14155770 Operation Research Project Transportation

7/30/2019 14155770 Operation Research Project Transportation

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Operations research projectON

Transportation Problem


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As any good work is incomplete without acknowledging the people whomade it possible, this acknowledgement is incomplete without thanking our

family, friends, and our faculty, without whose support this project wouldn't

have taken shape.

Since we have joined Jaipuria Institute of Management, LUCKNOW we

have gained so much knowledge, which has been possible due to the well-

managed education imparted to us under conditions, which are quite

conducive to learning, at our college.

We express our sincere gratitude toDr. 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 Bus Depot.

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 and encouragement.


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I ntroduction

Linear P rogramming

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

a linearobjective function, subject to linear equality and linear inequality

constraints. Informally, linear programming determines the way to achieve

the best outcome (such as maximum profit or lowest cost) in a given

mathematical model and given some list of requirements represented as

linear equations.

More formally, given apolytope (for example, apolygon or apolyhedron),

and a real-valued affine function

Defined on this polytope, a linear programming method will find a point in

the polytope where this function has the smallest (or largest) value. Such

points may not exist, but if they do, searching through the polytope vertices

is guaranteed to find at least one of them.

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

Maximize: ctxSubject to: Ax


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Linear Programming Assumptions

Linear programming requires linearity in the equations as shown in the

above structure. In a linear equation, each decision variable is multiplied by

a constant coefficient with no multiplying between decision variables and no

nonlinear functions such as logarithms. Linearity requires the following

assumptions:

1) Proportionality - a change in a variable results in a proportionate change

in that variable's contribution 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-integervalues, taking on fractional values.Integer programmingtechniques can be

used if the divisibility assumption does not hold.

In addition to these linearity assumptions, linear programming assumes

certainty; that is, that the coefficients are known and constant.

The Effect of Constraints

Constraints exist because certain limitations restrict the range of a variable's

possible values. A constraint is considered to be bindingif changing it alsochanges the optimal solution. Less severe constraints that do not affect the

optimal solution are non-binding.

Tightening a binding constraint can only worsen the objective function

value, and loosening a binding constraint can only improve the objective

function value. As such, once an optimal solution is found, managers can

seek to improve that solution by finding ways to relax binding constraints.


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Route planning

Network arises in numerous settings and in variety of guises. Transportation,

electrical networks pervade our daily lives. Many network optimization

models are special types of linear programming models. Route planning or

the shortest path problem is one of them. In this problem we consider an

undirected and connected network with 2 special nodes, called source and

destination. Associated with each link is nonnegative distance. The objective

is to find the shortest path from the source to the destination. A relatively

straightforward algorithm is available for this problem. The essence of this

procedure is that it fans out from the origin, identifying the shortest path to

each node of the network in the ascending order of their shortest distances

from the 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 of the simplex technique called transportation method.

Because of its major application in solving problems involving several

product sources and several destinations of products, this type of problem is

frequently called the transportation problem. It gets its name from its

application to problems involving transporting products from several sources

to several destinations. Although the formation can be used to represent

more general assignment and scheduling problems as well as transportationand distribution problems. The two common objectives of such problems are

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, destinations requirements and costs of material shipping from

each source to each destination are given constantly. The transportation

problem can be described using following linear programming mathematical

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 has four factories supplying four warehouses and its

management wants to determine the minimum-cost shipping schedule for its
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weekly output of chests. Factory supply, warehouse demands, and shipping

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 several different ways; the only requirement is that the destinationneeds be met within the constraints of source supply.

The Transportation Matrix

The transportation matrix is where supply availability at each factory is

shown in the far right column and the warehouse demands are shown in the

bottom row. The unit shipping costs are shown in the small boxes within the

cells. It is important at this step to make sure that the total supply

availabilities and total demand requirements are equal. Often there is an

excess supply or demand. In such situations, for the transportation method towork, a dummy warehouse or factory must be added. Procedurally, this

involves inserting an extra row (for an additional factory) or an extra column

(for an ad warehouse). The amount of supply or demand required by the

dummy equals the difference between the row and column totals.

It deals with sources where a supply of some commodity is available and

destinations where the commodity is demanded. The classic statement of the

transportation problem uses a matrix with the rows representing sources and

columns representing destinations. The algorithms for solving the 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 given source and destination, a large cost of M is

entered. This discourages the solution from using such 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.


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Matrix model of a transportation problem.

The network model of the transportation problem is shown in Fig. 10.

Sources are identified as the 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. All other parameters are set to the default values. The

network has a special form important in graph theory; it is called a bipartite

network since the nodes can be divided into two parts with all arcs going

from one part to the other.

On each supply node the positive external flow indicates supply flow

entering the network. On each destination node a demand is a negative fixed

external flow indicating that this amount must leave the network.


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Optimum solution, z = 46.

Variations of the classical transportation problem are easily handled by

modifications of the network 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 in units of

product, the gain factors can be used to represent transformation efficiency

at each source. If some minimal flow is required in certain links, arc lower

bounds can be set to nonzero values.


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Problems faced by Kaiserbagh Bus Depot

1). The Director of Roadways, Uttar Pradesh, Knows that the problem of existing

temporary bus stand in Kaiserbagh is the increased waiting cost on behalf of thecustomers. It is known that customers arrive at a Poisson process at the rate of 100 per

hour. The time required to deal with a customer has an exponential distribution with a

mean service time of 30 seconds. The director feels 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 facility is Rs. 4.56 crores. Its

been assumed that the new facility will be operatatble with an estimated life of 10

years. The new facility will reduce 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 resultin reduction in average service time to 10 seconds.

The director wants to evaluate the best proposal he should undertake so as to reduce the

total cost 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 hourMean service time () = 30 seconds=120 per hour

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

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

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.

Inter arrival time() = 100 per hour

Mean service time () = 15 seconds=240 per hour

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


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

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

= 100/240(240-100)

= .297Total cost of waiting = LqCw

= .297 600

= 178.57Initial cost =initial investment/10yrs36012=45600000/43200=1055.55

Total cost = Initial cost + Total cost of waiting

=1055.55 + 178.57=1234.12

Proposal 1.

Inter arrival time() = 100 per hourMean service time () = 10 seconds=360 per hour

Cost of waiting(Cw) = 10 per minute =600 per hourModel used --- (M/M/1):(/GD)

Total number of passenger waiting in que(Lq) = / (- )= 100/360(360-100)

= .107

Total cost of waiting = LqCw= .107 600

= 64.10

Initial cost =initial investment/10yrs36012=62500000/43200=1446.75

Total cost = Initial cost + Total cost of waiting

=1446.75 + 64.10=1510.85

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


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

Superfast bus service consisting of three new buses from Kaiserbagh Bus Stand. The

buses are of three categories namely Marcopolo, Tata SE202, Echier Super105.Thelocations are Barabanki, Sitapur and Gonda. The cost structur is as follows:-

Barabanki Sitapur GondaDistance 30 85 105

Proposed Ticket

Charge

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:-

Barabanki Sitapur Gonda

Marcopolo 900 2150 3150

Tata SE202 950 2225 3625

Echier Super105 1090 2555 3815

Solution and Senstivity Report On excel Sheet.

14155770 Operation Research Project Transportation

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