Exercise 3.5

Let P = {x ∈ℜ3∣x1 + x2 + x3 = 1,x ≥0} and consider the vector x = (0,0,1). Find the set of feasible directions at x.

Answers

By Definition 3.1, a vector d ∈ ℜn is feasible iff

∃ ⁡𝜃 > 0 : x+𝜃d = [0 0 1 ]+𝜃 [d1 d2 d3 ] = [ 𝜃d1 𝜃d2 1 + 𝜃d3 ] ∈ P.
(1)

By construction of P, this is the same as

∃ ⁡𝜃 > 0 : { 𝜃d1 + 𝜃d2 + (1 + 𝜃d3) = 1 𝜃d1 ≥ 0 𝜃d2 ≥ 0 1 + 𝜃d3 ≥ 0.
(2)

By canceling out 1 in the first constraint and factoring out 𝜃≠0, we see that the three components of d must sum up to zero. Considering that 𝜃 > 0, the second and the third constraints are equivalent to d1 ≥ 0 and d2 ≥ 0 respectively. The third constraint 𝜃d3 ≥−1 is always satisfied by tweaking 𝜃 ≈ 0 when d3≠0 and is thus redundant (in an extreme case when d3 = 0, we also must have d1,d2 = 0 and thus d = 0 - any 𝜃 > 0 works in that case). We thus got rid of 𝜃 and are left with an equivalent formulation:

d1 + d2 + d3 = 0 d1 ≥ 0 d2 ≥ 0.
(3)

The set of feasible directions at x ∈ P is thus

{ [d1 d2 d3 ] ∈ ℜ3| d3 = −d1 − d2 d1,d2 ≥ 0 }.

PIC

Figure 1: Feasible directions at x.
User profile picture
2022-02-17 11:45
Comments