5

I'm trying to prepare for my midterm and I was going over some problems out of my algorithm book but can't seem to figure out the following problem:

Find necessary and sufficient conditions on the reals a and b under which the linear program

max: x+y
ax + by <=1
x, y =>0

(a) is infeasible. (b) is unbounded. (c) has a finite and unique optimal solution.

here is what I've come up with: for (a), we can add another constraint: ax+by=>5

I'm not sure what to do about b and c.I'm not sure If I'm allowed to change the constraints I'm already given or add new ones.

Any help will be appreciated. Thanks so much advance.

sap
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  • This problem sounds to me like you are allowed to choose `a` and `b`, but may not add or otherwise modify any constraints of the program. Except the part about "necessary and sufficient" means you need to describe a way to determine which of the three cases (if any) applies no matter what `a` and `b` you're given. – aschepler Nov 15 '10 at 22:24
  • Just curious: is that a "linear program" or a "linear programming model"? You know correct nomenclature is key in this field. – R. Martinho Fernandes Nov 15 '10 at 22:25
  • Should be linear programming model but that's how it is written in my book. – sap Nov 15 '10 at 22:27

3 Answers3

4

a) I'm not sure if this is possible unless you add a constraint just like you did.
b) if a and b are both less than or equal to zero your problem will be unbounded
c) if a and b are both larger than zero, and they are not equal to each other you will have a unique optimal solution

Haluk
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  • More precisely: (a) is never possible without additional constraint, since `(0, 0)` is always feasible. (b) the problem is unbounded if and only if _either_ `a<=0` or `b<=0` (whatever the value of the second coefficient is) – Nicolas Grebille Oct 24 '14 at 19:32
1

a. This linear program never infeasible. No matter what value of a and b, there is always a feasible solution to satisfy ax + by <= 1

b. This linear program is unbounded when either a <= 0 or b <= 0.

c. Finite and unique optimal solution exist when a != b and both a > 0 and b > 0

kdebugging
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-1

For part (a): it is infeasible when either a=0 and b<0 OR a<0 and b=0