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Why is one optimal value greater than or equal to the other one here?
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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding the optimal value in an optimization problemWhen is a linear program feasible but not optimal?Why sharing optimal costs is so important in optimality?Why the definition for optimal value is the $inf{f_0(x)}$ rather than $min{f_0(x)}$?What is the optimal value of a quadratic program when there does not exist a solution?Prove the optimal value of this minimum is not smaller than the optimal value of this maximumWhy is the error here at most $frac{1}{3}$?Why is this the optimal value of the dual problem?How does this list of optimal values prove Farkas' lemma?How is this optimal value derived?
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Let the first program be $$min frac{c^Tx + d}{e^Tx+f} text{ subject to }{Gx le h, Ax = b}$$
It can be transformed to equivalent linear program:
$$min (c^Ty + dz) text{ subject to } {Gy - hz ge 0, Ay - bz = 0, e^Ty + fz = 1, z ge 0}$$
If $x$ is feasible in the first one, then $y = frac{x}{e^Tx + f}, z = frac{1}{e^Tx + f}$ is feasible in the second one. It follows that the optimal value of the original program is greater than or equal to the optimal value of the transformed program.
Can someone explain why the optimal value of the original program is greater than or equal to the optimal value of the transformed program?
proof-explanation convex-optimization
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add a comment |
$begingroup$
Let the first program be $$min frac{c^Tx + d}{e^Tx+f} text{ subject to }{Gx le h, Ax = b}$$
It can be transformed to equivalent linear program:
$$min (c^Ty + dz) text{ subject to } {Gy - hz ge 0, Ay - bz = 0, e^Ty + fz = 1, z ge 0}$$
If $x$ is feasible in the first one, then $y = frac{x}{e^Tx + f}, z = frac{1}{e^Tx + f}$ is feasible in the second one. It follows that the optimal value of the original program is greater than or equal to the optimal value of the transformed program.
Can someone explain why the optimal value of the original program is greater than or equal to the optimal value of the transformed program?
proof-explanation convex-optimization
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$begingroup$
This just reflects the fact that you're only proving one direction of the if-and-only-if equivalence. You're proving that every feasible point in the first problem leads to a feasible point in the second—but you're not saying anything about optimality. So given an optimal points for the first problem, you've only proved that leads to a feasible, but possibly suboptimal, point in the second. Without further proof, it's possible that the optimal point for the second problem cannot be mapped back to the first.
$endgroup$
– Michael Grant
Mar 25 at 3:25
add a comment |
$begingroup$
Let the first program be $$min frac{c^Tx + d}{e^Tx+f} text{ subject to }{Gx le h, Ax = b}$$
It can be transformed to equivalent linear program:
$$min (c^Ty + dz) text{ subject to } {Gy - hz ge 0, Ay - bz = 0, e^Ty + fz = 1, z ge 0}$$
If $x$ is feasible in the first one, then $y = frac{x}{e^Tx + f}, z = frac{1}{e^Tx + f}$ is feasible in the second one. It follows that the optimal value of the original program is greater than or equal to the optimal value of the transformed program.
Can someone explain why the optimal value of the original program is greater than or equal to the optimal value of the transformed program?
proof-explanation convex-optimization
$endgroup$
Let the first program be $$min frac{c^Tx + d}{e^Tx+f} text{ subject to }{Gx le h, Ax = b}$$
It can be transformed to equivalent linear program:
$$min (c^Ty + dz) text{ subject to } {Gy - hz ge 0, Ay - bz = 0, e^Ty + fz = 1, z ge 0}$$
If $x$ is feasible in the first one, then $y = frac{x}{e^Tx + f}, z = frac{1}{e^Tx + f}$ is feasible in the second one. It follows that the optimal value of the original program is greater than or equal to the optimal value of the transformed program.
Can someone explain why the optimal value of the original program is greater than or equal to the optimal value of the transformed program?
proof-explanation convex-optimization
proof-explanation convex-optimization
edited Mar 24 at 20:11
Oliver G
asked Mar 24 at 19:57
Oliver GOliver G
1,2761634
1,2761634
$begingroup$
This just reflects the fact that you're only proving one direction of the if-and-only-if equivalence. You're proving that every feasible point in the first problem leads to a feasible point in the second—but you're not saying anything about optimality. So given an optimal points for the first problem, you've only proved that leads to a feasible, but possibly suboptimal, point in the second. Without further proof, it's possible that the optimal point for the second problem cannot be mapped back to the first.
$endgroup$
– Michael Grant
Mar 25 at 3:25
add a comment |
$begingroup$
This just reflects the fact that you're only proving one direction of the if-and-only-if equivalence. You're proving that every feasible point in the first problem leads to a feasible point in the second—but you're not saying anything about optimality. So given an optimal points for the first problem, you've only proved that leads to a feasible, but possibly suboptimal, point in the second. Without further proof, it's possible that the optimal point for the second problem cannot be mapped back to the first.
$endgroup$
– Michael Grant
Mar 25 at 3:25
$begingroup$
This just reflects the fact that you're only proving one direction of the if-and-only-if equivalence. You're proving that every feasible point in the first problem leads to a feasible point in the second—but you're not saying anything about optimality. So given an optimal points for the first problem, you've only proved that leads to a feasible, but possibly suboptimal, point in the second. Without further proof, it's possible that the optimal point for the second problem cannot be mapped back to the first.
$endgroup$
– Michael Grant
Mar 25 at 3:25
$begingroup$
This just reflects the fact that you're only proving one direction of the if-and-only-if equivalence. You're proving that every feasible point in the first problem leads to a feasible point in the second—but you're not saying anything about optimality. So given an optimal points for the first problem, you've only proved that leads to a feasible, but possibly suboptimal, point in the second. Without further proof, it's possible that the optimal point for the second problem cannot be mapped back to the first.
$endgroup$
– Michael Grant
Mar 25 at 3:25
add a comment |
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$begingroup$
This just reflects the fact that you're only proving one direction of the if-and-only-if equivalence. You're proving that every feasible point in the first problem leads to a feasible point in the second—but you're not saying anything about optimality. So given an optimal points for the first problem, you've only proved that leads to a feasible, but possibly suboptimal, point in the second. Without further proof, it's possible that the optimal point for the second problem cannot be mapped back to the first.
$endgroup$
– Michael Grant
Mar 25 at 3:25