# 辅导案例-AMA 505-Assignment 1

AMA 505 Assignment 1 1st Sem, 2020 – 2021
Due date: Nov 1, 2020, 10:30 pm. No late submission will be accepted.
1. For the following functions, find all the stationary points and determine their nature, if possible.
(a) (5 points) f(x) = x61 + x
6
2 − 6x1x2;
(b) (5 points) f(x) = x1x2 + x
3
1 − x22.
2. (15 points) Consider the function
f(x) =
m∑
i=1
ln(ea
T
i x + e−a
T
i x) +
1
2
‖x‖22,
where ai ∈ IRn for each i = 1, . . . ,m. Suppose that ‖A‖2 = 2, where A ∈ IRm×n is the matrix with
the ith row being aTi . Argue that any accumulation point of the sequence generated by steepest
descent with constant stepsize α = 0.3985 is a stationary point of f .
3. Consider the function f : IR2 → IR defined by
f(x) = (x1 + 2x2 − 3)2 + 3‖x‖22.
(a) (5 points) Compute ∇f(x).
(b) Consider an iterate of the following form:
xk+1 = xk + αkd
k.
Let dk be the steepest descent direction and αk be chosen to satisfy the Wolfe’s condition.
Suppose that xk is nonstationary for all k.
i. (5 points) Show that the sequence {xk} is bounded.
ii. (10 points) Show that any accumulation point of {xk} is stationary.
4. (10 points) Consider Minimize
x∈IRn
f(x) with
f(x) =
1
2
xT
[
1 1
1 3
]
x− x1.
Perform two iterations of conjugate gradient method, starting with x0 = (2, 3). Write down x1 and
x2.
5. Discuss whether the following functions are convex.
(a) (5 points) f : IR2 → IR defined by f(x) = x1x2.
(b) (5 points) f : IR→ IR defined by
f(x) =
{
−x if x < 0,
x if x ≥ 0.
(c) (5 points) f : IRn → IR defined by f(x) = ∑mi=1(aTi x− bi)6 + ‖x‖1, where ai ∈ IRn and bi ∈ IR,
i = 1, . . . ,m, are nonzero vectors and numbers, n > m > 1.
(d) (5 points) f : IR→ IR defined by
f(x) =
{
−x3 if x < 0,
x4 if x ≥ 0.
(e) (5 points) f : IRn → IR defined by f(x) = (‖x‖∞)4.
6. (a) (10 points) Let A ∈ IRm×n and c ∈ IRn, where 1 ≤ m < n. Suppose that there exists y¯ ∈ IRm+
so that c = AT y¯. Argue that there exists x¯ ∈ IRn satisfying Ax¯ ≤ 0 and cT x¯ ≤ 0.
AMA 505 Assignment 1
(b) Solve the following linear programming problem using simplex method.
i. (5 points)
Maximize x1 − 3x2 + 2x3
Subject to 2x1 + x2 + x3 ≤ 12,
x1 + 2x2 − x3 ≤ 21,
3x2 − x3 ≤ 13,
x1, x2, x3 ≥ 0.
ii. (5 points)
Maximize 3x1 + 2x2 − 7x3
Subject to x1 + 2x2 + 4x3 ≤ 30,
2x1 + 3x2 + 2x3 = 30,
x1, x2, x3 ≥ 0.
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