mb0032 operations research assignments
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ASSIGNMENTS
Subject code: MB 0032
(2 credits)
Set 1
Marks 30
SUBJECT NAME : OPERATIONS RESEARCH
Note: Each Question carries 10 marks
1. Describe in details the different scopes of application of Operations Research.
Answer: Operations Research (OR) in the USA, South Africa and Australia, and Operational
Research in Europe and Canada, is an interdisciplinary branch of applied mathematics and
formal science that uses methods such as mathematical modeling, statistics, and algorithms to
arrive at optimal or near optimal solutions to complex problems. It is typically concerned with
optimizing the maxima (profit, assembly line performance, crop yield, bandwidth, etc) or minima
(loss, risk, etc.) of some objective function. Operations research helps management achieve its
goals using scientific methods. The terms operations research and management science are often
used synonymously.
When a distinction is drawn, management science generally implies a closer relationship to the
problems of business management. The field of operations research is closely related to
Industrial engineering. Industrial engineers typically consider Operations Research (OR)
techniques to be a major part of their toolset. Some of the primary tools used by operations
researchers are statistics, optimization, probability theory, queuing theory, game theory, graph
theory, decision analysis, and simulation. Because of the computational nature of these fields,
OR also has ties to computer science, and operations researchers use custom-written and off-the-
shelf software.
Operations research is distinguished by its frequent use to examine an entire management
information system, rather than concentrating only on specific elements (though this is often
done as well). An operations researcher faced with a new problem is expected to determine
which techniques are most appropriate given the nature of the system, the goals for
improvement, and constraints on time and computing power. For this and other reasons, the
human element of OR is vital. Like any other tools, OR techniques cannot solve problems by
themselves.
Scope of operation Research
Examples of applications in which operations research is currently used include:
Critical path analysis or project planning: identifying those processes in a complex project which
affect the overall duration of the project
1. Designing the layout of a factory for efficient flow of materials
2. constructing a telecommunications network at low cost while still guaranteeing QoS
(quality of service) or QoS (Quality of Experience) if particular connections become very
busy or get damaged
3. Road traffic management and 'one way' street allocations i.e. allocation problems.
4. Determining the routes of school buses (or city buses) so that as few buses are needed as
possible
5. designing the layout of a computer chip to reduce manufacturing time (therefore reducing
cost)
6. Managing the flow of raw materials and products in a supply chain based on uncertain
demand for the finished products
7. Efficient messaging and customer response tactics
8. Robotizing or automating human-driven operations processes
9. Globalizing operations processes in order to take advantage of cheaper materials, labor,
land or other productivity inputs
10. Managing freight transportation and delivery systems (Examples: LTL Shipping,
intermodal freight transport)
11. Scheduling:
Personnel staffing
Manufacturing steps
Project tasks
Network data traffic: these are known as queuing models or queueing systems.
sports events and their television coverage
12. blending of raw materials in oil refineries
13. determining optimal prices, in many retail and B2B settings, within the disciplines of
pricing science
Operations research is also used extensively in government where evidence-based policy is used.
2. What do you understand by Linear Programming Problem? What are the
requirements of L.P.P.? What are the basic assumptions of L.P.P.?
Answer: Linear programming problem (LPP):
The standard form of the linear programming problem is used to develop the procedure for
solving a general programming problem.
A general LPP is of the form
Max (or min) Z = c1x1 + c2x2 + … +cnxn
x1, x2, ....xn are called decision variable. subject to the constraints
c1, c2,…. Cn, a11, a12,…. amn are all known constants
Z is called the "objective function" of the LPP of n variables which is to be maximized or
minimized.
Requirements of L.P.P :
There are mainly four steps in the mathematical formulation of linear programming problem as a
mathematical model. We will discuss formulation of those problems which involve only two
variables.
Identify the decision variables and assign symbols x and y to them. These decision
variables are those quantities whose values we wish to determine.
Identify the set of constraints and express them as linear equations/inequations in terms
of the decision variables. These constraints are the given conditions.
Identify the objective function and express it as a linear function of decision variables. It
might take the form of maximizing profit or production or minimizing cost.
Add the non-negativity restrictions on the decision variables, as in the physical problems,
negative values of decision variables have no valid interpretation.
There are many real life situations where an LPP may be formulated. The following examples
will help to explain the mathematical formulation of an LPP.
Example-1. A diet is to contain at least 4000 units of carbohydrates, 500 units of fat and 300
units of protein. Two foods A and B are available. Food A costs 2 dollars per unit and food B
costs 4 dollars per unit. A unit of food A contains 10 units of carbohydrates, 20 units of fat and
15 units of protein. A unit of food B contains 25 units of carbohydrates, 10 units of fat and 20
units of protein. Formulate the problem as an LPP so as to find the minimum cost for a diet that
consists of a mixture of these two foods and also meets the minimum requirements.
Suggested answer:
The above information can be represented as
Let the diet contain x units of A and y units of B.
Total cost = 2x + 4y
The LPP formulated for the given diet problem is
Minimize Z = 2x + 4y subject to the constraints
Basic Assumptions of L.P.P:
Linear programming is applicable only to problems where the constraints and objective function
are linear i.e., where they can be expressed as equations which represent straight lines. In real
life situations, when constraints or objective functions are not linear, this technique cannot be
used.
Factors such as uncertainty, weather conditions etc. are not taken into consideration.
There may not be an integer as the solution, e.g., the number of men required may be a
fraction and the nearest integer may not be the optimal solution.
i.e., Linear programming techniques may give practical valued answer which is not desirable.
Only one single objective is dealt with while in real life situations, problems come with
multi-objectives.
Parameters are assumed to be constants but in reality they may not be so.
3. Describe the different steps needed to solve a problem by simplex method.
Answer: Simplex method
The simplex method is a method for solving problems in linear programming. This method,
invented by George Dantzig in 1947, tests adjacent vertices of the feasible set (which is a
polytope) in sequence so that at each new vertex the objective function improves or is
unchanged. The simplex method is very efficient in practice, generally taking 2m to 3m
iterations at most (where m is the number of equality constraints), and converging in expected
polynomial time for certain distributions of random inputs (Nocedal and Wright 1999, Forsgren
2002). However, its worst-case complexity is exponential, as can be demonstrated with carefully
constructed examples (Klee and Minty 1972). A different type of methods for linear
programming problems are interior point methods, whose complexity is polynomial for both
average and worst case. These methods construct a sequence of strictly feasible points (i.e., lying
in the interior of the polytope but never on its boundary) that converges to the solution. Research
on interior point methods was spurred by a paper from Karmarkar (1984). In practice, one of the
best interior-point methods is the predictor-corrector method of Mehrotra (1992), which is
competitive with the simplex method, particularly for large-scale problems.
Dantzig's simplex method should not be confused with the downhill simplex method (Spendley
1962, Nelder and Mead 1965, Press et al. 1992). The latter method solves an unconstrained
minimization problem in n dimensions by maintaining at each iteration n+1 points that define a
simplex. At each iteration, this simplex is updated by applying certain transformations to it so
that it "rolls downhill" until it finds a minimum.
The Simplex Method is "a systematic procedure for generating and testing candidate vertex
solutions to a linear program." (Gill, Murray, and Wright, p. 337) It begins at an arbitrary corner
of the solution set. At each iteration, the Simplex Method selects the variable that will produce
the largest change towards the minimum (or maximum) solution. That variable replaces one of
its compatriots that is most severely restricting it, thus moving the Simplex Method to a different
corner of the solution set and closer to the final solution. In addition, the Simplex Method can
determine if no solution actually exists. Note that the algorithm is greedy since it selects the best
choice at each iteration without needing information from previous or future iterations.
The Simplex Method solves a linear program of the form described in Figure 3. Here, the
coefficients cj represent the respective weights, or costs, of the variables x
i. The minimized
statement is similarly called the cost of the solution. The coefficients of the system of equations
are represented by aij
, and any constant values in the system of equations are combined on the
right-hand side of the inequality in the variables bi.
Combined, these statements represent a linear program, to which we seek a solution of minimum
cost.
Solving this linear program involves solutions of the set of equations. If no solution to the set of
equations is yet known, slack variables x n+1
, x n+2
, …., x n+m
, adding no cost to the solution,
are introduced. The initial basic feasible solution (BFS) will be the solution of the linear program
where the following holds:
Once a solution to the linear program has been found, successive improvements are made to the
solution. In particular, one of the nonbasic variables (with a value of zero) is chosen to be
increased so that the value of the cost function, , decreases. That variable is then
increased, maintaining the equality of all the equations while keeping the other nonbasic
variables at zero, until one of the basic (nonzero) variables is reduced to zero and thus removed
from the basis. At this point, a new solution has been determined at a different corner of the
solution set.
The process is then repeated with a new variable becoming basic as another becomes nonbasic.
Eventually, one of three things will happen. First, a solution may occur where no nonbasic
variable will decrease the cost, in which case the current solution is the optimal solution. Second,
a non-basic variable might increase to infinity without causing a basic variable to become zero,
resulting in an unbounded solution. Finally, no solution may actually exist and the Simplex
Method must abort. As is common for research in linear programming, the possibility that the
Simplex Method might return to a previously visited corner will not be considered here.
The primary data structure used by the Simplex Method is "sometimes called a dictionary, since
the values of the basic variables may be computed (‘looked up’) by choosing values for the
nonbasic variables." (Gill, Murray, and Wright, p. 337) Dictionaries contain a representation of
the set of equations appropriately adjusted to the current basis. The use of dictionaries provide an
intuitive understanding of why each variable enters and leaves the basis. The drawback to
dictionaries, however, is the necessary step of updating them which can be time-consuming.
Computer implementation is possible, but a version of the Simplex Method has evolved with a
more efficient matrix-oriented approach to the same problem. This new implementation became
known as the Revised Simplex Method.
The steps of the Simplex Method also need to be expressed in the matrix format of the Revised
Simplex Method. The basis matrix, B, consists of the column entries of A corresponding to the
coefficients of the variables currently in the basis. That is if x2
is the fourth entry of the basis,
then [ a12
a22
…am2
]T
is the fourth column of B. (Note that B is therefore an m X m matrix.)
The non-basic columns of A constitute a similar though likely not square, matrix referred to here
as V.
ASSIGNMENTS
Subject code: MB 0032
(2 credits)
Set 2
Marks 30
SUBJECT NAME : OPERATIONS RESEARCH
Note: Each Question carries 10 marks
1. What are the important features of Operations Research? Describe in details the
different phases of Operations Research.
Answer: Important features of OR are:
i. It is System oriented: OR studies the problem from over all point of view of organizations or
situations since optimum result of one part of the system may not be optimum for some other
part.
ii. It imbibes Inter – disciplinary team approach. Since no single individual can have a thorough
knowledge of all fast developing scientific knowhow, personalities from different scientific and
managerial cadre form a team to solve the problem.
iii. It makes use of Scientific methods to solve problems.
iv. OR increases the effectiveness of a management Decision making ability.
v. It makes use of computer to solve large and complex problems.
vi. It gives Quantitative solution.
vii. It considers the human factors also.
Phases of Operations Research
The scientific method in OR study generally involves the following three phases:
i) Judgment Phase: This phase consists of
a) Determination of the operation.
b) Establishment of the objectives and values related to the operation.
c) Determination of the suitable measures of effectiveness and
d) Formulation of the problems relative to the objectives.
ii) Research Phase: This phase utilizes
a) Operations and data collection for a better understanding of the problems.
b) Formulation of hypothesis and model.
c) Observation and experimentation to test the hypothesis on the basis of additional data.
d) Analysis of the available information and verification of the hypothesis using pre established
measure of effectiveness.
e) Prediction of various results and consideration of alternative methods.
iii) Action Phase: It consists of making recommendations for the decision process by those who
first posed the problem for consideration or by anyone in a position to make a decision,
influencing the operation in which the problem is occurred.
2. Describe a Linear Programming Problem in details in canonical form.
Answer: Linear Programming
The Linear Programming Problem (LPP) is a class of mathematical programming in which the
functions representing the objectives and the constraints are linear. Here, by optimization, we
mean either to maximize or minimize the objective functions. The general linear programming
model is usually defined as follows:
Maximize or Minimize
Z = c1 x1 + c2 x 2 +………….. +cn x n
subject to the constraints,
a11 x1 + a12 x2 + …………….+ a1n xn ~ b1
a21 x1 + a22 x2 +……………..+ a2n xn ~ b2
……………………………………………
……………………………………………
am1x1 + am2 x2 +……………. +amn xn ~ bm
and x1 > 0, x2 > 0, xn > 0.
Where cj, bi and aij (i = 1, 2, 3, ….. m, j = 1, 2, 3…….. n) are constants determined from the
technology of the problem and xj (j = 1, 2, 3 n) are the decision variables. Here ~ is either < (less
than), > (greater than) or = (equal). Note that, in terms of the above formulation the coefficient
cj, aij, bj are interpreted physically as follows. If bi is the available amount of resources i, where
aij is the amount of resource i, that must be allocated to each unit of activity j, the “worth” per
unit of activity is equal to cj.
Canonical forms:
The general Linear Programming Problem (LPP) defined above can always be put in the
following form which is called as the canonical form:
Maximise Z = c1 x1+c2 x2 + …….+cn xn
Subject to
a11 x1 + a12 x2 +…………….. +a1n xn < b1
a21 x1 + a22 x2 +…………….. +a2n xn < b2
……………………………………………
……………………………………………
am1x1+am2 x2 + …… + amn xn < bm
x1, x2, x3, … xn > 0.
The characteristics of this form are:
1) all decision variables are nonnegative.
2) all constraints are of < type.
3) the objective function is of the maximization type.
Any LPP can be put in the cannonical form by the use of five elementary transformations:
1. The minimization of a function is mathematically equivalent to the maximization of the
negative expression of this function. That is, Minimize Z = c1 x1 + c2x2 + ……. + cn xn is
equivalent to
Maximize – Z = – c1x1 – c2x2 – … – cnxn.
2. Any inequality in one direction (< or >) may be changed to an inequality in the opposite
direction (> or <) by multiplying both sides of the inequality by –1.
For example 2x1+3x2 > 5 is equivalent to –2x1–3x2 < –5.
3. An equation can be replaced by two inequalities in opposite direction. For example, 2x1+3x2
= 5 can be written as 2x1+3x2 < 5 and 2x1+3x2 > 5 or 2x1+3x2 < 5 and – 2x1 – 3x2 < – 5.
4. An inequality constraint with its left hand side in the absolute form can be changed into two
regular inequalities. For example: | 2x1+3x2 | < 5 is equivalent to 2x1+3x2 < 5 and 2x1+3x2 > –
5 or – 2x1 – 3x2 < 5.
5. The variable which is unconstrained in sign (i.e., > 0, < 0 or zero) is equivalent to the
difference between 2 nonnegative variables. For example, if x is unconstrained in sign then x
= (x + – x – ) where x + > 0, x – < 0.
3. What are the different steps needed to solve a system of equations by the simplex
method?
Answer: To Solve problem by Simplex Method
1. Introduce stack variables (Si’s) for “ < ” type of constraint.
2. Introduce surplus variables (Si’s) and Artificial Variables (Ai) for “>” type of constraint.
3. Introduce only Artificial variable for “=” type of constraint.
4. Cost (Cj) of slack and surplus variables will be zero and that of Artificial variable will be “M”
Find Zj Cj for each variable.
5. Slack and Artificial variables will form Basic variable for the first simplex table. Surplus
variable will never become Basic Variable for the first simplex table.
6. Zj = sum of [cost of variable x its coefficients in the constraints – Profit or cost coefficient of
the variable].
7. Select the most negative value of Zj - Cj. That column is called key column. The variable
corresponding to the column will become Basic variable for the next table.
8. Divide the quantities by the corresponding values of the key column to get ratios select the
minimum ratio. This becomes the key row. The Basic variable corresponding to this row will be
replaced by the variable found in step 6.
9. The element that lies both on key column and key row is called Pivotal element.
10. Ratios with negative and “a” value are not considered for determining key row.
11. Once an artificial variable is removed as basic variable, its column will be deleted from next
iteration.
12. For maximisation problems decision variables coefficient will be same as in the objective
function. For minimization problems decision variables coefficients will have opposite signs as
compared to objective function.
13. Values of artificial variables will always is – M for both maximisation and minimization
problems.
14. The process is continued till all Zj – Cj > 0.
ASSIGNMENTS
Subject code: MB 0032
(3 or 4 credits)
Set 1
Marks 60
SUBJECT NAME : OPERATIONS RESEARCH
Note: Each Question carries 10 marks
1. Describe the broad classification of Operations Research models in details. Name the
different steps needed in OR approach of problem solving?
Answer: A model is known as the representation of the reality. It is known as an idealized
representation or abstraction of a real life system. The main objective of this model is to identify
significant factors and their interrelationship. A model is helpful is decision making as it
provides a simplified description of complexities and uncertainties of a problem in logical
structure.
A broad classification of OR models:
a) Physical modes include all form of diagrams, graphs and charts. They are designed to deal
with specific problems. They bring out significant factors and inter-relationship in pictorial firm
so as to facilitate analysis.
There are two types:
1) Ieonic models and 2) Analog models
Iconic model is known as an image of an object or system that is represented on a small scale.
We can say that these models can simulate the actual performance of a product.
On the other hand analog models are small physical systems that have similar characteristics and
work like an object. For example- Toy.
b) Mathematical Model or Symbolic models represent the decision variable of the system. The
model employs a set of mathematical symbols also. The variables are related by mathematical
system also. For example - Allocation, sequencing, replacement models etc.
c) It is by nature of Environment
We have 1) Deterministic model in which every thing is defined and the results are certain. Eg:
EOQ model 2) Probabilistic models in which the input and output variables follow a probability
distribution Eg: Games Theory.
d) By the extent of Generality: The tow models belonging to this class are 1) General model can
be applied in general and does not pertain to one problem only. Eg: Linear Programming 2)
Specific model is applicable under specific condition only. For example - Sales can response
curve or equation which can be known as a function of advertising that is applicable in the
marketing function alone.
The scientific method translates a real given problem into a mathematical representation which is
solved and retransformed into the original context. The OR approach to problem solving consists
of the following steps:
1) Definition of the problem
The first and the most important requirement is that the root problem should be identified and
understood. The problem should be identified properly, this indicates three major aspects:
1) A description of the goal or the objective of the study, 2) an identification of the decision
alternative to the system, and 3) a recognition of the limitations, restrictions and requirements of
the system.
2) Construction of the model
Depending on the definition of the problem, the operations research team should decide on the
most suitable model for representing the system. Such a model should specify quantitative
expressions for the objective and the constraints of the problem in terms of its decision variables.
3) Solution of the model
Once an appropriate model has been formulated the next stage in the analysis calls for its
solution and the interpretation of the solution in the context of the given problem – A solution to
a model implies determination of a specific set of decision variables that would yield on
optimum solution. An optimum solution is one which maximizes or minimizes the performance
of any measure in a model subject to the condition and constraints imposed on the model.
4) Validation the model
A model is a good representative of a system, and then the optimal solution must improve the
system’s performance. A common method for testing the validity of a model is to compare its
performance with some post data available for the actual system.
5) About Implementation of the final result
The optimal solution obtained from a model should be applied practice to improve the
performance of the system and the validity of the solution should be verified under changing
conditions.
2. Describe the graphical method to solve the Linear Programming Problem. Use the
following example –
Maximize Z = 30x1 + 40x2
Subject to the constraints
1.5x1 + 1.9x2 600
0.3 x1 + 0.2x2 100
0.0x1 + 0.2x2 30
and x1 0, x2 0
Answer: A LPP with 2 decision variables x1 and x2 can be solved easily by graphical method.
We consider the x1 x2 – plane where we plot the solution space, which is the space enclosed by
the constraints. Usually the solution space is a convex set which is bounded by a polygon; since a
linear function attains extreme (maximum or minimum) values only on boundary of the region, it
is sufficient to consider the vertices of the polygon and find the value of the objective function in
these vertices. By comparing the vertices of the objective function at these vertices, we obtain
the optimal solution of the problem.
The method of solving a LPP on the basis of the above analysis is known as the graphical
method. The working rule for the method is as follows:
Working Rule:
Step I: Write down the equations by replacing the inequality symbols by the equality symbol in
the given constraints.
Step II: Plot the straight lines represented by the equations obtained in step I.
Step III: Identify the convex polygon region relevant to the problem. We must decide on which
side of the line, the halfplane is located.
Step IV: Determine the vertices of the polygon and find the values of the given objective
function Z at each of these vertices. Identify the greatest and least of these values. These are
respectively the maximum and minimum value of Z.
Step V: Identify the values of (x1, x2) which correspond to the desired extreme value of Z. This
is an optimal solution of the problem.
Solution by Graphical Method
Let the horizontal axis represent x1and vertical axis represent x2. plot the constraint lines,
feasibility region has been shown in fig.
Any point on the thick line or inside the shaded portion will satisfy all the restrictions of the
problem then ABCDE is the feasibility region carried out by the constraints operating on the
objective function. This depicts the limits within which the values of the decision variables are
permissible. The intersection points C and D can be solved by the linear equations 0.1x2 < 30,
0.1x1 + 1.5x2 = 600, and 0.2x1 + 0.2x2 < 100 i.e. C(150,300) and D(300,180).
We work out the revenues at different corners points as tabulated below
At Point Feasible Solution of the
product mix
Corresponding revenue Total
revenue
X1 X2 From X1 From X2
A
B
C
D
E
0
0
15
0
30
0
0
30
0
30
0
18
0
0
7500
1500
0
2500
0
2400
2400
0
1440
0
0
24000
31500
29400
25000
50
0
0
0
0 0
From the above table we find that revenue is maximum at Rs.31500 when 150 unit of X1 and
300 units of X2 are produced.
3. Describe the Penalty Cost method (Big M Method) for solving L.P.P.
Answer: Penalty Cost Method Or Big-M Method
Consider a L.P.P. when atleast one of the constraints is of the type > or = . While expressing in
the standard form, add a non negative variable to each of such constraints. These variables are
called artificial variables. Their addition causes violation of the corresponding constraints, since
they are added to only one side of an equation, the new system is equivalent to the old system of
constraints if and only if the artificial variables are zero. To guarantee such assignments in the
optimal solution, artificial variables are incorporated into the objective function with large
positive coefficients in a minimization program or very large negative coefficients in a
maximization program. These coefficients are denoted by ± M.
Whenever artificial variables are part of the initial solution X0, the last row of simplex table will
contain the penalty cost M. The following modifications are made in the simplex method to
minimize the error of incorporating the penalty cost in the objective function. This method is
called Big M method or Penalty cost method.
1) The last row of the simplex table is decomposed into two rows, the first of which involves
those terms not containing M, while the second involves those containing M.
2) The Step 1 of the simplex method is applied to the last row created in the above modification
and followed by steps 2, 3 and 4 until this row contains no negative elements. Then step 1 of
simplex algorithm is applied to those elements next to the last row that are positioned over zero
in the last row.
3) Whenever an artificial variable ceases to be basic, it is removed from the first column of the
table as a result of step 4, it is also deleted from the top row of the table as is the entire column
under it.
4) The last row is removed from the table whenever it contains all zeroes.
5) If non zero artificial variables are present in the final basic set, then the program has no
solution.
In contrast, zero valued artificial variables in the final solution may exist when one or more of
the original constraint equations are redundant.
4. Why Duality concept is important in OR? Describe the economic importance of the
Duality concept.
Answer: The Importance of Duality Concept Is Due To Two Main Reasons
i. If the primal contains a large number of constraints and a smaller number of variables, the
labour of computation can be considerably reduced by converting it into the dual problem and
then solving it.
ii. The interpretation of the dual variable from the loss or economic point of view proves
extremely useful in making future decisions in the activities being programmed.
Economic importance of duality concept
The linear programming problem can be thought of as a resource allocation model in which the
objective is to maximize revenue or profit subject to limited resources. Looking at the problem
from this point of view, the associated dual problem offers interesting economic interpretations
of the L.P resource allocation model. We consider here a representation of the general primal and
dual problems in which the primal takes the role of a resource allocation model.
Primal
Maximize
From the above resource allocation model, the primal problem has n economic activities and m
resources. The coefficient cj in the primal represents the profit per unit of activity j.
Resource i, whose maximum availability is bi, is consumed at the rate aij units per unit of
activity j.
Interpretation of Duel Variables –
For any pair of feasible primal and dual solutions,
(Objective value in the maximization problem) ≤ (Objective value in the minimization problem)
At the optimum, the relationship holds as a strict equation. Note: Here the sense of optimization
is very important. Hence clearly for any two primal and dual feasible solutions, the values of the
objective functions, when finite, must satisfy the following inequality.
The strict equality, z = w, holds when both the primal and dual solutions are optimal.
Consider the optimal condition z = w first given that the primal problem represents a resource
allocation model, we can think of z as representing profit in Rupees. Because bi represents the
number of units available of resource i, the equation z = w can be expressed as profit (Rs) = Σ
(units of resource i) x (profit per unit of resource i) This means that the dual variables yi,
represent the worth per unit of resource i [variables yi are also called as dual prices, shadow
prices and simplex multipliers]. With the same logic, the inequality z < w associated with any
two feasible primal and dual solutions is interpreted as (profit) < (worth of resources) This
relationship implies that as long as the total return from all the activities is less than the worth of
the resources, the corresponding primal and dual solutions are not optimal. Optimality is reached
only when the resources have been exploited completely, which can happen only when the input
equals the output (profit). Economically the system is said to remain unstable (non optimal)
when the input (worth of the resources) exceeds the output (return). Stability occurs only when
the two quantities are equal.
5. Describe the Matrix Minimum method of finding the initial basic feasible solution in the
transportation problem.
Answer: The Initial basic Feasible solution using Matrix Minimum Method
Let us consider a T.P involving m-origins and n-destinations. Since the sum of origin capacities
equals the sum of destination requirements, a feasible solution always exists.
Any feasible solution satisfying m + n – 1 of the m + n constraints is a redundant one and hence
can be deleted. This also means that a feasible solution to a T.P can have at the most only m + n
– 1 strictly positive component, otherwise the solution will degenerate.
It is always possible to assign an initial feasible solution to a T.P. in such a manner that the rim
requirements are satisfied.
This can be achieved either by inspection or by following some simple rules. We begin by
imagining that the transportation table is blank i.e. initially all xij = 0. The simplest procedures
for initial allocation discussed in the following section.
Matrix Minimum Method
Step 1:Determine the smallest cost in the cost matrix of the transportation table. Let it be cij ,
Allocate xij = min ( ai, bj) in the cell ( i, j)
Step 2: If xij = ai cross off the ith row of the transportation table and decrease bj by ai go to step
3.
if xij = bj cross off the ith column of the transportation table and decrease ai by bj go to step 3.
if xij = ai= bj cross off either the ith row or the ith column but not both.
Step 3: Repeat steps 1 and 2 for the resulting reduced transportation table until all the rim
requirements are satisfied whenever the minimum cost is not unique make an arbitrary choice
among the minima.
6. What do you understand by the Integer Programming Problem? Describe the Gomory’s
All-I.P.P. method for solving the I.P.P. problem.
Answer: Integer Programming Problem
The Integer Programming Problem I P P is a special case of L P P where all or some variables
are constrained to assume nonnegative integer values. This type of problem has lot of
applications in business and industry where quite often discrete nature of the variables is
involved in many decision making situations. Eg. In manufacturing the production is frequently
scheduled in terms of batches, lots or runs; In distribution, a shipment must involve a discrete
number of trucks or aircrafts or freight cars .
An integer programming problem can be described as follows:
Determine the value of unknowns x1, x2, … , xn so as to optimize z = c1x1 +c2x2 + . . .+ cnxn
subject to the constraints ai1 x1 + ai2 x2 + . . . + ain xn =bi , i = 1,2,…,m and xj > 0 j = 1, 2,
… ,n where xj being an integral value for j = 1, 2, … , k ≤ n.
If all the variables are constrained to take only integral value i.e. k = n, it is called an all (or pure)
integer programming problem. In case only some of the variables are restricted to take integral
value and rest (n – k) variables are free to take any non negative values, then the problem is
known as mixed integer programming problem.
Gomory’s All – IPP Method
An optimum solution to an I. P. P. is first obtained by using simplex method ignoring the
restriction of integral values. In the optimum solution if all the variables have integer values, the
current solution will be the desired optimum integer solution. Otherwise the given IPP is
modified by inserting a new constraint called Gomory’s or secondary constraint which represents
necessary condition for integrability and eliminates some non integer solution without losing any
integral solution. After adding the secondary constraint, the problem is then solved by dual
simplex method to get an optimum integral solution. If all the values of the variables in this
solution are integers, an optimum inter-solution is obtained, otherwise another new constrained is
added to the modified L P P and the procedure is repeated. An optimum integer solution will be
reached eventually after introducing enough new constraints to eliminate all the superior non
integer solutions. The construction of additional constraints, called secondary or Gomory’s
constraints, is so very important that it needs special attention.
The iterative procedure for the solution of an all integer programming problem is as follows:
Step 1: Convert the minimization I.P.P. into that of maximization, if it is in the minimization
form. The integrality condition should be ignored.
Step 2: Introduce the slack or surplus variables, wherever necessary to convert the inequations
into equations and obtain the optimum solution of the given L.P.P. by using simplex algorithm.
Step 3: Test the integrality of the optimum solution
a) If the optimum solution contains all integer values, an optimum basic feasible integer solution
has been obtained.
b) If the optimum solution does not include all integer values then proceed onto next step.
Step 4: Examine the constraint equations corresponding to the current optimum solution. Let
these equations be represented by
Where n’ denotes the number of variables and m’ the number of equations.
Choose the largest fraction of bis ie to find bii
Let it be [bk 1]
or write is as
f ko
Step 5: Express each of the negative fractions if any, in the k th row of the optimum simplex
table as the sum of a negative integer and a nonnegative fraction.
Step 6: Find the Gomorian constraint
and add the equation
to the current set of equation constraints.
Step 7: Starting with this new set of equation constraints, find the new optimum solution by dual
simplex algorithm. (So that Gsla (1) is the initial leaving basic variable).
Step 8: If this new optimum solution for the modified L.P.P. is an integer solution. It is also
feasible and optimum for the given I.P.P. otherwise return to step 4 and repeat the process until
an optimum feasible integer solution is obtained.
ASSIGNMENTS
Subject code: MB 0032
(3 or 4 credits)
Set 2
Marks 60
SUBJECT NAME : OPERATIONS RESEARCH
Note: Each Question carries 10 marks
1. Describe in details the OR approach of problem solving. What are the limitations of the
Operations Research?
Answer: The OR approach to problem solving consists of the following steps:
1. Definition of the problem.
2. Construction of the model.
3. Solution of the model.
4. Validation of the model.
5. Implementation of the final result.
1. Definition of the problem
The first and the most important requirement is that the root problem should be identified and
understood. The problem should be identified properly, this indicates three major aspects: (1) a
description of the goal or the objective of the study, (2) an identification of the decision
alternative to the system, and (3) a recognition of the limitations, restrictions and requirements of
the system.
2. Construction of the model
Depending on the definition of the problem, the operations research team should decide on the
most suitable model for representing the system. Such a model should specify quantitative
expressions for the objective and the constraints of the problem in terms of its decision variables.
A model gives a perspective picture of the whole problem and helps tackling it in a well
organized manner. If the resulting model fits into one of the common mathematical models, a
convenient solution may be obtained by using mathematical techniques. If the mathematical
relationships of the model are too complex to allow analytic solutions, a simulation model may
be more appropriate. There are various types of models which can be constructed under different
conditions.
3. Solution of the model
Once an appropriate model has been formulated, the next stage in the analysis calls for its
solution and the interpretation of the solution in the context of the given problem. A solution to a
model implies determination of a specific set of decision variables that would yield an Optimum
solution. An Optimum solution is one which maximize or minimize the performance of any
measure in a model subject to the conditions and constraints imposed on the model.
4. Validation the model
A model is a good representative of a system, then the Optimal solution must improve the
system’s performance. A common method for testing the validity of a model is to compare its
performance with some past data available for the actual system. The model will be valid if
under similar conditions of inputs, it can reproduce the past performance of the system. The
problem here is that there is no assurance that future performance will continue to duplicate past
behavior. Also, since the model is based on careful examination of past data, the comparison
should always reveal favorable results. In some instances this problem may be overcome by
using data from trial runs of the system. It must be noted that such a validation method is not
appropriate for nonexistent systems, since data will not be available for comparison.
5. Implementation of the final result
The optimal solution obtained from a model should be applied practice to improve the
performance of the system and the validity of the solution should be verified under changing
conditions. It involves the translation of these results into detailed operating instructions issued
in an understandable form to the individuals who will administer and operate the recommended
system. The interaction between the operations research team and the operating personnel will
reach its peak in this phase.
Limitations of OR
The limitations are more related to the problems of model building, time and money factors.
i) Magnitude of computation: Modern problem involve large number of variables and hence to
find interrelationship, among makes it difficult.
ii) Non – quantitative factors and Human emotional factor cannot be taken into account.
iii) There is a wide gap between the managers and the operation researches
iv) Time and Money factors when the basic data is subjected to frequent changes then
incorporation of them into OR models is a costly affair.
v) Implementation of decisions involves human relations and behaviour.
2. What are the characteristics of the standard form of L.P.P.? What is the standard form
of L.P.P.? State the fundamental theorem of L.P.P..
Answer: The Standard Form Of LPP
The characteristics of the standard form are:
1. All constraints are equations except for the non negativity condition which remain inequalities
(>, 0) only.
2. The right hand side element of each constraint equation is non-negative.
3. All variables are non-negative.
4. The objective function is of the maximization or minimization type.
The inequality constraints can be changed to equations by adding or subtracting the left hand
side of each such constraints by a nonnegative variable. The nonnegative variable that has to be
added to a constraint inequality of the form < to change it to an equation is called a slack
variable. The non-negative variable that has to be substracted from a constraint inequality of the
form ³ to change it to an equation is called a surplus variable. The right hand side of a constraint
equation can be made positive by multiplying both sides of the resulting equation by (-1)
wherever necessary. The remaining characteristics are achieved by using the elementary
transformations introduced with the canonical form.
The Standard Form Of The LPP
Any standard form of the L.P.P. is given by
Fundamental Theorem Of L.P.P.
Given a set of m simultaneous linear equations in n unknowns/variables, n > m, AX = b, with
r(A) = m. If there is a feasible solution X > 0, then there exists a basic feasible solution.
3. Describe the Two-Phase method of solving a linear programming problem with an
example.
Answer: Two Phase Method
The drawback of the penalty cost method is the possible computational error that could result
from assigning a very large value to the constant M. To overcome this difficulty, a new method
is considered, where the use of M is eliminated by solving the problem in two phases. They are
Phase I: Formulate the new problem by eliminating the original objective function by the sum of
the artificial variables for a minimization problem and the negative of the sum of the artificial
variables for a maximization problem. The resulting objective function is optimized by the
simplex method with the constraints of the original problem. If the problem has a feasible
solution, the optimal value of the new objective function is zero (which indicates that all artificial
variables are zero). Then we proceed to phase II. Otherwise, if the optimal value of the new
objective function is non zero, the problem has no solution and the method terminates.
Phase II : Use the optimum solution of the phase I as the starting solution of the original
problem. Then the objective function is taken without the artificial variables and is solved by
simplex method.
Examples:
7) Use the two phase method to
Maximise z = 3x1 – x2
Subject to 2x1 + x2 > 2
x1 + 3x2 < 2
x2 < 4,
x1, x2 > 0
Rewriting in the standard form,
Maximize z = 3x1 – x2 + 0S1 – MA1 + 0.S2 + 0.S3
Subject to 2x1 + x2 – S1 + A1 = 2
x1 + 3x2 + S2 = 2
x2 + S3 = 4,
x1, x2, S1, S2, S3, A1 > 0.
Phase I :
Consider the new objective,
Maximize Z* = – A1
Subject to 2x1 + x2 – S1 + A1 = 2
x1 + 3x2 + S2 = 2
x2 + S3 = 4,
x1, x2, S1, S2, S3, A1 > 0.
Solving by Simplex method, the initial simplex table is given by
x1 enters the basic set replacing A1.
The first iteration gives the following table:
Phase I is complete, since there are no negative elements in the last row.
The Optimal solution of the new objective is Z* = 0.
Phase II:
Consider the original objective function,
Maximize z = 3x1 – x2 + 0S1 + 0S2 + 0S3
Subject to
x2 + S3 = 4
x1, x2, S1, S2, S3 > 0
with the initial solution x1 = 1, S2 = 1, S3 = 4, the corresponding simplex table is
Proceeding to the next iteration, we get the following table:
Since all elements of the last row are non negative, the current solution is optimal.
The maximum value of the objective function
Z = 6 which is attained for x1 = 2, x2 = 0.
4. What do you understand by the transportation problem? What is the basic assumption
behind the transportation problem? Describe the MODI method of solving transportation
problem.
Answer: This model studies the minimization of the cost of transporting a commodity from a
number of sources to several destinations. The supply at each source and the demand at each
destination are known. The transportation problem involves m sources, each of which has
available ai (i = 1, 2, …..,m) units of homogeneous product and n destinations, each of which
requires bj (j = 1, 2…., n) units of products. Here ai and bj are positive integers. The cost cij of
transporting one unit of the product from the i th source to the j th destination is given for each i
and j. The objective is to develop an integral transportation schedule that meets all demands from
the inventory at a minimum total transportation cost.
It is assumed that the total supply and the total demand are equal.
The condition (1) is guaranteed by creating either a fictitious destination with a demand equal to
the surplus if total demand is less than the total supply or a (dummy) source with a supply equal
to the shortage if total demand exceeds total supply. The cost of transportation from the fictitious
destination to all sources and from all destinations to the fictitious sources are assumed to be zero
so that total cost of transportation will remain the same.
The Transportation Algorithm (MODI Method)
The first approximation to (2) is always integral and therefore always a feasible solution. Rather
than determining a first approximation by a direct application of the simplex method it is more
efficient to work with the table given below called the transportation table. The transportation
algorithm is the simplex method specialized to the format of table it involves:
i) finding an integral basic feasible solution
ii) testing the solution for optimality
iii) improving the solution, when it is not optimal
iv) repeating steps (ii) and (iii) until the optimal solution is obtained.
The solution to T.P is obtained in two stages. In the first stage we find Basic feasible solution by
any one of the following methods a) Northwest corner rale b) Matrix Minima Method or least
cost method c) Vogel’s approximation method. In the second stage we test the B.Fs for its
optimality either by MODI method or by stepping stone method.
Modified Distribution Method / Modi Method / U – V Method.
Step 1: Under this method we construct penalties for rows and columns by subtracting the least
value of row / column from the next least value.
Step 2: We select the highest penalty constructed for both row and column. Enter that row /
column and select the minimum cost and allocate min (ai, bj)
Step 3: Delete the row or column or both if the rim availability / requirements is met.
Step 4: We repeat steps 1 to 2 to till all allocations are over.
Step 5: For allocation all form equation ui + vj = cj set one of the dual variable ui / vj to zero and
solve for others.
Step 6: Use these value to find Δij = cij – ui – vj of all Δij >, then it is the optimal solution.
Step 7: If any Dij £ 0, select the most negative cell and form loop. Starting point of the loop is
+ve and alternatively the other corners of the loop are –ve and +ve. Examine the quantities
allocated at –ve places. Select the minimum. Add it at +ve places and subtract from –ve place.
Step 8: Form new table and repeat steps 5 to 7 till Δij > 0
5. Describe the North-West Corner rule for finding the initial basic feasible solution in the
transportation problem.
Answer: North West Corner Rule
Step1: The first assignment is made in the cell occupying the upper left hand (north west) corner
of the transportation table. The maximum feasible amount is allocated there, that is x11 = min
(a1,b1)
So that either the capacity of origin O1 is used up or the requirement at destination D1 is
satisfied or both. This value of x11 is entered in the upper left hand corner (small square) of cell
(1, 1) in the transportation table
Step 2: If b1 > a1 the capacity of origin O, is exhausted but the requirement at destination D1 is
still not satisfied , so that at least one more other variable in the first column will have to take on
a positive value. Move down vertically to the second row and make the second allocation of
magnitude
x21 = min (a2, b1 – x21) in the cell (2,1). This either exhausts the capacity of origin O2 or
satisfies the remaining demand at destination D1.
If a1 > b1 the requirement at destination D1 is satisfied but the capacity of origin O1 is not
completely exhausted. Move to the right horizontally to the second column and make the second
allocation of magnitude x12 = min
(a1 – x11, b2) in the cell (1, 2) . This either exhausts the remaining capacity of origin O1 or
satisfies the demand at destination D2 .
If b1 = a1, the origin capacity of O1 is completely exhausted as well as the requirement at
destination is completely satisfied. There is a tie for second allocation, An arbitrary tie breaking
choice is made. Make the second allocation of magnitude x12 = min (a1 – a1, b2) = 0 in the cell
(1, 2) or x21 = min (a2, b1 – b2) = 0 in the cell (2, 1).
Step 3: Start from the new north west corner of the transportation table satisfying destination
requirements and exhausting the origin capacities one at a time, move down towards the lower
right corner of the transportation table until all the rim requirements are satisfied.
6. Describe the Branch and Bound Technique to solve an I.P.P. problem.
Answer: The Branch And Bound Technique
Sometimes a few or all the variables of an IPP are constrained by their upper or lower bounds or
by both. The most general technique for the solution of such constrained optimization problems
is the branch and bound technique. The technique is applicable to both all IPP as well as mixed
I.P.P. the technique for a maximization problem is discussed below:
Let the I.P.P. be
Subject to the constraints
xj is integer valued , j = 1, 2, …….., r (< n) –––– (3)
xj > 0 …………………. j = r + 1, …….., n ––––––––––(4)
Further let us suppose that for each integer valued xj, we can assign lower and upper bounds for
the optimum values of the variable by
Lj ≤ xj ≤ Uj j = 1, 2, …. r ––––––––––––– (5)
The following idea is behind “the branch and bound technique”
Consider any variable xj, and let I be some integer value satisfying Lj £ I £ Uj – 1. Then clearly
an optimum solution to (1) through (5) shall also satisfy either the linear constraint.
x j > I + 1 ––––––––––––– ( 6)
Or the linear constraint xj ≤ I ………………...(7)
To explain how this partitioning helps, let us assume that there were no integer restrictions (3),
and suppose that this then yields an optimal solution to L.P.P. – (1), (2), (4) and (5). Indicating
x1 = 1.66 (for example). Then we formulate and solve two L.P.P’s each containing (1), (2) and
(4).
But (5) for j = 1 is modified to be 2 ≤ x1 ≤ U1 in one problem and L1 ≤ x1 ≤ 1 in the other.
Further each of these problems process an optimal solution satisfying integer constraints (3)
Then the solution having the larger value for z is clearly optimum for the given I.P.P. However,
it usually happens that one (or both) of these problems has no optimal solution satisfying (3), and
thus some more computations are necessary. We now discuss step wise the algorithm that
specifies how to apply the partitioning (6) and (7) in a systematic manner to finally arrive at an
optimum solution.
We start with an initial lower bound for z, say z (0) at the first iteration which is less than or
equal to the optimal value z*, this lower bound may be taken as the starting Lj for some xj.
In addition to the lower bound z (0) , we also have a list of L.P.P’s (to be called master list)
differing only in the bounds (5). To start with (the 0 th iteration) the master list contains a single
L.P.P. consisting of (1), (2), (4) and (5). We now discuss below, the step by step procedure that
specifies how the partitioning (6) and (7) can be applied systematically to eventually get an
optimum integer valued solution.