C.R.Krishna Prasad, BIT Bangalore-4
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.
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PHASES OF OPERATIONS RESEARCH STUDY
I Formulation The Problem: What are the objectives, controlled variables,
uncontrolled variable constraints
IIConstructing a mathematical method
A mathematical model should include a) decision variable and parameter,,objective functions&constraints.
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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
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Techniques of OR
Distribution/allocation Models Waiting line/queuing models Production/inventory model Competitive strategy model/games
theory Network analysis Job sequencing models
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Replacement models Markovian models Simulation models
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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
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Limitations Magnitude of computation Absence of quantification Distance between manager and OR
experts
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Linear Programming Problems
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- 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.
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Structure of Linear Programming model.
The general structure of the Linear Programming model
essentially consists
of three components.
i) The activities (Decision variables) and their
relationships
ii) The objective function and
iii) The constraints
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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 variablesAnd 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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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)
Source A Source B
Copper 40% 75%
Unfit Scrap 7.5% 10%
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
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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 man-
days 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.
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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.
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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.
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F1 F2 F3 F4
Production cost/unit 15 18 14 14
Raw material cost/unit 10 9 12 9
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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
3 9 5 4 Selling price.
34
Requirement 80
S2 1 7 4 5 32 120
S3 5 8 3 6 31 150
10 150 50 100
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Assignment Problems Hungarian method
Step 1Balance the problem if it is unbalancedPlace an M as the cost element if some assignment is prohibitedConvert 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.
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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.
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Programmers
A B C D
1 120 100 80 90
2 80 90 110 70
3 110 140 120 100
4 90 90 80 90
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Problem 2 Suggest optimum solution to the following assignment
problem and also the minimum cost:
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Markets/salesmen
I II III IV
A 44 80 52 60
B 60 56 40 72
C 36 60 48 48
D 52 76 36 40
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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.
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A typical transportation problem contains
Inputs:Sources with availability
Destinations with requirementsUnit cost of transportation from various sources to destinations
Objective:To determine schedule of transportation to minimize total transportation cost
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How to solve?1. Define the objective function to be
minimized with the constraints imposed on the problem.
2. Set up a transportation table with m rows representing the sources and n columns representing the destination
3. 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
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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.
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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.
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North West corner method
1. 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)
2. Adjust the supply and demand numbers in the respective rows and columns allocation
3. 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.
4. 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
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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.
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Problem 1: Obtain initial solution in the following transportation problem by using Northwest corner rule method
Origins D1 D2 D3 D4 Supply/capacity/availabili
ty
O1 1 2 1 4 3030
O2 3 3 2 1 5050
O3 4 2 5 9 2020Demand/Requirements
2020 4040 3030 1010
C.R.Krishna Prasad, BIT Bangalore-4
Least cost method:
1. Select the cell with the lowest transportation cost among all the rows or column of the transportation table
2. If the minimum cost is not unique, then select arbitrarily any cell with this minimum cost. (preferably where maximum allocation is possible)
3. 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 D1 D2 D3 D4 Supply/capacity/availabili
ty
O1 1 2 1 4 3030
O2 3 3 2 1 5050
O3 4 2 5 9 2020Demand/Requirements
2020 4040 3030 1010
C.R.Krishna Prasad, BIT Bangalore-4
Vegel’s Approximation Method (VAM)1. 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.
2. 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 D1 D2 D3 D4 Supply/capacity/availabili
ty
O1 1 2 1 4 3030
O2 3 3 2 1 5050
O3 4 2 5 9 2020Demand/Requirements
2020 4040 3030 1010
C.R.Krishna Prasad, BIT Bangalore-4
The Modified Distribution Method1. Determine an initial basic feasible solution consisting
of m + n -1 allocations in independent positions using any of the three methods
2. 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.
3. 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/From
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
To/fro
m
D1 D2 D3 D4 Available
P1 19 30 50 12 7
P2 70 30 40 60 10
P3 40 10 60 20 18
Requiremetn
s
5 8 7 15
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 price per
unit
Demand
1 15 120
2 14 90
3 16 50
C.R.Krishna Prasad, BIT Bangalore-4
Assignment Problems Hungarian method
Step 1Balance the problem if it is unbalancedPlace an M as the cost element if some assignment is prohibitedConvert 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
Programmers
A B C D
1 120 100 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/salesmen
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
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.
M1 M2 M3 M4 M5
302 302 303 302 301
303 304 301 305 302
304 305 304 304 304
301 305
302
303
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
Salesmen
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
What is the sequence of visit of the salesman, so that the cost is minimum.
From/to
A B C D E
A 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
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 hour in Rs.
No.of pages typed per hour
A 5 12
B 6 14
C 3 8
D 4 10
E 4 11
JOB No. of pages
P 199
Q 175
R 145
S 198
T 178
C.R.Krishna Prasad, BIT Bangalore-4
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 lay-over.
C.R.Krishna Prasad, BIT Bangalore-4
Hyderabad-Delhi Delhi -Hyderabad
Flight no.
Departure
Arrival Flight no. Departure
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 III IV
Annual sales
(in Rs)
60,000 50,000 40,000 30,000
C.R.Krishna Prasad, BIT Bangalore-4
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
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 and Development projects where times cannot be predicted
5.Suitable for civil constructions, installation, ship building etc.
C.R.Krishna Prasad, BIT Bangalore-4
Rules for drawing the network diagrams.
•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.
1 2 3 4
C.R.Krishna Prasad, BIT Bangalore-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.
1 2 3 4
C.R.Krishna Prasad, BIT Bangalore-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.
1 2 3 4
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.
1 2 3 4
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 represented by an arrows.
In this method activities 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
A
B
C
D
E
F
G
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
A
B
C
D
E
14
3
7
4
10
KeyJob
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
A
BC
D
E
F
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
A
B
C
D
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
A
B
C
D
E
F
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
A
B
E
D
C
G
H
F
J
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
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
Problem Presentation
C.R.Krishna Prasad, BIT Bangalore-4
1 2 5 6
3
4
A
B
C
D
E
F
Obtain the machine
Obtain the budget approval
Hire the operator
Install the machine
Train the operator
Produce the goods
C.R.Krishna Prasad, BIT Bangalore-4
1 2A
Obtain the budget approval
C.R.Krishna Prasad, BIT Bangalore-4
1 2
3
A
B
Obtain the machine
Obtain the budget approval
C.R.Krishna Prasad, BIT Bangalore-4
C.R.Krishna Prasad, BIT Bangalore-4
1 2
3
4
A
B
C
Obtain the machine
Obtain the budget approval
Hire the operator
C.R.Krishna Prasad, BIT Bangalore-4
1 2
3
4
A
B
C
D
Obtain the machine
Obtain the budget approval
Hire the operator
Install the machine
5
C.R.Krishna Prasad, BIT Bangalore-4
1 2 5
3
4
A
B
C
D
E
Obtain the machine
Obtain the budget approval
Hire the operator
Install the machine
Train the operator
C.R.Krishna Prasad, BIT Bangalore-4
1 2 5 6
3
4
A
B
C
D
E
F
Obtain the machine
Obtain the budget approval
Hire the operator
Install the machine
Train the operator
Produce the goods
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
•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
1
3
4
6
2
5
7
8
Determination of Critical Path
2
4
3
5
4
5
23
26
1
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 daysPath B : 4 + 5 + 4 = 13 daysPath C : 4 + 5 + 6 + 1 = 16 daysPath D : 4 + 2 + 3 + 1 = 10 daysPath 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
1
3
4
6
2
5
7
8
Determination of Critical Path
2
4
3
5
4
5
23
26
1
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
1
4
6
3
5
2
B
3
C
3
A14
D
7
E4
F
10G 0
Construction of the Network and Determination Critical Path
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B
3
C
3
A(0,14)14
D
7
E4
F
10G 0
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B
3
C
3
A(0,14)14
D(14,21)
7
E4
F
10G 0
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B
3
C
3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
100G
F
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B
3
C
3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
F(25,35)
10G 0
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C
3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
F(25,35)
100G
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C
3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
F(25,35)
10G
(14,
14)
0
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
10G
(14,
14)
0
F(25,35)
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14
D(14,21)
7
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14
D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14 (0,14) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3(22,25)
A(0,14)14 (0,14) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3(22,25)
A(0,14)14 (0,14) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3(19,22)
C(14,17)3(22,25)
A(0,14)14 (0,14) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
KeyJob (ES,EF)
Duration (LS,LF)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3(19,22)
C(14,17)3(22,25)
A(0,14)(0)14 (0,14)(0) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)
Determination of TF and FF
G(1
4,14
)
0(22
,22)
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)3(22,25)
A(0,14)(0)14 (0,14)(0) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)(8)3(22,25)(8)
A(0,14)(0)14 (0,14)(0) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)(8)3(22,25)(8)
A(0,14)(0)14 (0,14)(0) D(14,21)(0)
7(14,21)(0)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)(8)3(22,25)(8)
A(0,14)(0)14 (0,14)(0) D(14,21)(0)
7(14,21)(0)
E(2
1,25
)(0)
4(21
,25)
(0)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)(8)3(22,25)(8)
A(0,14)(0)14 (0,14)(0) D(14,21)(0)
7(14,21)(0)
E(2
1,25
)(0)
4(21
,25)
(0)
F(25,35)(0)
10(25,35)(0)0(
22,2
2)
G(1
4,14
)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
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
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Example:
Construct the Network for the following
Project and determine the following
i) Critical Path
ii) ES,EF,LS,LF
iii) TF,FF
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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
1 2
4 7
3
6
5
8
(0, 2)(0)
(ES,EF)(FF)
Duration (LS,LF)(TS)Key
2(0, 2)(0)
(2, 5)(0)
3(5, 8)(3)
(5 ,9)(0
)
4(8, 12)(3)
(2, 7)(0)5(2, 7)(0)
(5, 6)(7)1(12, 13)(7)
(7,13) (0)
6(7, 13)(0
)
(7, 9)(0)
2(14,16)(7)
(13, 20)(0)
7(13, 20)(0)
(9,1
3)(0
)4(
16, 2
0)(7
)
(9,17)(0)8(12,20)(3)
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 and Development projects where times cannot be predicted
5.Suitable for civil constructions, installation, ship building etc.
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 requirementsUnit 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. Define the objective function to be minimized with the constraints imposed on the problem.
2. Set up a transportation table with m rows representing the sources and n columns representing the destination
3. 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 method1. 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)
2. Adjust the supply and demand numbers in the respective rows and columns allocation
3. 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.
4. 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 D1 D2 D3 D4 Supply/capacity/availabili
ty
O1 1 2 1 4 3030
O2 3 3 2 1 5050
O3 4 2 5 9 2020Demand/Requirements
2020 4040 3030 1010
C.R.Krishna Prasad, BIT Bangalore-4
Least cost method:
1. Select the cell with the lowest transportation cost among all the rows or column of the transportation table
2. If the minimum cost is not unique, then select arbitrarily any cell with this minimum cost. (preferably where maximum allocation is possible)
3. 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 D1 D2 D3 D4 Supply/capacity/availabili
ty
O1 1 2 1 4 3030
O2 3 3 2 1 5050
O3 4 2 5 9 2020Demand/Requirements
2020 4040 3030 1010
C.R.Krishna Prasad, BIT Bangalore-4
Vegel’s Approximation Method (VAM)1. 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.
2. 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 counter-strategy 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
Y pays Rs.100 to X
X pays Rs.80 t0 Y
X pays Rs.80 t0 Y
Y pays Rs.100 to X
C.R.Krishna Prasad, BIT Bangalore-4
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. Develop the payoff matrix2. Identify row minimums and select the
largest of these as player one’s maximin strategy
3. Identify column maximums and select the smallest of these as the opponent’s minimax strategy.
4. If the maximin value equals the minimax value, the game is a pure strategy game and that value is the saddle point.
5. 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 B b1 b2
Player a1 a11 a12 A a2 a21 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 non-matching player (B) which one would you choose and what would be your strategy?
C.R.Krishna Prasad, BIT Bangalore-4
Problem 6
Company B
company 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 II
Player A 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
MARKET
WHAREHOUSE
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 Characteristicsa) Queue length
average number of customers in queue waiting to get service
b) System length average number of customers in the system
c) Waiting time in queue average waiting time of a customer to get service
d) Total time in system average time a customer spends in the system
e) Server idle time 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 fast-food 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 levelb. average waiting time of the customers at
the fast food jointc. 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 linef. probability that the counter clerk is idleg. Probability of finding the clerk busyh. chances that customer is required to wait
more than 30 minutes in the systemi. probability of having four customer in the
systemJ) 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 = 0.833h) 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 stationFor the system to be workable, λ ≤ μ
C.R.Krishna Prasad, BIT Bangalore-4
Model 2: Poisson-exponential single server model – finite populationHas 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 Characteristicsa) Queue length
average number of customers in queue waiting to get service
b) System length average number of customers in the system
c) Waiting time in queue average waiting time of a customer to get service
d) Total time in system average time a customer spends in the system
e) Server idle time 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 fast-food 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 levelb. average waiting time of th4e customers
at the fast food jointc. 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 linef. probability that the counter clerk is idleg. Probability of finding the clerk busyh. chances that customer is required to wait
more than 30 minutes in the systemi. probability of having four customer in the
systemJ) 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 6Customers 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) The probability that there is no customer at the counter (i.e. that the system is idle)
b) The probability that there are more than 20 customers at the counter
c) The probability that there is no customer waiting to be served
d) The probability that a customer is being served and no body is waiting.
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 2 3 4 5 6 7 8
Maintenace cost
100 250 400 600 900 1250 1600 2000
C.R.Krishna Prasad, BIT Bangalore-4
Problem 2
Year
1 2 3 4 5 6 7 8
Maintenace cost
1000
1200 1400 1800 2300 2800 3400 4000
Resale price
3000 1500 750 375 200 200 200 200
C.R.Krishna Prasad, BIT Bangalore-4
Replacement of machines that deteriorates with time (considering time value of money)
1. Write in a column the running/maintenance costs of machine or equipment for different years Rn
2. 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 investment.
10,000 12000 15000
Total annual cost
2000 1500 1200
life 10 10 10
Salvage value
500 1000 1200
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 2 3 4 5
% of bulbs failing by that month
10 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 2 3 4 5
%of failure at the end of week
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 2 3 4 5 6
%surviving at the end of month
100 97 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
1
4
6
3
5
2
B
3
C
3
A14
D
7
E4
F
10G 0
Construction of the Network and Determination Critical Path
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B
3
C
3
A(0,14)14
D
7
E4
F
10G 0
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B
3
C
3
A(0,14)14
D(14,21)
7
E4
F
10G 0
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B
3
C
3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
100G
F
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B
3
C
3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
F(25,35)
10G 0
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C
3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
F(25,35)
100G
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C
3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
F(25,35)
10G
(14,
14)
0
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
10G
(14,
14)
0
F(25,35)
Determination of ES and EF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14
D(14,21)
7
E(2
1,25
)4
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14
D(14,21)
7
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14
D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3
A(0,14)14 (0,14) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3(22,25)
A(0,14)14 (0,14) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3
C(14,17)3(22,25)
A(0,14)14 (0,14) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3(19,22)
C(14,17)3(22,25)
A(0,14)14 (0,14) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
KeyJob (ES,EF)
Duration (LS,LF)
Determination of LS and LF
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)
3(19,22)
C(14,17)3(22,25)
A(0,14)(0)14 (0,14)(0) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)
Determination of TF and FF
G(1
4,14
)
0(22
,22)
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)3(22,25)
A(0,14)(0)14 (0,14)(0) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)(8)3(22,25)(8)
A(0,14)(0)14 (0,14)(0) D(14,21)
7(14,21)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)(8)3(22,25)(8)
A(0,14)(0)14 (0,14)(0) D(14,21)(0)
7(14,21)(0)
E(2
1,25
)4(
21,2
5)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)(8)3(22,25)(8)
A(0,14)(0)14 (0,14)(0) D(14,21)(0)
7(14,21)(0)
E(2
1,25
)(0)
4(21
,25)
(0)
F(25,35)
10(25,35)G
(14,
14)
0(22
,22)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
1
4
6
3
5
2
B(0,3)(11)
3(19,22)(1
9)
C(14,17)(8)3(22,25)(8)
A(0,14)(0)14 (0,14)(0) D(14,21)(0)
7(14,21)(0)
E(2
1,25
)(0)
4(21
,25)
(0)
F(25,35)(0)
10(25,35)(0)0(
22,2
2)
G(1
4,14
)
Determination of TF and FF
TF( CA) = LS (CA) - ES (CA)
FF(CA) = ES(SA) – EF (CA)
IF = ( ES(SA) – LF(PA)) - Duration of CA
KeyJob (ES,EF)(FF)
Duration (LS,LF)(TS)
C.R.Krishna Prasad, BIT Bangalore-4
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
1 2
4 7
3
6
5
8
(0, 2)(0)
(ES,EF)(FF)
Duration (LS,LF)(TS)Key
2(0, 2)(0)
(2, 5)(0)
3(5, 8)(3)
(5 ,9)(0
)
4(8, 12)(3)
(2, 7)(0)5(2, 7)(0)
(5, 6)(7)1(12, 13)(7)
(7,13) (0)
6(7, 13)(0
)
(7, 9)(0)
2(14,16)(7)
(13, 20)(0)
7(13, 20)(0)
(9,1
3)(0
)4(
16, 2
0)(7
)
(9,17)(0)8(12,20)(3)
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 and Development projects where times cannot be predicted
5.Suitable for civil constructions, installation, ship building etc.
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
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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
2
3 5
4
6
2- 6 -10
4 - 8 - 12
2 - 3 - 4
2 -
4 -
6
0 - 0 - 0
6 - 4 - 10
3 - 6 - 91 - 3
- 5
1.Construction of the Network
C.R.Krishna Prasad, BIT Bangalore-4
1
2
3 5
4
6
2- 6 -10
4 - 8 - 12
2 - 3 - 4
2 -
4 -
6
0 - 0 - 0
6 - 4 - 10
3 - 6 - 91 - 3
- 5
2. Calculation of Expected time for all the activities
6
3
8
40
6
10
3
Expected Time ( te):‘te’ can be calculated by the following formula te = (to + 4tm + tp) / 6
C.R.Krishna Prasad, BIT Bangalore-4
1
2
3 5
4
6
2- 6 -10
4 - 8 - 12
2 - 3 - 4
2 -
4 -
6
0 - 0 - 0
6 - 4 - 10
3 - 6 - 91 - 3
- 5
3. Determination of Critical Path
6
3
8
40
6
10
3
Expected Duration of the project Te = 20 days
Keyto - tm - tp
te
C.R.Krishna Prasad, BIT Bangalore-4
Activity to tm tp Critical activities
σ2 = ((t p – t o) / 6)2
1-2 2 6 10 1-2 1.78
1-3 4 8 12 -
2-3 2 4 6 2-3 0.44
2-4 2 3 4 -
3-4 0 0 0 3-4 0
3-5 3 6 9 -
4-6 6 10 14 4-6 1.78
5-6 1 3 5 -
Σ σ2 = 4.00
C.R.Krishna Prasad, BIT Bangalore-4
σNetwork = √Sum of variances along the
Critical Path
= √ (σNetwork )2
= √ 4
= 2
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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 theirtime 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
3
2
1 6
4 5
6 -12 - 30
4 - 19 - 28
3 - 6 - 15 2 - 5 - 8
3 - 9 - 27
3 - 9 - 27 6 -12 - 3
0
1.Construction of the Network
C.R.Krishna Prasad, BIT Bangalore-4
3
2
1 6
4 5
6 -12 - 30
4 - 19 - 28
3 - 6 - 15 2 - 5 - 8
3 - 9 - 27
3 - 9 - 27 6 -12 - 3
0
2. Calculation of Expected time for all the activities
Expected Time ( te):‘te’ can be calculated by the following formula te = (to + 4tm + tp) / 6
14
18
7 5
11 14
1 - 4 - 7
4
11
C.R.Krishna Prasad, BIT Bangalore-4
3
2
1 6
4 5
6 -12 - 30
4 - 19 - 28
3 - 6 - 15 2 - 5 - 8
3 - 9 - 27
3 - 9 - 27 6 -12 - 3
0
14
18
7 5
11 14
1 - 4 - 7
4
11
3. Determination of Critical Path
Expected Duration of the project Te = 32 days
Keyto - tm - tp
te
C.R.Krishna Prasad, BIT Bangalore-4
Activity σ2 = ((t p – t o) / 6)2 σ2
1-2 ((30 – 6)/6)2 16
2-6 ((28 – 4)/6)2 16
As there are two Critical Paths, the path which gives more variance(σ2) is taken as Critical Path
Path A
Σ σ2 = 32.00
C.R.Krishna Prasad, BIT Bangalore-4
Activity σ2 = ((t p – t o) / 6)2 σ2
1-3 ((15 – 3)/6)2 4
3-5 ((27 – 3)/6)2 16
5-6 ((30 – 6)/6)2 16
Path B
Σ σ2= 36.00σ = √Σ σ2 =√ 36 = 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
TS = ? (to be calculated) ,TE = 32, σ = 6
i.e. 1.28 = (TS– 32) / 6
TS = 39.68 days
σ =
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
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
Differences between PERT & CPM
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 and Development projects where times cannot be predicted
5.Suitable for civil constructions, installation, ship building etc.
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
Crash Normal
Job duration
Dir
ect
cost
Shorter the duration higher will be the Direct expenses
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Job durationCrash Normal
In
dir
ect
cost
Shorter the duration lesser will be the Indirect expenses
C.R.Krishna Prasad, BIT Bangalore-4
C
ost
Job duration
NormalOptimumCrash
Indirect cost
Direct cost
Total cost
C.R.Krishna Prasad, BIT Bangalore-4
Find the lowest cost and optimum cost schedule for the following project, given the over head expenses as Rs.45/-day.
Activity Normal duration
Crashduration
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 2
3
4
1.Construction of the Network
3 -1
7 - 3
4 - 2
5 - 240
10
40
20
KeyNormal duration – Crash duration
Cost of crashing per day in Rs.
C.R.Krishna Prasad, BIT Bangalore-4
1 2
3
4
2.Determination of Critical path
3 -1
7 - 3
4 - 2
5 - 240
10
40
20
KeyNormal duration – Crash duration
Cost of crashing per day in Rs.
C.R.Krishna Prasad, BIT Bangalore-4
1 32 4
4
3 -1 4 - 2 5 - 2
7 - 3
40 40 20
1012 days
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
Step 1.
C.R.Krishna Prasad, BIT Bangalore-4
1 32 43 -1 4 - 2 3 - 2
7 - 3
40 40 20
1010 days
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
Step 2.
C.R.Krishna Prasad, BIT Bangalore-4
1 32 43 -1 4 - 2 2 - 2
6 - 3
40 40 20
109 days
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
Step 3.
C.R.Krishna Prasad, BIT Bangalore-4
1 32 41 -1 4 - 2 2 - 2
6 - 3
40 40 20
107 days
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
Step 4.
C.R.Krishna Prasad, BIT Bangalore-4
1 32 43 -1 2 - 2 2 - 2
4 - 3
40 40 20
105 days
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
Step 5.
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
Differences between PERT & CPM
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 and Development projects where times cannot be predicted
5.Suitable for civil constructions, installation, ship building etc.