Linear Programming
UNIT 1
LINEAR PROGRAMMING
Structure
NOTES
1.0 Introduction
1.1 Unit Objectives
1.2 Basic Structure of LP Problem
1.3 Properties of the LP Model
1.4 Application Areas of Linear Programming
1.5 General Mathematical Model of LPP
1.6 Formulation of LP Model
1.7 Examples on LP Model Formulation
1.8 Solution of LPP
1.8.1 Graphical LP Solution
1.9 Some Special Cases in LP Solution
1.10 Solution of LPP Using Simplex Method
1.11 Special Cases in Simplex Method
1.12 Sensitivity Analysis
1.13 Summary
1.14 Key Terms
1.15 Question & Exercises
1.16 Further Reading and References
1.0 Introduction
Linear programming (LP) can best be defined as a group of mathematical
techniques that can obtain the very best solution to problems which have many possible
solutions. Linear programming can be used to solve a variety of industrial problems. In
most of the situations, resources available to the decision maker are limited. Several
competing activities require these limited resources. With the help of linear
programming those scarce resources are allocated in an optimal manner on the basis of
a given criterion of optimality. In most of the situations, the criterion of
optimality is either maximization of profit, revenue or minimization of cost, time and
distance, etc.
1.1 Unit Objectives
After studying this unit, you should be able to
Understand Basic Structure of LP Problem
Know the properties of LP Model
Know the Application areas of Linear Progrmming
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Linear Programming
NOTES
Understand formulation of LP Model
Understand Graphical and Algebraic (Simplex) Methods of solution of LP
Models
Use of slack, surplus and artificial variables in LP solution
Use of Big-M and Two phase methods to handle artificial variables
Understand solution process to handle changes in constraints and changes in
objective function with sensitivity analysis.
1.2 Basic Structure of LP Problem
Structure of all LPP has three important components.
(1) Decision variables (activities) : These are activities for which we want to
determine a solution. These are usually denoted by x1, x2, ...., xn.
(2) The objective function (goal) : This is a function which is expressed in terms
of decision variables and we want to optimize (maximize or minimize) this function.
(3) The constraints : These are limiting conditions on the use of resources. The
solution of LPP must satisfy all these constraints.
1.3 Properties of the LP Model
In linear programming, the word linear refers to linear relationship among variables
in a model. The word “programming” refers to modelling and solving a situation
mathematically. In a LP model, the objective and the constraints are all linear function
and are expressed in terms of decision variables. Any LP model must satisfy following
basic properties :(1) Proportionality (Linearity) : The contribution of each activity (decision
variable) in both the objective function and the constraints to be directly
proportional to the value of the variable.
(2) Additivity : In LP models, the total contribution of all the activities in the
objective function and in the constraints to be the direct sum of the individual
contributions of each variable.
(3) Certainty : In all LP models, all model parameters such as availability of
resources, profit (or cost) contribution of a unit of decision variable and use
of resources by a unit of decision variable must be known and constant.
1.4 Application Areas of Linear Programming
LP is one of the most popular technique to find best solution in variety of situations.
Some of the common applications of LP are :-
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(1) Agricultural Applications : LP can be applied in agricultural planning. e.g.
allocation of limited resources such as acreage, labour, water supply and
working capital, etc. in such a way so as to maximize net revenue.
(2) Military Applications : LP can be applied to maximize the effect of military
operations as well as to minimize the travel distance and cost of operations.
(3) Production Management : Most of the examples of LPP are related to
develop a suitable product mix. A Company can produce several different
products, each of which requires the use of limited production resources.
Product mix is developed using LP, knowing marginal contribution and amount
of available resource used by different product. The objective is to maximize
the total contribution, subject to all constraints.
Linear Programming
NOTES
Similarly LP can be used in production planning to minimize total operation
costs, in assembly line balancing to minimize the total elapse time, in blending
problem to determine minimum cost blend and also to minimize the trim losses
in case of products of standard size.
(4) Financial Management : LP is used for deciding investment activity among
several other activities in such away which maximizes the total expected
return or minimize risk under certain conditions.
(5) Marketing Management : LP may be used in determing the media mix to
maximize the effective exposure within constraints of budget and circulation
/ reach of various media. LP can be used for determining location of
warehouses and other facilities to minimize cost of distribution of products.
(vi) Personnel Management :- LP is used to allocate available manpowers to
different shifts / duties to minimize overtime cost or total manpower.
LP has also find applications in capital budgeting, health care, diet- mix, cupala
charging, fleet utilization and many more such situations.
1.5 General Mathematical Model of LPP
Let us consider n decision variables x1, x2.... xn.
Contribution of each of these decision variable is c1, c2.... cn respectively. So as to
Optimize (Max. or Min.) Z = c1 x1+c2x2 + ......... + cnxn.
Let us consider m constraints.
Total Available
of resource 1
Constraints are expressed as follows :-
a11 x1 + a12 x2 + .... + a1n xn (< or = or >) b1
○
○
○
○
○
○
○
○
○
○
○
○
○
○
○
a21 x1 + a22x2 + ....+a2nxn (< or = or >) b2
○
This coefficient
represent amount of
resource 2 used by per
unit of activity 1, i.e. x1)
am1 x1+ am2 x2+ .... +amn xn (< or = or >) bm
Unless specified, in all LP models nonnegativity constraints are also included.
These constraints signify that decision variables can take only non negative values.
These are expressed as x1 > 0, x2 > 0 ..... xn > 0
The above formation can also be expressed in a compact form using summation
sign.
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Linear Programming
n
Optimize (Max. or Min.) Z = Σ cj xj
(objective function)
j=1
NOTES
Subject to constraints
n
Σ aijxj (< or = or >) bi ; i=1, 2 .... m (constraints)
j=1
and xj > 0 ; j=1, 2...n
(Non negativity Constraints)
In all LP situation, left hand side of constraints is either less than, or equal to or
greater than right hand side. In a particular case of a resource, only one possibility may
take place.
1.6 Formulation of LP Models
Usefulness of LP technique starts with modlling of a given situation in a standard
form as shown in section 1.5. Various steps involved in modelling of LPP are as follows.
(i)
Indentify the decision variables and express them in terms of algebraic
symbols. (Mostly x1, x2... xn).
(ii) Indentify contribution of each of these decision variable in objective which is
to be optimized (maximize or minimize). Express objective function as shown
in section 1.5.
(iii) Identify different resources or conditions which are to be satifsied. Develop
constraint inequality for each constraint. Be careful for the sign (less than,
equal to, greater than) in writing constraints.
Also add non negativity constraints for all decision variables.
1.7 Examples on LP Model Formulation
Example 1 : A manufacturer wishes to determine how to make products A and B
so as to realize the maximum total profit from the sale of the products. Both products
are made in two processes I and II. It takes 7 hours in process I and 4 hours in process
II to manufacture 100 units of product A. It requires 6 hours in process I and 2 hours in
process II to manufacture 100 units of product B. Process I can handle 84 hours of
work and process II can handle 32 hours of work in the scheduled period. If the profit
is Rs. 11 per 100 units for product A and Rs. 4 per 100 units for product B, then how
much of each of products A and B should be manufactured to realize the maximum
profit? It is assumed that whatever is produced can be sold and that the set-up time on
the two processes is negligible. Formulate the above problem as a linear programming
model.
Solution :
Step 1 : Let
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x1 = number of Product A (in 100's)
x2 = number of Product B (in 100's)
}
There are decision variables.
Step 2 : Now contribution of per 100 units of product A is Rs 11 and contribution of per
100 units of product B is Rs. 4
Linear Programming
Therefore total contribution is 11x1 + 4x2
We want to maximize this total contribution function, which is objective of this
model. Therefore, objective function becomes,
NOTES
Maximize Z = 11x1 + 4x2
Step 3 : Identification of resources : Here two resources are given, namely, process I
and process II.
For process I, total use should be less than or equal to maximum capacity to
handle, i. e. 84 hrs.
So Constraint for process I is 7x1+ 6x2 < 84
Similarly, for process II, 4x1 + 2x1 < 32
Also nonnegativity constraints will be x1, x2 > 0
Example 2 :
A company manufactures three grades of a product A, B and C. The plant operates
on a three shift basis and the following data are available from the production records.
Requirement of Resources
Grade
Availablility
A
B
C
(Capacity per month)
0.25
0.12
0.80
500 tonnes
Mixing (Kilolitres per machine shift) 2.0
3.0
5.0
100 machine shifts
Packing (Kilolitres per shift)
10.0
10.0
80 shifts
Special additive (kgs per litre)
10.0
There are no limitations on other resources. The particulars of sales forecasts and
estimated contribution to overheads and profits are given below.
A
B
C
Maximum possible sales per month (kilolitres)
100
400
600
Contribution (Rs. per kilolitre)
4000
3500
2000
Due to Commitments already made, a minimum of 150 kilolitres per month of `C`
has to be necessarily supplied during the next year.
Just as the company was able to finalise the monthly production programme for
the next 12 months, an offer was received from a nearby competitor for hiring 25
machine shifts per month of mixing capacity for grinding `B` product, that can be spared
for at least a year. However, profit margin of `B` product will reduce by Rs. 1 per litre,
for the amount of product produced on hired facility. Formulate the LP model for
determining the monthly production programme to maximize contribution.
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Linear Programming
Solution :
Step 1 : Identification of decision variables :
x1 = Monthly quantity of product A (kilolitres)
Let
NOTES
x2 = Monthly quantity of product B using conpany's facilities (Kilolitres)
x3 = Monthly quantity of product B using hired facilities (Kilolitres)
x4 = Monthly quantity of product C (Kilolitres)
Step 2 : Development of objective function :
Contribution per kilolitre
Decision Variable Total contribution
4000
x1
4000x1
3500
x2
3500x2
Working or Hired facility 3500 - 1x 1000 = 2500 x3
2500x3
2000
x4
2000x4
So complete objective function is
Maximize Z = 4000x1 + 3500x2 + 2500x3 + 2000x4
Step 3 : Development of constraints ;
(i)
Special additive constraint :
0.25x1 + 0.12x2 + 0.12x3 + 0.80x4 < 500 (tonnes)
(ii)
Own mixing facility constraint
x1
+
2.0
(iii)
x2
+
3.0
x4
< 100 (machine shifts)
5.0
Hired mixing facility constraint
x3
< 25 (machine shifts)
3.0
(iv)
Packing Constraint
x1
+
10.0
(v)
x2
10.0
x3
10.0
+
x4
10.0
Marketing constraint
< 100
For product A
x1
For product B
x2+x3 < 400
For product C
150
Step 4 : Non negativity constraints
x1, x2, x3, x4 > 0
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+
< x4 .... 600
< 80 (shifts)
Linear Programming
Example 3 : ABC limited has a flect of 4 different type of 120 coaches, which operates
along 4 routes. Details of capacity, daily trips of each coach and the number of customers
expected for each route during a certain period are as follows :
Type of Capacity
Number of
Coach (passengers) coaches Available
Number of daily trips on route
1
2
3
4
1
50
30
4
5
6
4
2
60
50
2
4
3
3
3
30
20
1
2
1
2
4
40
20
2
3
2
3
250
300
200
400
Daily Number of Customers
NOTES
x1
The operational cost / trip and penalty cost if a customer is not given service is
given below :
Type of Coach
Operation cost / Trip on given route (Rupees)
1
2
3
4
1
250
700
200
500
2
400
500
500
300
3
600
350
400
600
4
350
400
450
250
Penalty cost per customer
50
40
30
60
Formulate the above situation as a LPP
Solution :
Let xij denote the number of trips made by ith type of coach along route J.
i varies from 1 to 4 and j also varies from 1 to 4.
Note : It is assumed that during each trip, coach carry passengers to its full capacity.
Developing objective function
It is required to operate trips of each coach in order to minimize operational cost/
trip, penalty cost to maximize profit, i. e.
Minimize Z
=
250x11 + 700x12 + 200x13 + 500x14
+ 400x21 + 500x22 + 500x23 + 300x24
+ 600x31 + 350x32 + 400 x33 + 600 x34
}
Operational Cost
Component
+ 350 x41 + 400 x42 + 450 x43 + 250 x44
+ 50 (250-50x11-60x21-30x31-40x41)
+ 40 (300-50x12-60x22-30x32-40x42)
+30(200-50x13-60x23-30x33-40x43)
+60 (400-50x14-60x24-30x34-40x44)
}
Penalty cost
Component
Sample Calculation of penalty cost is as follows :
Penalty Cost on route 1 is
50x Number of Customers not given service on route1.
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Linear Programming
50x [ Daily number of customers on route1 -
50 x Number of trips made by first
type of coach on route 1
60 x Number of trips made by
second type of coach on route 1
NOTES
30 x Number of trips made by
third type of coach on route 1
40 x Number of trips made by
fourth type of coach on route 1
Threrore penalty cost on route is
50 [250 - 50x11 - 60x21 - 30x31 - 40x41]
Similarly penalty coast on route 2,3 and 4 is calculated
Developing Constraints :
(i)
Capacity Constraints :
The number of passengers carried along any route is the sum of passengers
carried by each type of coach is the total number of trips made, viz.
50x11 + 60x21 + 30x31 + 40 x41 < 250
50x12 + 60x22 + 30x32 + 40 x42 < 300
50x13 + 60x23 + 30x33 + 40 x43 < 200
50x14 + 60x24 + 30x34 + 40 x44 < 400
(ii) Maximum number of trips constraints
Total number of trips constraints.
Total number of trips made by a given type of coach along a route cannto exceed
the product of number of coaches available and number of daily trips on that route. It
will be expressed as follows :
X11< 30 x 4
X21 .. 50 x 2
X31 .. 20 x 1
X41 .. 20 x 2
X12< 30 x 5
X22 .. 50 x 4
X32 .. 20 x 2
X42 .. 20 x 3
X13< 30 x 6
X23 .. 50 x 3
X33 .. 20 x 1
X43 .. 20 x 2
X14< 30 x 4
X24 .. 50 x 3
X34 .. 20 x 2
X44 .. 20 x 3
Finally add non negativity constraints :
Xij > 0, for all i and j.
1.8 Solution of LPP
Once a LP model is ready, next step is to find solution of this model. Solution of LP
model can be done either graphically or using simplex methos.
1.8.1 Graphical LP Solution
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A LPP can easily be solved by graphical method when it involves two decision
variables.
Linear Programming
The process includes two steps :
1
Determination of the feasible solution space.
Any solution (values of decision variables) which satisfy the constraints and non
negativity restrictions of LP model is known as feasible solution.
NOTES
Within the feasible solution space, feasible solution correspond to the extreme (or
corner) points of the feasible solution space.
2
Determination of the optimum solution from among all the feasible points in the
solution space.
Any basic feasible solution which optimizes (maximize or minimize) the objective
function of a LP model is called an optimum solution.
The optimal feasible solution, if it exists, will occur at one, or more, of the extreme
points that are basic feasible solutions.
The procedure uses two examples to show how maximization and minimization
objective functions are handled.
Example 4 :
Solve graphically following LP model :
Maximize Z = 5x1 + 4x2
Subject to constraints :
6x1 + 4x2 < 24
(1)
x1 + 2x2 < 6
(2)
-x1 + x2 < 1
(3)
x2 < 2
(4)
x1, x2 > 0
(5) & (6)
Solution :
Step 1 : Consider a graphical plane. First, we account for the non negativity
constraints which are given in inequality (5) & (6) i. e. x1> 0 and x2 > 0. In figure no. 1,
the horizontal axis and the vertical axis represent the x1 and x2 variables respectively.
xx2
2
x1
x1
Fig. 1
Thus, the nonnegativity of the variables restricts the solution space area to the
first gradrant that lies above the x1 - axis and to the right of the x2 - axis.
Step 2 : Each of the four constraints expressed as inequalities will be treated as
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Linear Programming
equalities and, then their respective intercepts on both the axis can be determined.
(i)
NOTES
For first constraint, after replacing 6x1 + 4x2 < 24 with the straight line 6x1+ 4x2 =
24, we can determine two distinct points by first setting x1 = 0 to obtain x2 = 6 and
then setting x2 = 0 to obtain x1 = 4. Thus we get a line passed through the two
points (0, 6) and (4, 0) as shown by line (1) in fig. 2.
x2
5
5
(0, 6) 6
5
(1)
4
3
2
1
(4, 0)
0
1
2
3
4
6
5
x1
x1
6
Fig. 2
Next, consider the effect of inequality. The line 1 divides the (x1, x2) plane. into
two half- spaces, one on each side of the graphed line. Only one of these two halves
satisfies the inequality. To determine the correct side, choose (0, 0) as a reference
point. If it satisfies the inequality, then the side in which it lies is the feasible half-space,
otherwise the other side is. To illustrate the use of the reference point (0, 0), in this
inequality (6 x 0 + 4 x 0 = 0) it is less than 24, the half - space represnting the inequality
includes the origin. This is shown by direction of arrow for (1) in figure 2. It is convenient
computationally to select (0, 0) as the reference point, unless the line happens to pass
through the origin in which case any other point can be used
(ii) Similarly for constraint 2, x1 + 2x2 < 6, we get points (6, 0) and (0, 3). Using
reference point method, this inequality includes the origin, Figure 3 shows common
area which satisfies inequality 1, 2 and non negativity constraints.
x2
5
8
7
6
5
4
1
3
2
2
1
0
6
1
2
3
4
5
6
7
8
x1
Fig. 3
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(iii) Finally, figure 4 shows the feasible solution space of the given LPP. The feasible
solution space satisfies all six constraints simultaneously. Any point in or on the
boundary of the area ABCDEF is part of the feasible solution space. All points
outside this area are infeasible, as these points do not satisfy one or more than one
constraints.
Linear Programming
NOTES
Fig. 4
Step 3 : Since the value of x1 and x2 have to lie in the shaded area which contains
an infinite number of points would satisfy the constraints of the given LPP. But we are
confined only to those points which correspond to corners of solution space. Thus, as
shown in figure 4, the corner points of feasible region are A = (0, 0), B = (4, 0), C = (3,
1.5), D = (2, 2), E = (1, 2), F = (0, 1).
Step 4 : Value of objective junction at corner points
Corner
points
Coordinates
of corner points (x1, x2)
Objective Function
Z = 5x1 +4x2
Value
A
(0, 0)
5 (0) + 4 (0)
0
B
(4, 0)
5 (4) + 4 (0)
20
C
(3, 1.5)
5 (3) + 4 (1.5)
21
D
(2, 2)
5 (2) + 4 (2)
18
E
(1, 2)
5 (1) + 4 (2)
13
F
(0, 1)
5 (0) + 4 (1)
4
Here we see that maximum value of objective function is 21 which is obtained at
the point C = (3, 1.5), i. e. x1 = 3 and x2 = 1.5 are values of decision variables.
Example 5
: Minimize Z
= .3x1 + .9x2
Subject to
x1 + x2 > 800
.21x1 - .30x2 < 0
.03x1 - .01x2 > 0
x1, x2 > 0
Solution :
Figure 5 provides the graphical solution of the model.
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1500
Mi
nim
ize
0
Z=
.3x
1
2
1000
+.
9x
2
A
<0
.3x 2
1x1
.2
500
B
=
+x 2
x1
(Students are advised to select
a point on any side of a particular
constraint, and if that point satisfies
constraint then direction of constraint
is towards it otherwise it will be
directed opposite to that point.
Verify.)
x2
01x
>
It is also to be noted that (0, 0)
cannot be used as a reference point
to know the direction of constraint 2
and 3.
1
NOTES
Check that second and third constraints pass through origin. To plot these constraints
as straight lines, we need one additional point. (Student should take any value of x1,
such as 100 or 200 in constraint 2 and compute x2 Similarly assume one value of x1 for
constraint 3 and compute corresponding value of x2)
03x
-
Linear Programming
0
80
Another important point to
0
500
1000
1500 6 x
1
understand is about closed feasible
Fig. 5
solution space. Here objective
function is of minimization nature so we can get a solution otherwise it will become a
case of unbounded solution. (Unbounded solution will be discussed later in the chapter.)
Next we will calculate value of objective function at two corner points, A and B.
At point A, x 1 = 200,
x2 = 600 and
At point B, x 1 = 470.6,
x2 = 329.4 and Z = 437.64
Z = 600
So the minimum value of Z lies at B giving solution as x1 = 470.6 and x2 = 329.40.
1.9 Some Special Cases in LP Solution
(1) Multiple optimal solution or Alternative optima : In two earlier examples,
optimal value of objective function lies at one of the corner points of the feasible
solution space. But it is also possible in some situations, when optinal solution
obtained is not unique. It happens when objective function (line equation of objective
function) is parallel to one of the lines showing constraint in making the feasible
space. In this case infinite number of solutions will take place giving same value of
objective function but with different combinations of decision variables.
Students should verify case of multiple optimal solution with objective function
Z = 2x1 + 4x2 in example 4.
This objective function is parallel to constraint (2), represented by line segment
CD in feasible solution space. All points on this segment give Z = 12.
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(2) Unbounded Solution : In some LP models, the values of one or more decision
variables can be increased indefinitely without violating any constraint.
Correspondingly value of objective function is also increased indefinitely. Students
should verify for a LP model having objective function of maximize Z = 2x1 + x2,
Subjected to constraints of x1 - x2 < 10 and 2x1 < 40 with non negativity conditions.
They will find that the solution space is unbounded in the direction of x2, and the
value of Z can be increased indefinitely. Unboundedness normally occurs because
of poor construction of model.
Linear Programming
NOTES
(3) Infeasible Solution : In some cases, constraints are incosistent. Then, no common
feasible area can be obtained. This situation can only occur when some constraints
are of the type < and some are of the type >.
When no common feasible area is available, it means that solution is not possible
for this model.
Students should verify for a LP model having objective function of maximize Z
= 40x1+60x2 Subject to constraints of 2x1 + x2 > 70, x1 + x2 < 40 and x1 + 3x2 > 90
with non negativity conditions.
From the practical point, an infeasible solution space points to the possibility
that the model is not formulated correctly.
1.10 Solution of LPP Using Simplex Method
Graphical solution method discussed earlier is suitable for two decision variables.
Solution always occur at one of the corner points of feasible solution space. Values of
decision variables (x1, x2) can be determined for different corner points by solving
linear equations of constraints giving birth to a corner point. We test value of objective
function for different pairs of values of decision variable to decide optimum values of
decision variables. But in real world many decision variables are possible in LP models.
In case of more than two decision variables, simplex method is more useful. The simplex
method enumerates solution only at few selected corner points, if we compare with
graphical method.
To start simplex process of solving LPP, first we should be careful for following
two conditions :(i)
All the constraints (except non negativity constraints for decision variables)
are equations with non negative right hand side.
(ii) All the variables are non negative.
[Note : All computer programmes directly accept inequalities. Just we need to
keep positive RHS.]
How to convert inequalities into equations
Consider an inequality, 2x1 + 3x2 < 12
To convert this type of inequality into equation, we add a variable in LHS of this
inequality as follows.
2x1 + 3x2 + S = 12, with S > 0
This `S` is unused or slack amount of this resource. `S` here is known as slack
variable.
Now consider another inequality, 3x1 + x2 > 8.
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Linear Programming
Here we subtract a variable in LHS of the inequality.
3x1 + x2 - S = 8, with S > 0
NOTES
This `S` is amount by which LHS exceeds the minimum limit. It represents a
surplus. `S` here is known as surplus variable.
In case of negative RHS such as 2x1 - 3x2 < - 2
We can do 2x1 - 3x2 + S = -2 and then -2x1 + 3x2 - S = 2
Or we multiply both sides of inquality by -1. This will also change sign of inequality
from
(<) to (>)
So whenever (<) sign is there a slack variable is added and in case of (>)
a surplus variable is subtracted to convert inequalities into equations.
Important definitions for simplex method
In a simplex model where m equations and n variables (m < n) are there, only m
variables are part of solution. These m variables are known as basic variables.
Remaining n - m variables are non- basic variables. Solution obtained from basic
variables is known as basic solution of the model. In case all variables in basic solution
are non negative, it is a basic feasible solution.
One of the basic feasible solution will give best value of the objective function.
This best value (maximum in a maximization problem and minimum in a minimization
problem) is the optimum solution of given LPP.
Computation using Simplex Method
To explain the process of solving a LPP using simplex method, we will consider an
example of maximization type. Necessary steps will be discussed with the help of this
example.
Example 6 : Maximize Z = 5x1 + 4x2
Subject to
6x1 + 4x2 < 24
x1 + 2x2 < 6
-x1 + x2 < 1
x2 < 2
x1, x2 > 0
Solution
Step 1 : Convert constraint inequalities into equations :
6x1 + 4x2 + S1
= 24
1st constraint
x1 + 2x2
=6
2nd constraint
=1
3rd constraint
=2
4th constraint
-x1 + x2
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x2
+ S2
+ S3
+ S4
Linear Programming
x1, x2, s1, s2, s3, s4 > 0
Here variables S1, S2, S3 and S4 are the slacks associated with the respective
constraints.
We will also add these slack variables in objective function but with `0` (zero)
coefficients. So objective function will be
NOTES
Max Z = 5x1 + 4x2 + 0s1 + 0s2 +0s3 +0s4
For further computation, we write the objective equation as
Z - 5x1 - 4x2 = 0
Step 2 : Starting Simplex table :
With the help of objective equation and constraint equations, we will make initial
simplex table. The design of the table specifies the set of basic and non basic variables
as well as provides the solution associated with the starting iteration.
Standard desing of the simplex table used in iteration process is as follows :
Basic
Variables
Z
Decision, slack / surplus variables
Z ROW
1
Co efficient of various variables in
objective equation
0
0
0
Coefficients of different variables
↓
↓
Solution
RHS
according to row.
Fig. 6
To make starting simplex table, we need initial basic variables. In this model four
constraints are given and total variables are six (two decision variables x1 and x2 and
four slack variables S1, S2, S3 and S4). So only four variables will be basic variables and
two will be non basic variables.
Normally decision variables are given zero values and slack variables are initial
basic variables. In this example
Nonbasic (zero) variables
: (x1, x2)
Basic variables
: (s1, s2, s3, s4)
With this initial consideration
Z=0
S1 = 24, S2 = 6, S3 = 1, and S4 = 2
Now we will put this initial information in standard simplex table design as shown
in figure 6.
Quantitative Techniques in
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Linear Programming
Table I : Initial Table
NOTES
Basic
Z
X1
X2
S1
S2
S3
S4
Solution
Z
1
-5
-4
0
0
0
0
0
S1
0
6
4
1
0
0
0
24
S2
0
1
2
0
1
0
0
6
S3
0
-1
1
0
0
1
0
1
S4
0
0
1
0
0
0
1
2
Step 3 : To check for optimality
After preparing simplex table, we need to check optimality of this solution. In this
case objective function Z = 5x1 + 4x2 shows that the solution can be improved by
increasing x1 or x2. In simplex table we write objective function as Z - 5x1 - 4x2 = 0. If
any variable with negative coefficient in Z row is available, it means value of Z can be
improved and it is known as optimality condition. Optimality condition will be satisfied if
all variables in 2 row has non negative coefficients. So any variable with higher negative
coefficient will improve the value of Z faster.
This rule is referred to as the optimality condition. It means that currently nonbasic
variable x1 will enter into the solution. This is known as entering variable. If two or
more non basic variables in Z row are having same maximum negative coefficient, it
means a tie for entering variable. Tie is broken arbitrarily.
When a new variable will enter into the solution one of the existing basic variable
will become nonbasic variable. This is known as leaving variable.
To determine leaving variable : We compute the nonnegative ratios of right hand
side of the equations (Solution Column) to the corresponding constraint coefficients
under the entering variable, x1, as the following table shows :
Basic
s1
Entering
x1
Solution
Ratio
6
24
24
= 4 (Minimum Ratio)
6
s2
1
6
6
=6
1
s3
-1
1
1
= ignored
-1
s4
0
2
2
= ignored
0
Quantitative Techniques in
Management : 16
Decision : To enter x1 in the solution, s1 will be the leaving variable.
The minimum non negative, ratio identifies the current basic variable as leaving
variable. Value of x1 after becoming basic variable will be 4. As in case of entering
variable, two or more basic variable can have similar valid minimum ratio. It is a tie for
leaving variable. This tie can also be broken arbitrarily.
Linear Programming
NOTES
Now we will prepare table II in the solution process by swapping the entering
variable x1 and the leaving variable s1 in the simplex table to produce the following sets
of nonbasic and basic variables :
Nonbasic (zero) variables :
(s1, x2)
Basic variables
(x1, s2, s3, s4)
:
The swapping process is based on the Gauss - Jordan row operations. Process of
Gauss - Jordan row operations is as follows :
1.
Pivot Column : This is the column of entering variable.
2.
Pivot Row : This is the row of leaving variable.
3.
Pivot Element : The element in table obtainted as a result of intersection of
pivot row and column.
Let us reproduce our original table below :
Entering variable (x1 is most negative in z row)
Leaving variable (min. non
negative ratio of solution
and corresponding pivot
column element)
↓
↓
Basic
Z
x1
x2
s1
s2
s3
s4
Solution
Z
1
-5
-4
0
0
0
0
0
S1
0
6
4
1
0
0
0
24
S2
0
1
2
0
1
0
0
6
S3
0
-1
1
0
0
1
0
1
S4
0
0
1
0
0
0
1
2
Pivot row
Pivot column
Pivot element
To get new table from current table, Gauss - Jordan operation will be done in two
steps. First to get new row elements of pivot row and then get elements for all other
rows in table.
(1) To get new pivot row, we will divide all elements of existing pivot row by pivot
element. Thus new pivot row in this case will be
Z
x1
0
6
=0
X1
6
x2
4
=1
6
6
s1
2
=/
3
s2
1
s3
s4
0
0
=0
6
6
RHS
0
=0
=0
6
24
6
=4
6
Quantitative Techniques in
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Linear Programming
(2) To get all other rows including z row, we will perform following calculation :
New row = (Current row) - (Its pivot column coefficient) x (New pivot row)
e. g. to get new z row
NOTES
New z row = (current z row) - (-5) (New x1 row)
= (1 -5
-4
0 0 0 0 0)
-
(-5)
(0
2
3
= (1 0 -
1
2
1
3
6
0 0 0 4)
5
0 0 0 20)
6
Students are advised to do calculations for other rows of table.
The new basic solution has rows for z, x1, s2, s3 and s4.
The table will be as follows
Entering variable
↓
z
x1
x2
s1
s2
s3
s4
Solution
z
1
0
- 2/ 3
5
/6
0
0
0
20
Ratio
x1
0
1
2
/3
1
/6
0
0
0
4
s2
0
0
4
/3
- 1/ 6
1
0
0
2
6 Pivot
Column
3
/2
s3
0
0
5
/3
1
0
1
0
5
3
s4
0
0
1
0
0
1
2
2
↓
↓
Leaving variable
/6
0
↓
Basic
Pivot Column
Using the concept of optimality we find here that one of the variable in z row is
with negative coefficient. It implies that solution is not optinal and x2 will be our new
entering variable. To find leaving variable we will perform as usual following calculation.
Entering variable
Colum (x2)
RHS
(Solution Column)
Ratio
x1
2
/3
4
4
s2
4
/3
2
s3
5
/3
5
s4
1
Basic
2/
2
=6
/
3
=1.5 (minimum)
/
4/
3
2
5
5/
2
=3
/
/
1
3
=2
Leaving variable
Here minimum ratio is corresponding to S2 variable. So S2 is new leaving variable
making space for entering variable x2.
New rows for table will be computed as follows.
New pivot (x2) row
Quantitative Techniques in
Management : 18
Linear Programming
1
(current s2 row)
=
4/3
3
(0
=
0
4/3
-1/6
1
0
0
2)
NOTES
4
New z row
= current z row - (-2/3) (new x2 row)
Similarly new x1, s3 and s4 rows will be computed. Resulted table will be as
follows:
Basic
z
x1
x2
s1
s2
s3
s4
Solution
z
1
0
0
3
/4
1
/2
0
0
21
x1
0
1
0
1
/4
- 1 /2
0
0
3
x2
0
0
1
- 1 /8
3
/4
0
0
3
/2
s3
0
0
0
3
/8
- 5 /4
1
0
5
/2
s4
0
0
0
1
/8
- 3 /4
0
1
1
/2
Again using the condition of optimality, we see that all coefficients in z row are
non negative. Hence current table is optimal.
We read optimal solution as follows :
Z = 21 with x1=3 and x2 = 3/2
Since S3 = 5/2 and s4 =1/2 are also present in the final table, we conclude that
resource 3 and resource 4 are available in abundant. As S1 and S2 are not part of final
solution, we conclude that first two resourices are scarce.
Example 7 : Maximize Z - 2x1 +x2 - 3x3 + 5x4
Subject to
x1 + 2x2 + 2x3 + 4x4 < 40
2x1 - x2 + x3 + 2x4 < 8
4x1 - 2x2 + x3 - x4 < 10
x1, x2, x3, x4 > 0
Solution : Convert constraints into equations :
x1 + 2x2 + 2x3 + 4x4 + s1
= 40
2x1 - x2 + x3 + 2x4
=8
4x1 - 2x2 + x3 - x4
+ s2
+ s3
= 10
x1, x2, x3, x4, s1, s2, s3 > 0
Initial basic solution can be obtained by putting x1, x2, x3, x4 = 0
so initial basic feasible solution is
(s1, s2, s3) = (40, 8, 10)
Quantitative Techniques in
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Linear Programming
Simplex tables will be as follows : ↓
NOTES
↓
Basic z
x1
x2
x3
x4
s1
s2
s3
Solution
z
1
-2
-1
3
-5
0
0
0
0
Ratio
s1
0
1
2
2
4
1
0
0
40
10
s2
0
2
-1
1
2
0
1
0
8
4
s3
0
4
-2
1
-1
0
0
1
10
-10 (ignored)
Second table will have s1, x4, s3 as basic variables
↓
↓
Basic z
x1
x2
x3
x4
s1
s2
s3
solution
z
1
3
- 7 /2
11
0
0
5
/2
0
20
Ratio
s1
0
-3
4
0
0
1
-2
0
24
6
x4
0
1
- 1 /2
1
/2
1
0
1
/2
0
4
-8 (ignored)
s3
0
5
- 5 /2
3
/2
0
0
1
/2
1
14
-28/5 (ignored)
/2
Third table will have x2, x4 and s3 as basic variables.
Basic z
x1
x2
x3
x4
s1
s2
s3
Solution
z
1
3
/8
0
9
0
7
/8
3
/4
0
41
x2
0
- 3 /4
1
0
0
1
/4
- 1 /2
0
6
x4
0
5
0
1
1
1
/8
1
/4
0
7
s3
0
25
0
4
0
5
/8
- 3 /4
1
29
/8
/8
/2
Check with optimality condition. This last table is optimal one, giving z = 41 with x2
= 6 and x4 = 7. Two of the decision variables, x1 and x3 are non basic in final solution,
i. e. these are with zero values. Since S3 is also in the final table, it means that third
resource is in abundance.
Solution of a LPP using surplus variable :
As shown in Example 6 and 7, LPs in which all constraints are (<) with non
negative RHS can be handled by putting slack variables in constraints to convert these
inequalities into equations. But in case of (>) or (=) constraints, we need to do some
extra efforts to bring these inequalities into standard form of simplex processing.
If (=) constraints are involved, we will use artificial variables only in those
equations. If (>) constraints are involved, we will use surplus and artificial variables in
these inequalities.
Quantitative Techniques in
Management : 20
To handle artificial variables, two popular methods are the M-method and the two
- phase methos.
Purpose of artificial variable is to help in providing initial basic feasible solution of
the problem. Both these methods are developed in such a way that artificial variables
will disappear from final solution.
Note : Students can think that slack and surplus variables may be part of final
solution while artificial variables are not. There is a meaning of slack and surplus variable
while artificial variables have no physical meaning.
Linear Programming
NOTES
M-Method : When we use artificial variables in case of (=) or (>) type constraints,
we put a coefficient M with this variable in the objective function of the problem. M is
considered to be a very large quantity. The purpose of M is to assign heavy penalty for
this artifical variable. So as it does not remain in the solution. If problem is a maximization
one we use -M as artificial variable objective coefficient and if problem is a minimization
one, we use +M as coefficient.
Example 8 : Minimize Z = 3x1 + 8x2
Subject to
x1 + x2 = 200
x1 < 80, x2 > 60; x1, x2 > 0
Solution :
Conversion of inequalities into equations : First constraint will have artificial variable,
R1. Second constraint will have slack variable S1. Third constraint will have surplus
variable S2 and one artificial variable R2.
Since it is a minimization problem R1 and R2 will have +M as objective coefficients
while S1 snd S2 will have "0" objective coefficients.
Problem in standard form will be as follows :
Minimize Z = 3x1 + 8x2 + MR1 + MR2
Subject to
x1 + x2
x1
+ R1
+ s1
= 80
- s2
x2
= 200
+ R2 = 60
x1, x2, s1, s2, R1, R2 > 0
The associated starting basic solution is given by (R1, S1, R2) = (200, 80, 60)
The starting table will be as follows :
Basic
z
x1
x2
S1
S2
R1
R2
Solution
z
1
-3
-8
0
0
-M
-M
0
R1
0
1
1
0
0
1
0
200
S1
0
1
0
1
0
0
0
80
R2
0
0
1
0
-1
0
1
60
Quantitative Techniques in
Management : 21
Linear Programming
Be Careful!
Before we apply condition of optimality for minimization problems here, we need
to make z row consistent with rest of the table.
NOTES
Please check, that current basic solution is (R1, S1, R2) = (200, 80, 60) yielding z =
260M while table shows solution of Z as zero. This incosistency stems from the fact
that R1 and R2 have nonzero coefficients in the z row.
To eliminate this inconsistency, we will perform some row operations. In particular,
see the highlighted elements in the R1 - row and the R2 - row.
Multiplying each of R1-row and R2 - row by M and adding the sum to the current
Z row, we get new Z row.
(-3
-8
0
0
-M
-M
0)
+M
(1
1
0
0
1
0
200)
+M
(0
1
0
-1
0
1
60)
-3+M
-8+2M
0
-M
0
0
260M
Modified table thus becomes
Basic z
x1
Z
1
R1
x2
↓
s1
s2
R1
R2
Solution
(-3+M) (-8+2M) 0
-M
0
0
260M Ratio
0
1
1
0
0
1
0
200
200
S1
0
1
0
10
0
0
0
80
ignored
R2
0
0
1
0
-1
0
1
60
60
This Modified table is a consistent one and fit for applying condition of optimality
and Gauss - Jordon row operations.
Applying optimality condition, we will notice that most positive Z row coefficient
is (-8+2M). Thus x2 is entering variable. (Students are advised to compare (-3+M) and
(-8+2M) for most positivity.)
After comparing the ratios obtained from solution column and pilot column elements,
R2 is chosen as leaving variable.
After applying Gauss - Jordon row operations for new pivot row and for other
rows including Z row, table obtained after iteration is as follows :
↓
Quantitative Techniques in
Management : 22
Basic
z
x1
Z
1
R1
↓
x2
s1
s2
R1
R2
Solution
(-3+M) 0
0
M-8
0
+8-2M 140M + 480
0
1
0
0
1
1
-1
140
S1
0
1
0
1
0
0
0
80
x2
0
0
1
0
-1
0
1
60
New table, after x1 becomes entering variable and s1 is leaving variable, will be as
follows :
Basic
z
x1
x2
s1
s2
R1
R2
Solution
Z
1
0
0
-M+3
M-8
0
8-2M
60M+720
R1
0
0
0
-1
1
1
-1
60
x1
0
1
0
1
0
0
0
80
x2
0
0
1
0
-1
0
1
60
Linear Programming
NOTES
In this table, using condition of optimality S2 will be entering and R1 will be leaving
variable.
New table after iteration will be
Basic
z
x1
x2
s1
s2
R1
R2
Solution
z
1
0
0
-5
0
-8M
-M
1200
s2
0
0
0
-1
1
1
-1
60
x1
0
1
0
1
0
0
0
80
x2
0
0
1
-1
0
1
0
120
This table is optimal Verify!
Optimal solution is Minimum Z = 1200
with x1 = 80 and x2 = 120
Two phase Method :
In two phase method, problem is solved in two phases. Phase I attempts to find a
starting basic feasible solution, and, if one is found, phase II is involed to solve the
original problem.
Example 9 : We will use same problem of example 8 to show various steps of
two- phase method.
Solution :
Phase I
Minimize r = R1 + R2
(A new objective function is defined which is always a minimization irrespective
of nature of original problem. This objective function is sum of artificial variables used
in problem to convert (=) or (>) type constraints into standard form.)
Subject to
x1 + x2
x1
x2
+ R1
+ s1
= 200
= 80
- s2 + R 2
= 60
x1, x2, s1, s2, R1, R2 > 0
Quantitative Techniques in
Management : 23
Linear Programming
NOTES
The associated table is given as with R1, S1, and R2 as initial basic variables.
Basic
x1
x2
s1
s2
R1
R2
Solution
Y
0
0
0
0
-1
-1
0
R1
1
1
0
0
1
0
200
S1
1
0
1
0
0
0
80
R2
0
1
0
-1
0
1
60
To make this table consistent for use of Gauss - Jordan iteration process, as in the
M-Method, R1 and R2 are substituted out in the r-row by using the following row
operation:
New r-row = old r row + (1xR1 row + 1xR2 row)
As a result, modified table is :
Basic
x1
x2
s1
s2
R1
R2
Solution
r
1
2
0
-1
0
0
260
R1
1
1
0
0
1
0
200
S1
1
0
1
0
0
0
80
R2
0
1
0
-1
0
1
60
After two iterations, optimal table will be :
Basic
x1
x2
s1
s2
R1
R2
Solution
r
0
0
0
0
-1
-1
0
s2
0
0
-1
1
1
-1
60
x1
1
0
1
0
0
0
80
x2
0
1
-1
0
1
0
120
Bacause minimum r = 0, phase I produces the basic feasible solution x1 = 80 and
x2 = 120 and s2 = 60
Check that artificial variables (R1 and R2) are not part of this solution. These
variables have done their job of providing a feasible basic solution.
We are now ready for phase II computations.
Phase - II
(i)
Delete columns of artificial variables from optimal table of phase - I.
(ii) Rewrite the problem with original objective function and new constraints
derived from optimal table of phase - I.
Minimize Z = 3x1 + 8x2
Subject to
Quantitative Techniques in
Management : 24
-s1, +s2
= 60
x1 + s 1
= 80
r2 -s1
= 120
Linear Programming
NOTES
x1, x2, s1, s2 > 0
So, associated table with this set of objective function and constraint equations is
Basic
x1
x2
s1
s2
Solution
z
-3
-8
0
0
0
s2
0
0
-1
1
60
x1
1
0
1
0
80
x2
0
1
-1
0
120
In this table also we can see that two basic variables x1 and x2 have non zero
coefficients in z row. These coefficients need to be converted into zeros and associated
row operation required will be
New z row = old z row + (3xX1 row + 8xX2 row)
The initial table of phase II is thus given as
Basic
x1
x2
s1
s2
Solution
z
0
0
-5
0
1200
s2
0
0
-1
1
60
x1
1
0
1
0
80
x2
0
1
-1
0
120
In this table when we apply condition of optimality we realize that it is an optimal
solution. (In a minimization objective function, optimality is achieved when Z row has
no positive coefficient.)
So optimal (minimum) z = 1200 with x1 = 80 and x2 = 120.
1.11 Special Cases in Simplex Method
As mentioned earlier also, similar special cases can be seen during application of
simplex calculations.
(1) Alternative optima or multiple optimal Solution :Consider the following LP formulation :Max. z = 2x1 + 4x2
Subject to
Quantitative Techniques in
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Linear Programming
6x1 + 4x2 < 24
x1 + 2x2 < 6
-x1 + x2 < 1
NOTES
x2 < 2
x 1 , x2 > 0
After putting this LPP in tabular arrangement :
Iteration
Basic
x1
x2
s1
s2
s3
s4
Solution
0
z
-2
-4
0
0
0
0
0
s1
6
4
1
0
0
0
24
s2
1
2
0
1
0
0
6
s3
-1
1
0
0
1
0
1
s4
0
1
0
0
0
1
2
z
0
0
0
2
0
0
12
s1
0
0
1
-6
0
8
4
s3
0
0
0
1
1
-3
1
x2
0
1
0
0
0
1
2
x1
1
0
0
1
0
-2
2
5
z
0
0
0
2
0
0
12
Alternative
s4
0
0
0.13
-0.75
0
1
0.50
Optimal
s3
0
0
0.38
-1.25
1
0
2.50
(S4 enters
x2
0
1
-0.13
0.75
0
0
1.50
and S1 Leaves x1
1
0
0.25
-0.50
0
0
3.00
4 (optinal)
Here iteration 4 gives optimal solution with basic variables x1, x2, S1 and s3. In
optimal solution one of the nonbasic variable S4 is having 0 coefficient in z row. This
gives a chance of alternative optima condition. After forcing S4 into the solution, checking
for minimum ratio, S1 becomes leaving variable. Iteration 5 gives another optimal solution
without changing value of solution. In both iterations, 4 and 5, maximum value of Z
remains unchanged, i. e. 12. But in iteration 4, basic variables are x1 = 2, x2 =2, s1 = 4
and s3 = 1, while in iteration 5, basic variables are x1 = 3, x2 = 1.50, s3 = 2.50 and s4 =
0.50
In practical terms, alternative solutions are useful because we can choose from
many solutions without affecting the quality of the objective value.
(2) Unbounded solution :
Consider the following LP formulation
Quantitative Techniques in
Management : 26
Linear Programming
Maximize
Z = 2x1 + 3x2
Subject to
x1 - 2x2 < 15
NOTES
3x1 < 50
x 1 , x2 > 0
Let us make tabular arrangement for this LP.
Iteration
0
Basic
x1
x2
s1
s2
Solution
z
-2
-3
0
0
0
s1
1
-2
1
0
15
s2
3
0
0
1
50
This is a maximization problem and x2 is a candidate to enter into solution space.
But neither s1 nor s2 can leave the solution. This means that x2 can be increased
indefinitely without violating any of the constraints.
Situation of unboundedness is a result of poor model making. It is possible that a
constraint which restricts value of x2 to be increased infinitely is missing.
(3) Infeasible Solution :
Infeasible solution can occur when constraints are having opposite signs of
inequalities.
consider the following problem :
Maximize z = 4x1 + 3x2
Subject to
x1 + 2x2 < 5
2x1 + 3x2 > 10
x1, x2 > 0
Using slack variable S1 in first constraint and S2 surplus variable in second constraint
alongwith artificial variable R1 in second constraint and making this LP model in standard
form.
(We are using a value of 100 for M in the objective function.)
Lteration Basic
x1
x2
s1
s2
R1
RHS
0
Z
-204
-303
0
100
0
-1000
S1
1
2
1
0
0
5
R1
2
3
0
-1
1
10
Z
0
105
204
100
0
20
x1
1
2
-1
-
-
5
R1
0
-1
-2
-1
1
0
2
Quantitative Techniques in
Management : 27
Linear Programming
Second iteration is optimal. But this solution has x1 and artificial variable R1 as
basic variables. Artificial variable, if present in final solution, will give a case of infeasible
solution.
Infeasible solution is a result of poor model construction.
NOTES
(4) Degeneracy :
Sometime it is possible to have a situation of tie for deciding leaving variable.
Minimum ratio is same for two current basic variables. This tie can be broken arbitrasily.
In this case, at least one basic variable will be zero in the next iteration and the new
solution is said to be degenerate.
Consider the following problem.
Maximize z = 2x1 + 5x2
Subject to
x1 + 3x2 < 6
x1 + 2x2 < 4
x1, x2 > 0
Given the slack variables s1, and s2, the following tables provide the simplex
iterations of the problem :
Iteration
Basic
x1
x2
s1
s2
RHS
0
z
-2
-5
0
0
0
x2 enters, s1 or s2
s1
1
3
1
0
6
may leave. Take s1 s 2
1
2
0
1
4
z
- 1/ 3
0
5
/3
0
10
x2
1
/3
1
1
/3
0
2
s2
1
/3
0
- 2/ 3
1
0
as leaving variable.
1
See one of the current basic variable s2 is zero now.
Here x1 will enter and s2 will leave.
2
z
0
0
1
1
10
(Optimum)
x2
0
1
1
-1
2
x1
1
0
-2
3
0
In degeneracy situation, objective value of the function may not improve but
optimality condition remains unsatisfied. It is normally a case of overdetermined problems.
1.12 Sensitivity Analysis
Quantitative Techniques in
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Sensitivity analysis tells range of input parameter variation without changing optimal
solution. Following two cases will be discussed to explain the concept of sensitivity
Linear Programming
analysis :
(a) Changes in the RHS of the constraints
(b) Changes in the objective function.
(a) Changes in the RHS of the constraints :
NOTES
Consider the following problem :
Maximize z = 2x1 + x2 + 4x3
Subject to
x1 + 2x2 + x3 < 30
(Resource 1)
3x1 + 3x3 < 60
(Resource 2)
x1 + 4x2 < 20
(Resource 3)
x1, x2, x3 > 0
Here availbility of resources are 30 units, 60 units and 20 units respectively. We
can change availability of these resources within a limit without altering current optimum
solution. We want to determine limits of these changes.
First let us see the optimum table for original model using S1, S2 and S3 as slack
variables in three constraints, respectively.
1.13 Summary
1.14 Key Terms
Feasible Solution : A solution which satisfies all constraints including non-negativity
constraints.
Op[timum Solution : A feasible solution which gives either maximum or minimum
value of objective function.
Quantitative Techniques in
Management : 29
Linear Programming
Slack variable : It represents unused amount of a resource.
Surplus variable : It represents shortage of a resource.
NOTES
Artificial variable : These variables have no physical meaning, except to help in getting
a starting solution.
Basic variables : Decision, slack, surplus or artificial variables which are part of
solution are known as basic variables. In other words, non zero variables are
basic variables.
Non-basic variables : All zero variables are non-basic variables.
Pivot Column : In a simplex table, column of entering variable.
Pivot Row : In a simplex table, row of leaving variable.
1.15 Questions and Exercises
PART - A
Que. 1. A Paint company makes 3 grades of paint (α, β, γ ) from 3 raw materials
(A, B, C). Raw materials are used in making of these 3 grades of paint as follows :
Grade Specifications
Unit selling (Rs.)
price (in per litre)
α
8.0
Not less than 50% raw material `A`
Not less than 25% raw material `B`
β
Not less than 25% raw material `A`
6.5
γ
Not less than 50% raw material `B`
5.5
Within above restrictions, raw materials can be used in any grade of paint. There
are capacity limitations on the amounts of three raw materials that can be used :
Raw material
Capacity
Price / litre
A
500 litres
9.5
B
500 litres
9.5
C
300 litres
6.5
It is required to produce the maximum profit. Formulate a LP model for the problem.
Quantitative Techniques in
Management : 30
Que. 2. A manufacturing company has contracted to deliver home windows over
the next 4 months. The demands for each month are 100, 250, 190 and 110 units
respectively. Production cost per window varies from month to month depending on the
cost of labour, material and utilities. The company estimates the production cost per
window over the next 4 months to be Rs. 500, Rs. 450, Rs. 550 and Rs. 500 respectively.
To take advantage of the fluctuations in manufacturing cost, company may elect to
produce more than is needed in a given month and hold the excess units for delivery in
later months. This however, will incur holding cost at the rate of Rs. 30 per window per
month assessed on end- of - month inventory. Formulate a LP model to determine the
Linear Programming
optimum production schedule.
Que 3. Find solution of following problems using graphical method :
(a) Max
Z=
30x1 + 40x2
NOTES
subject to
x1< 20
x2 > 10
4x1 + 2x2 < 100
4x1 + 6x2 < 180
x1 + x2 < 40
x1, x2 > 0
and
(b) Max. Z = 150 x1 + 250 x2
Subject to
x1
/1500 + x2/4500 > 1
x1 1000 + x2 8000 > 1
/
/
x 2000 + x 4000 > 1
/
/
x 3000 + x 9000 > 1
/
/
and
(c) Min.
1
2
1
2
x1, x2 > 0
20x1 + 40x2
Z=
Subject to
36x1 + 6x2 > 108
3x1 + 12 x2 > 36
20x1 + 10x2 > 100
and
x1, x2 > 0
1.16 Further Reading and References
Quantitative Techniques in
Management : 31
Transportation Model
UNIT 2
NOTES
TRANSPORTATION MODEL
Structure
2.0 Introduction
2.1 Unit Objectives
2.2 Transportation Model
2.3 Steps in Solution Process
2.3.1 Initial Basic Feasible Solution
2.3.2 Method of Multiplier for Optimality Test
2.4 Summary
2.5 Key Terms
2.6 Question & Exercises
2.7 Further Reading and References
2.0 Introduction
Trasportation problems are a special class of linear programming problems.
Transportation problems deal with physical distribution of goods and services from
various supply sources to various demand centres. The structure of a transportation
problem involves a large number of shipping routes from several supply origins to
several demand destinations. The objective is to determine the number of units of an
item which should be shipped from an origin to a destination in order to satisfy the
required quantity of goods or services at each destination centre, within the limited
quantity of items available at each supply centre, at the minimum transportation cost
and/or time.
Transportation models are widely used in supply chain management.
2.1 Unit Objectives
After studying this unit, you should be able to
Quantitative Techniques in
Management : 32
Understand basic structure of transportation model as a special case of linear
programming model.
Understand north-west corner method, least coast method and vogel’s
approximation method to obtain initial basic feasible solution of
transportation model.
Understand process of checking optimality of basic feasible solution using
method of multipliers.
Transportation Model
2.2 Transportation Model
Sources
Destinations
c11 : x11
NOTES
1
b1
2
2
b2
a3
3
3
b3
am
m
n
bn
a1
1
a2
C 11 : X11
Cmn : Xmn
Cmn : Xmn
Fig : 2.1
Fig
: 2.1
Figure 2.1 gives a general type of transportation model. Here we have taken m
supply sources and n demand centres (destinations). Each source can supply to each
destination. If i is a source and j is a particular destination then xij units are shipped
from ith source to jth destination at unit cost of transportation cij. Objective of
transportation model is to ensure fullfillment of demand of each destination such as b1,
b2, .......... bn etc. But no supply centre can supply more than its capacity such as a1, a2
.... am.
Finally, distribution of units from sources to destination is designed in such a way
to incur minimum cost of distribution.
Take following case, before we move to develope transportation model.
Example : A company has two production facilities at Mumbai and Nagpur with
production capacity at 1000 and 800 products per day, respectively. Three warehouses
at the company are at Pune, Gondia and Chandrapur with daily requirements of 900,
400 and 500 products, respectively. Transportation cost (in rupees) per unit between
production facilities to warehouses is given in table 1.
Table 1
Warehouses →
Pune
Gondia
Chandrapur
Mumbai
4
8
7
Nagpur
6
4
3
→
Production facility
With the help of knowledge of liner programming of Unit 1, we can make
following linear programming model :
Consider following decision Variables
X11
→
Units to be transported from Mumbai to Pune
X12
→
Units to be transpoted from Mumbai to Gondia
Quantitative Techniques in
Management : 33
Transportation Model
NOTES
X13
→
Units to be transported from Mumbai to Chandrapur
X21
→
Units to be transported from Nagpur to Pune
X22
→
Units to be transported from Nagpur to Gondia
X23
→
Units to be transported from Nagpur to Chandrapur
Objective function will be
Minimize (total transportation cost) Z
= 4x11 + 8x12 + 7x13 + 6x21 + 4x22 +3x23
Subject to the constraints
(i) Capacity Constraints
x11 + x 12 + x 13 = 1000
x 21 + x 22 + x 23 = 800
(ii) Requirement Contraints
x11 + x 21 = 900
x12 + x 22 = 400
x13 + x 23 = 500
and x11, x12, x13, x21, x22, x23 > 0
The LP model can be solved by the simplex method. However, with the special
structure of the constraints, we can solve the problem more conveniently using the
trasportation table shown the table 2.
Table 2 : Trasportation model for examples 1
Destination →
Pune
Gondia
Chandrapur
Supply
→
Sources
4
8
x12
7
Mumbai
x11
Nagpur
x21
x22
x23
Demand
900
400
500
6
x13
4
1000
3
1000 800
In each cell of table 2, we write units to be transported from one source to a
destination (xij) in lower left side and unit cost of transportation between this pair of
source and destination in upper right hand corner.
Taking table 2 and figure 1 into account, we can develop general mathematical
model of trasportation problem :
m
Min Z =
n
Σ
Σ
i =1
j=1
Subject to the constraints
Quantitative Techniques in
Management : 34
cij xij
n
Σ
Transportation Model
xij = ai i = 1,2 ...... m (supply Constraint)
j =1
m
Σ
NOTES
xij = bj j = 1,2 ...... n (demand Constraint)
i =1
and xij > 0 for all i and j
A necessary and sufficient condition for the existence of a feasible solution to the
trasportation problem is
Total Supply = Total demond
m
n
Σ ai = Σ bj
i=1
j=1
If the problem is unbalanced, we can always add dummy source or a dummy
destination to make the original problem balanced. Transportation Costs in dummy
row or dummy column are always zero.
If in example 1, demand at Pune is 1200 instead of 900 given originally. This will
make total demond of 2100 units daily against a production of 1800 Units daily from
two plants. To solve such unbalanced model, we will add a new dummy source with a
capacity of 300 units. The capacity of new source is total demond - total supply available.
The revised balanced model with dummy source is given in Table 3.
Table 3 : Transportation model with dummy source
Source
Pune
Gondia
4
Chandrapur
8
x12
7
Mumbai
x11
Nagpur
x21
Dummy Source
x31
x32
x33
Demand
1200
400
500
6
x13
4
x22
0
Supply
1000
3
x23
0
800
0
300
Again consider example 1, and see if supply available at both these plants are 100
units daily.
This will produce 2000 units daily againts a total daily demand of 1800 units.
This is also unbalanced problem. We will add a new column (new destination) with a
supply requirement of 200 units. Demand of new destination is total supply- total
demand. The revised balanced model with dummy destination is given in table 4.
Quantitative Techniques in
Management : 35
Transportation Model
NOTES
Table 4 : Transportation model with dummy destination
Source
Pune
Gondia
Mumbai
x11
Nagpur
x21
x22
x23
x24
Demand
900
400
500
200
4
Chandrapur
8
x12
6
Dummy Destination
7
x13
4
Supply
0
x14
3
1000
0
1000
2.3 Steps in Solution Process
The transportation problem can be solved with exact steps of the simplex method.
But special structure of transportation problem gives an advantage. We can directly
solve transportation problem without going for complex computations of simplex
method.
Following steps are used in solving a transportation problem.
1.
Formulate the problem and arrange the data in matrix form as shown in table 2.
2.
Detemine a starting basic feasible solution. Following three methods can be
used to get initial basic feasible solution.
North- West Corner Method (N-W-C Method)
Least Cost Method (LC Method)
Vogel's Approximation Method (VAM)
3.
Test initial solution for optimality. If the optimality condition is satisfied stop,
otherwise determine the entering variable from among all the nonbasic variables
and go to step 4.
4.
Use the feasibility condition to determine the leaving variable from among the
current basic variables and find the new basic solution. Return to step 3, till
optimality is reached.
2.3.1 Initital Basic Feasible Solution : (IBFS)
The IBFS or any feasible solution must satisfy all the supply and demand
conditions. If m sources and n destination are shown in a model, it must have m +n -l
allocations, i.e. basic variables.
[Please note that all cells of transportation table will not have positive allocations
only m + n -1 cells will have allocations.]
1.
Northwest Corner Method
N-C method is the simplest method to get a starting solution. However it can not
provide a high quality starting solution. The initial value of objective function is relatively
higher as compared to other methods to get starting solution.
The method starts from the northwest- corner cell of the table.
Quantitative Techniques in
Management : 36
1
Consider table 2, we will all allocate
900 units in the Northwest- corner cell.
1
3
supply
4
8
7
1000
100
6
4
3
800
900
This will complete allocations of
column 1 and supply from row 1 is reduced
to 100 Units only.
2
Demand 900
Transportation Model
NOTES
400 500
Table 2 A
This is shown in table 2 A Now we
will Cross the column of zero demand. (In
case all supplies are exhausted, we will cross row)
Now we will move horizantally to
Cell (1, 2). In table 2 B, we allocated
100 units in cell (1,2) and crossed row
1 as supplies are exhausted. Demand
of column 2 are adjusted to 300 units.
2
1
2
1
4
900
2
6
Demand 900
3
Supply
8
7
100
1000 100
4
800
3
400 500
Moving in this fashion table 2 C
300
gives complete allocation and also path
Table 2 B
of making these allocations. Check that
only four Cells are allocated. Applying rule of m + n - 1 = 3 + 2 -1 = 4, the current
allocation is feasible one.
The associated cost of this allocation is
Z = 900 x 4 + 100 x 8 + 300 x 4 + 500 x 3
= 7100 (Rs.)
[Please note that if both a row and a column
net to zero simultaneously, cross out one only,
and leave a zero supply (demand) in the uncrossed
out row (column). ]
2.
Least Cost Method
The quality of solution with LC method is
better than the solution with N-W-C method.
1
1
2
4
900
2
6
900
3
Supply
8
7
100
1000
4
3
800
300 500
400 500
Table 2 C
The method starts with assigning as much as possible to the cell with the Smallest
Unit Cost (ties are broken arbitrarily). Next we cross satisfied row or column and the
amount of supply and demand are adjusted accordingly. As in case of N-W-C method,
here also, if both a row and a column are satisfied simultaneously, only one is crossed
out.
Next we check smallest cost cell from among the uncrossed rows and columns
and repeat the process till all supply/ demand conditions are satisfied.
Table 2 D gives final allocation with
L-C method. The order of allocation is also
given in cells in circled numbers.
First we allocated to cell (2, 3) 500 units
satisfying column 3 and adjusting supply
from row to 300 units. Now it was tie
between cell (1, 1) and (2, 2) as both these
1
2
1
2
3
4
900 3
8
100 4
7
6
4
300 2
3 800
500 1
900
400
1000
Table 2 D
500
Quantitative Techniques in
Management : 37
Transportation Model
cells had same unit transportation cost of Rs. 4.
We arbitrarily allocated to cell (2, 2) and process was completed in four allocations.
The assoiciated cost of this allocation is Rs. 7100. (Same as in N-W-C method.)
NOTES
3.
Vogel Approximation Method (VAM)
VAM is an improved method to get starting solution. It normally produces best
quality of starting solution. In this method, each allocation is made on the basis of the
opportunity (or penalty) cost that would have been incurred if allocation in certain
cells with minimum unit transportation cost were missed. The procedure is as
follows :
1.
Calculate row/ column penalties :- This is done by subtracting the smallest unit
cost element in the row (column) from the next smallest unit cost element in the
same row (column).
2.
Select the largest panalty in rows
or columns :- Ties are broken
arbitrarily. Now allocate in cell
with least cost in the selected or
column keeping rim conditions in
mind. Adjust the supply and
demand, and cross out the satisfied
row or column. If a row and a
column
are
satisfied
simulataneously. Only one of the
two is crossed out, and the
remaining row (column) is
assigned zero supply (demand).
3.
Row
Penalty
1
2
3
1
4
8
7 1000 7-4=3
2
6
4
3 800
900
400
500
6-4
8-4
7-3
=2
=4
=4
If only one row or column with
zero supply or demand remain
uncrossed out, stop the process. In
case of positive supply or demand
in place of zero supply or demand,
allocate this to least cost cell in the
row or column.
4-3=1
Table 2 E
New
Penalty
1
2
3
1
4
8
7 1000 8-4=4
2
6
4
900
400
500
New 6-4
Penalty =2
8-4
=4
-
3 800 6-4=2
500 300
Table 2 F
If all the uncrossed out rows and columns have zero supply and demand, determine
the zero basic variables by L-C method.
Consider again table 2, table 2 E gives penalties of rows and colums.
As seen in table 2 E Column 2 and 3 are having similar penalties of 4. We arbitratily
decide to select column 3. Here cell (2, 3) is
1
2
3
with least Cost, So we will allocate 500 units
1
4
8
7 1000
(follow rim conditions).
Table 2 F shows first allocation and
calculation of new penalties.
Quantitative Techniques in
Management : 38
Continuing in the same manner, we get
table 2G, allocating in four cells and
2
900
100
6
4
300
3 800
500
900
400
500
Table 2 G
Transportation Model
satisfying all rim conditions.
The associated objective function value is 7100 Rs. (Same as in case of N-W-Cmethod and L-C method.) It is a matter of chance that objective function value in case
of all three methods is same. Normally objective function value improves (decreases)
from Northwest corner method to least cost method to vogel’s approximation method.
NOTES
Practice Questions
1. Compare the starting solution obtained by the N-W-C, L-C and VAM for each
of the following models.
(a)
D1
D2
D3
D4
Supply
S1
21
16
15
3
11
S2
17
18
14
23
13
S3
32
27
18
41
19
Demand
6
10
12
15
1
2
3
Supply
1
1
2
6
7
2
0
4
2
12
3
3
1
5
11
10
10
10
(b)
Destinations →
→
Source
Demand
(c)
Destinations →
1
2
Supply
1
80
215
1000
2
100
108
1500
3
102
68
1200
Demand
1900
1400
→
Source
2.3.2 Method of multiplier for Optimality Test
Once a starting feasible solutions is obtained, we now test it for optimality.
1.
We determine the entering variable from among the current nonbasic variables
that can improve (reduce the total transportation cost) the solution. If optimality
condition is satisfied, stop otherwise go to step 2.
2.
If there is an entering variable, then we determine the leaving variable using the
simplex feasibility condition. After getting new basic solution, return to step 1.
Quantitative Techniques in
Management : 39
Transportation Model
Example 2 : Take the following model and solve for optimality
1
2
3
Supply
1
5
1
8
12
2
2
4
0
14
3
3
6
7
4
Demand
9
10
11
NOTES
Table 5
Solution :- Table 5A gives starting solution of model given in Table 5, with NorthWest Corner Method
(Students verify feasibility condition in table 5A) (Students should also develop
IBFS using LC Method and VAM)
To determine entering variable in table 5A,
we use method of multipliers. In this method, we
associate the multipliers Ui and Vj with row i and
column j of the transportation table 5A.
For each current basic variable xij, we can
write
1
1
2
3
2
3
5
1
900 (3) 300 (4)
8
12
3
4
0
00 (3) 700 (4) 700000 4
3
9990 (3)
ui + vj = cij for each basic xij
9
6
7
00 (4) 400000 4
10
11
Table 5 A
where cij is the unit transportation cost of
moving single unit from ith source to jth destination.
For table 5A, these are arranged as follows :
Basic Variable
(u,v) Equation
Solution
X11
u1 + v1 = 5
Take u1 =0, v1 = 5
x12
u1 + v2 = 1
v2 = 1
x22
u2 + v2 = 4
u2 = 3
x23
u2 + v3 = 0
v3 = -3
x 33
u3 + v3 = 7
u3 = 10
We have Six unknowns, namely u1, u2, u3, v1, v2 and v3 but 5 (u, v) equations. So
we have taken arbitrarily u1 = 0 to get values of other multipliers.
Now we will use these (u, v) values to evaluate other nonbasic variables by
computing
ui + vj - cij , for each nonbasic Xij
Quantitative Techniques in
Management : 40
The results of those evaluations are shown in the following table :
Nonbasic Variable ui + vj - cij
x13
u1 + v3 - C13 = 0-3-8 = -11
x21
u2 + v1 - C21 = 3 + 5 -2 = 6
x31
u3 + v1 - C31 = 10+ 5 - 3=12
x32
u3 + v2 - C32 =10 + 3 - 6=7
Transportation Model
NOTES
As transportation models are mostly cost minimization model, we take most
positive ui + vj - cij as our entering variable. If no current nonbasic variable is with
positive sign, the current solution is the optimal one.
Here x31 is the entering variable.
Now we determine leaving variable. One of the current basic variable will become
nonbasic. The selection of x31 as the entering variable means that we want to allocate to
this cell, but how much? If α is the quantity allocated to this cell, we always need to be
careful about rim conditions and non negativity constraints.
Based on these conditions, we determine maximum value of α for x31 cell.
The process is as follows :
(1) Construct a closed loop starting and ending at the entering cell; (3, 1) in this
case.
This loop consists of connected horizantal and vertical line segments only.
Diagonals are not allowed. Except for the entering cell, each corner of this loop must
have one of the current basic variable. Exactly one loop exists for a given entering
variable.
Table 5B shows the loop for x31 for current example.
1
2
3
Supply
1
5
9-α
1
3+α
8
12
2
2
4
7-α
0
7+α
14
3
6
7
4-α
4
3
Demand
α
9
10
11
Table 5 B
Loop represented by
lines starting from (3, 1) to (3, 3) to (2, 3) to (2, 2) to (1,
2) to (1, 1) and finally to (3, 1) is the desired and unique one.
If we are allocating α to (3, 1) cell, to balance the rim conditions, we will subtract
α from (3, 3), add α in (2, 3) and substract α in (2, 2) add α in (1, 2) and finally
subtract α in (1,1). Students can observe a pattern in adding and subtracting α at the
corner points of loop.
The loop can be traced clockwise or anti clockwise without affecting the solution.
Quantitative Techniques in
Management : 41
Transportation Model
Since α will have nonnegative value, the new values of the variables then remain
nonnegative if
x11 = 9 - α > 0
x22 = 7 - α > 0
NOTES
x33 = 4 - α > 0
The corresponding maximum value of α is 4, which occurs when x33 reaches zero
level. So x33 is the leaving variable.
So x31 is entering and x33 is leaving variable. This requires adjustments in table
5A. New table after this current iteration is shown as 5C.
1
2
1
1
3
8
12
4
0
14
7
4
5
5
7
2
2
3
3
3
4
9
11
6
10
11
Table 5 C
The associated cost is
5 x 5 + 7 x 1 + 3 x 4 + 11 x 0 + 4 x 3
Rs = 56
Ealier cost from table 5 A was Rs. 104. So solution has improved by Rs. 48.
This can also be verified as follows :
ui + vj - cij for x31 = 12, allocation in (3,1) = 4 units
So cost will reduce by 12 x 4 = Rs. 48
But we will check this solution of table 5 C for deciding a new entering variable
from current non basic variables.
Again make (U, V) equations for current basic variables, as follows : (This basic
variables are taken from Table 5C)
x11
u1 + v1 = 5
Take u1 = 0 v1 = 5
x12
u1 + v2 = 1
v2 =1
x22
u2 + v2 = 4
u2 =3
x23
u2 + v3 = 0
v3 = -3
x31
u3 + v1 = 3
u3= -2
Now make (u, v) equations for current nonbasic variables
x13
u1 + v3 - C13 = 0 - 3 -8 = -11
x21
u2 + v1 - C21 = 3 + 5 -2 = 6
x32
u3 + v2 - C32 = -2 + 1- 6 = - 7
x 33 u3 + v3 - C33 = -2 - 2 + 7 = 3
The most positive coefficient is coming for x21. So This becomes new entering
variable
Quantitative Techniques in
Management : 42
Table 5 D gives closed loop for (2, 1) as entering variable
If α is allocated to (2, 1) than
1
3-α >0
1
5-α >0
2
The corresponding maximum value
of α is 3 and (2, 2) becomes at zero
level.
3
So x21 is entering and x22 is leaving
variable. Table 5E is obtained after
iteration 2.
The associated cost is
2 x 5 + 10 x 1 + 3 x 2 + 11 x 0
3
8
5
1
5-α
7-α
α
2
4
0
3-α
11
3
6
7
4
9
10
11
Table 5 D
1
2
1
5
2
2
for x 21 u2 + v1 - C21 = 6
allocation = 3 Units; Cost Reduction = 6
x 3 = Rs. 18
14
NOTES
4
1
12
4
0
14
7
4
11
3
4
9
12
10
2
3
Transportation Model
3
8
3
+ 4 x 3 = Rs. 38
Verify it as follows :
2
6
10
11
Table 5 E
So new total cost = 56 - 18 = Rs. 38
We can see that solutions is improving. (transportation cost is reducing)
Again get values of (u, v) with the help of baisc variable in table 5E
Taking u1 = 0, v1 = 5, v2 = 1, v3 = 3, u2 = -3, u3= -2
Now we calculate coefficients for nonbasic variables of table 5E.
x13
u1 + v3 - C13 = 0 + 3 - 8 = -5
x 22
u2 + v2 - C22 = - 3 + 1 - 4 = -6
x 32
u3 + v2 - C32 = - 2 + 1 - 6 = -7
x33
u3 + v3 - C33 = - 2 + 3 - 7 = -6
These values of ui +vj - cij are now negative for all nobasic xij. Thus ths solution
in Table 5E is optimal.
2.4 Summary
Quantitative Techniques in
Management : 43
Transportation Model
NOTES
2.5 Key Terms
Node : Node represents a source or a destination.
Arc : Routes linking a source to a destination.
Dummy Source : A source added in transportation model to balance the structure in
a way that unit transportation cost from dummy source to all destinations
in zero.
Dummy Destination : A destination added in transportation model to balance the
structure in a way that unit transportation cost from all sources to dummy
destination is zero.
2.6 Questions and Exercises
(1) Determine optimal allocation for the following transportation model :
F1
F2
F3
Capacity
W1
5
8
12
300
W2
7
6
10
600
W3
13
4
9
700
W4
10
13
11
400
Demand
700
400
800
2)
A company making paint has three plants A, B and C, each with a capacity to
produce 250 kg, 150 kg and 450 kg, respectively per day. The production costs
per kg in plants A, B and C, respectively are Rs. 40, Rs. 35 and Rs. 38 Four
institutional customers have placed orders for the paint on the following basis :
Customer
kg. ordered
Price offered Rs. per kg.
α
350
100
β
225
100
γ
350
102
θ
150
103
Shipping cost per kg from plants to cumstomers are given in the table below :
Quantitative Techniques in
Management : 44
Transportation Model
Customers
Plant
α
β
γ
θ
A
2
4
4
6
B
7
10
9
12
C
4
6
2
8
NOTES
Develop an optimal solution for this situation.
(3) Consider following model of a transportation problem :
1
1 1
2 3
3 2
Demand 30
2
2
4
3
20
3 Supply
1
20
5
40
3
30
20
In this case, if a unit from a source is not shipped out, a storage cost is incurred at
the rate of Rs. 5, Rs. 4 and Rs. 3 per unit for sources 1, 2 and 3 respectively. Additionally,
all the supply at source 2 must be shipped out completely to make room for a new
product. Use VAM to get starting solution and determine all the iteration leading to the
optimum shipping schedule.
(4) Consider following table :
1
2
3
1
5
6
3
75
2
1
4
2
20
3
7
6
5
50
10
80
15
If penalty costs per unit of unsatisfied demand are Rs. 5, Rs. 3 and Rs. 2 for
destinations 1, 2 and 3 respectively. Using N-W-C method, get starting solution and
also determine optimal solution.
2.7 Further Reading and References
Quantitative Techniques in
Management : 45
Assignment Model
UNIT 3
NOTES
ASSIGNMENT MODEL
Structure
3.0 Introduction
3.1 Unit Objectives
3.2 General Model of Assignment Problems
3.3 The Hungarian Method
3.4 Unbalanced Assignment Problem
3.5 Travelling Salesman Problem
3.6 Summary
3.7 Key Terms
3.8 Question & Exercises
3.9 Further Reading and References
3.0 Introduction
''What jobs should a manager assign to whom?'' The answer is ''The best person
for the job.'' This is the appropriate answer of assignment model. A job that happens to
match a worker’s skill costs less than one in which the operator is not as skillful.
The assignment model is a special case of transportation problem involving
minimum cost assignment of workers to jobs.
3.1 Unit Objectives
After Studying this unit, you should be able to
Understand assignment problems as a special case of transportation problems,
and therefore special case of linear programming problems.
Handle unbalanced assignemnt problems.
Use Hungarian method of solving assignment problems
Use of Horizontal and vertical lines to get optimal assignments.
Identify optimal route fro a ravelling salesman going from one city to different
city and finally returning to original city.
This unit will discuss Hungarian method of assignment as well as maximisation
problems.
3.2 General models of Assignment Problems
Quantitative Techniques in
Management : 46
In a general assignment model with n machines waiting for n jobs, we have
following table :
n
Cn1
Cn2 .............Cnn
1
Assignment Model
1
1
.......
1
2
.......
Machine
1
C11
C21
Jobs
2................n
C12 .............C1n
C22 .............C2n
NOTES
1
1 .............. 1
Table 3.1
The elements inside matrix represented by Cij cost of assigning machine i to j (i,
j = 1, 2 ---- n). For getting a solution of assignment model, we should always have
equal number of machines and jobs. In case of different numbers of is and js, we can
always add fictitious. machines or fictitious jobs to solve the model.
As a special case supply from each source and demand at each destination are
equal to '1'.
Now we can directly solve assignment problem using transportation algorithm.
But special nature of assignment problem led to the development of a simple and fast
method known as Hungarian Method to solve assignment problems.
3.3 The Hungarian Method
Take following model of assignment, where these jobs are to be done by three
workers. Workers charge different payment for different jobs. Following table 3.2 gives
the wages of workers for a particular job.
Jobs
1
2
3
A
15
10
9
Workers B
9
15
10
C
10
12
8
Table 3.2
As a manager of the company, you identify combination of worker - job for
minimum total Cost of assignment.
Various Steps of the Hungarian Method are as follows :
Step 1 : From Table 3.2, identify each row's minimum, and subtract it from all the
entries of the row. This will result in Table 3.3
Jobs
1
A 15-9
Workers B
9-9
C 10-8
2
3
Row Min.
10-9
9-9
P1 = 9
15-9 10-9
P2 = 9
12-8
P3 = 8
Table 3.3
8-8
Quantitative Techniques in
Management : 47
Assignment Model
Step 2 : Now in Table 3.3, identify each column's minmum and subtract it from
all the entries of the column.
1
2
3
NOTES
Workers
A
6
1-1
0
B
0
6-1
1
C
2
4-1
0
Column Minimum v1 = 0
v2 = 1 v3 =0
Table 3.4
Step 3 : Identify the optimal solution as the feasible assignment associated with
the zero elements of the Table 3.4. These zero elements should be unique in row and
column. So the combination is (A, 2), (B, 1), and (C, 3).
The cost of this allocation is 10 + 9 + 8 = Rs. 27.
Example 1 : Take data presented in following Table 3.5 as cost of assignment for
a pair of job and machine Determine minimum cost of assignment.
Machine 1
2
3
4
36
32
40
39
31
40
38
35
17
12
18
16
Jobs
A
B
C
D
20
24
22
36
Table 3.5
Solution : First we identify row minimum and subtract each row with respective
row minima. This will yield Table 3.6
→
Machines → 1
Jobs
2
3
4
A
3
19
14
0
B
12
20
28
0
C
4
22
20
0
D
20
23
19
0
Table 3.6
Now identify column minimum in table 3.6 and subtract column entries with
respective column minima. This yields table 3.7
→
Jobs Machines
2
3
4
A
0
0
0
0
B
9
1
14
0
C
1
3
6
0
D
17
4
5
0
→
1
Quantitative Techniques in
Management : 48
Table 3.7
In Table 3.7 we have seven cells with zero opportunity cost. Job A can be assignned
to any machine 1, 2, 3 or 4. If we assign job A to machine 1, than job B can be assigned
to machine 4. Now no further assignment can be made in zero cells. As machine 4 can
not be assigned to job C or job D as it is already paired with job B.
Assignment Model
NOTES
In order to make optimal assignments, we must locate four cells with a zero
opportunity cost such that a complete assignment to these cells can be made with a
total opportunity cost of zero.
Now to get solution in this situation we present a new method of drawing horizontal
and vertical lines in reduced matrix.
We draw horizontal and vertical straight lines in such a way to have minimum
number of such lines covering all zeros.
1
2
3
4
A
0
0
0
0
B
9
1
14
0
C
1
3
6
0
D
17
4
5
0
Table 3.8
Table 3.8 shows two lines, one horizontal in first row and one vertical in last
column. In this table we have four rows but only two lines are available, So we can not
make an optimal solution. To have an optimal solution, we should have minimum as
many straight lines as we have rows (or columns) in the matrix.
Now we need to revise the Table 3.8 as follows:
Select the smallest number from uncovered cells and subtract it from entries of
each cell of the uncovered part. We need to add the same numbers lying at the intersection
of any two lines. This will give following Table 3.9
1
2
3
4
A
0
0
0
0+1=1
B
9-1=8
1-1 = 0
14-1=13
0
C
1-1=0
3-1=2
6-1=5
0
D
17-1=16
4-1=3
5-1=4
0
Table 3.9
When we draw Straight lines in Table 3.9, we get Table 3.10
1
2
3
4
A
0
0
0
1
B
8
0
13
0
C
0
2
5
0
D
16
3
4
0
Table 3.10
Quantitative Techniques in
Management : 49
Assignment Model
NOTES
The number of lines is 4 which is equal to number of rows (or column). Hence
optimal assignment can now be made.
First we select a row or column with only one zero in it. Column 3 and row D
have only single zero in them. Arbitrarily taking column 3, we make an assignment in
it.
1
2
3
4
A
0
0
0
1
B
8
0
13
0
C
0
2
5
0
D
16
3
4
0
Table 3.11
We have Crossed column 3 and row A in Table 3.11, as Job A is allocated to
machine 3.
Similarly we assign B to 2, C to 1 and D to 4. Allocations are shown with encircled
zeros in table 3.11
The Cost of this assignment is as follows :
31 + 32 + 22 + 16 = Rs. 101.
Maximization Case :
Generally assignment problems are cost minimization problems. But in some cases,
appropriate assignment of workers on jobs can be profit maximization case to the
organization.
In such a case of maximization, reduce the initial table to an opportunity loss
table by subtracting all numbers from the largest number and then proceed as in a
normal minimization case.
Example 2 : Consider following effectiveness matrix of different salseman of
marketing company in different territories. How to assign these salesman to get
maximum effectiveness?
Territory
1
2
3
4
A
42
35
28
21
B
30
25
20
15
C
30
25
20
15
D
24
20
16
12
Table 3.12
Solution : Step 1 : First we will convert table 3.12, which is effectiveness matrix
into standard assignment model by subtracting from the highest element (i.e. 42), all
the elements of the given table. This will yield table 3.13 as follows :
Quantitative Techniques in
Management : 50
1
2
3
4
A
0
7
14
21
B
12
17
22
27
C
12
17
22
27
D
18
22
26
30
Assignment Model
NOTES
Table 3.13
Table 3.13 is in standard minimization assignment model, where regular Hungarian
method can be applied. First row minimum is identified for each row. Then we subtract
each entry of rows with respective row minima, yielding Table 3.14
1
2
3
4
A
0
7
14
21
B
0
5
10
15
C
0
5
10
15
D
0
5
8
12
Table 3.14
Now we identify column minimum and subtract each entry of a column from
respective. column minima. This produces Table 3.15
1
2
3
4
A
0
3
6
9
B
0
1
2
3
C
0
1
2
3
D
0
0
0
0
Table 3.15
Now we draw straight lines to cover all zero cells.
1
2
3
4
A
0
3
6
9
B
0
1
2
3
C
0
1
2
3
D
1
0
0
0
Table 3.16
Table 3.16 has two straight lines, but we have four rows. Therefore, it cannot
produce optimal solution. The minimum entry in uncovered cells is 1. We will add this
at intersection cell of straight lines (i.e. D, 1) and subtract it from all other uncovered
cells. This results in table 3.17.
Quantitative Techniques in
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Assignment Model
NOTES
1
2
3
4
A
0
2
5
8
B
0
0
1
2
C
0
0
1
2
D
1
0
0
0
Table 3.17
Again we will not get optimal solution, as minimum number of straight lines is 3.
Repeating the process as above, we get Table 3.18
1
2
3
4
A
0
2
4
7
B
0
0
0
1
C
0
0
0
1
D
2
1
0
0
Table 3.18
In Table 3.18 we have used four lines to cover all zeros.
So assignments can be made in following two ways :
1
2
3
4
A
0
2
4
7
B
0
0
0
C
0
0
D
2
1
1
2
3
4
A
0
2
4
7
1
B
0
0
0
1
0
0
C
0
0
0
0
0
0
D
2
1
0
0
Table 3.19
Table 3.20
The associated effectiveness is
42 + 25 + 20 + 12
= 99 Units
The associated effectiveness is
42 + 20 + 25 + 12
= 99 Units
It is a case of altermative optima also.
3.4 Unbalanced Assignment Problem
The Hungarian method of assignment requires that the number of columns and
rows in the assignment matrix be equal, i.e. the matrix should be square one. However,
when the given cost matrix is not a square matrix, the assignment problem is called an
unbalanced problem.
To handle unbalanced assignment models we add dummy rows or dummy column
to make the model square one.
Quantitative Techniques in
Management : 52
Example 3 : Consider the assignment model of table M and allocate available
machines to different locations.
Jobs
Assignment Model
1
2
3
4
M1
9
11
15
10
Machine M2
12
9
--
10
M3
--
11
14
11
NOTES
Table 3.21
Solution : The given assignment matrix in Table 3.21 is not balanced, it is a 3 x 4
matrix.
To make it balanced (square), we will add a dummy row. It will become 4 x 4
square matrix.
All entries of new row (dummy) will be zero. Table 3.22 present updated matrix.
1
2
3
4
M1
9
11
15
10
M2
12
9
--
10
M3
--
11
14
11
0
0
0
0
M4 (Dummy)
Table 3.22
Note : Two entries, one in second row, third column and second in third row, first
column are '-'. It means these pairings are not allowed for some reason. To go ahead, we
can also write M in these cells. M is a large cost to prohibit us to pair in these cells.
After identifying row penalties and column penalties and subtracting entries of
row by row we get updated Table 3.23
1
2
3
4
M1
0
2
6
1
M2
3
0
M-9
1
M3
M-11
0
3
0
M4
0
0
0
0
Table 3.23
Table 3.23 presents allocations marked with 'X' signs.
The total cost is 9 + 9 + 11 = Rs. 29.
See dummy machine is at job C. Practically, this means that no machine is allocated
to job 3.
3.5 Travelling Salesman Problem
Many a times, it happen to start from one point (city) and go to many other points
Quantitative Techniques in
Management : 53
Assignment Model
(Cities) and return to same point (city). This is known as travelling salesman problem
where salesman visits clients (city) so that his travelling cost/ time/ distance is minimized.
This situation can be solved using assignment modelling with some modifications.
NOTES
Example 4 : An agent of a company starts from City A and then goes to City B, C
and D. The following table 3.24 shows distances between various cities on different
routes.
FromTo A
B
C
D
A
--
8
7
9
B
6
--
11
5
C
15
11
--
6
D
9
5
6
--
Table 3.24
The distance from A to B is 8 kms but from B to A is 6 kms because of one way
streets, and other restrictions on movement between points. As no movement is possible
from a destination to it self, these cells are marked with a dash.
Solution :Table 3.25 and 3.26 gives solution of the model using assignment method.
To
A
B
C
D
A
--
1
0
2
B
1
--
6
0
C
9
5
--
0
D
4
0
1
--
From
Table 3.25
(Reduced Matrix after row penalties subtraction)
A
B
C
D
A
--
1
0
2
B
0
--
6
0
C
8
5
--
0
D
3
0
1
--
Table 3.26
(Reduced Matrix after column penalties subtraction)
This suggests that agent should go from A- C- D- B- A.
The distance travelled will be
Quantitative Techniques in
Management : 54
7 + 6 + 5 + 6 = 24 kms.
Example 5 : A salesman has to visit five cities A, B, C, D and E. The distances
between these five cities are given in table 3.27.
If the salesman starts from city A and has to come back to City A, which route
should he select so that total distance travelled by him is minimized?
Assignment Model
NOTES
To City
From City
A
B
C
D
E
A
0
17
16
18
14
B
17
--
18
15
16
C
16
18
--
19
17
D
18
15
19
--
18
E
14
16
17
18
---
Table 3.27
Solution : Table 3.28 gives assignment solution of table 3.27 using method of
drawing straight lines.
A
B
C
D
E
A
0
3
0
4
0
B
3
--
1
0
1
C
0
1
--
2
0
D
4
0
2
--
3
E
0
1
0
3
---
Table 3.28
Optimally we should move from A to C, C to A and E to A. And we should move
from O to B and B to D. This is not the solution we are looking for. Starting from A we
must touch all points before returing to A. This means that we must include some other
cells in our route even though their opportunity cost is not zero. But we should try to do
this with minimal increase in cost.
We now calculate the opportunity cost (that is x the additional cost that we will in
cur for not using the zero cell and using the nest cost lier cell) of each of the zero cells.
The opportunity cost the sum of least number in the row and column of the zero cell
that we are considering. Table 3.29 gives calculation of opportunity cost based on table
3.28
Opportunity Cost Least of
Least of
Sum
of Cell
row (a)
Column (b)
(a) + (b)
AC
0
0
0
BD
1
2
3
CE
0
0
0
DB
2
1
3
EA
0
0
0
Table 3.29
Quantitative Techniques in
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Assignment Model
Now make assignment to the zero cell with the highest opportunity cost. We
arbitrarily take BD for this purpose to break tie between BD and DB. Movement from
D to B will not be permissible, so we will assign a very high cost (distance here), M, to
this cell.
NOTES
We will Climinate row and column of assigned cell.
Table 3.30 is updated table
A
B
C
E
A
0
3
0
0
C
0
1
-
0
D
4
M
2
3
E
0
1
0
=
Table 3.30
Table 3.30 needs only three straight lines to cover all zeros, so we need to do one
iteration for optimal allocation.
Table 3.31 is the revised table.
A
B
C
E
A
-
3
0
0
C
0
1
-
0
D
2
M-2
0
1
E
0
1
0
-
Table 3.31
Table 3.31 can not give optimal assignment. Doing further iteration,
A
B
C
E
A
-
2
0
0
B
0
0
-
0
C
2
M-3
0
1
D
0
0
0
-
Table 3.32
Table 3.32 has four lines. Now we can have allocation.
Allocations in table 3.32 are AE, CA, DC, EB. We have already done one allocation.
So complete allocations are as follows :
AE → EB → BD → DC → CA
Total distance to be travelled will be
14 + 16 + 15 + 19 + 16
Quantitative Techniques in
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= 80 kms.
In case an assignment without returning to intermediate points cannot be made,
Assignment Model
repeat the process of evaluating the opportunity cost of each zero, assign the highest
opportunity cost cell and proceed as discussed in the example 5.
NOTES
3.6 Summary
3.7 Key Terms
3.8 Questions and Exercises
(1) How would you deal with the assignment problems of maximization of objective
function?
(2)
How would you handle unbalanced assignment model ?
(3) How would you handle assignment problem, where some assignments are
prohibited?
Practice Questions
(4) Single Unit of resource from warehouses 1, 2, 3, 4, 5 and 6 is required to send to
Quantitative Techniques in
Management : 57
Assignment Model
warehouses 7, 8, 9, 10, 11 and 12. The distance (in km) between these cities are
given in following table :
NOTES
From
To
7
8
9
10
11
12
1
31
62
29
42
15
41
2
12
19
39
55
71
40
3
17
29
50
41
22
22
4
35
40
38
42
27
33
5
19
30
29
16
20
23
6
72
30
30
50
41
20
How to decied combination of cities (1, 2, 3, 4, 5, 6) to (7, 8, 9, 10, 11, 12) to minimize the total
travelling distance?
Unit : Assignment
Practice Questions
(5)
Five workers are to be assigned to six jobs. Associated costs of worker- assignment to job
combination is given in below table :-
Machines
Workers
M1
M2
M3
M4
M5
M6
W1
12
3
6
2
5
9
W2
4
11
--
5
--
8
W3
8
2
10
9
7
5
W4
--
7
8
6
12
10
W5
5
8
9
4
6
1
Determine Cost of allocation.
(6)
The expected time required to be taken by a salesman in travelling from one City to
another are as follows :-
To
From
C1
C2
C3
C4
C5
C1
--
10
13
11
--
C2
10
--
12
10
12
C3
14
13
--
13
11
C4
11
10
14
--
10
C5
12
11
12
10
--
How should the salesman plan his trip, so that he covers each of these cities no more than
once, and completes his trip in minimum possible time required in travelling.?
Quantitative Techniques in
Management : 58
Assignment Model
3.9 Further Reading and References
NOTES
Quantitative Techniques in
Management : 59
Queuing Theory
UNIT 4
NOTES
QUEUING THEORY
Structure
4.0 Introduction
4.1 Unit Objectives
4.2 Need for Queuing Analysis
4.3 Elements of a Queuing Models
4.4 Kendall Notations of Queuing System
4.5 Stedy State Queuing System Analysis
4.6 Analysis of M/M/1 Queuing System
4.7 Summary
4.8 Key Terms
4.9 Question & Exercises
4.10 Further Reading and References
4.0 Introduction
Standing in queues are a very familiar phenomenon in our daily life. We wait for
getting our scooters filled at petrol pump or at beauty parlours to get a make-up.
Customers, jobs, machines wait to get their turn. Queus are formed when service provider
is busy and is operating at full capacity.
The objective of queuing analysis is to offer a reasonably satisfactory service to
waiting customers. As compared to linear programming, transportation or assignment
problems, queuing theory is not an optimization technique. Rather, it determines the
measures of performance of waiting lines, such as the average waiting time in queue
and the productivity of the service facility, which can then be used to design the service
installation.
4.1 Unit Objectives
After studying this unit, you should be able to
Understand need of queuing analysis
Understand elements of queuing model with respect to queue size, queue
discipline, behaviour of customers, types of service facility.
Understand use of random distributions in queuing analysis
Understand steady state analysis of some of the popular queuing systems
such as M/M/I
4.2 Need for quening Analysis
Quantitative Techniques in
Management : 60
Everytime anyone waits for a service, the time spent in waiting results in costs-
they may be indirect or direct. One of the simplest methods of reducing waiting time
and the resultant waiting costs is to increase the rate at which the service is provided by
increasing the number of service facilities. But provision of service facilities entails
costs. Is the benefit of reducing waiting costs going to outweigh the cost of providing
additional facilities or not? Therefore a detailed analysis of queuing system is required
to facilitate decisions related to start a new service point or not.
Queuing Theory
NOTES
4.3 Elements of a Queuing model
A queue system has two important elements :
(i)
Customers
(ii) Servers
Customers are generated from a source. Once these customers arrive at a service
facility, service may start immediately if server is idle otherwise customer will wait in
a queue.
To analyze queues, the arrival of customers is represented by the interarrival time
between successive customers, and the service is described by the service time per
customer. Generally interarrival and service time are probabilistic in nature. Take
example of customers arriving at a petrol pump, their inter arrival time is probabilistic
while patients arriving at a clinic with prior appointments are deterministic in nature.
Queue size is also important in analysis. It can be finite or infinite when system
can accomodate only few customers in queue, it is a finite queue. When there is no
restriction on number of customers waiting for service, it is infinite queue size.
Queue discipline is an important factor in the alanysis of queuing models. It means
the order in which customers are selected from a queue for service. Following are
important types of discipline, generally used in a queue system :
(a) First Come First Served (FCFS) : Customers waiting at a bank counter.
(b) Priority : In this case, some customers would receive priority over others
and service would be provided to them before others irrespective of their
arrival times. Arrival of a cardiac arrest patient at a hospital may need priority attention.
(c) Last In First Out (LIFO) :- Luggage loaded in an aircraft's luggage section.
(d) Service In Random Order (SIRO) :- When queue is not well defined, such as
children gather arround an ice candy verdor.
Queuing Behaviour of Customers :
Behaviour of calling population is also an important part of our analysis. Customers
can exhibit different types of behaviour as follows :
(a) Jockey :- Customers move from one queue to another in the hope of reducing
waiting time.
(b) Balk :- Customers may avoid joining a queue altogether because of
anticipated long delay.
(c) Renage :- Customers leave a queue after waiting for sometime in it.
Quering theory does not take such members into consideration. The quering models
deal with the patient customers who join the quere and wait till service is provided to them.
Quantitative Techniques in
Management : 61
Queuing Theory
NOTES
Design of Service Facility :- Servers at facility can be in series or in parallel or
networked.
Fig.4.1 a presents simplest facility design where we have single channel and only
single server.
Customers Waiting for Service
Service facility
Served Customers
Type A
Fig. 4.1 (Single channel, single server)
Example of such facility is a petrol pump with single pertol bunk.
Figure 4.2 presents single channel with multi server arrangement.
Server facility
A
Server facility
B
Fig. 4.2 (Single Channel Multi Server)
In a small Government hospital we can have only single window for registration
and after that only single doctor to meet. This is a case of single channel multi server
arrangement.
Figure 4.3 presents multi channel single server arrangement
Server facility
A
Server Facility
A
Fig. 4.3 (Multi Channel Single Server)
In this arrangement parallel facilities are of similar type. This is normal situation
at a barber's shop.
Size of the calling population :- The source from which customers are generated
may be finite or infinite. A finite source limits the customers arriving for service. An
infinite source is forever abundant.
Exponential Distribution in Queuing analysis :- In a queuing system, the arrival
of customers occurs in a completly random manner. It means that the occurrence of an
event is not influenced by the length of time that has elapsed since the occurrence of
the last event.
Random interarrival and service times are described by exponential distribution
in a queuing model.
According to exponential distribution, distribution of time between successive
arrivals t, is
f (t) = λe −λt , t > o
Quantitative Techniques in
Management : 62
Where λ is the rate per unit time at which events are generated.
Probability distribution function describing the number of random arrivals during
a specified period is the Poisson distribution.
If x be the number of arrivals that take place during a specified time unit (e.g., a
minute or an hour). λ is knowns, the poisson pdf is
Queuing Theory
NOTES
λ e
ρ {x=k} = ------------, k = 0, 1, 2....
k1
k
-λ
4.4 Kendall Notations of Queuing System
Kendall notations are used to summarize the characteristics of the queuing
situations. Format of these notations is (a/b/c) : (d/e/f), where
a = Arrivals distribution
b= Departures distribution
c= number of parallel servers
d= Queue discipline
e= Maximum number of customers in the system
f = Size of the calling population
Arrivals and departures distribution can be markovian (or poisson) (M), Constand
time (D) and Erlang or gamma (EK).
Let us consider an example (M/M/3) : (FCFS /10/∞). This model says arrival and
departures are exponentially distrituted and 3 parallel servers. The queue discipline is
First Come First Served, and there is a limit of 10 customers on the entire system. The
size of the source from which customers arrive is infinite.
4.5 Steady State Queuing System Analysis
Steady state behaviour of queue is achieved after a system is in operation for
long- run. Steady state performance of queuing system is done on the basis of following
measures.
Ls = Expected number of customers in system
Lv=Expected number of customers in queue
Ws = Expected waiting time in system.
Wv = Expected waiting time in queue
Quantitative Techniques in
Management : 63
Queuing Theory
4.6 Analysis of M/M/1 Queuing System
(Case 1 ) : Size of the system is infinite
NOTES
If arrival occurs at the rate of λ customers per unit time and the service rate is µ customers
per unit time.
1
Ws = ----µ−λ
(time units)
λ
ρ
Wv = ---------- = ---------- (time units)
µ (µ−λ)
µ (1−ρ)
λ
Where ρ = ---------µ
λ
Ls = ---------(µ−λ)
λ2
Lv = ---------µ (µ−λ)
Note : The analysis of M/M/1 is having a condition of
λ
------ ρ < 1, or λ>m, 9 fλ >µ.
µ
9f λ > µ, Steady State queuing system will not exist. It is easy to understand that
unless the service rate is larger than the arrival rate, queue length will continually
increase and no steady state can be reached.
Utilisation Factor :The ratio λ/µ , represented by ρ (rho) is called the utilisation factor. It is also the
λ
probability that the system is busy = =−−−−−
µ
therefore, probability that the system is idle = 1-ρ
Quantitative Techniques in
Management : 64
To understand the effect of relative values of λ and µ, following table is
useful.
λ
µ
ρ
Lv
1
5
.2
.05
2
5
.4
.27
3
5
.6
.90
4
5
.8
3.20
4.5
5
.9
8.10
5
5
1.00
∞
Queuing Theory
NOTES
Data of table can be shown on a graph also as follows :
10
8
3
Queue length (Lv)
2
1
0
.2
.4
.6
.8
1.00
Utilization factor (ρ)
Therefore for analysis purpose ... should be less than 1.
Example 1 A garage repairman finds that the time spent on a car is exponentially
distributed with a mean of 30 minutes. If he repairs cars in the order of arrival and if the
arrival follows a Poisson distribution approximately with an average rate of 10 per 8hour day, what is the garage's expected idle time each day? How many cars are ahead of
the average car just brought in?
10
Solution Here λ = ---------- = 1.25 cars per hour
8
1 ) 60 = 2 cars per hour
µ = (-----30
Server Utilization factor (traffic intensity) ρ = ----
λ
µ
1.25
= -------2
1.25
(a) Now utilization of garage in 8 hrs day 8xρ = 8x ------2
= 5 hrs.
Quantitative Techniques in
Management : 65
Queuing Theory
Therefore garage is idle = 8-5 = 3 hrs in a day.
(b) Expected number of cars in the garage
λ
1.25
5
Ls = ------- = ------- = ------µ−λ 2-1.25
3
NOTES
≈ 2 Cars
Example 2 : Arrivals at a shop is Poisson with an average time of 10 minutes
between one arrival and the next. Service time of customers is assumed to be distributed
exponentially, with mean of 3 minutes.
(a) Determine the probability that a person arriving at the shop will have to
wait.
(b) Shopkeeper will keep a second salesman when convinced that an arrival
would expect waiting for at least 3 minutes for a purchase. By how much
should the flow of arrivals increase in order to justify a second booth?
(c) What is the average length of the queue that forms time to time?
(d) What is the probability that it will take a customer more than 10 minutes
altogether to wait for a purchase and complete the shopping?
1
Solution :- Here λ = -------- = 0.10 person per minute
10
µ = 1/3 = 0.33 person per minute
λ
.10
10
ρ =-------- = -------- =-------- = .3
µ
.33
3
(a) Probability that a person has to wait is
p (n>0) = 1 ρ0=
λ
µ
ρ = .3
(b) To have a second salesman will be justified only if the arrival rate is more than
the waiting time. Let λ be the increased arrival rate. Then expected waiting time in the
queue will be
λ1
Wv = -------------µ (µ −λ1)
3
λ1
=--------------------------or λ1 = 0.16
.33 (.33 - λ1)
Hence, the increase in the arrival rate is 0.16-0.10 = .06 arrivals per minute.
Quantitative Techniques in
Management : 66
c) Average length of non-empty queue
µ
0.33
L = -------- = -------- ≈ 2 Customers
µ−λ
0.23
(d) Probability of waiting for 10 minutes or more is given by ........
p(t >10) =
10
∫
∞
λ
-------
µ
Queuing Theory
NOTES
(µ−λ) e-(µ−λ)t dt
∞
∫ (0.3) (0.23) e
-0.23t
p(t >10) =
10
dt
= 0.03
i. e. Only 3 percent of customers on an average will have to wait for 10 minutes or
more before they shop.
Case II (M/M/1) : (N/FCFS), i. e. as contrast to case I, this queue model has a
finite size of N.
Here if N customers are already in the system, new customer will not enter the
system, and is lost. Some of the performance measures are as follows:
(1) Expected number of customers in the system
Ls
ρ
(N+1) ρN+1
= ------ - ----------------- ; ρ#1 (λ# µ)
1-ρ
1-ρN+1
N
/2
; ρ=1 (λ=µ)
(2) Expected queue length (Customers waiting in the system)
λ
Lv = Ls - -----µ
(3) Probability of a customer in the system for n=0, 1, 2 .....N are obtained as follows
λ
ρn= (------)ρ
µ 0
; n<N
1-ρ
and ρo = ----------- for ρ ≠1 and ρ < 1
1 - ρN+1
(4) Expected waiting time of a customer in the system
Ls
ws = ------------λ (1-ρN)
Quantitative Techniques in
Management : 67
Queuing Theory
NOTES
(5) Expected waiting time of a customer in the queue
1
Lv
wv = ws - ----- or -------------µ
λ (1 - ρN)
Where PN = ρo ρN
Effective arrival rate, λeff = λ (1-PN)
λe
Effective traffic intensity, ρeff = --------µ
Example 3 : In a single server queuing system with poisson input and exponential
service time, mean arrival time is 3 jobs per hour and expected service time is 0.25
hour. The maximum number of jobs in the system can be two. Calculate expected
number of jobs in the system.
Solution : Data is
λ = 3, µ=4, N=2
ρ = λ/µ = 0.75
1-ρ
ρn=ρo ρn = ( -----------) ρn
1-ρN+1
1- .75
=(--------------) (.75)n = (0.43) (0.75)n
1- .753
1-ρ
1- .75
and Po = ----------- = ----------- = .431
1-ρN+1
1- .753
The expected number of jobs in the system is
N
2
Ls = Σ = nρn = Σ
n (0.43) (0.75)n
n=1
n=1
=0.43 {(0.75) + 2 (0.75)2= 0.81
Example 4 : In a car washing facility four parking spaces are available. If the
parking lot is full, newly arriving cars balk to other facilities. The owner wishes to
determine the impact of the limited parking space on losing customers to the competition
Given is λ = 4,
N = 4+1=5,
Quantitative Techniques in
Management : 68
µ = 6,
no. of server = 1,
infinite source capacity.
Queuing Theory
Solution : Here
ρ5
= ρ0ρ5
1-ρ
= (----------) = P5
1-P6
NOTES
1- 2/3
= (--------------) (2/3)5 = 0.04812
1-(2/3)6
This is the proportion of lost customers.
Now on the basis of 8 hrs days, this is equivalent to losing -(λ P5 )X 8 = 4 x
0.4812 X 8 = 1.54 ≈ 2 cars per 8 hrs day.
A decision regarding increasing the size of the parking lot should be based on the
value of lost business.
Example 5. An AMC is to be singed for maintenance of water filters in your
office. At an average two filters per month get defected. The cost of a filter being
unavailable is Rs. 5000 per month. Two companies have quoted for the contract. ABC
quoted at Rs 2500 per month. Whereas xyz has quoted at Rs. 3500 per month for the
contract. Enquiries reveal that ABC has an average repair capability of 4 filters per
month and xyz can repair 5 filters per month at an average. Who should be given the
contract?
Solution :- The company loses (incur cost) on account of failure of filter as
company has to procure mineral water for employes.
To decide about the contract, let us compare as follows :ABC
XYZ
(a) Arrival rate of filters for repairs (λ)
2
2
(b) Service rate (µ)
4
5
(c) ρ = (λ/µ)
.5
.4
λ
(d) Ls (No. in System) =-------µ−λ
1
.67
(e) Cost of defected filters
1X5000 = 5000 .67 X 5000 = 3350
(f) Cost of Contract
2500
3500
(g) Total Cost
7500
6850
This alalysis shows that the contract should be given to XYZ. Its faster repair
capability will ensure greater availbility of filters in the company and hence the lesser
cost.
4.7 Summary
Quantitative Techniques in
Management : 69
Queuing Theory
NOTES
4.8 Key Terms
4.9 Questions and Exercises
Que 1 :
Define a waitine line. Give brief description of the various types of queues.
Que. 2 : What is traffic intensity? If traffic intensity is 0.30, what is the percent of
time a system remains idle?
Que. 3 : What is service discipline? Describe some forms of common service
disciplines and illustrate with examples.
Que. 4 : Explain different types of physical layouts of a service system and their
impact on waiting times.
Que. 5 : What is kendall notation? How queuing models are defined with these
notations?
Practice Questions :Que. 6 : Customers arrive at a booth in a Poisson distributed arrival rate of 20 per
hour. Service time is exponentially distributed with an average time of 1
minute. Calculate the mean number in the waiting line, the mean waiting
time, the mean numbers in the system, the mean time in the system and the
utilisation factor.
Quantitative Techniques in
Management : 70
Que. 7 : A company distributes its products by trucks loaded at its single facility.
Both the company's trucks and the contractor's trucks are used for this purpose.
It was found that on an average a truck arrived every 5 minutes and the
average unloading time was 3 minutes. 50% trucks belonged to the contractor.
Find out
(a) The probability that a truck has to wait.
Queuing Theory
NOTES
(b) The waiting time of a truck that waits.
(c) The expected waiting time of a contractor's trucks per day, assuming a
24 hours shift.
Que. 8 : In an office with single clerk, there are only two chairs for waiting customers.
On an average one customer arrives every 10 minutes and each customer
takes 5 minutes for getting served. Making suitable assumptions, find
(a) The probability that an arrival will get a chair to sit on.
(b) The probability that an arrival will have to stand.
(c) Expected waiting time of a customer.
Que. 9 : At a railway station, only one train is handled at a time. The railway yard is
sufficient only for two trains to wait while the other is given signal to leave
the station. Trains arrive at the station at an average rate of 6 per hour and the
railway station can handle them on an average of 12 per hour. Assuming
Poisson arrivals and exponential service distribution, find the steady-state
probabilities for the various waiting time of a new train coming into the
yard.
Que. 10 : A TV repair shop can repair TVs at the rate of 10 per day, the repair time
being exponentially distributed. TVs arrive at the shop following a Poisson
distribution at an average rate of 8 per day. The shop can accommodate only
a maximum of 3 TVs in the system due to space shortage. TVs, which arrive
when the system is full, are not taken up for service and the potential loss is
estimated at Rs. 50 per TV. The shop owner has two alternatives to reduce
the customer loss.
A-
He can take additional space on rent at a monthly rental of Rs. 300.
Which would enable increasing the capacity of the system to 4 TVs.
B-
He can appoint a trained mechanic who has to be paid an extra monthly
salary of Rs. 350 and the service rate will increase to 12 TVs per day.
Based on the cost - benefit evaluation, recommend the best alternative.
4.10 Further Reading and References
Quantitative Techniques in
Management : 71
Decision Theory
NOTES
UNIT 5
DECISION THEORY
Structure
5.0 Introduction
5.1 Unit Objectives
5.2 Zone of Decision- Making
5.3 Steps in Decision- Making Process
5.4 Example to Demonstrate Preparation of Pay off Table
5.5 Decision Making Under Uncertainty
5.5.1 The Maximax Criterion (Optimistic Criterion)
5.5.2 The Maximin Criterion (Pessimistic Approach)
5.5.3 The Maximax Regret Criterion (Opportunity Lost Decision Criterion)
5.5.4 The Realism Criterion
5.5.5 Criterion of Insufficient Reason
5.6 Decision Making Under Risk
5.7 Expected Value of Perfect Information (EVPI)
5.8 Minimising Expected Losses
5.9 Decision Tree
5.9.1 Bayesian Revision of Probabilities
5.10 Summary
5.11 Key Terms
5.12 Question & Exercises
5.13 Further Reading and References
5.0 Introduction
Decision making is an integral part of manager's job. The success or failures that
an individual or organization experiences, depends to a large extent on the ability of
making appropriate decisions. Decision theory provides an analytical and systematic
approach to depict the expected result of a situation when alternative managerial actions,
and out comes are compared.
Decision maker has the choice of adopting a qualitative or a quantitative approach.
In a quantitative technique. there must be alternatives to choose from, decision maker
must be rational and consistent and the problem should conform to mathematical
rationality.
5.1 Unit Objectives
This unit discusses steps of decision - making process, making decisions under
various decision - making enuironments and related computations.
After studying this unit, you should be able to
Quantitative Techniques in
Management : 72
Understand various zones of decision making from ignorance to certainly.
Understand steps in decision making process.
Understand decision making under different environments such as under
uncertainly, under risk.
Understand various criteria for decision making under uncertainly
Understand use of decision tree in decision making.
Decision Theory
NOTES
5.2 Zones of Decision - Making
As per available knowledge, spectrum of decision making ranges from ignorance
to certainty.
ignorance
uncertainty
risk
certainty
Fig. - A
Spectrum of decision making on the basis of increasing knowledge from left to right.
In case of decision making under certainty, decision maker has the complete
knowledge of every alternative with certainty. Here, decision maker will select an
alternative that yields the highest payoff for the known future. For example to purchase
a fixed deposit scheme or National saving Certificate (NSC) is a case of full knowledge
about future with certainty. Depending upon interest rate, we can determine exact
maturity value on a future date.
In a risk situation, decision maker has less than complete knowledge with certainty
of the consequences of each alternative. Decision maker makes an assumption of the
probability with which each state of nature will occur. A case of getting a tail in the toss
of a fair coin has a probability of 0.5, is an example of decision - making under risk.
In a case of uncertainty, decision maker is unable to specify the probabilities with
which the various state of nature will occur.
5.3 Steps in Decision- Making Process
Decision making involves following steps :
1.
Identify and define the problem
2.
Make inventory of all possible future events. These are states of nature. These
can occur in the context of the decision problem.
3.
Identify all courses of action (decision choices or alternatives) available to
the decision - maker.
Note : Decision maker has no control over states of nature but alternatives
are under control of decision - maker.
4.
Make a payoff matrix from each pair of alternatives and future events. These
are normally expressed in monetary value.
5.
Using a mathematical technique decide best alternative for resulting in optimal
payoff.
Quantitative Techniques in
Management : 73
Decision Theory
NOTES
5.4 Example to demonstrate preparation of payoff table
Example 1 :- A firm makes three types of products. The fixed and variable costs
are given below :
Fixed cost (Rs)
Variable Cost per Unit (Rs)
Product A :
20,000
15
Product B :
30,000
12
Product C :
50,000
8
The likely demand in units of these products can 3000, 6000 or 9000 depending
upon poor, moderate or high demand scenario.
Selling price of each type of product is Rs. 20, then prepare the payoff matrix.
Solution :- Three future events are poor demand scenorio (D1)
Moderate demand scenario (D2)
and High demand scenario (D3)
Payoff in this example = Sales revenue - Cost
Now for different pairs of future events (D1, D2 or D3) and alternative course
(product A, B or C), payoff will be as follows :
AD1 = (3000 X 20) - (20000 + 15 X 30000)
= -5000
AD2 = (6000X20) - (20000 + 15 X 6000)
= 10,000
AD3 = (9000 X 20) - (20000 + 15 X 9000)
= 25,000
Similarly for
BD1 = -6000, BD3 = 18000,
BD3
= 42000
CD1 = -14000, CD2 = 28000,
CD3
= 58000
The payoff values are finally summarized in table A.1.
Table A.1
D1
alternatives
Future states
D2
D2
A
-5000
10,000
25,000
B
-6000
18,000
42,000
C
-14,000
28,000
58,000
5.5 Decision Making Under Uncertainty
Quantitative Techniques in
Management : 74
Decision maker does not have enough knowledge to assign probabilities of any
state of nature. Different criteria can be used to make a decision.
5.5.1 The maximax Criterion (Optimistic Critesion)
We select best of the best option for each strategy or choice available. First we
choose the maximum payoff possible for each alternative and then choose the alternative
with the maximum payoff within this group.
Decision Theory
NOTES
Using this criterion in table 5.1, we will first select D3 scenario for each product
category and then would opt for introducing product C.
5.5.2 The Maximin criterion (Pessimistic Approach)
We select best of worst option. We try to maximise our minimum possible payoff.
First we list the minimum payoff possible for each alternative and then select the
alternative within this group that gives the maximum payoff.
Using this criterion in table 5.1, we will first select D1 scenario for each product
category and then would opt for introducing product A.
5.5.3 The minimax Regret Critesion (Opportunity lost decision
critesion)
In this approach the opportunity loss or regret for not taking the best decision
under each state of nature is calculated.
For instance in example 1, if the demand turned out to be high and decision
maker had chosen to introduce product B then decision maker would suffer a loss of
Rs. 16000 for having chosen product C which would have given him the maximum
profit of Rs. 58,000. Decision maker would select the maximum regret for each
alternative and would attempt to minimise the opportunity loss or regret by choosing
the alternative which gives the least regret. For each state of nature the highest value is
taken and all other values of payoff for that state of nature are subtracted from it. This
is also known as Savage's Rule.
Table 5.2 Regret Values for table 5.1
Alternatives
D1
D2
D3
A
0
18,000
33,000
B
2000
10,000
16,000
C
9000
0
0
State of Nature
The maximum regret for each alternative has been circled and the minimum of
these regrets is double enclosed. Decision maker should introduce product C.
5.5.4 The Realism Criterion :- (Hurwicz's Rule)
This takes a middle path between maximax and minimax, that is optimism and
pessimism, through the use of a coefficient or index of optimism, denoted by α This
index lies between 0 to 1. α zero means pessimism and 1 means optimism about nature.
First we identify maximum and minimum payoff for each alternative. Then for each
alternative, we calculate measure of realism = α (Maximum payoff) +
(1-α) (Minimum payoff)
For example 1, take α = .75, then measure of realism values for three products
(alternatives) are :
Quantitative Techniques in
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Decision Theory
Product A = .75 (25,000 + .25 (-5000) = 17,500
Product B = .75 (42,000) + .25 (-6000) = 30,000
Product C = .75 (58,000) + .25 (-14,000) = 40,000
NOTES
Decision maker would introduce product C.
5.5.5 Criterion of Insufficient Reason
(Laplace or equal probabilities criterion) :- The expected value of each alternative
is calculated by considering equal probability for each state of nature.
1
In our example 1, three states are possible, so equal probability will be = ------3
Expected payoff will be
1
1
1
Product A = ----- (-5000) + ----- (10000) + ----- (25000) = 10000
3
3
3
1
1
1
Product B = ----- (-6000) + ----- (18000) + ----- (42000) = 18000
3
3
3
1
1
1
Product C = ----- (-14000) + ----- (28000) + ----- (58000) = 24000
3
3
3
Decision maker will choose product C as it has the highest expected value.
It may be noted that different rules may give different results. Which is the best or
correct?
Normally, in a situation of uncertainty, the decision maker acts according to his
nature, style and thinking. It is generally observed that the decision maker tends to
adopt the same rule for most decisions under uncertainty.
5.6 Decision Making Under Risk
Decision making under risk, is a probabilistic decision situation, in which more
than one state of nature exists and the decision - maker has sufficient information to
assign probability values to the likely occurrence of each of these states. Knowing the
probability distritution of the states of nature, the best decision is to select that alternative
which has the largest expected pay off value.
If probabilities of scenario D1, D2 and D3 in example 1 are .6, .3 and .1 respectively,
then expected values are
Quantitative Techniques in
Management : 76
for product A = .6 (-5000) + .3 (10,000) + .1 (25,000)
= 2500
for product B = .6 (-6000) + .3 (18000) + .1 (42000)
= 6000
for product C = .6 (-14000) + .3 (28000) + .1 (58000)
= 5800
Decision maker should opt for product B as this option will give the maximum
expected value.
Decision Theory
The above table is referred to as the decision matrix.
It is not necessary that the same probability of occurrence be assigned to different
states of nature for all the alternatives. If it is felt that the probability of occurrence of
different states of nature will also vary with the alternative, it can be built into the
matrix and expected values are calculated accordingly.
NOTES
Let us reconsider example 1 and assign different probabilities for different
alternatives as shown in table 5.3
Table 5.3
D1
D2
D3
Probability
0.6
0.3
0.1
Product A
-5000
10,000
25,000
Probability
0.4
0.4
0.2
Product B
-6000
18000
42000
Probability
0.2
0.3
0.5
Product C
-14000
28000
58000
Expected value
2500
13200
34600
Decision maker will now opt for product C as this gives the highest expected
value.
Let us consider a typical situation where decisions are taken daily. Raju is a
newspaper boy at Nasik Railway Station. He buys newspapers from agency at Rs. 1.50
per paper and sells it for Rs. 2 per paper. Any unsold paper at the end of the day has to
be thrown away. (In some cases salvage value can also be provided.) Raju's demand
fluctuates between 51 to 55 newspapers a day. He losses customers if he does not have
sufficient stock, but runs the risk of having to throw away newspapers and lose money
on it, if overstocked. Raju has maintained data on his customers. His data reveals that
the newspaper demand in the last 50 days was as follows in table 5.4.
Table 5.4
Demand in numbers
51
52
53
54
55
No. of days
15
20
7
6
2
Probability of demand
.3
.4
.14
.12
.04
Table 5.5 gives a payoff table for Raju.
Table 5.5 payoff table for table 5.4
Demand
51
52
53
54
55
Stock option Probability .3
.4
.14
.12
.04
51
25.50
25.50
25.50
25.50
25.50
Quantitative Techniques in
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Decision Theory
NOTES
52
24.00
26.00
26.00
26.00
26.00
53
23.50
25.00
26.50
26.50
26.50
54
22.50
24.00
25.50
27.00
27.00
55
21.50
23.00
24.50
26.00
27.50
As the demand varies between 51 to 55 per day, Raju needs to consider the options
of stocking 51, 52, 53, 54 or 55 newspapers per day. If he stocks 51 and the demand is
55 or more, he will be able to make a profit of only Rs. 25.50 as he would have no more
papers to sell even if the demand were more than 51 pepers (His profit on each paper
sold is Rs. 0.50 and the loss on each pepre thrown away is Rs. 1.50). If Raju stocks 52
pepers and the demand is 51 papers then he earns Rs. 25.50 from the 51 papers sold but
loses Rs. 1.50 as he has to throw away one paper for which he has paid Rs. 1.50. If he
sells 52 papers, his profit is Rs. 26.00. Further increase in demand does not affect his
profit as he does not have any more paper to sell. Table 5.5 is also called the conditional
profit table.
The table reflects the profit that Raju would make for any combination of a stocking
policy and demand. On the basis of this table we can calculate expected values for
different stocking levels with the help of probalities assigned for different levels of
demand.
Expected value (profit) for 51 newspaper stock policy
= .3X25.50 + .4X25.50 + .14 X 25.50 + .12 X 25.50 + .04 X 25.50
= 25.50
Expected value for 52 newspapers stock policy
= .3 X 24 + (.4 + .14 + .12 + .04) X 26 = 25.40
For 53 papers = .3X 23.50 + .4 X 25.50 + (.14 + .12 + .04) X 26.50 = 25.20
For 54 papers = .3X 23.50 + .4 X 24.00 + .14 X 25.50 + (.12 + .04) X 27.00 = 24.24
For 55 papers = .3 X 21.50 + .04 X 23.00 + .14 X 24.50 + .12 X 26.00 + .04 X 27.50 =
23.30
Raju should now stock 51 papers every day as this gives him the highest expected
value.
5.7 Expected value of perfect Information (EVPI)
In decision making under risk each state of nature is associated with the probability
of its occurrence. However, if the decision - maker can acquire perfect information
about the occurrence of various states of nature, then he will be able to select a course
of action that yields the desired payoff for whatever state of nature that actually occurs.
Quantitative Techniques in
Management : 78
This does not mean that the demand will not vary from 51 to 55 papers per day.
Demand would still be 51 papers per day 30 percent of the time, 52 papers per day 40
percent of the time, 53 papers 14 percent of the time, 54 papers 12 percent of the time
and 55 papers 4 percent of the time. But with perfect information Raju will know in
advance what the demand would be for the following day and can stock accordingly.
When demand is 51 papers per day. Raju will stock 51 papers and will get a profit
of Rs. 25.50. Table 5.6 reflects the conditional profit table for Raju if he has perfect
information. He will be able to eliminate all losses, both due to unsold newspapers as
well as the loss for being out of stock. This will represent the maximum profit that he
can expect.
Decision Theory
NOTES
Table 5.6
Demand
Expected Value
Stock policy
51
52
53
54
55
Prob.
.3
.4
.14
.12
.04
51
25.50
52
53
54
7.65
26.00
10.40
26.50
3.71
27.00
55
3.24
27.50
1.10
Total
26.10
The maximum profit that Raju can expect with perfect information is Rs. 26.10.
In the absence of perfect information, Raju's expected value was Rs. 25.50 . The value
of perfect information is 26.10 - 25.50 = Rs. 0.60 Raju should, therefore, not spend
more than Rs. 0.60 everyday in gathering this perfect information else he will not
attain the maximum expected profit of Rs. 26.10
5.8 Minimising Expected Losses
Losses Earlier discussion was based on maximising expected profit, but as an
alternative approach of making a conditional loss table and then attenpting to minimise
expected loss.
Raju, the newspaper boy suffers two types of loss - loss due to overstocking absolescence loss and loss due to understocking.
Table 5.7 is a conditional loss table for Raju.
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Decision Theory
Table 5.7
Stock
NOTES
Demand
Policy
51
52
53
54
55
Probability
.3
.4
.14
.12
.04
51
0
.50
1.00
1.50
2.00
52
1.50
0
.50
1.00
1.50
53
3.00
1.50
0
.50
1.00
54
4.50
3.00
1.50
0
1.50
55
6.00
4.50
3.00
1.50
0
Opportunity
Losses
Diagonal mode
of Zeros
Table 5.7 has a diagonal set of zeros. The figures above the diagonal represent
opportunity losses and the figures below the diagonal represent the obsolescence losses.
Following is the calculation of the expected value of loss :Expected loss value for 51 paper stocking policy.
= .4X.50+ 1.00 X .14 + 1.50 X .12 + 2 X .04 = .60
For 52 papers stocking policy
= 1.50 X .3 + .50 X .14 + 1.00 X .12 + 1.50 X .04 = 0.70
For 53 papers stocking policy
= 3.00 X .3 + 1.50 X .4 + .50 X .12 + 1.00 X .04 = 1.60
For 54 papers stocking policy
= 4.50 X .3 + 3.00 X .4 + 1.50 X .14 + 0.50 X 0.04 = 1.78
For 55 papers stocking policy
= 6.00 X .3 + 4.50 X .4 + 3.00 X .14 + 1.50 X .12 = 4.20
Raju should choose the alternative which minimixes the loss, that is he should
stock 51 news paper every day.
ML
The ratio ---------------MP + ML
is a good measure to decide the units to stock. Here ML is marginal loss, which
happens when demand is less then stocked units and MP is marginal profit, which
happens when demand is more than stocked units.
ML
The ratio ---------------MP + ML
Quantitative Techniques in
Management : 80
is a probability to decide about stocking of additional unit. If by stocking an
additional unit expected profit increases is more than expected loss, we will stock
additional unit. Use of this formula will save efforts of calculation of expected profit or
loss.
For Raju's example, we arrange data as follows :
Demand level
Probability
51
.3
52
.4
53
.14
54
.12
55
.04
Decision Theory
NOTES
5.8 shows an additional column of cumulative probability. See here that cumulative
probability is calculated from bottom to up.
Table 5.8
Demand level
Probability
Cumulative Probability
51
.3
1.00
52
.4
.70
53
.14
.30
54
.12
.16
55
.04
.04
ML
Now we calculate .... = -------------MP+ML
1.50
1.50
Now we calculate .... = -------------- = -------------.50+1.50
2.00
=
0.75
We can now compare this value with the probability table above. We will select
demand level which should be corresponding to just higher than this value of 0.75.
According to table 5.8 this level is at papers.
5.9 Decision Tree
Decision - making problems discussed so far have been limited to a single decision
over one period of time, because the payoffs, states of nature, course of action and
probabilities associated with the occurrence of states of nature are not subject to change.
However situations may arise when a decision - maker needs to revise his previous
decisions on getting new information and make a sequence of several interrelated
decisions over several future periods.
These aspects can often be best displayed in a network, form called the decision
tree.
Quantitative Techniques in
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Decision Theory
NOTES
A decision tree represents sequential, multi-stage logic of a decision problem.
Decision trees uses two symbols- a box to represent a decision node and a circle to
represent a chance node. Outcomes of chance nodes are different states of nature.
Following examples illustrate use and construction of decision tree.
Example 2. ABC Ltd. has to decide about setting up of a new plant - large Vs
small. Large plant will cost Rs. 200 lakhs while small plant costs Rs. 80 lakhs. Probability
of high demand, low demand and moderate demand over the next 10 years is 0.5, 0.2
and 0.3 respectively.
(a) A large plant with high demand will yield an annual profit of Rs. 50 lakhs.
(b) A large plant with moderate demand will yield an annual profit of Rs. 40
lakhs.
(c) A large plant with low demand will result in loss annually of Rs. 10 lakhs.
(d) A small plant with high demand would yield Rs. 20 lakhs annually, taking
into account the cost of lost sales due to inability to meet demand.
(e) A small plant with moderate demand will yield Rs. 30 lakhs, as the losses
due to lost sales will be lower.
(f)
A small plant with low demand will yield Rs. 40 lakhs annually, as the plant
capacity and demand will match.
Solution :- The figure 5.2 A presents decision tree for the given situation
Large plant
2
1
Small plant
3
High demand (.5)
Moderate demand (.3)
Low demand (.2)
High demand (.5)
Moderate demand (.3)
Low demand (.2)
50
40
(-10)
20
30
40
Fig. 5.2
Now we will calculate expected values at node 2 and node 3 as follows.
E. V. at node 2 = .5 X 50 + .3 X 40 + .2 (-10) = 35
The expected value in 10 years is Rs. 350 lakhs. The cost of large plant is Rs. 200
lakhs. Hence net expected gains at the end of 10 years is 350-200 = Rs. 150 lakhs.
E. V. at node 3 = .5X20 + .3X30 + .2X40 = 27
The expected value in 10 years is Rs. 270 lakhs. The cost of small plant is Rs. 80
lakhs. Hence net expected gains at the end of 10 years is 270-80 = Rs. 190 lakhs.
We will write these expected values on decision tree as shown in figure B.
Quantitative Techniques in
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190
E . V.
Big plant =
150 2
High demand (0.5)
Moderate demand (0.3)
Low demand (0.2)
E.V. Small plant 100
3
1
High demand(0.5)
50
40
(-10)
20
High demand (0.3)
High demand (0.2)
Decision Theory
NOTES
30
40
Fig 5.3
Seeing figure 5.3, the decision in this case is to build a small plant and the expected
value is Rs. 190 lakhs.
Example 3 :
A company can setup a commercial plant or a pilot plant at present and setup
commercial plant at a later time. Cost of setting a commercial plant is Rs. 20 lakhs and
cost of setting a pilot plant is Rs. 2 lakhs. However, when the commercial plant is set up
six months later after observing the performance of the product it will cost Rs. 25
lakhs. It has also been estimated that the probability of the product giving a high yield
during the pilot stage is 0.8 and that of giving a low yield is 0.2. If the product is
introduced commercially without going through a pilot plant stage, it is expected that it
will give a high yield of profits with a probability of 0.75. If the pilot plant does show
a high yield then the probability that the commercial plant will also give a high yield is
0.85; but if the pilot plant gives a low yield, the probability that the commercial plant
will give a high yield is only 0.1. The estimated profits from high yield at the commercial
stage are Rs. 100 lakhs and if the yield is low, the company will suffer a loss of Rs. 10
lakhs.
Solution :- Figure 5.4 presents decision tree for given problem :
High (.75)
Commercial plant
2
Low (.25)
High (.85)
1
Commercial plant
High (.08)
6
Low (.15)
5
Stop
Pilot plant
3
Stop
Low (.2)
High (.10)
5
Commercial plant
7
Low (.90)
Fig. 5.4
Quantitative Techniques in
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Decision Theory
Values in ( ) with high or low are probabilities.
Calculation of expected values at node 2, 6 and 7 is as follows :
NOTES
EV at node
2 = .75X 100 + .25 (-10) = 72.50
EV at node
6. = .85 X 100 + .15 (-10) = 83.50
EV at node
7 = .10 X 100 + .90 (-10) = 1.00
Net gains at node 2, 6 and 7 will be obtained after subtracting cost of plant.
At node
2 = 72.50 - 20 = 62.50
At node
6 = 83.50 - 25 = 58.50
At node
7 = 1 - 25 = -24
At node 4 EV is 58.50 and node 5 EV will be zero.
So EV at node
3 = (58.50) X .8 + .2 X 0 = 46.80
Net gain at node 3 = 46.80 - 2 = 44.80
Going backward to node 1, we see that net gain is high at node 2. So we should
build a commercial plant.
5.9.1 BAYESIAN Revision of Probabilities
Take an example of three alternatives, A, B and C.
These three alternatives are three crops. You always devote the farm to grow one
crop at a time. You now need to make a decision as to which crop to grow during the
coming season. Following is the estimates of crop yields and net incomes per bushel
under various weather conditions.
Weather
Expected yield in bushels per acre
Crop A
Crop B
Crop C
Dry
20
15
30
Moderate
35
20
25
Damp
40
30
25
Net income per bushel
5000
7500
5000
Probabilities of weather condition to be dry, moderate or damp are .3, .5 and .2
respectively.
Crop A
2
Dry (.3)
Moderate (.5)
Damp (.2)
Dry (.3)
Crop B
1
3
Crop C
Quantitative Techniques in
Management : 84
4
Fig. 5.5
Moderate (.5)
Damp (.2)
Dry (.3)
Moderate (.5)
Damp (.2)
Payoff table will be
Decision Theory
Weather
Prob
Crop A
Crop B
Crop C
Dry
.3
20
15
30
Moderate
.5
35
20
25
Damp
.2
40
30
25
E.V. of yield
31.5
20.5
26.5
Income per bushel
5000
7500
5000
E.V. of profit
157500
153750
132500
NOTES
Here we select crop A as the expected value of profit for it is the highest.
Now we use Bayes decision rule with respect to the prior probabilities of moderate
weather and damp weather (without changing the prior probability of dry weather.) by
resolving when the prior probability of moderate weather is 0.2.
New payoff table will be
Weather
Prob.
CropA
Crop B
Crop
Dry
.3
20
15
30
Moderate
.2 See new
35
20
25
Damp
.5 probabilities
40
30
25
E. V. of yield
33
23.5
26.5
Income per bushel
5000
7500
5000
E.V. of profit
165000
176250
132500
Now we select Crop B if the prior probability of moderate weather is 0.2.
Application of posterior probabilities :Ram has a large piece of land near Nasik. A survay of EIL tells possibility of
crude in this land. EIL approached Ram with an offer of buying the land from his for
Rs. 100 lakhs with a gaurantee that if oil was found on the land he would receive
another Rs. 1000 lakhs. On the other hand, if he decides to explore for oil himself, it
would cost him Rs. 100 lakhs for the exploration. If he gets oil, he will make a profit of
Rs. 2000 lakhs. Probability of oil in the land is 0.7.
While he was still thinking, Earth exploration ltd. approached him for taking
soundings for oil exploration. The soundings can detect the presence of gas in the area.
However, there is no certainty that oil would be present if gas is detected. The company
will do sounding at a cost of Rs. 5 lakhs. If oil is present, the chances that gas will be
found are 70 percent but there is also a 10 percent chance of finding gas even if there is
no oil. It is also easy to understand that if the soundings reveal no gas, the chances are
that EIL will withdraw the offer. Should he run the risk of not only losing Rs. 5 Lakhs
on the soundings but also Rs. 100 lakhs which the EIL is offering him? What would
happen if the soundings found gas? Should he then explore for oil himself or still sell
the land to EIL? What if even after discovery of gas no oil is found?
The probabilities that are presently known to Ram are :
Probability of oil being present is 0.7.
Quantitative Techniques in
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Decision Theory
Probability of oil not being present is 0.3
Probability that gas will struck if oil is present is 0.7
Probability that gas will struck if oil is not present is 0.1
NOTES
1)
If he sells to EIL and no oil is found, his payoff is Rs. 100 lakhs. If oil is found,
EIL Will give him a total of Rs. 1100 lakhs.
2)
If he explores himself and strikes oil, he will gain Rs. 2000 lakhs but if oil is not
found, he will lose Rs. 100 lakhs.
3)
Cost of soundings is Rs. 5 lakhs.
Now Ram will calculate posterior probabilities from the given a prior probability
by using Baye's formula.
P(G/o)=.3
P(G/o)=.9
P(G/o)=.7
P(G/o)=0.1
P (0) = .7
P (O) = .3
Figure 5.6 Venn diagram based on prior probabilities
(i)
Probability of gas being struck
= Prob. of gas when oil is present + Prob. of gas when oil is not present
= 0.7X.7 + .1 X .3 = 0.52
(ii)
Probability of gas not being struck = 1 - .52 = .48
(iii)
Probability of oil given gas is struck
.7 X .7
= Prob. of gas when oil is present / Prob. of striking gas = ----------- = .942
.52
(iv)
Prob. of oil given that gas is not struck
= Prob. of no gas when oil is present / Prob. of no gas
.3x.7
= ---------- = .4375
.48
These probabilities are shown on decision tree in figure 5.7 in highlighted fonts.
Quantitative Techniques in
Management : 86
Oil (0.7)
800
6
Sell
No Soundings
1370
3
1370
7
Keep
1
No gas (.84)
818.75
Decision Theory
1100
No oil (0.3) 100
Oil (0.7)
2000
No oil (0.3) (-100)
537.5
Sell
8
Oil (.4375)
100
Oil (.4375) 2000
1369.664
Keep
9
No Oil (.5625)
Oil (.942)
Sell
Gas (.52)
1100
No Oil (.5625)
4
2
Sounding
1364.664
NOTES
1100
10
1878.20
No Oil (.058) 100
Oil (.912)
2000
5
Keep
(-100)
11
No Oil (.058)
(-100)
Figure5.7 decision tree
Expected value calculation at different nodes will be as follows :
EV at node 6 = 1100 X .7 + 100 X .3 = 800
EV at node 7 = 2000 X .7 + (-100) X .3 = 1370
These values are shown at node 6 and 7. The value at node 7 being greater is
carried to the decision node 3 and the decision from node 3 is to keep the land.
Similarly, the value at node 8 is
= 1100 X .4375 + 100 X .5625 = 537.5
at 9 is
= 2000 X .4375 + (-100) X .5625 = 818.75
at 10 is
= 1100 X .942 + 100 X .058 = 1042
at 11 is
= 2000 X .942 + (-100) X .058 = 1878.20
On the basis of EVs at node 8 and 9, EV at node 4 will be 818.75.
On the basis of EVs at node 10 and 11, EV at node 5 will be 1878.20.
Now EV at node 2 = 818.75 X .48 + 1878.20 X .52
= 1369.664
Net gain at node 2 = EV at node 2 - Cost of soundings
= 1369.664 - 5 = 1364.664
Quantitative Techniques in
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Decision Theory
NOTES
Comparing net gains at node 2 and 3, we decide not to take soundings and not to
sell but keep the land for carrying exploration on own.
5.10 Summary
5.11 Key Terms
5.12 Questions and Exercises
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(1)
What is the difference between decision making under risk and decision making
under uncertainty?
Decision Theory
(2)
Describe the steps involved in decision making?
(3)
Briefly explain the different decision rules usually adopted in context of decision
making under uncertainty.
NOTES
Practice Questions :
(4)
You have Rs. 5 lakhs for one of the three investment options in the stock market
- a growth, government bonds or a blue chip company. Three possible states are
listed below.
Type of investment
Stock Market Trend
Boom
Moderate growth
collapse
Growth fund
375000
150000
-400000
Government bonds
500000
100000
-500000
Blue chip company
250000
100000
-300000
Using the minimax regret criterion for decision making which option should you
adopt?
(5)
The following is a payoff (in rupees) table for three strategies and two states of
nature :
Strategy
State of Nature
N1
N2
s1
40
60
s2
10
-20
s3
-40
150
Select a strategy using each of the following decision criteria : (a) Maximax, (b)
Minimax regret, (c) Maximin, (d) Minimum risk, assuming equiprobable states.
(6)
The following matrix gives the payoff (in Rs.) of different strategies S1, S2, S3
against conditions N1, N2, N3 and N4
State of Nature
Strategy
N1
N2
N3
N4
S1
4000
-100
6000
18000
S2
20000
5000
400
0
S3
20000
15000
-2000
1000
Select a strategy under (i) Pessimistic; (ii) Optimistic, (iii) Equal probability, (iv)
Regret, (v) Hurwicz Critesion, the degree of optimism being 0.75.
(7)
The probability of the demand for lorries for hiring on any day are as
follows :
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Decision Theory
NOTES
Demand
0
1
2
3
4
Probability
.1
.2
.3
.2
.2
Lorries have a fixed cost of Rs. 90 each day to keep. The daily hire charges are
Rs. 200. If the lorry hire company owns 4 lorries, what is its daily expectations? If the
company is about to go into business and currently has no lorries, how many lorries
should it buy?
(8)
The demand pattern of the cakes made in a bakery is as follows :
Demand
0
1
2
3
4
5
Probability
.05
.10
.25
.30
.20
.10
If the preparation cost is Rs. 3 per unit and selling price is Rs 4 per unit and
unsold cakes have salvage value of Rs. 0.50 per unit, how many cakes should the baker
bake to maximize profit?
(9)
A company is considering the introduction of a new product. It has defined two
levels of sales as 'high' and 'low' on which to base its decision and has estimated
the changes that each market level will occur, together with their costs and
consequent profits or losses. The information is summarized below :
States of nature
Probability
Alternatives
Introduce
(Rs. 000)
Do not Intorduce
(Rs. 000)
High Sales
0.3
150
0
Low Sales
0.7
-40
0
The company's marketing manager suggests that a market research survey may
be undertaken to provide further information on which to base the decision. On past
experience with a certain market research organisation, the marketing manager assesses
its ability to give good information in the light of subsequent actual sales achievements
as follows :
Market Research
(Survey outcome)
Actual Sales
High Market
Low Market
'High' sales forecast
0.5
0.1
Indecisive Survey report
0.3
0.4
Low Sales forecast
0.2
0.1
The market research survey will cost Rs. 20,000, state whether or not there is a
case for employing the market research organization.
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5.13 Further Reading and References
Decision Theory
NOTES
Quantitative Techniques in
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Probability
NOTES
UNIT 6
PROBABILITY
Structure
6.0 Introduction
6.1 Unit Objectives
6.2 Important Terms
6.3 Calculating Probability
6.4 Theorems of Probability
6.5 Calculation of Probability of at least One Event
6.6 Conditional Probability Theorem
6.7 Bernoulli Theorem
6.8 Bayes Theorem : (Revising Prior Estimates)
6.9 Summary
6.10 Key Terms
6.11 Question & Exercises
6.12 Further Reading and References
6.0 Introduction
The word probability or chance is very commonly used in day- to- day
conversation. We come across statements such as
"Probably NDA will come to power in next general elections."
"Probably it may rain tomorrow"
"It is possible that we may not join you at dinner." In these statements, we have
some element of uncertainity about happening of the event in question.
A numerical measure of uncertainty is provided by "Probability".
Probability theory is extensively used in the quantitative analysis of business and
economic problems.
6.1 Unit Objectives
This unit covers same common terms related to prbability theory, important
therorems of probability such as addition and multiplication theorems, conditional
probability therorem, Bernoulli theorem, Bayes’ theorem etc. These theorems help in
calculationg probabilities in different situations.
6.2 Important terms
(a) Simple Experiment or Trial :- The term experiment refers to processes which
result in different possible outcomes or observations.
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(b) Outcome : Output of an experiment is called outcome. The number of outcomes
depends upon the nature of the experiment and may be finite or infinite.
(c) Random Experiment : If in an experiment all the possible outcomes are known in
advance and none of the outcomes can be predicted with certainty, then such an
experiment is called a random experiment.
(d) Sample Space : A set of all possible, equally likely outcomes of a random
experiment is known as the sample space and is denoted by 'S'.
Probability
NOTES
For example
(i) In tossing a coin, S = (H,T)
(ii) In rolling a die, S = (1, 2, 3, 4, 5, 6)
(iii) When two coins are tossed
S = (HH, TT, TH, HT)
(e) Events : A single outcome or a group of outcomes constitutes an event. Events are
denoted by capital letters A, B, C etc. Events can be of following types :
(i)
Simple and compound Events : In case of simple events we consider the
probability of the happening or not happening of single event. For examples,
we might be interested in finding out the probability of drawing a, white ball
from a bag containing 5 white and 5 red balls.
On the other hand, in case of compound events, We consider the joint
occurrence of two or more events. For example, if a bag has 5 white and 5
red balls, if two successive draws of 2 balls are made, we shall be finding out
the probability of getting 2 white balls in the first draw and 2 red balls in the
second draw, this way, we are dealing with a compound event.
(ii) Independent Events : Events are said to be independent of each other if
happening of one event is not affected by any one of others.
(iii) Dependent Events : Here occurrence or non- occurrence of one event in any
one trial affects the probability of other events in other trials.
For example, if a card is drawn from a pack of playing cards and is not
replaced, this will alter the probability of the second card.
(iv) Mutually Exclusive Events : Two or more events are called mutually exclusive
if the happening of any one of them excludes the happening of all others in
the same experiment. Thus, in a game of tossing the coin, at one time, we
can either get only head or only tail. Thus 'Head' or Tail' are mutually exclusive
events in this experiment.
(v) Equally likely Events : Events are equally likely if there is no reason for an
event to occur in preference to any other event. For example, when an unbiased
coin is rolled then the outcomes, head and tail, are equally likely.
(vi) Exhaustive Events : Exhaustive events are those event which include all
possible outcomes of an experiment. Thus, in toss of a single coin, we can
get head (H) or tail (T). Hence exhaustive number of cases is 2, viz (H,T).
6.3 Calculating Probability
To find probability, we should know following two things :
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Probability
(i)
Number of favourable cases;
(ii) Total number of equally likely cases.
Example 1: In tossing a coin, what is the probability of a head or a tail?
NOTES
Number of Favourable cases
Probability = ---------------------------------------------------------------Total Number of equally likely cases
Here, Number of Favourable case = 1
Total Number of equally likely cases = 2
1
Probability = ------2
Example 2 : In rolling a die, what is the probability of getting an even number?
Number of Faourable cases = To get 2,4, or6 = 3
Total Number of equally likely cases = Getting 1,2,3,4,5,or 6 = 6
3
1
Probability =------- = ------6
2
Example 3 : (i) A ball is drawn from a bag containing 8 red, 7 white and 5 blue balls.
Determine the probability that it is (a) red (b) white (c) blue.
(ii) Find the probability of selecting a vowel 'U' from Vowels chosen at random
from an English back?
Solution : (i) Total number of balls in the bag = 8 + 7 + 5 = 20
(a)
There are 8 red balls in the bag.
Therefore, the probability that the ball drawn is red
8
= --------20
=
2
--------5
(b) There are 7 white balls in the bag.
(c)
(ii)
7
Therefore, the probability that the ball drawn is white = --------20
There are 5 blue balls in the bag.
5
1
Therefore, the probability that the ball drawn is blue = --------- = --------20
4
Total number of vowels (A,E,I,O,U) = 5
'U' is one vowel out of the total vowels. Hence the probability of selecting a vowel 'U'
1
= -------5
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Example 4 : What is the probability that a leap year selected at random will contain 53
Mondays?
Solution : In a Leap year there are 366 days, i. e. 52 weeks and 2 days. These
remaining two days can come in following seven ways :
(i)
Probability
Sunday and Monday
(ii) Monday and Tuesday
NOTES
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
So probability of 53 mondays is possible in only first two cases out of seven
cases, so required probabily
2
= -----7
6.4 Theorems of Probability
There are two important theorems of probability, viz :
(1)
The Addition Theorem; and
(2)
The Multiplication Theorem.
(1)
The Addition Theorem : The addition theorem can be applied in following two
situations :
(a) In case of two or more mutually exclusive events.
If two events A and B are mutually exclusive the probability of the occurrence of
either A or B is the sum of the individual probability of A and B.
P(A or B) = P (A) + P (B)
The theorem can be extended to three or more mutually exclusive events :
P (A or B or C) = P (A) + P (B) + P (C)
(b) When events are not mutually exclusive :
In this case addition theorem is modified as follows :
P (A or B) = p (A) + p (B) - p (A and B)
where p (Aand B) = Probability of A and B happening together.
In case of three events :
P (A or B or C) = p (A) + p (B) + p (C) - p (AB) - p (BC) - p (AC) + p (ABC)
Example 5 : A card is drawn out of a pack of cards. Find the probability that it is a card
of heart or diamond.
Solution : Total number of cards in a pack of cards = 52
Total number of heart cards in this pack (A) = 13
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Probability
Total number of diamond cards in this pack (B) = 13
13
1
Therefore, p (A) --------- = ----- ,
52
4
NOTES
13
1
p (B) = --------- =--------52
4
Since the events are mutually exclusive, the probability that the card drawn is
either a heart or a diamond
1
1
1
p (A or B) = p (A) + p (B)= ----- + ----- = ----4
4
2
Example 6 : What is the probability of getting the following totals with two dice?
(a) 8 (b) at least 8 (c) more than 8
Solution : Chart showing the possible outcomes in the throw of two dice.
II dice
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
6
6,1
6,2
6,3
6,4
6,5
6,6
I dice
Therefore, total number of outcomes in the throw of two dice = 36
(a)
No. of favourable cases to get the total of 8
= (6,2), (5,3), (4,4), (3,5), (2,6) = 5
5
So probability to get the total of 8 = ------36
(b)
To get the total at least 8. (You can get totals of 8, 9, 10, 11 or 12).
Students can identify such cases. This number is 15.
15
5
So probability is = ----- = ----36 12
Same probability can also be obtained using addition theorem in following way:
Event A : total of 8, Favourable Cases = 5
(6,2), (5,3), (4,4), (3,5), (2,6)
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5
So p (A)= ------36
Probability
Event B : Total of 9, Favourable cases = 4
NOTES
4
So p (B)= ------36
Event C : total of 10, Favourable Cases = 3
(6,4), (5,5), (4,6)
3
So p (C)= ------36
Event D : Total of 11, Favourable Cases = 2
(6,5), (5,6)
2
So p (D)= ------36
Event E = Total of 12, Favourable Cases = 1
(6,6)
1
So p (E)= ------36
Hence, the probability to get the total at least 8.
= p (A) + p (B) + p (C) + p (D) + p (E)
5
4
3
2
1
15
5
= ------- + ------- + ------- + ------- + ------- = ------- =------36
36
36
36
36
36 12
(C) To get the total of more than 8 means total may be 9 or 10 or 11 or 12.
= p(B) + p(C) + p(D) + p(E)
4
3
2
1
10
5
= ------- + ------- + ------- + ------- = ------- = ------36
36
36
36
36
18
Example 7 : A card is drawn from a pack of cards. What is the probability that the
card drawn is
(i)
either a heart or a king,
(ii) either a queen or a black colour.
Solution : Total number of cards in a pack of cards = 52
(i)
Number of favourable cases to get a card of heart (A) = 13
Number of favourable cases to get a card of king (B) = 4
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Probability
NOTES
13
∴ p (A) = -------52
4
and p(B) = -------52
The events are not mutually exclusive because the king of heart entails the
occurrence of both the events.
1
Probability to get a card of king of heart, i.e. p (AB) = -------52
Hence, the probability that the card drawn is either a heart or a king
13
4
1
16
4
= p (A) + p (B) - p (AB) = -------- +-------- - -------- = -------- = -------52
52
52
52
13
(ii) Number of favourable cases to get a card of queen (A) = 4
4
∴ p (A) = -------52
Number of favourable cases to get a card of black color (B) = 26
26
p
(B)
=-------∴
52
The events are not mutually exclusive because the queen of black colour entails
the occurrence of both the events.
2
p (AB) = ------52
Hence, the probability that the card drawn is either a queen or a black colour
4
26
2
28
7
= p (A) + p (B) - p (AB) = -------- +-------- - -------- = -------- = -------52
52
52
52
23
(2)
The Multiplication Theorem :- This theorem states that if two events A and B are
independent, the probability that they both will occur is equal to the product of
their individual probabilities.
Here
p(A and B) = p(A) X p(B)
For three events A, B and C, theorem can be extended as
p (A and B and C) = p (A) X p (B) X p (C)
Example 8 : Two cards are drawn from a pack of cards in succession with
replacement. Find the probability that both are aces.
Probability that the first card drawn is an ace (A)
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4
p(A) = -----52
Probability
Probability that the second card drawn is an ace (B)
NOTES
4
p(B) = -----52
Probability that both cards are aces, i.e. p (A and B)
= p (A) X p (B)
4
4
16
1
Hence p(A and B) = -------- X -------- = -------- or -------52
52
2704
169
Example 9 : In a game, cards are thoroughly shuffled and distributed equally
among four players. What is the probability that a specific player gets all the four
kings?
Solution : As we know there are 52 cards,
52
hence each player will get -------- = 13 cards.
4
Number of ways of selecting 13 cards out of 52 cards
= 52C13
Number of ways of selecting 4 kings out of 4 kings = 4 c4
Number of ways of selecting remaining 9 cards out of 48 cards = 48C9
Hence the probability that a specific player gets all the four kings
4
c4 x48C9
13X12X11X10
11
= --------------------- = --------------------------- =--------------------52
C13
52X51X50X49
4165
Example : A bag contains 5 red and 4 green balls. 3 balls are drawn twice. Before
the second draw balls are replaced in the bag. Find the probability that in the first draw
3 red balls and in the second draw 3 green balls will be drawn?
Total number of balls = 5 + 4 = 9
Total number of ways of drawing three balls from the 9 balls
9!
C3 =---------------------= 84
6!x 3!
9
The number of ways in which three red balls can be drawn out of 5
5!
C3 =---------------------= 10
3! x 2!
5
The number of ways in which three green balls can be drawn out of 4
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Probability
4!
C3 =---------------------= 4
3! x 1!
4
There fore, the probability of first event (A);
NOTES
i. e., draw of 3 red balls
10
=------------84
Similarly the probability of second event (B);
4
i. e., draw of 3 green balls =-----------84
The probability of the compound event, i.e. of 3 red balls in the first draw and of
3 green balls in the second draw :
10
4
5
-------- x -------- = ------84
84
882
6.5 Calculation of probability of at Least One Event
If we are given n independent events E1, E2,E3,... En, with respective probabilities
of occurrences as P1, P2,P3, .... Pn, then probability that none of them happens, is (1-P1),
X(1-P2), X (1-P3) X .... X (1-Pn). Hence, the probability of occurrence of at least one of
the events can be determined as follows.
P (happening of at least one of the event) :
= 1-[(1-P1) X (1-P2) X (1- P3).... X (1- Pn)]
Example 11 : The probability of a cricket team winning match at Nagpur is 2/5
and losing match at Pune is 1/7. What is the probability of the team winning at least one
match?
Solution :
2
The probability of a team winning match at Nagpur = -------5
2
3
The probability of team losing at Nagpur = 1 - -------- = -------5
5
1
The probability of team losing at Pune
= -----7
Therefore, the probability of losing at Nagpur and Pune both.
3
1
3
= ------ x ------ = -----5
7
35
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Therefore, the probability of winning at least one match.
=
3
1 ----------35
6.6 Conditional Probability Theorem
Probability
NOTES
The multiplication theorem is not applicable in case of dependent events. Two
events A and B are said to be dependent when A can occur only when B is known to
have occurred (or vice versa). The probability attached to such an event is called the
conditional probability and is denoted by p (A/B) or p (B/A). p (A/B) means the probability
of A given that B has occurred and p (B/A) means the probability of B given that A has
occurred.
B
p (AB)
p ------ = ----------A
p (A)
;
A
p (AB)
p ------ = -----------B
p (B)
Conditional probability for three events :
B
C
p(ABC) = p(A) X p --------- X p --------A
AB
( ) ( )
Example 2 : A bag contains 6 white and 9 black balls. Two successive drawings
of 4 balls in each draw are made in such a way that (a) balls are replaced before the
second draw and (b) balls are not replaced before the second draw. Find the probability
of getting 4 white balls in the first draw and 4 black balls in the second draw.
Solution : (a) When the balls are replaced before the second trial the number of
ways in which 4 balls may be drawn is 15C4.
The number of ways in which 4 white balls may be drawn = 6C4.
The number of ways in which 4 black balls may be drawn = 9C4.
Therefore the probability of drawing 4 white balls at first trial
6
=
C4
6x5
4x3x2
1
--------= ----------------x --------------------------- = -------15
C4
2
15x14x13x12
91
Theprobability of drawing 4 black balls at eh second trial
9
=
C4
9x8x7x6
4!
6
--------= -------------------- x ------------------------ = -------15
C4
4!
15x14x13x12
65
Therefore, the probability of getting 4 white balls in the first draw and 4 black
balls in the second draw
1
6
6
= -------- x -------- = -------91
65
5915
(b) When the balls are not replaced :
At the first trial, 4 balls may be drawn in 15C4 ways and 4 white balls may be
Quantitative Techniques in
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Probability
NOTES
drawn in 6C4 ways.
Therefore, the probability of 4 white balls at first trail
6
C4
1
= -------- = -------- (as above)
15
C4
91
When 4 white balls have been drawn and removed the bag contains 2 white and 9
black balls.
Therefore at second trial
9
=
C4
9x8x7x6
4!
21
--------= ----------------x ------------------------ = -------11
C4
4!
11x10x9x8
5
Therefore the probability of getting 4 white balls in the first draw and 4 black
balls in the second draw
1
21
3
= -------- x -------- = -------91
55
715
6.7 Bernoulli Theorem
The probability of the happening of an event in one trial being known, it is required
to find the probability of its happening once, twice, thrice exactly in n trials. This can
be done by Bernoulli's Theorem. In such a case let p be the probability of its happening
and q of its failure then if the event will happen exactly r times in n trials the probability
must be (r+1)n terms in the expansion of (q+p)n.
Accordingly, the probability of the happening of an event exactly r times in n trial
is cr pr qn-r
n
Example 13 : If on an average one out of 10 ships sinks in the high seas. Find the
probability that out of 5 expected ships at least 4 will reach safely.
Solution :
9
The probability of the safe arrival of a ship (p) = ------10
1
and p (q) =------10
We want the probability of the safe arrival of at least four ships out of five which
means all the five ships or four ships.
The probability of the safe arrival of the four ships
p (r=4) =nc4 (p)r (q)n-r
9
= c4 -----10
5
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4
1 1 32805
------ =-------------10
1,00,000
( )( )
The probability of the safe arrival of all the five ships.
Probability
1
9 4 1
59049
p (r=5) = c4 ------ ------ =-------------10
10
1,00,000
5
( )( )
NOTES
Since the events are mutually exclusive, the required probability that at least 4
shipes will arrive safely is given by p(4) + p (5)
32805
59049
91854
=------------------------+ ----------------=------------------------ = 0.92
1,00,000
1,00,000
1,00,000
6.8 Bayes' Theorem : (Revising prior Estimates)
In many business situations, certain probabilities were altered after the people
involved got additional information. The new probabilities are known as revised, or
posterior, probabilities.
The basic formula for conditional probability under dependence (already discussed)
p(AB)
p(B/A) =-------------- is called
p (A)
Bayes’ Theorem.
Bayes' theorem offers a powerful statistical method of evaluating new information
and revising our prior estimates of the probability that things are in one state or another.
According to Bayes' theorem, ''an event is known to have proceeded from one of
n mutually exclusive causes whose probabilities are P1, P2, ... Pn. Further more lot P1,
P2, ... Pn be the respective probabilities that when one of the n causes exists the event
will then have followed.
The probability that event proceeded from the mth cause is then :
PmPm
P = ------------------------------------------------------P1 P1 + P2 P2 + P3P3+ ...... + PnPn
Example 14 : An insurance company insured 2000 scooter driver, 4,000 car drivers
and 6,000 truck drivers. The probability of an accident involving a scooter driver, car
driver and a truck driver is 0.01, 0.03 and 0.15 respectively. One of the inuerued drivers
meets with an accident. What is the probability that he is a car driver?
Solution :
Number of scooter drivers
= 2,000
Number of car drivers
= 4,000
Number of truck drivers
= 6,000
.... Total number of drivers
= 2000 + 4000 + 6000 = 12000
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Probability
1
Probability of being a scooter driver (p1) = 2000 / 12000 = ---------6
NOTES
Probability of being a car driver (p2) = 4000 /12000
1
= ---------3
Probability of being a truck driver (p3) = 6000/12000
1
= ---------2
It is given that the
1
Probability of an accident involving a scooter driver (p1) = 0.01 = ---------100
Probability of an accident imvolving a car driver (p2) = 0.03
3
= ---------100
Probability of an accident involving a truck driver (p3) = 0.15
15
= ---------100
By Bayes' rule, if one of the insured drivers meets with an accident, then the
probability that he was a car driver
P2p2
P(P2P2) = ----------------------------------------------P1p1 + P2P2 + P3P3
1
3
---------- x ---------3
100
= -----------------------------------------------------------------------------------------------1
1
1 3
1
15
------x ------ + ------x ------ + ------x -----6 100
3 100
2 100
(
) (
) (
)
1
1
--------------100
100
1
600
3
= --------------------------------------------------= ------------ = ------ x ------ = -----1
1 15
52
100
52
26
------ + ------ + ----------600
100 200
600
(
)
( )
= 0.11539
Example - 15 :
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(a)
In a bolt factory machines 'A', 'B' and 'C' manufacture respectively 25%, 35%
and 40% of the total production. Of their output 5, 4, 2 percents are defective
bolts. A bolt drawn at random from the production and is found to be defective.
What is the probability that it was manufactured by machine 'A'?
(b)
What is the probability that it was manufactured by machine 'B'?
Solution :
(a)
Probability
Probability that bolt drawn is manufactured by machine 'A',
25
(P1) = ---------100
Probability that bolt drawn is manufacturered by machine 'B',
35
(P2) = ---------100
Probability that bolt drawn is manufactured by machine 'C'
40
(P3) = ---------100
Probability that the bolt drawn is defective given that it is manufactured by machine
NOTES
'A'
5
(P1) = ---------100
Probability that the bolt drawn is defective given that it is manufactured by machine
'B'
4
(P2) = ---------100
Probability that the bolt drawn is defective given that is is manufactured by machine
'C'
2
(p3) = ---------100
By Bayes' rule, probability that the bolt drawn is manufactured by machine 'A'
given that the bolt drawn is defective.
25
5
------ x -----P1P1
100
100
P(P1P1) = ------------------------------------------------ = -------------------------------------------------------------------------------------------------------P1p1 + P2P2 + P3P3
25 5
35 4
40 2
------x ------ + ------x ------ + ------x -----6 100
3 100
2 100
(
) (
) (
)
1
1
-------------80
80
1
2000
25
= --------------------------------------------------= ------------ = ------ x ------ = -----1
7
1
69
80
69
69
------ + ------ + ------------80
500
125
2000
(
)
(
)
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Probability
= 0.3623 or 36.23%
(b)
NOTES
Probability that the bolt drawn is manufactured by machine 'B' given that the bolt
drawn is defective
35
4
------ x -----P1P1
100
100
P(P1P1) = -------------------------------------------------------= -----------------------------------------------------------------------------------------------P1P1 + P2P2 + P3P3
25
5
35 4
40
2
------x ------ + ------x ------ + ------x ------
(
) (
100
1
(-------500 )
=
-------- =
1
--------
(
)
100
) (
100 100
)
100 100
7
69
28
-------- x -------- = -------500
2000
29
500
= 0.4057 ro 40.57%
Probability that the bolt drawn is manufactur by machine 'C' given that the
bolt drawn is defective.
P 3 p3
p(P3p3) = ---------------------------------------P1p1 + P2p2 + P3p3
40
=
2
------100
100
-------------------------------------------------------------------------------------25
5
35
4
40
2
------- x ------- + -------x ------- + ------- x -------
(100
100
(------- x
)
) ( 100
100
1
(--------)
=
125
-------- =
69
-------200
(
=
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)
1
2000
16
-------- x -------- = -------125
69
69
0.2319 ro 23.19%
) (100
100
)
6.9 Summary
Probability
NOTES
6.10 Key Terms
6.11 Questions and Exercises
Short Answer Questions :
1.
Explain the concept of conditional probability.
2.
Explain the multiplication theorem of probbility with suitable examples.
3.
Define the following :
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Probability
(i)
Simple event
(ii) Compound event
(iii) Mutually exclusive events
(iv) Exhaustive events
NOTES
4.
What is 'Bayes theorem'? Explain with example.
5.
State the addition theorem of probability
(a) when two events are mutually exclusive and (b) when they are not mutually
exclusive.
Numerical Questions
1.
What is the probability of getting the following totals with two dice?
(a) 9 (b) at least 9 (c) More than 9
2.
Find the probability of obtaining a total of 2 or 9 or 11 in a single throw with
two dice.
3.
Find out the probability of getting a total of either 7 or 11 in a single throw with
two dice.
4.
Three six- faced dice are thrown what is the probability of getting the total of 7?
5.
A card is drawn from a pack of cards. Find the probability of getting a king or a
card of spade.
6.
One number is drawn from numbers 1 to 150. Find the probability that it is either
divisible by 3 or 5.
7.
A doctor is to visit a patient once in the month of November. Find the probability
that he visits on a date which is a multiple of 5 or 6.
8.
50 tickets are numbered serially from 1 to 50. One ticket is taken out at random.
Find the probability that the ticket. So drawn is a multiple of 3 or 4.
9.
Tickets numbered serially from 1 to 100 are thoroughly shiffied and one ticket is
taken out. What is the probability that the ticket so drawn is (a) an odd number,
(b) 5 or multiple of 5 (c) a number which is a square?
10. One ticket is drawn at random from a bag containing 100 tickets numbers from
1 to 100. What is the probability that the ticket drawn is a multiple of 2 or 3 or 10?
11) From a bag containing four red and six blue balls, two balls are drawn at random.
What is the probability that both the balls are of different colours?
12) In a bag there are 4 red and 3 black balls. What is the probability of drawing the
first ball red and second black, the third red and younth black and so on if they
are drawn one at a time?
13) A problem is given to three students A, B and C whose chances of solving are 1/
3, 1/4 and 1/5. Find the probability that the problem will be solved.
14) A bag contains 5 black balls and 4 white balls. 3 balls are drawn one by one
without replacement. Find the chance that all the three balls drawn are of the
same colour.
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15) In a railway reservation office, two clerk are engaged in checking reservation
forms. On an average the first clerk checks 55% of the forms while the second
doesthe remaining. The first clerk has an error rate of 0.03 and the second has an
error rate of 0.02. A reservation form is selected at random from the total number
of forms checked during a day and is found to have an error. Find the probability
that it was checked (i) by the first (ii) by the second clerk.
16) A TV manufacturing company produces TV sets in its three plants with monthly
production of 500, 1000 and 2000 sets respectively. According to past experience
it is known that the probability of defective TV sets produced by the three plants
are respectively 0.005, 0.008 and 0.010. If a set is selected from a month's
production and it is found to be defective, find out (i) from which plant the set
comes? (ii) What is the probability that it comes from the (a) first plant (b) second
plant?
Probability
NOTES
6.12 Further Reading and References
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Probaility Disribution
NOTES
UNIT 7 : PROBABILITY DISTRIBUTION
Structure
7.0 Introduction
7.1 Unit Objectives
7.2 Discret and Continuous Random Distributions
7.3 Probability Distribution
7.4 Expected Value of a Random Variable
7.5 Binomial Distribution
7.6 Measures of Central Tendency and Dispersion for the Binomial Distribution
7.7 Poisson Distribution
7.7.1 Mean and Variance of a Poisson Distribution
7.8 Normal Distribution
7.8.1 Probability Density function of a Normal Distributions
7.9 Summary
7.10 Key Terms
7.11 Question & Exercises
7.12 Further Reading and References
7.0 Introduction
In case of statistical outcomes based on chance, the outcomes are expected to
vary, or in other words, the outcomes occur randomly. This unit focuses on the
probabilities of various outcomes that can occur in a particular type of experiment. It
continues the discussion of probability by introducing the concept of random variable
and probability distribution.
7.1 Unit Objectives
The objective of this unit is to introduce discrete and contineous random probability
distributions, to use the concept of exptected value to make decisions, to show which
probability distribution to use and how to find its values and finally to understand the
limitations of each of the probability distributions we use.
7.2 Discrete and Continuous Random Distributions
A random variable is a variable which contains the outcome of a chance
experiment. A ramdom variable that assumes either a finite number of values or a
countable infinite number of values is termed as a discrete random variable. For
example, consider an experiment to measure the number of patients who arrive in a
clinic during a time interval of 15 minutes. The possible outcomes may vary from
patients to patients. These outcomes (0, 1, 2, ....n) are the values of the random variables.
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In most cases, a discrete random variables are non negative whole numbers.
Many experiments have outcomes which cannot be described as discrete random
variable. These random variables assume any numerical value in an interval, or can
take values at every point in a given interval. These random variables are called as
continuous random variables. Experimental outcomes which are based on
measurement scale such as time, distance, weight, and temperature can be explained
by continuous random variables.
Probaility Disribution
NOTES
7.3 Probability Distribution
Probability distributions are related to frequency distributions. A theoretical
frequency distribution is a probability distribution that describes how outcomes are
expected to vary.
To understand the concepts of probability distribution take a fair coin and we toss
this fair coin twice.
Table 7.1 possible out comes from two tosses of a fair coin
First Toss Second Toss
Number of tails
on two tosses
Probability of this outcome
T
T
2
.5 X .5 = .25
T
H
1
.5 X .5 = .25
H
H
0
.5 X .5 = .25
H
T
1
.5 X .5 = .25
1.00
Table 7.2 Probability distribution of the possible number of tails from two tosses
of fair coin
T=0
(H, H)
Probability = .25
T=1
(H,T), T,H)
Probability = .50
T=2
(T, T)
Probability = .25
Now corresponding probability distribution is shown in figure 1.
.50
Probability .25
0
1
2
Number of tails
Figure 7.1
7.4 Expected value of a Random variable
Expected value of a random variable is calculated by multiplying each outcome
with its probability and then summing these products.
Table 7.3 : Calculation of expected value of the discrete random variable no. of
newspaper sold daily"
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Probaility Disribution
NOTES
Possible values of the
Random variable
Probability that the Random
variable will take on these values
(1)
(2)
(1) X (2)
10
.10
1.0
11
.10
1.1
12
.15
1.8
13
.15
1.95
14
.25
3.50
15
.25
3.75
1.00
13.10
Expected value of random
variable "no. of newspapers
sold daily"
This expected value is also known as mean value.
Variance : The expected value provides the mean value of the random variable.
A measure of dispersion (variance and standard deviation) can also be obtained using
this mean of random variable.
Variance of a discrete distribution is given by
Var (x) - σ -2 = Σ [(x-µ)2 x p (x)]
Where x is an outcome,
P(x) the probability of that outcome, and
µ is the mean.
Steps for calculating variance in case of discrete distribution can be summarized
as follows :
(i) Calculate the deviation (x - µ) .
(ii) Square the deviation and multiply it by corresponding value of probability.
(iii) Take sum of products obtained in step (ii) above.
Standard deviation :
Standard deviation is the square root of the variance.
Standard deviation (σ) = √var (x)
7.5 Binomial Distribution
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Binomial distribution is the most widely known and most commonly used
distribution among all discrete distributions. It describes discrete data resulting from
an experiment known as Bernoulli process.
Probaility Disribution
Tossing a fair coin for a fixed number of times is a Bernoulli process and the
outcomes of such tosses can be represented by binomial distribution.
This process can be described as follows:
NOTES
1.
Each trial has only two possible outcomes : H or T.
2.
The probability of the outcome of any trial remains fixed overtime. (.5 in case of
fair coin)
3.
The trials are independent in nature.
Now binomial formula is for
n!
probability of r success in n trials = p(r) = -------------- pxqn-x
r ! (n-x) !
Where
p = probability of success
q = 1 - p = probability of failure
r = number of success desired
n = number of trials undertaken
3!
e. g. probability of getting 2 tails in 3 tosses of a fair coin = -------------- (.5)2 (.5)1
2 ! (3 - 2) !
=
.0375
7.6 Measures of Central Tendency and Dispersion for
the Binomial Distribution
Mean = µ = np
Where
n = number of trials
p = probability of success
Standard deviation (σ) = √ npv
where
v = probability of failure = 1 - p
Practice Questions :
Que 1. Find the mean and standard deviation of the following binomial distributions.
(a) n = 15, p = .50
(b) n = 12, p = .75
(c) n = 100, p = .15
Que. 2 For a binomial distribution with n = 6 and p = 0.3, Find
(a) p (r = 4)
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Probaility Disribution
(b) p (r>2)
(c) p (r >4)
NOTES
7.7 Poisson Distribution
The Poisson distribution focuses on the number of discrete occurrences over an
interval. It is used to describe a number of processes such as distribution of telephone
calls going through a switchboard system, the demand of patients for service at a health
institution, the arrival of trucks and cars at a tollbooth etc.
Some of the improtant properties of Poisson distribution are as follows :
(i)
Each occurrence of an event is independent of the occurrence of the other
event.
(ii) The probability of an occurrence is the same for any two intervals of equal
length.
(iii) It describes discrete occurrences over a specific time interval.
(iv) The expected number of occurrences must hold constant for all the time
intervals of the same size.
(v) In each time interval, occurrences can range from zero to infinity.
Poisson formula is
λx x e-λ
P (x) = ---------------x!
where p (x) = probability of exactly x occurrence
λ = Mean number of occurrence in an interval
e = base of natural logarithm system (2.71828)
Example : Past record of accidents at an important crossroad indicate an average
of 5 accidents per week. The number of accidents are Poisson distributed. Calculate
probability of exactly 0, 1, 2, 3 or 4 accidents in any month.
Solution :
Here λ is 5.
X = 0, 1, 2, 3, and 4.
By applying Poisson formula
(i)
For exactly Zero accidents
(5)0 X e-5
p (0) = ------------------ = .0067
0!
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(ii) For exactly one accident
Probaility Disribution
(5)1 X e-5
p (1) = ------------------------ = .033
1!
NOTES
(iii) For exactly two accidents
(5)2 X e-5
p (2) = ---------------------------- = .084
2!
(iv) For exactly three accidents
(5)3 X e-5
p (3) = ------------------ = .14
3!
(v)
For exactly four accidents
54 X e-5
p (4) = ------------------ = .175
4!
Further, calculate the probability of three or fewer accidents.
This will be P(0), + P(1) + P (2) + P (3)
= 0.0067 + .033 + .084 + .14
= .2637
7.7.1 Mean and variance of a Paisson Probability Distribution
The mean of the poisson distribution is given by λ. This indicates that for a Poisson
distribution over a long period of time, a long - run average can be determined. The
variance of the Poisson distribution is also λ and the standard deviation is given by √λ
Practice Questions :
Que. 1. Given λ = 5, for a Poisson distribution, find
(a) P (x < 2)
(b) P (x > 5)
(c) P (x = 8)
Que. 2. Solve the following problems using paisson formula :
(a) P (x = 5 / λ = 7.0)
(b) P (x < 5 / λ = 3.5)
(c) P (2 < x < 4 / λ = 3.5)
7.8 Normal Distribution
Normal distribution is a continuous probability distribution. In recognetion of
contribution of Karl Gauss, 18th century mathematician, normal distribution is also known
as Gaussian distribution.
Normal distribution has a wide range of practical appliction, for example, where
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Probaility Disribution
NOTES
the random variables are human characteristics such as height, weight, marks scored in
a class etc.
Normal distribution has a wide range of application in management also. Statistical
quality control (SQC) is based on normal distribution.
Normal probability distribution is defined with the help of a bell shaped curve,
known as normal curve shown in fig. 7.2. As shown in figure 7.2, normal curve is
symmetrical around the centre line. It has left
Fig. 7.2 Normal Curve
hand and right hand tail which extend indefinitely but never touch the horizontal axis.
Characteristics of the Normal Probability Distribution :Seeing the normal curve in fig. 7.2, following important characteristics can be
inferred.
1.
Bell shaped normal curve has a single peak, thus it is unimodal.
2.
Mean of normally distributed population lies at the centre of normal curve.
3.
Mean, median and mode have same value in normal distribution and are at
centre line of normal curve.
4.
Two tails (left side and right side) extend indefinitely but never touch the
horizontal axis.
5.
Irrespective of value of mean µ and standard dewation σ for normal probability
distribution the total area under the normal curve remains 1.
Following figure 7.3 shows
three normal probability
distributions each of which has
the same mean but a different
statnard deviation.
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This figure 7.3 shows the
effect of standard deviation on
shape of normal curve. Large
values of standard deviation
results in wider, flatter curves,
exhibiting more variability in the
data.
σ =1
σ =5
σ =10
Fig. 7.3 Normal probability distribution with
identical means but different standard
deviations.
Areas under the normal curve
Probaility Disribution
Figure 7.4 shows normal curve and area under it for three different standard
deviation limits.
NOTES
µ =3 σ µ =2 σ µ =σ
µ
µ + σ µ =2 σ µ =2σ
Fig. 7.4 : Area under normal curve
This figure 7.4 shows that approximately 68% of all the values in a normally
distributed population lie within ±1 standard deviation from the mean. If we increase
limits to ±2 standard deviation, than 97.5% of all the values will come in this limit
99.73% of all the values lie within ±3 standard deviation from the mean.
7.8.1 Probability Density function of a Normal Distribution
Mean (µ) and standard deviation (σ) are two parameters used to define normal
distribution PDF of normal distribution is
1
f (x) = -------------- e
σ√2π
1
x-µ
- ------- ------2
σ
(
)
Here π = 3.14159, and e = 2.71828.
We use normal distribution tables rather this formula for analysing normal
distribution problems.
Using the standard Normal Probability Distribution Table :
Every unique pair of mean (µ) and standard diviation (σ) describes a different
normal distribution. To simplify the process of analysis, standard normal probability
distribution is evolved. Normal random variable is standardized using following formula:
x-µ
z = ----------σ
Where z is the number of standard deviations from x to the mean of this distribution.
Figure 7.5 gives use of 'z'. It shows that the use of z is just a change of the scale of
measurement on the horizontal axis.
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Probaility Disribution
NOTES
The Z score can
be defined as the
number of standard
deviation that a value,
x, is above or below
the mean of the
distribution
As
x-µ
shown in figure 7.5,
Z = ---------if value of x is less
σ
than mean, z score is
Fig. d Compatability of Z values with normally distributed
'-'ve and if value of x
variable `x’
is more than mean, z
score is '+'ve. In case x is equal to mean, z score is zero.
Example2 :- A project requires mean time to complete of 50 hrs and this is normally
distributed with a standard deviation of 5 hrs.
(9)
What is the probability that a contractor 'A' elected at random will require more
than 50 hours to complete the project?
Solution
In figure 7.6, we see that half of the area under the curve is located on either side
of the mean of 50 hrs. Thus we can conclude that the probability that the random
variable will take on a value higher than 50 is the colored half, or 0.5
(b) What is the probability that a contractor selected at random will take between
50 and 65 hrs to complete the project?
Solution
In figure 7.7, this
condition is shown with
colored portion. Now z value
for x = 65 will be
µ = 50hrs
σ = 5hrs
65 - 50
15
z = ------------ = ------ = 3
5
5
If we look up z=3 in
Appendix Table 1, we find a
probability of 0.4987. Thus
the chance that the contractor
selected at random would
require between 50 to 65 hrs
to complete the project is
slightly less than 0.5.
(c)
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What is the probability
that a contractor
selected at random will
take more than 60 hrs.
50
Fig. 7.6
µ = 50hrs
σ = 5hrs
z = 3.0
Areo= .4987
50
Fig. 7.7
65
to complete the program?
Probaility Disribution
Z=2.0
Solution : Figure 7.8
gives the solution Space of this
problem
Area = .4772 (from
appendix table 1)
NOTES
60-50
Here z = --------- = 2
5
Table 1 in appendix gives
µ = 50 60
probability of z = 2.0 as .4772.
Fig. 7.8
This is the probability of
completing the task in 50 to 60
hrs. However, we are interested to know the probability of more than 60 hrs which is
given by colored space in figure 7.8
The total area under curve in right side of mean is 0.5. So area under colored port
is 0.5-0.4772 = 0.0228. Therefore, there are just over 2 chances in 100 that a contractor
chosen at random would take more than 60 hrs. to complete the project.
(d)
What is the probability that a contractor selected at random will require less than
55 hours to complete the
Z=1.0
project?
Solution :-
Figure H gives the
solution space by colored
portion.
Area = .3413 (from
appendix table 1)
p (Less than 55)
= .8413
To get probability we first
determine z value for x = 55
µ = 50
55-50
z = --------- = 1.0
5
55
Fig. 7.9
Appendix table 1 gives corresponding probability as .341. In figure 7.9, it is
shown clearly that area under left side of mean will come completly giving total
probability as 0.5 + 0.3413 = 0.8413.
Thus, the chances of a
contractor requiring less than
55 hrs to complete the project
are about 84 percent.
(e)
Z=2.0, Area= .4772
Z=1.0, Area= .3413
p (40 to 55)
µ = 50hrs
σ = 5hrs
What is the probability
that a contractor chosen
at random will take
between 40 to 55 hrs. to
complete the project?
Solution :Figure
7.10
40
gives
50
55
Fig. 7.10
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Probaility Disribution
solution space from
x = 40 to x = 55. Corresponding z values are
NOTES
40 - 50
z1 = ----------------= 2
5
corresponding probability from Appendix table I = .4772
55 - 60
z2 = ----------------= 1
5
corresponding probability from Appendix table I = .3413
5
So required probability is .
4772
+ .3413
----------------.8185
----------------
Therefore, the chances of a contractor will complete the project between 40 to 55
hrs are about 81 percent.
Limitations of the Normal Probability Distribution
Tails of normal probability distribution never touch horizontal axis. This means
that there is some probability that the random variable can take an enormous values.
Though these probabilities can be very very small. Consider a case where a contractor
of earlier example may finish the project in 300 hrs. Now this will be 50 standard
deviation on right side of mean. Associated probability will be very small. Here we do
not loose much accuracy by ignoring values for out in the tails. But in exchange for the
convenience, we accept the fact that it can assign impossible empirical values.
Normal Distribution as an Approximation of the Binomial Distribution
Normal distribution is a continuous distribution but it can sometimes be used to
approximate binomial distributions which is a discrete distribution. In a binomial
distribution, the experiment involves a sequence of n identical trials and for each trial,
there can be two possible outcomes, success and failure. In a binomial distribution
trials are independent and probability of success (p) and probability of failure (q) remain
constant throughout the experiment.
In case of large number of trials in a binomial distribution, determining probability
is a cumbersome activity. Also table of dinomial distribution (see in Appendix table 3)
it does not have values of n more than 20. In these cases normal approximation of
binomial distribution is useful.
Following example illustrates the process of use of normal distribution in place
of binomial distribution.
Example 3 :- In a quality control department a random sample of 100 finished
items has been taken. What is the probability of obtaining 15 defective items?
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Solution :
Probaility Disribution
In this case, p = .15, and q = .85, n = 100
Step 1.
For converting two parameters of binomial distribution into normal
distribution, following formula can be used.
NOTES
µ = E (x) = np and σ = √npv
So µ = E (x) = np = 100 x .15 = 15
σ = √ npv = √ 100x.15x.85 = 3.57.
Step 2. For converting a discrete distribution to a continuous distribution, a
correction of + 0.50 or - 0.50 or ± 0.50, depending the problem is required.
This correction factor is called continuity correction factor. This continuity
correction factor is introduced because a continuous probability
distribution is used to approximate a discrete probability distribution.
So for P (x = 15), the discrete binomial distribution is approximated by P (14.5 <
x < 15.5) which is a continuous normal probability distribution.
Step 3. Calculate z values for x = 14.5 and x = 15.5
14.5 - 15
x-µ
z =--------- = ------------------ = - 0.14
σ
3.75
15.5 - 15
x-µ
z =--------- = ------------------ = + 0.14
σ
3.75
From Appendix table 1, probability corresponding to z = 0.14 or -0.14 is .0557
So required probability is
0.0557
+ 0.0557
------------0.1114
So, the normal approximation of the probability of 15 defective items in a sample
of 100 is 11%
Practice Question :
Q. 1. Determine the probability for the portion of the normal distribution described as
below :(a) P ( Z > 1.95)
(b) P (1.25 < z < 2.20)
(c) P (-0.50 < z < 1.12)
Q. 2 Use the normal approximation to compute the binomial probabilities for following:(a) n = 15, p = .40
at most 5 successes.
(b) n = 15, p = .50,
between 15 and 20 successes.
(c) n = 20, p = .30, 20 or more successes.
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Probaility Disribution
NOTES
Q. 3 On the basis of past experience, QC supervisors in a plant have noticed that 5
percent of all cars coming in for their annual inspection fail to pass.
Using the normal approximation to the binomial, find the probability that
between 6 and 15 of the next 200 cars to enter the inspection station will fail the
inspection.
Q. 4 Students in a class score average 55 marks. Suppose that marks scored is normally
distributed with a standard deviation of 4. If a student is randomly selected, what
is the probability that the marks scored by him are more than 60? What is the
probability that he scored less than 40?
7.9 Summary
7.10 Key Terms
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7.11 Questions and Exercises
Probaility Disribution
NOTES
7.12 Further Reading and References
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Regression and Correlation
Alalysis
UNIT 8
REGRESSION AND
CORRELATION ANALYSIS
NOTES
Structure
8.0 Introduction
8.1 Unit Objectives
8.2 Regression Analysis
8.2.1 Estimation Using the Regression Line
8.3 The Standard Error of Estimate
8.4 Coefficient of Determination
8.5 Correlation Analysis
8.5.1 Coefficient of Determination
8.5.2 Coefficient of Correlation
8.6 Summary
8.7 Key Terms
8.8 Question & Exercises
8.9 Further Reading and References
8.0 Introduction
In day - to - day life we come across a large number of problems involving use of
two variables. We can have series of marks of individuals in two subjects in an
examination, series of ages of husbands and wives in a sample of selected married
couples, series of exports and imports of a specific product during the number of years
from 2000 to 2012, the series of sales revenue and advertising expenditure of different
firms in a particular year and so on.
In a bivariate distritution, we may be interested to examine whether a relationship
exists between the two variables under study.
If however, there exists a relationship between two variables under consideration,
we may be interested to know the strength of that relationship or dependence. Further, we
may also be eager to estimate the value of one variable from the known value of the other.
Regression and correlation analysis deal with this type of problem.
8.1 Unit Objectives
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This unit tells about use of scatter diagrams to visualize the relationship between
tow variables, use of regression analysis to estimate the relationship between two
variables, use of least square estimating equation to predict future values of the dependent
variable, use of correlation analysis in describing the degree to which tow variables are
linearly related to each other, and use of coefficient of determination as a measure of
the strength of the relationship between two variables.
Regression and Correlation
Alalysis
8.2 Regression Analysis
Meaning :- Regression analysis refers to the methods by which estimates are
made of the values of a dependent variable from the values of an independent variable.
It is a technique of predicting the unknown values on the basis of the "average
relationship".
NOTES
Regression analysis is done with the help of a regression line. The regression
line describes the average relationship existing between X and Y variables, more
precisely, it is a line which displays mean values of Y for given values of X.
Regression analysis may be of following types :1) Linear and Curvilinear regression :- If the given bivariate data are plotted
on a graph, the points so obtained on the scatter diagram will more or less concentrate
round a curve, called the 'curve of regression.'
If the regression curve obtained by plotting the values of two variables is a straight
line, it is called a linear regression. The relationship between X and Y can also take the
form of a curve. This relationship is called curvilinear.
The direction of straight line or curve can indicate whether the relationship is
direct or inverse.
Figure (8.1) and (8.2) show direct and inverse linear relationships.
Fig (8.1) Direct linear
Fig (8.2) Inverse linear
Figure (8.3) and (8.4) show direct and inverse curvilinear relationships
Fig (8.3) Direct Curvilinear
Fig (8.4) Inverse Curvilinear
Figure (8.5) shows no relationship between X and
Y; therefore knowledge of the post concerning one variable
does not allow us to predict future occurrences of the other.
(2) Simple and Multiple Regression :- In case of
simple regression, we take only one independent variable
to determine the value of dependent variable. In case of
multiple regression, we use more than one independent
variable to determine value of dependent variable.
Fig. (8.5) No
relationship
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NOTES
8.2.1 Estimation using the Regression Line
Regression lines can be determined by graphic or algebraic method.
In graphic method, the points are plotted on a graph paper representing various
pairs of values of the concerned variables. These points give a picture of a scatter
diagram with several points scattered around. A regression line may be drawn in between
those points either by free hand or by a scale rule in such a way that the squares of the
vertical or the horizontal distances (as the case may be) between the points and the line
of regression so drawn is the least.
In algebraic method a regression equation is developed algebraically.
Equation for a straight line where the
dependent variable Y is determined by the
independent variable X is
Y
Y=a+bx
Y = a + b x.................... (1)
Where a = Y - intercept
}a
b = slope of the line
X
Fig 8.6 shows a regression line represented
by equation (1)
Fig (8.6)
Now the task is to find values of a and b. To understand the generic process
consider a straight line shown in figure 8.7
Y
Just by seeing a = 2, This is the point
where line cuts Y axis.
To find b, which is slope we use following
formula
Y2 - Y1
b =-------------X2 - X1
4
3
2
.................... (2)
5-3 2
b = ------- = ------- slope of the line
5-2 3
2
So line becomes Y = 2 + ---- x ........... (3)
3
1
}a=2
(x2, y2)=(5,
5)
(x1, y1)=(1,
3)
X
1
2
3 4 5
Fig (8.7)
Using line represented by equation (3), we can find value of dependent variable
Y when X = 6.
Y = 2 + 2/3 (6) = 6
Similarly we can find value of Y corresponding to any given value of X.
Points which are estimated using a value of X, will be denoted by Y^ So our line of
estimation will be
^
Y
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= a + b x .................... (4)
Now it is important to understand that line of regression must minimize the sum
of the squares of the errors. These errors are difference between actual (Y) and estimated
values (Y^).
Slope of best - fitting regression line (line of loast square fit) is determined using
ΣxY-nxY
b = ------------------------ .................... (5)
Σ x2 - n x-2
Regression and Correlation
Alalysis
NOTES
and Y - intercept is determined using
a = Y - bx .................... (6)
Where
a - Y-intercept
b = Slope from formula (5)
x = values of the independent variable
Y = values of the dependent variable
X = mean of the values of the independent variable
Y = mean of the values of the dependent variable
n = number of data points.
Example 1 : Following table presents maintenance expenses on S. T. buses for
different aged buses :
Table 8.1
Bus No.
Age of Bus (x)
Repair Expenses
(in Years)
(Rs.) (Y)
MH01-8890
5
70000
MH02-7677
3
70000
MH04-5555
3
60000
MH03-6666
1
40000
Solution : Table 8.1 presents raw data of the problem. In table 8.2 we arrange it
in the required format to determine values of b and a.
Table 8.2
Age (x)
Repair expenses (Y)
XY
X2
(1)
(2)
5
70000
350000
25
3
70000
210000
9
3
60000
180000
9
1
40000
40000
1
ΣX = 12
ΣY= 240000
ΣXY= 780000
ΣX2 = 44
ΣX
X
= --------- =
n
12
--------- =3
4
ΣY
-
Y
240000
= --------- = -------------- = 60000
n
4
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ΣXY-nxy
Regression and Correlation
Alalysis
b
=---------------ΣX2 - nx2
780000 - (4) (3) (60000)
=---------------------------------------44 - (4) (3)2
=
NOTES
7500
and a will be
a = Y - bx
= 60000 - (7500) (3)
= 37500
So regression line is
^
Y
= 37500 + 7500 X
Now using this line equation we can setimate repair expenses for a four year old
bus.
Y = 37500 + 7500 (4)
= 67500
8.3 The Standard Error of Estimate
Measuring reliability of the estimating equation is an important part of regression
analysis.
Y
Y
X
Fig. 8.8
X
Fig 8.9
Figure 8.8 shows that regression line is a more accurate estimator of the relationship
between X and Y than the regression line of figure 8.9 where the points are farther
away from the line.
The standard error of the estimate is a measure of reliability of the regression
line. The standard error of the estimate (Se) measures the variability, or scatter, of the
observed values around the regression line.
Se =
√
Σ (Y-Y)2
---------------------- ...................... ( 7)
n-2
Here we need to compute Y^ for every given value of Y with the help of regression
equation by substituting different values of X.
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To save efforts we can use formula (8) also.
√
Σ Y2-aΣY-bΣXY
------------------------------ ........................ ( 8)
n-2
Se =
NOTES
The larger values of Se represents the greater scattering of points around the
regression line. If Se = 0, we expect the estimating equation to be a perfect estimater of
the dependent variable.
Example 2 : Obtain the regression line (Y = a + bx) from the following data :
x
:
8
6
4
7
5
y
:
9
8
5
6
2
Solution :
x
y
x2
y2
xy
8
9
64
81
72
6
8
36
64
48
4
5
16
25
20
7
6
49
36
42
5
2
25
4
10
ΣX = 30
ΣY= 30
ΣX2 = 190
ΣY2 = 210
ΣXY= 192
ΣX
X = --------- =
n
30
--------5
=6
ΣY
30
Y- = ---------- = --------n
5
=6
b =
ΣXY-nxy
192 - (5) (6) (6)
-------------------- =---------------------------------------- = 1.2
ΣX2 - nx2
190 - (5) (6)2
a= Y - bx = 6 - (1.2) (6) = 1.2
So regression line is
Y = -1.2 + 1.2 x
Example 3. From the data given in example 2, find the standard error of the
estimate of regression equation.
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Alalysis
Solution - Following table presents various steps involved in calculation of
standard error of the estimate.
NOTES
Y
Y
Y-Y
(Y-Y)2
9
8.4
0.6
.36
8
6.0
2.0
4.00
5
3.6
1.4
1.96
6
7.2
-1.2
+1.44
2
4.8
-2.8
+7.84
Σ (Y - Y)2 = 15.60
Se =
√
Σ (Y-Y)2
---------------------- =
n-2
√
15.60
---------------------5-2
or using formula (8)
Se =
√
Se =
√
ΣY2 - aΣY - bΣXY
---------------------------------------------n-2
210 - (-1.2) (30) - (1.2) (192)
----------------------------------------------------------------
= 2.28
5-2
8.4 Coefficient of Determination
Coefficient of determination is a very commonly used measure of fit for regression
models and is denoted by r2. If measures the proportion of variation in y that can be
attributed to the independent variable X. The values of r2 range from 0 to 1.
Σ (Y-Y)2
Y2 = -------------------- ............................... ( 9)
Σ (Y-Y)2
Where Y = Given (Actual) value of dependent variable.
Y = Average value of dependent variable
Y = Estimated value of dependent variable
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If we take example 2 again, and calculate r2, various steps will be as follows :Y
Y
Y
Y-Y
(Y-Y)2 Y-Y
(Y-Y)2
P-1
(Y-Y)2
9
6
8.4
3
9
0.6
.36
2.4
5.76
8
6
6.0
2
4
2.0
4.00
0
----
5
6
3.6
-1
+1
1.4
1.96
-2.4
+5.76
6
6
7.2
0
0
-1.2
+1.44
1.2
+1.44
2
6
4.8
-4
+16
-2.8
+7.84
-1.2
----
Σ (Y - Y)2 = 15.60
Regression and Correlation
Alalysis
NOTES
Σ (Y - Y)2 = 30
Σ(Y-Y)2
so Y2 = ------------------Σ(Y-Y)2
14.40
= ------------------- = 0.48
30
This indicates that only 48% of the variation in Y can be explained by the
independent variable X. It also explains that 52% of the variations in Y are explained
by factors other than X.
8.5 Correlation Analysis
Correlation analysis is used to describe the agree to which one variable is linearly
related to another.
There are two measures for describing the correlation between two variables :
(i)
Coefficient of determination
(ii) Coefficient of correlation
8.5.1 Coefficient of Determination
This is a primary measure to determine the strength of association between two
variables, X and Y. Here Y is taken as dependent variable. Now variations of the Y
values in a data set can be around
(a) the fitted regression line
(b) their own mean
Consider a regression line giving relation between Y and X as Y = a + bx
So variations of the Y values around the regression line = Σ (Y - Y)2 ............ (10)
The second variation, that of the Y values around their own mean is
Σ (Y - Y)2 ............ (11)
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Σ(Y-Y)2
Co efficient of determination,
2
r = 1 = ------------------Σ(Y-Y)2
............ (12)
Compare formula given as (9) and (12) Both these indicate same r2.
NOTES
If r2 = +1, regression line is a perfect estimater.
If r2 = 0, there is no correlation between these variables.
Alternatively, r2 can also be determined using
aΣY + bΣXY - nY2
r2 = ------------------------------- ..........................(13)
ΣY2 - nY2
Meaning of all these notations are already given earlier.
8.5.2 Coefficient of correlation
Sample co efficient of correlation is denoted by r and is square root of the sample
co efficient of determination.
Sample co efficient of correlation, r =√ r2
The sign of r indicates the direction of the relationship between the two variables
X and Y.
Take a case of r2 = .81
if r = +.9, figure 8.10 gives the positive relationship between variables, while r =
-.9, figure 8.11 shows the negative relationship between variables.
slope is `+’ve
slope is `-’ve
r2 = .81
r = +.9
Fig. 8.10
r2 = .81
r = -.9
Fig. 8.11
Apart from these two coefficients, Karl Pearson's co efficient of correlation is
another important measure of correlation. It is given as
K.P. Coefficient of correlation
nΣXY - (ΣX) (ΣY)
= --------------------------------------------------------------- - 14
√ n (ΣX2) - (ΣX)2 √ n(ΣY2) - (ΣY)2
Example 4 : Following table shows sales revenue and publicity expenses of a
company for the past 5 months. Find the coefficient Karl Pearson's coefficient of
correlation between sales revenue and publicity expenses.
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Month
Jan
Mar
May
July
Sept
Publicity expenses
10
12
11
9
11
110
115
137
150
120
Regression and Correlation
Alalysis
(in thousand rupees)
Sales (in thousand rupees)
NOTES
Solution :- Calculation for correlation co efficient is shown in the following table:Month
Sales (X)
Pub. Expenses (Y) XY
X2
Y2
Jan
110
10
1100
12100
100
Mar
115
12
1380
13225
144
May
137
11
1507
18769
121
July
150
9
1350
22500
81
11
1320
14400
121
September 120
Here
ΣX = 632
ΣY = 53
ΣXY
= 6657
ΣX2 = 80994
ΣY2 = 567
Co efficient of correlation according to formula (14)
nΣXY - (ΣX) (ΣY)
= ---------------------------------------------------------------√ n(ΣX2) - (ΣX)2 √ n(ΣY2)- (ΣY)2
5 (6657) - (632) (53)
=--------------------------------------------------------------- = .55
√ 5 (80994) - (632)2 √ 5(567) - (53) 2
Hence, K.P. Correlation coefficient between sales revenue and pubicity expenses
is - 0.55. This indicates that both these variables are negatively correlated to the extent
of - 0.55. We can conclude that an increase in the expenditure on publicity will not
result in an increase in sales.
8.6 Summary
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NOTES
8.7 Key Terms
8.8 Questions and Exercises
(1) What is the conceptual framework of simple linear regression and how can we
use it for business decision making?
(2) Explain the concept of coefficient of determination and standard error of the
estimate in a regression model.
(3) How can we use correlation coefficient for determining the statistical significance
of the relationship between two variables in a regression model?
Practice Questions :
(4) A company is looking for a regression model to predict the impact of advertisement
on sales. Data of 12 randomly selected months are given below. Develop estimation
equation for the company.
Months
1
2
3
4
Advertisement Exp. 92 94
97
98 100
Sales
1020 990 1100 1050 1150 1120 1130 1200 1250 1220
930 900
5
6
7
102 104
8
9
105 105
10
11
12
107
107
110
Expenses and sales are in thousand rupees.
(5)
For the following set of data :
(a) plot the scatter diagram
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(b) Develop the estimating equation that best describes the data
(c) Predict Y for X = 10, 15, 20.
(6)
X
13
16
14
11
17
9
13
17
18
12
Y
6.2
8.6
7.2
4.5
9.0
3.5
6.5
9.3
9.5
5.7
Calculate sample co efficient of determination and the sample co efficient of
correlation for question number 5.
Regression and Correlation
Alalysis
NOTES
8.9 Further Reading and References
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Testing of Hypothesis
NOTES
UNIT 9
TESTING OF HYPOTHESIS
Structure
9.0 Introduction
9.1 Unit Objectives
9.2 Introduction to Hypothesis Testing
9.3 Concept Behind Hypothesis Testing
9.4 Process of Hypothesis Testing
9.5 Type I and Type II Error
9.6 Two-Tailed and One- Tailed Tests of Hypothesis
9.7 Deciding of Distribution for Hypothesis Testing
9.8 Hypothesis Testing of Proportions : Large Sample
9.9 Hypothesis Testing of Means when the Popolation Standard Deviation is not
9.10 Meaning of Chi-Square Test
9.11 The Chi-Square Distribution
9.12 Useful Points while using the Chi-Square Test
9.13 Summary
9.14 Key Terms
9.15 Question & Exercises
9.16 Further Reading and References
9.0 Introduction
Hypothesis testing is a very important tool for business analysis purpose to arrive
at a meaningful conclusion. Hypothesis testing is the soul of inferential statistics.
Hypothesis testing begins with an assumption, known as hypothesis. Hypothesis testing
is accepting or rejecting a hypothesis about a population parameter. This can not be
done simply by intuition. Instead, we need to learn how to decide objectively, on the
basis of sample information, whether to accept or reject a hypothesis. This chapter is to
discuss how hypothesis testing can be conducted around population parameters.
9.1 Unit Objectives
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The objectives of this unit are to introduce hypothesis testing, concept of hypotehsis
testing, to learn how to use samples to decide whether a population possesses a particular
characteristic, to understand the tow types of errors possible when testing hypotheses
to learn when to use one tailed tests and when to use two tailed tests, to learn process of
hypothesis testing and to understand testing and to understand how and when to use the
normal and t-distributions for testing hypotheses about populatin means and proportions.
Finally, unit also describes chi-square test for hypothesis testing and as a test fo goodness
of fit.
9.2 Introduction to hypothesis testing
In different fields, may be business, social science, pshychology etc. decision
makers need might answers to certain questions in order to take optimum decisions.
Suppose a manager of a large shopping mall wants to know job satisfaction level of
employees. He will make decision on the basis of the available information and in most
cases, information is obtained through sampling. To answer this question, manager
needs to collect sample data, compute the sample statistic and use this information to
determine the correctness of the hypothesized population parameter. Decision maker
develops a "hypothesis" which can be studied and explored. Suppose you are a Cricket
Coach. You are organizaing a coaching camp for 500 players. Now organizers of the
camp want to determine improvement in the players on the basis of coaching camp. To
save efforts, we will contact only 50 players with an effectiveness measurement
questionnaire.
Testing of Hypothesis
NOTES
The result that is obtained would not be the result from the entire population but
only from the sample. We will then set an assumption that 'coaching has not enhanced
efficiency' and will accept or reject this assumption through a well- defined statistical
procedure known as hypothesis testing.
9.3 Concept Behind Hypothesis Testing
The sample taken from population is considered to be true representative of the
population. Therefore, the known sample statistic is used for estimating the unknown
population parameter. In setting a hypothesis, we assume that the sample statistic will
be close to the hypothesized population parameter. This is possible in cases where the
hypothesized population parameter is correct and the sample statistic is a good estimate
of the population parameter.
In statistical analysis, we use the concept of probability to specify a probability
level at which we conclude that the observed difference between the sample statistic
and the population parameter is not due to chance.
9.4 Process of Hypathesis Testing
Hypothesis testing is a well - defined procedure. A sample is selected for estimating
the population parameters.
The first step in this procedure is stating the assumed or hypothesized value of the
population parameter before we start sampling. The assumption we want to test is
called the null hypothesis (Ho)
Theoretically, a null hypothesis is set as no difference and considered true, until
and unless it is proved wrong by the collected sample data. The null hypothesis is
always expressed in the form of an equation, which makes a claim regarding the specific
value of the population. In symbol a null hypothesis is represented as
H0 : µ = µ0
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Testing of Hypothesis
NOTES
Where µ is the population mean and µ0 is the hypothesized value of the population
mean. Suppose we want to test whether a population mean is equal to 50, a null
hypothesis can be set as 'population mean is equal to 50'. Symbolically, H0 : µ = 50
The term null hypothesis arises from statistical applications in the field of
agriculture and medicines. In order to test the effectiveness of a new fertilizer or drug,
the tested hypothesis (the null hypothesis) was that it had no effect, that is, there was no
difference between treated and untreated samples.
If our sample results fail to support the null hypothesis, we must conclude that
something else is true. Whenever we reject the null hypothesis, the conclusion we do
accept is called the alternative hypothesis (H1).
for the null hypothesis Ho : µ = 50
We will consider three possible alternative hypothesis.
H1
:
µ ≠ 50 (The alternative hypothesis is that the population mean is not
equal to 50)
H1
:
µ > 50 (the alternative hypothesis is that the population mean is greater
than 50.)
H1
:
µ < 50 (The alternative hypothesis is that the population mean is less
than 50.)
After making the hypothesis, we decide on an appropriate statistical test that will
be used for statistical analysis. Data type, level and number may provide a platform for
decidng the statistical test.
After deciding the test, we set the level of significance. The level of significance
(α) is probability of rejecting a null hypotheses even it is true. It is important to note
that the level of significance must be determined before we draw samples, so that the
obtained results are free from the choice bias of a decision maker. The levels of
significance which are generally applied are 1 percent, 5 percent or 10 percent.
If we test a hypothesis at the 5 percent level of significance, it means that we will
reject the null hypothasis if the difference between the sample statistic and the
hypothesized population parameter is so large that it or a larger difference would occur,
on the average, only five or fewer times in every 100 samples when the hypothesized
population parameter is correct. If we assume the hypothesis is correct, then the
significance level will indicate the percentage of sample means that is outside certain
limits.
Figure 9.1 shows graphical interpretation of 5 percent level of significance.
Region where there is no
significant difference between
the sample statistic and the
hypothesized
population
parameter
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0.025 of area
µH0 - 1.965
µH0 + 1.965
µH0
Fig. A
0.025 of area
Figure 9.1 shows that 2.5 percent of the area under the curve is located in each
tail. Appendix table 1 can be used to determine extend of area on both side of mean
value. It tells that interval extendes 1.96σ on either side of the hypothesized mean in
case of 95 percent of all area is under consideration.
From appendix table I, Check value for z = 1.90 under .06. This is 0.4750. Taking
both sides, it is .475 + .475 = .950 Therefore for 95% area, we will take 1.96σx on
either side of the mean value.
Testing of Hypothesis
NOTES
Here in 95% of the area, there is no significant difference between the observed
value of the sample statistic and the hypothesized value of the population parameter. In
the remaining 5 percent (colored regions on both tails), a significant difference does
exist.
We would accept the null hypothesis if the sample statistic falls in 95% of the
region. If sample statistic falls in two colored tails, we would reject null hypothesis.
In selecting a significance level, we need to understand that the higher the
significance level we use for testing a hypothesis, the higher the probability of rejecting
a null hypothesis when it is true.
9.5 Type I and Type II Errors
While testing hypothesis, null hypothesis should be accepted when it is true and
it should be rejected when it is false.
Rejecting a null hypothesis when it is true is called a Type I error and its prabability
is significance level of the test (α) Accepting a null hypothesis when it is false is called
a Type II error and its probability is called β (Beta). In fact following four outcomes are
possible in case of statistical testing of hypothesis.
1.
Rejecting a true null hypothesis (Type I error)
2.
Accepting a false null hypothesis (Type II error)
3.
Accepting a true null hypothesis (Correct Decision)
4.
Rejecting a false null hypothesis (Correct Decision)
There is a trade- off between type I and type II errors. We cannot commit Type I
and Type II errors at the same time on the same hypothesis test. Generally α and β are
inversely related to each other. To deal with this trade- off, we decide the appropriate
level of significance by examining the costs or penalties attached to both types of
errors.
Following two examples explain preference of Type I or Type II error.
Case I :- Type I error involves the efforts of reworking some products that should
have been accepted. At the same time, making a Type II error means taking a chance
that an entire group of users of this product will be severly affected. In this case we will
prefer a Type I error to Type II error and, will set very high levels of significance in its
testing to get low βs
Case II :- In a case Type I error involves major disassembling an entire generator
Quantitative Techniques in
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Testing of Hypothesis
NOTES
at the factory, but making Type II error involves relatively inexpensive warranty services
by the dealers. Here the manufacturer is more likely to prefer a Type II error and will
get lower significance levels in its testing.
9.6 Two - Tailed and one- Tailed Tests of Hypothesis
There are two types of tests of hypothesis. These are two- tailed tests and onetailed tests of hypothesis. Hypothesis formulation provides a base for test selection.
(a) Two- tailed test of hypothesis : A two - tailed test of a hypothesis will reject
the null hypathesis if the sample mean is significantly higher than or lower than the
hypothesized population mean. Thus a two - tailed test contain the rejection region on
both the tails of the sample distribution of a test statistic. If the level of significance is
α then the rejection region will be on both the tails of the normal curve, consisting of
α/2 area on both the tails of the normal curve. As shown in figure 9.1, we have
colored tails on both sides.
A two tailed test is appropriate, when the null hypothesis is µ=µ0 and the alternative
hypothesis is µ ≠ µ0
(b) One tailed test of hypothesis :
Unlike the two tailed test, the one tailed test contains the rejection region on one
tail of the sampling distribution of a test statistic.
Consider following null and alternative hypathesis.
H0 : µ = µ0 and
H1 : µ < µ0
Left tailed test
{
}
As shown in here we reject the null hypothesis if the computed sample statistic
is significantly lower than the hypothesized population parameter.
Acceptance Region
Acceptance Region
Rejection region
µ = µ0
Fig. 9.3
Fig 9.3 Acceptance and rejection
regions for one tailed test (right)
µ = µ0
Fig. 9.2
Fig 9.2 Acceptance and rejection
regions for one failed test (left)
Now consider the following set of null and alternative hypothesis.
H0 : µ = µ0 and
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H1 : µ < µ0
Right tailed test
{
}
As shown in fig 9.3, here we reject the null hypothesis if the computed sample
statistic is significantly higher than the hypothesized population parameter.
In both these cases (either left or right tail rejection) the entire rejection area
corresponding to the level of significance (α) is located only in one tail of the sampling
distribution of the statistic.
Testing of Hypothesis
NOTES
Following table gives some important values at various significance levels for
test statistic Z.
Significance level Confidence Level
One tailed Region
Two tailed region
(α)
(1−α)%
1%
99%
± 2.33
± 2.575
5%
95%
± 1.645
± 1.96
10%
90%
± 1.28
± 1.645
9.7 Deciding of Distribution for Hypothesis Testing
We can choose between normal distribution and t distribution for making tests of
mean.
If sample size is more than 30 and population standard deviation is known or
unknown normal distribution (Z) is used. In case sample size is less than 30 and we
assume the population is normal or approximately so, we use normal distribution when
the population standard deviation is known but in case of unknown population standard
deviation we use t distribution.
Hypothesis Testing of Means when the Population Standard Deviation is known
(a) Two-Tailed Tests of Means :- Take a case of supplying LPG gas cylinders to
Indian Oil. IOC needs pressure bearing capacity of 100 N/m2, but an excessively strong
cylin.
Cylinder is also dangerous. We take standard deviation of 1 N/m2. Taking a sample
of 50 cylinders from production, tests them, and finds that the mean stress capacity of
the sample is 98.5 N/ m2. Written in symbols.
µ0 = 100
Population mean and standard deviation
σ=1
{
n = 50
sample size
x = 98.5
sample mean
}
If we use a significance level (α) of 5% in testing, will the cylinders fulfill the
pressure requirements? It can be stated as :
H0 : µ = 100 (Null Hypothesis)
H1 : µ ± 100 (Alternative Hypothesis)
α = .05 (Level of significance)
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Testing of Hypothesis
In this illustration, please note
* Size of population is large enough
* Population standard deviation is known.
NOTES
Therefore we will use normal distribution. With following formula, calculate
standard error of the mean
σx
σ
= --------n
(A)
1
= --------√ 50
= 0.141 N/m2 (Std error of the mean)
With our knowledge of normal distribution, we can see in figure 9.4 that 95%
acceptance region is
µ0+1.96 σx
µ0-1.96 σx
distributed in two equal
parts of .475 on each
side of mean value.
Now determine
limits of acceptance
region : Uper Limit =
µ0 + 1.9σx
.025 of area
.475 of area
.025 of area
µ0 = 100
= 100 + 1.96 (1)
= 101.96 N/m 2
and Lower Limit = µ0 1.96 σx
.475 of area
Fig. 9.4
Two tailed hypothesis test of the 5% signficance level
= 100 - 1.96 (1)
= 98.04 N/m2
We can see that sample mean of 98.5 lies between 98.04 to 101.96. Therefore we
can accept the null hypothesis because there is no significant difference between the
hypothesized mean of 100 and the observed mean of 98.5. On this basis, we accept the
production run as meeting the presure requirements.
Alternatively we can convert original values on Z values to use standardized
scale. We will use formula
x - µ0
Z =-----------σx
98.5 - 100
= -----------------1
=
-1.5
This observed value falls within ± 1.96, lower and upper limits of the acceptance
region on Z scale. Once again we conclude to accept null hypothesis.
Quantitative Techniques in
Management : 142
Either we use original values or Z values (Standardized scale), the two methods
will always lead to the same conclusion.
(b) One-tailed tests of Means :
Testing of Hypothesis
Take following illustration :
Ram is a doctor at a government hospital in Nasik. He is using large quantities of
a particular packaged drug. The individual dose of this drug is 50 mg. If excessive
intake of this drug is given to a patient, extra amount is harmlessly passed off from the
human body, but insufficient dose do not give the desired medical effect Standard
deviation in the quantity of a drug in a packed dose is 1 mg. Ram is procuring this drug
from a supplier for a number of years. He inspects 50 doses of this drug at random from
a large supply and finds the mean of these doses to be 49.80 mg.
NOTES
Symbolically
µ0
=
50
(Population mean)
σ
=
1
(Population standard deviation)
n
=
50
(sample size)
x
=
49.80 (sample mean)
If Ram sets a 5% significance level, what is your conclusion about these doses?
Here H0 : µ = 50
H1
: µ < 50
α
= 0.05
(The mean is less than 50 mg)
As population standard deviation is known and n is larger than 30, we will use
normal distribution. Again with the
help of appendix table I we
determine Z value, corresponding to
.45 of area. This is clear in figure 9.5
explaining total portion of rejection
on the left tail of the curve. Students
can see that acceptance region is .45
on left side of mean and complete
area of right side of mean giving 95%
0
z
area of normal curve except 5% area
Fig. 9.6
on left tail.
Now we calculate standard error of mean using formula (A).
σ
1
σx = - ------------- = ----------------- = .141mg
√
n
√
50
To Standardize the observed values, we use Z formula of normal distribution.
Z
x-µ0 49.80-50
=---------- = -------------- = .141mg
σx
.141
= -1.418
Quantitative Techniques in
Management : 143
Testing of Hypothesis
NOTES
Figure 9.6 shows placing of Z, i. e. 1.418 on standardized scale. This value lies
in acceptance region.
Therefore Ram accepts the null
hypothesis because the observed mean of
the sample is not significantly lower than
our hypothesized mean of 50 mg.
Fig. 9.6
9.9 Hypothesis Testing of Proportions : Large Samples
(a)
Two Tailed tests :
Consider following illustration :
In a stock of a product, manager of the warehouse tells that roughly 80 percent of
the products are usable. Owner of the company makes a committee to assess the usability
of all products. This committee checks 100 products and finds that only 70% of the
sample are usable.
In symbols
P0 = 0.8 (Hypothesized value of the population proportion of success, usable
products in the present case.)
q0 - 0.2 (Hypothesized value of the population proportion of failures, non- usable
prod in the present case)
n = 100 (Sample size)
P = .7 (Sample proportion of usables)
V = .3 (Sample proportion of non- usables)
We are interested to test at 1% significance level the hypothesis that 0.8 of the
products are usable.
Developing null and alternative hypothesis :Ho : P = 0.8
H1 : P ± 0.8
α = .01
Quantitative Techniques in
Management : 144
In order to determine
whether
the
true
proportion is larger or
smaller
than
the
hypothesized proportion, a
two - tailed test of a
proportion is appropriate.
-2.57
0
+2.57 z
Fig. 9.7
Two tailed hypothesis test of a proportion at the
0.01 level of significance.
As shown in figure G, two colored region on each tail correspond to significance
Testing of Hypothesis
level.
Here np - 70
and nq = 30.
NOTES
Both are larger than 5, we can use the normal approximation of the binomial
distribution. From appendix table 1, we can check the critical value of Z for .495 of the
area under the curve is 2.57.
Calculate the standard error of the proportion, using the hypothesized values of
Po and qo in formula (B).
Po Vo
-------------σπ =
(B)
n
√
=
.8 x .2
-------------- = .04 (Std. error of the proportion)
√
100
Now standardize the sample proportion using formula (C).
Z
.7 - .8
P - Po
= --------- = --------σp
.04
= -2.5
If we locate this 9.7 value on figure G, we can see that this sample is in range of
acceptance. Therefore, we accept the null hypothesis. We infer that the true proportion
of usable products in the warehouse is 80 percent.
(b) One- tailed test of proportions
A one- tailed test of proportion is very much similar to one- tailed test of a mean.
Take a case, "A new faculty member in the college says that less than 50 percent
of the students of the college are complying with discipline standards of the college.
Principal of the college believes that 50 percent of the students are following discipline
standards. Manager of the college decides to test that hypothesis at the 2% signifitcance
level.
In symbols Ho : P = .5
H1 : P < .5
λ = .02
A Committee closely observed 50 students from a population of over 2000 students
and finds that 28 are following discipline codes. Is the assertion by new faculty member
a valid one?
Again in symbols :
qo - 0.5 (Hypothesized value of the population proportion that is fillowing discipline
code.)
Vo - 0.5 (Hypothesized value of the population proportion that is not following
discipline code.)
Quantitative Techniques in
Management : 145
Testing of Hypothesis
n = 50
p = 28/50 or 0.56
q = 0.44
NOTES
This is a one- tailed test
Specifically this is a left tailed test.
Figure9.8 shows
hypothesis graphically.
this
Here, np = 50 X .56 = 28
nq = 50 X .44 = 22
-2.05
Fig. 9.8
One tail test at the 0.02 level of significance
both are larger than 5, we can
use normal approximation of binomial distribution
(Students should check critical value of Z for .48 area.)
Now determine standard error of the proportion using formula (B).
σp =
(.56) (.44)
------------= .070
50
√
Now we standardize the sample proportion using formula (C)
0.56 - 0.5
Z = ---------------------------- = 0.8571
.070
We can see that this Z value + 0.8571 lies in acceptance region of figure 9.8.
Therefore we accept the null hypothesis.
This means that 50% students follow discipline standards is correct.
9.9 Hypothesis Testing of Means when the Population
Standard Deviation is not Known
(a)
Two- Tailed Tests of Means using the t - distribution :
If the sample size n is 30 or less and population standard deviation σ is not known,
we should use t distribution. The appropriate t distribution has n-1 degress of freedom.
Take example of a class, where teacher thinks the average marks around 75. When
a sample of 10 students was taken, it was found that the mean score is 68 and the
standard deviation of this score was 7.
In symbols,
Quantitative Techniques in
Management : 146
µ0 = 75
(Hypothesized value of the population mean.)
n = 10
(Sample size)
x = 68
(Sample mean)
s=7
(Sample standard deviation)
If we want to test the hypothesis at the 5% level of significance, we will make
hypothesis ((null and alternative first.
Testing of Hypothesis
Ho : µ = 75
H1 : µ ≠ 75
NOTES
λ = .05
Since we are interested in knowing whether the true mean score is larger or smaller
than the hypothesized score, a two- tailed test is appropriate
Figure 9.9 shows two shaded region, each containing .025 of the area under the t
distribution. Because the
sample size is 10, the
degree of freedom is 9.
Checking Appendix table
2, under the 5% (.05)
column and corresponding
to 9 degress of freedom,
we get critical value of t,
2.262. These values are
t
also shown in figure 9.9
Fig. 9.9 :Two tailed test of hypothesis at the 5%
Here population
level of significance using t distribution.
standard deviation is not
known, we will estimate it
using formula (D) with the help of sample standard deviation.
σ^
=s
(D)
=7
With the help of population standard deviation, computed using formula (D) above,
we will further determine standard error of the mean using formula (E)
σ^
σ^ x = ---------- (E)
n
7
=------------= 2.21
√10
Further we will standardize the sample mean (x) using formula (F)
x - µ0
t = --------------------- (F)
σ^ x
68 - 75
t = --------------------2.21
= 3.16
Putting this calculated value of t, on t distribution curve of figure 9.9, we conclude
that this lies outside of acceptance range.
Therefore we reject the null hypothesis.
Quantitative Techniques in
Management : 147
Testing of Hypothesis
(b) One tailed test of Means using t distribution :
The process of a one tailed test using t distribution is similar to one tailed hypothesis
testing using the normal distribution.
NOTES
But this test causes some difficulty. As one can see Appendix table 2. The area
given in column heading is combined area under both tails. Thus t - distribution is
appropriate to use in a two tailed test with two rejection of rejection.
9.10 Review Questions :
1.
What do we mean we reject a hypothesis on the basis of a sample?
2.
Define Type I and Type II errors.
3.
Define the term significance level.
4.
What is the relationship between the significance level of a test and Type I
error?
5.
What do we mean by null and alternative hypotheses?
Practice Questions :
6.
7.
8.
If we want to accept a null hypothesis that ... = 30 with 95 percent certainty
when it is true, and our sample size is 100, graphically represent the acceptance
and rejection regions for the following alternative hypotheses :
(a)
µ ± 30
(b)
µ > 30
(c)
µ < 30
For the following cases, specify which probability distribution to use in a
hypothesis test :
(a)
H0 : µ = 10; H1 : µ ± 10, x = 9.8, σ^ = 2.0, n = 35
(b)
H0 : µ = 40, H1 : µ > 40, x = 42, σ^ = 4.0, n = 15
(c)
H0 : µ = 8, H1 : µ < 8, x = 7.8, σ^ 0.10, n= 20
As per records available with a company in Maharashtra average household
income of Maharashtra is Rs. 10,000.
You have recently joined this company. You doubt about the accuracy
of this data. You have taken a random sample of 200 households in Maharashtra
to verify the available data and found sample mean of Rs. 11,000. Assume
that the population standard deviation of the household income is Rs. 1200.
At α = .05, verify your doubt.
9.10 Meaning of Chi-Square Test
The Chi - Square test (χ2 test) is one of the simplest and most widely used non
parametric test in statistical work. Chi- square test enables us to test whether more than
two population proportions can be considered equal.
Quantitative Techniques in
Management : 148
Many a times, managers need to know whether the differences they observe among
several sample proportions are significant or only due to chance.
Testing of Hypothesis
Suppose the campaign manager of a company studies three different states and
finds that 30, 40 and 50 percent, respectively of the customers surveyed in the three
states recognize the company's name. If this difference is significant, the manager may
conclude that location will affect the way the company should act. But if the difference
is not significant (that is, if the manager concludes that the difference is solely due to
chance), then he may decide that the place chosen to make a particular policy- making
campaign will have no effect on its reception.
NOTES
Contingency table :
Suppose, that in four states, the CRPF samples its employee's attitude toward jobperformance reviews.
Two methods (M1 and M2) are under discussion Respondents were given a choice
to prefer one of these two methods. Table 9.1 illustrates the response to this question
from the sample polled. This table is called a contingency table.
Table 9.1
Chattisgarh
Bihar
Andhra
Pradesh
Orissa
Total
Number who prefer
method M1
68
75
57
79
279
Number who prefer
method M2
32
45
33
31
141
Total employees sampled
in each region
100
120
90
110
420
This table 9.1 is a 2 X 4 contingency table because it consists of two rows and four
columns.
Observed and Expected Frequencies :Now we symbolize the true proportions of the total population of personnels who
prefer the method M1 as
pC - proportion in Chattisgarh who prefer method M1
pB - proportion in Bihar who prefer method M1
pA - proportion in Andhra Pradesh who prefer method M1
pO - proportion in Orissa who prefer method M1
Using these symbols, we can state the null and alternative hypotheses as follows:
H0 : pC = pB = pA = p0 - Null Hypothesis
H1 : pC, pB, pA and p0 - are not all equal Alternative Hypothesis
If the null hypothesis is true, we can combine the data from the four samples and
then estimate the proportion of the total force (the total population) that prefers the method
M1 :
Quantitative Techniques in
Management : 149
Testing of Hypothesis
Combined proportion who prefer method M1 assuming the null hypothesis of no
difference is true
68 + 75 + 57 + 79
279
= ---------------------------- = ------420
420
NOTES
= 0.6643
It means .6643 is the estimate of the population proportion prefering method M1
and (1- .6643) = .3357 is the estimate of the population proportion prefering methos
M2
Now we can extend our calculation as follows :
Table 9.2
Chattisgarh
Bihar
A.P.
Orissa
Number of personnel
expected
to prefer method M1
100 X .6643
=66.43
120 X .6643
= 79.72
90 X .6643
= 59.79
110 X .6643
= 73.07
Number of personnel
expected
to prefer method M2
100 X .3357
= 33.57
120 X .3357
= 40.28
90 X .3357
= 30.21
110 X .3357
=36.93
Now we can combine Table 9.1 and Table 9.2 as follows in table 9.3. It shows both
the actual, or observed, frequency of the personnel sampled who prefer each method and
theoretical, or expected frequency of sampled employees personnel preferring each
method.
Table 9. 3 : Comparison of observed and expected frequencies of sampled personnels
Chattisgarh
Bihar
A.P.
Orissa
Frequency preferring method
M1 : Observed (Actual)
68
75
57
79
Expected (theoretical)
66.43
79.72
59.79
73.07
M2 :Observed (Actual)
32
45
33
41
Expected (theoretical)
33.57
40.28
30.21
36.93
Frequency preferring method
To test the null hypothesis, pC=pB=pA=pO we must compose the frequencies that
were observed with the frequencies we would expect if the null hypothesis is true.
The chi- Square Statistic :
(fo - fe)2
χ2 = Σ ---------------- .......... (G)
fe
Quantitative Techniques in
Management : 150
Where χ2 = chi-square
fo = Observed frequency
Testing of Hypothesis
fe = Expected frequency
Σ = Summation symbol
Calculation of χ2 (chi -Square) using formula (G) is shown in table 9.4.
NOTES
Table 9.4 : Calculation of Chi- square (χ2)
Step 1
Step2
Step 3
fo
fe
(fo-fe)
(fo-fe)2
(fo-fe)2/fe
68
66.43
1.57
2.46
.0370
75
79.72
-4.72
22.28
.2795
57
59.79
-2.79
7.78
.1301
79
73.07
5.93
35.16
.4812
32
33.57
-1.57
2.46
.0733
45
40.28
4.72
22.28
.5531
33
30.21
2.79
7.78
.2575
41
36.93
-5.93
35.16
.9521
(fo - fe )2
Σ -------------- = χ2
fe
= 2.764
(Note : χ2 values can never be negative, as difference term in step 1 is squared in
step 2.)
9.11 The Chi- Square Distribution
Figure 9.10 shown the chisquare distribution for three different
degree of freedoms.
With increase in degree of
freedom, distribution becomes
symmetrical. Chi-square distribution
is also a probability distribution,
therefore area under the curve is 1.
Chi-Square distribution can
approximate to normal distribution for
very high values of DoFs.
Fig. 9.10
Appendix table 4 illustrates only
the areas in the tail most commonly used in significance tests using the Chi-Square
Quantitative Techniques in
Management : 151
Testing of Hypothesis
distribution.
Degress of Freedom (DoF) :
DoF = (Number of rows - 1) (Number of Columns - 1) - H
NOTES
If we prepare a contingency table of r rows and c columns, then DOF for this
table will be determind using formula H as follows :
DOF = (r-1) (c-1)
Now in our illustration, contingency table 9.1 has 2 rows and 4 columns, so
appropriate DOF is = (2-1) (4-1)
=3
Using the Chi- Square Test :If we want to test null hypothesis set here at the 0.10 level of significance, figure
9.11 shows the hypothesis test at the 0.10 level of significance.
In appendix table 4, we can
look under the 0.10 column and
move down to the 3 DOF row.
Corresponding χ2 is 6.251. This
means that with 3 DOF, the region
to the right of a χ2 value of 6.251
contains 0.10 of the area under the
curve. Therefore acceptance region
for the null hypothesis as shown in
figure 9.11 goes from the left tail of
the curve to the χ2 value of 6.251.
χ2
Sample
χ2 Value
Fig. 9.11
Now we can see that calculated value of x2 for our illustration, 2.764 is within
acceptance region. Therefore, we accept the null hypothesis.
Contingency Tables with more than Two Rows :Take a situation where we assume that amount of per month scholarship affects
duration of completion of Ph.D. degree. On a random sample of 550 research scholars
of various universities of Maharashtra, data was collected as presented in table 9.5.
Table 9.5
Duration of Ph. D programme by a researther
Quantitative Techniques in
Management : 152
Scholarship per month
< 3 years
3-4 years
4-5 years
Total
Rs. 1000
40
20
20
80
Rs. 5000
30
40
160
230
Rs. 10000
40
50
150
240
Total
110
110
330
550
Table 9.5 gives observed frequencies in the nine different duration of research
program and amount of scholarship per month. Null hypothesis will be duration of
research program and scholarship are independent.
Testing of Hypothesis
Alternative hypothesis will be "length of Ph.D programme depends on amount of
scholarship” Level of significance is 10%.
To use chi- square test, we will need estimated frequencies.
NOTES
To get fe for cell (1, 1) i. e. scholarship is Rs. 1000 per month and a candidate
completes Ph.D. in less than 3 years; we let
A = the event "Scholarship is Rs. 1000 per month."
B = the event "Completion of research in less than 3 years."
Then,
p (first cell) = p (A and B)
= P (A) X p (B)
80
= --------- x
550
(
)
110
---------550
(
)
8
= -------275
beause is the expected proportion in the first cell, the expected frequency in that
cell is
8
(---------) X (550) = 16 observations
275
In general, we can calculate the expected frequency of any cell using
RT X CT
fe =---------------- .................... (I)
n
Where
RT
= row total for the row containing that cell.
CT
= Column total for the column containing that cell
n
= total number of observations.
Table 9.6 gives expected frequencies and the value of χ2 using formula (I) and
formula (G)
Table 9.6
Row
Column
fo
fe
(fo-fe)2/fe
1
1
40
16
36
1
2
20
16
1
1
3
20
48
16.33
2
1
30
46
5.56
2
2
40
46
0.78
Quantitative Techniques in
Management : 153
Testing of Hypothesis
NOTES
2
3
160
138
3.50
3
1
40
48
1.33
3
2
50
48
0.08
3
3
150
144
0.25
Degree of freedom = (3-1) X (31)
4
=4
Figure '9.12' shows a chi-square
distribution with 4 DOF, showing
10%significance level. Critical value of
χ2 from table in appendix 4 - is 7.7794
From this, we see that our
sampled value of χ2 is in region of nonacceptance. Therefore, we reject the
null hypothesis.
7.7794
64.83
Fig. 9.12
9.12 Useful points while using the Chi-Square Test
(1) Use large sample size - Expected frequency of less than 5 in one cell of a
contingency table is too small to use.
(2) If χ2 values are zero, we should be careful to question whether absolutely no
difference exists between observed and expected frequencies.
(3) Data should not be presented in percentage or ratio form, rather they should
be expressed in original units.
χ2 as a Goodness of fit Test
Chi-Square test can also be used to decide whether a particular probability
distribution, such as Poisson or normal, is the appropriate distribution.
The Chi-square tests whether there is a significant difference between an observed
frequency distribution and a theoretical frequency distribution. In this manner we
determine goodness of fit of theoretical distribution.
A Company is concerned about the increasing violent altercations between its
employees. The number of violent incidents recorded by the management during six
randomly selected months is given in table 9.7.
Table 7
Months
Jan
Feb
Mar
Apr
May
June
No. of violent incidents
55
65
68
72
80
85
Use α = 0.05 to determine whether the data fits a uniform distribution.
Quantitative Techniques in
Management : 154
Here our hypotheses are
H0 : An uniform distribution is a good description of number of violent altercations
over the months.
Testing of Hypothesis
H1 : Number of violent altercations are not uniformly distributed over the months.
Table 9.8 : Presents calculation of expected frequencies and χ2
fe
=
= Σ fo
-------n
NOTES
(fo - fe)2
---------------fe
Months
fo
Jan
55
70
3.2142
Feb
65
70
0.3571
Mar
68
70
0.0571
Apr
72
70
0.571
May
80
70
0.9142
June
85
70
2.0571
χ2 = 6.65
Σfo = 420
Here n - no. of periods = '6' in this case
Σfo
420
So, fe = -------------- = -------- = 70
n
6
Here degree of freedom will be one less than the total number of classes of data.
In general, first we employ the (k-1) rule, where k is number of classes and then
subtract an additional degree of freedom for each population parameter that has to be
estimated from the sample data.
In current illustration, DOF = 6-1 = 5
For a given level (here it is .05), rule of acceptance or rejection will be, if χ2 cal >
χ2 critical reject the
Figure 9.13 presents Chi-Square distribution for DOF = 5. As taken from appendix
table 4, critical value of χ2 is 11.07 which
is higher than sampled value of x 2.
acceptance region
Therefore we accept the null hypathesis.
It means uniform distribution is a
good description of number of violent
altercations.
It is worth mentioning that
expected frequency in this case is based
on uniform distribution.
Expected frequency calculation is
distribution based.
6.65
11.07
Fig. 9.13
If we are looking for goodness of fit of binomial distribution, we will check,
probability of success in the binomical distribution table (in Appendix table 3) for a
Quantitative Techniques in
Management : 155
Testing of Hypothesis
particular n and p.
9.13 Summary
NOTES
9.14 Key Terms
Hypothesis
: It is an assumption made about a population parameter.
Nll Hypothesis
: Hypothesis about a population parameter we want to test.
Alternative Hypothesis : Statement (Conclusion) we accept when we reject null
hypothesis.
Type I Error
: Rejecting a null hypothesis when it is true.
Type II Error
: Accepting a null hypothesis when it is false.
One tailed test
: A hypothesis test in which there is only one rejection region,
either in left or right side.
Two tailed test
: A hypothesis test in which tow rejection regions are there.
Here Null hypothesis is rejected if the sample value is
significantly higher or lower than the hypothesized value of
the population parameter.
Contigency table
: A talbe of R rows and C coloums. Each row is for a level of
one variable, each column is for a level of another variable.
Entries in he cells of table are the observed frequencies with
which variable combination occured.
Expected Frequencies: These frequencies are expected in contigency table on the basis
of theoretical calculations.
Quantitative Techniques in
Management : 156
Testing of Hypothesis
9.15 Questions and Exercises
(1)
(2)
What is the Chi-Square goodness of fit test and what are the applications in
decision making?
NOTES
Explain the conceptual framework of Chi-Square test with respect to expected
and observed frequencies.
Practice Questions
(3) In a survey of India, where four regions, norther eastern, southern and western
were surveged with a random sample of 1000 persons in each region to know
brand awareness. Following results were obtained.
Region
Northern Eastern Southern
Western
Aware about Green Marketing
400
550
450
500
Do not aware about green marketing
600
450
550
500
(i) Develop a table of observed and expected frequencies for this problem.
(ii) Calculate the sample χ2 value.
(iii) State the null and alternative hypotheses.
(iv) At α = .05, test whether awareness level is the same across the four regions.
4)
A company has developed a new product. The company test marketed it in a
particular geographic region. The consumer opinion (obtained through a randomly
selected sample of 511 consumers) of different age groups is given in following
table :
Opinion / Age group
Above 18 Above 30
Above 50
Row Total
liked new brand
95
85
70
250
Did not like new brand 35
55
72
162
Indifferent
30
34
35
95
Column Total
160
174
177
511
Examine whether the consumer opinion for a new brand is independent of age
groups. Use α = 0.10
(5) At the 0.10 level of significance, can we conclude that the following 400
observations follow a poisson distribution with λ = 3?
Number of arrivals per hour
0
1
2
3
4
5 or more
Number of hours
20
57
98
85
78
62
(6) At the level of .20 significance, test goodness of fit of a interview process in
which three executives give positive or negative ranking, with binomial
distribution.
Quantitative Techniques in
Management : 157
Testing of Hypothesis
NOTES
Possible positive ratings from
three executives
0
1
2
3
Take p = .40
Quantitative Techniques in
Management : 158
Number of candidates receiving
these ratings
18
47
24
11
--------100
Z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
0.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
0.4236
0.4370
0.4484
0.4582
0.4664
0.4732
0.4788
0.4834
0.4871
0.4901
0.4925
0.4943
0.4957
0.4968
0.4977
0.4983
0.4988
0.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
0.4251
0.4382
0.4495
0.4591
0.4671
0.4738
0.4793
0.4838
0.4875
0.4904
0.4927
0.4945
0.4959
0.4969
0.4977
0.4984
0.4988
0.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
0.4265
0.4394
0.4505
0.4599
0.4678
0.4744
0.4798
0.4842
0.4878
0.4906
0.4929
0.4946
0.4960
0.4970
0.4978
0.4984
0.4989
Normal Distributions
0.06
0.0239
0.0636
0.0126
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
0.4279
0.4406
0.4515
0.4608
0.4686
0.4750
0.4803
0.4846
0.4881
0.4909
0.4931
0.4948
0.4961
0.4971
0.4979
0.4985
0.4989
0.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4756
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4979
0.4985
0.4989
0.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4429
0.4535
0.4625
0.4699
0.4761
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
0.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
0.4986
0.4990
Areas under the Standard Normal
Probability Distribution between
the Mean and Positive Values of Z
Appendix Table 1
Appendix Table
0.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4982
0.4987
0.4750
area
0.4875 of
of area
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4864
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4987
Z=1.96
0.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
0.4192
0.4332
0.4452
0.4554
0.4641
0.4713
0.4772
0.4821
0.4862
0.4893
0.4918
0.4938
0.4953
0.4965
0.4974
0.4981
0.4987
Quantitative Techniques in
Management : 159
0.05 of area
t= -2.132
DoF = 4
Area = .10 in both tails
Quantitative Techniques in
Management : 160
0.05 of are
t = +2.132
t = + 1.729
Degree of
Freedom0.10
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
40
60
120
Normal Distribution
Appendix Table 2 `t’ distribution
Areas in Both Tails Combined for
Student's t Distribution
Area in Both Tails Combined
0.10
0.05
0.02
0.01
6.314
12.706
31.821
63.652
2.920
4.303
6.965
9.925
2.353
3.182
4.541
5.841
2.132
2.776
3.747
4.604
2.015
2.571
3.365
4.032
1.943
2.447
3.143
3.702
1.895
2.365
2.998
2.499
1.860
2.306
2.896
3.355
1.833
2.262
2.821
3.250
1.812
2.228
2.764
3.169
1.796
2.201
2.718
3.106
1.782
2.179
2.681
3.055
1.771
2.160
2.650
3.012
1.761
2.145
2.624
2.922
1.753
2.131
2.602
2.947
1.746
2.120
2.583
2.921
1.740
2.110
2.567
2.898
1.734
2.101
2.552
2.878
1.729
2.093
2.539
2.861
1.725
2.086
2.528
2.845
1.721
2.080
2.518
2.831
1.721
2.074
2.508
2.819
1.714
2.069
2.500
2.807
1.711
2.064
2.492
2.797
1.708
2.060
2.485
2.787
1.706
2.056
2.479
2.779
1.703
2.052
2.473
2.771
1.701
2.048
2.467
2.763
1.699
2.045
2.462
2.756
1.697
2.042
2.457
2.750
1.684
2.021
2.423
2.704
1.671
2.000
2.390
2.660
1.658
1.980
2.358
2.617
1.645
1.960
2.326
2.576
Appendix Table3 Binomial Probabilities
P
n
r
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.9
0.14
0.15
0.16
0.17
r
n
2
0
1
2
0.9801
0.0198
0.0001
0.9604
0.0392
0.0004
0.9409
0.0582
0.0009
0.9216
0.0768
0.0016
0.9025
0.0950
0.0025
0.8836
0.1128
0.0036
0.8649
0.1302
0.0049
0.8464 0.8281 0.8100 0.7921
0.1472 0.1638 0.1800 0.1958
0.0064 0.0081 0.0100 0.0121
0.10
0.7744 0.7569
0.2112 0.2262
0.0144 0.0169
0.7396
0.2408
0.0196
0.7225
0.2550
0.0225
0.7056
0.2688
0.0256
0.6889 0.6724
0.2822 0.2952
0.0289 0.0324
2
1
0
2
3
0
1
2
3
0.9703
0..0294
0.0003
0.0000
0.9412
0.0576
0.0012
0.0000
0.9127
0.0847
0.0026
0.0000
0.8847
0.1106
0.0046
0.0001
0.8574
0.1354
0.0071
0.0001
0.8306
0.1590
0.0102
0.0002
0.8044
0.1816
0.0137
0.0003
0.7787
0.2031
0.0177
0.0005
0.7536
0.2236
0.0221
0.0007
0.7290
0.2430
0.0270
0.0010
0.7050
0.2614
0.0323
0.0013
0.6815
0.2788
0.0380
0.0017
0.6585
0.2952
0.0441
0.0022
0.6361
0.3106
0.0506
0.0027
0.6141
0.3251
0.0574
0.0034
0.5927
0.3387
0.0645
0.0041
0.5718
0.3513
0.0720
0.0049
0.5514
0.3631
0.0797
0.0058
3
2
1
0
3
4
0
1
2
3
4
0.9606
0.0388
0.0006
0.0000
-
0.9224
0.0753
0.0023
0.0000
-
0.8853
0.1095
0.0051
0.0001
0.0000
0.8493
0.1416
0.0088
0.0002
0.0000
0.8145
0.1715
0.0135
0.0005
0.0000
0.7807
0.1993
0.0191
0.0008
0.0000
0.7481
0.2252
0.0254
0.0013
0.0000
0.7164
0.2492
0.0325
0.0019
0.0000
0.6857
0.2713
0.0402
0.0027
0.0001
0.6561
0.2916
0.0486
0.0036
0.0001
0.6274
0.3102
0.0575
0.0047
0.0001
0.5997
0.3271
0.0669
0.0061
0.0002
0.5729
0.3424
0.0767
0.0076
0.0003
0.5470
0.3562
0.0870
0.0094
0.0004
0.5220
0.3685
0.0975
0.0115
0.0005
0.4979
0.3793
0.1084
0.0138
0.0007
0.4746
0.3888
0.1195
0.0163
0.0008
0.4521
0.3970
0.1307
0.0191
0.0010
4
3
2
1
0
4
5
0
1
2
3
4
5
0.9510
0.0480
0.0010
0.0000
-
0.9039
0.0922
0.0038
0.0001
0.0000
-
0.8587
0.1328
0.0082
0.0003
0.0000
-
0.8154
0.1699
0.0142
0.0006
0.0000
-
0.7738
0.2036
0.0214
0.0011
0.0000
-
0.7339
0.2342
0.0299
0.0019
0.0001
0.0000
0.6957
0.2618
0.0394
0.0030
0.0001
0.0000
0.6591
0.2866
0.0498
0.0043
0.0002
0.0000
0.6240
0.3086
0.0610
0.0060
0.0003
0.0000
0.5905
0.3280
0.0729
0.0081
0.0004
0.0000
0.5584
0.3451
0.0853
0.0105
0.0007
0.0000
0.5277
0.3598
0.0981
0.0134
0.0009
0.0000
0.4984
0.3724
0.1113
0.0166
0.0012
0.0000
0.4704
0.3829
0.1247
0.0203
0.0017
0.0001
0.4437
0.3915
0.1382
0.0244
0.0022
0.0001
0.4182
0.3983
0.1517
0.0289
0.0028
0.0001
0.3939
0.4034
0.1652
0.0338
0.0035
0.0001
03707
0.4069
0.1786
0.0392
0.0043
0.0002
5
4
3
2
1
0
5
6
0
1
2
3
4
5
6
0.9415
0.0571
0.0014
0.0000
-
0.8858
0.1085
0.0055
0.0002
0.0000
-
0.8330
0.1546
0.0120
0.0005
0.0000
-
0.7828
0.1957
0.0204
0.0011
0.0000
-
0.7351
0.2321
0.0305
0.0021
0.0001
0.0000
-
0.6899
0.2642
0.0422
0.0036
0.0002
0.0000
-
06470
0.2922
0.0550
0.0055
0.0003
0.0000
-
0.6064
0.3164
0.0688
0.0080
0.0005
0.0000
-
0.5679
0.3370
0.0833
0.0110
0.0008
0.0000
-
0.5314
0.3543
0.0984
0.0146
0.0012
0.0001
0.0000
0.4970
0.3685
0.1139
0.0188
0.0017
0.0001
0.0000
0.4644
0.3800
0.1295
0.0236
0.0024
0.0001
0.0000
0.4336
0.3883
0.1452
0.0289
0.0032
0.0002
0.0000
0.4046
0.3952
0.1608
0.0349
0.0043
0.0003
0.0000
0.3771
0.3993
0.1762
0.0415
0.0055
0.0004
0.0000
0.3513
0.4015
0.1912
0.0486
0.0069
0.0005
0.0000
0.3269
0.4018
0.2057
0.0562
0.0086
0.0007
0.0000
0.3040
0.4004
0.2197
0.0643
0.0106
0.0009
0.0000
6
5
4
3
2
1
0
6
7
0
1
2
3
4
5
6
7
0.9321
0.0659
0.0020
0.0000
-
0.8681
0.1240
0.0076
0.0003
0.0000
-
0.8080
0.1749
0.0162
0.0008
0.0000
-
0.7514
0.2192
0.0274
0.0019
0.0001
0.0000
-
0.6983
0.2573
0.0406
0.0036
0.0002
0.0000
-
0.6485
0.2897
0.0555
0.0059
0.0004
0.0000
-
0.6017
0.3170
0.0716
0.0090
0.0007
0.0000
-
0.5578
0.3396
0.0886
0.0128
0.0011
0.0001
0.0000
-
0.5168
0.3578
0.1061
0.0175
0.0017
0.0001
0.0000
-
0.4783
0.3720
0.1240
0.0230
0.0026
0.0002
0.0000
-
0.4423
0.3827
0.1419
0.0292
0.0036
0.0003
0.0000
-
0.4087
0.3901
0.1596
0.0363
0.0049
0.0004
0.0000
-
0.3773
0.3946
0.1769
0.0441
0.0066
0.0006
0.0000
-
0.3479
0.3965
0.1936
0.0525
0.0086
0.0008
0.0000
-
0.3206
0.3960
0.2097
0.0617
0.0109
0.0012
0.0001
0.0000
0.2951
0.3935
0.2248
0.0714
0.0136
0.0016
0.0001
0.0000
0.2714
0.3891
0.2391
0.0816
0.0167
0.0021
0.0001
0.0000
0.2493
0.3830
0.2523
0.0923
0.0203
0.0027
0.0002
0.0000
7
6
5
4
3
2
1
0
7
n
r
0..99
0.98
0.97
0.96
0.95
0.94
0.93
0.92
0.91
0.90
0.89
0.88
0.87
0.86
0.85
0.84
0.83
0.82
r
n
P
0.11
0.12
0.13
0.18
Quantitative Techniques in
Management : 161
Quantitative Techniques in
Management : 162
P
n r
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.9
0.10
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
r n
8 0
1
2
3
4
5
6
7
8
0.9227
0.0746
0.0026
0.0001
0.0000
-
0.8508
0.1389
0.0099
0.0004
0.0000
-
0.7837
0.1939
0.0210
0.0013
0.0001
0.0000
-
0.7214
0.2405
0.0351
0.0029
0.0002
0.0000
-
0.6634
0.2793
0.0515
0.0054
0.0004
0.0000
-
0.6096
0.3113
0.0695
0.0089
0.0007
0.0000
-
0.5596
0.3370
0.0888
0.0134
0.0013
0.0001
0.0000
-
0.5132
0.3570
0.1087
0.0189
0.0021
0.0001
0.0000
-
0.4703
0.3721
0.1288
0.0255
0.0031
0.0002
0.0000
-
0.4305
03826
0.1488
0.0331
0.0046
0.0004
0.0000
-
0.3937
0.3892
0.1684
0.0416
0.0064
0.0006
0.0000
-
0.3596
0.3923
0.1872
0.0511
0.0087
0.0009
0.0001
0.0000
-
0.3282
0.3923
0.2052
0.0613
0.0115
0.0014
0.0002
0.0000
-
0.2992
0.3897
0.2220
0.0723
0.0147
0.0019
0.0002
0.0000
-
0.2725
0.3847
0.2376
0.0839
0.0185
0.0026
0.0002
0.0000
-
0.2479
0.3777
0.2518
0.0959
0.0228
0.0035
0.0003
0.0000
-
02252
0.3691
0.2646
0.1084
0.0277
0.0045
0.0005
0.0000
-
0.2044
0.3590
0.2758
0.1211
0.0332
0.0058
0.0006
0.0000
-
8
7
6
5
4
3
2
1
0 8
9 0
1
2
3
4
5
6
7
8
9
0.9135
0.0830
0.0034
0.0001
0.0000
-
0.8337
0.1531
0.0125
0.0006
0.0000
-
0.7602
0.2116
0.0262
0.0019
0.0001
0.0000
-
0.6925
0.2597
0.0433
0.0042
0.0003
0.0000
-
0.6302
0.2985
0.0629
0.0077
0.0006
0.0000
-
0.5730
0.3292
0.0840
0.0125
0.0012
0.0001
0.0000
-
0.5204
0.3525
0.1061
0.0186
0.0021
0.0002
0.0000
-
0.4722
0.3695
0.1285
0.0261
0.0034
0.0003
0.0000
-
0.4279
0.3809
0.1507
0.0348
0.0052
0.0005
0.0000
-
0.3874
0.3874
0.1722
0.0446
0.0074
0.0008
0.0001
0.0000
-
0.3504
0.3897
0.1927
0.0556
0.0103
0.0013
0.0001
0.0000
-
0.3165
0.3884
0.2119
0.0674
0.0138
0.0019
0.0002
0.0000
-
0.2855
0.3840
0.2295
0.0800
0.0179
0.0027
0.0003
0.0000
-
0.2573
0.3770
0.2455
0.0933
0.0228
0.0037
0.0004
0.0000
-
0.2316
0.3679
0.2597
0.1069
0.0283
0.0050
0.0006
0.0000
-
0.2082
0.3569
0.2720
0.1209
0.0345
0.0066
0.0008
0.0001
0.0000
-
0.1869
0.3446
0.2823
0.1349
0.0415
0.0085
0.0012
0.0001
0.0000
-
0.1676
0.3312
0.2908
0.1489
0.0490
0.0108
0.0016
0.0001
0.0000
-
9
8
7
6
5
4
3
2
1
0 9
10 0
1
2
3
4
5
6
7
8
9
10
0.9044
0.0914
0.0042
0.0001
0.0000
-
0.8171
0.1667
0.0153
0.0008
0.0000
-
0.7374
0.2281
0.0317
0.0026
0.0001
0.0000
-
0.6648
0.2770
0.0519
0.0058
0.0004
0.0000
-
0.5987
0.3151
0.0746
0.0105
0.0010
0.0001
0.0000
-
0.5386
0.3438
0.0988
0.0168
0.0019
0.0001
0.0000
-
0.4846
0.3643
0.1234
0.0248
0.0033
0.0003
0.0000
-
0.4344
0.3777
0.1478
0.0343
0.0052
0.0005
0.0000
-
0.3894
0.3851
0.1714
0.0452
0.0078
0.0009
0.0001
0.0000
-
0.3487
0.3874
0.1937
0.0574
0.0112
0.0015
0.0001
0.0000
-
0.3118
0.3854
0.2143
0.0706
0.0153
0.0023
0.0002
0.0000
-
0.2785
0.3798
0.2330
0.0847
0.0202
0.0033
0.0004
0.000
-
0.2484
0.3712
0.2496
0.0995
0.0260
0.0047
0.0006
0.0000
-
0.2213
0.3603
0.2639
0.1146
0.0326
0.0064
0.0009
0.0001
0.0000
-
0.1969
0.3474
0.2759
0.1298
0.0401
0.0085
0.0012
0.0001
0.0000
-
0.1749
0.3331
0.2856
0.1450
0.0483
0.0111
1.0018
0.0002
0.0000
-
0.1552
0.3178
0.2929
0.1600
0.0573
0.0141
0.0024
0.0003
0.0000
-
0.1374
0.3017
0.2980
0.1745
0.0670
0.0177
0.0032
0.0004
0.0000
-
10
9
8
7
6
5
4
3
2
1
0 10
n r
0.99
0.98
0.97
0.96
0.95
0.94
0.93
0.92
0.91
0.90
0.89
0.88
0.87
0.86
0.85
0.84
0.83
0.82
r n
P
Table A. 7 _
M
0.10
10
12
9.23635
0.0157907
0.210721
0.58438
1.06362
1.61031
2.20413
2.83311
3.48954
4.16816
4.86518
5.57779
6.30380
7.04150
7.78954
8.54675
9.31224
10.08518
10.86494
11.65091
12.44260
13.23960
14.04149
14.84795
15.65868
16.47341
17.29188
18.11389
18.93924
19.76774
20.59924
29.05052
37.68864
46.45888
55.32894
64.27784
73.29108
82.35813
X2
2.7055
4.6052
6.2514
7.7794
9.2363
10.6446
12.0170
13.3616
14.6837
15.9872
17.2750
18.5493
19.8119
21.0641
22.3071
23.5418
24.7690
25.9894
27.2036
28.4120
29.6151
30.8133
32.0069
33.1962
34.3816
35.5632
36.7412
37.9159
39.0875
40.2560
51.8050
63.1671
74.3970
85.5270
96.5782
107.5650
118.4980
Area in Upper Tail
0.9
0.1
Appendix Table 4
8
0.95
6
0.975
0.0039322
0.102586
0.35185
0.71072
1.14548
1.63538
2.16735
2.73263
3.32512
3.94030
4.57481
5.22603
5.8') 186
6.57063
7.26093
7.96164
8.67175
9.39045
10.11701
10.85080
11.59132
12.33801
13.09051
13.84842
14.61140
15.37916
16.15139
16.92788
17.70838
18.49267
26.50930
34.76424
43.18797
51.73926
60.39146
61.12602
77.92944
4
0.99
0.0009821
0.050636
0.21579
0.48442
0.83121
1.23734
1.68986
2.17972
2.70039
3.24696
3.81574
4.40378
5.00874
5.62872
6.26212
6.90766
7.56418
8.23074
8.90651
9.59077
10.28291
10.98233
11.68853
12.40115
13.11971
13.84388
14.57337
15.30785
16.04705
16.79076
24.43306
32.35738
40.48171
48.75754
57.15315
6564659
74.22188
2
0.995
0.0001571
0.020100
0.11483
0.29711
0.55430
0.87208
1.23903
1.64651
2.08789
2.55820
3.05350
3.57055
4.10690
4.66042
5.22936
5.81220
6.40774
7.01490
7.63270
8.26037
8.89717
9.54249
10.19569
10.85635
11.52395
12.19818
12.87847
13.56467
14.25641
14.95346
22.16420
29.70673
37.48480
45.44170
53.53998
61.75402
70.06500
0
0.0000393
0.010025
0.07172
0.20698
0.41175
0.67573
0.98925
1.34440
1.73491
2.15585
2.60320
3.07379
3.56504
4.07466
4.60087
5.14216
5.69727
6.26477
6.84392
7.43381
8.03360
8.64268
9.26038
9.88620
10.51965
11.16022
11.80765
12.46128
13.12107
13.78668
20.70658
27.99082
35.53440
43.27531
51.17193
59.19633
67.32753
The Chi-Square Table
Degree of
Freedom
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
40
50
60
70
80
90
100
0.05
0.025
3.8415
5.0239
5.9915
7.3778
7.8147
9.3484
9.4877 11.1433
11.0705 12.8325
12.5916 14.4494
14.0671 16.0128
15.5073 17.5345
16.9190 19.0228
18.3070 20.4832
19.6752 21.9200
21.0261 23.3367
22.3620
247356
23.6848 26.1189
24.9958 27.4884
26.2962 28.8453
27.5871 30.1910
28.8693 31.5264
30.1435 32.8523
31.4104 34.1696
32.6706 35.4789
33.9245 36.7807
35.1725 38.0756
36.4150 39.3641
37.6525 40.6465
38.8851 41.9231
40.1133 43.1945
41.3372 44.4608
42.5569 45.7223
43.7730 46.9792
55.7585 59.3417
67.5048 71.4202
79.0820 83.2977
90.5313 95.0231
101.8795 106.6285
113.1452 118.1359
124.3421 129.5613
6.6349
9.2104
11.3449
13.2767
15.0863
16.8119
18.4753
20.0902
21.6660
23.2093
247250
26.2170
276882
29.1412
30.5780
31.9999
33.4087
34.8052
36.1908
37.5663
38.9322
40.2894
41.6383
42.9798
44.3140
45.6416
46.9628
482782
49.5878
50.8922
63.6908
76.1538
88.3794
100.4251
112.3288
124.1162
135.8069
0.01
7.8794
10.5965
12.8381
14.8602
16.7496
18.5475
20.2777
21.9549
235893
25.1881
26.7569
28.2997
29.8193
31.3194
32.8015
34.2671
35.7184
37.1564
38.5821
39.9969
41.4009
42.7959
44.1814
45.5584
46.9280
48.2898
49.6450
50.9936
52.3355
53.6719
66.7660
79.4898
91.9518
104.2148
116.3209
128.2987
140.1697
0.005
Quantitative Techniques in
Management : 163