A dual scheme for solving linear countable semi-infinite fractional programming problems

Abstract

Using a dual method for solving linear fractional programming problems, we propose an approach to find optimal solutions of linear countable semi-infinite fractional programming problems via optimizing sequences. Duality theorems are established. A dual scheme for solving linear countable semi-infinite fractional problems is proposed. Examples are provided.

Appendix

In this part, we mention some duality theorems for linear fractional programming problems introduced in [18, 26] and refined in [24]. For convenience, the theorems are restated for $\mathrm{(P_k)}$ and $\mathrm{(D_k)}$ without proofs except Theorem A.2.

Theorem A.1

(Weak duality) One has

$$\begin{aligned} \sup \{F(x) \mid x \in X_k \} \le \inf \{I(u,v) \mid (u,v) \in Y_k\}. \end{aligned}$$

Theorem A.2

(Strong duality) If $\bar{x}_k$ is an optimal solution of $\mathrm{(P_k)}$ then there exists an optimal solution $(\bar{u}_k,\bar{v}_k)$ of $\mathrm{({D}_k)}$ and

$$\begin{aligned} F(\bar{x}_k)=I(\bar{u}_k,\bar{v}_k). \end{aligned}$$

(15)

Proof

Let $\bar{x}_k$ be an optimal solution of $\mathrm{(P_k)}$. By Dinkelbach transformation and by setting $\displaystyle {\bar{\lambda }=\frac{c^T\bar{x}_k+c_0}{d^T\bar{x}_k+d_0}}$, we have the following linear problem:

$$\begin{aligned} \mathrm{(LP_k)} \quad \mathrm{Max} \{c^Tx+c_0-\bar{\lambda }(d^Tx+d_0) \mid \mathcal {A}^{(k)}x \le b^{(k)}, x \in \mathbb {R}^n_+ \}. \end{aligned}$$

It is easy to verify that $\bar{x}_k$ is also an optimal solution of $\mathrm{(LP_k)}$ and its optimal value is 0, (see [3, Theorem 3.2]). According to dual scheme in linear programming, we obtain the dual problem of $\mathrm{(LP_k)}$:

$$\begin{aligned} \mathrm{(LD_k)} \quad \mathrm{Min} \{(b^{k})^Tv+c_0-\bar{\lambda }d_0 \mid (\mathcal {A}^{(k)})^Tv \ge c -\bar{\lambda }d, v \ge 0, v \in \mathbb {R}^k \}. \end{aligned}$$

Note that the optimal value of $\mathrm{(LD_k)}$ is 0 and $\mathrm{(LD_k)}$ must have an optimal solution $\bar{v}^*\ge 0$. Hence,

$$\begin{aligned} (b^{k})^T\bar{v}^*+c_0-\bar{\lambda }d_0 =0. \end{aligned}$$

(16)

Choose $\bar{u}_k= \bar{x}_k$ and $\bar{v}_k=(d^T\bar{x}_k+d_0) \bar{v}^*$. We claim that $(\bar{u}_k,\bar{v}_k)$ is an optimal solution of $\mathrm{(D_k)}.$ Obviously, the conditions (6) and (7) hold for ${\bar{u}}_k$ and ${\bar{v}}_k$. We need to verify (4) and (5). Since

$$\begin{aligned} \begin{array}{lll} (d^T\bar{u}_k)c-(c^T\bar{u}_k)d-{\mathcal {A}^{(k)}}^T\bar{v}_k&{}=&{}(d^T\bar{u}_k)c-(c^T\bar{u}_k)d-{\mathcal {A}^{(k)}}^T(d^T\bar{x}_k+d_0) \bar{v}^*\\ {} &{}\le &{}(d^T\bar{x}_k)c-(c^T\bar{x}_k)d-(c-\bar{\lambda }d)(d^T\bar{x}_k+d_0)\\ &{}=&{}(d^T\bar{x}_k)c-(c^T\bar{x}_k)d-(d^T\bar{x}_k+d_0)c-(c^T\bar{x}_k+c_0)d\\ &{}=&{}c_0d-d_0c, \end{array} \end{aligned}$$

the condition (4) holds. Next we verify (5). Multiplying both sides of (16) by $(d^T\bar{x}_k+d_0)>0$ and noting that $\bar{\lambda }(d^T\bar{x}_k+d_0)=(c^T\bar{x}_k+c_0),$ we get

$$\begin{aligned} (b^{k})^T\bar{v}^*(d^T\bar{x}_k+d_0)+(c_0-\bar{\lambda }d_0)(d^T\bar{x}_k+d_0) =0. \end{aligned}$$

Hence,

$$\begin{aligned} \begin{array}{lll} c_0d^T\bar{u}_k-d_0c^T\bar{u}_k+(b^k)^T\bar{v}_k &{}=&{}c_0d^T\bar{x}_k-d_0c^T\bar{x}_k+(b^k)^T(d^T\bar{x}_k+d_0)\bar{v}^*\\ &{}=&{}c_0d^T\bar{x}_k-d_0c^T\bar{x}_k-[c_0(d^T\bar{x}_k+d_0)-d_0(c^T\bar{x}_k+c_0)]\\ &{}=&{}0. \end{array} \end{aligned}$$

Thus, $({\bar{u}}_k, {\bar{v}}_k)$ is a feasible solution of $\mathrm{(D_k)}$. Since $F(\bar{x}_k)=F(\bar{u}_k)=I(\bar{u}_k, \bar{v}_k)$, the pair $(\bar{u}_k, \bar{v}_k)$ is an optimal solution of $\mathrm{(D_k)}$. The proof is complete. $\square$

Theorem A.3

(Converse duality) If $(\bar{u}_k,\bar{v}_k)$ is an optimal solution of $\mathrm{({D}_k)}$ then there exists an optimal solution $\bar{x}_k$ of $\mathrm{(P_k)}$ such that the equality (15) holds.

Theorem A.4

One has

$$\begin{aligned} \mathrm{Sol(P_k)}=\mathrm{Pr}_{\mathbb {R}^n}{( Y_k)}, \end{aligned}$$

where $\mathrm{Pr}_{\mathbb {R}^n}{({Y}_k)}= \{ u_k \in \mathbb {R}^n \mid (u_k,v_k)\in {Y}_k\},$ ${Y}_k$ is the feasible set of $\mathrm{(D}_k)$ and $\mathrm{Sol(P_k)}$ is the solution set of $\mathrm{(P_k)}.$

Corollary A.1

The solution set of $\mathrm{(P_k)}$ is nonempty if and only if the feasible set of $\mathrm{({D}_k)}$ is nonempty.