operations research project

7/27/2019 Operations research project

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ACKNOWLEDGEMENT

Operations research project

ON

Transportation Problem

---

As any good work is incomplete without acknowledging the people who made it possible, this

acknowledgement is incomplete without thanking our family, friends, and our faculty, withoutwhose support this project wouldn't have taken shape.

Since we have joined Jaipuria Institute of Management, LUCKNOW we have gained so much


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knowledge, which has been possible due to the well-managed education imparted to us underconditions, which are quite conducive to learning, at our college.

We express our sincere gratitude to Dr. Masood Siddiqui, our teacher of Operations Research, who has helped us clarify our concepts by sharing his valued experiences in his teaching,research and training which have thereby become an unconscious part of our ideas and thoughts

while analyzing the Operations Research project work on Management of Kaiserbagh BusDepot.

Without his sincere help and guidance the project report would have not been a possible.

We thank all our team members who had worked hard to make the report to its present form.

Lastly we would like to thank our families for their continuing support, blessings andencouragement.

Introduction

Linear Programming

In mathematics, linear programming (LP) is a technique foroptimization of a linearobjectivefunction, subject to linear equality and linear inequalityconstraints. Informally, linearprogramming determines the way to achieve the best outcome (such as maximum profit or lowestcost) in a given mathematical model and given some list of requirements represented as linear

equations.

More formally, given a polytope (for example, a polygon or a polyhedron), and a real-valuedaffine function

Defined on this polytope, a linear programming method will find a point in the polytope where thisfunction has the smallest (or largest) value. Such points may not exist, but if they do, searchingthrough the polytope vertices is guaranteed to find at least one of them.

Linear programs are problems that can be expressed in canonical form:

Maximize: ctx

Subject to: Ax


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Represents the vector of variables (to be determined), while and are vectors of (known)coefficients and is a (known) matrix of coefficients. The expression to be maximized or

minimized is called the objective function ( in this case). The equations are theconstraints which specify a convex polyhedron over which the objective function is to beoptimized.

Linear programming can be applied to various fields of study. Most extensively it is used inbusiness and economic situations, but can also be util ized for some engineering problems.Some industries that use linear programming models include transportation, energy,telecommunications, and manufacturing. It has proved useful in modeling diverse types ofproblems in planning, routing, scheduling, assignment, and design.

Linear Programming Assumptions

Linear programming requires linearity in the equations as shown in the above structure. In alinear equation, each decision variable is multiplied by a constant coefficient with no multiplyingbetween decision variables and no nonlinear functions such as logarithms. Linearity requires thefollowing assumptions:

1) Proportionality - a change in a variable results in a proportionate change in that variable'scontribution to the value of the function.

2) Additivity - the function value is the sum of the contributions of each term.

3) Divisibility - the decision variables can be divided into non-integer values, taking on fractionalvalues. Integer programmingtechniques can be used if the divisibil ity assumption does not hold.

In addition to these linearity assumptions, linear programming assumes certainty; that is, that thecoefficients are known and constant.

The Effect of Constraints

Constraints exist because certain limitations restrict the range of a variable's possible values. Aconstraint is considered to be bindingif changing it also changes the optimal solution. Lesssevere constraints that do not affect the optimal solution are non-binding.

Tightening a binding constraint can only worsen the objective function value, and loosening abinding constraint can only improve the objective function value. As such, once an optimalsolution is found, managers can seek to improve that solution by finding ways to relax bindingconstraints.

Route planning

Network arises in numerous settings and in variety of guises. Transportation, electrical networkspervade our daily lives. Many network optimization models are special types of linearprogramming models. Route planning or the shortest path problem is one of them. In this problem
http://en.wikipedia.org/wiki/Convex_set


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we consider an undirected and connected network with 2 special nodes, called source anddestination. Associated with each link is nonnegative distance. The objective is to find theshortest path from the source to the destination. A relatively straightforward algorithm is availablefor this problem. The essence of this procedure is that it fans out from the origin, identifying theshortest path to each node of the network in the ascending order of their shortest distances fromthe origin, thereby solving the problem when destination node is reached.

The Transportation Problem

There is a type of linear programming problem that may be solved using a simplified version ofthe simplex technique called transportation method. Because of its major application in solvingproblems involving several product sources and several destinations of products, this type ofproblem is frequently called the transportation problem. It gets its name from its application toproblems involving transporting products from several sources to several destinations. Althoughthe formation can be used to represent more general assignment and scheduling problems aswell as transportation and distribution problems. The two common objectives of such problemsare either (1) minimize the cost of shipping m units to n destinations or (2) maximize the profit ofshipping m units to n destinations.

Let us assume there are m sources supplying n destinations. Source capacities, destinationsrequirements and costs of material shipping from each source to each destination are givenconstantly. The transportation problem can be described using following l inear programmingmathematical model and usually it appears in a transportation tableau.

There are three general steps in solving transportation problems.

We will now discuss each one in the context of a simple example. Suppose one company hasfour factories supplying four warehouses and its management wants to determine the minimum-cost shipping schedule for its weekly output of chests. Factory supply, warehouse demands, andshipping costs per one chest (unit) are shown below.

Data for Transportation Problem

At first, it is necessary to prepare an initial feasible solution, which may be done in severaldifferent ways; the only requirement is that the destination needs be met within the constraints ofsource supply.

The Transportation Matrix

The transportation matrix is where supply availabili ty at each factory is shown in the far rightcolumn and the warehouse demands are shown in the bottom row. The unit shipping costs areshown in the small boxes within the cells. It is important at this step to make sure that the totalsupply availabilities and total demand requirements are equal. Often there is an excess supply ordemand. In such situations, for the transportation method to work, a dummy warehouse or factory
http://orms.czu.cz/text/TranspTableau.htmlhttp://orms.czu.cz/text/MatModelTransProb.html


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must be added. Procedurally, this involves inserting an extra row (for an additional factory) or anextra column (for an ad warehouse). The amount of supply or demand required by the dummyequals the difference between the row and column totals.

It deals with sources where a supply of some commodity is available and destinations where thecommodity is demanded. The classic statement of the transportation problem uses a matrix withthe rows representing sources and columns representing destinations. The algorithms for solvingthe problem are based on this matrix representation. The costs of shipping from sources to

destinations are indicated by the entries in the matrix. If shipment is impossible between a givensource and destination, a large cost of M is entered. This discourages the solution from usingsuch cells. Supplies and demands are shown along the margins of the matrix. As in the example,the classic transportation problem has total supply equal to total demand.

Matrix model of a transportation problem.

The network model of the transportation problem is shown in Fig. 10. Sources are identified asthe nodes on the left and destinations on the right. Allowable shipping links are shown as arcs,while disallowed links are not included.

Network flow model of the transportation problem.

Only arc costs are shown in the network model, as these are the only relevant parameters. Allother parameters are set to the default values. The network has a special form important in graphtheory; it is called a bipartite network since the nodes can be divided into two parts with all arcsgoing from one part to the other.

On each supply node the positive external flow indicates supply flow entering the network. Oneach destination node a demand is a negative fixed external flow indicating that this amountmust leave the network.


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Optimum solution, z = 46.

Variations of the classical transportation problem are easily handled by modifications of thenetwork model. If links have finite capacity, the arc upper bounds can be made finite. If suppliesrepresent raw materials that are transformed into products at the sources and the demands are inunits of product, the gain factors can be used to represent transformation efficiency at eachsource. If some minimal flow is required in certain links, arc lower bounds can be set to nonzerovalues.


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Problems faced by Kaiserbagh Bus Depot

1). The Director of Roadways, Uttar Pradesh, Knows that the problem of existing temporary busstand in Kaiserbagh is the increased waiting cost on behalf of the customers. It is known thatcustomers arrive at a Poisson process at the rate of 100 per hour. The time required to deal witha customer has an exponential distribution with a mean service time of 30 seconds. The directorfeels that the cost of loss in customer goodwill due to waiting in queue is Rs. 10 per minute.

The diector has been approached with the following two alternatives:-

Proposal 1 is to shift the entire operations to a new location i.e. Old Kaiserbagh Bus Depot.The cost of transfer and designing a new facili ty is Rs. 4.56 crores. Its been assumed thatthe new facility will be operatatble with an estimated life of 10 years. The new facili ty willreduce the average service time to 15 seconds.

Proposal 2 is to shift the entire operations to a less populated area and designing a newfacility with an estimated cost of Rs. 6.25 crores. The new facility will result in reduction inaverage service time to 10 seconds.

The director wants to evaluate the best proposal he should undertake so as to reduce the totalcost of operations.

(The Bus-Stand is operatable for 12 hours in a day for 360 days in a year).

Solution

Inter arrival time() = 100 per hour

Mean service time () = 30 seconds=120 per hour

Cost of waiting(Cw) = 10 per minute =600 per hour

Model used --- (M/M/1):(/GD)


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Total number of passenger waiting in que(Lq) = / (- )

= 100/120(120-100)

= 4.167

Total cost of waiting = LqCw

= 4.167 600

= 2500.2

Proposal 1.






= 100/240(240-100)

= .297


= .297 600

= 178.57

Initial cost =initial investment/10yrs36012=45600000/43200=1055.55

Total cost = Initial cost + Total cost of waiting

=1055.55 + 178.57=1234.12

Proposal 1.



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= 100/360(360-100)

= .107


= .107 600

= 64.10



=1446.75 + 64.10=1510.85

Hence cost is minimum in proposal therefore proposal 1 will be selected.

2). The management of the Kaiserbagh Bus Stand is thinking of inaugurating a new Superfast


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bus service consisting of three new buses from Kaiserbagh Bus Stand. The buses are of threecategories namely Marcopolo, Tata SE202, Echier Super105.The locations are Barabanki,Sitapur and Gonda. The cost structur is as follows:-

Barabanki Sitapur Gonda

Distance 30 85 105

Proposed TicketCharge

20 50 70

The Capacity of buses and the running cost per kilometer is as follows:-

Marcopolo Tata SE202 Echier Super105

Capacity 60 70 80

Cost of Running 10 15 17

Thus, the profit for the respective routes are as follows:-


Marcopolo 900 2150 3150

Tata SE202 950 2225 3625

Echier Super105 1090 2555 3815

Solution and Senstivity Report On excel Sheet.


11/21

ACKNOWLEDGEMENT

Jaipuria University

logo

Operations research project

ON

Transportation Problem


12/21

---

As any good work is incomplete without acknowledging the people who made it possible, this

acknowledgement is incomplete without thanking our family, friends, and our faculty, withoutwhose support this project wouldn't have taken shape.

Since we have joined Jaipuria Institute of Management, LUCKNOW we have gained so muchknowledge, which has been possible due to the well-managed education imparted to us underconditions, which are quite conducive to learning, at our college.

We express our sincere gratitude to Dr. Masood Siddiqui, our teacher of Operations Research, who has helped us clarify our concepts by sharing his valued experiences in his teaching,research and training which have thereby become an unconscious part of our ideas and thoughtswhile analyzing the Operations Research project work on Management of Kaiserbagh BusDepot.

Without his sincere help and guidance the project report would have not been a possible.

We thank all our team members who had worked hard to make the report to its present form.

Lastly we would like to thank our families for their continuing support, blessings andencouragement.

Introduction

Linear Programming

In mathematics, linear programming (LP) is a technique foroptimization of a linearobjective

function, subject to linear equality and linear inequalityconstraints. Informally, linearprogramming determines the way to achieve the best outcome (such as maximum profit or lowestcost) in a given mathematical model and given some list of requirements represented as linearequations.

More formally, given a polytope (for example, a polygon or a polyhedron), and a real-valued
http://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Polyhedronhttp://en.wikipedia.org/wiki/Polygonhttp://en.wikipedia.org/wiki/Polytopehttp://en.wikipedia.org/wiki/Mathematical_modelhttp://en.wikipedia.org/wiki/Constraint_%28mathematics%29http://en.wikipedia.org/wiki/Linear_inequalityhttp://en.wikipedia.org/wiki/Linear_equalityhttp://en.wikipedia.org/wiki/Objective_functionhttp://en.wikipedia.org/wiki/Linearhttp://en.wikipedia.org/wiki/Optimization_%28mathematics%29http://en.wikipedia.org/wiki/Mathematics


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affine function

Defined on this polytope, a linear programming method will find a point in the polytope where thisfunction has the smallest (or largest) value. Such points may not exist, but if they do, searchingthrough the polytope vertices is guaranteed to find at least one of them.

Linear programs are problems that can be expressed in canonical form:

Maximize: ctx

Subject to: Ax


14/21

Route planning

Network arises in numerous settings and in variety of guises. Transportation, electrical networkspervade our daily lives. Many network optimization models are special types of linearprogramming models. Route planning or the shortest path problem is one of them. In this problemwe consider an undirected and connected network with 2 special nodes, called source anddestination. Associated with each link is nonnegative distance. The objective is to find theshortest path from the source to the destination. A relatively straightforward algorithm is availablefor this problem. The essence of this procedure is that it fans out from the origin, identifying theshortest path to each node of the network in the ascending order of their shortest distances fromthe origin, thereby solving the problem when destination node is reached.

The Transportation Problem

There is a type of linear programming problem that may be solved using a simplified version ofthe simplex technique called transportation method. Because of its major application in solvingproblems involving several product sources and several destinations of products, this type ofproblem is frequently called the transportation problem. It gets its name from its application toproblems involving transporting products from several sources to several destinations. Althoughthe formation can be used to represent more general assignment and scheduling problems aswell as transportation and distribution problems. The two common objectives of such problemsare either (1) minimize the cost of shipping m units to n destinations or (2) maximize the profit of

shipping m units to n destinations.

Let us assume there are m sources supplying n destinations. Source capacities, destinationsrequirements and costs of material shipping from each source to each destination are givenconstantly. The transportation problem can be described using following l inear programmingmathematical model and usually it appears in a transportation tableau.

There are three general steps in solving transportation problems.

We will now discuss each one in the context of a simple example. Suppose one company hasfour factories supplying four warehouses and its management wants to determine the minimum-

cost shipping schedule for its weekly output of chests. Factory supply, warehouse demands, andshipping costs per one chest (unit) are shown below.

Data for Transportation Problem
http://orms.czu.cz/text/TranspTableau.htmlhttp://orms.czu.cz/text/MatModelTransProb.html


15/21

At first, it is necessary to prepare an initial feasible solution, which may be done in severaldifferent ways; the only requirement is that the destination needs be met within the constraints ofsource supply.

The Transportation Matrix

The transportation matrix is where supply availabili ty at each factory is shown in the far rightcolumn and the warehouse demands are shown in the bottom row. The unit shipping costs areshown in the small boxes within the cells. It is important at this step to make sure that the totalsupply availabilities and total demand requirements are equal. Often there is an excess supply ordemand. In such situations, for the transportation method to work, a dummy warehouse or factorymust be added. Procedurally, this involves inserting an extra row (for an additional factory) or anextra column (for an ad warehouse). The amount of supply or demand required by the dummyequals the difference between the row and column totals.

It deals with sources where a supply of some commodity is available and destinations where thecommodity is demanded. The classic statement of the transportation problem uses a matrix with

the rows representing sources and columns representing destinations. The algorithms for solvingthe problem are based on this matrix representation. The costs of shipping from sources todestinations are indicated by the entries in the matrix. If shipment is impossible between a givensource and destination, a large cost of M is entered. This discourages the solution from usingsuch cells. Supplies and demands are shown along the margins of the matrix. As in the example,the classic transportation problem has total supply equal to total demand.

Matrix model of a transportation problem.

The network model of the transportation problem is shown in Fig. 10. Sources are identified asthe nodes on the left and destinations on the right. Allowable shipping links are shown as arcs,while disallowed links are not included.

Network flow model of the transportation problem.


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Only arc costs are shown in the network model, as these are the only relevant parameters. Allother parameters are set to the default values. The network has a special form important in graphtheory; it is called a bipartite network since the nodes can be divided into two parts with all arcsgoing from one part to the other.

On each supply node the positive external flow indicates supply flow entering the network. Oneach destination node a demand is a negative fixed external flow indicating that this amountmust leave the network.

Optimum solution, z = 46.

Variations of the classical transportation problem are easily handled by modifications of thenetwork model. If links have finite capacity, the arc upper bounds can be made finite. If supplies

represent raw materials that are transformed into products at the sources and the demands are inunits of product, the gain factors can be used to represent transformation efficiency at eachsource. If some minimal flow is required in certain links, arc lower bounds can be set to nonzerovalues.


17/21

Problems faced by Kaiserbagh Bus Depot

1). The Director of Roadways, Uttar Pradesh, Knows that the problem of existing temporary busstand in Kaiserbagh is the increased waiting cost on behalf of the customers. It is known thatcustomers arrive at a Poisson process at the rate of 100 per hour. The time required to deal witha customer has an exponential distribution with a mean service time of 30 seconds. The directorfeels that the cost of loss in customer goodwill due to waiting in queue is Rs. 10 per minute.

The diector has been approached with the following two alternatives:-

Proposal 1 is to shift the entire operations to a new location i.e. Old Kaiserbagh Bus Depot.The cost of transfer and designing a new facili ty is Rs. 4.56 crores. Its been assumed thatthe new facility will be operatatble with an estimated life of 10 years. The new facili ty willreduce the average service time to 15 seconds.

Proposal 2 is to shift the entire operations to a less populated area and designing a new

facility with an estimated cost of Rs. 6.25 crores. The new facility will result in reduction inaverage service time to 10 seconds.

The director wants to evaluate the best proposal he should undertake so as to reduce the totalcost of operations.

(The Bus-Stand is operatable for 12 hours in a day for 360 days in a year).

Solution


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= 100/120(120-100)

= 4.167


= 4.167 600

= 2500.2

Proposal 1.






= 100/240(240-100)

= .297


= .297 600

= 178.57




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=1055.55 + 178.57=1234.12

Proposal 1.






= 100/360(360-100)

= .107


= .107 600

= 64.10



=1446.75 + 64.10=1510.85

Hence cost is minimum in proposal therefore proposal 1 will be selected.


20/21

2). The management of the Kaiserbagh Bus Stand is thinking of inaugurating a new Superfastbus service consisting of three new buses from Kaiserbagh Bus Stand. The buses are of threecategories namely Marcopolo, Tata SE202, Echier Super105.The locations are Barabanki,Sitapur and Gonda. The cost structur is as follows:-


Distance 30 85 105

Proposed TicketCharge

20 50 70

The Capacity of buses and the running cost per kilometer is as follows:-

Marcopolo Tata SE202 Echier Super105

Capacity 60 70 80

Cost of Running 10 15 17

Thus, the profit for the respective routes are as follows:-


Marcopolo 900 2150 3150

Tata SE202 950 2225 3625

Echier Super105 1090 2555 3815

Solution and Senstivity Report On excel Sheet.


21/21

operations research project

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