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LINEAR PROGRAMMING PROBLEM

INTRODUCTION

In a decision-making embroilment, model formulation is important because it represents theessence of business decision problem. The term formulation is used to mean the process of converting the verbal description and numerical data into mathematical expressions whichrepresents the relevant relationship among decision factors, objectives and restrictions on the useof resources. Linear Programming (LP) is a particular type of technique used for economicallocation of 'scarce' or 'limited' resources, such as labour, material, machine, time, warehousespace, capital, energy, etc. to several competing activities, such as products, services, jobs, newequipment, projects, etc. on the basis of a given criterion of optimally. The phrase scarceresources means resources that are not in unlimited in availability during the planning period. Thecriterion 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 inall functional areas of management, hospitals, airlines, agriculture, military operations, oil refining,education, energy planning, pollution control, transportation planning and scheduling, researchand 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-lifedecision 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 usbe 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 aresulting proportional change in another variable. For example, doubling the investment on acertain project will exactly double the rate of return. The word programming refers to modellingand 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 toachieve the desired objective.

A large number of computer packages are available for solving a mathematical LP model butthere is no general package for building a model. Model building is an art that improves withpractice. 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 ModelThe 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 toselect from, we would not need LP. The evaluation of various alternatives is guided by the natureof objective function and availability of resources. For this, we pursue certain activities usuallydenoted by x1, x2…xn. T h e v a lu eof these activities represent the extent to which each of these isperformed. For example, in a product-mix manufacturing, the management may use LP to decidehow many units of each of the product to manufacture by using its limited resources such aspersonnel, 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 limitedresources, require simultaneous solutions. All decision variables are continuous, controllableand 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:



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 maybe either random variables represented by a known distribution (general or 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 areassumed 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 arecontinuous. 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 aproblem 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 fractionalvalues, 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 thefractional values to the nearest integer values. However, when we do so, we should evaluate therevised solution to determine whether the solution represented by the rounded values is afeasible solution and also whether the solution is the best integer solution. Secondly, if theproblem relates to a continuum of time and it is designed to determine optimal solution for a giventime 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 unitsof B per week, can be adopted without any difficulty. The fractional amount of production wouldbe taken to be the work-in-progress and become a portion of the production of the followingweek. In this case an output of 17 units of A and 31 units of B over a three-week period wouldimply 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 programmingproblem, forcing the solutions to be in integers only.

Additivity: The value of the objective function for the given values of decision variables and thetotal sum of resources used, must be equal to the sum of the contributions (profit or cost) earnedfrom 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 beequal to the sum of the profits earned separately from A and B. Similarly, the amount of aresource consumed by A and B must be equal to the sum of resources used for A and Bindividually.This assumption implies that there is no interaction among the decision variables (interactionis 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 assumenegative values. Thus, only non-negative levels of activity are considered feasible. Thisassumption is indeed a realistic one. For instance, in the production problems, the outputcannot obviously be negative, because a negative production implies that we should be able toreverse 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 functionand 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 beproportional to its amount. For example, if production of one unit of a product uses 5 hours of aparticular 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 toadjust 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 isused 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 Industryand 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 asto 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 planover 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 beproduced from standard materials in order to minimise the trim loss.



Financial Management(i) Portfolio selection This deals with the selection of specific investment activityamong 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 approachConvex Polyhedron

TYPES OF SOLUTION

(a) Solution. Values of decision variables x j (j = 1, 2, 3, ….n) which satisfy the constraints of thegeneral 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). asolution obtained by setting (n - m) of the variables equal to zero and solving the remaining mequations in m unknowns is called a basic solution. Zero variables (n - m) are called non-basicvariables 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 solutionto 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 objectivefunction is same as that of the constraint forming the boundary of the feasiblesolutions region, and

(ii) The constraint should form a boundary on the feasible region in the direction of optimalmovement 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 beunbounded but LP problem not to be unbounded, that is an unbounded feasible region may yield



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 precisestatement of the objective function.

Convert the inequalities into equalities by the addition of nonnegative slack variables. Theseinequalities should be symmetric or balanced so that each slack variable appears in eachequation 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 topof 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 thezero or negative, the optimal solution has been obtained. Otherwise, the presence of anypositive 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- rowentry is the key column.

Find the key row and the key number. Divide the entries in the “quantity” column by thecorresponding nonnegative entries of the key column to form replacement ratios, and comparethese 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 eachcolumn ) the product of the corresponding key-row number and the corresponding fixed ratioformed by dividing the old row number in the key column by the key number. The result willgive 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 programRepeat steps 3 & 4 until a program has been derived.

[ Linear-programming problems involving the minimization of an objective function usually containstructural of the “greater than equal to” type. They can also be solved by the simplex method. Thesimplex procedure for solving a linear-programming problem in which the objective is to minimizerather than maximize a given function, although basically the same as above, requires sufficientmodifications 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 precisestatement 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 variableappears 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 columnvariable, 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 rowabove 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 theentries are zero or positive, the optimum solution has been obtained. Otherwise, the presenceof any negative entry in the net-evaluation row indicates that a better program can beobtained.

Revise the program.

Find the key column. The under which falls the largest negative net-evaluation-entry is the keycolumn.

Find the key row and the key number. Divide the entries in the “Quantity” column by thecorresponding nonnegative entries in the key column to form replacement ratios, and comparethese 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 tothe of the “Quantity” column by the key number. The resulting numbers form thecorresponding row of the next tableau.

Transform the non-key rows. Subtract from the old row the number of a given non-key row (ineach 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 rownumber. 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 minimizationproblems 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 maximizationproblem, 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 inthe 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 SimplexMethod, 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 feasiblesolution.

(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 thefeasible (optimal) solution is reached.

(4) To determine the entering variable take ratios of the coefficients of non-basic variables inthe objective function to the corresponding coefficients in the row associated with theleaving variable. Ignore the ratios with positive or zero denominators. The enteringvariable 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 toobtain 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 thisconstraint is not satisfied by the optimal solution, the problem remains optimal but it becomesinfeasible. 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 andconcurrent relationships, time scheduling becomes very effective. Identifying the criticalactivities keeps the executive alert and in a state of preparedness, with alternative plansready in case these are needed. Breaking down the project into the smaller componentspermits better and closed control.

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

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 nodesand connectors. It is often graphically represented by a numbered circle. All activities in thenetwork must commence from some event. Such events are called the tail event. Similarly allactivities 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 thehead 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; thearrowheads 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 whichrequires that an activity which depends on another activity is shown to emerge from the headevent 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 bestrictly 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 activityrepresented by the dangling arrow 9-11 is undertaken with no result.

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

1 2 3 4

6 7

8

9



1. And initial event is one which has arrow/arrows coming out of it and none of the arrowentering 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 initialevent.

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:

Here we proceed from left to right. The event with the least x coordinate is assigned the smallestinteger, say 1. Other events are assigned progressively higher integers with regard to xcoordinate. If two were more events such as 4 & 5 have the same x coordinate the one towardsthe arrow should have a higher number.It is better to number of the events as 10, 20, 30.... to afford insertions of more activities andevents omitted by oversight.

Concurrent activities

Activities may not always be discreet that is, they may be done in part allowing the subsequentactivities to commence before the preceding activity is fully completed. Activities of this kind are tobe frequently encountered in batch production. If, for example, a batch of 50 spindles is to beprocessed onto machines obviously it is not necessary to process all the items of the batch on thefirst machines and then transfer these to the next machines. A few items processed on the firstmachines may be transferred to the second machine before completion of the entire batch on thefirst 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 givenactivity can be scheduled. Earliest finish time for an activity indicates the time by which the activity

1 2 4

5

3

6 7 8

X

Y



can be completed, at the earliest. To compute these time estimates, we will first of all computethe 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 wecan 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 timeestimate with maximum value is retained as the earliest event time and other values are

1 2

3

4

5 6

E1= 0 E

2= 6

E3= 14

E4= 16

E5= 36 E

6= 52

6

8

10

20

6

016

L1= 0 L

2= 6

L3= 16

L4= 16

L5= 36 L

6= 52



discarded. This maximum value represents the completion of all the activities ending at the eventunder consideration. In the above example, the earliest event time for event 4 will be 16.

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 theduration 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 theearliest start and finish times of all the activities on the network. The earliest start time of anactivity 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 theearliest 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 projectterminal event must occur on or before some arbitrary scheduled time. Thus, the backwardpass computations are started rolling back by arbitrarily specifying the latest allowableoccurrence time for the project terminal event. If no schedule date for the completion of theproject is specified, then the convention of setting the latest allowable time for the terminal eventequal to its earliest time, determined in the forward pass, is usually followed i.e. L = E for theterminal event of the project. This convention is called the zero slack convention. Followingthis, one can also interpret the latest allowable activity finish time (LFT) as the time to whichthe completion of an activity can be delayed without directly causing any increase in thetotal time to complete the project.

To explain the computation, let us again consider the network diagram in figure 24. Theterminal 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. Sothere are two latest allowable occurrence times for the event 3 as given below:

L3 = L4 – D34 = 16 – 0 = 16L3 = L5 – D35 = 36 – 6 = 30

We retain the minimum value as the latest occurrence time for the event 3 and ignoreother values. Therefore, the latest allowable occurrence time for the event 3 is 16. Similarly

L2 = L3 – D23 = 16 – 8 = 8L2 = L4 – D24 = 16 – 10 = 6

The latest occurrence time for the event 2 is thus 6 and the latest occurrence time for theevent 1 is equal to its earliest time i.e. zero.



In general, latest allowable occurrence time of an event can be calculated by selecting anappropriate 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 occurrencetime estimates.

Latest start and finish times of an activity

After computing latest allowable occurrence time for various events, one can compute thelatest start and finish times of an activity. The latest finish time of an activity is equal to thelatest 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 activitiesthrough 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 days1-2-3-5-6 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 activitieslying 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 Pathand 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 starttimes are the same. This is so since the critical activities cannot be scheduled later than theearliest 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 canbe delayed for sometime without affecting the project duration. This flexibility is termed as slack incase 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 timeand the earliest event time.Mathematically it may be calculated using the formula Li – Ei where Li is the latest allowableoccurrence 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 TF ij = LFT – EFT

In other words, it is the difference between the maximum time available for the activity and theactual time it takes to complete. Thus, total float indicates the amount of time by which the actualcompletion of an activity can exceed its earliest expected completion time without causing anydelay in the project duration.

Free float

It is defined as that portion of the total float within which an activity can be manipulated withoutaffecting the float of the succeeding activities. It can be determined by subtracting the head eventslack 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 theearliest starting point without affecting the earliest start, and therefore the total float of the activitiesfollowing it.

Independent float

It is defined as that portion of the total float within which an activity can be delayed for start withoutaffecting float of the preceding activities. It is computed by subtracting the tail event slack from thefree 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 negativevalue 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 thefloat of the successor activities. In other words, it can be defined as the difference between thelatest finish time of the activity under consideration and the earliest start time of the followingactivity, or 0, whichever is larger. Thus, interfering float refers to that portion of the activity floatwhich cannot be consumed without affecting adversely the float of the subsequent activity or activities.



ExampleActivity Duration1.2 4 days1.3 12 days1.4 10 days2.4 8 days

2.5 6 days3.6 8 days4.6 10 days5.7 10 days6.7 0 days6.8 8 days7.8 10 days8.9 6 days

With the help of the activities given above draw a network. Determine its critical path, earliest starttime, earliest finish tine, latest start time, latest finish time, total float, free float and independentfloat.



PROGRAMME EVALUATION AND REVIEW TECHNIQUE

(a) The optimistic time estimate: this is the estimate of the shortest possible time in which anactivity 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 isdenoted by tp.

(c) The most likely time estimate: this is the time which lies between the optimistic and pessimistictime estimates. It assumes that things go in a normal way with few setbacks. It is represented bytm.

Beta distribution is found to give fairly satisfactory results for most of the activities. For distributionof this type the standard deviation is approximately 1/6 of the range.

i.e6

o p

t t t S −=

The variance, therefore is

2

2

6

−= o p

t

t t S

Expected time:

The expected time (te) is the average time taken for the completion of the job. By using betadistribution, the expected time can be obtained from the following formula.

++=6

4 pmo

e

t t t t

Probability of achieving completion date

Suppose we wish to find out the probability that the project will be completed within the scheduledcompletion time. The time te as determined by beta distribution after taking into account the timeestimates viz. to, tp, tm only represents a 50% chance that the activity will be completed within timete.

In general are project consisting of several activities will have a normal distribution, that is for theproject as a whole, the distribution curve will be a normal curve and the probability of competingproject 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 issymmetrical about the mean value te. Hence the area under the curve AC is 50% of the total areaunder 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 standarddeviation in the enclosed by ACD is 84.1%. If Ts is on the left of te at a distance of one standarddeviation 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

T T Z

cp−=

Where T1 denotes the duration in which we wish to complete the project and Tcp represents theduration of the critical path, S.D. stands for standard deviation of the earliest finish of a network.

Computed variance Vt (=2

t S ) of all the activity durations of the critical path. Sum up these andtake the square root. This yields the S.D. of the earliest finish time of a network. Let the criticalpath duration the designated by Tcp assuming normal distribution for the total duration, you shouldbe in a position to find the confidence interval for Tcp. Look at the standard normal probabilitydistribution 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. Withinhow many weeks we expect the project to be completed with the probability of 0.99? (That is theproject length that he would expect to be exceeded only by 1% of the time if the project wererepeated many times in an identical manner).Solution

Tcp = 60 S.D. = 9 = 360 + (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 theseassumptions 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. Limitationsof resources may invalidate the independence which exists by the very definition of anactivity.

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

6. In projects fraught with uncertainty it is natural that the existing alternatives with differingoutcomes. For example for particular hardness is not obtained in a metal, an alloy mighthave to be used that is more expensive and also inferior on certain technicalconsiderations. There have been theoretical developments in this regard, and it may beworthwhile 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 techniquescould be weak. The contractors reacts with this weakness shrewdly whilst bidding. If thereare 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 theengineers tend to be lax here; occasionally with the honest deliberation even, their guesses may be wide off the mark.



Let OA represent the normal duration of completing a job and OC the normal cost involved tocomplete 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 itgoes upto say OD (Crash Cost). This only amounts to saving that by reducing the time period byBA the cost has increased by the amount CD. The rate of increase in the cost of activity per unitdecrease in time is known as cost slope and is described as follows.

Activity cost slope =OBOA

OC OD

AB

CD

−

−=

=Crashtime Normaltime

NormalCost t Crash

−

−cos

Crashcost

Normalcost

CrashTime

NormalTime

COST

DURATION FOR THE JOB

D E

C

F

ABO



Optimum duration: the total project cost is the sum of the direct and indirectcosts. In case the direct cost varies with the project duration time, the total costwould have the shape as indicated in the above figure.

At Point A, the cost will be minimum. The time corresponding to this point Point A is called theoptimum duration and the cost as optimum cost for the project.

COST

A

O

TOTAL PROJECT COST

DIRECT COST

INDIRECT COST

Crash NormalOptimal

TIME



TRANSPORTATION1. 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 costof 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 relationshipsbetween the origin capacities and the goods to be received by different destinations. Theseare called capacity constraints. The other four constraints (one for each column) willrelationships between destination requirements and the goods to be shipped from differentorigins. 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 shipmentsbe permitted. The general correspondence between a typical linear programming problem andthe transportation problem is thus complete.Minimize

General Transportation Tableau

Destination (Di)

Origin(oj)

D 1

D 2 _ _ _ _ _ _ Dn Ai

O1 C11X11

C12X12

C1nX1n

A1

O2 C21X21

C22X22

C2nX2n

A2

|||

Om Cm1Xm1

Cm2Xm2

CmnXmn

Am

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

∑∑==

=n

j

j

m

i

i B A11

5.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 solutionAn initial basic feasible solution for a given transportation may be obtained by following thenorthwest corner rule, by the application of Vogel’s approximation method, or by simpleinspection.Test for step 1: A basic feasible solution must include shipments covering m+n-1 cells. That isnumber of occupied cells is 1 less than the number of rows and columns in the transportationmatrix.If number is more then recheck the data. If the number of occupied cells is less than m+n-1then this is a degenerate solution. To resolve the degeneracy, add one or more epsilons tosome “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 columnnumbers in such a manner that for each occupied cell the relationship cij =ui+vj holds. Tostart, 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 andcolumn values.

b) Calculate the implied cost of empty cells. Once all the row and column values have beenassigned, 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 isdetermined by subtracting the actual cost of the empty cell from the implied cost. In other words, opportunity cost, for each cell is given byOpportunity 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 programDesign a new program such that the empty cell having the largest opportunity cost is included inthe 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 thecell to be filled, passes to an occupied cell in the same row or column as the emptycell, and then, making a series of alternate horizontal and vertical turns throughoccupied cells returns to the empty cell.

b) Place the plus (+) sign in the empty cell to be filled. Then alternately, place minussigns 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 emptycell. This is accomplished by adding the same amount to all cells having plus signsand subtracting it from all cells having minus signs. The improved program shouldhave the same number of occupied cells as the preceding program. If the number isless then the problem becomes degenerate. In such a case , add epsilons to somerecently 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 an opportunity cost value which is either zero or negative. This program will be an optimalprogram.

B. Modified distribution method (maximization case)

Except for one transformation, a transportation objective is to maximize a given functioncan be solved by the MODI algorithm as presented above. The transformation is made bysubtracting all the cij’s from the highest cij (profit) of the given transportation matrix. Thetransformed cij’s give us the relative costs, and the problem then becomes a minimizationproblem. Once an optimal solution to this minimization problem has been found, the valueof 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.

6.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



ASSIGNMENT

2. Define assignment problem.

The name assignment problem originates from the classical problems where the objective is toassign a number of origins (jobs) to the equal number of destinations (persons) t a minimum costor at maximum profit. To explain the nature of assignment problem, suppose there are n jobs to beperformed and n persons are available for doing these jobs. Assume that each person can doeach job at a time, though with varying degree of efficiency. Let cij be the cost (payment) if the ithperson is assigned the jth job, the problem is to find the assignment (which job should be assignedto which person) so that the total cost for performing all jobs is minimum. Problems of this kind areknown as assignment problems.

Further these types of problems ma consist of assigning men to offices, classes to rooms, driversto trucks, trucks to delivery routes etc. The assignment problem can be stated in the form of a n xn 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



3. Explain the mathematical formation of an assignment problem.

Mathematically the assignment problem can be stated as:Minimize the total cost

n m n

Z= Σ Σ 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

Σ xij +1 (one job is done by the ith person , j= 1,2,……. n )j=1

n

and Σ xij =1 ( only one person should be assigned the jth job , j =1,2,3,……n )

i=1where xij denotes that the jth job is to assign to the ith person.

• MINIMISE ( TOTAL COST ) m m

Z = ∑ ∑ cij xiji= 1 j = 1

Subject ton

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

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

and xij ≥ 0 for all i & j

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

3(a) The minimasation case:

Step 1 Determine the total opportunity cost matrixa) 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 madea) Cover all the zeroes of the current total-opportunity-cost matrix with the minimum possiblenumber of horizontal and vertical lines.b) If the number of lines in step 2a equals the number of rows (or columns) of the matrix, theproblem can be solved. Make a complete assignment so that the total opportunity costinvolved 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 thematrix, proceed to step 3.



Step 3 Revise the total opportunity cost matrixa) Subtract the lowest entry in the uncovered cells of the current total opportunity cost matrixfrom 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 4Repeat steps 2 and 3 until an optimal assignment having a total opportunity cost of zero canbe made.

Maximisation case

Except for one transformation, an assignment problem in which the objective is to maximize thetotal payoff measure can be solved by the assignment algorithm presented above. Thetransformation involves subtracting all the entries of the original payoff matrix from the highestentry of the original payoff matrix. The transformed entries give us the “ relative costs “and theproblem then becomes a minimization problem. Once the optimal assignment for this transformedminimization 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 beenmade.

QT Combined 01

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