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I would like to know why here on the page 30, the author writes about absolute values, at all ?

Phase I: minimize the sum of the artificial variables, starting from the BFS wherethe absolute value of the artificial variable for each constraint, or of the slack variable in case there is no artificial variable, is equal to that of the right-handside.

He takes the aboslute values of both, left hand side and right hand side. The BFS however, may have a sign and he will then need to go back to that sign?

user122424
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  • The BFS contains non-negative values only. – callculus42 Feb 25 '19 at 18:45
  • But the right hand side of some equation can be -2 say, and the artificial variable on the left hand side $-x_3$? – user122424 Feb 25 '19 at 18:49
  • No, you always add (+) artificial variables. Otherwise they wouldn´t be a part of the BFS. – callculus42 Feb 25 '19 at 18:51
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    See page 30 here item 1. : "...add a new artificial variable that has the same sign as the right-handside" – user122424 Feb 25 '19 at 18:52
  • The RHS is non-negative as well. – callculus42 Feb 25 '19 at 18:53
  • Why cannot be among equations one with r.h.s. value $-2$ ? And yet, why would they consider absolute values of them if they were always positive? – user122424 Feb 25 '19 at 18:54
  • Sure you can have a negative RHS. But for the simplex-algorithm the RHS has to be non negative. Example: $x_2+2x_2=-9$. Add an artificial variable. $x_2+2x_2-a_1=-9$. Now we need a non-negative RHS. Thus we multiply the equation by (-1): $-x_2-2x_2+a_1=9$. In general the formulation at this document is not very comprehensible – callculus42 Feb 25 '19 at 19:00
  • Have a look to what I´ve written here . I hope it helps. – callculus42 Feb 25 '19 at 19:04
  • @callculus Thank you for your explanation. BUT whether we have $x_2+2x_3-a_1$ OR $x_2+2x_3+a_1$ depends on whether we had initially $x_2+2x_3\geq -9$ OR $x_2+2x_3\leq -9$, respectively. Is this true? Note that I've renamed your $2x_2$ to $2x_3$ because your first variable already is $x_2$. – user122424 Feb 25 '19 at 20:17
  • The best advice I can give is to multiply both inequalities by (-1). The inequality signs change the direction. $(1) \ x_2+2x_3\geq -9 \Rightarrow -x_2-2x_3\leq 9$. And $(2) \ x_2+2x_3\leq -9 \Rightarrow -x_2-2x_3\geq 9$. Now both RHS are non-negative. For (1) you just need a slack variable. For (2) your need a surplus variable and a artificial variable. – callculus42 Feb 25 '19 at 20:29
  • I almost understand BUT for the moment why do I need an artificial variable for (2) ? How and why do I insert it? – user122424 Feb 25 '19 at 20:34
  • I KNOW NOW! It is because r.h.s. and slack variable have opposite signs, right? – user122424 Feb 25 '19 at 20:38
  • What would be the solution of $-x_2-2x_3\geq 9$? If we add a slack variable we get $-x_2-2x_3-s_1= 9$ Since $s_1\geq 0$ the value of it must be $0$. Thus we need an artificial variable to obtain $a_1=9$. – callculus42 Feb 25 '19 at 20:40
  • Not if the inequality has a $\leq$-sign. But at a $\geq$-inequality you are right. At all cases I presume that the RHS is non-negative. – callculus42 Feb 25 '19 at 20:44

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