Prerequisite Qualification: Optimization :Introduction to Financial Engineering and Risk Management (Introduction to Financial Engineering and Risk Management) Answers 2025

Question 1

Correct statements:

❌ Alice does not need to specify short position, since long-only strategy will generate the same result as strategy allowing short positions.

✅ Alice should use the positions in each portfolio as variables.

✅ Alice should specify whether short position is allowed by constraints.

Primal:

$Ax≥b\min c^T x \quad \text{s.t. } Ax \ge b$

Dual objective is:

$max b^T u$

Here $b = [1,1]^T$

✅ $u_1 + u_2$

Dual constraints come from:

$ATu=c,u≥0A^T u = c,\quad u \ge 0$

Correct constraints:

✅ $u1≥0u_1 \ge 0$
✅ $u2≥0u_2 \ge 0$
❌ $u_1 + u_2 = -1$
❌ $u_1 + 2u_2 = 1$

From Question 3 constraints:

$u1≥0,u2≥0u_1 \ge 0,\quad u_2 \ge 0$

Using $A^T u = c$ , solution gives:

$u_1 = 1,\; u_2 = 0$

Optimal value:

$b^T u = 1$

✅ Answer: 1

Correct statements:

✅ If $\ge f(x^*)$ for any $\in \mathbb{R}^n$ , then $x^*$ is the global minimum.

❌ Local condition for all $r > 0$ implies global minimum, not “local but not global”

✅ Global minimum must also be local minimum.

✅ If inequality holds in a neighborhood $∥y−x∗∥≤r\|y-x^*\|\le r$ , then $x^*$ is a local minimum.

$f(x_1,x_2)=x_1^2+2x_1x_2+x_2^3$

Gradient = 0 ⇒ critical point at $(0, 0)$

$f (0, 0) = 0$

✅ Local minimum value = 0.00

Maximize:

$2×1+3×2=6\ln x_1 + \ln x_2 \quad \text{s.t. } 2x_1+3x_2=6$

Using Lagrange multiplier:

$x1=1.5,x2=1x_1=1.5,\quad x_2=1$ $f_{\max}=\ln(1.5)+\ln(1)=\ln(1.5)$ $ln⁡(1.5)≈0.41\ln(1.5)\approx 0.41$

✅ Global maximum value = 0.41