A balanced transportation problem to minimize the sum of shortage and shipping costs, and the basic feasible solution, using the northwest corner method. A balanced transportation problem to minimize the sum of shortage and shipping costs has been formulated, and the basic feasible solution has been determined, using the northwest corner method. Given: A company supplies goods to three customers, who each require 30 units. The company has two warehouses. Warehouse 1 has 40 units available, and warehouse 2 has 30 units available. The costs of shipping 1 unit from warehouse to customer are shown in the following table. There is a penalty for each unmet customer unit of demand. With customer 1, a penalty cost of $ 90 is incurred, with customer 2, $ 80 ; and with customer 3, $ 110 . From Customer 1 Customer 2 Customer 3 Warehouse 1 $ 15 $ 35 $ 25 Warehouse 2 $ 10 $ 50 $ 40 Concept used: The northwest corner method can be used to find a basic feasible solution of a balanced transportation problem as follows: Start by selecting the north-west corner cell of the transportation matrix. Evaluate the minimum of supply and demand for that cell and allocate this number of units to that cell. Subtract the allocated number of units from both the supply and demand corresponding to that cell to get the new supply and the new demand. If the new supply is equal to 0 , leave that row and move down to the northwest corner cell of the new transportation matrix of unassigned cells. If the new demand equals 0 , leave that column and move right to the northwest corner cell of the new transportation matrix of unassigned cells. If both new supply and new demand are 0 , then leave both column and rows and move diagonally to the northwest corner cell of the new transportation matrix of unassigned cells. Repeat these steps until all the supply and demand values are 0 . Calculation: It is given that a company supplies goods to three customers, who each require 30 units. This implies that total demand is 3 × 30 = 90 . It is also given that the company has two warehouses such that warehouse 1 has 40 units available, and warehouse 2 has 30 units available. This implies that total supply is 40 + 30 = 70 . This shows that the given problem is an unbalanced transportation problem, where the demand is greater than the supply. The given problem can be converted to a balanced problem by introducing a dummy supply row, where the values in cell would be the corresponding penalty incurred instead of transportation costs. The balanced problem is formulated as follows: From Customer 1 Customer 2 Customer 3 Supply Warehouse 1 $ 15 $ 35 $ 25 40 Warehouse 2 $ 10 $ 50 $ 40 30 Dummy Warehouse $ 90 $ 80 $ 110 20 Demand 30 30 30 Note that, the prices in the row corresponding to the Dummy Warehouse are the respective penalty incurred for unmet demand. Also note that, the supply corresponding to the Dummy Warehouse is 20 , which is the total shortage in supply with respect to the total demand. Now, the above can be considered as a balanced problem and hence, the northwest corner method can be applied. Now, the cell in the northwest corner is the cell having a value of $ 15 , corresponding supply of 40 and corresponding demand of 30 . The minimum of supply and demand is 30 . Assigning 30 units to this cell and subtracting 30 from the corresponding supply and demand, the new supply is 40 − 30 = 10 and the new demand is 30 − 30 = 0 . Striking the corresponding column as new demand is 0 , the next step of the problem is given as follows: From Customer 1 Customer 2 Customer 3 Supply Warehouse 1 $ 15 ( 30 ) $ 35 $ 25 40 − 30 = 10 Warehouse 2 $ 50 $ 40 30 Dummy Warehouse $ 80 $ 110 20 Demand 30 − 30 = 0 30 30 Now, the cell in the northwest corner is the cell having a value of $ 35 , corresponding supply of 10 and corresponding demand of 30 . The minimum of supply and demand is 10 . Assigning 10 units to this cell and subtracting 10 from the corresponding supply and demand, the new supply is 10 − 10 = 0 and the new demand is 30 − 10 = 20 . Striking the corresponding row as new supply is 0 , the next step of the problem is given as follows: From Customer 1 Customer 2 Customer 3 Supply Warehouse 1 $ 15 ( 30 ) $ 35 ( 10 ) 10 − 10 = 0 Warehouse 2 $ 50 $ 40 30 Dummy Warehouse $ 80 $ 110 20 Demand 0 30 − 10 = 20 30 Now, the cell in the northwest corner is the cell having a value of $ 50 , corresponding supply of 30 and corresponding demand of 20 . The minimum of supply and demand is 20 . Assigning 20 units to this cell and subtracting 20 from the corresponding supply and demand, the new supply is 30 − 20 = 10 and the new demand is 20 − 20 = 0 . Striking the corresponding column as new demand is 0 , the next step of the problem is given as follows: From Customer 1 Customer 2 Customer 3 Supply Warehouse 1 $ 15 ( 30 ) $ 35 ( 10 ) 0 Warehouse 2 $ 50 ( 20 ) $ 40 30 − 20 = 10 Dummy Warehouse $ 110 20 Demand 0 20 − 20 = 0 30 Now, the cell in the northwest corner is the cell having a value of $ 40 , corresponding supply of 10 and corresponding demand of 30 . The minimum of supply and demand is 10 . Assigning 10 units to this cell and subtracting 10 from the corresponding supply and demand, the new supply is 10 − 10 = 0 and the new demand is 30 − 10 = 20 . Striking the corresponding row as new supply is 0 , the next step of the problem is given as follows: From Customer 1 Customer 2 Customer 3 Supply Warehouse 1 $ 15 ( 30 ) $ 35 ( 10 ) 0 Warehouse 2 $ 50 ( 20 ) $ 40 ( 10 ) 10 − 10 = 0 Dummy Warehouse $ 110 20 Demand 0 0 30 − 10 = 20 Now, there are no further cells to strike out. The left-out cell with a value of $ 110 has equal supply and demand of 20 . Then, this cell would be allocated 20 units. The new supply and demand are calculated which brings all supplies and demand to 0 as follows: From Customer 1 Customer 2 Customer 3 Supply Warehouse 1 $ 15 ( 30 ) $ 35 ( 10 ) 0 Warehouse 2 $ 50 ( 20 ) $ 40 ( 10 ) 0 Dummy Warehouse $ 110 ( 20 ) 0 Demand 0 0 0 This is the basic feasible solution of the given problem. Conclusion: A balanced transportation problem to minimize the sum of shortage and shipping costs has been formulated, and the basic feasible solution has been determined, using the northwest corner method.

Chapter 7 Solutions