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[2]. Show that g(x) = k · f (x) is a concave function of S if k < 0. Let x, y ∈ S, and α ∈ [0, 1]. Since f is a convex function of S, then
f (αx + (1 − α)y) ≤ αf (x) + (1 − α)f (y) (1)
Multiply (1) by k, then we have
k · f (αx + (1 − α)y) ≥ k · αf (x) + k · (1 − α)f (y) = α(k · f (x)) + (1 − α)(k · f (y)) = α(k · f )(x) + (1 − α)(k · f )(y) = αg(x) + (1 − α)g(y)
Thus, we obtain
g(αx + (1 − α)y) ≥ αg(x) + (1 − α)g(y)
We conclude that g = k · f is a concave function of S.
[3.a.]. Let x, y ∈ S. Then Ax ≤ b, and Ay ≤ b. For z = αx + (1 − α)y with α ∈ [0, 1],
Az = A(αx + (1 − α)y) = αAx + (1 − α)Ay ≤ α · b + (1 − α) · b = b
Thus, z = αx + (1 − α)y ∈ S. Hence, S is a convex set.
[3.b.]. Derive the condition for the feasible direction p at x ∈ S Let Aˆ be the submatrix of A corresponding to the rows of the ”active” con- straints, and ˆb be the corresponding sub-vector of b. Then p should satisfy the following inequality to be a feasible direction at x ∈ S;
Aˆ(x + αp) ≤ ˆb
with a sufficiently small α > 0. Since Axˆ = ˆb, then p should satisfy
Apˆ ≤ 0.
[4.a.]. Find the sets of all feasible directions at points at xc = (1, 1 , 1)T^. We want to find p so that xc + αp, α > 0 satisfies all the constraints (equality and inequality constraints). (1) From the ”equality” constraint, we have a condition for p = (p 1 , p 2 , p 3 )T such as p 1 + p 22 + 3p 3 = 0
(2) We need to check if each inquality constraint is ”active” or ”inactive” at xc.
a. x 1 ≥ 0: since 1 (1st element of xc) > 0, then x 1 ≥ 0 is inactive at xc. Thus, p can be any direction.
b. x 2 ≥ 0: since 1 (2nd element of xc) > 0, then x 2 ≥ 0 is inactive at xc. Thus, p can be any direction.
c. x 3 ≥ 0: since 1 (3rd element of xc) > 0, then x 3 ≥ 0 is inactive at xc. Thus, p can be any direction.
From (1) and (2), we have only one condition (from ”active” constraint) for p = (p 1 , p 2 , p 3 )T^ such as p 1 + 2p 2 + 3p 3 = 0.
Then, using p 1 = − 2 p 2 − 3 p 3 , we have
p = (p 1 , p 2 , p 3 )T^ = (− 2 p 2 − 3 p 3 , p 2 , p 3 )T = p 2 · (− 2 , 1 , 0)T^ + p 3 · (− 3 , 0 , 1)T^.
To sum up, the set of all feasible direction at xc is
{p 2 · (− 2 , 1 , 0)T^ + p 3 · (− 3 , 0 , 1)T^ : p 2 , p 3 ∈ R},
or {(p 1 , p 2 , p 3 )T^ : p 1 + p 2 + p 3 = 0}
In addition, the set of all feasible direction at xa is
{(p 1 , p 2 , p 3 )T^ : p 1 + p 2 + p 3 = 0, p 1 ≥ 0 , p 2 ≥ 0 },
the set of all feasible direction at xb is
{(p 1 , p 2 , p 3 )T^ : p 1 + p 2 + p 3 = 0, p 2 ≥ 0 }.
[4.b.]. Verify p = (3, 0 , −1)T^ is a feasible direction at xc. By choosing p 2 = 0, and p 3 = −1, p = (3, 0 , −1)T^ = 0·(− 2 , 1 , 0)T^ − 1 ·(− 3 , 0 , 1)T belongs to the set in 4.a. which means that p = (3, 0 , −1)T^ is a feasible direction at xc.
[4.b.]. With a feasible direction p = (3, 0 , −1)T^ at xc, determine the upper bound for step length α so that xc + αp is a feasible point. (1) We want to find α so that xc + αp satisfies all the constraints. For that, we only need to check ”inactive” (inequality) constraints aT^ x ≥ b; only when
aT^ p < 0, we have a condition for α such as α ≤ b−a
T (^) x aT^ p.
- x 1 ≥ 0: since 3 (1st element of p) > 0, then α can be any positive value.
- x 2 ≥ 0: since 1 (2nd element of p) = 0, then α can be any positive value.
- x 3 ≥ 0: since -1 (3rd element of p) < 0, we have α ≤ b−a
T (^) x aT^ p =^
0 −(1) − 1 = 1.
Thus, we have the condition for α; α ≤ 1. (upper bound for α is 1).
(2) With x = (1, 1 , 1)T^ and p = (3, 0 , −1)T^ , we want to find α s.t. (x + αp) = (1 + 3α, 1 , 1 − α)T^ satisfies all the constraints. It’s sufficient to consider the ”inactive” (inequality) constraints;
