MB0032 Operations Research

Sikkim Manipal University: Master of Business Administration- MBA Semester 2

Assignment : MB0032 Operation Research

Set 1

Question 1Describe in details the different scopes of application of Operations Research. Answer:Operations research, also known as operational research, is an interdisciplinary branch of applied mathematics and formal science that uses advanced analytical methods such as mathematical modeling, statistical analysis, and mathematical optimization to arrive at optimal or near-optimal solutions to complex decision-making problems. It is often concerned with determining the maximum (of profit, performance, or yield) or minimum (of loss, risk, or cost) of some real-world objective.

Originating in military efforts before World War II, its techniques have grown to concern problems in a variety of industries. Operations research encompasses a wide range of problem-solving techniques and methods applied in the pursuit of improved decision-making and efficiency. Some of the tools used by operations researchers are statistics, optimization, probability theory, queuing theory, game theory, graph theory, decision analysis, mathematical modeling and simulation. Because of the computational nature of these fields, OR also has strong ties to computer science.The major sub-disciplines in modern operations research, as identified by the journal Operations Research are:

Computing and information technologies

Decision analysis

Environment, energy, and natural resources

Financial engineering

Manufacturing, service sciences, and supply chain management

Policy modeling and public sector work

Revenue management

Simulation

Stochastic models

Transportation

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2

Assignment : MB0032 Operation Research

Problems addressed with operations research

critical path analysis or project planning: identifying those

processes in a complex project which affect the overall duration

of the project

floorplanning: designing the layout of equipment in a factory or

components on a computer chip to reduce manufacturing time

(therefore reducing cost)

network optimization: for instance, setup of telecommunications

networks to maintain quality of service during outages

allocation problems

routing, such as determining the routes of buses so that as few

buses are needed as possible

supply chain management: managing the flow of raw materials

and products based on uncertain demand for the finished

products

efficient messaging and customer response tactics

Automation: automating or integrating robotic systems in

human-driven operations processes

Globalization: globalizing operations processes in order to take

advantage of cheaper materials, labor, land or other productivity

inputs

Transportation: managing freight transportation and delivery

systems (Examples: LTL Shipping, intermodal freight transport)

Scheduling:

o personnel staffing

o manufacturing steps

o project tasks

o network data traffic: these are known as queueing models

or queueing systems.

o sports events and their television coverage

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Assignment : MB0032 Operation Research

blending of raw materials in oil refineries

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.

In 1967 Stafford Beer characterized the field of management science

as "the business use of operations research".However, in modern times

the term management science may also be used to refer to the

separate fields of organizational studies or corporate strategy

Applications of management science

Applications of management science are abundant in industry as

airlines, manufacturing companies, service organizations, military

branches, and in government. The range of problems and issues to

which management science has contributed insights and solutions is

vast. It includes:

Scheduling airlines, including both planes and crew,

Deciding the appropriate place to site new facilities such as a

warehouse, factory or fire station,

Managing the flow of water from reservoirs,

Identifying possible future development paths for parts of the

telecommunications industry,

Establishing the information needs and appropriate systems to

supply them within the health service, and

Identifying and understanding the strategies adopted by

companies for their information systems

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2

Assignment : MB0032 Operation Research

Question 2What 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: The mathematical models which tells to optimise (minimize or maximise) the objective function Z subject to certain condition on the variables is called a Linear programming problem (LPP).During World War II, the military managements in the U.K and the USA engaged a team of scientists to study the limited military resources and form a plan of action or programme to utilise them in the most effective manner. This was done under the name 'Operation Research' (OR) because the team was dealing with research on military operation.

Linear Programming Problems (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 formMax (or min) Z = c1x1 + c2x2 + … +cnxn

x1, x2, ....xn are called decision variable.

Application Areas of Linear ProgrammingThe Application Areas of Linear Programming are:     1. Transportation Problem     2. Military Applications     3. Operation of System Of Dams     4. Personnel Assignment Problem     5. Other Applications: (a). manufacturing plants, (b). distribution centres, (c). production management and manpower management.

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Assignment : MB0032 Operation Research

Basic Concept of Linear Programming ProblemObjective Function: The Objective Function is a linear function of variables which is to be optimised i.e., maximised or minimised. e.g., profit function, cost function etc. The objective function may be expressed as a linear expression.Constraints: A linear equation represents a straight line. Limited time, labour etc. may be expressed as linear inequations or equations and are called constraints.Optimisation: A decision which is considered the best one, taking into consideration all the circumstances is called an optimal decision. The process of getting the best possible outcome is called optimisation.Solution of a LPP: A set of values of the variables x1, x2,….xn which satisfy all the constraints is called the solution of the LPP.. Feasible Solution: A set of values of the variables x1, x2, x3,….,xn which satisfy all the constraints and also the non-negativity conditions is called the feasible solution of the LPP.Optimal Solution: The feasible solution, which optimises (i.e., maximizes or minimizes as the case may be) the objective function is called the optimal solution. Important terms Convex Region and Non-convex Sets.

Mathematical Formulation of Linear Programming ProblemsThere 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.

1. Identify the decision variables and assign symbols x and y to them. These decision variables are those quantities whose values we wish to determine.2. Identify the set of constraints and express them as linear equations/inequations in terms of the decision variables. These constraints are the given conditions.3. 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.4. Add the non-negativity restrictions on the decision variables, as in the physical problems, negative values of decision variables have no valid interpretation.

Advantages of Linear Programmingi. The linear programming technique helps to make the best possible use of available productive resources (such as time, labour, machines etc.)ii. In a production process, bottle necks may occur. For example, in a factory some machines may be in great demand while others may

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Assignment : MB0032 Operation Research

lie idle for some time. A significant advantage of linear programming is highlighting of such bottle necks.

Limitations of Linear Programming(a). 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.

     (b). Factors such as uncertainty, weather conditions etc. are not taken into consideration.

Graphical Method of Solution of a Linear Programming ProblemThe graphical method is applicable to solve the LPP involving two decision variables x1, and x2, we usually take these decision variables as x, y instead of x1, x2. To solve an LPP , the graphical method includes two major steps.

a) The determination of the solution space that defines the feasible solution (Note that the set of values of the variable x1, x2, x3,....xn which satisfy all the constraints and also the non-negative conditions is called the feasible solution of the LPP).

     b) The determination of the optimal solution from the feasible region. There are two techniques to find the optimal solution of an LPP. Corner Point Method and ISO- PROFIT (OR ISO-COST).

Some Exceptional CasesWe may come across LPP which may have no feasible (infeasible) solution or may have unbounded solution.If the intersection of the constraints is empty and the problem has no feasible solution. Therefore the given L.P.P has no solution.

Conclusion     The graphical method of solving an LPP is possible only if there are two decision variables (say x and y). This method is not suitable if there are three or more decision variables. In this case, there is a powerful method called 'simplex method'. The wide usage of liner programming helps in business and economics, to use the resources available in a planned and economical way. We have just learnt the basics of LPP, there is in fact a lot to learn in higher classes. A lot of research work is carried all over the world which is based on LPP.

REQUIREMENTS OF L.P.P

1) Decision variables and their relationship

2) Well defined objective function

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Assignment : MB0032 Operation Research

3) Existence of alternative courses of action

4) Non negative conditions of decision variables

BASIC ASSUPMTIONS OF L.P.P

1) Linearity : both objective function and constraints must be

expressed as linear inequalities.

2) Deterministic :All coefficient of decision variables in the

objective constraints expressions should be known and finite.

3) Additivity: the value of objective function for the given values of

decision variables and the total sum of resources used, must be

equal to sum of the contributions earned from each decision

variable and the sum of resources used by the decisions

variables respectively.

4) Divisibility:The solution of decision variables and resources can

be any non-negative values including fractions.

Question 3Describe the different steps needed to solve a problem by simplex method. Answer:The Simplex Method is another algorithm for solving LP problems. You recall that the Algebraic Method provides all vertices even those which are not feasible. Therefore, it is not an efficient way of solving LP problems with large numbers of constraints. The Simplex Method is a modification of the Algebraic Method, which overcomes this deficiency. However, the Simplex Method has its own deficiencies. For example, it requires that all variables be non-negative (>=0); also, all other constraints must be in <= form with non-negative right-hand-side (RHS) values.

Like the Algebraic Method, the simplex method is also a tabular solution algorithm. However, each tableau in the simplex method corresponds to a movement from one basic variable set BVS (extreme

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Assignment : MB0032 Operation Research

or corner point) to another, making sure that the objective function improves at each iteration until the optimal solution is reached.

The presentation of the simplex method is not universal. In the U.S.A. professors on the West Coast enjoy solving the minimization problems, while in the East the maximization version is preferred. Even within each of these groups you will find differences in presenting the simplex rules. The following process describes all the steps involved in applying the simplex solution algorithm:

1. Convert the LP to the following form:

Convert the minimization problem into a maximization one (by multiplying the objective function by -1).All variables must be non-negative.All RHS values must be non-negative (multiply both sides by -1, if needed). All constraints must be in <= form (except the non-negativity conditions). No strictly equality or >= constraints are allowed.If this condition cannot be satisfied, then use the Initialization of the Simplex Method

2. Convert all <= constraints to equalities by adding a different slack variable for each one of them.

3. Construct the initial simplex tableau with all slack variables in the BVS. The last row in the table contains the coefficient of the objective function (row Cj).

4. Determine whether the current tableau is optimal. That is:If all RHS values are non-negative (called, the feasibility condition)If all elements of the last row, that is Cj row, are non-positive (called, the optimality condition).

If the answers to both of these two questions are Yes, then stop. The current tableau contains an optimal solution.Otherwise, go to the next step.

5. If the current BVS is not optimal, determine, which nonbasic variable should become a basic variable and, which basic variable should become a nonbasic variable. To find the new BVS with the better objective function value, perform the following tasks:

o Identify the entering variable: The entering variable is the one with the largest positive Cj value (In case of a tie, we

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Assignment : MB0032 Operation Research

select the variable that corresponds to the leftmost of the columns) .

o Identify the outgoing variable: The outgoing variable is the one with smallest non-negative column ratio (to find the column ratios, divide the RHS column by the entering variable column, wherever possible). In case of a tie we select the variable that corresponds to the upmost of the tied rows.

o Generate the new tableau: Perform the Gauss-Jordan pivoting operation to convert the entering column to an identity column vector (including the element in the Cj row).

Go to step 4.

A short discussion on the simplex method strategy: At the start of the simplex procedure; the set of basis is constituted by the slack variables. Remember that the first BVS has only slack variables in it. The row Cj presents the increase in the value of the objective function that will result if one unit of the variable corresponding to the jth column was brought in the basis. This row answers the question: Can we improve the objective function by moving to a new BVS? We will call it The Indicator Row (since it indicates if the optimality condition is satisfied).

Criterion for entering a new variable into the BVS will cause the largest per-unit improvement of the objective function. Criterion for removing a variable from the current BVS maintains feasibility (making sure that the new RHS, after pivoting remain non-negative).

Warning: Whenever during the Simplex iterations yet get a negative RHS, it means you have selected a wrong outgoing variable. The best remedy is to start all over again.

Notice that there is a solution corresponding to each simplex tableau. The numerical of basic variables are the RHS values, while the other variables (non-basic variables) are always equal to zero.

Notice also that variables can exit and enter the basis repeatedly during the simplex algorithm.

Numerical Recipes states that the Simplex algorithm is 'almost always' O(max(N,M)), which means that the number of iteration is a factor of number of variables or number of constraints, whichever is larger.

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2

Assignment : MB0032 Operation Research

Geometric Interpretation of the Simplex Method: The simplex method always starts at the origin (which is a corner point) and then jumps from a corner point to the neighboring corner point until it reaches the optimal corner point (if bounded). Therefore, at each one of the simplex iterations, we are searching for a better solution among the vertices of a Simplex. A simplex in an n-dimensional space is the simplest shape having n + 1 vertices. For example, a triangle is a simplex in 2-dimensional space while a pyramid is a simplex in 3-dimensional space. These movements can be seen when you correspond each simplex tableau with an specific corner point in the graphical method e.g.; the carpenter's problem, as shown next.

A Numerical Example: The Carpenter's Problem

Maximize 5X1 + 3X2

Subject to:2X1 + X2 = 40X1 + 2X2 = 50and both X1, X2 are non-negative.

After adding two slack variables S1 and S2 the problem is equivalent to:

Maximize 5X1 + 3X2

Subject to:2X1 + X2 + S1 = 40X1 + 2X2 + S2 = 50and variables X1, X2, S1, S2 are all non-negative.

The initial tableau is:

BVS

X1X2

S1

S2

RHS

Column Ratio (C/R)

S1[2]

1 1 0 40 40/2

S2 1 2 0 1 50 50/1Cj 5 3 0 0

The solution shown by this tableau is: S1 = 40, S2 = 50, X1 = 0, and X2 = 0. This solution is the origin, shown in our graphical method.

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Assignment : MB0032 Operation Research

This table is not optimal since some of Cj elements are positive. The incoming variable is X1 and the outgoing variable is S1 (by C/R test). The pivot element is in the bracket. After pivoting, we have:

BVS

X1

X2 S1S2

RHS

Column Ratio (C/R)

X1 1 1/2 1/2 0 20 20/(1/2)=40

S2 0[3/2

]-

1/21 30 30/(3/2)=10

Cj 0 1/2-

5/20

The solution to this tableau is: X1 = 20, S2 = 30, S1 = 0, and X2 = 0. This solution is the corner point (20, 0), shown in our graphical method.

This table is not optimal, since some of Cj elements is positive. The incoming variable is X2 and the outgoing variable is S2 (by C/R test). The pivot element is in the bracket. After pivoting, we have:

BVS

X1

X2

S1 S2RHS

X1 1 0 2/3-

1/310

X2 0 1-

1/32/3 20

Cj 0 0-

7/3-

1/3

The solution to this tableau is: X1 = 10, X2 = 20, S1 = 0, and S2 = 0. This solution is the corner point (10, 20), shown in our graphical method.

This tableau is optimal, since all Cj elements are non-positive and all RHS are non-negative. The optimal solution is X1 = 10, X2 = 20, S1 =0, S2 = 0. To find the optimal value, plug in this solution into the objective function 5X1 + 3X2 = 5(10) + 3(20) = 110.

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Assignment : MB0032 Operation Research

Question 4

Describe the economic importance of the Duality concept. AnswerThe importance of duality concept is due to two main reasons:

a) 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.

b) 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 interpretation of duality: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.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 bj, is consumed at the rate aij units per unit of activity j.Economic interpretation of dual 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 important.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. Optimally 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

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the input (worth of the resources) exceeds the output (return). Stability occurs only when the two quantities are equal.

Question 5How can you use the Matrix Minimum method to find the initial basic feasible solution in the transportation problem. Answer:MATRIX MINIMUM METHOD (LEAST COST METHOD)

Is the method of computing basic feasible solution of transportation

problem, where the basic variables are chosen according to unit cost of

transportation .This method is very useful because it reduces the

computation and time required to determine the optimal solution.

The following step summarizes the approach:

1) Identify the box having minimum transportation cost (Cij).

2) If the minimum cost is not unique, then you are at liberty to

choose any cell.

3) Choose the value of the corresponding Xij as much as possible

subject to the capacity and requirement constraints.

4) Repeat steps 1-3, until all restrictions are satisfied.

Example 1:

Factory Retail shop

Supply

1 2 3 4

1 3 5 7 6 50

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Assignment : MB0032 Operation Research

2 2 5 8 2 75

3 3 6 9 2 25

Demand 20 20 50 60

Solution: We observe that C21=2,which is minimum transportation

cost,so X21=20,demand of the first column is satisfied. Allocation is

shown in the following table1:

Factory Retail

shopSupply

1 2 3 4

1 3 5 7 6 50

2 220 5 8 2 75

3 3 6 9 2 25

Demand 20 20 50 60

Now we observe that C24=2,which is the minimum transportation cost,

so X24=55,The supply for the second row is exhausted.

Factory Retail

shopSupply

1 2 3 4

1 3 5 7 6 50

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2 220 5 8 255 75

3 3 6 9 2 25

Demand 20 20 50 60 5

Proceeding in this way ,we observe that

X34=5,X12=20,X13=30,X33=20.The resulting feasible solution is

shown in the following table:

Factory Retail shop

Supply

1 2 3 4

1 3 x 730 6 50

2 220 5 8 255 75

3 3 6 920 25 25

Demand 20 20 50 60 5

Number of basic variables =m+n-1=3+4-1=6

Initial total transportation cost associated with the solution is

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calculated as given below.

=20x2+20x5+30x7+55x2+20x9+5x2

=650.

Question 6

Describe the Integer Programming Problem. Describe the Gomory’s All-I.P.P. method for solving the I.P.P. problem. Answer:

In Linear programming the design variables considered are supposed to take any real value. However in practical problems like minimization of labor needed in a project, it makes little sense in assigning a value like 5.6 to the number of labourers. In situations like this, one natural idea for obtaining an integer solution is to ignore the integer constraints and use any of the techniques previously discussed and then round-off the solution to the nearest integer value. However, there are several fundamental problems in using this approach:

1. The rounded-off solutions may not be feasible.

2. The objective function value given by the rounded-off solutions (even if some are feasible) may not be the optimal one.

3. Even if some of the rounded-off solutions are optimal, checking all the rounded-off solutions is computationally expensive (possible round-off values to be considered for an variable problem)

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Assignment : MB0032 Operation Research

Types of Integer Programming

When all the variables in an optimization problem are restricted to take only integer values, it is called an all – integer programming problem. When the variables are restricted to take only discrete values, the problem is called a discrete programming problem. When only some variable values are restricted to take integer or discrete, it is called mixed integer or discrete programming problem. When the variables are constrained to take values of either zero or 1, then the problem is called zero – one programming problem.

Integer Linear Programming

Integer Linear Programming (ILP) is an extension of linear programming, with an additional restriction that the variables should be integer valued. The standard form of an ILP is of the form,

The associated linear program dropping the integer restrictions is called linear relaxation LR. Thus, LR is less constrained than ILP. If the objective function coefficients are integer, then for minimization, the optimal objective for ILP is greater than or equal to the rounded-off value of the optimal objective for LR. For maximization, the optimal objective for ILP is less than or equal to the rounded-off value of the optimal objective for LR.

For a minimization ILP, the optimal objective value for LR is less than or equal to the optimal objective for ILP and for a maximization ILP, the optimal objective value for LR is greater than or equal to that of ILP. If LR is infeasible, then ILP is also infeasible. Also, if LR is optimized by integer variables, then that solution is feasible and optimal for IP.

A most popular method used for solving all-integer and mixed-integer linear programming problems is the cutting plane method by Gomory (Gomory, 1957).

Gomory’s Cutting Plane Method for All – Integer Programming

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Assignment : MB0032 Operation Research

Consider the following optimization problem.

The graphical solution for the linear relaxation of this problem is shown below.

It can be seen that the solution is and the optimal value of The feasible solutions accounting the integer constraints are shown by red dots. These points are called integer lattice points. The original feasible region is reduced to a new feasible region by including some additional constraints such that an extreme point of the new feasible region becomes an optimal solution after accounting for the integer constraints.

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The graphical solution for the example previously discussed taking x1 and x2 as integers are shown below. Two additional constraints (MN and OP) are included so that the original feasible region ABCD is reduced to a new feasible region AEFGCD. Thus the solution for this ILP is x1=4, x2=3 and the optimal value is Z=15.

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Gomary proposed a systematic method to develop these additional constraints known as Gomory constraints.

Generation of Gomory Constraints:

Let the final tableau of an LP problem consist of n basic variables (original variables) and m non basic variables (slack variables) as shown in the table below. The basic variables are represented as xi (i=1,2,…,n) and the non basic variables are represented as y j (j=1,2,…,m).

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Choose any basic variable xi with the highest fractional value. If there is a tie between two basic variables, arbitrarily choose any of them as. Then from the ith equation of table,

------- (1)

Express both bi and cij as an integer value plus a fractional part.

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General procedure for solving ILP:

1. Solve the given problem as an ordinary LP problem neglecting the integer constraints. If the optimum values of the variables are integers itself, then there is nothing more to be done.

2. If any of the basic variables has fractional values, introduce the Gomory constraints as discussed in the previous section. Insert a new row with the coefficients of this constraint, to the final tableau of the ordinary LP problem (Table 1).

3. Solve this by applying the dual simplex method.

Since the value of yj=0 in Table 1, the Gomory constraint equation becomes si=-βi which is a negative value and thus infeasible. Dual

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simplex method is used to obtain a new optimal solution that satisfies the Gomory constraint.

4. Check whether the new solution is all-integer or not. If all values are not integers, then a new Gomory constraint is developed from the new simplex tableau and the dual simplex method is applied again.

5. This process is continued until an optimal integer solution is obtained or it shows that the problem has no feasible integer solution.

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

Set 2

Question 1What are the important features of Operations Research? Describe in details the different phases of Operations Research. Answer:Operations research, also known as operational research, is an interdisciplinary branch of applied mathematics and formal science that uses advanced analytical methods such as mathematical modeling, statistical analysis, and mathematical optimization to arrive at optimal or near-optimal solutions to complex decision-making problems. It is often concerned with determining the maximum (of profit, performance, or yield) or minimum (of loss, risk, or cost) of some real-world objective. Originating in military efforts before World War II, its techniques have grown to concern problems in a variety of industries.

IMPORTANT FEATURES OF OR are:

1. It is system oriented: OR studies the problem from over all points of view

of organizations or situations since optimum result of one part of the system

may not be optimum for the other part.

2.It imbibes inter-disciplinary team approach: since no single individual can

have a thorough knowledge of all fast developing scientific know-how,

personalities from different scientific and managerial cadre form a team to

solve a problem.

3. It makes use of scientific methods to solve problems.

4. OR increases the effectiveness of management decision ability.

5. It makes use of computer to solve large and complex problems.

6. It gives quantitative solution.

7. It considers the human factor also.

PHASES OPERATIONS RESEARCH

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

1. Judgment Phase:

This phase consists of:

Determining of the operation.

Establishment of the objectives and values related to the operation.

Determining the suitable measures of effectiveness and

Formulation of the problem relative to the objectives.

Decision making begins with a situation in which a problem is recognized.

The problem may be actual or abstract, it may involve current operations

or proposed expansions or contractions due to expected market shifts, it

may become apparent through consumer complaints or through employee

suggestions, it may be a conscious effort to improve efficiency or a

response to an unexpected crisis.

It is impossible to circumscribe the breadth of circumstances that might be

appropriate for this discussion, for indeed problem situations that are

amenable to objective analysis arise in every area of human activity.

At the formulation stage,

statements of objectives, constraints on solutions, appropriate

assumptions, descriptions of processes, data requirements, alternatives for

action and metrics for measuring progress are introduced.

Because of the ambiguity of the perceived situation, the process of

formulating the problem is extremely important. The analyst is usually not

the decision maker and may not be part of the organization, so care must

be taken to get agreement on the exact character of the problem to be

solved from those who perceive it. There is little value to either a poor

solution to a correctly formulated problem or a good solution to one that

has been incorrectly formulated.

2. Research Phase:

This phase utilizes

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

a) Operations and data collection for a better understanding of a

problem.

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 hypothesis

using pre-established measure of effectiveness.

e) Prediction of various results and consideration of alternate methods.

Construct a Model

A mathematical model is a collection of functional relationships by which

allowable actions are delimited and evaluated. Although the analyst would

hope to study the broad implications of the problem using a systems

approach, a model cannot include every aspect of a situation.

A model is always an abstraction that is, by necessity, simpler than the

reality.

Elements that are irrelevant or unimportant to the problem are to be

ignored, hopefully leaving sufficient detail so that the solution obtained

with the model has value with regard to the original problem.

The statements of the abstractions introduced in the construction of the

model are called the assumptions. It is important to observe that

assumptions are not necessarily statements of belief, but are descriptions

of the abstractions used to arrive at a model. The appropriateness of the

assumptions can be determined only by subsequent testing of the model’s

validity.

Models must be both tractable -- capable of being solved, and valid --

representative of the true situation. These dual goals are often

contradictory and are not always attainable. We have intentionally

represented the model with well-defined boundaries to indicate its relative

simplicity.

Find a Solution(1)

The next step in the process is to solve the model to obtain a solution to

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the problem. It is generally true that the most powerful solution methods

can be applied to the simplest, or most abstract, model.

Some methods can prescribe optimal solutions while other only evaluate

candidates, thus requiring a trial and error approach to finding an

acceptable course of action.

It may be necessary to develop new techniques specifically tailored to the

problem at hand. A model that is impossible to solve may have been

formulated incorrectly or burdened with too much detail. Such a case

signals the return to the previous step for simplification or perhaps the

postponement of the study if no acceptable, tractable model can be found.

Find a Solution (2)

Of course, the solution provided by the computer is only a proposal. An

analysis does not promise a solution but only guidance to the decision

maker.

Choosing a solution to implement is the responsibility of the decision

maker and not the analyst. The decision maker may modify the solution to

incorporate practical or intangible considerations not reflected in the

model.

Establish the Procedure (1)

Once a solution is accepted a procedure must be designed to retain control

of the implementation effort.

Problems are usually ongoing rather than unique. Solutions are

implemented as procedures to be used repeatedly in an almost automatic

fashion under perhaps changing conditions.

Control may be achieved with a set of operating rules, a job description,

laws or regulations promulgated by a government body, or computer

programs that accept current data and prescribe actions.

Establish the Procedure(2)

Once a procedure is established (and implemented), the analyst and

perhaps the decision maker are ready to tackle new problems, leaving the

procedure to handle the required tasks.

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

But what if the situation changes?

An unfortunate result of many analyses is a remnant procedure designed

to solve a problem that no longer exists or which places restrictions on an

organization that are limiting and no longer appropriate. Therefore, it is

important to establish controls that recognize a changing situation and

signal the need to modify or update the solution.

3. Action Phase:

It consist of making recommendations for the decision process by those who

first pose the problem for consideration or by anyone in a position to make a

decision,influencing the operation in which the problem is occurs.

Implement the Solution

A solution to a problem usually implies changes for some individuals in the

organization. Because resistance to change is common, the implementation

of solutions is perhaps the most difficult part of a problem solving exercise.

Some say it is the most important part. Although not strictly the

responsibility of the analyst, the solution process itself can be designed to

smooth the way for implementation.

The persons who are likely to be affected by the changes brought about by

a solution should take part, or at least be consulted, during the various

stages involving problem formulation, solution testing, and the

establishment of the procedure.

The OR Process

Combining the steps we obtain the complete OR process.

In practice, the process may not be well defined and the steps may not be

executed in a strict order. Rather there are many loops in the process, with

experimentation and observation at each step suggesting modifications to

decisions made earlier.

The process rarely terminates with all the loose ends tied up. Work

continues after a solution is proposed and implemented. Parameters and

conditions change over time requiring a constant review of the solution

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and a continuing repetition of portions of the process.

Question 2

Describe a Linear Programming Problem in details in canonical form.

Answer:

Canonical form of LPP

Canonical form of standard LPP is a set of equations consisting of the

‘objective function’and all the ‘equality constraints’ (standard form of LPP)

expressed in canonical form.

Maximise Z=C1X1+C2X2+-------------+CnXn

Subject to

a11X1+a12X2+……..+a1nXn < b1

a21X1+a22X2+……..+a2nXn < b2

……………………………………….

……………………………………….

am1X1+am2X2+……..+amnXn < bm

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X1,X2,X3………….Xn> 0

The characteristics of this form are :

1. All decision variables are non negative.

2. All constrains are of < type.

3. The objective function is of maximum type.

Any LPP can be put in the canonical form by the use of five elementary

transformations:

1. The minimization of the function is mathematically equivalent to the

maximization of the negative expression of this function .That is ,

Minimise Z=C1X1+C2X2+-------------+CnXn is equivalent to

Maximise -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,< or zero) is

equivalent to the difference between 2 non negative variables. For example,

if x is unconstrained in sign then X=(X+-X-) where X+ > 0,X-<0.

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

Question 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 variable (Si’s) for < type of constraint.

2. Introduce surplus variables (Si’s) and artificial variable (Ai) for > type

of constraint.

3. Introduce artificial variable of( =)type of constraint.

4. Cost Cj of slack and surplus variables will be zero and that of artificial

variable will be M.

5. Find Zj-Cj for each variable.

6. Slack and artificial variable will form basic variable for the first simplex

table.

7. Zj=sum of [cost of variable x its coefficients in the constraints –Profit or

cost coefficient in the variable]

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8. 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.

9. 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.

10. The element that lies both on the key column and key row is

called pivotal element.

11. Ratios with the negative and α value are not considered for

determining key row.

12. Once a artificial variable is removed as basic variable, its column

will be deleted from next iteration.

13. For maximization problems decision variables coefficient will be

same as in the objective function. For minimization problems decision

variable coefficients will have opposite signs as compared to objective

functions.

14. Values of artificial variables will always is –M for both

maximization and minimization problems.

15. The process is continued till Zj-Cj > 0.

Simplex algorithm: To test for optimality of the current basic feasible solution

of the LPP,simplex algorithm has to be used.

Assume that no artificial variables are existing in the program.

Steps:

1. Locate the most negative number in the last (bottom) row of the

simplex table, excluding that of the last column and call the column in

which this number appears as the work column.

2. Form ratios by dividing each positive number in the work

column,excluding that of the last row into the element in the same row

and last column.designate the element in the work column that yields

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

the smallest ration as the pivot element .If more than one element

yields the same smallest ratio choose arbitrarily one of them.If no

element in the work column is non negative the programme has no

solution.

3. Use elementary row operations to convert the pivot element to unity

(1) and then reduce all other elements in the work column to zero.

4. Replace the x-variable in the pivot row and first column by x-variable in

the first row pivot column. The variable which is to be replaced is

called the outgoing variable and the variable that replaces is called the

incoming variable .This new first column is the current set of basic

variables.

5. Repeat steps 1 through 4 until there are no negative numbers in the

last row excluding the last column.

6. The optimal solution is obtained by assigning to each variable in the

first column that value in the corresponding row and last column. All

other variables are considered as non basic and have assigned value

zero. The associated optimal value of the objective function is the

number in the last row and last column for a maximization program but

the negative of this number for a maximization program.

Question 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:

A scooter production company produces scooters at the units situated at

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various places (called origins) and supplies them to the places where the

depot (called destination) are situated. Here the availability as well as

requirements of the various

Depots are finite and constitute the limited resources. This type of problem is

known as distribution or transportation problem in which the key idea is to

minimize the cost or the time of transportation.

Transportation problem 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 destination ,each of

which requires bj (j=1,2……,n) units of products.Here ai and bj are positive

integers. The cost cij of transportating one unit of the products from the ith

source to the jth destination is given for 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.

i.e.

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 fictious sources are

assumed to be zero so that the total cost of transportation will remain the

same.

The solution of transportation problem is obtained in two stages. In the first

stage basic feasible solution is found by one of the following method:

1) North West corner method

2) Matrix minima method or Least cost method

3) Vogel’s approximation method

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In the second stage the BFs 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 /requirement

is met.

Step 4: We repeat step 1 to 2, till 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 this value to find ∆ij=cij-ui-vj.If all ∆ij> 0, then it is the optimal

solution.

Step 7: If any ∆ij< 0, select the most negative cell and form loop. Starting

point of the loop is +ve and alternately 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 ∆ij>0.

Question 5

Describe the North-West Corner rule for finding the initial basic feasible

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solution in the transportation problem.

Answer:

INITIAL BASIC FEASIBLE SOLUTION:

Consider a Transportation problem involving m-origins and n-destinations.

Since the sum of origin capabilities equals the sum of destination

requirements, a 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 TP can have at the most only m+n-1 strictly positive components,

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.

There are three different methods to obtain the initial basic

feasible solution viz.

(I) North-West corner rule

(II) Lowest cost entry method

(III) Vogel’s approximation method

In the light of above problem let us discuss one by one.

NORTH -WEST CORNER RULE

In this method we distribute the available units in rows and column in such a

way that the sum will remain the same. We have to follow the steps given

below.

(a) Start allocations from north-west corner, i.e., from (1,1) position.

Here min (a1, b1), i.e., min (8,6)=6 units.

Therefore, the maximum possible units that can be allocated to this position

is 6, and

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write it as 6(2) in the (1,1) position of the table. This completes the allocation

in the first column and cross the other positions, i.e., (2,1) and (3,1) in the

column. (see Table 3)

Table 3

DEPOT UNIT B1 B2 B3 B4 STOCK

A1 6(2) 8-6=2

A2 X 10

A3 X 10

REQUIREMENT 6-6=0 8 9 15 32

(b) After completion of step (a), come across the position (1,2).Here min (8–

6,8)=2 units can be allocated to this position and write it as 2(3). This

completes the allocations in the

first row and cross the other positions, i.e., (1,3) and (1,4) in this row (see

Table 4).

Table 4

DEPOT UNIT B1 B2 B3 B4 STOCK

A1 6(2) 2(3) X X 2-2=0

A2 X 10

A3 X 10

REQUIREMENT 0 8-2=6 9 15 30

(c) Now come to second row, here the position (2,1) is already been struck

off, so consider the position (2,2). Here min (10,8–2)=6 units can be

allocated to this position and write it as 6(3). This completes the allocations

in second column so strike off the position (3,2) (see Table 5)

Table 5

DEPOT UNIT B1 B2 B3 B4 STOCK

A1 6(2) 2(3) X X 0

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

A2 X 6(3) 10-6=4

A3 X X 20

REQUIREMENT 0 0 9 15 24

(d) Again consider the position (2,3). Here, min (10–6,9)=4 units can be

allocated to this position and write it as 4(4). This completes the allocations

in second row so struck off the position (2,4) (see Table 6).

Table 6

DEPOT UNIT B1 B2 B3 B4 STOCK

A1 6(2) 2(3) X X 0

A2 X 6(3) 4(4) X 0

A3 X X 20

REQUIREMENT 0 0 9-4=5 15 20

(e) In the third row, positions (3,1) and (3,2) are already been struck off so

consider the position (3,3) and allocate it the maximum possible units, i.e.,

min (20,9–4)=5 units and write it as 5(7). Finally, allocate the remaining

units to the position (3,4), i.e., 15 units to this position and write it as 15(2).

All the allocations done in the above method complete the table as follows:

Table 7

DEPOT UNIT B1 B2 B3 B4 STOCK

A1 6(2) 2(3) X X 8

A2 X 6(3) 4(4) X 10

A3 X X 5(7) 15(2) 20

REQUIREMENT 6 8 9 15 38

From the above table calculate the cost of transportation as

6×2 + 2×3 + 6×3 + 4×4 + 5×7 + 15×2

= 12 + 6 + 18 + 16 + 35 + 30

= 117

i.e., Rs. 11700.

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

Question 6

Describe the Branch and Bound Technique to solve an I.P.P. problem.

Ans:

INTEGER PROGRAMMING (BRANCH AND BOUND METHOD)This method is applicable to both, all IPP as well as mixed IPP. Some times a few or all thevariables of an IPP are constrained by their upper or lower bounds. The most general method forthe solution of such constrained optimization problems is called Branch and Bound method.This method first divides the feasible region into smaller subsets and then examines each ofthem successively until a feasible solution that gives optimal value of objective function isobtained.Let the given IPP beMax Z=CXSubject tob AX ≤0 X ≥ are integers.In this method, we first solve the problem by ignoring the integrality condition.(i) If the solution is in integers, the current solution is optimum to the given IPP.(ii) If the solution is not in integer, say one of the variable Xr is not an integer, then where

are consecutive non-negative integers.Hence any feasible integer value of Xr must satisfy one of the two conditions.

These two conditions are mutually exclusive (both can't be true simultaneously). By addingthese two conditions separately to the given LPP we form different sub-problems.

Thus, we have branched or partitioned the original problem into two subproblems. Each of thesesub-problems is then solved separately as LPP.

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If any sub-problem yields an optimum integer solution it is not further branched. But if any subproblemyields a non-integer solution it is further branched into two sub-problems. Thisbranching process is continued until each problem terminates with either integer optimal solutionor there is an evidence that it cannot yield a better solution. The integer valued solution amongall the sub-problems which gives the most optimal value of the objective function is thenselected as the optimum solution.Note: For minimization problem the procedure is the same except that upper bounds are used.The sub-problem is said to be fathomed and is dropped from further consideration if it yields avalue of the objective function lower than that of the best available integer solution and i' isuseless to explore the problem any further.Example:

Use Branch and Bound technique to solve the following:

Since all coefficients are positive, an optimum solution is obtained. xt =0,x2 =7/4. and Max Z=7

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Sikkim Manipal University: Master of Business Administration- MBA Semester 2Assignment : MB0032 Operation Research

Since x2 = 4, this problem should be branched- into two sub-problems

Applying these two conditions separately in the given LPP we get two sub problems

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