient of the following distribution

x f12 114 416 618 1020 722 2

X f xf (x-m) (x-m)2 (x-m)2f (x-m)4f12 1 12 -5.6 31.36 31.36 983.4514 4 56 -3.6 12.96 51.84 671.8516 6 96 -1.6 2.56 15.36 39.3218 10 180 .4 0.16 1.60 0.25620 7 140 2.4 5.76 40.32 232.2422 2 44 4.4 19.36 38.72 749.62

30 528 179.20 2,676.74

M =528

30 = 17.6

σ2 = 179.20

30 = 5.973

σ4 = 35.677

M4 =

4 x m f

f =

2,676.74

30 = 89.22

Moment coefficient of Kurtosis =89.22

35.677 = 2.5

Note Coefficient of kurtosis can also be found using the method of assumed mean.

3.4 Indices An index number is an attempt to summarize a whole mass of data into one figure. The singlefigure shows how one year differs from another year.It is a statistical devise used to measure the change in the level of prices, wages output and other variables at given times, relative to their level at an earlier time which is taken as the base forcomparison purposes



Lesson Three 109


A simple price index =Pn

Po × 100 (an unweighted price index)

A simple quantity index =Qn

Qo × 100 (an unweighted quantity index)

Where pn is the price of a commodity in the current year (the year for which the price index tobe calculated)

Where po is the price of the same commodity in the base year (the year for comparisonpurposes)

Similarly Qn and Qo are defined in the same way

AGGREGATE PRICE INDEX NUMBERS AND QUANTITY INDEX NUMBERSPRICE INDEX QUANTITY INDEX

LASPEYRE‟S INDEX n o

o o

p q P q

× 100 n o

o o

q pq p

× 100

PAASCHE‟S INDEX n n

o n

p q

P q × 100 n n

o n

q q

q p × 100

Value index = n n

o o

p q

P q × 100

MODIFIED FORM OF THE LASPEYRE’S PRICE INDEX NUMBER

Laspeyre’s Price index 100

n

o

p

o p

o

w

w

Where w 0 are the proportions of the total expected in the basic period. This formula isfrequently used to calculate retail price index.

CHANGING THE BASE OF THE INDEXFor comparison purposes if two series have different base years, it is difficult to compare themdirectly. In such cases, it is necessary to change the base year of one of the series (or both) sothat both have the same base.It is also necessary to keep the index relevant to current conditions hence the need to change thebase from time to time.

Example; Year 1985 1986 1987 1988 1989 1990 1991 1992Price index 100 104 108 109 112 120 125 140

Suppose we wish to change the base year to 1989 We recalculate each index by expressing it as a percentage of 1989





Previous index Recalculated index1985 100 100

112 × 100 = 89.3

1986 104 104

112 × 100 = 92.9

1987 108 108

112 × 100 = 96.4

1988 109 109

112 × 100 = 97.3

1989 (new base year) 112 112

112 × 100 = 100

1990 120 120

112 × 100 = 107.1

1991 125 125

112 × 100 = 111.6

1992 140 140

112 × 100 = 125.0

When changing the base year, it is advisable to update the weights used in the base year.

CHAIN BASED INDEX NUMBERS A chain based index is one where the index is calculated every year using the previous year as thebase year. This type of index measures rate of change from year to year. This method is suitable where weights are changing rapidly and items are constantly beingbrought into the index and unwanted items taken out. It can be a price or quantity index

Previous index Recalculated chainbasedindex fixedbased index1985 100 100 100(1985 base year1986 104 104

100 × 100 = 104

104

100 × 100 = 104

1987 108 108

104 × 100 = 103.8

108

100 × 100 = 108

1988 109 109

108 × 100 = 100.9

109

100 × 100 = 109

1989 112 112

109

× 100 = 102.8112

100

× 100 = 112

1990 120 120

112 × 100 = 107.1

120

100 × 100 = 120

1991 125 125

120 × 100 = 104.2

125

100 × 100 = 125

1992 140 140

120× 100 = 112

140

100 × 100 = 140



Lesson Three 111


The Fisher’s index The Fisher‟s index acts as a compromise between Laspeyre‟s index and Paasche‟ index. It iscalculated as a geometric mean of the two indexes.

Retail price indexIt is weighted average of price relatives based upon an average household in the base year. The

items consumed are divided into groups such as food, housing, transport, alcoholic drinks,footwear, fuel, light, water, household goods, services e.t.c. each item included in the index isgiven a weighting and a price relative to the base is calculated. Modified form of laspeyre‟s priceindex formula is used as a weighted arithmetic mean of price relatives.

I.e. Retail Price index 0

0

100

n

o

p

pW

W

The index is used by the Government as a guide in determining the minimum wages, pensionrates unemployed benefits (in UK e.t.c). Trade unions use it as a basis for their wages claims.

DeflationIndexes may be used to deflate time series so that comparisons between periods may be made in

real termsIt is a process of reducing a value measured in current period prices to its equivalent in the baseperiod prices. The deflated value is what would have been necessary to purchase the sameamount of goods as the present value can purchase in the current period

Deflation Factor = n n

0 n

p q

p q × 100

Deflation of a time series Year Average monthly earnings (shs) Retail index Real earnings1 5,000 100 5000 = 50002 5,500 120

5,500 ×100

120 = 4,583.3

3 6,000 1406,000 ×

100

140 = 4,285.7

4 6,500 1706,500 ×

100

170 = 3,823.5

5 7,200 2007,200 ×

100

200 = 3,600.0

The technique of index number construction When preparing index numbers it is important to define

a)

The exact purpose of the indexb) How the items are to be selectedc) The choice of the weightsd) The choice of the basee) The type of average to be used

The base year should be as close to the normal trend as possible. The best methods should beused for collection of data. The items should be selected in such a way that they are a fairrepresentation of all the relevant items.Due consideration should be given to the weighting of all items selected





The index of industrial productionIt is a quantity index compiled by the government. It measures changes in the volume ofproduction in major industries. The index is a good indication of the state of national economy.It covers the following major industries in the UK

i. Mining and quarrying

ii. Manufacturing such as food, drinks and tobacco, chemicals, metal manufacture,engineering e.t.ciii. Textileiv. Construction v. Gas electricity, water e.t.c

It excludes agriculture, fishing, trade, transport, finance and other such industries.Each industries order is given a weighting. The weighting is based on average monthlyproduction in each industry in a fixed base year. It gives each item its relative importanceamongst all other items and thus gives a better estimate of the index for comparison purposes.

The Geometric Index (Industrial Share index) This index is an index of 30 selected top industrial companies. It is calculated by taking an

unweighted geometric mean of the price relatives of the selected shares.Example The share prices of ordinary shares of four companies on 1st January 1990 and 1st January 1991 were as follows.

Share Price on 1.1.1990 Price on 1.1.1991Company A Shs 10 Shs 12Company B Shs 12 Shs 15Company C Shs 20 Shs 25Company D Shs 5 Shs 6

Using an unweighted geometric index, calculate the index of share prices at 1.1.1991 if 1.1.1990is the base date, index 100

Solution

14

1 1

4 412 15 25 6 270002.25

10 12 20 5 12000

1.225

percentage increase = 22.5% index = 122.5

Inflation The inflation rate for a given period can be calculated using the following formula;

Inflation = Current retail price indexRetail price index in the base year

× 100

Marshal Hedge Worth Index

Marshal Hedge worth index = n o n

o o n

p p q

p q q × 100



Lesson Three 113


Tests For An Ideal Index Number1. Factor Reversal Test This test indicates that when the price index is multiplied with a quantity index i.e. factors arereversed), it should result in the value index.2. The time reversal testIf we reverse the time subscripts of a price or quantity index, the result should be reciprocal of

the original index.





LESSON 3 REINFORCING QUESTIONS

QUESTION ONE

a) Distinguish between discrete and continuous data.b) What is dispersion and what is the formula for the standard deviation?c) What is the measure of relative dispersion?d) Draw diagrams showing positive and negative skewness

QUESTION TWO

The managers of an import agency are investigating the length of time that customers take topay their invoices, the normal terms for which are 30 days net. They have checked the paymentrecord of 100 customers chosen at random and have compiled the following table:

Payment in Number of customers5 to 9 days 410 to 14 days 1015 to 19 days 1720 to 24 days 2025 to 29 days 2230 to 34 days 1635 to 39 days 840 to 44 days 3

Required:a) Calculate the arithmetic mean.b) Calculate the standard deviationc) Construct a histogram and insert the modal value.d) Estimate the probability that an unpaid invoice chosen at random will be between 30 and

39 days old.

QUESTION THREE

The price of the ordinary 25p shares of Manco PLC quoted on the stock exchange, at the closeof the business on successive Fridays is tabulated below

126 120 122 105 129 119 131 138125 127 113 112 130 122 134 136128 126 117 114 120 123 127 140124 127 114 111 116 131 128 137127 122 106 121 116 135 142 130

Requireda) Group the above date into eight classes. (4 marks)b) Calculate cumulative frequency, the median value, quartile values and the

semi-quartile range. (4 marks)c) Calculate the mean and standard deviation of your frequency distribution. (7 marks)d) Compare and contrast the values that you have obtained for:

i) The median and meanii) The semi-interquartile range and the standard deviation (5 marks)

(Total: 20 marks)



Lesson Three 115


QUESTION FOUR

Define the coefficient of variation.

The following table gives profits (in ten thousands of shillings) of two supermarkets over aduration of one year.

Month Supermarket A Supermarket B January 65 28February 48 33March 15 20 April 28 23May 41 69 June 59 45 July 41 53 August 10 15September 24 35

October 56 57November 92 99December 120 136

Required:i) Compute the coefficient of variation for each supermarket.ii) Indicate for which supermarket the variability of profits is relatively greater.

QUESTION FIVE

Prodco PLC manufactures an item of domestic equipment which requires a number ofcomponents which have varied as various modifications of the model have been used. Thefollowing table shows the number of components required together with the price over the last

three years of production.

COMPONENT 1981 1982 1983Prices Quantity Prices Quantity Prices Quantity

A 3.63 3 4.00 2 4.49 2B 2.11 4 3.10 5 3.26 6C 10.03 1 10.36 1 12.05 1D 4.01 7 5.23 6 5.21 5

Required:a) Establish the base weighted price indices for 1982 and 1983 based on

1981 for the item of equipment. (8 marks)

b)

Establish the current weighted price indices for 1982 and 1983 based on1981 for the item of equipment. (8 marks)c) Using the results of (a) and (b) as illustrations, compare and contrast

Laspeyre‟s and Paasche price index numbers. (4 marks)(Total: 20 marks)





QUESTION SIX

a) A company manufacturing a product known as 257 uses five components in its assembly.

The quantities and prices of the components used to produce a unit of K257 in 1982, 1983and 1984 are tabulated as follows:

COMPONENT 1982 1983 1984Quantity Prices Quantity Prices Quantity Prices

A 10 3.12 12 3.17 14 3.20B 6 11.49 7 11.58 5 11.67C 5 1.40 8 1.35 9 1.31D 9 2.15 9 2.14 10 2.63E 50 0.32 53 0.32 57 0.32

Required:i) Calculate Laspyere‟s type price index number for the cost of one unit of K257

for 1983 and 1984 based on 1982. (6 marks)

ii) Calculate Paasche type price index numbers for the cost of one unit of K257for 1983 and 1984 based on 1982. (6 marks)

iii) Compare and contrast the Laspeyre and Paasche price-index numbers youhave obtained in (i) and (ii) (3 marks)

A number of employers manufacturing plastic components used in plumbing have formedthemselves into an association for the purpose of negotiating with the trade union for thisindustrial sector.

The negotiations cover pay and contributions in this sector.

Required:Explain the usefulness of an index of Industrial Production and an index of retail prices to bothsides in a series of pay negotiations. (5 marks)

(Total: 20 marks)

QUESTION SEVEN

The data given below indicates the prices and production of some horticulatural products inCentral Territory:

Produce Production(1000 boxes)

Price per box (Shs)

1980 1990 1980 1990

Cabbages 48,600 62,000 100 150 Tomatoes 22,000 37,440 220 310Onions 47,040 61,430 180 200Spinach 43,110 55,720 130 170



Lesson Three 117


Required:Calculate the increase or decrease in prices from 1980 on the basis of the following indices:

a) Mean relativesb) Laspeyres index

c) Paasche indexd) Marshall – Hedgeworth indexe) Fishers index.




118 Measures of Relationships and Forecasting


LESSON FOUR

Measures of Relationships and Forecasting

- Correlation- Regression analysis- Multiple Linear Regression- Time series analysis and forecasting



Lesson Four 119


4.1 Correlation and Regression

Correlation This is an important statistical concept which refers to interrelationship or association between variables.

The purpose of studying correlation is for one to be able to establish a relationship, plan andcontrol the inputs (independent variables) and the output (dependent variables)In business one may be interested to establish whether there exists a relationship between the

i. Amount of fertilizer applied on a given farm and the resulting harvestii. Amount of experience one has and the corresponding performanceiii. Amount of money spent on advertisement and the expected incomes after sale of

the goods/service There are two methods that measure the degree of correlation between two variables these aredenoted by R and r.

(a) Coefficient of correlation denoted by r, this provides a measure of the strength ofassociation between two variables one the dependent variable the other the independent

variable r can range between +1 and – 1 for perfect positive correlation and perfectnegative correlation respectively with zero indicating no relation i.e. for perfect positivecorrelation y increase linearly with x increament.

(b) Rank correlation coefficient denoted by R is used to measure association between twosets of ranked or ordered data. R can also vary from +1, perfect positive rankcorrelation and -1 perfect negative rank correlation where O or any number near zerorepresenting no correlation.

SCATTER GRAPHS- A scatter graph is a graph which comprises of points which have been plotted but are

not joined by line segments- The pattern of the points will definitely reveal the types of relationship existing between

variables- The following sketch graphs will greatly assist in the interpretation of scatter graphs.





Perfect positive correlation

y

Dependant variable x

x

x

x

x

x

x

x

Independent variable

NB: For the above pattern, it is referred to as perfect because the points may easily berepresented by a single line graph e.g. when measuring relationship between volumes of salesand profits in a company, the more the company sales the higher the profits.

Perfect negative correlation

y x

Quantity sold x

X

x

x

x

x

x

x

10 20 Price X

This example considers volume of sale in relation to the price, the cheaper the goods the biggerthe sale.



Lesson Four 121


High positive correlation

y

Dependant variable xx

xx

xxxx

xxxx

xxx

xxxx

x

independent variable x

High negative correlation

y

quantity sold x

x

xxx

xxx

x

xx

xxx

price





No correlation

y

600 x x x x x

x x x

400 x x x x x

x x x x

200 x x x x x

x x x x0

10 20 30 40 50 x

h) Spurious Correlations- in some rare situations when plotting the data for x and y we may have a group showing

either positive correlation or – ve correlation but when you analyze the data for x and yin normal life there may be no convincing evidence that there is such a relationship. This implies therefore that the relationship only exists in theory and hence it is referredto as spurious or non sense e.g. when high passrates of student show high relation withincreased accidents.

Correlation coefficient- These are numerical measures of the correlations existing between the dependent and

the independent variables- These are better measures of correlation than scatter graphs (diagrams)- The range for correlation coefficients lies between +ve 1 and – ve 1. A correlation

coefficient of +1 implies that there is perfect positive correlation. A value of – ve showsthat there is perfect negative correlation. A value of 0 implies no correlation at all

- The following chart will be found useful in interpreting correlation coefficients



Lesson Four 123


__ 1.0 } Perfect +ve correlation

} High positive correlation

__ 0.5 }} Low positive correlation

__0 }

} Low negative correlation

__-0.5}

} High negative correlation

__-1.0} Perfect – ve correlation

There are usually two types of correlation coefficients normally used namely;-

Product Moment Coefficient (r)It gives an indication of the strength of the linear relationship between two variables.

r =2 22 2

n xy x y

n x x n y y

note that this formula can be rearranged to have different outlooks but the resultant is alwaysthe same.

Example The following data was observed and it is required to establish if there exists a relationshipbetween the two.X 15 24 25 30 35 40 45 65 70 75

Y 60 45 50 35 42 46 28 20 22 15





Solution Compute the product moment coefficient of correlation (r)X Y X 2 Y 2 XY15 60 225 3,600 90024 45 576 2,025 1,08025 50 625 2,500 1,250

30 35 900 1,225 1,05035 42 1,225 1,764 1,47040 46 1,600 2,116 1,84045 28 2,025 784 1,26065 20 4,225 400 1,30070 22 4,900 484 1,54075 15 5,625 225 1,125

424 X 363Y 2 21,926 X 2 15,123Y 12,815 XY

r =2 22 2

n xy x y

n x x n y y

r =2 2

10 12,815 424 363

10 21,926 424 10 15,123 363

=25,762

0.9339,484 19,461

The correlation coefficient thus indicates a strong negative linear association between the two variables.

Interpretation of r – Problems in interpreting r values

NOTE:

A high value of r (+0.9 or – 0.9) only shows a strong association between the two variablesbut doesn‟t imply that there is a causal relationship i.e. change in one variable causes changein the other it is possible to find two variables which produce a high calculated r yet theydon‟t have a causal relationship. This is known as spurious or nonsense correlation e.g. highpass rates in QT in Kenya and increased inflation in Asian countries. Also note that a low correlation coefficient doesn‟t imply lack of relation between variablesbut lack of linear relationship between the variables i.e. there could exist a curvilinearrelation. A further problem in interpretation arises from the fact that the r value here measures therelationship between a single independent variable and dependent variable, where as aparticular variable may be dependent on several independent variables (e.g. crop yield maybe dependent on fertilizer used, soil exhaustion, soil acidity level, season of the year, type ofseed etc.) in which case multiple correlation should be used instead.

The Rank Correlation Coefficient (R) Also known as the spearman rank correlation coefficient, its purpose is to establish whetherthere is any form of association between two variables where the variables arranged in a rankedform.



Lesson Four 125


R = 1 -2

2

6

1

d

n n

Where d = difference between the pairs of ranked values.

n = numbers of pairs of rankings

Example A group of 8 accountancy students are tested in Quantitative Techniques and Law II. Theirrankings in the two tests were.Student Q. T. ranking Law II ranking d d2 A 2 3 -1 1B 7 6 1 1C 6 4 2 4D 1 2 -1 1E 4 5 -1 1F 3 1 2 4

G 5 8 -3 9H 8 7 1 12 22d

d = Q. T. ranking – Law II ranking

R = 1 -2

2 2

6 6 221

1 8 8 1

d

n n

= 0.74

Thus we conclude that there is a reasonable agreement between student‟s performances in thetwo types of tests.

NOTE: in this example, if we are given the actual marks then we find r. R variesbetween +1 and -1.

Tied Rankings A slight adjustment to the formula is made if some students tie and have the same ranking theadjustment is

3

12

t t where t = number of tied rankings the adjusted formula becomes

R = 1 -

32

12

2

6

1

t t d

n n

Example Assume that in our previous example student E & F achieved equal marks in Q. T. and weregiven joint 3rd place.





Solution

Student Q. T. ranking Law II ranking d d2 A 2 3 -1 1B 7 6 1 1C 6 4 2 4

D 1 2 -1 1E 3 ½ 5 -1 ½ 2 ¼F 3 ½ 1 2 ½ 6 ¼G 5 8 - 3 9H 8 7 1 1

2 1226d

R = 1-

32

12

2

6

1

t t d

n n = 1 -

32 212 12

2

6 26 since 2

8 8 1t

= 0.68NOTE: It is conventional to show the shared rankings as above, i.e. E, & F take up the 3rd and

4th rank which are shared between the two as 3½ each.

ii. Coefficient of Determination This refers to the ratio of the explained variation to the total variation and is used to measure thestrength of the linear relationship. The stronger the linear relationship the closer the ratio will beto one.

Coefficient determination = Explained variation Total variation

Example (Rank Correlation Coefficient)

In a beauty competition 2 assessors were asked to rank the 10 contestants using the professionalassessment skills. The results obtained were given as shown in the table below

Contestants 1st assessor 2nd assessor A 6 5B 1 3C 3 4D 7 6E 8 7F 2 1G 4 8H 5 2

J 10 9K 9 10



Lesson Four 127


REQUIREDCalculate the rank correlation coefficient and hence comment briefly on the value obtained

d d2

A 6 5 1 1B 1 3 -2 4C 3 4 -1 1D 7 6 1 1E 8 7 1 1F 2 1 1 1G 4 8 -4 16H 5 2 3 9 J 10 9 +1 1K 9 10 -1 1

Σd2 = 36

∴ The rank correlation coefficient R

R = 1 -

2

2

6

1

d

n n

= 1 -2

6 36

10 10 1

= 1 -216

990

= 1 – 0.22

= 0.78

Comment: since the correlation is 0.78 it implies that there is high positive correlation betweenthe ranks awarded to the contestants. 0.78 > 0 and 0.78 > 0.5

ExampleContestant 1st

assessor2nd assessor d d2

A 1 2 -1 1B 5 (5.5) 3 2.5 6.25C 3 4 -1 1D 2 1 1 1E 4 5 -1 1F 5 (5.5) 6.5 -1 1G 7 6.5 -0.5 0.25H 8 8 0 0

Σd2 = 11.25

Required: Complete the rank correlation coefficient

∴ R = 1 -2

2

6

1

d

n n = 1 -

6 11.25

8 63

= 1 – 67.5

504





= 1 – 0.13

= 0.87

This implies high positive correlation

Example (Rank Correlation Coefficient)

Sometimes numerical data which refers to the quantifiable variables may be given after which arank correlation coefficient may be worked out.Is such a situation, the rank correlation coefficient will be determined after the given variableshave been converted into ranks. See the following example;

Candidates Math r Accounts r d d2P 92 1 67 5 -4 16Q 82 3 88 1 2 4R 60 5(5.5) 58 7(7.5) -2 4S 87 2 80 2 0 0 T 72 4 69 4 0 0U 60 5(5.5) 77 3 -2.50 6.25

V 52 8 58 7(7.5) 0.5 0.25 W 50 9 60 6 3 9X 47 10 32 10 0 0 Y 59 7 54 9 -2 4

Σd2 = 43.5

∴ Rank correlation r = 1 -2

2

6

1

d

n n

= 1 -2

6 43.5

10 10 1 = 1 –

261

990

= 0.74 (High positive correlation between mathematicsmarks and accounts)

Example(Product moment correlation) The following data was obtained during a social survey conducted in a given urban arearegarding the annual income of given families and the corresponding expenditures.

Family (x)Annualincome £ 000

(y)Annualexpenditure £ 000

xy x2 Y 2

A 420 360 151200 176400 129600B 380 390 148200 144400 152100C 520 510 265200 270400 260100

D 610 500 305000 372100 250000E 400 360 144000 160000 129600F 320 290 92800 102400 84100G 280 250 70000 78400 62500H 410 380 155800 168100 144400 J 380 240 91200 144400 57600K 300 270 81000 90000 72900 Total 4020 3550 1504400 1706600 1342900



Lesson Four 129


RequiredCalculate the product moment correlation coefficient briefly comment on the value obtained The produce moment correlation

r =2 22 2

n xy x y

n x x n y y

Workings:

X =4020

10= 402

3550355

10Y

r =22

10 1, 504, 400 4020 3550

10 1,706,600 4020 10 1,342,900 3550

= 0.89

Comment: The value obtained 0.89 suggests that the correlation between annual income andannual expenditure is high and positive. This implies that the more one earns the more one

spends.

4.2 REGRESSION- This is a concept, which refers to the changes which occur in the dependent variable as

a result of changes occurring on the independent variable.- Knowledge of regression is particularly very useful in business statistics where it is

necessary to consider the corresponding changes on dependant variables wheneverindependent variables change

- It should be noted that most business activities involve a dependent variable and eitherone or more independent variable. Therefore knowledge of regression will enable abusiness statistician to predict or estimate the expenditure value of a dependant variable when given an independent variable e.g. consider the above example for annual incomes

and annual expenditures. Using the regression techniques one can be able to determinethe estimated expenditure of a given family if the annual income is known and vice versa

- The general equation used in simple regression analysis is as followsy = a + bx

Where y = Dependant variablea= Interception y axis (constant)b = Slope on the y axisx = Independent variablei. The determination of the regression equation such as given above is normally

done by using a technique known as “the method of least squares‟. Regression equation of y on x i.e. y = a + bx





y x x Line of best fit

x x

x x

x x

x x

x x

x

The following sets of equations normally known as normal equation are used to determine theequation of the above regression line when given a set of data.

Σy = an + bΣx Σxy = aΣx + bΣx2

Where Σy = Sum of y values Σxy = sum of the product of x and yΣx = sum of x values Σx2= sum of the squares of the x valuesa = The intercept on the y axisb = Slope gradient line of y on x

NB: The above regression line is normally used in one way only i.e. it is used to estimate the y

values when the x values are given.Regression line of x on y i.e. x = a + by- The fact that regression lines can only be used in one way leads to what is known as a

regression paradox- This means that the regression lines are not ordinary mathematical line graphs which

may be used to estimate the x and y simultaneously- Therefore one has to be careful when using regression lines as it becomes necessary to

develop an equation for x and y before doing the estimation. The following example will illustrate how regression lines are used

Example An investment company advertised the sale of pieces of land at different prices. The following

table shows the pieces of land their acreage and costs



Lesson Four 131


Piece of land (x)Acreage Hectares (y) Cost £ 000 xy x2 A 2.3 230 529 5.29B 1.7 150 255 2.89C 4.2 450 1890 17.64D 3.3 310 1023 10.89

E 5.2 550 2860 27.04F 6.0 590 3540 36G 7.3 740 5402 53.29H 8.4 850 7140 70.56 J 5.6 530 2969 31.36

Σx =44.0 Σy = 4400 Σxy= 25607 Σx2 = 254.96

RequiredDetermine the regression equations of

i. y on x and hence estimate the cost of a piece of land with 4.5 hectaresii. Estimate the expected average if the piece of land costs £ 900,000

Σy = an + bΣxy

Σxy = a∑x + bΣx2

By substituting of the appropriate values in the above equations we have4400 = 9a + 44b …….. (i) 25607 = 44a + 254.96b ……..(ii)

By multiplying equation …. (i) by 44 and equation …… (ii) by 9 we have193600 = 396a + 1936b …….. (iii) 230463 = 396a + 2294.64b ……..(iv)

By subtraction of equation …. (iii) from equation …… (iv) we have36863 = 358.64b102.78 = b

by substituting for b in …….. (i)

4400 = 9a + 44( 102.78)4400 – 4522.32 = 9a – 122.32 = 9a-13.59 = a

Therefore the equation of the regression line of y on x is Y = 13.59 + 102.78x

When the acreage (hectares) is 4.5 then the cost(y) = -13.59 + (102.78 x 4.5)= 448.92= £ 448, 920

Note that Where the regression equation is given by

y= a + bx Where a is the intercept on the y axis and

b is the slope of the line or regression coefficientn is the sample size

then,

intercept a = y b x

n





Lesson Four 133


4.3 Multiple Linear Regression Models There are situations in which there is more than one factor which influence the dependent variableExampleCost of production per week in a large department depends on several factors;

i. Total numbers of hours worked

ii. Raw material used during the weekiii. Total number of items produced during the weekiv. Number of hours spent on repair and maintenance

It is sensible to use all the identified factors to predict department costsScatter diagram will not give the relationship between the various factors and total costs The linear model for multiple linear regression if of the type; (which is the line of best fit).

y = α + b1x1 +b2x2 +………… + bnxn We assume that errors or residuals are negligible.In order to choose between the models we examine the values of the multiple correlationcoefficient r and the standard deviation of the residuals α. A model which describes well the relationship between y and x‟s has multiple correlationcoefficient r close to ±1 and the value of α which is small.

ExampleOdino chemicals limited are aware that its power costs are semi variable cost and over the lastsix months these costs have shown the following relationship with a standard measure of output.

Month Output (standard units) Total power costs £ 0001 12 6.22 18 8.03 19 8.64 20 10.45 24 10.26 30 12.4

Requiredi. Using the method of least squares, determine an appropriate linear relationship

between total power costs and outputii. If total power costs are related to both output and time (as measured by the number

of the month) the following least squares regression equation is obtainedPower costs = 4.42 + (0.82) output + (0.10) month Where the regression coefficients (i.e. 0.82 and 0.10) have t values 2.64 and 0.60respectively and coefficient of multiple correlation amounts to 0.976Compare the relative merits of this fitted relationship with one you determine in (a).Explain (without doing any further analysis) how you might use the data to forecasttotal power costs in seven months.

Solutiona)

Output (x) Power costs (y) x2 y2 xy12 6.2 144 38.44 74.4018 8.0 324 64.00 144.0019 8.6 361 73.96 163.4020 10.4 400 108.16 208.0024 10.2 576 104.04 244.8030 12.4 900 153.76 372.00Σx = 123 Σ y = 55.8 Σx2 = 2705 Σy2 = 542.36 Σxy= 1,206.60





Lesson Four 135


The value of b and a will turn out to be 3.11 and 9.6 i.e. relationship will be of the formOutput = 9.6 + 3.11 × month

For this equation forecast for 7th month will beOutput = 9.6 + 3.11 × 7

= 9.6 + 21.77= 31.37 units

Using the equation , Power costs = 2.29 + 0.34 × output= 2.29 + 0.34 × 31.37= 2.29 + 10.67= 12.96 i.e. £ 12,960

Non Linear RelationshipsIf the scatter diagram and the correlation coefficient do not indicate linear relationship, then therelationship may be non – linear Two such relationships are of peculiar interest

x b y ab and y ax

Both of these can be reduced to linear model. Simple or multiple linear regression methods arethen used to determine the values of the coefficients

i. Exponential model x y ab

Take log of both sideslog y = log a + log bx log y = log a + xlog b

Let log y = Y and log a = A and log b = B

Thus we get Y = A + Bx. This is a linear regression model

ii. Geometric modelb y ax

using the same technique as abovelog y = log a + blog x Y = A + bX

Where Y = log y A = log aX = log x

Using linear regression technique (the method of least squares), it is possible to calculate the value of a and b

TIME SERIES AND ANALYSIS This is the mathematical or statistical analysis on past data arranged in a periodic sequence.Decision making and planning in an organization involves forecasting which is one of the time

series analysis.

Impediments in time series analysis Accuracy of data in reflectinga) Drastic changes e.g. in the advent of a major competitor, period of war or sudden change of

taste.b) For long term forecasting internal and external pressures makes historical data less effective.





1. Moving AveragePeriodical data e.g. monthly sales may have random fluctuation every month despite a generaltrend being evident. Moving average helps in smoothing away these random changes.

A moving average is the forecast for a period that takes the average of the previous periods.

Example: The table below represents company sales, calculate 3 and 6 monthly moving averages, for thedata

Months Sales January 1200February 1280March 1310 April 1270May 1190 June 1290 July 1410

August 1360September 1430October 1280November 1410December 1390

Solution. These are calculated as follows

April‟s forecast = Jan + Feb + Mar

3 =

1200+1280+1310

3

May‟s forecast =Feb + Mar + Apr

3 =

1280+1310+1270

3

And so on…

Similarly for 6 monthly moving average

July forecast = Jan + Feb + Mar+ Apr+ May + Jun

6 =

1200+1280+1310+1270+1190+1290

6

And so on… 3 months moving average 6 months moving average

April 1263May 1287 June 1257 July 1250 1257 August 1297 1292September 1353 1305October 1400 1325November 1357 1327December 1373 1363

Note: When plotting moving average on graphs the points are plotted as the midpoint of the period ofthe average, e.g. in our example the forecast for April (1263) is plotted on mid Feb.



Lesson Four 137


Characteristics of moving average1) The more the number of periods in the moving average, the greater the smoothing

effect.2) Different moving averages produce different forecasts.3) The more the randomness of data with underlying trend being constant then the

more the periods should be involved in the moving averages.

Limitations of moving averages.1) Equal weighing with disregard to how more recent data is more relevant.2) Moving average ignores data outside the period of the average thus it doesn‟t fully

utilise available data.3) Where there is an underlying seasonal variation, forecasting with unadjusted moving

average can be misleading.

2. Exponential smoothing This is a weighted moving average technique, it is given by:

New forecast = Old forecast + (Latest Observation – Old forecast) Where = Smoothing constant

This method involves automatic weighing of past data with weights that decrease exponentially with time.

ExampleUsing the previous example and smoothing constant 0.3 generate monthly forecasts

Months Sales Forecasts: = 0.3 January 1200February 1280 1200March 1310 1224 April 1270 1250May 1190 1256 June 1290 1233

July 1410 1250 August 1360 1283September 1430 1327October 1280 1358November 1410 1335December 1390 1357

SolutionSince there were no forecasts before January we take Jan to be the forecast for February.

Feb – 1200For March;

March forecast = Feb forecast + 0.3 ( Feb sales – Feb forecast)= 1200 + 0.3 (1280 – 1200)=1224

Note:

The value lies between 0 and 1. The higher the value, the more the forecast is sensitive to the current status.

Characteristics of exponential smoothing

More weight is given to the most recent data.









Lesson Four 141


13 27 52.66 5114 39 54.50 72

Year 4 15 92 56.34 16316 53 58.18 91

Trend estimates and percentage variations table.

Step 4

Average the percentage variations to find the average seasonal variations.Q1 Q2 Q3 Q4% % % %65 99 181 8055 106 180 7151 83 157 9451 72 163 91222 360 681 336

4 = 56% 90% 170% 84%

These then are the average variations expected from the trend for each of the quarters; forexample, on average the first quarter of each year will be 56% of the value of the trend. Becausethe variations have been averaged, the amounts over 100% (Q3 in this example). This can bechecked by adding the average and verifying that they total 400% thus:

56% + 90% + 170% + 84% = 400%.

On occasions, roundings in the calculations will make slight adjustments necessary to theaverage variations.

Step 5

Prepare final forecasts based on the trend line estimates from “trend estimates and percentages variation table” (i.e. 30.58, 32.42, etc) and the averaged seasonal variations from the table above.(i.e. 56%, 90%, 170% and 84%)

The seasonally adjusted forecast is calculated thus:

Seasonally adjusted forecast = Trend estimate × Seasonal variation%

X (quarters) Y (sales) Seasonally adjustedforecast

1 20 17.12 Year 1 2 32 29.18

3 62 58.244 29 30.32

5 21 21.24 Year 2 6 42 35.80

7 75 70.758 31 36.51





9 23 25.37 Year 3 10 39 42.43

11 77 83.2712 48 42.69

13 27 29.49

Year 4 14 39 49.0515 92 95.7816 53 48.87

Seasonally adjusted forecasts The forecasts are compared with the actual data to get some idea of how good extrapolatedforecasts might be. With further analysis they enable us to quantify the residual variations.

Extrapolation using the trend and seasonal factorsOnce the formulae above have been calculated, they can be used to forecast (extrapolate) futuresales. If it is required to estimate the sales for the next year (i.e. Quarters 17, 18, 19 and 20 in ourseries) this is done as follows:

Quarter 17 Basic trend = 28.74 + 1.84 (17)= 60.02

Seasonal adjustment for a first quarter = 56% Adjusted forecast = 60.02 × 56%

= 33.61

A similar process produces the following figures:

Adjusted forecasts Quarter 18 = 55.6719 = 108.2920 = 55.05

Notes: a) Time series decomposition is not an adaptive forecasting system like moving averages

and exponential smoothing.b) Forecasts produced by such an analysis should always be treated with caution.

Changing conditions and changing seasonal factors make long term forecasting adifficult task.

c) The above illustration has been an example of a multiplicative model. This is theseasonal variations were expressed in percentage or proportionate terms. Similar steps would have been necessary if the additive model had been used except that the variations from the trend would have been the absolute values. For example, the firsttwo variations would have been

Q1: 20 – 30.58 = absolute variation = -10.58Q2: 32 – 32.42 = absolute variation = - 0.42

And so on.

The absolute variations would have been averaged in the normal way to find theaverage absolute variation, whether + or -, and these values would have been used tomake the final seasonally adjusted forecasts.



Lesson Four 143


2. Trend and seasonal variation using moving averages When the correlation coefficient is low the method of calculating the regression linethrough the actual data points should not be used. This is because the regression line istoo sensitive to changes in the data values.

In such circumstances, calculating a regression line through the moving average trend

points is more robust and stable.

Example 1 is reworked below using this method and, because there are many similaritiesto the earlier method, only the key stages are shown.

x y 3 point movingaverage (1)

Trend line (2) Actual%

Trend 1 20 34.38 582 32 38 35.70 903 62 41 37.02 1674 29 37.3 38.34 765 21 30.7 39.66 53

6 42 46 40.98 1027 75 49.3 42.30 1778 31 43 43.62 719 23 31 44.94 5110 39 46.3 46.26 8411 77 54.7 47.58 16212 48 50.7 48.90 9813 27 38 50.22 5414 39 52.7 51.54 7615 92 61.3 52.86 17416 53 54.18 98

Trend estimates and percentage variations utilizing moving averages

The first three moving average is calculated as follows20 32 62

= 38 which is entered opposite period 23

The next calculated:

32 62 29 = 41, and so on

3 The regression line y = a + bx of the moving average values is calculated in the normal mannerand results in the following:

y = 33.06 + 1.32x

This is used to calculate the trend line:

e.g. For Period 1:y = 33.06 + 1.32(1) = 34.38For Period 2:y = 33.06 + 1.32 (2) = 35.70

The percentage variations are averaged as previously shown, resulting in the following values:





Q1 Q2 Q3 Q4 Average seasonal variation % 54 89 170 86

The trend line and the average seasonal variations are then used in a similar manner to thatpreviously described.

For example, to extrapolate future sales for the next year (i.e. quarters 17, 18, 19 and 20) is asfollows:

Quarter 17Forecast sales = (33.06 + 1.32(17)) × 0.54 = 29.97

A similar process produces the following figures:

Quarter 18 = 50.5719 = 98.8420 = 51.13

Forecast errors

Differences between actual results and predictions may arise from many reasons. They may arisefrom random influences, normal sampling errors, choice of the wrong forecasting system oralpha value or simply that the future conditions turn out to be radically different from the past. Whatever the cause(s) management wish to know the extent of the forecast errors and variousmethods exist to calculate these errors.

A commonly used technique, appropriate to time series, is to calculate the mean squared error of thedeviations between forecast and actual values then choose the forecasting system and/orparameters which gives the lowest value of mean squared errors, i.e. akin to the „least squares‟method of establishing a regression line.

Longer- term forecasting

Moving averages, exponential smoothing and decomposition methods tend to be used for shortto medium term forecasting. Longer term forecasting is usually less detailed and is normallyconcerned with forecasting the main trends on a year to year basis. Any of the techniques ofregression analysis described in the preceding chapters could be used depending on theassumptions about linearity or non- linearity, the number of independent variables and so on. The least squares regression approach is often used for trend forecasting.

Forecasting using least squares

Example 2Data have been kept of sales over the last seven years

Year 1 2 3 4 5 6 7Sales (in „000 units 14 17 15 23 18 22 27

It is required to forecast the sales for the 8th year



Lesson Four 145


Solution Years (x) Sales (y) xy x2

1 14 14 12 17 34 43 15 45 94 23 92 165 18 90 256 22 132 367 27 189 49

x=28 y = 136 xy=596 x2= 140

136 = 7a + 28b

596 = 28a + 140b

b = 1.86 And substituting in one of the equations we obtain

a = 12 Regression line = y = 12 + 1.86x

Or, Sales in (‘000s of units) = 12.00 + 1.86 (no of years)

We use this expression for forecasting, for 8th year sales = 12 + 1.86 (8)=26.88 i.e. 26,888 units






QUESTION ONE

a) What is meant by correlation?b) Why is the co-efficient of determination calculated?c) Define R. (coefficient of rank correlation)

QUESTION TWO

Explain the difference between Linear model, exponential model and geometric model, and write down their formulas

QUESTION THREE

An analysis of representatives‟ car expenses shows that the expenses are dependent on the milestravelled (x) and the type of journey (x). the general form is:

y = a + b1x1 + b2x2

Calculations have produced the following values (where y is expenses per month)

y = £86 + 0.37x1 + 0.08x2

r2x1 = 0.78

r2x2 = 0.16

R = 0.88

Interpret these values

QUESTION FOUR

Month Actual Sales (units) January 450February 440March 460 April 410May 380 June 400 July 370 August 360

September 410October 450November 470December 490 January 460

Required.Provide 3 month, 6 month and 12 month moving average.



Lesson Four 147


QUESTION FIVE

The manager of a company is preparing revenue plans for the last quarter of 1993/94 and forthe first three quarters of 1994/5. The data below refer to one of the main products:

Revenue April-June July-Sept Oct-Dec Jan-MarchQuarter 1 Quarter 2 Quarter 3 Quarter 4

£‘000 £‘000 £‘000 £‘000 £‘000 1990/91 49 37 58 671991/92 50 38 59 681992/93 51 40 60 701993/94 50 42 61 -

Required:a) Calculate the four-quarterly moving average trend for this set of data.b) Calculate the seasonal factors using either the additive model or the multiplicative model,

but not both.c) Explain, but do not calculate how you would use the results in parts (a) and (b) of this

question to forecast the revenue for the last quarter of 1993/4 and for the first three

quarters of 1994/95.





Lesson Four 149


c) Using the least square technique, calculate the values of a an b in the equation y = a + bx inorder to predict costs given the output, and explain the meaning of the calculated values.

QUESTION EIGHT

Your company has been selling data base and spreadsheets for the last four years and has foundthe business to vary with season. The quarterly sales figures for the last four years are shown intable 6b1 and table 6b2 shows the deviation from the trend at the appropriate periods

Table 6b1

Quarterly sales in £ 000s

Year Q1 Q2 Q3 Q41983 360 530 3541984 304 430 750 3951985 340 500 660 5091986 374 590 710 5211987 440

Table 6b2

Seasonal deviation from trend in £ 000s

Year Q1 Q2 Q3 Q41983 -421984 -128 -37 276 -931985 -145 12 153 -151986 -165 43 153

Requiredi. establish the trend figures from the data in the two tables

ii. establish the seasonal variations for the four year periodiii. using your results from parts (i) and (ii) forecast sales for 1987 quarter 2

QUESTION NINE1. The directors of your company wish to make a serious study of the heating costs of the ****

block. The data for the last sixteen quarterly periods are tabulated as follows.Heating costs in £

Quarter

Year Q1 Q2 Q3 Q41980 15601981 1730 1554 1504 16301982 1950 1595 1540 17001983 1860 1709 1574 17901984 1910 1721 1640





Requireda) Assuming the additive model calculate the trend of heat costs using the method of

moving averagesb) Estimate the seasonal deviations from trendc) Estimate the heating costs for quarter IV of 1984 and comment on any factors

affecting the reliability of your forecast




Lesson Four 151


COMPREHESIVE ASSIGNMENT TWOWork out these question for three hours (exam condition) then hand them in to DLC for marking


SECTION I

QUESTION ONE

a) In the just concluded higher education seminar at Shoppers Paradise, Nairobi, the College ofBusiness Administration of Highland University states in some of its promotional materialthat the average graduate of the college earns over Sh.3 million a year. Assume, forsimplicity, that only four people have graduated to date; Sam, Tom, Jackie and Mary whoearn Sh.1.6 million, Sh.1.8 million, Sh.1.8 million and Sh.2 million respectively in a year.

Required:Compute the mean, median and the mode. Is the college‟s claim correct? (3 marks)

b) Let us change our assumption about the number of graduates in (a) above and insteadassume that five people have graduated. They consist of the four listed above and Suki whoearns Sh.5.3 million per year.

Required:Compute the mean, median and the mode for the five graduates. Is the college‟s claimcorrect? (3 marks)

c) Changing our assumption one more time about the number of graduates, let us assume thatsix people have graduated. They consist of the four original ones, Suki, who earns Sh.5.3million a year; and Bob who earns Sh.6.7 million a year.

Required:i) Compute the mean, median and mode for the six graduates. Is the college‟s claim correct?

(2 marks)ii) Comment on what happened to the mean, median and mode as you moved from part (a)

to (b) to (c) of this problem. (2 marks)iii) What do the results in (ii) above suggest about the relative stability of the mean, median

and the mode? (2 marks)iv) How do you feel about the ethics of this college in claiming that their average graduates

earn over Sh.3 million a year? (2 marks)

d) Genuine athletic Company Ltd., manufactures weight-lifting equipment. The company‟stop-of-the-line equipment are used in events such as the Olympics and other prestigious

professional weight-lifting competitions. Consequently, it is very important that if a barbellplate is stamped say, “50 kilogrammes”, it weighs very close to 50 kilogrammes. In addition,a barbell plate must have a hole just slightly larger than 1 centimeter in diameter so that it will slip onto the 1 centimeter diameter bar easily but fit smoothly when it is in place.

A recent sampling of barbell plates of 10 and 50 kilogrammes revealed the followinginformation:





i) Weights of 10 kilogram plates had an arithmetic mean of 10.013 kilogrammes and astandard deviation of 0.124 kilogrammes.

ii) Weights of 50 kilogram plates had an arithmetic mean of 50.032 kilogrammes and astandard deviation of 0.465 kilogrammes.

iii) Diameters of holes in the 10 kilogram plates had an arithmetic mean of 1.22 centimetersand a standard deviation of 0.187 centimeters.

iv) Diameter of holes in the 50 kilogram plates had an arithmetic mean of 1.20 centimetersand a standard deviation of 0.183 centimeters.

Required:Determine whether the production process associated with one size of barbell plantproduced more variable results than the production process associated with the other size. (6 marks)

(Total: 20 marks)

QUESTION TWO

a) The index of industrial production in the Utopian country by July 2001 is given below:

July 2001 IndexSector Weight (1994 = 100)Mining and quarrying 41 361Manufacturing:Food, drink and tobacco 77 106Chemicals 66 109Metal 47 72Engineering 298 86 Textiles 67 70Other manufacturing 142 91Construction 182 84- Gas, electricity and water 80 115

Required:i) Calculate the index of industrial production for all industries and manufacturing industries.

(6 marks)ii) Comment on your results. (4 marks)

b) Explain some of the uses of index numbers. (5 marks)c) What are some of the limitations of index numbers? (5 marks)

(Total: 20 marks)QUESTION THREE

Leisure Publishers Ltd. recently published 20 romantic novels by 20 different authors. Salesranged from just over 5,000 copies for one novel about 24,000 copies for another novel. Before

publishing, each novel had been assessed by a reader who had given it a rating between 1 and 10. The managing director suspects that the main influence on sales is the cover of the book. Theillustrations on the front covers were drawn either by artist A or artist B. the short descriptionon the back cover of the novel was written by either editor C or editor D.

A multiple regression analysis was done using the following variables: Y Sales (million of shillings)X 1 1 if front cover is by artist A

2 if front cover is by artist B



Lesson Four 153


X 2 readers‟ rating X 1 1 if the short description of the novel is by editor C

2 if the short description of the novel is by editor D

The computer analysis produced the following results:

Correlation coefficient r = 0.921265Standard error of estimate = 2.04485

Analysis of varianceDegrees of freedom Sum of squares Mean square F ratio

Regression 3 375.37 125.12 29.923Residue 16 66.903 1.1814

Individual analysis of variables Variable Coefficient Standard error F ValueConstant 15.7588 2.54389 38.3751 -6.25485 0.961897 42.284

2 0.0851136 0.298272 0.0814283 5.86599 0.922233 40.457741

Correlation coefficients1 -0.307729 0 -0.674104

1 0.123094 0.3108381 0.627329

1

Required:a) The regression equation. (3 marks)b) Does the regression analysis provide useful information? Explain. (3 marks)c) Explain whether the covers were more important for sales than known quality of the novels.

(4 marks)d) State with 95% confidence the difference in sales of a novel if its cover illustrations were

done by artist B instead of artist A. (5 marks)e) State with 95% confidence the difference in sales of a novel if its short description was by

editor D and not editor C. (5 marks)(Total: 20 marks)

QUESTION FOUR

a) Explain the difference between regression and correlation analysis . (4 marks)

b) Explain why the existence of a significant correlation does not imply causation. (2 marks)

c) A bakery bakes cakes under the brand name „super cakes‟. Irene Juma, the manageress does

not know the cost of each cake. She therefore gathers data on the total cost of each day‟sproduction for the last 10 days. The results are shown in the table below;





Day Number of cakes(‘00’ units)

Total cost (Sh.‘000’)

1 22.5 23.02 21.0 21.6

3 27.5 23.34 21.5 24.05 30.0 28.26 20.0 22.47 24.0 23.18 26.5 25.39 18.3 20.110 17.0 16.5

Required:i) Estimate the total cost function using the ordinary least squares method. State the fixed

cost and unit cost. (11 marks)

ii) If each cake is sold at Sh.10, determine the break even number of cakes. (3 marks)(Total: 20 marks)

QUESTION FIVE

Differentiate between additive model and the multiplicative model as used in time series analysis. (4 marks)

The sales data of XYZ Ltd. (in millions of shillings) for the years 2001 and 2004 inclusive are asgiven below:

Quarter Year 1 2 3 4

2001 40 64 124 582002 42 84 150 622003 46 78 154 962004 54 78 184 106

Required:i) The trend in the data using the least squares method. (8 marks)ii) The estimated sales for each quarter of year 2004. (4 marks)iii) The percentage variation of each quarter‟s actual sales for year 2004. (4 marks)

(Total: 20 marks)SECTION II

QUESTION SIX

a) Explain the following terms as used in index numbers:i) Price index (2 marks)ii) Quantity index (2 marks)iii) Composite index (2 marks)iv) Value index (2 marks)

b) The following prices and quantities reflect the average weekly consumption pattern of acertain family for the years 2001 and 2002.



Lesson Four 155


Year 2001 Year 2002Price (p0 ) Quantity (q 0 ) Price (p1 ) Quantity (q 1 )

Item Sh. Sh.Oranges (kg) 15 2 25 1Milk (Litres) 30 2 35 2

Bread (Loafs) 30 3 40 3Eggs (Dozens) 50 1 65 1

Required:i) Price relatives for each item (4 marks)ii) Laspeyres price index (4 marks)iii) Paasche price index (4 marks)

(Total: 20 marks)QUESTION SEVEN

Explain three methods of fitting a trend in time series analysis. (6 marks)

The quarterly sales data for Chuce hardware are given below:

Quarter Year 1 2 3 4

(Sh. Million) (Sh. Million) (Sh. Million) (Sh. Million)2000 8.5 10.4 7.5 11.82001 9.5 12.2 8.8 13.62002 10.4 13.5 9.7 13.12003 9.5 11.7 8.4 12.92004 10.9 13.7 10.1 15.0

Required:(a) The centred four-quarter moving averages. (6 marks)

(b) The specific seasonal variation for each quarter (3 marks)(c) The typical seasonal indices (3 marks)(d) Explain the third quarter typical seasonal index (2 marks)

(Total: 20 marks)QUESTION EIGHT a) A machine produces circular bolts and as a quality control test, 250 bolts were selected

randomly and the diameter of their heads measured as follows:

Diameter of head (cm) Number of components0.9747 - 0.9749 20.9750 - 0.9752 60.9753 - 0.9755 8

0.9756 - 0.9758 150.9759 - 0.9761 420.9762 - 0.9764 680.9765 - 0.9767 490.9768 - 0.9770 250.9771 - 0.9773 180.9774 - 0.9776 120.9777 - 0.9779 40.9780 - 0.9782 1





Required:b)

i) Determine whether the customer is getting reasonable value if the label on the circularbolt advertises that the average diameter of the head is 0.97642 cm. (8 marks)

ii) In what situation would weighted mean be used? (3 marks)iii) Describe briefly how to estimate the median on a grouped frequency distribution

graphically? (3 marks)iv) Why is the mode not used extensively in statistical analysis? (3 marks)“The standard deviation is the natural partner to the mean”. Explain (3 marks)

(Total: 20 marks)



Probability 157


LESSON FIVE

Probability

Contents- Probability theory- Bayes Theorem and conditional probability- Permutations and combinations- Discrete probability distributions- Continuous probability distribution



158 Lesson Five


5.1 PROBABILITY- Probability is a very popular concept in business management. This is because it covers

the risks which may be involved in certain business situations. It is a fact that when abusiness investment is being arranged, the outcome is usually uncertain. Therefore theconcept of probability may be used to describe the degree of uncertainty of a

particular business outcome Probability may therefore be defied as the chances of a given event occurring.Numerically, probability values range between 0 and 1. a probability of 0 implies thatthe event cannot occur at all. A probability of 1 implies that the event will certainlyoccur. Therefore other events have their probabilities with values lying between 0 and 1

- The formular used to determine probability is as follow

Probability (x) =outcomes Total

outcomesFavourable

n

r

Application of Probability in Business1. Business games of chance e.g. Raffles Lotteries e.t.c.

2. Insurance firms: this is usually done when a new client or property is being insured. Thecompany has to be certain about the chances of the insured risks occurring.

3. Business decision making regarding viability of projects thus the projects with a greaterprobability has greater chances.

Example A bag contains 80 balls of which 20 are red, 25 are blue and 35 are white. A ball is picked atrandom what is the probability that the ball picked is:(i) Red ball(ii) Black ball(iii) Red or Blue ball.

Solution

(i) Probability of a red ball = R Pbag theinballsof number Total

bag theinballsredof Number

=4

1

80

20

(ii) Probability of black ball = BPballsof number Total

bag theinballsblack of Number

= 080

0

(iii) P(R or B) =80

25

80

20

80

25or

80

20

=16

9

Note: in probability or is replaced by a plus (+) sign. See addition rule.

Common termsEvents: an event is a possible outcome of an experiment or a result of a trial or an observation.



Probability 159


Mutually exclusive events A set of events is said to be mutually exclusive if the occurance of any one of the eventsprecludes the occurrence of any of the other events e.g. when tossing a coin, the events are ahead or a tail these are said to be mutually exclusive since the occurrence of heads for instanceimplies that tails cannot and has not occurred.It can be represented in venn diagram as.

E1

E2

= Ø

E1 E2 ≠ Ø

Consider a survey in which a random sample of registered voters is selected. For each voterselected their sex and political party affiliation are noted. The events “KANU” and “woman”are not mutually exclusive because the selection of KANU does not preclude the possibly thatthe voter is also a woman.

Independent EventsEvents are said to be independent when the occurance of any of the events does not affect theoccurrence of the other(s).

e.g. the outcome of tossing a coin is independent of the outcome of the preceeding orsucceeding toss.

ExampleFrom a pack of playing cards what is the probability of;(i) Picking either a „Diamond‟ or a „Heart‟ → mutually exclusive (ii) Picking eigher a „Flower‟ or an „Ace‟ → indepent events

Solutions.(i) P(Diamond or Heart)

= P(Diamond) + P(Heart)

E1 E 2

E1 E 2

Non-mutually exclusive events(independent events)



160 Lesson Five


=52

26

52

13

52

13

= 0.5

(ii) P(Flower or Ace)= P(Flower) + P(Ace) – P(Flower and Ace)

=52

1

52

4

52

13

=52

4 = 0.31

Note: that the formula used incase of independent events is different to the one ofmutually exclusive.

Rules of Probability(a) Additional Rule – This rule is used to calculate the probability of two or more mutually

exclusive events. In such circumstances the probability of the separate events must be

added.

Example What is the probability of throwing a 3 or a 6 with a throw of a die?

Solution

P(throwing a 3 or a 6) =3

16

16

1

(b) Multiplicative rule This is used when there is a string of independent events for which individualprobability is known and it is required to know the overall probability.

Example What is the probability of a 3 and a 6 with two throws of a die?

SolutionP(throwing a 3) and P(6)

= P(3) and P(6) =36

16

16

1

Note: 1) In probability „and‟ is replaced by „x‟ – multiplication.2) P(x) and P(y) ≠ P(x and y) note that these two are different. The first

implies P(x) happening and P(y), but if the order of which happenedfirst is unimportant then we have p(x and y).

In the example above:P(3) and P(6) =

361

butP(3 and 6) = P(3 followed by 6) or P(6 followed by 3)

= [P(3) P(6)] or [P(6) P(3)]

=18

136

136

1





162 Lesson Five


P(CSGP) = 0.9 × 0.95 = 0.855P(CSBP) = 0.9 × 0.05 = 0.045P(ISGP) = 0.1 × 0.3 = 0.03P(ISBP) = 0.1 × 0.7 = 0.07

1.00

- Probability of getting a good part (GP) = CSGP or ISGP= CSGP + ISGP= 0.855 + 0.03 = 0.885

Note: Good parts may be produced when the machine is correctly set up and also when itsincorrectly setup. In 1000 trials, 855 occasions when its correctly setup and good parts produced(CSGP) and 30 occasions when its incorrectly setup and good parts produced (ISGP).

- Probability that the machine is correctly set up after getting a good part.

= 0.966

0.885

0.855

P(GP)

P(CSGP)

outcomespossible Total

outcomesfavourableof Number

Or

= P(CS|GP) = 0.9660.885

0.855

P(GP)

P(CSGP)

ExampleIn a class of 100 students, 36 are male and studying accounting, 9 are male but not studyingaccounting, 42 are female and studying accounting, 13 are female and are not studyingaccounting.Use these data to deduce probabilities concerning a student drawn at random.

Solution: Accounting A Not accounting A Total

Male M 36 9 45Female F 42 13 55 Total 78 22 100

P(M) = 0.45100

45

P(F) = 0.55100

55

P(A) = 0.78100

78

P A = 0.22100

22

P(M and A) = P(A and M) =100

36 = 0.36

P(M and A ) = 0.09

P(F and A ) = 0.13



Probability 163


These probabilities can be express differently as;

P(M) = P(M and A) or P(M and A )= 0.36 + 0.09 = 0.45

P(F) = P(F and A) or P(F and A )

= 0.42 + 0.13 = 0.55P(A) = P(A and M) + P(A and F) = 0.36 + 0.42 = 0.78

P A = P( A and M) + P( A and F) = 0.09 + 0.13 = 0.22

Now calculate the probability that a student is studying accounting given that he is male.

This is a conditional probability given as P(A|M)

P(A|M) = 0.800.45

0.36

P(M)

M)andP(A

From the formula above we get that,P(A and M) = P(M) P(A|M) ……………….. (i) Note that P(A|M) ≠ P(M|A)

Since P(M|A) =P(A)

Mand AP this is known as the Bayes‟ rule.

Bayes‟ rule/Theorem

This rule or theorem is given by

P(A|B) =P(B)

ABP AP

It‟s used frequently in decision making where information is given the in form of conditionalprobabilities and the reverse of these probabilities must be found.

Example Analysis of questionnaire completed by holiday makers showed that 0.75 classified their holidayas good at Malindi. The probability of hot weather in the resort is 0.6. If the probability ofregarding holiday as good given hot weather is 0.9, what is the probability that there was hot weather if a holiday maker considers his holiday good?

Solution

P(A|B) =P(B)

ABP AP

Let H = hot weatherG = GoodP(G) = 0.75 P(H) = 0.6 and P(G|H) = 0.9 (Probability of regard holiday asgood given hot weather)

Now the question requires us to get



164 Lesson Five


P(H|G) = Probability of (there was) hot weather given that the holiday has been ratedas good).

=0.75

0.90.6

P(G)

HGPHP

= 0.72.

Worked examples on probability1. A machine comprises of 3 transformers A, B and C. The machine may operate if at least 2transformers are working. The probability of each transformer working are given as shownbelow;

P(A) = 0.6, P(B) = 0.5, P(C) = 0.7 A mechanical engineer went to inspect the working conditions of those transformers. Find theprobabilities of having the following outcomes

i. Only one transformer operatingii. Two transformers are operatingiii. All three transformers are operatingiv. None is operating v. At least 2 are operating vi. At most 2 are operating

Solution

P(A) =0.6 P( A ) = 0.4 P(B) = 0.5 P(~B) = 0.5

P(C) = 0.7 P( C ) = 0.3

i. P(only one transformer is operating) is given by the following possibilities1st 2nd 3rd

P (A B C ) = 0.6 x 0.5 x 0.3 = 0.09

P ( A B C ) = 0.4 x 0.5 x 0.3 = 0.06

P ( A B C) = 0.4 x 0.5 x 0.7 = 0.14∴ P(Only one transformer working)

= 0.09 + 0.06 + 0.14 = 0.29

ii. P(only two transformers are operating) is given by the following possibilities.1st 2nd 3rd

P (A B C ) = 0.6 x 0.5 x 0.3 = 0.09

P (A B C) = 0.6 x 0.5 x 0.7 = 0.21

P ( A B C) = 0.4 x 0.5 x 0.7 = 0.14

∴ P(Only two transformers are operating)= 0.09 + 0.21 + 0.14 = 0.44

iii. P(all the three transformers are operating).

= P(A) x P(B) x P(C)

= 0.6 x 0.5 x 0.7

= 0.21

iv. P(none of the transformers is operating).



Probability 165


= P( A ) x P( B ) x P( C )

= 0.4 x 0.5 x 0.3

= 0.06

v. P(at least 2 working).= P(exactly 2 working) + P(all three working)

= 0.44 + 0.21

= 0.65

vi. P(at most 2 working).= P(Zero working) + P(one working) + P(two working)= 0.06 + 0.29 + 0.44= 0.79

5.3 Permutations and Combinations

Definition

Permutation- This is an order arrangement of items in which the order must be strictly observed

Example

Let x, y and z be any three items. Arrange these in all possible permutations

1st 2nd 3rd X Y Z

X Z Y Y X Z Six different permutations Y Z XZ Y XZ X Y

NB: The above 6 permutations are the maximum one can ever obtain in a situation where thereare only 3 items but if the number of items exceeds 3 then determining the no. of permutationsby outlining as done above may be cumbersome. Therefore we use a special formula todetermine such permutations. The formula is given below

The number of permutations of „r‟ items taken from a sample of „n‟ items may be provided as

n

Pr = !r -n

!nwhere; ! = factorial

e.g.

i. 3P3 =!33

!3

=!0

123 note; 0! = 1



166 Lesson Five


=1

6= 6

ii. 5P3 =!3-5

!5

=21

12345

= 60

iii. 7P5 =!5-7

!7

=12

1234567

=2

5040

= 2520

Example There are 6 contestants for the post of chairman secretary and treasurer. These positions can befilled by any of the 6. Find the possible no. of ways in which the 3 positions may be filled.SolutionChairman Secretary Treasurer

6 5 4 Therefore the no of ways of filing the three positions is 6 x 5 x 4 = 120

6P3 =!3-6

!6

=123

123456

=6

720

= 120

Combinations

Definition A combination is a group of times in which order is not important.For a combination to hold at any given time it must comprise of the same items but if a newitem is added to the group or removed from the group then we have a new combination

Example3 items x, y and z will have 6 different permutations but only one combination.

The following formular is usually used to determine the no. of combinations in a given situation.!

! !

n

r

nC

r n r

Example

i. 8

7

8!

7! 8 7 !C



Probability 167


8! 8 7!

7!1! 1 7!

= 8

ii. 6

4

6!

4! 6 4 !C

6! 6 5 4!

4!2! 4! 2 1

= 15

iii. 8

3

8!

3!5!C

8 7 6 5!

3 2 1 5!

= 56

Example There is a committee to be selected comprising of 5 people from a group of 5 men and 6 women. If the selection is randomly done. Find the possibility of having the followingpossibilities (combinations)

i. Three men and two womenii. At least one man and at least one woman must be in the committeeiii. One particular man and one particular woman must not be in the committee (one

man four women)

Solutioni. The committee size = 5 people

The group size = 5m + 6w

∴ assuming no restrictions the committee can be selected in 11C5 the committee has to consist of 3m & 2w

∴ these may be selected as follows.5C3 × 6C2 P( committee 3m and 2w)

5 6

3 2

115

C C

C note that this formula can be fed directly to your scientific calculator and attain a solution.

5! 6!

3!2! 4!2!

11!5!6!

5 4 3 2 1 6 5 4! 5 4 3 2 1 6!

3 2 1 2 1 2 1 4! 11 10 9 8 7 6!

=77

27



168 Lesson Five


ii. P( at least one man and at least one woman must be in the committee ) The no. of possible combinations of selecting the committee without any woman = 5C5 The probability of having a committee of five men only

5

5

11

5

1

462

C

C

the probability of having a committee of five women only

6

5

11

5

6!

5!1!11!

5!6!

C

C

6 5! 5!6!

5!1! 11 10 9 8 7 6!

=77

1

∴ P( at least one man and at least one woman )= 1 – {P( no man ) + P( no woman )}

= 1 – 462

1

77

1

= 1 – 462

16

= 1 – 462

7

=465

455

iii. P(one particular man and one particular woman must not be in the committee would be determined as follows

The group size = 5m + 6wCommittee size = 5 people

Actual groups size from which toSelect the committee = 4m + 5wCommittee = 1m + 4w

The committee may be selected in 9C5 The one man may be selected in 4C1 ways The four women may be selected in 5C4 ways

∴ P( committee of 4w1man ).5 4

4 1

9

5

C C

C

5! 4!

4!1! 1!3!9!

4!5!



Probability 169


5 4! 4 3! 4!5!

1!4! 1 3! 9 8 7 6 5!

=63

10

5.4 DISCRETE PROBABILITY DISTRIBUTIONS

BINOMIAL PROBABILITY DISTRIBUTIONBinomial probability distribution is a set of probabilities for discrete events. Discrete events arethose whose results or outcomes can be counted. Binomial probabilities are commonlyencountered in business situations e.g. in quality control activities the binomial probabilities arefrequently used especially when determining the probability of having a certain no. of defectiveitems in a given consignment.

- The binomial probability distribution is usually characterized by the fact that thebinomial events have to fulfill the following properties

i. Each event has 2 possible outcomes only known as success or failureii. The probability of each outcome is independent of the previous outcomesiii. The sample size is generally fixediv. The probabilities of success and failure tend to approach 0.5 if the sample size

increases (in the event when an unbiased coin is thrown a number of times) v. The probabilities are given by the following equation

9

5 1 n r r P r C p p

!1

!

n r r n p p

r n r

Where p = Probability of successr = no. of successesn = sample size

q = 1 – P = Probability of failure

Example 1 A medical survey was conducted in order to establish the proportion of the population which was infected with cancer. The results indicated that 40% of the population were suffering fromthe disease. A sample of 6 people was later taken and examined for the disease. Find the probability that thefollowing outcomes were observed

a) Only one person had the diseaseb) Exactly two people had the diseasec) At most two people had the diseased) At least two people had the disease

e)

Three or four people had the disease

SolutionP(a persona having cancer) = 40% = 0.4 = PP(a person not having cancer) = 60% = 0.6 = 1 – p = qa) P(only one person having cancer)

= 6C1 (0.4)(0.6)5

=!1!5

!6(0.4)1(0.6)5



170 Lesson Five


= 0.1866Note that from the formula

nCrprq n-r: n = sample size = 6p = 0.4r = 1 = only one person having cancer

b) P( 2 people had the disease )

= 6C2 (0.4)2 (0.6)4 =

!2!4

!6(0.4) 2 (0.6)5

=12!4

!456 (0.4) 2 (0.6)5

= 15 × (0.4) 2 (0.6)5

= 0.311

c) P( at most 2 ) = P(0) + P(1) + P(2) = P(0) or P(1) or P(2)

So we calculate the probability of each and add them up.

P(0) = P(nobody having cancer)= 6C0 (0.4) 0(0.6)6

=!6!0

!6(0.4) 0(0.6)6

= (0.6)6 = 0.0467

The probabilities of P(1) and P(2) have been worked out in part (a) and (b)

Therefore P( at most 2 ) = 0.0467 + 0.1866 + 0.311 = 0.5443

d) P( at least 2 )

= P(2) + P(3) + P(4) + P(5) + P(6)

= 1 – [P(0) + P(1)] This is a shorter way of working out the solution since

[P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1]

= 1 – (0.0467 + 0.1866)

= 0.7667e) P( 3 or 4 people had the disease )

= P(3) +P(4)

= 6C3(0.4)3(0.6)3 + 6C4(0.4)4(0.6)2

=!3!3

!6(0.4) 3(0.6)3 +

!4 !2

!6 (0.4) 4(0.6)2

= 6 × 5 × 4 × 3! (0.4) 3(0.6)3 + 6 × 5 × 4! (0.4) 4(0.6)2

3 × 2 × 1 × 3! 2 × 1 × 4!

= 20(0.4)3(0.6)3 + 15(0.4)4(0.6)2

= (20 × 0.013824) + (15 × 0.009216)

= 0.27648 + 0.13824

= 0.41472



Probability 171


Example 2 An insurance company takes a keen interest in the age at which a person is insured.Consequently a survey conducted on prospective clients indicated that for clients having the

same age the probability that they will be alive in 30 years time is3

2 . This probability was

established using the actuarial tables. If a sample of 5 people was insured now, find the

probability of having the following possible outcomes in 30 yearsa) All are aliveb) At least 3 are alivec) At most one is alived) None is alivee) At least 1 is alive

Sample size = 52 13 3

where as P alive p P not alive q

a)

5 05 2 15 3 3

5 02 13 3

523

5

5!5!0!

32

243

P all alive P r

C

b)

3 25 2 13 3 3

5 0 3 22 1 2 13 3 3 3

3 3

3 4 5 3 4 5

3

5! 10

3!2!

80

243

325

243

80 80 32

3 243 243 243

192

243

P atleast alive P r

P orP orP P P P

P C

P

P

4 15 2 14 3 3

4 1 42 1 2 13 3 3 3

4

5! 5 4! 4!1! 4! 1

80

243

P C



172 Lesson Five


c)

0 5 1 45 52 1 2 10 13 3 3 3

5 41 2 13 3 3

1 1

0 1

5! 5!

0!5! 1!4!

1 10

243 243

11

243

P atmost is alive P r

P P

C C

d)

e)

0 55 2 10 3 3

0

1

243

P none is alive P r

C

1 1

1

11

243

242

243

P atleast alive P r

P none alive

POISSON PROBABILITY DISTRIBUTION

- This is a set of probabilities which is obtained for discrete events which are described asbeing rare. Occasions similar to binominal distribution but have very low probabilitiesand large sample size.

Examples of such events in business are as follows:i. Telephone congestion at midnightii. Traffic jams at certain roads at 9 o‟clock at night iii. Sales boomiv. Attaining an age of 100 years (Centureon)- Poisson probabilities are frequently applied in business situations in order to determine

the numerical probabilities of such events occurring.- The formula used to determine such probabilities is as follows

!

xe P x

x Where x = No. of successes

⋋ = mean no. of the successes in the sample ( ⋋ = np)e = 2.718

Example 1 A manufacturer assures his customers that the probability of having defective item is 0.005. Asample of 1000 items was inspected. Find the probabilities of having the following possibleoutcomes

i. Only one is defectiveii. At most 2 defectiveiii. More than 3 defective

P(x) =!x

λe xλ

( ⋋ = np = 1000 × 0.005) = 5i. P(only one is defective) = P(1) = P(x = 1)

=1!

52.718 15

Note that 2.718-5=52.718

1



Probability 173


=52.718

5

=33.148

5

= 0.0337

ii.

P(at most 2 defective) = P(x ≤ 2) = P(0) + P(1) + P(2)

P(x = 0) =!0

505e

= 2.718-5

=5718.2

1

=336.148

1

= 0.00674

P(1) = 0.0337

P(2) =!2

5718.2 25

=148.3362

25

= 0.08427

P(x≤2) = 0.00674 + 0.0337 + 0.08427

= 0.012471

iii. P(more than 3 defective) = P(x > 3)

= 1 – 3P2P1P0P

BINOMIAL MATHEMATICAL PROPERTIES1. The mean or expected value = n × p = np

Where; n = Sample Sizep = Probability of success

2. The variance = npqWhere ; q = probability of failure = 1 - p

3. The standard deviation = npq

Example A firm is manufacturing 45,000 units of nuts. The probability of having a defective nut is 0.15Calculate the following

i. The expected no. of defective nutsii. The variance and standard deviation of the defective nuts in a daily consignment of

45,000

SolutionSample size n = 45,000P(defective) = 0.15 = pP(non defective) = 0.85 = q

i. ∴ the expected no of defective nuts= 45,000 × 0.15 = 6,750

ii. The variance = npq= 45000 × 0.85 × 0.15



174 Lesson Five


= 5737.50

The standard deviation = npq

= 5737.50 = 75.74

POISSON MATHEMATICAL PROPERTIES1. The mean or expected v alue = np = λ

Where ; n = Sample Sizep = Probability of success

2. The variance = np = ⋋

3. Standard deviation = np =

Example The probability of a rare disease striking a given population is 0.003. A sample of 10000 wasexamined. Find the expected no. suffering from the disease and hence determine the varianceand the standard deviation for the above problem

SolutionSample size n = 10000P( a person suffering from the disease ) = 0.003 = p

∴ expected number of people suffering from the diseaseMean = λ = 10000 × 0.003

= 30

= np = ⋋ variance = np = 30

Standard deviation = np = ⋋

= 30

= 5.477

5.5 PROBABILITY DISTRIBUTION FOR CONTINUOUS RANDOM VARIABLES.In a continuous distribution, the variable can take any value within a specified range, e.g. 2.21 or1.64 compared to the specific values taken by a discrete variable e.g 1 or 3. The probability isrepresented by the area under the probability density curve between the given values. The uniform distribution, the normal probability distribution and the exponential distributionare examples of a continuous distribution

- The normal distribution is a probability distribution which is used to determineprobabilities of continuous variables

Examples of continuous variables are

o Distanceso Timeso Weightso Heightso Capacity e.t.c

- Usually continuous variables are those, which can be measured by using the appropriateunits of measurement.

- Following are the properties of the normal distribution



Probability 175


1. The total area under the curve is = 1 which is equivalent to the maximum valueof probability

Normal probability

Distribution curve

Line of symmetry

2. The line of symmetry divides the curve into two equal halves3. The two ends of the normal distribution curve continuously approach the

horizontal axis but they never cross it4. The values of the mean, mode and median are all equal

NB: The above distribution curve is referred to as normal probability distribution curve becauseif a frequency distribution curve is plotted from measurements of a given sample drawn from anormal population then a graph similar to the normal curve must be obtained.

- It should be noted that 68% of any population lies within one standard deviation, ±1σ - 95% lies within two standard deviations ±2σ - 99% lies within three standard deviations ±3σ

Where σ = standard deviation

0 Z

A e Yrs

Tail endTail end



176 Lesson Five


STANDARDIZATION OF VARIABLES- Before we use the normal distribution curve to determine probabilities of the

continuous variables, we need to standardize the original units of measurement, byusing the following formular.

Z =σ

μ

Where χ = Value to be standardizedZ = Standardization of xµ = population meanσ = Standard deviation

Example A sample of students had a mean age of 35 years with a standard deviation of 5 years. A student was randomly picked from a group of 200 students. Find the probability that the age of thestudent turned out to be as follows

i. Lying between 35 and 40ii. Lying between 30 and 40iii. Lying between 25 and 30

iv. Lying beyond 45 yrs v. Lying beyond 30 yrs vi. Lying below 25 years

Solution(i). The standardized value for 35 years

Z =σ

=

5

35-35 = 0

The standardized value for 40 years

Z =σ

=

5

35-40 = 1

∴ the area between Z = 0 and Z = 1 is 0.3413 (These values are checked from the normaltables see appendix) The value from standard normal curve tables. When z = 0, p = 0 And when z = 1, p = 0.3413Now the area under this curve is the area between z = 1 and z = 0

= 0.3413 – 0 = 0.3413

∴ the probability age lying between 35 and 40 yrs is 0.3413(ii). 30 and 40 years

Z = σ

= 5

3530

= 5

5

= -1

Z =σ

=

5

3540 = 1

∴ the area between Z = -1 and Z = 1 is= 0.3413 (lying on the positive side of zero) + 0.3413 (lying on the negative side ofzero)



Probability 177


P = 0.6826

∴ the probability age lying between 30 and 40 yrs is 0.6826(iii). 25 and 30 years

Z =σ

=

5

3525 =

5

10 = -2

Z =σ

=

5

3530 = -1

∴ the area between Z = -2 and Z = -1Probability area corresponding to Z = -2

= 0.4772 (the z value to check from the tables is 2)Probability area corresponding to Z = -1

= 0.3413 (the z value for this case is 1)

∴ the probability that the age lies between 25 and 30 yrs= 0.4772 – 0.3413 (The area under this curve)

P = 0.1359

iv). P(beyond 45 years) is determined as follow = P(x > 45)

Z =σ

=

5

3545 =

5

10 = + 2

Probability corresponding to Z = 2 = 0.4772 = probability of between 35 and 45

∴ P(Age > 45yrs) = 0.5000 – 0.4772= 0.0228

The exponential distribution The exponential distribution is of particular importance because of the wide ranging nature ofthe practical situations in which it is used.

Examples1. The length of time until an electronic device fails2. The time required to wait for the first emission of a particle from a radio active

source3. The length of time between successive accidents in a large factory

Assume that a probability density function f(x) is valid between the values a and b, then

(i).. b

a

( ) 1 i.e. The area under the curve is equal to 1 f x dx

(ii).The mean of the distributionb

a

E x xf x dx

(iii) The variance of the distribution = E(x2 ) – [E(x)]2

Where 2 2

b

a

E x x f x dx



178 Lesson Five


Example of continuous probability distribution function The distribution of a random variable x has a probability density function f(x) given by

f(x) = kx for 0 ≤ x ≤1f(x) = 0 elsewhereWhere k is constant

Required.

i. Show the value of k is 2ii. Find the mean of f(x)iii. Find the variance of f(x)

Solutioni) 1

0

112

2 00

2

1

. 1

1 0 1

2

k

k

f x dx

kx dx x

k

ii) 1

0

11

2 323 0

0

2 23 3

2

0

Mean E x xf x dx

x dx x

iii) 22

22

122 2

3

0

141 42 90

1 42 9

2

1

18

b

a

Variance E x E x

x f x dx Mean

x x dx

x

Variance

Exponential distribution

Example The mean life of an electrical component is 100 hours and its life has an exponentialdistribution.Find

a. The probability that it will last less than 60 hoursb. The probability that it will last more than 90 hours

Solution A continuous random variable X has an exponential distribution, if for some constant k >0 ithas the probability density function

. for 0

0 elsewhere

kxk e x f x



Probability 179


The function f(x) is positive for all values of x and the area under the curve

0 0

1kx f x dx ke dx

The mean of an exponential distribution with parameter k is 1k

and its variance is 2

1

k

Example The mean of an exponential distribution is 100, find;a) P(x<60)b) P(x>90)

solution.

a) 1

100

60

1100

0

160 100

100

x P x e dx mean thus k

100

600.6

0

1

0.45

x

e e

b) 90 1 90 P x P x

100

90

1100

0

900.9 0.9

0

1

1

0.41

x

e dx

e e

The students t distribution The students t distribution was presented by W. S. Gosset in 1908 under the pen name of

„student‟. The t distribution is of great importance in the so called small sample tests and isprofoundly used in statistical inference The t distribution has a single parameter, known as the number of degrees of freedom. It is

denoted by the Greek symbol ℧ (read as nu). It can be interpreted as the number of useful itemsof information generated by a sample of given size. The degrees of freedom are sample size lessone (v = n-1)

Properties of t distribution1. The t distribution ranges from – ∞ to ∞ first as does the normal distribution2. The t distribution like the standard normal distribution is bell shaped and

symmetrical around mean zero3. The shapes of the t distribution changes as the number of degrees of freedom

changes4. The t distribution is more platykurtic that the normal distribution5. The t distribution has a greater dispersion than the standard normal

distribution. As n gets larger the t distribution approaches the normaldistribution when n = 30 the difference is very small



180 Lesson Five


Relation between the t distribution and standard normal distribution is shown in the followingdiagram

Standard normal distribution

T distribution n = 15

T distribution n = 5

- 4 - 3 - 2 -1 0 1 2 3 4

Note that the t distribution has different shapes depending on the size of the sample. When thesample is quite small the height of the t distribution is shorter than the normal distribution andthe tails are wider.

Assumptions of t distribution1. The sample observations are random2. Samples are drawn from normal distribution3. The size of sample is thirty or less n ≤ 30

Application of t distribution- Estimation of population mean from small samples- Test of hypothesis about the population mean- Test of hypothesis about the difference between two means

Chi Square distributionChi square was first used by Karl Pearson in 1900. It is denoted by the Greek letter χ2. itcontains only one parameter, called the number of degrees of freedom (d-f), where the termdegree of freedom represents the number of independent random variables that express the chisquare

Properties1. Its critical values vary with the degree of freedom. For every increase in the numberof degrees of freedom there is a new χ 2 distribution.

2. This possesses additional property so that when χ 21 and χ 22 are independent andhave a chi square distribution with n1 and n2 degrees of from χ 21 + χ 22 will also bedistributed as a chi square distribution with n1 + n2 degrees of freedom

3. Where the degrees of freedom is 3.0 and less the distribution of χ 2 is skewed. But,for degrees of freedom greater than 30 in a distribution, the values of χ 2 arenormally distributed



Probability 181


4. The χ 2 function has only one parameter, the number of degrees of freedom.

P(x) ℧ = 1

℧ = 2

℧ = 3

℧ = 4

℧ = 5

0 1 2 3 4 5 6 7 8 9 10

. χ 2

5. χ 2 distribution is a continuous probability distribution which has the value zeroat its lower limit and extends to infinity in the positive direction. Negative valueof χ2 is not possible because the differences between the observed and expectedfrequencies are always squared

F distribution or Variance ratio distributionIt was developed by R. A Fisher in 1924 and is usually defined in terms of the ratio of the variances of two normally distributed populationsIt is used to test the hypothesis that the two normally distributed populations have two equal variancesF distribution ratio of the variances between two normally distributed population may be

expressed as2

2

2

2

2

1

2

1

ds

ds

With ℧1 = n1 – 1 and ℧2 = n2 – 1 degrees of freedom

Where normal population means are unknownn1 – sample size of independent random 1n2 – sample size of independent random 2



182 Lesson Five


2

1 s - Sample variance of 12

2 s – sample variance of 22

1d - Population variance of 12

2d Population variance of 2

2

1 s and 2

2 s are given by

2

112

1

1 1

x x s

nas the unbiased estimator of d12

2

222

2

1 1

x x s

n as the unbiased estimator of d22

if 2

1d = 2

2d then the statistic F =22

21

S

S Larger estimate of variance

Smaller estimate of variance

F – Distribution with n1 – 1 and n2 – 1 degrees of freedom. F distribution depends on the degrees

of freedom ℧1 for the numerator and ℧2 for the denominator. It has parameters ℧1 and ℧2

such that for different values of ℧1 and ℧2 will have different distributions.

Properties1. The shape of the f distribution depends upon the number of degrees of

freedom2. The mean and variance of the f distribution are

Mean = ℧1 for ℧2 >2-v2 - 2

1

2 1 2

2

1 2 2

2 2

2 4

v v vVariance

v v v for ℧2 > 4

3. The f distribution is positively skewed and its skewness decreases with increases

in ℧1 and ℧24. The value of f must be positive or zero since variances are squares and can

never assume negative values

Assumptionsa) All sample observations are randomly selected and independentb) The total variance of the various sources of variance should be additive.c) The ratio of S12 to S22 should be equal to or greater than 1d) The population for each sample must be normally distributed with identical mean of

variancee) F value can never be negative



Probability 183



QUESTION ONE

The quality controller, Mr. Brooks, at Queensville Engineers has become aware of the need foran acceptance sampling programme to check the quality of bought-in components. This is ofparticular importance for a problem the company is currently having with batches of pumpshafts bought in from a local supplier. Mr. Brooks proposes the following criteria to assess whether or not to accept a large batch of pump shafts from this supplier.

From each batch received take a random sample of 50 shafts, and accept that batch if no morethan two defectives are found in the sample.

Mr. Brooks needed to calculate the probability of accepting a batch P a , when the proportion ofdefectives in the batch, p, is small (under 10%, say)

Required:a) Explain why the Poisson distribution is appropriate to invstigate this situation.b) Using the Poisson distribution, determine the probability of accepting a batch P a,

containing p=2% defectives if the method is used.Determine P a, for p = 0%, 2%, 5%, 10%, 15%

QUESTION TWO

A woven cloth is liable to contain faults and is subjected to an inspection procedure. Any faulthas a probability of 0.7 that it will be detected by the procedure, independent of whether anyother fault is detected or not.

Required:a) If a piece of cloth contains three faults, A, B and C,

i) Calculate the probability that A and C are detected, but that B is undetected;ii) Calculate the probability that any two of A, B and C be detected, the other fault

being undetected;iii) State the relationship between your answers to parts (i) and (ii) and give reasons for

this.

b) Suppose now that, in addition to the inspection procedure given above, there is a secondarycheck which has a probability of 0.6 of detecting each fault missed by the first inspectionprocedure. This probability of 0.6 applies independently to each and every fault undetectedby the first procedure.

i) Calculate the probability that a piece of cloth with one fault has this faultundetected by both the inspection procedure and the secondary check;

ii) Calculate the probability that a piece of cloth with two faults has one of these faultsdetected by either the inspection procedure or the secondary check, and one fault

undetected by both;iii) Of the faults detected, what proportion are detected by the inspection procedureand what proportion by the secondary check?



184 Lesson Five


QUESTION THREE

A company has three production sections S1,S2 and S3 which contribute 40%,35% and 25%,respectively, to total output. The following percentages of faulty units have been observed:

S1 2% (0.02)S2 3% (0.03)S3 4% (0.04)

There is a final check before output is dispatched. Calculate the probability that a unit foundfaulty at this check has come from section 1, S1

QUESTION FOUR

Assuming a Binomial Distribution what is the probability of a salesman making 0,1,2,3,4,5 or 6sales in 6 visits if the probability of making a sale on a visit is 0.3?

Do not use tables for this question.

QUESTION FIVE

Records show that 60% of students pass their examinations at first attempt. Using the normalapproximation to the binomial, calculate the probability that at least 65% of a group of 200students will pass at the first attempt.

QUESTION SIX

A batch of 5000 electric lamps has a mean life of 1000 hours and a standard deviation of 75hours. Assume a normal distribution.

a) How many lamps will fail before 900 hours?b) How many lamps will fail between 950 and 1000 hours?c) What proportion of lamps will fail before 925 hours?

d)

Given the same mean life, what would the standard deviation have to be to ensure thatno more than 20% of lamps fail before 916 hours?





186 Lesson Six


Systematic Sampling This sampling is a part of simple random sampling in ascending or descending orders. Insystematic sampling a sample is drawn according to some predetermined object. Suppose apopulation consists of 1000 units, then every tenth, 20th or 50th item is selected. This method is very easy and economical. It also saves a lot of time

Multistage sampling This is similar to stratified sampling except division is done on geographical/location basis, e.g. acountry can be divided into provinces and then survey is done in 4 towns in each province. Thishelps to cut traveling costs for a surveyor.

Cluster Sampling This is where a few geographical regions e.g. a location, town or village are selected at randomand say every single household or shop in that area is interviewed. This again cuts on costs.

Judgment SamplingHere the interviewer selects whom to interview believing that their view is more fundamental

since they might be directly affected e.g. to find out effects of public transport one may chose tointerview only people who don‟t own cars and travel frequently to work.

6.2 THE CENTRAL LIMIT THEOREM The theory was introduced by De Moivre and according to it; if we select a large number ofsimple random samples, say from any population and determine the mean of eachsample, the distribution of these sample means will tend to be described by the normal

probability distribution with a mean µ and variance σ2/n. This is true even if the populationitself is not normal distribution. Or the sampling distribution of sample means approaches to anormal distribution irrespective of the distribution of population from where the sample is takenand approximation to the normal distribution becomes increasingly close with increase in samplesizes

Types of distribution

Population distributionIt refers to the distribution of the individual values of population. Its mean is denoted by „µ‟

Sample distributionIt is the distribution of the individual values of a single sample. Its mean is generally written as“ x ”. it is not usually the same as µ

Distribution of Sample Means or sampling distribution A sample of size n is taken from the parent population and mean of the sample is calculated. This is repeated for a number of samples so that we have a distribution of sample means, which

approaches a normal distribution.

Standard errors of the mean

The series of sample means 1X , 2X , 3X …….. is normally distributed or nearly so (accordingto the central limit theorem). It can be described by its mean and its standard deviation. Thisstandard deviation is known as the standard error.



Sampling and Estimation 187


Standard error of the mean =n

sS

x

Note: this formula is satisfactory for larger samples and a large population i.e. n > 30 and n >5% of N.- The word „error‟ is in place of „deviation‟ to emphasize that variation among sample means

is due to sampling errors.- The smaller the standard error the greator the precision of the sample value.

6.3 Statistical inferenceIt is the process of drawing conclusions about attributes of a population based upon informationcontained in a sample (taken from the population).It is divided into estimation of parameters and testing of hypothesis. Symbols for statistic ofpopulation parameters are as follows.

Sample Statistic Population Parameter Arithmetic mean x µStandard deviation s σ Number of items n N

Statistical estimationIt is the procedure of using statistic to estimate a population parameterIt is divided into point estimation (where an estimate of a population parameter is given by asingle number) and interval estimation (where an estimate of a population is given by a range in which the parameter may be considered to lie) e.g. a bus meant to take a class of 100 students(population N) for trip has a limit to the maximum weight of 600kg of which it can carry, theteacher realizes he has to find out the weight of the class but without enough time to weigheveryone he picks 25 students selected at random (sample n = 25). These students are weighedand their average weight recorded as 64kg ( X - mean of a sample) with a standard deviation (s),now using this the teacher intends to estimate the average weight of the whole class (µ –

population mean) by using the statistical parameters standard deviation (s), and mean of thesample ( x ).

Characteristic of a good estimator(i) Unbiased: where the expected value of the statistic is equal to the population

parameter e.g. if the expected mean of a sample is equal to the population mean(ii) Consistency: where an estimator yields values more closely approaching the

population parameter as the sample increases(iii) Efficiency: where the estimator has smaller variance on repeated sampling.(iv) Sufficiency: where an estimator uses all the information available in the data

concerning a parameter

Confidence Interval The interval estimate or a „confidence interval‟ consists of a range (an upper confidence limit andlower confidence limit) within which we are confident that a population parameter lies and weassign a probability that this interval contains the true population value The confidence limits are the outer limits to a confidence interval. Confidence interval is theinterval between the confidence limits. The higher the confidence level the greater theconfidence interval. For example A normal distribution has the following characteristic

i. Sample mean ± 1.960 σ includes 95% of the population



188 Lesson Six


ii. Sample mean ± 2.575 σ includes 99% of the population

1. LARGE SAMPLES These are samples that contain a sample size greater than 30(i.e. n>30)

(a) Estimation of population mean

Here we assume that if we take a large sample from a population then the mean of thepopulation is very close to the mean of the sampleSteps to follow to estimate the population mean includes

i. Take a random sample of n items where (n>30)

ii. Compute sample mean ( X ) and standard deviation (S)iii. Compute the standard error of the mean by using the following formular

Sx

=n

s

where Sx

= Standard error of mean

S = standard deviation of the samplen = sample size

iv.

Choose a confidence level e.g. 95% or 99% v. Estimate the population mean as under

Population mean µ = χ ± (appropriate number) ×Sx

„ Appropriate number‟ means confidence level e.g. at 95% confidence level is 1.96this number is usually denoted by Z and is obtained from the normal tables.

Example The quality department of a wire manufacturing company periodically selects a sample of wirespecimens in order to test for breaking strength. Past experience has shown that the breakingstrengths of a certain type of wire are normally distributed with standard deviation of 200 kg. Arandom sample of 64 specimens gave a mean of 6200 kgs. Find out the population mean at 95%level of confidence

Solution

Population mean = χ ± 1.96 Sx

Note that sample size is alredy n > 30 whereas s and x are given thus step i), ii) and iv) areprovided.Here: X = 6200 kgs

Sx

=n

s =

64

200 = 25

Population mean = 6200 ± 1.96(25)= 6200 ± 49= 6151 to 6249

At 95% level of confidence, population mean will be in between 6151 and 6249

FINITE POPULATION CORRECTION FACTOR (FPCF)If a given population is relatively of small size and sample size is more than 5% of thepopulation then the standard error should be adjusted by multiplying it by the finite populationcorrection factor





FPCF is given by =1n

n N

where N = population sizen = sample size

Example

A manager wants an estimate of sales of salesmen in his company. A random sample 100 out of500 salesmen is selected and average sales are found to be Shs. 75,000. if a sample standarddeviation is Shs. 15000 then find out the population mean at 99% level of confidence

Solution

Here N = 500, n = 100, X = 75000 and S = 15000NowStandard error of mean

= Sx

=n

s x

1n

n N

=100

15000 x1500

100500

=10

15000 x

499

400

=10

15000(0.895)

Sx

= 1342.50 at 99% level of confidence

Population mean = X ± 2.58 S x =shs 75000 ± 2.58(1342.50)=shs 75000 ± 3464= Shs 71536 to 78464

b) Estimation of difference between two means We know that the standard error of a sample is given by the value of the standard deviation

( σ)divided by the square root of the number of items in the sample ( n ).But, when given two samples, the standard errors is given by

B A X X S =

B

B

A

A

n

S

n

S 22

Also note that we do estimate the interval not from the mean but from the difference betweenthe two sample means i.e. B A X X . The appropriate number of confidence level does not change Thus the confidence interval is given by;

B A X X ± Confidence level B A X X

S

= B A X X ± Z B A X X

S



190 Lesson Six


Example

Given two samples A and B of 100 and 400 items respectively, they have the means 1X = 7 ad

2X = 10 and standard deviations of 2 and 3 respectively. Construct confidence interval at 70%

confidence level?

SolutionSample A B

1X = 7 2X = 10

n1 = 100 n2 = 400S1 = 2 S2 = 3

The standard error of the samples A and B is given by

B A X X S =

4 9

100 400

=25

400 =

5

20

= ¼ = 0.25

At 70% confidence level, then appropriate number is equal to 1.04 (as read from the normaltables)

1 2 X X = 7 – 10 = - 3 = 3

We take the absolute value of the difference between the means e.g. the value of X = absolute

value of X i.e. a positive value of X.Confidence interval is therefore given by

= 3± 1.04 (0.25 ) From the normal tables a z value of 1.04 gives a value of 0.7.

= 3± 0.26= 3.26 and 2.974

Thus 2.974 ≤ X ≤ 3.26

Example 2 A comparison of the wearing out quality of two types of tyres was obtained by road testing.Samples of 100 tyres were collected. The miles traveled until wear out were recorded and theresults given were as follows Tyres T1 T2

Mean 1X = 26400 miles 2X = 25000 miles

Variance S21= 1440000 miles S

22= 1960000 milesFind a confidence interval at the confidence level of 70%

Solution

1X = 26400

2X = 25000Difference between the two means

21 X X = (26400 – 25000)





192 Lesson Six


= P ± 1.96 Sp where 1.96 = Z.= 0.70 ± 1.96 (0.016)

= 0.70 ± 0.03

= 0.67 to 0.73

= between 67% to 73%

Example 2 A sample of 600 accounts was taken to test the accuracy of posting and balancing of accounts where in 45 mistakes were found. Find out the population proportion. Use 99% level ofconfidence

SolutionHere

n = 600; p =45

600 = 0.075

q = 1 – 0.075 = 0.925

Sp = pq

n =

0.075 0.925

600

= 0.011

Population proportion= P ± 2.58 (Sp )= 0.075 ± 2.58 (0.011)

= 0.075 ± 0.028

= 0.047 to 0.10

= between 4.7% to 10%

d) Estimation of difference between population proportionsLet the two proportions be given by P1 and P2, respectively Then the difference (absolute) between the two proportions is given by (P1 – P2 ) The standard error is given by

21 P P S =

1 2

pq pq

n n where p = 1 1 2 2

1 2

p n p n

n n and q = 1 - p

Then given the confidence level, the confidence interval between the two populationproportions is given by

(P1 – P2 ) ± Confidence level21 P P

S

= (P1 – P2 ) ± Z1 2

pq pq

n n





Where P = 1 1 2 2

1 2

p n p n

n n always remember to convert P1 & P2 to P.

2. SMALL SAMPLES(a) Estimation of population mean

If the sample size is small (n<30) the arithmetic mean of small samples are not normallydistributed. In such circumstances, students t distribution must be used to estimate thepopulation mean.In this case

Population mean µ = X ± x

ts

X = Sample mean

xS =

s

n

S = standard deviation of samples =

2

1

x x

nfor small samples.

n = sample size v = n – 1 degrees of freedom. The value of t is obtained from students t distribution tables for the required confidence level

Example A random sample of 12 items is taken and is found to have a mean weight of 50 grams and astandard deviation of 9 grams What is the mean weight of population

a) with 95% confidenceb) with 99% confidence

Solution

50; X S = 9; v = n – 1 = 12 – 1 = 11;9

12 x

sS

n

µ = x‟ ± x

ts

At 95% confidence level

µ = 50 ± 2.2629

12

= 50 ± 5.72 grams

Therefore we can state with 95% confidence that the population mean is between 44.28 and55.72 grams At 99% confidence level

µ = 50 ± 3.259

12

= 50 ± 8.07 grams



194 Lesson Six


Therefore we can state with 99% confidence that the population mean is between 41.93 and58.07 grams

Note: To use the t distribution tables it is important to find the degrees of freedom ( v = n – 1).

In the example above v = 12 – 1 = 11

From the tables we find that at 95% confidence level against 11 and under 0.05, the value of t =2.201

6.4 Hypothesis Testing

Definition- A hypothesis is a claim or an opinion about an item or issue. Therefore it has to be testedstatistically in order to establish whether it is correct or not correct- Whenever testing an hypothesis, one must fully understand the 2 basic hypothesis to be testednamely

i. The null hypothesis (H0 )ii. The alternative hypothesis(H1 )

The null hypothesis This is the hypothesis being tested, the belief of a certain characteristic e.g. Kenya Bureau ofStandards (KBS) may walk to a sugar making company with an intention of confirming that the2kgs bags of sugar produced are actually 2kgs and not less, they conduct hypothesis testing withthe null hypothesis being: H0 = each bag weighs 2kgs. The testing will set out to confirm this orto refute it.

The alternative hypothesis While formulating a null hypothesis we also consider the fact that the belief might be found tobe untrue hence we will reject it. We therefore formulate an alternative hypothesis which is acontradiction to the null hypothesis, thus when we reject the null hypothesis we accept thealternative hypothesis.

In our example the alternative hypothesis would beH1 = each bag does not weigh 2kg

Acceptance and rejection regions All possible values which a test statistic may either assume consistency with the null hypothesis(acceptance region) or lead to the rejection of the null hypothesis (rejection region or criticalregion) The values which separate the rejection region from the acceptance region are called critical values

Type I and type II errors While testing hypothesis (H0 ) and deciding to either accept or reject a null hypothesis, there are

four possible occurrences.a) Acceptance of a true hypothesis (correct decision) – accepting the null hypothesis and ithappens to be the correct decision. Note that statistics does not give absolute information,thus its conclusion could be wrong only that the probability of it being right are high.

b) Rejection of a false hypothesis (correct decision).c) Rejection of a true hypothesis – (incorrect decision) – this is called type I error, with

probability = α. d) Acceptance of a false hypothesis – (incorrect decision) – this is called type II error, with

probability = β.





Levels of significance A level of significance is a probability value which is used when conducting tests of hypothesis. A level of significance is basically the probability of one making an incorrect decision after thestatistical testing has been done. Usually such probability used are very small e.g. 1% or 5%

0.5000 0.4900

1% provision for errors

0

Critical value

NB: If the standardized value of the mean is less than – 1.65 we reject the null hypothesis (H0 )and accept the alternative Hypothesis (H1 ) but if the standardized value of the mean is morethan – 1.65 we accept the null hypothesis and reject the alternative hypothesis

The above sketch graph and level of significance are applicable when the sample mean is < (i.e.less than the population mean)

0.45

5% = 0.05

Crititical value = -1.65

0

Critical re ion



196 Lesson Six


The following is used when sample mean > population mean

Acceptance region

Critical region (rejection region)

5% = 0.05

0 Z = 1.65 (critical value)

NB: If the sample mean standardized value < 1.65, we accept the null hypothesis but reject thealternative. If the sample mean value > 1.65 we reject the null hypothesis and accept thealternative hypothesis The above sketch is normally used when the sample mean given is greater than the populationmean

Accept null hyp( reject Alternative hyp)

Reject null hyp (accept alt hyp) Reject null hyp (accept alt hyp)

0.05% = 0.05 0.495 0.495 0.5% = 0.05

-2.58 +2.58

NB: if the standardized value of the sample mean is between – 2.58 and +2.58 accept the nullhypothesis but otherwise reject it and therefore accept the alternative hypothesis





TWO TAILED TESTS A two tailed test is normally used in statistical work(tests of significance) e.g. if a complaintlodged by the client is about a product not meeting certain specifications i.e. the item willgenerate a complaint if its measurements are below the lower tolerance limit or above theupper tolerance limit

Region of acceptance for

H0

Critical region Critical region

15cm 17 ½ cm

NB: Alternative hypothesis is usually rejected if the standardized value of the sample mean liesbeyond the tolerance limits (15cm and 17 ½ cm).

ONE TAILED TEST This is a test where the alternative hypothesis (H1:) is only concerned with one of the tails of thedistribution e.g. to test a business complaint if the complaint is above the measurements of item

being shorter than is required.E.g. a manufacturer of a given brand of bread may state that the average weight of the bread is500 gms but if a consumer takes a sample and weighs each of the pieces of bread and happensto have a mean of 450 gms he will definitely complain about the bread which is underweight. The statistical analysis to be done will concentrate on the left tail of the normal distribution in which one will have to establish whether 450 gms being less than 500g is statistically significant.Such a test therefore is referred to as one tailed test.



198 Lesson Six


left

On the other hand the test may compuliate on the right hand tail of the normal distribution when this happens the major complaint is likely to do with oversize items bought. Therefore thetest is known as one tailed as the focus is on one end of the normal distribution.

Number of standard errors Two tailed test One tailed test

5% level of significance 1.96 1.651% level of significance 2.58 2.33

HYPOTHESIS TESTING PROCEDURE Whenever a business complaint comes up there is a recommended procedure for conducting astatistical test. The purpose of such a test is to establish whether the null hypothesis oralternative hypothesis is to be accepted. The following are steps normally adopted

1. Statement of the null and alternative hypothesis2. Statement of the level of significance to be used.3. Statement about the test statistic i.e. what is to be tested e.g. the sample mean, sample

proportion, difference between sample means or sample proportions4. Type of test whether two tailed or one tailed.5. Statement on critical values using the appropriate level of significance6. Standardizing the test statistic7. Conclusion showing whether to accept or reject the null hypothesis

STANDARD HYPOTHESIS TESTSIn principal, we can test the significance of any statistic related to any probability distribution.

However we will be interested in a few standard cases. The sample statistics mean, proportionand variance, are related to the normal, t, F, and chi squared distributions Thus





1. Normal test

Test a sample mean ( X ) against a population mean (µ) (where samples size n > 30 andpopulation variance σ2 is known) and sample proportion, P(where sample size np >5 and nq>5 since in this case the normal distribution can be used to approximate the binomialdistribution

2. t test

Tests a sample mean ( X ) against a population mean and especially where the population variance is unknown and n < 30.

3. Variance ratio test or f testIt is used to compare population variances and it is used with samples of any size drawnfrom normal populations.

4. Chi squared testIt can be used to test the association between attributes or the goodness of fit of anobserved frequency distribution to a standard distribution

Example 1 A certain NGO carried out a survey in a certain community in order to establish the average at which the girls are married. The results of the survey indicated that the marriage age for the girlsis 19 yearsIn order to establish the validity of the mean marital age, a sample of 50 women was interviewedand the average age indicated that they got married at the age of 16 years. However the differentages at which they were married differed with the standard deviation of 2.1years The sample data indicates that the marital age is less 19 years. Is this conclusion true or not ?

RequiredConduct a statistical test to either support the above conclusion drawn from the sample statistics

i.e. the marriage age is less than 19 years, use a level of significance of 5%

Solution1. Null hypothesis

H0: μ (mean marital age) = 19 years Alternative hypothesis H1: μ (mean marital age) < 19 years

2. The level of significance is 5%

3. The test statistics is the sample mean age, X = 16 years4. The critical value of the one tailed test (one tailed because the alternative hypothesis is

an inequality) at 5% level of significance is – 1.65







3. The test statistics is the sample mean age, x‟ = 30.24. The critical value of the one tailed test at 5% level of significance is + 2.33

0.4900

1% = 0.01

2.33

5. The standardized value of the sample mean is

Z = x

X

S =

4

70

30.2 28 = 4.6

6. Since 4.6 > 2.33, we reject the null hypothesis but accept the alternative hypothesis at1% level of significance i.e. the new sample mean life span is statistically significanthigher than the population mean Therefore the research undertaken was worth while or justified

Example 3

A construction firm has placed an order that they require a consignment of wires which have amean length of 10.5 meters with a standard deviation of 1.7 m The company which produces the wires delivered 90 wires, which had a mean length of 9.2 m., The construction company rejected the consignment on the grounds that they were differentfrom the order placed.

RequiredConduct a statistical test to indicate whether you support or not support the action taken by theconstruction company at 5% level of significance.

SolutionNull hypothesis µ = 10.5 m Alternative hypothesis µ ≠ 10.5 mLevel of significance be 5%

The test statistics is the sample mean X = 9.2m The critical value of the two tailed test at 5% level of significance is ± 1.96 (two tailed test).



202 Lesson Six


- 1.96 +1.96

The standardized value of the test Z =

Z =X

X -μ

S =

1.7

90

9.2 10.5= - 7.25

Since 7.25 < 1.96, reject the null hypothesis but accept the alternative hypothesis at 5% levelof significance i.e. the sample mean is statistically different from the consignment ordered bythe construction company. Therefore support the action taken by the construction company

TESTING THE DIFFERENCE BETWEEN TWO SAMPLE MEANS (LARGESAMPLES) A large sample is defined as one which contains 30 or more items (n≥30) Where n is the sample

sizeIn a business those involved are constantly observant about the standards or specifications ofthe item which they sell e.g. a trader may receive a batch of items at one time and another batchat a later time at the end he may have concluded that the two samples are different in certainspecifications e.g. mean weight mean lifespan, mean length e.t.c. further it may becomenecessary to establish whether the observed differences are statistically significant or not. If thedifferences are statistically significant then it means that such differences must be explained i.e.there are known causes but if they are not statistically significant then it means that thedifference observed have no known causes and are mainly due to chanceIf the differences are established to be statistically significant then it implies that the complaints, which necessitated that kind of test, are justified

Let X 1 and X 2 be any two samples whose sizes are n1 and n2 and mean X 1 and X 2. Standard

deviation S1 and S2 respectively. In order to test the difference between the two sample means, we apply the following formulas

Z = 1 2

1 2

X X

S X X where 1 2S X X =

2 2

1 2

1 2

S S

n n

Example 1 An agronomist was interested in the particular fertilizer yield output. He planted maize on 50equal pieces of land and the mean harvest obtained later was 60 bags per plot with a standard





204 Lesson Six


Example 2 An observation was made about reading abilities of males and females. The observation lead to aconclusion that females are faster readers than males. The observation was based on the timestaken by both females and males when reading out a list of names during graduation ceremonies.In order to investigate into the observation and the consequent conclusion a sample of 200 men

were given lists to read. On average each man took 63 seconds with a standard deviation of 4seconds A sample of 250 women were also taken and asked to read the same list of names. It was foundthat they on average took 62 seconds with a standard deviation of 1 second.

RequiredBy conducting a statistical hypothesis testing at 1% level of significance establish whether thesample data obtained does support earlier observation or not

SolutionH0: µ1 = µ2 H1: µ1 ≠ µ2

Critical values of the two tailed test is at 1% level of significance is 2.58.

Z =1 2

1 2

X X

S X X

Z =2 24 1

200 250

63 62 = 3.45

Acceptance region

Rejection region

- 2.58 0 +2.58 +3.45

Since 3.45 > 2.33 reject the null hypothesis but accept the alternative hypothesis at 1% levelof significance i.e. there is a significant difference between the reading speed of Males andfemales, thus females are actually faster readers.

TEST OF HYPOTHESIS ON PROPORTIONS This follows a similar method to the one for means exept that the standard error used in thiscase:





Sp = Pq

n

Z score is calculated as, Z = P

Sp Where P = Proportion found in the sample.

Π – the hypothetical proportion.

Example A member of parliament (MP) claims that in his constituency only 50% of the total youthpopulation lacks university education. A local media company wanted to acertain that claim thusthey conducted a survey taking a sample of 400 youths, of these 54% lacked universityeducation.

Required: At 5% level of significance confirm if the MP‟s claim is wrong.

Solution.Note: This is a two tailed tests since we wish to test the hypothesis that the hypothesis is

different ( ≠) and not against a specific alternative hypothesis e.g. < less than or > morethan.

H0 : π = 50% of all youth in the constituency lack university education.H1 : π ≠ 50% of all youth in the constituency lack university education.

Sp = pq

n=

0.5 0.5

400

x= 0.025

Z =0.54 0.50

0.025= 1.6

at 5% level of significance for a two-tailored test the critical value is 1.96 since calculated Z value

< tabulated value (1.96).i.e. 1.6 < 1.96 we accept the null hypothesis.

Thus the MP‟s claim is accurate.

HYPOTHESIS TESTING OF THE DIFFERENCE BETWEEN PROPORTIONS

ExampleKen industrial manufacturers have produced a perfume known as “fianchetto .” In order to testits popularity in the market, the manufacturer carried a random survey in Back rank city where10,000 consumers were interviewed after which 7,200 showed preference. The manufactureralso moved to area Rook town where he interviewed 12,000 consumers out of which 1,0000showed preference for the product.

RequiredDesign a statistical test and hence use it to advise the manufacturer regarding the differences inthe proportion, at 5% level of significance.

SolutionH0 : π1 = π2 H1 : π1 ≠ π2



206 Lesson Six


The critical value for this two tailed test at 5% level of significance = 1.96.

Now Z = 1 2 1 2

1 2

P P

S P P

But since the null hypothesis is π1 = π2, the second part of the numerator disappear i.e.π1 - π2 = 0 which will always be the case at this level.

Then Z = 1 2

1 2

P P

S P P

Where;

Sample 1 Sample 2Sample size n1 = 10,000 n2 = 12,000Sample proportion of success P1 =0.72 P2 = 0.83

Population proportion of success. Π1 Π2

Now 1 2S p p =1 2

pq pq

n n

Where P = 1 1 2 2

1 2

p n p n

n n

And q = 1 – pin our case

P =10,000(0.72) 12,000(0.83)

10,000 12,000

=84,000

22,000

= 0.78q = 0.22

1 2

0.78 0.22 0.78 0.22

10,000 12,000S P P

= 0.00894

Z =0.72 0.83

0.00894 = 12.3

Since 12.3 > 1.96, we reject the null hypothesis but accept the alternative. the differencesbetween the proportions are statistically significant. This implies that the perfume is muchmore popular in Rook town than in Back rank city.

HYPOTHESIS TESTING ABOUT THE DIFFERENCE BETWEEN TWOPROPORTIONSIs used to test the difference between the proportions of a given attribute found in two randomsamples.





The null hypothesis is that there is no difference between the population proportions. It meanstwo samples are from the same population.HenceH0 : π1 = π2 The best estimate of the standard error of the difference of P1 and P2 is given by pooling thesamples and finding the pooled sample proportions (P) thus

P = 1 1 2 2

1 2

p n p nn n

Standard error of difference between proportions

1 2

1 2

pq pqS p p

n n

And Z = 1 2

1 2

P P

S p p

ExampleIn a random sample of 100 persons taken from village A, 60 are found to be consuming tea. Inanother sample of 200 persons taken from a village B, 100 persons are found to be consumingtea. Do the data reveal significant difference between the two villages so far as the habit oftaking tea is concerned?

SolutionLet us take the hypothesis that there is no significant difference between the two villages as faras the habit of taking tea is concerned i.e. π1 = π2 We are given

P1 = 0.6; n1 = 100P2 = 0.5; n2 = 200

Appropriate statistic to be used here is given by

P = 1 1 2 2

1 2

p n p n

n n

=0.6 100 0.5 200 60 100

100 200 300

= 0.53q = 1 – 0.53

= 0.47

1 2S P P =1 2

pq pq

n n

=0.53 0.47 0.53 0.47

100 200

= 0.0608

Z =0.6 0.5

0.0608

= 1.64



208 Lesson Six


Since the computed value of Z is less than the critical value of Z = 1.96 at 5% level ofsignificance therefore we accept the hypothesis and conclude that there is no significantdifference in the habit of taking tea in the two villages A and B

t distribution (student’s t distribution) tests of hypothesis (test for small samples n < 30)

For small samples n < 30, the method used in hypothesis testing is exactly similar to the one forlarge samples exept that t values are used from t distribution at a given degree of freedom v,instead of z score, the standard error Se statistic used is also different.Note that v = n – 1 for a single sample and n1 + n2 – 2 where two sample are involved.

a) Test of hypothesis about the population mean When the population standard deviation (S) is known then the t statistic is defined as

t = X

X

S where

X

S S

n

Follows the students t distribution with (n-1) d.f. where

X = Sample mean

μ = Hypothesis population mean n = sample size

and S is the standard deviation of the sample calculated by the formula

S =

2

1

X X

n for n < 30

If the calculated value of t exceeds the table value of t at a specified level of significance, the nullhypothesis is rejected.

Example Ten oil tins are taken at random from an automatic filling machine. The mean weight of the tinsis 15.8 kg and the standard deviation is 0.5kg. Does the sample mean differ significantly from theintended weight of 16kgs. Use 5% level of significance.

Solution

Given that n = 10; x = 15.8; S = 0.50; μ = 16; v = 9H0 : μ = 16 H1 : μ ≠ 16

=0.5

10 X

S

t =0.5

10

15.8 16

=0.2

0.16

= -1.25 The table value for t for 9 d.f. at 5% level of significance is 2.26. the computed value of t issmaller than the table value of t. therefore, difference is insignificant and the null hypothesis isaccepted.





b) Test of hypothesis about the difference between two means The t test can be used under two assumptions when testing hypothesis concerning the differencebetween the two means; that the two are normally distributed (or near normally distributed)populations and that the standard deviation of the two is the same or at any rate not significantlydifferent.

Appropriate test statistic to be used is

t =21

1 2

X X

X X

S at (n1 + n2 – 2) d.f.

The standard deviation is obtained by pooling the two sample standard deviation as shownbelow.

Sp =2 2

1 1 2 2

1 2

1 1

2

n S n S

n n

Where S1 and S2 are standard deviation for sample 1 & 2 respectively.

Now1 X

S =1

Sp

n and

2 X S =

2

Sp

n

1 2 X X S =

21

2 2

X X S S

Alternatively1 2 X X

S = Sp 1 2

1 2

n n

n n

Example Two different types of drugs A and B were tried on certain patients for increasing weights, 5persons were given drug A and 7 persons were given drug B. the increase in weight (in pounds)

is given below

Drug A 8 12 16 9 3Drug B 10 8 12 15 6 8 11

Do the two drugs differ significantly with regard to their effect in increasing weight? (Given that

v= 10; t0.05 = 2.23)

SolutionH0 : μ1 = μ2 H1 : μ1 ≠ μ2

t =1 2

1 2

X X

X X

S

Calculate for 1 X , 2 X and S







A BNo. of sales 20 18 Average sales in shs 170 205Standard deviation in shs 20 25

SolutionH0 : μ1 = μ2 H1 : μ1 ≠ μ2 Where

Sp =2 2

1 1 2 2

1 2

1 1

2

n S n S

n n

1 2 X X S = Sp

1 2

1 2

n n

n n

Where: 1 X =170, 2 X = 205, n1 = 20, n2 = 18, S1 = 20, S2 = 25, V = 36

Sp =2 219 20 17 25

20 18 2

= 22.5

1 2

3822.5

360 X X S

= 7.31

t = 170 2057.31

= 4.79t0.05(36) = 1.9 (Since d.f > 30 we use the normal tables)

The table value of t at 5% level of significance for 36 d.f. when d.f. >30, that t distribution is thesame as normal distribution is 1.9. since the value computed value of t is more than the table value, we reject the null hypothesis. Thus, we conclude that there is significant difference in theaverage sales between the two salesmen

Testing the hypothesis equality of two variances The test for equality of two population variances is based on the variances in two independentlyselected random samples drawn from two normal populationsUnder the null hypothesis 2

2

2

1 σσ

F =

2

2

2

2

2

1

2

1

σ

s

σ

s

Now under the H0 :2

2

2

1 σσ it follows that



212 Lesson Six


F =2

1

22

S

S which is the test statistic.

Which follows F – distribution with V 1 and V 2 degrees of freedom. The larger sample variance isplaced in the numerator and the smaller one in the denominatorIf the computed value of F exceeds the table value of F, we reject the null hypothesis i.e. the

alternate hypothesis is accepted

ExampleIn one sample of observations the sum of the squares of the deviations of the sample valuesfrom sample mean was 120 and in the other sample of 12 observations it was 314. test whetherthe difference is significant at 5% level of significance

Solution

Given that n1 = 10, n2 = 12, Σ(x1 – 1 X )2 = 120

Σ(x2 – 2 X )2 = 314

Let us take the null hypothesis that the two samples are drawn from the same normal population

of equal varianceH0 :

2

2

2

1 σσ

H1:2

2

2

1 σσ

Applying F test i.e.

F =2

1

2

2

S

S

=

211

1

222

2

1

1

X X

n

X X

n

=120

9

31411

=13.33

28.55

since the numerator should be greater than denominator

F =28.55

2.113.33

The table value of F at 5% level of significance for V 1 = 9 and V 2 = 11. Since the calculated value of F is less than the table value, we accept the hypothesis. The samples may have beendrawn from the two population having the same variances.

Chi square hypothesis tests (Non-parametric test)(X 2 ) They include amongst others

i. Test for goodness of fitii. Test for independence of attributesiii. Test of homogeneityiv. Test for population variance





The Chi square test ( χ2 ) is used when comparing an actual (observed) distribution with ahypothesized, or explained distribution.

It is given by; χ2 =

2O E

E Where O = Observed frequency

E = Expected frequency The computed value of χ2 is compared with that of tabulated χ2 for a given significance level anddegrees of freedom.

i. Test for goodness of fit This tests are used when we want to determine whether an actual sample distribution matches aknown theoretical distribution The null hypothesis usually states that the sample is drawn from the theoretical populationdistribution and the alternate hypothesis usually states that it is not.

ExampleMr. Nguku carried out a survey of 320 families in Ateka district, each family had 5 children andthey revealed the following distributionNo. of boys 5 4 3 2 1 0No. of girls 0 1 2 3 4 5No. of families 14 56 110 88 40 12

Is the result consistent with the hypothesis that male and female births are equally probable at5% level of significance?

SolutionIf the distribution of gender is equally probable then the distribution conforms to a binomialdistribution with probability P(X) = ½. Therefore

H0 = the observed number of boys conforms to a binomial distribution with P = ½

H1 = The observations do not conform to a binomial distribution.On the assumption that male and female births are equally probable the probability of a malebirth is P = ½ . The expected number of families can be calculated by the use of binomialdistribution. The probability of male births in a family of 5 is given byP(x) = 5cX Px q 5-x (for x = 0, 1, 2, 3, 4, 5,)

= 5cX ( ½ )5 (Since P = q = ½ ) To get the expected frequencies, multiply P(x) by the total number N = 320. The calculations areshown below in the tables

x P(x) Expected frequency = NP(x)0 5c0 ( ½ )5 = 1

32 320 × 1

32 = 10

1 5c1 ( ½ )5 = 532

320 × 532

= 50

2 5c2 ( ½ )5 =1032

320 × 1032

= 100

3 5c3 ( ½ )5 =1032

320 × 1032

= 100

4 5c4 ( ½ )5 = 532

320 × 532

= 50







ExampleIn a sample of 200 people where a particular devise was selected, 100 were given a drug and theothers were not given any drug. The results are as follows

Drug No drug TotalCured 65 55 120Not cured 35 45 80

Total 100 100 200 Test whether the drug will be effective or not, at 5% level of significance.

SolutionLet us take the null hypothesis that the drug is not effective in curing the disease. Applying the χ2 test The expected cell frequencies are computed as follows

E11 = 1 1 R C

n=

120 100

200 = 60

E12 = 1 2 R C

n

=120 100

200

= 60

E21 = 2 1 R C

n=

80 100

200 = 40

E22 = 2 2 R C

n=

80 100

200 = 40

The table of expected frequencies is as follows60 60 12040 40 80

100 100 200O E (O – E) 2 (O – E) 2 /E65 60 25 0.41755 60 25 0.62535 40 25 0.41745 40 25 0.625

Σ(O – E) 2 /E = 2.084

Arranging the observed frequencies with their corresponding frequencies in the following table we get

χ2 =

2

O E E

= 2.084

V= (r – 1) (c-1) = (2 – 1) (2 – 1) = 1; 2

)05.0(tabulated = 3.841







SUMMARY OF FORMULAE IN HYPOTHESIS

Testing(a) Hypothesis testing of mean

For n>30

Z = X

X

S Where X

S S n at level of significance.

For n < 30

t = X

X

S where

X

S S

n

at n – 1 d.flevel of significance

(b) Difference between means (Independent samples)For n > 30

Z =1 2

1 2

X X

X X S

Where1 2

2 2

1 2

1 2 X X

S S S

n n

At = level of significanceFor n < 30

t =1 2

1 2

X X

X X

S at n1 + n2 – 2 d.f

where 1 2

1 2

1 2

p X X

n nS S n n

and2 2

1 1 2 2

1 2

1 1

2 p

n S n S S

n n

(c) Hypothesis testing of proportions

Z = p

p

S

Where: Sp = pq

n

p = Proportion found in sampleq = 1 – p

= hypothetical proportion

(d) Difference between proportions

Z =1 2

1 2

P P

P P

S



218 Lesson Six


Where:

1 2

1 2

P P

pq pqS

n n

p = 1 1 2 2

1 2

p n p n

n n

q = 1 – P(e) Chi-square test

X 2 =

2O E

E

Where O = observed frequency

E =Column total × Row total

Sample Size= expected frequency

(f) F – test (variance test)

F =2

1

2

2

S

S

here the bigger value between the standard deviations make the numerator.






QUESTION ONE

A firm purchases a very large quantity of metal off-cuts and wishes to know the average weightof an off-cut. A random sample of 625 off-cuts is weighed and it is found that the mean sample weight is 150 grams with a sample standard deviation of 30 grams. What is the estimate of thepopulation mean and what is the standard error of the mean? What would be the standard errorif the sample size was 1225?

QUESTION TWO

A sample of 80 is drawn at random from a population of 800. The sample standard deviation was found to be 6 grams.

- What is the finite population correction factor?- What is the approximation of the correction factor?- What is the standard error of the mean?

QUESTION THREE

State the Central Limit Theorem

QUESTION FOUR

a) What is statistical inference?b) What is the purpose of estimation?c) What are the properties of good estimators?d) What is the standard error of the mean?e) What are confident limits?f) When is the Finite Population Correction Factor used? What is the formula?g) How are population proportions estimated?

h)

What are the characteristics of the t distribution?QUESTION FIVE

A market research agency takes a sample of 1000 people and finds that 200 of them know ofBrand X. After an advertising campaign a further sample of 1091 people is taken and it‟s foundthat 240 know of Brand X.It is required to know if there has been an increase in the number of people having an awarenessof Brand X at the 5% level.

QUESTION SIX

The monthly bonuses of two groups of salesmen are being investigated to see if there is a

difference in the average bonus received. Random samples of 12 and 9 are taken from the twogroups and it can be assumed that the bonuses in both groups are approximately normallydistributed and that the standard deviations are about the same. The same level of significance isto be used.

The sample results were n1=12 n2=9 x1=£1060 x2=£970

s1=£63 s2=£76



220 Lesson Six


QUESTION SEVEN

Torch bulbs are packed in boxes of 5 and 100 boxes are selected randomly to test for thenumber of defectives

Number ofDefectives

Numberof boxes

Totaldefectives

0 40 01 37 372 17 343 5 154 1 45 0 0

100 90 The number of any individual bulb being a reject is

905 0.18

100

and it is required to test at the 5% level whether the frequency of rejects conforms to a binomialdistribution.

QUESTION EIGHT

a) Define type I and type II errors.b) What is a two-tail test?c) What is the best estimate of the population standard deviation when the two samples are

taken

QUESTION NINE

Express Packets guarantee 95%of their deliveries are on time. In a recent week 80 deliveries were made and 6 were late and the management says that, at the 95%level there has been asignificant improvement in deliveries.

Can the MD‟s statement be supported? If not, at what level of confidence can it be supported? A batch of weighing machines has been purchased and one machine is selected at random fortesting. Ten weighing tests have been conducted and the errors found are noted as follows:

Test Errors (gms)1 4.62 8.23 2.14 6.35 5.06 3.6

7 1.48 4.19 7.010 4.5

The purchasing manager has previously accepted machines with a mean error of 3.8 gms andasserts that these tests are below standard. Test the assertions at 5% level.

Compare your answers with those given in lesson 9 of the study pack





COMPREHESIVE ASSIGNMENT THREEWork out these question for three hours (exam condition) then hand them in to DLC for marking


Time allowed: Three hoursSECTION I

QUESTION ONE

a) Explain what is meant by the following terms as used in statistical inference:

i) Statistical hypothesis; (2 marks)ii) Test of a hypothesis; (2 marks)iii) Type I error; (2 marks)iv) Type II error; (2 marks) v) Level of significance (2 marks)b) Cross Lines Group (CLG) has two factories in different parts of the country. Their

Resources, including the labour force skills are regarded as identical and both factories were built at the same time.

A random sample of output data during a given period has been taken from each factoryand converted to standard hours of output per employee. The data are given below:

Factory 1 42 50 43 39 41 49 52 41 46 48Factory 2 39 45 36 42 52 37 43 41 40 39

You are given that for factory 1 mean = 45.1 and variance = 20.10 and that for factory 2mean = 41.4 and variance = 21.16.

Required:i) Test the hypothesis that the mean of standard hours for employees in the two factories

is the same. (7 marks)ii) Comment briefly on the conditions of the test and interpret the outcome. (3 marks)

(Total: 20 marks)

QUESTION TWO

a) State clearly what is meant by two events being statistically independent. (2 marks)

b) In a certain factory which employs 500 men, 2% of all employees have a minor accident in agiven year. Of these, 30% had safety instructions whereas 80% of all employees had no

safety instructions.

RequiredFind the probability of an employee being accident-free given that he had:i) No safety instructions (5 marks)ii) Safety instructions (5 marks)

c) An electric utility company has found that the weekly number of occurrences of lightningstriking the transformers is a Poisson distribution with mean 0.4.



222 Lesson Six


Required:i) The probability that no transformer will be struck in a week. (3 marks)ii) The probability that at most two transformers will be struck in a week (5 marks)

(Total: 20 marks)

QUESTION THREEExplain the difference between the paired t-test and the two-sample t-test (4 marks)

Trendy Tyres Ltd. Has introduced a new brand of tyres which in their advertisements claim tobe superior to their only competitor brand. The Roadmaster Tyres. The brand manager ofRoadmaster Tyres disputes this claim which he says is an advertisement gimmick. The brandmanagers of the two companies agree to run a road test for the brands. Ten (10) saloon cars ofuniform weight and identical specifications are to be used for the test. Each car is fitted withboth brands of tyres: One brand at the front the other brand at the rear. The cars cover adistance of 5,000 kilometers and the trend wear is recorded as follows:

Trend tyres Roadmaster tyres

centimeters Centimeters1 1.08 1.122 1.06 1.093 1.24 1.164 1.20 1.245 1.17 1.236 1.21 1.257 1.18 1.208 1.10 1.159 1.22 1.1910 1.60 1.13

Required:i) Determine whether Trendy Tyres Ltd.‟s claim is true using α = 0.01 (15 marks)ii) What are the assumptions you have made in (i) above? (1 mark)

(Total: 20 marks)

QUESTION FOUR

Kenear Commercial bank Ltd. commissioned a research whose results indicated that automaticteller machines (ATM) reduces the cost of routine banking transactions.

Following this information, the bank installed an ATM facility at the premises of Joy ProcessingCompany Ltd., which for the last several months has exclusively been, used by Joy‟S 605employees. Survey on the usage of the ATM facility by 100 of the employees in a month

indicated the following:





Number of times ATM used

Frequency

0 201 322 20

3 134 105 5

Required:a) An estimate of the proportion of Joy‟s employees who do not use the ATM facility in a

month (2 marks)b) i) Determine the 95% confidence interval for the estimate in (a) above (5 marks)ii) Can the bank be certain that at least 40% of Joy‟s employees will use the ATM facility?

(1 mark)c) The number of ATM transactions on average an employee of Joy makes per month

(3 marks)

d)

Determine the 95% confidence interval of the mean number of transactions made by anemployee in a month. (6 marks)e) Is it possible that the population mean number of transactions is four? Explain. (3 marks)

(Total: 20 marks)

QUESTION FIVE

State any five problems encountered in the construction of the consumer price index. (5 marks)

An investment analyst gathered the following data on the 91-day Treasury bill rates for the years2003 and 2004.

Month Treasury bill rates (%)

2003 2004 January 3.2 5.5February 3.0 5.2March 2.8 4.3 April 2.5 3.6May 2.9 3.3 June 3.4 2.7 July 3.7 2.4 August 4.0 2.0September 3.8 2.3October 4.2 2.8November 4.5 3.1December 5.1 3.7

The analyst would like to determine if on average there was a significant change in the Treasurybill rates over the two years.

Required:i) The mean and variance of the Treasury bill rates for each year. (10 marks)ii) Determine if there is a significant difference in the average Treasury bill rates (use a

significance level of 1%) (5 marks)



224 Lesson Six


Note:

2 2

1 1 2 22

1 2

1 1

2

n S n S S

n n

(Total: 20 marks)

SECTION II

QUESTION SIXa) Describe the characteristics of the following distributions:

i) Binomial distribution (3 marks)ii) Poisson distribution. (3 marks)

b) High Grade Meat Ltd. Produces beef sausages and sells them to various supermarket. Inorder to satisfy the industry‟s requirement, the firm may only produces 0.2 per cent ofsausages below a weight of 80 grammes. The sausage producing machine operates with astandard deviation of 0.5 grammes. The weights of the sausages are normally distributed.

The firm‟s weekly output is 300,000 sausages and the sausage ingredients cost Sh.5.00 per

100 grammes. Sausages with weights in excess of 82 grammes require additional ingredientscosting Sh.2.50 per sausage.

Required:i) The mean weight at which the machine should be set. (4 marks)ii) The firm‟s weekly cost of production (10 marks)

(Total: 20 marks)

QUESTION SEVEN

a) The past records of Salama Industries indicate that about 4 out of 10 of the company‟sorders are for export. Further, their records indicate that 48 per cent of all orders are forexport in one particular financial quarter. They expect to satisfy about 80 orders in the next

financial quarter.

Required:Determine the probability that they will break their previous export record. (7 marks)Explain why you have used the approach you have chosen to solve part (i) above. (2 marks)

b) Gear Tyre Company has just developed a new steel-belted radial tyre that will be soldthrough a national chain of discount stores. Because the tyre is a new product, thecompany‟s management believes that the mileage guarantee offered with the tyre will be animportant factor in the consumer acceptance of the product. Before finalizing the tyremileage guarantee policy, the actual road test with the tyres shows that the mean tyre mileageis μ = 36,500 kilometers and the standard deviation is σ = 5,000 kilometers. In addition, the

data collected indicate that a normal distribution is a reasonable assumption.Required:

i) Gear Tyre Company will distribute the tyres if 20 per cent of the tyres manufactured can beexpected to last more than 40,000 kilometres. Should the company distributed the tyres?

(4 marks)ii) The company will provide a discount on a new set of tyres if the mileage on the original

tyres does not exceed the mileage stated on the guarantee.





What should the guarantee mileage be if the company wants no morethan 10% of the tyres to be eligible for the discount? (4 marks)

c) Explain briefly some of the advantages of the standard normal distribution. (3 marks)(Total: 20 marks)

QUESTION EIGHTa) Explain the following terms used in statistical inference:

i) Null hypothesis (2 marks)ii) Parametric test (2 marks)iii) Coefficient of correlation (2 marksiv) Rank correlation coefficient (2 marks)

b) State four areas that the chi-square distribution is used (4 marks)

c) In an analysis of the results of telecommunication students, the examining board classifiedthe results as either credit, pass or discontinued. Further, the board analyzed the students‟method of study which was either full-time, part-time or private. An employee of the board

cross-classified the examination results and the method of study of 300 students. He thencomputed a test statistic of 42.28

Required:i) State the null and alternative hypotheses that should be tested. (4 marks)ii) What conclusion can be drawn from the results of the data? (use α = 0.05) (4 marks)

(Total: 20 marks)



226 Decesion Theory


LESSON SEVEN

Decision Theory

- Decision theory- Decision trees and sequential decisions- Game theory

7.1 Decision Theory

Types of decisions There are many types of decision making

1. Decision making under uncertainty These refer to situations where more than one outcome can result from any single decision

2. Decision making under certainty Whenever there exists only one outcome for a decision we are dealing with this category e.g.linear programming, transportation assignment and sequencing e.t.c.

3. Decision making using prior dataIt occurs whenever it is possible to use past experience (prior data) to develop probabilities forthe occurrence of each data

4. Decision making without prior dataNo past experience exists that can be used to derive outcome probabilities in this case thedecision maker uses his/her subjective estimates of probabilities for various outcomes

Decision making under uncertaintySeveral methods are used to make decision in circumstances where only the pay offs are knownand the likelihood of each state of nature are known

a) Maximin Method This criteria is based on the “conservative approach‟ to assume that the worst possible is goingto happen. The decision maker considers each strategy and locates the minimum pay off foreach and then selects that alternative which maximizes the minimum payoff

IllustrationRank the products A B and C applying the Maximin rule using the following payoff tableshowing potential profits and losses which are expected to arise from launching these threeproducts in three market conditions(see table 1 below)



Lesson Eight 227


Pay off table in £ 000‟s Boom condition Steady state Recession Mini profits row

minimaProduct A +8 1 -10 -10Product B -2 +6 +12 -2

Product C +16 0 -26 -26

Table 1Ranking the MAXIMIN rule = BAC

b) MAXIMAX method This method is based on „extreme optimism‟ the decision maker selects that particular strategy which corresponds to the maximum of the maximum pay off for each strategy

IllustrationUsing the above example

Max. profits row maxima

Product A +8Product B +12Product C +16

Ranking using the MAXIMAX method = CBA

c) MINIMAX regret method This method assumes that the decision maker will experience „regret‟ after he has made thedecision and the events have occurred. The decision maker selects the alternative whichminimizes the maximum possible regret.

Illustration

Regret table in £ 000‟s Boom condition Steady state Recession Mini regret rowmaxima

Product A 8 5 22 22Product B 18 0 0 18Product C 0 6 38 38

A regret table (table 2) is constructed based on the pay off table. The regret is the „opportunityloss‟ from taking one decision given that a certain contingency occurs in our example whetherthere is boom steady state or recession The ranking using MINIMAX regret method = BAC

d) The expected monetary value method The expected pay off (profit) associated with a given combination of act and event is obtainedby multiplying the pay off for that act and event combination by the probability of occurrenceof the given event. The expected monetary value (EMV) of an act is the sum of all expectedconditional profits associated with that act

Example A manager has a choice between



228 Decesion Theory


i. A risky contract promising shs 7 million with probability 0.6 and shs 4 million withprobability 0.4 and

ii. A diversified portfolio consisting of two contracts with independent outcomes eachpromising Shs 3.5 million with probability 0.6 and shs 2 million with probability 0.4

Can you arrive at the decision using EMV method?

Solution The conditional payoff table for the problem may be constructed as below.

(Shillings in millions)Event Ei Probability (Ei) Conditional pay offs decision Expected pay off decision

(i) Contract (ii) Portfolio(iii) Contract (i) x (ii) Portfolio (i) x (iii)Ei 0.6 7 3.5 4.2 2.1E2 0.4 4 2 1.6 0.8

EMV 5.8 2.9

Using the EMV method the manager must go in for the risky contract which will yield him ahigher expected monetary value of shs 5.8 million

e) Expected opportunity loss (EOL) method This method is aimed at minimizing the expected opportunity loss (OEL). The decision makerchooses the strategy with the minimum expected opportunity loss

f) The Hurwiz method This method was the concept of coefficient of optimism (or pessimism) introduced by L.Hurwicz. The decision maker takes into account both the maximum and minimum pay off foreach alternative and assigns them weights according to his degree of optimism (or pessimism). The alternative which maximizes the sum of these weighted payoffs is then selected

g) The Laplace method This method uses all the information by assigning equal probabilities to the possible payoffsfor each action and then selecting that alternative which corresponds to the maximumexpected pay off

Example A company is considering investing in one of three investment opportunities A, B and Cunder certain economic conditions. The payoff matrix for this situation is economic condition

Investmentopportunities

1 £ 2 £ 3 £

A 5000 7000 3000B -2000 10000 6000C 4000 4000 4000

Determine the best investment opportunity using the following criteriai. Maximinii. Maximaxiii. Minimaxiv. Hurwicz (Alpha = 0.3



Lesson Eight 229


SolutionEconomic condition

Investmentopportunities

1 £ 2 £ 3 £ Minimum £ Maximum £

A 5000 7000 3000 3000 7000B -2000 10000 6000 -2000 10000

C 4000 4000 4000 4000 4000i. Using the Maximin rule Highest minimum = £ 4000

Choose investment Cii. Using the Maximax rule Highest maximum = £ 10000

Choose investment Biii. Minimax Regret rule

1 2 3 Maximumregret

A 0 3000 3000 3000B 7000 0 0 7000C 1000 6000 2000 6000

Choose the minimum of the maximum regret i.e. £3000Choose investment Aiv. Hurwicz rule: expected values

For A (7000 x 0.3) + (3000 x 0.7) = 2100 + 2100 = £4200For B (10000 x 0.3) + (-2000 x 0.7) = 3000 + 1400 = £ 1600For C (4000 x 0.3) + (4000 x 0.7) = 1200 + 2800 = £ 4000Best outcome is £ 4200 choose investment A

Value of perfect informationIt relates to the amount that we would pay for an item of information that would enable us toforecast the exact conditions of the market and act accordingly.

The expected value of perfect information EVPI is the expected outcome with perfectinformation minus the expected outcome without perfect information namely the maximumEMV

ExampleFrom table 1 above and given that the probabilities are Boom 0.6, steady state 0.3 and recession0.1 then When conditions of the market are; boom launch product C: profit = 16 When conditions of the market are; steady state launch product B: profit = 6 When conditions of the market are; recession launch product B: profit = 12 The expected profit with perfect information will be

(16 x 0.6) + (6 x 0.3) + (12 x 0.1) = 12.6

our expected profit choosing product C is 7the maximum price that we would pay for perfect information is 12.6 – 7 = 5.6

7.2 DECISION TREES AND SUB SEQUENTIAL DECISIONS A decision tree is a graphic display of various decision alternatives and the sequence of events asif they were branches of a tree.



230 Decesion Theory


- The symbol and indicates the decision point and the situation of uncertainty

or event respectively. The node depicted by a square is a decision node while outcome

nodes are depicted by a circle.

- Decision nodes: points where choices exist between alternatives and managerial decisions ismade based on estimates and calculations of the returns expected.

- Outome nodes are points where the events depend on probabilities

Illustration of a tree diagram

Event

111

event B1

ACT E1 D2 B2 112

A1 E2 C1

D1 D3

A2 C2 121

For example 111 represents the payoff of the act event combination A1 – E1 – B1 When probabilities of various events are known they are written along the correspondingbranches. Joint probabilities are obtained by multiplying the probabilities along the branches

ExampleKauzi Agro mills ltd (KAM) is considering whether to enter a very competitive market. In caseKAM decided to enter this market it must either install a new forging process or pay overtime wages to the entire workers. In either case, the market entry could result in

i. high salesii. medium salesiii. low salesiv. no sales

a) Construct an appropriate tree diagramb) Suppose the management of KAM has estimated that if they enter the market

there is a 60% chance of their stakeholders approving the installation of thenew forge. (this means that there is a 40% chance of using overtime) a random

sample of the current market structure reveals that KAM has a 40% chance ofachieving high sales, a 30% chance of achieving medium sales, a 20% chance ofachieving low sales and a 10% chance of achieving no sales. Construct theappropriate probability tree diagram and determine the joint probabilities for various branches

c) Market analysts of KAM have indicated that a high level of sales will yield shs1,000,000 profit; a medium level of sales will result in a shs 600000 profit a lowlevel of sales will result in a shs 200000 profit and a no sales level will causeKAM a loss of shs 500000 apart from the cost of any equipment. Entering the

131

122



Lesson Eight 231


market will require a cash outlay of either shs 300000 to purchase and install aforge or shs 10000 for overtime expenses should the second option be selected.Draw the appropriate decision tree diagram

Solutiona) The tree diagram for this problem is illustrated as follows

The 1st stage of drawing a tree diagram is to show all decision points and outcome points donefrom left to right, concentrate first on the logic of the problem and on probabilities or valuesinvolved. This is called forward pass. The resultant is the figure below:

The entire sample space of act event choices is available to KAM are summarized in the tableshown belowPath Summary of alternative Act event sequence

0 – 1 – 3 – 5 Enter market, install forge, high sales0 – 1 – 3 – 6 Enter market, install forge, medium sales0 – 1 – 3 – 7 Enter market, install forge, low sales0 – 1 – 3 – 8 Enter market, install forge, no sales0 – 1 – 4 – 9 Enter market, use overtime, high sales0 – 1 – 4 – 10 Enter market, use overtime, medium sales0 – 1 – 4 – 11 Enter market, use overtime, low sales0 – 1 – 4 – 12 Enter market, use overtime, no sales0 – 2 Do not enter the market

Outcome/event

0

1

2

3

4

5

6

7

8

9

10

11

12

Act Act/event

Install forge

Use overtime

High sales

Medium sales

Low sales

No sales

High sales

Medium sales

Low sales

No sales

stop

Do not enter market

Tree diagram



232 Decesion Theory


b) The appropriate probability tree is shown in the figure below. The alternatives availableto the management of KAM are identified. The joint probabilities are the result of thepath sequence that is followed. For example, the sequence „enter market install forge,low sales‟ yields (0.6) (0.2) = 0.12 = probability to install forge and get low sales.

(c) The overall decision is determined after analysis of the expected values at various pointsso the correct decision (with the highest expected value is made. The stage is worked

from right to left and is known as the backward pass. - The expected value for a decision is the highest pay off value where as the E.V

for an outcome is the summation of probability x pay off value of each branch.In both cases any expenditure incurred due to the selection of the said option isdeducted.

- In our caseNode 3 = 000,501.0000,2002.0000,6003.0000,000,14.0

- 300,000E.V. = 615,000 – 300,000 = 315,000

Node 4 = 000,501.0000,2002.0000,6003.0000,000,14.0 - 10,000

E.V. = 615,000 – 10,000 = 605,000

Node 1 = (0.6 × 315,000) + (0.4 × 605,000)E.V. = 431,000

Node 0 = The highest of (0;431,000)Since not entering the market has a 0 expected value= 431,000 = thus the decision should be to enter the market.

0

1

2

3

4

Use overtime

Don‟t enter market

Enter Market 0.6

0.4

Install forge

(10,000)

(300,000)

0.4

0.3

0.2

0.1

0.4

0.3

0.2

0.1

HS = 0.24 = 1,000,000

MS = 0.18 = 600,000

LS = 0.12 = 200,000

NS = 0.06 = - 500,000

HS = 0.16 = 1,000,000

MS = 0.12 = 600,000

LS = 0.08 = 200,000

NS = 0.04 = - 500,000

Pay offs



Lesson Eight 233


This is represented as below in a tree diagram.

BAYES THEORY AND DECISION TREESIt makes an application of Bayes‟ Theorem to solve typical decision problems. This is examined

a lot so it is important to clearly understand it.

Example:Magana Creations is a company producing Ruy Lopez brand of cars. It is contemplatinglaunching a new model, the Guioco. There are several possibilities that could be opted for.- Continue producing Ruy Lopez which has profits declining at 10% per annum on a

compounding basis. Last year its profit was Shs. 60,000.- Launch Guioco without any prior market research. If sales are high annual profit is put at

Shs. 90,000 with a probability which from past data is put at 0.7. Low sales have 0.3probability and estimated profit of Shs. 30,000.

- Launch Guioco with prior market research costing Shs. 30,000 the market research willindicate whether future sales are likely to be „good‟ or „bad.‟ If the research indicates „good‟

then the management will spend Shs. 35,000 more on capital equipment and this willincrease annual profits to Shs. 100,000 if sales are actually high. If however sales are actuallylow, annual profits will drop to Shs. 25,000. Should market research indicate „good‟ andmanagement not spend more on promotion the profit levels will be as for 2nd scenarioabove.

- If the research indicate „bad‟ then the management will scale down their expectations to giveannual profit of Shs. 50,000 when sales are actually low, but because of capacity constrints ifsales are high profit will be Shs. 70,000.Past history of the market research company indicated the following results.

0

1

3

4

Use overtime

Don‟t enter market

Enter Market

EV = 431,000

Install forge

EV = 605,000

EV = 315,000

0.4

0.3

0.2

0.1

0.4

0.3

0.2

0.1

1,000,000

600,000

200,000

- 500,000

1,000,000

600,000

200,000

- 500,000

0

0.6

0.4



234 Decesion Theory


Actual salesHigh Low

Predicted saleslevel

Good 0.8* 0.1Bad 0.2 0.9

*When actual sales were high the market research company had predicted good sales level80% of the time.

Required:Use a time horizon of 6 years to indicate to the management of the company which optiontheory should adopt (Ignore the time value of money).

Solution(a) First draw the decision tree diagram

Computations; note how probability figures are arrived at.- The decision tree dictates that the following probabilities need to be calculated.

P(G)P(B)

2 A

B

C

1

D

E

Ruy Lopez(option 1)

GUIOCO(option 2)

MarketResearch(option 3)

Good

Bad

Extra 35,000

No extra

High 0.7

Low 0.3

P H G

P L G

0.95

0.05P H G

0.95

P L G

0.05P H B

0.34

P L B

0.66

60,000 (declining)

90,000

30,000

100,000

25,000

90,000

30,000

70,000

50,000

For market research



Lesson Eight 235


P(H|G)P(L|G)P(H|B)P(L|B)

P(G|H) = 0.8P(B|H) = 0.2P(G|L) = 0.1P(B|L) = 0.9P(H) = 0.7P(L) = 0.3

Good P(G&H) = P(H) × P(G|H)0.7 × 0.8 = 0.56

P(G&L) = P(L) × P(G|L)0.3 × 0.1 = 0.03

Bad B&H = P(H) × P(B|H)0.7 × 0.2 = 0.14

P(B&L) = P(L) × P(B|L)0.3 × 0.9 = 0.27

High 0.7 Low 0.3

P(G) = P(G and H) + P(G and L)= 0.56 + 0.03 = 0.59

P(B) = P(B and H) + P(B and L)= 0.14 + 0.27 = 0.41Note that P(G) + P(B) = 0.59 + 0.41 = 1.00

From Bayes‟ rule;

G| 0.56|G 0.95

0.59

G| 0.03|G 0.05

0.59

B| 0.14|B 0.34

0.41

B| 0.27|B 0.66

0.41

P H P H P H

P G

P L P L P L

P G

P H P H P H

P B

P L P L P L

P B

Evaluating financial outcome:Option 1:Last year Shs. 60,000 profits

Year Shs.1 = 60,000 × 0.9 = 54,000.02 = 60,000 × 0.92 = 48,000.03 = 60,000 × 0.93 = 43,740.04 = 60,000 × 0.94 = 39,366.05 = 60,000 × 0.95 = 35,429.56 = 60,000 × 0.96 = 31,886.5

253,022.0

For sales outcome;

Given





Lesson Eight 237


the action of others. Both competing sides face a similar problem. Hence game theory is ascience of conflictGame theory does not concern itself with finding an optimum strategy but it helps to improvethe decision process.Game theory has been used in business and industry to develop bidding tactics, pricing policies,advertising strategies, timing of the introduction of new models in the market e.t.c.

RULES OF GAME THEORYi. The number of competitors is finiteii. There is conflict of interests between the participantsiii. Each of these participants has available to him a finite set of available courses of

action i.e. choicesiv. The rules governing these choices are specified and known to all players

While playing each player chooses a course of action from a list of choices availableto him

v. the outcome of the game is affected by choices made by all of the players. The choicesare to be made simultaneously so that no competitor knows his opponents choice untilhe is already committed to his own

vi.

the outcome for all specific choices by all the players is known in advance andnumerically defined When a competitive situation meets all these criteria above we call it a game

NOTE : only in a few real life competitive situation can game theory be applied because all therules are difficult to apply at the same time to a given situation.

Example Two players X and Y have two alternatives. They show their choices by pressing two types ofbuttons in front of them but they cannot see the opponents move. It is assumed that bothplayers have equal intelligence and both intend to win the game. This sort of simple game can be illustrated in tabular form as follows:

Player YButton R Button t

Player X Button m X wins 2 points X wins 3 pointsButton n Y wins 2 points X wins 1 point

The game is biased against Y because if player X presses button m he will always win. Hence Y will be forced to press button r to cut down his losses

Alternative examplePlayer Y

Button R Button tPlayer X Button m X wins 3 points Y wins 4 points

Button n Y wins 2 points X wins 1 point

In this case X will not be able to press button m all the time in order to win(or button n).similarly Y will not be able to press button r or button t all the time in order to win. In such asituation each player will exercise his choice for part of the time based on the probability



238 Decesion Theory


Standard conventions in game theoryConsider the following table

Y3 -4

X -2 1

X plays row I, Y plays columns I, X wins 3 pointsX plays row I, Y plays columns II, X looses 4 pointsX plays row II, Y plays columns I, X looses 2 pointsX plays row II, Y plays columns II, X wins 1 points

3, -4, -2, 1 are the known pay offs to X(X takes precedence over Y)here the game has been represented in the form of a matrix. When the games are expressed inthis fashion the resulting matrix is commonly known as PAYOFF MATRIX

STRATEGYIt refers to a total pattern of choices employed by any player. Strategy could be pure or a mixed

oneIn a pure strategy, player X will play one row all of the time or player Y will also play one of thiscolumns all the time.In a mixed strategy, player X will play each of his rows a certain portion of the time and player Y will play each of his columns a certain portion of the time.

VALUE OF THE GAME The value of the game refers to the average pay off per play of the game over an extendedperiod of time

Example

3 4

6 2

Player Y

Player X

in this game player X will play his first row on each play of the game. Player y will have to playfirst column on each play of the game in order to minimize his loosesso this game is in favour of X and he wins 3 points on each play of the game. This game is a game of pure strategy and the value of the game is 3 points in favour of X

ExampleDetermine the optimum strategies for the two players X and Y and find the value of the gamefrom the following pay off matrix

3 -1 4 2

-1 -3 -7 0

4 -7 3 -9

Player Y

Player X

Strategy assume the worst and act accordinglyif X plays first



Lesson Eight 239


if X plays first with his row one then Y will play with his 2nd column to win 1 point similarly if Xplays with his 2nd row then Y will play his 3rd column to win 7 points and if x plays with his 3rd row then Y will play his fourth column to win 9 pointsIn this game X cannot win so he should adopt first row strategy in order to minimize losses This decision rule is known as „maximum strategy‟ i.e. X chooses the highest of these minimumpay offs

Using the same reasoning from the point of view of yIf Y plays with his 1st column, then X will play his 3rd row to win 4 pointsIf Y plays with his 2nd column, then X will play his 1st row to lose 1 pointIf Y plays with his 3rd column, then X will play his 1st row to win 4 pointsIf Y plays with his 4th column, then X will play his 1st row to win 2 points

Thus player Y will make the best of the situation by playing his 2nd column which is a „Minimaxstrategy‟ This game is also a game of pure strategy and the value of the game is – 1(win of 1 point pergame to y) using matrix notation, the solution is shown below

Row Minimum

column maximum

3 -1 4 2 1

-1 -3 -7 0 7

4 -7 3 -9 9

4 -1 4 2

Player Y

Player X

In this case value of the game is – 1Minimum of the column maximums is – 1Maximum of the row is also – 1i.e. X‟s strategy is maximim strategy

Y‟s strategy is Minimax strategy Saddle Point The saddle point in a pay off matrix is one which is the smallest value in its row and the largest value in its column. It is also known as equilibrium point in the theory of games.Saddle point also gives the value of such a game. In a game having a saddle point, the optimumstrategy for both players is to pay the row or column containing the saddle point.Note: if in a game there is no saddle point the players will resort to what is known as mixedstrategies.

Mixed StrategiesExampleFind the optimum strategies and the value of the game from the following pay off matrixconcerning two person game

1 4

5 3

Player Y

Player X

In this game there is no saddle pointLet Q be the proportion of time player X spends playing his 1 st row and 1-Q be the proportionof time player X spends playing his 2nd row



240 Decesion Theory


SimilarlyLet R be the proportion of time player Y spends playing his 1st column and 1-R be theproportion of time player Y spends playing his second row The following matrix shows this strategy

1

1 4

1 5 3

Player Y

R R

Q Player X

Q

X’s strategyX will like to divide his play between his rows in such a way that his expected winning or loses when Y plays the 1st column will be equal to his expected winning or losses when y plays thesecond column

Column 1Points Proportion played Expected winnings1 Q Q5 1-Q 5(1-Q)

Total = Q + 5(1 – Q)Column 2

Points Proportion played Expected winnings4 Q 4Q3 1-Q 3(1-Q)

Total = 4Q + 3(1 – Q) Therefore Q + 5(1-Q) = 4Q +3(1-Q)

Giving Q =5

2 and (1-Q) =5

3

This means that player X should play his first row5

2 th of the time and his second row5

3 th of

the timeUsing the same reasoning1×R + 4(1-R) = 5R +3(1-R)

Giving R =5

1 and (1-R) =5

4

This means that player Y should divide his time between his first column and second column inthe ratio 1:4

1 45 5

25

35

1 4 5 3

Player Y

Player X

Short cut method of determining mixed matrices

1 4

5 3

Player Y

Player X



Lesson Eight 241


Step ISubtract the smaller pay off in each row from the larger one and smaller pay off in each columnfrom the larger one

1 4 4-1 3

5 3 5-3 2

5 1 4 4 3 1

Step IIInterchange each of these pairs of subtracted numbers found in step I

1 4 2

5 3 3

1 4

Thus player X plays his two rows in the ratio 2: 3 And player Y plays his columns in the ratio 1:4 This is the same result as calculated before

To determine the value of the game in mixed strategiesIn a simple 2 x 2 game without a saddle point, each players strategy consists of two probabilitiesdenoting the portion of the time he spends on each of his rows or columns. Since each playerplays a random pattern the probabilities are listed under

Pay off Strategies which produce this pay off Joint probability1 Row I column I

252

51

52

4 Row I column II25

85

45

2

5 Row II column I25

35

15

3

3 Row II column II 2512

54

53

Expected value (or value of the game)Pay off Probability p(x) Expected value x (p(x)1

252

252

425

8 25

32

525

3 25

15

325

12 25

36

Ƹx p(x) = 85/25 = 17/5 = 3.43.4 is the value of the game

DominanceDominated strategy is useful for reducing the size of the payoff tableRule of dominance



242 Decesion Theory


i. If all the elements in a column are greater than or equal to the correspondingelements in another column, then the column is dominated

ii. Similarly if all the elements in a row are less than or equal to the correspondingelements in another row, then the row is dominatedDominated rows and columns may be deleted which reduces the size of the gameNB always look for dominance and saddle points when solving a game

ExampleDetermine the optimum strategies and the value of the game from the following 2xm pay offmatrix game for X and Y

6 3 1 0 3

3 2 4 2 1

Y

X

In this columns I, II, and IV are dominated by columns III and V hence Y will not play thesecolumnsSo the game is reduced to 2×2 matrix, hence this game can be solved using methods alreadydiscussed

1 3

4 1

Y

X

GRAPHICAL METHODGraphical methods can be used in games with no saddle points and having pay off m×2 or 2×nmatrix The aim is to substitute a much simpler 2×2 matrix for the original m×2 or 2×m matrix

Example IDetermine the optimum strategies and the value of the game from the following pay off matrix

game.

6 3 1 0 3

3 2 4 2 1

Y

X

Draw two vertical axes and plot two pay offs corresponding to each of the five columns. Thepay off numbers in the first row are plotted on axis I and those in second row on axis II



Lesson Eight 243


Axis I Axis II 2 K 2

6 A 6 1 1

5 5 0 0

4 4 -1 L -1

3 B 3 -2 -2

2 2 -3 -3

1 1 -4 -4

0 0 -5 -5

-1 -1 -6 -6

-2 T -2 -7 -7

-3 K -3 -8 -8

-4 L -4 -9 -9

Example I Example II

Thus the two pay off number 6 and 3 in the first column are shown respectively by point A onaxis I and point B on axis IIOn the two intersecting lines at the very bottom thicken them from below to the point ofintersection i.e. highest point on the boundary.

The thick lines on the graph KT and LT meet at T The two lines passing through T identify the two critical moves of Y which combined with Xyield the following 2 × 2 matrix

1 3

4 1

Y

X

The value of the game and the optimum strategies can be calculated using the methodsdescribed earlier

Example IIDetermine the optimum strategies and the value of the game from the following pay off matrix

concerning two person 4 × 2 game

6 2

3 4

2 9

7 1

Y

X



244 Decesion Theory


The method is similar to the previous example, except we thicken the line segments which bindsthe figure from the top and taken the lowest point on the boundary The segments KP, PM and ML drawn in thick lines bind the figure from the top and theirlowest intersection M through which the two lines pass defines the following 2 × 2 matrixrelevant to our purpose

3 4

7 1

Y

X

The optimal strategies and the value of the game can now be calculated

Non Zero Sum GamesUntil recently there was no satisfactory theory either to explain how people should play non zerogames or to describe how they actually play such gamesNigel Howard (1966) developed a method which describes how most people play non zero sumgames involving any number of persons

ExampleEach individual farmer can maximize his own income by maximizing the amount of crops thathe produces. When all farmers follow this policy the supply exceeds demand and the prices fall.On the other hand they can agree to reduce the production and keep the prices high This creates a dilemma to the farmer This is an example of a non zero sum gameSimilarly marketing problems are non zero sum games as elements of advertising come in. insuch cases the market may be split in proportion to the money spent on advertising multipliedby an effectieness factor

Prisoners DilemmaIt is a type of non zero sum game and derives its name from the following story

The district attorney has two bank robbers in separate cells and offers each a chance ofconfession. If one confesses and the other does not then the confessor gets two years and theother one ten years. If both confess they will get eight years each. If both refuse to confess thereis only evidence to ensure convictions on a lesser charge and each will receive 5 years

Another example The table below is a pay off matrix for two large companies A and B. initially they both have thesame prices. Each consider cutting their prices to gain market share and hence improve profit

Corporation B

Corporation A

The entries in the pay off matrix indicate the order of preference of the players i.e. first A thenB. We may suppose that if both player study the situation, they will both decide to play row Icolumn I(3,3).

Maintain prices Decrease pricesmaintain prices 3,3 status quo 1 , 4 B gets market share and

profitDecrease prices 4, 1, A gains market

share and profit(2,2) Both retain market sharebut lose profit



Lesson Eight 245


HoweverSuppose A‟s reasoning is as follows If B plays column I then I should play row 2 because I will increase my gain to 4In the same way B‟s reasoning may be as follows If A plays row I then I should play column 2 to get pay off 4 per playIf both play 2(row 2 column 2) each two receives a pay off of 2 only

In the long run pay off forms a new equilibrium point because if either party departs from it without the other doing so he will be worse off before he departed from itGame theory seems to indicate that they should play (2,2) because it is an equilibrium point butthis is not intuitively satisfying. On the other hand (3,3) is satisfying but does not appear toprovide stability. Hence the dilemma.

Theory of Metagames This theory appears to describe how most people play non zero sum games involving a numberof personsPrisoners dilemma is an example of this. The aim is to identify points at which players actuallytend to stabilize their play in non zero sum games. This theory not only identifies equilibrium points missed by traditional game theory in games

that have one or more such points but also does so in games in which traditional theory finds nosuch pointIts main aim is that each player is trying to maximize the minimum gain of his opponent

ADVANTAGES AND LIMITATIONS OF GAME THEORY

AdvantageGame theory helps us to learn how to approach and understand a conflict situation and toimprove the decision making process

LIMITATIONS1. Businessmen do not have all the knowledge required by the theory of games. Most often

they do not know all the strategies available to them nor do they know all the strategies

available to their rivals2. there is a great deal of uncertainty. Hence we usually restrict ourselves to those games with

known outcomes3. The implications of the Minimax strategy is that the businessman minimizes the chance of

maximum loss. For an ambitious business man, this strategy is very conservative4. the techniques of solving games involving mixed strategies where pay off matrices are rather

large is very complicated5. in non zero sum games, mathematical solutions are not always possible. For example a

reduction in the price of a commodity may increase overall demand. It is also not necessarythat demand units will shift from one firm to another





Lesson Eight 247


The managers believe that if they employ specialist-engineering consultants, their chancesof finishing the building on time will be trebled. But if the building is still late, it wouldonly be one or two months late, with equal probability.

Requireda) To draw a tree diagram to represent this decision problem, using squares for

decision points, circles for random outcomes, and including probabilitiesrevenues and penalties;

b) To analyse the tree using expected value techniques:

c) To write a short report for the managers, with reasons and comments,recommending which decision to make.

QUESTION THREE

Define minimax and maximax decision rules

QUESTION FOUR

A has two ammunition stores, one of which is twice as valuable as the other. B is an attacker who can destroy an undefended store but he can only attack one of them. A can onlysuccessfully defend one of them.

What would A do so as to maximize his return from the situation no matter what B may do?

QUESTION FIVE

Determine the optimum strategies and the value of the game for the following pay off matrix.

X Y

1 2 -1-2 1 1

2 0 1Compare your answers with those given in lesson 9 of the study pack





Lesson Eight 249


Solution:1. Formulating a linear programming model

Step 1: Identifying variables: The variables here are the number of units of Benko and Benoni lager produced by Longcastling breweries per day; we can represent them as:

X 1= a unit of Benko lager.

X 2= a unit of Benoni lager.

Step 2: Identify the objective:

Definition : An objective is the desired result i.e. optimization of a function dependent ondecision variable and subject to some constraints.

The objective of Long Castling Breweries is to maximize daily contribution. Objective functionis the formula that will give us the total contribution in a day for both Benko lager and Benonilager. The information above can thus be represented in a tabular form as:

PRODUCT Maximum available(PER DAY) X 1 X 2 hours/dayMachine hours 0.5 0.33 12Labor hours 0.5 0.5 14Contribution 4 3

Objective function = 4X 1 + 3X 2 The objective is to maximize 4X 1 + 3X 2

Step 3: Identifying constraints (constraints formulation)

Definition : Constraints are circumstances that govern achievement of an objective.Limitations must be quantified mathematically and they must be linear.

For Long castling breweries we have limited machine hours (12hrs/day), which must be shared

among production of Benko and Benoni lagers. Therefore production must be such that the numbers of machine hours required is less than orequal to 12 hours per day.

0.5X 1+0.33X 2 ≤ 12hrs

Similarly for labour hours we have:0.5X 1+0.5X 2 ≤ 14hrs

Non-Negativity: It is logical assumption to assume that the company cannot manufacturenegative amounts of a product, thus it can only manufacture either zero product or more. Therefore we have: X 1≥0

X 2≥0 or X 1, X 2≥0

Thus the complete linear programming model is;Maximize 4X 1 + 3X 2

Subject to the constraints;0.5X 1 + 0.33X 2 ≤ 12 0.5X 1 + 0.5X 2 ≤ 14 X 1, X 2 ≥ 0



250 Decesion Theory


Solving linear programming problems The question requires us to optimize (in our case, maximize) the objective (the contributionfunction), or in simple terms we are required to solve the linear programming model.Solving linear programming model entails finding the values of variables that satisfy allinequalities simultaneously and optimize the objective.

Graphical solution This method is used to solve LP models in case where only two variables are involved. For morethan two variables (multivariable) then the simplex technique (algebraic method may be used).Now in solving the problem above we first draw the axis, taking X 1 to be the y axis and X 2 to bethe X axis.

X1

0

X2

Next we plot the scales on each axis to approximate the scales to use them, we consider eachconstraint equation. We get the value of one of variables putting the other variable to be zero

and by substituting the inequality ≤or ≥ with equality sign (=).

For: 0.5X 1 + 0.33X 2 ≤ 12 When X 1 = 0

0.5(0) + 0.33X 2 = 120.33X 2 = 12X 2 = 12/0.33 = 36

Therefore point to plot is (36, 0) … Implying that when X 2 = 36 then X 1 = 0

When X 2 = 00.5X 1 + 0.33(0) = 120.5X 1 = 12X 1 = 12/0.5 = 24

Therefore the point is (0, 24)

For 0.5X 1 + 0.5X 2 ≤ 14 When X 1 = 0

0.5(0) + 0.5X 2 = 140.5X 2 = 14X 2 = 14/0.5 = 28




Lesson Eight 251


When X 2 = 00.5X 1 + 0.5(0) = 140.5X 1 = 14X 1 = 14/0.5 = 28


Comparing these values we see that X 2 ranges between 0 – 28, therefore we can have the graphplotted as:

Next draw each limitation (constraint) as separate line on the graph.For 0.5X 1 + 0.33X 2 ≤ 12 The two points that represent this line are (36, 0) and (0, 24). This is plotted as a straight linefrom 36 on X 2 axis to 24 on X 1 axis.



252 Decesion Theory


Now including the Non-Negativity constraints since no negative product can be produced;X 1 ≥ 0; x2 ≥ 0



Lesson Eight 253


We must now consider how to choose the production which will maximize contribution. This we do by plotting a line representing the objective function (4x1 + 3x2 ).First choose a convenient point inside the feasible region

eg X 2(10) +3X 1(20) = 40 +60= Sh 100

All of the other product mixes that give a contribution of Sh.100 lies on the line:

100 = 4X 1 + 3X 2 ..................................................................................... (i)<<This line is called a contribution line>>

Picking another point, say X 2 =10 ad X 1 = 20Its contribution value is SH 110, thus give a contribution line of

110 = 4X 1 + 3X 2 ........................................................................................ (ii)

Plotting these two contribution lines to our graph we get two parallel lines.

Until we reach the last feasible solution(s) before the line moves entirely out of the feasibleregion.



254 Decesion Theory


Point X is the last feasible solution. Coordinates of this point give a combination of the twolager‟s production volumes that fetches the highest contribution. Coordinates of point X can be read from the graph, but for precision they are calculated bysolving simultaneously the equations of the two lines that intersect at point X.

The two constraints are called binding or limiting constraints. They are the resources being fullyused thus preventing daily contribution from increasing further. Therefore to get point x we solve:

0.5X 1 + 0.33X 2 = 12 ….(i) 0.5X 1 + 0.5X 2 = 14 ….(ii)

Since X is the intersection of these two constraints, solving by deducting (i) from (ii) we get0.17X 2=2X 2 =11.76

And substituting X 2 = 11.76 to equation (i) we getX 1 = 16.24

Therefore 11.76 units of Benko lager and 16.24 units of Benoni lager need to be produced formaximum contribution.

Contribution = 4(16.24) + 3(11.76) = 100.24

Assumption made in linear programming Assumptions that are made to solve these types of problems are that:.

Proportionality: all activities in linear programming problems are proportional to thelevel of decision variables.



Lesson Eight 255


Divisibility: the solution to a linear programming problem does not have to be aninteger but for strictly whole number solutions, use integer programming.Non-negativity: no decision variable can be negative. Additivity: the total of all activities in linear programming problems are assumed toequal to the sum of individual activities.

Special cases in linear programmingInfeasibility: This is w hen all constraints don‟t satisfy a particular point thus there is nofeasible solution.Redundancy: A constraint is considered redundant if it does not affect the feasibleregion. This happens in cases of excess resources since it does not limit attainment ofthe objective.Multiple optimal solutions: This occurs when the objective function has the same slopeas a binding constraint.

Minimizing problem

Example

A manufacturing company has acquired new machine for producing product P at a rate of 25units per hour with a 98% rate of efficiency. The company requires to produce atleast 1800 unitsof P per day. The 10 old machines that the company has, produce 15 units of P with a 95%efficiency. The cost of operating the new machine is Sh. 4 per hour and Sh. 3 per hour for the old ones. The cost incurred due to inefficiency is Sh. 2 per unit; It is government policy that at least 2 ofthe new machines must be indulged into production. The company wishes to optimally allocate the machines in order to minimize the totalmanufacturing cost if the total available hours for production in a day are 8 hours.

SolutionIn minimizing problems we use ≥ (greater or equal to) type inequality.

Step 1: Identifying variables.Since the problem requires us to appropriately allocate the machines in order to minimize thecosts thus our variables are the new machines and the old machines, we can let;

X 1 =new machinesX 2=old machines

Step 2: Identify objectives: The objective is to minimize manufacturing costs. Total manufacturing cost per machine=Operating cost + (inefficiency rate × number of units ×cost of loss) Therefore cost for new machines

Cost = 4+(0.02 × 25 × 2) = Sh. 5 per hour=5 × 8= Sh. 40 per day]Similarly for old machinesCost = 3+(0.05×15×2) = Sh. 4.5 per hour

=4.5×8=Sh. 36 per day Therefore the objective function is to minimize 40X 1 + 36X 2 Step 3: Identifying constraint functions.

X 1 ≤ 8 X 2 ≤ 10



256 Decesion Theory


(25 × 8) X 1 +(8 × 15) X 2 ≥1800 (This can be simplified further as) 200 X 1 +120 X 2 ≥1800 (dividing through by 40) 5X 1 +3 X 2 ≥45 X 1 ≥2 X 1, X 2 ≥0

Thus the LP model is:Minimize 40X 1 + 36X 2 Subject to: X 1 ≤ 8

X 2 ≤ 10 5X 1 +3 X 2 ≥45 X 1 ≥2 X 1, X 2 ≥0

Plotting this on a graph we get:

The line X 1 ≥2 does not affect the feasible region (doesn‟t cause reduction of the feasibleregion), this constraint doesn‟t limit attainment of the objective, thus its kno wn as a redundantconstraint.Now picking a convenient point inside the feasible region, say (6, 10)

We get a total cost of 600 = (6(40)+10(36)) Thus the objective function line of 40X 1 + 36X 2 = 600

Moving this line parallel toward the origin to locate the last apex before the line completely falloff the feasible region, we get:

X1

X2

8



Lesson Eight 257


Point X is the point of optimal solution The binding constraints here are:

X 1 ≤ 8 And 5X 1 +3 X 2 ≥45 Solving this to get coordinates of point X

5X 1 +3 X 2 = 45 (when X 1 = 8)40 + 3 X 2 = 45

X 2 = 5/3 Thus the solution is X 1 = 8

X 2 = 5/3

Shadow or dual pricesDefinition : A shadow price or a dual price is the amount increase (or decrease) of the objectivefunction when one more (or one less) of the binding constraints is made available.Consider example 1 .

Maximize 4X 1 + 3 X 2Subject to: 0.5X 1 + 0.33X 2 ≤ 12 (Machine hours)

0.5X 1 + 0.5X 2 ≤ 14 (labor hours)

Starting with machine hours; let‟s assume that one more machine hour is available (with laborhours remaining constant) We get:

0.5X 1 + 0.33X 2 = 130.5X 1 + 0.5X 2 = 14

Solving this simultaneously we get the values of X 1 and X 2 as0.17 X 2 = 1X 2 = 5.88X 1 = 22.12



258 Decesion Theory


Thus the contribution is4(22.12) + 3(5.88) = Sh.106.12

Comparing this with its original contribution of Sh.100.24 (see example 1) we see increasingmachine hours by one unit has increased contribution by Sh.5.88, which is the shadow price permachine hour.Note: This figure is also arrived at if we assume that machine hours are reduced by 1 unit ie 12-

1.Similarly assuming that one more labor hour is made available, then contribution change is:0.5X 1 + 0.33X 2 = 120.5X 1 + 0.5X 2 = 15

Solving this simultaneously gives:0.17 X 2 =3X 2 = 17.65X 1 = 12.35

Which give a contribution of:4(12.35) + 3(17.65) = Sh.102.35

The contribution change is Sh.2.11 which is the shadow price per labor hour.

Note: The shadow prices apply in so far as the constraint is binding for example if more and morelabor hours are available it will reach a point where labor hours are no longer scarce thus laborhours cease to be a binding constraint and its shadow price becomes a zero.(All non-bindingconstraints have zero shadow price). Logically its senseless to pay more to increase a resource, which is already abundant.

Interpretation of shadow prices A shadow price of a binding constraint indicates to management how much extra contribution will be gained by increasing a unit of the scarce resource.In the example above Sh.2.11 is the shadow price for labor hours. This implies that managementis ready to pay up to Sh.2.11 extra per hour for the extra hours i.e. say an employee is paid sh.5per hour and one day he works for two hours extra (overtime), the management is prepared to

pay up to sh.7.11 per hour for the two hours overtime worked.

Sensitivity AnalysisDefinition : Sensitivity analysis is the test of how certain changes in resources affect the optimalsolution.

In sensitivity analysis we consider the effect of additional limiting or non-limiting constraints. We already know that adding more non-limiting constraints does not change the optimalsolution. We also know that adding more binding constraints affects the objective function.It is very important for the management to know how much of a limit resource can be madeavailable until it has no effect on the objective function (ie ceases to be a binding resource)

SIMPLEX METHOD When analyzing linear programming problems with three or more variables the graphicalmethod becomes enadequate, in such cases we employ simplex method . Simplex method is analgebraic procedure for solving systems of equations requiring optimization of the objectivefunction.. This method can be applied to any number of variables, the more they are the more complex itbecomes to workout a solution on paper. Computer programs e.g. Tora are used to solve themost intricate problems.





260 Decesion Theory


Thus the tableu will appear as follows.Solution Variable

Products Slack Variables SolutionQuantity

1 x 2 x 3 x 4 x

2 x 14

1 120

0 20

4 x 10 15 0 1 450

Z 45 80 0 0 0

5. next we conduct row operations that aim to reduce elements falling in the same columnas the previously marked element to zero. These row operations may sometimenecessitate multiplying or deviding the selected row with an arbitrary number.

Therefore:Row 2 – 15×Row 1

4

15 152 4 20

* 31

4 4 4

10 15 0 1 450

15 15 0 300

6 0 1 150

x

x

new x

note that the aim was to attain the zero.

2

*

45 80 0 0 0

80 20 80 4 0 1600

25 0 4 0 1600

Z

x

new Z

on replacing the new rows to the tableu we get2nd tableu.

Solution Variable


1 x 2 x 3 x 4 x

2 x 14

1 120

0 20

4 x 14

6 0 34

1 150

Z 25 0 -4 0 -1600

Since in the Z row under products column we still have values greator than zero, weconduct another operation.

Taking the column with a Z value of 25, we repeat the process in the same manner.14

14

20 80

150 6 24

thus we pick the x4 row and mark the element 14

6 , the row solution variable is changed to

x1 and we devide the row by 14

6 to convert the marked element to 1.

Therefore; x1 1 0 325 4

25 = 24

Next we do the row operations



Lesson Eight 261


1 12 4 20

31 41 4 100 100

* 1 12 50 25

1 0 20

4 0 6

0 1 24

x

x

new x

1

*

25 0 4 0 160025 25 0 3 4 600

0 0 7 4 2200

Z x

new Z

the tableu becomes, 3rd TableuSolution Variable


1 x 2 x 3 x 4 x

2 x 0 1 150

125

24

1 x 1 0 325

425

24

Z 0 0 -7 -4 -2200 This is the final tableu since the Z row has no values greator than zero thus we have theoptimal solution.

Interpretation

to maximise Z we need to produce 24 units x2 and 24 units of x1, we obtain these values from the solutions quantity column

thus, Z = 24(45) + 24(80) = 3000

we have zero slack (unused quantities of constraints). Assume tableu 2 is the final tableu and let the constraint with variable x3 be labour

hours and x4 be raw materials, the slack wouldhve been interpreted as:o 150 units of raw materials were unusedo to maximize Z we produce 20units of x2 and none of x1.

The values represnts in Z row under slack vaiable column represents shadow prizes. Thus the shadow prize for the first constraint with x3 is 7 and the shadowprize for the second constraint with the vaiable x4 is 4.

INTERPRETATION OF COMPUTER GENERATED SOLUTION

Example

Maximize 1 2 225 20 24 x x x

1 2 3: , ,where x Xtragrow x Youngrow x Zupergrow

Subject to 1 30.3 0.2 500 x x

2 3

1 2 3

2 3

1

1 2 3

0.5 0.4 1000

0.2 0.1 0.1 800

0.4 0.3 600

1500

0, 0, 0

x x

x x x

x x

x

x x x



262 Decesion Theory


The computer generated solution for this problem is as follows;Objective value = 71666.7

Variable Value Obj. Coeff Obj Value ContributionX1: Xtragrow 1666.7 25 4166.7X2: Youngrow 1500 20 30000X3: Zupergrow 0 24 0

Constraint RHS Slack-/Surplus+1(<) 500 02(<) 1000 250-3(<) 800 316.67-4(<) 600 05(>) 1500 166.7+

Sensitivity Analysis Variable Current obj coeff Min obj coeff Max Obj Coeff Reduced costX1: Xtragrow 25 13.50 Infinity 0X2: Youngrow 20 9.78 Infinity 0X3: Zupergrow 24 -Infinity 31.67 7.67

Constraint Current RHS Min RHS Max RHS Dual price1(<) 500 450 975 83.332(<) 1000 750 Infinity 03(<) 800 483.3 Infinity 04(<) 600 0 800 505(>) 1500 -Infinity 1666.7 0

RequiredInterpret the data generated by the the computer.

Solution. Table 1:Objective value, is the solution to objective function (e.g the solution to this example is 71,436) The four columns of table 1 are to be interpreted as follows;

Variable: these are the variables of the model. In our example we have x 1 = Xtragrow , x 2 = Youngrow and x 3 = Zupergrow Value: this is value that the variables assume at optimal solution (to optimize theobjective function one needs to produce this amounts of the variables). In our example weare required to produce 1,666.67 of x1 and 1,750 of x2 and none of x3 Objective coefficient: these are the coefficients of the objective functionObjective value contribution: this is the value contributed by each variable to theobjective function (for x1=25×1,666.67), the total of this is equal to our objective value(i.e 41,666.67+35,000=76,666.67).

The 3 columns of the second part of table1 can be interpreted as follows;Constraints: this is constraints of the model representing the limited resources.RHS: the Right hand side value is the limiting value of the constraint. E.g for the firstconstraint the maximum amount of material A is 500 tons. Slack/surplus: at optimal production not all the materials for some of the constrants willbe fully utilized, slack is the amount of material that is left over after production. Forconstraint 1 and 4 no material remained, this also implies that these are the binding constraints i.e theiradjustment directly affects the objective solution



Lesson Eight 263


Sensitivity Analysis This is the analysis of the effect of adjusting variables or constraint, whether te objectivesolution will be affected. How much of the objective coefficient (or the maximum availableamount of a constraint) can be reduced or increased without affecting the objective solution.

The columns of this table can be interpreted as follows; Variable: as explained above

Current objective coefficient: this is the value of coefficients of the objective function

Minimum objective coefficient: this is how low the coefficient can be reduced withoutaffecting the optimal basis . The coefficient for x1 can be reduced from 25 to 13.50 (the prize for

Xtragrow fall to as low as $13.50 from $25) but the optimal solution will remain the same

Maximum objective coefficient: this is how high the coefficient can be increased withoutaffecting the optimal basis

Reduced cost: this is amount by which the coefficient of the variable has to be adjusted

with for it to become a basic variable (included to the objective optimal solution). X1and x2 have 0 reduced costs implying that they already make part of the optimal solution, x3 willrequire to be increased by 7.67 for it to make part of the basic variable.

The second part of the table is interpreted as follows;Constraints: as described above Current Rihgt Hand Side: the limiting value of the constraint. E.g for the first constraint themaximum amount of material A is 500 tons. Minimum RHS: the lowest the available amount of the constraint can be reduced without affecting the optimal basis Maximum RHS: the highest the available amount of the constraint can be increased without affecting the optimal solution

Dual price: this is amount increase to the objective contribution due to a unit increaseof the available constraint. Since there are only 500 tons of material A if management decides toincrease it by a unit to 501tons then the objective optimal solution will be 76,666.67+83.33.

TRANSPORTATION A transportation problem deals with a number of sources of supply (e.g a manufacturingcompany, warehouse) and a number of destinations (e,g shops, houses). The usual objective isminimizing transportation costs of supplying items from a set of source points to a set ofdestinations. A major characteristic of this problem is the linearity requirement, i.e. transport cost fom onepoint to another must be clearly defined, if it will cost sh.50 to transport a bag from a warehouseto shop A then it will cost sh.250 to transport 5 bags.

Assumptions

The model assumes a homogeneous commodity, one type of commodity Total supply is equal to total demand

Example 164 chambers a computer support firm has three branches at different parts of the city, it receivesorders for a total of 15 desktop computers from four customers. In total in the three branches



264 Decesion Theory


there are 15 machines available. The management wish to minimise delivery costs by dispatchingthe computers from the appropriate branch for each customer.

Details of the availabilities, 'requirements, and transport costs per computer are given in thefollowing table.

Table 1Customer Customer Customer Customer Total

A B C DComputers 3 3 4 5 15

transportationcost

per unit

Branch X. 2 Sh.13 11 15 20 Available Branch Y 6 Sh.17 14 12 13

Branch Z 7 Sh.18 18 15 12 Total 15

SolutionStep 1 Make an initial feasible allocation of deliveries by selecting the cheapest route first, and

allocate as many as possible then the next cheapest and so on. The result of such an

allocation is as follows.Table 2

Requirement A B C D3 3 4 5

X 2 Units 2(1) Available Y 6 Units 1(4) 1(3) 4(2)

Z 7 Units 2(5) 5(2)

Note: the number in the table represent deliveries of computers and the number in the brackets (1), (2), etcrepresent the sequence in which they are inserted, lowest cost first i.e.

Sh.

1. 2 units X → B sh.11/unit Total cost 222. 4 units Y → C sh.12/unit Total cost 485 units Z → D sh.12/unit Totals cost 60

3. The next lowest cost move which is feasible i.e. doesn‟t exceed row or column totals is 1unit Y → B sh.14/unit 14

4. similarly the next lowest feasible allocation 1 unit Y→ Ash.17/unit 17

5. finally to fulfill the row /column totals 2 units Z → A sh.18/unit __36197

Step 2. Check solution obtained to see if it represents the minimum cost possible. This is doneby calculating „shadow costs‟ (i.e. an imputed cost of not using a particular route) andcomparing these with the real transport costs to see whether a change of allocation isdesirable.

This is done as follows:Calculate a nominal 'dispatch' and 'reception' cost for each occupied cell by making anassumption that the transport cost per unit is capable of being split between dispatch andreception costs thus:



Lesson Eight 265


D(X) + R(B) = 11D(Y) + R(A) = 17D(y) + R(B) = 14D(Y) + R(C) = 12D(Z) + R(A) = 18D(Z) + R(D) = 12

Where D(X), D(Y) and D(Z) represent Dispatch cost from depots X, Y and Z, and R(A) R(B),R(C) and R(D) represent Reception costs at customers A, B, C, D.

By convention the first depot is assigned the value of zero i.e. D(X) = 0 and this value issubstituted in the first equation and then all the other values can be obtained thus

R(A) = 14 D(X) = 0R(B) = 11 D(Y) = 3R(C) = 9 D(Z) = 4R(D) = 8

Using these values the shadow costs of the unoccupied cells can be calculated. The unoccupiedcells are X : A, X : C, X : D, Y : D, Z : B, Z : C.

Shadowcosts

D(X) + R(A) = 0 + 14 = 14D(X) + R(C) = 0 + 9 = 9D(X) + R(D) = 0 + 8 = 8D(Y) + R(D) = 3 + 8 = 11D(Z) + R(B) = 4 + 11 = 15D(Z) + R(C) = 4 + 9 = 13

These computed 'shadow costs' are compared with the actual transport costs (from Tab- I), Where the actual costs are less than shadow costs, overall costs can be reduced by allocating unitsinto that cell.

Actual Shadow + Cost increasecost - cost - Cost reduction

CellX:A 13 - 14 = -1X : C 15 - 9 = +6X : D 20 - 8 = + 12 Y: D 13 - 11 = +2Z : B 18 - 15 = +3Z : C 15 - 13 = +2

The meaning of this is that total costs could be reduced by sh.1 for every unit that can betransferred into cell X : A. As there is a cost reduction that can be made the solution , Table 2is not optimum.

Step 3: Make the maximum possible allocation of deliveries into the cell where actual costs areless than shadow costs using occupied cells i.e.

Cell X : A from Step 2, The number that can be allocated is governed by the need to keep withinthe row and column totals. This is done as follows:



266 Decesion Theory


Table 3Requirement

A B C D3 3 4 5

X 2 Units + 2 -

Available Y 6 Units 1 - 1 + 4Z 7 Units 2 5

Table 3 is a reproduction of Table 2 with a number of + and - inserted. These were inserted forthe following reasons.Cell X : A + indicates a transfer in as indicated in Step 2Cell X : B - indicates a transfer out to maintain Row X total.Cell Y : B + indicates a transfer in to maintain Column B totalCell Y : A - indicates a transfer out to maintain Row Y and Column A totals.

The maximum number than can be transferred into Cell X : A is the lowest number in the

Minus cells i.e. cells Y : A, and X : B which is 1 unit. Therefore 1 unit is transferred in the + and - sequence described above resulting in the followingtable

Table 4Requirement

A B C D3 3 4 5

X 2 Units 1 1 Available Y 6 Units 2 4

Z 7 Units 2 5

The total cost of this solution is

Sh.Cell X:A 1 unit @ sh.13 = 13Cell X:B 1 Unit @ sh.11 = 11Cell Y:B 2 Units @ sh.14 = 28Cell Y:C 4 Units @ sh.12 = 48Cell Z:A 2 Units @ sh.18 = 36Cell Z:D 5 Units @ sh.12 = 60

196

The new total cost is sh.1 less than the total cost established in Step 1. This is the result expected

because it was calculated in Step 2 that sh.1 would be saved for every unit we were able totransfer to Cell X : A and we were able to" transfer 1 unit only.

Notes: Always commence the + and - sequence with a + in the cell indicated by the (actual cost -shadow cost) calculation. Then put a - in the occupied cell in the same row which has anoccupied cell in its column. Proceed until a - appears in the same column as the original +.

Step 4. Repeat Step 2 i.e. check that solution represents minimum cost. Each of the processes in



Lesson Eight 267


Step 2 are repeated using the latest solution (Table 4) as a basis, thus: Nominal dispatchand reception costs for each occupied cell.

D(X) + R(A) = 13D(X) + R(B) = 11D(y) + R(B) = 14

D(Y) + R(C) = 12DZ) + R(A) = 18D(Z) + R(D) = 12

On setting D(X) to be 0, the rest of the values are found to be

R(A) = 13 D(X) = 0R(B) = 11 D(Y) = 3R(C) = 9 D(Z) = 5R(D) = 7

Using these values the shadow costs of the unoccupied cells are calculated. The unoccupied cellsare X:C , X:D, Y:A, Y:D, Z:B, and Z:C

Therefore;D(X) + R(C) = 9D(X) + R(D) = 7D(Y) + R(A) = 16D(Y) + R(D) = 10D(Z) + R(B) = 16

D(Z) + R(C) = 14

The computed shadow costs are compared with actual costs to see if any reduction in cost ispossible.

Actual Shadow+ Cost

increasecost - cost - Cost reduction

Cell X :C 15 - 9 = +6X:D 20 - 7= +13 Y:A 17 - 16 = +1 Y:D 13 - 10 = +3Z:B 18 - 16 = +2Z:C 15 - 14 = +1

It will be seen that all the answers are positive, therefore no further cost reduction is possibleand optimum solution has been reached. Thus the optimal solution is represented by table 4

UNEQUAL SUPPLY AND DEMAND QUANTITIESConsider the following example.Example 2 Wanjiru books supplies in a firm dealing with import of books and it has three storesstrategically situated around the country. Yesterday the company received orders to supply 100books from 4 schools, of the books ordered the firm has 110 books in stock. The firm wishes to



268 Decesion Theory


minimize cost and its seeking your advice, advise the firm.Below is a table of availability and requirement;

RequiredSch. A Sch. B Sch. C Sch. D Total

Books 25 25 42 8 100

Store I 40 Sh.3 16 9 transportcosts perBook

Store II 20 Sh.1 9 3 8 Available Store III 50 Sh.4 5 2 5

Total 110

Solution

Step 1: add a dummy destination to table 5 with zero transport costs and requirements equal tothe surplus availability.

RequiredSch. A Sch. B Sch. C Sch. D Dummy Total

Books 25 25 42 8 10 100Store I 40 Sh.3 16 9 0 transport

costs perBook

Store II 20 Sh.1 9 3 8 0 Available Store III 50 Sh.4 5 2 5 0

Total 110

Step 2. Now that the quantity available equals the quantity required (because of insertion of thedummy) the solution can proceed in exactly the same manner described in the firstexample. First set up an initial feasible solution

Requirement A B C D Dummy25 25 42 8 10

I 40 5(4) 17(6) 8(3) 10(7) Available II 20 20(1)

III 50 8(5) 42(2)

The numbers in the table represent the allocations made and the numbers in brackets representthe sequence they were inserted based on lowest cost and the necessity to maintain row/columntotals. The residue of 10 was allocated to the dummy. The cost of this allocation is

Sh. Sh.I→A 5 units @ 3 15

I→B 17 units @ 16 272I→D 8units @ 2 16

I→Dummy 10 units @ zero costII→A 20 units @ 1 20III→B 8 units @ 5 40III→C 42 units @ 2 84

447



Lesson Eight 269


Step 3. Check solution to see if it represents the minimum cost possible in the same manner aspreviously described i.e.

Dispatch & Reception Costs of used routes:

D(I) + R(A) = 3D(I) + R(B) = 16

D(I) + R(D) = 2D(I) + R(Dummy) = 12D(II) + R(A) = 1D(III) + R(B) = 5D(III) + R(C) = 2

Setting D(I) at zero the following values are be obtained

R(A) =3 D(I) =0R(B) =16 D(I) =-2R(C) =13 D(III) =-11R(D) =2R(Dummy) =0

Using these values the shadow costs of the unused routes can be calculated .The unused routesare I:C,II:B,II:C,II:D,II:Dummy,III:D,and Dummy

ShadowCosts£

D (I) + R(C) = 0+13 =13D (II). + R (B) = -2+16 =14D (II). + R(C) = -2+13 =11D (II) + R (D) = -2+ 2 =0D (II) + R (Dummy) = -2+0 =-2D (III) + R (A) = -11+3 =-8D (III) + R (D) = -11+2 =-9D (III) + R (Dummy) = -11+0 =-11

The shadow costs are then deducted from actual costs

It will be seen that total cost can be reduced by £8 per unit for every unit that can be transferredinto Cell II:C

Step4.Make the maximum possible allocation of deliveries into Cell II:C.This is done byinserting a sequence of +and -,maintaining row and column totals.



270 Decesion Theory


Requirements A B C D Dummy25 25- 42 8 10

I 40 5+ 17- 8 10 Available II 20 20-

III 50 8+ 42-

The maximum transferable number is the lowest number in the minus cell, i.e. 17. after thetransfer is made we get;

A B C D Dummy25 25- 42 8 10

I 40 22 0 8 10 Available II 20 3 17

III 50 25 25

Step 3 is repeated again to check if the cost is minimum after setting D(I) = 0.

In our case after deducting shadow costs from actual costs we find that there are no morenegative numbers thus we deduce from the last table that the minimum transportation cost is,

(22×3) + (8×2) + (10×0) + (3×1) + (17×3) + (25×5) + (25×2) = Sh.311

Maximization using Transportation Transportation problems are usually minimizing problems, on occasions problems are framed sothat the objective is to make the allocations from sources to destinations in a manner whichmaximizes contribution or profit. These problems are dealt with similar to minimizing problemsbut the reverse of it. i.e.a) Make initial feasible allocation on basis of maximum contribution first, then next highest

and so on.b) For optimum, the differences between actual and shadow contributions for the unused

routes should be all negative. If not, make allocation into cell with the largest positive difference.c) In case there are more items available than are required, a dummy destination with zero

contribution should be introduced and the maximizing procedure in a). followed

8.2 Assignment Models The following example will be used as a basis of the step-by-step explanation.

Example 1 A company employs services engineers based at various locations throughout the country toservice and repair their equipment installed in customer‟s premises. Four requests for serviceshave been received and the company finds that four engineers are available. The distances each

of the engineers is from the various customers, is given in the following table and the company wishes to assign engineers to customers to minimise the total distances to be travelled.

Customers W X Y Z

Alf 25 18 23 14Bill 38 15 53 23Charlie 15 17 41 30Dave 26 28 36 29



Lesson Eight 271


Step 1. Reduce each column by the smallest figure in that column. The smallest figures are 15, 15,23 and 14 and deducting these values from each element in the columns produces the followingtable.

Table 2

W X Y Z

A 10 3 0 0B 23 0 30 9C 0 2 18 16D 11 13 13 15

Step 2 Reduce each row by the smallest figure in that row. The smallest figures are 0, 0, 0 and 11 and deducting these values gives the following table.

Table 3 W X Y Z

A 10 3 0 0B 23 0 30 9

C 0 2 18 16D 0 2 2 4

Note : Where the smallest value in a row is zero (i.e. as in rows A, B and C above) the row is, ofcourse, unchanged.

Step 3 Cover all the zero in the table 3 by the minimum possible number of lines. The lines may behorizontal or vertical.

Table 4

W X Y Z A 10 3 0 0

B 23 0 30 9C 0 2 18 16D 0 2 2 4

Note: Line 3, covering Row B, could equally well have been drawn covering column X.

Step 4.Compare the number of lines with the number of assignments to be made (in this examplethere are 3 lines and 4 assignments).If the number of line equals the number of assignments to bemade go to step 6.

If the number of lines is less than the number of assignments to be made (i.e. as in this example which has three lines and four assignments) then

a) Find the smallest uncovered element from step 3, called X (in Table 4 this value is 2).b) Subtract X to every element in the matrix.c) Add back to every element covered by a line. If an element is covered by two lines, for

example, cell A: W in Table 4, X is added twice.

Note: The effect of these steps is that X is subtracted from all covered by one line remainunchanged, and elements covered by two lines are increased by X.



272 Decesion Theory


Note: The effect of these steps is that X is subtracted from all uncovered elements, elementscovered by one line remains unchanged, and elements covered by two lines are increased byX.

Carrying out this procedure on Table 4 produces the following results:In Table 4 the smallest elements is 2. New table is

Table 5

W X Y Z

A 12 3 0 0B 25 0 30 9C 0 0 16 14D 0 0 0 2

Note : It will be seen that cells A: W and B: W have been increased by 2; cells A : X, A : Y,A :Z, B:X,B:Y, B:Z, C:W and D:W are unchanged, and all other cells have been reduced by 2.

Step 5. Repeat steps 3 and step 4 until the number of lines covering the zero equals the numberof assignments without any further repetition, thus:

Table 6

W X Y Z

A 12 3 0 0 Line 1B 25 0 30 9 Line 2C 0 0 16 14 Line 3D 0 0 0 2 Line 4

Step 6 when the number of lines equals the number of assignments to be made, use thefollowing rules:

a) Assign to any zero which is unique to both a column and a row.b) Assign to any zero which is unique to a column or a row.c) Ignoring assignments already made repeat rule (b) until all assignments are

made.

Carrying out this procedure for our example results in the following:a) (Zero unique to both a column and a row). None in this example.b) (Zero unique column or row). Assign B to X and A to Z. The position is

now as follows.

Table 7

W X Y Z

A Row Satisfied Column satisfiedB Row Satisfied Column satisfiedC 0 Column Satisfied 16 Column SatisfiedD 0 Column Satisfied 0 Column Satisfied



Lesson Eight 273


c) Repeating rule (b) results in assigning D to Y and C to W.Notes:a) Should the final assignment not be to a zero, then more lines than necessary were used in

step 3.b) If a block of 4 or more zero‟s is left for the final assignment, then a choice of assignment

exits with the same mileage.

Step 7 Calculate the total mileage of the final assignment. A to Z Mileage 14B to X 15C to W 15D toY 36

80 Miles

The assignment technique for maximising A maximising assignment problem typically involves making assignments so as to maximisecontribution. To maximise only one step 1 from above differs-the columns are reduced by thelargest number in each column. From then on the same rules apply that are used for minimising.

Maximising example

Example 2 The previous example No.1 will be used with the changed assumptions that the figuresrelate to contribution and not mileage and that it is required to maximise contribution.The solution would be reached as follows.(In each case the step number corresponds tothe solution given for Example No 1.)

Original data

Table 8

W X Y Z A 25 18 23 14 ContributionsB 38 15 53 23 to be gainedC 15 17 41 30D 26 28 36 29

Step 1: Reduce each column by the largest figure in that column and ignore the resulting signs.

Table 9

W X Y Z

A 13 10 30 16B 0 13 0 7C 23 11 12 0D 12 0 17 1



274 Decesion Theory


Step 2. Reduce each row by smallest figures in that row.

Table 10

W X Y Z

A 3 0 20 6B 0 13 0 7C 23 11 12 0D 12 0 17 1

Step 3.Cover zeros by minimum possible number of lines.

Table 11

W X Y Z A 3 0 20 6B 0 13 0 7

C 23 11 12 0D 12 0 17 1

Step 4. If a number of lines equals the number of assignments to be made go to step 6.If less, (asin this example), carry out the „uncovered element‟ procedure previously described. This resultsin the following table:

Table 12

W X Y Z A 0 0 17 6B 0 16 0 10C 20 11 9 0

D 9 0 14 1

Table 13

W X Y Z A 0 0 17 6B 0 16 0 10C 20 11 9 0D 9 0 14 1

Step 6. Make assignment in accordance with the rules previously described which result in thefollowing assignment:

C to ZD to X A to WB to Y



Lesson Eight 275


Step 7.Calculate contribution to be gained from the assignments.

C to Z 30D to X 28 A to W 25

B to Y 53 Total 136

Notes:a) It will be apparent that maximising assignment problems can be solved in virtually the

same manner as minimising problems.b) The solution methods given are suitable for any size of matrix. If a problem is as small

as the illustration used in this chapter, it can probably be solved merely by inspection.

Unequal sources and destinations5. To solve assignments problems in the manner described the matrix must be square, i.e. thesupply must equal the requirements. Where the supply and requirements are not equal, an

artificial source or destination must be created to square the matrix. Thecost/mileage/contributions etc for the fictitious column or row should be zero throughout.

Solution methodHaving made the sources equal the destinations, the solutions method will be as normal, treatingthe fictitious elements as though they were real. The solution method will automatically assign asource or destination to the fictitious row or column and the resulting assignment will incur zerocost or gain zero contribution.

Points to notea) The assignment technique can be used for allocating type of problems, e.g. taxis to

customers, jobs to personnel.b) Most practical problems of size illustrated could be solved fairly readily using

nothing more than commonsense. However, the technique illustrated can be usedto solve much larger problems.

Exercises with answers1. A foreman has four fitters and has been asked to deal with five jobs. The times for each job

are estimated as follows:

Fitters

Alf Bill Charlie Dave

Job 1 6 12 20 12

Job 2 22 18 15 20 Job 3 12 16 18 15 Job 4 16 8 12 20 Job 5 18 14 10 17

Allocate the men to the jobs so as to minimise the total time taken and identify the job which will not be dealt with.



276 Decesion Theory


2. A company has four salesmen who have to visit four clients. The profits records fromprevious visits are shown in the table and it is required to maximise profits by the bestassignments.

A B C D

Customer 1 6 12 20 1222 18 15 2012 16 18 1516 8 12 20

Answers to exercises

1. Dummy fitter inserted to square matrix A B C D DUMMY

1 6 12 20 12 02 22 18 15 20 03 12 16 18 15 0

4 16 8 12 15 05 18 14 10 17 0

Reduce columns by the smallest element and cover by lines

0 4 10 0 0

16 10 5 8 06 8 8 3 010 0 2 8 012 6 0 5 0

4 lines so not optimum, smallest element 3

Therefore reduce uncovered elements by 3 and increase elements crossed by 2 lines by 3

0 4 10 0 313 7 2 5 03 5 5 0 010 0 2 8 312 6 0 5 3

5 Lines so optimum.

AssignmentsB to 4C to 5 A to 1Dummy to 2

8.3 NETWORK ANALYSIS This is a system of interrelationship between jobs and tasks for planning and control ofresources of a project by identifying critical path of the project.



Lesson Eight 277


Dangling activity

Terminology Activity. Task or job of work, which takes time and resources e.g building a bridge. It is

represented by an arrow which indicates where the task begins and ends

Event (node). This is a point in time and it indicates the start or finish of an activity e.g in buildinga bridge, rails installed. It is represented by a circle.

Dummy activity. An activity that doesn‟t consume time or resources, it is merely to show logicaldependencies between activities so as abide by rules of drawing a network, it isrepresented by a dotted arrow

Network. This is a combination of activities and events (including dummy activities)

Rules for Drawing a Networka) A network should only have one start point and one finish point ( start event and finish

event )b) All activities must have at least one preceding event ( tail event ) and at least one

succeeding event ( head event), but an activity may not share the same tail event and headevent.

c) An activity can only start after its tail event has been reachedd) An event is only complete after all activities leading to it are complete.e) Activities are identified by alphabetical or numeric codes i.e. A,B,C; 1,2,3 or

identification by head or tail events 1-2, 2-4, 3-4,1-4…

f) Loops (a series of activities leading back to the same event) and danglers (activities which do not link to the overall project) are not allowed

Dummy Events This is an event that does not consume time or resources, it is represented by dotted arrow.Dummies are applied when two or more events occur concurrently and they share the same

head and tail events e.g. when a car goes to a garage tires are changed and break pads as well,instead of representing this as;

Loop

A- Tires Chan ed

B- Break pads Changed

Car Arrives (CA) Car ready (CR)



278 Decesion Theory


B

ACRCA

These events are represented as;

Example of a network.

Activities1-2 - where 1 is the preceding event where as 2 is the succeeding event of the activity1-32-42-53-54-54-65-6

6-7

8.4 Network Analysis-Time Analysis Assessing the time

a) After drawing the outline of the network time durations of the activities are theninserted.a) Time estimates. The analysis of the projects time can be achieved by using :

i. Single time estimates for each activity. These estimates would be based on thejudgment of the individual responsible or by technical calculations using datafrom similar projects

ii. Multiple time estimates for each activity. the most usual multiple timeestimates are three estimates for each activity , i.e. optimistic (O), Most Likely

(ML), and Pessimistic (P). These three estimates are combined to give anexpected time and the accepted time formula is:

Expected time =6

4MLPO

For example assume that the three estimates for an activity areOptimistic 11 daysMost likely 15 daysPessimistic 18 days

1

2

3

4

5

6 7



Lesson Eight 279


B2

A

1

E

1

C

3

F

2

D4

B

2

A

1

E

1

C

3

F

2

D

4

Expected time =6

1541811

= 14.8 daysb) Use of time estimates. as three time estimates are converted to a single time

estimate. There is no fundamental difference between the two methods asregards the basic time analysis of a network. However, on completion of thebasic time analysis, projects with multiple time estimates can be furtheranalyzed to give an estimate of the probability of completing the project by ascheduled date.

c) Time units. Time estimates may be given in any unit, i.e. minutes , hours, daysdepending on the project. All times estimates within a project must be in thesame units otherwise confusion is bound to occur.

Basic time analysis – critical pathb) The critical path of a network gives the shortest time in which the whole project can

be completed. It is the chain of activities with the longest duration times. There maybe more than one critical path which may run through a dummy.

Earliest start times (EST)– Forward pass , Once the activities have been timed we can

assess the total project time by calculating the ESTs for each activity. The EST isthe earliest possible time at which a succeeding activity can start. Assume the following network has been drawn and the activity times estimated indays.

The ESTs can be inserted as follows.

EST

The method used to insert the ESTs is also known as the forward pass, this is obtainedby;

EST = The greater of [EST (tail event) + Activity duration]

2

0 1 3 4 5

2

3

0

0

1

1

3

4

4

7

5



280 Decesion Theory


B

2

A1

E1

C3

F2

D

4

B

2

A1

E

1

C

3

F

2

D

4

a) Start from the start event giving it 0 values,b) For the rest of the events EST is obtained by summing the EST of the tail event and the

activity durationc) Where two or more routes converge into an activity, calculate individual EST per route

and then select the longest route (time)

d) The EST of the finish event is the shortest time the whole project can be completed.

Latest Start Times (LST) – Backward pass. this is the latest possible time with which apreceding activity can finish without increasing the project duration. After this operationthe critical path will be clearly defined.

From our example this is done as follows;

LST

LST = Lowest of [LST (head event) – activity duration]

a) Starting at the finish event, insert the LST (i.e. 9 for our example) ,and work backwardsthrough the network.

b) deduct each activity duration from the previously calculated LST (i.e. head LST).

c)

Where the tails of activities join an event, the lowest number is taken as the LST for thatevent

Critical Path. . This is the chain of activities in a network with the longest duration Assessmentof the resultant network shows that one path through the network (A, B, D, F) hasEST's and LST's which are identical this is the critical path.

The critical path can be indicated on the network either by a different colour or bytwo small transverse lines across the arrows along the path, thus in our example wehave;

2

3 3

0

0 0

1

1 1

3

4 6

4

7 7

5

9 9

23 3

0

0 0

1

1 1

3

4 6

4

7 7

5

9 9



Lesson Eight 281


A B

10

C

Section of the network

Activities along the critical path are vital activities which must be completed by theirEST's/LST's otherwise the project will be delayed.

Non critical activities (in the example above, C and E) have spare time or float

available. C and/ or E could take up to an additional 2 days in total withoutdelaying the project duration. If it is required to reduce the overall project durationthen the time of one or more of the activities on the critical path must be reducedperhaps by using more labour, or better equipment to reducing job times.

FLOAT

Float or spare time can only be associated with activities which are non-critical. By definition,activities on the critical path cannot have float. There are three types of float, Total Float, Free

Float and Independent Float. To illustrate these types of float we use the following example.

a) Total float . Amount of time by which a path of activities could be delayed withoutaffecting the overall project duration. The path in this example consists of one activityonly i.e. B

Total Float = Latest Finish time (LFT) - Earliest Start time(EST) time – Activity

Duration Total Float = 50 - 10 - 10= 30 days

b) Free float Amount of time an activity can be delayed without affecting thecommencement of a subsequent activity at its earliest start time, but may affect float ofa previous activity.

Free Float = Earliest Finish Time(EFT) - EST - Activity Duration

Free Float = 40-10-10= 20 days

c) Independent float. Amount of time an activity can be delayed when all preceding activities

are completed as late as possible and all succeeding activities commenced as early apossible. Independent float therefore does not affect the float of either preceding orsubsequent activities.

Independent float = EFT- Latest Start time (EST) - Activity Duration

Independent float = 40 - 20 - 10= 10 days

5

10 20

6

40 50



282 Decesion Theory


A4 C

5

B 7

D

6

E

2

F3

G

5

H11 I 7

J

4

K

3

L

4

Note: for examination purposes, float always refers to total float

The total float can be calculated separately for each activity but it is often useful to find the total float over

chains of non-critical activities between critical events

Example.

The following represents activities of a network. Activity Preceding Activity Duration Days

A - 4B A 7C A 5D A 6E B 2F C 3G E 5H B,F 11I G,H 7 J C 4

K D 3L I,J,K 4

Required:

a) Draw the network diagram and find the critical pathb) Calculate the floats of the network in question

Solution. (a)

1

02

4

3

11

6

9

4

13

930

8

23

5

12

1034

7

10



Lesson Eight 283


A4 C

5

B 7

D

6

E2

F 3

G

5

H11

I 7

J

4

3

L4

First we draw the network structure ensuring it fits the data above We then label all activities from 1 to 12 and indicate activity durationConduct a forward pass operation (to obtain the diagram above)Operate backward pass to establish the critical path, thus we have…

Therefore we get the critical path to be, A- C- F- H- I- L b) The floats of the network,

Activity

Duration

Total Float Free Float Independent

Float

Activity EST LST EFT LFT D LFT -EST- D EFT-EST-D EFT-LST-D

*A 0 0 4 4 4 - - -B 4 4 11 15 7 4 - -

*C 4 4 9 9 5 - - -D 4 4 10 22 6 12 - -

E 11 15 13 21 2 8 - -*F 9 9 15 15 3 - - -G 13 21 23 23 5 5 5 -*H 12 12 23 23 11 - - -*I 23 23 30 30 7 - - - J 9 9 30 30 4 17 17 17K 10 22 30 30 3 17 17 5*L 30 30 34 34 4 - - -

0 0

4 4

11 15

9 9

13 18

30 30

23 23

12 12

34 34

10 27



284 Decesion Theory


The total float on the non-critical chains are;

Non-critical

chain

Time required

(sum of duration)

Time available

(LFT of last activity-EST of 1st activity)

Total Float over

chain

B,E,G 14 19 5B,Dummy 7 8 1

D,K 9 26 17 J 4 21 17

Slack

This is the difference between the EST and LST for each event. Strictly it does not apply toactivities but on occasions the terms are confused in examination questions and unless thecontext makes it abundantly clear that event slack is required, it is likely that some form ofactivity float is required. Events on the critical path have zero slack.

8.5 Cost Scheduling

This is done by calculating the cost of various project durations, cost analysis seeks to find the

cheapest way of reducing the overall cost duration of a project by increasing labour hours,equipment e.t.c.

Terminologies

Normal cost . The costs associated with a normal time estimate for an activity. Often the normaltime estimate is set at the point where resources (labour, equipment, etc.) are usedin the most efficient manner.

Crash cost . The costs associated with the minimum possible time for an activity. Crash costs,because of extra wages, overtime premiums, extra facility costs are always higherthan normal costs.

Crash time . The minimum possible time that an activity is planned to take. . The minimum time isinvariably brought about by the application of extra resources, e.g. more labour ormachinery.

Cost slope . This is the average cost of shortening an activity by one time unit (day, week, month asappropriate). The cost slope is generally assumed to be linear and is calculated asfollows:

Cost slope = Crash cost – Normal costNormal time – Crash time

Example A project has the following activities and costs. You are required to prepare the least costschedules for all possible durations from normal time – normal cost to crash time – crash cost.



Lesson Eight 285


A

4

B

8

D

9

C

5 E5

Activity Preceding

Activity

Duration

days

Crash

time

Cost

(Shs).

Crash cost Cost slope

A - 4 3 360 420 60B - 8 5 300 510 70C A 5 3 170 270 50D A 9 7 220 300 40E B,C 5 3 200 360 80

Project duration and costs

(a) Normal duration = 14 daysCritical path = A,C,EProject cost (cost of all activities in normal time) = Shs. 1,250.

(b) Reduce by 1 day the activity on the critical path with the lowest cost slope. Thus wereduce C at extra cost of Shs. 50.

NowProject duration = 13 daysProject cost = Shs. 1,300

Note: that all activities are now critical.

(c) Further reducing the critical path by 1 day will require that more than one activity isaffected because there exist several critical paths.

Reduce by 1 day Extra cost Activities critical A and B 60 + 70 = 130 All

D and E 40 + 80 = 120 AllB, C and D 70 + 50 + 40 = 160 All A and E 60 + 80 = 140 A, D, B, E

From this we realize that reducing D and E is the cheapest.

However closer examination of the fourth alternative reveals that C is now non-criticaland has 1 day float. Since we earlier reduced C for Shs. 50, if we reduce A and E andincrease C by a day which will save Shs. 50.

0

0 0

1

4 4

2

9 9

314 14



286 Decesion Theory


Then the net cost for 12 day duration = 1,300 + (140 – 50) = 1,390.

The network becomes………

(d) Next we reduce D & EProject duration = 11 daysProject cost = 1,510Critical activities = All

(e) Final reduction possible is by reducing B, C & D for Shs. 160 the network thenbecomes.

Duration = 10 daysCost = Shs. 1,670Critical activities = All.

Note: only critical activities affect project duration.: Always look for a possibility of increasing the duration of a previously

crashed activity.

1

3 3

0

0 0

2

7 7

3

10 10

3 crash

C

4

B

7

D

7 Crash

E

3 (crash)

A

1

3 3

0

0 0

2

7 7

3

12 12

3 crash

C

5

B

8

D

9

E

4

A



Lesson Eight 287


SCHEDULING RESOURCES AND GANTT CHART

Apart from time, cost network analysis also help in controlling and planning of resources.

Example A project has the following activity durations and resource requirements.

Activity Preceding activity Duration (days) Resource requirement (man power) A - 6 3B - 3 2C - 2 2D C 2 1E B 1 2F D 1 1

Requiredi) What is the networks critical pathii) Draw a gantt chart diagram indicating activity times, using their estimate.

iii) Show resource requirement on a day to day basis assuming all events commence at theirestimates.iv) Assuming that only six employees are available, how will the activities be planned for?

Solutioni)

Activities Duration EST LST Man power A 6 0 0 3B 3 0 0 2C 2 0 0 2D 2 2 3 1E 1 3 5 2

F 1 4 5 1



288 Decesion Theory


ii) A gantt chart or a bar chart. This is a diagram indicating a resource scalednetwork.

iii) Resource requirements on a day to day basis.

iv) When on 6 manpower resources are available then we adjust the activities toaccommodate this and still end at the given critical time duration i.e.



Lesson Eight 289


Node Networks This network also known as a procedure diagram is represented with the same information as anetwork diagram.Its characteristics are;

i) Activities are shown in boxes instead of arrowsii) Events are not represented.iii) The arrows linking boxes indicate the sequence precedence of activities.iv) Dummies aren‟t necessary.

E.g.

Would appear as



290 Decesion Theory


A full activity node network is represented as;

This is represented as;

Note:i) EST and LST are calculated by the same process we learnt earlier.ii) EFT and LFT are calculated by adding the activity time duration to EST and LST

respectively.

iii) Critical path is similarly identified by identifying equal EST and LST throughout thepath.



Lesson Eight 291



QUESTION ONE

Regal Investments has just received instructions from a client to invest in two shares; one anairline share, the other an insurance share. The total maximum appreciation in share value over

the next year is to be maximized subject to the following restrictions:- the total investment shall not exceed Sh.100,000- at most Sh.40,000 is to be invested in the insurance sharesquarterly dividends must total at least Sh.2,600

The airline share is currently selling for Sh.40 per share and its quarterly dividend is Sh.1pershare. The insurance share is currently selling for Sh.50 per share and the quarterly dividend isSh.1.50 per share. Regal‟s analysts predict that over the next year, the value of the airline share will increase by Sh.2 per share and the value of the insurance share will increase by Sh.3 pershare. A computer software provided the following part solution output:

Objective Function Value = 5,400 Variable Number Reduced cost

Airline shares 1,500 0.000Insurance shares 800 0.000

Constraint Slack/Surplus Dual prices Total investment 0.000 0.050Investment in insurance 0.000 0.010Dividends 100.000 0.000





Lesson Eight 293


ii) Within what limits must the unit profits lie for each of the frames for this solution toremain optimal?

b) Explain the limitations of the technique you have used to solve part (a) above.(Q 6 Dec 2000)

QUESTION FOURa) Define the following terms as used in linear programming:

i) Feasible solutionii) Transportation problemiii) Assignment problem

b) The TamuTamu products company ltd is considering an expansion into five new salesdistricts. The company has been able to hire four new experienced salespersons. Uponanalysing the new salesperson‟s past experience in combination with a personality test which was given to them, the company assigned a rating to each of the salespersons foreach of the districts .These ratings are as follows:

c) Districts

Salespersons

1 2 3 4 5 A 92 90 94 91 83B 84 88 96 82 81C 90 90 93 86 93D 78 94 89 84 88

The company knows that with four salespersons, only four of the five potential districts canbe covered.

Required:

i) The four districts that the salespersons should be assigned to in order to maximize thetotal of the ratingsii) Maximum total rating. (Q 6 June 2002)

QUESTION FIVE

a) Explain the value of sensitivity analysis in linear programming problems and show howdual values are useful in identifying the price worth paying to relax constraints.

b) J.A Computers is a small manufacturer of personal computers. It concentrates onproduction of three models- a Desktop 386, a Desktop 286, and a Laptop 486, eachcontaining one CPU Chip. Due to its limited assembly facilities JA Computers are unable toproduce more than 500 desktop models or more than 250 Laptop models per month. It has

one hundred and twenty 80386 chips (these are used in Desktop-386) and four hundred80286 chips (used in desktop 286 and Laptop 486) for the month. The Desktop 386 modelrequires five hours of production time, the Desktop 286 model requires four hours ofproduction time, and the Laptop 486 requires three hours of production time. J.AComputers have 2000 hours of production time available for the coming month. Thecompany estimates that the profit on Desktop 386 is Sh. 5,000. for a desktop 286 the profitis Sh.3,400 and Sh.3,000 profit for a laptop 486.



294 Decesion Theory


Required:Formulate this problem as a profit maximization problem and mention the basicassumptions that are inherent in such models.

c) An extract of the output from a computer package for this problem is given below:

Output solutionX 1=120, X 2 = 200, X 3 = 200Dual values Constraints 3 150

Constraints 4 90Constraints 5 20

Sensitivity analysis of objective function coefficients:

Variable Lowerlimit

Original value

Upperlimit

X 1 100 250 No limitX 2 150 170 200X 3 127.5 150 170

Sensitivity analysis on R.H.S ranges.

Constraints Lowerlimit

Original value

Upperlimit

1 320 500 No limit2 200 250 No limit3 80 120 1304 350 400 412.55 1950 2000 2180

X 1=Monthly production level for Desktop 386.

X 2 =Monthly production level for Desktop 286.X 3=Monthly production level for Laptop 486.

Required:i) Interpret the output clearly, including optimum product mix, monthly profit, unused

resources and dual valuesii) Explain the purpose of upper limits and lower limits for the variables X 1,X 2,X 3 and

constraints 1 to 5.iii) Calculate the increase in profit if the company is able to produce a further 10 CPU

80386 chips. (Q7 July 2000 Pilot paper)

QUESTION SIX

Preface Retailers is a high-technology retailer and mail order business. In order to improve itsprocess the company decides to install a new microcomputer system to manage its entireoperation (i.e. payroll, accounts, inventory).

Terminals for each of its many stores will be networked for fast, dependable service. Thespecific activities that Preface will need to accomplish before the system is up and running arelisted below. The table also includes the necessary increased staffing to undertake the project.



Lesson Eight 295


Activity Preceding Activities

Duration(Days)

IncreasedStaff

A. Build insulated enclosureB. Decide on computer systemC. Electrical wiring of room

D. Order and collect computerE. Install air conditioningF. Install computerG. Staff testingH. Install softwareI. Staff training

--

A

B AD, E

BC, FG, H

413

242523

132

122111

Required: a) Draw a network diagram for the project and determine the critical path and its

duration.b) Assuming that all activities start as soon as possible, draw a progress chart for the

project, showing the times at which each activity takes place and the manpower

requirements.c) The union has decided that any staff employed on the project must be paid for theduration of the project whether they work or not, at a rate of £500 per day.

Assuming that the same staff is employed on the different activities, determine the work schedule that will minimise labour costs though not necessarily the projecttime. What is the cost associated with this schedule?

Comment on the validity of the assumption.



296 Decesion Theory


COMPREHENSIVE ASSIGNMENT FOURWork out these question for three hours (exam condition) then hand them in to DLC for marking



QUESTION ONE

a) Define the following terms as used in game theory:i) Dominance. (2 marks)ii) Saddle point. (2 marks)iii) Mixed strategy. (2 marks)iv) Value of the game (2 marks)

b) Consider the two person zero sum game between players A and B given the following pay-off table:

Player B Strategies1 2 3 4

Player A Strategies 1 2 2 3 -12 4 3 2 6

Required:i) Using the maximin and minimax values, is it possible to determine the value of the game?

Give reasons. (3 marks)ii) Use graphical methods to determine optimal mixed strategy for player A and determine

the value of the game. (9 marks)(Total: 20 marks)

QUESTION TWO

Central and Eastern Industries is planning to introduce a new mobile phone service. To do so,the following activities are necessary:

Preceding Expected Standard Activity Activity Time (weeks) Deviation (weeks) A - 6 1.0B - 3 0.5C A 5 1.0D A 4 1.0E A 3 0.5F C 3 0.5G D 5 1.0H B.D.E 5 1.0I H 2 0.5 J F.G.I 3 1.0

The costs of the project are estimated to be Sh.10 million. If the projects is completed within 24 weeks the expected net revenue will be about Sh.100 million but if the deadline of 24 weeks is



Lesson Eight 297


not met, the service will fail to penetrate the market and a net revenue of Sh.2 million isexpected.

Required:a) Determine how long the project would take. (8 marks)b) If the start of activities B, E and G are respectively delayed by 3,2 and 2 weeks,

how would this affect the total project time? (6 marks)c) Determine a 95% confidence interval for the expected time of the projectand explain your answer. Ignore the delays referred to in (b) above. (3 marks)

d) Determine the expected profit on this project. Again ignore the delays referredin (b) above (3 marks)

(Total: 20 marks)

QUESTION THREE

a) Explain the value of sensitivity analysis in linear programming problems and show how dual values are useful in identifying the price worth paying to relax constrains. (4marks)

b) J.A Computers is a small manufacturer of personal computers. It concentrates productionon three models, a Desktop 286,and a laptop 486,each containing one CPU Chip. Due to

its limited assembly facilities J.A Computers are unable to produce more than 500 desktopmodel or more than 250 Laptop models per month. It has one hundred and twenty 80386chips (these are used in Desktop – 386) and four hundred 80286 chips(used in desktop 286model requires four hours of production time, and the laptop 486 requires three hours ofproduction time. J.A Computers have 2000hours of production time available for thecoming month. The company estimates that the profit on a desktop 386 is Sh.5000, for adesktop 286 the profit is Sh.3400 and Sh.3000 profit for a Laptop 486.

Required:Formulate this problem as a profit maximization problem and mention the basicassumptions that are inherent in such models. (7 marks)

a)

An extract of the output from a computer package for this problem is given below:Output solution

X 1 = 120, X 2 = 200, X 3 = 200

Dual values Constraints 3 150Constraints 4 90Constraints 5 20

Sensitivity analysis of objective function coefficients Variable Lower limit Original value Upper limitX 1 100 250 No limitX 2 150 170 200X 3 127.5 150 170

Sensitivity analysis on R. H. S rangesConstraints Lower limit Original value Upper limit1 320 500 No limit2 200 250 No limit3 80 120 1304 350 400 412.5



298 Decesion Theory


5 1950 2000 2180

X 1 = Monthly level for Desktop 386.X 2 = Monthly production level for Desktop 286X 3 = monthly production level for Laptop 486

Required:i) Interpret the output clearly, including optimum product mix, monthly profit,

unused resources and dual values. (3 marks)ii) Explain the purpose of upper limits for the variables X 1, X 2 ,X 3 and

constraints 1 to 5. (3 marks)iii) Calculate the increase in profit if the company is able to produce a further

10 CPU 80836 chips. (3 marks)(Total: 20 marks)

QUESTION FOUR

a) Give two applications of simulation in business. (2 marks)

b) Collins Simiyu recently acquired a piece of land in Kitale. A property development companyhas offered him Sh.300,000 for the piece of land. He has to make a decision on whether tocultivate the piece of land or to sell it to the property development company. if he decidesto cultivate the land, there is a probability of getting a high, medium or low harvest. Theexpected net income for each of the above states of harvest is shown below:

State of harvest Net income (Sh.)High 500,000Medium 100,000Low (200,000)

From past experience, there is a 10 per cent probability that the harvest will be low, a 30 pr cent

probability that the harvest will be medium and a 60 per cent probability that the harvest will behigh. Collins Simiyu can engage an agricultural expert to carry out a survey on the productivityof the land, which will cost him Sh.30,000. The agricultural expert gives the followinginformation as to the reliability of such surveys (prior probabilities).

Results of survey State of harvestHigh Medium Low Total

Accurate 0.35 0.10 0.05 0.5Not accurate 0.25 0.10 0.15 0.5

0.60 0.20 0.20 1.0

Required:i) Construct a decision tree for the above problem. (6 marks)ii) The expected monetary value for each decision (10 marks)iii) The decision that you would recommend (2 marks)

(Total: 20 marks)

QUESTION FIVE

a) Explain the difference between assignment and transportation problems. (4 marks)

b) State the assumptions made in solving a transportation problem. (4 marks)



Lesson Eight 299


c) Umoja Engineering Works Ltd. Has a network of branches all over Kenya. The branchesare used to service, repair and install equipment for their clients. Currently, the Nairobibranch has four clients who require installation of equipment. Each client requires theservices of one engineer.

There are four engineers who are not engaged at the moment and can be assigned any one of the

tasks. However, these engineers have to travel from different locations and the Nairobi branchhas to meet their travel and subsistence allowances. The allowances vary from one engineer toanother and according to the client the engineer has been assigned to work for.

The table below shows the costs (in thousands of shillings) associated with each engineer.

ClientEngineer 1 2 3 4 A 37.0 27.0 34.0 21.0B 57.0 22.0 79.0 34.0C 22.0 25.0 61.0 45.0D 39.0 42.0 54.0 43.0

Required:i) The assignments to be made in order to minimize the total cost of the

engineers. (10 marks)ii) The minimum cost of using engineers. (12 marks)

(Total: 20 marks)

SECTION IIQUESTION SIX

a) Define the following terms as used in the network analysis:

i) Crash time (2 marks)

ii)

Optimistic time (2 marks)iii) Forward pass (2 marks)iv) Dummy activity (2 marks) v) Slack (2 marks)

b) James Mutiso is a computer engineer in an information technology firm. The firm hasdecided to install a new system to be used by the firm‟s help desk. James Mutiso hasidentified some activities required to complete the installation.

The table below provides a summary of the activities‟ durations and the required number oftechnicians:

Activity Duration (Weeks) Required number of technicians1-2 3 21-3 1 42-4 3 42-5 2 23-4 2 43-6 4 44-5 2 25-6 2 26-7 2 2



300 Decesion Theory


Required:i) Draw a gantt chart for the project. (6 marks)ii) Mr. Mutiso would like to reschedule activities so that not more than 6 technicians

are required each week.

Determine if this is possible and how it can be achieved by rescheduling

the activities. (4 marks)(Total: 20 marks)

QUESTION SEVEN

Regal investments has just received instructions from a client to invest in two shares; one anairline share, the other an insurance share. The total maximum appreciation in share value overthe next year is to be maximized subject to the following restrictions:

The total investment shall not exceed Sh.100,000 At most Sh.40,000 is to be invested in the insurance sharesQuarterly dividends must total at least Sh.2,600

The airline share is currently selling for Sh.40 per share and its quarterly dividend is Sh.1 pershare. The insurance share is currently selling for Sh.50 per share and the quarterly dividend isSh.1.50 per share. Regal‟s analysts predict that over the next year, the value o f the airline share will increase by Sh.2 per share and the value of the insurance share will increase by Sh.3 pershare.

A computer software provided the following part solution output:Objective Function Value = 5,400

Variable Number Reduced cost Airline shares 1,500 0.000Insurance shares 800 0.000

Constraint Slack/Surplus Dual prices Total investment 0.000 0.050Investment in insurance 0.000 0.010Dividends 100,000 0.000

Objective coefficient Ranges Variable Lower limit Current value Upper limitInsurance share 2.5000 3.000 No upper limit Airline share 0.000 2.000 2.400

Right-hand Side RangesConstraint Lower limit Current value Upper limit

Total investment 96,000.00 100,000 No upper limitInvestment in insurance 20,000.00 40,000 100,000.00Dividends No lower limit 2,600 2,700.00

Required:i) Formulate the above problem. (7 marks)ii) Explain what reduced cost and dual prices columns above mean. (5 marks)iii) How should the client‟s money be invested to satisfy the restrictions? (4 marks)iv) Suppose Regal‟s estimate of the airline shares appreciation is in error, within



Lesson Eight 301


what limits must the actual appreciation lie for the answer in (c) above toremain optimal? (4 marks)

(Total: 20 marks)

QUESTION EIGHT

The linear programming model and output model below represent a problem whose solution will tell a road side kiosk owner how many of the four different types of household goods tostock in order to maximize profits. It is assumed that every item stocked will be sold. The variables measure the packets of Unga, Spaghetti, Rice and Sugar to stock respectively. Theconstraints measure storage space in units, special display racks, demand and a marketingrestriction, respectively.

Maximize Z = 4X 1+ 6X 2+ 5X 3+ 3.5X 4

Subject to :2X 1 + 3X 2+ 3X 3+ X 4 ≤ 120 (1)1.5X 1 + 2X 2 ≥ 54 (2)

2X 2 + X 3 + X 4 ≤ 72 (3)

X 2 + X 3 ≥ 12 (4)

Where X 1 = packets of UngaX 2 = packets of SpaghettiX 3 = packets of RiceX 4 = packets of Sugar

Optimal solution Variable Value Reduced costX 1 12.00 -X 2 0.00 0.50X 3 12.00 -

X 4 60.00 -

Constraint Stack/surplus Dual/shadow price1 - 2.002 - -3 - 1.504 - -2.50

Objective Coefficient Ranges Variable Lower limit Current value Upper limitX 1 1.50 - 5.00X 2 No limit - 6.50X 3 4.50 - 7.50X 4 3.00 - No limit

Right Hand Side RangesConstraint Lower limit Current value Upper limit1 96 - 1682 18 - No limit3 24 - 964 0 - 24



302 Decesion Theory


Required:a) Determine the retailer‟s optimal profit level. (2 marks)b) Determine and interpret the missing values under:

i) Reduced cost column. (2 marks)ii) Slack/surplus column, indicate whether the value is a slack or a surplus.

(2 marks)iii) Dual/shadow price column. (2 marks)iv) Current value column under objective coefficient ranges. (2 marks) v) Current value column under right hand side ranges. (2 marks)

c) Interpret the value 0.50 under the reduced cost column and values; 2.00, 1.50and -2.50 under the dual/shadow price column (2 marks)

d) Determine whether the current, optimal solution would change if thecurrent profit of packets of Unga is increased by Sh.2.00. (2 marks)

e) Determine by how much the amount of space would increase before thereis a change in the dual/shadow price. (2 marks)

f) The above problem could have been solved manually. Explain how the

optimal solution can be determined using the manual approach. (2 marks)(Total: 20 marks)



Revision Aid 303


LESSON NINE

Revision Aid

SOLUTIONS

LESSON ONE

QUESTION ONE

1. ( – 3, 12) 2. (12, 0) 3. (10, 4) 4. ( 8 , 8 )

QUESTION TWO

a) 0, 4; b) 3, 3; c) 10.5, 36 x y x y x y

QUESTION THREE

Equillibrium will be attained

Population23.33

46.67

north

south

P

P

QUESTION FOUR

1.

The vector for final demand =

60

6060

The input/output coefficient matrix (also called the matrix for technical coefficients). Call it A

80 100 100 1 1 14 4 3320 400 300

80 200 60 1 1 14 2 5320 400 300

80 100 100 1 1 14 4 3320 400 300

A

let X be the total output vector.

Thus; X = AX + YX = (I – A ) -1 Y



304 Lesson Nine


31 1 1 1 14 4 3 4 4 3

1 1 1 1 1 14 2 5 4 2 5

1 1 1 1 1 24 4 3 4 4 3

1 0 0

0 1 0

0 0 1

I A

Inverse (I - A) -1 =

17 13160 4 60

13 5 760 12 30

3 5116 4 16

68 60 52240 152 100 56

23 2345 60 75

Therefore X =

68 60 52 60 10,8001 1

52 100 56 60 17,48023 23

45 60 75 60 10,800

=

469.57

542.60

469.57

in Shs

QUESTION FIVE

i) Express each percentage as a decimal and the matrix equation becomes,

1 1

2 2

0.4 0.6

0.5 0.25

t b

t b

ii)

Put t 1= 400 and t 2= 700 and the matrix equation becomes.

1

2

0.4 0.6 400

0.5 0.25 700

b

b

That is b1 = (0.4 400) (0.6 700) = 580 kilos

b2 = (0.5 400) (0.25 700) = 375 kilos

iii) To establish the value of t 1 and t 2 which correspond to b 1 and b 2 it is necessary to formthe inverse of the matrix.

0.4 0.6

0.5 0.25

which is



Revision Aid 305


1.25 3.0

2.5 2.0

the equation to determine t 1 and t 2 now becomes,

1

2

1.25 3.0 600

2.5 2.0 450

t t

Thus;t 1 = - 1.25×600 + 3.0 × 450

= 600 kilos and

t 2 = - 2.5×600 - 2 .0 × 450= 600 kilos

QUESTION SIX

i)

2 2 2 2 10 -2 A3 = =

3 -3 3 -3 -3 15

2 2 10 -2 14 26 A2 = =

3 -3 -3 15 39 -51

ii)

F(A) = A3

– 3A2

– 2A + 4114 26 10 -2 2 2 1 0

= -3 -2 +439 -51 -3 15 3 -3 0 1

14 26 30 6 4 4 4 0= - - +

39 -51 -9 45 6 -6 0 4

-16 28=

42 -86

iii) -3 -2 1 1

1_ = 4 6 A-1 = -12 -3 2 1 -1

4 6



306 Lesson Nine


QUESTION SEVEN

i) 12

10 15 6 0 8 348

15 20 6 4 18 = 536

16 24 8 6 22 660

Cost of Standard set = 348 pence = £3.48Cost of Deluxe set = 536 pence = £5.36Cost of super set = 660 pence = £6.60

ii) 10 15 16 3,000

100 10 15 6 015 20 24 4,300

80 = or (100 80 50)15 20 6 46 6 8 1,48050 16 24 8 6

0 4 6 620

Squares 3,000, triangles 4,300 hexagons 1,480, octagons 620

iii) 12

8(3,000 4,300 1,480 620)

18

20= 110,680 in pence

= £1,106.80

QUESTION EIGHT

a) A B C storage maintenance

10 12 50 2 0.5 156 48NP = 60 0 20 3 1.5 = 160 40

2 0.5

Cost of storing in warehouse Y = 156p = £1.56Cost of storing in warehouse W =160p = £1.60Cost of maintaining warehouse Y = 48p = £0.48Cost of maintaining warehouse W = 40p = £0.40

b) A B C storage maintenance



Revision Aid 307


Day I Y 10 10 50 2 0.5 150 45 cost in W 40 0 20 3 1.5 = 120 30 pence

2 0.5

A B C storage maintenance

Day II 10 10 60 2 0.5 191 60.5 cost in40 15 20 3 1.5 = 165 52.5 pence

2 0.5

For total cost add Day I and Day II values

c) A B C

10 150 45 191 60.5+3 × Cost in pence

120 30 165 52.5



308 Lesson Nine


LESSON TWO

QUESTION ONE

a) 6

b)

c)

d)

QUESTION TWO

2 30dc QdQ

Therefore Q = 15 at minimum

Note: which is positive indicating a minimum value

QUESTION THREE

n(є) = 250

P + 12 + 59 = 147 giving P = 76Q + 59 + 37 = 102 giving Q = 6

i) Those who did not vote= 250 – (76 + 12 + 14 + 59 + 6 + 37)= 250 – 204 = 46

ii) x = 76 + 12 + 14 = 102

32

x

1

1 2 x

32

1 1,

2

dy y x

dx x

2

22

d c

dQ



Revision Aid 309


x = 12 + 59 + 6 = 77

z = 37 + 14 + 6 = 57

iii) x won the election.

QUESTION FOURi) To find maximum or minimum value we use differential calculus as follows:

3

R 14 8112

x x

23

8112

dR x

dx

2

2

60

12 2

d R x x

dx

i.e. 81 – x = 0 Which gives x = 18 or = -18thus x =18 or – 18

Check for a maximum or a minimum

2

2

dx

R d =

2

x

when x = 18,2

2

dx

R d = -9 which is negative

Therefore at x = 18, the value of R is maximumSimilarly at x = -18, the value of R is minimum.

Therefore, the number of units that maximize the revenue = 18 units

ii) The maximum revenue is given by

R = 14 + 81 × 18 – 12

18 3

= Ksh.986

iii) The price per unit to maximize the revenue is

18

986 = 45.78 i.e. Ksh.54.78

QUESTION FIVE

i) Profit is maximized when

2

81 04

x



310 Lesson Nine


Marginal Revenue = Marginal cost

Note: Profit = Revenue – Costπ = R – C

grad of π = Grad of R – grad of C= (marginal Revenue – marginal Cost)

Grad of π = 0 for maximum and minimum.

PRODUCTION DEPARTMENT

Marginal Revenue = Revenue per unit = 6 + 6 × 0.8= Shs.10.8 per unit

Cost per unit =Q

20,000 + 6 + 0.0002 (marginal cost)

ThereforeQ

20,000 + 6.0002 = 10.8

Giving Q =4.7998

20,000 = 4,166.84

= 4,167 units

ii) SALES DEPARTMENT

Profit = Revenue – CostRevenue = Sales price x Quantity = p(40,000 – 2,000p)

= 40,000p – 2,000p2

Cost = 2 X q + 6,000 + 6q + 80% of 6q= 2q + 6p + 4.8q + 6,000= 12.8q + 6,000= 12.8 (40,000 – 2,000p) + 6,000

Profit = (40,000p – 2,000p2 ) – [6,000 + 12.8(40,000 – 2,000p)]

π = 40,000p – 2,000p2 – 6,000 – 512,000 + 25,600p

π = -2,000p2 + 65,600p – 518,000

dp

d = -4,000p + 65,600



Revision Aid 311


2

2

dp

d = -4,000p + 65,600 When

dp

d = 0

P =

400

65,600 = 16.4

Selling price = Shs.16.4

iii) Quantity 4,167 (as in i)Selling price Shs 16.4 (as in ii)

Firms Profit = Total Revenue – Total CostRevenue = 4,167 × 16.4 = Shs.68,338.8

Cost of production

Production Department 20,000 + 6.0002 × 4,167

Sales Department 6,000 + 2 × 4,167

Total = 26,000 + 25,002.833 + 8,334

= 59,336.83

= Shs.59,336.85

Profit = 68,338.8 – 59,336.85

= Shs.9,002

iv) Quantity and sales price that maximizes the shop‟s profit.

Revenue = (40,000 – 2,000p)p [Q = 40,000 – 2,000p]

Cost = (40,000 – 2,000p) (6 + 0.0002 + 2) + 20,000 + 6,000

Profit = (40,000p – 2,000p2 ) – [(40,000 – 2,000p) (8.0002) + 26,000]

π = 40,000p – 2,000p2 – 320,008 + 16,000.4p - 26,000

= -2,000p2 + 56,000.4p – 340,008

For maximum profit

dp

d = 0 and

2

2

dp

d is – ve

dp

d = -4,000p + 56,000.4



312 Lesson Nine


Therefore -4,000p + 56,000.4 = 0 giving p = 1.0001 = Shs.14

2

2

dp

d = -4,000 which is – ve

for maximum profit = -2,000 (14) + 56,000.4 × 14 – 340,008

= -392,000 + 784,005.6 – 340,008

= 47,998

Quantity Q = 40,000 – 2,000 × 14

= 12,000 units.

QUESTION SIX

a) Quadratic functions in decision making.Due to economies of scale, the cost of production is usuall dependent upon volume of sales.

Total revenue = sale price per unit × quantity sold (say x )Sales price is usually a function of x; f(x).Hence total revenue = f(x) × x ; which may be a quadratic function

Using calculus techniques (i.e. maxima and minima) we can calculate the optimum valueof x, which might give maximum profits or minimum costs.

b) Demand function p = 400 – q (price p in shillings and q in Kg)

Average total cost of prodcing the quantity =

Hence total cost

Revenue = Price × Quantity = p× q = (400 – q ) q = 400q – q 2

Profit π = (400q – q 2 ) – (1000 + 100q – 5q 2 + q 3 )= 300q + 4q 2 – q 3 – 1000

i.

= average fixed cost/unit

as quantity sold gradually increases, average fixed cost per unitdecreases

ii. for maximum profit

2100100 5q q

q

2 2 31000 100 5 1000 100 5q q q q q qq

1000 Fixed cost (100 is fixed cost)

Quantity produced and soldq



Revision Aid 313


2 3

2

2

2

2

2

300 4 1000

300 8 3 (puting =0 )

8 60.53we get 11.422 (note that the negative part of q is invalid)

6

8 6

for 11.422 ; is negative, it gives the mximumvalue of

q q q

d d q q

dq dq

q Kg

d qdq

d q

dq2 3maximum profit 300(11.42) 4(11.42) (11.42) 1000

Sh.1458.31

iii.

2

2

13 2 0 (Product demand function)

13 2 0 (Product demand function)

13 2 and 13 2

total revenue for (13 2 ) 13 2

total revenue for (13 2 ) 13 2

total revenue

x

y

x y

p x y x

p x y y

p x y p x y

x x y x x x xy

y x y y y xy y2 2

2 2

2 2

2 2

for and combined (13 2 ) 13 2

13 13 2 2 2

total cost

Profit Total revenue - total cost

(13 13 2 2 2 ) ( )

12 12 2 2 2

x y x x xy y xy y

x y x y xy

x y

x y x y xy x y

x y x y xy

2 2

2 2

2

13 2 0 (Product

Using partial differentiation and standard calculus techniques for maxima and minima

12 4 2 , 12 2 4

4 (-ve), 4 (-ve)

2

x p x y

x y x y x y

x y

x y



314 Lesson Nine


22 2 2

2 2

:

For maximum or minimum values 0 , 0

For maximum value

In this case, 12 2 4 0 ; 12 4 2 0

which gives; 2 , 2

other conditions are a

Rules

x y

x y x y

x y x y

x units y units

lso certified.

Price P 13 4 2 7

Price P 13 2 4 7

x

y

QUESTION SEVEN

i. The required matrix equation is

5 8 640

4 12 820

that 640,000 has been written as 640 (in thousands)

5 8multiplying both sides with an inverse matrix of we get;

4 12

12 81

4 560 32

x

y

Note

5 8 12 8 6401

4 12 4 5 82060 32

28 0 11201 1

0 28 154028 28

40

55

40, 55

x

y

x

y

x

y

thus x y ii.

1. Marginal productivity

2

2

6010 (this is the rate of change)

total productivity P is given by the function

6010

6010 constant

x

P dx x

x C where C is a x



Revision Aid 315


For the values; 5 and 62

60 ; 62 10(5)

5

24

60 10 24

10

6010(10) 24 118

10

x P

we have C

thus C

Hence P x x

when x

P furnaces

2. Marginal productivity is the increase in output of electric furnaces per week if thecapitalization is increased by Sh.1 million.



316 Lesson Nine


LESSON THREE

QUESTION ONE

a) Discrete data have distinct values with no intermediate points, whereas continuous data can haveany values over a range either a whole number or any fraction.

b) Dispersion is the variation or scatter of a set of values.

Standard deviation is represented by;

2 2

;1

x x x x s

n n

Where s: is for a sample and σ: is for a population

c) This is the coefficient of dispersion of a distribution that is used in comparing dispersion betweendistributions. It is given by;

variation 100%coefficient of x .

d) See Text

QUESTION TWO

a) Payment in days Mid-point Number of customers

x f f fx2

5 – 9 7 4 28 19610 – 14 12 10 120 1,440

5 – 19 17 17 289 4,91320 – 24 22 20 440 9,68025 – 29 27 22 594 16,03830 -34 32 16 512 16,38435 – 39 37 8 296 10,95240 – 44 42 3 123 5,292

100 2,405 64,895

fx 2,405 24.05

f 100 Arithmetic Mean days



Revision Aid 317


b) 2

2

2

fx fxStandard deviation

f f

64,895 2, 405

100 100

8.4

c) Histogram to show payment record of 100 customers

Number ofCustomers

25

22

20 2017

15 16

10 108

5 4

3

4.5 9.5 14.5 19.5 24.5 29.5 34.5. 39.5 44.5Days taken to settle debt

d) Out of 100, 16 i.e. in the class 30 to 34 days and 8 lie in the class 35 to 39 days. Therefore, the

best estimate that an unpaid invoice chosen at random will be between 30 and 39 days old is

64

2,478 = 0.24

QUESTION THREE

a) The smallest value in the distribution is 105, the largest in the distribution is 142. The range tobe spanned is 142 – 105, i.e. 37. The following grouping is a suggestion.

The classes should be of equal width.Group Tally Frequency

105 but less than 110 I I 2110 115 I I I I 5115 120 I I I I 4120 125 I I I I I I I 8125 130 I I I I I I I I 10130 135 I I I I 5135 140 I I I I 4140 145 II 2

40



318 Lesson Nine


b) Calculate the cumulative frequency.

Group Frequency CumulativeFrequency

105 but less than 110 2 2

110 115 5 7115 120 4 11120 125 8 19125 130 10 29130 135 5 34135 140 4 38140 145 2 40

Summary Median value = 125 pFirst quartile = 119 p Third quartile = 131 p

The semi-interquartile range is given by the formula ½ (third quartile value – first quartile value)

Thus the semi-interquartile range = ½(131 – 119)= ½ × 12= 6p

c)

Group Mid-point f fx105 but less than 110 107.5 2 215.0110 115 112.5 5 562.5

115 120 117.5 4 470.0120 125 122.5 8 980.0125 130 127.5 10 1,275.0130 135 132.5 5 662.5135 140 137.5 4 550.0140 145 142.5 2 285.0

40 5,000

Calculate the mean value first using the

Formula x = Σfx Σf

Thus x = 5,000 ÷ 40

x = 125p

Continuing now with x = 125 we can calculate the standard deviation





320 Lesson Nine


Mean =12

613 = 51.08

= 51.1

Standard Dev = 34.25

Coefficient of Variation Supermarket A

=49.92

30.5 × 100

= 61.1%

Coefficient of Variation Supermarket B

= 34.25 × 100

51.1= 67.02

Hence variability of supermarket B is relatively greater than supermarket A

QUESTION FIVE

The objective of this question is to test the candidate‟s knowledge of: Use of base and current weighting index numbers and contrast their construction.

a) To establish the base weighted indices, the weights are the quantities used in 1981. thefollowing tabulation leads to the solution.

For 1981Competent Weight Price (£) Price X weight (£) A 3 3.63 10.89B 4 2.11 8.44C 1 10.03 10.03D 7 4.01 28.07

57.4For 1982Competent Weight Price (£) Price X weight (£)

A 3 4.00 12.00B 4 3.10 12.40C 1 10.36 10.36

D 7 5.23 36.6171.37For 1983Competent Weight Price (£) Price X weight (£)

A 3 4.49 13.47B 4 3.26 13.04C 1 12.05 12.05D 7 5.21 36.47

75.03



Revision Aid 321


The base weighted price indices are therefore for

1981, 100

1982, 57.43

71.37

× 100 = 124.27

198357.43

75.03 × 100 = 130.65

b) The current weighted indices use the weights of the components in 1982 to establish the1982 index, then the weights of the components in 1983 for the 1983 index. The followingtabulation leads to the solution.

The 1981 index as before 100

For 1982Component Weight

19821981

Price (£)1981

Price × Weight (£)

1982Price (£)

1982Price ×

Weight (£) A 2 3.63 7.26 4.00 8.00B 5 2.11 10.55 3.10 15.50C 1 10.03 10.03 10.36 10.36D 6 4.01 24.06 5.23 31.38

51.90 65.24

The 1982 current weighted price index is therefore

51.90

65.24 × 100 that is 125.70

Component Weight 1981Price (£)

1981Price ×

Weight (£)

1982Price (£)

1982Price ×

Weight (£) A 2 3.63 7.26 4.49 8.98B 6 2.11 12.66 3.26 19.56C 1 10.03 10.03 12.05 12.05D 5 4.01 20.05 5.21 26.05

50.00 66.64

The 1983 current weighted price index is therefore

50.00

66.64 × 100 that is 133.28

c) Laspeyres price indices use weights at the base period, whereas Paasche price indices use weights from the current period.



322 Lesson Nine


The weights at the base period will always be available whereas the current weights may notalways be available. In application like Retail or Consumer Price Index establishing thecurrent weights will be much more difficult and expensive than establishing the currentprices. These arguments give a preference for the Laspeyres type of index number.

It can be argued that the use of current weights reflects the present situation moreaccurately, giving a preference for the Paasche type of index number. However, incomputing the series of index numbers, it is clearly demonstrated in part (a) that thedenominator need only be calculated once in the series for base weighted indices, whereas arecomputation is needed for current weighted indices. This leads to a favouring of theLaspeyres type index.

QUESTION SIX

a) i) The Laspeyres Price index is often summarized by the formula

100n o

o o

P Q

P Q

That is the weighting factor in the calculation is the quantity at the base period.

For 1983

Component Quantity Price (1982) Quantity 1982 × Price1982

Price 1983 Quantity1982 × Price1982

A 10 3.12 31.20 3.17 31.70B 6 11.49 68.94 11.58 69.48

C 5 1.40 7.00 1.35 6.75D 9 2.15 19.35 2.14 19.26E 50 0.32 16.00 0.32 16.00

142.49 143.19

Thus the Laspeyres Price Index for 1983

=143.19

100142.49

= 100.49

For 1984

Component Quantity Price Quantity 19821982 1984 × Price 1984

A 10 3.2 32.00B 6 11.67 70.20C 5 1.31 6.55D 9 2.63 23.67



Revision Aid 323


E 50 0.32 16.00148.24

Thus the Laspeyres Price Index for 1984

= 142.49

148.24

× 100

= 104.04

ii) The Paasche Price Index is often summarized by the formula

100n n

o n

P Q

P Q

That is the weighting factor in the calculation is the quantity at the current year.

For 1983



A 12 3.12 37.44 3.17 38.04B 7 11.49 80.43 11.58 81.06C 8 1.40 11.20 1.35 10.80D 9 2.15 19.35 2.14 19.26E 53 0.32 16.96 0.32 16.96

165.38 166.12

Thus the Paasche Price Index for 1983=

166.12100

165.38

= 100.45

For 1984



A 14 3.12 43.68 3.20 44.80

B 5 11.49 57.45 11.67 58.35C 9 1.40 12.60 1.31 11.79D 10 2.15 21.50 2.63 26.30E 57 0.32 18.24 0.32 18.24

153.47 159.48

Thus the Paasche Price Index for 1984



324 Lesson Nine


=153.47

159.48 × 100

= 103.92

iii) The following is a comparison between the index numbers.

Year Laspeyres Paasche1983 100.49 100.451984 104.03 103.92

There is little to choose between the two measures in 1983 as the weightings, that is, thequantity for 1982 and 1983 are close. The situation is different in 1984 where the weightingshave increased in three cases by a significant amount over the 1982 figures resulting in alarge increase in the Laspeyres index than the Paasche index.

b) An index of industrial production would probably by calculated on a month by month basisby central government, indicating in percentage terms by how much production in theindustrial sector has either grown or declined over the previous month or year. The full

index would be made up from components which apply to particular sectors and so it ispossible for an employer to measure increase or decrease in production in that sector andincrease in production may thus be rewarded and a decrease would be looked upon with lessfavour.

An index of retail prices would also probably be calculated on a month by month basis bycentral government, and is an indication in percentage terms of the increase or decrease inretail prices. This measure is often used to quantify inflation. A trade union may thereforeargue its case for an increase in pay to be greater than or equal to the rate of inflation tokeep up with the cost of living and not decrease the living standard s of its members. Anindex of wages may also be published by central government, and may be available fordifferent industrial sectors.

QUESTION SEVEN

Note : The production for 1980 and 1990 is given in 1,000 boxes. As long as units are kept thesame throughout the problem, the rates will not change.

Produce Production1,000 boxes

Price perbox

Po(Shs)Pn

Pn

Po Po Qo

n o p q

1980 oQ 1990 nQ 1980 1990

Cabbages 48,600 62,000 100 150 1.5 4,860,000 7,290,000 Tomatoes 22,000 37,440 220 310 1.4091 4,840,000 6,820,000

Onions 47,040 61,430 180 200 1.1111 8467,200 9,408,000Spinach 43,110 55,720 130 170 1.3077 5,604,300 7,328,700Σ 5.3279 23,71,500 30,846,700

a) Mean Relative Index of prices =4

5.3279 133.20



Revision Aid 325


Hence there is a 33.2% increase in average price of the four horticultural products from1980 to 1990.

b) Laspeyre‟s Price Index =23,771,500

30,846,700 × 100 = 129.76

% increase 29.76%

c) Paasche‟s Price Index = Σpnq n × 100ΣpOq n

And

d) Marshall Hedge-worth index = ΣPn(q n + q n ) × 100ΣPO (q n + q n )

pnq n poq n pn + q n pn (q n + pn ) pn(pn + q n )Cabbages 9,300,000 6,200 110,600 16,590,000 11,060,000 Tomatoes 11,606,400 8,236,800 59,400 18,426,400 13,076,800Onions 12,286,000 11,057,400 108,470 21,694,000 19,524,600Spinach 9,472,400 7,43,600 98,830 16,801,100 12,847,900

Σ 42,664,800 32,737,800 73,511,500 56,509,300

e) Paasche‟s index =32,737,800

42,664,800 × 100 = 130.32

% increase 30.32%

f) Marshall Hedge-worth Index =56,509,300

73,511,500 × 100 = 130.09

% increase = 30%

g) Fisher‟s Price Index = (129.76 × 130.32)1/2 = 130.4

This gives a percentage increase = 30%



326 Lesson Nine


LESSON FOURQuestion 1 and 2See text

QUESTION THREE

r2x1 = 0.78

This is the coefficient of determination of miles traveled to cost and means that 78% of totalcost is attributable to mileage

r2x1 = 0.16 i.e. 16% of cost is accounted for by the type of journey

R 2 = 0.88 is the overall coefficient of determination and indicates that the multiple regressionequation accounts for 88% of the total variation in costs. The coefficient in the equation are:

a = 86 = fixed costsb1 = 0.37 = amount per mile

b2 = 0.08 = influence of the type of journey

QUESTION FOUR

Forecasts produced by3 monthlyMoving average

6 monthly moving average 12 monthly moving average

450437417397 423383 410377 397

380 388407 395443 410470 425 424



Revision Aid 327


QUESTION FIVE

a) Year Qtr Sum of

four qtrsSum of two

qtrs Trend Actual

minustrend

Actual/

trend1990/91 1 49

2 37 2113 58 212 423 52.875 5.125 1.0974 67 213 425 53.125 13.875 1.2611 50 214 427 53.375 -3.375 0.9372 38 215 429 53.625 -15.625 0.7093 59 216 431 53.875 5.125 1.0954 68 218 434 54.25 13.75 1.2531 51 219 437 54.625 -3.625 -.9342 40 221 440 55 -15 0.7273 60 220 441 55.125 4.875 1.0884 70 222 442 55.25 14.75 1.267

1 50 223 445 55.625 -5.625 0.8992 423 61

b) Either

c) An explanation of the forecasting method:i) Plot a graph of the trend.

Additive modelQuarter

Year Q1 Q2 Q3 Q41990/91 5.125 13.8751991/1992 -3.375 -15.625 5.125 13.751992/93 -3.625 -15 4.875 14.751993/94 -5.625 ______ _____ ______

Total -12.625 -30.625 15.125 42.375 Average -4.208 -15.313 5.042 14.125 = -0.354Seasonal variation -4 -15 5 14 = 0

Or

Multiplicative modelQuarter

Year Q1 Q2 Q3 Q41990/91 1.097 1.2611991/1992 0.937 0.709 1.095 1.2531992/93 0.934 0.727 1.088 1.267

1993/94 0.899 Total 2.770 0.718 3.280 3.781 Average 0.923 0.718 1.093 1.260 = 3.994 Adjustment factor* 1.0015 1.0015 1.0015 1.0015Seasonal variation 0.924 0.719 1.095 1.262 = 4.000* adjustment factor = 4/(3.994) = 1.0015



328 Lesson Nine


ii) By eye, establish an appropriate forecast of the trend for the last qurter of 1993/94 andthe first three quarters of 1994/5. (Note: linear regression or the high/low method maybe appropriate methods to establish the forecast.)

iii) Adjust the forecast trend for these quarters for the seasonal variations:

Additive model

Estimated data value = forecast trend value + appropriate seasonal variation value.

Multiplicative methodMultiply each point by the appropriate seasonal factor.

QUESTION SIX

a) A = y – bx

b =2

xx

y y xx

A B A × Bx x x - x y y y - y (x - x )2

2 6 -4 60 104.8 -44.8 179.2 168 6 2 132 104.8 27.2 54.4 46 6 0 100 104.8 -4.8 0 08 6 2 120 104.8 15.2 30.4 4

10 6 4 150 104.8 45.2 180.8 164 6 -2 84 104.8 -20.8 41.6 44 6 -2 90 104.8 -14.8 29.6 42 6 -4 68 104.8 -36.8 147.2 06 6 0 104 104.8 -0.8 0 16

10 6 4 140 104.8 35.2 140.8 1660 1,048 804.0 80

x10

60 y = 104.8

10

1,048

Therefore b =80

804 = 10.05

a = 104.8 – (10.05 × 6) = 44.5

where y = total cost (x 10)x = age in years

b) Maintenance costs using formula from (a):

Age of vehicles Cost(years) (£ = 10)

1 54.552 64.60



Revision Aid 329


3 74.654 84.705 94.756 104.807 114.858 124.90

9 134.9510 145.00

c) 12-year-old vehicle would have an estimated maintenance cost of:

44.5 + (10.05 × 12) = 165.1 (£ x 10), or £ x 10), or 1,651.00

This forecast is an extrapolation beyond the data and consequently is less sound.

QUESTION SEVEN

a) Two possible reasons for the large variation in output each month are:

Seasonal variationProduction problems in some months

b) Graph showing relationship between output and costs.

The graph shows there is a strong positive relationship between output and costs. This meansthat output may be used to predict costs.



330 Lesson Nine


x y xy x2 y2

16 170 2,720 256 28,90020 240 4,800 400 57,60023 260 5,980 529 67,60025 200 7,500 625 90,00025 280 7,000 625 78,400

19 230 4,370 361 52,90016 200 3,200 256 40,00012 160 1,920 144 25,60019 240 4,560 361 57,60025 290 7,250 625 84,10028 350 9,800 784 122,50012 200 2,400 144 40,000

240 2,920 61,500 5,110 745,200

b =2240-5.11012

2,920240-61,50012 =

3,720

37,200 = 10

a =12

2,920 – 10 ×

12

240 = 43.333

y = 43.333 = 10x

This means that when output is zero, costs are zero £43.333, and that for every one unit increasein output, costs will rise by £10. This assumes linearity.

QUESTION EIGHT

i. The tabulation of the trend pattern is as follows and has been computed using the formula

Trend = Sales – Seasonal deviation

Year Quarter Sales Seasonaldeviation

Trend

£ 000 £ 000 £ 0001983 2 360

3 5304 354 -42 396

1984 1 304 -128 4322 430 -37 4673 750 276 4744 395 -93 488

1985 1 340 -145 485

2 500 12 4883 660 153 5074 509 -15 524

1986 1 374 -165 5392 590 43 5473 710 153 5574 521

1987 1 440



Revision Aid 331


ii. The seasonal variations are established from the following table

Quarter Year 1 2 3 41983 -421984 -128 -37 276 -93

1985 -145 12 153 -151986 -165 43 153 Total -438 18 582 -150 Average -146 6 194 -50 Total 4 Adjustedaverage

-147 5 193 -51 Total 0

Adjustment to each average – 4/4 that is – 1

To the nearest integer the seasonal variations are

Quarter 1 -1472 5

3 193

4 -51

iii. Forecast sales 1987 quarter II =

590 + 5 (£ 000)

= 595 (£ 000)

QUESTION NINE

The object of this question was to test the candidates knowledge of the time series and the

ability to present data on a labeled diagram

Year Quarter Costs £ Four quartertotal

Four quarter total inpairs

Trend Deviation

1980 IV 15601981 I 1730 6348

II 1554 6418 12766 1595.77 -41.75III 1504 6638 13056 1632.00 -128.00IV 1630 6679 13317 1664.63 -34.63

1982 I 1950 6715 13394 1674.25 275.75II 1595 6785 13500 1687.50 -92.50III 1540 6695 13480 1685.00 -145.00IV 1700 6809 13504 1688.00 12.00

1983 I 1860 6843 13652 1706.50 153.50II 1709 6933 13776 1722.00 -13.00III 1574 6983 13916 1739.50 -165.50IV 1790 6995 13978 1747.25 42.75

1984 I 1910 7061 14056 1757.00 153.00II 1721III 1640



332 Lesson Nine


a) The calculations for the trend figures and the deviations are summarized in the above table

The seasonal effect is removed from the data by first totaling four quarter figures, then

totaling the four quarter figures in pairs and finally dividing by eight and centering the trend

figure at the middle point

Deviation is the difference between the costs and the corresponding trend figureb) The seasonal deviation is another calculation produced from a table

SeasonalQuarter

Deviations

Year I II III IV1983 -41.75 -128.00 -34.631984 275.75 -92.50 -145.00 12.001985 153.50 -13.00 -165.50 42.751986 153.00 Total 582.25 -147.25 -438.50 20.12 Average 194.08 -49.08 -146.17 6.71 Total = 5.54 Adjustedaverage

192.70 -50.46 -147.55 5.33 Total = 0.02

Seasonaldeviation

193.00 -50.00 -148.00 5

The adjustment is obtained by reducing each figure by 1.38 (that is 5.54∸4)

c) From the trend the forecast trend value of Quarter IV of 1984 is £ 1790. the seasonal

deviation is £ 5, hence the forecast heating costs for quarter IV for 1984 are £ 1795

It can be seen that the trend is a little variable and the seasonal deviations are not very

regular, possibly due to the weather. Another possible reason for the irregular fluctuation in the heating cost trend line is an

uneven increase in the price of the heating medium (for example, oil or gas or

electricity.)



Revision Aid 333


LESSON FIVE

QUESTION ONE

a) The Poisson distribution is appropriate because there is a small probability of an eventoccurring these are discrete values and the average of these events (i.e. m = np) is below

10 (50 × 0.1 = 5)

b) The Poisson formula is

P( )!

xe x

In this exmple m = np = 50(0.02) =1

P ( x 2) = P(x=0) +P ( x=1 ) + p (x=2 )1

2

0! 1! 2!

e e e

1 1 2 12718 2718 1 2718

1 1 2 1

0.3679 0.3679 0.1840 0.92

c) Similarly , the following can be drawn up

P Pa0 1.0000.02 0.9200.05 0.5440.10 0.1250.15 0.020

QUESTION TWO

(a) (i) Probability = 0.7 × 0.3× 0.7 = 0.147

(ii) Probability =3C2 (0.7)2

(0.3) = 0.441;

(iii) The answer to (ii) is 3× the answer to (i)

This is because either A or B could be detected, A and C, or B and C. Each of these situations isequally probable, with the probability given in part (i).

(b) (i) The probability that the defect is undetected by the inspection procedure is0.3.



334 Lesson Nine


The probability that it is undetected by the secondary check is 0.4. The probability that it is undetected by both is 0.3 × 0.4 = 0.12

(ii) Call the two faults A and B.

We can have the possibilities:

A is found by inspection procedure, B not found by either procedure.

B is found by inspection procedure, A not found by either procedure.

A is found by secondary check, A and B not found by inspection procedure.

B is found by secondary check, A and B not found by inspection procedure.

Adding together the probabilities corresponding these four possibilities, the required probabilityis

(0.7×0.3×0.4) + (0.3×0.7×0.4) + (0.3×0.3×0.6×0.4) + (0.3×0.3×0.4×0.6)=0.0840 + 0.0840 + 0.0216 + 0.0216 + 0.1680 + 0.0432

=0.2112

(iii) Probability that a fault is detected by the inspection procedure is 0.7.Probability that a fault is detected by secondary check is 0.3 × 0.6 =0.18

Therefore the proportion of faults detected by the inspection procedure and secondary check,respectively, is

0.7 0.18 and

0.7 0.18 0.7 0.18

70 18that is and

88 88

QUESTION THREE

Let F represent a unit which has been found to be faulty.Let P(S1 ) = probability that a unit chosen at random comes from S1

Let P(S2 ) = probability that a unit chosen at random comes from S2

Let P(S3 ) = probability that a unit chosen at random comes from S3

P(S1 ) = 0.40P(S2 ) = 0.35

P(S3 ) =0.25

1.00

The percentages of faulty unit are as follows:



Revision Aid 335


1

2

3

P(F S ) 0.02

P(F S ) 0.03

P(F S ) 0.04

The required probability may be expressed as;

1P(S F) the unknown probability is P(F) to be slotted into the formula

Note that

1 1

1

2 2

2

3 3

3

P(S ) P(F S )P(S F)

P(F)

P(S ) P(F S )P(S F)

P(F)

P(S ) P( F S )P(S F)

P(F)

the faulty part can only have come from S1 or S2 or S3 and so

1 2 3P(S F)+P(S F)+P(S F)=1.0

since P(F) is a denominator and the sum equals unity then the expression

1 1 2 2 3 3P(S ) P( F S )+P(S ) P(F S )+P(S ) P(F S )

must be equal to P(F)

thus P(F)= (0.4 0.02)+(0.35 0.03)+(0.25 0.04)

0.0080 0.0105 0.0100 = 0.0285

Substitution into 1 1

1

P(S ) P( F S )P(S F)

P(F)

Gives1

0.4 0.02P(S F)

0.0285

= 0.2807

QUESTION FOUR

0 66

0

1 56

1

2 46

2

3 36

3

0 0.3 0.7

1 0.3 0.7

2 0.3 0.7

3 0.3 0.7

P C

P C

P C

P C



336 Lesson Nine


4 264

5 165

6 066

4 0.3 0.7

5 0.3 0.7

6 0.3 0.7

P C

P C

P C

QUESTION FIVE

P = 0.6 q = (1-p) = 1-.06 = 0.4

0.035200

0.40.6

n

pq σ

0.65 0.601.43

0.035 z

Which gives 0.4236 (42.36%)

This means there is a (0.5-0.4236) 0.0764(7.64%) chance of 65% or more passing the first

attempt. This is graphically shown below.

Probability

density

42.36%

7.64%

0.60 0.65 X proportions

i.e. the mean

QUESTION SIX

a) The left-hand tail of the distribution below 900 hours represents the number of lamps that will fail before 900 hours. Accordingly, if the probability of the distribution above 900 hoursis found and deducted from 0.5, the required number can be found thus:

The probability of which is 0.40821000 900

1.3375

z



Revision Aid 337


Lamps failing before 900 hours =5000(0.5- 0.4082)=459

b)

1000 950 0.6775

z The probability of which is 0.2486

Lamps failing between 1000 and 950 hours = 5000 (0.2486)=1243

c)

1000 9251

75 z

The probability of which is 0.3413

Proportion failing between before 925 hours = 0.5 – 0.3413 = 15.87%

d) A probability of 0.3(0.5-02) is found in the tables with a Z score of 0.84

Thus 0.84=1000-916 84

z= =1 s.d.= =100s.d. 0.84



338 Lesson Nine


LESSON SIX

QUESTION ONE

The sample mean is 150 grams so that the estimate of the population mean is 150 grams.

i.e. 150 grams 150 grams x Where x means best estimate of .

When n= 625Standard error of the mean

s 30= = =1.2grams

625n

When n =1225

x

30s = =0.857 grams

1225

QUESTION TWO

Correction factor =

n 80Approximation to correction factor= 1- 1 0.9486

N 800

It will be seen that to three significant figures, it is accurate enough for all practical purposes, thetwo formulae give the same result, i.e. 0.949.

Standard deviation error of the means xS =

1

60.949

80

0.637 grams

n

N n

Note: The standard error without correction is6

0.671 grams80

Thus the precision of the sample estimate, measured by the standard error, is determined notonly by the absolute size of the sample but also to some extent by the proportion of thepopulation sampled.

N-n 800 800.9493

N-1 800 1



Revision Aid 339


QUESTION THREE

The Central Limit Theorem states that the means of samples (and the medians and standarddeviations) tend to be normally distributed almost regardless of the shape of the originalpopulation.

QUESTION FOUR

See text

QUESTION FIVE

H0 : 2 = 1

H1 : 2 1 (one-tail test)

1

200P = 0.2

1000 2

240 p = =0.22

1091

Pooled sample proportion200+240

p= 0.211000+1091

and 1- 1- 0.21 0.79q p

0.01781091

0.790.21

1000

0.790.212P1PS

0.20-0.22z= =1.120.0178

The critical value of one-tailed test t the 5% level is 1.64 so that as the calculated value islower than this value we conclude there is insufficient evidence to reject the nullhypothesis.

QUESTION SIX

0 1 2

1 2

H :μ -μ =0

H1:μ -μ =0

It will be seen that this is a two-tailed test. The common standard deviation is calculated first.

2 2

1 1 2 2

1 2

(n -1)s (n 1)ssp=

n +n 2

2 2(12 1)63 (9 1)76

12 9 2

68.77



340 Lesson Nine


Px1

1

x2

2

2 21 2 x1 x2

S 68.77s = 19.85

n 12

sp 68.77s = 22.92

n 9s(x -x )= s +s =30.32

and, finally the t score can be calculated

1 2

1 2

x -x 1060 970t= 2.97

s(x -x ) 30.32

At 5% level with (n1+n2-2)= (12+9-2) = 19 degrees of freedom value is 2.093.

Since the calculated value is greater than 1.96, the null hypothesis can be rejected at the 5% level,i.e. we conclude that there is a significant difference between the mean monthly incomes.

QUESTION SEVEN

We are testing whether the observed number of defects fits a binomial distribution, thus;

H0: the observed number of defects conforms to a binomial distribution of the form(p+q) 5 where p=0.18

H1 :that the observations do not conform.

The observed frequencies are already given so its only necessary to calculate the frequenciesexpected from a binomial distribution to the power 5 i.e. (p+q)5, where p=0.18 and q=1-0.18=0.82.

The probabilities of the various values of p and q can be found from binomial probability tablesif available. Alternatively they can be calculated from the binomial expansion. I.e.

P5 +5 (p4 q)+ 10(p3 q2)+ 10(p2 q3)+ 5(p q4)+q5

This shows the probabilities for 5, 4, 3, 2, 1 and 0 defectives and when p = 0.18 (the probabilityof a bulb being defective) and q=0.82 (the probability of not being defective) the probabilitiesrange from 0.0002 (i.e. 0.185 ) for five defectives to 0.3711 (0.825 ) for no defectives. When theprobabilities are known they are multiplied by 100 boxes to find the expected frequencies whichare used in the normal X 2 procedures.

The table below summarises the calculations:



Revision Aid 341


Defectives No. of boxesObserved

BinomialProbabilities

ExpectedFrequency

(O - E) 2 (O - E) 2 E

0 40 0.3711 37.11 8.35 0.221 37 0.4069 40.69 13.62 0.332 17 0.1786 17.86 0.74 0.04

3 51 0 }

0.0392 3.920.40.02 }4 0.0040 2.76 0.64

5 0.0002 1.23

Note: Because of the very small values of the expected frequencies for 3, 4 and 5 defectives theyhave been combined into one but it makes little difference to the results if they are notcombined.

The calculated 2 value is compared with the 2 value for the appropriate degrees of freedom.

Because the last three classes have been combined there are four classes remaining, i.e. for 1, 2

and the combined class for 3-5 rejects thus V =n – 2 = 4 - 2=2.

The X 2 value for two degrees of freedom at the 5% level is 5.991 and, as the calculated value of1.23 is well below this, we accept the null hypothesis and conclude that the observed values fit abinomial distribution to the power 5 when p = 0.18.

Note: If the last three classes had not been combined, the calculated X2 score would have been1.8 and there would have been 4 degrees of freedom. At 4 degrees of freedom the score is 9.488,so the conclusion would be the same, i.e. we accept the Null Hypothesis.

QUESTION EIGHT

See textQUESTION NINE

1. a)

S.e. = 0.03380

0.90.1

n

pq

Where p =proportion late

Actual p =

80

6 = 0.075

0.1 0.0750.75 standard errors

0.033

This is less than the standard errors at the 5% level ± 1.96 so we conclude there is nosignificant improvement in deliveries.



342 Lesson Nine


b) 0.75 standard errors from the mean would cover ± 0.2734 i.e. 54.68%, say 55% of thepopulation, so the MD‟s claim could be accepted at any level of confidence

2. Calculated mean of sample = 4.68 gmsCalculated sample s.d. = 1.968

n 10 best estimate of population s.d.= 1.986 2.093

n-1 9

2.093std. error of mean= 0.66

10

There are 10-1 =9 d.f and it is a one-tailed test. The 5%value for a one-tailed test is1.833.

The sample mean should be within 1.833*0.66 gms = 1.21 gms.

The actual difference is 4.68-3.8 gms= 0.88 gms so the figures do not support thePurchasing Manager‟s assertion.



Revision Aid 343


LESSON SEVEN

QUESTION ONE

a)

b).Drilling Immediately:

£ 0.55 120+0.45 -40 £ 66-18 £48mm m

Tests:

Successful and drill, £ 0.8 100+0.20 -50 £ 80-10 £48mm m

Fail and drill, £ 20 100+0.8 -50 £ 20-40 £20mm m

There fore EV = £ 0.7 70+0.3 15 £ 49+4.5 £53.5mm m

Overall EV = £53.5

Thus course of action is: first carry out tests, if successful proceed to drill if tests fail, sell theexploitation rights.

Fail 0.45

Succeed 0.55

Fail 0.2

Succeed 0.8

Sell70

15

Fail 0.8

NPV

45

120

-40

65

100

-50

15

100

-50

Succeed 0.2Drill

Fail 0.3

Succeed 0.7

53.5Tests

Drill immediatel53.5

1

2

2

3

3

4

1

Sell immediatel

Sell

-20

70

48

Drill



344 Lesson Nine


QUESTION TWO

a) tree diagram

b) EV calculationsEV(present arrangement) = £(0.3×1000) + (0.23×900) + (0.23×800) + (0.23×700)

= 860

EV (with consultants) = 785

c) on strict EV calculations the firm should not use consultants, however management may

consider to use consultants in order to improve the chances of completing on time thussafeguarding their reputation.

QUESTION THREE

See text

QUESTION FOUR

Let the value of the small store be = 1 And let the value of the large store be = 2

If both survive, A loses nothing, if only larger store survives, A loses 1 and if smaller store

survives, A will lose 2.

Continue as present

Employ consultants

(Cost = £200,000)

On time (p = 0.3)

1 month delay (p = 0.23)

2 months delay (p = 0.23)

3 months delay (p = 0.23)

1 month delay (p = 0.05)

months delay (p = 0.05)

n time

1,000

1,000 – 100 = 900

1,000 – 200 = 800

1,000 – 300 = 700

1,000 – 200 = 800

1,000 – 200 - 100 = 700

1,000 – 200 - 200 = 600

Profits before construction

costs (£’000)



Revision Aid 345


GAIN MATRIX FOR A

DEFENDER

A

B I II Row

Minimum

I

II

0 -2

-1 0

-2

-1

ColumnMaximum 0 0

There is no saddle point.

Hence this is a problem of mixed strategy.

Using method givenI II

I

II

0 -2

-1 0

0-(-2)=2

0-(-1)=1

0-(-1)=1 0-(-2)=2

The final strategy is given by the matrix

A

A plays his first row 1/3 rd of the time (randomly)

A plays his second row 2/3 rd of the time.

Similarly

B plays his first column 2/3 rd of the time

B plays his second column 1/3 rd of the time.

Attack the smaller storeI

Attack the larger storeII

Defend the smaller storeI

Both survive0

The larger store destroyed-2

Defend the larger storeII The smaller storedestroyed-1

Both survive0

I III

II

0 -2

-1 0

1/3

2/3

2/3 1/3



346 Lesson Nine


The values of the game

=0×1/3×2/3 +(-2)×(1/3×1/3) +(-1)×2/3×2/3 +0×1/3×2/3

=0-2/9-4/9+0

= -6/9

= -2/3

QUESTION FIVE

YRowMinimum

X1 2 -1 -1-2 1 -2 -22 2 1 0

ColumnMaximum

2 2 1

There is no saddle effect.

Let the probabilities of y be p, q, r.

Then the three payoffs to y corresponding to each of the three counter moves of his opponent xmust all be equal to the optimal value v of the game.

Y‟s payoffs against the three moves of x are

1p+2q+(-1r) -2p+1q+1r 2p+0q+1r

We obtain three equations by equating each of the three payoffs to v

1p+2q-1r = -2p+1q+1r = 2p+0q+1r = v

Also p+q+r = 1 (Total probability)

1p+2q+1r= -2p+1q+1r….. (1) 1p+2q+1r= 2p+0q+1r……(2) p+q+r = 1 …..(3)

Solving these equations simultaneously, we get

P=2/17, q=8/17, and r=7/17Similarly using the same reasoning as before

Let the three probabilities of x be p‟, q‟, r‟

We get 1p‟+2q‟+1r‟ = 2p‟+1q‟+0r‟ = -1p‟+1q‟+1r‟

Also p+q+r = 1



Revision Aid 347


Solving them simultaneously, we get

P‟=3/17, q‟=5/17, and r‟= 9/17

Hence x should play his rows in the ratio 3: 5: 9

And y should play his columns in the ratio 2: 8: 7

( Note: Rows and columns should be played at random)

Σ [(payoff * joint probability of the payoff]

2 3 8 3 7 3i.e. 1 2 ( 1)

17 17 17 17 17 17

+6 other values calculated in the same way as above which amount to 11/17

Alternatively

Value of the game is1 3 (-2) 5 2 9 11

3 5 9 17



348 Lesson Nine


LESSON EIGHT

QUESTION ONE

a) Let x1 and x2 be the number of shares invested in airline and insurance shares respectively. Then the objective function will be as follows:Objective function (maximize)

z = 2x1 + 3x2 Share appreciation.

Subject to the following restrictions:1. 40x1 + 50x2 ≤ 100,000 sh. Total investment.2. 50x1 ≤ 40,000 sh. Investment in insurance.3. x1 + 1.5x2 ≥ 2600 sh. Dividends.4. x1, x2 ≥ 0 Non-negativity of number.

b) Reduced cost represents the amount that objective function coefficient of a non-basicdecision variable must improve (increase in this case) to be put into the basic. In this case

since reduced cost is equal to zero for both variables, it means they are in the basic.

Dual prices on the other hand mean the amount that share appreciation will improve incase any of the limiting constraints is increased by one unit. This occurs for constraints thatthe slack is exactly zero. In this case total investment and investment in insurance do havepositive dual prices while Dividends does have zero dual price (can not increase shareappreciation if increased by one unit).

c) From the computer solution, the client‟s money should be invested as follows, to satisfy therestrictions:1500 shares to be invested in Airline shares, and800 shares to be invested in insurance shares, to give an optimum quarterly share

appreciation of sh. 5400.d) For the optimal decision to remain the airline shares appreciation should not be lower than

2.5, but can be any higher amount. That is, the optimum solution is insensitive to increasein the share appreciation.

QUESTION TWO

a) Formulation of the problem Take x1 to be the number of large loaves and x2 to be the number of small loaves.

Objective function

z = 5x1 + 3x2 Profit

Constraints.Line 1 x1 ≤ 280 Maximum number of large loaves.Line 2 x2 ≤ 400 Maximum number of small loaves.Line 3 10x1 + 8x2 ≤ 4000 Space (1000-3 ) m3 Line 4 25x1 + 12.5x2 ≤ 8000 Hours (1000-3 )

5 x1, x2 ≥ 0 Non-negativity.



Revision Aid 349


Using graphical method.

Small Vs Large loaves

-100

0

100

200

300

400

500

600

700

0 100 200 300 400 500

x1

x 2

3 4 1 2

The feasible area is that enclosed by corner points ABCDEFCorner points are where the optimum feasible solution exists. The upper points are to be

considered for maximization problem. At corner point A, Line 2 and x1=0 intersectSo x2= 400Putting these value in the objective function gives the followingz = 5x1 + 3x2=5 0 + 3 400=1200 Profit A

At corner point B, Line 2 and Line 3 intersect

x2= 400 and x1= 8010

40084000

Putting these value in the objective function gives the followingz = 5x1 + 3x2=5 80 + 3 400=1600 ProfitB

At corner point C, Line 3 and Line 4 intersect

The equations for the lines are:Line 3 10x1 + 8x2 = 4000 (1)Line 4 25x1 + 12.5x2 = 8000 (2)Multiplying equation (1) by 25 and equation (2) by 10 gives the following

(10x1 + 8x2 = 4000) 25(25x1 + 12.5x2 = 8000) 10

250x1 + 200x2 = 100,000 (3)

BA

C

D

EF



350 Lesson Nine


250x1 + 125x2 = 80,000 (4)Deducting equation (4) from Equation (3) gives:

75x2 =20,000x2 =266.7

x1 = 7.18610

266.78-4000

Putting these values in the objective function gives the followingz = 5x1 + 3x2=5 186.7 + 3 266.7=1733 ProfitC

At point D, Line 1 and Line 4 intersect.x1 = 280

x2 = 8012.5

252808000

The profit at this point is then equal to:z = 5x1 + 3x2=5 280 + 3 80=1640 ProfitD

At point E, Line 1 and x2 = 0 intersectx2 = 0 x1 = 280

z = 5x1 + 3x2=5 280 + 3 0=1400 ProfitE

Comparing these profits, it is at point C that profit is maximized.So the solution is that:

x1 = 186 No. of large loaves produced.x2 = 266 No. of small loaves produced.

And the maximum profit is Profit c = Shs. 1,733 where x 1 = 186.7, x 2 = 266.7

NOTE: Two methods can be used to solve the problem. It is easily solved using thegraphical rather than the simplex method, since it is just two variables and sensitivityanalysis is not required.

b) To solve this kind of problem (linear programming problem) the following procedure isfollowed:- First, the problem has to be formulated. That is, the objective function and constraints

are determined.Objective function is that which is to be optimised.Constraints are the limitations in resources.

- Secondly, the method of solving is determined. In this case, of a two-variable problem,the better method to use is graphical method, rather than simplex method.

- Thirdly, the constraints are taken as equalities and a line graph drawn. The unwantedregions are shaded out. Resulting region indicates the feasible region. The optimumpoint exists where there are corner points, which show extreme amounts. Formaximization it is the outer ones to the right and up. For minimization it is the lowerside.

- Lastly, the profit is determined at those points where there is maximum profit, is thepoint to be used.

NOTE: This part simply asks for the procedure followed.



Revision Aid 351


QUESTION THREE

a) i) Simplex method will be appropriate.

Formulation of problem.Objective function.

Let x1 and x2 be the number of Deluxe and Professional bicycle frames producedrespectively per week.

z = 1000x1 + 1500x2 Profit sh.

Constraints:2x1 + 4x2 ≤ 100 Aluminum alloy3x1 + 2x2 ≤ 80 Steel alloyx1, x2 ≥ 0

In standard form:0 = z – 1000x1 – 1500x2 + 0s1 + 0s2 100 = 2x1 + 4x2 + s1 + 0s2

80 = 3x1 + 2x2 + 0s1 + s2

Table 1x1 x2 s1 s2 Solution Ratio

s1 2 4 1 0 100 25s2 3 2 0 1 80 40z -1000 -1500 0 0 0

Table 2x2 1/4 1 1/4 0 25 50s2 2 0 -1/2 1 30 15z -250 0 375 0 37,500

Table 3x2 0 1 3/8 -1/4 17.5x1 1 0 -1/4 1/2 15z 0 0 312.5 125 41,250

Stop here The optimal weekly production schedule is as follows:Deluxe bicycle Frame = 17.5 ≈17 Professional bicycle Frame = 15

ii) Let Δ1 be the change in profit from Deluxe bicycle frame.Δ2 be the change in profit from Professional bicycle frame. So

C1 = 1000 + Δ1 and C2 = 1500 + Δ2 limit of profit.From the final table: To avoid entry ofs1 312.5 – 1/4Δ1 > 0 Δ1 < 1250s2 125 + 1/2Δ1 > 0 Δ1 > -250

From the two conditions:-250 < ∆1 < 1250 and



352 Lesson Nine


750 < C1 < 2250

To avoid entry ofs1 312.5 + 3/8Δ2 > 0 Δ2 > -833.33s2 125 – 1/4Δ2 > 0 -Δ2 > -500 Δ2 < 500

So from the two conditions:-833.33 < ∆2 < 500 And C2 varies as follows666.7 < C2 < 2000

NOTE: This problem could be solved graphically with part (i) Easily determined. Part(ii) Limits will be determined from equating slopes of the objective function which hascoefficients with constraints nearest to it.For part (ii), accurate drawings will be required. Intuition will have to be followed andthere will be an assumption that fractions are possible.

b) The technique is really involving.

Assumes fractions are possible, which is not really the case like here where we cannot make½ a bicycle frame.

QUESTION FOUR

a) i) A feasible solution is one that satisfies the objective function and given constraintsii) Transportation problem is a special linear programming problem where there a

number of sources and destinations and an optimum allocation plan is required. Totaldemand equal total supply

iii) Assignment problem is a special kind of transportation problem where the number ofsources equals the number of destinations. That means for every demand there is onesupply.

b) This is a case of assignment problem. Assignment problems usually require that the number of sources equal the number ofsupply. Here there are 5 districts and only 4 salespersons. A dummy salesperson E isintroduced with zero ratings.

Districts1 2 3 4 5

A 92 90 94 91 83Sales persons B 84 88 96 82 81

C 90 90 93 86 93D 78 94 89 84 88E 0 0 0 0 0

By following the Hungarian method:Firstly:For each row, the lowest rating is reduced from each rating in the particular row. Thisresults to a row reduced rating table. Then all the zeroes are to be crossed by the leastnumber of vertical and horizontal lines. If the number of lines equal the number of rows(or columns = 5 in this case) then the final assignment has been determined. Otherwise thefollowing steps are followed.



Revision Aid 353


1 2 3 4 5 A 9 7 11 8 0B 3 7 15 1 0C 4 4 7 0 7D 0 16 11 6 10

E 0 0 0 0 0

Secondly, for each column, the lowest rating is reduced from every rating in the particularcolumn. In this case the table will remain the same since the dummy salesperson has ratingsof zero for every district. Thirdly a revision of the opportunity-rating table is done. The smallest rating in the table not covered by the lines is taken (in this case it is one). Thisis reduced from all the uncrossed ratings and added to the ratings at the intersection of thecrossings. Then all the zeroes are to be crossed by the least number of vertical andhorizontal lines. If the number of lines equal the number of rows (or columns = 5 in thiscase) then the final assignment has been determined.Otherwise the following steps are followed.

1 2 3 4 5 A 8 6 10 8 0B 2 6 14 0 0C 4 4 7 0 8D 0 16 11 6 11E 0 0 0 0 1

Third step is repeated as follows:1 2 3 4 5

A 6 4 8 8 0B 0 4 12 0 0C 2 2 5 0 8D 0 16 11 8 13E 0 0 0 2 3

Still the optimal solution has not been reached. Third step is again repeated to give thefollowing table:

1 2 3 4 5 A 6 2 6 8 0B 0 2 10 0 0C 2 0 3 0 8D 0 14 9 8 13E 0 0 0 4 5

An optimal assignment can now be determined since the number of lines crossing the

ratings is equal to 5.Lastly, the assignment procedure is that a row or column with only one zero is identifiedand assigned. This row or column is now eliminated. The other zeroes are then assigneduntil the last zero is assigned. This step-by-step assignment is shown on the following tablefrom the first one to the fifth one.



354 Lesson Nine


District

1 2 3 4 5

A 6 2 6 8

Sales person B 0 2 10 0

C 2 3 0 8

D 14 9 8 13

E 0 0 4 5

The assignment is as follows

Salesperson District Rating A 5 83B 4 82C 2 90D 1 78 Total rating 333

The total rating is 333.

QUESTION FIVE

a) Sensitivity analysis measures how sensitive a linear programming solution is to changes inthe values of parameters. These parameters include the coefficients of objective function,

limiting resources and non-limiting resources.

So sensitivity analysis involves changing any of these parameters and showing how thelinear programming problem is affected.Dual values indicate the additional improvement of the solution due to additional unit oflimiting resource. In that way, the additional improvement of solution is the price worthpaying to release a constraint

b) Let x1, x2 and x3 be the units of desktop 386, Desktop 286 and laptop 486Maximize profitZ = 5000x1 + 3400x2 + 3000x3

Subject to

x1+x2 500 limit of desktop modelsx3 250 limit of laptop modelx3 120 limit of 80386 chipsx2+x3 400 limit of 80286 chips5x1+4x2+3x2 2000 hours available Assumptions x1, x2, x3 0

Linearity/proportionality

01

02

03

05

04



Revision Aid 355


DivisibilityDeterministic Additive

c) i) The optimum product mix is that the numbers of units to produce are

Desktop 386-120Desktop 286-200Laptop 486-200Maximum profit isZ=5000 120+3400 200+3000 200=Sh1,880,000Unused resources include the following JK computers can still produce 180 more desktop models (500-120-200) and 50 laptopmodels (250-200)For used up resources the prices to pay for any additional unit are as followsSh150 for 80386 chipSh90 for 80286 chipSh20 for any hour

ii) The range for the variables x1, x2 and x3 are to indicate where the number of units canchange without affecting the basic solution

The range for the constraints indicate the extent the resources can be changed withoutaltering the basic solution of the linear programming problem

iii) The dual value of 80386 chip is Sh 150. That is the addition increase in profit due toincrease of one chip. So if the company increases the number of chips by 10, theadditional profit will be 10 150=Sh1,500.

QUESTION SEVEN

a) Network

The critical path is A, E, F, H, I. The duration of the critical path is 15 days.

b) Assuming all the activities start as soon as possible, the following chart shows whenactivities will start and finish.

1 6

4 4

8 8 10 10 12 12 15 15

EST LST



356 Lesson Nine


Consequently the following resource allocation is required on a day to day basis.

Day Activities Number of increased staff1 A, B 1+3 = 42, 3 A, D, G 1+1+1 = 34 A, G 1+1 = 25, 6 C, E, G 2+2+1 = 57 C, E 2+2 = 4

8 E 2 = 29, 10 F 2 = 211, 12 H 1 = 113, 14, 15 I 1 = 1

Current costs with 5 staff = 15 days 5 £500 = £37500 If activity C is delayed to start on day 7 only 4 staff are required and the project durationunchanged. The cost is

15 days 4 £500 = £30000

However if the duration is increased by starting B on Day 1 and delaying activities A, E, F, H, I

by one day and C until day 12 only a maximum of 3 staff are required.

Critical

activities

Time scale

(days)

G

DB

A E F H I

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

C

Other

activities



Revision Aid 357


MOCK EXAMINATION

Work out these question for three hours (exam condition) then hand them in to DLC for marking

Instructions: Answer any THREE questions from SECTION I and TWO questions from SECTION II.

Marks allocated to each question are shown at the end of the question. Show all your workings


QUESTION ONE

a) Explain the importance of set theory in business (4marks)b) By use of matrix algebra, develop the leontief inverse matrix. (8marks)c) Digital ltd manufactures and sells floppy disks at Nairobi industrial area. The average

revenue (AR)(in thousands of shillings )of producing x floppy disks are given by thefollowing functions

2 5 5012 2

50 x

ATC x x

And AR=800-2X 2 Where: x is the number of floppy disks produced

Required:i. The profit function (3marks)ii. The number of floppy disks required to maximize profit (3marks)iii. The maximum profit (2marks)

(Total: 20marks)

QUESTION TWOa) State any five problems encountered in the construction of the consumer price index

(5marks)b) An investment analyst gathered the following data on the 91-day Treasury bill rates for the

years 2003and 2004Month Treasury bill rates (%)

2003 2004 January 3.2 5.5February 3.0 5.2March 2.8 4.3 April 2.5 3.6May 2.9 3.3

June 3.4 2.7 July 3.7 2.4 August 4.0 2.0September 3.8 2.3October 4.2 2.8November 4.5 3.1December 5.1 3.7

The analyst would like to determine if on average there was a significant change in the Treasury bill rates over the two years.



358 Lesson Nine


Required:i. The mean and variance of the Treasury bill rates for each year (10marks)ii. Determine if there is a significant difference in the average Treasury bill rates (use a

significant level of 1%). (5marks)

Note: S2

= 2

11

21

2

22

2

11

nn

S nS n

QUESTION THREE

a) Describe the characteristics of the following statistical distributionsi. Binomial distribution (3marks)ii. Poisson distribution (3marks)

b) High Grade Meat Ltd produces beef sausages And sells them to various supermarkets .Inorder to satisfy the industry‟s requirements ,the firm may only produce 0.2percent ofsausages below a weight of 80 grammes .The sausage producing machine operates with astandard deviation of 0.5 grammes .The weights of the sausages are normally distributed

The firm‟s weekly output is 300,000sausages and the sausage ingredients cost shs5.00

per 100 grammes ,sausages with weights in excess of 82 grammes require additionalingredients costing sh 2.50 per sausage

Requiredi. The mean weight at which the machine should be set (4marks)ii. The firm‟s weekly cost of production (10marks)

(Total: 20marks)

QUESTION FOUR

a) A survey of undergraduate students at High Fliers University (HFU)showed the followingresults regarding gender and the fields of specialization in their studies

Field of specializationGender Business Science Arts TotalMale 100 250 100 450Female 200 50 100 350 Total 300 300 200 800

Requiredi. Determine if the field of specialization in the studies is dependent on gender (use

significance level of 5%) (10marks)ii. An earlier survey showed that the proportion of female students taking science was only

10%of the total student population taking science .Does the data above show anysignificant improvement in the proportion of female students taking science (use a

significance level of 5%) (6marks)

b) Charles Nzioka who is a barber has found out that he can shave on average 4 customers perhour .The arrival rate of customers averages 3customers per hour

Requiredi. The proportion of time that Charles Nzioka is idle (1 mark)ii. The probability that a customer receives immediate service upon arrival (1mark)iii. Average number of customers in the queuing system (1 mark)



Revision Aid 359


iv. Average time a customer spends in the queuing system (1 mark)(Total: 20marks)

QUESTION FIVE

a. Differentiate between the additive model and the multiplicative model as used in time seriesanalysis (4marks)

b. The sales data of XYZ Ltd (in millions of shillings) for the year 2001 to 2004 inclusive aregiven below.

Quarter Year 1 2 3 42001 40 64 124 582002 42 84 150 622003 46 78 154 962004 54 78 184 106

Requiredi. The trend in the data using the least square method (8marks)ii. The estimated sales for each quarter of the year 2004 (4marks)

iii. The percentage variation of each quarter‟s actual sales for the year 2004 (4marks)(Total: 20marks)

SECTION IIQUESTION SIX

a. Give two applications of simulation in business (2marks)b. Collins Simiyu recently acquired a piece of land in Kitale .A property development company

has offered him 300,000 for the piece of land .He has to make a decision on whether tocultivate the piece of land or to sell it to the property development company If he decides tocultivate the land ,there is a probability of getting a high ,medium ,or low harvest .Theexpected net income for each of the above states of harvest is shown below:

State of harvest Net income (sh)High 500,000Medium 100,000Low (20,000)

From past experience there is a 10percent probability that the harvest will be low, a 30 percent probability that the harvest will be medium and a 60percent probability that the harvest will be high .Colins Simiyu can engage an agricultural expert to carry out a survey on theproductivity of the land which will cost him sh30, 000. The agricultural expert gives thefollowing information as to the reliability of such surveys (prior probabilities)

Results of survey state of harvestHigh Medium Low Total

Accurate 0.35 0.10 0.05 0.5Not accurate 0.25 0.10 0.15 0.50.60 0.20 0.20 1.0

Requiredi. Construct a decision tree for the above problem (6marks)ii. The expected monetary value for each decision (10marks)iii. The decision that you would recommend (2marks)

(Total: 20marks)



360 Lesson Nine

QUESTION SEVEN

a. Explain the difference between assignment and transportation problems (4marks)b. State the assumptions made in solving a transportation problem (4marks)c. Umoja Engineering Works Ltd has a network of branches all over Kenya .The

branches are used to service, repair and install equipment for their clients .Currently,the Nairobi branch has four clients who require installation of equipment .Eachclient requires the services of one engineer There are four engineers who are not engaged at the moment and can be assignedany one of the tasks .However, these engineers have to travel from differentlocations and the Nairobi branch has to meet their travel and subsistenceallowances. The allowances vary from one engineer to another and according to theclient the engineer has been assigned to work for. The table below shows the costs (in thousand of shillings) associated with eachengineer

ClientEngineer 1 2 3 4 A 37.0 27.0 34.0 21.0B 57.0 22.0 79.0 34.0

C 22.0 25.0 61.0 45.0D 39.0 42.0 54.0 43.0

Requiredi. The assignments to be made in order to minimise the total cost of

engineers (10marks)ii. The minimum cost of using the engineers (2marks)

(Total 20marks)

QUESTION EIGHT

a. Define the following terms as used in network analysis:i. Crash time (2marks)

ii. Optimistic time (2marks)iii. Forward pass (2marks)iv. Dummy activity (2marks) v. Slack (2marks)

b. James Mutiso is a computer engineer in an information technology firm .The firm hasdecided to install a new computer system to be used by the firm‟s helpdesk .JamesMutiso has identified nine activities required to complete the installation. The table below provides a summary of the activities durations and the required numberof technicians

Activity Duration(weeks) Required number of technicians

1-2 3 21-3 1 42-4 3 42-5 2 23-4 2 43-6 4 44-5 2 2

Documents

Quantitative Analysis Book-revision Kit