Sensitivity Ranges Can Be Computed Only for the Right-hand Sides of Constraints.
Vector on Right Hand Side (RHS) associated with the constrains of a Linear Programming model may have dissimilar practical interpretations such as the availability of inputs for the industry of certain products, limiting of chapters, demand requirements, amidst other. As a result it is interesting to analyze the impact of the alter of ane or more coefficients of the vector on right hand side over the original results of the model without the need of optimizing again. In item if nosotros verify that:
It can exist said that the current optimal base is preserved. This implies that the basic variables of the model will remain so nether this new scenario and therefore you will notice the new optimal solution of the trouble through the resolution of the original system of equations (the original active constrains are preserved). Now, if any of the coefficients in the calculation of the basic variables vector takes a negative value , we have an infeasible basic solution, which forces u.s. to make an update of the results of the model to find the new solution, optimal base and optimal value, but non needing to passing through the reoptimization of it.
Changes in the Right Hand Side (RHS) of the Constraint
Consider the post-obit Linear Programming model:
When solving this Linear Programming model with Simplex Method you lot accomplish the next final tableau, where s1, s2, and s3 are the slack variables of the constrains 1, 2, and 3, respectively:
The bones variables are x=100, s2=400, y=350, all of which meet the conditions of not negativity (i.e. is a basic viable solution) and also the reduced cost of the non-basic variables (s1 y s3) are bigger or equal to zero, the necessary and sufficient condition to ensure that nosotros have the optimal solution of the problem (optimal bones viable solution). In addition and related with the previous suggestion we can confirm the results we got:
At present lets consider that the correct hand side of the constrain ane changes from its original value i.600 to 1.650. Does information technology change the current optimal basis? To practise this we recalculate the vector of the bones variables:
Y'all can see that all the coefficients of the vector of basic variables (Xb) are bigger or equal to goose egg, i.due east. the optimum base of operations (aforementioned basic variables) is preserved but the optimal solution changes to x=125, s2=250, y=350. Additionally the optimal value now is V(P)=3.175. Yet, it is non necessary to continue the iterations of the Simplex Method (as we are facing an optimal basic feasible solution) and preventing united states from doing a reoptimization.
The natural question is:What happens if, when computing the vector of the bones variables, at to the lowest degree one of the variables gets a negative value? Now let´due south alter simultaneously the correct sides of constrains 1 and 2 to 2.000 and 1.500, respectively. The new basic variables vector is defined in the following way:
Detect that now the basic variable s2=-ane.000 takes a value that does not satisfy the status of non negativity for the determination variables. To address this situation ofinfeasibility information technology is necessary to update the last tableau of the Simplex Method with the value of the bones variables and the objective function value:
In social club to find the optimal solution of this problem from the above table, the Dual Simplex Methodcan be applied. The basic variable given by the base is s2 (basic variable associated with row ii where we detect the negative 'right hand side' ). In order to make up one's mind the variable that goes to the base, we calculate the minimum quotient: Min{-three/2/-three}=1/2 ==> s1 goes to the base. Nosotros update the tableau of the Simplex Method with the post-obit:
You can run across that it was just necessary to practise an additional iteration to get the optimal solution for the new scenario (ten=400/3, s1=ane.000/3, y=350) with an optimal value of V(P)=3.200. The post-obit chart fabricated with Geogebra, allows usa to see the new optimal solution and structure of the problem, where now the optimal solution finds the 2 and three active constrains (the original problem in its optimal solution considered ane and 3 as active constraints):
Source: https://www.linearprogramming.info/changes-in-the-right-hand-side-rhs-of-the-constraint-sensitivity-analysis-in-linear-programming/
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