Exercise · ZW-WOLFE-CONDITIONS-007 · ★★☆ Intermediate
Let \(f(x)=x^2\), \(x_k=1\), \(p_k=-1\), \(c_1=0.1\), \(c_2=0.9\). Which condition fails with \(\alpha=2\)?
Choose one option.
Show answer
- C. Armijo, because \(1>0.6\)
Explanation
With \(\alpha=2\), \(f(-1)=1\) while the right side is \(1+0.1\cdot 2\cdot(-2)=0.6\), so Armijo fails. The curvature condition still holds.
Review the concept: Wolfe conditions
Two inequalities that a line-search step \(\alpha\) must satisfy to guarantee sufficient decrease of the function and that the step is not too short.
\[f(x_k+\alpha p_k)\le f(x_k)+c_1\alpha\nabla f(x_k)^\top p_k\]
More practice problems: wolfe conditions
- ZW-WOLFE-CONDITIONS-001 Which condition requires the function to decrease enough relative to its initial value?
- ZW-WOLFE-CONDITIONS-002 Which Wolfe condition prevents the step from being too short?
- ZW-WOLFE-CONDITIONS-003 In the Wolfe conditions, what relation must the constants c_1 and c_2 satisfy?
- ZW-WOLFE-CONDITIONS-004 If a step α satisfies both Wolfe conditions, what is guaranteed?
- ZW-WOLFE-CONDITIONS-005 What is the name of the variant that bounds the slope at the new point in absolute value?
- ZW-WOLFE-CONDITIONS-006 Let f(x)=x^2, x_k=1, p_k=-1, c_1=0.1. Does α=0.5 satisfy the Armijo condition?
- ZW-WOLFE-CONDITIONS-008 A step α that is too short barely changes the gradient. Which Wolfe condition rejects it?
- ZW-WOLFE-CONDITIONS-009 Let f(x)=x^2, x_k=1, p_k=-1, c_1=0.1, c_2=0.9. What happens with α=0.01?