Consider the following optimization problem:Minimizex∈IR3 2x21 + x22 + 2x23Subject to x1 + 2x2 + x3 ≤ −3.(a) (10 points) For each c > 0, defineqc(x) := 2x21 + x22 + 2x23 + c2 (x1 + 2x2 + x3 + 3)2+Argue that qc is convex and find the global minimizer of qc.(b) (10 points) For each µ > 0, defineℓµ(x) := 2x21 + x22 + 2x23 − µ ln (−3 − x1 − 2x2 − x3) .Argue that ℓµ is convex and find the global minimizer of ℓµ. You may use without proof thefact that the function t 7 → − ln(−3 − t) is convex (as an extended real-valued function).
Question
Consider the following optimization problem:Minimizex∈IR3 2x21 + x22 + 2x23Subject to x1 + 2x2 + x3 ≤ −3.(a) (10 points) For each c > 0, defineqc(x) := 2x21 + x22 + 2x23 + c2 (x1 + 2x2 + x3 + 3)2+Argue that qc is convex and find the global minimizer of qc.(b) (10 points) For each µ > 0, defineℓµ(x) := 2x21 + x22 + 2x23 − µ ln (−3 − x1 − 2x2 − x3) .Argue that ℓµ is convex and find the global minimizer of ℓµ. You may use without proof thefact that the function t 7 → − ln(−3 − t) is convex (as an extended real-valued function).
Solution
(a) The function qc(x) is a sum of two convex functions, hence it is convex. The first part 2x21 + x22 + 2x23 is a quadratic function which is convex. The second part c2 (x1 + 2x2 + x3 + 3)2 is also a quadratic function which is convex.
To find the global minimizer of qc, we take the derivative of qc with respect to x and set it equal to zero. This gives us the system of equations:
4x1 + c2 (2x1 + 4x2 + 2x3 + 6) = 0, 2x2 + c2 (x1 + 4x2 + x3 + 3) = 0, 4x3 + c2 (x1 + 2x2 + 2x3 + 3) = 0.
Solving this system of equations will give us the global minimizer of qc.
(b) The function ℓµ(x) is a difference of a convex function and a concave function. The first part 2x21 + x22 + 2x23 is a quadratic function which is convex. The second part − µ ln (−3 − x1 − 2x2 − x3) is a concave function because the negative logarithm function is concave.
To find the global minimizer of ℓµ, we take the derivative of ℓµ with respect to x and set it equal to zero. This gives us the system of equations:
4x1 - µ / (−3 − x1 − 2x2 − x3) = 0, 2x2 - 2µ / (−3 − x1 − 2x2 − x3) = 0, 4x3 - µ / (−3 − x1 − 2x2 − x3) = 0.
Solving this system of equations will give us the global minimizer of ℓµ.
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