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Methods for Constrained Optimization - MIT Certificate Error

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so that( x) CT λu = ∇f+ ,x min= x 0+ ku,where k is the value that minimizes( k) f ( x ku)F =0+ .We can confirm that.u( x min) = 0x minis the minimum by checking that the gradient projection vector3.6 Gradient Projection Method <strong>for</strong> Inequality ConstraintsThe gradient projection method can also be extended to solve optimization problems withlinear inequality constraints of the <strong>for</strong>m:minimize f ( x)subject to Dx − e ≤ 0 .The algorithm that I implemented <strong>for</strong> applying gradient projection to problems withinequality constraints is as follows:1. Find the global minimum x 0.2. Check if x0satisfies the inequality constraints. If it does, we’re done, i.e. xmin= x0.If not, continue.3. Find any vector x′ that is on the active set of constraints, i.e. some boundary pointthat satisfies all constraints.4. Given that the vector x′ is on the active set, check if this vector is the minimum bytesting if u( x′ ) ≈ 0 . If it is, then we’re done, i.e. x = x′min. If not, use gradientprojection to update x′ .5. If at some point the vector x′ goes outside the constraints, i.e. the gradient projectionreached beyond another constraint, then find the point of intersection of the newconstraint and the old. Set this point as the new x′ , and add the new constraint to theactive set. Make a new projection from this point and update x′ to the projectedvector.6. If no such point of intersecting constraints was found, it indicates that x′ was alreadyon an optimal corner, so the old x′ is optimal.7. Return to Step 4.4 Per<strong>for</strong>mance of Different <strong>Methods</strong> on Toy ProblemFor the purpose of demonstration, I chose a toy problem of the <strong>for</strong>m:22minimize f ( x)= x + x x + 2x− 6x− 6x1521 1 2 2 1 2+

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