Is this a trick question? Standard form

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Homework Statement



[PLAIN]http://img191.imageshack.us/img191/7440/unledtev.png


The Attempt at a Solution



There are like 2 other problems in my book similar to this one.

I thought problems posed in this manner are already in standard form. They say

"max [obj f]

s.t.

constraints, for variables positive "
 
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flyingpig said:

Homework Statement



[PLAIN]http://img191.imageshack.us/img191/7440/unledtev.png


The Attempt at a Solution



There are like 2 other problems in my book similar to this one.

I thought problems posed in this manner are already in standard form. They say

"max [obj f]

s.t.

constraints, for variables positive "

I hope your book does not say that variables are positive, for often they are not: they can be ZERO as well, and often are in an optimal solution. So, you should say non-negative, not positive. Problems with positive variables may not have any optimal solutions; the simplest example of this is min x, subject to x > 0.

Does the problem above satisfy ALL the requirements of a "standard" problem?

RGV
 
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Oh I have change 3x_1 +3x_2 + x_3 \geq 2 to -3x_1 - 3x_2 - x_3 \leq -2

And for x_1 + 2x_3 = -4, I have to change it to -x_1 - 2x_3 \leq 4 because x_1 and x_3 are nonnegative ?

Also what does u.r.s. mean...? Because I just assumed it meant it can be positive..

EDIT:

x_1 + 2x_3 = -4

Could also say

x_1 + 2x_3 \geq -4 and x_1 + 2x_3 \leq -4

Then

-x_1 -2x_3 \leq 4 and x_1 + 2x_3 \leq -4 would make the requirements for constraints in standard form.
 
flyingpig said:
Oh I have change 3x_1 +3x_2 + x_3 \geq 2 to -3x_1 - 3x_2 - x_3 \leq -2

And for x_1 + 2x_3 = -4, I have to change it to -x_1 - 2x_3 \leq 4 because x_1 and x_3 are nonnegative ?

Also what does u.r.s. mean...? Because I just assumed it meant it can be positive..

EDIT:

x_1 + 2x_3 = -4

Could also say

x_1 + 2x_3 \geq -4 and x_1 + 2x_3 \leq -4

Then

-x_1 -2x_3 \leq 4 and x_1 + 2x_3 \leq -4 would make the requirements for constraints in standard form.

Different authors have different definitions of "standard form". For example, the standard form in https://netfiles.uiuc.edu/angelia/www/ge330fall09_stform4.pdf is max or min cx, st AX = b, x >= 0 (obtained by using slack or surplus variables if necessary). In others sources the standard is a minimization, in some others a maximization, in some others the constraints must all be <=, etc. Myself, I prefer the form max cx st Ax=b, x >= 0 form, because that is the form you need to get started on the simplex method. However, *ALL sources agree that 'x >= 0' is part of the standard*.

In your problem, x_3 urs means, I think, that x_3 is unrestricted in sign; that is, x_3 can be < 0 or >= 0. That makes your problem non-standard, and you are asked to do something to it to put it into standard form. More than that I cannot say without solving your problem for you.

RGV
 
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Ray Vickson said:
Different authors have different definitions of "standard form". For example, the standard form in https://netfiles.uiuc.edu/angelia/www/ge330fall09_stform4.pdf is max or min cx, st AX = b, x >= 0 (obtained by using slack or surplus variables if necessary). In others sources the standard is a minimization, in some others a maximization, in some others the constraints must all be <=, etc. Myself, I prefer the form max cx st Ax=b, x >= 0 form, because that is the form you need to get started on the simplex method. However, *ALL sources agree that 'x >= 0' is part of the standard*.

It does say (max) in parenthesis, let's go with mine!

In your problem, x_3 urs means, I think, that x_3 is unrestricted in sign; that is, x_3 can be < 0 or >= 0. That makes your problem non-standard, and you are asked to do something to it to put it into standard form. More than that I cannot say without solving your problem for you.

RGV

Oh that's easy, I can just make it into positive as I have and erase my new inequality!

Thanks
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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