**Transcription of the Question:** --- Consider the standard-form LP: \[ \begin{align*} \text{Minimize} \quad & 4y_1 - 3y_2 + 5y_3 \\ \text{Subject to} \quad & y_1 - 0.5y_3 = 0.5 \\ & y_1 + y_2 = 2 \\ & y_1, y_2, y_3 \geq 0 \end{align*} \] Conduct the Affine Scaling Search at \( y^{(0)} = (1, 1, 1) \) and determine the next feasible solution \( y^{(1)} \). --- # 1. **Complete Solution with All Calculations** --- ### **Step 1: Express the Problem in Matrix Form** Let \[ c = \begin{bmatrix} 4 \\ -3 \\ 5 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 0 & -0.5 \\ 1 & 1 & 0 \end{bmatrix}, \quad b = \begin{bmatrix} 0.5 \\ 2 \end{bmatrix} \] The initial point is: \[ y^{(0)} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \] --- ### **Step 2: Affine Scaling Direction** Affine scaling direction is given by: \[ d = -D^2 A^T (A D^2 A^T)^{-1} (A y^{(0)} - b) - D^2 \left[c - A^T \lambda\right] \] But for the first step, use the feasible affine scaling direction: \[ d = -D^2 \left[c - A^T \lambda\right] \] where \[ D = \text{diag}(y^{(0)}) \] So, \[ D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] Therefore, \[ D^2 = D = I \] So, \[ d = -\left[c - A^T \lambda\right] \] --- ### **Step 3: Compute \(\lambda\)** \[ \lambda = (A D^2 A^T)^{-1} A D^2 c \] Since \(D^2 = I\): \[ A A^T = \begin{bmatrix} 1 & 0 & -0.5 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ -0.5 & 0 \end{bmatrix} = \begin{bmatrix} 1^2 + 0^2 + (-0.5)^2 & 1*1 + 0*1 + (-0.5)*0 \\ 1*1 + 1*0 + 0*(-0.5) & 1*1 + 1^2 + 0^2 \end{bmatrix} \] Compute each entry: - (1,1): \(1^2 + 0^2 + (-0.5)^2 = 1 + 0 + 0.25 = 1.25\) - (1,2): \(1*1 + 0*1 + (-0.5)*0 = 1 + 0 + 0 = 1\) - (2,1): \(1*1 + 1*0 + 0*(-0.5) = 1 + 0 + 0 = 1\) - (2,2): \(1^2 + 1^2 + 0^2 = 1 + 1 + 0 = 2\) So, \[ A A^T = \begin{bmatrix} 1.25 & 1 \\ 1 & 2 \end{bmatrix} \] Now, compute \(A c\): \[ A c = \begin{bmatrix} 1 & 0 & -0.5 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 4 \\ -3 \\ 5 \end{bmatrix} = \begin{bmatrix} 1*4 + 0*(-3) + (-0.5)*5 \\ 1*4 + 1*(-3) + 0*5 \end{bmatrix} = \begin{bmatrix} 4 - 2.5 \\ 4 - 3 \end{bmatrix} = \begin{bmatrix} 1.5 \\ 1 \end{bmatrix} \] Now, \[ \lambda = (A A^T)^{-1} (A c) \] First, find \((A A^T)^{-1}\): The determinant: \[ \det = (1.25)(2) - (1)(1) = 2.5 - 1 = 1.5 \] The inverse: \[ (A A^T)^{-1} = \frac{1}{1.5} \begin{bmatrix} 2 & -1 \\ -1 & 1.25 \end{bmatrix} = \begin{bmatrix} 1.3333 & -0.6667 \\ -0.6667 & 0.8333 \end{bmatrix} \] Multiply: \[ \lambda = \begin{bmatrix} 1.3333 & -0.6667 \\ -0.6667 & 0.8333 \end{bmatrix} \begin{bmatrix} 1.5 \\ 1 \end{bmatrix} \] Compute each element: - First row: \(1.3333 \times 1.5 + (-0.6667) \times 1 = 2.0 - 0.6667 = 1.3333\) - Second row: \(-0.6667 \times 1.5 + 0.8333 \times 1 = -1.0 + 0.8333 = -0.1667\) Thus, \[ \lambda = \begin{bmatrix} 1.3333 \\ -0.1667 \end{bmatrix} \] --- ### **Step 4: Compute \(c - A^T \lambda\)** First, compute \(A^T \lambda\): \[ A^T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ -0.5 & 0 \end{bmatrix} \] \[ A^T \lambda = \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ -0.5 & 0 \end{bmatrix} \begin{bmatrix} 1.3333 \\ -0.1667 \end{bmatrix} = \begin{bmatrix} 1*1.3333 + 1*(-0.1667) \\ 0*1.3333 + 1*(-0.1667) \\ -0.5*1.3333 + 0*(-0.1667) \end{bmatrix} = \begin{bmatrix} 1.3333 - 0.1667 \\ -0.1667 \\ -0.66665 \end{bmatrix} = \begin{bmatrix} 1.1666 \\ -0.1667 \\ -0.6667 \end{bmatrix} \] Now, \[ c - A^T \lambda = \begin{bmatrix} 4 \\ -3 \\ 5 \end{bmatrix} - \begin{bmatrix} 1.1666 \\ -0.1667 \\ -0.6667 \end{bmatrix} = \begin{bmatrix} 2.8334 \\ -2.8333 \\ 5.6667 \end{bmatrix} \] --- ### **Step 5: Compute the Direction \(d\)** \[ d = - D^2 [c - A^T \lambda] = -I [c - A^T \lambda] = -[c - A^T \lambda] \] \[ d = - \begin{bmatrix} 2.8334 \\ -2.8333 \\ 5.6667 \end{bmatrix} = \begin{bmatrix} -2.8334 \\ 2.8333 \\ -5.6667 \end{bmatrix} \] --- ### **Step 6: Compute Maximum Feasible Step Size \(\alpha\)** The new iterate is \(y^{(1)} = y^{(0)} + \alpha d\), with \(y^{(1)} \geq 0\). Find \(\alpha_{max}\) so that \(y^{(0)} + \alpha d \geq 0\): For each component: \[ y^{(1)}_i = y^{(0)}_i + \alpha d_i \ge