MB0032 Set 1 and Set 2 completed

8/14/2019 MB0032 Set 1 and Set 2 completed

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Name: Falguni Pandit Registration No.: 520966021MBA II SEM

OPERATIONS RESEARCH - MB0032

MB0032

Registration No.: 520966021

Page No: 1 Name: Falguni Pandit |Registration No.: 520966021


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Set 11. Describe the broad classification of Operations Research

models in details. Name the different steps needed in ORapproach of problem solving?Answer: A model is known as the representation of the reality. It isknown as an idealized representation or abstraction of a real life system.

The main objective of this model is to identify significant factors and theirinterrelationship. A model is helpful is decision making as it provides asimplified description of complexities and uncertainties of a problem inlogical structure.

A broad classification of OR models:a) Physical modes include all form of diagrams, graphs and charts. They

are designed to deal with specific problems. They bring out significantfactors and inter-relationship in pictorial firm so as to facilitate analysis.There are two types:1) Ieonic models and 2) Analog modelsIconic model is known as an image of an object or system that isrepresented on a small scale. We can say that these models can simulatethe actual performance of a product.On the other hand analog models are small physical systems that havesimilar characteristics and work like an object. For example- Toy.b) Mathematical Model or Symbolic models represent the decision variableof the system. The model employs a set of mathematical symbols also.

The variables are related by mathematical system also. For example -Allocation, sequencing, replacement models etc.c) It is by nature of Environment We have 1) Deterministic model in whichevery thing is defined and the results are certain. Eg: EOQ model 2)Probabilistic models in which the input and output variables follow aprobability distribution Eg: Games Theory.d) By the extent of Generality: The tow models belonging to this class are1) General model canbe applied in general and does not pertain to oneproblem only. Eg: Linear Programming2) Specific model is applicable under specific condition only. For example -

Sales can response curve or equation which can be known as a function of advertising that is applicable in the marketing function alone.

The scientific method translates a real given problem into a mathematicalrepresentation which is solved and retransformed into the originalcontext. The OR approach to problem solving consists of the followingsteps:1) Definition of the problem

The first and the most important requirement is that the root problemshould be identified and understood. The problem should be identifiedproperly, this indicates three major aspects:1) A description of the goal or the objective of the study,2) an identification of the decision alternative to the system, and



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OPERATIONS RESEARCH - MB00323) a recognition of the limitations, restrictions and requirements of thesystem.

2) Construction of the model

Depending on the definition of the problem, the operations research teamshould decide on the most suitable model for representing the system.Such a model should specify quantitative expressions for the objectiveand the constraints of the problem in terms of its decision variables.3) Solution of the modelOnce an appropriate model has been formulated the next stage in theanalysis calls for its solution and the interpretation of the solution in thecontext of the given problem A solution to a model implies

determination of a specific set of decision variables that would yield onoptimum solution. An optimum solution is one which maximizes orminimizes the performance of any measure in a model subject to thecondition and constraints imposed on the model.4) Validation the modelA model is a good representative of a system, and then the optimalsolution must improve the systems performance. A common method fortesting the validity of a model is to compare its performance with somepost data available for the actual system.5) About Implementation of the final result

The optimal solution obtained from a model should be applied practice toimprove theperformance of the system and the validity of the solution should beverified under changing conditions.

2. Describe the graphical method to solve the LinearProgramming Problem. Use the following example Maximize Z = 30x 1 + 40x 2Subject to the constraints1.5x 1 + 1.9x 2 6000.3 x 1 + 0.2x 2 100

0.0x 1 + 0.2x 2

30and x 1 0, x 2 0

Answer: A LPP with 2 decision variables x1 and x2 can be solved easilyby graphical method.We consider the x1 x2 plane where we plot the solution space, which isthe space enclosed by the constraints. Usually the solution space is aconvex set which is bounded by a polygon; since a linear function attainsextreme (maximum or minimum) values only on boundary of the region, itis sufficient to consider the vertices of the polygon and find the value of the objective function in these vertices.



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OPERATIONS RESEARCH - MB0032By comparing the vertices of the objective function at these vertices, weobtain theoptimal solution of the problem.

The method of solving a LPP on the basis of the above analysis is knownas the graphicalmethod. The working rule for the method is as follows:Working Rule:Step I: Write down the equations by replacing the inequality symbols bythe equality symbol in the given constraints.Step II : Plot the straight lines represented by the equations obtained instep I.Step III: Identify the convex polygon region relevant to the problem. Wemust decide on which side of the line, the halfplane is located.

Step IV: Determine the vertices of the polygon and find the values of thegiven objectivefunction Z at each of these vertices. Identify the greatest and least of these values. These are respectively the maximum and minimum value of Z.Step V: Identify the values of (x1, x2) which correspond to the desiredextreme value of Z. This is an optimal solution of the problem.Solution by Graphical Method

Let the horizontal axis represent x1and vertical axis represent x2. plot theconstraint lines,feasibility region has been shown in fig.

Any point on the thick line or inside the shaded portion will satisfy all the restrictions of the problem then ABCDE is the feasibility region carried out by the constraints operating on theobjective function. This depicts the limits within which the values of the decision variablesare permissible. The intersection points C and D can be solved by the linear equations 0.1x2< 30, 0.1x1 + 1.5x2 = 600, and 0.2x1 + 0.2x2 < 100 i.e. C(150,300) and D(300,180).

We work out the revenues at different corners points as tabulated below



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From the above table we find that revenue is maximum at Rs.31500 when150 unit of X1 and 300 units of X2 are produced.

3. Describe the Penalty Cost method (Big M Method) for solvingL.P.P.Answer: Penalty Cost Method Or Big-M MethodConsider a L.P.P. when atleast one of the constraints is of the type > or= . While expressing in the standard form, add a non negative variable toeach of such constraints. These variables are called artificial variables.

Their addition causes violation of the corresponding constraints, sincethey are added to only one side of an equation, the new system isequivalent to the old system of constraints if and only if the artificialvariables are zero. To guarantee such assignments in the optimal solution,artificial variables are incorporated into the objective function with largepositive coefficients in a minimization program or very large negativecoefficients in a maximization program. These coefficients are denoted by M.

Whenever artificial variables are part of the initial solution X0, the last rowof simplex table will contain the penalty cost M. The followingmodifications are made in the simplex method to minimize the error of incorporating the penalty cost in the objective function. This method iscalled Big M method or Penalty cost method.1) The last row of the simplex table is decomposed into two rows, the firstof which involves those terms not containing M, while the second involvesthose containing M.2) The Step 1 of the simplex method is applied to the last row created inthe above modification and followed by steps 2, 3 and 4 until this rowcontains no negative elements. Then step 1 of simplex algorithm isapplied to those elements next to the last row that are positioned overzero in the last row.



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OPERATIONS RESEARCH - MB00323) Whenever an artificial variable ceases to be basic, it is removed fromthe first column of the table as a result of step 4, it is also deleted fromthe top row of the table as is the entire column under it.4) The last row is removed from the table whenever it contains all zeroes.5) If non zero artificial variables are present in the final basic set, then theprogram has nosolution. In contrast, zero valued artificial variables in the final solutionmay exist when one or more of the original constraint equations areredundant.

4. Why Duality concept is important in OR? Describe the economicimportance of the Duality concept.Answer: The Importance of Duality Concept Is Due To Two MainReasons

i. If the primal contains a large number of constraints and a smallernumber of variables, thelabour of computation can be considerablyreduced by converting it into the dual problem and then solving it.ii. The interpretation of the dual variable from the loss or economic pointof view provesextremely useful in making future decisions in the activities beingprogrammed.Economic importance of duality concept

The linear programming problem can be thought of as a resourceallocation model in which the objective is to maximize revenue or profitsubject to limited resources. Looking at the problem from this point of view, the associated dual problem offers interesting economicinterpretations of the L.P resource allocation model. We consider here arepresentation of the general primal and dual problems in which theprimal takes the role of a resource allocation model.PrimalMaximize

From the above resource allocation model, the primal problem has neconomic activities and m resources. The coefficient cj in the primalrepresents the profit per unit of activity j. Resource i, whose maximumavailability is bi, is consumed at the rate aij units per unit of activity

j. Interpretation of Duel Variables For any pair of feasible primal and dual solutions,(Objective value in the maximization problem) (Objective value in theminimization problem) At the optimum, the relationship holds as a strict

equation. Note: Here the sense of optimization is very important. Hence



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OPERATIONS RESEARCH - MB0032clearly for any two primal and dual feasible solutions, the values of theobjective functions, when finite, must satisfy the following inequality.

The strict equality, z = w, holds when both the primal and dual solutionsare optimal.Consider the optimal condition z = w first given that the primal problemrepresents a resource allocation model, we can think of z as representingprofit in Rupees. Because bi represents the number of units available of resource i, the equation z = w can be expressed as profit (Rs) = (units of resource i) x (profit per unit of resource i) This means that the

dual variables yi,represent the worth per unit of resource i [variables yi are also called asdual prices, shadow prices and simplex multipliers]. With the same logic,the inequality z < w associated with any two feasible primal and dualsolutions is interpreted as (profit) < (worth of resources) This relationshipimplies that as long as the total return from all the activities is less thanthe worth of the resources, the corresponding primal and dual solutionsare not optimal. Optimality is reached only when the resources have beenexploited completely, which can happen only when the input equals theoutput (profit). Economically the system is said to remain unstable (nonoptimal) when the input (worth of the resources) exceeds the output(return). Stability occurs only when the twoquantities are equal.

5. Describe the Matrix Minimum method of finding the initialbasic feasible solution in the transportation problem.Answer: The Initial basic Feasible solution using Matrix MinimumMethodLet us consider a T.P involving m-origins and n-destinations. Since thesum of origin capacities equals the sum of destination requirements, afeasible solution always exists.

Any feasible solution satisfying m + n 1 of the m + n constraints is aredundant one and hence can be deleted. This also means that a feasiblesolution to a T.P can have at the most only m + n 1 strictly positive component, otherwise the solution will degenerate. It isalways possible to assign an initial feasible solution to a T.P. in such amanner that the rim requirements are satisfied.

This can be achieved either by inspection or by following some simplerules. We begin by imagining that the transportation table is blank i.e.initially all xij = 0. The simplest procedures for initial allocation discussedin the following section.Matrix Minimum Method



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OPERATIONS RESEARCH - MB0032Step 1: Determine the smallest cost in the cost matrix of thetransportation table. Let it be cij , Allocate xij = min ( ai, bj) in the cell ( i,

j)Step 2: If xij = ai cross off the ith row of the transportation table anddecrease bj by ai go to step 3. if xij = bj cross off the ith column of thetransportation table and decrease ai by bj go to step 3.if xij = ai= bj cross off either the ith row or the ith column but not both.Step 3: Repeat steps 1 and 2 for the resulting reduced transportationtable until all the rimrequirements are satisfied whenever the minimum cost is not uniquemake an arbitrary choice among the minima.



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6. What do you understand by the Integer Programming Problem?Describe the Gomorys All-I.P.P. method for solving the I.P.P.problem.Answer: Integer Programming Problem

The Integer Programming Problem I P P is a special case of L P P where allor some variables are constrained to assume nonnegative integer values.

This type of problem has lot of applications in business and industrywhere quite often discrete nature of the variables is involved in manydecision making situations. Eg. In manufacturing the production isfrequently scheduled in terms of batches, lots or runs; In distribution, ashipment must involve a discrete number of trucks or aircrafts or freightcars .An integer programming problem can be described as follows:

Determine the value of unknowns x1, x2, , xn so as to optimize z =c1x1 +c2x2 + . . .+ cnxnsubject to the constraints ai1 x1 + ai2 x2 + . . . + ain xn =bi , i = 1,2,,mand xj > 0 j = 1, 2, ,n where xj being an integral value for j = 1, 2, , k n.If all the variables are constrained to take only integral value i.e. k = n, itis called an all (or pure) integer programming problem. In case only someof the variables are restricted to take integral value and rest (n k)variables are free to take any non negative values, then the problem isknown as mixed integer programming problem.Gomorys All IPP MethodAn optimum solution to an I. P. P. is first obtained by using simplexmethod ignoring therestriction of integral values. In the optimum solution if all the variableshave integer values, the current solution will be the desired optimuminteger solution. Otherwise the given IPP is modified by inserting a newconstraint called Gomorys or secondary constraint which representsnecessary condition for integrability and eliminates some non integersolution without losing any integral solution. After adding the secondaryconstraint, the problem is then solved by dual simplex method to get anoptimum integral solution. If all the values of the variables in this solution

are integers, an optimum inter-solution is obtained, otherwise anothernew constrained is added to the modified L P P and the procedure isrepeated. An optimum integer solution will be reached eventually afterintroducing enough new constraints to eliminate all the superior noninteger solutions. The construction of additional constraints, calledsecondary or Gomorysconstraints, is so very important that it needs special attention.

The iterative procedure for the solution of an all integer programmingproblem is as follows:Step 1: Convert the minimization I.P.P. into that of maximization, if it is in

the minimization form. The integrality condition should be ignored.



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OPERATIONS RESEARCH - MB0032Step 2: Introduce the slack or surplus variables, wherever necessary toconvert the inequations into equations and obtain the optimum solution of the given L.P.P. by using simplex algorithm.Step 3: Test the integrality of the optimum solutiona) If the optimum solution contains all integer values, an optimum basicfeasible integer solution has been obtained.b) If the optimum solution does not include all integer values then proceedonto next step.Step 4: Examine the constraint equations corresponding to the currentoptimum solution. Let these equations be represented by

Where n denotes the number of variables and m the number of equations.Choose the largest fraction of bis ie to find {bi}iLet it be [bk 1]or write is asf koStep 5: Express each of the negative fractions if any, in the k th row of the optimum simplextable as the sum of a negative integer and a nonnegative fraction.Step 6: Find the Gomorian constraint

and add the equation

to the current set of equation constraints.Step 7: Starting with this new set of equation constraints, find the newoptimum solution by dual simplex algorithm. (So that Gsla (1) is the initialleaving basic variable).Step 8: If this new optimum solution for the modified L.P.P. is an integersolution. It is also feasible and optimum for the given I.P.P. otherwise



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OPERATIONS RESEARCH - MB0032return to step 4 and repeat the process until an optimum feasible integersolution is obtained.



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OPERATIONS RESEARCH - MB00321. Describe in details the OR approach of problem solving. Whatare the limitations of theOperations Research?Answer: The OR approach to problem solving consists of the followingsteps:1. Definition of the problem.2. Construction of the model.3. Solution of the model.4. Validation of the model.5. Implementation of the final result.1. Definition of the problem

The first and the most important requirement is that the root problemshould be identified andunderstood. The problem should be identified properly, this indicates

three major aspects: (1) adescription of the goal or the objective of the study, (2) an identification of the decisionalternative to the system, and (3) a recognition of the limitations,restrictions and requirements of the system.2. Construction of the modelDepending on the definition of the problem, the operations research teamshould decide on themost suitable model for representing the system. Such a model shouldspecify quantitativeexpressions for the objective and the constraints of the problem in termsof its decision variables.A model gives a perspective picture of the whole problem and helpstackling it in a wellorganized manner. If the resulting model fits into one of the commonmathematical models, aconvenient solution may be obtained by using mathematical techniques. If the mathematicalrelationships of the model are too complex to allow analytic solutions, asimulation model may be

more appropriate. There are various types of models which can beconstructed under differentconditions.3. Solution of the modelOnce an appropriate model has been formulated, the next stage in theanalysis calls for itssolution and the interpretation of the solution in the context of the givenproblem. A solution to amodel implies determination of a specific set of decision variables thatwould yield an Optimumsolution. An Optimum solution is one which maximize or minimize the

performance of any



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OPERATIONS RESEARCH - MB0032measure in a model subject to the conditions and constraints imposed onthe model.4. Validation the modelA model is a good representative of a system, then the Optimal solutionmust improve thesystems performance. A common method for testing the validity of amodel is to compare itsperformance with some past data available for the actual system. Themodel will be valid if undersimilar conditions of inputs, it can reproduce the past performance of thesystem. The problemhere is that there is no assurance that future performance will continue toduplicate past behavior.Also, since the model is based on careful examination of past data, the

comparison should alwaysreveal favorable results. In some instances this problem may be overcomeby using data from trialruns of the system. It must be noted that such a validation method is notappropriate fornonexistent systems, since data will not be available for comparison.5. Implementation of the final result

The optimal solution obtained from a model should be applied practice toimprove theperformance of the system and the validity of the solution should beverified under changingconditions. It involves the translation of these results into detailedoperating instructions issued inan understandable form to the individuals who will administer and operatethe recommendedsystem. The interaction between the operations research team and theoperating personnel willreach its peak in this phase.Limitations of OR

The limitations are more related to the problems of model building, timeand money factors.

i) Magnitude of computation: Modern problem involve large number of variables and hence tofind interrelationship, among makes it difficult.ii) Non quantitative factors and Human emotional factor cannot be takeninto account.iii) There is a wide gap between the managers and the operationresearchesiv) Time and Money factors when the basic data is subjected to frequentchanges thenincorporation of them into OR models is a costly affair.v) Implementation of decisions involves human relations and behaviour.

2. What are the characteristics of the standard form of L.P.P.?What is the standard form



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OPERATIONS RESEARCH - MB0032of L.P.P.? State the fundamental theorem of L.P.P..Answer: The Standard Form Of LPP

The characteristics of the standard form are:1. All constraints are equations except for the non negativity conditionwhich remain inequalities(>, 0) only.2. The right hand side element of each constraint equation is non-negative.3. All variables are non-negative.4. The objective function is of the maximization or minimization type.

The inequality constraints can be changed to equations by adding orsubtracting the left hand sideof each such constraints by a nonnegative variable. The nonnegativevariable that has to be added

to a constraint inequality of the form < to change it to an equation iscalled a slack variable . Thenon-negative variable that has to be substracted from a constraintinequality of the form tochange it to an equation is called a surplus variable . The right hand sideof a constraint equationcan be made positive by multiplying both sides of the resulting equationby (-1) wherevernecessary. The remaining characteristics are achieved by using theelementary transformationsintroduced with the canonical form.The Standard Form Of The LPPAny standard form of the L.P.P. is given by

Fundamental Theorem Of L.P.P.Given a set of m simultaneous linear equations in n unknowns/variables, n> m, AX = b, with r(A)= m . If there is a feasible solution X > 0 , then there exists a basic feasiblesolution.3. Describe the Two-Phase method of solving a linearprogramming problem with anexample.Answer: Two Phase Method

The drawback of the penalty cost method is the possible computationalerror that could result



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OPERATIONS RESEARCH - MB0032from assigning a very large value to the constant M. To overcome thisdifficulty, a new method isconsidered, where the use of M is eliminated by solving the problem intwo phases. They arePhase I: Formulate the new problem by eliminating the original objectivefunction by the sum of the artificial variables for a minimization problem and the negative of thesum of the artificialvariables for a maximization problem. The resulting objective function isoptimized by thesimplex method with the constraints of the original problem. If theproblem has a feasiblesolution, the optimal value of the new objective function is zero (whichindicates that all artificial

variables are zero). Then we proceed to phase II. Otherwise, if the optimalvalue of the newobjective function is non zero, the problem has no solution and themethod terminates.Phase II : Use the optimum solution of the phase I as the starting solutionof the originalproblem. Then the objective function is taken without the artificialvariables and is solved bysimplex method.Examples:7) Use the two phase method toMaximise z = 3x1 x2Subject to 2x1 + x2 > 2x1 + 3x2 < 2x2 < 4,x1, x2 > 0Rewriting in the standard form,Maximize z = 3x1 x2 + 0S1 MA1 + 0.S2 + 0.S3Subject to 2x1 + x2 S1 + A1 = 2x1 + 3x2 + S2 = 2x2 + S3 = 4,

x1, x2, S1, S2, S3, A1 > 0.Phase I :Consider the new objective,Maximize Z* = A1Subject to 2x1 + x2 S1 + A1 = 2x1 + 3x2 + S2 = 2x2 + S3 = 4,x1, x2, S1, S2, S3, A1 > 0.



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OPERATIONS RESEARCH - MB0032Solving by Simplex method, the initial simplex table is given by

x1 enters the basic set replacing A1.

The first iteration gives the following table:

Phase I is complete, since there are no negative elements in the last row.

The Optimal solution of the new objective is Z* = 0.Phase II:Consider the original objective function,Maximize z = 3x1 x2 + 0S1 + 0S2 + 0S3Subject to

x2 + S3 = 4x1, x2, S1, S2, S3 > 0



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OPERATIONS RESEARCH - MB0032with the initial solution x1 = 1, S2 = 1, S3 = 4, the corresponding simplextable is

Proceeding to the next iteration, we get the following table:

Since all elements of the last row are non negative, the current solution isoptimal.

The maximum value of the objective functionZ = 6 which is attained for x1 = 2, x2 = 0.4. What do you understand by the transportation problem? Whatis the basic assumptionbehind the transportation problem? Describe the MODI method of solving transportationproblem.Answer: This model studies the minimization of the cost of transporting acommodity from anumber of sources to several destinations. The supply at each source andthe demand at eachdestination are known. The transportation problem involves m sources,each of which hasavailable ai (i = 1, 2, ..,m) units of homogeneous product and ndestinations, each of whichrequires bj (j = 1, 2., n) units of products. Here ai and bj are positiveintegers. The cost cij of

transporting one unit of the product from the i th source to the j thdestination is given for each i



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OPERATIONS RESEARCH - MB0032and j. The objective is to develop an integral transportation schedule thatmeets all demands fromthe inventory at a minimum total transportation cost.It is assumed that the total supply and the total demand are equal.

The condition (1) is guaranteed by creating either a fictitious destinationwith a demand equal tothe surplus if total demand is less than the total supply or a (dummy)source with a supply equalto the shortage if total demand exceeds total supply. The cost of

transportation from the fictitiousdestination to all sources and from all destinations to the fictitious sourcesare assumed to be zeroso that total cost of transportation will remain the same.The Transportation Algorithm (MODI Method)

The first approximation to (2) is always integral and therefore always afeasible solution. Ratherthan determining a first approximation by a direct application of thesimplex method it is moreefficient to work with the table given below called the transportationtable. The transportationalgorithm is the simplex method specialized to the format of table itinvolves:i) finding an integral basic feasible solutionii) testing the solution for optimalityiii) improving the solution, when it is not optimaliv) repeating steps (ii) and (iii) until the optimal solution is obtained.

The solution to T.P is obtained in two stages. In the first stage we findBasic feasible solution byany one of the following methods a) Northwest corner rale b) MatrixMinima Method or least

cost method c) Vogels approximation method. In the second stage wetest the B.Fs for itsoptimality either by MODI method or by stepping stone method.Modified Distribution Method / Modi Method / U V Method .Step 1 : Under this method we construct penalties for rows and columnsby subtracting the leastvalue of row / column from the next least value.Step 2 : We select the highest penalty constructed for both row andcolumn. Enter that row /column and select the minimum cost and allocate min (ai, bj)Step 3 : Delete the row or column or both if the rim availability /

requirements is met.Step 4 : We repeat steps 1 to 2 to till all allocations are over.



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OPERATIONS RESEARCH - MB0032Step 5 : For allocation all form equation ui + vj = cj set one of the dualvariable ui / vj to zero andsolve for others.Step 6 : Use these value to find ij = cij ui vj of all ij >, then it is theoptimal solution.Step 7 : If any Dij 0, select the most negative cell and form loop.Starting point of the loop is+ve and alternatively the other corners of the loop are ve and +ve.Examine the quantitiesallocated at ve places. Select the minimum. Add it at +ve places andsubtract from ve place.Step 8 : Form new table and repeat steps 5 to 7 till ij > 05. Describe the North-West Corner rule for finding the initial basicfeasible solution in the

transportation problem.Answer: North West Corner RuleStep1: The first assignment is made in the cell occupying the upper lefthand (north west) cornerof the transportation table. The maximum feasible amount is allocatedthere, that is x11 = min(a1,b1)So that either the capacity of origin O1 is used up or the requirement atdestination D1 is satisfiedor both. This value of x11 is entered in the upper left hand corner (smallsquare) of cell (1, 1) inthe transportation tableStep 2: If b1 > a1 the capacity of origin O, is exhausted but therequirement at destination D1 isstill not satisfied , so that at least one more other variable in the firstcolumn will have to take on apositive value. Move down vertically to the second row and make thesecond allocation of magnitudex21 = min (a2, b1 x21) in the cell (2,1). This either exhausts thecapacity of origin O2 or

satisfies the remaining demand at destination D1.If a1 > b1 the requirement at destination D1 is satisfied but the capacityof origin O1 is notcompletely exhausted. Move to the right horizontally to the secondcolumn and make the secondallocation of magnitude x12 = min(a1 x11, b2) in the cell (1, 2) . This either exhausts the remainingcapacity of origin O1 orsatisfies the demand at destination D2 .If b1 = a1, the origin capacity of O1 is completely exhausted as well asthe requirement at

destination is completely satisfied. There is a tie for second allocation, Anarbitrary tie breaking



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OPERATIONS RESEARCH - MB0032choice is made. Make the second allocation of magnitude x12 = min (a1 a1, b2) = 0 in the cell(1, 2) or x21 = min (a2, b1 b2) = 0 in the cell (2, 1).Step 3: Start from the new north west corner of the transportation tablesatisfying destinationrequirements and exhausting the origin capacities one at a time, movedown towards the lowerright corner of the transportation table until all the rim requirements aresatisfied.6. Describe the Branch and Bound Technique to solve an I.P.P.problem.Answer: The Branch And Bound TechniqueSometimes a few or all the variables of an IPP are constrained by theirupper or lower bounds or

by both. The most general technique for the solution of such constrainedoptimization problems isthe branch and bound technique. The technique is applicable to both allIPP as well as mixedI.P.P. the technique for a maximization problem is discussed below:Let the I.P.P. be

Subject to the constraints

xj is integer valued , j = 1, 2, .., r (< n) (3) xj > 0 . j = r + 1, .., n (4)Further let us suppose that for each integer valued xj, we can assign lowerand upper bounds forthe optimum values of the variable byLj xj Uj j = 1, 2, . r (5)

The following idea is behind the branch and bound techniqueConsider any variable xj, and let I be some integer value satisfying Lj I Uj 1 . Then clearlyan optimum solution to (1) through (5) shall also satisfy either the linearconstraint.

x j > I + 1 ( 6)Or the linear constraint xj I ...(7)

To explain how this partitioning helps, let us assume that there were nointeger restrictions (3),and suppose that this then yields an optimal solution to L.P.P. (1), (2),(4) and (5). Indicating x1= 1.66 (for example). Then we formulate and solve two L.P.Ps eachcontaining (1), (2) and (4).

But (5) for j = 1 is modified to be 2 x1 U1 in one problem and L1 x1 1 in the other.



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OPERATIONS RESEARCH - MB0032Further each of these problems process an optimal solution satisfyinginteger constraints (3)

Then the solution having the larger value for z is clearly optimum for thegiven I.P.P. However, itusually happens that one (or both) of these problems has no optimalsolution satisfying (3), andthus some more computations are necessary. We now discuss step wisethe algorithm thatspecifies how to apply the partitioning (6) and (7) in a systematic mannerto finally arrive at anoptimum solution.We start with an initial lower bound for z, say z (0) at the first iterationwhich is less than or equalto the optimal value z*, this lower bound may be taken as the starting Lj

for some xj.In addition to the lower bound z (0) , we also have a list of L.P.Ps (to becalled master list)differing only in the bounds (5). To start with (the 0 th iteration) themaster list contains a singleL.P.P. consisting of (1), (2), (4) and (5). We now discuss below, the step bystep procedure thatspecifies how the partitioning (6) and (7) can be applied systematically toeventually get anoptimum integer valued solution.

Set 2

Q1: Compare and contrast NPV with IRR.

ANS 1.

Net present value method The cash inflow in different years are discounted (reduced) to their present valueby applying the appropriate discount factor or rate and the gross or total presentvalue of cash flows of different years are ascertained. The total present value of cash inflows are compared with present value of cash outflows (cost of project)and the net present value or the excess present value of the project and thedifference between total present value of cash inflow and present value of cashoutflow is ascertained and on this basis, the various investments proposals areranked.



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OPERATIONS RESEARCH - MB0032Cash inflow = earnings / profits of an investment after taxes but beforedepreciation

The present value of cash outflows = initial cost of investment and the commentof project at various points of time ^

Decision rule After ranking various investments proposals on basis on netpresent value, projects with negative net present value (net present value of cash inflows less than their original costs) are rejected and projects with positiveNPV are considered acceptable. In case of mutually exclusive alternativeprojects, projects with higher net present value are selected. Net present valuemethod is suitable for evaluating projects where cash flows are uneven.

Merits

1. The most significant advantage is that it explicitly recognizes the time value of money, e.g., total cash flows pertaining to two machines are equal but the netpresent value are different because of differences of pattern of cash streams.

The need for recognizing the total value of money is thus satisfied.

2. It also fulfills the second attribute of a sound method of appraisal. In that itconsiders the total benefits arising out of proposal over its life time.

3. It is particularly useful for selection of mutually exclusive projects.4. This method of asset selection is instrumental for achieving the objective of financial management, which is the maximization of the shareholder's wealth. Inbrief the present value method is a theoretically correct technique in the

selection of investment proposals.

Demerits

1. It is difficult to calculate as well as to understand and use, in comparison withpayback method or average return method.

2. The second and more serious problem associated with present value methodis that it involves calculations of the required rate of return to discount the cashflows. The discount rate is the most important element used in the calculation of the present value because different discount rates will give different presentvalues. The relative desirability of a proposal will change with the change of discount rate. The importance of the discount rate is thus obvious. But thecalculation of required rate of return pursuits serious problem. The cost of capitalis generally the basis of the firm's discount rate. The calculation of cost of capitalis very complicated. In fact there is a difference of opinion even regarding theexact method of calculating it.

3. Another shortcoming is that it is an absolute measure. This method will acceptthe project which has higher present value. But it is likely that this project mayalso involve a larger initial outlay. Thus, in case of projects involving differentoutlays, the present value may not give dependable results.

4. The present value method may also give satisfactory results in case of twoprojects having different effective lives. The project with a shorter economic life



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OPERATIONS RESEARCH - MB0032is preferable, other things being equal. It may be that, a project which has ahigher present value may also have a larger economic life, so that the funds willremain invested for longer period while the alternative proposal may haveshorter life but smaller present value. In such situations the present valuemethod may not reflect the true worth of alternative proposals. This method issuitable for evaluating projects whose capital outlays or costs differ significantly.

Internal rate of return method The technique is also known as yield oninvestment, marginal efficiency value of capital, marginal productivity of capital,rate of return, time adjusted rate of return and so on. Like net present value,internal rate of return method also considers the time value of money fordiscounting the cash streams. The basis of the discount factor however, isdifficult in both cases. In the net present value method, the discount rate is therequired rate of return and being a predetermined rate, usually cost of capitaland its determinants are external to the proposal under consideration. Theinternal rate of return on the other hand is based on facts which are internal tothe proposal. In other words, while arriving at the required rate of return forfinding out the present value of cash flows, inflows and outflows are notconsidered. But the IRR depends entirely on the initial outlay and cash proceedsof project which is being evaluated for acceptance or rejection. It is thereforeappropriately referred to as internal rate of return. The IRR is usually, the rate of return that a project earns. It is defined as the discount rate which equates theaggregate present value of net cash inflows (CFAT) with the aggregate presentvalue of cash outflows of a project. In other words it is that rate which gives thenet present value zero. IRR is the rate at which the total of discounted cashinflows equals the total of discounted cash outflows (the initial cost of investment). It is used where the cost of investment and its annual cash inflows

are known but the rate of return or discounted rate is not known and is requiredto be calculated.

Accept / Reject decision The use of IRR as a criterion to accept capital investment decision involves acomparison of actual IRR with required rate of return, also known as cut off rateor hurdle rate. The project should qualify to be accepted if the internal rate of return exceeds the cut off rate. If the internal rate of return and the required rateof return be equal, the firm is indifferent as to accept or reject the project. Incase of mutually exclusive or alternative projects, the project which has thehighest IRR will be selected provided its IRR is more than the cut off rate. In casethere are budget constraints, the projects are ranked in descending order of theirIRR and are selected subject to provisions.

Evaluation of IRR

1. Is a theoretically correct technique to evaluate capital expenditure decision. Itpossesses the advantages which are offered by the NPV criterion such as, itconsiders the time value of money and takes into account the total cash inflowsand outflows.

2. In addition, the IRR is easier to understand. Business executives and non-

technical people understand the concept of IRR much more readily than theyunderstand the concept of NPV. For instance, Business X will



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OPERATIONS RESEARCH - MB0032understand the investment proposal in a better way if it is said that the total IRRof Machine B is 21% and cost of capital is 10% instead of saying that NPV of Machine B is Rs. 15,396.3. It itself provides a rate of return which is indicative of profitability of proposal.

The cost of capital enters the calculation later on.

4. It is consistent with overall objective of maximizing shareholders wealth.According to IRR, the acceptance / rejection of a project is based on acomparison of IRR with required rate of return. The required rate of return is theminimum rate which investors expect on their investment. In other words, if theactual IRR of an investment proposal is equal to the rate expected by theinvestors, the share prices will remain unchanged. Since, with IRR, only suchprojects are accepted which have IRR of the required rate, therefore, the shareprices will tend to rise. This will naturally lead of maximization of shareholderswealth. ^

The IRR suffers from serious limitations:

1. It involves tedious calculations. It involves complicated computation problems.

2. It produces multiple rates which can be confusing. This situation arises in thecase of non-conventional projects.

3. In evaluating mutually exclusive proposals, the project with highest IRR wouldbe picked up in exclusion of all others. However in practice it may not turn out tobe the most profitable and consistent with the objective of the firm i.e.,maximization of shareholders wealth.

4. Under IRR, it is assumed that all intermediate cash flows are reinvested at theIRR. It is rather ridiculous to think that the same firm has the ability to reinvestthe cash flows at different rates. The reinvestment rate assumption under theIRR is therefore very unrealistic. Moreover it is not safe to assume always thatintermediate cash flows from the project may be reinvested at all. A portion of cash inflows may be paid out as dividends, a portion may be tied up with currentassets such as stock, cash, etc. Clearly, the firm will get a wrong picture of theproject if it assumes that it invests the entire intermediate cash proceeds.

Further it is not safe to assume that they will be reinvested at the same rate of return as the company is currently earning on its capital (IRR) or at the currentcost of capital (k).

NPV versus IRR NPV indicates the excess of the total present value of futurereturns over the present value of investments. IRR (or DFC rate) indicates on theother hand the rate at which the cash flows (at present values) are generated inthe business by a particular project.Both NPV and IRR iron out the difference due to interest factor or say higherreturns in earlier years and higher returns in later years (though the total returnsin absolute terms may be around the same for several projects).Between the two, IRR or DFC rate is the more sophisticated method a popularas well, since:

(a) IRR method -mostly subjective decision regarding discounting rate. ^



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(b) Whilst under NPV the main basis of comparison is between different NPV's of different projects, under IRR or DFC rate approach a number of basis is available.For example

DFC rate Vs Discount rate of return (on normal operations) ^ DFC rate Vs Cutoff rate of the company DFC rate Vs Borrowing rate (on cost of capital) DFCrates between different projects

(c) The results under DFC rate approach are simpler for the management tounderstand and appreciate. We should however be very careful in applying thedecision rules properly when NPV and IRR calculation shows divergent results.

The rules are

(i) the projects be the basis of decision when mutually exclusive in character;

(ii) there is capital rationing situation

(d) IRR should be a better guide when there are plenty of project situations (as itis there in a long enterprise) and no major constraints (for example, in respect of macro projects).



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OPERATIONS RESEARCH - MB0032Q2: Zodiac Ltd is considering purchase of investment worth Rs. 40lakhs. The estimated life and the new cash flow fir 3 years are as under.

MachinesA

B C

Estimated life 3 Years 3 Years 3yearsCash inflows(in lakhs)1 Year 27 06 122 Year 18 21 803 Year 55 33 30

Ans 2

A) Payback period

Machine A. Rs.27 lakhs will be recovered in 1st year & the balance 13 lakhs (40 27) will be recovered in 2nd year of 18 lakhs

Payback period = 1year +(13/18*12month) or =1year +13/181 year 8.6 month = 1year + 0.72

1.72year

machine B. Rs.06 lakhs will be recovered in 1st year & the balance 34 lakhs(40 6) will be recovered in 2nd year of 21 lakhspayback period = 1year +(35/21*12month) or =1year + 35/21

= 1year 20 month = 1year +1.662.66year

machine c. Rs.12 lakhs will be recovered in 1st year & the balance 28 lakhs(40 12) will be recovered in 2nd year of 80 lakhspayback period = 1year + (28/80*12month) or =1year + 28/80= 1year + 4.2 month = 1year + 0.35

= 1.35year



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machine A. Rs. 35.4546 will be recovered in 2nd year & balance 4.5454(40-35.4546) will be recovered in 3rd year out of 39.149

=2year + (4.5454/39.149)=2year + 0.1161

=2.11year

machine A. Rs. 22.0968 will be recovered in 2nd year & balance 17.9032(40-22.0968) will be recovered in 3rd year out of 23.489

=2year + (17.9032/23.489)=2year +0.7621= 2.76year

machine A. Rs. 74.4936 will be recovered in 2nd year & balance -34.4936(40- 74.4936) will be recovered in 3rd year out of 95.8436

=2year + (-34.4936/95.8436)=2year + -0.3598= 1.64year



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OPERATIONS RESEARCH - MB0032Q3: The cash flow stream of Nanotech Ltd, is as follows:

years 0 1 2 3 4 5 6Cash flows(in millions)

-120 -100 40 60 80 100 130

The cost of capital is 13% find MIRR.

Ans 3:

Present value of cost = 120 * 100/1.13= 194.69

Terminal value of cash flow4 3 2

= 40(1.13) + 60(1.13) +80(1.13)+100(1.13)+130= 40*1.6305 + 60*1.4429 + 80*1.2769 + 113 + 130=496.95

MIRR is obtain on solving the following equation.6194.69 = 496.95/(1+mirr)6(1+mirr) = 496.95/194.696(1+mirr) = 2.5525

Q3: Elaborate different sources of risk in a projectANS. Risk in the project are many. It is possible to identify three separate anddistinct type of risk in any project.1. Stand alone risk : it is measured by the variability of expected returns of theproject2. portfolio risk : a firm can be viewed as portfolio of projects having a certaindegree of risk. When new project is added to the existing portfolio of project, therisk profile of the firm will alter. The degree of the change in the risk depends onthe existing portfolio of theprojects. If the return from the new project is negatively correlated with thereturn from portfolio, the risk of the firm will be further diversified away.3. market or beta risk: it is measured by the effect of the project on the beta of the firm. The market risk for a project is difficult to estimate. Stand alone risk of a project when the project is considered in isolation. Corporate risk is theprojects risks of the firm. Market risk is systematic risk. The market risk is themost important risk because of the direct influence it has on stock prices.

Source of risk: the source of risk are1. project specific risk2. competitive or competition risk

3. industry - specific risk4. international risk



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OPERATIONS RESEARCH - MB00325. market risk

1. project specific risk: the source of this risk could be traced to somethingquite specific to the project. Managerial deficiencies or error in estimation of cash flow or discount rate may lead to a situation of actual cash flow relisedbeing less than that projected.2. competitive risk or competition risk: unanticipated of a firms competitors willmaterially affect the cash flows expected from a project. Because of this theactual cash flow from a project will be less than that of the forecast.3. industry- specific : industry specific risks are those that affect all the firms inthe industry. It could be again grouped in to technological risk, commodity riskand legal risk. All these risks will affect the earnings and cash flows of theproject. The changes in technology affect all the firms not capable of adaptingthemselves to emerging new technology. The best example is the case of firmmanufacturing motors cycles with two stroke engines. When technologicalinnovation replaced the two stroke engines by the four stroke engines thosefirms which could not adapt to new technology had to shut down theiroperations. Commodity risk is the arising from the affectof price changes on goods produced and marketed. Legal risk arise fromchanges in laws and regulations application to the industry to which the firmbelongs. The best example is the imposition of service tax on apartments by thegovernment of India when the total numberof apartments built by a firm engaged in that industry exceeds a prescribed limit.Similarly changes in import export policy of the government of India have ledto the closure of some firms or sickness of some firms.4. international risk : these types of risk are faced by firms whose businessconsists mainly of exports or those who procure their main raw material from

international markets. For example, rupee dollar crisis affected the software andBPOs because it drastically reduce their profitability. Another best example isthat of the textile units in Tirupur in Tamilnadu, exporting their major part of thegarments produces. Rupee gaining and dollar weakening reduced theircompetitiveness in the global markets.



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Q4: The expected cash flow of Tejaswini Ltd, are an follow. (5 Marks)

Year Cash flow0 (50000)1 90002 80003 70004 120005 21000

The certainty equivalent factor balance as per the following equation t=1-05.05t. Calculate theNPV of the project if the risk free rate of return is 9%.

Ans 4:



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Q5: From the following information prepare cash budgets for VSI Co. Ltd.:

Particulars Jan Feb March AprilOpening cash balanceCollection from customer Payments :Raw materials purchaseSalary and WagesOther expensesIncome TaxMachinery

20,0001,30,00025,0001,00,00015,0006,000---

1,60,00045,0001,05,00010,000--------

1,65,00040,0001,00,00015,000----20,000

2,30,000632001,14,20012,000------

The firm wants to maintain a minimum cash balance of Rs.25000 foreach month. Creditors are allowed one-month credit. There is no lag inpayment of salary, other expenses.

Ans 5:

For JAN:Bank over Draft 21,000 for maintain minimum cash balance for

month of

February 21000+4000=25000.For FEB:

No need to take bank over Draft because closing balance 25,000.For MARCH:

Bank over Draft 10,000 for maintain minimum cash balance formonth of April 15,000+10,000=25000.

At last closing balance month of April 65,600 so return to bank 31,000form month

of april , 65,600 31,000= 34,400



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Case StudyAssume you are an external financial consultant. You have beenapproached by a client M/s Technotron Ltd to give a presentation onCredit Policy adopted by Information technology companies. You arespecifically asked to deal with credit standards, credit period, cashdiscounts and collection programme. For the sake of simplicity take anytwo information technology companies and analyze the credit policyfollowed by them.

ANS:-Credit policy Variables1. Credit standards.2. Credit period.3. Credit discount and

4. Collection programme.1. Credit standards : The term credit standards refer to the criteria forextending credit to customers. The bases for setting credit standards are.a. Credit ratingb. Referencesc. Average payment periodd. Ratio analysis

There is always a benefit to the company with the extension of credit to itscustomers but with the associated risks of delayed payment or non payment,funds blocked in receivables etc. The firm may have light credit standards. It

may sell on cash basis and extend credit only to financially strong customers.Such strict credit standards will bring down bad debt losses andreduce the cost of credit administration. But the firm may not be able to increaseits sales. The profit on lost sales may be more the costs saved by the firm. Thefirm should evaluate the trade off between cost and benefit ofany creditstandards.

2. Credit period: Credit period refer to the length of time allowed to itscustomers by a firm to make payment for the purchase made by customers of the firm. It is generally expressed in days like 15 days or 20 days. Generally,firms give cash discount if payment are made within the specified period. If afirm follows a credit period of net 20 it means that it allows to itscustomers 20 days of credit with no inducement for early payments. Increasingthe credit period will bring in additional sales from existing customers and newsales from new customers. Reducing the credit period will lower sales, decreaseinvestments in receivables and reduce the bad debt loss. Increasing the creditperiod increases the incidence of bad debt loss. The effect of increasing thecredit period on profits of the firm are similar to that of relaxing the creditstandards.

3. Cash discount: Firms offer cash discount to induce their customer to makeprompt payments. Cash discount have implications on sales volume, averagecollection period, investment in receivables, incidence of bad debt and profits. A

cash discount of 2/10 net 20 means that a cash discount of 2% is offered if the



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OPERATIONS RESEARCH - MB0032payment is made by the tenth day; other wise full payment will have to be madeby 20 th day.4. Collection programme : The success of a collection programme depends onthe collection policy pursued by the firm. The objective of a collection policy is toachieve. Timely collection of receivable, there by releasing funds locked inreceivables and minimize the incidenceof bad debts. The collection programmes consists of the following.

1. Monitoring the receivables2. Reminding customers about due date of payment3. On line interaction through electronic media to customers about the paymentsdue around the due date.4. Initiating legal action to recover the amount from overdue customer as thelast resort the dues from defaulted customers. Collection policy formulated shallnot lead to bad relationship with customers.

MB0032 Set 1 and Set 2 completed

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