BF 360
Operations Research
Lecturer: Mwale Richard
July 31, 2022
Operations Research
BF 360
Contents
1 INTRODUCTION TO OPERATIONS RESEARCH
5
1.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.2
Phases and Process of O.R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.3
Tools and Techniques of O.R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.4
Application of O.R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1.5
Limitation of O.R
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1.6
Tutorial sheet 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2 INTRODUCTION TO LINEAR PROGRAMMING-GRAPHICAL METHOD
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2.1
Introduction to Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2.2
Linear Programming Problem Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2.3
Solving the LP problem graphically . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2.4
Types of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2.5
Tutorial sheet 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 LINEAR PROGRAMMING-SIMPLEX METHOD
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3.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3.2
Big M Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3.3
Two Phase Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3.4
Tutorial Sheet 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 DUALITY OF A LINEAR PROGRAMMING PROBLEM
4.1
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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4.2
Dual Problem Formulation and Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4.3
Procedure for converting a primal into a dual and vice versa . . . . . . . . . . . . . . . . . . .
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4.4
Tutorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5 TRANSPORTATION PROBLEM
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5.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
5.2
Transportation Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.3
North West Corner Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.4
Least Cost Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.5
Vogel Approximation Method (VAM)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.6
Unbalanced Transportation Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
5.7
Transshipment Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.8
Maximization case of Transportation Problem . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.9
Tutorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6 ASSIGNMENT PROBLEM
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6.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6.2
Assignment Problem Structure and Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6.3
Unbalanced Assignment problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6.4
Maximization in an Assignment Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6.5
Diagonal Rule of the Assignment Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6.6
Tutorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7 NETWORK PROBLEM
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7.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.2
Project Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.2.1
Component of a Project Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.2.2
Construction of a Project Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Shortest Path Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.3.1
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7.3
Steps in Shortest path Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4
Critical Path Method
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.5
Project Evaluation & Review Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.6
Earliest and Latest Times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.6.1
Earliest Times of an Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.6.2
Latest Times of an Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.6.3
Total Float of an Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.6.4
Expected Times of an Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Tutorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.7
8 QUEUING THEORY
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8.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.2
Queuing System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.2.1
Input (arrival pattern) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.2.2
Inter-arrival times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.2.3
Service Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.2.4
Queuing discipline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.2.5
Customer’s behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.3
Classification of Queuing Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.4
Model 1: (M/M/1) : (GD/∞/∞) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.5
Model 2: (M/M/C) : (GD/ ∞/∞) Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8.6
Tutorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9 DECISION THEORY
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9.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.2
Decision under condition of Certainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.3
Decision under condition of Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.4
Decision under condition of Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.4.1
Optimistic (Maxmax) Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.4.2
Pessimistic (Maxmin) Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.4.3
Minmax Regret Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.4.4
Laplace (Equal Probability) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.4.5
Hurwicz (Realism) Criterion
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Tutorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9.5
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INTRODUCTION TO OPERATIONS RESEARCH
Objectives
At the end of the session, you should be able to:
• Understand the meaning, purpose, and tools of Operations research.
• Describe the stages of O.R.
• Explain the application of operations research
• Describe the limitation of O.R.
1.1
Introduction
Operation Research is a relatively new field. The contents and the boundaries of the field are not
fixed yet. Therefore, to put up a standard definition of the term Operations Research is quite difficult of
a task. The OR begins when mathematical and quantitative techniques are used to support the decision
being taken. The main job of a manager or CEO is the decision making. In our days to day life we make
the decisions even without paying much attention to them. The decisions are made simply by the use of
common sense, intuition ,judgement and expertise without the use of any mathematical or any statistical
model in simple situations. But the decision we are concerned here with are complex and huge responsible.
Examples are public transportation network planning in a city having its own layout of factories, residential
blocks or finding the appropriate product mix when there exists a large number of products with different
profit contributions and production requirement etc.
Operations research can be regarded as science in the because of its description, understanding and prediction
of the systems behaviour, especially man-machine system. The OR experts basically undergo the following
three classical aspect of science,these are:
(i) Determining the system behaviour
(ii) Analysing the system behaviour by developing appropriate models
(iii) Predict/estimating the future behaviour using the developed models
The significance on the analysis of operations entirely distinguishes O.R. from other research and engineering.
Operations research is an interdisciplinary discipline which provided solution to problem of the military
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operations during the World War II, and also successful in other operations.
As highlighted earlier, defining O.R. is a difficult task. The definition given by some experts and societies
on the subject together to help us understand O.R. According to the Operational research of Great Britain,
they defined O.R as follows:
Definition 1.1. Operations research is defined as a management science which deals with complex problems
arising in the direction and management of large system of men, machine, material and money in industry,
business, government and defense.
Operations Research also can be defined as the use of quantitative methods to assist analysts and
decision-makers in designing, analyzing, and improving the performance or operation of systems.
1.2
Phases and Process of O.R.
The phases and process of O.R. are also called the stages of development of O.R. There are basically
six important steps to follow in O.R. These are;
Step I: Observe the problem Environment
The first step in the process of O.R. development is observing the environment of the problem. This step
involves, site visit, research, observation and so on.... . These activities helps to provide sufficient information
to the O.R. specialist to formulate the problem.
Step II: Analysis and Defining of the problem
This step deals with the analysis and definition of the problem. Under this stage in addition to the problem
definition, the objectives,uses and limitation of O.R. study of the problem are also defined. The results of
this step is to have a clear picture of the solution and its nature.
Step III: Develop a Model
This step deals with the development of a model representing some real situation. The model are basically
mathematical model which helps to describe the system or process in form of equations. The different
activities in the stage are variable definition, formulating equations etc. The developed model is then tested
in the field under different environmental constraints and modelled in order to work. In some case the model
is modified to satisfy the management with the results.
Step IV: Select appropriate data input
With appropriate data inputs, the model works perfectly. Therefore selecting appropriate input data is very
cardinal step in O.R. development process This step includes the following activities; internal/external data
analysis, fact analysis, and collection of opinions an also the use of computer data banks. The objective of
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the stage is to provide sufficient data input to operate and test the model developed in the previous step.
Step V: Provide a solution and test its reasonableness
In this step we get a solution with the help of the model and data input. The solution is not implemented
immediately, but instead the solution is used to test the model and find out if there are any limitations.
If the solution is found to be unreasonable or the behaviour of the model is not perfect, then then model
is updated and modified at this stage. The results of this step is the solution(s) that supports the current
organizational objectives.
Step I: Implement the solution
The final stage in the O.R. process involve implementing the solution obtained in step V.
1.3
Tools and Techniques of O.R.
Operations research uses tools and techniques to find the solutions of the problems. The commonly
used tools/techniques are mathematical procedures, cost analysis, electronic computation. However there
are many tools/techniques that exist. Below are some of the tool and techniques used in O.R.
Linear Programming
This is a constrained optimisation technique, which optimise some criterion within some constraints. In
Linear programming the objective function (profit, loss or return on investment) and constraints are linear.
There are different methods available to solve linear programming.
Game Theory
This is used for making decisions under conflicting situations where there are one or more players/opponents.
In this the motive of the players are divided. The success of one player tends to be at the cost of other
players and hence they are in conflict.
Decision Theory
Decision theory is concerned with making decisions under conditions of complete certainty about the future
outcomes and under conditions such that we can make some probability about what will happen in future.
Queuing Theory:
This is used in situations where the queue is formed (for example customers waiting for service, aircrafts
waiting for landing, jobs waiting for processing in the computer system, etc). The objective here is minimizing
the cost of waiting without increasing the cost of servicing.
Inventory Models: Inventory model make a decisions that minimize total inventory cost. This model
successfully reduces the total cost of purchasing, carrying, and out of stock inventory.
Simulation
Simulation is a procedure that studies a problem by creating a model of the process involved in the problem
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and then through a series of organized trials and error solutions attempt to determine the best solution.
Some times this is a difficult/time consuming procedure. Simulation is used when actual experimentation is
not feasible or solution of model is not possible.
Non-linear Programming: This is used when the objective function and the constraints are not linear
in nature. Linear relationships may be applied to approximate non-linear constraints but limited to some
range, because approximation becomes poorer as the range is extended. Thus, the non-linear programming
is used to determine the approximation in which a solution lies and then the solution is obtained using linear
methods.
Dynamic Programming:
Dynamic programming is a method of analyzing multistage decision processes. In this each elementary
decision depends on those preceding decisions and as well as external factors.
Integer Programming:
If one or more variables of the problem take integral values only then dynamic programming method is used.
For example number or motor in an organization, number of passenger in an aircraft, number of generators
in a power generating plant, etc.
Markov Process:
Markov process permits to predict changes over time information about the behaviour of a system is known.
This is used in decision making in situations where the various states are defined. The probability from one
state to another state is known and depends on the current state and is independent of how we have arrived
at that particular state.
Network Scheduling:
This technique is used extensively to plan, schedule, and monitor large projects (for example computer
system installation, R & D design, construction, maintenance, etc.). The aim of this technique is minimize
trouble spots (such as delays, interruption, production bottlenecks, etc.) by identifying the critical factors.
The different activities and their relationships of the entire project are represented diagrammatically with
the help of networks and arrows, which is used for identifying critical activities and path. There are two main
types of technique in network scheduling, they are: Program Evaluation and Review Technique (PERT) – is
used when activities time is not known accurately/ only probabilistic estimate of time is available. Critical
Path Method (CPM) – is used when activities time is know accurately.
Information Theory:
This analytical process is transferred from the electrical communication field to O.R. field. The objective
of this theory is to evaluate the effectiveness of flow of information with a given system. This is used
mainly in communication networks but also has indirect influence in simulating the examination of business
organizational structure with a view of enhancing flow of information.
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Application of O.R.
Today, almost all fields of business and government utilize the benefits of Operations Research.
There are many of applications of Operations Research. However it is impossible to cover all applications of
O.R. in brief. The following are the abbreviated set of typical operations research applications to show how
widely these techniques are used today:
Accounting:
Assigning audit teams effectively, Credit policy analysis, Cash flow planning, Developing standard costs,
Establishing costs for byproducts, and Planning of delinquent account strategy.
Construction:
Project scheduling, monitoring and control, Determination of proper work force, Deployment of work force,
and Allocation of resources to projects.
Facilities Planning:
Factory location and size decision, Estimation of number of facilities required, Hospital planning, International logistic system design, Transportation loading and unloading, and Warehouse location decision.
Finance:
Building cash management models, Allocating capital among various alternatives, Building financial planning models, Investment analysis, Portfolio analysis, and Dividend policy making.
Manufacturing:
Inventory control , Marketing balance projection, Production scheduling,and Production smoothing.
Marketing:
Advertising budget allocation, Product introduction timing, Selection of Product mix, and Deciding most
effective packaging alternative.
Organizational Behaviour / Human Resources:
Personnel planning, Recruitment of employees, Skill balancing Training program scheduling, and Designing
organizational structure more effectively.
Purchasing:
Optimal buying, Optimal reordering, and Materials transfer
Research and Development:
R & D Projects control, R & D Budget allocation, and Planning of Product introduction.
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Limitation of O.R
Operations Research has number of applications; in the same way it also has certain limitations.
The limitations are mostly associated with the the model building and money and time factors problems
involved in its application. Some of them are as given below:
• Distance between O.R. specialist and Manager: Operations Researchers job needs a mathematician or statistician, who might not be aware of the business problems. Similarly, a manager is unable
to understand the complex nature of Operations Research. Thus there is a big gap between the two
personnel.
• Money and Time Costs: The basic data are subjected to frequent changes, incorporating these
changes into the operations research models is very expensive. However, a fairly good solution at
present may be more desirable than a perfect operations research solution available in future or after
some time.
• Non-quantifiable Factors: When all the factors related to a problem can be quantifiable only then
operations research provides solution otherwise not. The non-quantifiable factors are not incorporated
in O.R. models. Importantly O.R. models do not take into account emotional factors or qualitative
factors.
• Implementation: Once the decision has been taken it should be implemented. The implementation
of decisions is a delicate task. This task must take into account the complexities of human relations
and behavior and in some times only the psychological factors.
• Magnitude of Calculations: The aim of the O.R. is to find out optimal solution taking into consideration all the factors. In this modern world these factors are enormous and expressing them in
quantitative model and establishing relationships among these require voluminous calculations, which
can be handled only by machines.
1.6
Tutorial sheet 1
1. Define Operations Research.
2. Explain the various steps in the O.R. development process.
3. Discuss the applications of O.R
4. Discuss the limitation of O.R.
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5. Describe different techniques of O.R.
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INTRODUCTION TO LINEAR PROGRAMMING-GRAPHICAL
METHOD
Objectives
At the end of the session, you should be able to:
• Formulate Linear Programming Problem.
• Identify the characteristics of linear programming problem.
• Solve the problem graphically
• Identify the various types of solutions
2.1
Introduction to Linear Programming
Linear programming is a versatile technique which can be applied to variety of management prob-
lem, these are; Advertising, Distribution, Production, Refinery operation, and Transport analysis. Linear
programming is useful not only in industry and but also in non-profit organisations such as Education, Government, Hospital, and Libraries. The linear programming method is applicable in problems characterized
by decision variables. The objective function and the constraint can be expressed as linear function of the
decision variables.
Terminologies used in Linear programming
Decision variables can be defined as a quantity that a decision marker controls. It represents the level of
achievement. The solution of the linear programming will give the optimal value for each and every variable
of the model, and can be denoted by x1 , x2 .
An objective function represents some principal objective criterion or goal that measures the effectiveness
of the system such as maximizing profits or productivity, or minimizing cost or consumption. The general
format of the objective function is;
Z = c1 x1 + c2 x2 + · · · + cn xn
Objective Function Coefficient is a constant representing the cost or profit per unit of carrying out an
activity. E.g. Let c1 , c2 , . . . , cn are the cost per unit of products a1 , a2 , . . . , an respectively.
Technological Coefficient (aij ) is the amount of resource i required for activity j, where i ranges from 1
to m and j ranges from 1 to n.
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Set of constraint: A constraint is some kind of restriction/limitation on the total amount of a particular resource required to carry out the activities at various level. There is always some practical limitation/restriction on the availability of resources these are; man, material, machine, or time for the system.
These constraints are expressed as linear equations involving the decision variables. Solving a linear programming problem means determining actual values of the decision variables that optimize the objective
function subject to the limitation imposed by the constraints. General format of constraints is;
a11 x1 + a12 x2 + · · · + a1n xn ≥, ≤, or = b1
The main important feature of linear programming model is the presence of linearity in the problem. Thus
the amount of resources required fro a given activity is directly proportional to the level of that activity.
Non-negativity is another assumption made under linear programming. The non-negativity assumption permits the decision variables to have positive values only.
2.2
Linear Programming Problem Formulation
The linear programming problem formulation is demonstrated through a product mix problem.
The product mix problem occurs in an industry where it is possible to manufacture a variety of products.
A product has a certain margin of profit per unit, and uses a common pool of limited resources. In this
case the linear programming technique identifies the products combination which will maximize the profit
subject to the availability of limited resource constraints.
Example 2.1. Suppose an industry is manufacturing two types of products P1 and P2 . The profits per Kg
of the two products are K30 and K40 respectively. These two products require processing in three types of
machines. The following table shows the available machine hours per day and the time required on each
machine to produce one Kg of P1 and P2 . Formulate the problem in the form of linear programming model.
Profit/kg
P1
P2
Total machine hour/day
Machine 1
3
2
600
Machine 2
3
5
800
Machine 3
5
6
1100
Solution
The procedure for linear programming problem formulation is as follows:
Variable definition
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Let x1 and x2 be the amount of product 1 P1 and product 2 P2
Objective: Maximize the profit, thus objective function
Z = 30x1 + 40x2
Constraints:
Since one Kg of P1 requires 3 hours of processing time in machine 1 while the corresponding requirement of
P2 is 2 hours. So, the first constraint can be expressed as
3x1 + 2x2 ≤ 600
In the similar way the constraints corresponding to machine 2 and 3 are;
3x1 + 5x2 ≤ 800
3x1 + 6x2 ≤ 1100
Furthermore there is no negative production, which may be represented algebraically as
x1 ≥ 0
x2 ≥ 0
Therefore, the product mix problem in the linear programming model is as follows:
Maximize : Z = 30x1 + 40x2
Subject to :
3x1 + 2x2 ≤ 600
3x1 + 5x2 ≤ 800
5x1 + 6x2 ≤ 1100
x1 ≥ 0
x2 ≥ 0
Example 2.2. A company owns two flour mills viz. A and B, which have different production capacities
for high, medium and low quality flour. The company has entered a contract to supply flour to a firm every
month with at least 8, 12 and 24 quintals of high, medium and low quality respectively. It costs the company
K2000 and K1500 per day to run mill A and B respectively. On a day, Mill A produces 6, 2 and 4 quintals
of high, medium and low quality flour, Mill B produces 2, 4 and 12 quintals of high, medium and low quality
flour respectively. Formulate the LPP so that each mill is operated in order to meet the contract order most
economically.
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Solution
Let x1 and
2
be the number of days of operating the mills A and B.
Here the objective is to minimize the cost of the machine runs and to satisfy the contract order.
The linear programming problem is given by
Minimize : Z = 2000x1 + 41500x2
Subject to :
6x1 + 2x2 ≥ 8
2x1 + 4x2 ≥ 12
4x1 + 12x2 ≥ 24
x1 ≥ 0
2.3
x2 ≥ 0
Solving the LP problem graphically
A two variable linear programming problem can be easily solved using graphical method. The
method is simple but the principle of solution is depends on certain analytical concepts, they are:
Convex Region:
A region R is convex if and only if for any two points on the region R the line connecting those points lies
entirely in the region R.
Extreme Point:
The extreme point E of a convex region R is a point such that it is not possible to locate two distinct points
in R, so that the line joining them will include E. The extreme points are also called as corner points or
vertices.
Thus, the following result provides the solution to the linear programming model:
”If the minimum or maximum value of a linear function defined over a convex region exists, then it must be
on one of the extreme points”.
Feasible Region:
The collection of all the feasible solution is called as the feasible region.
Example 2.3. Consider the maximization problem described in example 2.1.
Maximize : Z = 30x1 + 40x2
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Subject to :
3x1 + 2x2 ≤ 600
3x1 + 5x2 ≤ 800
5x1 + 6x2 ≤ 1100
x1 ≥ 0
x2 ≥ 0
Solve the LP problem graphically
Solution
Let
3x1 + 2x2 = 600
2x1 + 5x2 = 800
5x2 + 6x2 = 1100
For 3x1 + 2x2 = 600
x1
0
200
x2
300
0
For 3x1 + 5x2 = 800
x1
0
266.7
x2
160
0
x1
0
220
x2
183.3
0
And for 5x1 + 6x2 = 1100
We graph the constraint and shade the feasible region
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Figure 2.1
The following Table 1 shows the calculation of maximum value of the objective function
Extreme points
x1
x2
Objective function Z = 30X1 + 40x2
A
0
0
0
B
0
160
6400
C
100
100
7000
D
175
37.5
6750
E
200
0
6000
Table 2.1
The value of Z is 7000, which is the objective function maximum value. The optimum value variables are
x1 = 100 and X2 = 100.
Interpretation:We therefore require the industry to manufacture 100kg and 100kg of product P1 and P2
respectively in order to maximize the profits of the two products
Example 2.4. Consider the minimization problem described in example 2.2.
Minimize : Z = 2000x1 + 1500x2
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Subject to :
6x1 + 2x2 ≥ 8
2x1 + 4x2 ≥ 12
4x1 + 12x2 ≥ 24
x1 ≥ 0
x2 ≥ 0
Solve the LP problem graphically
Solution
In a similar way we have;
Figure 2.2
The following Table 2 shows the calculation of minimum value of the objective function
Extreme points
x1
x2
Objective function Z = 2000X1 + 15000x2
A
0
4
6000
B
0.4
2.8
5000
C
6
0
12000
Table 2.2
The value of Z is 500, which is the objective function minimum value. The optimum value variables are
x1 = 0.4 and X2 = 2.8.
Interpretation:We therefore require mill A and mill B to operate 0.4 days and 2.8 days respectively in order
to minimize cost of the machine runs and to satisfy the contract order
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Types of Solutions
Multiple Optimal Solutions
When the objective function passed through only the extreme point located at the intersection of two half
planes, then the linear programming problem possess unique solutions. As we saw in example 2.3 and
example 2.4 (which possessed unique solutions). However, the objective function coincides with one of the
half planes generated by the constraints in the problem, will possess multiple optimal solutions. Consider
the example below.
Example 2.5. A company purchasing scrap material has two types of scarp materials available. The first
type has 30% of material X, 20% of material Y and 50% of material Z by weight. The second type has 40%
of material X, 10% of material Y and 30% of material Z. The costs of the two scraps are K120 and K160
per kg respectively. The company requires at least 240 kg of material X, 100 kg of material Y and 290 kg of
material Z. Find the optimum quantities of the two scraps to be purchased so that the company requirements
of the three materials are satisfied at a minimum cost.
Solution
Firstly we formulate the linear programming model. Let us introduce the decision variables x1 and x2
denoting the amount of scrap material to be purchased. Here the objective is to minimize the purchasing
cost. So, the objective function here is
Minimize : Z = 120x1 + 160x2
Subject to :
0.3x1 + 0.4x2 ≥ 240
0.2x1 + 0.1x2 ≥ 100
0.5x1 + 0.3x2 ≥ 290
x1 ≥ 0
x2 ≥ 0
By graphing the constraints we have
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Figure 2.3
The following Table 2 shows the calculation of minimum value of the objective function
Extreme points
x1
x2
Objective function Z = 120X1 + 160x2
A
0
1000
160000
B
100
800
140000
C
400
300
96000
D
800
0
96000
Table 2.3
The extreme points are A, B, C, and D. One of the objective functions 120x1 + 160x2 = Z family coincides
with the line CD at the point C with value Z = 96000, and the optimum value variables are x1 = 400, and
x2 = 300.
And at the point D with value Z = 96000, and the optimum value variables are x1 = 800, and x2 = 0.
Thus, every point on the line CD minimizes objective function value and the problem contains multiple
optimal solutions.
Unbounded Solution
When the feasible region is unbounded, a maximization problem may not have optimal solution, since the
values of the decision variables may be increased arbitrarily. This is illustrated with the help of the following
problem.
Example 2.6.
Maximize : Z = 3x1 + x2
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Subject to :
x1 + x2 ≥ 6
−x1 + x2 ≤ 6
−x1 + 2x2 ≥ −6
x1 ≥ 0
x2 ≥ 0
Solution
By graphing the constraints we have
Figure 2.4
Figure 2.4 shows the unbounded feasible region and demonstrates that the objective function can be made
arbitrarily large by increasing the values of x1 and x2 within the unbounded feasible region.
In this case, we do not have a point (x1 , x2 ) that is optimal because there are always other feasible points
for which objective function is larger.
Note that it is not the unbounded feasible region alone that prevents an optimal solution. The minimization
of the function subject to the constraints shown in the figure 2.4 would be solved at one the extreme point (A
or B). The unbounded solutions typically come into existence because some real constraints, which represent
a practical resource limitation, have been missed from the linear programming formulation. In such situation
the problem needs to be reformulated and re-solved.
Infeasible Solution
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A linear programming problem is said to be infeasible if no feasible solution of the problem exists. This
section describes infeasible solution of the linear programming problem with the help of the following.
Example 2.7.
Minimize : Z = 200x1 + 300x2
Subject to :
0.4x1 + 0.6x2 ≥ 240
0.2x1 + 0.2x2 ≤ 80
0.4x1 + 0.3x2 ≥ 180
x1 ≥ 0
x2 ≥ 0
Solution
By graphing the constraints we have
Figure 2.5
The region purple includes all the solutions which satisfy the first constraint (4x1 + 6x2 ≥ 2400) and the
third in pink contains all solution which satisfy the constraint (4x1 + 3x2 ≥ 1800). The region in green
contains all solutions which satisfy the second constraint (2x1 + 2x2 ≤ 800).
Hence, there is no solution satisfying all the three constraints (first, second, and third). Thus, the linear
problem is infeasible
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Tutorial sheet 2
1. Suppose an industry is manufacturing tow types of products P1 and P2 . The profits per Kg of the
two products are K30 and K40 respectively. These two products require processing in three types of
machines. The following table shows the available machine hours per day and the time required on each
machine to produce one Kg of P1 and P2 . Formulate the problem in the form of linear programming
model.
Profit/Kg
P1 (K30)
P2 (K40)
Total available Machine hours/day
Machine 1
3
2
600
Machine 2
3
5
800
Machine 3
5
6
1100
2. A juice company has its products viz. canned apple and bottled juice with profit margin K4 and K2
respectively pre unit. The following table shows the labour, equipment, and ingredients to produce
each product per unit.
Canned Apple
Bottle Juice
Total available Machine hours/day
Labour
3.0
2.0
12.0
Equipment
3.2
1.0
8.0
Ingredients
2.4
2.0
9.0
Formulate the linear programming problem (model) specifying the product mix which will maximize
the profit without exceeding the levels of resources.
3. A company owns two flour mills viz. A and B, which have different production capacities for high,
medium and low quality flour. The company has entered a contract to supply flour to a firm every
month with at least 8, 12 and 24 quintals of high, medium and low quality respectively. It costs the
company K2000 and K1500 per day to run mill A and B respectively. On a day, Mill A produces 6,
2 and 4 quintals of high, medium and low quality flour, Mill B produces 2, 4 and 12 quintals of high,
medium and low quality flour respectively. How many days per month should each mill be operated
in order to meet the contract order most economically.
4. A company purchasing scrap material has two types of scarp materials available. The first type has 30%
of material X, 20% of material Y and 50% of material Z by weight. The second type has 40material
X, 10% of material Y and 30% of material Z. The costs of the two scraps are K120 and K160 per kg
respectively. The company requires at least 240 kg of material X, 100 kg of material Y and 290 kg
of material Z. Find the optimum quantities of the two scraps to be purchased so that the company
requirements of the three materials are satisfied at a minimum cost.
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5. An organization is interested in the analysis of two products which can be produces from the idle time
of labour, machine and investment. It was notified on investigation that the labour requirement of the
first and the second products was 4 and 5 units respectively and the total available man hours was
48. Only first product required machine hour utilization of one hour per unit and at present only 10
spare machine hours are available. Second product needs one unit of byproduct per unit and the daily
availability of the byproduct is 12 units. According to the marketing department the sales potential of
first product cannot exceed 7 units. In a competitive market, first product can be sold at a profit of K6
and the second product at a profit of K10 per unit. Formulate the problem as a linear programming
model. Also determine graphically the feasible region. Identify the redundant constraints if any.
6. Find graphically the feasible region of the linear programming problem given in Q2.
7. A bed mart company is in the business of manufacturing beds and pillows. The company has 40
hours for assembly and 32 hours for finishing work per day. Manufacturing of a bed requires 4 hours
for assembly and 2 hours in finishing. Similarly a pillow requires 2 hours for assembly and 4 hours
for finishing. Profitability analysis indicates that every bed would contribute K80, while a pillow
contribution is K55 respectively. Find out the daily production of the company to maximize the
contribution (profit).
8. Solve the following LP problems graphically
(a) Maximize:
1170x1 + 1110x2
Subject to:
9x1 + 5x2 ≥ 500
7x1 + 9x2 ≥ 300
5x1 + 3x2 ≤ 1500
7x1 + 9x2 ≤ 1900
2x1 + 4x2 ≤ 1000
x1 , x2 ≥ 0
(b) Minimize:
2x1 + 1.7x2
Subject to:
0.15x1 + 0.10x2 ≥ 1.0
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0.75x1 + 1.70x2 ≥ 7.5
1.30x1 + 1.10x2 ≤ 10
x1 , x2 ≥ 0
(c) Maximize:
2x1 + 3x2
Subject to:
x1 − x2 ≤ 1
x1 + x2 ≥ 3
x1 , x2 ≥ 0
(d) Maximize:
3x1 + 2x2
Subject to:
9x1 + 5x2 ≥ 500
2x1 − 3x2 ≥ 0
3x1 + 4x2 ≤ −12
x1 , x2 ≥ 0
(e) Maximize:
4x1 + 4x2
Subject to:
9x1 + 5x2 ≥ 500
−2x1 + x2 ≤ 1
x1
≤2
x1 + x2 ≤ 3
x1 , x2 ≥ 0
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LINEAR PROGRAMMING-SIMPLEX METHOD
Objectives
• Understand the basics of simplex method.
• Explain the simplex calculations.
• Understand two phase and Big M methods.
3.1
Introduction
Linear Programming with two variables can be solved graphically. The graphical method of solving
linear programming problem is of limited application in the business problems as the number of variables
increases. If the linear programming problem has larger number of variables, the suitable method for solving
is Simplex Method. The simplex method is an iterative process, through which it reaches an optimal solution
to either minimum or maximum value of the objective function.
The simplex method also helps the decision maker/manager to identify the following:
• Redundant Constraints.
• Multiple Solutions.
• Unbounded Solution.
• Infeasible Problem.
Consider the following problem solved using the simplex method
Example 3.1. Solve the following LPP problem using simplex method
Maximize :
12x1 + 16x2
10x1 + 20x2 ≤ 120
Subject to:
8x1 + 8x2 ≤ 80
x1 ≥ 0,
x2 ≥ 0
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Solution:
Procedure
Step 1: Write the LPP problem in standard form : A standard LP problem has ” ≤ ” and the constraints
are positive and then we proceed by introducing the slack variables that is :
Maximize : 12x1 + 16x2 + 0s1 + 0s2
Subject to:
10x1 + 20x2 + s1 +
8x1 + 8x2 +
+s2
= 120
= 80
x1 , x2 , s1 , s2 ≥ 0
Step 2: Construct the simple table.
The simplex table and the following the component:
Cj is the coefficient of the j th term of the objective function and CBi is the coefficient of the ith basic
variable
The value at the intersection of the key row and key column is called the key element. Fro the maximization
problems, the maximum value in Cj − Zj forms the key column. The smallest ratio obtained from division
of the solution and the key column values form the key row. The value of Zj is computed using the formula:
Zj =
X
(CBi )(aij )
Optimality condition: For maximization problems we have that
C j − Zj ≤ 0
or Zj − Cj ≥ 0
then optimality is reached otherwise select the variable with the maximum Cj − Zj as the entering variable.
For minimization problems we have that
C j − Zj ≥ 0
or Zj − Cj ≤ 0
then optimality is reached otherwise select the variable with the minimum Cj − Zj as the entering variable.
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Initial table
Table 3.1
The key column in table 3.1 is the column highlighted in blue, whilst the key row is the one highlighted in
yellow. Thus the key element is highlighted in red.
1st Iteration
Computations: We divide every element in the key row by 20 to get a new row 1. i.e.
N ewR1 =
1
OldR1
20
To find the new elements of row 2 we use the formula below.
New element = Old element −
Corresponding key column element × Corresponding key row element
Key element
As a sample calculation, the computation of the new value of row 2 and column X1 is shown below:
New element = 8 −
8 × 10
=4
20
we proceed in the same way for the rest of the elements in the second row. In this case s1 is the leaving
variable and x2 is the entering variable, thus we obtain the table below
Table 3.2
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We check if the optimality condition Cj − Zj ≤ 0, from the table 3.2 we seen that this condition is not
satisfied. We proceed with the iterations.
From table 3.2, the key column highlighted in blue, whilst the key row is the one highlighted in yellow and
the key element is highlighted in red.
2nd Iteration
Computations: We divide every element in the key row by 4 to get a new row 2. i.e.
N ewR2 =
1
OldR2
4
To find the new elements of row 1 we use the formula below.
New element = Old element −
Corresponding key column element × Corresponding key row element
Key element
we have table 3.3
Table 3.3
Since Cj − Zj ≤ 0, then optimality has been reached, hence we have
x1 = 8,
x2 = 2
and Z = 128
Example 3.2. Solve the following minimization LP problem using simplex method:
Minimize : Z = 2x1 − 3x2 + 6x3
Subject to :
3x1 − x2 + 2x3 ≤ 7
2x1 + 4x2 ≥ −12
−4x1 + 3x2 + 8x3 ≥ 10
x1 , x3 , x2 ≥ 0
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Solution:
Minimize : Z = 2x1 − 3x2 + 6x3
Subject to :
3x1 − x2 + 2x3 ≤ 7
−2x1 − 4x2 ≤ 12
−4x1 + 3x2 + 8x3 ≥ 10
x1 , x3 , x2 ≥ 0
By rewriting in standard form and introducing the sack variables we have:
Minimize : Z = 2x1 − 3x2 + 6x3 + 0s1 + 0s2 + 0s3
Subject to :
3x1 − x2 + 2x3 + s1 +
−2x1 − 4x2 + 0x3 +
−4x1 + 3x2 + 8x3 +
=7
+s2 +
= 12
+s3 = 10
x1 , x3 , x2 , s1 , s2 , s3 ≥ 0
Initial table
Table 3.4
Here the key column is highlighted in blue, and the key row is in yellow and the key element is in red.
1st Iteration
Computations:
N ewR3 =
30
1
OldR3
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Using the formula below to find the new row 1 and new row 2
New element = Old element − [Corresponding key column element × Corresponding New element]
Table 3.5
2nd Iteration
Computations: We get a new row one by dividing through
N ewR1 =
3
5
i.e.
3
OldR1
5
Using the formula below to find the new row 2 and new row 3
New element = Old element − [Corresponding key column element × Corresponding New element]
Table 3.6
Since Cj − Zj ≥ 0, then optimality has been reached, therefore we have
x1 =
31
,
5
x2 =
58
5
and x3 = 0
31
so that
Z=−
112
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Big M Method
When some constraints are of
0
=0 , or
0
≥0 type, then they will not contain any basic variables.
In order to have a basic variables in each of them, we need to introduced a new variable called artificial
variable, If the objective function is a maximization type, then the coefficient of the variable in the objective
function should be −M ; otherwise it should be +M , where M is a large value.
If the constraint contains 0 ≤0 we add a slack variable (+S) and if the constraint has the sign 0 ≥0 then we
subtract the slack variable (−S) and add the artificial variable (+M )
Consider the following example on how to solve a LPP using big M method.
Example 3.3.
Minimize : Z = 7x1 + 15x2 + 20x3
Subject to :
2x1 + 4x2 + 6x3 ≥ 24
3x1 + 9x2 + 6x3 ≥ 30
x1 , x2 , x3 ≥ 0
Solution:
Minimize : Z = 7x1 + 15x2 + 20x3 + 0s1 + 0s2 + 0s3 + M (A1 + A2 )
Subject to :
2x1 + 4x2 + 6x3 − s1 + A1 = 24
3x1 + 9x2 + 6x3 − s2 + A2 = 30
x1 , x2 , x3 , s1 , s2 , s3 , A1 ,
and A2 ≥ 0
Initial table
Table 3.4
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From table 3.4, the key column is highlighted in blue and the key row is the one in yellow, with a key element
in red. In this case our leaving variable is A2 and entering variable is x2 . 1st Iteration
Computations:
We find the new R2 by dividing every element in the old row R2 by the key element i.e.
1
OldR2
9
N ewR2 =
And new R1 is obtained by using this formula
New element = Old element − [Corresponding key column element × Corresponding New element]
Thus we have
Table 3.5
From table 3.5, the key column is highlighted in blue and the key row is the one in yellow, with a key element
in red. In this case our leaving variable is A1 and entering variable is x3 . 2nd Iteration
Computations:
We find the new R1 by dividing every element in the old row R2 by the key element i.e.
N ewR1 =
3
OldR1
10
And new R2 is obtained by using this formula
New element = Old element − [Corresponding key column element × Corresponding New element]
We have;
Table 3.6
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Since Cj − Zj ≥ 0, then optimality has been reached, therefore we have
x1 = 0,
3.3
x2 =
6
5
and x3 =
16
5
so that
Z = 82
Two Phase Simplex Method
When the Big M method is computerized, the value of M is assumed to be very large number. As a
result, we suffer from the truncation/rounding off of the coefficient value while performing different iterations
which will give inaccurate results. In order to overcame this difficulty, two phase simplex method is used
to solve LP problem involving 0 ≤0 or 0 ≥0 type of constraints.
The following is the procedure we follow when using the two phase simplex method;
Phase 1:
1. Form a modified problem for phase 1 from the original one. This is done by replacing the objective
function with the sum of only the artificial variable along the same set of constraints of the original
problem.
2. Prepare the initial table for phase 1
3. Apply the usual simplex method till the optimality is reached.
4. Check whether the objective function value is zero in the optimal table of phase 1. If yes, then go to
phase 2, otherwise conclude that the original problem has no feasible solution and stop.
Phase 2:
1. Obtain a modified table using the following steps;
1.1. Drop the columns in the optimum table of phase 1 corresponding to the artificial variables which
are currently non-basic
1.2. If some artificial variable(s) is (are) present at zero level in the basic solution of the optimal table
of phase 1, substitute its objective function coefficient(s) with zero.
1.3. Substitute the coefficient of the original objective function in the optimal table of the phase 1 for
the remaining variable and make the necessary changes in CB column
2. Perform further iterations till optimality condition is reached and then stop.
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Example 3.4. Consider the following LP model and solve it using the two phase method.
Minimize : Z = 10x1 + 6x2 + 2x3
Subject to :
−x1 + x2 + x3 ≥ 1
3x1 + x2 − x3 ≥ 2
x1 , x2 , x3 ≥ 0
Solution:
Minimize : Z = 10x1 + 6x2 + 2x3 + 0s1 + 0s2 + M (A1 + A2 )
Subject to :
−x1 + x2 + x3 − s1 + A1 = 1
3x1 + x2 − x3 − s2 + A2 = 2
x1 , x2 , x3 , s1 , s2 , s3 , A1 ,
and A2 ≥ 0
Phase 1:
Minimize : Z = A1 + A2
Subject to :
−x1 + x2 + x3 − s1 + A1 = 1
3x1 + x2 − x3 − s2 + A2 = 2
x1 , x2 , x3 , s1 , s2 , s3 , A1 ,
and A2 ≥ 0
Initial table
Table 3.7
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In the table 3.7, the key column is highlighted in blue and the key row is in yellow with a key element
highlighted in red. Thus we have that the leaving variable is A1 and x2 is our entering variable. 1st
Iteration
Computations:
New R1 is just equal to the old R1 i.e.
N ewR1 = OldR1
We can obtain by using new R2 by using the formula
New element = Old element − [Corresponding key column element × Corresponding New element]
so that;
Table 3.8
In the table 3.8, the key column is highlighted in blue and the key row is in yellow with a key element
highlighted in red. Thus we have that the leaving variable is A2 and x1 is our entering variable. 2nd
Iteration
Computations: We obtain the new R2 by dividing every element in the key row by the key element that
is;
N ewR1 =
1
OldR1
4
We can obtain by using new R1 by using the formula
New element = Old element − [Corresponding key column element × Corresponding New element]
so that;
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Table 3.9
Since Cj − Zj ≥ 0, then optimality has been reached, therefore the objective function value is zero in the
optimal table then we go to phase two.
Phase 2:
We replace back the coefficient if the original objective function and check if the optimality is reached. If it
is reached then we end the iteration and conclude otherwise we continue with the iteration till optimality is
reached.
Initial table
Table 3.10
Thus since Cj − Zj ≥ 0, then optimality has been reached, therefore we have
x1 =
3.4
1
,
4
x2 =
5
4
and x3 = 0
so that
Z = 10
Tutorial Sheet 3
1. A soft drinks company has two products these are; Coco-cola and Pepsi with profit of K2 an K1 per
unit. The following table illustrates the labour, equipment and materials to produce per unit of each
product. Determine suitable product mix which maximizes the profit using simplex method.
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2. A factory produces three using three types of ingredients viz. A, B and C in different proportions. The
following table shows the requirements o various ingredients as inputs per kg of the products.
The three profits coefficients are 20, 20 and 30 respectively. The factory has 800 kg of ingredients A,
1800 kg of ingredients B and 500 kg of ingredient C. Determine the product mix which will maximize
the profit and also find out maximum profit.
3. Solve the following LP problems using the simplex method
(a)
Minimize : Z = 2x1 − 3x2 + 6x3
Subject to :
3x1 − x2 + 2x3 ≥ 7
2x1 + 4x2 ≥ −12
−4x1 + 3x2 + 8x3 ≥ −12
x1 , x2 , x3 ≥ 0
(b)
Maximize : P = 3x1 + 4x2 + x3
Subject to :
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x1 + 2x2 + x3 ≤ 6
2x1
+ 2x2 ≤ 4
3x1 + x2 + x3 ≤ 9
x1 , x2 , x3 ≥ 0
3. Solve the following linear programming problem using two phase simplex method and Big M method.
Maximize : Z = 12x1 + 15x2 + 9x3
Subject to :
8x1 + 16x2 + 12x3 ≤ 250
4x1 + 8x2 + 10x3 ≥ 80
7x1 + 9x2 + 8x3 = 105
x1 , x2 , x3 ≥ 0
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DUALITY OF A LINEAR PROGRAMMING PROBLEM
Objectives:
• Introduction
• Dual Problem Formulation and Properties
• Simple Way of Solving Dual Problem
4.1
Introduction
Every linear programming problem has a corresponding linear programming problem called the
dual. Suppose the original problem is a maximization problem then the dual problem is minimization
problem and vice-versa. In either case the final table of the dual problem will contain both the solution
to the dual problem and the solution to the original problem. The solution of the dual problem is readily
obtained from the original problem solution if the simplex method is used.
The formulation of the dual problem also sometimes called duality and is helpful for the understanding
of the linear programming. The original linear programming problem is known as primal problem, and
the derived problem is known as dual problem.The concept of the dual problem is important for several
reasons. Most important are
(i) The variables of dual problem can convey important information to managers in terms of formulating
their future plans and
(ii) In some cases the dual problem can be instrumental in arriving at the optimal solution to the original
problem in many fewer iterations, which reduces the labour of computation.
The variable of the dual problem is known as the dual variables or shadow price of the various
resources. The solution to the dual problem leads us to the solution of the original problem and thus efficient
computational techniques can be developed through the concept of duality. Finally, in the competitive
strategy problem solution of both the original and dual problem is necessary to understand the complete
problem.
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4.2
BF 360
Dual Problem Formulation and Properties
Suppose the original problem is in the standard form then the dual problem can be formulated
using the following rules:
1. The number of constraints in the original problem is equal to the number of dual variables. The number
of constraints in the dual problem is equal to the number of variables in the original problem.
2. The objective function coefficients of the prima appear as right-hand side numbers in the dual and vice
versa.
3. The right hand side elements of the primal appear as objective function coefficients in the dual and
vice versa
4. If the original problem is a maximization problem then the dual problem is a minimization problem.
Similarly, if the original problem is a minimization problem then the dual problem is a maximization
problem.
5. If original problem has less than or equal (≤) to type of constraints then the dual problem will have
greater than or equal (≥) to type constraints.
6. The coefficients of the constraints of the original problem which appear from left to right are placed
from top to bottom in the constraints of the dual problem and vice versa. In other words te input output coefficient matrix of the dual is the transpose of the input – output coefficient matrix of the
primal and vice versa.
The Dual Linear Programming Problem is explained with the help of the following Example 4.1
Example 4.1. The doctor advises a patient visited him that the patient is weak in his health due to shortage
of two vitamins, i.e., vitamin X and vitamin Y. He advises him to take at least 40 units of vitamin X and 50
units of Vitamin Y everyday. He also advises that these vitamins are available in two tonics A and B. Each
unit of tonic A consists of 2 units of vitamin X and 3 units of vitamin Y. Each unit of tonic B consists of 4
units of vitamin X and 2 units of vitamin Y. Tonic A and B are available in the medical shop at a cost of
K3 per unit of A and K2, 50 per unit of B. The patient has to fulfil the need of vitamin by consuming A and
B at a minimum cost.
Solution:
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Example 4.2. A company manufactures two products X and Y on three machines Turning, Milling and
finishing machines. Each unit of X takes, 10 hours of turning machine capacity, 5 hours of milling machine
capacity and 1 hour of finishing machine capacity. One unit of Y takes 6 hours of turning machine capacity,
10 hours of milling machine capacity and 2 hours of finishing machine capacity. The company has 2500 hours
of turning machine capacity, 2000 hours of milling machine capacity and 500 hours of finishing machine
capacity in the coming planning period. The profit contribution of product X and Y are K23 per unit and
K32 per unit respectively. Formulate the linear programming problems and write the dual.
Solution:
4.3
Procedure for converting a primal into a dual and vice versa
Case 1 - Maximization Problem:
The objective function of primal is of maximization type and the structural constraints are of ≤ type. Now
if the basis variables are x, y and z, give different name for variables of dual. Let them be a, b, and c etc.
Now write the structural constraints of dual reading column wise. The coefficients of variables in objective
function will now become the left hand side constants of structural constraints. And the left hand side
constants of primal will now become the coefficients of variables of objective function of dual.
Example 4.3. Consider the LPP given below:
Case 2 - Minimization Problem The procedure is very much similar to that explained in case
1.
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Example 4.4. Consider the LPP given below:
Case 3: When the problem has got both ≥ and ≤ constraints, then depending upon the objective
function convert all the constraints to either ≥ or≤ type. That is, if the objective function is of minimization
type, then see that all constraints are of ≥ type and if the objective function is of maximization type, then
see that all the constraints are of ≤ type. To convert and ≥ to and ≤ or and ≤ to and ≥, simply multiply
the constraint by −1. Once you convert the constraints, then write the dual as explained in case 1 and 2.
Consider the example:
Example 4.5. Consider the LPP given below:
Case 4: When one of the constraint is an equation, then we have to write two versions of the
equation, that is remove equality sign and write ≥ and ≤ sign for each one of them respectively and then
write the dual as usual. Consider the example given below:
Example 4.6. Consider the LPP given below:
4.4
Tutorial
1. Construct the dual problem and find the solution for both the primal and the dual problem of the
following LP problems.
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(a)
Maximize
Z = 5x1 + 2x2 + 6x3 + 4x4
x1 + x2 + x3 + x4 ≤ 140
s.t.
2x1 + 5x2 + 6x3 + x4 ≥ 200
x1 + 3x2 + x3 + 2x4 ≤ 150
x1 , x2 , x3 , x4 ≥ 0
(b)
Maximize
Z = 4x + 3y
2x + 9y ≤ 100
s.t.
3x + 6y ≤ 120
x + y = 50
x, y ≥ 0
(c)
Minimize
s.t.
Z = y1 + 2y2 + 3y3
2y1 + 3y2 − y3 ≥ 20
y1 + 3y2 + 3y3 ≤ 15
y2 + 2y3 = 10
y1 , y2 , y3 ≥ 0
2. The Nitrogen Chemicals of Zambia manufactures 2 brands of fertilizers, Sulpha–X and Super–Nitro.
The Sulpher, Nitrate and Potash contents (in percentages) of these brands are 10–5–10 and 5–10–10
respectively. The rest of the content is an inert matter, which is available in abundance. The company
has made available, during a given period, 1050 tons of Sulpher, 1500 tons of Nitrates, and 2000 tons
of Potash respectively. The company can make a profit of K200/– per tone on Sulpha – X and K300/–
per ton of Super– Nitro. If the object is to maximise the total profit how much of each brand should
be procured during the given period?
(a) Formulate the above problem as a linear programming problem and solve it.
(b) Write the dual of the above problem and Determine it solution.
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5
BF 360
TRANSPORTATION PROBLEM
Objectives:
• Formulation of a Transportation Problem
• Determine basic feasible solution using various methods
• Make unbalanced Transportation Problem into balanced one using appropriate method.
• An extension to the transportation problem.
• Describe suitable method for maximizing the objective function instead of minimizing
5.1
Introduction
A Transportation Problem is a special class of linear programming problem, where the objective
is focused on minimizing the cost of distributing a product from a number of sources (e.g. factories) to a
number of destinations (e.g. warehouses) while satisfying both the supply limits and the demand requirement.Simplex Method is unsuitable for solving Transportation Problem because of the special structure of
the Transportation Problem. The model assumes that the distributing cost on a given route is directly
proportional to the number of units distributed on that route. Generally, the transportation model can be
extended to areas other than the direct transportation of a commodity, including among others, inventory
control, employment scheduling, and personnel assignment.
Consider the example below which illustrate special feature of the transportation problem.
Example 5.1. Suppose a manufacturing company owns three factories (sources) and distribute his products
to five different retail agencies (destinations). The following table shows the capacities of the three factories,
the quantity of products required by the various retail agencies and the cost of shipping one unit of the product
from each of the three factories to each of the five retail agencies.
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The above table is referred as Transportation Table, which provides the basic information regarding the transportation problem. The quantities inside the table are known as transportation cost per
unit of product. The capacity (Supply) of the factories 1, 2, 3 is 50, 100 and 150 respectively. The requirement (Demand) of the retail agency 1, 2, 3, 4, 5 is 100,60,50,50, and 40 respectively.
In this case, the transportation cost of one unit
from factory 1 to retail agency 1 is 1
from factory 1 to retail agency 2 is 9,
from factory 1 to retail agency 3 is 13, and so on.
A transportation problem can be formulated as linear programming problem using variables with two subscripts. Let
x11 = Amount to be transported from factory 1
to retail agency 1
x12 = Amount to be transported from factory 1
to retail agency 2
······
······
······
x35 = Amount to be transported from factory 3
to retail agency 5
Let the transportation cost per unit be represented by C11 , C12 , . . . C35 that is C11 = 1, C12 = 9, and so on.
Let the capacities of the three factories be represented by a1 = 50, a2 = 100, a3 = 150, and also we let the
requirement of the retail agencies are b1 = 100, b2 = 60, b3 = 50, b4 = 50, and b5 = 40
Then the LPP is given by:
Minimize
C11 x11 + C12 x12 + · · · + C35 x35
Subject to
Since the sum must be less than or equal to the available capacity, then we have
x11 + x12 + x13 + x14 + x15 ≤ a1
x21 + x22 + x23 + x24 + x25 ≤ a2
x13 + x32 + x33 + x34 + x35 ≤ a3
Now because the sum must be greater than or equal to the demand off the market. We cannot send less
than what is required
x11 + x21 + x31 ≥ b1
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x12 + x22 + x32 ≥ b2
x13 + x23 + x33 ≥ b3
x14 + x24 + x34 ≥ b4
x15 + x25 + x35 ≥ b5
x11 + x21 + · · · + x35 ≥ 0
However, the problem has 8 constraints and 15 variables. So, it is not possible to solve such a
problem using simplex method. This is the reason for the need of special computational procedure to solve
transportation problem. There are varieties of procedures, which are described in the next section.
5.2
Transportation Algorithm
The steps of the transportation algorithm are exact parallels of the simplex algorithm, they are:
Step 1: Determine a starting basic feasible solution, using any one of the following three methods.
1. North West Corner Method.
2. Least Cost Method.
3. Vogel Approximation Method.
Step 2: Determine the optimal solution using the following method
1. MODI (Modified Distribution Method) or UV Method.
The special structure of the transportation problem allows securing a non artificial basic feasible solution
using one the following three methods. North West Corner Method, Least Cost Method and Vogel
Approximation Method.
The difference among these three methods is the quality of the initial basic feasible solution they produce,
in the sense that a better initial solution yields a smaller objective value. Generally the Vogel Approximation
Method produces the best initial basic feasible solution, and the North West Corner Method produces the
worst, but the North West Corner Method involves least computations.
5.3
North West Corner Method
The method starts at the North West (upper left) corner cell of the tableau (variable x11).
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Step 1: Allocate as much as possible to the selected cell, and adjust the associated amounts of capacity
(supply) and requirement (demand) by subtracting the allocated amount.
Step 2: Cross out the row (column) with zero supply or demand to indicate that no further assignments
can be made in that row (column). If both the row and column becomes zero simultaneously, cross out one
of them only, and leave a zero supply or demand in the uncrossed out row (column).
Step 3: If exactly one row (column) is left uncrossed out, then stop. Otherwise, move to the cell to the
right if a column has just been crossed or the one below if a row has been crossed out. Go to step 1.
Example 5.2. Consider the problem discussed in Example 5.1 to illustrate the North West Corner Method
of determining basic feasible solution.
Solution
50
1
9
1
3
36
5
1
50
24
50 1
2
1
6
20
1
1
4
10
3
3
100
60
50
10
50
1
50
50
23
50
40
50
2
6
40
150
100
50
140
90
40
300
The un-cancelled bold numbers indicates the order in which the allocated amounts are generated. The
starting basic solution is given as X11 = x21 = x22 = 50 x32 = 10 x33 = x34 = 50 and x35 = 40
Total cost = (50 × 1) + (50 × 24) + (50 × 12) + (10 × 33) + (50 × 1) + (50 × 23) + (40 × 26) = 4420
5.4
Least Cost Method
The least cost method is sometimes called matrix minimum method is a method which looks
for the row and the column corresponding to minimum Cij . This method finds a better initial basic feasible
solution by concentrating on the cheapest routes. Unlike in the North West Corner Method, instead of
starting the allocation with the northwest cell, we start by allocating as much as possible to the cell with
the smallest unit cost. If there are two or more minimum costs then we should select the row and the
column corresponding to the lower numbered row. If they appear in the same row we should select the
lower numbered column. We then cross out the satisfied row or column, and adjust the amounts of capacity
(supply) and requirement(demand) accordingly. If both a row and a column is satisfied simultaneously, only
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one is crossed out. Next, we look for the uncrossed-out cell with the smallest unit cost and repeat the process
until we are left at the end with exactly one uncrossed-out row or column.
Example 5.3. The least cost method of determining initial basic feasible solution is illustrated with the help
of Example 5.1.
Solution
50
1
60
2
4
50
1
4
100
9
12
33
60
13
36
16
20
50
1
50
50
5
1
40
23
50
50
1
2
6
40
100
150
60
50
100
300
50
The un-cancelled bold numbers indicates the order in which the allocated amounts are generated. The
starting basic solution is given as X11 = x31 = 50 X22 = 60 X25 = 40 and x33 = x34 = 50.
Total cost = (50 × 1) + (60 × 12) + (50 × 14) + (40 × 1) + (50 × 1) + (50 × 23) = 2710
5.5
Vogel Approximation Method (VAM)
VAM is an improved version of the least cost method that generally produces better solutions.
The steps involved in this method are:
Step 1: For each row (column) with strictly positive capacity (requirement), determine a penalty by subtracting the smallest unit cost element in the row (column) from the next smallest unit cost element in the
same row (column).
Step 2: Identify the row or column with the largest penalty among all the rows and columns. If the penalties
corresponding to two or more rows or columns are equal we select the topmost row and the extreme left
column.
Step 3: We select Xij as a basic variable if Cij is the minimum cost in the row or column with largest
penalty. We choose the numerical value of Xij as high as possible subject to the row and the column constraints. Depending upon whether ai or bj is the smaller of the two ith row or j th column is crossed out.
Step 4: The Step 2 is now performed on the uncrossed-out rows and columns until all the basic variables
have been satisfied.
Example 5.4. Consider the following transportation problem
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Determine the total transportation cost using VAM
Solution:
5.6
Unbalanced Transportation Problem
From the previous section we discussed the balanced transportation problem i.e. when the total
supply (capacity) at the origins is equal to the total demand (requirement) at the destination. In this section
we are going to focus our attention on the the unbalanced transportation problems i.e. when the total supply
is not equal to the total demand, these are called unbalanced transportation problem.
In the unbalanced transportation problem if the total supply is more than the total demand then we introduce
an additional column known as a dummy column which will indicate the surplus supply with transportation
cost zero. Similarly, if the total demand is more than the total supply an additional row known as a dummy
row is introduced in the transportation table which indicates unsatisfied demand with zero transportation
cost.
Example 5.5. Consider the following unbalanced transportation problem
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Solution:
5.7
Transshipment Problem
An extension to the transportation problem is referred as the transshipment problem. Trans-
shipment Problem is a special kind of transportation problem in which commodities are transported from
the sources through some intermediate or trans-shipment points before it reaches its destination.
There are two types of transshipment problem these are;
• Transshipment problem with source and destination acting as transient nodes (Model I)
• Transshipment problem with some transient nodes between sources and destination (Model 2)
Mode I
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Example 5.6. Consider the following transshipment problem involving 4 sources and 2 destinations. The
supply values of the sources S1 ; S2 ; S3 and S4 are 100 units, 200 units, 150 units and 350 units, respectively.
The demand values of the destinations D1 and D2 are 350 units and 450 units, respectively. The transportation cost per unit between different sources and destinations are summarized in the table below. Solve the
transshipment problem.
Steps to solve transshipment problem
• Check whether it is balanced or unbalanced
B =⇒
supply = demand
If the problem is unbalanced we add a dummy row or dummy column.
• Add the value if 0 B 0 to all the rows and the columns
• Find the total transportation cost by using any of these methods
– Least cost method
– Northwest corner
– VAM
• Draw the shipping patterns
Solution:
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5.8
BF 360
Maximization case of Transportation Problem
Basically, the transportation problem is a minimization problem, as the objective function is to
minimize the total cost of transportation. Hence, when we would like to maximize the objective function.
Given a maximization problem, we can subtract all the elements in the matrix from the highest element in
the matrix. Then the problem becomes minimization problem.
Example 5.7. A company has three factories located in three cities viz. X, Y, Z. These factories supplies
consignments to four dealers viz. A, B, C and D. The dealers are spread all over the country. The production
capacity of these factories is 310, 100 and 290 units per month respectively. The net return per unit product
is given in the following table:
Determine a suitable allocation to maximize the total return
Solution:
5.9
Tutorial
1. Four companies. W , X, Y and Z supply the requirements of three warehouses. A, B and C respectively.
The companies’ availability, warehouses requirements and the unit cost of transportation are given in
the following table. Find an initial basic feasible solution using
(a) North-West Corner Method
(b) Least Cost Method
(c) Vogel Approximation Method
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2. The ABT transport company ships truckloads of food grains from three sources X, Y , Z to four mills
A, B, C, D respectively. The supply and the demand together with the unit transportation cost per
truckload on the different routes are described in the following transportation table. Assume that the
unit transportation costs are in hundreds of dollars. Determine the optimum minimum shipment cost
of transportation using VAM
3. An organization has three plants at X, Y , Z which supply to warehouses located at A, B, C, D,
and E respectively. The capacity of the plants is 800, 500 and 900 per month and the requirement of
the warehouses is 400, 400, 500, 400 and 800 units respectively. The following table shows the unit
transportation cost.
Determine an optimum distribution for the organization in order to minimize the total cost of transportation using VAM.
4. Consider the following transportation problem where the origins are plants and destinations are depots
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In addition to the above, suppose that the unit cost of transportation from plant to plant and from
depot to depot are as:
And hence
Find out minimum transshipment cost of the problem and also compare this cost with the corresponding
minimum transportation cost.
5. According to the following type and quantities of dresses
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Four different dress makers are submitted the tenders, who undertake to supply not more than the
quantities indicated below:
The store estimates that its profit per dress will vary according to the dress maker as indicates in the
following table:
Determine how should the orders to be places for the dresses so as to maximize the profit.
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6
BF 360
ASSIGNMENT PROBLEM
Objectives:
• Introduction
• Assignment Problem Structure and Solution
• Unbalanced Assignment Problem
• Infeasible Assignment Problem
• Maximization in an Assignment Problem
6.1
Introduction
The Assignment Problem can define as follows: Given n facilities, n jobs and the effectiveness of
each facility to each job, here the problem is to assign each facility to one and only one job so that the
measure of effectiveness if optimized. Here the optimization means Maximized or Minimized.
There are many management problems have an assignment problem structure.
For example, the head of the department may have 6 people available for assignment and 6 jobs to fill. Here
the head may like to know which job should be assigned to which person so that all tasks can be accomplished
in the shortest time possible.
Another example ca be a marketing set up by making an estimate of sales performance for different salesmen
as well as for different cities one could assign a particular salesman to a particular city with a view to
maximize the overall sales. Note that with n facilities and n jobs there are n! possible assignments. The
simplest way of finding an optimum assignment is to write all the n! possible arrangements, evaluate their
total cost and select the assignment with minimum cost. But this method leads to a calculation problem of
formidable size even when the value of n is moderate. For n = 10 the possible number of arrangements is
3268800.
6.2
Assignment Problem Structure and Solution
The Assignment problem structure is similar to that of the transportation problem, and is as follows.
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The element cij represents the measure of effectiveness when ith person is assigned j th job. Assume that the
overall measure of effectiveness is to be minimized. The element xij represents the number of ith individuals
assigned to the j th job.
Types of assignment Problems
There are some types in assignment problem. They are:
(i) Balanced Assignment problem
(ii) Unbalanced Assignment problem
Assignment Problem Procedure(Hungarian Method)
The Hungarian Method is discussed in the form of a series of computational steps as follows, when the
objective function is that of minimization type.
Phase 1: Row and Column Reductions
• Step 1: Subtract the minimum value of each row from all the entries of that row.
• Step 2: Subtract the minimum value of each column from all the entries of that column.
Phase 2: Optimization of the Problem
• Step 1: Draw a minimum number of lines to cover all the zeros of the matrix
(a) Row Scanning:
Starting from the first row see if there is a single zero. If yes marks a square around it and draw
a vertical line passing through that zero, otherwise skip that row. After scanning the last row
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check whether all the zeros are covered with line. If yes then proceed go to Step 2 , otherwise,
do column scanning.
(b) Column Scanning:
Starting from the first column see if there is a single zero. If yes marks a square around it and
draw a horizontal line passing through that zero, otherwise skip that column. After scanning the
last Scanning check whether all the zeros are covered with line.
• Step 2: Check whether the number of squares marked is equal to the number of rows of the matrix
(Optimality conditions. If yes proceed to Step 5, otherwise go to Step 3
• Step 3: Identity the minimum values of the undeleteted cell values.
(i) Add the minimum undeleted cell value at the intersection point of the present matrix.
(ii) Subtract the minimum undeleted cell value from all the undeleted cell value.
(iii) The rest of the entries remain the same.
• Step 4: Go to Step 1
• Step 5: Take the solution as marked by the squares as the optimal solution.
Example 6.1. A company has five jobs J1 , J2 , J3 , J4 , and J5 and five machines M1 , M2 , M3 , M4 and M5 .
The given matrix shows the procession time in hours required to process the jobs on the machines. Assign
the jobs to the machines so as to minimize the total required time.
Solutions:
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6.3
BF 360
Unbalanced Assignment problem
In the previous section we assumed that the number of persons to be assigned and the number of
jobs were same. Such kind of assignment problem is called balanced assignment problem.
Suppose if the number of person is different from the number of jobs then the assignment problem is called
unbalanced assignment problem. If the number of persons is more than the number of jobs , then we
need to introduce one or more dummy jobs of zero duration to make the unbalanced assignment problem into
balanced assignment problem. Then the balanced assignment problem can be solved by using the Hungarian
Method as discussed in the previous section. The persons to whom the dummy jobs are assigned are left out
of assignment.
Similarly, if the number of persons is less than number of jobs then we have introduce one or more dummy
persons with zero duration to modify the unbalanced into balanced and then the problem is solved using the
Hungarian Method. Here the jobs assigned to the dummy persons are left out.
Example 6.2. A marketing company wants to assign three employees A, B, and C to four offices located
at W , X, Y and Z respectively. The assignment cost for this purpose is given in following table.
Solutions:
6.4
Maximization in an Assignment Problem
There are situations where certain facilities have to be assigned to a number of jobs so as to
maximize the overall performance of the assignment. In such cases the problem can be converted into a
minimization problem and can be solved by using Hungarian Method. Here the conversion of maximization
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problem into a minimization can be done by subtracting all the elements of the cost table from the highest
value of that table.
Example 6.3. Consider the problem of five different machines can do any of the required five jobs with
different profits resulting from each assignment as illustrated below: enter
Find out the maximum profit through optimal assignment
6.5
Diagonal Rule of the Assignment Problem
The following is the procedure will follow when using the diagonal rule in an assignment problem.
Phase 1: Row and Column reduction
Phase 2: Optimization of the problem
Step 1: Draw a minimum number of lines to cover all the zeros of the matrix
(a) Row scanning
(b) Column scanning
Check whether all zeros are covered with lines. If yes proceed to the next step, Otherwise select the zeros
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diagonally opposite with each other.
Step 2: Check if the number of square boxes is equal to the number of rows. If yes, report the solution as
marked by the square as the optimal solution, otherwise go to step 3. Step 3: Identify the minimum value
of the undeleted cell values, then
• Copy the table
• Add the minimum value to the element at the intersection.
• Subtract the minimum value to the undeleted element.
• Go to step 1.
Example 6.4. Solve the following assignment problem using Hungarian method, The matrix entries represents the processing times in hours.
Operators/Job
A
B
C
D
1
5
3
2
8
2
7
9
2
6
3
6
4
5
7
4
5
7
7
8
Solution:
6.6
Tutorial
1. Four different jobs are to be done on four machines, one job on each machine, as set up costs and times
are too high to permit a job being worked on more than one machine. The matrix given below gives
the times of producing jobs on different machines. Assign the jobs to machine so that total time of
production is minimized.
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2. A tourist company owns a one car in each of the five locations viz. L1 , L2 , L3 , L4 , L5 and a passengers
in each of the five cities C1 , C2 , C3 , C4 , C5 respectively. The following table shows the distant between
the locations and cities in kilo meter. How should be cars be assigned to the passengers so as to
minimize the total distance covered.
3. A manager has 4 jobs on hand to be assigned to 3 of his clerical staff. Clerical staff differs in efficiency.
The efficiency is a measure of time taken by them to do various jobs. The manager wants to assign
the duty to his staff, so that the total time taken by the staff should be minimum. The matrix given
below shows the time taken by each person to do a particular job. Help the manager in assigning the
jobs to the personnel.
4. The city post office has five major counters namely, Registration (R), Savings (S), Money – Order (M),
Postal stationary (P) and Insurance / license (I). The postmaster has to assign five counters o five clerks
A, B, C, D and E one for each counter. Considering the experience and ability of these clerks he rates
their suitability on a certain 10 - point scale of effectiveness of performance for accomplishing different
counter duties, as listed below. Assign the counters to the clerks for maximum effective performance.
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5. A marketing manager wants to assign salesman to four cities. He has four salesmen of varying experience. The possible profit for each salesman in each city is given in the following table. Find out an
assignment which maximizes the profit.
6. Banda’s three wife, Rani, Brinda, and Fathima want to earn some money to take care of personal expenses during a school trip to the local beach. Mr. Banda has chosen three chores for his wife:washing,
cooking, sweeping the cars. Mr. Banda asked them to submit bids for what they feel was a fair pay for
each of the three chores. The three wife of Shiva accept his decision. The following table summarizes
the bid received
Assign the chores to Mr Banda’s wives so that he can minimize the cost of paying them for the assigned
chores.
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7
BF 360
NETWORK PROBLEM
Objectives:
• Introduction
• Understand the definitions of important terms
• Understand the development of project network diagram
• Understand and solve a shortest path problem
• Solve Critical Path Method (CPM) problems
• Solve Program Evaluation Review Technique (PERT) Problem
• Determining the Earliest and Latest times of an activity
7.1
Introduction
A network consists of several destinations or jobs which are linked with one another. A manager
will have occasions to deal with some network or other. Certain problems pertaining to networks are taken
up for consideration in this unit.
7.2
Project Network
A project network (or project activity network) is a graphical depiction (very similar to a flow
chart) that shows the sequence in which the project’s terminal elements must be completed. In other words
a project activity network displays the order in which a project’s activities are to be completed.
7.2.1
Component of a Project Network
Certain key concepts/terms pertaining to a project network are described below:
Definition 7.1. An activity means a work. A project consists of several activities. An activity takes time.
It is represented by an arrow in a diagram of the network.
For example, an activity in house construction can be flooring. This is represented as follows:
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Construction of a house involves various activities. Flooring is an activity in this project. We can say that
a project is completed only when all the activities in the project are completed.
Definition 7.2. An event is the beginning or the end of an activity. Events are represented by circles in a
project network diagram. The events in a network are called the nodes.
Example:
Starting a punching machine is an activity. Stopping the punching machine is another activity.
Definition 7.3. An event just before another event is called the predecessor event.
Definition 7.4. An event just following another event is called the successor event
Example: Consider the following
From the above diagram:
Event 1 is predecessor for the event 2.
Event 2 is successor to event 1.
Event 2 is predecessor for the events 3, 4 and 5.
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Event 4 is predecessor for the event 6.
Event 6 is successor to events 3, 4 and 5
Definition 7.5. A network is a series of related activities and events which result in an end product or
service.
The activities shall follow a prescribed sequence. For example, while constructing a house, laying
the foundation should take place before the construction of walls. Fitting water tapes will be done towards
the completion of the construction. Such a sequence cannot be altered.
Definition 7.6. A dummy activity is an activity which does not consume any time.
Sometimes, it may be necessary to introduce a dummy activity in order to provide connectivity to
a network or for the preservation of the logical sequence of the nodes and edges.
7.2.2
Construction of a Project Network
A project network consists of a finite number of events and activities, by adhering to a certain
specified sequence. There shall be a start event and an end event (or stop event). All the other events shall
be between the start and the end events. The activities shall be marked by directed arrows. An activity
takes the project from one event to another event. An event takes place at a point of time whereas an
activity takes place from one point of time to another point of time.
Example 7.1. Construct the network diagram for a project with the following activities:
Solution:
Example 7.2. Develop a network diagram for the project specified below:
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Solution:
7.3
Shortest Path Problem
The shortest path method aims to find how a person can travel from one location to another,
keeping the total distance travelled to the minimum. In other words, it seeks to identify the shortest route
to a series of destinations.
7.3.1
Steps in Shortest path Method
The procedure consists of starting with a set containing a node and enlarging the set by choosing
a node in each subsequent step.
Step 1
Begin, by locating the origin. Then, find the node nearest to the origin. Mark the distance between the
origin and the nearest node in a box by the side of that node. In some cases, it may be necessary to check
several paths to find the nearest node.
Step 2
Repeat the above process until the nodes in the entire network have been accounted for. The last distance
placed in a box by the side of the ending node will be the distance of the shortest route. We note that the
distances indicated in the boxes by each node constitute the shortest route to that node. These distances
are used as intermediate results in determining the next nearest node
Example 7.3. A leather manufacturing company has to transport the finished goods from the factory to the
store house. The path from the factory to the store house is through certain intermediate stations as indicated
in the following diagram. The company executive wants to identify the path with the shortest distance so as
to minimize the transportation cost.Determine the shortest distance
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Solution:
7.4
Critical Path Method
Critical Path Method (CPM) is the network scheduling technique used when activities time is know
accurately. CPM aims at the determination of the time to complete a project and the important activities
on which a manager shall focus attention. The following are the component of the CPM. Assumption: In
CPM, it is assumed that precise time estimate is available for each activity.
Project Complexion Time: From the start event to the end event, the time required to complete all the
activities of the project in the specified sequence is known as the project completion time.
Path in a Project: A continuous sequence, consisting of nodes and activities alternatively, beginning with
the start event and stopping at the end event of a network is called a path in the network.
Critical Path & Critical activities: Consider all the paths in a project, beginning with the start event
and stopping at the end event. For each path, calculate the time of execution, by adding the time for the
individual activities in that path.
The path with the largest time is called the critical path and the activities along this path are called the
critical activities or bottleneck activities.
The activities are called critical because they cannot be delayed. However, a non-critical activity may be
delayed to a certain extent. Any delay in a critical activity will delay the completion of the whole project.
However, a certain permissible delay in a non –critical activity will not delay the completion of the whole
project. It shall be noted that delay in a non-critical activity beyond a limit would certainly delay the
completion the whole project. Sometimes, there may be several critical paths for a project. A project
manager shall pay special attention to critical activities.
Example 7.4. The following details are available regarding a project:
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Determine the critical path, the critical activities and the project completion time.
Solution:
Example 7.5. Determine the completion time and the critical activities for the following project:
Solution:
7.5
Project Evaluation & Review Technique
Program Evaluation and Review Technique (PERT) is a network scheduling technique used when
activities time is not known accurately/ only probabilistic estimate of time is available. PERT is a tool that
helps a project manager in project planning and control. It enables him in continuously monitoring a project
and taking corrective measures wherever necessary. This technique involves statistical methods.
Assumption
Note that in CPM, the assumption is that precise time estimate is available for each activity in a project.
However, in PERT, we assume that it is not possible to have precise time estimate for each activity and
instead, probabilistic estimates of time alone are possible.
A multiple time estimate approach is followed here. In probabilistic time estimate, the following 3 types of
estimate are possible:
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(i) Pessimistic time estimate ( tp )
(ii) Optimistic time estimate ( to )
(iii) Most likely time estimate ( tm )
Definition 7.7. The optimistic estimate of time is based on the assumption that an activity will not
involve any difficulty during execution and it can be completed within a short period
Definition 7.8. A pessimistic estimate of time is made on the assumption that there would be unexpected
problems during the execution of an activity and hence it would consume more time
Definition 7.9. The most likely time estimate is made in between the optimistic and the pessimistic
estimates of time
Thus, the three estimates of time have the relationship
t0 ≤ tm ≤ tp
In practical, neither the pessimistic nor the optimistic estimate may hold in reality, but the most likely
time estimate is expected to prevail in almost all cases. Therefore, it is preferable to give more weight to
the most likely time estimate. Thus we give a weight of 4 to most likely time estimate and a weight
of 1 each to the pessimistic and optimistic time estimates. In doing so, we get the time estimate (te )
as the weighted average of these estimates as follows:
te =
t0 + 4tm + tp
6
Since we have taken 6 units i.e. 1 for tp , 4 for tm and 1 for to , hence we divide the sum by 6. With this
time estimate, we can determine the project completion time as applicable for CPM. Since PERT involves
the average of three estimates of time for each activity, this method is very practical and the results from
PERT will be have a reasonable amount of reliability.
Measure of Certainty
The 3 estimates of time are such that
t0 ≤ tm ≤ tp
So that the range for the time estimate is tp − to . The time taken by an activity in a project network follows
a distribution with a standard deviation which is approximately
σ=
tp − to
6
So that the variance is
σ2 =
tp − to
6
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The certainty of the time estimate of an activity can be analysed with the help of the variance. The greater
the variance, the more uncertainty in the time estimate of an activity.
Example 7.6. Two experts A and B examined an activity and arrived at the following time estimates
Determine which expert is more certain about his/her estimates of time
Solution:
Example 7.7. Find out the time required to complete the following project and the critical activities:
Solution:
Example 7.8. Determine the time, variance and standard deviation of the project with the following time
estimates in weeks:
Solution:
Example 7.9. A project consists of seven activities with the following time estimates. Find the probability
that the project will be completed in 30 weeks or less.
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Solution:
7.6
Earliest and Latest Times
A project manager has the responsibility to make sure that a project is completed by the stipulated
date, without delay. Therefore much attention should be focused on this aspect in what follows.
7.6.1
Earliest Times of an Activity
Here we can consider two type of earliest time
(i) Earliest Start Time of an activity
(ii) Earliest Finish Time of an activity.
Definition 7.10. Earliest Start Time of an activity (ES ) is the earliest possible time of starting that
activity on the condition that all the other activities preceding to it were began at the earliest possible times.
Definition 7.11. Earliest Finish Time of an activity (EF ) is the earliest possible time of completing
that activity. It is given by the formula.
The Earliest Finish Time of an activity = The Earliest Start Time of the activity + The
estimated duration to carry out that activity.
EF = ES + TE
7.6.2
Latest Times of an Activity
Here we consider two types of latest times
(i) Latest Finish Time of an activity
(ii) Latest Start Time of an activity
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Definition 7.12.
BF 360
Latest Finish Time of an activity((LF ) is the latest possible time of completing
that activity on the condition that all the other activities succeeding it are carried out as planned by the
management and without delaying the project beyond the stipulated time.
Definition 7.13. Latest Start Time of an activity (LS ) is the latest possible time of beginning that
activity.
It is given by the formula:
Latest Start Time of an activity = The Latest Finish Time of the activity - The estimated duration to carry
out that activity.
LS = LF − TE
7.6.3
Total Float of an Activity
Float seeks to measure how much delay is acceptable. It sets up a control limit for delay
Definition 7.14. The total float of an activity is the time by which that activity can be delayed without
delaying the whole project. It is given by the formula
Total Float of an Activity = Latest Finish Time of the activity - Earliest Finish Time of that activity.
TF = LF − EF
It is also given by the formula:
Total Float of an Activity = Latest Start Time of the activity - Earliest Start Time of that activity.
TF = LS − ES
Since a delay in a critical activity will delay the execution of the whole project, the total float of a
critical activity must be zero.
7.6.4
Expected Times of an Activity
An event occurs at a point of time. We can consider
(i) Earliest Expected Time of Occurrence of an event
(ii) Latest Allowable Time of Occurrence an event
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The Earliest Expected Time of Occurrence of an event is the earliest possible time of expecting that
event to happen on the condition that all the preceding activities have been completed.]
The Latest Allowable Time of Occurrence of an event is the latest possible time of expecting that
event to happen without delaying the project beyond the stipulated time.
Procedure to Determine the Earliest Expected Times of an Event
1. Take the Earliest Expected Time of Occurrence of the Start Event as zero
2. For an event other than the Start Event, find out all paths in the network which connect the Start
node with the node representing the event under consideration
3. In the “Forward Pass” (i.e., movement in the network from left to right), find out the sum of the time
duration of the activities in each path identified in Step 2
4. The path with the longest time in Step 3 gives the Earliest Expected Time of Occurrence of the even
The Rule for finding the earliest expected time of an event: For an event under consideration,
locate all the predecessor events and identify their earliest expected times. With the earliest expected time of
each event, add the time duration of the activity connecting that event to the event under consideration. The
maximum among all these values gives the Earliest Expected Time of Occurrence of the event.
Procedure to Determine the Latest Expected Times of an Event
We consider the “Backward Pass” (i.e., movement in the network from right to left). The latest allowable
time of occurrence of the End Node must be the time of completion of the project. Therefore it shall be
equal to the time of the critical path of the project
1. Identify the latest allowable time of occurrence of the End Node
2. For an event other than the End Event, find out all paths in the network which connect the End node
with the node representing the event under consideration.
3. In the “Backward Pass” (i.e., movement in the network from right to left), subtract the time durations
of the activities along each such path.
4. The Latest Allowable Time of Occurrence of the event is determined by the path with the longest time
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in Step 3. In other words, the smallest value of time obtained in Step 3 gives the Latest Allowable
Time of Occurrence of the event.
The Rule for finding the latest allowable time of an event: For an event under consideration, locate
all the successor events and identify their latest allowable times. From the latest allowable time of each
successor event, subtract the time duration of the activity that begins with the event under consideration.
The minimum among all these values gives the Latest Allowable Time of Occurrence of the event
Definition 7.15. The allowable time gap for the occurrence of an event is known as the slack of that
event. It is given by the formula:—
Slack of an event = Latest Allowable Time of Occurrence of the event - Earliest Expected
Time of Occurrence of that event.
Definition 7.16. The slack of an activity is the float of the activity
Example 7.10. The following details are available regarding a project:
Determine the earliest and latest times, the total float for each activity, the critical activities and the project
completion time.
Solution:
7.7
Tutorial
1. Explain the terms: event, predecessor event, successor event, activity, dummy activity,
network.
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2. Construct the network diagram for the following project:
3. The dependency relationships between the activities of a project are shown in the table below. Draw
a project diagram for these activities.
Activity:
A
B
C
D
E
Predecessor
–
–
–
A,B B
F
B,C
Activity:
4. The following are the details of the activities in a project:
Activity:
A
B
C
D
E
F
G
H
I
J
K
Predecessor
–
A
A
A
B
B
D
C,F,G
C,F,G
E,H
I,J
8
13
9
12
14
8
7
12
9
10
7
Activity:
Duration
(weeks)
(a) Construct a project network diagram.
(b) Determine shortest path method
(c) Determine the critical activities and hence the critical paths
5. The following are the time estimates and the precedence relationships of the activities in a project
network:
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Draw the project network diagram. Determine the critical path and the project completion time.
6. Draw the network diagram for the following project. Determine the time, variance and standard
deviation of the project.:
7. Consider the following project with the estimates of time in weeks:
Find the probability that the project will be completed in 27 weeks
8. The following are the details of the activities in a project:
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Activity
Predecessor
Duration
Activity
(Weeks)
A
-
15
B
A
17
C
A
21
D
B
19
E
B
22
F
D,E
18
G
C,F
15
Calculate the earliest and latest times, the total float(slack) for each activity and the project completion
time.
9. The following are the details of the activities in a project:
Activity:
A
B
C
D
E
F
G
H
I
J
K
Predecessor
–
A
A
A
B
B
D
C,F,G
C,F,G
E,H
I,J
8
13
9
12
14
8
7
12
9
10
7
Activity:
Duration
(weeks)
Determine the earliest and latest times, the total float(slack) for each activity and the project completion time.
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8
8.1
BF 360
QUEUING THEORY
Introduction
A flow of customers from finite or infinite population towards the service facility forms a queue
(waiting line) an account of lack of capability to serve them all at a time. In the absence of a perfect
balance between the service facilities and the customers, waiting time is required either for the service
facilities or for the customers arrival. In general, the queuing system consists of one or more queues and one
or more servers and operates under a set of procedures. Depending upon the server status, the incoming
customer either waits at the queue or gets the turn to be served. If the server is free at the time of arrival of
a customer, the customer can directly enter into the counter for getting service and then leave the system.
In this process, over a period of time, the system may experience “ Customer waiting” and /or “Server
idle time
8.2
Queuing System
A queuing system can be completely described by;
(1) Input (arrival pattern)
(2) Service mechanism (service pattern)
(3) Queue discipline and
(4) Customer’s behaviour
8.2.1
Input (arrival pattern)
The input described the way in which the customers arrive and join the system. In general,
customers arrive in a more or less random manner which is not possible for prediction. Thus the arrival
pattern can be described in terms of probabilities and consequently the probability distribution for
8.2.2
Inter-arrival times
(the time between two successive arrivals) must be defined. We deal with those Queuing system in
which the customers arrive in Poisson process. The mean arrival rate is denoted by λ
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8.2.3
BF 360
Service Mechanism
This means the arrangement of service facility to serve customers. If there is infinite number of
servers, then all the customers are served instantaneously or arrival and there will be no queue. If the number
of servers is finite then the customers are served according to a specific order with service time a constant
or a random variable. Distribution of service time follows Exponential distribution defined by
f (t) = λe−λt
8.2.4
t>0
Queuing discipline
It is a rule according to which the customers are selected for service when a queue has been formed.
The most common disciplines are;
(i) First In First out (FIFO) or First come first served – (FCFS)
In this system, the element that came first ise served first, so that every element has a fair chance
of getting service. Moreover it is understood that it gives a good morale and discipline in the queue.
When the condition of FIFO is violated, there arises the trouble and the management is answerable
for the situation.
(ii) Last in first out – (LIFO)
In this system, the element arrived last will have a chance of getting service first. In general, this
does not happen in a system where human beings are involved. But this is quite common in Inventory
system. This is what we call Last come first served). This can also be written as First In Last Out
(FILO).
(iii) Service in random order (SIRO)
In this case the items are called for service in a random order. The element might have come first or
last does not bother; the servicing facility calls the element in random order without considering the
order of arrival. This may happen in some religious organisations but generally it does not followed in
an industrial / business system.
(iv) Service by Priority
Priority disciplines are those where any arrival is chosen for service ahead of some other customers
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already in queue. In the case of Pre-emptive priority the preference to any arriving unit is so high that
the unit is already in service is removed / displaced to take it into service. A non- pre-emptive rule
of priority is one where an arrival with low priority is given preference for service than a high priority
item
8.2.5
Customer’s behaviour
(a) Generally, it is assumed that the customers arrive into the system one by one. But in some cases,
customers may arrive in groups. Such arrival is calledBulk arrival.
(b) If there is more than one queue, the customers from one queue may be tempted to join another queue
because of its smaller size. This behaviour of customers is known as jockeying.
(c) If the queue length appears very large to a customer, he/she may not join the queue. This is known
as Balking of customers
(d) Sometimes, a customer who is already in a queue will leave the queue in anticipation of longer waiting
line. This kind of departure is known as reneging.
(e) Collusion: In this case several customers may collaborate and only one of them may stand in the
queue. One customer represents a group of customer. Here the queue length may be small but service
time for an individual will be more. This may break the patience of the other customers in the waiting
line and situation may lead a lot of confusion.
8.2.6
Variables
The following are the variable that we will be making use in our study of queuing systems;
n - Number of customers in the system.
c - Number of servers in the system.
Pn (t) – Probability of having n customers in the system at time t.
Pn - Steady state probability of having customers in the system.
P0 - Probability of having zero customer in the system.
Lq - Average number of customers waiting in the queue.
Ls - Average number of customers waiting in the system(in the queue and in the service counters).
Wq - Average waiting time of customers in the queue.
Ws - Average waiting time of customers in the system(in the queue and in the service counters).
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δ- Arrival rate of customers.
µ - Service rate of server.
φ - Utilization factor of the server.
δef f - Effective rate of arrival of customers.
M - Poisson distribution.
N - Maximum numbers of customers permitted in the system. Also, it denotes the size of the calling source
of the customers.
GD - General discipline for service. This may be first in first – serve (FIFS), last-in-first serve (LIFS),
Service in Random Order(SIRO) etc.
Utilization factor of the server: The ultilization factor in a queuing system is given by;
φ=
δ
Mean arrival time
=
Mean service time × number of servers
cµ
Utilization factor is often seen as a measure of productivity in other words it is the fraction of time where
the servers are busy and therefore it is considered desirable for it to be high.
8.3
Classification of Queuing Model
Generally, queuing models can be classified into six categories using Kendall’s notation with six
parameters to define a model. The parameters of this notation are;
P - Arrival rate distribution i.e probability law for the arrival /inter – arrival time.
Q - Service rate distribution, i.e probability law according to which the customers are being served.
R - Number of Servers (i.e number of service stations)
X - Service discipline.
Y - Maximum number of customers permitted in the system.
Z - Size of the calling source of the customers.
A queuing model with the above parameters is written as (P/Q/R : X/Y /Z).
8.4
Model 1: (M/M/1) : (GD/∞/∞)
In this model
(i) the arrival rate follows Poisson (M ) distribution.
(ii) Service rate follows Poisson distribution (M )
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(iii) Number of servers is 1
(iv) Service discipline is general disciple (i.e. GD)
(v) Maximum number of customers permitted in the system is infinite (∞)
(vi) Size of the calling source is infinite (∞)
Under this model the ultilization factor is given by
φ=
Mean arrival time
δ
=
Mean service time × number of servers
cµ
Since the number of server is 1
φ=
δ
µ
The steady state equations to obtain, Pn the probability of having customers in the system and the values
for Ls , Lq , Ws and Wq are given below.
Pn : the probability of having customers in the system
Pn = φn (1 − φ),
where φ =
n = 0, 1, 2, 3 . . . , ∞
δ
µ
Ls – Average number of customers waiting in the system (i.e waiting in the queue and in the service station)
Ls =
φ
1−φ
Lq – Average number of customers waiting in the queue
Lq = Ls −
δ
φ
φ2
=
−φ=
µ
1−φ
1−φ
Lq =
φ2
1−φ
Ws - Average waiting time of customers in the system
Ws =
Ls
φ
φ
1
=
=
×
δ
(1 − φ)δ
1 − φ µφ
N.B (δ = µφ)
Ws =
1
µ−δ
Wg = Average waiting time of customers in the queue
Lq
1
Wq =
=
δ
δ
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since δ = µφ then
1
Wq =
µφ
φ2
1−φ
φ
µ − φµ
Wq =
Wq =
φ
µ−δ
Example 8.1. The arrival rate of customers at a banking counter follows a Poisson distribution with a
mean of 30 per hours. The service rate of the counter clerk also follows Poisson distribution with mean of
45 per hour.
(a) What is the probability of having zero customer in the system?
(b) What is the probability of having 8 customer in the system?
(c) What is the probability of having 12 customer in the system?
(d) Find Ls , Lq , Ws and Wq
Solution:
Example 8.2. At one-man barber shop, customers arrive according to Poisson distribution with mean arrival
rate of 5 per hour and the hair cutting time was exponentially distributed with an average hair cut taking
10 minutes. It is assumed that because of his excellent reputation, customers were always willing to wait.
Calculate the following:
(i) Average number of customers in the shop and the average numbers waiting for a haircut.
(ii) The percentage of time arrival can walk in straight without having to wait.
(iii) The percentage of customers who have to wait before getting into the barber’s chair.
Solution:
8.5
Model 2: (M/M/C) : (GD/ ∞/∞) Model
The parameters of this model are as follows:
(i) Arrival rate follows Poisson distribution
(ii) Service rate follows Poisson distribution
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(iii) Number of servers is c.
(iv) Service discipline is general discipline.
(v) Maximum number of customers permitted in the system is infinite
Under this model the ultilization factor is given by
φ=
Mean arrival time
δ
=
Mean service time × number of servers
cµ
Since the number of server is c
δ
cµ
φ=
Then the steady state equation to obtain the probability of having n customers in the system is
Pn =
φ n P0
,
n!
0≤n≤c
Where as
P0 =

X φn n!

+
φc
−1

h
i
c! 1 − φc 
Under this model , Lq – the average number of customers waiting in queue the is given by
φc−1
Lq =
P0
(c + 1)!(c − φ)2
Ls )-Average number of customers waiting in the system is
Ls = Lq + φ
Wq - Average waiting time of customers in the queue is
Lq
δ
Wq =
and Ws - Average waiting time of customers in the system are given by
Ws = Wq +
1
µ
The probability that all servers are busy, or the probability that an arrival has to wait is
P (n ≥ c) =
φc
h
i
c! 1 − φc
Example 8.3. At a central warehouse, vehicles are at the rate of 24 per hour and the arrival rate follows
Poisson distribution. The unloading time of the vehicles follows exponential distribution and the unloading
rate is 18 vehicles per hour. There are 4 unloading crews. Find
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(i) Po and P3
(ii) Lq , Ls , Wq and Ws
Solution:
Example 8.4. A supermarket has two sales girls at the sales counters. The service time for each customer
is exponential with a mean of 4 minutes, and the people arrive in a Poisson fashion at the rate of 10 an
hour.
(a) Calculate the probability that a customer has to wait for service.
(b) Find the expected percentage of idle time for each sales girl?
(c) If a customer has to wait, what is the expected length of his waiting time?
Solution:
8.6
Tutorial
1. The arrival rate of customers at a banking counter follows a Poisson distribution with a mean of 45
per hour. The service rate of the counter clerk also Poisson distribution with a mean of 60 per hours.
(a) What is the probability of having zero customer in the system (p0 )
(b) What is the probability of having 5 customer in the system (p5 )
(c) What is the probability of having 10 customer in the system (p10 ).
(d) Find Ls , Lq , Ws and Wq and interpret the values
2. A self-service store employs one cashier at its counter. Nine customers arrive on an average every 5
minutes whiles the cashier can serve 10 customers in 5 minutes. Assuming the arrival rate follows a
Poison distribution and the service time follows an exponential distribution. Determine the following
(a) Utilization factor
(b) Average number of customers in the system.
(c) Average number of customers in queue or average length.
(d) Average time a customers spend in the system.
(e) Average time a customers wait before being served.
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3. At a central warehouse, vehicles arrive at the rate of 9 per hour and the arrival rate follows a Poison
distribution. The unloading time of the vehicles follows an exponential distribution and the unloading
rate is 3 vehicles per hour. There are 4 unloading crews. Determine the following
(a) p0 and p3
(b) Lq and Ls
(c) Wq and Ws
4. A petrol station has two pumps. The service time follows the exponential distribution with mean 4
minutes and cars arrive for service in a Poisson process at the rate of 10 cars per hour. Find the
probability that a customer has to wait for service. What proportion of time the pump remains idle?
5. A T.V. Repairman finds that the time spent on his jobs have an exponential distribution with mean
of 30 minutes. If he repairs sets in the order in which they come in, and if the arrival of sets is
approximately Poisson with an average rate of 10 per 8 hour day, what is repairman’s expected idle
time each day? How many jobs are ahead of the average set just brought in?
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9
BF 360
DECISION THEORY
Objectives:
• Introduction
• Decision making under condition of Certainty (Deterministic Decision)
• Decision Making under condition of Risk
• Decision making under condition of Uncertainty
9.1
Introduction
Decision making is an integral part of any business organization. The process involves selecting the
best among several decisions through a proper evaluation of the parameters of each decision environment.
Definition 9.1. Decision theory is mainly concerned with determining which decision is optimal from a
set of possible alternatives for a specific condition.
Basically decision theory is comprised of three component or element these are;
(i) Decision alternatives- are different possible strategies the decision maker adopts.
(ii) States of nature- refer to as the future event, not under the control of the decision marker, which
may or may not occur. The states of nature should be defined so that they are mutually exclusive and
collectively exhaustive.
(iii) Payoff -is defines as the consequence resulting from a specific combination of a decision alternatives
and states of nature.
A table showing payoffs for all possible combination of the decision alternative and states of nature is called
a payoff table. Payoff can be expressed in terms of profit and cost, time, distance or any other measure
appropriate for decision problem being analysed. A decision can then be made after having a numerical
value to each possible alternative course of action. The types of decisions can be classified into the following
categories;
(a) Decision condition of Certainty.
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(b) Decision under condition of Risk.
(c) Decision under Condition of Uncertainty
9.2
Decision under condition of Certainty
Decision making under certainty assumes that all relevant information required to make decision
is certain in nature and is well known. It uses a deterministic model, with complete knowledge, stability
and no ambiguity. To make decision, the manager will have to be quite aware of the strategies available and
their payoffs and each strategy will have unique payoff resulting in certainty. The decision-making may be
of single objective or of multiple objectives.
Example: In the economic evaluation of alternatives, if the annual revenues of the different alternatives
are known in advance and which are not subject to any variation in future, these quantities are deterministic
quantities and the decision on the selection of the best alternative is an example of decision under certainty
9.3
Decision under condition of Risk
Decision-making under risk describes a situation in which each strategy results in more than one
outcomes or payoffs and the manager attaches a probability measure to these payoffs. This model covers the
case when the manager projects two or more outcomes for each strategy and he or she knows, or is willing
to assume, the relevant probability distribution of the outcomes. The following assumptions are to be made:
(1) Availability of more than one strategies.
(2) The existence of more than one states of nature.
(3) The relevant outcomes.
(4) The probability distribution of outcomes associated with each strategy.
The optimal strategy in decision making under risk is identified by the strategy with highest expected
utility (or highest expected value) , if the payoffs are in terms of profits. In the case were he payoff are in
terms of cost we select the optimal strategy as the one having the least expected value. The expected value
formula is given by:
EV (di ) =
N
X
P (sj )Vij
j=1
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Where
N is the number of states of nature.
P (sj ) is the probability of the j th state of nature.
Vij is the payoff of the ith decision alternative and the j th state of nature.
Example 9.1. A marketing manager of an insurance company has kept complete records of the sales effort
of the sales personnel. These records contain data regarding the number of insurance policies sold and net
revenues received by the company as a function of four different sales decision strategies. The manager has
constructed the conditional payoff matrix given below, based on his records. (The state of nature refers to the
number of policies sold). The number within the table represents utilities. Suppose you are a new salesperson
and that you have access to the original records as well as the payoff matrix. Which decision strategy would
you follow?
Solution:
9.4
Decision under condition of Uncertainty
Decision making under uncertainty is formulated exactly in the same way as decision making under
risk, only difference is that no probability to each strategy is attached. In decision making under uncertainty,
there are no probabilities attached to set of the states of nature. Sometimes we may have only positive
elements in the given matrix, indicating that the company under any circumstances will have profit only.
Sometimes, we may have negative elements, indicating potential loss. While solving the problem of decision
making under uncertainty. There are five approaches that we can use to make a decision under the condition
of uncertainty these are;
(a) Optimistic (Maxmax) Criterion
(b) Pessimistic (Maxmin) Criterion
(c) Minmax Regret Criterion.
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(d) Laplace (Equal Probability) Criterion
(e) Hurwicz (Realism) Criterion
9.4.1
Optimistic (Maxmax) Criterion
Under thus criterion, we determine the best possible outcome in each strategy, and then identify
the best of the best outcome in order to select the optimal strategy.
Example 9.2. The management of Trade Kings company is considering the use of a newly discovered chemical which, when added to detergents, will make the washing stet, thus eliminating the necessity of adding
softeners. The management is considering at present time, these three alternative strategies;
S1 = Add the new chemical to the currently marketed detergent DETER and sell it under label ’NEW
IMPROVED DETER’.
S2 = Introduce a brand new detergent under the name of ’SUPER SOFT’.
S3 = Develop a new product and enter the softener market under the name ’EXTRA WASH’.
The management has decided for the time being that only one of the three strategies is economically feasible
(under given market condition). The marketing research department is requested to develop a conditional
payoff matrix for this problem. After conducting sufficient research, based on personal interviews and anticipating the possible reaction of the competitors, the marketing research department submits the payoff matrix
given below. Select the optimal strategy.
Solution:
9.4.2
Pessimistic (Maxmin) Criterion
When criterion of pessimistic is applied to solve the problem under uncertainty, first determine
worst possible outcome in each strategy (row minimums), and select the best of the worst outcome
in order to select the optimal strategy.
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Example 9.3. Consider example (9.2), determine which decision strategy would you follow using pessimistic
criterion.
Solution:
9.4.3
Minmax Regret Criterion
In this case, we have to determine the regret matrix or opportunity loss matrix. To find the
opportunity loss matrix (column opportunity loss matrix), subtract all the elements of a column from the
highest element of that column. The obtained matrix is known as regret matrix. To select the optimal
strategy we first determine the maximum regret that the decision maker can experience for each strategy,
then select the minimum of the maximum regrets. The choice process can be known as minimax
regret. Suppose two strategies have same minimax element, then the manager needs additional factors that
influence his selection.
Example 9.4. Consider example (9.2), determine which decision strategy would you follow using minmax
regret criterion.
Solution:
9.4.4
Laplace (Equal Probability)
Since do not have any objective evidence of a probability distribution for the states of nature, one
can use subjective criterion. Not only this, as there is no objective evidence, as a result we can assign equal
probabilities to each of the state of nature. This subjective assumption of equal probabilities is known as
Laplace criterion, or criterion of insufficient reason in management literature. Once equal probabilities
are attached to each state of nature, we revert to decision making under risk. In short this is more like finding
the average payoff then selecting the highest as the optimal strategy.
Example 9.5. Consider example (9.2), determine which decision strategy would you follow using laplace
criterion.
Solution:
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9.4.5
BF 360
Hurwicz (Realism) Criterion
This is also known as weighted average criterion, it is a compromise between the maximax and
maximin decisions criteria. It takes both of them into account by assigning them weights in accordance with
the degree of optimism or pessimism. The alternative that maximizes the sum of these weighted payoffs is
then selected.The decision maker’s degree of optimism is denoted by α, so that
Payoff = α × maximum payoff + (1 − α) × minimum payoff
Example 9.6. Consider example (9.2), determine which decision strategy would you follow using Hurwicz
criterion with α = 0.75
Solution:
9.5
Tutorial
1. The following matrix gives the payoff of different strategies(alternatives) in profits S1 , S2 and S3 against
condition(events)or States of nature N1 ,N2 , N3 and N4
N1
N2
N3
N4
S1
4000
−100
6000
18000
S2
20000
5000
400
0
S3
20000
15000
−2000
1000
Probability
0.05
0.25
0.30
0.40
Indicate the decision under the following approach
(i) Pessimistic(maxmin) criterion.
(ii) Optimistic(maxmax) criterion.
(iii) Equal probability criterion.
(iv) Minmax Regret criterion.
(v) Huricz criterion take α = 0.60
(vi) Expected value criterion.
2. The following matrix gives the payoff of different strategies (alternatives) A, B, and C against conditions
(events) W , X, Y and Z. Identify the decision taken under the following approaches:
(i) Pessimistic
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(ii) Optimistic
(iii) Equal probability
(iv) Regret
(v) Hurwicz criterion. The decision maker’s degree of optimism (α) being 0.7.
3. Mikalile tradings is considering three options for managing its data processing operation: continue with its own staff, hire an outside vendor to do the managing (referred to as outsourcing), or use a combination of its own staff and an outside vendor. The profit of the
operation depends on future demand. The annual profit of each option (in thousands of Zambian
kwacha (ZMW)) depends on demand as follows
Determine which staffing option Makalile trading should use in order to maximize the profit by
using;
(i) Decision making under condition of risk.
(ii) Optimistic (Maximax) approach
(iii) Pessimistic (Maxmin) approach
(iv) Minimax Regret approach
(v) Laplace (Equal probability) approach
(vi) Huricz(Realism) approach take α = 0.8
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