This question is about the “Pivoting” step in the Simplex algorithm procedure. T

Question

This question is about the “Pivoting” step in the Simplex algorithm procedure. The step updates the Simplex tableau by pivoting on the intersection of the entering-variable column and the leaving-variable row, i.e. perform EROs on the tableau to get a 1 in the pivot position, and 0s above and below it. We know that one ERO type is “Add a multiple of one row to another row.” Consider that we are trying to make a nonzero element above or below the pivot position become 0. Please provide detailed explanations for the following questions.
(a) Will adding a multiple of the reduced cost row (row 0) to the row in which the nonzero element exists induce a wrong solution?
(b) Will adding a multiple of a constraint row other than the leaving-variable row to the row in which the nonzero element exists induce a wrong solution?
(c) Is it always most efficient to add a multiple of the leaving-variable row to the row in which the nonzero element exists?

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Explanation / Answer

While using simplex method to solve a Linear Programming Problem, We make the intersection element of leaving row and entering colum as PIVOT element, now our goal is to make this pivot element 1 and all the other elements in that colum 0.

So, Firstly we divide the entire row containing the pivot element by the pivot element, thus making the pivot element 1.

Next, in order to make the other elements in that column 0, we multiply the row containing the pivot element by the value of the element that has to be made 0 and subtract it from the row containing pivot element.

We continue this untill we get the desirable 0s and 1s in the column.

This is the method that I find easy and follow it while using Simplex method.

I hope this helps.

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