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Chapter 6.1.0 TRANSPORTATION PROBLEM The transportation problem is a special class of the linear programming problem. Itdeals with the situation in which a commodity is transported from Sources toDestinations. The objective is to determine the amount of commodity to be transportedfrom each source to each destination so that the total transportation cost is minimum.EXAMPLE 1.1A soft drink manufacturing firm has m plants located in m different cities. The totalproduction is absorbed by n retail shops in n different cities. We want to determine thetransportation schedule that minimizes the total cost of transporting soft drinks fromvarious plants to various retail shops. First we will formulate this as a linearprogramming problem.MATHEMATICAL FORMULATIONLet us consider the m-plant locations (origins) as O1 , O2 , …., Om and the n-retail shops(destination) as D1 , D2 , ….., Dn respectively. Let ai ≥ 0, i= 1,2, ….m , be the amountavailable at the ith plant Oi . Let the amount required at the jth shop Dj be bj ≥ 0, j=1,2,….n.Let the cost of transporting one unit of soft drink form ith origin to jth destination be Cij , i= 1,2, ….m, j=1,2,….n. If xij ≥ 0 be the amount of soft drink to be transported from ith origin to jth destination , then the problem is to determine xij so as to Minimize m nz = ∑∑ xij cij i =1 j =1Subject to the constraint and xij ≥ 0 , for all i and j. n∑xj =1 ij = ai , i = 1,2,...m m ∑ xij = b j , j =1,2,...n.i =1This lPP is called a Transportation Problem. 1
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THEOREM 1.1A necessary and sufficient condition for the existence of a feasible solution to thetransportation problem is that m n∑a = ∑bi =1 i j =1 jRemark. The set of constraints m n ∑ x = b and ∑ x = a ij j ij ii =1 j =1Represents m+n equations in mn non-negative variables. Each variable xij appears inexactly two constraints, one is associated with the origin and the other is associated withthe destination.Note. If we are putting in the matrix from, the elements of A are either 0 or 1.THE TRANSPORTATYION TABLE: D1 D2 …… Dn supply O1 c11 c12 ….. c1n a1 O2 c21 c22 ….. …. c2n a2 … …… ….. ….. ….. ….. : Om cm1 cm2 …. … cmn am Requirement b1 b2 … …. bnDefinition. (Loop). In a transportation table, an ordered set of four or more cells is saidto form a loop if :(I) Any two adjacent cells in the ordered set lie in the same row or in the same column.(II) Any three or more adjacent cells in the ordered set do not lie in the same row or in the same column.RESULT:A feasible solution to a transportation problem is basic if and only if the correspondingcells in the transportation table do not contain a loop. To find an initial basic feasiblesolution we apply:(1) The North-West corner rule(2) Vogel`s Approximation method. 2
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1.1 THE NORTH-WEST CORNER RULEStep (1). The first assignment is made in the cell occupying the upper left-hand (NorthWest) corner of the transportation table. The maximum feasible amount is allocated there,i.e; x11 = min( a1, b1 ) .Step (2). If b1 > a1, the capacity of origin O 1 is exhausted but the requirement at D 1is not satisfied. So move downs to the second row, and make the second allocation:x21 = min ( a2 , b1 – x11 ) in the cell ( 2,1 ).If a1 > b1 , allocate x12 = min ( a1 - x11 , b2 ) in the cell ( 1,2) .Continue this until all the requirements and supplies are satisfied.EXAMPL 1.1.1Determine an initial basic feasible solution to the following transportation problem usingthe North-West corner rule: D1 D2 D3 D4 Availability O1 6 4 1 5 14 O2 8 9 2 7 16 O3 4 3 6 2 5Requirement 6 10 15 4Solution to the above problem is: 6 8 6 4 1 5 2 14 8 9 2 7 1 4 4 3 6 2Now all requirements have been satisfied and hence an initial basic feasible solution tothe transportation problem has been obtained. Since the allocated cells do not form aloop, the feasible solution is non-degenerate. Total transportation cost with this allocationis: 3
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Z = 6*6 + 4*8 + 2*9 +14*2 + 1*6 +4* = 128.VOGEL’S APPROXIMATION METHOD (VAM ).Step 1. For each row of the transportation table, identify the smallest and the next to-smallest costs. Determine the difference between them for each row. Display themalongside the transportation table by enclosing them in parenthesis against the respectiverows. Similarly compute the differences for each column.Step 2. Identify the row or column with the largest difference among all the rows andcolumns. If a tie occurs, use any arbitrary tie breaking choice. Let the greatest differencecorrespond to ith row and the minimum cost be Cij . Allocate a maximum feasible amount xij = min ( ai , bj ) in the ( i, j )th cell, and cross off the ith row or jth column.Step 3. Re compute the column and row differences for the reduced transportation tableand go to step 2. Repeat the procedure until all the rim requirements are satisfied.Remark. VAM determines an initial basic feasible solution, which is very close to theoptimum solution.PROBLEM 1.1.2Obtain an initial basic feasible solution to the following transportation problem usingVogels approximation method. I II III IV A 5 1 3 3 34 B 3 3 5 4 15 C 6 4 4 3 12 D 4 -1 4 2 19 21 25 17 171.2 MOVING TOWARDS OPTIMALITY(1) DETERMINE THE NET-EVALUATIONS ( U-V METHOD) 4
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Since the net evaluation is zero for all basic cells, it follows thatzij - cij = ui +vj - cij , for all basic cells (i, j). So we can make use of this relation to findthe values of ui and vj . Using the relation ui +vj = cij , for all i and j which (i,j) is abasic cell, we can obtain the values of ui `s and vj `s. After getting the values of ui `s andvj `s, we can compute the net-evaluation for each non-basic cell and display them inparenthesis in the respective cells.(2) SELECTION OF THE ELECTING VARIABLESChoose the variable xrs to enter the basis for which the net evaluation zrs -crs = max { zij - cij > 0} .After identifying the entering variable xrs , form a loop which starts at the non-basic cell(r,s) connecting only basic cells . Such a closed path exists and is unique for any non-degenerate solution. Allocate a quantity θ alternately to the cells of the loop starting +θto the entering cell. The value of θ is the minimum value of allocations in the cellshaving -θ.Now compute the net-evaluation for new transportation table and continue the aboveprocess till all the net-evaluations are positive for non-basic cells.1.3 DEGENCY IN TRANSPORTATION PROBLEM Transportation with m-origins and n-destinations can have m+n-1 positive basicvariables, otherwise the basic solution degenerates. So whenever the number of basiccells is less than m + n-1, the transportation problem is degenerate.To resolve the degeneracy, the positive variables are augmented by as many zero-valuedvariables as is necessary to complete m +n –1 basic variables.UNBALANCED TRANSPORTATION PROBLEMIf m n , ∑ a ≠ ∑b i =1 i j =1 jThe transportation problem is known as an unbalanced transportation problem. Thereare two casesCase(1). 5
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m n∑ a > ∑bi =1 i j =1 j Introduce a dummy destination in the transportation table. The cost of transporting tothis destination is all set equal to zero. The requirement at this destination is assumed tobe equal to m n∑ a − ∑bi =1 i j =1 j.Case (2) . m n∑ a < ∑bi =1 i j =1 jIntroduce a dummy origin in the transportation table, the costs associated with are setequal to zero. The availability is n m∑b − ∑aj =1 j i =1 i1.4 THE ASSIGNMENT PROBLEMSuppose there are n-jobs for a factory and has n-machines to process the jobs. A job i(i=1,2,…n ) when processed by machine j ( j=1,2,…n) is assumed to incur a cost c ij .Theassignment is to be made in such a way that each job can associate with one and only onemachine. Determine an assignment of jobs to machines so as to minimize the overall cost.1.4.1 MATHEMATCAL FORMULATIONWe can define xij = 0, if the ith job not assigned to jth machine. = 1 , if the ith job is assigned to jth machine.We can assign one job to each machine, n n∑xi =1 ij = 1, and ∑ xij = 1 j =1The total assignment cost is given by n nz = ∑∑ cij xij j =1 i =1 6
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1.4.2 THE ASSIGNMENT ALGORITHMStep (1). Determine the effectiveness matrix. Subtract the minimum element of each rowof the given cost matrix from all of the elements of the row. Examine if there is at leastone zero in each row and in each column. If it is so, stop here, otherwise subtract theminimum element of each column from all the elements of the column. The resultingmatrix is the starting effectiveness matrix.Step (2). Assign the zeroes:(a). Examine the rows of the current effective matrix successively until a row withexactly one unmarked zero is found. Mark this zero, indicating that an assignment will bemade there. Mark all other zeroes lying in the column of above encircled zero. The cellsmarked will not be considered for any future assignment. Continue in this manner untilall the rows have taken care of.(b). Similarly for columns.Step (3). Check for Optimality. Repeat step 2 successively till one of the followingoccurs.(a). There is no row and no column without assignment. In such a case, the currentassignment is optimal.(b). There may be some row or column without an assignment. In this case the currentsolution is not optimal. Proceed to next step.Step (4). Draw minimum number of lines crossing all zeroes as follows. If the number oflines is equal to the order of the matrix, then the current solution is optimal, otherwise itis not optimal. Go to the next step>Step (5). Examine the elements that do not have a line through them. Select the smallestof these elements and subtract the same from all the elements that do not have a linethrough them, and add this element to every element that lies in the intersection of thetwo lines.Step (6). Repeat this until an optimal assignment is reached.PROBLEM 1.4.1Consider the problem of assigning five jobs to five persons. The assignment costs aregiven as follows: 7
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Jobs Person 1 2 3 4 5 A 8 4 2 6 1 B 0 9 5 5 4 C 3 8 9 2 6 D 4 3 1 0 3 E 9 5 8 9 5 1.4.4.UNBALANCEED ASSIGNMENT PROBLEM When the cost matrix of an assignment problem is not a square matrix, i.e; numberof sources is not equal to the number of destinations, the assignment problem is called anunbalanced assignment problem. In such problems, dummy rows or columns are addedin the matrix so as to complete it to form a square matrix. 8
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