INTRODUCTION

Transportation models play an important role in logistics and supply chain management for reducing cost and improving service. Therefore, the goal is to find the most cost effective way to transport the goods. Transportation problems are among the most pressingstrategic developmentproblems in many cities, often a major constraint for long-term urban development ingeneral, and very closely related to land development, economicstructure,energypolicies, and environmental quality. Since all citizens are either enjoying thetransportation system or, and often at the same time, suffering from it, it is an important element of the urban quality of life. The transportation problems is generally to be solved ,deals with inefficiency of urban transportation systems and underlying land use patterns, whichnegatively affect quality of life, economic efficiency, and the environment; the high(and often hidden) costs of urban transportation in both socio-economic andenvironmental terms; and in particular the environmental consequences both in termsof physical aspects that include land and resource use, ecological aspects, andhuman health problems.Efficient tools for comprehensive strategic analysis that are directly useful to cityadministrations are lacking. New strategies for sustainable mobility require wellbalanced combinations of measures with impacts on

improved land-use/economic development planning; improved planning, management and use of transport

infrastructures andfacilities; incorporation of the real costs of both infrastructure and environment in investment policies and decisions and also in user costs;

development of public transport and improvement of its competitive position ,continued technical improvement of vehicles and fuels.

incentives for the use of less polluting fuels; promotion of a more environmentally rational use of the private car, including behavioural changes.

These problems can only be addressed with a consistent and comprehensiveapproach and planning methodology that helps to design strategies for sustainablecities. This has to include an integration of socio-economic, environmental andtechnological concepts including the development, integration,and demonstration ofmethodologies to improve forecasting, assessment and strategic policy level decision

Mathematical Model of Transportation Problem Mathematically a transportation problem is nothing but a special linear programming problem in which the objective function is to minimize the cost of transportation subjected to the demand and supply constraints. Let ai = quantity of the commodity available at the origin i, bj = quantity of the commodity needed at destination j, cij = transportation cost of one unit of a commodity from origin I to destination j, and xij = quantity transported from origin I to the destination j. Mathematically, the problem is Minimize z = ΣΣ xij cij

S.t. Σxij = ai, i= 1,2,…..m Σxij = bj, j= 1,2,…..,n

and xij ≥ 0 for all i and j .

Let us consider an example to understand the formulation of mathematical model of transportation problem of transporting single commodity from three sources of supply to four demand destinations. The sources of supply can be production facilities, warehouse or supply point, characterized by available capacity. The destination are consumption facilities, warehouse or demand point, characterized by required level of demand.

FORMULATION OF TRANSPORATATION PROBLEM AS A LINEAR PROGRAMMING MODEL Let P denote the plant (factory) where the goods are being manufactured & W denote the warehouse (godown) where finished products are stored by the company before shipping to various destinations. Further let, xij = quantity (amount of goods) shipped from plant Pi to the warehouse Wj, and Cij = transportation cost per unit of shipping from plant Pi to the Warehouse Wj.

Objective-function. The objection function can be represented as:

Minimize Z = c11x11 + C12x12 + C13x13 (i.e. cost of shipping + c21x21 + c22x22 + c23x23 from a plant + c31x31 + c32x32 + c33x33 to the ware house)

Supply constraints. x11 + x12 + x13 = S1

x21 + x22 + x23 = S2

x31 + x32 + x33 = S3

Demand constraints.

x11 + x21 + x31 = D1

x21 + x22 + x23 = D2

x31 + x32 + x33 = D3

Either, xij ≥ for all values of I and j (ie; x11, x12, … all such values are ≥ 0) .It is further assumed that: S1 + S2 + S3 = D1 + D2 + D3

The total supply available at the plants exactly matches the total demand at the destinations. Hence, there is neither excess supply nor excess demand. Such type of problems where supply and demand are exactly equal are known as Balanced Transportation Problem. In general, if a transportation problem has m rows an n column, then the problem is solvable if there are exactly (m + n –1) basic variables. A transportation problem is said to be unbalanced if the supply and demand are not equal. If Supply < demand, a dummy supply variable is introduced in the equation to make it equal to demand ,Likewise, if demand < supply, a dummy demand variable is introduced in the equation to make it equal to supply.

points to note:-

1) Total supply = total demand then it is a balanced transportation problem, otherwise it is a unbalanced problem.

2) The unbalanced problem can be balanced by adding a dummy supply center (row) or a dummy demand center (column) as the need arises.

3) When the number of positive allocation at any stage of feasible solution is less than the required number (row +Column – 1) the solution is said to be degenerate otherwise non-degenarete.

The solution algorithm to a transportation problem can be summarized into following steps: Step 1. Formulate the problem and set up in the matrix form.

The formulation of transportation problem is similar to LP problem formulation. Here the objective function is the total transportation cost and the constraints are the supply and demand available at each source and destination, respectively. Step 2. Obtain an initial basic feasible solution. This initial basic solution can be obtained by using any of the following methods:

i. North West Corner Rule

It is a simple and an efficient method to obtain an initial solution.This method does not take into account the cost of transportation on any route of transportation. This method can be summarized as follows :

Step 1: Start with the cel at the upper left (north-west) corner of the transportation matrix and allocate as much as possible equal to the minimum of the rim values for the first row and first column, i.e. min (a1, b1).

Step 2: (a) If allocation made in step 1 is equal to the capacity of the first source (a1, in first row), then move vertically downward to the cell (2,1) in the second row and first column and apply step1 again, for next allocation.

(b) If alcation made in step1 is equal to the first destination (b1, in first column), then move horizontally to the cell (1,2) in the first row and second column and apply step1 again for next allocation.

(c) If a1 = b1, allocate x11 = a1, or b1 and move diagonally to the cell (2,2).

Step 3: Continue the procedure step by step till an allocation is made ion the south-east corner cell of the transportation table.

ii. Vogel Approximation Method

Vogel’s approximation Method (penalty or regret method) is a heuristic method and is preferred to the other two methods described above. In this method each allocation is made on the

basis of the opportunity (or penalty or extra) cost that would have incurred if allocation in certain cells with minimum unit transportation cost were missed. In this method allocations are made so that the penalty cost is minimized. The advantage of this method is that it gives an initial solution which is nearer to an optimum solution or is the optimum solution itself.The steps in VAM are as follows:

Step1: calculate penalties for each (column) by taking the difference between the smallest and next smallest unit transportation cost in the same row (column).This difference indicates the penalty or extra cost which has to be paid if one fails to allocate to the cell with the minimum unit transportation cost.

Step2: Select the row or column with the largest penalty and allocate as much as possible in the cell having the least cost in the selected row or column satisfying the rim conditions. If there is a tie in the values of penalties then it can be broken by selecting the cell where maximum allocation can be made.

Step3: Adjust the supply and demand and cross out the satisfied row or column. If a row and a column are satisfied simultaneously, only one of them is crossed out and the remaining row (column) is assigned a zero supply (demand). Any row or column with zero supply or demand should not be used in computing future penalties.

Step4: Repeat step1 to 3 until the entire available supply at various sources and demand at various destinations are satisfied.

(iii) Least cost method

Since our abjective is to minimize the total transportation cost , we musttry to transport as must as possible through those routes (cells)where the unit transportation cost is minimum.

This method takes in to account the minimum unit cost and can be summarized as follows:

Step1:select the cell with the smallest unit cost in the entire transportation table and allocate as much as possible to this cell and eliminate (line out )that row or column in which either supply or demand is exhausted.if both a row and column are satisfied simultaneously only one may be crossed out .In case the smallest unit cost cell is not unique ,the then select the cell where maximum can be made.

Step2:after adjusting the supply and demand for all uncrossed – out rows and column repeat the procedure with smallest unit cost among the remaining all rows and column of the transportation and allocate as much as possible to this cell and eliminate (line out )that row and column in which either supply and demand is exhausted.

Step3:Repeat the procedure available until the entire available supply at various sources and demand at various destination is satisfied. The solution so abtained need not to be non degenerate.

The solution obtained by any of the above methods must fulfill following conditions:

i. The solution must be feasible, i.e., it must satisfy all the supply and demand constraints. This is called RIM CONDITION.

ii. The number of positive allocations must be equal to m + n – 1, where, m is number of rows and n is number of columns

The solution that satisfies both the above mentioned conditions are called a non-degenerate basic feasible solution.

Step 3. Test the initial solution for optimality. Using any of the following methods can test the optimality of obtained initial basic solution: i. Stepping Stone Method ii. Modified Distribution Method (MODI)

If the solution is optimal then stop, otherwise, determine a new improved solution.

Step 4. Updating the solution repeat Step 3 until the optimal solution is arrived at.

Case : sharma milk products pvt ltd , gujurat , india

A dairy firm has three plants located through out a state .the daily milk production at each plant is as fol;lows :

Plant 1: 6 hundred litres

Plant 2: 1 hundred litres

Plant 3: 10 hundred litres

Each day the firm must fufil the needs of its four distribution centres .minimum requirement at each center is as follows :

Distribution centre 1`: 7 hundred litres

Distruibution centre 2: 5 hundred litres

Distruibution centre 3: 3 hundred litres

Distruibution centre 4: 2 hundred litres

Cost of shipping one hundred litres from each plant to distribution centre is given in the folloing tablr in hundreds rupees:

Distribution centres

PLANT D1 D2 D3 D4

P1 2 3 11 7

P2 1 0 6 1

P3 5 8 15 9

In the above problem we are going to minimize the cost by using 3 methods which are as follows:

A)North West Corner Rule

B)Least Cost Method

C)Vogel’s Approximation Method

Solution : (A) North West Corner Rule

Plant D1 D2 D3 D4 SUPPLY

P1 2

3

11 7

6=a1

P2 1

0

6

1

1=a2

P3 5

8 15 9 10=a3

DEMAND 7=B1 5=B2 3=B3 2=B4

1)Comparing a1 and b1. Since a1 < b1; allocate x11 = 6. This exhaust the supply at p1 and left 1 unit as unsatisfied demand at d1.

2)Move to cell (p2, d1). Comparing a2 and b1 (i.e. 1 and1). Since a2 = b1, therefore allocate x21 =1.

3)Move to cell (p3,d2). Since supply at p3= the demand at d2,d3 and d4, therefore x32 =5,x33 and x34=2.

It may be noted the number of allocated cells (also called basic cells) are 5 which is 1 less than the required number m+m-1(3+1-1=6). Thus this solution is the degenerate solution.The transportation cost associasted with this solution is :

Total cost = Rs. (2x6+1x1+8x5+15x3+9x2)x100

=RS 11600

(B) Least Cost Method

D1 D2 D3 D4 supply

6

1

5 3 2

2

3 11 7 6

1 0

6 1 1

5 8 15 9 10

Demand 7 5 3 2

1)The lowest unit cost 0 is in cell (p2,d2), therefore maximum possible allocation which can be made here is 1. This exhausts the supply at planned p2, therefore row2 is crossed out.

2)The next lowest unit cost is 2 in cell (p1,d1). The maximum possible allocation which can be made here is 6. This exhausts the supply at planned p1, therefore row p1 is crossed out.

3)Since the total supply at p3 is now equal to the unsatisfied demand at all the four distribution centres, therefore maximum possible allocation satisfying the supply and demand conditions are made in cells (p3,d1), (p3,d2), (p3,d3) and (p3,d4).

The number of allocated cells in this case are 6 which is equal to the required number, m+n-1(3+4-1=6). Thus this solution is non degenerate. The transportation cost associated with this solution is

Total cost= Rs. (2x6+5x1+8x4+15x3+9x2)x100 = Rs. 11200.

(C) Vogal’s Approximation Method

D1 D2 D3 D4 Supply row penalty

6

1

4 3 21

2

3 11 7 6 1 1 5 -

1 0

6

1 1 1 - - -

5 8 15

910 3 3 4 4

7 5 3 21 3 5 6

3 5 4 2

3 - 4 2

5 - 15 9

The number of allocated cell in this case is 6 which is equal to the required number m+n-1(3+4-1=6), therefore this solution is nondegenerate. The transportation cost associated with this solution is Total cost = Rs. ( 2x1+3x5+1x1+5x6+15x3+9x1)x100 =Rs 10200

It can be seen that the total transportation cost found by VAM is lower than the cost of transportation determined by the other two methods. Therefore it is of advantage to use this method in order to find an initial basic feasible solution.

CONCLUSION

At last we can conclude that the transportation cost is an important element of the total cost structure for any business. In this modern era of hypercompetitive competition, the key criteria for any business is to achieve high profits and success by reducing its cost. Hence transportation cost being an important and major element of the cost structure need to be reduced (taken care off) to the extent possible.

1 5

3 16

1

The above problem of minimizing the transportation cost has been duly shown in the study, through the use of three different matters which are NORTH-WEST CORNER RULE, LEAST COST METHOD and VOGAL’S APPROXIMATION METHOD, among which Vogal’s Approximation Method is considered the most efficient one in minimizing cost. Through the use of these models a business can identify easily and efficiently how to plan out its transportation, so that it can not only minimize the cost of transporting goods and services but also create time utility by reaching the goods and services at the right place and at the right time.