# Simplex integer solution

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I'm bit confused about integer solution from the rational solution. I understood that we have to enforce constraints to get integer solution.

So when we select v= floor(x0) = floor(-1/5) =  -1 . why we have selected x0?  can we select any variables from both basic and non-basic?

If floor did not work for the selected  variable, should we select the same variable with ceil function? I mean, if values of y0 is not an integer, should we repeat the step? When can we terminate? I am not sure that, we will end up in an infinite loop. Could you please help on this?

## 1 Answer

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We have to strengthen the constraints to the next integer values for all solutions that are not yet integers. x0 was just the first, one could also choose y1 or y2.

If floor does not work, the next step would be to try the ceil-value with the same variable as explained on page 118 of the related chapter. If that does also not work, there is no solution, but it may also happen that the recursion does not terminate as explained on 117.
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Okay.. If I select a variable and tried with ceil function, should i just check one-iteration? When should i stop the iteration for floor function ?
Well, you must continue with the strengthening until an integer solution is found, or it turns out that there is no solution with the strengthened constraints. In the first case, it terminates with a solution, in the second case, you backtrack and tray with floor. It that should also fail, there is no solution at all.
Yeah..got it. Thanks for the clarification.

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