KENYATTA UNIVERSITY

SCHOOL OF BUSINESS

DEPARTMENT: MANAGEMENT SCIENCE DEPARTMENT

UNIT CODE & NAME: BMS 840 –QUANTITATIVE TECHNIQUES

WRITTEN BY: Ms. Gladys Kimutai and Mr. Paul Sang

Copyright © Kenyatta University, 2014
All Rights Reserved
Published By:
KENYATTA UNIVERSITY PRESS

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ii

MODULE SUMMARY

During the last two decades, there has been a dramatic change in the business environment. The trend towards complexity has increased the risk associated with business decisions, making it more important to have a sound information base. To do well in such an environment, one will need to understand how to identify quality information and recognize the solid, reliable research on which one‟s high-risk decisions as a manager can be based. One also needs to know how to analyze the research findings. The study of quantitative techniques provides one with the knowledge and skills needed to solve the problems and the challenges of a fast-paced decisionmaking environment.
Managers make decisions on a day to day basis and it is necessary for them to be able to analyze the data so as to be able to make optimal decisions. This module has ten lesson which cover matrix algebra, markov analysis, Linear programming, differentiation, applications of differentiation to cost, revenue and profit functions, integral calculus, inventory models, sampling and estimation theory, hypothesis testing and chi-square tests.

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MODULE OBJECTIVES

By the end of the course, the student should be able to:-

1. Perform various operations on matrices matrix algebra,
2. Apply the concept of matrices in solving simultaneous equations, input-output analysis and markov analysis,
3. Formulate and solve Linear programming using the graphical and simplex method
4. Differentiate various functions and apply to cost, revenue and profit functions
5. Apply integral calculus to profit and cost functions
6. Determine the optimal levels of inventory that an organization can hold
7. Explain the various sampling techniques
8. Calculate the point and interval estimates of population mean and proportions
9. Test hypothesis about means, proportions and variance
10. Carry out the chi-square tests.

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TABLE OF CONTENTS
MODULE SUMMARY ................................................................................................... iii
MODULE OBJECTIVES ............................................................................................... iv
TABLE OF CONTENTS ................................................................................................ iv
LESSON ONE: MATRIX ALGEBRA ........................................................................... 1
1.0
Introduction ..................................................................................................................... 1
1.1
Expected Learning Outcomes ......................................................................................... 1
1.2
Definition of a Matrix ..................................................................................................... 2
1.3
Types of matrices ............................................................................................................ 2
1.4
Matrix Operations ........................................................................................................... 4
1.4.1 Matrix addition and subtraction .................................................................................. 4
1.4.2 Matrix Multiplication .................................................................................................. 4
1.4.3 Inverse of a matrix ...................................................................................................... 5
1.5
Solution of Simultaneous Equations ............................................................................... 8
1.6
Input Output Analysis ................................................................................................... 11
1.7
Assumptions of input – output models ......................................................................... 11
1.8
Types of input –output models ..................................................................................... 12
1.9
Purpose of input output analysis ................................................................................... 12
1.10 Activities ....................................................................................................................... 15
LESSON TWO: MARKOV ANALYSIS ...................................................................... 19
2.0
Introduction ................................................................................................................... 19
2.1
Lesson Objectives ......................................................................................................... 19
2.2
Assumptions of Markov analysis .................................................................................. 20
2.3
Terms used in Markov Analysis ................................................................................... 20
2.4
Information Flow in the Markov Analysis.................................................................... 21
2.5
Absorbing States ........................................................................................................... 31
2.6
Activities ....................................................................................................................... 35
LESSON THREE: LINEAR PROGRAMMING ........................................................ 40
3.0
Introduction ............................................................................................................. 40
3.1
Expected Learning Outcomes ....................................................................................... 40
3.2
Definition of Linear Programming ............................................................................... 40
3.3
Basic parts of a linear programming problem............................................................... 41
3.4
Assumptions of Linear programming ........................................................................... 41
3.5
Applications of Linear Programming ........................................................................... 41
3.6
Mathematical formulation of Linear Programming problems ...................................... 42
3.7
Graphical solution method ............................................................................................ 43
3.8
The Simplex method ..................................................................................................... 45
3.9
Advantages of Linear Programming ............................................................................. 54
3.10 Limitations of Linear Programming ............................................................................. 54
3.11 Activities ....................................................................................................................... 54
LESSON FOUR: DIFFERENTIATION ...................................................................... 59
4.0
Introduction ................................................................................................................... 59
4.1
Expected Learning Outcomes ....................................................................................... 59
4.2
Rules of differentiation ................................................................................................. 59
4.3
Second derivative of a function .................................................................................... 63
4.4
Partial derivatives.......................................................................................................... 63 v 4.5
Finding the gradient ...................................................................................................... 64
4.6
Finding a stationary point ............................................................................................. 64
4.7
Turning points (Maximum and Minimum) ................................................................... 65
4.8
Activities ....................................................................................................................... 65
LESSON FIVE: APPLICATIONS DIFFERENTIATION TO .................................. 68
REVENUE, COST AND PROFIT FUNCTIONS ........................................................ 68
5.0
Introduction ................................................................................................................... 68
5.1
Lesson Objectives ......................................................................................................... 68
5.2
Revenue Application ..................................................................................................... 68
5.3
Cost Application ........................................................................................................... 70
5.4
Profit Application.......................................................................................................... 72
5.5
Applications of partial derivatives ................................................................................ 74
5.6
Marginal Approach to Profit Maximization ................................................................. 75
5.6.1 Marginal Revenue ..................................................................................................... 76
5.6.2 Marginal Cost............................................................................................................ 76
5.6.3 Marginal Profit Analysis ........................................................................................... 76
5.7
Activities ....................................................................................................................... 79
LESSON SIX: INTERGRAL CALCULUS ................................................................. 83
6.0
Introduction ................................................................................................................... 83
6.1
Lesson Objectives ......................................................................................................... 83
6.2
Rules of Integration....................................................................................................... 83
6.3
Application of Integral Calculus ................................................................................... 87
6.3.1 Finding the area between a curve and the x-axis ...................................................... 87
6.3.2 Finding the area between two curves ........................................................................ 87
6.3.3 Application of integration to cost, revenue and profit functions .............................. 88
6.4
Activity ......................................................................................................................... 93
LESSON SEVEN: INVENTORY CONTROL ............................................................ 97
7.0
Introduction ................................................................................................................... 97
7.1
Expected Learning Outcomes ....................................................................................... 97
7.2
Definition of Inventory ................................................................................................. 98
7.3
Reasons for holding stock ............................................................................................. 98
7.4
Alternative reasons for the existence of stocks ............................................................. 98
7.5
Inventory / Stock Costs ................................................................................................. 99
7.6
Factors Affecting Inventory .......................................................................................... 99
7.7
Objective of inventory control .................................................................................... 100
7.8
ECONOMIC ORDER QUANTITY MODEL (EOQ) MODEL ................................. 100
7.9
The derivation of the EOQ model ............................................................................... 101
7.10 EOQ model for production runs ................................................................................. 109
7.11 Inventory model with planned shortages .................................................................... 110
7.12 Selective Approaches to Inventory Control ................................................................ 113
7.13 Activities ..................................................................................................................... 116
LESSON EIGHT: SAMPLING AND ESTIMATION THEORY ............................ 119
8.0
Introduction ................................................................................................................. 119
8.1
Expected Learning Outcomes ..................................................................................... 119
8.2
Definition of sampling ................................................................................................ 119
8.3
Reasons for sampling .................................................................................................. 120 vi 8.4
Sampling Procedures .................................................................................................. 120
8.5
Types of Probability sampling methods ..................................................................... 120
8.6
Non - Probability Sampling Methods ......................................................................... 122
8.7
Estimation Theory....................................................................................................... 123
8.8
Confidence interval for the population mean.............................................................. 124
8.9
Confidence interval for a population proportion ........................................................ 126
8.10 Confidence interval of the difference between two population means ( 1   2 ) ...... 128
8.11 Confidence interval of the difference between two population proportions............... 129
8.12 Qualities of a good estimator ...................................................................................... 130
8.13 Selecting the sample size to estimate a population mean ........................................... 131
8.14 Selecting the sample size to estimate a population proportion ................................... 133
8.15 Exercises ..................................................................................................................... 134
LESSON NINE: HYPOTHESIS TESTING.............................................................. 136
9.0
Introduction ................................................................................................................. 136
9.1
Expected Learning Outcomes ..................................................................................... 136
9.2
Definition of Hypothesis Testing ................................................................................ 136
9.3
Procedure for testing a hypothesis .............................................................................. 137
9.4
One-tailed and Two-tailed tests .................................................................................. 138
9.5
Testing the Population Mean When the Population Variance Is Known ................... 138
9.6
Testing the Population Mean when the Population Variance is Unknown ................ 141
9.7
Testing the population proportion ............................................................................... 144
9.8
Inference about a Population Variance ....................................................................... 147
9.9
Summary ..................................................................................................................... 150
9.10 Activities ..................................................................................................................... 150
LESSON TEN: CHI-SQUARE TESTS ...................................................................... 152
10.0 Introduction ................................................................................................................. 152
10.1 Lesson objectives ........................................................................................................ 152
10.2 Chi-square test of a multinomial experiment (Goodness-of-fit test) .......................... 153
10.3 Chi-Square Test of a Contigency Table ...................................................................... 156
10.4 Activities ..................................................................................................................... 159

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LESSON ONE: MATRIX ALGEBRA
1.0 Introduction

This lesson covers matrix algebra, the definition, matrix operations, determinants, inverse and finally its application in making business decisions.

1.1

Expected Learning Outcomes

By the end of the lessson, the learner should able to:
 Describe the different types of matrices
 Perform various operations on matrices
 Solve simultaneous equations using matrix approach.
 Apply matrices to input-output problems

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1.2

Definition of a Matrix

A matrix is a rectangular (or square) array of numbers in rows and columns. Every matrix is characterized by: Order or dimension: It is defined by the number of rows (m) and number of columns (n) and denoted (m*n) and read as m by n.
 Membership: Matrices are denoted by capital letters while their members / elements are designated by subscripted lower case letters of the alphabet. The first subscript refers to the row to which the element belongs while the second the column.

 a11

 a 21
A=  .

 .
a
 m1

1.3

a12 . ... ain 

a 22 .... . a 2 n 
.
. .
. 

.
. .
.  a m 2 . . a mn 


Types of matrices

The following are the different types of matrices:-

(a) Null matrix : It is a matrix whose all elements are equal to zero.

0 0 0
e.g. A( 2*3)  

0 0 0
(b) Rectangular matrix: A matrix in which the number of rows is not equal to the number of columns (m  n)

1 2 3
e.g. A  

 4 5 6
i.

Row matrix: It is a rectangular matrix with one row and more than one columns (m  1 and n  1) e.g. B  7 2 5 9

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ii.

Column matrix: It is a rectangular matrix with one matrix and more than one rows

2
(n  1 and m  1) e.g.  4 
 
6 
 
(c) Square matrix: A matrix in which the number of rows is equal to the number of columns
(m  n)

(d) Transpose matrix: It is a matrix derived by interchanging the rows(Columns) with the columns
(rows).

1 4 
1 2 3


A
 The transpose of A is A'  2 5
 4 5 6
3 6


(e) Identity Matrix (unit matrix): It is a square matrix having all the elements of the leading diagonal (that diagonal running from top left to bottom right) being equal to one, while the others

1 0 0
1 0

 are predominantly equal to zero i.e. I  
 , I  0 1 0 
0 1 
0 0 1 


(f) Diagonal matrix: It‟s a square matrix having all its elements being equal to zero except the main

5 0 0  diagonal consisting of numbers other than one e.g. C  0 2 0


0 0 7 


(g) Scalar matrix: It is a diagonal matrix whose all diagonal elements are equal

5 0 0 
e.g. D  0 5 0


0 0 5 


(h) Symmetric matrix: it is a square matrix whose elements above the main diagonal are mirror images of the elements below the main diagonal e.g.

2 5 6
1 2 


E
 , F  5 3 7 
 2 3
6 7 4


(i) Sub-matrix: It is a matrix derived by deleting some rows and / or columns from a matrix.

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2 5 6 
9 10
G  0 9 10 H  
 H is a sub-matrix of G


7 4 
3 7 4 



1.4

Matrix Operations
1.4.1 Matrix addition and subtraction

Two or more matrices may be added / subtracted if they are of the same order, corresponding elements being added / subtracted. Elements correspond in two or more matrices if and only if they occupy the same position as defined by the row and column that define their positions. The order of the sum / difference between two or more matrices is equal to the order of the two or more matrices being added / subtracted.

Example

1 3 
Given that A = 
 4  2 B =




  3 2

 0 4




Find (a) A+B (b) B – A

Solution

 1 3    3 2  1  3 3  2    2 5 
A+B = 
 4  2 +  0 4 =  4  0  2  4 =  4 2
 
 
 


 
 

 
  3  1 2  3   4  1
B-A = 
 0  4 4  (2)    4 6 


 


1.4.2 Matrix Multiplication
Given two matrices, A( m1*n1) and B( m 2*n 2) , then Am1*n1 * Bm 2*n 2   Cm1*n 2  . A( m1*n1) is called a premultiplier and B( m 2*n 2) is called a post multiplier. Two matrices may be multiplied if the number of columns in the pre-multiplier is equal to the number of rows in the post multiplier with the

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product having an order defined by the rows in the pre-multiplier and the columns of the post multipliers Example

6 2 
 2 6
1  2 
2 1 7 


Given that A  
 , B  7 2  , C  3 4 5 , D  1 3
 3 5




5 8 


Find : (a) AB (b) BC (c) CD

Solution

2 6 1  2 (2 * 1)  (6 * 7) (2 * 2)  (6 * 2) 44  14
(a) AB = 



4 
3 5 7 2   (3 *1)  (5 * 7) (3 * 2)  (5 * 2)  38

1  2 2 1 7  4  7  3
(b) BC  



7 2  3 4 5  20 15 59 
6 2 
2 1 7  
  48 63
(c) CD  
 1 3  47 58
 3 4 5  5 8  




1.4.3 Inverse of a matrix
Any matrix A has another matrix A-1 derived from it such that A * A1  I where I is an identity matrix. Both matrices A and A-1 are square matrices of the same order.

Procedure for finding an inverse
i. Find the cofactor matrix template which is a matrix of signs obtained by raising -1 to the power of the sum of the row and column position e.g.

  111
 21
 1

 11 2   


 12 2    

ii. Find the cofactor matrix. A cofactor of an element a ij is the product of the minor of a ij and

1i j
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iii. Find the determinant iv. Find the adjoint matrix which is the transpose of the cofactor matrix
v. Find the inverse i.e. Inverse 

1
* Adjo int
Determinant

Example
Find the inverse of the following matrices

 2 5
(a) A  

1 6 

6 5 3
(b) B  8 4 5


7 6 3


Solution

 2 5
(a) A  

1 6 
The cofactor matrix template is given by

  
  


The cofactor matrix

 6  1 
  5  2


The determinant is (2 * 6)  (5 * 1)  7
Inverse 

(b)

1  6  1
7  5 2 



6 5 3
B  8 4 5


7 6 3


Finding the cofactors of the coefficient matrix:

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 4

 6
 5

 6
 5
 4


5
3
3
3
3
5

8
7
6

7
6

8


5
3
3
3
3
5

8
7
6

7
6

8



4

6
 18 11 20 
5 
 3 1 
 3

6
 13  6  16

5 
4


The determinant  (6 * 18)  (5 *15)  (3 * 20)  7

13 
 18 3
 11  3  6 
The inverse = 1 

7
 20  1  16



Example
The national office of a car rental company is planning its maintenance for the next year. The company‟s management is interested in determining the company‟s needs for certain repair parts.
The company rents saloon cars, station wagons and double cab pick ups. The matrix as shown below indicates the number of each type of vehicle available for renting in the four regions of the country:Saloon Station wagons

Double cabs

Region

150

300

400

Coast

110

200

300

Central

90

100

100

Western

120

300

200

Eastern

Four repair parts of particular interest, because of their cost and frequency of replacement are fan belts, plugs, batteries and tyres. On the basis of studies of maintenance records in different parts of the country, the management have determined the average number of repair parts needed per car during a year. These are summarized below
Saloons Station wagons

Double cabs

17

16

15

Fan belts

12

8

5

Plugs

9

7

5

Batteries

4

7

6

Tyres

7

Required:i.

The total demand for each type of car

ii.

The total number of each repair part required for the fleet

iii.

The total cost for all repair parts

Solution

(a) (i) Total demand for each type of car saloon central

Western

Eastern

150  110   90  120   470 
Station Wagons 300  200  100  300   900 

 
   
 

Double Cabs 400 300 100 200 1000

 
   
 

Saloon

(ii) The total number of each repair part required for the fleet

17 16 15
37,390 Fan Belts
12 8 5   470  17,840  Plugs

  900   

 15,530  Batteries
Batteries  9 7 5  

 1000 
 14,180  Tyres
Tyres  4 7 6  

 fan belts
Plugs

(iii) The total cost for all repair parts

37,390
17,840 
  275,394,500
1250 800 6500 8000
15,530 


14,180 

1.5

Solution of Simultaneous Equations

Procedure
i. Model the simultaneous equation as a matrix ii. Find the inverse of the coefficient matrix iii. Pre-multiply the inverse on both sides of the equation
8

iv. Solve for the unknowns

Example
A movie theatre charges Sh. 80 for each adult admission and Sh. 50 for each child. One
Saturday, 525 tickets were sold, bringing in a total of 32,550. How many of each type of ticket were sold? (Use the matrix method)

Solution
Let x and y denote the number of the adult and children tickets sold respectively
The set of equations will be x  y  525
80 x  50 y  32550

Rewriting in matrix form

 1 1   x   525 
80 50  y   32550

  

The cofactor matrix is given by

 50  1
 80 1 


The determinant is (50 *1)  (80 * 1)  30
The inverse is 

1  50  1
30  80 1 



Pre-multiplying the inverse on the right hand side we have:-

 x
1  50  1  525  210
 y    30  80 1  32550  315
 


 


Therefore the number of tickets that must be bought is 210 and 315 for adults and children respectively Example

9

A firm manufactures three types of pans; round bottom, square bottom and triangular bottom.
These products use three available inputs from material; man hour and machine hour as shown in the table below:

Inputs

Per unit utilization of inputs

Availability

Round

Square

Triangular

bottom

bottom

bottom

Raw material

6

5

3

110

Man-hour

8

4

5

130

Machine hour

7

6

3

125

Using the matrix method, determine the number of units of each product to produce in order to utilize completely the available resources.

Solution
Let x, y and z denote units of the round bottom, square bottom and triangular bottom pans respectively. 6 x  5 y  3z  110
8 x  4 y  5 z  130
7 x  6 y  3z  125
Re-writing the equations in matrix form:

6 5 3  x  110
8 4 5  y   130

   
7 6 3  z  125

   
Finding the cofactors of the coefficient matrix:

 4

 6
 5

 6
 5
 4


5
3
3
3
3
5

8
7
6

7
6

8



5
3
3
3
3
5

8
7
6

7
6

8



4

6
 18 11 20 
5 
 3 1 
 3

6
 13  6  16

5 
4


The determinant  (6 * 18)  (5 *15)  (3 * 20)  7

10

13 
 18 3
 11  3  6 
The inverse = 1 

7
 20  1  16


13   5 
 x
 18 3
Therefore  y   1  11  3  6   10
 
  
7
z
 20  1  16 10
 

  
Therefore, the numbers of units of each product to produce in order to utilize completely the available resources are 5, 10 and 10 units of the round bottom, square bottom and triangular bottom pans respectively.

1.6

Input Output Analysis

Input output analysis is used to analyze the interdependence of various segments of an economy.
According to the technique, an economy is divided into two broad sectors, the sources or producers sector and the destinations or inputs or users‟ sectors.
The producers‟ sectors includes:

The section that manufactures goods



The section that provides services

These goods and services are intended for:

Industrial usage and are called industrial goods and services



Final consumption and these are termed consumer goods and services

The output of each industry from each industry is either for industrial use or for final consumption. The model aims at indicating how products produced from different sectors of the economy are consumed by the industrial sector and the final consumption sector. It indicates how the independent sectors of economy interact. The interaction of these sectors is necessary information for planning purposes so as adequate stock of goods can be produced by the economy. A device to assist in such planning is the technical coefficients.

1.7

Assumptions of input – output models

 All inputs into an economy emerges as outputs i.e. input equals output
11

 Industries / sectors interdependently or interactively produce outputs with sectoral output acting as inputs into the same or other sectors.

Types of input –output models

1.8

Input output models are of two types: Closed and open models. Closed models are those where all that is produced by an industry / sector is consumed by sectors interacting in the output production process. Open models are those in which some of the production is consumed by those who produce it and the rest of the production is consumed by external bodies.

1.9

Purpose of input output analysis


Planning for economic development of a region or state



Shows how resources in an economy are shared among contributing sectors



Helps in determination of GDP



Reveals impact of decision made by the contributing sectors on the economy

Input output equations
Consider a two sector economy with total outputs of X1 and X2 respectively. The total output equations will be given by

a11 X 1  a12 X 2  d1  X 1 a 21X 1  a 22 X 2  d 2  X 2
Rewriting in matrix form, we have

 a11 a12   X 1   d1   X 1 
a
       
 21 a 22   X 2  d 2   X 2  a12 
a
 d1 
 X1 
Let A   11
 D  d  X   X 
a 21 a 22 
 2
 2
Therefore,

12

AX  D  X
D  X  AX

D  X I  A
X 

D
1
 I  A D
I  A

Where X= Total output, A = Technical coefficient matrix ,

I  A1 = Leontiff inverse
Example 1
Given the Input-output table below: P

Q

Final demand

Total

P

60

140

300

500

Q

110

90

150

350

(a)

Find the Technical Coefficient Matrix.

(b) Determine the Total Output given that the final demand for output from industry P increased by 10% while that from Q went down by 5%.
Solution
(a) The technical coefficient matrix will be given by:-

 60

A   500
110

 500

140 
350   0.12 0.4 
90  0.22 0.26

 
350 

(b)

1 0 0.12 0.4   0.88  0.4
I A



0 1 0.22 0.26  0.22 0.74 
Determinant  (0.88 * 0.74) - (-0.22 * -0.4)  0.5632
Evaluating the changes in the final demand , we have
P  1.1 * 300  330
Q  0.95 *150  142.5

X  I  A D
1

X 

1 0.74 0.4   330   534.8 

0.5632 0.22 0.88 142.5 351.5625


 

13

Example 2

An economy consists of three industries X, Y and Z and each produces one product. The interaction of the use of X, Y and Z production over some fixed period of time is as shown below: X

Y

Z

Final

Total

demand

production

X

100

40

80

140

360

Y

40

60

40

180

320

Z

60

40

40

100

240

Suppose demand is expected to increase by 30% for X and Y and reduce by the same percentage for Z industries. What must the total output be to meet this projected demand?

Solution
The technical coefficient matrix is;-

40
80
100

360
320
240  0.28 0.13 0.33

60
40
 40
   0.11 0.19 0.17

360
320
240 
 60
0.17 0.13 0.17
40
40

 360
320
240 



1 0 0 0.28 0.13 0.33  0.72  0.13  0.33
I  A  0 1 0   0.11 0.19 0.17    0.11 0.81  0.17

 
 

0 0 1 0.17 0.13 0.17  0.17  0.13 0.83 

 
 











0.18
 0.13
 0.13
 0.13
 0.13
0.81

 0.17
0.83
 0.33
0.83
 0.33
 0.17

 0.11
 0.17
0.72

 0.17
0.72

 0.11



 0.17
0.83
 0.33
0.83
 0.33
 0.17

 0.11
 0.17
 0.72

 0.17
0.72

 0.11



14

0.81 

 0.13 
0.6502 0.1202 0.152 
 0.13  

  0.1508 0.5415 0.1157
 0.13  
0.2894 0.1157 0.5689

 0.13  
0.81 


The determinant is  (0.72 * 0.6502)  (0.1202 * 0.13)  (0.152 * 0.33)  0.402358

The inverse is 1

0.6502 0.1508 0.2894
0.1202 0.5415 0.1157

0.402358 
 0.152 0.1157 0.5689



The new values of the new demand are:
X  1.3 *140  182
Y  1.3 *180  234
Z  0.7 *100  70

The total output will be:

0.6502 0.1508 0.2894 182  432.17
0.1202 0.5415 0.1157 234  389.43
1

 

0.402358 
 0.152 0.1157 0.5689  70   234.73


 


1.10 Activities

1.

An investor has Sh. 20,000 to invest. If part is invested at 8% and the rest at 12%, how much should be invested at each rate to yield 11% on the total amount invested? (Use the matrix method) 2.

A company produces three products P, Q and R using raw materials A, B and C. One unit of
P requires 1, 2 and 3 units of A, B and C respectively. One unit of Q requires 2, 3 and 2 units of A, B and C respectively. One unit of R requires1, 2 and 2 units of A, B and C respectively. The number of units available for raw material A, B and C are 8, 14 and 13 units respectively. Using the matrix method, find how many units of each product to poduce in order to utilize completely the available resources.
15

3.

A sales man has the following record of sales during three months for three items A, B and
C which have different rates of commission
Months

Sales in units
A

B

Total
C

commission
(KSh.)

January

90

100

20

800

February

130

50

40

900

March

60

100

30

850

Using the matrix method, find out the rates of commission in Sh. on items A, B and C.
4.

Food Perfect Corporation manufactures three models of the Perfect Food processors X, Y and Z. Each model of X processor requires 30, 40 and 30 minutes of electrical assembly, mechanical assembly and testing respectively. Each model of Y requires 20, 50 and 30 minutes of electrical assembly, mechanical assembly and testing respectively. Each model of
Z requires 30, 30 and 20 minutes of electrical assembly, mechanical assembly and testing respectively. If 2500 minutes of electrical assembly, 3500 minutes of mechanical assembly and 2400 minutes of testing are used in one day, how many of each model will be produced?(Use the matrix method)

5.

A citrus company completes the preparation of its products by cleaning, filling and labeling bottles. Each case of orange juice requires 10, 4 and 2 minutes in the cleaning, filling and labeling machine respectively. Each case of mango juice requires 12, 4 and 1 minute in the cleaning, filling and labeling machine respectively. Each case of pineapple juice requires 9, 6 and 1 minute in the cleaning, filling and labeling machine respectively. If the company runs the cleaning machine for 398 minutes, the filling machine for 164 minutes and the labeling machine for 58 minutes, how many cases of each type of juice are prepared? (Use the matrix method)

6.

The following is the input-output table for three industries X1, X2 and X3.

Users
X1

X2

X3

16

F

Total Output

Producers

X1

8

10

10

4

32

X2

8

20

6

6

40

X3

8

10

10

2

30

Required:
i.

Determine the technology matrix.

ii.

If the final demand for the outputs of industries X1, X2 and X3 becomes 6, 4 and 6 respectively, find the outputs required of the industries.

7.

An economy consists of three industries R, S and T. the interrelationships between the production of this industries is as shown below. Units in “000” tonnes.

R

S

T

Consumer

Total

R

50

20

40

70

180

S

20

30

20

90

160

T

30

20

20

50

120

Required:
i.

Determine the Input-output matrix.

ii.

If the demand is expected to change for R, S and T and it becomes 60, 110 and 60 respectively. What must the total output be for each industry to satisfy this demand?

8.

Consider a hypothetical economy consisting of three sectors A, B and C with inputs and outputs for a particular period. The balanced transaction table has the form shown in the following table. The figures given are in billions of shillings.
INPUT TO
A

B

External
C

Total

Demand

Output

Output from A

18

30

45

15

108

Output from B

27

30

60

3

120

Output from C

54

40

60

26

180

Other input

9

20

15

Total input

108

120

180

Required:17

a) Derive the input output matrix from the above data.
b) If the external demand was to change from 15, 3 and 26 billion shillings to 10, 2 and 20 billion, calculate the total output for sector A, B and C.
9.

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 vehicles output requires 0.2 units of steel, 0.4 units from motor vehicles and 0.2 units from construction. A unit of construction output requires 0.3 units of steel, 0.4 units from motor vehicles and 0.1 units from construction. The final demand is 20 million units from steel, 50 million units from motor vehicles and 30 million units from construction.
Required:a) The technical coefficient matrix
b) If the final demands from steel drops by 2 million units and that from motor vehicles increases 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?

10. An economy consists of three interdependent sectors; agriculture, mining and manufacturing. The flow of inputs and outputs between the industries is represented in the table below:Inputs (Tonnes)

Final

Output(Tonnes)

Agriculture

Mining

Manufacturing

demand

Agriculture

20,000

32,500

37,500

10,000

Mining

30,000

65,000

37,500

30,000

Manufacturing

40,000

32,500

12,500

40,000

Required:a) The technical coefficient matrix
b) The leontiff inverse matrix
c)

The output level for each sector of the final demand for the agricultural sector increases by
1000 tons and that of the manufacturing sector falls by 500 tons and the final demand for the mining sector remains unchanged.
18

LESSON TWO: MARKOV ANALYSIS
2.0

Introduction

Markov analysis is a descriptive tool designed to predict the behaviour of a system over time. This lesson covers the concept of Markov analysis, its assumptions, computation of transition probabilities, state conditions steady states and analysis of absorbing chains.

2.1

Lesson Objectives

By the end of the lesson, the students should be able to;
i. State the assumptions of markov analysis ii. Compute the state probabilities and steady states iii. Analyze absorbing chains

Markov analysis is a descriptive tool designed to predict the behaviour of a system over time. A markov analysis is a procedure that can be used to describe the behaviour of a system in a dynamic situation. Specifically, it describes and predicts the movement of a system, among different system states, as time passes. This movement is done in a probabilistic (stochastic)
19

environment. To be able to predict the future movements and conditions of such a system would be of value to the management. Markov makes prediction such as:
 The probability of finding a system in any particular state at any given time.
 The long run probabilities of being in each state.

2.2

Assumptions of Markov analysis

The following are the assumptions of markov analysis
 The system has a finite number of discrete states, none of which is „absorbing‟
(a state that, once entered, cannot be left).
 The system‟s condition (state) in any given period depends only on its condition in the preceding period and on the transition probabilities.
 The transition probabilities are constant over time.
 Changes in the system may occur once and only once each period.
 The transition period occurs with regularity.

2.3


Terms used in Markov Analysis
Stochastic process:-it arises whenever we have a series of events in which each event is determined by chance.



Transition matrix: - it is a matrix containing probabilities of a process moving from a certain condition (or state) in the current stage (time period) to one of the possible states in the next stage. The elements in the matrix are transitional probabilities.



Recurrent state: it‟s a state in which the probability is one that will be reentered at some time in future but not necessarily in the next immediate time.



Steady state:-It is a time reached by the process where the probabilities no longer change with time i.e. the process is in equilibrium. The systems state probabilities become independent of time.

20

2.4

Information Flow in the Markov Analysis

The Markov model is based on two sets of input data, the transition matrix and the existing
(initial) conditions. From these inputs, the model makes two predictions, usually expressed as vectors:  The probabilities of the system being in any state, at any given future time.
 The long-run (Steady state) probabilities.

The transitional probabilities
The markov process describes the movement of a system from a certain condition (or state) in the current stage (time period) to one of n possible states in the next stage. The system moves in an uncertain environment. All that is known is the probability associated with any possible move
(transition). This probability termed the transition probability, Pij is the likelihood that the system currently in state i, will move to state j in the next period. The transitional probabilities can be found by undertaking a marketing research- asking people about their preferences and seeing how they change over time, monitoring their actions over time and directly monitoring the flow of items over time.
Initial conditions:-It describes the situation the system is currently in. It is always described by a row matrix
State probabilities:-These are the probabilities of a given system being in a certain state at any given time period. The initial probabilities and the transition probabilities are used to derive the state probabilities.
Steady state (equilibrium) conditions
One of the major properties of markov chains is that, in the long run, the process usually stabilizes. A stabilized system is said to approach steady state or equilibrium when the system‟s state probabilities have become independent of time

Example One
The manufacturer of Tamu Soft drinks has been facing stiff competition on its main brand TamuCola soda. The management is considering an extensive advertising and rebranding campaign for

21

Tamu-Cola soda. If the current branding remains, the transition matrix of consumers between tamu-Cola and other brands will be as follows:To
Tamu -Cola
Tamu –Cola

0.85

0.15

Others

From

Others

0.25

0.75

After the advertising and rebranding campaign, the transition matrix is expected to change as follows:To
Tamu -Cola
Tamu –Cola

0.9

0.1

Others

From

Others

0.3

0.7

The advertising and rebranding campaign is expected to cost Sh. 20 million each year. There are 40 million consumers of soft drinks in the market and for each consumer, the average profitability is Sh. 5 annually. Required:a)

The equilibrium state proportion of consumers using tamu-Cola
i.

Before the advertising campaigns

ii.

After the advertising campaign

b) The expected annual profit increase or decrease after the advertising campaign. Would you recommend the advertising campaign?
Solution
(a) The equilibrium state proportion before the advertising campaign

0.85 0.15
1  x 
  x 1  x 
0.25 0.75
0.85 x  0.25  0.25 x  x

x

0.25  0.4 x
0.25
x 
 0.625
0.4
1  x  1  0.625  0.375
 The equillibri um state proportion before advertising campaigns is 0.625 0.375

22

The equilibrium state proportion before the advertising campaign

0.9 0.1
1  x 
  x 1  x 
0.3 0.7
0.9 x  0.3  0.3x  x

x

0.3  0.4 x
0.3
x 
 0.75
0.4
1  x  1  0.75  0.25
 The equillibri um state proportionafter advertising campaigns is 0.75 0.25

(b) The expected annual profit increase or decrease after the advertising campaign.
Profits before advertising campaigns  (0.625 * 40,000,000 * 5)  125,000,000
Profits after advertising campaigns  (0.75 * 40,000,000 * 5)  20,000,000  130,000,000
Annual profit increase  130,000,000  125,000,000  5,000,000
Yes, I would you recommend the advertising campaign

Example Two
In a small town with three advocates X, Y and Z, each advocate knows that some clients switch back and forth, depending on which advocate is available at the time the client needs one. There are no new clients in the current legal market; however, none of the old clients are leaving the area. During a slack period, the three advocates collected data which identified the number of clients each advocate had seen during the preceding year. Table 1 summarizes the results of this study and Table 2 summarizes the manner in which the clients were gained or lost.
TABLE 1: Data Summary – Client Activity
Advocate

Clients

as

at Change during the year

Clients

as

at

January 1, 2008

Gain

Loss

January 1, 2009

X

400

75

50

425

Y

500

50

150

400

23

Z

500

100

25

575

TABLE 2: Gain – loss Summary
Advocate Clients as Gains

at From

losses

Clients

From

To

To

To

as

Y

January X

From

Z

X

Y

Z

January

1, 2008

at

1, 2009

X

400

0

50

25

0

50

0

425

Y

500

50

0

0

50

0

100

400

Z

500

0

100

0

25

0

0

575

Construct the state transition matrix that describes the above problem.

Solution
The transition matrix will be given by
To

X Y
X  350
 400
Y  50
From 
 500
Z  25
 500


Z
50
400
350
500
0


0 
0 
0.875 0.125
100  
  0.1
0.7
0.2 


500 
0
0.95
475   0.05



500 

Example Three
Habib‟s food liner, Ngugi‟s Supermarket and Quick Stop groceries are the only three grocery stores in sand town, a small town in Northern Kenya. A market research involving shoppers over a ten week period indicated that Habib retains 85% of its shoppers while 10% and 5% shift to
Ngugi‟s and Quick Stop respectively each subsequent week. Ngugi‟s lose 20% to Habib and 5% to Quick stop. Of those who shop in Quick stop, 15% and 10% shift to habib and Ngugi respectively. The total grocery store customers in sand town are estimated at 10,000. Each shopper generates revenue of about KSh. 500 per week. What are the projected weekly revenues for each grocery store at equilibrium?
24

(a)

The transitional matrix is given by

0.85 0.1 0.05
0.20 0.75 0.05
0.15 0.10 0.75

Let p, q and r denote the proportion of customers who shop in H, N and Q respectively.
0.85 0.1 0.05
 p q r 0.20 0.75 0.05   p q r 


0.15 0.10 0.75


0.85 p  0.20q  0.15r  p..............(i )
0.1 p  0.75q  0.10r  q................(ii )
0.05 p  0.05q  0.75r  r...................(iii ) p  q  r  1 p  1  q  r...................(iv ) substituti ng (iv ) int o(ii )
0.1(1  q  r )  0.75q  0.10r  q
0.1  0.1q  0.1r  0.75q  0.1r  q
 0.35q  0.1
 q  10

35

 0.2857

substituti ng (iv ) int o(iii )
0.05(1  q  r )  0.05q  0.75r  r
0.05  0.05q  0.05r  0.05q  0.75r  r
0.3r  0.05
r  5

 0.1667
30
p  1  q  r  1  0.2857  0.1667  0.5476

The total number of shoppers in each grocery n the long run will be:

0.5476 5476 
10,0000.2857  2857

 

0.1667 1667 

 

The weekly revenues for each grocery at equilibrium will be:-

25

5476  2,738,000
5002857  1,428,000 

 

1667   833,500 

 


Example Four
A university‟s service department is considering leasing one of two possible computers. A computer can be found in operating condition (O) or nonoperating condition (NO). The daily transition matrix of the two computers under identical maintainance follows:
Computer A
O

Computer B
NO

O

NO

O

0.95

0.05

O

0.98

0.02

NO

0.90

0.10

NO

0.85

0.15

Which computer should the university lease if the leasing charges are the same?

Solution

In order to find which computer the university should lease if the leasing charges are the same, we find the long-run probabilities
Computer A

0.95 0.05 q 
   p q
0.90 0.10
0.95 p  0.90q  p
0.05 p  0.10q  q p  q  1 p  1  q substituti ng ;
0.05(1  q )  0.10q  q
0.05  0.05q  0.10q  q
 0.95q  0.05 q  5  0.05
95
 p  1  0.05  0.95

p

Computer B

26

0.98 0.02 q 
   p q
0.85 0.15
0.98 p  0.85q  p
0.02 p  0.15q  q p  q  1 p  1  q substituti ng ;
0.02(1  q )  0.15q  q
0.02  0.02q  0.15q  q
 0.87 q  0.02 q  2  0.02
87
 p  1  0.02  0.98

p

Therefore, since computer B will be operational most of the time (98%), the university should lease it.

Example Five
The following table indicates the customers buying behaviour for a given product Produced by
ABC Company with two brands. Currently, the market share for Brand A is 27.5%, Brand B is
37.5% and Competitor‟s Brand C is 35%.
Next months brand
Choice

Brand A

Brand B

Current months

Competitors
Brand C

Choice

Brand A

0.2

0.2

0.6

Brand B

0.1

0.5

0.4

Competitors Brand C

0.5

0.2

0.3

Determine
(a) The market shares one month and two months from now.
(b) The long run probabilities
(c) ABC Company is considering two alternative policies for promoting its products
27

1.

Promote Brand A only. This will cost Sh. 150,000 invested in a lump sum and is expected to change the transition matrix to:
To

A

B

C

A

0.6

0.2

0.2

B

0.4

0.4

0.2

C

0.6

0.1

0.3

From

2.

Promote Brand B only. This will cost Sh. 280,000 invested in a lump sum and is expected to change the transition matrix to:
To

A

B

C

A

0.1

0.5

0.4

B

0.2

0.8

0

C

0.3

0.5

0.2

From

Required:
i.

Which policy will bring larger increases in ABC‟s total share of the market in the long run? ii.

Which policy will be more efficient in terms of the gain per one thousand shilling invested in the long run?

iii.

Assume that each percentage of increased share in the total market is worth Sh.10,000 to ABC‟s; which policy (if any) should ABC take?

Solution
(a) The market shares one month from now will be:-

0.2 0.2 0.6
0.275 0.375 0.350.1 0.5 0.4  0.2675 0.3125 0.42


0.5 0.2 0.3



(b) In the long run
28

0.2 0.2 0.6
x y 1  x  y  0.1 0.5 0.4  x


0.5 0.2 0.3


0.2 x  0.1 y  0.5  0.5 x  0.5 y  x

y 1  x  y

0.5  1.3 x  0.4 y............................(i )
0.2 x  0.5 y  0.2  0.2 x  0.2 y  x
0.2  0.7 y
0.2
 0.2857
0. 7
0.5  (0.4 * 0.2857) x  0.2967
1. 3
1  x  y  1  0.2857  0.2967  0.4176

y 

 The long run probabilities are 0.2967 0.2857 0.4176

(c) Long run Probabilities
(i) Long run probability if brand A only is promoted

0.6 0.2 0.2
x y 1  x  y 0.4 0.4 0.2  x


0.6 0.1 0.3


0.6 x  0.4 y  0.6  0.6 x  0.6 y  x

y 1  x  y

0.6  x  0.2 y.................(i )
0.2 x  0.4 y  0.1  0.1x  0.1 y  y
0.1  0.1x  0.7 y................(ii )
Solving equation (i) and (ii) simultaneously, we have
0.6  x  0.2y
 0.6  x  0.2 y
10(0.1  -0.1x  0.7y)  1   x  7 y
1.6 

7.2y

1.6
 0.2222
7.2
x  0.6  (0.2 * 0.2222)  0.5556

y 

1  x  y  1  0.5556  0.2222  0.2222
 The long run probabilities if Brand A only is promoted is 0.5556 0.2222 0.2222

Long run probability if brand B only is promoted

29

x

 0.1 0.5 0.4 y 1  x  y 0.2 0.8 0   x


0.3 0.5 0.2



y 1  x  y

0.1x  0.2 y  0.3  0.3 x  0.3 y  x
0.3  1.2 x  0.1 y.................(i )
0.5 x  0.8 y  0.5  0.5 x  0.5 y  y
0.5  0.7 y
0.5
 0.7143
0.7
0.3  (0.1 * 0.7143) x  0.1905
1.2
1  x  y  1  0.1905  0.7143  0.0952

y 

 The long run probabilities if Brand b only is promoted is 0.1905 0.7143 0.0952

The market share for ABC : Before promotion  0.2967  0.2857  0.5824
 If Brand A only is promoted  0.5556  0.2222  0.7778
 If Brand B only is promoted  0.1905  0.7143  0.9048
Therefore the policy which will bring a larger increase in ABC‟s total share of the market in the long run is promoting Brand B only

(ii) Increase in the market shares if : If Brand A only is promoted  77.78% - 58.24%  19.54%
 If Brand B only is promoted  90.48% - 58.24%  32.24%

Brand A only

150000  19.54%
1000  ?
1000

*19.54%  0.13%
150000

Brand B Only

30

280000  32.24%
1000  ?
1000

* 32.24%  0.12%
280000

Therefore promoting Brand A only is more efficient in the long run

(iii)

Profit = Total Revenue – Total Cost



If Brand A only is promoted, Profit  (19.54 *10000)  150000  45400



If Brand B only is promoted, Profit  (32.24 *10000)  280000  42400

Therefore Promoting Brand A is better

2.5

Absorbing States

If there are n states, out of which m are absorbing states, and n  m transient states, the matrix is expressed as follows:

Q R 
P

O I 
In the above matrix, rows and columns of P correspond to states 1,2,......, n  m ; Q represents the square matrix of the order n  m of transient probabilities representing transitions between transient states: R is an n  m by m matrix representing transitions from transient to absorbing states ; I is an identity matrix of the order m , depicting the fact that it is not possible to go from one absorbing state to another, and O is a null matrix consisting of all zeros. The matrix

I  Q1 is

known as the fundamental matrix of the markov chains. The ij th element of this

matrix gives the expected number of periods that will be spent in the transient state i before reaching an absorbing state. Similarly, if the chain begins in a given transient state i , the probability that it shall be eventually in absorbing state j would be given by the ij th element of the matrix I  Q R .
1

31

Example One
A company has two production departments X and Y and three service departments A, B and C.
The direct cost allocated to each of the departments and percentage of total cost of each service department apportioned to various departments are given below:Department

Direct

Cost Percentage allocation of total cost of

(Sh.)

the department
A

B

C

X

120,000

40

30

30

Y

290,000

30

30

50

A

25,000

0

30

10

B

124,000

20

0

10

C

77,000

10

10

0

Draft this problem as absorbing chain and determine the total cost (allocated and apportioned) for each production department
Solution
The matrix P will be as given below:A
P

B
C
X
Y

A
0

B C
X
Y
0.2 0.1 0.4 0.3

0.3 0 0.1 0.3 0.3
0.1 0.1 0 0.3 0.5
0
0
0
1
0
0
0
0
0
1

 0 0.2 0.1
Q  0.3 0 0.1


0.1 0.1 0 



 0.4 0.3
R  0.35 0.3


0.15 0.4



32

 0.2  0.1
1 0 0  0 0.2 0.1  1
0 1 0  0.3 0 0.1   0.3
I Q  
1
 0.1
 
 

0 0 1 0.1 0.1 0    0.1  0.1
1 

 
 


Finding the cofactors, we have











1
 0.1
 0.2
 0.1
 0.2
1

 0.1
 0.3

1
 0.1
 0.1
1

1
 0.1
 0.1
1

 0.1
 0.3

 0.1
1
 0.1
1
 0.1
 0.1

 0.3
 0.1
1

 0.1
1

 0.3


1 

 0.1 
0.99 0.31 0.13
 0.2  

   0.21 0.99 0.12
 0.1 
0.12 0.13 0.94

 0.2  
1 


Determinant = (1* 0.99)  (0.2 * 0.31)  (0.1* 0.13)  0.915

I  Q

1

0.99 0.21 0.12  0.4 0.3 0.54098 0.45902
 0.31 0.99 0.13 0.35 0.3  0.50273 0.49727
R 1

 

0.915 
0.13 0.12 0.94 0.15 0.4 0.40437 0.59563


 


The cost apportioned will be

0.54098 0.45902
25000 124000 770000.50273 0.49727  106999.51 119000.49


0.40437 0.59563


The total cost allocated and apportioned will be
X  120000  106999.51  226999.51
Y  290000  119000.49  409000.49

Example Two
ABC Company has three service departments X, Y and Z and two production departments A and
B. Overhead is allocated to the production departments for inclusion in the stock valuation. The

33

analysis of benefits received by each department during the last quarter and the overhead expense incurred by each department were as follows:
Service departments

Percentage to be allocated to departments
X

Y

Z

A

B

X

0

25

5

40

30

Y

10

0

25

35

30

Z

30

15

0

15

40

Direct overhead expense (Sh.‟000)

60

25.5

60.5

26

84

Formulate the above as an absorbing chain and determine the total overhead to be allocated from each of the service departments X, Y and Z to the production departments.

Solution
The matrix P will be as given below:X
P

X
0

Y
Z
0.25 0.05

A
0.4

Y
Z
A
B

0.1 0
0.25 0.35 0.3
0.3 0.15
0
0.15 0.4
0
0
0
1
0
0
0
0
0
1

 0 0.25 0.05
Q  0.1 0
0.25


0.3 0.15
0 



B
0.3

 0.4 0.3
R  0.35 0.3


0.15 0.4



 0.25  0.05
1 0 0  0 0.25 0.05  1
0 1 0  0.1 0
    0.1
I Q  
0.25 
1
 0.25
 

0 0 1 0.3 0.15
0   0.3  0.15
1 

 
 


Finding the cofactors, we have
34











1
 0.15
 0.25
 0.15
 0.25
1

 0.25
1
 0.05
1
 0.05
 0.25

 0.1
 0.3
1

 0.3
1

 0.1


 0.25
1
 0.05
1
 0.05
 0.25

 0.1
 0.3
1

 0.3
1

 0.1



1 

 0.15 
0.9625 0.175 0.315
 0.25  

  0.2575 0.985 0.225
 0.15 
0.1125 0.255 0.975

 0.25  
1 


Determinant = (1* 0.9625)  (0.25 * 0.175)  (0.05 * 0.315)  0.903

I  Q

1

0.9625 0.2575 0.1125  0.4 0.3 0.5449 0.4551
 0.175 0.985 0.255  0.35 0.3  0.5017 0.4983
R 1

 

0.903 
 0.315 0.225 0.975  0.15 0.4 0.3887 0.6113


 


The total cost apportioned will be:-

0.5449 0.4551
60000 25500 605000.5017 0.4983  69003.7 79996.3


0.3887 0.6113


The total cost allocated and apportioned will be
A  26000  69003.7  95003.7
B  84000  79996.3  163996.3

2.6

Activities

1. One half of a country‟s population lives in the city and one half in the suburbs. There is an
80% chance that a suburban resident will remain in the suburbs and a 20% chance that he or she will move to the city within the next month. A city dweller has a 50-50% chance of staying in the city or moving to the suburbs. Determine:

35

(a) The percentage of the population who will be in the suburb and the city one and two months from now.
(b) The steady state probabilities.

2. Joe Fundi, the caretaker manager of sky towers is in the process of deciding which of two elevator maintenance companies to select for providing maintenance and repair services for the buildings elevators. The guarantees for promptness and quality of service form one week to the next for the two companies are as follows.
Quick response service Ltd

JIT Elevator Service Ltd

To

To

From

Operation

Breakdown

From

operation

breakdown

Operation

0.85

0.15

Operation

0.95

0.05

Breakdown 0.95

0.05

Breakdown 0.85

0.15

Advise Joe Fundi on which of the two elevator companies to select.

3. Continental Corporation operates a large fleet of cars for which an extensive preventive maintainance program is utilized. The cars can be classified in one of the three states: Good
(G), Fair (F) and poor (P). The transition matrix of these cars is as follows:
TO

G

F

P

G

0.6

0.3

0.1

F

0.2

0.6

0.2

P

0.1

0.4

0.5

FROM

Required:
i.

Assume that currently there are 100 cars in good shape, 60 in fair shape and 20 in poor shape. How many cars will be found in each condition next week?

ii.

How many cars will be found in each condition once the process stabilizes?

4. The market for homeowners insurance in a particular city is dominated by two companies: national property and United family. Currently, national property insures 50% of the homes in the city, united family insures 20% and the remainder are insured by a collection of
36

smaller companies. United family decides to offer rebates to increase its market share. This has the following effects on the insurance purchases for the next several years; each of 30% of national property‟s customers switch to united family and 10% switch to other companies,
20% of united family customers switch to national property and 5% switch to other companies, 30% of the customers of other companies switch to national property and 40% switch to united family. Determine the market share of the companies after two years and also in the long run.
5. A manufacturer uses Machine Z in one of its production processes. Due to its constant usage, the machine is inspected for maintenance purposes on a monthly basis. On inspection, the condition of the machine is classified into four possible states as follows:State 1: Good
State 2: Minor repairs
State 3: Major repairs
State 4: Write off
Thereafter, the manufacturer adopts either of the following maintainance policies:Policy 1: Minor repairs if the machine is in state 2, overhaul if the machine is in state 3 and replace if the machine is in state 4
Policy 2: Minor repairs if the machine is in state 1, 2 or 3 and replace the machine if it is in state 4

The costs of maintainance associated with each policy are shown in the table below:State

Policy 1 (Sh.)

Policy 2 (Sh.)

1

0

0

2

100,000

100,000

3

400,000

300,000

4

600,000

600,000

The transition probability matrices for each policy are as given below:Policy 1
To
1

Policy 2
To

2

3

4

1

37

2

3

4

1 0
0.8 0.1 0.1
 0.1 0.6 0.2 0.1
2
 from 
3 0.05 0.3 0.45 0.2


4 1
0
0
0

1 0 0.8 0.1 0.1
2 0 0.5 0.4 0.1

From 
3 0 0 0.4 0.6


0
0
4 1 0

Required:a)

The long-run proportion of times the machine would be expected to be in each state if Policy
1 is adopted

b) The long run proportion of times the machine would be expected to be in each state if policy
2 is adopted
c)

The expected long run average cost for each policy

d) The policy you would recommend to the manufacturer

6. A company has two production departments X and Y and three service departments A, B and
C. The direct cost allocated to each of the departments and percentage of total cost of each service department apportioned to various departments are given below:Department

Direct Cost

Percentage allocation of total cost

(Sh.)

A

B

C

X

80,000

40

30

30

Y

90,000

20

40

20

A

10,000

0

20

30

B

20,000

20

0

20

C

50,000

20

10

0

Draft this problem as absorbing chain and determine the total cost (allocated and apportioned) for each production department

7. A credit card company is attempting to determine a more effective set of credit control policies. It has traditionally classified all its accounts receivables into one of the five categories listed below:-

Accounts receivable category

Status of accounts receivable
0 – 30 days late

1
38

2

31 – 90 days late

3

91 + days late

4

Paid in full

5

Defaulted

Based on the historical data, the company has derived the following transition matrix describing the behaviour of various categories on a weekly basis
From

To category

category

1

2

3

4

5

1

0.4

0.2

0.1

0.2

0.1

2

0.3

0.4

0.1

0.1

0.1

3

0.2

0.4

0.1

0.1

0.2

4

0.0

0.0

0.0

1.0

0.0

5

0.0

0.0

0.0

0.0

1.0

Currently, the company has the following accounts receivables in various categories:
Category

Amount (Sh)

1

2,000,000

2

4,000,000

3

3,000,000

Required: Calculate the expected amounts which may eventually be:
(a) Collected
(b)Defaulted

39

LESSON THREE: LINEAR PROGRAMMING
3.0

Introduction

Linear programming is a mathematical technique that deals with the optimization of a linear function of variables known as objective function to a set of linear inequalities known as constraints. It is a method of determining an optimum programme of inter-dependent activities in view of available resources.

3.1

Expected Learning Outcomes

By the end of the lesson the students should be able to:
 Define the term linear programming
 Describe the assumptions of linear programming
 Formulate and solve linear programming problems using both the graphical and simplex method
 Identify the various applications of linear programming
 Discuss the advantages and disadvantages of linear programming

3.2

Definition of Linear Programming

 It is a mathematical technique that deals with the optimization of a linear function of variables known as objective function to a set of linear inequalities known as constraints.
 It is a method of determining an optimum programme of inter-dependent activities in view of available resources.

40

3.3
i.

Basic parts of a linear programming problem
The objective function: it describes the primary purpose of the formulation i.e. to maximize profit or to minimize cost.

ii.

The constraint set: It is a system of equalities and / or inequalities describing the conditions under which optimization is to be accomplished e.g. machine hours, manhours, materials etc.

3.4

Assumptions of Linear programming
 Linearity: Costs, revenues or any physical properties, which form the basis of the problem, vary in direct proportion with the quantities or numbers of components produced.  Divisibility: Quantities, revenues and costs are infinitely divisible i.e. any fraction or decimal answer is valid.
 Certainty: The technique makes no allowance for uncertainty in the estimate made.
 Positive solutions: Non-negativity constraints are introduced to ensure only positive values are considered.
 Interdependence between demand products is ignored, products may be complementary or a substitute for one another.
 Time factors are ignored. All production is assumed to be instantaneous.
 Costs and benefits which cannot be quantified easily, such as liquidity, good will and labour stability are ignored.

3.5

Applications of Linear Programming
 Business and industry e.g. in the petroleum industry
 Food processing industry: to determine the optimal mix of feeds
 Paper and textile industry: To determine the optimal cutting method to minimize trim losses  Transport industry: To determine the best route
 Financial institutions – to determine investment plans.
41

 Advertising media: assigning advertising expenditures to different media plans.
 Politics- campaign strategies
 Auditing – to find the number of financial audits
 Agriculture- amount of fertilizer to apply per acre
 Hospital scheduling- number of nurses to employ
 Marketing – determining the best marketing strategy
 Crude oil refining

3.6

Mathematical formulation of Linear Programming problems

If there are n decision variables and m constraints in the problem, the mathematical formulation of the LP is:Optimization (Max or Min) Z  c1 x1  c2 x2  ........  cn xn

Subject to the constraints a11x1  a12 x2  ............  a1n xn  b1 a21x1  a22 x2  ............  a2n xn  b2 am1 x1  am2 x2  ............  a1n xn  bm

and x1 , x2 ,........xn  0

where
X i  decision var iable
Cj  constant representing per unit contribution of the objective function of the

jth decision variable

ai j  constant representing exchange coefficient of the jth decision variable in the ith constraint bi  constant representing ith constraint requirement or availability

42

Example

XYZ chemical company is producing two products A and B. the processing times are 3 hrs and 4 hrs per unit for A on operations one and two respectively and 4 hrs and 5 hrs per unit for B on operations one and two respectively. The available time is 18 hrs and 21 hrs for operations one and two respectively. The product A can be sold at Sh. 3 profit per unit and B at Sh.8 profit per unit. Formulate the above as a LP problem.

Solution
Let x = one unit of product A and y = one unit of product B.
The constraints are the available times on machine one and two

X

y

Availability

Operation one

3

4

18

Operation two

4

5

21

Therefore the linear programming problem would be

maximize Z  3x  8y subject to :
3 x  4 y  18
4 x  5 y  21 x  0, y  0

Solutions to LP problems
LP problems can be solved by the help of the following methods:
i.

Graphical method

ii.

Simplex method

3.7

Graphical solution method

In order to solve a LP problem graphically, the following procedure is adopted:-

43

i.

Formulate the appropriate LP problem

ii.

Graph the constraint inequalities as follows: treat each inequality as though it were an equality and for each equation, arbitrarily select two sets of points. Plot each of the points and connect them with appropriate lines.

iii.

Identify the solution space or feasible region which satisfies all the constraints simultaneously. For  constraints, this region is below the lines and for  constraints, the region is above the lines.

iv.

Locate the solution points on the feasible region. These points always occur at the corner points of the feasible region.

v.

Evaluate the objective function at each of the corner points.

vi.

Identify the optimum value of the objective function.

Example

M ax Z  50 x  18 y
s.t.
2 x  y  100 x  y  80 x, y  0
Identifying the points to plot
From equation (i) 2 x  y  100  x 0

100

y

50

0

From equation (i) x  y  80  x 0

40

y

40

0

Drawing the graph and identifying the feasible area we have

44

Evaluating the objective function at the corner points
Corner Point

X

Y

Max Z  50 x  18 y

A

0

80

18*80 = 1440

B

20

60

(50*20) + (60*18) = 2080

C

50

0

50 * 50 = 2500

Optimal solution : X = 50 and Y = 0
Maximum value of the objective function is 2500

3.8

The Simplex method

The simplex method is an iterative algorithm for efficiently solving L.P problems. It was first developed in 1947 by G.B. dantzig and his associates in the U.S. Department of the air force.

The simplex process
1. Standardize the problem into a linear programming tableau.
2. Generate the initial solution, called a basis.
3. Test the solution for optimality. If the solution is not optimal, improve it (go to step 4); otherwise go to step 6
4. Identify one variable that will leave the basis and one variable that will enter the basis.
45

5. Generate an improved solution. The improved solution is checked for optimality. If it is not optimal, then steps 4 and 5 are repeated. If it is optimal, step 6 is undertaken.
6. Find if more than one optimal solution exists.

Standardizing the problem
Since only corner points of the feasible solution space are to be checked, and since these points are defined by the intersection of equations, it is necessary to convert the inequalities in the problem statement into equations in order to find the co-ordinates of the intersecting points. Such a conversion depends on the type of constraints involved.
Constraints with inequalities of the smaller- than- or-equal-to type
A slack variable is added to each of these type constraints.
Example
Maximize Z = 300 x1  250 x2
Subject to:
2 x1  1x 2  40
1x1  3x 2  45
1x1  0 x 2  12

Step 1: Standardizing the problem
Maximize Z = 300 x1  250 x2  0s1  0s2  0s3
Subject to:
2 x1  1x 2  1s1  0s 2  0s3  40
1x1  3x 2  0s1  1s 2  0s3  45
1x1  0 x 2  0s1  0s 2  1s3  12

Basic variables and a basis
In the above example, there exists five unknown variables; x1, x2, s1, s2 and s3. and on the other hand, there are only three equations. A solution to a system of linear equations requires that the number of equations equal the number of variables, otherwise, there will be an infinite number of solutions. To overcome this problem, two of the five variables are set to zero. Then it is possible to solve the system of linear equations for the three remaining variables. In general, two cases are distinguished: 46

1. Solutions with exactly m variables, where m is the number of constraints, are called basic solutions. The m variables in these solutions are called basic variables and they constitute the basis of the solution. The remaining variables have values of zero.
2. Solutions that include more than m nonzero variables are called nonbasic solutions.

The basic solutions can be further divided into two groups: feasible and infeasible. The feasible solutions are those that satisfy all the constraints i.e. they are in the feasible area and the value of all their variables must be nonnegative. The simplex algorithm searches for basic feasible solutions only.

Step 2: Generate the initial solution
2 x1  1x 2  1s1  0s 2  0s3  40
1x1  3x 2  0s1  1s 2  0s3  45
1x1  0 x 2  0s1  0s 2  1s3  12

A natural solution to this set equations would be to set x1 and x2 as the nonbasic variables with s1, s2 and s3 as the basic variables. This is because x1 and x2 appear in more than one equation, whereas s1, s2 and s3 each appear only once and in different equations and can thus be solved easily. Setting x1 = x2 = 0 and solving s1, s2 and s3 ;

s1  40, s 2  45 and s3  12
The resulting profit is found from the objective function as:-

max imizeZ  300 x1  250 x2  0s1  0s 2  0s3
 300(0)  250(0)  0(40)  0(45)  0(12)  0

The structure of a tableau
List of basic

Coefficients

Quantity

Decision

variables

of objective

in solution

variables

Slack (surplus)

function
Basis

Unit profits

variables
Quantity

X1

47

X2

S1

S2

S3

S1

0

40

2

1

1

0

0

S2

0

45

1

3

0

1

0

S3

0

12

1

0

0

0

1

Cj row (coefficients of the objective function)

300

250

0

0

0

Zj row (new coefficient: per unit “losses”)

0

0

0

0

0

Cj – Zj row (The evaluator)

300

250

0

0

0

The Zj row: this row shows the amount of profits the objective function will be reduced by when one unit of the variable, in each column is brought into the basis. Each value in the row is found by: Multiply the elements in the “unit profits” column by the corresponding elements of the main body, on a variable-by-variable basis and total the results. E.g. for column 2: 0(1) + 0(3) + 0(0) =
0.
The evaluator row: Cj – Zj. This row shows the net impact on the value of the objective function of bringing one unit of each of the column variables into the basis. The Cj‟s indicate how much will be gained while the Zj‟s indicate how much will be simultaneously lost. If the difference between the two is positive, then the value of the objective function can be increased by introducing one unit of the variable into the solution. The value of cj – zj represents the opportunity cost of not having one unit of the corresponding variable in the solution.

Step 3: Test for optimality
Rule for optimality: For an optimal solution of a maximization problem, it is sufficient if all the coefficients of the Cj – zj row are nonpositive (either zero or negative).
Since the solution is not optimal, it can be improved by exchanging one variable in the solution.
The improvement is done by:
 Identifying the incoming variable
 Identifying the outgoing variable
 Building an improved tableau

Step 4: Identifying the incoming and the outgoing variable
48

The incoming variable: an incoming variable (currently nonbasic, to be changed to a basic variable) is chosen by inspecting the Cj – Zj row of the current solution for that variable with the largest positive coefficient. In our example X1 = 300. This is also known as the pivot column.
Rule for determining the outgoing variable: Look down the column of the incoming variable and consider only positive elements (coefficients). Then, divide the quantity of each constraint by the corresponding coefficient in the column of the incoming variable. The row with the smallest ratio (it may be zero) is selected as the outgoing row.
The coefficient of the incoming variable in the outgoing (pivot) row is called the pivot element.
Incoming

Basi

Unit profits

Quantity

X1

X2

S1

S2

S3

Ratio

S1

0

40

2

1

1

0

0

20

S2

0

45

1

3

0

1

0

45

S3

0

12

1

0

0

0

1

12

Cj (unit profits)

300

250

0

0

0

Zj (unit losses)

0

0

0

0

0

300

250

0

0

0

s

Outgoing

Cj – Zj

Step 5: Generating an improved solution
The solution is improved by introducing the incoming variable into the basis and removing the outgoing variable. This is done row by row on the old the old tableau, forming, in the process, a new one. The procedure starts with transforming the row of the outgoing variable, then the other basis rows, and finally transforming the Cj, Zj and Cj- Zj rows.
Rule for Transformation of the outgoing row:
Divide the main body elements and the „quantity” of the outgoing row by the pivot element.
Transformation of the other rows
1. Identify the coefficient at the intersection of the row to be transformed and the incoming column. 2. Multiply this number, in turn, by every element of the transformed outgoing row
49

3. Subtract the result of the second activity from the old row to derive the new transformed row.

From our example, the first row will be
Quantity

X1

X2

S1

S2

S3

The new third row

12

1

0

0

0

1

The row to be transformed

40

2

1

1

0

0

Minus: 2 * new third row

-24

-2

0

0

0

-2

Result: new first row

16

0

1

1

0

-2

Quantity

X1

X2

S1

S2

S3

Second row

45

1

3

0

1

0

Minus: 1 * new third row

-12

-1

0

0

0

-1

Result : new second row

33

0

3

0

1

-1

The second row will be:

The new tableau will be
Inco
ming
Basi

Unit profits

Quantity

X1

X2

S1

s

S

S3

Ratio

2

S1

16

0

1

1

0

-2

16

S2

0

33

0

3

0

1

-1

11

X1

Outgoing

0

300

12

1

0

0

0

1

Limit
=

Cj (unit profits)

300

250

0

0

0

Zj (unit losses)

300

0

0

0

300

0

250

0

0

-300

Cj – Zj

Step 3: Test for optimality
50

The new solution must now be tested for optimality. It can be seen that the variable X2 in the new Cj – Zj row has a positive coefficient, therefore the current solution is not optimal.

Step 4(repeat): Identifying the incoming and the outgoing variable
X2 has the largest positive coefficient in the Cj – Zj row (250) and therefore it is the incoming variable. S2 is the outgoing variable since it has the smallest nonnegative ratio.

Step 5: (repeat) Generate an improved solution
The rows of the tableau of the old solution are transformed one at a time to a new solution.

Basis

Unit

Quantit

X1

X2

S1

S2

S3

S1

profits

y

0

5

0

0

1

1

5

250

11

0

1

0

300

12

1

0

0

0

1

Cj (unit profits)

300

250

0

0

0

Zj (unit losses)

300

250

0

250 3

650 3

0

0

0

 250 3

X2
X1

Cj – Zj

1

3

3

1

3
3

 650 3

Interpretation of the Results
X1 = 12, x2 = 11,
S1 = 5: constraint one has a slack of 5, i.e. only 35 hours of labour out of 40 hours will be utilized. The value of the objective function (profit):
0(5) + 250 (11) + 300 (12) = 6, 350.

Example
A firm uses three machines in the manufacture of three products. Each unit of product A requires
3, 2 and 1 hour on machine I, II and III respectively. Each unit of product B requires 4, 1 and 3 hour on machine I, II and III respectively while each unit of product C requires 2hours each on
51

the three machines. The contribution margin of the three products is Sh. 30, Sh. 40 and Sh. 35 per unit respectively. The machine hours available on the three machines are 90, 54 and 93 respectively. Required:a) Formulate the above problem as a linear programming problem.
b) Obtain an optimal solution to the problem using the simplex method. Which of the three products shall not be produced by the firm? Why?

Solution
(a) Let x1 , x2 , x3 represent the outputs of product A, B and C respectively.

MaxZ  30 x1  40 x 2  35 x3
s.t.
3 x1  4 x 2  2 x3  90
2 x1  x 2  2 x3  54 x1  3 x 2  2 x3  93 x1 , x 2 , x3  0

(b) Standardizing the problem,

MaxZ  30 x1  40 x 2  35 x3  0s1  0s 2  0s3
s.t.
3x1  4 x 2  2 x3  s1  90
2 x1  x 2  2 x3  s 2  54 x1  3x 2  2 x3  s3  93 x1 , x 2 , x3 , s1 , s 2 , s3  0
To find the initial solution, we let x1  x2  x3  0  s1  90, s2  54, s3  93
Initial tableau
Basis Unit

Quantity

x1

x2

x3

S1

S2

S3

Ratio

profits

S1

0

90

3

4

2

1

0

0

22.5

S2

0

54

2

1

2

0

1

0

54

0

93

1

3

2

0

0

1

31

30

40

35

0

0

0

S3
Cj

52



Zj

0

0

0

0

0

0

Cj-Zj

30

40

35

0

0

0



Tableau 2
Basis Unit

Quantity

x1

x2

x3

S1

S2

S3

Ratio

3

1

1

1

0

0

45

0

3

1

0

21

0

1

0

1

51

profits

x2

40

45

S2

0

63

S3

0

51

2
2
2

5

4
4

5

4

2

4

1

2

3

2

4
4

Cj

30

40

35

0

0

0

Zj

30

40

20

10

0

0

Cj-Zj

0

0

15

-10

0



0



Tableau 3
Basis Unit

Quantity

x1

x2

x3

S1

S2

1

0

1

1

0

1

1

0

0

2

40

35

0

40

35

15

0

0

 15

S3

Ratio

0

45

0

21

1

51

profits

x2

40

12

1

x3

35

21

5

S3

0

15

5

3
6
3

Cj

30

Zj

85

Cj-Zj

 25

2
2

3
6
3

2
1

3
3
3

0
10

2

2

0
0

-10

0

The optimal solution is x1  0, x2  12, x3  21, s3  15 . The objective function = 1215.

53

This indicates that product A should not be produced by the firm. This is because each unit of it produced would reduce the profit by 12.50.

3.9

Advantages of Linear Programming

The advantages of linear programming are: Helps in attaining the optimum use of productive factors
 Improves the quality of decisions
 Improves the knowledge and skill of tomorrow‟s executives
 It highlights the bottlenecks in the production process
 It gives insight and perspective into problem situations
 Enables one to consider all possible solutions to problems.
 Enables one to come up with better and more successful decisions
 It‟s a better tool for adjusting to meet changing conditions.

3.10 Limitations of Linear Programming
The limitations of linear programming are:
 It treats all relationships as linear
 It is assumed that any activity is infinitely divisible
 It takes into account single objective only i.e. profit maximization or cost minimization
 It can be adopted only under the condition of certainty i.e. that the resources available, per unit contribution, costs etc are known with certainty. This does not hold in real situations. 3.11 Activities

1.

A company makes two products (1 and 2). Each product requires time on two machines A and B. the specifications for each product are as follows
Product 1
54

Product 2

Processing time on machine A (hrs/unit)

2

2

Processing time on machine B (hrs/unit)

1

2

Material and labour cost (Sh./ unit)

14

15

Selling price (Sh./ unit)

16

18

Maximum possible sale (units)

130

150

The amount of time available on machine A is 360 hrs and on machine B is 260 hrs.
Formulate the above as an LP problem, and determine the number of units product 1 and
2, which should be sold in order to maximize total contribution for the company.
2.

ABC Company produces two products. Unit profit for product A is Sh. 60 and for product
B is Sh. 50. Each product must pass through two machines P and Q. Product A requires 10 minutes on machine P and 8 minutes on machine Q. Product B requires 20 minutes on machine P and 5 minutes on Machine Q. machine P is available 200 minutes a day while machine Q is available 80 minutes a day. The company must produce at least two units of product A and five units of product B each day. Units that are not completed in a given day are finished the following day. Required:
i.

Formulate the above as a linear programming problem

ii.

Solve using the graphical method and determine the most profitable daily production plan.

3.

A firm makes two product X and Y and has a total production capacity of nine tonnes per day, X and Y requiring the same production capacity. The firm has a permanent contract to supply at least two tonnes of X and at least three tonnes of Y per day to another company.
Each tonne of X requires twenty machine hours production time and each tonne of Y requires fifty machine hours production time. The daily maximum possible number of machine hours is three hundred and sixty. The entire firm‟s output can be sold and the profit made is Sh. 80 per tonne of X and Sh. 120 per tonne of Y.
Required:
i.

Formulate the above as a linear programming problem.

ii.

Using the graphical method, determine the number of tonnes of X and Y that the company should produce, hence the maximum profit.

55

4.

Reddy Mikks Company produces both interior and exterior paints from two raw materials
M1 and M2. The following table provides the basis data of the problem:-

Tonnes of raw material per tonne of

Maximum daily

Exterior paint

availability (Tonnes)

Interior Paint

Raw Material M1

6

4

24

Raw Material M2

1

2

6

Profit per tonne

5

4

(Sh. 000)

A market survey indicates that the daily demand for interior paint cannot exceed that of exterior paint by more than one tonne. Also the maximum daily demand of interior paint is two tonnes. Reddy Mikks Company wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit.
Required:i.

Formulate the above as a linear programming problem.

ii.

Using the graphical method, determine the optimum product mix hence the optimum profit.

5.

A chicken farmer wishes to produce a chicken feed which is a blend of two inputs coded P and Q. the table below shows the percentage of ingredients in each blend input into the blend and the minimum daily requirements. Each unit of input P and Q costs KSh. 40 and
KSh. 10 respectively.
Proportion of input per unit of blend

Minimum

Ingredients

P

Q

requirements

Protein

30%

25%

400g

Carbohydrates

45%

15%

80g

Salt

15%

30%

120g

Water

5%

10%

200g

Required:

56

(a)

Define the decision variables and the constraints for the above problem hence formulate the problem as a linear programming model given the farmer‟s problem objective is to produce the blend that meets the minimum requirement at least cost.

(b)

Solve the above problem graphically hence indicate the optimal solution and the optimal objective value.

6.

An electronics firm is undecided as to the most profitable mix for its products. The products now manufactured are transistors, resistors and electron tubes with a profit (per 100 units) of Sh. 100, Sh. 60 and Sh. 40 respectively. To produce a shipment of transistors containing
100 units requires one hour of engineering services, ten hours of direct labour and two hours of administrative services. To produce 100 resistors are required one hour, four hours and two hours of engineering, direct labour and administrative time respectively. To produce one shipment of the electron tubes (100 units) requires one hour, five hours and six hours of engineering, direct labour and administrative time respectively. There are 100 hours of engineering services, 600 hours of direct labour and 300 hours of administration available with the firm. What is the most profitable mix?

7.

A farmer has 1,000 acres on which he can grow corn, wheat or soyabeans. Each acre of corn costs Sh. 100,000 for preparation, requires 7 man-days of work and yields a profit of
Sh. 30,000. An acre of wheat costs Sh. 120,000 for preparation, requires 10 man-days of work and yields a profit of Sh. 40,000. An acre of soyabeans costs Sh. 70,000 for preparation, requires 8 man-days of work and yields a profit of Sh. 20,000. If the farmer has Sh.100, 000,000 for preparation and can count on 8,000 man-day of work, how many acres should be allocated to each crop to maximize profits?

8.

An automobile company manufactures three models of cars. There is a back log of orders with the company. Model A requires 60, 100 and 80 worker-days in three production processes I, II and III respectively. Model B requires 100, 240 and 100 worker-days. Model
C requires 200, 360 and 160 worker-days, respectively in three production processes. The numbers of workers employed in the three production processes are 15, 30 and 15 respectively and an average worker is on the job for 200 working days a year. The expected

57

profit for each model is Sh. 7500, 15000 and 30000 respectively. With this capacity, determine the company‟s optimum product mix and total profit.

58

LESSON FOUR: DIFFERENTIATION
4.0 Introduction

Differentiation is the process of finding the derivative of a function. The derivative represents the instantaneous rate of change in the dependent variable given a change in the independent variable.

4.1

Expected Learning Outcomes

By the end of the lesson, the students should be able to:
1. Differentiate any given function, using the rules of differentiation.
2. Find the second derivative of any given function
3. Find the partial derivatives of a given function
4. Find the gradient of a given curve at a given point
5. Identify maximum and minimum turning points

4.2

Rules of differentiation
(a) Power Rule
If y  x n ,

dy
 nx n 1 . This holds at all points except at x  0 dx Example
If y  x 5 ,

dy
 5 x 51  5 x 4 dx 59

(b) Constant Function
If y  c a constant, then

dy
0
dx

Example
If y  6,

dy
0
dx

(c) Constant Times a Function
If y  cu where u  f (x) ,

dy du c dx dx

Example
If y  4 x 3 ,

dy
 4 * 3x 2  12 x 2 dx (d) Sum or Difference of Functions
If y  u  v where u  f (x) and v  g (x) ,

dy du dv
.


dx dx dx

Example
If y  8 x 3  10 x,

dy
 24 x 2  10 dx If y  3x 3  5 x 2 ,

dy
 9 x 2  10 x dx (e) Product Rule
If y is a product of two functions i.e. y  uv where u and v are functions of x , then

dy dv du
u v dx dx dx Example 1
Find the derivative of the function y  2 x 2 5x  3
Solution
Let

u  2x 2



du
 4 x and dx v  5x  3



dv
5
dx

60

dy dv du
u v
 2 x 2 * 5  5 x  35 dx dx dx 2
2
 10 x  20 x  12 x  30 x 2  12 x

Example 2





Find the derivative of the function y  x  5 x 2  3
Solution
u  x5



du
 1 and dx v  x2  3

Let



dv
 2x dx 



dy dv du
u v
 x  52 x  x 2  3 1  3x 2  10 x  3 dx dx dx (f) Quotient Rule
If y is a quotient of two functions i.e. y 

dy

dx

v

u where u and v are functions of x , then v du dv u dx dx
2
v

Example 1

2x 2  3
Find the derivative of the function y  x Solution u  2x 2  3

dy

dx

v



du
 4 x and dx vx

Let



dv
1
dx

du dv u
2
2 dx dx  x4 x   2 x  3 1  2 x  3 v2 x2 x2 



Example 2

61

Find the derivative of the function y 

x2  3 x2 Solution u  x2  3

dy

dx

v



du
 2 x and dx v  x2

Let



dv
 2x dx du dv u
2
2
3
3 dx dx  x 2 x   x  3 2 x  2 x  2 x  6 x  6 x  6
2
v2 x4 x4 x3 x2 



 

(g) Chain rule
Chain rule is used to differentiate compound functions. If y  u  where u is a function of x , n then

dy du dy dy du or  nu n 1

* dx dx dx du dx

Example 1
Find the derivative of the function y  x 3  5x  3
Solution
The above function can be re-written as y  x 3  5x  3





Let

yu

1

2

du
 3x 2  5 dx u  x 3  5x  3

Therefore

1

dy 1  12
 u du 2

2

dy dy du 1  12
1
1

*
 u * (3x 2  5)  (3x 2  5)( x 3  5 x  3) 2 dx du dx 2
2
2
3x  5

2 x 3  5x  3
Example 2





Find the derivative of the function y  3x 2  6 x  5
Solution
62

3

2

Let

u  3x 2  6 x  5

Therefore

yu

3

2

du
 6x  6 dx dy 3 12
 u du 2

dy dy du 3 12

*
 u * (6 x  6) dx du dx 2
1
3
 (3x 2  6 x  5) 2 (6 x  6)  9 x  9 3x 2  6 x  5
2



4.3



1

2

Second derivative of a function

The second derivative of y  f (x) is denoted by f ' ' ( x) or

d2y
.
dx 2

Example
Find the second derivative of the function y  5x 3  4 x 2  10 x  25
Solution

dy
 15 x 2  8 x  10 dx d2y
 30 x  8 dx 2

4.4

Partial derivatives

If a function has more than one variable, partial derivatives can be calculated for each variable, by assuming all the other variables to be constants.
Example
Find the partial derivatives the following function with respect to x and y.
Z  10 x  15 y 2  8x 2 y  20

Solution

Z
 10  16 xy
x
Z
 30 y  8 x 2
y
63

4.5

Finding the gradient

When finding the gradient of a curve at a given point, the function is differentiated and the value of x is inserted at the required points in the derivative function.
Example
Find the gradient of the curve Y  5x 2  3x  5 when x = 2.
Solution
dy
 10 x  3 dx Therefore when x = 2, the gradient of the curve  10(2)  3  23

4.6

Finding a stationary point

A stationary point is that point where the gradient is zero. It may be a maximum or a minimum point. The following steps are taken to find out a stationary point
 Find the derivative of the function
 Equate the derivative to Zero
 Solve the resulting equation
Example
Find the point at which the curve Y  2 x 2  3x  10 will have a stationary point.
Solution
dy
 4x  3 dx  4x  3  0 x 3
 0.75
4

There will be a stationery point when x  0.75

64

4.7

Turning points (Maximum and Minimum)

The turning points show the maximum or minimum values of a given function. The gradient at both maximum and minimum points is zero. In order to find out the maximum or minimum points, the second derivative is calculated. The following rules apply:
First derivative

Second derivative

Maximum

Zero

Negative

Minimum

Zero

Positive

Example
Find the point at which the curve Y  x 3  3x  2 will have a maximum or minimum point.
Solution
dy
 3x 2  3  3x 2  3  0 dx x 2  1, x  1 or x  1

There will be stationary points when x = +1 or x = -1
Finding the second derivative

d2y
 6x dx 2 d2 y if x   1 then 2  6 *1  6 dx d2 y if x  - 1 then 2  6 * 1  6 dx Therefore there is a minimum point when x = +1 and a maximum point when x = -1.

4.8

Activities

1. Find the derivatives of the following functions
65

(a) y  50
(b) y  6 x 3  15x 2  3x  10





(c) y  9 x 2  23x  5

5

(d) Y  3x 5  14 x 3  16 x  10





(e) Y  2 x 2  3 4 x 3  10

Y



3  x2 x2 (f)
2. Differentiate the following functions
i.

y  x10

ii.

y  15x 2  10 x  23

iii.

y  4 x 4  5x 3 2 x 2  5

iv.

y

v.

y  4 x 3  3x  5







x2  x 1 x2  x 1





7

3. Find the second derivative of the function
(a) Y  3x 5  14 x 3  16 x  10
(b) y  x 3  4 x 2  5x  50
(c) y  6 x 3  15x 2  3x  10
(d) y  15x 3  6 x 2  x  8
4. Find the partial derivatives the following function with respect to x and y.
Z  15xy  5 y 2 x 2  8x  9 y  16

5. Find the gradient of the curve

Y  2 x 2  3x  10 when x = -1.
6. Find the point at which the following curves will have a stationary point.
66

(a) Y  6 x 2  3x  15
7. Find the points at which the following curves will have a maximum or minimum point.
(a) Y  120 x  x 2  0.02 x 3
(b) Y  25  5x  2 x 2

67

LESSON FIVE: APPLICATIONS DIFFERENTIATION TO

REVENUE, COST AND PROFIT FUNCTIONS
5.0

Introduction

Various mathematical concepts are used to solve business problems. The main objective of a business enterprise is to maximize its profits. In order to achieve this objective, costs must be minimized and revenue maximized. A business firm is said to be at equilibrium when its marginal revenue is equal to marginal cost. The techniques of calculus can be used to analyse revenue, cost and profit functions so as to determine the points of maximum revenue, maximum profit and minimum cost.
.

5.1

Lesson Objectives

By the end of the lesson, the students should be able to: 1. Apply the concepts of differentiation to revenue, profit and cost functions
2. Determine the number of units to be produced and sold to maximize revenue and profit and to minimize costs.
3. Determine the maximum amount of revenue, maximum profit or minimum costs 4. Apply marginal cost analysis in solving business problems

5.2

Revenue Application

Total Revenue = (Price per unit)(Quantity sold) = P*Q
Total revenue would be maximized when

d 2TR dTR 0
 0 and dq dq 2
68

Example 1:
The demand for the product of a firm varies with the price that the firm charges for the product
.The firm estimates that annual total revenue R (stated in 1000”s) as a function of the price P is given by R  f ( p)  50 p 2  500 p
(a) Determine the price, which should be charged in order to maximize total revenue.
(b) What is the maximum value of annual total revenue?

Solution:
(a) The maximum value of R will occur at the vertex of the graph of the function R  f ( p)  50 p 2  500 p

dR
 f ' ( p)  100 p  500 dp At the maximum point, the gradient is equal to zero
 100 p  500  0 p 500
5
100

Second order derivations should be found to determine whether it is a relative maximum or relative minimum.

d 2R
 f ' ' ( p)  100  0 hence the point is a maximum. dp 2
Therefore the price which should be charged in order to maximize total revenue = 5.

(b) The maximum value of annual total revenue is
R  f ( p)  50 p 2  500 p  50(5) 2  500(5)  1250  1250  1250

Example 2
A public transportation company has been experimenting on a possibility of developing a system of charging fares .The demand functions, which expresses the ridership as a function of fare charged is given below:
Q  10,000  125 p

69

where Q equals the average number of riders per hour and p equals the fare in shillings.
(a) Determine the fare, which should be charged in order to maximize hourly bus fare revenue. (b) What is the expected maximum revenue?
(c) How many riders per hour are expected under this figure?

Solution:
(a) The general expression for total revenue is
R  p * q  p * (10,000  125 p)  10,000 p  125 p 2

The first derivative is

dR
 10,000  250 p dp If the derivative is set to zero
10,000  250 p  0
 p  40

The second derivative is found to determine the nature of the critical point.

d 2R
 f ' ' ( p)  250  0 hence the point is a maximum dp 2
The fare, which should be charged in order to maximize hourly bus fare revenue, is
P = 40
(b) The expected maximum revenue is
R  10,000 p  125 p 2  10,000(40)  125(40) 2  200,000

(c) The average number of rider‟s expected each hour will be given by the demand function
i.e.
Q  10,000  125 p  10,000  125(40)  5000 riders per hour

5.3

Cost Application

Cost represents cash outflows for an organization. Most organizations seek ways to minimize this outflow i.e. inventory management. A common problem in organizations is determining how
70

much of a needed item should be kept on hand. For retailers, the problem may relate to how many units of each product should be kept in stock. For producers, the problem may involve how much of each raw material should be kept available. Concerning the question of how much
“Inventory” to keep on hand, there may be costs associated with having too little or too much inventory on hand.
Total Cost  Fixed Cost  Variable Cost

Fixed costs are costs, which do not change with the level of production while variable costs are costs, which change with a change in the level of production.
Total cost would be minimized when

d 2TC dTC 0
 0 and dq dq 2

Example:
A retailer of motorized bicycles has examined cost data and has determined a cost function which expresses the annual cost of purchasing, owning, and maintaining inventory as a function of the size (number of units) of each order it places for the bicycles. The cost of function is,

C  f (q) 

4,860
 15q  750,000 q Where C equals annual inventory cost, stated in dollars and q equals the number of cycles ordered each time the retailer replenishes the supply.
a) Determine the order size, which minimizes annual inventory cost.
b) What is minimum annual inventory cost expected to equal?

Solution
a) The first derivative is:

dC
 f ' (q)  4,860q  2  15 dq 4,860
- 4,860q -2  15  0  2  15 q q 2  324  q  18

71

The value q  18 is meaningless in this application (negative order quantities are not possible) .The nature of the only meaningful critical point q  18 is checked by finding f"

d 2C
9,720 9720
 f ' ' ( p)  9,720q 3 

 1.667  0 . hence the point is a minimum.
2
dp q3 183

i.e.

Annual inventory cost will be minimized when 18 bicycles are ordered each time the retailer replenishes the supply.

b) Minimum annual inventory costs are determined by calculating f (18) or
C

5.4

4,860
 15(18)  750,000  750,540
18

Profit Application

Profit  Total Revenue - Total Cost

Total profit would be maximized when

d 2P dP 0
 0 and dq dq 2

Example 1:
A major cosmetic and beauty supply firm, which specializes in door-to-door sales approach, has found that the response of sales to the allocation of additional sales representation behaves according to the law of diminishing returns .For one region sales district, the company estimates that annual profit P , stated in hundreds of shillings, is a function of the number of sales representations x assigned to the district .The function these two variables is
P  f ( x)  12.5x 2  1,375x  1,500

a) What number of representations will result in maximum profit for the district?
b) What is the expected maximum profit?

Solution
72

a) The derivative of the profit function is dP  25 x  1375 dx 1375
25 x  1375  x 
 55
25
f ' ( x) 

Checking the nature of the critical point we find f ' ' ( x)  25  0 hence the point is a maximum point.
The number of representations that will result in maximum profit for the district is 55.

b) The expected maximum profit is
P  f ( x)  12.5(55) 2  1,375(55)  1,500  36,312.5

We conclude that annual profit will be maximized at a value of Sh.36, 312.5(100s), or 3,631,250, if 55 representatives are assigned to the district.

Example 2:
A manufacturer has developed a new design for solar collection panels. Marketing studies have indicated that annual demand for the panels will depend on the price charged. The demand function for the panels has been estimated as

q  100,000  200 p
Where q equals the number of units demanded each year and p equals the price in shillings.
Engineering studies indicate that the total cost of producing q panels is estimated well by the function C  150,000  100q  0.003q . Formulate the profit function p = f (q) which states the annual profit p as a function of the number of units q which are produced and sold.

Solution
R  p*q from q  100,000 - 200p 200p  100,000 - q  p  500 - 0.005q
 R  (500 - 0.005q) * q  500q - 0.005q 2

Now that the revenue and cost function have been stated in terms of q, the profit function can be defined as

73

P  R  C  500q  0.005q 2  [150,000  100q  0.003q 2 ]
 0.008q 2  400q  150,000

5.5

Applications of partial derivatives

If a function has more than one variable, partial derivatives can be calculated for each variable, by assuming all the other variables to be constants.
Procedure for determining the optimal values (Maximum and Minimum) for partial derivatives
i.

Find the first partial derivatives of the dependent variable with respect to all the independent variables.

z z

0
x y

ii.

Simultaneously equate these to zero i.e.

iii.

Determine the second order partial differentials at the critical points. If


2z
2z
and
 0 , then the function is minimum at the critical points.
yy
xx



2z
2z
and
 0 , then the function is maximum at the critical points.
yy
xx

Example
The total profit per acre on a wheat farm, has been found to be related to the expenditure per acre for (a) Labour and (b) soil conditioners and fertilizers. If x represents the shillings per acre spent on labour and Y represents the shillings per acre spent on soil improvement:
Profit   48 X  60Y  10 XY  10 X 2  6Y 2
What are the values of X and Y that maximize profits?
Solution

74


 48  10 y  20 x
x

 60  10 x  12 y
y
48  10 y  20 x  0 ..............(i)
60  10 x  12 y  0...............(ii) solving equation (i) and (ii) simultaneously :
- 20x  10y  -48
10x - 12y  -60 multiplying equation (ii) by 2
- 20x  10y  -48
20x - 24y  -120
 14 y  168
168
y 
 12
14
10 x  12(12)  60
10 x  60  144
84
x
 8. 4
10
The values of x and y that will maximize profits are y = 12 and x = 8.4.

5.6

Marginal Approach to Profit Maximization

An alternative approach to finding the profit maximization point involves marginal analysis.
Popular among economist, marginal analysis examines incremental effects on profitability.
Given that a firm is producing a certain number of units each year, marginal analysis would be concerned with the effect on the profit if one additional unit is produced and sold.
Requirements for using the marginal approach
i.

It must be possible to identify the total revenue function and the total cost function separately. ii.

The revenue and the cost function must be stated in terms of the level of output or number of units produced and sold.

75

5.6.1 Marginal Revenue
It is the additional revenue derived from one more units of a product or service. If each unit of a product sells at the same price, the marginal revenue is always equal to the price. For example, the linear revenue function
R = 10q
Represent a situation where each unit sells for Sh.10. The marginal revenue for selling one additional unit is Sh. 10 at any level of output q.
For a total revenue function R(q) , the derivative R' (q) represent the instantaneous rate of change in total revenue given a change in the number of units sold R´ also represent a general expression for the slope of the graph of the total revenue function. For purposes of marginal analysis, the derivative is used to represent the marginal revenue i.e. MR  R' (q)

5.6.2 Marginal Cost
It is the additional cost incurred as a result of producing and selling one more unit of a product or service. Linear cost function assume that the variable cost per unit is constant. For such a functions the marginal cost is the same at any level of output e.g. C  105,000  3.5q
Where variable cost per unit is constant at Sh.3.50 the marginal cost for this cost function is always Sh.3.50.

For a total cost function C (q) , the derivative C ' (q) represents the instantaneous rate of change in the total cost given a change in the number of units produced. C ' (q) also represents a general expression for the slope of the graph of the total cost function.
For purposes of marginal analysis, the derivative is used to represent the marginal cost, or
MC  C ' (q)

As with R´, C´ can be used to approximate the marginal cost associated with producing the next unit. 5.6.3 Marginal Profit Analysis
This is concerned with the effect on profit if one additional unit of a product is produced and sold. As long as the additional revenue brought in by the next unit exceeds the cost of producing
76

and selling that unit, there is a net profit from producing and selling that unit and total profit increases. If, however, the additional revenue from selling the next unit is exceeded by the cost of producing and selling the additional unit, there is a net loss from that next unit and total profit decreases. NOTE:
 If MR  MC , produce the next unit.
 If MR  MC , do not produce the next unit.
 Profit will be maximized when MR  MC i.e. production should continue up to the point where R' (q)  C ' (q)
 Given a level of output q* where R' (q)  C ' (q) (Or MR  MC ), producing q* will result in profit maximization if R' ' (q*)  C ' ' (q*)

Example
ABC Ltd employed a cost accountant who developed two functions to describe the operations of the firm. He found the marginal revenue function to be MR  25  5x  2 x 2 and the marginal cost function to be MC  15  2 x  x 2 where x is the level of output. Determine the profit maximizing output and the total profit at that point.
Solution
For profit maximization MR = MC therefore

77

25  5 x  2 x 2  15  2 x  x 2
25  15  5 x  2 x  2 x 2  x 2  0
10  3 x  x 2  0 solving the above quadratic equation
- x 2  2 x  5 x  10  0 x( x  2)  5( x  2)  0
 x  2  0 x  2 x  5  0  x  -5 for profit maximisati on

d2P
0
d 2x

d 2P
 3  2 x d 2x d2P  when x  2 2  3  (2 * 2)  7 d x d2P when x  5
 3  (2 * 5)  7 d 2x
Therefore, the profit maximising output is two units.
The total profit at that point would be

 (10  3x  x

2

)dx  10 x  3 x 2  1 x 3
2
3

when x  2
P  10(2) - 3 (4)  1 (8)  20  6  2.67  11.33
2
3

78

5.7

Activities

1. The annual revenue form a firm depends upon the number of units produced specifically the functions which describes total revenue R  0.01x 2  5,000 x  25,000 where

x is the

number of units produced.
(a)

Determine the number of units x , which will result in maximum revenue.

(b) What is the expected maximum Revenue?

2.

The total cost of producing q units of a certain products is described by the function.
C  350,000  7,500q  0.25q 2

Where C is the total cost stated in shillings
a) Determine how many units q should be produced in order to minimize the average cost per unit.
b) What is the minimum average cost per unit?
c) What is total cost of production at this level of output?

3. A company estimates that the demand for its products fluctuates with the price it charges
.The demand function is q  280,000  400 p
Where q equals the number of units demanded and p equals the price in dollars. The total cost of producing q units of the product is estimated by the function.
C  350,000  300q  0.0015q 2

(a)

Determine how many units q should be produced in order to maximize annual profit.

(b) What price should be charged?
(c)

What is the annual profit expected to equal?

4. The total cost and total revenue function for a product are
C (q)  500  100q  0.5q 2 and R(q)  500q

(a)

Using the marginal approach, determine the profit-maximizing level of output.

(b) What is the maximum profit?
79

5. The total cost and total revenue functions for a product are
C (q)  40,000  25q  0.025q 2

(a)

R(q)  75q  0.008q 2

and

Using the marginal approach, determine the profit-maximizing level of output.

(b) What is the maximum profit?

6. A firm producing jackets estimates that the total cost of making X jackets is given by TC =
600 + 60X and the sales revenue per unit is given by R = 200 – 4X.
i.

How many jackets should the firm produce to breakeven?

ii.

What is the possible maximum profit for this firm?

7. The cost accountant of a firm producing colour television has worked out the total cost function for the firm as TC  120Q  Q 2  0.02Q 3 . A sales manager has provided the sales forecasting function as P  114  0.25Q where P is price and Q the quantity sold. Required:
i.

Find the level of production that will yield minimum average cost per unit and determine whether this level of output maximizes profit for the firm.

ii.

Determine the price that will maximize profit for the firm.

iii.

Determine the maximum revenue for this firm.

8. A firm has analyzed their operating conditions, prices and costs and have developed the following functions:
Revenue = 400Q  4Q 2 and cost = Q 2  10Q  30 where Q is the number of units sold. The firm wishes to maximize profit and wishes to know:
(a) What quantity should be sold?
(b) At what price?
(c) What will be the amount of profit?
(d) Find the point of maximum value of the revenue function?

9. ABC Ltd supply tool kits for the home handyman. Each tool kit comprises a standard plastic box which contains a variable number of tools depending on the type of tools, the market and

80

the wholesalers requirements. The firm has derived a profit function which shows that their profits are dependent both on the number of tool kits supplied and the number of tools in each kit. The profit function is as follows: P = 8K  0.0001K 2  0.05KT  77.5T 2  10,000
Where P = Profit in KSh. K = Number of Kits and T = Number of tools in each kit. How many tool kits containing how many tools should be sold?

10. A car costs Sh.75, 000 in the market and the running cost for x kilometers is given by
V .C ( x)  x  30 x( x  1). Find, when is the total average cost minimum.

11. A firm‟s demand function is given by

P  100  2 x

and its cost function is

C ( x)  20 x  3000 . Determine the optimum level of output for profit maximization.

12. A factory produces x calculators per day. The total daily cost in shillings incurred is
5x 2  700 x  500. if the calculators are sold for Sh. (100 –10x) each, find the number of

calculators that would maximize the daily profit.

13. A firm has analyzed its operating conditions and has developed the following functions
Total Revenue =  10q 2  200q
Total Cost = q 2  20q  1000
Where q is the number of units produced and sold.
Determine the value of q that:i.

Maximizes revenue and hence the Maximum Revenue

ii.

Minimizes Cost and hence the Minimum Cost

iii.

Maximizes profit and hence the Maximum Profit.

14. A manufacturer has determined a cost function, which expresses the annual cost of purchasing, owning and maintaining its raw material inventory as a function of the size of each order. The cost function is-

C

51,200
 80q  750,000 q 81

Where q equals the size of each order (in tons) and C equals the annual inventory cost.
a) Determine the order size q , which minimizes annual inventory cost.
b) What are minimum inventory costs expected to equal?
15. A manufacturer has found that if he wants to increase his output, he must lower his price. His total revenue (TR) from an output x, is given by the expression TR  x148  x  . His production costs are Shs. 1,000 fixed and Shs. 36 per unit variable and the total cost
TC  1,000  36 x . Required:

Find:
(a) The output that would maximize revenue.
(b) The maximum total revenue.
(c) The profit P in terms of the number of units (x).
(d) The output x, that would maximize profit.

16. A firm selling a trade directory has developed a profit function as follows:
P  9D  0.0005D 2  0.06DA  80 A2  5,000

Where D = Number of directories sold and A = number of advertising pages.
How many directories containing how many advertising pages should be sold to maximize profits?

17. Given Marginal Cost (MC) = 36 + (q – 8)2 and Marginal Revenue (MR) = 100 – 2q where q is quantity of units. Find: i.

The profit maximizing output of the firm.

ii.

The total cost and total revenue function assuming that fixed cost and fixed revenue is equal to zero.

iii.

The value of maximum profit.

82

LESSON SIX: INTERGRAL CALCULUS
6.0

Introduction

Integration is the branch of calculus concerned with the theory and applications of integrals. While differential calculus focuses on rates of change, such as slopes of tangent lines and velocities, integral calculus deals with total size or value, such as lengths, areas, and volumes. The two branches are connected by the fundamental theorem of calculus, which shows how a definite integral is calculated by using its antiderivative (a function whose rate of change, or derivative, equals the function being integrated). For example, integrating a velocity function yields a distance function, which enables the distance traveled by an object over an interval of time to be calculated. As a result, much of integral calculus deals with the derivation of formulas for finding antiderivatives.

.

6.1

Lesson Objectives

By the end of the lesson, the students should be able to: 1. Integrate various functions by using the rules of integration
2. Differentiate between integration and differentiation
3. Apply integration in solving business problems and
4. Find the area of a curve between two points.

6.2

Rules of Integration

83

Integration is the reverse of differentiation. Given the derivative of a function, the process of finding the original function is the reverse of differentiation, f is said to be an antiderivative of f ' . Given f ' ( x)  4 then f ( x)  4 x  C

Example 1:
Find the antiderivative of any constant of f ' ( x)  0 .
Solution.
The derivative of any constant function is 0. Therefore, the antiderivative is f ( x)  C
Example 2:
Find the derivative of f ' ( x)  2 x  5
Solution.
f ( x)  x 2  5 x  C

Given that f is a continuous function

 f ( x)dx  F ( x)  C
If F´(x) = f(x) ,C is the constant of integration.

RULE 1: Constant Functions:

 kdx  kx  C , Where k is any constant.
Examples:
a)  (2)dx  2 x  C .

b)

 3 2 dx  3 2 x  C .

c)



d)

2dx  2 x  C .

 0dx  (0) x  C  C .
84

RULE 2: Power Rule n  x dx 

x n 1
 C provided that n  1 n 1

Examples

x2
C
2

a)

 xdx 

b)

2
 x dx 

x3
C
3
3

3 x 2 x dx   x dx 
C  2 x 2 C
3
3
2
1

c)



d)

1 x 2
1
dx   x 3 dx 
C   2 C
 x3
2
2x

2

RULE 3: Constant Times a Function

 Kf ( x)dx  k  f ( x)dx.
Examples

a)

b)

c)

5 xdx  5 xdx 



Where k is any constant.

5x 2
C
2

x2 x3 dx  1  x 2 dx  1 *  C  1 x 3  C
2
2
2 3
6



3 x dx  3

1 x 1

dx  3 x

1

2

1 x 2 dx  3 *
 6x 2  C  6 x  C
1
2

RULE 4: Addition or Difference Rule

If

 f (x) dx

and

 g ( x)dx exist, then   f ( x)  g ( x)dx   f ( x)dx   g ( x)dx
85

Example
a)

3x
 3x  6dx   3xdx   6dx 

2

2

 C1  (6 x  C 2 ) 

3x 2
 6x  C
2

N.B Even though the two integrals technically results in separate constants of integration, these constants may be considered together as one.

RULE 5: Power Rule Exception

x

1

dx  ln x  C

RULE 6: Exponential Function



x

dx   x  C

RULE 7:



 f ( x)n1  C, n  1
 f ( x) . f ' ( x)dx  n n 1

Example

5x  3
3
 5x  3 5dx 

4

4

C

RULE 8:

 f ' x

f x

dx   f  x   C

Examples:
(a) Evaluate



2

2 x x dx

Solution
86

 2x

x2

dx   x  C
2



(b) Evaluate

3

x 23 x dx

Solution f ( x)  3x 3 and f ' ( x)  9 x 2



6.3

3

x 23 x dx =

9 2 3x3
1
1 3x3
2 3x3
3 3x3
 x  dx = 9  9 x  dx =  x  dx  9   C
9

Application of Integral Calculus

Integration can be used to determine the area between curves and also to solve business and problems. 6.3.1 Finding the area between a curve and the x-axis
For a function f that is continuous over [a, b], the area between y  f (x) and the x axis from x
= a to x = b can be found using definite integrals as follows: b 

For f ( x)  0 over [a, b]: Area =

 f ( x)dx a b



For f ( x)  0 over [a, b]: Area =  [ f ( x)]dx a Example
Find the area bounded by f ( x)  6 x  x 2 and y  0 for 1  x  4 .
Solution
4



x3 
(4) 3  
(1) 3 
 3(1) 2 
Area =  (6 x  x )dx  3x 2    3(4) 2 

  48  24  24
3 1 
3  
3 

1
4

2

6.3.2 Finding the area between two curves
If

f and g are continuous and f ( x)  g ( x) over the interval [a, b] , then the area bounded by

y  f (x) and y  g (x) for a  x  b is given by

87

b

 [ f ( x)  g ( x)]dx

Area =

a

Example
Find the area bounded by the graphs of f ( x)  1 x  3, g ( x)   x 2  1, x  -2 and x  1
2
Solution

 x


 2 x

2
2[ f ( x)  g ( x)]dx  2  2  3   x  1dx  2 x  2  2 dx






1

Area =

1

1

1

 x3 x2

1 1
  8 4
 33
 
 2     2   
  4 
 8.25
4
  3 4
 4
3
 2  3 4

6.3.3 Application of integration to cost, revenue and profit functions Examples:
1. The function describing the marginal cost of producing a product is

MC  x  100

Where x equals the number of units produced .It is also known that total cost equals Sh.
40,000 when x  100 . Determine the total cost function.
Solution:
To determine the total cost, we must first find the antiderivative of the marginal cost function, or C ( x) 

x2
 100 x  C
2

Given that C (100) = 40,000, we can solve for the value of C, which happens to represent the fixed cost.

100 2
 100(100)  C
2
40,000  5,000  10,000  C
40000 

c  25,000
The specific function representing the total cost of producing the product is

C ( x) 

x2
 100 x  25,000
2

2. The marginal revenue function for a company‟s product is

88

MR  50,000  x

Where x equals the number of units produced and sold. If total revenue equals 0 when no units are sold, determine the total revenue function for the producer.
Solution:
Because the marginal revenue function is the derivative of the total function is the antiderivative of MR.

R( x)  50,000 x 

x2
C
2

Since we are told that R(0)  0 , substitution of x  0 and R  0 into the above equation yields 0  50,000(0) 

02
C
2

C0

Thus the total revenue function for the company‟s product is R( x)  50,000( x) 

x2
2

3. An automobile manufacturer estimates that the annual rate of expenditure r (t ) for maintenance on one of its model is represented by the function r (t )  100  10t 2 .
(a)

What is the expected maintenance expenditures during the automobile first 5 years?

(b) Of the above expenditures, what is expected to be incurred during the fifth year?

Solution
(a)



5

0

(100  10t 2 )dt  [100t 

10t 3
3

5
0

= 100(5) 

10(5)3
 500  416.67
3

=$916.67

(b) Of these expenditures, those expected to be incurred during the firth year are estimated as

89

5

2
 (100  10t )dt  100t 
4

10t 3
3


10(5)3  
10(4)3 
 100(4) 
= 100(5) 


3  
3 


5
4

=916.67- 400  213.33)

=Sh. 303.34
4. A state civic organization is conducting its annual fund-raising campaign for its summer camp program for the disadvantaged. Campaign expenditures will be incurred at a rate of
$10,000 per day. From past experience it is known that contribution will be high during the early stages of the campaign and will tend to fall off as the campaign continues. The function describing the rate at which contribution are received is C (t )  100t 2  20000
Where t represents the day of the campaign, and C (t ) equals the rate at which contributions are received, measured in shillings per day. The organization wishes to maximize the net proceeds from the campaign.
a) Determine how long the campaign should be conducted in order to maximize net proceeds. b) What are the total campaign expenditures expected to equal?
c) What are the total contributions expected to equal?
d) What are the net proceeds (total contribution less total expenditures) expected to equal.

Solution
a) The function which describes the rate at which expenditures e(t ) are incurred is e (t)=10,000
C (t ) = e(t )

-100t2+20,000=10,000 t2 = 100 t = 10 days negative root is meaningless
b) Total campaign expenditures are represented by
E = e (t)
=(Sh. 10,000 per day)(10 days)
= Sh.100,000
90

c) Total contribution during the 10 days are represented by

C=

10



0

(100t 2  20,000)dt

t3
-100
 20,000t
3

=

10
0

 100(10)3
 20,000(10)
=
3
= -33,333.33+200,000 = $166,666.67
d) Net Proceeds = Costs – Expenditures = Sh.166, 666.67 - Sh.100, 000
=Sh. 66,666.67

Example
Gatheru and Kabiru Certified Public Accountants have recently started to give business advice to their clients. Acting as consultants, they have estimated the demand curve of a clients firm to be:
AR  200  8Q

Where AR is average revenue in millions of shillings and Q is the output in units.
Investigators of the client firm‟s cost profile shows that marginal cost (MC) is given by
MC  Q 2  28Q  211(In millions of shillings)

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

The total cost function

ii.

The total revenue function

iii.

The level of output that maximize profit

Solution
(b)

91

i.

The total cost function

TC   MCdq   (q 2  28q  211)dq  1 q 3  14q 2  211q  10
3
ii.

The total revenue function
TR  AR * Q  (200  8q)q  200q  8q 2

iii.

The level of output that maximize profit

Pr ofit  TR  TC  200q  8q 2  1 q 3  14q 2  211q  10   1 q 3  6q 2  11q  10
3
3
2
d d 
To max  ,
 0, 2  0 dq dq d  q 2  12q  11  0 dq  12  12 2  (4 * 1 * 11)  12  10
q 

2
2
 12  10
 12  10 q  1 or q 
 11
2
2 d 2
 2q  12 dq 2 d 2
When q  1, 2  2(1)  12  10 dq d 2
When q  11, 2  2(11)  12  10  0 hence maximum dq  The level of output tha will maximize profit  11 units t Example
The cost function of a company is given by C ( x)  x 3  60 x 2  1200 x  1000 . If the product sells for Ksh. 588 each,
i. What level of production will maximize the profits? ii. Find the total cost, total revenue and total profit at the optimal production.

Solution
i. What level of production will maximize the profits?

92

  TR - TC
  Price * Quantity  588x
  588 x  x 3  60 x 2  1200 x  1000
   x 3  60 x 2  612 x  1000
To max  ,

d d 2
 0, 2  0 dx dx

d
 3 x 2  120 x  612  0 dx  120  120 2  (4 * 3 * 612)  120  84 x 
2(3)
6
 120  84
120  84 x  6 or x 
 34
6
6 d 2
 6 x  120 dx 2 d 2
When x  6, 2  6(6)  120  84 dx d 2
When x  34, 2  6(34)  120  84  0, hence maximum dx  The level of production that will maximize profit  34 units ii. Find the total cost, total revenue and total profit at the optimal production.

TC  34 3  60(34 2 )  1200(34)  1000  11,744
TR  588(34)  19,992
Profit   34 3  60(34 2 )  612(34)  1000  8248

6.4

Activity

1. Evaluate the following integrals
(a)

 60dx

(b)

 10  6 x  15x dx

(c)



2

5

x dx
93

 20
(d) 
3 2
 x

(e)

 x

(f)

 3x 

2

2


dx



 2x

x3

 x  1dx
5

dx

2. Find the area bounded by the graphs of f ( x)  x 2  x, g ( x)  2 x, for -2  x  3
3. Find the area bounded by the graphs of f ( x)  2 x 2 , g ( x)  4  2 x, for -2  x  2
4. Find the area between the graph of f ( x)  x 2  9 and the x axis over the indicated intervals
(i) [0, 2]

(ii) [2, 4]

5. Find the area bounded by f ( x)  x 2  1 and y  0 for  1  x  3 .

6. Find the area bounded by the graphs of f ( x)  x 2  1, g ( x)   1 x  3, x  -1 and x  2
2

7. A company specializing in mail order sales approach is beginning a promotional campaign.
Advertising expenditures will cost the firm $5,959 per day. Marketing specialists estimates that the rate at which profit (exclusive of advertising costs) will be generated from the promotion campaign decreases over the length of the campaign. Specifically, the rate r (t ) for this campaign is estimated by the function r (t )  50t 2  10,000

Where t represents the day of the campaign and r (t ) is measured in dollars per day.
In order to maximize net profit, the firm should conduct the campaign as long as r (t ) exceeds the daily advertising cost.

a) Graph the function r (t) and the function c (t)=5,950, which describes the rate at which advertising expenses are incurred.
b) How long should the campaign be conducted?
c) What are the total advertising expenditures expected to equal during the campaign?
d) What net profit will be expected
94

8. A hospital blood bank conducts an annual blood drive to replenish its inventory of blood. The hospital estimates that blood will be donated at a rate of r(t) pints per day, where

d (t )  300 0.1t and t equals the length of the blood drive in days. If the goal for the blood drive is 2,000 pints, when will the hospital reach its goal?

9. A manufacturer of jet engines estimates that the rate at which maintenance costs are incurred on its engines is a function of the number of hours of operation of the engine. For one engine used on commercial aircraft, the function is r ( x)  60  0.040 x 2 where x equals the number of hours of operation and r (x) equals the rate at which repairs cost are incurred in dollars per hour of operation.
(a) Determine the rate at which costs are being incurred after 100 hours of operation.
(b) What are the total costs expected to equal during the first 100 hours of operation.

10. The rate w(t ) at which solid waste is being generated in Nairobi city is described by the function w(t )  0.5 0.025t where w(t ) is stated billions of tons per year and t equals time measured in years (t = 0 corresponds to January 1 2005)
a) Determine the rate at which solid waste is expected to be generated at the beginning of year 2015.
b) What total tonnage is expected to be generated during the 20-year period from 2005 through 2025?
11. The demand for commercial forest timber has been increasing rapidly over the past three to four decades. The function describing the rate of demand for timber is d (t )  20  0.003t 2 where d (t ) is stated in billions of cubic feet per year and t equals time

in year (t = 0 corresponds to January 1 2005)
a) Determine the rate of demand at the beginning of 2010
b) Determine the rate of demand at the beginning of 2020
c) Determine the total demand for timber during the period 2005 through 2024
{Hint: integrate between t = 0 and t = 20}

95

12. The learning curve for a particular job has been defined as h(x) =

20
 4, x >0 x Where h (x) equals the production rate measured in hours per unit and x equals the unit produced. a) Determine the production rate h (x) at the time of 10th unit (x = 10)
b) Integrating the learning curve over a specified interval provides an estimate of the total number of production hours required over the corresponding range of output. Determine the total number of hours expected for producing the first 20 units by integrating h (x) between x = 1 and x = 20
c) Sketch h.
d) Is there any limit suggested as to how efficient an employee can become at this job

96

LESSON SEVEN: INVENTORY CONTROL
7.0

Introduction

Inventory management is a science primarily about specifying the shape and percentage of stocked goods. It is required at different locations within a facility or within many locations of a supply network to precede the regular and planned course of production and stock of materials. The scope of inventory management concerns the fine lines between replenishment lead time, carrying costs of inventory, asset management, inventory forecasting, inventory valuation, inventory visibility, future inventory price forecasting, physical inventory, available physical space for inventory, quality management, replenishment, returns and defective goods, and demand forecasting. Balancing these competing requirements leads to optimal inventory levels, which is an on-going process as the business needs shift and react to the wider environment.
Markov analysis is a descriptive tool designed to predict the behaviour of a system over

7.1 This lesson covers the concept of Markov analysis, its assumptions, computation of time. Expected Learning Outcomes transition probabilities, state conditions steady states and analysis of absorbing chains.

By the end of the lesson, the students should be able to;
1. Explain the various reasons why organizations hold stock
2. Describe the various costs associated with holding inventory
3. Derive the economic order quantity model
4. Explain the various techniques used to manage inventory

97

7.2

Definition of Inventory

 It‟s a quantity of items held in storage so as to meet some future demand.
 It‟s a stock of physical assets having some economic value, which can either be in the form of materials, labour or money.
 They are goods, which are procured, stored and used for the day to day functioning of the organisation. Inventory Control are the techniques of maintaining stock items at desired levels i.e. correct quality and quantity.

Inventory decisions
 How much to order?
 When should the order be placed?
 How much safety stock should be kept?

7.3

Reasons for holding stock

 To ensure that sufficient goods are available to meet anticipated demand
 To absorb variations in demand and production
 To provide a buffer between production processes
 To take advantage of bulk purchasing discounts
 To meet possible shortages in the future
 To absorb seasonal fluctuations in usage or demand
 To enable production processes to flow smoothly and efficiently
 As a necessary part of the production process
 As deliberate investment policy particularly in times of inflation or possible shortage.

7.4

Alternative reasons for the existence of stocks

 Obsolete items are retained in stock
 Poor or non existent inventory control resulting in over-large orders, replenishment orders being out of phase with production
 Inadequate or non-existent stock records
98

 Poor liaison between the production control, purchasing and marketing departments
 Sub-optimal decision-making e.g. the production department might increase W-I-P stocks unduly so as to ensure long production runs.

7.5

Inventory / Stock Costs

Whether as a result of deliberate policy or not, stock represents an investment by the organisation. There are four major types of costs associated with holding inventory. They include: Unit costs, Ordering costs, Holding / carrying costs and Shortage / stock – out costs
1. Unit costs: it‟s the price paid for one unit of the commodity under consideration.
2. Ordering costs: includes all the expenses of placing orders. Is assumed to be a fixed cost per order. Examples – clerical and paperwork expenses of purchasing, inspection, receiving, book-keeping and data processing that are directly related to ordering as well as the expenses of delivery, postage and telephone charges.
3. Holding / carrying costs:
 Cost of capital – interest paid on capital invested.
 Storage costs – rental fees, light, security.
 Store keeping operations.
 Insurance and taxes
 Obsolescence and deterioration of the items stored.
4. Shortage / stock – out costs
 Loss of goodwill, potential future sales and loss of reputation.
 Disruption in production, spoilage of idle machines, cost of idle machines, idle labour, spoilage of materials and delays in shipment.

7.6

Factors Affecting Inventory

The following factors affect how much to order and when to order:1. Economic parameters
 Purchase price / production cost
99

 Selling price
 Procurement costs
 Carrying / holding costs
 Shortage / stock-out costs
 Cost of operating the information processing systems.
2. Demand.
3. Ordering cycle i.e. time between two successive placements of orders. Can be continuous or periodic. 4. Delivery lag / lead time i.e. time between the placement of the requisition for an item and its receipt for actual use.
5. Time horizon i.e. planning period over which inventory is to be controlled e.g. annually.
6. Number of supply echelons
7. Number of stages of inventory
8. Number of items which can be affected by the floor space or limited total capital.
9. Availability of items.
10. Governments / company‟s policy e.g. on imports like explosives and inflammables.

7.7

Objective of inventory control

It is to maintain stock levels so that the combined costs are at a minimum. This can be achieved by establishing two factors, when to order and how many to order.

7.8

ECONOMIC ORDER QUANTITY MODEL (EOQ) MODEL

Assumptions:
 The demand (D) for the item is certain, continuous and constant over time.
 The unit costs (CU), the order costs (CO) and holding cost (CH) are known while the stockout costs (CS) are so large that no demand goes unmet.
 The lead time is known and fixed.
 The replenishment is scheduled in such a way that shipments arrive exactly when the inventory level reaches zero i.e. no shortages or surplus.
100

 Orders for different items are independent of each other.

7.9

The derivation of the EOQ model

Total cost  Total unit cost  Total order costs  Total Carrying costs

Total cost  Price per unit * Quantity purchased per year  Cost per order * Number of orders per year
 carrying cost per unit * Average inventory
Let D  Demand per year, Cu  Price per unit, Co  Cost per order, Ch  Carrying cost per unit
TC  CuD  D Co  Q Ch
Q
2
In order to M in TC,

dTC d 2TC
 0,
0
dQ dQ 2

TC  CuD  DQ 1Co  Q Ch
2
dTC
DCo Ch
  DQ  2 Co  1 Ch  0  2 
2
dQ
2
Q

 2DCo  ChQ 2

2 DCo
2 DCo
 Q* 
Ch
Ch
2
d TC 2 DCo

 0 since Q will always be positive dQ 2
Q3

Q2 

Example One
The demand for an item is constant at 45 units per month. The unit cost is Ksh. 100 and the cost per order is Ksh. 120 and a carrying cost is Ksh. 15 per unit per year. Determine the EOQ.

Solution

D  45 * 12  540
EOQ 

2CoD

Ch

2 * 120 * 540
 92.95  93 units
15

Example Two
A manufacturing company has determined from an analysis of its accounting and production data for a certain part that its demand is 9,000 units per annum and is uniformly distributed over the year. Its cost price is Sh. 2 per unit, ordering cost is Sh. 40 per order and the inventory
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carrying charge is 9% of the inventory value. Further, it is known that the lead time is uniform and equals 8 working days and that the total working days in a year are 300. Determine:a)

The EOQ

b)

The optimum number of orders per annum.

c)

The total ordering and holding costs associated with the policy of ordering and amount equal to EOQ.

d)

The re-order level

e)

The number of day‟s stock at re-order level

f)

The length of the inventory cycle

g)

The amount of savings that would be possible by switching to the policy of ordering EOQ determined in (a) from the present policy of ordering the requirements of this part thrice a year and

h)

The increase in the total cost associated with ordering (i) 20% more and (ii) 40% less than the EOQ.

Solution

a)

The EOQ

EOQ 

2CoD

Ch

2 * 40 * 9,000
 2,000 units
0.09 * 2

D 9000

 4.5 .
Q * 2000

b)

The optimum number of orders per annum 

c)

The total ordering and holding costs associated with the policy of ordering and amount equal to EOQ.
Co D

Q

*

 Ch Q

*

2

 9000
  2000


* 40   
* 0.18   360
 2000
  2

9000
 240 units
300

d)

The re-order level = Lead time in days*demand per day  8 *

e)

The number of day‟s stock at re-order level = lead time = 8 days

102

f)

The length of the inventory cycle

Q * 2000

 0.222 year or 0.222 * 300  66.67 days
D
9000
Alternatively



g)

Q*
2000

 66.67 days
Demand per day
30

The amount of savings that would be possible by switching to the policy of ordering EOQ determined in (a) from the present policy of ordering the requirements of this part thrice a year Ordering thrice a year means that Q 

9000
 3000 units
3

 9000
  3000


* 40   
* 0.18   390
Q
 3000
  2

 Savings  390 - 360  Sh. 30 per year
Co D

h)

*

 Ch Q

*

2

The increase in the total cost associated with ordering
 20% more than the EOQ
Q  1.2 * 2000  2400
 9000
  2400


* 40   
* 0.18   366
2  2400
Q
  2

 Savings  366 - 360  Sh. 6 per year
Co D

*

 Ch Q

*

 40% less than the EOQ
Q  0.6 * 2000  1200
 9000
  1200


* 40   
* 0.18   408
2  1200
Q
  2

 Savings  408 - 360  Sh. 48 per year
Co D

*

 Ch Q

*

Example Three
Yogesh keeps his inventory in special containers. Each container occupies 10 sq ft of store space.
Only 5,000 sq ft of the storage space is available. The annual demand for the inventory item is
9,000 containers, priced at Sh. 8 per container. The ordering cost is estimated at Sh. 40 per order

103

and the annual carrying cost amount of 25% of the inventory value. Would you recommend to
Yogesh to increase his storage space? If so, how much should be the increase?

Solution

EOQ 

2CoD

Ch

2 * 40 * 9,000
 600 units
2

With EOQ = 600 units, the maximum level of inventory would be 600 units and the space required to hold it equals 600*10 = 6000 square feet. The present capacity being 5000 square feet, he should increase the storage capacity by 1000 square feet

Example Four
A company uses 8,000 units of a product as raw material, costing Sh. 10 per unit. The administrative cost per purchase is Sh. 40. The holding costs are 28% of the average inventory.
The company is following an optimal purchase policy and places orders according to the EOQ. It has been offered a quantity discount of one per cent if it purchases its entire requirement only four times a year. Should the company accept the offer of quantity discount of one per cent?

Solution
Evaluation of whether the company accept the offer of quantity discount of one per cent

The EOQ is
EOQ 

2CoD

Ch

TC  Cu D  Co D

Q

2 * 40 * 8000
 478 units
0.28 *10

*

 Ch Q

*

2

Total cost without the discount will be : 8000
  478

TC  10 * 8000  
* 40   
*10 * 0.28   81340
 478
  2


Evaluating the total cost with the discount
104

Q

8000
 2000
4

Cu  0.99 *10  9.9

 8000
  2000

TC  9.9 * 8000  
* 40   
* 0.28 * 9.9   82132
 2000
  2

Since the total cost with the discount is greater that without, the company should not accept the offer Example five
A hardware store procures and sells hardware items. Information on an item is given below:Expected annual sales

=

8000 units

Ordering costs

=

Sh. 180 per order

Holding costs

=

10% of the average inventory value

The item can be purchased according to the following schedule
Lot size

Unit price (Sh.)

1 – 999

22

1000 – 1499

20

1500 – 1999

19

2000 and above

18.50

Determine the best order size

Solution
Finding the EOQ at the various price breaks , we‟ve :-

105

2CoD
Ch

EOQ 
Q1* 

2 *180 * 8000
 1248 units
0.1 * 1850

 Not feasible

Q* 
2

2 *180 * 8000
 1231 units
0.1 *19

 Not feasible

Q* 
3

2 * 180 * 8000
 1200 units
0.1 * 20

 Feasible

Evaluating the total cost, we have:-

TC  Cu D  C o D

Q*

 Ch Q

*

2

 8000
  1200

TC(1200)  20 * 8000  
* 180   
* 0.1 * 20   162,400
 1200
  2

 8000
  1500

TC(1500)  19 * 8000  
*180   
* 0.1 *19   154,385
 1500
  2

 8000
  2000

TC( 2000)  18.50 * 8000  
*180   
* 0.1 *18.50   150,570
 2000
  2


Since the total cost is minimum at 2000 units, the best order size is 2000 units

Example six
An aircraft uses high tensile bolts at an approximately constant rate of 50,000 units per year. The bolts cost Sh. 20 each and the purchase department estimated the cost at Sh. 200 to place an order. The holding cost per unit is 20% of the unit cost. No shortages are allowed.
Required:
i.

What is the economic order quantity?

ii.

How frequently should the orders be placed?

106

iii.

If orders could be executed only once in two months, the ordering quantity would be higher than the optimal quantity. By this what would be the percentage change in the total relevant cost? iv.

The company finds, to its error, that the cost of placing an order was Sh. 5,000 and carrying cost was 15% per year and not the earlier data. How much was the company losing per year on inventory because of imperfect information?

v.

Working on the new ordering cost and carrying cost, the company receives the following offer from the supplier:
Quantity

Price per unit

Up to 20,000 pieces per order

20.00

Above 20,000 pieces up to 30,000

19.50

Above 30,000 pieces up to 45,000

19.25

Above 45,000

19.00

Should they make use of this offer? vi. If the entire requirements have to be bought in a single order, what should be the justifiable unit price offer to the company?

Solution
i.

the economic order quantity
EOQ =

ii.

2CoD

Ch

2 * 20 * 50,000
 2,236 units
0.2 * 200

How frequently should the orders be placed?
Length of the order cycle =

iii.

EOQ 2,236

 0.045 years  16 days
D
50,000

If orders could be executed only once in two months, the ordering quantity would be higher than the optimal quantity. By this what would be the percentage change in the total relevant cost? Number of orders = 6
Order quantity =

50,000
 8,333 units
6

107

Total relevant cost = ordering cost + Holding cost  (6 * 200) 
Total

relevant

cost

under

policy

8,333
* 4  17,866
2

of

ordering

EOQ

 2Co DCh  2 * 200 * 50,000 * 4  8,944
% change in the total relevant cost =

iv.



17,866  8,944
*100  99.75%
8,944

The company finds, to its error, that the cost of placing an order was Sh. 5,000 and carrying cost was 15% per year and not the earlier data. How much was the company losing per year on inventory because of imperfect information?(4 marks)
With the revised values the EOQ=

2CoD

Ch

2 * 5,000 * 50,000
 12,910 units
0.15 * 200

Total relevant cost  2Co DCh  2 * 5,000 * 50,000 * 3  38,730
Actual total cost under policy in (i) above = 

50,000
2236
* 5,000 
* 0.15 * 20  115,161
2236
2

Therefore loss on account of imperfect information = 115,161 – 38,730 = 76,431.

v.

Working on the new ordering cost and carrying cost, the company receives the following offer from the supplier:
Quantity

Price per unit

Up to 20,000 pieces per order

20.00

Above 20,000 pieces up to 30,000

19.50

Above 30,000 pieces up to 45,000

19.25

Above 45,000

19.00

Should they make use of this offer?
From (iv) it is clear that EOQ = 12,910 units s feasible with a cost price of Sh. 20 per unit.
Since the price break is available at a quantity of 20,000 only, it is evident that lower prices would not yield feasible results. Therefore we determine the total cost of EOQ = 12910 and at each price break.

108

TC (12,910)  50,000 * 20 

50,000
12,910
* 5,000 
* 0.15 * 20  1,038,730
12,910
2

 50,000
  20,000

TC (20,000)  50,000 *19.5  
* 5,000   
* 0.15 *19.5   1,016,750

 20,000
  2
 50,000
  30,000

TC (30,000)  50,000 *19.25  
* 5,000   
* 0.15 *19.25   1,014,146

 30,000
  2
 50,000
  45,000

TC (45,000)  50,000 *19.00  
* 5,000   
* 0.15 *19.00   1,019,681

 45,000
  2
Since the total cost is lowest at quantity of 30,000 units, the order should be placed for a quantity just above 30,000.

vi.

If the entire requirements have to be bought in a single order, what should be the justifiable unit price offer to the company?
If the requirement has to be bought in a single lot, the price may be obtained by setting the total cost with that price equal to the total cost which is available with the offer in (v) above. Thus with the unit price set to x
50,000
50,000
* 5,000 
* 0.15 x  1,014,146
50,000
2
53,750 x  1,009,146
 50,000 x 

x

1,009,146
 18.77
53,750

Therefore a unit price offer of 18.77 would be justifiable as an alternative to the proposal given.

7.10 EOQ model for production runs
In the classical EOQ model, it is assumed that the entire quantity of the item ordered for is received in a single lot. However, there are cases when the goods are received for inventory at a constant rate over time and they are also being consumed at a constant rate. This is particularly relevant for situations when the item in question is being produced internally rather than being procured from external suppliers. When the production begins, a constant number of units are
109

supposed to be added to the inventory each day till the time the production run is completed.
Simultaneously, the items would be demanded at a constant rate. The rate at which they are produced has to be higher than the rate of their depletion, for inventory to build up.

7.11 Inventory model with planned shortages
In general inventory situations, a shortage or stock out is considered undesirable and is avoided, if possible. This is because shortages may and are likely to mean loss of customer good will, reduction in future orders, unfavourable change in the market share etc. while in some situations the customers shift to other sources for their requirements and so may be lost forever, in some others the customers may not withdraw the orders and wait until the next shipment arrives (back ordering situation). The EOQ model assumes that the inventory is replenished precisely when the inventory level falls off to zero. With the assumption of back ordering, shortages may in fact be deliberately planned to occur. It may be advisable on economic considerations, especially when the value of the item in question is very high with consequent high holding cost. It is a question of setting off the cost of shortage against the saving in the holding cost.

Example One
A manufacturing company needs 2500 units of a particular component every year. The company buys it at the rate of Sh. 30 per unit. The order processing cost for this part is estimated at Sh. 15 and the cost of carrying a part in stock comes to about Sh.4 per year. The company can manufacture this part internally. In that case, it saves 20% of the price of the product. However, it estimates a set-up cost of Sh. 250 per production run. The annual production rate would be 4800 units. However, the inventory holding costs remain unchanged. i. Determine the EOQ and the optimal number of orders placed in a year. ii. Determine the optimum production lot size and the average duration of the production run. iii. Should the company manufacture the component internally or continue to purchase it from the supplier?

Solution
110

i. Determine the EOQ and the optimal number of orders placed in a year.

D  2500, A  15, h  4
EOQ 

2AD

h

2 *15 * 2500
 137 units
4

Optimal number of orders

D
2500

 18
EOQ 137

ii. Determine the optimum production lot size and the average duration of the production run. D  2500, S  250 per set up, h  4, p  4800, d  2500
Economic Lot Size(ELS) 

2DS h p

pd

Average duration of the production run 

2 * 2500 * 250
4800
 808 units
4
4800  2500

ELS
808

 0.32 year
Demand 2500

iii. When the item is purchased from outside
TC  DC 

D
EOQ
* A
*h
EOQ
2

 2500
  137 
TC  2500 * 30  
*15   
* 4   75548
 137
  2


When item is produced internally
Cost per unit  0.8 * 30  24
Set up cost = 250 per set up

TC  DC 

D
ELS p  d
*S 
*
*h
ELS
2 p  2500
  808 2300 
TC  2500 * 24  
* 250   
*
* 4
 808
  2 4800 
TC  61548

The company should manufacture the product internally

Example Two

111

The annual demand for a component Z is 208,000 units at a steady weekly rate of 4,000 units.
An appropriate formula for calculating the economic batch quantity for production of a component which is used (at a rate of s and produced (at a rate r of per week) at the same time is EBQ 

2 AC o
. The cost of installing the line of producing Z was Sh. 6,000 for a
(1  s / r )ip

maximum production capacity of 8,000 per week. The operating costs at full capacity are Sh. 100 per week for labour, Sh.600 per week for material, Sh. 300 per week for variable overhead and
Sh. 250 per week for fixed overhead. The cost of preparing the production order, producing drawings and so on is Sh. 40 each time production is required. Storage costs including interest have been calculated as Sh. 2 per unit per annum.
(a) Calculate the most economic quantity that should be produced each time the line is set up.
(b) Advise the management if it now thinks that there is an opportunity to produce a special one-off order for 50,000 Z‟s delivery in six months‟ time.

Solution
(a) Set up cost = 40, storage cost = Sh. 2 per unit year
The other costs are not relevant to determine the batch size as they are not related to and do not affect it
Annual demand (A) = 208,000
Production Rate (r) = 8000 units / week
Usage rate (s) = 4000 units per week

EBQ 

2 * 208000 * 40
 4079 units
 4000 
1 
*2
 8000 

Therefore the production pattern would be to have a production run each week to produce the weekly requirement

(b) At present the variable costs associated with this component are:Labour

100

Material

600
112

Variable overhead

300
1000 per week

The maximum capacity being 8000 units per week, the cost per unit at this level =

 1000

8000

 0.125 . However, the storage costs is Sh. 2 per unit per year which is excessively

high in relation to the unit production cost. As such, the production should be deferred. With the extra capacity of 4000 units per week over and above the current level of production the extra order would take

50000
 12.5 weeks to produce so that the production of this special order
4000

should be delayed for 26 – 12.5 = 13.5 weeks

7.12 Selective Approaches to Inventory Control
All items of inventory cannot and need not be controlled with equal attention. some approaches to inventory control include:

(a) ABC analysis
The inventory items in an organization are classified on the basis of their usage in monetary terms. It is very common to observe that usually a small number of items account for a large share of the total cost of materials and a comparatively large number of items involve an insignificant share. Based on the criterion, the items are divided into three categories as follows:A: High consumption value items
B: Moderate consumption value items and
C: Low consumption value items
The division of the items into various categories is accomplished by plotting the usage value of items to obtain the ABC distribution curve which is also called the Pareto curve. The mechanics of the ABC analysis is given in a step wise manner:Step 1: Obtain a list of items along with information on their unit cost and the periodic (Annual) consumption. Step 2: Determine the annual usage for each of the items by multiplying unit cost with number of units and rank them in descending order on the basis of their respective usage values.

113

Step 3: express the value for each item as a percentage of the aggregate usage value, then cumulate the percent of annual usage values.
Step 4: Obtain the percentage value for each of the items. For n items, each item shall represent

100 per cent. n Step 5: Using the data on cumulated values of items and the cumulated percentage usage values, plot the curve by showing these respectively on X and Y axes.
Step 6: Determine appropriate divisions for the A, B and C categories. The curve would rise steeply up to a point. This point is marked and the items up to that point constitute the type A items. Similarly, the curve would only be moderately sloped towards upright. The point beyond which the slope is negligible is marked and the items covered beyond that point are classed as C type items because they cause only a negligible increase in the cost. The other items are the B type items for which the curve depicts a gradual upward rise. After the items are so classified, the inventory decisions are made on the basis of this classification. “A” items would call for a strict control and should be delivered near the time of use. The protection against their stock outs may be set as high as 98%. The “C” items might be kept in open storage available and as they are demanded they might be issued without formalities. For these, the re-order quantities might be larger than their respective EOQ‟s. Periodic review system may be invoked for the procurement of such items. In regard to the class B items, the policy would be of a fairly tight control. Example
Perform ABC analysis using the following data
Item

Units

Unit price (Sh.) Item

Units

Unit price (Sh.)

1

700

5

7

6000

0.20

2

2400

3

8

300

3.50

3

150

10

9

30

8.00

4

60

22

10

2900

0.40

5

3800

1.50

11

1150

7.10

6

4000

0.50

12

410

6.20

(b) VED analysis
The items are classified on the basis of their criticality to the production process or other services. In the VED classification of materials, V stands for vital items without which the
114

production process would come to a stand still. E in the system denotes essential items whose stockout would adversely affect the efficiency of the production system. Although the system would not altogether stop for want of these items, yet their non-availability might cause temporary losses in, or dislocation of production. The D items are the desirable items which are required but do not immediately cause a loss of production. VED analysis is done mainly in respect of spare parts.
(c) HML analysis
This is similar to the ABC analysis except that in this analysis, the items are classified on the basis of unit cost rather than their usage value. The items are classified accordingly as their cost per unit is H-High, M-Medium and L-Low. This analysis is useful for keeping control over materials consumption at the departmental level.
(d) SDE analysis
This uses the criterion of the availability of items. S stands for Scarce items which are in short supply, D refers to difficult items (items that might be available in the indigenous market but cannot be procured easily), while E represents easily available items from the local markets.
(e) S-OS analysis
This is based on the nature of supplies wherein S represents the Seasonal items and OS the Offseasonal items. This classification of items is done with the aim of determining proper procurement strategies.
(f) FSN analysis
Based on the consumption pattern of the items, the FSN classification of items as fast moving, slow moving and Non-moving. This speed classification helps in the arrangement of stocks in the stores and in determining the distribution and handling patterns.
(g) XYZ analysis
XYZ analysis is based on the closing inventory value of different items. Items whose inventory values are high are classified as X items while those with low investment in them are termed as Z items. Other items are the Y items whose inventory value is neither too high nor too low.

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7.13 Activities

1. The city council of Nairobi uses 100 replacement lamps a month for its street lights. Each lamp costs the city Ksh. 8. Ordering costs are estimated at Sh. 27 per order and the holding costs are 25%. The city council currently orders according to the EOQ. The supplier has now offered the city a 2% discount if the city will buy 600 lamps at a time. Should the city council accept the offer?

2. Given an inventory system where
Yearly demand = 120 units
Ordering cost = Sh. 45
Price per unit = Sh. 200
Annual carrying cost = 24% of the unit cost
i.

Determine The economic order Quantity

ii.

The supplier offers a 1% discount on the unit price if the items are purchased in lots of 100 at a time. Should the management accept the offer?

iii.

Determine the minimum percentage discount that will make the offer attractive?

3. Kenyatta university health unit buys a certain antibiotic from a large supplier. The drug can be bought at the following prices:
Quantity

Price per unit (Sh.)

1 – 4,999

2.75

5,000 – 9,999

2.60

Over 10,000 units

2.50

The demand for the drug in the health unit is 50,000 units a year. There is an ordering cost of Sh. 50 per order and a holding cost of 20% of the cost of the item per unit per year. Find the optimal purchasing policy for the hospital.

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4. ABC departmental store sells 25,000 type A shirts a year. The supplier offers a generous quantity discount. The price list is given below: Quantity

Price per shirt ($)

0 –999

2.50

1,000 – 1749

2.00

1750 – 2, 499

1.50

Over 2,500

1.00

Given that order cost is $ 20. Inventory carrying cost is 20% of the value of the item.
Determine the best inventory policy for ABC.
5. The purchasing manager of a distillery company is considering three sources of supply for oak barrels. The first supplier offers any quantity of barrels at Sh. 150 each. The second supplier offers barrels in lots of 150 or more at Sh. 125 per barrel. The third supplier offers barrels in lots of 250 or more at Sh. 100 each. The distillery uses 1,500 barrels a year at constant rate. Carrying costs are 40 per cent and it costs the purchasing agent Sh. 400 to place an order. Calculate the total annual cost for the orders placed to the probable suppliers and find out the supplier to whom orders should be placed.
2. A dealer supplies you the following information pertaining to an item of inventory:
Annual demand

800 units

Buying cost

Sh. 150 per order

Inventory carrying cost

Sh. 3 per unit per year

Back ordering cost

Sh. 20 per unit per year

i. What will the optimal number of units of the inventory item he should buy in one lot? ii. What quantity he should allow to be back ordered? iii. What will be the cost savings, if any, resulting from back ordering? iv. What would the maximum inventory of the item at any time of the year?
v. If the dealer wants that no more than 25% of the units can be back ordered, should the policy of back ordering be adopted?

3. The demand for an item is deterministic and constant over time at 600 units per year. The item costs Sh. 50 per unit and the cost of placing an order is estimated to be Sh. 5. The inventory carrying cost is 20% and the shortage cost is Sh. 1 per unit per month. Find the
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optimal ordering quantity if stock outs are permitted and the units can be back ordered at the shortage cost indicated. What will the company lose if no stock outs are permitted?
4. A dealer supplies you the following information with regard to a product dealt in by him.
Annual demand:

10,000 units

Ordering cost:

Sh. 10 per order

Inventory carrying cost:

20% of the value of inventory per year

Price:

Sh. 20 per unit

The dealer is considering the possibility of allowing some back order (stock-out) to occur. He has estimated that the annual cost of backordering will be 25% of the value of inventory.
Required:
i. What should the optimum number of units of the product he should buy in one lot? ii. What quantity of the product should be allowed to be back ordered if any? iii. What should the maximum quantity of inventory at any time of the year? iv. Would you recommend to allow back ordering? If so what would be the annual cost saving by adopting the policy of back ordering?

118

LESSON EIGHT: SAMPLING AND ESTIMATION THEORY
8.0

Introduction

Sampling is that part of statistical practice concerned with the selection of a subset of individuals from within a population to yield some knowledge about the whole population, especially for the purposes of making predictions based on statistical inference.

8.1

Expected Learning Outcomes

By the end of the lesson, the students should be able to;
1. Explain the various reasons for sampling
2. Describe the probabilistic and non-probabilistic sampling techniques
3. Differentiate between point and interval estimators
4. Estimate the confidence intervals for population means and proportions

8.2

Definition of sampling

Sampling is that part of statistical practice concerned with the selection of a subset of individuals from within a population to yield some knowledge about the whole population, especially for the purposes of making predictions based on statistical inference.
Population: It‟s a complete set of individuals, cases or objects with some observable characteristics. A census is a count of all the elements in a population.
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Sample: A sample is a subset of a particular population. The target population is that population to which a researcher wants to generalize the results of the study. There must be a rationale for defining and identifying the accessible population from the target population.
Sampling: It‟s the process of selecting a sample from a population.

8.3

Reasons for sampling

The following are some of the reasons why researchers sample and not take the whole population: Cost
 Time: Greater speed of data collection
 Destructive nature of certain tests
 Greater accuracy of results
 Physical impossibility of checking all items in the population.
 Availability of population elements.

8.4

Sampling Procedures

There are two major ways of selecting samples;
 Probability sampling methods
 Non - Probability sampling methods

Probability Sampling Methods
Samples are selected in such a way that each item or person in the population has a known
(Nonzero) likelihood of being included in the sample.

8.5

Types of Probability sampling methods

1. Simple Random Sampling:
A sample is selected so that each item or person in the population has the same chance of being included.
Advantages
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 Easy to implement with automatic dialing and with computerized voice response systems. Disadvantages
 Requires a listing of population elements.
 Takes more time to implement
 Uses larger sample sizes
 Produces larger errors
 Expensive

2. Systematic Random Sampling:
The items or individuals of the population are arranged in some manner. A random starting point is selected and then every kth member of the population is selected for the sample.
Advantages
 Simple to design
 Easier to use than the simple random.
 Easy to determine sampling distribution of mean or proportion.
 Less expensive than simple random.
Disadvantages
 Periodicity within the population may skew the sample and results.
 If the population list has a monotonic trend, a biased estimate will result based on the start point.

3.

Stratified Random Sampling:
A population is divided into subgroups called strata and a sample is selected from each stratum. After the population is divided into strata, either a proportional or a nonproportional sample can be selected. In a proportional sample, the number of items in each stratum is in the same proportion as in the population while in a non-proportional sample, the number of items chosen in each stratum is disproportionate to the respective numbers in the population.
Advantages
 Researcher controls sample size in strata
121

 Increased statistical efficiency
 Provides data to represent and analyze subgroups.
 Enables use of different methods in strata.
Disadvantages
 Increased error will result if subgroups are selected at different rates
 Expensive especially if strata on the population have to be created.

4. Cluster Sampling:
The population is divided into internally heterogeneous subgroups and some are randomly selected for further study. It is used when it is not possible to obtain a sampling frame because the population is either very large or scattered over a large geographical area. A multi-stage cluster sampling method can also be used.
Advantages
 Provides an unbiased estimate of population parameters if properly done.
 Economically more efficient than simple random.
 Lowest cost per sample, especially with geographic clusters.
 Easy to do without a population list.
Disadvantages
 More error (Lower statistical efficiency) due to subgroups being homogeneous rather the heterogeneous.

8.6

Non - Probability Sampling Methods

It is used when a researcher is not interested in selecting a sample that is representative of the population. 1. Convenience or Accidental Sampling
Involves selecting cases or units of observation as they become available to the researcher
e.g. asking a question to the radio listeners, roommates or neighbours.
2. Purposive Sampling
Allows the researcher to use cases that have the required information with respect to the objectives of his or her study e.g. educational level, age group, religious sect etc.

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3. Quota Sampling
The researcher purposively selects subjects to fit the quotas identified e.g.
 Gender: Male or Female.
 Class Level: Graduate or Undergraduate
 School: Humanities, Science or human resource development.
 Religion: Muslim, Protestant, catholic, Jewish.
 Fraternal affiliation: member or nonmember.
 Social economic class: Upper, middle or lower.
Advantage
 Widely used by pollsters, marketers and other researchers.
Disadvantages
 It gives no assurance that the sample is representative of the variables being studied.
 The data used to provide controls may be outdated or inaccurate.
 There is a practical limit on the number of simultaneous controls that can be applied to ensure precision.
 Since the choice of subjects is left to field workers, they may choose only friendly looking people.
4. Snow ball sampling
It is used when the population that possesses the characteristics under study is not well known and can be best located through referral networks. Initial subjects are identified who in turn identify others. Commonly used in drug cultures, teenage gang activities,
Mungiki sect, insider trading, Mau Mau etc.

8.7

Estimation Theory

Sample information is used to shade some light on the population characteristics i.e. we infer population properties based on findings on the sample. Statistical inference falls into two main areas i.e. statistical estimation and hypothesis testing.
Statistical Estimation: The characteristics of the sample (sample statistic) are used to estimate or approximate some unknown population characteristics.
Hypothesis testing: The population characteristics are known or assumed. The sample characteristics are used to verify or ascertain this assumed or known population characteristic.
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The assignment of values to a population parameter is based on a sample is called estimation.
The values assigned to a population parameter based on the value of a sample statistic is called an estimate of the population parameter. The sample statistic used to estimate a population parameter is called an estimator.
Estimation can be undertaken in two forms namely:
 Point estimation
 Interval estimation
Point estimation: A single value is calculated from sample data to estimate the unknown population parameter. The value of a sample statistic that is used to estimate a population parameter is called a point estimate and the procedure is called point estimation. Thus a sample mean ( x ) can be used as a point estimate of the population mean  .
Interval estimation: a point estimate may be right or wrong. It does not indicate the degree of error or uncertainty or precision in the estimate. In estimating business parameters an error is an obvious expectation. Therefore, uncertainty should be accommodated in business estimations.
Interval estimation includes the degree of error or uncertainty in the estimate. In interval estimation, an interval is constructed around a point estimate and it is stated that this interval is likely to contain the corresponding population parameter. Each interval is constructed with regard to a given confidence level and is called a confidence interval.

8.8

Confidence interval for the population mean

This is computed by

X Z

n or

X Z S

n Z depends on the level of confidence.

The standard error of the mean when:

population standard deviation known is given by  X 



based on the sample standard deviation is given by

124

 n SX 

S n Examples
1. A study involves selecting a random sample of 256 sales representatives under the age of 35.
One item of interest is their annual income. The sample mean is Sh. 55,420 and the sample standard deviation is 2050.
i.

What is the point estimate?

ii.

What is the 95% confidence interval for the population mean rounded to the nearest Sh.
10?

Solution
i. The point estimate is 55,420 ii. What is the 95% confidence interval for the population mean rounded to the nearest Sh. 10?

  X Z S

n

  55420  1.96 2050


  55420  251.125



256 

55170    55670

2. A bank wishing to determine the average amount of time a customer must wait to be served took a random sample of 100 customers and found that the mean waiting time was 7.2 minutes. Assuming that the population standard deviation is known to be 15 minutes, find the
90% confidence interval estimate of the mean waiting time for all the bank‟s customers.

Solution

  X Z

n

  7.2  1.64515


  7.2  2.4675



100 

4.7325    9.6675

125

The width of a confidence interval depends on the size of the maximum error which depends on the values of z,  and n . However the value of  is not within the control of the researcher, hence the width of CI depends on:
i.

The value of Z which depends on the confidence level

ii.

The sample size n

The confidence level determines the value of Z which in turn determines the size of the maximum error. Z value increases as the CL increases and vice versa. The higher the CL, the wider / larger the width of the confidence interval, other things remaining the same. An increase in the sample size decreases the width of the confidence interval. Therefore, if we want to decrease the width of the confidence interval, there are two choices:i.

Lower the confidence level

ii.

Increase the sample size

However, lowering the confidence level is not a good choice because a lower confidence may give less reliable results. Therefore, we prefer to increase the sample size if we wish to decrease the width of a confidence interval.

8.9

Confidence interval for a population proportion

A point estimate for the population proportion is found by dividing the number of successes in the sample by the number sampled.

ˆ
The confidence interval for a sample proportion is p  Z

ˆ p Where

ˆ
ˆ
p(1  p) n sample proportion

Z

Standard normal value for the degree of confidence selected

n

is the sample size

Examples

1.

A market survey was conducted to estimate the proportion of homemakers who would recognize the brand name of a cleanser based on the shape and colour of the container. Of
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the 1400 homemakers, 420 were able to identify the brand name. Using the 99% degree of confidence, calculate the confidence interval of the population proportion.
Solution
ˆ p 420
 0.3
1400

P  0.3  2.575

0.3 * 0.7
1400

P  0.3  0.032
0.268  P  0.332

2.

Suppose the Nation TV network is considering replacing one of its prime time dramas with a new family oriented comedy. Before a final decision is made a random sample of 400 prime time viewers is conducted. After seeing a preview of the comedy, 250 indicated that they would watch it.
i.

What is the point estimate of the proportion of viewers in the population who will watch the new show?

ii.

Develop a 95% confidence interval for the proportion of viewers who will watch the new show.

Solution
i.

What is the point estimate of the proportion of viewers in the population who will watch the new show?

ˆ p ii.

250
 0.625
400

Develop a 95% confidence interval for the proportion of viewers who will watch the new show.
P  0.625  1.96

0.625 * 0.375
400

P  0.625  0.047
0.578  P  0.672

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8.10 Confidence interval of the difference between two population means
( 1   2 )
1   2  ( x1  x2 )  Z 

2

2

2

S1
S
 2 n1 n2

Example
A manufacturing company operates two factories A and B. The company wishes to establish the difference between the average daily production levels of the two factories. The following data was obtained for this purpose:Factory

xi

Si

ni

A

2600

182

38

B

1980

150

32

Estimate 1   2 at : i.

90% Confidence level

ii.

95% Confidence level

iii.

99% Confidence level

Solution
i.

90% Confidence level

1   2  ( x1  x 2 )  Z 

2

2

2

S1
S
 2 n1 n2

1   2  2600  1980  1.645

182 2 150 2

38
32

1   2  620  65.28
554.72  1   2  685.28 ii. 95% Confidence level

128

1   2  ( x1  x 2 )  Z 

2

2

2

S1
S
 2 n1 n2

182 2 150 2
1   2  2600  1980  1.96

38
32
1   2  620  77.78
542.22  1   2  697.78 iii. 99% Confidence level

1   2  ( x1  x 2 )  Z 

2

2

2

S1
S
 2 n1 n2

1   2  2600  1980  2.575

182 2 150 2

38
32

1   2  620  102.19
517.81  1   2  722.19

8.11 Confidence interval of the difference between two population proportions ˆ
ˆ
P1  P2  ( p1  p 2 )  Z

ˆ ˆ
ˆ ˆ p1q1 p 2 q 2

n1 n2 Example
In a public opinion survey, 60 out of a sample of 100 high-income voters and 40 out of a sample of 75 low-income voters supported a decrease in sales tax. Estimate the differences in proportions with:i.

90% Confidence level

ii.

95% Confidence level

iii.

99% Confidence level

Solution

129

60
 0.6
100
40
ˆ
p2 
 0.53
75
ˆ p1 

i.

90% Confidence level

ˆ
ˆ
P1  P2  ( p1  p 2 )  Z

ˆ ˆ
ˆ ˆ p1 q1 p 2 q 2

n1 n2 P1  P2  0.6  0.53  1.645

0.6 * 0.4 0.53 * 0.47

100
75

P1  P2  0.07  0.124
 0.054  P1  P2  0.194

ii.

95% Confidence level

ˆ
ˆ
P1  P2  ( p1  p 2 )  Z

ˆ ˆ
ˆ ˆ p1 q1 p 2 q 2

n1 n2 P1  P2  0.6  0.53  1.96

0.6 * 0.4 0.53 * 0.47

100
75

P1  P2  0.07  0.148
 0.078  P1  P2  0.218

iii.

99% Confidence level

ˆ
ˆ
P1  P2  ( p1  p 2 )  Z

ˆ ˆ
ˆ ˆ p1 q1 p 2 q 2

n1 n2 P1  P2  0.6  0.53  2.575

0.6 * 0.4 0.53 * 0.47

100
75

P1  P2  0.07  0.195
 0.125  P1  P2  0.265

8.12 Qualities of a good estimator
1. Unbiased: an unbiased estimator of a population parameter is an estimator whose expected value is equal to that parameter i.e. if you were to take an infinite number of samples,
130

calculate the value of the estimator in each sample, and then average these values, the average value would equal the parameter.
2. Consistency: an unbiased estimator is said to be consistent if the difference between the estimator and the parameter grows smaller as the sample size grows large.
3. Should be efficient: an efficient estimator should have the least variance or least standard error. 4. Should be sufficient: an estimator is said to be sufficient if it extracts from the sample such an amount of information as no other estimator does.

8.13 Selecting the sample size to estimate a population mean
One of the most common questions asked of statisticians is, how large should the sample taken in a survey be? The answer to this question depends on three factors:i.

The parameter to be estimated

ii.

The desired confidence level of the interval estimator

iii.

The maximum error of estimation, where error of estimation is the absolute difference between the point estimator and the parameter e.g. the point estimator of  is x so that the error of estimation  x  

The maximum error of estimation is also called the error bound and is denoted B.
Suppose the parameter of interest in an experiment is the population mean  . The confidence interval estimator (assuming a normal population, with the population variance known) is

x  z / 2

 n . If we want to estimate  to within a certain specified bound B, we will want the

confidence interval estimator to be x  B . As a consequence, we have z / 2

z   n , we get the following result n    / 2 
 B 

2

131

 n  B . Solving for

A popular method of approximating  is to begin by approximating the range of the random variable. A conservative estimate of  is the range divided by 4 i.e.  

Range
. This produces
4

a larger value of  , which results in a larger value of n , which then estimates  with an interval at least as good as was specified.

Examples
1. A production manager would like to estimate the mean time required for workers to complete a task on an assembly line. Assume that she knows that  is 80 seconds. How large a sample should she draw to estimate  to within 5 seconds with (i) 90% confidence (ii) 95% confidence (iii) 99% confidence
90% confidence

z  
1.645 * 80  n    /2   
  692.7424  693
5


 B 
2

2

95% confidence

z  
1.96 * 80  n    /2   
  983.4496  984
5


 B 
2

2

99% confidence

z  
 2.575 * 80  n    /2   
  1697.44  1698
5


 B 
2

2

2. Find n , given that we want to estimate  to within 10 units with 95% confidence, assuming that   100

z  
1.96 *100  n    /2   
  384.16  385
 10

 B 
2

2

3. The operations manager of a large production plant would like to estimate the average amount of time a worker takes to assemble a new electronic component. After observing a
132

number of workers assembling similar devices, she noted that the shortest time taken was 10 minutes and the longest time taken was 22 minutes. How large a sample of workers should she take if she wants to estimate the mean assembly time to within 20 seconds? Assume that the confidence level is to be 99%.



Range
22  10

 3 * 60  180 seconds
4
4

z  
 2.575 * 180  n    /2   
  537.08  538
20


 B 
2

2

8.14 Selecting the sample size to estimate a population proportion
z
ˆˆ pq 
The sample size necessary to estimate p is n    / 2

B





2

1. The manager of a bank feels that 35% of branches will have enhanced yearly collection of deposits after introducing a hike in interest rate. Determine the sample size such that the mean proportion is within plus or minus 0.06 at a confidence level of (i) 90% (ii) 95% and
(iii) 99%.
Solution
90% confidence level
2

z
1.645 0.35 * 0.65 
ˆˆ
pq  n    /2
 
  171
B
0.06






2

95% confidence level
2

z
1.96 0.35 * 0.65 
ˆˆ
pq  n    /2
 
  243
B
0.06






2

99% confidence level

133

2

z
 2.575 0.35 * 0.65 
ˆˆ
pq  n    /2
 
  419
B
0.06






2

2. How large a sample should be taken in order to estimate p to within 0.01 with 95% confidence? assume that
a) You have no information about the value of p
b) p is believed to be approximately 0.10
c)

p is believed to be approximately 0.90

Solution
a) You have no information about the value of p
2

z
1.96 0.5 * 0.5 
ˆˆ
pq  n    /2
 
  9604
B
0.01






2

b) p is believed to be approximately 0.10
2

z
1.96 0.1 * 0.9 
ˆˆ
pq  n    /2
 
  3458
B
0.01






c)

2

p is believed to be approximately 0.90
2

z
1.96 0.9 * 0.1 
ˆˆ
pq  n    /2
 
  3458
B
0.01






2

8.15 Exercises

134

1.

The results from a sample taken n  25,  X i  500,  X i2  12400 , find the point estimates of the following:
(i) The mean 
(ii) The variance  2
(iii)The standard deviation
(iv) The standard error of the sample mean

2.

Determine the sample size necessary to estimate  to within 10 units with 99% confidence.
We know that the range of the population is 200 units.

3.

The director of a management school feels that 55% of students will have enhanced performance if additional input is given to them. Determine the sample size such that the mean proportion is within plus or minus 0.10 at a confidence level of 95%.

135

LESSON NINE: HYPOTHESIS TESTING
9.0

Introduction

This lesson gives an overview of the concepts in hypothesis testing. It describes the procedure of testing a hypothesis, differentiates between one-tailed and twotailed tests and type I and Type II errors. Examples of testing hypothesis about a single population mean when the population variance is given and not given are discussed.

9.1

Expected Learning Outcomes

By the end of the lesson, the students should be able to;
 Define the term hypothesis
 Differentiate between one-tailed and two-tailed tests
 Describe the procedure for testing hypothesis
 Test hypothesis about the mean when the population variance is known


9.2

Test hypothesis about the mean when the population variance is unknown

Definition of Hypothesis Testing

Hypothesis: It‟s a statement about a population parameter developed for the purpose of testing.
Hypothesis testing: It‟s a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.

136

9.3

Procedure for testing a hypothesis

The following are the steps that are followed when testing hypothesis
1.

State the null and alternate hypothesis

2.

Select a level of significance.

3.

Identify the test statistic

4.

Formulate a decision rule and identify the rejection region

5.

Compute the value of the test statistic

6.

Make a conclusion.

State the null hypothesis (HO) and alternate hypothesis (HA)
 The null hypothesis is a statement about the value of a population parameter. It should be stated as “ There is no significant difference between ……………”. It should always contain an equal sign.
 The alternate hypothesis is a statement that is accepted if sample data provide enough evidence that the null hypothesis is false.

Select a level of significance
A level of significance is the probability of rejecting the null hypothesis when it is true. It is designated by  and should be between 0 –1.
Types of errors that can be committed
i.

Type I error: it is rejecting the null hypothesis, when it is true.

ii.

Type II error: It is not rejecting the null hypothesis, when it is false.
Null hypothesis

Do not reject HO

Reject HO

HO is True

Correct decision

Type I error

HO is false

Type II error

Correct decision

Identify the test statistic

137

A test

statistic is

the statistic that will be used to

test the hypothesis

e.g.

, , Fand 2 (chi  square)

Formulate a decision rule
A decision rule is a statement of the conditions under which the null hypothesis is rejected and the conditions under which it is not rejected. The region or area of rejection defines the location of all those values that are so large or so small that the probability of their occurrence under a true null hypothesis is rather remote.
Compute the value of the test statistic and make a conclusion.
The value of the test statistic is determined from the sample information, and is used to determine whether to reject the null hypothesis or not.

9.4

One-tailed and Two-tailed tests

 A test is one tailed when the alternate hypothesis states a direction e.g.
Ho:

The mean income of women is equal to the mean income of men

HA: The mean income of women is greater than the mean income of men
 A test is two tailed if no direction is specified in the alternate hypothesis
Ho:

There is no difference between the mean income of women and the mean income of men

HA:

There is a difference between the mean income of women and the mean income of men

9.5

Testing the Population Mean When the Population Variance Is Known

When the population variance is known and the population is normally distributed, the test statistic for testing hypothesis about  is Z 

x

 n when  2 is known is

x  Z 
2

n

138

. The confidence interval estimator of 

Example One
A study by the Coca-Cola Company showed that the typical adult Kenyan consumes 18 gallons of Coca-Cola each year. According to the same survey, the standard deviation of the number of gallons consumed is 3.0. A random sample of 64 college students showed they consumed an average (mean) of 17 gallons of cola last year. At the 0.05 significance level, can we conclude that there is a significance difference between the mean consumption rate of college students and other adults?
Solution
1.

Stating the null and alternate hypothesis

H 0 :   18
H A :   18
2.

Level of significance:   0.05

3.

Test statistic Z 

X 

 n 4.

Rejection region
Z  / 2  Z 0.025  1.96

5.

Value of the test statistic

Zc 

X 

 n 6.

If Z c  1.96 or Zc  1.96, Reject H o



17  18
 2.67  1.96
3
64

Conclusion

Reject H0. Yes, there is a significance difference between the mean consumption rate of college students and other adults

Example Two

The manager of a departmental store is thinking about establishing a new billing system for the stores credit customers. After a thorough financial analysis, she determines that the new system
139

will not be cost effective if the average monthly account is less than 70,000. A random sample of
200 monthly accounts is drawn, for which the mean monthly account is Sh. 66,000. With  =
0.05, is there sufficient evidence to conclude that the new system will not be cost effective?
Assume that the population standard deviation is Sh. 30,000.

Solution
1.

Stating the null and alternate hypothesis

H 0 :   70,000
H A :   70,000
2.

Level of significance:   0.05

3.

Test statistic Z 

X 

 n 4.

Rejection region
Z   Z 0.05  1.645

5.

Value of the test statistic
Zc 

X 

 n 6.

If Z c  1.645, Reject H o



66000  70000
 1.89  1.645
30000
200

Conclusion

Reject H0. Yes, there is sufficient evidence to conclude that the new system will not be cost effective Example Three
Past experience indicates that the monthly long distance telephone bill per household in a particular community is normally distributed, with a mean of Sh. 1012 and a standard deviation of Sh. 327. After an advertising campaign that encouraged people to make long distance telephone calls more frequently, a random sample of 57 households revealed that the mean

140

monthly long distance bill was Sh. 1098. Can we conclude at the 10% significance level that the advertising campaign was successful?
Solution
1.

Stating the null and alternate hypothesis

H 0 :   1012
H A :   1012
2.

Level of significance:   0.1

3.

Test statistic Z 

X 

 n 4.

Rejection region
Z   Z 0.1  1.28

5.

Value of the test statistic
Zc 

X 

 n 6.

If Zc  1.28, Reject H o



1098  1012
 1.99  1.28
327
57

Conclusion
Reject H0. Yes, there is sufficient evidence to conclude that the advertising campaign was successful 9.6

Testing the Population Mean when the Population Variance is Unknown

When the population variance is unknown and the population is normally distributed, the test statistic for testing hypothesis about  is t 

x which has a student t distribution with s n

n  1 degrees of freedom.

141

We now have two different test statistic for testing the population mean. The choice of which one to use depends on whether or not the population variance is known.
 If the population variance is known, the test statistic is Z 

x

 n  If the population variance is unknown, the test statistic is t 

The confidence interval estimator of  when  2 is unknown is

x s n

x  t

d. f  n  1

s
2

n

d. f .  n  1

Example One
A manufacturer of automobile seats has a production line that produces an average of 100 seats per day. Because of new government regulations, a new safety device has been installed, which the manufacturer believes will reduce average daily output. A random sample of 15 days output after the installation of the safety device is shown below:
93, 103, 95, 101, 91, 105, 96, 94, 101, 88, 98, 94, 101, 92, 95
Assuming that the daily output is normally distributed, is there sufficient evidence at the 5% significance level, to conclude that average daily output has decreased following the installation of the safety device?
Solution
1.

Stating the null and alternate hypothesis

H 0 :   100
H A :   100
2.

Level of significance:   0.05

3.

Test statistic t 

4.

Rejection region

X  s n

t , n1  t 0.05,14  1.761

5.

If t c  1.761, Reject H o

Value of the test statistic
142

 X  1447  X
X 

X n 

 139917

2

1447
 96.47
15

 X 


2

S tc 
6.

X

2

n 1

n

1447 2
139917 
15  4.85

14

X   96.47  100

 2.82  1.761 s 4.85 n 15

Conclusion
Reject H0. Yes, there is sufficient evidence to conclude that average daily output has decreased following the installation of the safety device

Example Two
A courier service advertises that its average delivery time is less than six hours for local deliveries. A random sample of the amount of time this courier takes to deliver packages to an address across town produced the following times (rounded to the nearest hour).
7, 3, 4, 6, 10, 5, 6, 4, 3, 8
Is there sufficient evidence to support the courier‟s advertisement at the 5% level of significance? Solution
1.

Stating the null and alternate hypothesis

H0 :   6
HA :  6
2.

Level of significance:   0.05

3.

Test statistic t 

4.

Rejection region

X  s n

143

t , n1  t 0.05,9  1.833

5.

Value of the test statistic

 X  56  X
X 

X n 

2

 360

56
 5.6
10

 X 


2

S tc 

6.

If t c  1.833, Reject H o

X

2

n 1

n



56 2
360 
10  2.27
9

X 
5.6  6

 0.56  1.833 s 2.27 n 10

Conclusion
Do not Reject H0. No, there is no sufficient evidence to conclude that the advertising campaign was successful

9.7

Testing the population proportion

ˆ
The sample proportion p is a approximately normally distributed, with mean p and standard deviation pq
ˆ
, provided that n is large ( np  5 and nq  5 ). Since p is approximately normal, n it follows that the standardized variable Z 

ˆ p p pq n

is approximately standard normally

distributed. The null and alternate hypotheses of tests of proportions are set up in the same way as the hypothesis of tests about mean and variance. The test statistic for p is Z 

ˆ
Confidence interval estimator of p is p  Z 

2

144

ˆˆ pq n

ˆ p p pq n

Examples:
1. An inventor has developed a system that allows visitors to museums, zoos and other attractions to get information at the touch of a digital code. For example, zoo patrons can listen to an announcement (recorded on a microchip) about each animal they see. It is anticipated that the device would rent for $3.00 each. The installation cost for the complete system is expected to be about $400,000. The ABC zoo is interested in having the system installed, but the management is uncertain about whether to take the risk. A financial analysis of the problem indicates that if more than 10% of the zoo visitors rent the system, the zoo will make a profit. To help make the decision, a random sample of 400 zoo visitors is given details of the systems capabilities and cost. If 48 people say that they would rent the device, can the management of the zoo conclude at the 5% significance level that the investment would result in a profit?
Solution
1.

Stating the null and alternate hypothesis

H 0 : P  0.1
H A : P  0.1
2.

Level of significance:   0.05

3.

Test statistic Z 

4.

Rejection region

ˆ p p pq n

Z   Z 0.05  1.645

5.

If Z c  1.645, Reject H o

Value of the test statistic
ˆ
p

Z

48
 0.12
400
ˆ p p pq n



0.12  0.1
0.1 * 0.9
400

 1.33  1.645

145

6.

Conclusion
Do not Reject H0. No, there is no sufficient evidence to conclude that the investment would result in a profit.

2. In a random sample of 100 units from an assembly line, 22 were defective. Does this provide sufficient evidence at the 10% significance level to allow us to conclude that the defective rate among all units exceeds 10%?

Solution
1.

Stating the null and alternate hypothesis

H 0 : P  0.1
H A : P  0.1
2.

Level of significance:   0.1

3.

Test statistic Z 

4.

Rejection region

ˆ p p pq n

Z   Z 0.1  1.28

5.

Value of the test statistic
ˆ
p

Z

6.

If Zc  1.28, Reject H o

22
 0.22
100
ˆ p p pq n



0.22  0.1
0.1 * 0.9
100

 4  1.28

Conclusion
Reject H0. Yes, there is sufficient evidence to conclude that the defective rate among all units exceeds 10%
146

9.8

Inference about a Population Variance

Variance can be used to describe a number of different situations e.g.
 In Quality control, engineers must ensure that products coming out of a production line meets specifications such as size, weight, volume etc.
 In finance, the variance of the returns on a portfolio of investments is a measure of the uncertainty and risk inherent in a portfolio.
The sample variance is an unbiased, consistent estimator of the population variance.

Chi-square sampling distribution
In repeated sampling from a normal population whose variance is  2 , the variable chi-square distributed with (n-1) degrees of freedom. The variable

n  1s 2
2

n  1s 2
2

is

is called the chi-

square statistic and is denoted by  2 . The  2 variable can equal any value between 0 and x.

Chi-square notation
The value of  2 such that the area to its right under the chi-square curve is equal to x and is
2
denoted by   . The value  12 is the point such that the area to its right is 1   . Hence , the

area to its left is  .

Testing the population variance
Test statistic for  2
The test statistic used to test hypothesis about 

2

is 

2

n  1s 2

2

which is chi-square

distributed with (n-1) degrees of freedom, provided the population random variable is normally distributed. Estimating the population variance
The confidence interval estimator of  2 is
147

LCL 

n  1s 2
2


UCL 

n  1s 2
 12

2

2

Examples
1. A manufacturer of a bottle-filling machine claims that the standard deviation of the fills from his machine is less than 2 cc. In a random sample of 10 fills, the sample standard deviation was 1.19 cc. Is this sufficient evidence at the 5% level of significance to support the manufacturer‟s claim? (Assume a normal population).
Solution
1. Stating the null and alternate hypothesis

H0 : 2  4
H A : 2  4
2. Level of significance:   0.05
3. Test statistic:  2 

n  1s 2
2

4. Rejection region :  2   21  ,n1   .2 ,9  3.325, If  2  3.325, Reject Ho
95
5. Value of test statistic

2 

n  1s 2
2



9 *1.19 2
 3.186  3.325
4

6. Conclusion
Reject H0. Yes, there is sufficient evidence to support the manufacturer‟s claim.
Example
Some traffic experts claim that the variability of automobile speeds is a critical factor in determining how many accidents are likely to occur on a highway. The greater the variability, the more the accidents. Suppose a random sample of 101 cars reveals that the mean and variance of their speeds are 57.3 Km/h and 88.7 (Km/h)

2

respectively. Can we conclude at the 5%

significance level that the variance of all cars speeds exceeds 50 (Km/h) 2?

Solution
1. Stating the null and alternate hypothesis
148

H 0 :  2  50
H A :  2  50
2. Level of significance:   0.05
3. Test statistic:  2 

n  1s 2
2

4. Rejection region :  2   2 ,n1   .2 ,100  124.342
05
5. Value of test statistic



2

n  1s 2



2



100 * 88.7
 177  124.342
50

6. Conclusion
Reject H0. Yes, there is sufficient evidence to conclude that the variance of all cars speeds exceeds 50 (Km/h) 2

149

9.9

Summary

Hypothesis testing is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement. The steps that are followed when testing hypothesis are state the null and alternate hypothesis, select a level of significance, identify the test statistic, formulate a decision rule and identify the rejection region, compute the value of the test statistic and make a conclusion.

9.10 Activities

1. The owner of a large fleet of taxis is trying to estimate his costs for next years operations.
One major cost is for fuel purchases. Because of the high cost of gasoline, the owner has converted his taxis to operate on propane. He needs to know what the average consumption will be, so he decides to take a random sample of eight taxis and measures the miles per gallon achieved. The results are as follows:
28.1, 33.6, 42.1, 37.5, 27.6, 36.8, 39.0, 29.4
Estimate with 95% confidence the mean propane mileage for all taxis in his fleet? (Assume that the distribution of mileage is normal).
2. A highway patrol officer believes that the average speed of cars traveling over a certain stretch of highway exceeds the posted limit of 55 km/ hr. A random sample of 10 cars had their speeds measured by radar. The results in Km / hr are as follows:
71, 53, 62, 49, 59, 52, 58, 61, 85, 55

150

Do these data provide sufficient evidence to support the highway patrol belief, at the 10% level of significance?
3. A factory produces a component that is used in manufacturing computers. Each component is tested prior to shipment to determine whether or not it is defective. In a random sample of
250 units, 20 were found to be defective. Estimate with 99% confidence the true proportion of defective components produced by the factory. (0.036, 0.124)
4. A manufacturer of computer chips claims that more than 90% of his products conform to specifications. In a random sample of 1,000 chips drawn from a large production run, 75 were defective. Do the data provide sufficient evidence at the 1% level of significance to enable us to conclude that the manufacturer‟s claim is true? What is the P-value of the test?
5. In 2009, 15% of the households in Kisumu City indicated that they owned a sewing machine.
In 2011, there was a reason to believe that there was some increase in this percentage. A survey based on a random sample of 900 households was taken and it was found that 189 households had a sewing machine. Can we conclude at the 5% significance level that there has been a significant increase in the sale of sewing machines
6. A company manufactures steel shafts for use in engines. One method of judging inconsistencies in the production process is to determine the variance of the lengths of the shafts. A random sample of 10 shafts produced the following measurements of their lengths in centimeters
20.5, 19.8, 21.1, 20.2, 18.9, 19.6, 20.7, 20.1, 19.8, 19.0
Find a 90% confidence interval estimate for the population variance  2 assuming that the lengths of the steel shafts are normally distributed.
7. One important factor in inventory control is the variance of the daily demand for a product.
Management believes that demand is normally distributed with the variance equal to 250. In an experiment to test this belief about  2 , the daily demand was recorded for 25 days. The data has been summarized as follows:

x  50.6
S 2  500
Do the data provide sufficient evidence to show that management‟s belief about the variance is untrue? (Use   0.01 ).

151

LESSON TEN: CHI-SQUARE TESTS
10.0 Introduction

This lesson covers the chi-square test for multinomial experiment and contingency table

10.1 Lesson objectives

By the end of the lesson, the students should be able to perform the following tests:1. Chi-square test of a multinomial experiment
2. Chi-square test for contingency table

A chi-squared test is any statistical hypothesis test in which the sampling distribution of the test statistic is a chi-squared distribution when the null hypothesis is true. Also considered a chisquared test is a test in which this is asymptotically true, meaning that the sampling distribution
(if the null hypothesis is true) can be made to approximate a chi-squared distribution as closely as desired by making the sample size large enough. The chi-square test is used to determine whether there is a significant difference between the expected frequencies and the observed frequencies in one or more categories

152

10.2 Chi-square test of a multinomial experiment (Goodness-of-fit test)
A multinomial experiment is a generalized version of a binomial experiment that allows for more than two possible outcomes on each trial of the experiment. The following are the properties of a multinomial experiment
 The experiment consists of a fixed number n of trials.
 The outcome of each trial can be classified into exactly one of k categories called cells
 The probability P1 that the outcome of a trial will fall into a cell i remains constant for each trial, for i  1, 2, 3, .........k. moreover, P  P2 ........Pk  1.
1
 Each trial of the experiment is independent of the other trials.

Test statistic k oi  ei 2

i 1

ei

 
2

Rejection region

 2   2 , k-1
Example one
Two companies A and B have recently conducted aggressive advertising campaigns in order to maintain and possibly increase their respective shares of the market for a particular product.
These two companies enjoy a dominant position in the market. Before advertising campaigns began, the market share for Company A was 45% while Company B had a market share of 40%.
Other competitors accounted for the remaining market share of 15%. To determine whether these market shares changed after the advertising campaigns, a marketing analyst solicited the preferences of a random sample of 200 consumers of this product. Of the 200 consumers, 100 indicated a preference for Company‟s A‟s product, 85 preferred Company‟s B product and the remainder preferred one or another of the products distributed by other competitors. Conduct a test to determine at the 5% level of significance, whether the market shares have changed from the levels they were at before the advertising campaigns occurred.
Solution
1. Stating the null and alternate hypothesis
153

Ho: P1= 0.45, P2 = 0.4, P3 = 0.15
HA: At least one of the Pi is not equal to its specified value.
2. Level of significance:   0.05 k 3. Test statistic:  2   i 1

(oi  ei ) 2 ei 4. Rejection region :  2   2 ,k 1   .2 , 2  5.99147
05
5. Value of the test statistic: assuming that the null hypothesis is correct, we can calculate the expected number of consumers who prefer A, B and others using the formula

ei  np .
Company

Observed frequency oi  ei 2

Expected

oi  ei 2

frequency

ei

A

100

90

100

1.11

B

85

80

25

.31

Others

15

30

225

7.50

Total

200

200

k

Therefore  2   i 1

8.92

(oi  ei ) 2
 8.92 ei 6. Conclusion: Reject Ho
There is sufficient evidence at the 5% level of significance to allow us to conclude that the market shares have changed from the levels they were at before the advertising campaigns occurred.

Example Two
To determine if a single die, is balanced, or fair, the die was rolled 600 times. The observed frequencies with which each of the six sides of the die turned up are recorded in the following table: Face

1

2

3
154

4

5

6

Observed frequency

114 92

84

101

107

102

Is there sufficient evidence to conclude at the 5% level of significance, that the die is not fair?
Solution
1.

Stating the null and alternate hypothesis

H o : p1  p 2  p3  p 4  p5  p6  1

6
H A : At least one of the Pi s is not equal ot its specified value
2.

Level of significance:   0.05

3.

Test statistic:  2  

k

i 1

(oi  ei ) 2 ei 4.

Decision Rule :  2   2 ,k 1   .2 ,5  11.0705,
05

5.

If  2  11.0705, Reject Ho

Value of the test statistic:
Assuming that the null hypothesis is correct, we can calculate the expected number of consumers who prefer A, B and others using the formula ei  np .
Observed

Expected

oi  ei 2

frequency

Face

frequency

ei

1

114

100

1.96

2

92

100

0.64

3

84

100

2.56

4

101

100

0.01

5

107

100

0.49

6

102

100

0.04

Total

600

600

5.7

(oi  ei ) 2
 5.7  11.0705
Therefore    ei i 1 k 2

6.

Conclusion:

155

Do not Reject Ho. There is no sufficient evidence at the 5% level of significance to allow us to conclude that that the die is not fair.
Rule of five
For the discrete distribution of the test statistic  2 to be adequately approximated by the continuous chi-square distribution, the conventional rule is to require that the expected frequency for each cell be at least 5. Where necessary, cells should be combined in order to satisfy this condition. The choice of cells to be combined should be made in such a way that meaningful categories result from the combination.

10.3 Chi-Square Test of a Contigency Table
A contingency table is a rectangular table which items from a population are classified according to two characteristics. The objective is to analyze the relationship between two qualitative variables i.e. to investigate whether a dependence relationship exists between two variables or whether the variables are statistically independent. The number of degrees of freedom for a contingency table with r rows and c columns is d . f .  r - 1 c  1 .
Example One
A sample of employees at a large chemical plant was asked to indicate a preference for one of three pension plans. The results are given in the following table:Pension Plan
Job Class

Plan A

Plan B

Plan B

Supervisor

10

13

29

Clerical

19

80

19

Labourer

81

57

22

At the 1% significance level, determine whether there is a relationship between the pension plan selected and the job classification of employees?
Solution
Pension Plan

Total

Job Class

Plan A

Plan B

Plan B

Supervisor

10

13

29

156

52

Clerical

19

80

19

118

Labourer

81

57

22

160

Total

110

150

70

330

We need to conduct a chi-square of the contingency table to determine whether the classifications are statistically independent.
Ho: The two classifications are independent
HA: the two classifications are dependent k Test statistic:  2   i 1

(oi  ei ) 2 ei 2
Rejection region :  2   2 ,( r 1)( c1)   0.01, 4  13.2767

The value of the test statistic
To compute the expected values for each cell, multiply the row total by the column total and divide by the total number of shirts sampled.

Observed frequency

Expected frequency

o  e2

o

e

e

1

10

17.33

3.1003

2

13

23.64

4.7889

3

29

11.03

29.2766

4

19

39.33

10.5087

5

80

53.64

12.9539

6

19

25.03

1.4527

7

81

53.33

14.3564

8

57

72.73

3.4021

9

22

33.94

4.2005

Cell i

Total

k

Value of the test statistic :  2   i 1

84.0401

(oi  ei ) 2
 84.0401 ei Conclusion: Reject Ho.

157

There is enough evidence at the 1% significance level to conclude that the two classifications are dependent. Example Two
The Coca Cola Company sells four brands of sodas in East Africa. To help determine if the same marketing approach used in Kenya can be used in Uganda and Tanzania, one of the firm‟s marketing analysts wants to ascertain if there is an association between the brand of Soda preferred and the nationality of the consumer. She first classifies the population according to the brand of soda preferred i.e. Fanta, Sprite, Coke and Krest. Her second classification consists of the three nationalities; Kenyan, Tanzanian and Ugandan. The marketing analyst then interviews a random sample of 250 Soda drinkers from the three countries, classifies each according to the two criteria and records the observed frequency of drinkers falling into each of the cells as shown in the table below.
Soda preference
Nationality

Coke

Krest

Sprite

Fanta

Total

Kenyan

72

8

12

23

115

Ugandan

26

10

16

33

85

Tanzanian

7

10

14

19

50

Total

105

28

42

75

250

Based on the above sample data, can we conclude at the 1% level of significance that there is a relationship between the preference of the soda drinkers and their nationality?
Solution
We need to conduct a chi-square of the contingency table to determine whether the classifications are statistically independent.
Ho: The two classifications are independent
HA: the two classifications are dependent

(oi  ei ) 2
Test statistic:    ei i 1 k 2

2
Rejection region :  2   2 ,( r 1)( c1)   0.01,6  16.8119

The value of the test statistic
To compute the expected values for each cell, multiply the row total by the column total and divide by the total number of respondents sampled.
158

Observed frequeny

Expected frequency

o  e2

o

e

e

1

72

48.30

11.63

2

26

35.70

2.64

3

7

21.00

9.33

4

8

12.88

1.85

5

10

9.52

0.02

6

10

5.60

3.46

7

12

19.32

2.77

8

16

14.28

0.21

9

14

8.40

3.73

10

23

34.50

3.83

11

33

25.50

2.21

12

19

15.00

1.07

Cell i

k

Value of the test statistic:  2   i 1

(oi  ei ) 2
 42.75 ei Conclusion: Reject Ho.
Based on the sample data, we can conclude at the 1% significance level that there is a relationship between preferences of soda drinkers and their nationality.

10.4 Activities

1. Grades assigned by an economics instructor have historically followed a symmetrical distribution. 159

Grade

A

B

C

D

F

Percentage

5

25

40

25

5

A sample of 150 grades revealed the following
Grade

A

B

C

D

F

Number

11

32

62

29

16

Can we conclude at the 1% level of significance that this year‟s grades are distributed differently than they were in the past?
2. The trustee of a company‟s pension plan has solicited the opinions of a sample of the company‟s employees regarding a proposed revision of the plan. A breakdown of the responses is shown in the table below: Response

Lower level

Middle

Top

management

management

management

For

67

32

11

Against

63

18

9

Is there sufficient evidence at the 5% significance level, to conclude that the responses differ among the three groups of employees?
3. The operations manager at a shirt manufacturing plant has been concerned about the large number of defects that the company‟s three shifts have been producing. They appear to be three types of defects: Improper stitching, buttons not aligned with button holes and inconsistent colouring. The manager decides to investigate the problem. As a first step to improving the quality, she wants to know if the number and type of defects are the same for all three shifts. A random sample of one day‟s shirt production is taken. The number of each type of defect and the number of perfect shirts for each are shown in the following table.
Shift
Shirt condition

1

2

3

Total

Perfect

224

249

238

711

Improperly stitched

15

19

21

55

Unaligned buttons

8

12

12

32

Inconsistent colour

17

16

11

44

Total

264

296

282

842

160

Do these results allow the operations manager to conclude that at the 10% significance level, there are differences in quality among the three shifts?

161