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OPTIMIZING A LINEAR FRACTIONAL FUNCTION OVER AN INTEGER EFFICIENT SET
Leila YOUNSI-ABBACI •
University of Bejaia, Department of Electrical Engineering, Faculty of Technology, Research Unit LaMOS, 06000 Bejaia, Algeria [email protected]
Abstract
Over recent decades, significant advancements have been made in optimization over the efficient set. This paper introduces a novel exact algorithm designed to optimize a linear fractional objective function over the integer efficient set of a multi-objective linear programming problem (MOILP). Without enumerating all efficient solutions, our method employs a selection strategy to iteratively improve the primary objective while progressively refining the feasible region and excluding dominated points. By exploring edge connections within the truncated feasible space, the proposed algorithm ensures convergence to the global optimal value in a finite number of iterations. A numerical example demonstrates the algorithm's effectiveness and practical application. This approach addresses critical challenges in multi-objective integer programming, particularly the nonconvexity of the efficient set and the absence of explicit feasible set descriptions.
Keywords: multiple objective programming, integer programming, linear fractional programming, efficient solutions.
1. Introduction
Multi-objective integer programming (MOIP) is an important research area as many practical situations require discrete representations by integer variables and many decision makers have to deal with several objectives. Some note-worthy practical environments where the MOIP problems find their applications are supply chain design, logistics planning, scheduling and financial planning.
In the past two decades, researchers and practitioners have shown increased interest in the problem of optimizing a linear function on the efficient set of multiple objective linear programming problem (MOLP). Several methods and algorithmic ideas have been developed-in general, these approaches can be classified and grouped according to the methodological concepts-which include, among others, adjacent vertex search technique ([16, 9,10], nonadjacent methods [7], dual approach [19], etc. An overview of these approaches can be found in Yamamoto [21].
In addition to the continuous case, few algorithms have been suggested for solving the problem involving discrete decision variables. For the first time in [15] made an attempt to optimize on the integer efficient set, where only an upper bound value for the main objective is proposed. Jorge [13] developed approach that defines a sequence of progressively more constrained single-objective integer problems that successively eliminates undesirable points.
Fractional programming is an optimization problem in which ratio of two linear functions is optimized subject to some constraints [5, 14]. Integer Linear Fractional Programming problem is
an important class of problems arising in criteria Decision Making when some or all the model variables represent discrete decisions.
In preparing this paper, a special effort has been made do make certain that it is self-contained and that it is suitable both a as a text and as a reference. within we developed an algorithm that optimized linear fractional function ever the efficient set of a MOILP without explicitly having to enumerate all the efficient solutions. Given a Integer Linear Programming problem with Multiple Objective (MOILP):
P)i "max" Zi = CiX, i £{1,...,p} ^
1 s.t. x £ DD
Where Ci £ Rn, for each i £ {l,..., p}, A £mxn, b £m and D is a polyhedral set of n defined as D={x £n | Ax = b, x > 0, integer}. To avoid the technicality we assume throughout the paper that D is nonembounded.
The search of specific methods for solving (l) that provide the decision maker with his/her preferred efficient solution without having to explicitly determine the set of all efficient solutions of (l)denoted by E(Pd), efficiency and non-dominance are defined as follows (see [17, 23, 24]) is doubtless a very difficult task that can be tackled in many different ways. One of such approaches, that has been studied successfully by Philip [16], in which an algorithm based on moving to adjacent efficient vertices is outlined when O(x) is a linear function, and lots of papers followed his work [22]. Our aim in this study is to provide one approach in the discrete case, consists of optimizing O(x) a Linear Fractional function representing the preferences of the decision-maker over the efficient set of (1). Formally, the problem under consideration can be defined as:
. Ux + a
max °(x) = Vx+p (2)
s.t. x £ (Pd )
Where a, p are scalars; p, q £ Rn.
The main difficulty of the problem arises from the nonconvexity of the efficient set (E(Pd)), which is the union of several faces of X. This problem was first considered by [16], in which an algorithm based on moving to adjacent efficient vertices is outlined when 1 is a linear function, and lots of papers followed his work.
It is worth noting that solving (2) involves several difficulties since its feasible set, (E(PD)), is not explicitly known, nor a convenient implicit description (say, e.g., integer linear) is available. As a consequence, (2) is a global optimization problem, frequently with multiple local (not necessarily global) optima [[22], [11]]. However, some particular instances of problem (1) can be solved straightforwardly, due to their special characteristics. More precisely, when the multi objective problem IP is completely efficient [2].
Generally, E(PD) = D. Otherwise, if (D) is completely efficient, E(PD) can be substituted by D and, in such cases, solving (Pe) is equivalent to solving the following program:
^ , . Ux + a
/t, \ i max O(x) = 77-7 .„,
(PE-relaxed) { K ' Vx + p (3)
s.t. x £ D
2. The main results
Definition 1. A point x0 £ D is said to be efficient of (1) if and only if there does not exist another point x1 £ D such that Zi(x1) > Zi(x0) for all i £ {1,..., p} and Zi(x1) > Zi(x0) for at least one i £ {1,...,p}.
2.1. Testing Efficiency
The following result (see [12]) is used in various steps of the algorithm to test the efficiency of a given feasible solution of problem (1).
Theorem 1. Let x* be an arbitrary element of the region D. x* E EFF if and only if the optimal
value of the objective ty is null in the following mixed integer linear programming problem:
P
max ty(x) = ^Yj
(Px*){ i=i (4)
V x M s.t. Cx - IY = Cx*
x E D, Yi E R+; Vi E {1,..., p}.
C is a matrix (p,n) of which her ieme line corresponds to ci, i = 1,2,..., p, I is the matrix identity
(p, p) and Y = (Yi)i=1.....p. The problem (Px*) .
Is often used to test the efficiency of a given point. (Px*) can be also used to generate an efficient point even x* is not efficient ([9]).
2.2. Notation and Definitions
• xk = x^j is one optimal integer solution obtained in Dkat step k.
• Bk is the basis associated with solution xk;
• akj E Rmk x1 is the activity vector of x^j with respect to the current truncated region D^
• Ik = {j I the vector akj is a column of the basis Bk} (indices of basic variables);
• Nk = {j I the vector a^j is not a column of the basis Bk} (indices of non-basic variables);
• ykj = (ykij) = (Bk)-1 aKj, where yKj E Rmkx1;
• Uj = the jth component of vector U;
• Vj = the jth component of vector V;
• pkj = E piykij
iElk
— E Why
ie lk
Zk,i _ Uxk + a
• Zl (xk ) — K2 vxk+a
• Ykj — Zk,2(Pj — Pk,j) — Zki(qj — qk,j) , the updated value of the jth component of the reduce gradient vector Yk
Definition 2. Assume that jk e Nk An edge Ejk incident to a solution Xk is defined as the set
Ek — < xi e R1l+NkI
Xi — xKi — OjkVkijk for i e Ik xj,k — Qj,k
xa — 0 for all a e Nk \ {jk}
xk i
Where 0 < d < min{—— lykj > 0}, djk is a positive integer and djk x ykj for i E Ik are
iElk yk,ijk
integers for all i E Ik if such integer values exist.
Theorem 2. [14] Let Xi be an optimal solution of problem (3) All integer feasible solutions of problem (3) alternate to Xi on an edge Ej1 of region D (or truncated region Di ) emanating from it, in the direction of vector a1/j1, j1 G /1 with /1 = {j G N1 | Yi j = 0} lie in the open half space
E xj <1
jGNi\{ji}
Theorem 3. [4] The point Xi of D is an optimal solution of problem (3) if and only if the reduce gradient vector y = pp — ixq is such that Yj < 0 for all j Gfc.
Theorem 4. [17] x* G E(PD if, and only if, {(x* + C>) n D = x*}.
3. Development of the algorithm and theoretical results
The proposed algorithm provides a global optimal solution of (PE) without specifying all efficient solutions of (P(D)).
Initially, we solve the relaxed problem (3) associated to problem (PE). Obviously, only in a reduced number of special cases would the solution of (3) provide the optimal solution of (PE). So if it were not the case, a new efficient solution dominating the previous one is then obtained. The efficient solution xl issued from the efficiency test is considered as a first efficient solution.
Assuming that all coefficients of matrix C are integers, at iteration k, the feasible set D is reduced gradually by eliminating all dominated solutions by C(x)1 (see Sylva and Crema, 2004, 2007). The resolution of the following problem enables us to perform this elimination:
(Rfl) : max{|x G D — Uls=iDs} (5)
{xs;s = 1,...,l — 1} are solutions of (PD) obtained at iterations 1,2,...,l — 1 respectively. Where Ds = {x Gn ICx < Cxs} and {Cxs}S=1 is a subset of nondominated criteria vectors for problem
(Pd ).
D — uS=i Ds
cix > (cix = 1)ys + Mi(1 — ys, i = 1,2,...p s = 1,2,...,l.)
Eys > 1, s = 1,2.....l
i=1
y* G {0,1}, i = 1,2,...p s = 1,2,...,l
xi D
where Mi is a lower bound for any feasible value of the ith objective function. The associate variables ys i = 1,2, ...p of Cxs and additional constraints are added to impose an improvement on at least one objective function. Note that when yis = 0, the constraint is not restrictive and when y\ = 1a strict improvement is forced in the ith objective function evaluated at Cxs.
We start exploring all edges incident to xl corresponding to /1 until an efficient solution is found to improve Op We solve the problem (Rfl ). The optimal solution obtained, xl, produces a minimum value of the criterion O(x) in the reduced domain. The process continue in this manner until the current feasible space becomes empty or O(xl) > O0pt.
Proposition 1. [6] Let x1 x2,...,xl be efficient solutions to problem (PD) and Ds = {x Gn ICx <
Cxs}. Let x* be an efficient solution to the multi-objective integer problem Pk = "max"{Cx,x G D — us=iDs}. Then x* is an efficient solution to the problem (PD ).
3.1. Theoretical Results
Proposition 2. Let x1, x2, ...,xl be efficient solutions to problem P(D) and Ds = {x Gn ICx < Cxs}.
Let xl be an alternative solution of xl of the problem (Rfl ) with Ux. , + a > max { Ux + .
F v Jl' Vxl+1 + p jGi.....lXVxs +
Leila YOUNSI-ABBACI RT&A, No 1 (82)
OPTIMIZING A LINEAR FRACTIONAL_Volume 2°, March 2025
If xl is an efficient solution to problem (P(D)) then is an optimal solution of (Pe).
if problem (Rfi) is unfeasible then {Cxs}S—1 is the entire set of non-dominated criterion
vectors for problem (PD).
Proof. Suppose on the contrary that xl is not an optimal solution of (PE). Then a feasible solution exists x e E(PD) such that with the value of the function main to the x point superior a
Uxl
V ! + a • As x is an alternative solution for x ( ©9, — 0)to (Rfl) because xs
Uxl + a Uxl + a
—-- — —--. Thus x G Ul 1 Ds therefore xx G Ds for some s G il,..., l} and, accordingly
vxl+a Vxl + a s—1
Ux + a uxs + a
to the definition of Ds, Cx < Cxs. As x e E(PD we have that —-- < —--.
v D vxe+a vxs+a
, Ux1 + a Ux1 + a Ux + a Uxs + a ,. . . , , , consequently —;-- — —;-- < 777-- < —-- who is contradicting with the hy-
4 y vxl+a vxl+a vxe+a vxs+a 6 y
. Ux1 + a r Uxs + a -, pothesis —;-- > max { —-- }.
F vxl + a jel.....l vxs + a
If (Rfl ) is unfeasible then E(PD ) Ç Uls—1 Ds and for any x e E(PD ) there exists an xs such that Cx < Cxs. In this case we must proceed as follows: let xx e E(PD) for the reason there is an 3s e 1,..., l with Cxs > Cx then Cxs — Cx (and Cx is a dominated vector).
3.2. Algorithm
The algorithm used to obtain an integer optimal solution to our main problem (PE) is can be summarized as follows:
Algorithm 1: part 1 input :
A(mxn): matrix of constraits,
b(mxn),
RHS vector,
C(px n): matrix of criteria. U(1xn), V(1xn): main criterion vector, a, ft: are scalars. output :
X0pt:optimal solution of the problem (PE), ^opt:optimal value of the main criterion O initialization: for i ^ 1 to p do
solve Mi= min{Cix, x E D} set the lower bounds;
Oopt := -inf, l := 1,
Ei := , D := D, optimal := false, alternative := false, explore := true.
Algorithm 2: part 2
while optimal:=false do
solve P{F = max{, x G D};
if P1RF is infeasible then
X0pt an optimal solution of (PE); optimal : = true,'Terminate;
else
let xl bean optimal solution of PRF;
efficiency test: solve (p(xl)), Y is the optimal solution criteria; if Y = 0 then
xl an efficient solution; X0pt an optimal solution of (PE) and Oopt = O(xl);
else
xl is not efficient solution, x1 an optimal solution of (p(xl )) is efficient; solve Q(xl) = max{, x G D, Cx = Cx1};
let xl bean optimal solution of Q(xl ) if 0(xl) > Oopt then
_ Xopt = xl, Oopt = Ux+ß, let El+i = El U {xl};
l = l + 1 and D := D uS= Ds; Ds = {x G Zn/Cx < Cxl, xl G El_1} solvePl = {maxUx+ß, x G DD}, let xl an optimal of Pi ;
if D = or O(xl) < Oopt then
Xopt an optimal solution of (PE) and Oopt the optimal value of (PE) optimal := trwe,Terminate;
else
optimal:=False ; solve (P(xl) if Y = 0 then
xl an efficient solution X0pt = xl; optimal:=True; El+1 = El U {x1}; Terminate; else
xl is an optimal solution of P(xl) is efficient Ej+i = El U {x1}; construct the set r = {j G Ni/yj = 0}; if r = then i:=1;
while rl = and explore:= true do
(search xi integer efficient solution for xl ) calculate ©^the integer
part of miniGik { -±L /y}ih > 0};
xl, ih 'n
if ®0l
0 then
rl = rl \{Jl (i)};
else
© := ©0l;
while © > 0 and alternative:=False do
searching for a efficient integer solution on edge Ejl
corresponding to ©0l and test for efficiency, solve P(D)
if Y = 0 then
alternative:=true; O0pt := Oxexpl; optimal:=True;
El+1 = El U {xexpl}; Terminate;
else
L © := © — 1
i:=i+1;
l:=l+1;
Proposition 3. The algorithm terminates in a finite number of iterations.
Proof. By hypothesis provided D is non-empty and D is bounded, {Cxs }S=1 is finite. With the progression of to advance in the algorithm, the domain of feasibility becomes more and more is strictly reduced by the theorem (sylva [18],[6]) or O0pt strictly increases.
The theorem (3) guarantees that we can obtain an optimal solution integer of (PfR) if it exists, and the theorem (Testing efficiency 1 with [9]) one gets an optimal solution for the problem (2) having in mind that at least one new efficient solution is generated at each iteration since for an arbitrary l none of the previously generated efficient point is feasible, the proof is thus complete. ■
4. Numerical illustration
To illustrate the use of this algorithm, we consider the following integer linear program with tow objectives:
(P(D))
max Z1 = x1 - 2x2 max Z2 = —x1 + 4x2 s.t. -2x1 + x2 < 0, 6x1 + x2 < 21, - 2x1 + 4x2 < 6, x1, x2 E
(6)
Figure 1: Space of the decisions
In this example, it is easy to see that D contains 11 feasible points (see Figure 1). Using the characterization of efficiency presented in Theorem (4), it can be shown that seven of them are efficient. Particularly, the efficient set E(PD) is given by: FF = {(2,0), (2,1), (2,2), (3,0), (3,1), (3,2), (3,3)}. With the aim of illustrating how our algorithm works, we will solve the problem (2) given by
(Pf(E))
x1 + x2 - 1
max O = ---
5x1 + x2 - 1
s.t. x1, x2 eff
step 0: Initialization We take Oinf = -to, O;
sup
l = 1.
After solving {min Cix, x E D} i = 1,2, the lower bounds of the objective functions are
M1 = -3, M12 = -3
(Pf(R))
Xi + X2 - 1
max O = ---
5x1 + x2 - 1
s.t. —2x1 + x2 < 0,
6x1 + x2 < 21, -2x1 + 4x2 < 6, x1, x2 E
(8) is solved, yielding the optimal solution x1 = (0,0), Let Z(xl) = (0,0). • Iteration 1.
• Step 1. In order to test the efficiency of x1 we solve the problem (9), that is:
(8)
(P(x1))
max 0 = Y1 + Y2 s.t. (x1, x2) G D
x1 — 2x2 — Y1 = 0 —x1 + 4x2 — Y2 = 0 Yi > 0, i = 1,2.
(9)
The optimal value of (9) is 2 , which is achieved at the point x1 — (2,1). Thus, x1 eFF and x1 eFF, since Cx1 > Cx1 We set Gsup — ^(x1 ) — 1
Step 2. When (10) defined as:
(Tfi )
xi + x2 — 1
max O = ---
5xi + x2 — 1
s.t. (x1, x2) G D
x1 — 2x2 = 0
—x1 + 4x2 = 2
(10)
is solved, x1 — x11 — (2,1) is obtained as the optimal solution. Let z1 — Cx1 — (0,2) ^(x1 ) — 1/5 > ®inf — —œ, put ®inf — 1/5 et Xopt — x1 Ginf — ®suV, go to step 3
Step 3. The optimal solution of (11)
max
O
x1 + x2 — 1
(RF )
5x1 + x2 — 1 s.t. x1, x2 G D
x1 — 2x2 > y1 — 3(1 — y1) (1) —x1 + 4x2 > 3y1 — 3(1 — y1 ) (2)
y1 + y\ > 1, y\, y1 G {0,1}
(11)
is x2 = (1,2), y = (0,1), being Z(x2) = (-3,7) and Y = (0,2). In order to test the efficiency of x1 we solve the problem (9), that is:
(P(x2))
max 0 = Y1 + Y2 s.t. (x1, x2) G D
x1 — 2x2 — Y1 = —3 —x1 + 4x2 — Y2 = 7 Yi > 0, i = 1,2.
(12)
The optimal value of (12) is 2 , which is achieved at the point x2 — (3,3); Y — (0,2).
1
Thus, x2 eFF and x2 eFF, We set Gsup — G(x2) — 3 go to step 4.
Figure 2: The reduced regionD1
Step 4. J2 = {j G N2 | Hh
0} = 0, go to step 2.
Iteration 2.
• Step 2. When (10) defined as:
xx + x2 - 1
max O = ---
5xi + x2 - 1
(Tf1 ) ^ s.t. (x1, x2) G D
x1 - 2x2 = -3 -x1 + 4x2 = 9
(13)
is solved, xc2 = x2 = (3,3) is obtained as the optimal solution. Let z1 = Cx1 = (0,2). ^(x1 ) = 5/17 > Oinf = 1/5, put Oinf = 5/17 et Xopt = x2 Oinf = Osup, go to step 3
Step 3.The optimal solution of (11)
(RF1 )
x1 + x2 - 1
max O = ---
5x1 + x2 - 1
(x1, x2 ) G D
s.t.
x1 - 2x2 > y11 - 3(1 - y1 ) (1) —x1 + 4x2 > 3y1 - 3(1 - y1) (2) y1 + y1 > 1, yj, y2 G {0,1}
x1 - 2x2 > -2y2 - 3(1 - y1) —x1 + 4x2 > 10y2 - 3(1 - y2) y2 + y2 > 1, y1,y2 G {0,1}
(3) (4)
(14)
The problem (14) )is note feasible. Terminate, X0pt = x2 = (3,3) is an optimal solution of (PE) with O(x2) = 5/17.
The set of all solutions efficient of this problem is: FF = {(2,0), (2,1), (2,2), (3,0), (3,1), (3,2), (3,3)}. However, our algorithm optimizes the linear fractional function O = xi + x2-- without having
to determine all these solutions but only {(2,1), (3,3)}.
5x1 + x2 - 1
Figure 3: The reduced region D2
5. Conclusion
The proposed algorithm optimizes a linear fractional function over the integer set of a multi-objective linear program (Pe) by using classical strategies of fractional programming and cutting plane techniques without having to enumerate all the efficient solutions. The main advantage of the proposed solution methodology is that no nonlinear optimization is required.
Although the research themes addressed is difficult, it is hoped that this article motivate the researchers to develop better solution procedures.
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