AE 6310 Assignment 2

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Feb 20, 2024

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AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B Q.1) Consider the function f (x) = (x − 3)2x3(x − 6)4 on the interval x ∈ [0, 7]. Based on the starting point x = 2.2 do the following. (a) Consider search direction p = −f (x)/|f (x)|, plot the line search function on the interval α ′ ′ ∈ [0, 4] and report the intervals in α that satisfy the strong Wolfe conditions for the constants c1 = 0.1, c2 = 0.1. Hint: You should generate a single plot that has four curves: φ(α), φ(0) + c1αφ (0), |φ (α)|, and c2|φ ′ ′ (0)|, ′ and use these curves to identify the intervals that satisfy the strong Wolfe conditions. The given figure plots the four curves. To satisfy the strong Wolfe’s condition, the value of alpha should be considered where green line is below red line given that the blue line is already below the yellow line. From the figure it can be assumed that the value of alpha that satisfies Wolfe’s condition is 1. From 0.65 to 0.85 2. From 3.4 to 4 1

AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B Q.2) Consider the Rosenbrock function From the starting point x = (1, 0), answer the following questions. (a) Discuss the condition for a descent direction at x and identify which of the following directions are descent directions. • p1 = (0, −1) • p2 = (0, 1) • p3 = (−1, 1) • p4 = (1, −1) The given Rosenbrock equation is: The required condition to check the descent condition is: Evaluating ∇ ' f ( x ) = ¿ ¿ ¿ = Substituting for all values of p at (1 , 0) 1. = = +200 > 0 ( Does not satisfy ) 2. = = -200 < 0 ( Satisfy ) 3. = = -600 < 0 ( Satisfy ) 4. = = +600 > 0 ( Does not satisfy ) 2

AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B Figure-1) For p = (0,-1): As the value of > 0 , no value of c1 will satisfy the required alpha. Figure-2) For p = (0,1): As the value of > 0 , we will have alpha that satisfies the sufficient decrease condition c1 Alpha 0.01 0 to 1.98 0.1 0 to 1.80 0.5 0 to 1 3

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AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B Figure-3) For p = (-1,1): As the value of > 0 , we will have alpha that satisfies the sufficient decrease condition c1 Alpha 0.01 0 to 0.98 0.1 0 to 0.75 0.5 0 to 0.35 4

AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B Figure-4) For p = (0,-1): As the value of > 0 , no value of c1 will satisfy the required alpha. Q-3 Verify your implementation by solving the optimization problem from xinit = (0, 0) with tolerance ε. Plot the convergence criterion || ∇ f (x)||2 v.s. iteration number. Use logarithmic scale for y-axis. Show that your final solution satisfies the convergence criterion || ∇ f (x)||2 < ε. The tolerance used for the code is 10 -3 Output: Converged after 6296 iterations. Optimal solution: [0.9992102 0.9984224] Objective value at optimal solution: 6.239781604883472e-07 5

AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B (b) Pick 10 random initial points (using np.random.rand(...)) from xinit ∈ [−1.5, 1.5]2, run the opti- mization, plot the Euclidean distance r = ||x − x ∗ ||2 v.s. iteration number for 10 cases in one figure. Use logarithmic scale for y-axis. Report the average number of line searches and function evaluations. 6

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AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B  Average number of line searches: 5219.6  Average number of funtion evaluations: 46400.7 (Q-4) (a) Output: Converged after 217 iterations. Optimal solution: [0.99999997 0.99999994] Objective value at optimal solution: 1.1076508791145594e-15 7

AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B (Q-4) (b) 8

AE 6310 Assignment 2 Spring 2024 | Bhanavadiya, Aman B  Average number of line searches: 85.2  Average number of funtion evaluations: 651.7 9

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