Based upon the results and key operations given in the above sections, the new branch and bound algorithm for solving the EMFP problem, and hence the MFP problem, can be described as follows.

**Step 0**. (Initialization) Choose convergence tolerance *ϵ* > 0, set iteration counter *k* := 0 and the initial active nod as Ω_{0} = *X*^{0} × *D*^{0}.

Solve the initial convex relaxation problem (RMFP) over region *X*^{0} × *D*^{0}, if the (RMFP) is not feasible then there is no feasible solution for the (MFP). Else, denote the optimal value and solution as *f*_{0} and
${x}_{opt}^{0}$,
respectively, then we can obtain the initial upper and lower bound of the optimal value for the MFP problem, that is, *v* :=
$f({x}_{opt}^{0}),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{\_}{v}:={f}_{0},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{where}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}f(x)={\displaystyle max\{\frac{{n}_{i}(x)}{{d}_{i}(x)}\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}i=1,2,\cdots ,\phantom{\rule{thinmathspace}{0ex}}p\}.}$
And then, if v − *v* < *ϵ*, the algorithm can stop, and
${x}_{opt}^{0}$ is the optimal solution of the (MFP), otherwise proceed to step 1.

**Step 1**. (Reduction) Substitute the interval end points
${l}_{n+1}^{k}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{n+1}^{k}$ by utilizing the reduction technique (6).

**Step 2**. (Branching) Partition *D*^{k} into two new sub-intervals according to the adapted rule described in section 4.1. Add the new nods into the active nods set and denote the set of new partitioned intervals as
${\stackrel{~}{D}}^{k}$.

**Step 3**. (Bounding) For each subregion still of interest *X*^{0} × *D*^{k, μ} ⊆ *X*^{0} × *D*, *μ* = 1, 2, …, *s*_{k}, obtain the optimal solution and value for the RMFP problem corresponding outcome interval *D*^{k, μ} by solving some convex programming, if *LB*(*D*^{k, μ}) > *v*, or *f*(*x*^{k, μ}) < *v* then delete *D*^{k, μ} from
${\stackrel{~}{D}}^{k}$. Otherwise, let *LB*^{k} = min{*LB*(*D*^{k, μ}) | *μ* = 1, 2, ⋯, *s*_{k}} and *UB*^{k} = min{*f*(*x*^{k, μ}), *v*}, then we can update the lower and upper bounds as follows
$$\underset{\_}{v}=:max\{\underset{\_}{v},\phantom{\rule{thinmathspace}{0ex}}L{B}^{k}\},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{\_}{v}=:min\{\overline{v},\phantom{\rule{thinmathspace}{0ex}}U{B}^{k}\},$$

and let
$$\mathrm{\Omega}}_{k}=({\mathrm{\Omega}}_{k}\mathrm{\setminus}X)\bigcup {\stackrel{~}{D}}^{k$$

**Step 4**. (Termination) Let
$${\mathrm{\Omega}}_{k+1}={\mathrm{\Omega}}_{k}\mathrm{\setminus}\{X\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}f(X)-LB(X)\le \u03f5,\phantom{\rule{thinmathspace}{0ex}}X\in {\mathrm{\Omega}}_{k}\}.$$

If Ω_{k+1} = ∅, the algorithm can be stopped, *v* is the global optimal value for (MFP). Otherwise, set *k* : = *k* + 1, select *X*^{k} from Ω_{k} with
${X}^{k}=\underset{X\in {\mathrm{\Omega}}_{k}}{\text{argmin}}LB(X),$ then return to Step 3.

#### Theorem 4.1

*The proposed algorithm either terminates within finite iterations with an optimal solution for the MFP is found*, *or generates an infinite sequence of iterations such that along any infinite branches of the branch*-*and-bound tree*, *any accumulation point of the sequence* {*x*^{k}} *will be the global optimal solution of the (MFP)*.

#### Proof

(1) If the proposed algorithm is finite, it will terminate in some iteration *k*, *k* ≥ 0. And it can be known, by the termination criteria, that
$$\overline{v}-\underset{\_}{v}\le \u03f5.$$

From Step 0 and Step 3, it implies that
$$f({x}^{k})-L{B}_{k}\le \u03f5.$$

Let *v*_{opt} be the optimal value of the MFP problem. Then by section 3 and section 4.1, we known that
$$f({x}^{k})\ge {v}_{opt}\ge L{B}_{k}.$$

Hence, taken together, it implies that
$${v}_{opt}+\u03f5\ge L{B}_{k}+\u03f5\ge f({x}^{k})\ge {v}_{opt}.$$

And thus the proof of part (1) is completed.

(2) If the algorithm is infinite and generates an infinite feasible solution sequence
$\{({x}^{k},\phantom{\rule{thinmathspace}{0ex}}{x}_{n+1}^{k})\}$ via solving the (RMFP). Let
${x}_{n+1}^{k}=f({x}^{k}),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\{({x}^{k},\phantom{\rule{thinmathspace}{0ex}}{x}_{n+1}^{k})\}$ is a feasible solution sequence for the (EMFP). Since the sequence {*x*^{k}} is bounded, it must have accumulations, without loss of generality, assume
$\underset{k\to \mathrm{\infty}}{lim}{x}^{k}={x}^{\ast}.$ On the other hand, by the continuity of *n*_{i}(*x*) and *d*_{i}(*x*), we can get
$$\underset{k\to \mathrm{\infty}}{lim}{x}_{n+1}^{k}=\phantom{\rule{thinmathspace}{0ex}}\underset{k\to \mathrm{\infty}}{lim}f({x}^{k})=f({x}^{\ast}).$$(7)

Also, according to the branching regulation described before, we can see that
$$\underset{k\to \mathrm{\infty}}{lim}{l}_{n+1}^{k}=\phantom{\rule{thinmathspace}{0ex}}\underset{k\to \mathrm{\infty}}{lim}{u}_{n+1}^{k}={x}_{n+1}^{\ast}.$$(8)

What’ more, note that
${l}_{n+1}^{k}\le f({x}^{k})\le {u}_{n+1}^{k},$ taken (7) and (8) together, we can come to the conclusion that
$${x}_{n+1}^{\ast}=\phantom{\rule{thinmathspace}{0ex}}\underset{k\to \mathrm{\infty}}{lim}f({x}^{k})=f({x}^{\ast})=\phantom{\rule{thinmathspace}{0ex}}\underset{k\to \mathrm{\infty}}{lim}{\overline{x}}_{n+1}^{k}.$$

Therefore
$({x}^{\ast},\phantom{\rule{thinmathspace}{0ex}}{x}_{n+1}^{\ast})$ is also a feasible solution for the (EMFP). Further more, since the lower bound sequence *LB*^{k} for the optimal value is increasing and lower bounded by the optimal value *v*_{opt} in the algorithm, combining the continuity of *x*_{n+1}, we have
$$\underset{k\to \mathrm{\infty}}{lim}L{B}^{k}={x}_{n+1}^{\ast}\le {v}_{opt}\le \phantom{\rule{thinmathspace}{0ex}}\underset{k\to \mathrm{\infty}}{lim}{x}_{n+1}^{k}={x}_{n+1}^{\ast}.$$

That is,
$({x}^{\ast},\phantom{\rule{thinmathspace}{0ex}}{x}_{n+1}^{\ast})$ is an optimal solution for the (EMFP), and of course *x*^{*} is an optimal solution for the (MFP) according to the equivalence of the problems (MFP) and (EMFP), the proof is completed.□

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