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The instructions and solutions for an online midterm exam in the optimization methods course offered in spring 2022. The exam covers topics such as linear programming, graphical method, and convex functions. Students are required to apply the simplex method, evaluate gradients and hessian matrices, and sketch feasible sets to find optimal solutions.
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Online Midterm Examination.Quaranteed Success.Attained A+> Time allowed: 1.75 hours Time: 9:30-11:15am, March 11, 2022 Policy:
x 1 ,x 2 ,x 3 ,s 1 ,s 2 ,s 3 ,z s.t. z + x 1 − 2 x 2 − 3 x 3 = 0 − x 1 + x 2 + s 1 = 3 − x 2 + 2 x 3 + s 2 = 2 x 1 − 3 x 3 + s 3 = 6 , x 1 ,x 2 ,x 3 ,s 1 ,s 2 ,s 3 ≥ 0. It is clear that we can take s 1 ,s 2 ,s 3 as the initial basis which will lead to a feasible CPF solution. We thus form the simplex table: BV z x 1 x 2 x 3 s 1 s 2 s 3 RHS z 1 1 -2 -3 0 0 0 0 s 1 0 -1 1 0 1 0 0 3 s 2 0 0 -1 2 0 1 0 2 s 3 0 1 0 -3 0 0 1 6 1 We select x 3 as the entering variable and s 2 as the leaving variable. Observe the following row opera- tions: − 0 2 −
Resulting in the table:
(^0 1 0 0 3) 1 0 1 / 2 0 1 0 0 3 / 2 1 9 B V z x 1 x 2 x 3 s 1 s 2 s 3 R HS z 1 1 -7/2 0 0 3/2 0 3 s 1
x 3
Solution: (a) Notice that the constraint max{ x 1 ,x 2 } ≤ 1 is equivalent to x 1 ≤ 1, and x 2 ≤ 1 simultaneously. The feasible region is thus sketched as: Applying the graphical method shows that the only optimal solution is (1 , 1) with an optimal objective value of 2. (b) The feasible region is sketched as: ✗✗ 2 I - ¥EEz >×' Applying the graphical method shows that the only optimal solution is (2 , 0) with an optimal objective value of 2. (c) Notice that satisfying the constraints min 0 is equivalent to having max{ x 1 ,x 2 } ≤ 1 ,x 1 ≥ 0 ,x 2 ≥ 0 OR. In other words, the feasible region is given by the union of the feasible sets in part (a) and (b). This can be sketched as: ✗✗ 2 ✗ ✗ 2 " % £ 1 "^ "
' "' Applying the graphical method shows that there are two optimal solutions: (1 , 1) AND (2 , 0). Both have an optimal objective value of 2. Problem 3. (25%) Answer the following question on convex functions. (a) (5%) Let a ∈ R and b ≥ 0 be given. Consider the following set X = { x,y ∈ R : y > 0} Consider the function f : X → R , Evaluate the gradient vector ∇ f ( x,y ) and the Hessian matrix ∇^2 f ( x,y ). (b) (10%) Show that f ( x,y ) in part (a) is a convex function for all ( x,y ) ∈ X. (c) (10%) Set b > 0 and consider the function g : X → R , Show that g ( x,y ) is not a convex function. (Hint: check the non-PSD-ness of the Hessian matrix by fixing ( x,y ).) Solution. (a) The gradient and Hessian are given as: . (b) Our idea is to check the PSD-ness of ∇^2 f ( x,y ). Observing that which can be simplified as
(a) The decision variables are: xA – amt. of time on EGGG3100 , xB – amt. of time on FCEC1200 , xC – amt. of time on FCEC The constraints are: . and The objective is to maximize w.r.t. xA,xB,xC for Thus, the overall optimization problem is given as: s.t. xA + xB + xC ≤ 30 , xA ≥ 0 , xB ≥ 0 , xC ≥ 0 , xC ≥ 15 / 4. (b) To rewrite the above optimization problem, we introduce the slack variables zA,zB,zC and observe the following equivalent form s.t. Replacing zA ≤ min{16 , 4 xA } by zA ≤ 16 , zA ≤ 4 xA , ..., lead to s.t. This is a standard form LP as desired. (c) We set the following as decision variables: xA – amt. of time on EGGG3100 , xB – amt. of time on FCEC1200 , xC – amt. of time on FCEC zA – GP of EGGG3100 , zB – GP of FCEC1200 , zC – GP of FCEC