Finding and recognizing infeasible Lagrange multiplier points

In summary, the conversation discusses the use of Lagrangian and KKT conditions in solving optimization problems. It is mentioned that the assumptions made in solving the problem may lead to infeasible points, and the question is raised whether obtaining 3M = 0 in one such scenario would be considered an infeasible point. The conversation also highlights the importance of understanding the problem and not relying solely on traditional methods.
  • #1
clustro
Maximize: 3*v*m
subject to:
L - m - v >= 0
V - v >= 0
m - 6 >= 0
M - m >= 0

Where L, M, and V are positive integers.

Lagrangian (call it U):

U = 3vm + K1(L - m - v) + K2(V - v) + K3(m - 6) + K4(M - m)

Where K1-K4 are the slack variables/inequality Lagrange multipliers.
Which yield the KKT conditions:

dU/dv = 3m - K1 - K2 = 0
dU/dm = 3v - K1 + K3 - K4 = 0
K1(L - m - v) = 0
K2(V - v) = 0
K3(m - 6) = 0
K4(M - m) = 0
K1-K4 >= 0

Now, suppose we assume K1 = K2 = 0, K3 and K4 != 0.

This yields:

m - 6 = 0
M - m = 0
(so m = M)

3(M) - K1 - K2 = 0

but we assumed K1 = K2 = 0, and plugging into the second KKT condition yields:

3(M) - K1 - K2 = 0, 3M = 0, which is not true.

I do not understand if I have made an error, or if this result is to be interpreted in some fashion. Does this simply mean that the point is infeasible? It just seems strange to obtain the result that way; other times I can solve for all variables and clearly see that K1-K4 are not all positive, or that another constraint is being violated.
 
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  • #2
A good place to start is K3(m - 6) = 0, which means either K3 = 0 or m = 6.
If K3 != 0, the first equation says K1 + K2 = 18.

You don't say why you chose to assume K1 = K2 = 0, K3 and K4 != 0, but that assumption isn't consistent with the equations you are trying to solve.
 
  • #3
AlephZero said:
A good place to start is K3(m - 6) = 0, which means either K3 = 0 or m = 6.
If K3 != 0, the first equation says K1 + K2 = 18.

You don't say why you chose to assume K1 = K2 = 0, K3 and K4 != 0, but that assumption isn't consistent with the equations you are trying to solve.

Well the reason I assumed it, is because from what I read, that is how these types of problems are solved. You do not know a priori which K's will be zero, and which will be nonzero. The only way to find out is to assume some of them are zero, and then plug in and see the results. If it leads to constraint violations, then the point is infeasible.

Its just several times when I discovered infeasible points, I solved for v and m, but discovered that one of the K values with be negative (all K's must be >= 0 for a correct solution). In this case, I found 3M = 0. It was different, and I wasn't sure if that result was correctly interpreted as an "infeasible point."

edit:

By K1 - K4 >= 0, I meant "K1 thru K4 >=0". Sorry!
 
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  • #4
Does nobody know? :(
 
  • #5
These problems are rarely straightforward.

First of all, you shouldn't solve a problem in a certain way just because it's how it's usually done. That usually means you don't understand what you're doing. If there is a solution to this optimization problem then there must be a set of {m,v,K1,K2,K3,K4}, which satisfy the conditions.

Second, I doubt anyone's going to solve this for you. From what I can see you made no mistake except for starting with the wrong set of assumptions. Try a different set until you eventually reach the solution.

There are obviously many assumptions to make, and you've failed to try them all, so just keep going.
 
  • #6
AntsyPants said:
These problems are rarely straightforward.

First of all, you shouldn't solve a problem in a certain way just because it's how it's usually done. That usually means you don't understand what you're doing. If there is a solution to this optimization problem then there must be a set of {m,v,K1,K2,K3,K4}, which satisfy the conditions.

Second, I doubt anyone's going to solve this for you. From what I can see you made no mistake except for starting with the wrong set of assumptions. Try a different set until you eventually reach the solution.

There are obviously many assumptions to make, and you've failed to try them all, so just keep going.

I have not asked anyone to solve the problem for me. That is not what I am looking for. I probably won't solve it all the way myself, because there are too many permutations and the solution is not direly important. The question I am asking is more theoretical.
I am simply asking if this would be considered an infeasible point, since you obtain 3M = 0, which contradicts the assumptions made.

I am assuming it is, but maybe someone who knows more/has more experience with the KKT conditions/Lagrange multipliers has more insight.

First of all, you shouldn't solve a problem in a certain way just because it's how it's usually done. That usually means you don't understand what you're doing.
What leads you to believe that this is what I am doing? Indeed, I read that you make various assumptions about the Lagrange multipliers, but that makes sense/seems like the only real way to solve these problems, e.g. by assuming that various constraints are either slacking or riding, and trying to find the optimum over all points that obey the constraints.
 

Related to Finding and recognizing infeasible Lagrange multiplier points

1. What are Lagrange multiplier points and why are they important?

Lagrange multiplier points are points in a function where the gradient of the function is equal to the gradient of the constraint. They are important because they help us find the maximum or minimum value of a function subject to certain constraints.

2. How do you find infeasible Lagrange multiplier points?

Infeasible Lagrange multiplier points can be found by setting up the Lagrange multiplier formula and solving for the Lagrange multipliers. If the resulting Lagrange multipliers are negative, there are no feasible solutions and the point is considered to be infeasible.

3. What is the significance of infeasible Lagrange multiplier points?

Infeasible Lagrange multiplier points indicate that there is no feasible solution to the optimization problem. This could be due to the constraints being too restrictive or the function being too complex.

4. Can infeasible Lagrange multiplier points be used in practical applications?

No, infeasible Lagrange multiplier points are not useful in practical applications. They do not represent feasible solutions and therefore cannot be implemented in real-world situations.

5. How do you recognize infeasible Lagrange multiplier points graphically?

Infeasible Lagrange multiplier points can be recognized graphically by plotting the function and the constraint on a graph and looking for points where the gradients of the two curves are equal. If there are no points of intersection, then the point is infeasible.

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