Question: QUESTION 1 This Question worth 0.5 mark) represents the Continous Assessement (Total 3 Marks The variable to remove from the current basis is the variable

QUESTION 1 This Question worth 0.5 mark)
QUESTION 1 This Question worth 0.5 mark)
QUESTION 1 This Question worth 0.5 mark)
QUESTION 1 This Question worth 0.5 mark) represents the Continous Assessement (Total 3 Marks The variable to remove from the current basis is the variable with the biggest positive e-val. the variable with the smallest positive 9-4 value. the variable with the smallest positive ratio value the variable with the biggest positive ratio value QUESTION 2 This Question worth 0.5 mark) represents the Continous Assessement (Total 3 Marks Using a Simplex table, we know we have reached the optimal solution for a minimization problem when the grow, has to numbers in it has no nonzero numbers in it none of the answers Oh no positive numbers in it O has no negative numbers in it QUESTION 3 (This Question worth 0.5 mark) represents the Continous Assessement Total 3 Marks) Artificial variables are added to the left hand side of the constraint, wher: A feasible solution is available O A feasible solution is not available O A basic feasible solution is available A basic feasible solution is not available QUESTION 4 (This Question worth 0.5 mark) represents the Continous Assessement (Total 3 Marks Which of the following is true in the case of the simplex method of L.R? The simplex algorithm is an iterative procedure. Inequalities are not converted into equations It is limited for two variables problems It cannot be used for two variables problems QUESTIONS K This Question worth 0.5 mark) represents the Continous Assessement (Total 3 Marks) A tie (equality for leaving variable in simplex proceduer implies: cycling No solution Optimality O Degeneracy QUESTIONG Markel

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