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TechWidget Incorporated maintains three warehouses located across the United States in order to service to four customers. Warehouse has a supply of 35 units, warehouse 2 has a supply of 50 units, and warehouse 3 has a supply of 40 units. Customer 1 has a demand of 45 units, customer 2 has a demand of 20 units, customer 3 has a demand of 30 units, and customer 4 has a demand of 30 units. Provided below is a balance transportation tableau that includes the transportation cost between each warehouse and customer. TechWidget desires to minimize the total transportation cost while satisfying customer demand. Additionally, an initial BFS, whose values shown in red, has been provided in the tableau. Customer 1 Customer 2 Customer 3 Customer 4 6 10 9 Warehouse 1 35 12 13 Warehouse 2 10 20 20 14 9 16 Warehouse 3 s3 40 10 30 d1 45 d4-30 (a) Use the transportation simplex method to find an optimal solution- show and label all steps. Specify the allocation of supply used to satisfy customer demand. Determine the total transportation cost. (b) Setup and solve this problem in LINGO. Provide the model and corresponding output.

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Answer #1

(a) Optimal solution is determined using transportation simplex method as below:

Customer Supply 35 50 40 0 4 Wh 1 Wh 2 Wh 3 35 Cu2 10 20 20 14 16 10 30 Demand 45 20 30 30 Total Cost1180 Optimisation using transportation simplex method Basic Variables Iteration 1 Calculate reduced cost (Wij) for Non-Basic Variables x12 ultv2 60+11 -65 x13 ul+v3- 10012 10 2 x14 ul+4 901 9 x24 u244- 711 x31 u3+- 144+8 142 x32|u2+v2-91= 4 + 11-9 = 61most positive value, entering variable Wij corresponding to x32 is >0, therefore it is the new basic variable. In the next iteration, we use stepping stone method to assign x32 Customer V2 2 V3 Wh 1 Wh2 Wh 3 4 Supply check 35 50 40 35 12 13 10 10 30 14 16 30 30 10 Demand45 20 30 Total Cost1120 Basic Variables Iteration 2 Calculate reduced cost (Wij) for Non-Basic Variables x12 u1+v2 60 11 -65 most posive value, entering variable x13 u1+v3- 10012 10 2 x14 ul+y4 90 7-9 2 x24 u244 717 x31 u3+1 142+8 14-8 X33 u3+3 16 2+ 12 166 Wij corresponding to x12 is >0, therefore it is the new basic variable. In the next iteration, we use stepping stone method to assign x12

(b) Solution using LINGO is following:

Optimal result:


Variable Value Reduced Cost
X11 0.000000 2.000000
X12 10.00000 0.000000
X13 25.00000 0.000000
X14 0.000000 7.000000
X21 45.00000 0.000000
X22 0.000000 3.000000
X23 5.000000 0.000000
X24 0.000000 2.000000
X31 0.000000 5.000000
X32 10.00000 0.000000
X33 0.000000 3.000000
X34 30.00000 0.000000

Total cost = 1020

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