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OPERATION RESEARCH UNITED COLLEGE OF ENGG & RESEARCH FOR MBA IST YEAR 1 FROM GAURAV SONKAR OPERATION RESEARCH UNITED COLLEGE OF ENGG & RESEARCH (Q1) Write short notes on “Transportation Problem”. Describe methods to obtain an initial feasible solution for a transportation problem. (03-04; 05-06) (A1) The transportation problem is a particular class of linear programming in which the objective is to transport a commodity from several supply origin to different demand destinations at a minimum total cost. A general transportation problem looks like: Destination D1 D2 Dn Availability …….. Origin C11 C12 C1n O1 x11 x12 …….. x1n a1 C21 C22 C2n O2 x21 x22 …….. x2n a2 . . . . . . . . …….. . . . . . . . Cm1 Cm2 Cmn Om xm1 xm2 …….. xmn am Requirement b1 b2 …….. bn Σai=Σbj General Transportation Table Where ai amount of commodity available at origin to Oi bj amount of commodity required at destination Dj Cij cost of transporting one unit of commodity from origin Oi to destination Dj xij amount of commodity to be transported from origin Oi to destination Dj Mathematical formulation of Transportation Problem: Min z = ΣΣ cij xij subject to constraints Σxij = ai (i = 1, 2,……., m) 2 Σcij = bj (j = 1, 2,……., n) 3 And xij ≥ 0 FOR MBA IST YEAR 2 FROM GAURAV SONKAR OPERATION RESEARCH UNITED COLLEGE OF ENGG & RESEARCH Constraints 2 and 3 are called availability constraints and requirement constraints respectively. Methods to obtain an Initial Feasible Solution NORTH WEST LEAST COST VOGEL’S CORNER APPROXIMATION METHOD METHOD METHOD (I)NORTH WEST CORNER METHOD: This method does not take into account the cost of transportation on any route of transportation. This method consists of following steps: Step I: Start with north-west corner (uppermost left hand corner) of the transportation table allocate maximum possible amount to it, so that either the capacity of row is exhausted or the destination requirement of the first column is satisfied. Step II: (a) If the supply for the first row is exhausted. Then, move down vertically to the first cell in the second row. (b) If the demand for the first column is satisfied. Then, move right horizontally to the second cell in the first row. (c) If both supply in the row and demand in the column is satisfied. Then, move diagonally. Step III: Continuing in this way till an allocation is made in the south-east corner cell of the transportation table. (II) LEAST COST METHOD: This method takes into account the minimum unit cost of transportation for obtaining initial basic feasible solution. This method consists of following steps: Step I: Select the cell with lowest cost and allocate maximum possible amount to it; if there is a tie make an arbitrary selection. Step II: Delete the row or column or both which are satisfied by the allocation. Step III: Again search for the next lowest cost empty cell for which demand and supply are not exhausted and make the allocations. Repeat step I and II until all supply and demand conditions are satisfied. FOR MBA IST YEAR 3 FROM GAURAV SONKAR OPERATION RESEARCH UNITED COLLEGE OF ENGG & RESEARCH (III) VOGEL’S APPROXIMATION METHOD: This is the most preferred method as initial feasible solution obtained by this method is either optimal or very close to the optimal solution. Basic steps in VAM are as follows: Step I: Calculate the penalty cost i.e. the difference between the lowest and the second lowest cost values for each row and each column. Step II: Enter the penalty cost of each row to the right of corresponding row and the penalty cost of each column under the corresponding column. Step III: Select the row or column with the largest penalty and allocate maximum possible amount to the cell with the minimum cost in the selected row or column. If there are more than one largest penalty rows or columns we select any of them arbitrarily. Step IV: Cross out the particular row or column on which the requirement is satisfied. Step V: Recompute the penalty cost for each row and column for the reduced transportation table and go to the previous steps. Step VI: Repeat steps I to V until all the rim requirements are satisfied. (Q2) Explain the term ‘Degeneracy’ in the context of transportation problem. (03-04) (A2) Degeneracy: Degeneracy in an m x n transportation problem occurs whenever the number of positive allocations in a BFS is less than m+n-1. The degeneracy in the transportation problems may occur at the two stages: (a) At initial stages when the number of positive allocations in the initial BFS is less than m + n -1. (b) At any stage while moving towards optimal solution. This happens when two or more occupied cells with the same minimum allocation become unoccupied simultaneously. Necessity to remove degeneracy: (a) In such cases, the current solution can not be improved because it is not possible to draw a closed path (loop) for every occupied cell. (b) Also the values of dual variables ui and vj which are used to test the optimality can not be computed. Resolution of Degeneracy: To resolve degeneracy, we select as many unoccupied cells as is necessary to have exactly m + n – 1 allocation. Now we allocate an extremely small positive amount to these unoccupied cells, denoted by ‘ε’ and is assumed to satisfy following conditions: (a) 0+ε=ε (b) Xij ± ε = Xij The cells containing ε’s are kept in the transportation table until either degeneracy is removed or an optimal solution is attained, whichever occurs first. Then we set ε equals to zero. FOR MBA IST YEAR 4 FROM GAURAV SONKAR OPERATION RESEARCH UNITED COLLEGE OF ENGG & RESEARCH (Q3)What is meant by unbalanced transportation problem? Explain the method for solving such a problem with an example (03-04) (A3) A transportation problem is said to be unbalanced if total supply is not equal to total demand i.e. m n ∑ ai ≠ ∑ bj i=1 j=1 In this situation there are two cases arises. Case I: If the total supply is more than total demand i.e. ∑ ai > ∑ bj In this case, we introduce a dummy destination in the transportation table. The requirement at this destination is assumed to be equal to ∑ ai - ∑ bj. The unit transportation cost is taken to be zero. Case II: If the total supply is less than total demand i.e. ∑ ai < ∑ bj In this case, we introduce a dummy origin in the transportation table. The availability at this origin is assumed to be equal to ∑bj - ∑ai. The unit transportation cost is taken to be zero. Thus an unbalanced transportation problem is converted into a balanced transportation problem which can be solved by either of North-West Corner Method, Least Cost Method or Vogel’s Approximation Method. EXAMPLE: Consider the following transportation problem: Plant Market Supply A B C D E 11 20 7 8 50 F 21 16 10 12 40 G 8 12 18 19 70 Demand 30 25 35 40 Solution: Here Total Supply (∑ai) = 160 and Total Demand (∑bj) = 130 Since, ∑ ai > ∑ bj , the given problem is unbalanced. To convert it into balanced transportation problem we introduce a dummy market E, with demand 160- 130 = 30, such that the unit transportation cost from each plant to market is equal to zero. Thus, we obtain the following transportation problem. Plant Market Supply A B C D E X 11 20 7 8 0 50 Y 21 16 10 12 0 40 Z 8 12 18 19 0 70 Demand 30 25 35 40 30 160 FOR MBA IST YEAR 5 FROM GAURAV SONKAR OPERATION RESEARCH UNITED COLLEGE OF ENGG & RESEARCH (Q4) Define the assignment problem. (02-03; 03-04; 05-06) (A4) The assignment problem is a particular case of transportation problem in which the objective is to assign a number of resources to an equal number of activities so as to minimize total cost or maximize total profit. Typical examples of decision making situations where assignment model can be used are: workers to machines sales personnel to different sales areas products to factories jobs to machines classes to rooms A general assignment problem looks like: Jobs Supply Person 1 2 ……….. n 1 C12 ………….. C1n 1 C11 2 C22 …………… C2n 1 C21 : : . . . . : . . . . Cn2 ……………. Cnn 1 n Cn1 Demand 1 1 This matrix is called cost matrix where Cij is the cost ……… 1 of assigning ith person to jth job The cost matrix is same as that of a transportation problem except that availability of the resource and the demand at each of the destination is unity. Let xij denotes the assignment of resource i to the activity j, such that 1, if resource i is assigned to activity j Xij = 0, otherwise Then the mathematical model of assignment problem can be represented as: n n Minimize Z = ∑ ∑ Cij . xij i=1 j=1 Subject to constraints n ∑ xij = 1, for all i (resource availability) j=1 n and ∑ xij = 1, for all j (activity requirement) i=1 and xij = 0 or 1, for all i and j FOR MBA IST YEAR 6 FROM GAURAV SONKAR OPERATION RESEARCH UNITED COLLEGE OF ENGG & RESEARCH Thus, it is clear that the assignment problem is a special case of transportation problem with the characteristics: (a) The cost matrix is a square matrix, and (b) The optimal solution for the problem would always be such that there would be only one assignment in a given row or column of the cost matrix. (Q5)Show assignment problem is a special case of transportation problem. (07-08) (A5) Same as above. (Q6) Explain Hungarian Method for solving an assignment problem. (01-02) (A6) Step I: Row Operation: Subtracting smallest element in each row from all other elements of that row. Column Operation: Subtracting smallest element in each column from all other elements of that column. Step II: MAKING ASSIGNMENTS: Search for an optimal assignment in the finally modified cost matrix as follows: Examine the first row. If there is only one zero in it, then enclose this zero in a box ( ) cross (X) all the zero of that column passing through the enclosed zero. Similarly done with other rows. If any row has more than one zero, then do not touch that row & passes on the next row. Repeat the same procedure with columns. Step III: If each row and each column of the reduced matrix has exactly one enclosed zero, then the enclosed zero yield an optimal assignment. If not, then go to next step. Step IV: Draw minimum number of horizontal and/or vertical lines to cover all the zeros of the reduced matrix. Since the assignment is not optimal, the number of lines will be less than n (i.e. order of matrix). In order to move towards optimality, generates more zeros as follows: i. Find the smallest of the elements of the reduced matrix not covered by any of the lines. Let this element be α. ii. Subtract α from each of the element not covered by the lines and add α to the element at the intersection of these lines. Do not change the remaining elements. Step V: Go to step II and repeat the procedure till an optimal assignment is achieved. FOR MBA IST YEAR 7 FROM GAURAV SONKAR