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(a) Draw a valid loop starting and finishing at X31. (b) Identify the donor and recipient cells. (c) Determine the leaving basic variable.

(a) Draw a valid loop starting and finishing at X31. (b) Identify the donor and recipient cells. (c) Determine the leaving basic variable.

Practical Management Science
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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### Problem Statement: **4.** Reconsider the previous problem, assume that we have determined that the current basic feasible (BF) solution is not an optimal solution, and assume that \( x_{31} \) is the entering basic variable. #### Tasks: (a) Draw a valid loop starting and finishing at \( x_{31} \). (b) Identify the donor and recipient cells. (c) Determine the leaving basic variable. ### Table Explanation: - The table represents a transportation problem with three sources (rows labeled 1, 2, 3) and four destinations (columns labeled Destination 1, Destination 2, Destination 3, Destination 4). - Supply values for each source are given in the last column, while demand values for each destination are listed in the last row. - The numbers inside the cells represent the cost associated with transporting from a source to a destination. - Some cells contain circled numbers, which represent the current allocation or transported units. ### Key Features: - **Highlighted Entry**: The cell \( x_{31} \) in row 3, column 1 is marked with "Enter" indicating it is the entering basic variable. - **Current Allocations**: - \( x_{11} = 10 \) - \( x_{12} = 2 \) - \( x_{21} = 8 \) - \( x_{22} = 9 \) - \( x_{24} = 1 \) - \( x_{34} = 10 \) ### Task Solutions: - **(a) Draw a valid loop**: Start and finish the loop at the cell marked "Enter" (x₃₁), connecting through the cells with allocations to form a closed loop for potential reallocation. - **(b) Identify the donor and recipient cells**: The donor cell is the cell within the loop with the smallest allocation, while recipient cells will be adjusted positively along the loop. - **(c) Determine the leaving basic variable**: The leaving basic variable corresponds to the donor cell where the smallest allocation in the loop is reduced to zero. This exercise reinforces understanding of the transportation model in linear programming and optimization strategies involving basic feasible solutions, candidate loops, and pivoting.
Transcribed Image Text:### Problem Statement: **4.** Reconsider the previous problem, assume that we have determined that the current basic feasible (BF) solution is not an optimal solution, and assume that \( x_{31} \) is the entering basic variable. #### Tasks: (a) Draw a valid loop starting and finishing at \( x_{31} \). (b) Identify the donor and recipient cells. (c) Determine the leaving basic variable. ### Table Explanation: - The table represents a transportation problem with three sources (rows labeled 1, 2, 3) and four destinations (columns labeled Destination 1, Destination 2, Destination 3, Destination 4). - Supply values for each source are given in the last column, while demand values for each destination are listed in the last row. - The numbers inside the cells represent the cost associated with transporting from a source to a destination. - Some cells contain circled numbers, which represent the current allocation or transported units. ### Key Features: - **Highlighted Entry**: The cell \( x_{31} \) in row 3, column 1 is marked with "Enter" indicating it is the entering basic variable. - **Current Allocations**: - \( x_{11} = 10 \) - \( x_{12} = 2 \) - \( x_{21} = 8 \) - \( x_{22} = 9 \) - \( x_{24} = 1 \) - \( x_{34} = 10 \) ### Task Solutions: - **(a) Draw a valid loop**: Start and finish the loop at the cell marked "Enter" (x₃₁), connecting through the cells with allocations to form a closed loop for potential reallocation. - **(b) Identify the donor and recipient cells**: The donor cell is the cell within the loop with the smallest allocation, while recipient cells will be adjusted positively along the loop. - **(c) Determine the leaving basic variable**: The leaving basic variable corresponds to the donor cell where the smallest allocation in the loop is reduced to zero. This exercise reinforces understanding of the transportation model in linear programming and optimization strategies involving basic feasible solutions, candidate loops, and pivoting.
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