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Engineering Computer Science OPERATIONS RESEARCH >INTERNATIONAL EDITI

Program plan: To formulate a balanced transportation problem to minimize the sum of shortage and shipping costs using the transportation simplex method. Explanation: The transportation problem is used to reduce the cost of delivering a good from a variety of sources or origins to a variety of destinations. An additional supply row is used in the formulation of balanced transportation by adding it for the shortage of demand. The costs for the dummy row are the penalty of each customer. Total demand for both warehouses can be calculated as: 3 customers× 30 units= 90 units Then, the total supply for both warehouses would be: 40 + 30 = 70 units As total demand is greater than the total supply, in this case, means it is an unbalanced problem that needs to be balanced, therefore, an additional supply row is added to the transportation problem makes the Shortage of demand. As to make the total transportation problem balanced it is required to make the demand and supply equation is. Therefore, dummy storage would be added to the total supply of warehouses to make the supply equal to the demand. Dummy storage supply is determined as: 90 − 40 + 30 = 20 Similarly, dummy storage would be included for each customer that is given in the statement as penalty cost with customer 1 is $90, with customer 2, $80, and with customer 3, $110. This cost is the penalty for each customer in warehouses. According to the transportation simplex method, first, it is needed to balance the transportation problem which is shown in the below table. To From Customer 1 Customer 2 Customer 3 Supply Warehouse 1 $15 $35 $25 40 Warehouse 2 $10 $50 $40 30 Dummy Storage 90 80 110 20 Demand 30 30 30 90 Now, penalties need to be calculated by determining the difference between the greatest and the smallest cost for each customer in warehouses. Penalty for customer 1: Greatest cost − Lowest cost = $ 15 − $ 10 = $ 5 Penalty for customer 2: Greatest cost − Lowest cost = $ 50 − $ 35 = $ 15 Penalty for customer 3: Greatest cost − Lowest cost = $ 40 − $ 25 = $ 15 Penalty for warehouse 1: Greatest cost − Lowest cost = $ 35 − $ 15 = $ 20 Penalty for warehouse 2: Greatest cost − Lowest cost = $ 50 − $ 10 = $ 40 Penalty for dummy storage: Greatest cost − Lowest cost = $ 110 − $ 80 = $ 30 From Customer 1 Customer 2 Customer 3 Supply Penalty Warehouse 1 $15 $35 $25 40 20 Warehouse 2 $10 $50 $40 30 40 Dummy Storage 90 80 110 20 30 Demand 30 30 30 90 Penalty 5 15 15 Therefore, warehouse 1 has a penalty cost of $20, and warehouse 2 has a penalty cost of $40 whereas dummy storage involves a penalty of $30. Then, the allocation of maximum units is needed which is determined with the highest penalty and the lowest cost, i.e., the one denoted by an asterisk. It would find out as: For customer 1: Greatest penalty − Lowest cost = 90 − 10 = 80 For customer 2: Greatest penalty − Lowest cost = 80 − 35 = 45 For customer 3: Greatest penalty − Lowest cost = 110 − 25 = 85 From Customer 1 Customer 2 Customer 3 Supply Penalty Warehouse 1 $15 $35 $25* 40 20 Warehouse 2 10 50 40 30 40 Dummy Storage 90 80 110 20 30 Demand 30 30 30 90 Penalty 80 45 85 As the maximum penalty is with customer 3 and warehouse 1, therefore, the asterisk-marked numbers are eliminated. Now, the new penalties would be as follows: From Customer 1 Customer 2 Customer 3 Supply Penalty Warehouse 1 $15* $35* $25* 10 20 Warehouse 2 10 50 40* 30 40 Dummy Storage 90 80 110* 20 10 Demand 30 30 0* 90 Penalty 80 45 Conclusion: Here, a penalty of customer 3 is eliminated, and therefore, its demand becomes zero. The maximum number of units would be assigned to 10 in column 1. Due to this, new dummy storage will become 10 which would be the lowest and it cut down the cost of shortage and shipping.

Program plan: To formulate a balanced transportation problem to minimize the sum of shortage and shipping costs using the transportation simplex method. Explanation: The transportation problem is used to reduce the cost of delivering a good from a variety of sources or origins to a variety of destinations. An additional supply row is used in the formulation of balanced transportation by adding it for the shortage of demand. The costs for the dummy row are the penalty of each customer. Total demand for both warehouses can be calculated as: 3 customers× 30 units= 90 units Then, the total supply for both warehouses would be: 40 + 30 = 70 units As total demand is greater than the total supply, in this case, means it is an unbalanced problem that needs to be balanced, therefore, an additional supply row is added to the transportation problem makes the Shortage of demand. As to make the total transportation problem balanced it is required to make the demand and supply equation is. Therefore, dummy storage would be added to the total supply of warehouses to make the supply equal to the demand. Dummy storage supply is determined as: 90 − 40 + 30 = 20 Similarly, dummy storage would be included for each customer that is given in the statement as penalty cost with customer 1 is $90, with customer 2, $80, and with customer 3, $110. This cost is the penalty for each customer in warehouses. According to the transportation simplex method, first, it is needed to balance the transportation problem which is shown in the below table. To From Customer 1 Customer 2 Customer 3 Supply Warehouse 1 $15 $35 $25 40 Warehouse 2 $10 $50 $40 30 Dummy Storage 90 80 110 20 Demand 30 30 30 90 Now, penalties need to be calculated by determining the difference between the greatest and the smallest cost for each customer in warehouses. Penalty for customer 1: Greatest cost − Lowest cost = $ 15 − $ 10 = $ 5 Penalty for customer 2: Greatest cost − Lowest cost = $ 50 − $ 35 = $ 15 Penalty for customer 3: Greatest cost − Lowest cost = $ 40 − $ 25 = $ 15 Penalty for warehouse 1: Greatest cost − Lowest cost = $ 35 − $ 15 = $ 20 Penalty for warehouse 2: Greatest cost − Lowest cost = $ 50 − $ 10 = $ 40 Penalty for dummy storage: Greatest cost − Lowest cost = $ 110 − $ 80 = $ 30 From Customer 1 Customer 2 Customer 3 Supply Penalty Warehouse 1 $15 $35 $25 40 20 Warehouse 2 $10 $50 $40 30 40 Dummy Storage 90 80 110 20 30 Demand 30 30 30 90 Penalty 5 15 15 Therefore, warehouse 1 has a penalty cost of $20, and warehouse 2 has a penalty cost of $40 whereas dummy storage involves a penalty of $30. Then, the allocation of maximum units is needed which is determined with the highest penalty and the lowest cost, i.e., the one denoted by an asterisk. It would find out as: For customer 1: Greatest penalty − Lowest cost = 90 − 10 = 80 For customer 2: Greatest penalty − Lowest cost = 80 − 35 = 45 For customer 3: Greatest penalty − Lowest cost = 110 − 25 = 85 From Customer 1 Customer 2 Customer 3 Supply Penalty Warehouse 1 $15 $35 $25* 40 20 Warehouse 2 10 50 40 30 40 Dummy Storage 90 80 110 20 30 Demand 30 30 30 90 Penalty 80 45 85 As the maximum penalty is with customer 3 and warehouse 1, therefore, the asterisk-marked numbers are eliminated. Now, the new penalties would be as follows: From Customer 1 Customer 2 Customer 3 Supply Penalty Warehouse 1 $15* $35* $25* 10 20 Warehouse 2 10 50 40* 30 40 Dummy Storage 90 80 110* 20 10 Demand 30 30 0* 90 Penalty 80 45 Conclusion: Here, a penalty of customer 3 is eliminated, and therefore, its demand becomes zero. The maximum number of units would be assigned to 10 in column 1. Due to this, new dummy storage will become 10 which would be the lowest and it cut down the cost of shortage and shipping.

OPERATIONS RESEARCH >INTERNATIONAL EDITI

OPERATIONS RESEARCH >INTERNATIONAL EDITI

4th Edition

ISBN: 9780534423629

Author: WINSTON

Publisher: CENGAGE L

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OPERATIONS RESEARCH >INTERNATIONAL EDITI

Ch. 7.1 - Prob. 1P Ch. 7.2 - Prob. 1P Ch. 7.3 - Prob. 1P Ch. 7.5 - Prob. 1P Ch. 7 - Prob. 1RP Ch. 7 - Prob. 5RP Ch. 7 - Prob. 6RP Ch. 7 - Prob. 10RP Ch. 7 - Prob. 14RP Ch. 7 - Prob. 23RP

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