Activity
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Operations research
C.R.Krishna Prasad, BIT Bangalore-4
Origin and Development
The main origin of OR was during the II world war
Military management in England called upon a team
of scientists to study the Strategic* and Tactical*
problems related to air and land defence of the
country.
Since they were having very limited resources, it
was necesent osary to decide upon the most
effective utilization of them., eg the efficient ocean
transport, effective bombing, etc.
(* premeditated)
C.R.Krishna Prasad, BIT Bangalore-4
Contd….
Their mission was to formulate the specific
proposals and plans for aiding the military
commands to arrive at the decisions on
optimal utilization of scarce military resources
and efforts and
also to implement the
decision effectively.
Scientific and systematic approaches
involved in OR provided a good intellectual
support to the strategic initiatives of the
military commands.
C.R.Krishna Prasad, BIT Bangalore-4
i
One group in britain came to known as
Blacket circus.(radar OR Unit in gun site).
The US military team named operational
analysis ,operational evaluation, operational
research,
system
analysis
,system
evaluation, system research & management.
But military team were dealing with research
on (military) operation, the work of this team
of scientists named as Operation Research
in England.
C.R.Krishna Prasad, BIT Bangalore-4
Introduction
End of the war, the success of military teams
attracted the attention of industrial managers
Who were seeking their complex executive
type problems.
The most common problem was : what
method should be adopted so that the total
cost is minimum or total profit is
maximum.
C.R.Krishna Prasad, BIT Bangalore-4
The first mathematical technique in this field
(called the simplex method of linear
programming) was developed by american
mathematician,George B Damtzig.
Since then techniques & application have
been developed through the effort and
cooperation of interested individual in
academic institutions and industry both.
C.R.Krishna Prasad, BIT Bangalore-4
Definitions of OR
OR is the systematic method oriented
study of the basic structure, characteristic,
functions
and
relationships
of
an
organisation to provide the executive with
a sound, scientific and quantitative basis
for decision making.
OR is concerned with scientifically
deciding how to best design and operate
man-machine systems usually requiring the
allocation of resources.
C.R.Krishna Prasad, BIT Bangalore-4
Contd..
OR is the art of giving bad answers to the
problems to which otherwise worse answers
Are given.
An art of winning the war without actually
fighting it.
C.R.Krishna Prasad, BIT Bangalore-4
Characteristics of OR
OR approaches problem solving and decision making
from a total system’s perspective
It is interdisciplinary model
Model building and mathematical manipulation
provide the methodology .
OR is for operations economy
Primary focus on decision making.
C.R.Krishna Prasad, BIT Bangalore-4
PHASES OF OPERATIONS RESEARCH
STUDY
I Formulation The Problem:
What
are
the
objectives,
uncontrolled variable constraints
controlled
variables,
IIConstructing a mathematical method
A mathematical model should include a) decision
variable and parameter,,objective functions&constraints.
C.R.Krishna Prasad, BIT Bangalore-4
III Deriving the solutions from the model
OR models include LPP, Transportation, Assignment,
Queuing models, Network analysis, Job sequencing,
Replacement models, simulation models
IV Testing the model and
solution(updating the model)
V: Controlling the solution
VI: Implementing the solution
C.R.Krishna Prasad, BIT Bangalore-4
Techniques of OR
Distribution/allocation Models
Waiting line/queuing models
Production/inventory model
Competitive strategy model/games
theory
Network analysis
Job sequencing models
C.R.Krishna Prasad, BIT Bangalore-4
Replacement models
Markovian models
Simulation models
C.R.Krishna Prasad, BIT Bangalore-4
Scope of O.R.
In agriculture
In Finance- BEP,Capital budgeting,
SAPM, cash flow,financial planning
In Industry
In Marketing
In Personnel management
In Production management
In Life Insurance
C.R.Krishna Prasad, BIT Bangalore-4
Limitations
Magnitude of computation
Absence of quantification
Distance between manager and OR
experts
C.R.Krishna Prasad, BIT Bangalore-4
Linear Programming Problems
C.R.Krishna Prasad, BIT Bangalore-4
- Formulation
Linear Programming is a mathematical technique for optimum
allocation of limited or scarce resources, such as labour, material,
machine, money, energy and so on , to several competing activities
such as products, services, jobs and so on, on the basis of a given
criteria of optimality.
C.R.Krishna Prasad, BIT Bangalore-4
Contd….
The term ‘Linear’ is used to describe the proportionate relationship of
two or more variables in a model. The given change in one variable
will always cause a resulting proportional change in another variable.
The word , ‘Programming’ is used to specify a sort of planning that
involves the economic allocation of limited resources by adopting a
particular course of action or strategy among various alternatives
strategies to achieve the desired objective.
C.R.Krishna Prasad, BIT Bangalore-4
Structure of Linear Programming model.
The general structure of the Linear Programming model
essentially consists
of three components.
i)
The
activities
(Decision
variables)
relationships
ii) The objective function and
iii) The constraints
C.R.Krishna Prasad, BIT Bangalore-4
and
their
i) The activities are represented by
X1, X2, X3 ……..Xn. These are known as Decision variables.
ii) The objective function of an LPP (Linear Programming Problem)
is a mathematical representation of the objective in terms a
measurable quantity such as profit, cost, revenue, etc.
Optimize (Maximize or Minimize) Z=C1X1 +C2X2+ ………..Cn Xn
Where Z is the measure of performance variable
X1, X2, X3, X4…..Xn are the decision variables
And C1, C2, …Cn are the parameters that give contribution to
decision variables.
iii) Constraints
are the set of linear inequalities and/or equalities
which impose restriction of the limited resources
C.R.Krishna Prasad, BIT Bangalore-4
General Mathematical Model of an LPP
Optimize (Maximize or Minimize) Z=C1 X1 + C2 X2 +……+CnXn
Subject to constraints,
a11X1+ a 12X2+………………+ a 1nXn (<,=,>) b1
a21X1+ a 22X2+………………+ a 2nXn (<,=,>) b2
a31X1+ a 32X2+………………+ a 3nXn (<,=,>) b3
am1X1+ a m2X2+………………+ a mnXn (<,=,>) bm
and X1, X2 ….Xn > 0
C.R.Krishna Prasad, BIT Bangalore-4
Guidelines for formulating Linear Programming model
i) Identify and define the decision variable of the problem
ii) Define the objective function
iii) State the constraints to which the objective function should be
optimized (i.e. either Maximization or Minimization)
iv) Add the non-negative constraints from the consideration that the
negative values of the decision variables do not have any valid physical
interpretation
C.R.Krishna Prasad, BIT Bangalore-4
Example 1.
A firm is engaged in producing two products. A and B. Each unit of
product A requires 2 kg of raw material and 4 labour hours for
processing, where as each unit of B requires 3 kg of raw materials
and 3 labour hours for the same type. Every week, the firm has an
availability of 60 kg of raw material and 96 labour hours. One unit of
product A sold yields Rs.40 and one unit of product B sold gives
Rs.35 as profit.
Formulate this as an Linear Programming Problem to determine as
to how many units of each of the products should be produced per
week so that the firm can earn maximum profit.
C.R.Krishna Prasad, BIT Bangalore-4
i) Identify and define the decision variable of the problem
Let X1 and X2 be the number of units of product A and product
B produced per week.
ii) Define the objective function
Since the profits of both the products are given,
the objective function is to maximize the profit.
MaxZ = 40X1 + 35X2
C.R.Krishna Prasad, BIT Bangalore-4
iii) State the constraints to which the objective function should be
optimized (i.e. Maximization or Minimization)
There are two constraints one is raw material constraint and the other
one is labour constraint..
The raw material constraint is given by
2X1 + 3X2 < 60
The labour hours constraint is given by
4X1 + 3X2 < 96
C.R.Krishna Prasad, BIT Bangalore-4
Finally we have,
MaxZ = 40X1 + 35X2
Subject to constraints,
2X1 + 3X2 < 60
4X1 + 3X2 < 96
X1,X2 > 0
C.R.Krishna Prasad, BIT Bangalore-4
Example 2.
The agricultural research institute suggested the farmer to spread out
atleast 4800 kg of special phosphate fertilizer and not less than 7200 kg of
a special nitrogen fertilizer to raise the productivity of crops in his fields.
There are two sources for obtaining these – mixtures A and mixtures B.
Both of these are available in bags weighing 100kg each and they cost
Rs.40 and Rs.24 respectively. Mixture A contains phosphate and nitrogen
equivalent of 20kg and 80 kg respectively, while mixture B contains these
ingredients equivalent of 50 kg each. Write this as an LPP and determine
how many bags of each type the farmer should buy in order to obtain the
required fertilizer at minimum cost.
C.R.Krishna Prasad, BIT Bangalore-4
i) Identify and define the decision variable of the problem
Let X1 and X2 be the number of bags of mixture A and mixture B.
ii) Define the objective function
The cost of mixture A and mixture B are given ;
the objective function is to minimize the cost
Min.Z = 40X1 + 24X2
C.R.Krishna Prasad, BIT Bangalore-4
iii) State the constraints to which the objective function should be
optimized.
The above objective function is subjected to following constraints.
20X1 + 50X2 >4800
Phosphate requirement
80X1 + 50X2 >7200
Nitrogen requirement
X1, X2 >0
C.R.Krishna Prasad, BIT Bangalore-4
Finally we have,
Min.Z = 40X1 + 24X2
is subjected to three constraints
20X1 + 50X2 >4800
80X1 + 50X2 >7200
X1, X2 >0
C.R.Krishna Prasad, BIT Bangalore-4
Example 3.
A manufacturer produces two types of models M1 and M2.Each
model of the type M1 requires 4 hours of grinding and 2 hours of
polishing; where as each model of M2 requires 2 hours of grinding
and 5 hours of polishing. The manufacturer has 2 grinders and 3
polishers. Each grinder works for 40 hours a week and each
polisher works 60 hours a week. Profit on M1 model is Rs.3.00 and
on model M2 is Rs.4.00.Whatever produced in a week is sold in the
market. How should the manufacturer allocate his production
capacity to the two types of models, so that he makes maximum
profit in a week?
C.R.Krishna Prasad, BIT Bangalore-4
i)
Identify and define the decision variable of the problem
Let X1 and X2 be the number of units of M1 and M2 model.
ii) Define the objective function
Since the profits on both the models are given, the objective function
is to maximize the profit.
Max Z = 3X1 + 4X2
C.R.Krishna Prasad, BIT Bangalore-4
iii) State the constraints to which the objective function should be
optimized (i.e. Maximization or Minimization)
There are two constraints one for grinding and the other for polishing.
The grinding constraint is given by
4X1 + 2X2 < 80
No of hours available on grinding machine per week is 40 hrs. There
are two grinders. Hence the total grinding hour available is 40 X 2 = 80
hours.
C.R.Krishna Prasad, BIT Bangalore-4
The polishing constraint is given by
2X1 + 5X2 < 180
No of hours available on polishing machine per week is 60
hrs. There are three grinders. Hence the total grinding hour
available is 60 X 3 = 180 hours.
C.R.Krishna Prasad, BIT Bangalore-4
Finally we have,
Max Z = 3X1 + 4X2
Subject to constraints,
4X1 + 2X2 < 80
2X1 + 5X2 < 180
X1,X2 > 0
C.R.Krishna Prasad, BIT Bangalore-4
Example 4.
A firm can produce 3 types of cloth, A , B and C.3 kinds of wool
are required Red, Green and Blue.1 unit of length of type A
cloth needs 2 meters of red wool and 3 meters of blue wool.1
unit of length of type B cloth needs 3 meters of red wool, 2
meters of green wool and 2 meters of blue wool.1 unit type of C
cloth needs 5 meters of green wool and 4 meters of blue wool.
The firm has a stock of 8 meters of red,10 meters of green and
15 meters of blue. It is assumed that the income obtained from
1 unit of type A is Rs.3, from B is Rs.5 and from C is
Rs.4.Formulate this as an LPP.( December2005/January 2006)
C.R.Krishna Prasad, BIT Bangalore-4
i) Identify and define the decision variable of the problem
Let X1, X2 and X3 are the quantity produced of cloth type A,B and C
respectively.
ii) Define the objective function
The incomes obtained for all the three types of cloths are given;
the objective function is to maximize the income.
Max Z = 3X1 + 5X2 + 4X3
C.R.Krishna Prasad, BIT Bangalore-4
iii) State the constraints to which the objective function should be
optimized.
The above objective function is subjected to following three
constraints.
2X1 + 3X2 < 8
2X2 + 5X3 < 10
3X1 + 2X2 + 4X3 < 15
X1, X2 X3 >0
C.R.Krishna Prasad, BIT Bangalore-4
Finally we have,
Max Z = 3X1 + 5X2 + 4X3
is subjected to three constraints
2X1 + 3X2 < 8
2X2 + 5X3 < 10
3X1 + 2X2 + 4X3 < 15
X1, X2 X3 >0
C.R.Krishna Prasad, BIT Bangalore-4
Example 5.
A Retired person wants to invest upto an amount of Rs.30,000 in fixed
income securities. His broker recommends investing in two Bonds: Bond
A yielding 7% and Bond B yielding 10%. After some consideration, he
decides to invest atmost of Rs.12,000 in bond B and atleast Rs.6,000 in
Bond A. He also wants the amount invested in Bond A to be atleast equal
to the amount invested in Bond B. What should the broker recommend if
the investor wants to maximize his return on investment? Solve
graphically. (January/February 2004)
C.R.Krishna Prasad, BIT Bangalore-4
i) Identify and define the decision variable of the problem
Let X1 and X2 be the amount invested in Bonds A and B.
ii) Define the objective function
Yielding for investment from two Bonds are given; the
objective function is to maximize the yielding.
Max Z = 0.07X1 + 0.1X2
C.R.Krishna Prasad, BIT Bangalore-4
iii) State the constraints to which the objective function
should be optimized.
The above objective function is subjected to following three
constraints.
X1 + X2 < 30,000
X1 > 6,000
X2 < 12,000
X1 -- X2 >0
X1, X2 >0
C.R.Krishna Prasad, BIT Bangalore-4
Finally we have,
MaxZ = 0.07X1 + 0.1X2
is subjected to three constraints
X1 + X2 < 30,000
X1 > 6,000
X2 < 12,000
X1 -- X2 >0
X1, X2 >0
C.R.Krishna Prasad, BIT Bangalore-4
Minimization problems
Example 6.
A person requires 10, 12, and 12 units chemicals A, B and C
respectively for his garden. A liquid product contains 5, 2 and 1
units of A,B and C respectively per jar. A dry product contains 1,2
and 4 units of A,B and C per carton.
If the liquid product sells for Rs.3 per jar and the dry product sells
for Rs.2 per carton, how many of each should be purchased, in
order to minimize the cost and meet the requirements?
C.R.Krishna Prasad, BIT Bangalore-4
i) Identify and define the decision variable of the problem
Let X1 and X2 be the number of units of liquid and dry products.
ii) Define the objective function
The cost of Liquid and Dry products are given ;
The objective function is to minimize the cost
Min. Z = 3X1 + 2X2
C.R.Krishna Prasad, BIT Bangalore-4
iii) State the constraints to which the objective function should be
optimized.
The above objective function is subjected to following three
constraints.
5X1 + X2 >10
2X1 + 2X2 >12
X1 + 4X2 >12
X1, X2 >0
C.R.Krishna Prasad, BIT Bangalore-4
Finally we have,
Min. Z = 3X1 + 2X2
is subjected to three constraints
5X1 + X2 >10
2X1 + 2X2 >12
X1 + 4X2 >12
X1, X2 >0
C.R.Krishna Prasad, BIT Bangalore-4
Example 7.
A Scrap metal dealer has received a bulk order from a customer for a
supply of atleast 2000 kg of scrap metal. The consumer has specified
that atleast 1000 kgs of the order must be high quality copper that can
be melted easily and can be used to produce tubes. Further, the
customer has specified that the order should not contain more than
200 kgs of scrap which are unfit for commercial purposes. The scrap
metal dealer purchases the scrap from two different sources in an
unlimited quantity with the following percentages (by weight) of high
quality of copper and unfit scrap.
C.R.Krishna Prasad, BIT Bangalore-4
Source A
Source B
Copper
40%
75%
Unfit Scrap
7.5%
10%
The cost of metal purchased from source A and source B are Rs.12.50
and Rs.14.50 per kg respectively. Determine the optimum quantities of
metal to be purchased from the two sources by the metal scrap dealer
so as to minimize the total cost (February 2002)
C.R.Krishna Prasad, BIT Bangalore-4
i) Identify and define the decision variable of the problem
Let X1 and X2 be the quantities of metal to be purchased from the two
sources A and B.
ii) Define the objective function
The cost of metal to be purchased by the metal scrap dealer are given;
the objective function is to minimize the cost
Min. Z = 12.5X1 + 14.5X2
C.R.Krishna Prasad, BIT Bangalore-4
iii) State the constraints to which the objective function should be
optimized.
The above objective function is subjected to following three constraints.
X1 + X2 >2,000
0.4X1 + 0.75X2 >1,000
0.075X1 + 0.1X2 + 4X3 < 200
X1, X2 >0
C.R.Krishna Prasad, BIT Bangalore-4
Finally we have,
Min. Z = 12.5X1 + 14.5X2
is subjected to three constraints
X1 + X2 >2,000
0.4X1 + 0.75X2 >1,000
0.075X1 + 0.1X2 + 4X3 < 200
X1, X2 >0
C.R.Krishna Prasad, BIT Bangalore-4
Example 8.
A farmer has a 100 acre farm. He can sell all tomatoes, lettuce or
radishes and can raise the price to obtain Rs.1.00 per kg. for
tomatoes , Rs.0.75 a head for lettuce and Rs.2.00 per kg for
radishes. The average yield per acre is 2000kg.of tomatoes, 3000
heads of lettuce and 1000 kgs of radishes. Fertilizers are available
at Rs.0.50 per kg and the amount required per acre is 100 kgs for
each tomatoes and lettuce and 50kgs for radishes. Labour
required for sowing, cultivating and harvesting per acre is 5 mandays for tomatoes and radishes and 6 man-days for lettuce. A total
of 400 man-days of labour are available at Rs.20.00 per man-day.
Formulate this problem as LP model to maximize the farmers profit.
C.R.Krishna Prasad, BIT Bangalore-4
i) Identify and define the decision variable of the problem
Let X1 and X2 and X3 be number acres the farmer grows
tomatoes, lettuce and radishes respectively.
ii) Define the objective function
The objective of the given problem is to maximize the profit.
The profit can be calculated by subtracting total expenditure
from the total sales
Profit = Total sales – Total expenditure
C.R.Krishna Prasad, BIT Bangalore-4
The farmer produces 2000X1 kgs of tomatoes, 3000X2
heads of lettuce, 1000X3 kgs of radishes.
Therefore the total sales of the farmer will be
= Rs. (1 x 2000X1 + 0.75 x 3000X2 + 2 x 100X3)
Total expenditure (fertilizer expenditure) will be
= Rs.20 ( 5X1 + 6X2 + 5X3 )
Farmer’s profit will be
Z = (1 x 2000X1 + 0.75 x 3000X2 + 2 x 100X3) –
{ [0.5 x 100 x X1+0.5 x 100 x X2 + 50xX3]+ [20 x 5 x X1+20
x 6 x X2 + 20 x 5 x X3]}
=1850X1 + 2080X2 + 1875X3
C.R.Krishna Prasad, BIT Bangalore-4
Therefore the objective function is
Maximise Z = 1850X1 + 2080X2 + 1875X3
iii) State the constraints to which the objective function should be
optimized.
The above objective function is subjected to following constraints.
Since the total area of the firm is 100 acres
X1 + X2 + X3 < 100
The total man-days labour is 400 man-days
5X1 + 6X2 + 5X3 < 400
C.R.Krishna Prasad, BIT Bangalore-4
Finally we have,
Maximise Z = 1850X1 + 2080X2 + 1875X3
is subjected to three constraints
X1 + X2 + X3 < 100
5X1 + 6X2 + 5X3 < 400
X1, X2 X3 >0
C.R.Krishna Prasad, BIT Bangalore-4
Assumptions of Linear Programming
Certainty.
In all LP models it is assumed that, all the model parameters such as
availability of resources, profit (or cost) contribution of a unit of decision
variable and consumption of resources by a unit of decision variable
must be known with certainty and constant.
Divisibility (Continuity)
The solution values of decision variables and resources are assumed to
have either whole numbers (integers) or mixed numbers (integer (digit)or
fractional (partial/part)). However, if only integer variables are desired,
then Integer programming method may be employed.
C.R.Krishna Prasad, BIT Bangalore-4
Additivity
The value of the objective function for the given value of decision
variables and the total sum of resources used, must be equal to the
sum of the contributions (Profit or Cost) earned from each decision
variable and sum of the resources used by each decision variable
respectively. /The objective function is the direct sum of the
individual contributions of the different variables
Linearity
All relationships in the LP model (i.e. in both objective function and
constraints) must be linear.
C.R.Krishna Prasad, BIT Bangalore-4
Problem : A company has four factories F1,F2,F3,F4 and
manufacturing the same product. Production and raw
material costs differ from factory to factory and are given in
the following table in the first two rows. The transportation
costs are from the factories to sales depots, S1,S2,S3,and
S4are also given. The last two columns in the table give the
sales price and the total requirement at each depot. The
production capacity of each factory is given in the last row.
C.R.Krishna Prasad, BIT Bangalore-4
F1
F2
F3
F4
Production cost/unit
15
18
14
14
Raw material cost/unit
10
9
12
9
C.R.Krishna Prasad, BIT Bangalore-4
Determine the most profitable production and
distribution schedule and the corresponding profit.
The surplus production should be taken to yield zero
profit.
Transp.cost
S1
S2
S3
3
9
5
4
Selling price. Requirement
80
34
1
7
4
5
32
120
5
8
3
6
31
150
10
150
50
100
C.R.Krishna Prasad, BIT Bangalore-4
Assignment Problems
Hungarian method
Step 1
Balance the problem if it is unbalanced
Place an M as the cost element if some
assignment is prohibited
Convert into equivalent min problem if it is a max
problem
In a given matrix subtract the smallest element in
each row from every element of that row and do the
same in the column.
C.R.Krishna Prasad, BIT Bangalore-4
Step2 In the reduced matrix obtain from step 1, subtract
the smallest element in each column from every element
of that column
Step 3 Make the assignment for the reduced matrix
obtained from step 1 and step 2
(all the zeros in rows/columns are either marked (□) or (x)
and there is exactly one assignment in each row and each
column. In such a case optimum assignment policy for the
given problem is obtained.
If there is row or column with out an assignment go to the
next step.
C.R.Krishna Prasad, BIT Bangalore-4
Step 4 Draw the minimum number of vertical and horizontal
lines necessary to cover all the zeros in the reduced matrix
obtained from step 3 by adopting the following procedure.
(i) mark(√) all rows that do not have assignments
(ii) Mark (√) all columns (not already marked) which have
zeros in the marked rows
(iii) Mark (√) all rows (not already marked) that have
assignments in marked columns
(iv) Draw straight lines through all unmarked rows and
marked columns
C.R.Krishna Prasad, BIT Bangalore-4
Step 5 If the number of lines drawn are equal to
the number of rows or columns, then it is an optimum
solution ,otherwise go to step 6
Step 6 Select the smallest element among all the
uncovered elements. Subtract this smallest element from all
the uncovered elements an add it to the element which
lies at the intersection of two line. Thus we obtain another
reduced matrix for fresh assignments.
Step 7 go the step 3 and repeat the procedure until the
umber of assignment become equal to the number of rows
or columns. In such a case, we shall observe that
row/column has an assignment. Thus, the current solution is
an optimum solution.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1
A company centre has got four expert
programmers. The centre needs four
application programmes to be developed. The
head of the computer centre, after studying
carefully the programme’s to be developed,
estimate the computer time in minutes required
by the respective experts to develop the
application programmes as follows.
C.R.Krishna Prasad, BIT Bangalore-4
Program
mers
1
A
B
C
D
120
100
80
90
2
80
90
110
70
3
110
140
120
100
4
90
90
80
90
C.R.Krishna Prasad, BIT Bangalore-4
Problem 2
Suggest optimum solution to the following assignment
problem and also the minimum cost:
C.R.Krishna Prasad, BIT Bangalore-4
Markets/s
alesmen
I
II
III
IV
A
44
80
52
60
B
60
56
40
72
C
36
60
48
48
D
52
76
36
40
C.R.Krishna Prasad, BIT Bangalore-4
C.R.Krishna Prasad, BIT Bangalore-4
Transportation
Transportation models deals with the transportation of a
product manufactured at different plants or factories
supply origins) to a number of different warehouses
(demand destinations). The objective to satisfy the
destination requirements within the plants capacity
constraints at the minimum transportation cost.
C.R.Krishna Prasad, BIT Bangalore-4
A typical transportation problem contains
Inputs:
Sources with availability
Destinations with requirements
Unit cost of transportation from various sources
to destinations
Objective:
To determine schedule of transportation to minimize
total transportation cost
C.R.Krishna Prasad, BIT Bangalore-4
How to solve?
1.
2.
3.
Define the objective function to be minimized
with the constraints imposed on the problem.
Set up a transportation table with m rows
representing the sources and n columns
representing the destination
Develop an initial feasible solution to the
problem by any of these methods a) The
North west corner rule b) Lowest cost entry
method c)Vogel’s approximation method
C.R.Krishna Prasad, BIT Bangalore-4
4. Examine whether the initial solution is feasible or not.( the
solution is said to be feasible if the solution has allocations
in ( m+n-1) cells with independent positions.
5. Test wither the solution obtained in the above step is
optimum or not using a) Stepping stone method b)
Modified distribution (MODI) method.
6.If the solution is not optimum ,modify the shipping schedule.
Repeat the above until an optimum solution is obtained.
C.R.Krishna Prasad, BIT Bangalore-4
Applications
To minimize shipping costs from factories
to warehouses or from warehouses to
retails outlets.
To determine lowest cost location of a
new factor, warehouse or sales office
To determine minimum cost production
schedule that satisfies firm’s demand and
production limitations.
C.R.Krishna Prasad, BIT Bangalore-4
North West corner method
1.
2.
3.
4.
Select the northwest corner cell of the transportation
table and allocate as many units as possible equal to the
minimum between availability supply and demand
requirements i.e.(min (s1,d1)
Adjust the supply and demand numbers in the respective
rows and columns allocation
A. If the supply for the first row is exhausted ,then move
down to the first cell in the second row and first column
and go to step 2.
If the demand for the first column is satisfied, then move
horizontally to the next cell in the second column and first
row and go to step 2
C.R.Krishna Prasad, BIT Bangalore-4
5. If for any cell, supply equals demand, then the next
allocation can be made in cell either in the next row or
column.
6. Continue the procedure until the total available quantity is
fully allocated to the cells as required.
Advantages; it is simple and reliable. Easy to compute
,understand and interpret.
Disadvantages: This method does not take into considerations
the shipping cost, consequently the initial solution obtained
b this method require improvement.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1: Obtain initial solution in the following transportation
problem by using Northwest corner rule method
Origins
Supply/c
apacity/a
vailability
D1
D2
D3
D4
O1
O2
1
3
2
3
1
2
4
1
30
50
O3
4
20
2
40
5
30
9
10
20
Demand/R
equirement
s
C.R.Krishna Prasad, BIT Bangalore-4
Least cost method:
1.
2.
3.
Select the cell with the lowest transportation cost among
all the rows or column of the transportation table
If the minimum cost is not unique, then select arbitrarily
any cell with this minimum cost. (preferably where
maximum allocation is possible)
Repeat steps 1 and 2 for the reduced table until the
entire supply at different factories is exhausted to satisfy
the demand at different warehouses.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1: Obtain initial solution in the following transportation
problem by using Least cost method
Origins
Supply/c
apacity/a
vailability
D1
D2
D3
D4
O1
O2
1
3
2
3
1
2
4
1
30
50
O3
4
20
2
40
5
30
9
10
20
Demand/R
equirement
s
C.R.Krishna Prasad, BIT Bangalore-4
1.
2.
Vegel’s Approximation Method (VAM)
Compute a penalty for each row and column in the
transportation table. The penalty for a given row and
column is merely the difference between the smallest
cost and next smallest cost in that particular row or
column.
Identify the row or column with the largest penalty. In this
identified row or column, choose the cell which has the
smallest cost and allocate the maximum possible
quantity to the lowest cost cell in that row or column so
as to exhaust either the supply at a particular source or
satisfy demand at warehouse.( If a tie occurs in the
penalties, select that row/column which has minimum
cost. If there is a tie in the minimum cost also, select the
row/column which will have maximum possible
assignments)
C.R.Krishna Prasad, BIT Bangalore-4
3.Reduce the row supply or the column demanded by the
assigned to the cell
4.If the row supply is now zero, eliminate the row, if the column
demand is now zero, eliminate the column, if both the
row, supply and the column demand are zero, eliminate
both the row and column.
5. Recompute the row and column difference for the
reduced transportation table, omitting rows or columns
crossed out in the preceding step.
6. Repeat the above procedure until the entire supply at
factories are exhausted to satisfy demand at different
warehouses.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1: Obtain initial solution in the following transportation
problem by using VAM
Origins
Supply/c
apacity/a
vailability
D1
D2
D3
D4
O1
O2
1
3
2
3
1
2
4
1
30
50
O3
4
20
2
40
5
30
9
10
20
Demand/R
equirement
s
C.R.Krishna Prasad, BIT Bangalore-4
1.
2.
3.
The Modified Distribution Method
Determine an initial basic feasible solution consisting of m
+ n -1 allocations in independent positions using any of
the three methods
Determine a set of number for each row and each
column. Calculate Ui ( i= 1,2,..m)and Vj (j = 1,2..n)for
each column, and Cij = (Ui + Vj ) for occupied cells.
Compute the opportunity cost
Δij = Cij - (Ui + Vj ) for each unoccupied cells.
C.R.Krishna Prasad, BIT Bangalore-4
4. Check the sign of each opportunity cost: if all the
Δij are positive or zero, the given solution is optimum. If one of
the values is zero there is another alterative solution for the
same transportation cost. If any value is negative the given
solution is not optimum. Further improvement is possible.
5. Select the unoccupied cell with the largest negative
opportunity cost as the cell to be included in the next
solution.
6. Draw a closed path or loop for the unoccupied cell
selected in step 5.It may be noted that right angle turns in
this path are permitted only a occupied cells and at the
original unoccupied cell
C.R.Krishna Prasad, BIT Bangalore-4
7. Assign alternative plus and minus signs at the unoccupied
cells on the corner points of the closed path with a plus sign
at the cell being evaluated.
8.Determine the maximum number of units that should be
shipped to this unoccupied cell. The smallest one with a
negative position on the closed path indicates the number
of units that can be shipped to the entering cell. This
quantity is added to all the cells on the path marked with
plus sign and subtract from those cells mark with minus
sign. In this way the unoccupied cell under consideration
becomes an occupied cell making one of the occupied
cells as unoccupied cell.
9.Repeat the whole procedure until an optimum solution is
attained i.e. Δij is positive or zero. Finally calculate new
transportation cost.
C.R.Krishna Prasad, BIT Bangalore-4
Problem: 4 A distribution system has the following
constraints.
Factory
capacity (in units)
A
45
B
15
C
40
Warehouse
Demand (in units)
I
25
II
55
III
20
The transportation costs per unit( in rupees) allocated with
each route are as follows.
C.R.Krishna Prasad, BIT Bangalore-4
To/Fro
m
I
II
III
A
10
7
8
B
15
12
9
C
7
8
12
C.R.Krishna Prasad, BIT Bangalore-4
Find the optimum transportation schedule and the
minimum total cost of transportation.
Problem 3
A company is spending Rs.1,000 on transportation of its
units from these plants to four distribution centres. The
supply and demand of units, with unity cost of
transporataion are given in the table.
What can be the maximum saving by optimum scheduling.
C.R.Krishna Prasad, BIT Bangalore-4
D1
D2
D3
D4
Avail
able
P1
19
30
50
12
7
P2
70
30
40
60
10
P3
40
10
60
20
18
Requir
emetn
s
5
8
7
15
To/fro
m
C.R.Krishna Prasad, BIT Bangalore-4
Special cases in Transportation
Unbalanced transportation
Maximisation
Restricted routes
C.R.Krishna Prasad, BIT Bangalore-4
A product is produced by 4 factories F1,
F2,F3 and F4. Their unit production cost
are Rs.2,3,1,and 5 only. Production
capacity of the factories are 50,70,40
and 50 units respectively. The product is
supplied to 4 stores S1,S2,S3 and S4., the
requirements of which are 25,35,105 and
20 respectively. Unit cost of transportation
are given below
C.R.Krishna Prasad, BIT Bangalore-4
S1
S2
S3
S4
F1
2
4
6
11
F2
10
8
7
5
F3
13
3
9
12
F4
4
6
8
3
C.R.Krishna Prasad, BIT Bangalore-4
Find the optimal
transportation plan such
that total production and
transportation cost is
minimum.
PROBLEM
A particular product is
manufactured in factories
A,B ,C and D: it is sold at
centres 1,2,and 3. the
cost in rupees of product
per unit and capacity of
each plant is given below
Factory
Cost Rs
Per unit
Capacity
A
12
100
B
15
20
C
11
60
D
13
80
C.R.Krishna Prasad, BIT Bangalore-4
The sales prices is Rs
per unit and the
demand are as
follows.
Find the optimal
solution
Sales
centers
Sales
Demand
price per
unit
1
15
120
2
14
90
3
16
50
C.R.Krishna Prasad, BIT Bangalore-4
Assignment Problems
Hungarian method
Step 1
Balance the problem if it is unbalanced
Place an M as the cost element if some
assignment is prohibited
Convert into equivalent min problem if it is a max
problem
In a given matrix subtract the smallest element in
each row from every element of that row and do the
same in the column.
C.R.Krishna Prasad, BIT Bangalore-4
Step2 In the reduced matrix obtain from step 1, subtract
the smallest element in each column from every element
of that column
Step 3 Make the assignment for the reduced matrix
obtained from step 1 and step 2
(all the zeros in rows/columns are either marked (□) or (x)
and there is exactly one assignment in each row and each
column. In such a case optimum assignment policy for the
given problem is obtained.
If there is row or column with out an assignment go to the
next step.
C.R.Krishna Prasad, BIT Bangalore-4
Step 4 Draw the minimum number of vertical and horizontal
lines necessary to cover all the zeros in the reduced matrix
obtained from step 3 by adopting the following procedure.
(i) mark(√) all rows that do not have assignments
(ii) Mark (√) all columns (not already marked) which have
zeros in the marked rows
(iii) Mark (√) all rows (not already marked) that have
assignments in marked columns
(iv) Draw straight lines through all unmarked rows and
marked columns
C.R.Krishna Prasad, BIT Bangalore-4
Step 5 If the number of lines drawn are equal to
the number of rows or columns, then it is an optimum
solution ,otherwise go to step 6
Step 6 Select the smallest element among all the
uncovered elements. Subtract this smallest element from all
the uncovered elements an add it to the element which
lies at the intersection of two line. Thus we obtain another
reduced matrix for fresh assignments.
Step 7 go the step 3 and repeat the procedure until the
umber of assignment become equal to the number of rows
or columns. In such a case, we shall observe that
row/column has an assignment. Thus, the current solution is
an optimum solution.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1
A company centre has got four expert
programmers. The centre needs four
application programmes to be developed. The
head of the computer centre, after studying
carefully the programme’s to be developed,
estimate the computer time in minutes required
by the respective experts to develop the
application programmes as follows.
C.R.Krishna Prasad, BIT Bangalore-4
Program
mers
1
A
B
C
D
120
100
80
90
2
80
90
110
70
3
110
140
120
100
4
90
90
80
90
C.R.Krishna Prasad, BIT Bangalore-4
Problem 2Suggest optimum solution to the following
assignment problem and also the minimum cost:
C.R.Krishna Prasad, BIT Bangalore-4
Markets/s
alesmen
I
II
III
IV
A
44
80
52
60
B
60
56
40
72
C
36
60
48
48
D
52
76
36
40
C.R.Krishna Prasad, BIT Bangalore-4
Problem: 3
A company has taken the third floor of a multi-storied
building for rent with a view to locate one of their zonal
offices. There are five main rooms in this to be assigned
to five managers. Each room has it sown advantages
and disadvantages. Some have windows, some are
closer to the wash rooms or to the canteen or secretarial
pool. The rooms are of all different sizes and shapes.
Each of the five managers were asked to rank their room
preference amongst the rooms 301,302,303,304 and 305.
Their preferences were recorded in a table as indicated
below
C.R.Krishna Prasad, BIT Bangalore-4
M1
M2
M3
M4
M5
302
302
303
302
301
303
304
301
305
302
304
305
304
304
304
301
305
302
303
Most of the managers did not list al the five rooms since they
were not satisfied with some of these rooms and they have
left these from the list. Assuming that their preferences can
be quantified by numbers, find out as to which manager
should be assigned to which room so that their total
preference ranking is a minimum.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 4 A company plans to assign 5
salesmen to 5 districts in which it
operates. Estimates of sales revenue in
thousands of rupees for each salesman in
different districts are given in the
following table. In your opinion, what
should be the placement of the salesmen
if the objective is to maximize the
expected sales revenue?.
C.R.Krishna Prasad, BIT Bangalore-4
Expected sales data district wise
Sales
men
D1
D2
D3
D4
D5
S1
40
46
48
36
48
S2
48
32
36
29
44
S3
49
35
41
38
45
S4
30
46
49
44
44
S5
37
41
48
43
47
C.R.Krishna Prasad, BIT Bangalore-4
Problem5 : A traveling salesman has to
visit 5 cities. He wishes to start from a
particular city, visit each city once and
then return to his starting point. The
traveling cost for each city from a
particular city is given below
C.R.Krishna Prasad, BIT Bangalore-4
From
/to
A
A
B
C
D
E
X
4
7
3
4
B
4
X
6
3
4
C
7
6
X
7
5
D
3
3
7
X
7
E
4
4
5
7
X
What is the sequence of visit of the salesman,
so that the cost is minimum.
C.R.Krishna Prasad, BIT Bangalore-4
Problem6. A solicitors firm employs typists on
hourly piece rate basis for their work . There
are five typists for service and their charges
and speeds are different. According to an
earlier understanding only one job is given to
one typist and the typist is paid for full hour
even if he works for a fraction of an hour. Find
the least cost allocation for the following data.
C.R.Krishna Prasad, BIT Bangalore-4
typist
Rate per No.of
hour in
pages
Rs.
typed
per hour
A
5
12
B
6
14
C
3
8
D
4
10
E
4
11
C.R.Krishna Prasad, BIT Bangalore-4
JOB
No. of
pages
P
199
Q
175
R
145
S
198
T
178
Problem 7ABC airline, operating 7 days a
week, has given the following time-table. The
crews must have a minimum lay-over of 5
hours between flight. Obtain the pairing of
flights that minimizes la-overtime away from
home . For any given pairing the crew will be
based at the city that results the smallest layover.
C.R.Krishna Prasad, BIT Bangalore-4
Hyderabad-Delhi
Flight
no.
Departur
e
Arrival
Delhi -Hyderabad
Flight no. Departur
e
Arrival
A1
6 AM
8 AM
B1
8 AM
10 AM
A2
8 AM
10 AM
B2
9 AM
11AM
A3
2 PM
4 PM
B3
2 PM
4 PM
A4
8 PM
10 PM
B4
7 PM
9 PM
C.R.Krishna Prasad, BIT Bangalore-4
Problem 8.
A company has four territories and four salesmen available
for assignment. The territories are not equally rich in their
sales potential. It is estimated that a typical salesman
operating I each territory would bring the following weekly
sales.
Territory
I
II
Annual sales
(in Rs)
60,000
50,000
C.R.Krishna Prasad, BIT Bangalore-4
III
40,000
IV
30,000
The four salesmen are also considered to
differ in ability. It is estimated that working
under the same conditions their yearly sales
would be proportionately as follows
Salesman
A
B
C
D
Proportion
7
5
5
4
C.R.Krishna Prasad, BIT Bangalore-4
If the criteria is to maximize expected sales,
the intuitive answer is to assign the best
salesman to the richest territory, the next best
salesman to the second richest and so on.
Verify this answer by the assignment
technique.
C.R.Krishna Prasad, BIT Bangalore-4
Thank you
Give feedback to
krishmandya@rediffmail.com
C.R.Krishna Prasad, BIT Bangalore-4
Network Analysis
•Network is a graphical representation of all the
Activities and Events arranged in a logical and
sequential order.
•Network analysis plays an important role in project
management.
•A project is a combination of interrelated activities
all of which must be executed in a certain order for its
completion.
C.R.Krishna Prasad, BIT Bangalore-4
Activity: Activity is the actual performance of the
job. This consumes resources (Time, human
resources, money, and material)
Event: An event refers to start or completion of a
job. This does not consume any resources.
C.R.Krishna Prasad, BIT Bangalore-4
•Analyzing network, the planning, scheduling and
control of a project becomes easier.
•PERT and CPM are the two most popular network
analysis technique used to assist managers in
planning and controlling large scale projects.
•PERT- (Programme Evaluation Review Technique)
•CPM - (Critical Path Method)
C.R.Krishna Prasad, BIT Bangalore-4
Applications: Construction of a Residential complex,
Commercial complex,
Petro-chemical complex
Ship building
Satellite mission development
Installation of a pipe line project etc...
C.R.Krishna Prasad, BIT Bangalore-4
Historical Evolution.
Before 1957 there was no generally accepted procedure
to aid the management of a project. In 1958 PERT was
developed by team of engineers working on a Polaris
Missile programmer of the navy. This was a large
project involved 250 prime contractors and about 9000
job contractors. It had about 19 million components. In
such projects it is possible that a delay in the delivery
of a small component might hold the progress of entire
project. PERT was used successfully and the total time
of completion was reduced from 7 years to 5 years.
C.R.Krishna Prasad, BIT Bangalore-4
In 1958 Du Pont Company used a technique called
Critical Path Method (CPM) to schedule and control a
very large project like overhauling of a chemical plant,
there by reducing the shutdown period from 130hrs to
90 hrs saving the company 1 million dollar.
Both of these techniques are referred to as project
scheduling techniques.
C.R.Krishna Prasad, BIT Bangalore-4
PERT
CPM
1. It is a technique for
planning scheduling &
controlling of projects
whose activities are
subject to uncertainty in
the performance time.
Hence it is a probabilistic
model
1. It is a technique for
planning scheduling &
controlling of projects
whose activities not
subjected to any
uncertainty and the
performance times are
fixed. Hence it is a
deterministic model
2.It is an Event oriented
system
2. It is an Activity oriented
system
C.R.Krishna Prasad, BIT Bangalore-4
3.Basically does not
differentiate critical and
non-critical activities
Differentiates clearly the
critical activities from the
other activities.
4. Used in projects where
resources (men, materials,
money) are always
available when required.
4. Used in projects where
overall costs is of
primarily important.
Therefore better utilized
resources
5. Suitable for Research
5.Suitable for civil
and Development projects constructions, installation,
where times cannot be
ship building etc.
predicted
C.R.Krishna Prasad, BIT Bangalore-4
Rules for drawing the network diagrams.
1
2
3
4
•In a network diagram, arrows represent the
activities and circles represent the events.
•The tail of an arrow represents the start of an
activity and the head represent the completion
of the activity.
C.R.Krishna Prasad, BIT Bangalore-4
1
2
3
4
•The event numbered 1 denotes the start of the
project and is called initial event.
• Event carrying the highest number in the network
denotes the completion of the project and is called
terminal event.
C.R.Krishna Prasad, BIT Bangalore-4
1
3
2
4
•Each defined activity is represented by one and only
arrow in the network.
•Determine which operation must be completed
immediately before other can start.
•Determine which other operation must follow the
other given operation.
C.R.Krishna Prasad, BIT Bangalore-4
1
2
3
4
•The network should be developed on the basis of
logical, analytical and technical dependencies
between various activities of the project.
C.R.Krishna Prasad, BIT Bangalore-4
The basic network construction – Terminology used.
Network representation: There are two types of
systems –
AOA system
(Activity
on Arrow system)
AON system
(Activity on Node system
)
In this activities are
In this method activities
represented by an arrows. are represented in the
circles.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1.
Construct an arrow diagram for the following project.
Activities
Relationship
A
Precedes B,C
B
Precedes D,E
C
Precedes D
D
Precedes F
E
Precedes G
F
Precedes G
C.R.Krishna Prasad, BIT Bangalore-4
E
B
G
A
F
C
D
C.R.Krishna Prasad, BIT Bangalore-4
Problem 2.
Construct an arrow diagram for the following project.
Job
Immediate
predecessor
Duration
A
-
14 Days
B
A
3 Days
C
A
7 Days
D
C
4 Days
E
B,D
10 Days
C.R.Krishna Prasad, BIT Bangalore-4
E
10
B
14
D
3
A
4
C
7
Key
Job
Duration
C.R.Krishna Prasad, BIT Bangalore-4
Problem 3.
Construct an arrow diagram for the following project.
Job
Immediate predecessor
A
-
B
-
C
A,B
D
A
E
D
F
C,E
C.R.Krishna Prasad, BIT Bangalore-4
C
B
F
A
E
D
C.R.Krishna Prasad, BIT Bangalore-4
Problem 4.
Construct an arrow diagram for the following project.
Activity
Predecessor
A
-
B
-
C
-
D
A,B
E
B,C
C.R.Krishna Prasad, BIT Bangalore-4
D
A
B
C
E
C.R.Krishna Prasad, BIT Bangalore-4
Problem 5.
Construct an arrow diagram for the following project.
Activity
Predecessor
A
-
B
-
C
-
D
A,B
E
B,C
F
A,B,C
C.R.Krishna Prasad, BIT Bangalore-4
D
A
B
F
C
E
C.R.Krishna Prasad, BIT Bangalore-4
Problem 6.
Draw the PERT network for the following project
Event A is followed by events B & C
Event D is preceded by events B & C
Event H is the successor to event E
Event E is the successor to event B
Event F is the successor to event D & G
Event C is the predecessor to event G
Event J is preceded by events F,G, & H
C.R.Krishna Prasad, BIT Bangalore-4
E
B
H
A
J
D
F
C
G
C.R.Krishna Prasad, BIT Bangalore-4
Network Analysis
•Network is a graphical representation of all the
Activities and Events arranged in a logical and
sequential order.
•Network analysis plays an important role in project
management.
•A project is a combination of interrelated activities
all of which must be executed in a certain order for its
completion.
C.R.Krishna Prasad, BIT Bangalore-4
Project situation (An example)
A new machine is required by a department for which
budget approval is needed. The new machine
necessitates employment of an operator who would be
trained for operating this machine. The operator can be
hired as soon as the proposal of buying machine is
cleared, and trained on the same machine in the
training division of the company. Once the machine is
installed and worker is trained , the trail production can
commence.
C.R.Krishna Prasad, BIT Bangalore-4
Problem Presentation
Activity
Description
Duration (weeks)
Immediate
Predecessor(s)
A
Obtain the budget
approval
2
-
B
Obtain the machine
5
A
C
Hire the operator
1
A
D
Install the machine
1
B
E
Train the operator
6
C
F
Produce the goods
1
D,E
C.R.Krishna Prasad, BIT Bangalore-4
Obtain the
machine
Obtain the
budget approval
Install the
machine
3
D
B
1
A
F
2
5
C
Hire the
operator
Produce the
goods
6
E
4
C.R.Krishna Prasad, BIT Bangalore-4
Train the
operator
Obtain the
budget approval
1
A
2
C.R.Krishna Prasad, BIT Bangalore-4
Obtain the
machine
Obtain the
budget approval
3
B
1
A
2
C.R.Krishna Prasad, BIT Bangalore-4
C.R.Krishna Prasad, BIT Bangalore-4
Obtain the
machine
Obtain the
budget approval
3
B
1
A
2
C
Hire the
operator
4
C.R.Krishna Prasad, BIT Bangalore-4
Obtain the
machine
Obtain the
budget approval
Install the
machine
3
D
B
1
A
2
5
C
Hire the
operator
4
C.R.Krishna Prasad, BIT Bangalore-4
Obtain the
machine
Obtain the
budget approval
Install the
machine
3
D
B
1
A
2
5
C
Hire the
operator
E
4
C.R.Krishna Prasad, BIT Bangalore-4
Train the
operator
Obtain the
machine
Obtain the
budget approval
Install the
machine
3
D
B
1
A
F
2
5
C
Hire the
operator
Produce the
goods
6
E
4
C.R.Krishna Prasad, BIT Bangalore-4
Train the
operator
Network Analysis
•Network is a graphical representation of all the
Activities and Events arranged in a logical and
sequential order.
•Network analysis plays an important role in project
management.
•A project is a combination of interrelated activities
all of which must be executed in a certain order for its
completion.
C.R.Krishna Prasad, BIT Bangalore-4
•The network should be developed on the
basis of logical, analytical and technical
dependencies between various activities
of the project.
C.R.Krishna Prasad, BIT Bangalore-4
CRITICAL PATH
Meaning: The longest path in a project network
which determine the duration of the project is
known as critical path.
C.R.Krishna Prasad, BIT Bangalore-4
Determination of Critical Path
3
2
2
5
4
1
8
5
4
6
5
3
1
6
2
2
3
4
7
Step 1.List all the possible sequences from start to finish
Step 2.For each sequence determine the total time required from start to
finish.
Step 3.Identify the longest path (Critical Path)
C.R.Krishna Prasad, BIT Bangalore-4
Step 1. List all the possible sequences from start to finish
Path A : 1 – 2 – 5 – 8
Path B : 1 – 3 – 5 – 8
Path C : 1 – 3 – 6 – 7 – 8
Path D : 1 – 3 – 4 – 7 – 8
Path E : 1 – 3 – 4 – 6 – 7 – 8
Step 2.For each sequence determine the total time required from
start to finish.
Path A : 2 + 3 + 4 = 9 days
Path B : 4 + 5 + 4 = 13 days
Path C : 4 + 5 + 6 + 1 = 16 days
Path D : 4 + 2 + 3 + 1 = 10 days
Path E : 4 + 2 + 2 + 6 + 1 = 10 days
C.R.Krishna Prasad, BIT Bangalore-4
Step 2.For each sequence determine the total time
required from start to finish.
Path A : 2 + 3 + 4 = 9 days
Path B : 4 + 5 + 4 = 13 days
Path C : 4 + 5 + 6 + 1 = 16 days
Path D : 4 + 2 + 3 + 1 = 10 days
Path E : 4 + 2 + 2 + 6 + 1 = 10 days
Step 3.Identify the longest path (Critical Path)
Path C : 4 + 5 + 6 + 1 = 16 days
Path C : 1 – 3 – 6 – 7 – 8
C.R.Krishna Prasad, BIT Bangalore-4
Determination of Critical Path
3
2
2
5
4
1
8
5
4
6
5
3
1
6
2
2
3
4
7
Step 1.List all the possible sequences from start to finish
Step 2.For each sequence determine the total time required from start to
finish.
Step 3.Identify the longest path (Critical Path)
C.R.Krishna Prasad, BIT Bangalore-4
Float (Slack)
•Float (Slack ) refers to the amount of time by which
a particular event or an activity can be delayed
without affecting the time schedule of the network.
•Float (Slack)
Float (Slack) is defined as the difference between
latest allowable and the earliest expected time.
Event Float/Slack = LS – ES
Where LS = Latest start time
ES = Early start time.
C.R.Krishna Prasad, BIT Bangalore-4
Earliest start : Denoted as ‘ES’
Earliest start time is the earliest possible time by
which the activity can be started.
Early finish time : Denoted as ‘EF’
Early finish time is the earliest possible time by
which the activity can be completed.
C.R.Krishna Prasad, BIT Bangalore-4
Latest start time : Denoted as ‘LS’
Latest start time is the latest possible time by
which the activity can be started
Late finish time : Denoted as ‘LS’
Late finish time is the latest possible time by which
the activity can be completed.
C.R.Krishna Prasad, BIT Bangalore-4
Total float (TF) / Total slack (TS)
Total float of the job is the differences between its
Late start and Early start ‘or’ Late finish and Early
finish
i.e.
TF( CA) = LS (CA) - ES (CA)
Or
TF( CA) = LF (CA) - EF (CA)
CA = Current activity
C.R.Krishna Prasad, BIT Bangalore-4
Free float (FF)
Free float is the amount of time a job can be
delayed without affecting the Early start time of any
other job.
FF(CA) = ES(SA) – EF (CA)
CA = Current Activity
SA = Succeeding Activity
C.R.Krishna Prasad, BIT Bangalore-4
Independent Float (IF)
Independent Float is the amount of time that can
be delayed without affecting either predecessor
or successor activities.
IF = ES(SA) – LF(PA)
- Duration of CA
ES = Early Start
LF = Late Finish
SA = Succeeding Activity
PA = Preceding Activity
CA = Current Activity
C.R.Krishna Prasad, BIT Bangalore-4
Example:
Construct the Network for the following
Project and determine the following
i) Critical Path
ii) ES,EF,LS,LF
iii) TF,FF
C.R.Krishna Prasad, BIT Bangalore-4
Activity
Duration
1-2
14
1-4
3
2-3
7
2-4
0
3-5
4
4-5
3
5-6
10
C.R.Krishna Prasad, BIT Bangalore-4
Construction of the Network and Determination Critical Path
4
F
5
1
G 0
D
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F
5
1
G 0
D
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F
5
1
G 0
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F
5
1
G 0
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F(25,35)
5
1
G 0
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F(25,35)
5
1
G 0
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F(25,35)
5
0
1
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F(25,35)
5
0
1
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7(14,21)
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7(14,21)
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7(14,21)
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7(14,21)
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
10(25,35)
1
D(14,21)
2
7(14,21)
3
Key
Job (ES,EF)
Duration (LS,LF)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)
2
7(14,21)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)
2
7(14,21)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)
2
7(14,21)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)(0)
2
7(14,21)(0)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)(0)
2
7(14,21)(0)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)(0)
5
10(25,35)(0)
1
D(14,21)(0)
2
7(14,21)(0)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Activity Duration
ES
EF
LS
LF
TF
FF
1-2
14
0
14
0
14
0
0
1-4
3
0
3
19
22
19
11
2-3
7
14
21
14
21
0
0
2-4
0
14
14
22
22
0
0
3-5
4
21
25
21
25
0
0
4-5
3
14
17
22
25
8
8
5-6
10
25
35
25
35
0
0
C.R.Krishna Prasad, BIT Bangalore-4
Example:
Construct the Network for the following
Project and determine the following
i) Critical Path
ii) ES,EF,LS,LF
iii) TF,FF
C.R.Krishna Prasad, BIT Bangalore-4
Activity
Duration
1-2
2
2-3
3
2-4
5
3-5
4
3-6
1
4-6
6
4-7
2
5-8
8
6-8
7
7-8
4
C.R.Krishna Prasad, BIT Bangalore-4
5
3
8
(0, 2)(0)
1
2
6
2(0, 2)(0)
(7, 9)(0)
4
Key
7
2(14,16)(7)
(ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
Activity
Duration
ES
EF
LS
LF
TF
FF
1-2
2
0
2
0
2
0
0
2-3
3
2
5
2
8
3
0
2-4
5
2
7
5
7
0
0
3-5
4
5
9
8
12
3
0
3-6
1
5
6
12
13
7
7
4-6
6
7
13
7
13
0
0
4-7
2
7
9
14
16
7
0
5-8
8
9
17
12
20
3
0
6-8
7
13
20
13
20
0
0
7-8
4
9
13
16
20
7
0
C.R.Krishna Prasad, BIT Bangalore-4
PERT Model
Historical Evolution.
Before 1957 there was no generally accepted procedure
to aid the management of a project. In 1958 PERT was
developed by team of engineers working on a Polaris
Missile programme of the navy. This was a large
project involved 250 prime contractors and about 9000
job contractors. It had about 19 million components. In
such projects it is possible that a delay in the delivery
of a small component might hold the progress of entire
project. PERT was used successfully and the total time
of completion was reduced from 7 years to 5 years.
C.R.Krishna Prasad, BIT Bangalore-4
Differences between PERT & CPM
PERT
CPM
1. It is a technique for
planning scheduling &
controlling of projects
whose activities are
subject to uncertainty in
the performance time.
Hence it is a probabilistic
model
1. It is a technique for
planning scheduling &
controlling of projects
whose activities not
subjected to any
uncertainty and the
performance times are
fixed. Hence it is a
deterministic model
2.It is an Event oriented
system
2. It is an Activity oriented
system
C.R.Krishna Prasad, BIT Bangalore-4
3.Basically does not
differentiate critical and
non-critical activities
Differentiates clearly the
critical activities from the
other activities.
4. Used in projects where
resources (men, materials,
money) are always
available when required.
4. Used in projects where
overall costs is of
primarily important.
Therefore better utilized
resources
5. Suitable for Research
5.Suitable for civil
and Development projects constructions, installation,
where times cannot be
ship building etc.
predicted
C.R.Krishna Prasad, BIT Bangalore-4
Thank you
Give feedback to
krishmandya@rediffmail.com
C.R.Krishna Prasad, BIT Bangalore-4
Transportation
Transportation models deals with the transportation of a
product manufactured at different plants or factories
supply origins) to a number of different warehouses
(demand destinations). The objective to satisfy the
destination requirements within the plants capacity
constraints at the minimum transportation cost.
C.R.Krishna Prasad, BIT Bangalore-4
A typical transportation problem contains
Inputs:
Sources with availability
Destinations with requirements
Unit cost of transportation from various sources
to destinations
Objective:
To determine schedule of transportation to minimize
total transportation cost
C.R.Krishna Prasad, BIT Bangalore-4
How to solve?
1.
2.
3.
Define the objective function to be minimized
with the constraints imposed on the problem.
Set up a transportation table with m rows
representing the sources and n columns
representing the destination
Develop an initial feasible solution to the
problem by any of these methods a) The
North west corner rule b) Lowest cost entry
method c)Vogel’s approximation method
C.R.Krishna Prasad, BIT Bangalore-4
4. Examine whether the initial solution is feasible or not.( the
solution is said to be feasible if the solution has allocations
in ( m+n-1) cells with independent positions.
5. Test wither the solution obtained in the above step is
optimum or not using a) Stepping stone method b)
Modified distribution (MODI) method.
6.If the solution is not optimum ,modify the shipping schedule.
Repeat the above until an optimum solution is obtained.
C.R.Krishna Prasad, BIT Bangalore-4
Applications
To minimize shipping costs from factories
to warehouses or from warehouses to
retails outlets.
To determine lowest cot location of a
new factor, warehouse or sales office
To determine minimum cost production
schedule that satisfies firm’s demand and
production limitations.
C.R.Krishna Prasad, BIT Bangalore-4
North West corner method
1.
2.
3.
4.
Select the northwest corner cell of the transportation
table and allocate as many units as possible equal to the
minimum between availability supply and demand
requirements i.e.(min (s1,d1)
Adjust the supply and demand numbers in the respective
rows and columns allocation
A. If the supply for the first row is exhausted ,then move
down to the first cell in the second row and first column
and go to step 2.
If the demand for the first column is satisfied, then move
horizontally to the next cell in the second column and first
row and go to step 2
C.R.Krishna Prasad, BIT Bangalore-4
5. If for any cell, supply equals demand, then the next
allocation can be made in cell either in the next row or
column.
6. Continue the procedure until the total available quantity is
fully allocated to the ells as required.
Advantages; it is simple and reliable. Easy to compute
,understand and interpret.
Disadvantages: This method does not take into considerations
the shipping cost, consequently the initial solution obtained
b this method require improvement.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1: Obtain initial solution in the following transportation
problem by using Northwest corner rule method
Origins
Supply/c
apacity/a
vailability
D1
D2
D3
D4
O1
O2
1
3
2
3
1
2
4
1
30
50
O3
4
20
2
40
5
30
9
10
20
Demand/R
equirement
s
C.R.Krishna Prasad, BIT Bangalore-4
Least cost method:
1.
2.
3.
Select the cell with the lowest transportation cost among
all the rows or column of the transportation table
If the minimum cost is not unique, then select arbitrarily
any cell with this minimum cost. (preferably where
maximum allocation is possible)
Repeat steps 1 and 2 for the reduced table until the
entire supply at different factories is exhausted to satisfy
the demand at different warehouses.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1: Obtain initial solution in the following transportation
problem by using Northwest corner rule method
Origins
Supply/c
apacity/a
vailability
D1
D2
D3
D4
O1
O2
1
3
2
3
1
2
4
1
30
50
O3
4
20
2
40
5
30
9
10
20
Demand/R
equirement
s
C.R.Krishna Prasad, BIT Bangalore-4
1.
2.
Vegel’s Approximation Method (VAM)
Compute a penalty for each row and column in the
transportation table. The penalty for a given row and column is
merely the difference between the smallest cost and next
smallest cost in that particular row or column.
Identify the row or column with the largest penalty. In this
identified row or column, choose the cell which has the
smallest cost and allocate the maximum possible quantity to
the lowest cost cell in that row or column so as to exhaust either
the supply at a particular source or satisfy demand at
warehouse.( If a tie occurs in the penalties, select that
row/column which has minimum cost. If there is a tie in the
minimum cost also, select the row/column which will have
maximum possible assignments)
C.R.Krishna Prasad, BIT Bangalore-4
3.Reduce the row supply or the column demanded by the
assigned to the cell
4.If the row supply is now zero, eliminate the row, if the column
demand is now zero, eliminate the column, if both the
row, supply and the column demand are zero, eliminate
both the row and column.
5. Recompute the row and column difference for the
reduced transportation table, omitting rows or columns
crossed out in the preceding step.
6. Repeat the above procedure until the entire supply at
factories are exhausted to satisfy demand at different
warehouses.
C.R.Krishna Prasad, BIT Bangalore-4
Theory of games
Game theory may be defined as “ a body of
knowledge that deals with making decisions
when two or more intelligent and rational
opponents are involved under conditions of
conflict and competition.
The approach of the game theory is to seek to
determine a rival’s most profitable counterstrategy to one’s own ‘best moves to formulate
the appropriate defensive measures.
C.R.Krishna Prasad, BIT Bangalore-4
Basic terminology
Game:
Let us consider a two person coin tossing game
between two player X and Y . Each player tosses
an unbiased coin simultaneously. If the two
simultaneous tosses match i.e. the toss of each
player given either heard up or the tail up
simultaneous ( eg. Either (H,H ) or (T,T) the player
Y pays Rs.100 to player X. But if the toss of player
x gives Head up and toss of the player Y gives
tail up or vice versa, then the player X pays Rs.80
to Y. This can be put up in the tabular form as
follows
C.R.Krishna Prasad, BIT Bangalore-4
Y
X
X pays Rs.80 t0
Y pays Rs.100 to Y
X
X pays Rs.80 t0
Y
C.R.Krishna Prasad, BIT Bangalore-4
Y pays Rs.100 to
X
In this game of coin tossing, the
outcome of the toss of each player is
only two i.e. Hear or Tail, which are
called the moves or strategies of the
players. Here the strategies of both the
players are the same but they may be
different. The results or termination of
each play., is expressed in terms of X’ s
payoff with the help of payoff matrix as
shown in the above table.
C.R.Krishna Prasad, BIT Bangalore-4
A competitive situation is thus called a
‘game’ if it has the following properties.
The number of competitors called players is
finite
The players at rationally and intelligently
Each player has available to him a finite
number of choices or possible course of
actions called strategies. The number of
choices need not be the same for each
player.
C.R.Krishna Prasad, BIT Bangalore-4
All relevant information, i.e., the different strategies of each
player and the amount of gain or loss on an individual's
move( strategy) are known to each player in advance.
The players select their respective courses of action (
strategies) simultaneously.
The player make individual decisions without direct
communication
The maximizing player attempt to maximize his gains and the
minimizing player tries to minimize his losses
The pay-off is fixed and determined in advance.
C.R.Krishna Prasad, BIT Bangalore-4
Two persons zero sum game: a game
of two persons, in which the gains of
one player are the losses of the other
player is called a two person zero sum
game, i.e., in a two person zero sum
game, the algebraic sum of the gains
to both the players after a play is
bound to be zero. These are also
known as rectangular games.
C.R.Krishna Prasad, BIT Bangalore-4
Pay off matrix: A strategy is a course of action
taken by one of the participants in a game,
and the payoff is the result or outcome of the
strategy.
Maximin Principle: A player adopts a pessimistic
attitude and plays safe. i.e. his strategy is always
that which result in the best out of the worst
outcomes. HE decides to play that strategy
which corresponds to the maximum of the
minimum gains for different course of actions.
C.R.Krishna Prasad, BIT Bangalore-4
Minimax Principle: The minimizing player
would also like to play safe and he selects
the strategy which corresponds to the
minimum of the maximum losses for his
different course of action.
Optimal strategy: A course of action or play
which puts the player in the most preferred
position, irrespecctive of the strategy of his
competitors is called an optimal strategy.
C.R.Krishna Prasad, BIT Bangalore-4
Value of the game: It is the expected pay-off of play
when all the players of the game follow their optimal
strategies. The game is called fair if the value of the
game is zero and unfair if it is non-zero.
Solutions methods of pure strategy games
(with saddle point)
in case of pure strategy game, the maximizing player
arrives at his optimal strategy o the basis of the miximin
criterion, while the minimizing player’s strategy is based
on the minimax criterion. The game is solved when the
maximin value equals minimax value( which is known as
saddle point)
C.R.Krishna Prasad, BIT Bangalore-4
1.
2.
3.
4.
5.
Develop the payoff matrix
Identify row minimums and select the largest
of these as player one’s maximin strategy
Identify column maximums and select the
smallest of these as the opponent’s minimax
strategy.
If the maximin value equals the minimax
value, the game is a pure strategy game and
that value is the saddle point.
The value of the game of player one is the
maximin value, and to player two, the value is
the negative of the minimax value.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1: Solve the following games
whose payoff matrix is given by
Player B
Player
A
B1
B2
B3
B4
A1
1
7
3
4
A2
5
6
4
5
A3
7
2
0
3
C.R.Krishna Prasad, BIT Bangalore-4
Firm B
Firm A
B1
B2
B3
B4
B5
A1
3
-1
4
6
7
A2
-1
8
2
4
12
A3
16
8
6
14
12
A4
1
11
-4
2
1
C.R.Krishna Prasad, BIT Bangalore-4
PRINCIPLE OF DOMINANCE
If a strategy is inferior to another, it is said to be
dominated
1. If all elements in a row are less than or equal to the
corresponding elements in another row, then that row is
dominated and can be deleted from the matrix.
2. If all elements in a column are grater than or equal to
the corresponding elements in another column, then
that column is dominated and can be deleted from the
matrix.
3. A pure strategy may be dominated if if is inferior to
average of two or more other pure strategies.
C.R.Krishna Prasad, BIT Bangalore-4
Note: dominance property is used when the saddle
point does not exist
Problem. Apply the rule of dominance for the following
matrix .
Player B
Player
A
B1
B2
B3
A1
9
8
-7
A2
3
-6
4
A3
6
7
7
C.R.Krishna Prasad, BIT Bangalore-4
Solution methods of strategy games
(games without saddle point)
C.R.Krishna Prasad, BIT Bangalore-4
If it is a 2 × 2 game
Player
A
a1
a2
Player B
b1
a11
a21
b2
a12
a22
If A plays a1 with probability x and a2 with probability 1-x, and B
plays b1 with probability y and b2 with probability 1-y, then
a22 – a21
a22 – a12
x = ----------------------------y = --------------------------(a11 + a22) – (a21 + a12)
(a11 + a22) – (a21 + a12)
and
(a11 × a22) – (a21 × a12)
v = ----------------------------(a11 + a22) – (a21 + a12)
C.R.Krishna Prasad, BIT Bangalore-4
Pure strategy: If a player knows exactly what the
other player is going to do, a deterministic situation is
obtained and objective function is to maximize the
gain. Therefore, the pure strategy is a decision rule
always to select a particular course of action.
Mixed strategy: If a player is guessing as to which
activity is to be selected by the other on any
particular occasion, a probabilistic situation is
obtained and objective function is to maximize the
expected gain. Thus, the mixed strategy is a selection
among pure strategies with fixed probabiities.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 4 Solve the following game
Player B
Player
A
I
II
III
IV
I
-5
2
1
20
II
5
5
4
6
III
4
-2
0
-5
C.R.Krishna Prasad, BIT Bangalore-4
Problem 5:Two players A and B without showing
each other, put on a table a coin, with head or
tail up. A wins Rs.8 when both the coins show
head and Rs.1 When both are tails. B wins Rs.3
when the coins do not match. Given the
choice of being matching player ( A) or nonmatching player (B) which one would you
choose and what would be your strategy?
C.R.Krishna Prasad, BIT Bangalore-4
Problem 6
Company B
compa
ny
A
I
II
III
IV
I
3
2
4
0
II
3
4
2
4
III
4
2
4
0
IV
0
4
0
8
C.R.Krishna Prasad, BIT Bangalore-4
Problem 7 Two firms A and B make colour and black& white T.V.
sets. Firm A can make either 150 colour sets in a week or an equal
number of B & W T.V. sets, and make a profit of Rs.400 per colour
set and Rs.300 for B & W set. Firm B can, on the other hand make
either 300 colour sets, or 150 colour and 150 B & W sets, or 300 B &
W sets and manufacturers would share market in the proportion in
which they manufacture a particular type of set. Write the pay off
matrix of A per week. Obtain A’s and B’s optimum strategies and
value of the game.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 8 Graphical method
Solve the following (2X 3) game graphically
Player B
Player
A
b1
b2
b3
a1
1
3
11
a2
8
5
2
C.R.Krishna Prasad, BIT Bangalore-4
Problem 9
Solve the following game graphically whose e payoff
matrix for the player A is given in the table
Player B
I
Player A
II
I
2
4
II
2
3
III
3
2
IV
-2
6
C.R.Krishna Prasad, BIT Bangalore-4
Revision: Transportation
Given the following transportation problem
WHA
REH
OUS
E
MARKET
A
B
C
SUPPLY
W1
10
12
7
180
W2
14
11
6
100
W3
9
5
13
160
W4
11
7
9
120
Demand
240
200
220
C.R.Krishna Prasad, BIT Bangalore-4
Obtain the initial basic feasible solution
by VAM. Is the optimal solution obtained
by you unique? If not, what is the other
optimal solution?
C.R.Krishna Prasad, BIT Bangalore-4
Assignment: Assign the mechanics to the jobs .
JOB
M
E
C
H
A
NIC
1
2
3
4
5
A
10
3
3
2
8
B
9
7
8
2
7
C
7
5
6
2
4
D
9
10
9
6
10
C.R.Krishna Prasad, BIT Bangalore-4
Model: 1Operating Characteristics
Queue length
a)
average number of customers in queue waiting to get
service
System length
b)
average number of customers in the system
Waiting time in queue
c)
average waiting time of a customer to get service
Total time in system
d)
average time a customer spends in the system
Server idle time
e)
relative frequency with which system is idle
C.R.Krishna Prasad, BIT Bangalore-4
Measurement parameters
λ= mean number of arrivals per time period (eg. Per
hour)
μ = mean number of customers served per time period
Probability of system being busy/traffic intensity
ρ= λ/μ
Average waiting time system Ws = 1/(μ-
λ)
Average waiting time in queue
Wq= λ/ μ(μ- λ)
Average number of customers in the system
Ls = λ/ (μ- λ)
C.R.Krishna Prasad, BIT Bangalore-4
Average number of customers in the queue
Lq = λ2/ μ(μ- λ)
Probability of an empty facility/system being
idle
P(0) = 1– P(w)
Probability of being in the system longer than
time (t)
P(T>t)= e –(μ- λ)t
Probability of customers not exceeding k in the
system
P (n.≥k) = ρk
P( n>k) = ρ(k+1)
Probability of exactly N customers in the system
P(N) = ρN (1-ρ)
C.R.Krishna Prasad, BIT Bangalore-4
Problem : At a service counter of fastfood joint, the customers arrive at the
average interval of six minutes whereas
the counter clerk takes on an average 5
minutes for preparation of bill and
delivery of the item. Calculate the
following
a. counter utilisation level
b. average waiting time of the customers
at the fast food joint
c. Expected average waiting time in the
line
C.R.Krishna Prasad, BIT Bangalore-4
d. Average number of customers in the service
counter area
e. average number of customer in the line
f. probability that the counter clerk is idle
g. Probability of finding the clerk busy
h. chances that customer is required to wait more
than 30 minutes in the system
i. probability of having four customer in the system
J) probability of finding more than 3 customer in
the system
C.R.Krishna Prasad, BIT Bangalore-4
Solutions: Given λ = 60/10 = 10 customer/hr
μ = 12 customer/hr
A) traffic intensity ρ = λ / μ = 10/12 = 0.833
b) waiting time in the system
Ws = 1/ μ- λ = 1/12-10 = 0.5 hr
C) waiting time in the queue
Wq = λ/ μ (μ- λ) = 10/12(12-10) = 0.416 hr
D) number of customer in the system
Ls= λ/ (μ- λ) = 10/12-10 = 5 customers
E) Number of customer in the queue
Lq = λ2 / μ (μ- λ) = 102 /12(12-10) = 4.167 customers
C.R.Krishna Prasad, BIT Bangalore-4
f) probability that the counter clerk is idle
1- ρ = 1- λ / μ = 1- 10/12 = 0.167
g. Probability of finding the clerk busy
ρ = λ / μ = 10/12 = 0.833
h) chances of probability that customer
wait more than 30min = 30/60 = 0.5 hrs
P (T>t) = e – (μ- λ) t
P (T>0.5) = e – (12- 10) 0.5 = 0.368
C.R.Krishna Prasad, BIT Bangalore-4
I) probability of having four customer in the system
P (N) = ρN (1-ρ)
P (4) = ρ4 (1-ρ) = (0.833)4(1-0.833) = 0.0806
j) probability of finding more than 3 customer in
the system
P (n>k) = ρ (k+1)
P (n>3) = ρ (3+1) = (λ / μ) 4= (10/12) 4
= 0.474
C.R.Krishna Prasad, BIT Bangalore-4
Queuing Theory
Queuing System: General Structure
Arrival Process
According to source
According to numbers
According to time
Service System
Single server facility
Multiple, parallel facilities with single queue
Multiple, parallel facilities with multiple queues
Service facilities in a parallel
C.R.Krishna Prasad, BIT Bangalore-4
Queue Structure
First come first served
Last come first served
Service in random order
Priority service
C.R.Krishna Prasad, BIT Bangalore-4
Model 1: Poisson-exponential single
server model – infinite population
Assumptions:
Arrivals are Poisson with a mean arrival rate of, say λ
Service time is exponential, rate being μ
Source population is infinite
Customer service on first come first served basis
Single service station
For the system to be workable, λ ≤ μ
C.R.Krishna Prasad, BIT Bangalore-4
Model 2: Poisson-exponential single
server model – finite population
Has same assumptions as model 1, except that
population is finite
C.R.Krishna Prasad, BIT Bangalore-4
Model 3: Poisson-exponential multiple server
model – infinite population
Assumptions
Arrival of customers follows Poisson law, mean rate λ
Service time has exponential distribution, mean service
rate μ
There are K service stations
A single waiting line is formed
Source population is infinite
Service on a first-come-first-served basis
Arrival rate is smaller than combined service rate of all
service facilities
C.R.Krishna Prasad, BIT Bangalore-4
Model: 1Operating Characteristics
Queue length
a)
average number of customers in queue waiting to get
service
System length
b)
average number of customers in the system
Waiting time in queue
c)
average waiting time of a customer to get service
Total time in system
d)
average time a customer spends in the system
Server idle time
e)
relative frequency with which system is idle
C.R.Krishna Prasad, BIT Bangalore-4
Measurement parameters
λ= mean number of arrivals per time period (eg. Per
hour)
μ = mean number of customers served per time period
Probability of system being busy/traffic intensity
ρ= λ/μ
Average waiting time system Ws = 1/(μ-
λ)
Average waiting time in queue
Wq= λ/ μ(μ- λ)
Average number of customers in the system
Ls = λ/ (μ- λ)
C.R.Krishna Prasad, BIT Bangalore-4
Average number of customers in the queue
Lq = λ2/ μ(μ- λ)
Probability of an empty facility/system being idle
P(0) = 1– P(w)
Probability of being in the system longer than time (t)
P(T>t)= e –(μ- λ)t
Probability of customers not exceeding k in the system
P (n.≥k) = ρk
P( n>k) = ρ(k+1)
Probability of exactly N customers in the system
P(N) = ρN (1-ρ)
C.R.Krishna Prasad, BIT Bangalore-4
Problem 1. Customers arrive at a booking
office window, being manned by a single
individual a a rate of 25per hour. Time
required to serve a customer has
exponential distribution with a mean of
120 seconds. Find the mean waiting time
of a customer in the queue.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 2: A repairman is to be hired to repair
machines which breakdown at a n average
rate of 6 per hour. The breakdowns follow
Poisson distribution. The non-production time of
a machine is considered to cost Rs. 20 per hour.
Two repairmen Mr. X and Mr.Y have been
interviewed for this purpose. Mr. X charges Rs.10
per hour and he service breakdown machines
at the rate of 8 per hour. Mr. Y demands Rs.14
per hour and he services at an average of 12
per hour. Which repairman should be hired? (
Assume 8 hours shift per day)
C.R.Krishna Prasad, BIT Bangalore-4
Problem 3: A warehouse has only one loading
dock manned by a three person crew. Trucks
arrive at the loading dock at an average rate
of 4 trucks per hour and the arrival rate is
Poisson distributed. The loading of a truck takes
10 minutes on an average and can be
assumed to be exponentially distributed . The
operating cost of a truck is Rs.20 per hour and
the members of the crew are paid @ Rs.6 each
per hour. Would you advise the truck owner to
add another crew of three persons?
C.R.Krishna Prasad, BIT Bangalore-4
Problem 4; At a service counter of fastfood joint, the customers arrive at the
average interval of six minutes whereas
the counter clerk takes on an average 5
minutes for preparation of bill and
delivery of the item. Calculate the
following
a. counter utilisation level
b. average waiting time of th4e customers
at the fast food joint
c. Expected average waiting time in the
line
C.R.Krishna Prasad, BIT Bangalore-4
d. Average number of customers in the service
counter area
e. average number of customer in the line
f. probability that the counter clerk is idle
g. Probability of finding the clerk busy
h. chances that customer is required to wait more
than 30 minutes in the system
i. probability of having four customer in the system
J) probability of finding more than 3 customer in
the system
C.R.Krishna Prasad, BIT Bangalore-4
Problem 5: Customers arrive at a one-window drive-in bank
according to a Poisson distribution with mean 10 per hour.
Service time per customer is exponential with mean 5
minutes. The space in front of the window including that
for the serviced car accommodate a maximum of 3 cars.
Other cars can wait outside the space. Calculate
A) what is the probability that an arriving customer can
drive directly to the space in front of the window.
B) what is the probability that an arriving customer will have
to wait outside the indicated space
C) How long is arriving customer expected to wait before
stating the service.
C.R.Krishna Prasad, BIT Bangalore-4
D) How many spaces should be provided in front of the
window so that all the arriving customers can wait in front
of the window at least 20% of the time.
Problem 6
Customers arrive at the first class ticket counter of a theatre at
a rate of 12 per hours. There is one clerk serving the
customers at a rate of 30 per hour. Assuming the
conditions for use of the single channel queuing model,
evaluate
C.R.Krishna Prasad, BIT Bangalore-4
a)
b)
c)
d)
The probability that there is no customer
at the counter (i.e. that the system is
idle)
The probability that there are more than
20 customers at the counter
The probability that there is no customer
waiting to be served
The probability that a customer is being
served and no body is waiting.
C.R.Krishna Prasad, BIT Bangalore-4
Thank you
Give feedback to
krishmandya@rediffmail.com
C.R.Krishna Prasad, BIT Bangalore-4
Replacement decisions
Replacement theory is concerned with
the problem of replacement of
machines, electricity bulbs etc due to
their deteriorating efficiency ,failure or
breakdown. Replacement is generally
carried out in the following situation.
When the existing items outlived, when
existing items destroyed
C.R.Krishna Prasad, BIT Bangalore-4
Failure mechanism of items
A) Gradual failure( increase in expenditure for
operating costs, decreased in productivity of
the equipment
B) Sudden failure
i) Progressive failure
ii) Retrogressive failure
iii) Random failure
C.R.Krishna Prasad, BIT Bangalore-4
Problem1: The cost of a machine is Rs 6100 &
its scrap value is only Rs 100. Maintenance
costs are followed from the experience. When
should the machine be replaced
Year
1
Main 100
tena
ce
cost
2
3
4
5
6
7
8
250
400
600
900
1250
1600
2000
C.R.Krishna Prasad, BIT Bangalore-4
Problem 2
Year
1
Mai 100
nten 0
ace
cost
2
3
4
1200
1400
1800
2300
2800
3400
4000
750
375
200
200
200
200
Res 3000 1500
ale
pric
e
5
C.R.Krishna Prasad, BIT Bangalore-4
6
7
8
1.
2.
Replacement of machines that
deteriorates with time (considering time
value of money)
Write in a column the
running/maintenance costs of machine
or equipment for different years Rn
In the next column write the discount
factor indicating the present value of a
rupee received after (i-1)year., v n-1
=(1/1+r ) n-1
C.R.Krishna Prasad, BIT Bangalore-4
3.The two column values are multiplied to get
present value of the maintenance costs i.e. Rn v
n-1
4.These discounted maintenance costs are then
cumulated to the ith year to get
∑Rn v n-1
5.The cost of the machine or equipment is added
to the values obtained in step 4 above to
obtain C + ∑Rn v n-1
6. The discount factors are then cumulated to get
∑ v n-1
C.R.Krishna Prasad, BIT Bangalore-4
7. The total costs obtained in (step 5) are
divided by the corresponding value of
the accumulated discount factor for
each of the years
8. Now compare the column of
maintenance costs which is constantly
increasing, with the last column. Replace
the machine in the latest year that the
last column exceeds the column of
maintenance cost.
C.R.Krishna Prasad, BIT Bangalore-4
Problem. The cost of a new machine is
Rs.5000. The maintenance cost during the
nth year is given by Rn = 500 X (n-1),
n=1,2,…Suppose that the discount rate
per year is 0.05 . After how many years it
will be economical to replace the
machine by a new one?
C.R.Krishna Prasad, BIT Bangalore-4
Problem
An entrepreneur is considering to purchase a machine
for his factory. Relevant data about alternative
machines are as follows.
As a advisor to the buyer, you have been asked to
select the best machine considering 12%normal rate of
return.
Single payment present worth factor at 12%for 10 years
0.322
Annual series present worth factor at 12% for 10 years
5.650
C.R.Krishna Prasad, BIT Bangalore-4
machine
A
B
C
Present
10,000
investment.
Total annual 2000
cost
life
10
12000
15000
1500
1200
10
10
Salvage
value
1000
1200
500
C.R.Krishna Prasad, BIT Bangalore-4
Problem 4: The management of a large hotel is
considering the periodic replacement of light
bulbs fitted in ins rooms. There are 500 rooms in
the hotel and each room has 6 bulbs. The
management is now following the policy of
replacing the bulbs as they fail at a total cost of
Rs.3 per bulb. The management feels that this
cost can be reduced to Rs.1 by adopting the
periodic replacement method. On the basis of
the information give below, evaluate the
alternative and make a recommendation to
the management.
C.R.Krishna Prasad, BIT Bangalore-4
Months of use
1
% of bulbs failing 10
by that month
2
3
4
5
25
50
80
100
C.R.Krishna Prasad, BIT Bangalore-4
Assumptions 1. Bulbs failing during a
month are replaced just before the end
of the month.
The actual percentage of failures during
a month as for a sub population of bulbs
with same age is the same as the
expected percentages of failures during
the month for that sub population
C.R.Krishna Prasad, BIT Bangalore-4
Problem 5:Following mortality rates have
been observed for a certain type of fuses
Week
1
%of failure at the 5
end of week
2
3
4
5
15
35
75
100
C.R.Krishna Prasad, BIT Bangalore-4
There are 1,000 fuses in use and it costs Rs.5 to
replace an individual fuse. If all fuses were
replaced simultaneously if would cost Rs.1.25
per fuse. It is proposed to replace all fuses at
fixed intervals of time, whether or not they have
burnt out, and to continue replacing burnt-out
fuses as they fail. At what intervals, the group
replacement should be made? Also prove that
this optimal policy is superior to the straight
forward policy of replacing each fuse only
when it fails.
C.R.Krishna Prasad, BIT Bangalore-4
Problem 6: A computer contains 10,000
resistors. When any of the register fails it is
replaced. The cost of replacing a single
resistor is Rs.10 only. If all the resistors are
replaced at the same time, the cost per
resistor would be reduced to Rs.3.50. The
per cent surviving by the end of the
month t is as follows. What is the optimum
plan.
C.R.Krishna Prasad, BIT Bangalore-4
month
0
1
%survi 100 97
ving at
the end
of
month
2
3
4
5
6
90
70
30
15
0
C.R.Krishna Prasad, BIT Bangalore-4
Float (Slack)
•Float (Slack ) refers to the amount of time by which
a particular event or an activity can be delayed
without affecting the time schedule of the network.
•Float (Slack)
Float (Slack) is defined as the difference between
latest allowable and the earliest expected time.
Event Float/Slack = LS – ES
Where LS = Latest start time
ES = Early start time.
C.R.Krishna Prasad, BIT Bangalore-4
Earliest start : Denoted as ‘ES’
Earliest start time is the earliest possible time by
which the activity can be started.
Early finish time : Denoted as ‘EF’
Early finish time is the earliest possible time by
which the activity can be completed.
C.R.Krishna Prasad, BIT Bangalore-4
Latest start time : Denoted as ‘LS’
Latest start time is the latest possible time by
which the activity can be started
Late finish time : Denoted as ‘LS’
Late finish time is the latest possible time by which
the activity can be completed.
C.R.Krishna Prasad, BIT Bangalore-4
Total float (TF) / Total slack (TS)
Total float of the job is the differences between its
Late start and Early start ‘or’ Late finish and Early
finish
i.e.
TF( CA) = LS (CA) - ES (CA)
Or
TF( CA) = LF (CA) - EF (CA)
CA = Current activity
C.R.Krishna Prasad, BIT Bangalore-4
Free float (FF)
Free float is the amount of time a job can be
delayed without affecting the Early start time of any
other job.
FF(CA) = ES(SA) – EF (CA)
CA = Current Activity
SA = Succeeding Activity
C.R.Krishna Prasad, BIT Bangalore-4
Independent Float (IF)
Independent Float is the amount of time that can
be delayed without affecting either predecessor
or successor activities.
IF = ES(SA) – LF(PA)
- Duration of CA
ES = Early Start
LF = Late Finish
SA = Succeeding Activity
PA = Preceding Activity
CA = Current Activity
C.R.Krishna Prasad, BIT Bangalore-4
Example:
Construct the Network for the following
Project and determine the following
i) Critical Path
ii) ES,EF,LS,LF
iii) TF,FF
C.R.Krishna Prasad, BIT Bangalore-4
Activity
Duration
1-2
14
1-4
3
2-3
7
2-4
0
3-5
4
4-5
3
5-6
10
C.R.Krishna Prasad, BIT Bangalore-4
Construction of the Network and Determination Critical Path
4
F
5
1
G 0
D
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F
5
1
G 0
D
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F
5
1
G 0
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F
5
1
G 0
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F(25,35)
5
1
G 0
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F(25,35)
5
1
G 0
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F(25,35)
5
0
1
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of ES and EF
4
F(25,35)
5
0
1
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7(14,21)
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7(14,21)
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7(14,21)
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
1
D(14,21)
2
7(14,21)
3
C.R.Krishna Prasad, BIT Bangalore-4
10(25,35)
6
Determination of LS and LF
4
F(25,35)
5
10(25,35)
1
D(14,21)
2
7(14,21)
3
Key
Job (ES,EF)
Duration (LS,LF)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)
2
7(14,21)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)
2
7(14,21)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)
2
7(14,21)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)(0)
2
7(14,21)(0)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)
5
10(25,35)
1
D(14,21)(0)
2
7(14,21)(0)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Determination of TF and FF
4
F(25,35)(0)
5
10(25,35)(0)
1
D(14,21)(0)
2
7(14,21)(0)
3
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
Key
IF = ( ES(SA) – LF(PA)) - Duration of CA
Job (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
6
Activity Duration
ES
EF
LS
LF
TF
FF
1-2
14
0
14
0
14
0
0
1-4
3
0
3
19
22
19
11
2-3
7
14
21
14
21
0
0
2-4
0
14
14
22
22
0
0
3-5
4
21
25
21
25
0
0
4-5
3
14
17
22
25
8
8
5-6
10
25
35
25
35
0
0
C.R.Krishna Prasad, BIT Bangalore-4
Example:
Construct the Network for the following
Project and determine the following
i) Critical Path
ii) ES,EF,LS,LF
iii) TF,FF
C.R.Krishna Prasad, BIT Bangalore-4
Activity
Duration
1-2
2
2-3
3
2-4
5
3-5
4
3-6
1
4-6
6
4-7
2
5-8
8
6-8
7
7-8
4
C.R.Krishna Prasad, BIT Bangalore-4
5
3
8
(0, 2)(0)
1
2
6
2(0, 2)(0)
(7, 9)(0)
4
Key
7
2(14,16)(7)
(ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
PERT Model
Historical Evolution.
Before 1957 there was no generally accepted procedure
to aid the management of a project. In 1958 PERT was
developed by team of engineers working on a Polaris
Missile programme of the navy. This was a large
project involved 250 prime contractors and about 9000
job contractors. It had about 19 million components. In
such projects it is possible that a delay in the delivery
of a small component might hold the progress of entire
project. PERT was used successfully and the total time
of completion was reduced from 7 years to 5 years.
C.R.Krishna Prasad, BIT Bangalore-4
Differences between PERT & CPM
PERT
CPM
1. It is a technique for
planning scheduling &
controlling of projects
whose activities are
subject to uncertainty in
the performance time.
Hence it is a probabilistic
model
1. It is a technique for
planning scheduling &
controlling of projects
whose activities not
subjected to any
uncertainty and the
performance times are
fixed. Hence it is a
deterministic model
2.It is an Event oriented
system
2. It is an Activity oriented
system
C.R.Krishna Prasad, BIT Bangalore-4
3.Basically does not
differentiate critical and
non-critical activities
Differentiates clearly the
critical activities from the
other activities.
4. Used in projects where
resources (men, materials,
money) are always
available when required.
4. Used in projects where
overall costs is of
primarily important.
Therefore better utilized
resources
5. Suitable for Research
5.Suitable for civil
and Development projects constructions, installation,
where times cannot be
ship building etc.
predicted
C.R.Krishna Prasad, BIT Bangalore-4
In PERT model, 3 time values are associated
with each activity. They are
i)Optimistic time = to
ii)Pessimistic time = tp
iii)Most likely time = tm
These three times provide a measure of uncertainty
associated with that activity
C.R.Krishna Prasad, BIT Bangalore-4
Optimistic Time: This is the shortest possible time
in which the activity can be finished. It assumes
that every thing goes well.
Pessimistic Time: This is the longest time that an
activity could take. It assumes that every thing
goes wrong.
Most likely Time: It is the estimate of the normal
time that an activity would take. This assumes
normal delays.
C.R.Krishna Prasad, BIT Bangalore-4
Expected Time ( te):
‘te’ can be calculated by the following formula
te = (to + 4tm + tp) / 6
Example.
If a job has to = 5 days, tp = 17 days, tm = 8 days
Then Expected time for the job would be
te = (to + 4tm + tp) / 6
= (5 + 4 x 8 + 17) / 6
= 9 days
C.R.Krishna Prasad, BIT Bangalore-4
Variability of activity times
•Standard deviation and Variance are
commonly used in statistics to measure the
variability of number.
In PERT model, to measure the variability of an
activity time duration standard deviation and
variance are used.
A large standard deviation represents high
variability and vice-versa.
C.R.Krishna Prasad, BIT Bangalore-4
•Calculation of Standard Deviation and Variance
Variance = (Standard deviation )2
Standard deviation =(t p – t o) / 6
•Expected length of the Critical Path = te of all the
activities along the Critical Path
C.R.Krishna Prasad, BIT Bangalore-4
Probability of completing the project within a given date
Z = (TS – TE ) / σ
Where TS = Scheduled time for project completion
TE = Expected time for the project completion
σ = Standard deviation for the Network
C.R.Krishna Prasad, BIT Bangalore-4
σNetwork = √Sum of variances along the
Critical Path
= √ (σNetwork )2
C.R.Krishna Prasad, BIT Bangalore-4
Example:
Construct the Network for the following project and
calculate the probability of completing the project in 25 days
Activity
to
tm
tp
1-2
2
6
10
1-3
4
8
12
2-3
2
4
6
2-4
2
3
4
3-4
0
0
0
3-5
3
6
9
4-6
6
10
14
5-6
1
3
5
C.R.Krishna Prasad, BIT Bangalore-4
1.Construction of the Network
2-3-4
2
4
1
6
3-6-9
3
5
C.R.Krishna Prasad, BIT Bangalore-4
2. Calculation of Expected time for all the activities
2-3-4
2
4
3
1
6
3-6-9
3
5
6
Expected Time ( te):
‘te’ can be calculated by the following formula
te = (to + 4tm + tp) / 6
C.R.Krishna Prasad, BIT Bangalore-4
3. Determination of Critical Path
2-3-4
2
4
3
1
6
3-6-9
3
5
6
to - tm - tp
Key
te
Expected Duration of the project Te = 20 days
C.R.Krishna Prasad, BIT Bangalore-4
Activity
to
tm
tp
Critical
activities
1-2
2
6
10
1-2
1-3
4
8
12
-
2-3
2
4
6
2-3
2-4
2
3
4
-
3-4
0
0
0
3-4
3-5
3
6
9
-
4-6
6
10
14
4-6
5-6
1
3
5
-
σ2 = ((t p – t o) / 6)2
1.78
0.44
0
1.78
Σ σ2 = 4.00
C.R.Krishna Prasad, BIT Bangalore-4
σNetwork = √Sum of variances along the
Critical Path
= √ (σNetwork )2
=√
4
=2
C.R.Krishna Prasad, BIT Bangalore-4
Probability of completing the project within a given date
Z = (TS – TE ) / σ
Where TS = Scheduled time for project completion
TE = Expected time for the project completion
σ = Standard deviation for the Network
= (25 – 20) / 2
= + 2.5
C.R.Krishna Prasad, BIT Bangalore-4
From the Normal distribution Table, we get the
probability of completing the project in 25 days is
99.4%
C.R.Krishna Prasad, BIT Bangalore-4
Example.
The following table lists the jobs of a network along with their
time estimates.
Activity
to
tm
tp
1-4
3
9
27
1-3
3
6
15
1-2
6
12
30
4-5
1
4
07
3-5
3
9
27
3-6
2
5
08
5-6
6
12
30
2-6
4
19
28
C.R.Krishna Prasad, BIT Bangalore-4
a) Draw the project network.
b) What is the probability that the job will be
completed in 35 days?
c) What due date has 90% chance of being met?
C.R.Krishna Prasad, BIT Bangalore-4
1.Construction of the Network
2
1
3 - 6 - 15
3
4
2-5-8
5
C.R.Krishna Prasad, BIT Bangalore-4
6
2. Calculation of Expected time for all the activities
2
1
3 - 6 - 15
2-5-8
3
7
5
4
1-4-7
4
5
Expected Time ( te):
‘te’ can be calculated by the following formula
te = (to + 4tm + tp) / 6
C.R.Krishna Prasad, BIT Bangalore-4
6
3. Determination of Critical Path
2
1
3 - 6 - 15
2-5-8
3
7
5
4
1-4-7
4
5
to - tm - tp
Key
te
Expected Duration of the project Te = 32 days
C.R.Krishna Prasad, BIT Bangalore-4
6
As there are two Critical Paths, the path which
gives more variance(σ2) is taken as Critical Path
Path A
Activity
σ2 = ((t p – t o) / 6)2
σ2
1-2
((30 – 6)/6)2
16
2-6
((28 – 4)/6)2
16
Σ σ2 = 32.00
C.R.Krishna Prasad, BIT Bangalore-4
Path B
σ2 = ((t p – t o) / 6)2
Activity
σ =
σ2
1-3
((15 – 3)/6)2
4
3-5
((27 – 3)/6)2
16
5-6
((30 – 6)/6)2
16
√ Σ σ2
=
√ 36
Σ σ2= 36.00
=6
Therefore the Critical Path is 1 - 3 - 5 - 6
C.R.Krishna Prasad, BIT Bangalore-4
b)
Probability of completing the project within a given date
Z = (TS – TE ) / σ
Where TS = Scheduled time for project completion
TE = Expected time for the project completion
σ = Standard deviation for the Network
= (35 – 32) / 6
= + 0.5
C.R.Krishna Prasad, BIT Bangalore-4
From the Normal distribution Table, we get the
probability of completing the project in 35 days is
69.15%
C.R.Krishna Prasad, BIT Bangalore-4
c)
The due date for 90% chance of being met.
Probability of completing the project within a given date
Z = (TS – TE ) / σ
The value of Z from the table for a 90% probability is +1.28
σ =
=6
TS = ? (to be calculated) ,TE = 32, σ
i.e. 1.28 = (TS– 32) / 6
TS = 39.68 days
C.R.Krishna Prasad, BIT Bangalore-4
CPM Model
In 1958 Du Pont Company used a technique called
Critical Path Method (CPM) to schedule and control a
very large project like overhauling of a chemical plant,
there by reducing the shutdown period from 130hrs to
90 hrs saving the company 1 million dollar.
Both of these techniques are referred to as project
scheduling techniques.
C.R.Krishna Prasad, BIT Bangalore-4
Differences between PERT & CPM
PERT
CPM
1. It is a technique for
planning scheduling &
controlling of projects
whose activities are
subject to uncertainty in
the performance time.
Hence it is a probabilistic
model
1. It is a technique for
planning scheduling &
controlling of projects
whose activities not
subjected to any
uncertainty and the
performance times are
fixed. Hence it is a
deterministic model
2.It is an Event oriented
system
2. It is an Activity oriented
system
C.R.Krishna Prasad, BIT Bangalore-4
3.Basically does not
differentiate critical and
non-critical activities
Differentiates clearly the
critical activities from the
other activities.
4. Used in projects where
resources (men, materials,
money) are always
available when required.
4. Used in projects where
overall costs is of
primarily important.
Therefore better utilized
resources
5. Suitable for Research
5.Suitable for civil
and Development projects constructions, installation,
where times cannot be
ship building etc.
predicted
C.R.Krishna Prasad, BIT Bangalore-4
Cost considerations in PERT / CPM
The total cost of any project comprises of two costs.
•Direct cost - material cost, manpower loading
•Indirect cost - overheads such as managerial services,
equipment rent, building rent etc.
C.R.Krishna Prasad, BIT Bangalore-4
Direct cost
Crash
Normal
Job duration
Shorter the duration higher will be the Direct expenses
C.R.Krishna Prasad, BIT Bangalore-4
Indirect cost
Crash
Normal
Job duration
Shorter the duration lesser will be the Indirect expenses
C.R.Krishna Prasad, BIT Bangalore-4
Cost
Total cost
Direct cost
Crash
Optimum
Job duration
C.R.Krishna Prasad, BIT Bangalore-4
Normal
Find the lowest cost and optimum cost schedule for
the following project, given the over head expenses
as Rs.45/-day.
Activity
Normal
duration
Crash
duration
Cost of
crashing per
day
1-2
3
1
Rs.40
2-3
4
2
Rs.40
2-4
7
3
Rs.10
3-4
5
2
Rs.20
C.R.Krishna Prasad, BIT Bangalore-4
1.Construction of the Network
4
3 -1
1
2
40
3
Normal duration – Crash duration
Key
Cost of crashing per day in Rs.
C.R.Krishna Prasad, BIT Bangalore-4
2.Determination of Critical path
4
3 -1
1
2
40
3
Normal duration – Crash duration
Key
Cost of crashing per day in Rs.
C.R.Krishna Prasad, BIT Bangalore-4
Step 1.
3 -1
1
4-2
5-2
2
3
40
40
7-3
12 days
4
20
4
10
Activity
crashed
Days
saved
Project
duration
Cost of crashing
Total
cost of
crashing
Over Head
cost
Total
cost
None
0
12
-Nil-
-Nil-
45 x 12
540
3-4
2
10
20 x 2 =40
40
45 x 10
490
3-4 &2-4
1
9
20x1+10x1 =30
70
45 x 9
475
1-2
2
7
40 x 2 =80
150
45 x 7
465
2-3&2-4
2
5
40x2+10x2 =100
250
45 x 5
475
C.R.Krishna Prasad, BIT Bangalore-4
Step 2.
3 -1
1
4-2
3-2
2
3
40
4
20
40
7-3
10 days
10
Activity
crashed
Days
saved
Project
duration
Cost of crashing
Total
cost of
crashing
Over Head
cost
Total
cost
None
0
12
-Nil-
-Nil-
45 x 12
540
3-4
2
10
20 x 2 =40
40
45 x 10
490
3-4 &2-4
1
9
20x1+10x1 =30
70
45 x 9
475
1-2
2
7
40 x 2 =80
150
45 x 7
465
2-3&2-4
2
5
40x2+10x2 =100
250
45 x 5
475
C.R.Krishna Prasad, BIT Bangalore-4
Step 3.
3 -1
1
4-2
2-2
2
3
40
4
20
40
6-3
9 days
10
Activity
crashed
Days
saved
Project
duration
Cost of crashing
Total
cost of
crashing
Over Head
cost
Total
cost
None
0
12
-Nil-
-Nil-
45 x 12
540
3-4
2
10
20 x 2 =40
40
45 x 10
490
3-4 &2-4
1
9
20x1+10x1 =30
70
45 x 9
475
1-2
2
7
40 x 2 =80
150
45 x 7
460
2-3&2-4
2
5
40x2+10x2 =100
250
45 x 5
475
C.R.Krishna Prasad, BIT Bangalore-4
Step 4.
1 -1
1
4-2
2-2
2
3
40
4
20
40
6-3
7 days
10
Activity
crashed
Days
saved
Project
duration
Cost of crashing
Total
cost of
crashing
Over Head
cost
Total
cost
None
0
12
-Nil-
-Nil-
45 x 12
540
3-4
2
10
20 x 2 =40
40
45 x 10
490
3-4 &2-4
1
9
20x1+10x1 =30
70
45 x 9
475
1-2
2
7
40 x 2 =80
150
45 x 7
460
2-3&2-4
2
5
40x2+10x2 =100
250
45 x 5
475
C.R.Krishna Prasad, BIT Bangalore-4
Step 5.
3 -1
1
2-2
2-2
2
3
40
4
20
40
4-3
5 days
10
Activity
crashed
Days
saved
Project
duration
Cost of crashing
Total
cost of
crashing
Over Head
cost
Total
cost
None
0
12
-Nil-
-Nil-
45 x 12
540
3-4
2
10
20 x 2 =40
40
45 x 10
490
3-4 &2-4
1
9
20x1+10x1 =30
70
45 x 9
475
1-2
2
7
40 x 2 =80
150
45 x 7
460
2-3&2-4
2
5
40x2+10x2 =100
250
45 x 5
475
C.R.Krishna Prasad, BIT Bangalore-4
Differences between PERT & CPM
PERT
CPM
1. It is a technique for
planning scheduling &
controlling of projects
whose activities are
subject to uncertainty in
the performance time.
Hence it is a probabilistic
model
1. It is a technique for
planning scheduling &
controlling of projects
whose activities not
subjected to any
uncertainty and the
performance times are
fixed. Hence it is a
deterministic model
2.It is an Event oriented
system
2. It is an Activity oriented
system
C.R.Krishna Prasad, BIT Bangalore-4
3.Basically does not
differentiate critical and
non-critical activities
Differentiates clearly the
critical activities from the
other activities.
4. Used in projects where
resources (men, materials,
money) are always
available when required.
4. Used in projects where
overall costs is of
primarily important.
Therefore better utilized
resources
5. Suitable for Research
5.Suitable for civil
and Development projects constructions, installation,
where times cannot be
ship building etc.
predicted
C.R.Krishna Prasad, BIT Bangalore-4
