AC - 3 August 2012 QE
Part 1 =
$ \begin{align} 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, \end{align} $
where N − 1 is the number of steps performed in the uncertainty range reduction process.
Solution:
The reduction factor is (1 − ρ1)(1 − ρ2)(1 − ρ3)...(1 − ρN − 1) Since $ 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, $ we have $ 1- \rho_{N-2} = \frac{F_{3}}{F_{4}} $ and so on.
Then, we have $ (1- \rho_{1})(1- \rho_{2})(1- \rho_{3})...(1- \rho_{N-1}) = \frac{F_{N}}{F_{N+1}} \frac{F_{N-1}}{F_{N}} ... \frac{F_{2}}{F_{3}} = \frac{F_{2}}{F_{N+1}} $ Therefore, the reduction factor is $ \frac{2}{F_{N+1}} $
Solution 2:
The uncertainty interval is reduced by $ (1- \rho_{1})(1- \rho_{2})(1- \rho_{3})...(1- \rho_{N-1}) = \frac{F_{N}}{F_{N+1}} \frac{F_{N-1}}{F_{N}} ... \frac{F_{2}}{F_{3}} = \frac{F_{2}}{F_{N+1}} $
$ \color{blue} \text{In this specific case, we use only N-1 iterations. } $
(ii)(10 pts) It is known that the minimizer of a certain function f(x) is located in the interval[-5, 15]. What is the minimal number of iterations of the Fibonacci method required to box in the minimizer within the range 1.0? Assume that the last useful value of the factor reducing the uncertainty range is 2/3, that is
$ 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, $
Solution:
Final Range: 1.0; Initial Range: 20.
$ \frac{2}{F_{N+1}} \le \frac{1.0}{20} $, or $ F_{N+1} \ge 40 $
So, N + 1 = 9
Therefore, the minimal iterations is N-1 or 7.
Solution 2: Since the final range is 1 and the initial range is 20, we can say $ \frac{2}{F_{N+1}} \le \frac{1.0}{20} \text{or equivalently } F_{N+1} \ge 40 $
From the inequality, we get $ F_{N+1} \ge 40 , \text{ so } N+1=9 $. Therefore the minimum number of iteration is N-1=7
$ \color{blue} F_9=55 \text{ and } F_8=34 $
Problem 2. (20pts) Employ the DFP method to construct a set of Q-conjugate directions using the function
$ f = \frac{1}{2}x^TQx - x^Tb+c $ $ =\frac{1}{2}x^T \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}x-x^T\begin{bmatrix} 2 \\ 1 \end{bmatrix} + 3. $
Where x(0) is arbitrary.
Solution:
$ f = \frac{1}{2}x^TQx - x^Tb+c $
Use initial point x(0) = [0,0]T</sub> and H0 = I2
In this case
$ g^{(k)} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} x^{(k)} - \begin{bmatrix} 2 \\ 1 \end{bmatrix} $
Hence $ g^{(0)} = \begin{bmatrix} -2 \\ -1 \end{bmatrix}, $ $ d^{(0)} = -H_0g^{(0)} =- \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -2 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} $
Because f is a quadratic function
$ \alpha_0 = argminf(x^{(0)} + \alpha d^{(0)}) = - \frac{g^{(0)^T}d^{(0)}}{d^{(0)^T}Qd^{(0)}} = - \frac{\begin{bmatrix} -2 & -1 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix}}{\begin{bmatrix} 2 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix}} = \frac{1}{2} $
$ x^{(1)} = x^{(0)} + \alpha d^{(0)} = \frac{1}{2} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix} $
$ \Delta x^{(0)} = x^{(1)}- x^{(0)} = \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix} $ $ g^{(1)} =\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} x^{(1)} - \begin{bmatrix} 2 \\ 1 \end{bmatrix}= \begin{bmatrix} -\frac{1}{2} \\ 1 \end{bmatrix} $
$ \Delta g^{(0)} = g^{(1)} - g^{(0)} = \begin{bmatrix} -\frac{3}{2} \\ 2 \end{bmatrix} $
Observe that $ \Delta x^{(0)} \Delta x^{(0)^T} = \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{4} \end{bmatrix} $ $ \Delta x^{(0)^T} \Delta g^{(0)} = \begin{bmatrix} 1 & \frac{1}{2} \end{bmatrix}\begin{bmatrix} \frac{3}{2} \\ 2 \end{bmatrix} = \frac{5}{2} $ $ H_0 \Delta g^{(0)} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{3}{2} \\ 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} \\ 2 \end{bmatrix}, $ $ (H_0 \Delta g^{(0)})(H_0 \Delta g^{(0)})^T = \begin{bmatrix} \frac{9}{4} & 3 \\ 3 & 4 \end{bmatrix} $ $ \Delta g^{(0)^T}H_0 \Delta g^{(0)} = \begin{bmatrix} \frac{3}{2} & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{3}{2} \\ 2 \end{bmatrix} = \frac{25}{4} $ Using the above, now we have $ H_1 = H_0 + \frac{\Delta x^{(0)} \Delta x^{(0)^T}}{\Delta x^{(0)^T} \Delta g^{(0)}} - \frac{(H_0 \Delta g^{(0)})(H_0 \Delta g^{(0)})^T}{\Delta g^{(0)^T}H_0 \Delta g^{(0)} } = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} \frac{2}{5} & \frac{1}{5} \\ \frac{1}{5} & \frac{1}{10} \end{bmatrix} - \frac{25}{4}\begin{bmatrix} \frac{9}{4} & 3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{26}{25} & -\frac{7}{25} \\ -\frac{7}{25} & \frac{23}{50} \end{bmatrix} $
T'hen we have, $ d^{(1)} = -H_1 g^{(0)} = - \begin{bmatrix} \frac{26}{25} & -\frac{7}{25} \\ -\frac{7}{25} & \frac{23}{50} \end{bmatrix} \begin{bmatrix} -\frac{1}{2} \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix} $
$ \alpha_1 = argminf(x^{(1)} + \alpha d^{(1)}) = - \frac{g^{(1)^T}d^{(1)}}{d^{(1)^T}Qd^{(1)}} = - \frac{\begin{bmatrix} -2 & 1 \end{bmatrix}\begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix}}{\begin{bmatrix} \frac{4}{5} & -\frac{3}{5}\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix}} = \frac{5}{2} $
$ x^{(2)} = x^{(1)} + \alpha_1 d^{(1)} = \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix} + \frac{5}{2}\begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix} $
$ \Delta x^{(1)} = x^{(2)} - x^{(1)} = \begin{bmatrix} 2 \\ -\frac{3}{2} \end{bmatrix} $ $ g^{(2)} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} x^{(0)} - \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $
Note that we have $ d^{(0)^T}Qd^{(0)} = 0; $ that is, $ d^{(0)} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} $ and $ d^{(1)} = \begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix} $ are Q-conjugate directions.
Solution 2:
$ \text{Let the initial point be } x^{(0)}= \begin{bmatrix} 0\\ 0 \end{bmatrix} \text{and initial Hessian be } H_0=\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} $
$ g^{(k)} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} x^{(k)} - \begin{bmatrix} 2 \\ 1 \end{bmatrix} , \text{so} $
$ g^{(0)} = \begin{bmatrix} -2 \\ -1 \end{bmatrix}, $ $ d^{(0)} = -H_0 g^{(0)} =- \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} -2 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} $
$ \alpha_0 = - \frac{g^{(0)^T}d^{(0)}}{d^{(0)^T}Qd^{(0)}} = - \frac{\begin{bmatrix} -2 & -1 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix}}{\begin{bmatrix} 2 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix}} = \frac{1}{2} $
$ x^{(1)} = x^{(0)} + \alpha d^{(0)} = \frac{1}{2} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix} $
$ \Delta x^{(0)} = x^{(1)}- x^{(0)} = \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix} $
$ g^{(1)} =\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} x^{(1)} - \begin{bmatrix} 2 \\ 1 \end{bmatrix}= \begin{bmatrix} -\frac{1}{2} \\ 1 \end{bmatrix} $
$ \Delta g^{(0)} = g^{(1)} - g^{(0)} = \begin{bmatrix} -\frac{3}{2} \\ 2 \end{bmatrix} $
$ \text{If we plug in the above numbers in the formula, we can get} $
$ H_1 = H_0 + \frac{\Delta x^{(0)} \Delta x^{(0)^T}}{\Delta x^{(0)^T} \Delta g^{(0)}} - \frac{(H_0 \Delta g^{(0)})(H_0 \Delta g^{(0)})^T}{\Delta g^{(0)^T}H_0 \Delta g^{(0)} } = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} \frac{2}{5} & \frac{1}{5} \\ \frac{1}{5} & \frac{1}{10} \end{bmatrix} - \frac{25}{4}\begin{bmatrix} \frac{9}{4} & 3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{26}{25} & -\frac{7}{25} \\ -\frac{7}{25} & \frac{23}{50} \end{bmatrix} $
$ d^{(1)} = -H_1 g^{(1)} = - \begin{bmatrix} \frac{26}{25} & -\frac{7}{25} \\ -\frac{7}{25} & \frac{23}{50} \end{bmatrix} \begin{bmatrix} -\frac{1}{2} \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix} $
$ \alpha_1 = - \frac{g^{(1)^T}d^{(1)}}{d^{(1)^T}Qd^{(1)}} = - \frac{\begin{bmatrix} -2 & 1 \end{bmatrix}\begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix}}{\begin{bmatrix} \frac{4}{5} & -\frac{3}{5}\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix}} = \frac{5}{2} $
$ x^{(2)} = x^{(1)} + \alpha_1 d^{(1)} = \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix} + \frac{5}{2}\begin{bmatrix} \frac{4}{5} \\ -\frac{3}{5} \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix} $
$ \Delta x^{(1)} = x^{(2)} - x^{(1)} = \begin{bmatrix} 2 \\ -\frac{3}{2} \end{bmatrix} $
$ g^{(2)} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} x^{(2)} - \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $
$ \text{When the gradient is 0, we reach the minimum point, which is } x^{(2)}=\begin{bmatrix} 3 \\ -1 \end{bmatrix} $
Problem 3. (20pts) For the system of linear equations,$ Ax=b $ where
$ A = \begin{bmatrix} 1 & 0 &-1 \\ 0 & 1 & 0 \\ 0 & -1& 0 \end{bmatrix}, b = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} $
Find the minimum length vector $ x^{(\ast)} $ that minimizes $ \| Ax -b\|^{2}_2 $
Problem 4. (20pts) Use any simplex method to solve the following linear program.
Maximize x1 + 2x2 S'ubject to $ -2x_1+x_2 \le 2 $ $ x_1-x_2 \ge -3 $ $ x_1 \le -3 $ $ x_1 \ge 0, x_2 \ge 0. $
Problem 5.(20pts) Solve the following problem:
Minimize $ -x_1^2 + 2x_2 $ Subject to $ x_1^2+x_2^2 \le 1 $ $ x_1 \ge 0 $ $ x_2 \ge 0 $
(i)(10pts) Find the points that satisfy the KKT condition.
(ii)(10pts)Apply the SONC and SOSC to determine the nature of the critical points from the previous part.
