Need help with # 6 **Transportation Problem and Algorithms****6. Transportation Problem:**- **Objective:** Find solutions using different methods for the transportation problem.- **Table Details:** - **Stores (Columns 1 to 3):** Lists the costs of transportation from each warehouse to each store. - **Warehouses (Rows 1 to 3):** Contains: - Costs to transport goods to each store. - Supplies available at each warehouse. - **Demands:** Lists the demand at each store.**Problems:** a) **Northwest Corner Solution:** Find an initial feasible solution using the northwest corner rule. b) **Set of Prices:** Based on the initial solution, determine a set of prices for transportation. c) **Unused Entry:** Identify an unused entry in the transportation table that could lower the solution's cost. d) **Spanning-tree Solution:** Develop a new spanning-tree solution using the entry identified in part (c).**7. Other Algorithm and Game Theory Concepts:** a) **Symmetry Representation:** Understanding why setting \(x_i\) equal to 2 in the cycle structure affects the count of 2-colorings left fixed by a symmetry \(\pi\). b) **Augmenting Flow Algorithm Termination:** Explanation of the termination condition for the Augmenting Flow Algorithm when no more iterations are possible. c) **Grundy Value and Kernel of Game:** Exploration of why positions with a Grundy value of 0 represent the kernel in game theory.This content provides insights into solving transportation problems and understanding key concepts in algorithms and game theory, useful for students and practitioners in operations research and computer science.

Hi! You have posted multiple questions. We will be solving only the first question. If you need an answer to others then kindly repost the question by specifying it. For (6), For (a), Let D1, D2, D3 denotes the stores and S1, S2, S3 denotes Warehouse for simplicity. TOTAL number of supply constraints : 3TOTAL number of demand constraints : 3The problem Table is, D1 D2 D3 Supplies S1 6 4 6 20 S2 8 9 7 40 S3 6 7 5 30 Demand 30 20 40 So, basically, we need to minimize the transportation cost.The rim values for S1=20 and D1=30 are compared.The smaller of the two i.e. min(20,30) = 20 is assigned to S1 D1.This exhausts the capacity of S1 and leaves 30 - 20=10 units with D1. Table-1: D1 D2 D3 Supply S1 6(20) 4 6 0 S2 8 9 7 40 S3 6 7 5 30 Demand 10 20 40 The rim values for S2=40 and D1=10 are compared.The smaller of the two i.e. min(40,10) = 10 is assigned to S2 D1.This meets the complete demand of D1 and leaves 40 - 10=30 units with S2. Table-2: D1 D2 D3 Supply S1 6(20) 4 6 0 S2 8(10) 9 7 30 S3 6 7 5 30 Demand 0 20 40 The rim values for S2=30 and D2=20 are compared.The smaller of the two i.e. min(30,20) = 20 is assigned to S2 D2.This meets the complete demand of D2 and leaves 30 - 20=10 units with S2.Table-3: D1 D2 D3 Supply S1 6(20) 4 6 0 S2 8(10) 9(20) 7 10 S3 6 7 5 30 Demand 0 0 40 The rim values for S2=10 and D3=40 are compared.The smaller of the two i.e. min(10,40) = 10 is assigned to S2 D3. This exhausts the capacity of S2 and leaves 40 - 10=30 units with D3.Table-4: D1 D2 D3 Supply S1 6(20) 4 6 0 S2 8(10) 9(20) 7(10) 0 S3 6 7 5 30 Demand 0 0 30 The rim values for S3=30 and D3=30 are compared.The smaller of the two i.e. min(30,30) = 30 is assigned to S3 D3.Table-5: D1 D2 D3 Supply S1 6(20) 4 6 0 S2 8(10) 9(20) 7(10) 0 S3 6 7 5(30) 0 Demand 0 0 0 The initial feasible solution is, D1 D2 D3 Supply S1 6 (20) 4 6 20 S2 8 (10) 9 (20) 7 (10) 40 S3 6 7 5 (30) 30 Demand 30 20 40 Thus, the minimum total transportation cost = 6×20+8×10+9×20+7×10+5×30=600Here, the number of allocated cells = 5 is equal to m + n - 1 = 3 + 3 - 1 = 5∴ This solution is non-degenerate