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I was asked to re-write this linear programming question with an equation less.

MAX 7X1+5X2

S.t :

4X1+3X2 <= 2400
2X1+0.5X2 <= 750
X1 >= 100
X1,X2 >= 0

What I did was I used the simplex method and I found that the maximum profit is 4030 with X1 = 100 and X2=666. Can I use that and say to obtain the maximum profit, X1 has always to be 100, then the third equation is an extra?

dfrib
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Zok
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3 Answers3

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Since we only consider a simple 2-dimensional problem, we can solve this graphically. First note that the gradient of the objective function is

∇f_obj = (7, 5)

From this points and onwards, we'll denote your variable X1 by x, and X2 by y.

The constraints describe the polytope (a) below, and the level curves for the objective function is given in (b) (brighter contour: increased objective function value).

enter image description here

The optimal value is marked by the red dot in (b) above, (x^*, y^*) = (262.5, 450).

It's apparent that the inequality constraints 4x+3y <= 2400 and 2x+0.5y <= 750 are both active, as the optimum is given in the intersection of these two.

The constraint x >= 100 (X1 >= 100), is, however, not active, and hence redundant.

dfrib
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  • I can not remove the non-negativity equations. And I do not think so since we don't know the value of X2. I tried to many methods but it seems at the end i get many different answers . – Zok Feb 03 '16 at 01:08
  • Yes I get it but in the question I was asked not to touch the non-negativity constraint. I forgot to mention it above sorry – Zok Feb 03 '16 at 01:17
  • The question is correct. In that case, can I say that since x1=262.5, then X1 >100 is redun.? And I wear it says that it is prohibited to touch the non-neg. equation.. – Zok Feb 03 '16 at 02:01
  • @Zok: Since we have a linear program, the optimum will always be in one of the extreme points of the polytype defined by the constraints of our program. Since LP:s are always convex, and we're considering a maximization problem, the optimum extreme point can be found by traversing the corners of the polytope as long as each traversal moves in the direction of the gradient of the objective function. From the level curves above, it's apparent that, for any point in the positive polytope: the gradient is directed away from the border `X1=100`. Hence, `X1>=100` is redundant. – dfrib Feb 03 '16 at 02:10
  • Okay thank you so much ! I appreciate your help. One last thing out of topic that I dont understand.. In another question, I have the last table of simplex method with the law row of 0 0 1 0 2 0 1 256 X1 = 0 X2 =0 X3=1 U=0 V=2 H=0 and Z=1 with Maximum profit of 256. Is this a unique solution? – Zok Feb 03 '16 at 02:17
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    @Zok Happy to help! Regarding your other question: I can't really help out without sitting down with the full problem and a paper and a pen (simplex iterations are a bit messy to perform by hand), and I'm off to bed shortly. Try asking (including the problem as well as your best try) at math.stackexchange.com, pure math questions are better suited there. Good luck with your optimising! – dfrib Feb 03 '16 at 02:20
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[1] 2x1 + 0.5x2 ≤ 750  
[2] 2x1 + 0.5x2 ≤ 4500 / 6
[3] 6 * (2x1 + 0.5x2) ≤ 4500
[4] 12x1 + 3x2 ≤ 4500
[5]
    12x1 + 3x2 ≤ 4500
 -   4x1 + 3x2 ≤ 2400
  ---------------------
     8x1 ≤ 2100
[6] x1 ≥ 2100 / 8
[7] x1 ≥ 262,5

That 6 in step [2] refers to how much times 3x2 in the first constraint is greater than 0.5x2 in the second constraint, in short, 3x2 / 0.5x2 = 6.

So, the third constraint x1 >= 100 can be eliminated, because, in reality, x1 must be greater than or equal 262,5, considering the fourth constraint x1,x2 >= 0.

Felypp Oliveira
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  • You analysis up to and including step 5 simply computes for which value of `x1` that the two lines `2x1 + 0.5x2 = 750` and `4x1 + 3x2 = 2400` cross. You cannot, however, go from step [5] to step [6] (changed sign?). Note that values both larger and smaller than `x1=262.5` are allowed (see the polytope of the feasible region in my answer above). Also, note that you cannot infer that constraint `x1 >= 100` is redundant **without including the objective function in your analysis**. If e.g. obj. function is `f(x1,x2) = -7x1-5x2`, then constraint `x1 >= 100` is very relevant. – dfrib Feb 03 '16 at 15:49
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Okay so the answer is the following:-

X1 >= 100. <=> X1-100 >= 0 X1 - 100 = y

Or X1 = y+100 Replace X1 by (y+100) in the first 2 equations. Replace X1 in the non negativity equation by y, remove the third equation .

Zok
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  • This is, to be blunt, incorrect. You cannot infer that constraint `X1 >= 100` is redundant without considering the objective function. If the objective function would be, as an example, `f(X1, X2) = -7X1-5X2`, then the constraint `X1 >= 100` is very much relevant. The only _truly_ redundant constraint (w.r.t. to the feasible region of the problem, and _not_ w.r.t. the gradient direction of the objective function) is `X1>=0`: this constraint is never active (or needed) as `X1>=100` ensures that the former is true, by default. – dfrib Feb 05 '16 at 23:34