1 2 We studying these special cases to: 1- Present a theoretical explanation of these situations. 2- Provide a practical interpretation of what these special results could mean in real life. Degeneracy It is a term used for a basic feasible solution having one or more basic variable at value zero (0) in the next iteration and the new solution is said to be degenerate. This is in itself not a problem, but making simplex iterations from a degenerate solution may give rise to cycling (circling), meaning that after a certain number of iterations without improvement in objective value the method may turn back to the point where it started. From practical standpoint, The condition reveals that the model has at least one redundant constraint. Degeneracy (Example) Maximize: Subject to: (Solution) The constraints: Where and are slack variables. (Solution) In iteration 0, and tie for the leaving variable, leading to degeneracy in iteration 1 because the basic variable assume a zero value. The optimum reached in one additional iteration. From theoretical standpoint, degeneracy has two implications: 1- Cycling (circling): Same objective no change and improve. It is possible to have no improve and no termination for computation 2- in previous example, Both iteration, though differing I the basic- nonbasic categories of the variable, yield identical values for all variables and objective value------ Is it possible then to stop the computation at iteration 1 (when degeneracy first appear) even though it is not optimum? Answer: NO .. Why??? Answer: Because the solution may be temporarily degenerate demonstrate When the objective function is parallel to nonredundant constraint( a constraint that satisfied as an equation at the optimal solution), the objective function can assume the same optimal value at more than one solution point, thus giving rise to alternative optima. Alternative Optima Alternative Optima (Example) Maximize: Subject to: Solution Any point on the line segment BC, in the following figure, represents an alternative optimum with the same objective value z=10 Solution In first iteration, Optimal solution is 10 when x2=5/2, x1=0 The coefficient for x1 is 0, which indicates that x1 can enter the basic solution without changing the value of z. In second iteration, The new optimal solution is 10 when x1=3, x2=1 Unbounded Solution It occurs when nonbasic variables are zero or negative in all constraints coefficient (max) and variable coefficient in objective is negative Unbounded Solution (Example) Maximize: Subject to: Solution Maximize: Subject to: Solution Both and have negative z equation coefficients. Hence either one can improve the solution Select as entering variable (has most negative coefficient) All constraint coefficient under are negative or zero, so there is no leaving variable and that can be increased indefinitely without violating any constraint. each unit increase in will increase z by 1, an infinite increase in leads to an infinite increase in z. We ca say, if all value of (nonbasic variable) either zero or negative, the solution space will be unbounded Solution Infeasible solution LP models with inconsistent constraint have no feasible solution. If all constraint are of the type with nonneagtive RHS, there is a feasible solution. Other type of constraints, we use artificial variables. Infeasible solution (Example) Subject to: Maximize: Solution The following tableaux provides the simplex iterations : The optimum iteration 1 shows that R is positive (=4), which indicates that the problem is infeasible. Solution By allowing the artificial variable to be positive, the simplex method, in essence, has reversed the direction of the inequality from This result is called pseudo-optimal solution 24