LINEAR PROGRAMMING PROBLEM

INTRODUCTION

In a decision-making embroilment, model formulation is important

because it represents the essence of business decision problem. The

term formulation is used to mean the process of converting the verbal

description and numerical data into mathematical expressions which

represents the relevant relationship among decision factors, objectives

and restrictions on the use of resources. Linear Programming (LP) is a

particular type of technique used for economic allocation of 'scarce' or

'limited' resources, such as labour, material, machine, time, warehouse

space, capital, energy, etc. to several competing activities, such as

products, services, jobs, new equipment, projects, etc. on the basis of a

given criterion of optimally. The phrase scarce resources means

resources that are not in unlimited in availability during the planning

period. The criterion of optimality, generally is either performance, return

on investment, profit, cost, utilily, time, distance, etc.

George B Dantzing while working with US Air Force during World War

II, developed this technique, primarily for solving military logistics

problems. But now, it is being used extensively in all functional areas of

management, hospitals, airlines, agriculture, military operations, oil

refining, education, energy planning, pollution control, transportation

planning and scheduling, research and development, etc. Even though

these applications are diverse, all I.P models consist of certain common

properties and assumptions. Before applying linear programming to a

real-life decision problem, the decision-maker must be aware of all these

properties and assumptions, which are discussed later in this chapter.

Before discussing in detail the basic concepts and applications of

linear programming, let us be clear about the two words, linear and

programming. The word linear refers to linear relationship among

variables in a model. Thus, a given change in one variable will always

cause a resulting proportional change in another variable. For example,

doubling the investment on a certain project will exactly double the rate

of return. The word programming refers to modelling and solving a

problem mathematically that involves the economic allocation of limited

resources by choosing a particular course of action or strategy among

various alternative strategies to achieve the desired objective.

A large number of computer packages are available for solving a

mathematical LP model but there is no general package for building a

model. Model building is an art that improves with practice. To illustrate,

how to build I.P models, a variety of examples are given in this

chapter.

STRUCTURE OF LINEAR PROGRAMMING

General Structure of LP Model

The general structure of LP model consists of three components.

Decision variables (activities): We need to evaluate various

alternatives (courses of action) for arriving at the optimal value of

objective function. Obviously, if there are no alternatives to select from,

we would not need LP. The evaluation of various alternatives is guided

by the nature of objective function and availability of resources. For this,

we pursue certain activities usually denoted by x1, x2…xn. The value of

these activities represent the extent to which each of these is performed.

For example, in a product-mix manufacturing, the management may use

LP to decide how many units of each of the product to manufacture by

using its limited resources such as personnel, machinery, money,

material, etc.

These activities are also known as decision variables because they

arc under the decision-maker's control. These decision variables, usually

interrelated in terms of consumption of limited resources, require

simultaneous solutions. All decision variables are continuous,

controllable and non-negative. That is, x1>0, x2>0, ....xn>0.

The objective function: The objective function of each L.P problem is a

mathematical representation of the objective in terms of a measurable

quantity such as profit, cost, revenue, distance, etc. In its general form,

it is represented as:

Optimise (Maximise or Minimise) Z = c1x1 + c2X2. … cnxn

where Z is the mcasure-of-performance variable, which is a function of

x1, x2 ..., xn. Quantities c1, c2…cn are parameters that represent the

contribution of a unit of the respective variable x1, x2 ..., xn to the

measure-of-performance Z. The optimal value of the given objective

function is obtained by the graphical method or simplex method.

The constraints: There are always certain limitations (or constraints)

on the use of resources, e.g. labour, machine, raw material, space,

money, etc. that limit the degree to which objective can be achieved.

Such constraints must be expressed as linear equalities or inequalities in

terms of decision variables. The solution of an L.P model must satisfy

these constraints.

The linear programming method is a technique for choosing the best

alternative from a set of feasible alternatives, in situations in which the

objective function as well as the constraints can be expressed as linear

mathematical functions. In order to apply linear programming, there are

certain requirements to me met.

(a) There should be an objective which should be clearly identifiable

and measurable in quantitative terms. It could be, for example,

maximisation of sales, of profit, minimisation of cost, and so on.

(b) The activities to be included should be distinctly identifiable and measurable in quantitative terms, for instance, the products included in a production planning problem.

(c) The resources of the system which arc to be allocated for the attainment of the goal should also be identifiable and measurable quantitatively. They must be in limited supply. The technique would involve allocation of these resources in a manner that would trade off the returns on the investment of the resources for the attainment of the objective.

(d) The relationships representing the objective as also the resource

limitation considerations, represented by the objective function and

the constraint equations or inequalities, respectively must be linear in

nature.

(e) There should be a series of feasible alternative courses of action

available to the decision makers, which are determined by the

resource constraints.

When these stated conditions are satisfied in a given situation, the

problem can be expressed in algebraic form, called the Linear

Programming Problem (LPP) and then solved for optimal decision. We

shall first illustrate the formulation of linear programming problems and

then consider the method of their solution.

ASSUMPTIONS OF LINEAR PROGRAMMING

The following four basic assumptions are necessary for all linear

programming models.

Certainty: In all LP models, it is assumed, that all model parameters

such as availability of resources, profit (or cost) contribution of a unit of

decision variable and consumption of resources by a unit of decision

variable must be known and is constant. In some cases, these may be

either random variables represented by a known distribution (general or

Additional info:

A large number of decision problems faced by a business manager involve allocation of

resources to various activities, with the objective of increasing profits or decreasing

costs, or both. When resources arc in abundance no difficulty is experienced. But such

cases are very rare. Practically in all situations, the management is confronted with the

problem of scarce resources. Normally, there are several activities to perform with

limitations of either of the resources or their use prevents each activity from being

performed to the desired level. Thus, the manager has to take a decision as to how

best the resources be allocated among the various activities.

The decision problem becomes complicated when a number of resources are required

to be allocated. There are several activities to perform. Rule of thumb, even for an

experienced manager, in all likelihood may not provide the right answer in such cases.

The decision problems can be formulated, and solved using mathematical

programming techniques.

Mathematical programming involves optimisation of a certain function, called the

objective functions subject to certain constraints. For example, a manager may be

faced with the problem of deciding the appropriate product mix of the four products.

With the profitability of the products along with the requirements of raw materials,

labour etc. known, his problem can be formulated as a mathematical programming

problem taking the objective function as the maximisation of profits obtainable from the

resources keeping in view the various constraints—the availability of raw materials,

labour supply, market and The methods of mathematical programming can be divided

into three groups: linear, integer, and non-programming.

may be statistical) or may tend to change, then the given problem can

be solved by a stochastic LP model or parametric programming. The

linear programming is obviously deterministic in nature.

Divisibility (or continuity): The solution values of decision variables

and resources are assumed to have either whole numbers (integers) or

mixed numbers (integer and fractional). However, if only integer variables

are desired, e.g. machines, employees, etc. the integer programming

method may be applied to get the desired values.

It is also an assumption of a linear programming model that the

decision variables are continuous. As a consequence, combinations of

output with fractional values, in the context of production problems, are

possible and obtained frequently. For example, the best solution to a

problem might be to produce 5 2/3 units of product A and 10 1/3 units

of product B per week.

Although in many situations we can have only integer values, but we can

deal with the fractional values, when they appear, in the following ways.

Firstly, when the decision is a one-shot decision, that is to say, it is not

repetitive in nature and has to be taken only once, we may round the

fractional values to the nearest integer values. However, when we do so,

we should evaluate the revised solution to determine whether the

solution represented by the rounded values is a feasible solution and

also whether the solution is the best integer solution. Secondly, if the

problem relates to a continuum of time and it is designed to determine

optimal solution for a given time period only, then the fractional values

may not be rounded. For instance in the context of a production

problem, a solution like the one given earlier to make 5 2/3 units of A

and 10 units of B per week, can be adopted without any difficulty. The

fractional amount of production would be taken to be the work-in-

progress and become a portion of the production of the following week.

In this case an output of 17 units of A and 31 units of B over a three-

week period would imply 5 2/3 units of A and 10 units of B per week.

Lastly, if we must insist on obtaining only integer values of the decision

variables, we may restate the problem as an integer programming

problem, forcing the solutions to be in integers only.

Additivity: The value of the objective function for the given values of

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

the sum of the contributions (profit or cost) earned from each decision

variable and the sum of the resources used by each decision variable,

respectively. For example, the total profit earned by the sale of two

products A and B must be equal to the sum of the profits earned

separately from A and B. Similarly, the amount of a resource consumed

by A and B must be equal to the sum of resources used for A and B

individually.

This assumption implies that there is no interaction among the

decision variables (interaction is possible when, for example, some

product is a by-product of another one).

Finite choices A linear programming model also assumes that a

limited number of choices are available to a decision-maker and the

decision variables do not assume negative values. Thus, only non-

negative levels of activity are considered feasible. This assumption is

indeed a realistic one. For instance, in the production problems, the

output cannot obviously be negative, because a negative production

implies that we should be able to reverse the production process and

convert the finished output back into the raw materials!

Linearity (or proportionality): All relationships in the LP model (i.e. in

both objective function and constraints) must be linear. In other words,

for any decision variable j, the amount of particular resource say i used

and its contribution to the cost one in objective function must be

proportional to its amount. For example, if production of one unit of a

product uses 5 hours of a particular resource, then making 3 units of that

product uses 3 x 5 = 15 hours of that resource.

ADVANTAGES OF LINEAR PROGRAMMING

Following are certain advantages of linear programming.

1.Linear programming helps in attaining the optimum use of productive resources. It also indicates how a decision-maker can employ his productive factors effectively by selecting and distributing (allocating) these resources.

2.Linear programming techniques improve the quality of decisions. The decision-making approach of the user of this technique becomes more objective and less subjective.

3.Linear programming techniques provide possible and practical

solutions since there might be other constraints operating outside the problem which must be taken into account. Just because we can produce so many units docs not mean that they can be sold. Thus, necessary modification of its mathematical solution is required for the sake of convenience to the decision-maker.

4.Highlighting of bottlenecks in the production processes is the most significant advantage of this technique. For example, when a bottleneck occurs, some machines cannot meet demand while other remains idle for some of the time.

5.Linear programming also helps in re-evaluation of a basic plan for changing conditions. If conditions change when the plan is partly carried out, they can be determined so as to adjust the remainder of the plan for best results.

LIMITATIONS OF LINEAR PROGRAMMING

In spite of having many advantages and wide areas of applications, there arc some limitations associated with this technique. These are given below. Linear programming treats all relationships among decision variables as linear. However, generally, neither the objective functions nor the constraints in real-life situations concerning business and industrial problems are linearly related to the variables.

1. While solving an LP model, there is no guarantee that we will get integer valued solutions. For example, in finding out how many men and machines would be required lo perform a particular job, a non-integer valued solution will be meaningless. Rounding off the solution to the nearest integer will not yield an optimal solution. In such cases, integer programming is used to ensure integer value to the decision variables.

2. Linear programming model does not take into consideration the effect of time and uncertainty. Thus, the LP model should be defined in such a way that any change due to internal as well as external factors can be incorporated.

3. Sometimes large-scale problems can be solved with linear programming techniques even when assistance of computer is available. For it, the main problem can be fragmented into several small problems and solving each one separately.

4. Parameters appearing in the model are assumed to be constant but in real-life situations, they are frequently neither known nor constant.

It deals with only single objective, whereas in real-life situations we may come across conflicting multi-objective problems. In such cases, instead of the LP model, a goal programming model is used to get satisfactory values of these objectives.

APPLICATION AREAS OF LINEAR PROGRAMMING

Linear programming is the most widely used technique of decision-

making in business and Industry and in various other fields. In this

section, we will discuss a few of the broad application areas of linear

programming.

Agricultural Applications

These applications fall into categories of farm economics and farm management. The former deals with agricultural economy of a nation or region, while the latter is concerned with the problems of the individual farm.

The study of farm economics deals with inter-regional competition and optimum allocation of crop production. Efficient production patterns can be specified by a linear programming model under regional land resources and national demand constraints.

Linear programming can be applied in agricultural planning, e.g. allocation of limited resources such as acreage, labour, water supply and working capital, etc. in a way so as to maximise net revenue.

Military Applications

Military applications include the problem of selecting an air weapon system against enemy so as to keep them pinned down and at the same time minimising the amount of aviation gasoline used. A variation of the transportation problem that maximises the total tonnage of bombs dropped on a set of targets and the problem of community defence against disaster, the solution of which yields the number of defence units that should be used in a given attack in order to provide the required level of protection at the lowest possible cost.

Production Management

(i) Product mix A company can produce several different products, each of which requires the use of limited production resources. In such cases, it is essential to determine the quantity of each product to be produced knowing its marginal contribution and amount of available resource used by it. The objective is to maximise the total contribution, subject to all constraints.

(ii) Production planning This deals with the determination of minimum cost production plan over planning period of an item with a fluctuating demand, considering the initial number of units in inventory, production capacity, constraints on production, manpower and all relevant cost factors. The objective is to minimise total operation costs.

(iii) Assembly-line balancing This problem is likely to arise when an item can be made by assembling different components. The process of assembling requires some specified sequcnce(s). The objective is to minimise the total elapse time.

(iv) Blending problems These problems arise when a product can be made from a variety of available raw materials, each of which has a particular composition and price. The objective here is to determine the minimum cost blend, subject to availability of the raw materials, and minimum and maximum constraints on certain product constituents.

(v) Trim loss When an item is made to a standard size (e.g. glass, paper sheet), the problem that arises is to determine which

combination of requirements should be produced from standard materials in order to minimise the trim loss.

Financial Management

(i) Portfolio selection This deals with the selection of specific investment activity among several other activities. The objective is to find the allocation which maximises the total expected return or minimises risk under certain limitations.

(ii) Profit planning This deals with the maximisation of the profit margin from investment in plant facilities and equipment, cash in hand and inventory.

Marketing Management

(i) Media selection Linear programming technique helps in determining the advertising media mix so as to maximise the effective exposure, subject to limitation of budget, specified exposure rates to different market segments, specified minimum and maximum number of advertisements in various media.

(if) Travelling salesman problem The problem of salesman is to find the shortest route from a given city, visiting each of the specified cities and then returning to the original point of departure, provided no city shall be visited twice during the tour. Such type of problems can be solved with the help of the modified assignment technique.

Cm) Physical distribution Linear programming determines the most economic and efficient manner of locating manufacturing plants and distribution centres for physical distribution.

Personnel Management

a) Staffing problem Linear programming is used to allocate optimum manpower to a particular job so as to minimise the total overtime cost or total manpower.

b) Determination of equitable salaries Linear programming technique has been used in determining equitable salaries and sales incentives.

c) Job evaluation and selection Selection of suitable person for a specified job and evaluation of job in organisations has been done with the help of linear programming technique.

Other applications of linear programming lie in the area of administration, education, fleet utilisation, awarding contracts, hospital administration and capital budgeting, etc.

GRAPHICAL SOLUTION

Extreme point enumeration approach

Convex Polyhedron

TYPES OF SOLUTION

(a) Solution. Values of decision variables xj (j = 1, 2, 3, ….n) which satisfy the constraints of the general L. P. P., is called the solution to that L. P. P.

(b) Feasible solution. Any solution that also satisfies the nonnegative restrictions of the general L. P. P. is called a feasible solution.

(c) Basic Solution. For a set of m simultaneous equations in n unknowns (n> m). a solution obtained by setting (n - m) of the variables equal to zero and solving the remaining m equations in m unknowns is called a basic solution. Zero variables (n - m) are called non-basic variables and remaining m are called basic variables and constitute a basic solution.

(d) Basic Feasible Solution. A feasible solution to a general L.P.P. which is also basic solution is called a basic feasible solution.

(e) Optimum Feasible Solution. Any basic feasible solution which optimizes (maximizes or minimizes) the objective function of a general L.P.P. is known as an optimum feasible solution to that L.P.P.

(f) Degenerate Solution. A basic solution to the system of equations is called degenerate if one or more of the basic variables become equal to zero.

SPECIAL CASES IN LINEAR PROGRAMMING

Alternative (or Multiple) Optimal Solutions

So far we have seen that the optimal solution of any linear programming problem occurs at an extreme point of the feasible region and the solution is unique, i.e. no other solution yields the same value of the objective function. However, in certain cases, a given LP problem may have more than one solution yielding the same optimal objective function value. Each of such optimal solutions is termed as alternative optimal Solution.

There are two conditions that should be satisfied for an alternative optimal solution to exist:

(i) The given objective function is parallel to a constraint that forms the boundary (or edge) of the feasible solutions region. In other words, the slope of the objective function is same as that of the constraint forming the boundary of the feasible solutions region, and

(ii) The constraint should form a boundary on the feasible region in the direction of optimal movement of the objective function. In other words, the constraint should be an active constraint.

Remark: The constraint is said to be active or binding or tight, if at optimality, the left-hand side of a constraint equals the right-hand side. In other words, an equality constraint is always active. An inequality constraint may or may not be active.

Geometrically, an active constraint is one that passes through one of the extreme points of the feasible solution space.

Unbounded Solution

Sometimes an LP problem will not have a finite solution. This means when one or more decision variable values and the value of the objective function (maximisation case) are permitted to increase infinitely without violating the feasibility conditions, then the solution is said to be unbounded. It is important to note that there is the difference between a feasible region being unbounded and an LP problem being unbounded. It is possible for a feasible region to be unbounded but LP problem not to be unbounded, that is an unbounded feasible region may yield some definite value of the objective function. The general cause for an unbounded LP problem is a mistake in mathematical model formulation.

Infeasible solution

It has already been stated that a solution is called feasible if it satisfies

all the constraints and the non-negativity conditions. Sometimes it is

possible that the constraints may be inconsistent so that there is no

feasible solution to the problem. Such a situation is called infeasibility.

This means, there is no unique (single) feasible region. Such a problem

arises due to wrong model formulation with conflicting constraints.

Infeasibility depends solely on the constraints as has nothing to do with

the objective function.

In the graphic approach to the solution to an LPP, the infeasibility is

evident if its feasible region is empty so that there is no feasible region

in which all the constraints may be satisfied simultaneously.

Infeasibility vs Unboundedness

Both infeasibility and unboundedness have a similarity in that there is no optimal solution in either case. But there is a striking difference between the two, while in infeasibility there is not a single feasible solution, in unboundedness there are infinite feasible solutions but none of them can be termed as the optimal.

Redundancy

A redundant constraint is one that does not affect the feasible solutions region (or space) and thus redundancy of any constraint does not cause any difficulty in solving an LP problem graphically. In other words constraint appears redundant when it may be more binding than another.

SIMPLEX

Define l.p.p. (mathematical)

Explain the summary procedure for the maximization case of the simplex method. Step 1 Formulate the problem Translate the technical specifications of the problems into

inequalities, and make a precise statement of the objective function. Convert the inequalities into equalities by the addition of nonnegative

slack variables. These inequalities should be symmetric or balanced so that each slack variable appears in each equation with a proper coefficient.

Modify the objective function to include the slack variables. Step 2 Design an initial program (A basic feasible solution) Calculate the net evaluation row: To get a number in the net

evaluation row under a column, multiply the entries in that column by the corresponding numbers in the objective column, and add the products. Then subtract this sum from the number listed in the objective row at the top of the column. Enter the result in the net evaluation row under the column.

Test : Examine the entries in the net evaluation row for the given simplex tableau. If all the zero or negative, the optimal solution has been obtained. Otherwise, the presence of any positive entry in the net-evaluation row indicates that a better program can be obtained.

Step 3 Revise the program Find the key column. The column under which falls the largest

positive net-evaluation- row entry is the key column. Find the key row and the key number. Divide the entries in the

“quantity” column by the corresponding nonnegative entries of the key column to form replacement ratios, and compare these ratios. The row in which falls the smallest replacement ratio is the key row. The number which lies at the intersection of the key row and the key column is the key number.

Transform the key row. Divide all the numbers in the key row (starting with and to right of the “quantity” column) by the key number. The resulting numbers form the corresponding row of the next tableau.

Transform the non-key rows. Subtract from the old row number of a given key row (in each column ) the product of the corresponding key-row number and the corresponding fixed ratio formed by dividing the old row number in the key column by the key number. The result will give the corresponding new row number. Make this transformation for all the non-key rows.

Enter the results of (3) and (4) above in a tableau representing the revised program.

Step 4 Obtain the optimal program Repeat steps 3 & 4 until a program has been derived. [ Linear-programming problems involving the minimization of an objective function usually contain structural of the “greater than equal to” type. They can also be solved by the simplex method. The simplex procedure for solving a linear-programming problem in which the objective is to minimize rather than maximize a given function, although basically the same as above, requires sufficient modifications to deserve the listing of a separate summary. ] 9. Summary procedure for the simplex method (minimization case) Step 1 Formulate the problem Translate the technical specification of the problem into inequalities,

and make a precise statement of the objectivity function. Convert the equalities into inequalities by the subtraction of

nonnegative slack variables. Then modify these equations by the addition of nonnegative artificial slack variables. These equations should be asymmetric or balanced so that each slack and artificial slack variable appears in each equation with a proper coefficient.

Modify the objective functions to include all the slack artificial slack variables.

Step 2 Design an initial program (a basic feasible solution). Design the first program so that only the artificial slack variables are

included in the solution. Place the program in a simplex tableau. In the objective row, above each column variable, place the corresponding coefficient of that variable from step 1.c.In particular, place a zero above each column containing an artificial slack variable.

Step 3 Test and revise the program. Calculate the net evaluation row. Toto get a number in the net

evaluation row under a column, multiplying the entries in that column by the corresponding number in the objective column, and add the products. Then subtract the sum from the number listed in the objective row above the column. Enter the result in the net-evaluation row under the column.

Test. Examine the entries in the net-evaluation row for the given simplex tableau. If all the entries are zero or positive, the optimum solution has been obtained. Otherwise, the presence of any negative entry in the net-evaluation row indicates that a better program can be obtained.

Revise the program. Find the key column. The under which falls the largest negative net-

evaluation-entry is the key column. Find the key row and the key number. Divide the entries in the

“Quantity” column by the corresponding nonnegative entries in the key column to form replacement ratios, and compare these ratios. The row in which the smallest replacement ratio falls is the key row. The number which lies at the intersection of the key row and the key column is the key number.

Transform the key row. Divide all the numbers in the key row (staring with and to the right to the of the “Quantity” column by the key number. The resulting numbers form the corresponding row of the next tableau.

Transform the non-key rows. Subtract from the old row the number of a given non-key row (in each column) the product of the corresponding fixed ratio formed by dividing the old row number in the key column by the number. The result will give the corresponding new row number. Make the transformation for all rows.

Enter the results of (3) and (4) above in a tableau representing the revised program.

Step 4 Obtain the optimal program. Repeat steps 3and 4 until an optimal program has been derived.

We repeat the following comments comparing the maximization and minimization problems as solved by the simplex method.

The procedure for calculating the net-evaluation row is the same in

both cases. However, whereas the largest positive value is chosen to

identify the incoming product in a maximization problem, the most

negative value is chosen in a minimization problem. The rest of the

mechanics, namely, the transformation of the key and the non-key rows,

is exactly the same. The decision rule identifying the optimal solution is

the absence of any positive value in the non-evaluation row in the

maximization problem, and the absence of any negative value in the

minimization problem.

Algorithm, degeneracy

DUAL SIMPLEX

In ordinary simplex method we start with feasible but non-optimal

solution while in Dual Simplex Method, we start with infeasible but

optimal solution. Successive iterations will maintain optimality to remove

infeasibility of the solution. The following steps are followed to arrive at

optimal feasible solution.

(1) Write down objective function as maximization and all constraints as ≤ or =.

(2) Construct first dual simplex table from the given problem in usual manner.

(3) The leaving variable is the basic variable having the most negative value (Break ties, if any, arbitrarily). If all the basic variables are non-negative, the process ends and the feasible (optimal) solution is reached.

(4) To determine the entering variable take ratios of the coefficients of non-basic variables in the objective function to the corresponding coefficients in the row associated with the leaving variable. Ignore the ratios with positive or zero denominators. The entering variable is the one with the smallest absolute value of the ratio. (Break ties, if any, arbitrarily). If all the denominators are zero or positive, the problem has no feasible solution.

After selecting the entering and leaving variables, row operations

are applied as usual to obtain the next table.

Application of this Dual Simplex Method is useful in Sensitivity

Analysis. For example, suppose a new constraint is added to the

problem after the optimal solution is reached. If this constraint is not

satisfied by the optimal solution, the problem remains optimal but it

becomes infeasible. The Dual Simplex Method is then used to clear the

infeasibility in the problem.

Example : Minimize Z = 2x1 + x2 subject to 3x1 + x2 ≥ 3, 4xx +

3x2 ≥ 6,

x1 + 2x2 ≤ 3, x1, x2 ≥ 0.

CPM and PERT

CRITICAL PATH ANALYSIS

Advantages of critical path analysis

1. It allows for a comprehensive view of the entire project. Because of the sequential and concurrent relationships, time scheduling becomes very effective. Identifying the critical activities keeps the executive alert and in a state of preparedness, with alternative plans ready in case these are needed. Breaking down the project into the smaller components permits better and closed control.

2. Critical path analysis offers economic and an effective system of control based on the principle of management by exception. That is need for corrective action arises only in exceptional situations and in most other cases; performance is in conformity with the plans.

3. It is a dynamic pool of management which calls for a constant review, early formulation of the network, and finding the current path of relevance and optimum resources allocation.

Events

The beginning and ending points of an activity or a group of activities are

called events, aka nodes and connectors. It is often graphically

represented by a numbered circle. All activities in the network must

commence from some event. Such events are called the tail event.

Similarly all activities in the network must our terminal point called the

head event.

In the network, symbol "i" is used for the tail event, also called the

preceding event and "j" for the head event also called the succeeding

event of an activity.

The activity is then denoted by "i-j".

Conventions adopted in drawing networks

There are two conventions normally adopted while drawing networks.

They are

Time flows from left to right. Head events always have a number higher than that of tail events.

Graphical representation of events and activities

Events are represented by numbers within circles. Activities are

represented by arrows; the arrowheads represent the completing of the

activities. Lengths and orientation of the arrow are of no significance and

are only chosen for convenience.

Fundamental properties governing the representation of events and

activities

The representation of events and activities is governed by one simple

dependency rule which requires that an activity which depends on

another activity is shown to emerge from the head event of the activity

upon which depends and that only dependent activities are drawn in this

way.

Errors and logical sequence: two types of errors in logic may arise when

drawing network, particularly when it is a complicated one. These are

known as looping and dangling.

Looping: Normally in the network, the arrow points from left to right. This

convention is to be strictly adhered, as this would avoid the illogical

looping, as shown wrongly below.

Dangling: The situation represented by the following diagram is also at

fault, since the activity represented by the dangling arrow 9-11 is

undertaken with no result.

1 2 3 4

To overcome problems arising due to dangling arrows, we must make

sure that

All events except the first and the last must have at least one activity entering and one activity leaving them.

All activities must start to finish within event.

Duplicate activities: consider the following figure

Here A and B are known as duplicate activities because they have the

same head and tail events. A remedy the situation is introduction of a

dummy activity.

Dummy activity: it is a hypothetical activity which consumes known

source or time. It is represented by a dotted line and inserted in the

network to clarify activity that turn on the following situations

(i) it is created to make equities with, starting and finishing events

distinguishable,

(ii) to identify and maintain the proper precedence relationship between

activities that are not connected by events

(iii) to bring all "loose ends" to single initial and a single terminal event in

each network using dummies if necessary.

6 7

8

9

6 7 8

A

B

For example the problem of duplicate activities may be circumvented as

shown:

Numbering the events

The event numbers in the network should in some respects reflects the

logical sequences. When a complicated network has been drawn when

the problem of assigning numbers to the events involved in the network

arises. A rule devised by D. R. Fulkerson, involving the following steps

may be followed to resolve the problem of numbering events.

1. And initial event is one which has arrow/arrows coming out of it and none of the arrow entering it. In a network there will be only one such event. Call it 1.

2. Delete all arrows coming of from the event 1. This will give us at least one more initial event.

3. Numbers these events as 2,3… 4. delete all emerging arrows from these numbered events which will

create new initial events. Then follow steps 3. 5. Continue the above steps the last event is obtained which has no

arrows coming out of it.

Consider the numbering of events in the following figure:

6 8

A

B 7

1 2 4

5

3

6 7 8

X

Y

Here we proceed from left to right. The event with the least x coordinate

is assigned the smallest integer, say 1. Other events are assigned

progressively higher integers with regard to x coordinate. If two were

more events such as 4 & 5 have the same x coordinate the one towards

the arrow should have a higher number.

It is better to number of the events as 10, 20, 30.... to afford insertions of

more activities and events omitted by oversight.

Concurrent activities

Activities may not always be discreet that is, they may be done in part

allowing the subsequent activities to commence before the preceding

activity is fully completed. Activities of this kind are to be frequently

encountered in batch production. If, for example, a batch of 50 spindles

is to be processed onto machines obviously it is not necessary to

process all the items of the batch on the first machines and then transfer

these to the next machines. A few items processed on the first machines

may be transferred to the second machine before completion of the

entire batch on the first machine. Concurrent activities are to be

encountered in sewage work, trenching, laying pipe, welding pipe, and

backfilling all going on simultaneously with suitable lags on construction

work.

Forward pass computations

As stated above, the purpose of the forward pass is to compute the

earliest start (EST) and finish time (EFT) for each activity. The EST time

indicates the earliest time that a given activity can be scheduled.

Earliest finish time for an activity indicates the time by which the activity

can be completed, at the earliest. To compute these time estimates, we

will first of all compute the earliest allowable occurrence time for various

events of the network.

It is a convention to keep the earliest allowable occurrence time of the

START event as zero.

To understand, how this time estimate for other events is computed, let

us consider the following network diagram.

Earliest allowable occurrence time

In the network shown above, event 1 stands for the beginning of the

activity 1 – 2 and we can say that it occurs at the time zero i.e. E1 = 0.

Event 2 stands for the finish of the activity 1 – 2 , thus event 2 can occur

at the earliest time E2 which is computed as

E2 = 0 + D 12 = 0 + 6 = 6

Where D12 stands for the duration of activity 1 – 2

Event 3 stands for finish of the activity 2 –3 and its earliest time is

E3 = E2 + D23 = 6 + 8 = 14

The event 4 can occur either at the end of the activity 3-4 or at the

finish of activity 2 – 4. In this case, there will be two time estimates as

follows:

E4 = E2 + D34 = 14 + 0 = 14

E4 = E2 + D24 = 6 + 10 = 16

In case, two or more time estimates exist for a particular

event, then the time estimate with maximum value is retained as the

earliest event time and other values are discarded. This maximum

value represents the completion of all the activities ending at the event

under consideration. In the above example, the earliest event time for

event 4 will be 16.

1 2

3

4

5 6

E1= 0 E2= 6

E3= 14

E4= 16

E5= 36 E6= 52

6

8

10

20

6

0 16

L1= 0 L2= 6

L3= 16

L4= 16

L5= 36 L6= 52

A general rule can also be given here for determining the earliest

event time as below:

E1 = Max (Ei + Dii)

When E1 is the earliest time for even j, Ei is the earliest time for

event I and Dii is the duration of the activity i-j.

Earliest start and finish times of an activity

After computing the earliest event time of various events, one can easily

compute the earliest start and finish times of all the activities on the

network. The earliest start time of an activity is given by the

earliest allowable occurrence time of the tail even of that activity.

Thus, in our example, the earliest start time of the activity 1-2 will be

given by the earliest time of the event 1 i.e. it will be 0. The earliest start

time for the activities 2-3 and 2-4 will be given by the earliest time of

event 2 which is equal to 6. The earliest time for the activities 3-4 and 3-

5 will be 14 which is the earliest time for the event 3 and so on.

The earliest finish time of an activity will be simply equal to the

earliest start time of the activity plus the duration of that activity.

Hence, in our example, earliest finish time of activity 1-2 will be 0 + 6 =

6, for activity 2-3, it will be 6 + 8 = 14 and for activity 2-4, it will be 6 + 10

= 16 and so on.

The complete computations for all the activities are shown in columns

(3) and (4) of Table 1.

1.4.3 Backward pass computations

The purpose of the backward pass is to compute latest start and finish

times for each activity. These computations are precisely a “mirror

image” of the forward pass computations. The term “latest allowable

occurrence” of an even (denoted by L1) is used in the sense that the

project terminal event must occur on or before some arbitrary scheduled

time. Thus, the backward pass computations are started rolling

back by arbitrarily specifying the latest allowable occurrence time

for the project terminal event. If no schedule date for the completion

of the project is specified, then the convention of setting the latest

allowable time for the terminal event equal to its earliest time,

determined in the forward pass, is usually followed i.e. L = E for the

terminal event of the project. This convention is called the zero slack

convention. Following this, one can also interpret the latest allowable

activity finish time (LFT) as the time to which the completion of an

activity can be delayed without directly causing any increase in the

total time to complete the project.

To explain the computation, let us again consider the network diagram in

figure 24. The terminal even is 6 so we set L6 = E6 = 52 and we start

rolling back. The latest allowable occurrence time for the events 5 and 4

are L5 = 52 – 16 = 36 and L4 = 36 – 20 = 16 respectively. It may be

noted here that can roll back to event 3 via activity 3 –5 as well as

activity 3 – 4. So there are two latest allowable occurrence times for the

event 3 as given below:

L3 = L4 – D34 = 16 – 0 = 16

L3 = L5 – D35 = 36 – 6 = 30

We retain the minimum value as the latest occurrence time for the

event 3 and ignore other values. Therefore, the latest allowable

occurrence time for the event 3 is 16. Similarly

L2 = L3 – D23 = 16 – 8 = 8

L2 = L4 – D24 = 16 – 10 = 6

The latest occurrence time for the event 2 is thus 6 and the latest

occurrence time for the event 1 is equal to its earliest time i.e. zero.

In general, latest allowable occurrence time of an event can be

calculated by selecting an appropriate formula among the following two:

Li = Li – Dii or

Li = minimum (Li – Dii)

The second formula is used for the event having two or more latest

allowable occurrence time estimates.

Latest start and finish times of an activity

After computing latest allowable occurrence time for various

events, one can compute the latest start and finish times of an activity.

The latest finish time of an activity is equal to the latest allowable

occurrence time of the head event of that activity.

i.e. LFT (i-j) = Lj

The latest start time of an activity is equal to its latest finish

time minus its duration.

i.e. LST (i-j) = LFT (i-j) – Dij

These computations are shown in column (4) and column (6) of

Table 1, given below:-

Table 1

Start Finish

Activity Duration Earliest

time

Latest

time

Earliest

time

Latest

time

(1) (2) (3) (4) (5) (6)

1-2 6 0 0 6 6

2-3 8 6 8 14 16

2-4 10 6 6 16 16

3-4 0 14 16 14 16

3-5 6 14 30 20 36

4-5 20 16 16 36 36

5-6 16 36 36 52 52

The critical path determination

After having computed various time estimates, we are now interested in

finding the critical Path of the network. A network will consist of a

number of parts. A path is a continuous series of activities through the

network that leads from the initial event (or node) of the network to its

terminal event. For finding the critical Path, we list out all possible paths

through a network along with their duration. In the network under

consideration, various paths have been listed as follows:

Path length in days

1-2-3-5-6 36

1-2-4-5-6 52

1-2-3-4-5-6 50

Critical Path: a path in a project network is called critical if it is the

longest Path. The activities lying on the critical Path are called the critical

activities.

In the above example, the Path 1-2-4-5-6 with the longest duration of 52

days is the critical Path and the activities 1-2,2-4, 4-5 and 5-6 are the

critical activities.

Calculation of floats

It may be observed that for every critical activities in a network, the

earliest start and latest start times are the same. This is so since the

critical activities cannot be scheduled later than the earliest scheduled

time without delaying the total project duration, they do not have any

flexibility in scheduling. However, noncritical activities do have some

flexibility. That is these activities can be delayed for sometime without

affecting the project duration. This flexibility is termed as slack in case of

an event and as floats in case of an activity.

Some people do not make any distinction between a slack and a float.

Slack time for an event

The slack time or slack of an event in a network is the difference

between the latest event time and the earliest event time.

Mathematically it may be calculated using the formula Li – Ei where Li is

the latest allowable occurrence time and Ei is the earliest allowable

occurrence time of an event i.

Total float of an activity

The total activity float is equal to the difference between the earliest and

latest allowable start or finish times for the activity in question. Thus, for

an activity (i-j), the total float is given by:

TFij = LST – EST or TFij = LFT – EFT

In other words, it is the difference between the maximum time available

for the activity and the actual time it takes to complete. Thus, total float

indicates the amount of time by which the actual completion of an

activity can exceed its earliest expected completion time without causing

any delay in the project duration.

Free float

It is defined as that portion of the total float within which an activity can

be manipulated without affecting the float of the succeeding activities. It

can be determined by subtracting the head event slack from the total

float of an activity.

i.e. FFij = TFij – (slack of event j)

The free float indicates the value by which an activity in question can be

delayed beyond the earliest starting point without affecting the earliest

start, and therefore the total float of the activities following it.

Independent float

It is defined as that portion of the total float within which an activity can

be delayed for start without affecting float of the preceding activities. It is

computed by subtracting the tail event slack from the free float.

i.e. IFij = FFij – (slack of event i)

The independent float is always either equal to a less than the free float

of an activity. If a negative value is obtained, the independent float is

taken to be 0.

Interfering float

Utilisation of the float of an activity can affect the float of subsequent

activities in the network. Thus, interfering float can be defined as that

part of the total float which causes a reduction in the float of the

successor activities. In other words, it can be defined as the difference

between the latest finish time of the activity under consideration and the

earliest start time of the following activity, or 0, whichever is larger. Thus,

interfering float refers to that portion of the activity float which cannot be

consumed without affecting adversely the float of the subsequent activity

or activities.

Example

Activity Duration

1-2 4 days 1-3 12 days 1-4 10 days 2-4 8 days 2-5 6 days 3-6 8 days 4-6 10 days 5-7 10 days 6-7 0 days 6-8 8 days 7-8 10 days 8-9 6 days

With the help of the activities given above draw a network. Determine its

critical path, earliest start time, earliest finish tine, latest start time, latest

finish time, total float, free float and independent float.

PROGRAMME EVALUATION AND REVIEW TECHNIQUE

(a) The optimistic time estimate: this is the estimate of the shortest

possible time in which an activity can be completed on the ideal

conditions. For this estimate, no provision for delays of setbacks are

made. We shall denote this estimate by to.

(b) The pessimistic time estimate: this is the maximum possible time

equity to accomplish the job. If everything went long and normal

situations prevailed, this would be the time estimate. It is denoted by tp.

(c) The most likely time estimate: this is the time which lies between the

optimistic and pessimistic time estimates. It assumes that things go in a

normal way with few setbacks. It is represented by tm.

Beta distribution is found to give fairly satisfactory results for most of the

activities. For distribution of this type the standard deviation is

approximately 1/6 of the range.

i.e 6

op

t

ttS

The variance, therefore is 2

2

6

op

t

ttS

Expected time:

The expected time (te) is the average time taken for the completion of

the job. By using beta distribution, the expected time can be obtained

from the following formula.

6

4 pmo

e

tttt

Probability of achieving completion date

Suppose we wish to find out the probability that the project will be

completed within the scheduled completion time. The time te as

determined by beta distribution after taking into account the time

estimates viz. to, tp, tm only represents a 50% chance that the activity will

be completed within time te.

In general are project consisting of several activities will have a normal

distribution, that is for the project as a whole, the distribution curve will

be a normal curve and the probability of competing project in time equal

to the mean value Te which is ½.

A standardised normal curve has an area equal to 1 and the standard

deviation of 1. Further, it is symmetrical about the mean value te. Hence

the area under the curve AC is 50% of the total area under the curve

ACB. The area under the curve ACD depends on the location of Ts along

the time axis. The point te can be taken as the reference point and the

distance (te- Ts) can be expressed in terms of standard deviation. For

example, if Ts is on the right of te at a distance of one standard deviation

in the enclosed by ACD is 84.1%. If Ts is on the left of te at a distance of

one standard deviation then the area enclosed is 15.9%.

A distance of +1 corresponds to 84.1% probability and a distance of -1

corresponds to 15.9% probability.

We calculate the value of the standard normal variate (Z) as follows:

..

1

DS

TTZ

cp

Where T1 denotes the duration in which we wish to complete the project

and Tcp represents the duration of the critical path, S.D. stands for

standard deviation of the earliest finish of a network.

Computed variance Vt (= 2

tS ) of all the activity durations of the critical

path. Sum up these and take the square root. This yields the S.D. of the

earliest finish time of a network. Let the critical path duration the

designated by Tcp assuming normal distribution for the total duration, you

should be in a position to find the confidence interval for Tcp. Look at the

standard normal probability distribution tables for the probability of

competing project within the given duration of T1.

Examples 10. PERT calculations yield a project length of 60 weeks with

the variance of 9. Within how many weeks we expect the project to be

completed with the probability of 0.99? (That is the project length that he

would expect to be exceeded only by 1% of the time if the project were

repeated many times in an identical manner).

Solution

Tcp = 60 S.D. = 9 = 3

60 + (3 x 2.3) = 67

A few comments on assumptions of PERT AND CPM

1. Beta distribution may not always be applicable. 2. The formula for the expected duration and S. D. are simplifications.

Maccrinnon and Ryavec reached the conclusion that in certain cases the errors, because of these assumptions may even be to the tune of 33 percent.

3. The errors owing to the aforesaid simplification and assumptions may be compounded or may cancel each other to an extent.

4. In computing the S. D. of the critical path, independence of activities is implied. Limitations of resources may invalidate the independence which exists by the very definition of an activity.

5. It may not always be possible to sort of completely identifiable activities and state where they begin and where they end.

6. In projects fraught with uncertainty it is natural that the existing alternatives with differing outcomes. For example for particular hardness is not obtained in a metal, an alloy might have to be used that is more expensive and also inferior on certain technical considerations. There have been theoretical developments in this regard, and it may be worthwhile to incorporate the concept of decision tree analysis depending on the situation.

7. Time estimates have an element of subjectiveness and, to that extent, the techniques could be weak. The contractors reacts with this weakness shrewdly whilst bidding. If there are cost plus contracts they would deliberately underestimate the time for chances of being awarded the contract. Incentive type contracts might lead to an opposite bias.

8. Cost-time trade-offs, for deriving the cost curve slopes, are subjective again and call for ability of expertise of the technology as well as genuine effort to estimate. Often the engineers tend to be lax here; occasionally with the honest deliberation even, their guesses may be wide off the mark.

Distinctions between PERT and CPM

The PERT and CPM models are similar in terms of their basic structure,

rationale and mode of analysis. However, there are certain distinctions

between pert and CPM networks which are ennumerated below.

1. CPM is activity oriented that is CPM network is built on the basis of activities. Also results of various calculations are considered in

terms of activities of the project. On the other hand, PERT is event oriented.

2. CPM is a deterministic model that is it does not take into account the uncertainties involved in the estimation of time for execution of a job or an activity. It completely ignores the probabilistic element of the problem. Pert, however is the probabilistic model. It uses three estimates of the activity time; optimistic, pessimistic and most likely; with a view to take into account time uncertainty. Thus, the expected duration of each activity is probabilistic and expected duration indicates that there is 50% probability of getting the job done within that time.

3. CPM places dual emphasis on time and cost and evaluates the trade-off between project cost and project time. By deploying additional resources, it allows the critical path project manager to manipulate project duration within certain limits so that project duration can be shortened at an optimal cost. On the other hand, pert is primarily concerned with time. It helps the manager to schedule and coordinate various activities so that the project can be completed on schedule time.

4. CPM is commonly used for those projects which are repetitive in nature and where one has prior experience of handling similar projects. What is generally used for those projects with time required to complete various activities are not known before hand. Thus, pert is widely used for planning and scheduling research and development projects.

PROJECT CRASHING

In some cases, there are compelling reasons to complete a project

earlier than the originally estimated time duration of the critical path

computed on the basis of normal activity times, by employing extra

resources. An example would be introduction of a new project. The

motives in hastening the project might be to ensure that the competitors

do not steal a march. Increase or decrease in the total duration of the

completion time for project is closely associated with cost

considerations. In such cases when the total time duration is reduced,

the project cost increases, but in some exceptional cases project cost is

reduced as well. Production cost occurs in the cases of those projects

which make use of a certain type of resources for example a machine

and whose time is more valuable than the operator’s time.

Some definitions:

Activity cost: it is defined as the cost of performing and completing a

particular activity or task.

Crash cost, Cc: this is the direct cost that is anticipated in completing an

activity within the crash time.

Crash time, Ct: This is the minimum time required to complete an

activity.

Normal cost Nc: this is the lowest possible direct cost required to

complete an activity.

Normal time Nt: this is the minimum time required to complete an activity

at normal cost.

Activity costs slope: the costs slope indicates the additional cost incurred

per unit of time saved in reducing the duration of an

activity.

Let OA represent the normal duration of completing a job and OC the

normal cost involved to complete the job. Assume that the management

wish to reduce the time of completing the job to OB from normal time

OA. Therefore under such a situation the cost of the project increases

and it goes upto say OD (Crash Cost). This only amounts to saving that

by reducing the time period by BA the cost has increased by the amount

CD. The rate of increase in the cost of activity per unit decrease in time

is known as cost slope and is described as follows.

Activity cost slope = OBOA

OCOD

AB

CD

= CrashtimeNormaltime

NormalCosttCrash

cos

Optimum duration: the total project cost is the sum of the direct and

indirect costs. In case the direct cost varies with the project duration

time, the total cost would have the shape as indicated in the above

figure.

t Point A, the cost will be minimum. The time corresponding to this point

Point A is called the optimum duration and the cost as optimum cost for

the project.

Crash

cost

Normal

cost

Crash

Time

Normal

Time

COST

DURATION FOR THE JOB

D E

C

F

A B O

TRANSPORTATION

What is the transportation problem?

The transportation problem consists of three components. First we can formulate a linear objective function, which is to be minimized. This function will represent the total shipping cost of all the goods to be sent to all the destinations. Second we can write a table of constraints. Of the seven constraints of this problem, three (one for each row) will give the relationships between the origin capacities and the goods to be received by different destinations. These are called capacity constraints. The other four constraints (one for each column) will relationships between destination requirements and the goods to be shipped from different origins. These are called as requirement constraints. Third we can specify a set of non-negativity constraints for the structural variables xij. They will state that no negative shipments be permitted. The general correspondence between a typical linear programming problem and the transportation problem is thus complete. Minimize

COST

A

O

TOTAL PROJECT COST

DIRECT COST

INDIRECT COST

Crash Normal Optimal

TIME

General Transportation Tableau

Destination (Di)

Origin

(oj)

D 1

D 2

_ _ _ _ _

_

Dn

Ai

O1

C11

X11

C12

X12

C1n

X1n

A1

O2

C21

X21

C22

X22

C2n

X2n

A2

|

|

|

Om

Cm1

Xm1

Cm2

Xm2

Cmn

Xmn

Am

Req Bj B1 B2 -------- Bn

n

j

j

m

i

i BA11

What are the methods to solve a transportation problem?

A. Procedure summary for the modified distribution method

(minimization case)

Step 1 Obtain a basic feasible solution

An initial basic feasible solution for a given transportation may be

obtained by following the northwest corner rule, by the application of

Vogel’s approximation method, or by simple inspection.

Test for step 1: A basic feasible solution must include shipments

covering m+n-1 cells. That is number of occupied cells is 1 less than

the number of rows and columns in the transportation matrix.

If number is more then recheck the data. If the number of occupied

cells is less than m+n-1 then this is a degenerate solution. To resolve

the degeneracy, add one or more epsilons to some “suitable” empty

cells so that the number of occupied cells becomes equal to m+n-1.

Step 2 Determine the opportunity costs of the empty cells

(Opportunity costs = Implied costs –Actual costs)

a) Determine a complete set of row and column numbers (values). When in a given program, the number of occupied cells equals m+n-1, proceed to assign row and column numbers in such a manner that for each occupied cell the relationship cij =ui+vj holds. To start, a value of zero can be assigned to any row having an occupied cell. For each occupied cell, the actual shipping cost per unit should equal the sum of its row and column values.

b) Calculate the implied cost of empty cells. Once all the row and column values have been assigned, the implied cost of a given empty cell can be calculated as follows :

Implied cost = row value + column value.

c) Determine the opportunity costs of empty cells. The opportunity cost of an empty cell is determined by subtracting the actual cost of the empty cell from the implied cost. In other words, opportunity cost, for each cell is given by Opportunity cost = ui+ vj –cij.

If the opportunity costs of all the cells are non-positive, an optimal

solution has been obtained else a better program can be

designed. Thus step 2 serves as a test for optimality.

Step 3 Designed an improved program

Design a new program such that the empty cell having the largest

opportunity cost is included in the solution. This is accomplished in the

following manner:

a) draw a loop of horizontal and vertical arrows in such a manner that it starts from the cell to be filled, passes to an occupied cell in the same row or column as the empty cell, and then, making a series of alternate horizontal and vertical turns through occupied cells returns to the empty cell.

b) Place the plus (+) sign in the empty cell to be filled. Then alternately, place minus signs and plus signs at the beginnings and ends of the connecting ends of the loops.

c) Examine those occupied cells in which the minus signs have been placed. Of these, the cell having the least number is vacated by transferring these units to the empty cell. This is accomplished by adding the same amount to all cells having plus signs and subtracting it from all cells having minus signs. The improved program should have the same number of occupied cells as the preceding program. If the number is less then the problem becomes degenerate. In such a case , add epsilons to some recently vacated cells such that the number of occupied cells again equals m+n-1.

Step 4: Repeat steps 2 and 3 until a program is achieved in which

each empty cell has a n opportunity cost value which is either zero

or negative. This program will be an optimal program.

B. Modified distribution method (maximization case)

Except for one transformation, a transportation objective is to

maximize a given function can be solved by the MODI algorithm

as presented above. The transformation is made by subtracting all

the cij’s from the highest cij (profit) of the given transportation

matrix. The transformed cij’s give us the relative costs, and the

problem then becomes a minimization problem. Once an optimal

solution to this minimization problem has been found, the value of

the objective function can calculated by inserting the original

values of the cij’s for those routes, which form the basis (occupied

cells) in the optimal solution.

What is degeneracy? How can it be resolved?

The solution for non-degenerate basic feasible solution with exactly

m+n-1 strictly positive allocations in independent positions has been

discussed so far. However, sometimes it is not possible to get such

initial feasible solution to start with. Thus degeneracy occurs in the

transport problem whenever the number of occupied cells is less than

m+n-1.

Degeneracy in transportation can occur in two ways:

Basic feasible solutions may be degenerate from the initial stage

onward.

They may become degenerate at any intermediate stage.

Resolution of degeneracy during the initial stage.

To resolve degeneracy, allocate an extremely small amount of goods

(close to zero) to one or more of the empty cells so that the number of

occupied cells becomes m+n-1. The containing this small allocation, is of

course, considered to be an occupied cell.

Rule: The extremely small quantity usually denoted by the Greek letter

(delta) is introduced in the least cost independent cell subject to

following rule. If the necessary, two or more ’s can be introduced in the

least and second least cost independent cells.

Resolution of degeneracy during the solution stages

The transportation problem may also become degenerate during

solution stages. This happens when most favorable quantity is allocated

to the empty cell having the largest negative cell evaluation resulting in

simultaneous vacation of two or more of the currently occupied cells. To

solve degeneracy allocate (delta) to one or more of recently vacated

cells so that the no. of occupied cells is m+n-1 in the new solution.

What is an unbalanced Transportation problem? Explain the

procedure for balancing an unbalanced TP.

If in a TP the sum of all available quantities is not equal to the sum of

requirements, that is, m n

ai <> bj

i=1 j=1 then such a problem is called an unbalanced

transportation problem. An unbalanced transportation problem can be

modified to a balanced problem by introducing a fictitious sink in the first

case and a fictitious source in the second. The inflow from the fictitious

source to the sink represents the unfilled demand at that sink. Similarly

the inflow from the source to the fictitious sink represents the surplus at

that source. For convenience the cost of transportation of a unit from the

fictitious source or to the fictitious sink as the case maybe is assumed

zero. Following is the procedure to balance a TP.

Balancing the given transportation problem

To solve a given transportation problem we must establish equality

between the total capacities of the origins and the total requirements of

the destinations. Three cases can arise.

Case 1 : bi = dj In this case, the total capacity of the origins equals

the total requirements of the destinations. The problem can be arranged

in the form of a matrix, along with relevant cost data, and the

transportation algorithm may be applied directly to obtain a solution.

Case 2 : bi > dj In this case, the total capacity of the origins exceeds

the total requirement of the destinations. A “dummy” destination can be

added to the matrix to absorb the excess capacity. The cost of shipping

from each origin to this dummy destination is assumed to be zero. The

adding of a dummy destination establishes equality between the total

origin capacities and the total destination requirements. The problem is

then amenable to solution by the transportation algorithm.

ASSIGNMENT

Define assignment problem- The name assignment problem originates from the classical problems where the objective is to assign a number of origins (jobs) to the equal number of destinations (persons) t a minimum

cost or at maximum profit. To explain the nature of assignment problem, suppose there are n jobs to be performed and n persons are available for doing these jobs. Assume that each person can do each job at a time, though with varying degree of efficiency. Let cij be the cost (payment) if the ith person is assigned the jth job, the problem is to find the assignment (which job should be assigned to which person) so that the total cost for performing all jobs is minimum. Problems of this kind are known as assignment problems. Further these types of problems ma consist of assigning men to offices,

classes to rooms, drivers to trucks, trucks to delivery routes etc. The

assignment problem can be stated in the form of a n x n matrix [cij] of

real numbers as given in the table below.

JOBS

1 2 ……………………. j

……………… n

1 C11 C12 ….. C1j

…… C1n

2 C21 C22 ….. C2j

…… C2n

. . .

. . . . .

. PERSONS i Ci1 Ci2

…... Cij …… Cin

. .

. . . . .

. .

n Cn1 Cn2 …… Cnj

…… Cnn

Explain the mathematical formation of an assignment problem. Mathematically the assignment problem can be stated as:

Minimize the total cost

n m n

Z= cij xij , i= 1,2,3,….. xij = 1 for all I

(workers available)

i=1j=1 j = 1

n

subject to restrictions of the form: xij = 1

for all j (job req)

xij = 1 : if ith person is assigned the jth job .

= 0 : if not

n

xij +1 (one job is done by the ith person , j=

1,2,……. n )

j=1

n

and xij =1 ( only one person should be assigned the jth

job , j =1,2,3,……n )

i=1

where xij denotes that the jth job is to assign to the ith person.

MINIMISE ( TOTAL COST ) m m

Z = cij xij

i= 1 j = 1

Subject to

n

xij = ai j i= 1,2 …… m (capacity constraints)

j = 1

m

xij = bj ; j = 1,2 ….. n (req. constraint)

and xij 0 for all i & j

Explain the procedure summary for the assignment method for (a) minimization case & (b) maximization case.

(a) The minimasation case:

Step 1 Determine the total opportunity cost matrix

a) Arrive at a column opportunity cost matrix by subtracting the lowest

entry of each column of the given payoff matrix from all the entries in

the column.

b) Then subtract the lowest entry of each row of the matrix obtained

in (a) from all the entries in its row.

The result of step 1b gives the total-opportunity-cost matrix.

Step 2 Determine whether an optimal assignment can be made

a) Cover all the zeroes of the current total-opportunity-cost matrix with

the minimum possible

number of horizontal and vertical lines.

b) If the number of lines in step 2a equals the number of rows (or

columns) of the matrix, the problem can be solved. Make a complete

assignment so that the total opportunity cost involved in the

assignment is zero.

c) if the number of lines drawn in step 2a is less than the number of

rows (or columns ) of the matrix, proceed to step 3.

Step 3 Revise the total opportunity cost matrix

a) Subtract the lowest entry in the uncovered cells of the current total

opportunity cost matrix from all the uncovered cells.

b) Add the same lowest entry to only those cells in which the covering

lines of step 2 cross.

The result of steps 3a and 3b is a revised total opportunity cost

matrix.

Step 4

Repeat steps 2 and 3 until an optimal assignment having a total

opportunity cost of zero can be made.

Maximisation case

Except for one transformation, an assignment problem in which the

objective is to maximize the total payoff measure can be solved by the

assignment algorithm presented above. The transformation involves

subtracting all the entries of the original payoff matrix from the highest

entry of the original payoff matrix. The transformed entries give us the “

relative costs “and the problem then becomes a minimization problem.

Once the optimal assignment for this transformed minimization problem

has been identified, the total value of the original payoff measure can be

found by adding the individual original entries for those cells to which the

assignments have been made.