Solving Linear Integer Models with Variable Bounding

Abstract

We present a technique to solve the linear integer model with variable bounding. By using the continuous optimal solution of the linear integer model, the variable bounds for the basic variables are approximated and then used to calculate the optimal integer solution. With the variable bounds of the basic variables known, solving a linear integer model is easier by using either the branch and bound, branch and cut, branch and price, branch cut and price, or branch cut and free algorithms. Thus, the search for large numbers of subproblems, which are unnecessary and common for NP Complete linear integer models, is avoided.

1. Introduction

It is challenging to solve the NP Complete linear integer programming (LIP) model using the available methods [1,2,3]. Linear integer models are used when the continuity assumption in linear programming (LP) does not hold. For discrete quantities such as the number of computer machines, aircraft, automobiles, personnel, warehouses, or stations, continuity is not a strong assumption. Several approaches can be used to solve this model, including branch and bound (BB) [4,5], branch and price (BP) [6], branch and cut (BC) [7], branch cut and price (BCP) [8], and branch cut and free (BCF) [9] algorithms. In addition, reformulation techniques that can help reduce the complexity of LIP models exist [10]. The LIP models can also be solved by using a characteristic equation [1]. A hypermetaheuristic approach to solving mixed integer optimization problems was proposed by [11]. Data envelopment analysis (DEA)-related problems were solved to prove the efficiency and validity of the method; however, proving the validity of DEA-related problems only limited the study. A new strategy to solve linear integer programming problems with simplex directions was presented by [12]. To prove the validity of the method, the results were compared with CPLEX software results. With this method, the authors managed to compute accurate solutions for 34 out of 100 problems. Mixed integer linear programming was used by [13] to solve the station capacity allocation problem of a star-tree pipe network. The results of the model proved the validity of the method. Integer linear programming was applied in [14] to optimize energy-aware high-level synthesis and reliability. The results were compared with a genetic algorithm for several problems, and the integer linear programming approach resulted in optimal solutions. Mixed integer models were applied in [15] to optimize a vanadium redox flow battery with electrolyte maintenance, variable efficiencies, and capacity fade. The results proved that the problem was solved in a more accurate way when using this method as compared with other simpler methods. The branch and bound approach was enhanced in [16] to solve the knapsack linear integer problem. The problems were solved as subproblems that were less complex and required fewer iterations to obtain the optimal solution.

We present a way to find a bound such that a predetermined number of integers is accommodated in a calculated variable range. Using the continuous optimal solution of the linear integer model, the integer bounds for the basic variables are approximated and then used to calculate the optimal integer solution. The search for large numbers of subproblems, which are unnecessary and common for NP Complete linear integer models, is avoided. Linear integer models are NP-hard problems; thus, computing the optimal solution in a reasonable computational time is remarkably difficult. Combining metaheuristics and exact solution approaches is an efficient and effective solution. The variable-bounding method that we propose helps one to determine the variable bounds for linear integer models and thus reduces the number of iterations when applying exact-solution approaches such as branch and bound and branch and cut, among others, to solve the linear integer models. The approximated integer bounds for the variables are utilized when calculating the optimal integer solutions. The exact methods require fewer iterations when the integer bounds for the variables are approximated first; thus, the search for a large number of subproblems, which are unnecessary, is avoided. Essentially, the main contribution of this research is the development of the variable-bounding method, which can be used to approximate the variable integer bounds for the variables; the approximated bounds will be utilized to reduce the complexity and the number of iterations when applying the exact-solution approaches.

2. The Linear Integer Programming (LIP)

$$\begin{gathered} \mathrm{M}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{z}\mathrm{e} Y_o=c_1{y}_1+c_2{y}_2+\dots +c_n{y}_n, \\ \mathrm{Such} \mathrm{that}: \\ {a}_{11}{y}_1+a_{12}{y}_2+\dots +a_{1n}{y}_n\le b_1, \\ {a}_{21}{y}_1+a_{22}{y}_2+\dots +a_{2n}{y}_n\le b_2, \\ \dots \\ {a}_{m1}{y}_1+a_{m2}{y}_2+\dots +a_{mn}{y}_n\le b_m, \end{gathered}$$

(1)

where $a_{ij}, b_j \mathrm{a}\mathrm{n}\mathrm{d} c_i \mathrm{a}\mathrm{r}\mathrm{e} \mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{r}\mathrm{s}, Y_o \mathrm{i}\mathrm{s} \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{o}\mathrm{b}\mathrm{j}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e} \mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}, y_i\ge 0 \mathrm{a}\mathrm{n}\mathrm{d} \mathrm{a}\mathrm{n} \mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{r}, i=1,2,\dots ,m \mathrm{a}\mathrm{n}\mathrm{d} j=1,2,\dots ,n$.

3. The Continuous Optimal Table

The integer restrictions on the LIP model, which restrict integers that are difficult to use, can be relaxed. This is performed by changing the LIP model into an ordinary linear program (LP), as the LP is easier to solve than the LIP version. The relaxed LP form is solved to obtain a continuous optimal solution, which is provided in

Table 1. LP continuous optimal table.

	${\mathit{y}}_1$	${\mathit{y}}_2$	…	${\mathit{y}}_{\mathit{m}}$	${\mathit{s}}_1$	${\mathit{s}}_2$	…	${\mathit{s}}_{\mathit{m}}$	R.H.S
$Y_o$	0	0	…	0	${\lambda }_1$	${\lambda }_2$	…	${\lambda }_m$	R
$y_1$	1	0	…	0	${\pi }_{11}$	${\pi }_{12}$	…	${\pi }_{1m}$	${\beta }_1$
$y_2$	0	1	…	0	${\pi }_{21}$	${\pi }_{22}$	…	${\pi }_{2m}$	${\beta }_2$
…	…	…	…	…	…	…	…	…	…
$y_m$	0	0	…	1	${\pi }_{m1}$	${\pi }_{m2}$	…	${\pi }_{mm}$	${\beta }_m$

where ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m,{\beta }_1,{\beta }_2,\dots ,{\beta }_m$ are non-negative constants, whereas ${\pi }_{ij} \mathrm{a}\mathrm{n}\mathrm{d} R$ are constants. The variables $y_1,y_2,\dots ,y_m$ are made basic and $s_1,s_2,\dots ,s_m$ are made nonbasic for convenience. Any other arrangement is also possible.

.

4. Determining the Expressions for the Basic Variable Limits

$$\mathrm{Let}{\mathsf{\lambda }}_1{\mathrm{s}}_1+{\mathsf{\lambda }}_2{\mathrm{s}}_2+\dots +{\mathsf{\lambda }}_{\mathrm{m}}{\mathrm{s}}_{\mathrm{m}}\ge ∆{\mathrm{Y}}_0$$

(2)

where $∆Y_0$ is a reduction in the optimal objective value R. By introducing a slack variable $s_{m+1}$, the constraint becomes Equation (3)

$${-\mathsf{\lambda }}_1{\mathrm{s}}_1-{\mathsf{\lambda }}_2{\mathrm{s}}_2-\dots -{\mathsf{\lambda }}_{\mathrm{m}}{\mathrm{s}}_{\mathrm{m}}+{\mathrm{s}}_{\mathrm{m}+1}=-∆{\mathrm{Y}}_0$$

(3)

By adding this extra row to the current optimal solution, it becomes

Table 2. Adding a new constraint.

	${\mathit{y}}_1$	${\mathit{y}}_2$	…	${\mathit{y}}_{\mathit{m}}$	${\mathit{s}}_1$	${\mathit{s}}_2$	…	${\mathit{s}}_{\mathit{m}}$	${\mathit{s}}_{\mathit{m}+1}$	R.H.S
$Z$	0	0	…	0	${\lambda }_1$	${\lambda }_2$	…	${\lambda }_m$	0	R
$y_1$	1	0	…	0	${\pi }_{11}$	${\pi }_{12}$	…	${\pi }_{1m}$	0	${\beta }_1$
$y_2$	0	1	…	0	${\pi }_{21}$	${\pi }_{22}$	…	${\pi }_{2m}$	0	${\beta }_2$
…	…	…	…	…	…	…	…	…	…	…
$y_m$	0	0	…	1	${\pi }_{m1}$	${\pi }_{m2}$	…	${\pi }_{mm}$	0	${\beta }_m$
$s_{m+1}$	0	0	…	0	$-{\lambda }_1$	$-{\lambda }_2$	…	$-{\lambda }_m$	1	$-∆Y_0$

.

We then applied the dual simplex procedure, and n possible solutions were found. The first basic variable in the first solution is of the form provided in Equation (4):

$${\mathrm{y}}_1={\mathsf{\beta }}_1+{\mathrm{h}}_1^1∆{\mathrm{Y}}_0,\forall \mathrm{i}.$$

(4)

The second basic variable in the first solution is of the form provided in Equation (5):

$${\mathrm{y}}_2={\mathsf{\beta }}_2+{\mathrm{h}}_2^1∆{\mathrm{Y}}_0,\forall \mathrm{i}.$$

(5)

The mth basic variable in the mth solution is of the form provided in Equation (6):

$${\mathrm{y}}_{\mathrm{m}}={\mathsf{\beta }}_{\mathrm{m}}+{\mathrm{h}}_{\mathrm{m}}^1∆{\mathrm{Y}}_0,\forall \mathrm{i}.$$

(6)

where $h_i^{\stackrel{-}{i}}$ is a constant and $\stackrel{-}{i}=1,2,...,m$

Thus,

$$\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ \dots \\ {\mathrm{x}}_{\mathrm{m}}\end{array}\right]=\left[\begin{array}{c}{\mathsf{\beta }}_1+{\mathrm{h}}_1^1∆Z\\ {\mathsf{\beta }}_2+{\mathrm{h}}_2^1∆Z\\ \dots \\ {\mathsf{\beta }}_{\mathrm{m}}+{\mathrm{h}}_{\mathrm{m}}^1∆Z\end{array}\right],\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ \dots \\ {\mathrm{x}}_{\mathrm{m}}\end{array}\right]=\left[\begin{array}{c}{\mathsf{\beta }}_1+{\mathrm{h}}_1^2∆Z\\ {\mathsf{\beta }}_2+{\mathrm{h}}_2^2∆Z\\ \dots \\ {\mathsf{\beta }}_{\mathrm{m}}+{\mathrm{h}}_{\mathrm{m}}^2∆Z\end{array}\right],\dots ,\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ \dots \\ {\mathrm{x}}_{\mathrm{m}}\end{array}\right]=\left[\begin{array}{c}{\mathsf{\beta }}_1+{\mathrm{h}}_1^{\mathrm{m}}∆Z\\ {\mathsf{\beta }}_2+{\mathrm{h}}_2^{\mathrm{m}}∆Z\\ \dots \\ {\mathsf{\beta }}_{\mathrm{m}}+{\mathrm{h}}_{\mathrm{m}}^{\mathrm{m}}∆Z\end{array}\right]$$

(7)

From these m possible solutions, the bounds for the variables that are nonbasic can be determined using the formula provided in Equation (8):

$${\mathsf{\beta }}_{\mathrm{i}}+{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0\le {\mathrm{y}}_{\mathrm{i}}\le {\mathsf{\beta }}_{\mathrm{i}}+{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0$$

(8)

where

$${\ell }_{\mathrm{i}}^{\mathrm{L}}=\min \left[{\mathrm{h}}_{\mathrm{i}}^1,{\mathrm{h}}_{\mathrm{i}}^2,\dots ,{\mathrm{h}}_{\mathrm{i}}^{\mathrm{m}}\right]$$

(9)

and

$${\ell }_{\mathrm{i}}^{\mathrm{U}}=\max \left[{\mathrm{h}}_{\mathrm{i}}^1,{\mathrm{h}}_{\mathrm{i}}^2,\dots ,{\mathrm{h}}_{\mathrm{i}}^{\mathrm{m}}\right]$$

(10)

5. Arranging Variables in the Order of Their Restrictions

Arranging the variables in an order that begins with the most restricted variables will reduce the number of iterations during the searching process. The approaches that are branch-and-bound-related converge fast if the first iteration begins with the most-restricted variables.

5.1. Ordering Branching Variables

Branching must always begin with the most-restricted variables. In this case, the variable with the most restrictions is defined as the variable with the lowest number of possible integral values.

Proof. 

Let the number of possible integer values for variable $x_i$ be provided by $I_i$.

Thus,

$$\mathrm{n}\left({\mathrm{x}}_{\mathrm{i}}\right)={\mathsf{\omega }}_{\mathrm{i}}$$

(11)

Thus, the number of possible integers for the various variables are provided in Equation (12):

$$\begin{gathered} \mathrm{n}\left({\mathrm{x}}_1\right)={\mathsf{\omega }}_1, \\ \mathrm{n}\left({\mathrm{x}}_2\right)={\mathsf{\omega }}_2, \\ \dots \\ \mathrm{n}\left({\mathrm{x}}_{\mathrm{m}}\right)={\mathsf{\omega }}_{\mathrm{m}}. \end{gathered}$$

(12)

Additionally, let the number of branches in ascending order be Equation (13):

$${\mathsf{\omega }}_1\le {\mathsf{\omega }}_2\le \dots \le {\mathsf{\omega }}_{\mathrm{m}}$$

(13)

Thus, variable $y_1$ is the most restricted variable, followed by $y_2$ and then variable $y_3$ in that order up to variable $y_m$. The following number of nodes are visited when constructing the search tree. In this type of search process, the worst-case scenario is assumed. □

5.2. Starting with the Most Restricted Variable

Step 1: $\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s} \mathrm{t}\mathrm{a}\mathrm{k}\mathrm{e}\mathrm{n}={\mathsf{\omega }}_1$

Step 2: $\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s} \mathrm{t}\mathrm{a}\mathrm{k}\mathrm{e}\mathrm{n}={\mathsf{\omega }}_1{\mathsf{\omega }}_2$

Step m: $\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s} \mathrm{t}\mathrm{a}\mathrm{k}\mathrm{e}\mathrm{n}={\mathsf{\omega }}_1{\mathsf{\omega }}_2\dots {\mathsf{\omega }}_{\mathrm{m}}$

Let the total number of iterations $(\sum \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\left(\mathrm{A}\right)) \mathrm{b}\mathrm{e} \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{v}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{d} \mathrm{b}\mathrm{y} \mathrm{E}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n} (14)$:

$$\sum \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\left(\mathrm{A}\right)={\mathsf{\omega }}_1+{\mathsf{\omega }}_1{\mathsf{\omega }}_2+\dots +{\mathsf{\omega }}_1{\mathsf{\omega }}_2\dots {\mathsf{\omega }}_{\mathrm{m}}$$

(14)

5.3. Starting with the Variable with the Least Restriction

Step 1: $\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s} \mathrm{t}\mathrm{a}\mathrm{k}\mathrm{e}\mathrm{n}={\omega }_m$

Step 2: $\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s} \mathrm{t}\mathrm{a}\mathrm{k}\mathrm{e}\mathrm{n}={\omega }_m{\omega }_{m-1}$

Step m: $\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r} \mathrm{o}\mathrm{f} \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s} \mathrm{t}\mathrm{a}\mathrm{k}\mathrm{e}\mathrm{n}={\omega }_m{\omega }_{m-1}\dots {\omega }_1$

Let the total number of iterations $(\sum \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\left(\mathrm{B}\right)) \mathrm{b}\mathrm{e} \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{v}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{d} \mathrm{b}\mathrm{y} \mathrm{E}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n} (15)$:

$$\sum \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\left(\mathrm{B}\right)={\mathsf{\omega }}_{\mathrm{m}}+{\mathsf{\omega }}_{\mathrm{m}}{\mathsf{\omega }}_{\mathrm{m}-1}+\dots +{\mathsf{\omega }}_{\mathrm{m}}{\mathsf{\omega }}_{\mathrm{m}-1}\dots {\mathsf{\omega }}_1$$

(15)

Therefore,

$$\sum \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\left(\mathrm{A}\right)\le \sum \mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\left(\mathrm{B}\right)$$

(16)

More information on the search process can be found in [17], and in this search process, the worst-case scenario is assumed. Let the restriction index on variable $y_i$ be provided by ${\gamma }_i$; then, we have Equations (17) and (18):

$${\mathsf{\gamma }}_{\mathrm{i}}={\ell }_{\mathrm{i}}^{\mathrm{U}}∆\mathrm{Z}-{\ell }_{\mathrm{i}}^{\mathrm{L}}∆\mathrm{Z}$$

(17)

$${\mathsf{\gamma }}_{\mathrm{i}}={\mathsf{\phi }}_{\mathrm{i}}∆\mathrm{Z}$$

(18)

where ${\phi }_i={\ell }_i^U-{\ell }_i^L$. Thus, the variables can be arranged in their restricted order by arranging their indices in ascending order. The smaller the value of ${\phi }_i$, the more restricted the variable $y_i$ is. We present a way to find a bound such that a predetermined number of integers is accommodated in a calculated variable range.

6. Calculating the Basic Variable Integral Bounds

The decrease in the objective value $(∆Y_0)$ can be obtained by solving the following single variable LP:

$$\begin{gathered} \mathrm{M}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{z}\mathrm{e}=∆Y_0, \\ \mathrm{Such} \mathrm{that}: \\ {\phi }_1∆Y_0\ge \omega , \\ {\phi }_2∆Y_0\ge \omega , \\ \dots \\ {\phi }_m∆Y_0\ge \omega \end{gathered}$$

(19)

where $\omega$ is a range that accommodates at least an integer. The optimal solution to Equation (19) is readily available and is provided by Equation (20):

$$∆{\mathrm{Y}}_0=\mathrm{M}\mathrm{a}\mathrm{x}\left[\frac{\mathsf{\omega }}{{\mathsf{\phi }}_1},\frac{\mathsf{\omega }}{{\mathsf{\phi }}_2},\dots ,\frac{\mathsf{\omega }}{{\mathsf{\phi }}_{\mathrm{m}}}\right]$$

(20)

6.1. Proof

The LP model in Equation (19) can be rewritten as Equation (21):

$$\begin{gathered} \mathrm{M}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{z}\mathrm{e}=∆{\mathrm{Y}}_0, \\ \mathrm{such} \mathrm{that} \\ ∆{\mathrm{Y}}_0\ge \frac{\mathsf{\omega }}{{\mathsf{\phi }}_1}, \\ {\mathrm{Y}}_0\ge \frac{\mathsf{\omega }}{{\mathsf{\phi }}_2}, \\ \dots \\ {\mathrm{Y}}_0\ge \frac{\mathsf{\omega }}{{\mathsf{\phi }}_{\mathrm{m}}} \end{gathered}$$

(21)

The dual variable of Equation (21) becomes that provided in Equation (22):

$$\begin{gathered} \mathrm{M}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{z}\mathrm{e}\frac{\omega }{{\phi }_1}{x}_1+\frac{\omega }{{\phi }_1}{x}_2+\dots +\frac{\omega }{{\phi }_1}{x}_m, \\ \mathrm{such}\mathrm{that} \\ {x}_1+x_2+\dots +x_m\le 1 \end{gathered}$$

(22)

where $x_i$ is the dual variable.

6.2. Determining the Value of $\omega$

The number $\omega$ can be determined in many ways. One of these ways is to use the interval provided in Equation (23):

$$1<\mathsf{\omega }$$

(23)

This interval is valid because the shortest distance between any two integers is one. If the distance is less than one, then we may not find an integer in the interval:

$$2\le \mathsf{\omega }$$

(24)

It may seem logical to start from two onwards:

$$\mathsf{\omega }=2,3,4,\dots$$

(25)

The range $\omega$ is not necessarily an integer; in this case, we are using integers for convenience. Any value of $\omega$ greater than one may be used. Once the restriction expressions for the basic variables are known, the variable bounds can be determined. From the most restricted variable $y_i,$ the decrease in the objective value can be calculated given that at least two integers are accommodated in the bounds as provided in Equation (26):

$$∆{\mathrm{Y}}_0=\mathrm{M}\mathrm{i}\mathrm{n}\left[\frac{1+\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{U}}} \mathrm{o}\mathrm{r} \frac{2-\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{U}}},\frac{1+\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{L}}} \mathrm{o}\mathrm{r} \frac{2-\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{L}}}\right]$$

(26)

where $\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]$ is the fractional part of ${\mathsf{\beta }}_{\mathrm{i}}$. $\frac{1+\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{U}}}$ and $\frac{1+\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{L}}}$ are used when ${\ell }_{\mathrm{i}}^{\mathrm{U}}<0$ and ${\ell }_{\mathrm{i}}^{\mathrm{L}}<0$, respectively, and $\frac{2-\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{U}}}$ and $\frac{2-\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{L}}}$ are used when ${\ell }_{\mathrm{i}}^{\mathrm{U}}>0$ and ${\ell }_{\mathrm{i}}^{\mathrm{L}}>0$, respectively.

Proof. 

The expression ${\mathsf{\beta }}_{\mathrm{i}}+{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0\le {\mathrm{y}}_{\mathrm{i}}\le {\mathsf{\beta }}_{\mathrm{i}}+{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0,$ provided in Equation (8), can be written as Equation (27):

$$\stackrel{-}{{\mathsf{\beta }}_{\mathrm{i}}}+\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0\le {\mathrm{y}}_{\mathrm{i}}\le \stackrel{-}{{\mathsf{\beta }}_{\mathrm{i}}}+\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0$$

(27)

where $\stackrel{-}{{\mathsf{\beta }}_{\mathrm{i}}}$ is the integer and $\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]$ is the fractional part of ${\mathsf{\beta }}_{\mathrm{i}}$.

For the bounds $\stackrel{-}{{\mathsf{\beta }}_{\mathrm{i}}}+\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0$ and $\stackrel{-}{{\mathsf{\beta }}_{\mathrm{i}}}+\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0$ to be integers, $\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0$ and $\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0$ must be integers. This is possible as per Equations (28) and (29).

$$\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0=1,2,3,\dots$$

(28)

$$\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0=1,2,3,\dots$$

(29)

For ${\ell }_{\mathrm{i}}^{\mathrm{U}}>0$ and ${\ell }_{\mathrm{i}}^{\mathrm{L}}>0$, respectively, at least two integers are within the bounds of the most restricted variable provided in Equation (30):

$$\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0=2$$

(30)

and

$$\mathrm{f}\left[\mathsf{\beta }\right]+{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0=2$$

(31)

Thus, $∆{\mathrm{Y}}_0=\frac{2-\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{U}}}$ and $∆{\mathrm{Y}}_0=\frac{2-\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{L}}}$.

Similarly,

$$\mathrm{f}\left[\mathsf{\beta }\right]-{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0=0,-1,-2,\dots$$

(32)

$$\mathrm{f}\left[\mathsf{\beta }\right]-{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0=0,-1,-2,\dots$$

(33)

because ${\ell }_{\mathrm{i}}^{\mathrm{U}}<0$ and ${\ell }_{\mathrm{i}}^{\mathrm{L}}<0.$

Additionally, at least two integers are present in between the bounds provided in Equations (34) and (35):

$$\mathrm{f}\left[\mathsf{\beta }\right]-{\ell }_{\mathrm{i}}^{\mathrm{L}}∆{\mathrm{Y}}_0=-1$$

(34)

$$\mathrm{f}\left[\mathsf{\beta }\right]-{\ell }_{\mathrm{i}}^{\mathrm{U}}∆{\mathrm{Y}}_0=-1$$

(35)

Thus, $∆{\mathrm{Y}}_0=\frac{1+\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{U}}}$ and $∆{\mathrm{Y}}_0=\frac{1+\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{L}}}$.

The value of $∆{\mathrm{Y}}_0$ is substituted to obtain the bounds for all the basic variables, and once these are available, they can easily be changed to integers, as shown in Equation (36):

$${\mathrm{I}}_{\mathrm{i}}^{\mathrm{L}}\le {\mathrm{y}}_{\mathrm{i}}\le {\mathrm{I}}_{\mathrm{i}}^{\mathrm{U}}$$

(36)

where ${\mathrm{I}}_{\mathrm{i}}^{\mathrm{L}} \mathrm{a}\mathrm{n}\mathrm{d} {\mathrm{I}}_{\mathrm{i}}^{\mathrm{U}}$ are the lower and upper integer limits, respectively. □

7. Determining the Bound for the Liner Integer Model

A linear integer model can usually be solved with the approximated integer bounds with fewer subproblems than in the case where no variable bounds are present. If no feasible solution exists, then ${\mathsf{\omega }}_1{\mathrm{s}}_1+{\mathsf{\omega }}_2{\mathrm{s}}_2+\dots +{\mathsf{\omega }}_{\mathrm{n}}{\mathrm{s}}_{\mathrm{n}}\ge ∆{\mathrm{Y}}_0$ or Equation (37):

$${\mathrm{Y}}_0\le \mathrm{R}-∆{\mathrm{Y}}_0$$

(37)

Equation (37) depicts a strong bound for the linear integer model. This is then used as an upper bound for the branch and bound, branch and cut, branch and price, branch cut and price, and branch cut and free algorithms, or any selected suitable procedure. If a feasible solution exists, then the upper bound is optimal.

Justification

The whole area

$$\mathrm{R}-∆{\mathrm{Y}}_0\le {\mathrm{Y}}_0\le \mathrm{R}$$

(38)

is searched for integer points, and the feasible optimal integer solution for this small region is the global optimal one.

7.1. Procedure for Finding a Strong Bound

Step 1: Relax and solve the model so that a continuous optimal solution can be obtained.

Step 2: Express the basic variables in terms of their approximated integer variable bounds.

Step 3: Solve the LIP model now with the approximated variable bounds. When branching, start with the most restricted variables.

Step 4: If the solution is feasible, then it is optimal; otherwise, an upper bound is obtained as ${\mathrm{Y}}_0\le \mathrm{R}-∆{\mathrm{Y}}_0$ and is used in the selected approach for the LIP model.

7.2. Numerical Illustration

Consider the three-variable linear integer model provided in Equation (39) that is used to prove the validity and applicability of the variable bounding approach. Without the introduction of the variable bounds, the branch and bound method requires 9123 iterations to solve the problem, and only 3 iterations are required when variable bounds are computed first. This numerical illustration clearly shows the importance and advantages of approximating the integer variable bounds.

$$\begin{gathered} \mathrm{M}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{z}\mathrm{e}{\mathrm{Y}}_0=17{\mathrm{y}}_1+19{\mathrm{y}}_2+22{\mathrm{y}}_3, \\ \mathrm{Such} \mathrm{that} \\ 9{\mathrm{y}}_1+15{\mathrm{y}}_2-15{\mathrm{y}}_3\le 87, \\ 10{\mathrm{y}}_1-12{\mathrm{y}}_2+6{\mathrm{y}}_3\le 45, \\ -7{\mathrm{y}}_1+13{\mathrm{y}}_2+15{\mathrm{y}}_3\le \mathrm{99,008}. \end{gathered}$$

(39)

where ${\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_3\ge 0 \mathrm{a}\mathrm{n}\mathrm{d} \mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{r}$.

The continuous optimal table is provided in

Table 3. Continuous optimal table.

	${\mathit{y}}_1$	${\mathit{y}}_2$	${\mathit{y}}_3$	${\mathit{s}}_1$	${\mathit{s}}_2$	${\mathit{s}}_3$	R.H.S
$Y_o$	0	0	0	1.1918	1.9440	1.8809	186,410.8961
$y_1$	1	0	0	0.0438	0.0713	0.0153	1519.3595
$y_3$	0	0	1	−0.0078	0.0377	0.0438	4336.3971
$y_2$	0	1	0	0.0326	−0.0051	0.0346	3430.5815

.

Whereby the variables are expressed in terms of $∆{\mathrm{Y}}_0$. From the continuous optimal table provided in Table 3, we have Equation (40):

$$1.1918{\mathrm{y}}_1+1.944{\mathrm{y}}_2+1.8809{\mathrm{y}}_3\ge ∆{\mathrm{Y}}_0$$

(40)

Because $\mathrm{n}=3$, the possible solutions in terms of $∆{\mathrm{Y}}_0$ are provided as follows:

When variable ${\mathrm{s}}_1$ enters the basis:

$${\mathrm{y}}_1=1519.3595-0.0368∆{\mathrm{Y}}_0,$$

$${\mathrm{y}}_2=3430.5815-0.0274∆{\mathrm{Y}}_0,$$

$${\mathrm{y}}_3=4336.3971-0.0065∆{\mathrm{Y}}_0.$$

When variable ${\mathrm{s}}_2$ enters the basis:

$${\mathrm{y}}_1=1519.3595-0.0367∆{\mathrm{Y}}_0,$$

$${\mathrm{y}}_2=3430.5815-0.0026∆{\mathrm{Y}}_0,$$

$${\mathrm{y}}_3=4336.3971-0.0194∆{\mathrm{Y}}_0.$$

When variable ${\mathrm{s}}_3$ enters the basis:

$${\mathrm{y}}_1=1519.3595-0.0081∆{\mathrm{Y}}_0,$$

$${\mathrm{y}}_2=3430.5815-0.0184∆{\mathrm{Y}}_0,$$

$${\mathrm{y}}_3=4336.3971-0.0233∆{\mathrm{Y}}_0.$$

By arranging the variables in the order of their descending restriction, the variable bounds become that provided in Equation (41):

$$\begin{gathered} 1519.3595-0.0368∆{\mathrm{Y}}_0\le {\mathrm{y}}_1\le 1519.3595-0.0081∆{\mathrm{Y}}_0 \\ 3430.5815-0.0274∆{\mathrm{Y}}_0\le {\mathrm{y}}_2\le 3430.5815-0.0026∆{\mathrm{Y}}_0 \\ 4336.3971-0.0233∆{\mathrm{Y}}_0\le {\mathrm{y}}_3\le 4336.3971-0.0065∆{\mathrm{Y}}_0 \end{gathered}$$

(41)

The variable ranges can be calculated as provided in Equation (42):

$$\begin{gathered} \mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e} y_1,{\tau }_1=-0.0081∆Y_0+0.0368∆Y_0=0.0287∆Y_0, \\ \mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e} y_2,{\tau }_2=0.0026∆Y_0+0.0274∆Y_0=0.03∆Y_0, \\ \mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e} y_3,{\tau }_3=0.0065∆Y_0+0.0233∆Y_0=0.0298∆Y_0. \end{gathered}$$

(42)

By arranging the variables in their approximated descending order of restriction, we have the following:

Descending order $\stackrel{{\mathrm{y}}_1{\mathrm{y}}_3{\mathrm{y}}_2}{\to }$

Calculating the basic variable limits

The most restricted basic variable is ${\mathrm{y}}_1$; thus,

$$1519.3595-0.0368∆{\mathrm{Y}}_0\le {\mathrm{y}}_1\le 1519.3595-0.0081∆{\mathrm{Y}}_0$$

(43)

Because $∆{\mathrm{Y}}_0=\mathrm{M}\mathrm{i}\mathrm{n}\left[\frac{1+\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{U}}} \mathrm{o}\mathrm{r} \frac{2-\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{U}}},\frac{1+\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{L}}} \mathrm{o}\mathrm{r} \frac{2-\mathrm{f}\left[{\mathsf{\beta }}_{\mathrm{i}}\right]}{{\ell }_{\mathrm{i}}^{\mathrm{L}}}\right]$, the smallest $∆{\mathrm{Y}}_0$ (i.e., $∆{\mathrm{Y}}_1)$ is calculated by using the equation provided in Equation (44):

$$∆{\mathrm{Y}}_0=\mathrm{M}\mathrm{i}\mathrm{n}\left[\frac{1.3595}{0.0368}=36.9428,\frac{1.3595}{0.0081}=167.8395\right]=36.9428$$

(44)

By replacing $∆{\mathrm{Y}}_0$ with $∆{\mathrm{Y}}_1$, we have Equation (45):

$$\begin{gathered} 1518.00\le {\mathrm{y}}_1\le 1519.06 \\ 3429.57\le {\mathrm{y}}_2\le 3430.68 \\ 4335.53\le {\mathrm{y}}_3\le 4336.63 \end{gathered}$$

(45)

After adjusting to the integer bounds, we have Equation (46):

$$\begin{gathered} 1518\le {\mathrm{y}}_1\le 1519 \\ 3430\le {\mathrm{y}}_2\le 3430 \\ 4336\le {\mathrm{y}}_3\le 4336 \end{gathered}$$

(46)

The upper bound is provided by ${\mathrm{Y}}_0\le \mathrm{186,410.8961}-36.9428;$ thus,

$${\mathrm{Y}}_0\le \mathrm{186,373.9533}$$

(47)

Solving the LIP model with the approximated integer bounds

By selecting an automated branch and bound algorithm and solving the LIP model with the approximated integer bounds, a feasible solution is obtained and verified in only three iterations. The optimal solution obtained is

$$\begin{gathered} {\mathrm{Y}}_0=\mathrm{186,368}, \\ {\mathrm{y}}_1=1518, \\ {\mathrm{y}}_2=3430, \\ {\mathrm{y}}_3=4336 \end{gathered}$$

(48)

Without the bounds, the branch and bound method (automated version) requires 9123 iterations to obtain the same optimal solution provided in Equation (48).

8. Conclusions

We solved linear integer models by using the variable bounding method. Finding a bound is important and is supposed to be the first attack method that is required for any linear integer programming problem. The computational complexity and many iterations that are required to search the NP Complete LIP problems may be significantly reduced by using bounds. One can independently generate n possible solutions, which are used to calculate limits. These independent calculations allow one to use parallel processors [18]. Using parallel processing alongside the available approaches for linear integer programming has been unsuccessful. A stronger bound can be found by increasing the number of integers accommodated within the limits of the most restricted variable. However, there should be a balance between the strength of the bound and the resulting number of iterations required to generate it.