Transportation-Problem
Question 1 |
A basic feasible solution of an mXn Transportation-Problem is said to be non-degenerate, if basic feasible solution contains exactly____number of individual allocation in ___positions
m+n+1, independent | |
m+n-1, independent | |
m+n-1, appropriate | |
m-n+1, independent |
Question 1 Explanation:
The initial solution of a transportation problem is said to be non-degenerate basic feasible solution if it satisfies:
→The solution must be feasible, i.e. it must satisfy all the supply and demand constraints.
→The number of positive allocations must be equal to m+n-1, where m is the number of rows and n is the number of columns.
→All the positive allocations must be in independent positions
A few terms used in connection with transportation models are defined below.
1. Feasible solution: A feasible solution to a transportation problem is a set of non-negative allocations, xij that satisfies the rim (row and column) restrictions.
2. Basic feasible solution: A feasible solution to a transportation problem is said to be a basic feasible solution if it contains no more than m + n – 1 non – negative allocations, where m is the number of rows and n is the number of columns of the transportation problem.
3. Optimal solution: A feasible solution (not necessarily basic) that minimizes (maximizes) the transportation cost (profit) is called an optimal solution.
4. Non-degenerate basic feasible solution: A basic feasible solution to a (m x n) transportation problem is said to be non – degenerate if, the total number of non-negative allocations is exactly m + n – 1 (i.e., number of independent constraint equations), and these m + n – 1 allocations are in independent positions.
5. Degenerate basic feasible solution: A basic feasible solution in which the total number of non-negative allocations is less than m + n – 1 is called degenerate basic feasible solution.
Note: It is very standard and regular question and asked in UGC-NET Dec-2015 paper-3
Ref-https://solutionsadda.in/ugc-net-cs-2015-dec-paper-3/
→The solution must be feasible, i.e. it must satisfy all the supply and demand constraints.
→The number of positive allocations must be equal to m+n-1, where m is the number of rows and n is the number of columns.
→All the positive allocations must be in independent positions
A few terms used in connection with transportation models are defined below.
1. Feasible solution: A feasible solution to a transportation problem is a set of non-negative allocations, xij that satisfies the rim (row and column) restrictions.
2. Basic feasible solution: A feasible solution to a transportation problem is said to be a basic feasible solution if it contains no more than m + n – 1 non – negative allocations, where m is the number of rows and n is the number of columns of the transportation problem.
3. Optimal solution: A feasible solution (not necessarily basic) that minimizes (maximizes) the transportation cost (profit) is called an optimal solution.
4. Non-degenerate basic feasible solution: A basic feasible solution to a (m x n) transportation problem is said to be non – degenerate if, the total number of non-negative allocations is exactly m + n – 1 (i.e., number of independent constraint equations), and these m + n – 1 allocations are in independent positions.
5. Degenerate basic feasible solution: A basic feasible solution in which the total number of non-negative allocations is less than m + n – 1 is called degenerate basic feasible solution.
Note: It is very standard and regular question and asked in UGC-NET Dec-2015 paper-3
Ref-https://solutionsadda.in/ugc-net-cs-2015-dec-paper-3/
Question 2 |
Consider the following transportation problem:
is degenerate solution | |
is optimum solution | |
needs to improve | |
is infeasible solution |
Question 2 Explanation:
Step1: In vogel’s approximation method first find out the row difference(by calculating the difference between the two smallest values of that row) and column difference(by calculating the difference between the two smallest values of that row)
Then select the row having highest row difference and after selecting row choose the column/cell of that row having minimum value.
After that fulfill the demand by using supply from the selected row and column.
After that if demand becomes zero then delete that column and if the row become zero then delete that row and then repeat step 1.
Then select the row having highest row difference and after selecting row choose the column/cell of that row having minimum value.
After that fulfill the demand by using supply from the selected row and column.
After that if demand becomes zero then delete that column and if the row become zero then delete that row and then repeat step 1.
There are 2 questions to complete.