Operation Research Project Transportation

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, 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 wellmanaged education imparted to us under conditions, 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 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.

Introduction Linear Programming In mathematics, linear programming (LP) is a technique for optimization of a linear objective 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 a polytope (for example, a polygon or a polyhedron), 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: ctx Subject to: Ax<=b 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 the constraints which specify a convex polyhedron over which the objective function is to be optimized. Linear programming can be applied to various fields of study. Most extensively it is used in business and economic situations, but can also be utilized 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 of problems in planning, routing, scheduling, assignment, and design.

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-integer values, taking on fractional values. Integer programming techniques 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 binding if changing it also changes 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.

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 transportation and 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

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 destination needs 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 to work, 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.

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.

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.

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 the customers. 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.

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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 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(λ) = 100 per hour Mean service time (µ) = 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) = λ²/ µ(µ- λ) = 100²/120(120-100) = 4.167 Total cost of waiting = Lq×Cw = 4.167 ×600 = 2500.2 Proposal 1. Inter arrival time(λ) = 100 per hour Mean service time (µ) = 15 seconds=240 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) = λ²/ µ(µ- λ) = 100²/240(240-100) = .297 Total cost of waiting = Lq×Cw = .297 ×600 = 178.57 Initial cost =initial investment/10yrs×360×12=45600000/43200=1055.55 Total cost = Initial cost + Total cost of waiting =1055.55 + 178.57=1234.12 Proposal 1. Inter arrival time(λ) = 100 per hour Mean service time (µ) = 10 seconds=360 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) = λ²/ µ(µ- λ) = 100²/360(360-100) = .107 Total cost of waiting = Lq×Cw = .107 ×600 = 64.10 Initial cost =initial investment/10yrs×360×12=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.

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.The locations are Barabanki, Sitapur and Gonda. The cost structur is as follows:Distance Proposed Ticket Charge

Barabanki 30 20

Sitapur 85 50

Gonda 105 70

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

Marcopolo 60 10

Tata SE202 70 15

Echier Super105 80 17

Thus, the profit for the respective routes are as follows:Marcopolo Tata SE202 Echier Super105

Barabanki 900 950 1090

Sitapur 2150 2225 2555

Solution and Senstivity Report On excel Sheet.

Gonda 3150 3625 3815