11 The Transportation Problem Question 1 The initial allocation of a transportation problem, alongwith the unit cost of transportation from each origin to destination is given below. You are required to arrive at the minimum transportation cost by the Vogel s Approximation method and check for optimality. (Hint : Candidates may consider u 1 = 0 at Row 1 for initial cell evaluation) Requirement 8 6 4 11 2 8 6 2 18 10 9 9 12 9 6 10 8 7 6 3 7 7 8 2 2 9 3 5 6 11 4 Availability 12 8 8 8 4 40 (May 2007, 6 Marks) The concept tested in this problem is Degeneracy with respect to the transportation problem. Total of rows and columns = (4 + 5) = 9. Hence, the number of allocations = 9 1 = 8. As the actual number of allocation is 7, a zero allocation is called for. To resolve this, an independent cell with least cost should be chosen. R4C2 has the least cost (cost = 3), but this is not independent. The next least cost cell R4C3 (cost = 5) is independent.
The Transportation Problem 11.2 9 C1 2 C2 5 C3 6 C4 2 C5 Total 8 6 4 0R1 11 2 8 6 2 18 10 0R2 9 9 12 9 6 10 8 2R3 7 6 3 7 7 8 2 0 2 0R4 9 3 5 6 11 4 Total 12 8 8 8 4 40 Forming Equations through allocated cells Basic equation Setting R1 = 0 other values R1 + C2 = 2 Setting R1 = 0, C2 = 2 R1 + C4 = 6 C4 = 6 R1 + C5 = 2 C5 = 2 R2 + C1 = 9 R2 = 0 R3 + C3 = 3 R3 = 2 R4 + C1 = 9 C1 = 9 R4 + C3 = 5 C3 = 5 R4 + C4 = 6 R4 = 0 Evaluate unallocated cells R1C1 = 11 0 9 = 2 R3C1 = 7 + 2 9 = 0 R1C3 = 8 0 5 = 3 R3C2 = 6 + 2 2 = 6 R2C2 = 9 0 2 = 7 R3C4 = 7 + 2 6 = 7 R2C3 = 12 0 5 = 7 R3C5 = 7 + 2 2 = 7 R2C4 = 9 0 6 = 3 R4C2 = 3 0 2 = 1 R2C5 = 6 0 2 = 4 R4C5 = 11 0 2 = 9 Since all the evaluation is 0 or +ve, the optimal solution is obtained. Optimal cost = (8 2) + (6 6) + (4 2) + (10 9) + (8 3) + (2 9) + (0 5) + (2 6) =
11.3 Advanced Management Accounting 16 + 36 + 8 + 90 + 24 + 18 + 10 + 12 = Rs. 204. Note: As regards allocation of the zero values, the solution to the above problem is also obtained by allocating the zero value in other independent cells such as R1C3, R2C2, R2C3, R3C1, R3C2, R3C4, R3C5. In such situation there will be one more iteration. Question 2 Goods manufactured at 3 plants, A, B and C are required to be transported to sales outlets X, Y and Z. The unit costs of transporting the goods from the plants to the outlets are given below: Plants Sales outlets A B C Total Demand X 3 9 6 20 Y 4 4 6 40 Z 8 3 5 60 Total supply 40 50 30 120 You are required to: (i) Compute the initial allocation by North-West Corner Rule. (ii) Compute the initial allocation by Vogel s approximation method and check whether it is optional. (iii) State your analysis on the optionality of allocation under North-West corner Rule and Vogel s Approximation method. (10 Marks) (May, 2008) 20 20 3 9 6 20 20 40 4 4 6 30 30 60 8 3 5 40 50 30 120 (i) Initial allocation under NW corner rule is as above. Initial cost : 20 3 = 60 20 4 = 80 20 4 = 80
The Transportation Problem 11.4 (ii) 30 3 = 90 30 5 = 150 460 Initial solution by VAM: 20 20 3 3 9 6 20 20 40 0 0 2 4 4 6 50 10 60 2 2 2 8 3 5 40 50 30 1 1 1 4 1 1 1 1 Initial solution: 20 3 = 60 20 4 = 80 50 3 = 150 20 6 = 120 10 5 = 100 460 Checking for optimality 3 u 1 = 0 4 6 u 2 = 1 3 5 u 3 = 0 V1 = 3 V2 = 3 V3 = 5 u i + v j
11.5 Advanced Management Accounting 3 5 0 4 1 3 0 3 3 5 ij = c ij ( u i + v j ) ij 0 5 6 1 0 Solution is optimal Conclusion: The solution under VAM is optimal with a zero in R 2 C 2 which means that the cell C 2 R 2 which means that the cell C 2 R 2 can come into solution, which will be another optimal solution. Under NWC rule the initial allocation had C 2 R 2 and the total cost was the same Rs. 460 as the total cost under optimal VAM solution. Thus, in this problem, both methods have yielded the optimal solution under the 1 st allocation. If we do an optimality test for the solution, we will get a zero for ij in C 3 R 2 indicating the other optimal solution which was obtained under VAM. Question 3 State the methods in which initial feasible solution can be arrived at in a transportation problem. (3 Marks) (Nov., 2008) The methods by which initial feasible solution can be arrived at in a transportation model are as under: (i) North West Corner Method. (ii) Least Cost Method (iii) Vogel s Approximation Method (VAM) Question 4 The cost per unit of transporting goods from the factories X, Y, Z to destinations. A, B and C, and the quantities demanded and supplied are tabulated below. As the company is working out the optimum logistics, the Govt.; has announced a fall in oil prices. The revised unit costs are exactly half the costs given in the table. You are required to evaluate the minimum transportation cost. (6 Marks)(June, 2009) Destinations Factories A B C Supply
The Transportation Problem 11.6 X 15 9 6 10 Y 21 12 6 10 Z 6 18 9 10 Demand 10 10 10 30 The problem may be treated as an assignment problem. The solution will be the same even if prices are halved. Only at the last stage, calculate the minimum cost and divide it by 2 to account for fall in oil prices. A B C X 15 9 6 Y 21 12 6 Z 6 18 9 Subtracting Row minimum, we get A B C X 9 3 0 Y 15 6 0 Z 0 12 3 A B C Subtracting Column minimum, No of lines required to cut Zeros = 3 Cost / u Units Cost Revised Cost Allocation: X B 9 10 90 45 Y C 6 10 60 30 Z A 6 10 60 30 210 105 Minimum cost = 105 Rs. Alternative Solution I Least Cost Method
11.7 Advanced Management Accounting X B Y C Z A Test for optimality No. of allocation = 3 No. of rows m =3, no. of column = 3 m + n 1 = 3 + 3 1 = 5 2 very small allocation are done to 2 cells of minimum costs, so that, the following table is got: A B C X 15 10 9 e 6 Y 21 12 10 6 Z 10 6 18 e 9 m + n 1 = 5 Now testing for optimality
The Transportation Problem 11.8 9 e 6 e v j 6 9 6 u i + v j for unoccupied cells A B C X 6 - - Y 6 9 - Z - 9-6 u i 0 0 0 Diff = Cij (u i + v j ) A B C X 9 - - Y 15 3 - Z - 9 - All Δ ij > 0, Hence this is the optimal solution. Original Costs Reduced Costs due to Oil Price X B 9 4.5 10 45 Y C 6 3 10 30 Z A 6 3 10 30 Total cost of transportation is minimum at Rs.105 Qty. Cost 105
11.9 Advanced Management Accounting Alternative Solution II No. of rows + no. of column 1 m + n 1 = 5 No. of allocation = 3 Hence add e to 2 least cost cells so that
The Transportation Problem 11.10 Now m + n 1 = 5 Testing for optimality, u i, v j table A B C u i X 4.5 e 0 Y 3 Z 3 e v j 3 4.5 3 u i + v j for unoccupied cells 0 0 3 - - 3 4.5 - - 4.5 - Cij 7.5 - - 3 - - 11.5 6-3 4.5 - - 9 - - 4.5 - Δ ij = C ij (u i + v j ) 4.5 - - 11.5 1.5-8.5 4.5 - u i +v j
11.11 Advanced Management Accounting All Δ ij > 0. Hence the solution is optimal. Qty. Cost/u Total Cost X B 10 4.5 45 Y C 10 3 30 Z A 10 3 30 Total minimum cost at revised oil prices 105 Question 5 How do you know whether an alternative solution exists for a transportation problem? (4 Marks)(Nov., 2009) The Δ i j matrix = Δ i j = Ci j (u i + v j ) Where c i is the cost matrix and (u i + v j ) is the cell evaluation matrix for allocated cell. The Δ i j matrix has one or more Zero elements, indicating that, if that cell is brought into the solution, the optional cost will not change though the allocation changes. Thus, a Zero element in the Δ i j matrix reveals the possibility of an alternative solution. Question 6 A company has three plants located at A, B and C. The production of these plants is absorbed by four distribution centres located at X, Y, W and Z. the transportation cost per unit has been shown in small cells in the following table: Factories Distribution Centres X Y W Z Supply (Units) A 6 9 13 7 6000 B 6 10 11 5 6000 C 4 7 14 8 6000 Demand (Units) 4000 4000 4500 5000 18000 17500 Find the optimum solution of the transportation problem by applying Vogel s Approximation Method. (8 Marks)(Nov., 2010)
The Transportation Problem 11.12 Step 1 : Initial Allocation based on Least cost cells corresponding to highest differences X Y W Z Dummy Total A 2,000 3,500 500 6,000 B 1,000 5,000 6,000 C 4,000 2,000 6,000 TOTAL 4,000 4,000 4,500 5,000 500 18,000 Step 2 : Δij Matrix values for Unallocated cells X Y W Z Dummy A 0 0 B 2 3 2 C 3 3 2 All Δij values > 0. Therefore initial allocation is optimal. Step 3 : Optimal Transportation Cost Units Costs involved Total A to Y 2,000 9 18,000 A to W 3,500 13 45,500 B to W 1,000 11 11,000 B to Z 5,000 5 25,000 C to X 4,000 4 16,000 C to Y 2,000 7 14,000 Total minimum cost 1,29,500 Note : Since there are zeroes in the Δij Matrix alternate solutions exist. Question 7 Will the initial solution for a minimization problem obtained by Vogel s Approximation Method and the Least Cost Method be the same? Why? (4 Marks) (May, 2011) The initial solution need not be the same under both methods.
11.13 Advanced Management Accounting Vogel s Approximation Method uses the differences between the minimum and the next minimum costs for each row and column. This is the penalty or opportunity cost of not utilising the next best alterative. The highest penalty is given the 1 st preference. This need not be the lowest cost. For example if a row has minimum cost as 3, and the next minimum as 2, penalty is 1; whereas if another row has minimum 4 and next minimum 6, penalty is 2, and this row is given preference. But least cost given preference to the lowest cost cell, irrespective of the next cost. Vogel s Approximation Method will to result in a more optimal solution than least cost. They will be the same only when the maximum penalty and the minimum cost coincide Question 8 The following matrix is a minimization problem for transportation cost. The unit transportation costs are given at the right hand corners of the cells and the ij values are encircled. D1 D2 D3 Supply 3 4 4 F 1 500 9 6 7 F 2 300 300 88 88 4 6 5 F 3 200 200 88 88 Deamnd 300 400 300 1000 Find the optimum solution (s) and the minimum cost. (5 Marks) (May, 2011) Δ ij values are given for unallocated cells. Hece, no. of allocated cells = 5, which = 3 + 3 1 = no. of columns + no of rows 1. Allocating in other than Δ ij cells. Factory S1 D2 D3 Supply 3 4 4 300 100 100 500 9 6 7 88 300 88 300
The Transportation Problem 11.14 4 6 5 200 88 88 200 300 400 300 1000 This solution is optional since Δ ij are non-ve. For the other optional solution, which exists since Δ ij = 0 at R 3 C 1, this cell should be brought in with a loop : R 3, C 1 R 1 C 1 R 1 C 3 R 3 C 3. Working Notes: Step I : R 1 C 1 (Minimum of 300, 500) Step II : R 2 C 2 (Minimum of 300, 400) Step III : R 1 C 2 balance of C 2 total : 100, R 1 Total = 100 Step IV : R 1 C 3 100 (balance of C 3 total = 200) Step V : R 3 C 3 200 Solution I -100 +100 300 100 100 300 +100 88 200-100 Solution II 200 100 200 300 100 100 Solution I Solution II Cost: 3 х 300 = 900 3 х 200 = 600 4 х 100 = 400 4 х 100 = 400 4 х 100 = 400 4 х 200 = 800 6 х 300 = 1800 6 х 300 = 1800 5 х 200 = 1000 5 х 100 = 500 4 х 100 = 400 Minimum Cost 4500 4500
11.15 Advanced Management Accounting Question 9 The following table gives the unit transportation costs and the quantities demanded/supplied at different locations for a minimization problem: Supply Demand C 1 C 2 C 3 C 4 Total Units R 1 100 120 200 110 20000 R 2 160 80 140 120 38000 R 3 180 140 60 100 16000 Total Units 10000 18000 22000 24000 You are required to find out which cell gets the 3rd allocation in the initial basic feasible solution under each of the following methods and to give the cell reference, cost per unit of that cell and the quantity allocated to that cell : (i) North West Corner Rule (ii) Vogel's Approximation Method (iii) Least Cost Method (Candidates may use the standard notation of C i R j for cell reference.( e.g. C 2 R 3 means the cell at the intersection of Column 2 and Row 3 ) (Note: The full solution is not required to be worked out). (5 Marks)(May, 2012) Sl. No Method Cell Reference Cost/unit Quantity I II III i) North West Corner Rule C 2 R 2 80 8,000 ii) Vogel s method C 3 R 2 140 6,000 C 1 R 1 100 10,000 iii) Least Cost Method C 1 R 1 100 10,000 Question 10 In a transportation problem for cost minimization, there are 4 rows indicating quantities demanded and this totals up to 1,200 units. There are 4 columns giving quantities supplied. This totals up to 1,400 units. What is the condition for a solution to be degenerate? (3 Marks)(May, 2012) or
The Transportation Problem 11.16 The condition for degeneracy is that the number of allocations in a solution is less than m+n-1. The given problem is an unbalanced situation and hence a dummy row is to be added, since the Column quantity is greater than that of the Row quantity. The total number of Rows and Columns then = 9 i.e. (5+4). Therefore, m+n-1 = 8, i.e. if the number of allocations is less than 8, then degeneracy would occur. Question 11 Explain the term 'Degeneracy' in the context of a transportation problem. How can this be solved? (5 Marks)(Nov., 2012) A transportation problem s solution has m+n-1 basic variables, (where m,n are the number of rows and columns) which means that the number of occupied cells in such a solution is one less than the number of rows and number of columns. When the number of occupied cells in a solution is less than m+n-1, the solution is called a degenerate solution. Such a situation is handled by introducing an infinitesimally small allocation e in the least cost and independent cell. If the number of occupied cells < m+n-1 by one, then only one e needs to be introduced. If the number of occupied cells is less by more than one, to the extent of shortage, e s will have to be introduced till the condition that no. of occupied cells = m+n-1. For e.g. if no. of occupied cells in a solution is 7 and we have m+n-1 = 9, then, we have to introduce two quantities of e, say e 1 and e 2 in 2 of the least cost independent cells. Degeneracy occurs because in any particular allocation (earlier than the last allocation), the row and column totals get simultaneously fulfilled. (In the last allocation, it is always that row and column get fulfilled). Then, we have a degeneracy by one number, i.e. no. of occupied cells +1= m+n-1. We need to put one e. In the subsequent allocation, if again row and column totals get fulfilled simultaneously, again there will be a shortage of occupied cells and another e will be required. Due to this concept, an assignment problem, solved by transportation technique taking demand quantity = supply quantity = 1 in every row and column will require an e for each allocation other than the last one. For e.g. in a 5 x 5 assignment problem, there are 4 allocations other than the last one.therefore, 4 e s will be required.i.e. m + n -1 will be 5+5-1, =9, whereas, the no. of occupied cells will be 5.To resolve the degeneracy, we will need 4 e s. The e has to be placed in the least cost independent cell, for arriving at the optimal solution as early as possible. If, by mistake, we place e in the second least cost but independent cell, after the u i, v j step, the e will be shifted to the least cost independent cell, thereby necessitating one more iteration. This is similar to the simplex table. If we bring in a wrong
11.17 Advanced Management Accounting variable by mistake, it will go out in the next iteration. The only thing is that the solution will be reached later. Question 12 XYZ Company has three plants and four warehouses. The supply and demand in units and the corresponding transportation costs are given. The table below shows the details taken from the solution procedure of the transportation problem : WAREHOURSES I II III IV Supply A 5 101 10 4 5 10 Plants B 20 6 8 7 5 2 25 C 5 4 10 2 5 5 7 20 Demand 25 10 15 5 the following questions. Give brief reasons: (i) Is this solution feasible? (ii) Is this solution degenerate? (iii) Is this solution optimum? (8 Marks)(May,2013) (i) Is this solution feasible? A necessary and sufficient condition for the existence of a feasible solution to the transportation problem is that (ii) m n a i b j i 1 j 1 Where a i = quantity of product available at origin i. b j = quantity of product available at origin j. In other words, the total capacity (or supply) must equal total requirement (or demand) As the supply 55 units (10+25+20) equals demand 55 units (25+10+15+5), a feasible solution to the problem exists. Is this solution degenerate?
The Transportation Problem 11.18 When the number of positive allocations at any stage of the feasible solution is less than the required number (rows + columns -1), the solution is said to be degenerate solution. In given solution total allocated cells are 6 which are equal to 4+3-1 (rows + columns -1). Therefore, the initial basic solution is not a degenerate solution. (iii) Is this solution optimum? Test of Optimality: (u i +v j ) matrix for allocated cells u i 4-1 6 2 2 4 2 5 0 v j 4 2 5 0 (u i +v j ) matrix for unallocated cells u i 3 1-1 -1 4 7 2 0 0 v j 4 2 5 0 ij = C ij (u i +v j ) matrix 2 9 6 4 0 Since, all cells values in ij = C ij (u i +v j ) matrix are non- negative, hence the solution provided by XYZ Company is optimum. It may be noted that zero opportunity cost in cell (B, III) indicates a case of alternative optimum solution. Question 13 Define the following terms in relation to a transportation problem: (i) Degeneracy (ii) Prohibited routes (4 Marks)(Nov., 2013) 7
11.19 Advanced Management Accounting (i) Degeneracy: A transportation problem s solution has m+n 1 basic variables, (where m,n are the number of rows and columns) which means that the number of occupied cells in such a solution is one less than the number of rows and number of columns. When the number of occupied cells in a solution is less than m+n 1, the solution is called a degenerate solution. Such a situation is handled by introducing an infinitesimally small allocation e in the least cost and independent cell. (ii) Prohibited Routes: Sometimes in a given transportation problem, some routes may not be available. There could be several reasons for this such as bad road conditions or strike etc. In such situations, there is a restriction on the route available for transportation. To handle such type of a situation, a very large cost (or a negative profit for the maximization problem) represented by or M is assigned to each of such routes which are not available. Due to assignment of very large cost, such routes would automatically be eliminated in the final solution. The problem is the solved in its usual way. Question 14 Will the solution for a minimization problem obtained by Vogel's Approximation Method and Least Cost Method be the same? Why? (4 Marks) (May, 2014) The initial solution need not be the same under both methods. Vogel s Approximation Method (VAM) uses the differences between the minimum and the next minimum costs for each row and column. This is the penalty or opportunity cost of not utilising the next best alternative. The highest penalty is given the 1 st preference. This need not be the lowest cost. For example if a row has minimum cost as 2, and the next minimum as 3, penalty is 1; whereas if another row has minimum 4 and next minimum 6, penalty is 2, and this row is given preference. But Least Cost Method gives preference to the lowest cost cell, irrespective of the next cost. Solution obtained using Vogel s Approximation Method is more optimal than Least Cost Method. Initial solution will be same only when the maximum penalty and the minimum cost coincide. Question 15 In a 3 x 4 transportation problem for minimizing costs, will the R 2 C 1 cell (at the intersection of the 2 nd row and 1 st column) always figure in the initial solution by the North West Corner Rule? Why? (4 Marks) (May, 2014)
The Transportation Problem 11.20 The Initial solution obtained by the North-West Corner Rule in transportation need not always contain the R 2 C 1 cell. In the North-West Corner Rule the first allocation is made at R 1 C 1 cell and then it only moves towards R 2 C 1 cell when the resources at the first row i.e. R 1 is exhausted first than the resources of first column i.e. C 1. On the contrary if resources at first column i.e. C 1 is exhausted first then the next allocation will be at R 1 C 2. For example the resource availability at first row (R 1 ) is 1,500 units and the demand in first column (C 1 ) is 1,000 units. In this case resource availability of first row (R 1 ) will be exhausted to the extent of the demand in first column (C 1 ) first and then the remaining resource availability at first row (R 1 ) will be used to meet the demand of the second column (C 2 ). In this example cell R 2 C 1 will not come in initial solution obtained by the North-West Corner Rule. Question 16 In a transport problem for cost minimization, there are 4 rows indicating quantities demanded and these totals up to 1800 units. There are 4 columns giving quantities supplied and these totals up to 2,100 units. What is the condition for a solution to be degenerate? (4 Marks) (November, 2014) The condition for degeneracy is that the number of allocations in a solution is less than m+n-1. The given problem is an unbalanced situation and hence a dummy row is to be added, since the column quantity is greater than that of the row quantity. The total number of rows and columns will be 9 i.e. (5 rows and 4 columns). Therefore, m+n-1 = 8, i.e. if the number of allocations is less than 8, then degeneracy would occur.