Comparison of Optimization Algorithms Problem A: Consider the following three optimization problems: Numerically find the minimum (=optimal) feasible design vector x for each of the three problems using a gradient-based search technique of your choice  For each run (2runs) record the starting point you used, the iteration history (objective value on y-axis and iteration number on x-axis), the final point at which the algorithm terminated and whether or not the final solution is feasibleProblem B: Repeat problem a but this time using a heuristic technique of your choice Explain how you “tuned” the heuristic algorithm. compare convergence history of both method and frequency at which the technique gets trapped in a local maximum1. The Rosenbrock Function This function is known as the “banana function” because of its shape; it is described mathematically in Equation 1. In this problem, there are two design variables with lower and upper limits of [−5,5]. The Rosenbrock function has a known global minimum at [1,1] with an optimal function value of zero. Minimize f(x) = 100 (x2 −(x1)^2)^2  + (1−x1) ^2 2. The Eggcrate Function: This function is described mathematically in Equation 2. In this problem, there are two design variables with lower and upper bounds of [−2π,2π]. The Eggcrate function has a known global minimum at [0,0] with an optimal function value of zero. Minimize f(x) = (x 1 )^2 + (x2)^2 +25 ((sin^2) x1 +(sin^2) x2)3.Golinski’s Speed Reducer :This hypothetical problem represents the design of a simple gearbox such as might be used  in a light airplane between the engine and propeller to allow each to rotate at its most efficient speed. The objective is to minimize the speed reducer’s weight while satisfying the 11 constraints imposed by gear and shaft design practices. A known feasible solution obtained by a sequential quadratic programming (SQP) approach (Matlab’s fmincon) is a 2994.34 kg gearbox with the following values for the seven design variables: [3.5000,0.7000,17,7.3000,7.7153,3.3502,5.2867]. This is a feasible solution with four active constraints, but is it an optimal solution?
Comparison of Optimization Algorithms Problem A: Consider the following three optimization problems: Numerically find the minimum (=optimal) feasible design vector x for each of the three problems usi
g) In a binary encoded GA with two design variables, one with 4-bits, and one with 16 bits, what is the total number of possible population members? (a) 2 18 (b) 2 20 (c) 2 24 (d) 2 32 (a-2) Your objective is to design the cheapest possible bridge to span a highway. The total span of the bridge (across both halves of the highway) must be L= 30 meters, and it must support its own weight and a load q= 33 10 4 N/m along its span (a total load of about 1 10 6 N plus its own weight). The bridge span will be supported by between one and four I-beams. In the gure above, the I-beams would be parallel to each other going into the page, for example it could be one I- beam in the middle of the bridge, or one I-beam on both sides of the bridge, etc. and this will be represented by the design variable, n Ibeams . The shape of the I-beams will be represented by three continuous design variables, the height, h, the ange width, b, and thickness, t. The middle support will be rectangular (when viewed from above), and will have two design variables, the width, w, and depth, d. Let, r Ibeams be the density of the material used for the I-beams, then the mass of the I-beams can be computed using, MIbeams = [ 2bt + ( h 2t) t] L r Ibeams n Ibeams : Let r Support be the density of the material used for the support, and H=5m be the height of the bridge above the ground, then the mass of the middle support can be computed using MSupport = wd H r Support : There is a constraint that the stress of the I-beam is less than the material failure stress for the I-beam, s Failure-Ibeams . (Note: gis the gravitational constant (9.81 m/s 2 ).) s Ibeams =q L 2 2 + M Ibeams L 4 g 8 I Ibeam n Ibeams h 2 s Failure-Ibeams ; where I I beam is the moment of inertia for the I-beam given by IIbeam =( h 2t) 3 t 12 + 2″ t3 b 12 + t b h 2 t 2 2# : In addition, there is a constraint that the shear stress in the I-beams is less than the material failure stress: tIbeams = M Ibeams g + qL 4 [2 bt + ( h 2t) t] n Ibeams s Failure-Ibeams : For the middle support there are two constraints, the column cannot buckle and the stress must be less than the material failure stress. Buckling is based on a requirement that the applied load is less than a critical load: PApplied =M Ibeams g + qL 2 P Crit ; where the critical load is a function of th slowest moment of inertia of the support and the modulus of elasticity of the support material, E support , P Crit =p 2 E Support minn w3 d 12 ;wd 3 12 o 4 H 2 : The stress requirement is that the applied stress is less than the support material failure stress, sSupport =P Applied wd s Failure-Support : The bridge span (I-beams) can be made from A1 6061, A36 Steel, A514 Steel, or Titanium; how- ever, the support can be made from A1 6061, A36 Steel, A514 Steel, or Concrete. The reason for the difference is that concrete cannot be loaded in tension. The material properties and prices are listed in the Table: Material Density (kg/m 3 ) Modulus of Elasticity (GPa = 10 9 N/m 2 ) Failure Stress (MPa = 10 6 N/m 2 ) Cost($/kg) A1 6061 2700 70 270 2.05 A36 Steel 7850 210 250 0.62 A514 Steel 7900 210 700 0.90 Titanium 4500 120 760 16.00 Concrete 2400 31 70 0.04 Your objective is to nd the dimensions of the I-beams, number of I-beams, and material type for the I-beams, as well as the dimensions of the support and material type for the support to minimize cost of the bridge. Where c Ibeams and c Support are the cost per kilogram of the materials used for the I-beams and Support. The total bridge cost is then: C= c Ibeams M Ibeams + c Support M Support : (a) Please explain what optimization algorithm you chose to nd the cheapest possible bridge and why. (b) What are the design variables and cost for the cheapest bridge?

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