TRANSPORTATION PROBLEM

TRANSPORTATION MODELS INTRODUCTION A transportation transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n demand points using the capacities of m supply points. The transportation problem is one of the subclasses of linear programming problem where the objective is to transport various quantities of a single homogeneous product that are initi initial ally ly stored stored at vario various us origi origins ns,, to diffe differen rentt dest destina inati tions ons in such such a way way that that the the total total transportation cost is minimum. F.I. Hitchaxic developed the basic transportation problem in 1941. However it could be solved for optimally as an answer to complex business problem only in 1951, when George B. Dantzig applied the concept of Linear Programming in solving the Transportation models. Transportation models or problems are primarily concerned with the optimal (best possible) way in which a product produced produced at different factories 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. 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 transportation so as to increase the profit on sales. Transportation Transportation problems arise in all such cases. It aims at providing assistance to the top management in ascertaining how many units of a 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. Mathematical Model of Transportation Problem Mathematically Mathematically a transportation transportation problem is nothing but a special linear programming 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, b j = 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 x ij cij S.t. Σxij = ai, i= 1,2,…..m Σxij = b j, j=   j= 1,2,…..,n and xij ≥ 0 for all i and j . Let Let us consid consider er an exam example ple to unders understan tand d the formu formulat lation ion of math mathem emat atica icall model 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 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 thefinished products are stored by the company before shipping to various destinations. Further let, x ij = quantity (amount of goods) shipped from plant Pi to the warehouse W j, and Cij = transportation cost per unit of shipping from plant Pi to the Warehouse W j. Objective-function. The objection function can be represented as: Minimize Z = c 11x11 + C12x12 + C13x13 + c21x21 + c22x22 + c23x23+ c31x31 + c32x32 + c33x33 from a plant 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

(i) 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. Example 1 : A firm has 3 factories located at A, E, and K which produce the same product. There are four major product district centers situated at B, C, D, and M. Average daily product at A, E, K is 30, 40, and 50 units respectively. The average daily requirement of this product at B, C, D, and M is 35, 28, 32, 25 units respectively. The cost in Rs. of transportation per unit of product from each factory to each district centre is given in table 1 Factories A E K Demand

B 6 5 8 35

C 8 11 9 28

D 8 9 7 32

M 5 7 13 25

Supply 30 40 50

The problem is to determine the name of product, no. of units of product to be transported from each factory to various district centers at minimum cost.

Factories A E K Demand

B X11 X21 X31 35

C X12 X22 X32 28

D X13 X23 X33 32

M X14 X24 X34 25

Supply 30 40 50

Xij = No. of unit of product transported from ith factory to jth district centre. Total transportation cost: Minimize = 6x 11 + 8x12 + 8x13 + 5x14 + 5x21 + 11x22 + 9x23 + 7x24 + 8x31 + 9x32 + 7x33 + 13x34 subject to : x 11 + x12 + x13 + x14 = 30 x21 + x22 + x23 + x24 = 40 x31 + x32 + x33 + x34 = 50 x11 + x21 + x31 = 35 x12 + x22 + x32 = 28 x13 + x23 + x33 = 32 x14 + x24 + x34 = 25 xij ≥ 0 Since number of variables is very high, simplex method is not applicable. Feasible condition: Total supply = total demand. Or Σ a i = Σ b j = K(say) i= 1,2,…..,n and j = 1,2,….,n Things to know: 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. 4) Cell in the transportation table having positive allocation will be called occupied cells, otherwise empty or non occupiedcell .

Some Definitions The following terms are to be defined with reference to thetransportation problems: (A) Feasible Solution (F.S.) A set of non-negative allocations xij ³ 0 which satisfies therow and column restrictions is known as feasible solution. (B) Basic Feasible Solution (B.F.S.)A feasible solution to a m-origin and n-destination problem issaid to be basic feasible solution if the number of positiveallocations are (m+n–1).If  the number of allocations in a basic feasible solutions areless than (m+n–1), it is called degenerate basic feasible solution(DBFS) (otherwise non-degenerate). (C) Optimal Solution A feasible solution (not necessarily basic) is said to be optimalif it minimizes the total transportation cost

(A) Initial Basic Feasible Solution : There are three different methods to obtain the initial basicfeasible solution viz. (I) North-West corner rule (II) Lowest cost entry method  (III) Vogel’s approximation method  Solution for a transportation problem The solution algorithm to a transpiration 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 ii. Matrix Minimum Method iii. Vogel Approximation Method 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.

Example 1: Powerco Formulation  . Powerco has three electric power plants that supply the electric needs of four cities.  The associated supply of each plant and demand of each city is given in the table 1.  The cost of sending 1 million kwh of electricity from a plant to a city depends on the distance the electricity must travel.  A transportation problem is specified by the supply, the demand, and the shipping costs. Relevant data can be summarized in a transportation tableau.

From

To

Plant 1

City 1 City 2 City 3 City 4 Supply   (Million kwh) Rs.8 Rs.6 Rs.10 Rs.9 35

Plant 2 

Rs.9

Plant 3

Rs.14 Rs.9

Demand  45 (Million kwh)

Rs.12 20

Rs.13

Rs.7

50

Rs.16

Rs.5

40

30

30

Example 1: Solution  Decision Variables  Powerco must determine how much power is sent from each plant to each city so x ij = Amount of electricity produced at plant i and sent to city j    x 14 = Amount of electricity produced at plant 1 and sent to city 4  Constraints  A supply constraint ensures that the total quality produced does not exceed plant capacity. Each plant is a supply point.  A demand constraint ensures that a location receives its demand. Each city is a demand point.  Since a negative amount of electricity can not be shipped all  x ij’s must be non negative  LP Formulation of Powerco’s Problem Min Z = 8 x 11+6 x 12+10 x 13+9 x 14+9 x 21+12 x 22+13 x 23+7 x 24 +14 x 31+9 x 32+16 x 33+5 x 34 S.T.:  x 11+ x 12+ x 13+ x 14 <= 35 (Supply Constraints)  x 21+ x 22+ x 23+ x 24 <= 50  x 31+ x 32+ x 33+ x 34 <= 40  x 11+ x 21+ x 31 >= 45  x 12+ x 22+ x 32 >= 20  x 13+ x 23+ x 33 >= 30  x 14+ x 24+ x 34 >= 30  x ij >= 0 (i= 1,2,3; j= 1,2,3,4)   

 

(Demand Constraints)

  j =n i =m If   si = ∑d   j ∑demand, then total supply equals to total the problem is said to be a balanced i=   j = transportation problem. If total supply exceeds total demand, we can balance the problem by adding dummy demand point. Since shipments to the dummy demand point are not real, they are assigned a cost of zero. If a transportation problem has a total supply that is strictly less than total demand the problem has no feasible solution. No doubt that in such a case one or more of the demand will be left unmet. 1

1



Generally in such situations a penalty cost is often associated with unmet demand and as one can guess the total penalty cost is desired to be minimum.

METHODS FOR SOLVING TRANSPORTATION PROBLEM There are five methods to determine the solution for balanced transportation problem: 1. Northwest Corner method 2. Minimum cost method 3. Vogel’s approximation method 4. Row Minimum Method 5. Column Minimum Method The five methods differ in the "quality" of the starting basic solution they produce and better  starting solution yields a smaller objective value.We present the five methods and an illustrativeexample is solved by these five methods. 1. North West-Corner Method The method starts at the northwest-corner cell (route)of the tableau (variable x11) (i)Allocate as much as possible to the selected cell and adjust the associated amounts of  supply and demandby subtracting the allocated amount. (ii)Cross out the row or Column with zero supply or demand to indicate that no further  assignments can be made in that row or column. If both a row and a column net to zero simultaneously, cross out one only and leave a zero supply (demand in the uncrossed-out row (column). (iii) If exactly one row or column is left uncrossed out, stop .otherwise, move to the cell to the right if a column has just been crossed out or below if a row has been crossed out .Go to step (i). 2. Minimum-Cost Method The minimum-cost method finds a better starting solution by concentrating on the cheapest routes. The method starts by assigning as much as possible to the cell with the smallest unit cost .Next, the satisfied row or column is crossed out and the amounts of supply and demand are adjusted accordingly. If both a row and acolumn are satisfied simultaneously, only one is crossed out, the same as in the northwest –corner method .Next ,look for the uncrossed-out cell with the smallest unit cost and repeat the process until exactly one row or column is left uncrossed out . 3. Vogel’s Approximation Method (VAM) Vogel’s Approximation Method  is an improved version of the minimum-cost method that generally produces better starting solutions. (i) For each row (column) determine a penalty measure by subtracting the smallest unit cost element in the row(column) from the next smallest unit cost element in the same row (column). (ii) Identify the row or column with the largest penalty.Break ties arbitrarily. Allocate as much as possible to the variable with the least unit cost in the selected row orcolumn .Adjust the supply and demand and cross out the satisfied row or column. If a row and column are satisfied simultaneously, only one of the two is crossed out, and the remaining row (column) is assigned zero supply (demand).

(iii) (a) If exactly one row or column with zero supply or demand remains uncrossed out, stop. (b) If one row (column) with positive supply(demand) remains uncrossed out, determine the basic variables in the row (column) by the least cost method .stop. (c) If all the uncrossed out rows and columns have(remaining) zero supply and demand, determine the zero basic variables by the least-cos tmethod .stop. ). (d) Otherwise, go to step (i). 4. Row Minimum Method Row minimum method start with first row and choose the lowest cost cell of first row so that either the capacity of the first supply is exhausted or the demand at jth distribution center is satisfied or both. Three cases arise: (i)If the capacity of the first supply is completely exhausted, cross off the first row and proceed to the second row. (ii)If the demand at jth distribution center is satisfied,cross off the jth column and reconsider  the first row with the remaining capacity. (iii)If the capacities of the first supply as well as the demand at jth distribution center are completely satisfied,make a zero allocation in the second lowest cost cell of the first row .cross off the row as well as the jth column and move down to the second row.Continue the process for the resulting reduced transportation table until all the rim conditions (supply and demand condition) are satisfied. 5. Column Minimum Method Column minimum method starts with first column and chooses the lowest cost cell of first column so that either the demand of the first distribution center is satisfied or the capacity of  the ith supply is exhausted or both .three cases arise: (i)If the demand of the first distribution center is satisfied,cross of the first column and move right to the second column. (ii)If the capacity of ith supply is satisfied, cross off ith row and reconsider the first column with the remaining demand. (iii)If the demands of the first distribution center as well as the capacity of the ith supply are completely satisfied,make a zero allocation in the second lowest cost cell of thefirst column. Cross of the column as well as the ith rowand move right to the second column. Continue the process for the resulting reduced transportation table until all the rim conditions are satisfied. ILLUSTRATIVE EXAMPLE (SUNRAY TRANSPORTATION) Sun Ray Transportation Company ships truckloads of grain from three silos to four mills The supply (in truckloads) and the demand (also in truckloads) together with the unit transportation costs per truckload on thedifferent routes are summarized in the transportation model in table.1. Table

Table 1. Transportation model of example (SunRay Transportation) The model seeks the minimum-cost shipping schedule between the silos and the mills. This is equivalent to determining the quantity xij shipped from silo i to mill j (i=1, 2, 3; j=1, 2, 3, 4) 1. Northwest-Corner method The application of the procedure to the model of the example gives the starting basic solution in table.2.

Table 2. The starting solution using Northwest-corner method  The Starting basic Solution is given as

X11=5, x12=10, x22=5, x23=15, x24=5, x34=10 The objective function value is Z=5×10+10×2+5×7+15×9+5×20+10×18=Rs.520 2. Minimum Cost Method The minimum-cost method is applied to Example(SunRay Transportation) in the following manner: 1. Cell (1,2) has the least unit cost in the tableau (=Rs.2).the most that can be shipped through (1,2) is x12=min(15,15)=15 which happens to satisfy both row 1 and column 2 simultaneously, we arbitrarily cross out column2 and adjust the supply in row 1 to 0. 2. Cell (3,1) has the smallest uncrossed-out unit cost(=Rs.4).Assign x31=5, cross out column 1 because it is satisfied and adjust the demand of row 3 to 10-5=5 truckloads. 3.continuing in the same manner ,we successively assign15 truckloads to cell (2,3),0 truckloads to cell (1,4),5 truckloads to cell(3,4) and 10 truckloads to cell (2,4).The resulting starting solution is summarized in this table.3. The arrows show the order in which the allocations are made. Table

Table 3. The starting solution using minimum-cost method  The starting solution (consisting of 6 basic variables) is x12=15, x14=0, x23=15, x24=10, x31=5, x34=5 The associated objective function value is Z =15×2+0×11+15×9+10×20+5×Σ+5×18=Rs.475 The quality of the least cost starting solution is better than of the northwest –corner method because it yields a smaller value of Z (Rs.475 versus Rs.520 in the northwest cornermethod.

3. Vogel’s Approximation Method (VAM) VAM is applied to Example (SunRay Transportation)in the following manner:-

1. We computes from row 3 because row 3 has the largest penalty =10 and cell (3,1) has the smallest unit cost in the row, the amount 5 is assigned to x31.Column 1 is now satisfied and must be crossed out. Next, new penalties are recomputed in the table. 6 2.Shows that row 1 has the highest penalty(=9).hence ,we assign the maximum amount possible to cell (1,2),whichyields x12=15and simultaneously satisfies both row 1 and column 2 .we arbitrarily cross out column 2 and adjust the supply in row 1 to zero as table.7.

Table 4. Step1 to determine the starting solution using (VAM)

Table 5. Step2 to determine the starting solution using (VAM)

Table 6. Step3 to determine the starting solution using (VAM)

Table 7. Step4 to determine the starting solution using (VAM)

4. Row Minimum Method Row minimum method is applied to Example (SunRayTransportation) in table. 8. The associated objective function value is

Z =15×2+12×5+9×15+20×5+10×18=Rs.505 This cost is less than northwest-corner method

Table 8. The starting solution using row minimum method  5. Column Minimum Method Column minimum method is applied to Example (SunRayTransportation) in table. 9. The associated objective function value is Z =4×5 +15×2 +9×15 +10×20 +18×5 =Rs. 475.This cost is less than northwest-corner method

Table 9.the starting solution using column minimum method 

COMPARISON BETWEEN THE FIVE METHODS

North-west corner method  is used when the purpose of completing demand No. 1 and then the next and is used when the purpose of completing the warehouse No. 1 and then the next. Advantage of northwest corner method is quick solution because computations take short time but yields a bad solution because it is very far from optimal solution. Vogel's approximation method and Minimum-cost method  is used to obtain the shortest road. Advantage of  Vogel’sapproximation method and Minimum-cost method  yields the best starting basic solution because gives initial solution very near to optimal solution but the solution of Vogel’s approximation methods is slow because computations take long time. The cost of transportation with Vogel's approximation method and Minimum-cos tmethod  is less than north-west corner method. Row-minimum method  is used when the purpose of completing the warehouse No. 1 and then the next. Row minimum cost is useful in small number ofsupply and when the cost of  transportation on supply.The cost of transportation is less than North-west corner  method.Column minimum method is used when the purpose of completing demand No. 1 and then the next .Column minimum cost is useful in small number of demand and when the cost of transportation on demand .The cost of transportation is less than North-west corner  method. (B) Test for Optimization : In part (A) of this section we have learnt how to obtain an initial basic feasible solution. Solutions so obtained may be optimal or may not be optimal, so it becomes essential for us totest for optimization. For this purpose we first define non-degenerate basic feasiblesolution. Definition: A basic feasible solution of an (m × n) transportation problem is said to be nondegenerate if it has followingtwo properties : (a) Initial basic feasible solution must contain exactly m+n–1number of individual allocations. (b) These allocations must be in independent positions.Independent positions of a set of  allocations means that it is always impossible to form any closed loop through these allocations. See fig. given below.

Closed

Non - independent positions

Independent

positons

DEGENCY IN TRANSPORTATION PROBLEM Transportation with m-origins and n-destinations can have m+n-1 positive basic variables, otherwise the basic solution degenerates. So whenever the number of basic cells is less than m + n-1, the transportation problem is degenerate. To resolve the degeneracy, the positive variables are augmented by as many zero-valued variables as is necessary to complete m +n –1 basic variables

SUBMITTED TO MANISH MISHRA

SUBMITTED BY RICHA SHARMA

M.B.A. III SEM