Transportation Problem Introduction The basic transportation problem was developed in 1941 by F.I. Hitchaxic. However it could be solved for optimally as an answer to complex business problem only in 1951,when Geroge B. Dantzig applied the concept of Linear Programming in solving the Transportation models. Transportation problems are primarily concerned with the optimal (best possible) way in which a product produced at different factories or plants (called supply origins) can be transported to a number of warehouses (called demand destinations). The objective in a transportation problem is:-To fully satisfy the destination requirements within the operating production capacity constraints at the minimum possible cost. Introduction Whenever there is a physical movement of goods from the point of manufacture to the final consumers through a variety of channels of distribution (wholesalers, retailers, distributors etc.), there is a need to minimize the cost of transportation so as to increase the profit on sales. Transportation problems arise in all such cases. It aims at providing assistance to the top management in ascertaining how many units of a Introduction particular product should be transported from each supply origin to each demand destinations to that the total prevailing demand for the company’s product is satisfied, while at the same time the total transportation costs are minimized. Formulation of Transportation Problem as a Linear Programming Model A Transportation problem is simply a special case of LPP. Hence to convert into a LP model, some basic steps are:1. P denote the plant (factory) where the goods are being manufactured & 2. W denote the warehouse (godown) where the finished products are stored by the company before shipping to various destinations. Formulation ….. Xij = no. of units of the product to be transported from factory i to warehouse j or from plant Pi to the warehouse Wj, and Cij = transportation cost per unit of shipping from plant Pi to the Warehouse Wj. Formulation …. i.e.; 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. • Supply (from various sources) are written in the rows, while a column is an expression for the demand of different warehouses. In general, if a transportation problem has m rows an n columns, then the problem is solvable if there are exactly (m + n –1) basic variables. Unbalanced Transportation Problem : A transportation problem is said to be unbalanced if the supply and demand are not equal. Two situations are possible:- 1. If Supply < demand, a dummy supply variable is introduced in the equation to make it equal to demand. 2. If demand < supply, a dummy demand variable is introduced in the equation to make it equal to supply. Example : Formulate the following transportation problem as a LP model: Example Source Designation W1 P1 W2 11 X11 P2 X31 6 15 8 X22 22 X14 X23 13 23 X24 8 X33 22 Supply (capacity) 3 21 4 17 X32 20 W4 X13 X12 X21 Requirement (Demand) 6 7 P3 W3 31 29 X34 25 73 Formulating as a Linear Programming model, we have: Minimise Z = 11X11 + 6X12 + 15X13 + 3X14+ 7X21 + 8X22 +4X23 + 13X24 +22X31 + 17X32 + 8X33 + 31X34 Subject to: Capacity constraints: X11 + X12 + X13 + X14 =21 X21 + X22 + X23 + X24 = 23 X31 + X32 + X33 + X34 = 29 Notation used: Xij = no. of units transported from source i to destination j Requirement constraints: X11 + X21 + X31 = 6 X12 + X22 + X32 = 20 X13 + X23 + X33 = 22 X14 + X24 + X34 = 25 Hence the transportation problem is successfully formulated as a LP model. The various steps for solving a transportation problem are as follows: • From the given problem, express the objective function which must be minimized along with the relevant constraints. • Set up a transportation table, where supply or sources (factories, plants) are denoted as rows and demand or destinations (warehouses, markets) are denoted by columns. • Find out an initial feasible solution by using one of the three rules (discussed later in the chapter). • A solution so obtained is aid to be feasible if and only if, such a solution has allocations in the total number of (m + n – 1) cells. • By ascertaining the opportunity costs, the solution obtained above can be checked for optimality. Opportunity cost may be defined as the cost of sacrifice. Here, it denoted the reduction in cost due to the inclusion of a particular cell in the solution. An optimum solution is obtained when there is a positive opportunity cost for each of the empty cells. Where the solution so obtained is not found to be optimal, that empty cell which results in the largest saving is included and the above steps are again repeated. Initial Feasible Solutions Most commonly used methods are: 1. North-West corner Rule (NWC) 2. Least Cost method (LCM) 3. Vogel’s Approximation method (VAM) Least Cost Method As the name suggests, of all the routes select the one where shipping cost is the least. Now, consider the supply available at the corresponding source and demand at the corresponding destination and put the lower of the two as the quantity to be transported through that route. After that delete the source or destination, whichever is satisfied. Least Cost method….. Consider the remaining routes and again choose the one with the smallest cost and make assignments. Continue in this manner until all the units are assigned. If there is a tie in the min. cost, so that 2 or more routes have the same least cost of shipping, then conceptually either of them may be selected. However a better initial solution is obtained if the route chosen is the one where largest quantity can be transported. Least Cost Method… Thus if there are three cells for which the least cost is equal, then consider all of these one by one and determine the quantity which can be dispatched, and choose the all with the largest quantity. If there is still a tie, then either of them may be selected. Vogel’s Approximation Method (VAM) Like the previous method, VAM also considers the shipping cost, but in a relative sense, when making allocations. First consider the each row of the matrix individually & find the difference between two least cost cells in it. Then repeat this for each of the column. Identify the column or row with the max. difference value. Vogel’s Approximation Method (VAM) Now, consider the cell with min. cost in that row or column and assign the max. units possible. In case of a tie in the largest cost difference, although either of them may be chosen but it is preferable to choose the cost difference corresponding to which the largest no. of units may be assigned or corresponding to which the cell chosen has min. cost. Delete the column or row which is satisfied. Vogel’s Approximation Method (VAM) Again find out the differences and proceed in the same manner. Continue until all units have been assigned. Testing of Optimality The Modified Distribution Method (MODI) This method is an efficient method of testing the optimality of a transportation solution. Step1: Add to the transportation table a column on the RHS titled ui and a row in the bottom of it labelled vj. Step2: • Assign any value arbitrarily to a row or column variable ui or vj. Generally, u1=0 is assigned. The Modified Distribution Method (MODI) • Consider every occupied cell in the first row individually and assign the column value vj (when the occupied cell is in the jth column of the row) which is such that the sum of the row and the column values is equal to the unit cost value in the occupied cell. With the help of these values, consider other occupied cells one by one and determine the appropriate values if ui’s & vj’s, taking in each case ui+vj=cij The Modified Distribution Method (MODI) Step3: • After determining all the values of ui & vj, calculate for each unoccupied cell Dij=ui+vj-cij. The Dij’s represents the opportunity costs of various cells. • If all the empty cells have –ve opportunity cost, the solution is optimal & unique. • If some empty cells has a zero opportunity cost but if none of the other empty cells have +ve opportunity cost, then it implies that the given The Modified Distribution Method (MODI) solution is optimal but that is not unique there exists other solutions that would be as good as this solution. • If the solution contains a +ve opportunity cost for one or more of the empty cells, the solution is not optimal. In such a case, the cell with the largest opportunity cost value is selected, a closed loop traced & transfers of units along with the route are made. The Modified Distribution Method (MODI) Note: This method is workable only when solution is non-degenerate, i.e. for a matrix of the order m*n, there are exactly m+n-1 non zero (occupied) cells. Special Topics of Transportation • Unbalanced transportation Problem • Prohibited Routes • Unique vs multiple solutions • Degeneracy • Maximization problem Unbalanced Transportation Problem • For solving the transportation problem it is required that the aggregate supply is equal to the aggregate demand. Practically situation may arise when the two are unequal. • Two possibilities are: when the total supply exceeds total demand and when total demand exceeds total supply. Such problems are called unbalanced transportation problems. • Then before solving we must balance the demand & supply. Unbalanced Transportation Problem • When supply exceeds demand, the excess supply is assumed to go to inventory. A column of slack variables is added to the transportation table which represents dummy destination with a requirement equal to the amount of excess supply and the transportation cost equal to zero. • When demand exceeds supply, balance is restored by adding a dummy origin. The row representing it is added with an assumed total availability equal to the difference between total demand & supply and Unbalanced Transportation Problem with each cell having a zero unit cost. • After balancing the transportation problem, solution proceeds in exactly same manner as discussed earlier. Prohibited routes • Sometimes in a given transportation problem some routes may not available. This can happen due to many reasons like the weather condition, strike etc. • There is a restriction on the routes available for transportation. To handle a situation of this type, we assign a very large cost represented by M to each of such routes which are not available. • The effect of adding a large cost element would be that such routes would automatically be eliminated in the final solution. Unique vs Multiple Solutions • The optimal solution of a given problem may or may not be optimal. • If all Dij values are –ve, then the solution is unique. • If some Dij=0, then multiple optimal solutions are indicated so that there exist transportation patterns other than the one obtained which can satisfies all the rim requirements. • To obtain an alternate solution, trace a closed loop starting with cell Dij=0, and get the revised soluition

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