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On the mixed integer signomial programming problems
Applied Mathematics and Computation 170 (2005) 1436–1451
www.elsevier.com/locate/amc
On the mixed integer signomial programming problems Ching-Ter Chang Department of Information Management, Changhua University of Education, Paisa Village, Changhua 50058, Taiwan, ROC
Abstract This paper proposes an approximate method to solve the mixed integer signomial programming problem, for which the objective function and the constraints may contain product terms with exponents and decision variables, which could be continuous or integral. A linear programming relaxation is derived for the problem based on piecewise linearization techniques, which first convert a signomial term into the sum of absolute terms; these absolute terms are then linearized by linearization strategies. In addition, a novel approach is included for solving integer and undefined problems in the logarithmic piecewise technique, which leads to more usefulness of the proposed method. The proposed method could reach a solution as close as possible to the global optimum. 2005 Elsevier Inc. All rights reserved. Keywords: Piecewise linear function; Signomial programming; Linearization technique
E-mail address: [email protected] 0096-3003/$ - see front matter 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.01.034
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1. Introduction This paper addresses the mixed integer signomial programming (MISP) problem, which seeks a global optimum to a mixed integer signomial objective function subject to a set of mixed integer signomial constraint functions, each decision variable could be continuous or integral. Such optimization problems occur quite frequently in engineering design, inventory control, scheduling, etc. [1,3,4,10]. Some approaches in the literature for solving the related problem called the polynomial programming (PP) problem are discussed as follows. Horst and Tuy [6], Hanson et al. [5] derived outer approximation techniques and interval analysis based on sufficient conditions for convergence and then proposed a variety of methods for solving PP problems. These methods, however, are only convergent in the absence of blocking sub-problems. In addition, these methods exploit the pure 0–1 structure of the problem, and are therefore not directly applicable for MISP problems. Sherali and Tuncbilek [11] derived a generic algorithm called Reformulation Linearization Technique (RLT) to solve PP problems, their algorithm generates non-linear implied constraints by taking the products of bounding terms in the constraints set to a suitable order. Defining new variables, one for each polynomial term appearing in the problem, subsequently linearizes the resulting problem. Incorporating appropriate bound factor products in their RLT scheme, and employing a suitable partitioning technique have developed a convergent branch and bound algorithm. Although the RLT algorithm is very promising in solving a PP problem, however, there are some difficulties for applying that algorithm to solve a MISP Problem, as listed below: (i) RLT algorithm cannot be directly applied to solve problems containing integer variables. (ii) There are various approaches in designing a suitable RLT process for solving a PP problem. There is no adequate understanding of how best to choose and implement an approach. (iii) RLT algorithm required generating several types of implied constraints at the expense of a huge increase in problem size, which might cause a heavy computation burden. Li and Chou [7] proposed a method for solving the PP problem, based on the multi-level single linkage technique of Rinnooy and Timmer [9], to obtain a global optimum at a pre-specified confidence level. The difficulty of their method is that in order to reach a global optimum at a sufficiently high confidence level, their method the solving of a huge amount of non-linear optimization problems based on various starting points. This always causes a heavy computational burden. Michelon and Maculan [15] proposed a lagrangean decomposition technique for solving SP problems. Li and Chang [8] derived a 0–1 linearization approach for solving PP problems. However, it can only solve a small size of PP problems. Suppose all variables in the problem are continuous, then a PP problem can be reformulated as a geometric programming problem [2,3]. For some special cases it is simpler computationally to reformulate a PP problem as a geometric program. However, current
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geometric programming methods only exploit the continuous structure of the problem, and are incapable of treating MISP problems. In addition, only if the degree of difficulty is less than two could a geometric programming method be applied to solve a PP problem. Tsai et al. [13] proposed some convexification techniques for solving discrete signomial engineering design problems. The mathematical formulation of MISP problem is given below. 1.1. MISP problem Minimize
ZðxÞ ¼
n X p¼1
Subject to
m X q¼1
hq
Y
cp
Y
xai
i2J
xbi 6 lk ;
k ¼ 1; 2; . . . ; K;
i2J
where 0 < xi 6 xi 6 xi (xi and xi are, respectively, the upper and lower bound of xai ); cp, hq, and lk are constants and unrestricted in sign; xi could be continuous or integral variables indexed by some set J; a and b could be an integer or noninteger exponent. This paper proposes a novel method for solving the MISP problem. What follows are the advantages of the proposed approach: (i) The proposed method can treat a MISP problem where a and b are allowed to be non-integer, and xi could be continuous or integral variables. In addition, there is no restriction on the degree of difficulty. (ii) The proposed method can handle a MISP problem without solving a huge amount of non-linear programs. (iii) The proposed method could reach a solution as close as possible to the global optimum, in which the error can be pre-specified by a decision maker. This study first converted the MISP problem into a piecewise linear function (PLF) problem. The non-linear integer formulation strategy is then derived for solving the problem containing the integer variable. Following that, some numerical examples of generalized SP and engineering design problems are solved to demonstrate the usefulness of the proposed method.
2. Formulation of MISP problem 2.1. Formulation of the non-linear real term in MISP problem By refereeing to Chang [14], any signomial term, xai , in MISP can be represented by the following PLF.
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
f ðxÞ ¼ a0 þ s1 ðx a1 Þ þ
n1 X si si1 ðjx ai j þ x ai Þ; 2 i¼2
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ð1Þ
where ai, i = 1, 2, . . . , n are break points of f(x) and si, i = 1, 2, . . . , n are slopes of line segments between ai and ai+1. For more concise expression of (1), let us consider a simple SP problem below. Example 1 Minimize
1:3 z ¼ x1 x1:7 2 x3
Subject to X 2 F
ðF is a feasible setÞ:
We choose break points at a1 = 1, a2 = 2, a3 = 3, and a4 = 4. By referring to (1), Example 1 can be approximated as follows: Minimize z Subject to ln z ¼ ln x1 þ 1:7 ln x2 þ 1:3 ln x3 ; ln x1 ¼ 0:69314ðx1 1Þ 0:14383ðjx1 2j þ x1 2Þ 0:0589ðjx1 3j þ x1 3Þ; ln x2 ¼ 0:69314ðx2 1Þ 0:14383ðjx2 2j þ x2 2Þ 0:0589ðjx2 3j þ x2 3Þ; ln x3 ¼ 0:69314ðx3 1Þ 0:14383ðjx3 2j þ x3 2Þ 0:0589ðjx3 3j þ x3 3Þ; z ¼ eln z ; X 2 F ðF is a feasible setÞ: If x1 = 2, x2 = 3, and x4 = 4 in this instance, the tolerance error approximated below ln x1 ¼ 0:69314ð2 1Þ ¼ 0:69314; ln x2 ¼ 0:69314ð3 1Þ 0:14383ðj3 2j þ 3 2Þ ¼ 1:09862; ln x3 ¼ 0:69314ð4 1Þ 0:14383ðj4 2j þ 4 2Þ 0:0589ðj4 3j þ 4 3Þ ¼ 1:3863; ln z ¼ ln x1 þ 1:7 ln x2 þ 1:3 ln x3 ¼ 4:362984; z ¼ e4:362984 ¼ 78:49100247; error ¼ 0:00104073: 1:3 Now, we really transform a SP term x1 x1:7 2 x3 into an approximate PLF form successful. The error of this approximation is 0.00104073.
By referring to [14], all absolute terms in (1) can be linearized by the following programs P1–P3.
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Program P1 (for linearizing an absolute term with non-positive coefficient in the objective function/constraints). Function Subject to
z ¼ 2g þ 2uai Mðu 1Þ < x ai 6 Mu; x þ ðu 1ÞM 6 g 6 x þ ð1 uÞM; g 6 Mu; g P 0;
ð2Þ ð3Þ ð4Þ ð5Þ
where u is a 0–1 variable; g is a continuous variable. Program P2 (for linearizing an absolute term with positive coefficient in the objective function/constraints). Function
z ¼ 2g 2uai
Subject to
ð2Þ–ð5Þ;
where all variables are as defined in P1. Program P3 (for linearizing an absolute term with positive coefficient in the objective function). ! j m X X Minimize z¼2 x aj þ dk j¼1
Subject to
xþ
m X
k¼1
d k P am ;
k¼1
0 6 d 1 6 a1 ; 0 6 d k 6 ak ak1
for k ¼ 2; 3; . . . ; m;
x 2 F ðF is a feasible setÞ: Notably, the accuracy of the PLF heavily depends upon the number of break points for each variable. However, when the number of break points is increased, the number of 0–1 variables in the PLF is also increased. This will cause a heavy computation burden in the solution process of the problem. In order to solve this problem, a way of choosing the optimal break points has been proposed by Chang [14]. 2.2. Formulation of the non-linear integer term in MISP problem a a a For an be expressed as P integer variable xi ; 0 < xi 6 xi 6 xi ; xi can P xai ¼ xi þ Kk¼0 2k uk , where K is a minimal integer satisfying Kk¼0 2k P xi xi , and uk is a 0–1 variable, then ln xai can be expressed as
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
ln xai ¼ ln xi þ
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K K X Ki X X ½lnð2k þ xi Þ ln xi uk þ ½lnð2k þ 2i Þ þ ln xi k¼0
i¼0
k¼0
lnð2k þ xi Þ lnð2i þ xi Þ uik : For instance, consider an integer x, 2 6 x 6 9, denoting x as x = 2 + u0 + 2u1 + 4u2, ln x can then be expressed as ln x ¼ ln 2 þ ðln 3 ln 2Þu0 þ ðln 4 ln 2Þu1 þ ðln 6 ln 2Þu2 þ ðln 5 þ ln 2 ln 3 ln 4Þu01 þ ðln 7 þ ln 2 ln 3 ln 6Þu02 þ ðln 8 þ ln 2 ln 4 ln 6Þu12 ; where u0, u1, u2 2 {0, 1}, and u01, u02, and u12 are additional continuous variables satisfied the following inequalities: ðu0 þ u1 2Þ þ 1 6 u01 6 ð2 u0 u1 Þ þ 1; u0 6 u01 6 u0 ; u1 6 u01 6 u1 ; ðu0 þ u2 2Þ þ 1 6 u02 6 ð2 u0 u2 Þ þ 1; u0 6 u02 6 u0 ; u2 6 u02 6 u2 ; ðu1 þ u2 2Þ þ 1 6 u12 6 ð2 u1 u2 Þ þ 1; u1 6 u12 6 u1 ; u2 6 u12 6 u2 : If x = 5 in this instance, this leads to u0 = u1 = 1, u2 = 0, u01 = 1, and then ln 5 ¼ ln 2 þ ðln 3 ln 2Þ þ ðln 4 ln 2Þ þ ðln 5 þ ln 2 ln 3 ln 4Þ ¼ 1:609437912: Based on the above strategy, the MISP with non-integer exponent variable can be easily solved. 2.3. The undefined problem of the logarithmic piecewise technique Assume that x1, x2, and x3 are a positive value in Example 1; otherwise (value zero or negative), the logarithmic LPF is undefined. Chang [14] proposed a bounded method to solve the problem of a logarithm of zero. However, the problem of the logarithm being negative can still not be solved by the current published methods. It can be observed that the problem can be a solved if the problem with a negative variable can be converted into the equivalent problem with a positive variable. For more simplicity, but without loss of generality, the following example is given an idea on how to solve the logarithm being negative problem in the LPF. Example 2 Minimize Subject to
x1 x2 x3 X 2 F ðF is a feasible setÞ; 4 6 x1 6 4;
3 6 x2 6 3;
3 6 x3 6 3:
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Example 2 cannot be solved using logarithmic PLF because the problem contains three negative undefined variables as x1 2 [4, 4], x2 2 [3, 3], and x3 2 [3, 3], In order to overcome the undefined problem, the following equivalent program with positive variables is then introduced. Program P4 Minimize
ð Þz
Subject to
y 1 ¼ jx1 j;
y 2 ¼ jx2 j;
ð6Þ
y 3 ¼ jx3 j;
ln u ¼ ln y 1 þ ln y 2 þ ln y 3 ;
ð7Þ
ln y 1 ¼ 0:69314ðy 1 1Þ 0:14383ðjy 1 2j þ y 1 2Þ 0:0589ðjy 1 3j þ y 1 3Þ;
ð8Þ
ln y 2 ¼ 0:69314ðy 2 1Þ 0:14383ðjy 2 2j þ y 2 2Þ;
ð9Þ
ln y 3 ¼ 0:69314ðy 3 1Þ 0:14383ðjy 3 2j þ y 3 2Þ;
ð10Þ
z ¼ eln u ;
ð11Þ
X 2 F ðF is a feasible setÞ;
ð12Þ
0 6 y 1 6 4;
0 6 y 2 6 3;
4 6 x1 6 4;
0 6 y 3 6 3;
3 6 x2 6 3;
3 6 x3 6 3;
ð13Þ ð14Þ
where (•) is a sign determinator. Assume negative variable, xi, as ÔoddÕ numbers and positive variable, xi, as ÔevenÕ numbers. The rules of multiplying follows the same kind of pattern: 2 negatives gives you a positive product, 2 positives gives you a positive product, and one of each gives you a negative product and so on. Thus, the behavior of the sign determinator (•) can be expressed as follows: þ; number of negative variables; xi ; are even numbers; ð Þ ¼ ; number of negative variables; xi ; are odd numbers: If x1 = 2, x2 = 3, and x3 = 2 in this instance, the problem can be approximated below. y 1 ¼ 2;
y 2 ¼ 3;
y 3 ¼ 2;
ln y 1 ¼ 0:69314ð2 1Þ ¼ 0:69314; ln y 2 ¼ 0:69314ð3 1Þ 0:14383ðj3 2j þ 3 2Þ ¼ 1:09862; ln y 3 ¼ 0:69314ð2 1Þ ¼ 0:69314; ln u ¼ 0:69314 þ 1:09862 þ 0:69314 ¼ 2:4849; z ¼ e2:4849 ¼ 11:99992;
error ¼ 0:00008:
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(•) is a negative in sign (2 positives (i.e., x2, x3) and 1 negative (i.e., x1) give you a negative product). Clearly, the objective value is approximated 11.99992. From P4 we can see that there have two major elements in solving the negative undefined problem: (1) yi = jxij; (2) a sign determinator (•). Use the following auxiliary constraints to reach yi = jxij. y i ¼ ð2si 1Þxi ;
ð15Þ
ðsi 1Þxi P 0; si xi P 0;
ð16Þ ð17Þ
where si is a 0–1 variable. Proposition 1. (15)–(17) and yi = jxij are equivalent in the sense that they have the same optimal solutions. Proof. If xi > 0 then si = 1 (from (16)). This forces yi = xi (from (15)). If xi < 0 then si = 0 (from (17)). This forces yi = xi (from (15)). It can be observed that (15)–(17) are equivalent to yi = jxij, This completes the proof of this proposition. h In order to determine the behavior of sign determinator (•), the logical relationships of P5 are listed in Table 1. Program P5 Minimize
ð Þz
Subject to
ð15Þ–ð17Þ;
for i ¼ 1; 2; 3;
ð18Þ
ð7Þ–ð14Þ; where s1, s2, and s3 are 0–1 variables.
Table 1 Logical relationships of sign determinator (•) in P5 s1
s2
s3
(•) (sign determinator)
0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
Negative (i.e., x1 < 0, x2 < 0, and x3 < 0) Positive (i.e., x1 < 0, x2 < 0, and x3 > 0) Positive (i.e., x1 < 0, x2 > 0, and x3 < 0) Negative (i.e., x1 < 0, x2 > 0, and x3 > 0) Positive (i.e., x1 > 0, x2 < 0, and x3 < 0) Negative (i.e., x1 > 0, x2 < 0, and x3 > 0) Negative (i.e., x1 > 0, x2 > 0, and x3 < 0) Positive (i.e., x1 > 0, x2 > 0, and x3 > 0)
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From Table 1 we can see that all cases of negative in sign of (•) can be expressed as: (1 s1)(1 s2)(1 s3) or (1 s1)s2s3 or s1(1 s2)s3 or s1s2(1 s3). Substituting (•) in the objective function of P5 by these logic relationships to obtain the following equivalent objective function. 2½ð1 s1 Þð1 s2 Þð1 s3 Þ ð1 s1 Þs2 s3 s1 ð1 s2 Þs3 s1 s2 ð1 s3 Þ z þ z;
ð19Þ
By referring to the approach of Chang [16], all cubic mixed integer terms s1s2s3 in (19) can be easily linearized as follows: s1 s2 s3 ¼ w;
ð20Þ
where s1, s2, s3, and w satisfy the following inequalities: ðs1 þ s2 þ s3 3Þ þ 1 6 w 6 ð3 s1 s2 s3 Þ þ 1;
ð21Þ
s1 6 w 6 s1 ;
ð22Þ
s2 6 w 6 s2 ;
ð23Þ
s3 6 w 6 s3 ;
ð24Þ
where w is a continuous variable. The above inequalities can be checked as follows: (i) if
Q3
(ii) if
Q3
i¼1 si
¼ 1 then this forces w to be one (from (21)).
i¼1 si
¼ 0 then this forces w to be zero (from (22)–(24)).
The same idea can be applied to quadratic mixed integer terms si sj in (19). Without loss of generality, (19) can be checked by the following cases. (i) If x1 = 2, x2 = 3, and x3 = 1 in P5 then s1 = 0, s2 = 1, and s3 = 0 (from (18)). This forces (19) to be z (i.e., 2[0]z + z). (ii) If x1 = 2, x2 = 3, and x3 = 2 in P4 then s1 = 0, s2 = 1, and s3 = 1 (from (18)). This forces (19) to be z (i.e., 2[1]z + z). Consider the following problem with negative variables. Example 3 Minimize
z ¼ x1 x2 x3 þ x1 x2
Subject to
2x1 þ x2 þ x3 P 0;
ð25Þ
x1 þ 2x2 þ x3 6 5;
ð26Þ
2 6 x1 6 2;
3 6 x2 6 3;
2 6 x3 6 3:
ð27Þ
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This problem contains variables with negative range, which cannot be solved by logarithmic PLF approach. In contrast, this problem can be converted into the following equivalent problem using the above strategies. Program P6 Minimize z ¼ 2½ð1 s1 Þð1 s2 Þð1 s3 Þ ð1 s1 Þs2 s3 s1 ð1 s2 Þs3 s1 s2 ð1 s3 Þ u1 þ u1 þ 2½ð1 s1 Þs2 s1 ð1 s2 Þ u2 þ u2 Subject to ð15Þ–ð17Þ for i ¼ 1;2; 3 Use to replace y 1 ¼ jx1 j; y 2 ¼ jx2 j; y 3 ¼ jx3 j 3 ln u1 ¼ ln y 1 þ ln y 2 þ ln y 3 ; 7 7 ln u2 ¼ ln y 1 þ ln y 2 ; 7 7 7 ln y 1 ¼ :69314ðy 1 1Þ; 7 7 7 Use to approximate ln y 2 ¼ :69314ðy 2 1Þ 7 7 :14383ðjy 2 2j þ y 2 2Þ; 7 7 z ¼ x1 x2 x3 þ x1 x2 ; 7 7 ln y 3 ¼ :69314ðy 3 1Þ 7 7 :14383ðjy 3 2j þ y 3 2Þ; 7 5 u1 þ u2 ¼ elnu1 þ elnu2 ; 0 6 y 1 6 2; 0 6 y 2 6 3; 0 6 y 3 6 3 Corresponding toð27Þ; ð25Þ–ð27Þ:
All absolute terms and mixed integer terms in P6 can be easily linearized by P1 and P2. Solve this problem by LINDO [12] to obtain the approximate global optimal solutions as (x1, x2, x3, y1, y2, y3, ln y1, ln y2, ln y3) = (2, 3, 3, 2, 3, 3, .69324, 1.09861, 1.09861), objective value = 23.99987, error = 0.00013.
3. Numerical examples Example 4. Solve the following problem with a signomial objective function and linear constraints. The tolerable errors are specified as E1 6 0.3 and E3 6 0.03. Minimize Subject to
1:2 x1 x0:5 2 x3 þ 2x1 2x1 þ x2 þ x3 P 8;
ð28Þ
x2 þ 2x3 P 10:5; x1 þ 2x3 6 10;
ð29Þ ð30Þ
1 6 x1 6 5;
1 6 x2 6 5;
1 6 x3 6 5:
ð31Þ
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It is noticeable that because the objective function contains a signomial term which has non-integer exponents, this problem can not be solved by RLT method (Sherali and Tuncbilek [11]). Since all coefficients in the objective function are positive and all constraints are linear, this example can be converted 1:2 into PLF problem. Denote z ¼ x1 x0:5 2 x3 and approximate z by (1) where three best initial break points are chosen as follows, by referring to Chang [14] 77:13 1 e2:86 1 ¼ 2:86; b1 ¼ ln ¼ 1:75; ln 77:13 ln 1 2:86 ln 1 77:13 e2:86 b3 ¼ ln ¼ 3:69; ln 77:13 2:86
b2 ¼ ln
where 77.13 and 1 is the upper and lower bound of z, respectively. The associated slope values are computed as s1 ¼
e1:75 e0 ¼ 2:72; 1:75 0
s3 ¼
e3:69 e2:86 ¼ 27:21; 3:69 2:86
s2 ¼
e2:86 e1:75 ¼ 10:55; 2:86 1:75
s4 ¼
e4:35 e3:69 ¼ 56:72: 4:35 3:69
z is then expressed approximately as 10:55 2:72 ðjy 1:75j þ y 1:75Þ 2 27:21 10:55 56:72 27:21 þ ðjy 2:86j þ y 2:86Þ þ 2 2 ðjy 3:69j þ y 3:69Þ;
z ffi 1 þ 2:72y þ
ð32Þ
where y = y1 + 1.2y2 + 1.2y3 = ln x1 + 1.2 ln x2 + 1.2 ln x3. Since ln xi is a concave function of xi, only one break point is chosen for linearizing ln xi in order to save the number of new added 0–1 variables. This break point is chosen following, by referring to Chang [14] ai1 ð0Þ ¼
51 ¼ 2:49; ln 5 ln 1
for i ¼ 1; 2; 3:
The associated slope values becomes si1 ¼
ln 2:49 ln 1 ¼ :61; 2:49 1
si2 ¼
ln 5 ln 2:49 ¼ :28: 5 2:49
Denote yi as yi = ln xi, which is approximately expressed as y i ffi :61ðxi 1Þ þ
:28 :61 ðjxi 2:49j þ xi 2:49Þ; 2
i ¼ 1; 2; 3:
ð33Þ
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Expression (32) and (33) can then be linearized by P1 through P3 as follows: z ffi 1 þ 2:72y þ ð10:55 2:72Þðy 1:75 þ d 1 Þ þ ð27:21 10:55Þ ðy 2:86 þ d 1 þ d 2 Þ þ ð56:72 27:21Þ ðy 3:69 þ d 1 þ d 2 þ d 3 Þ; y þ d 1 þ d 2 þ d 3 P 3:69; d 1 6 1:75; d 2 6 1:11; d 3 6 :83; y ¼ y 1 þ :5y 2 þ 1:2y 3 ; y i ¼ :61ðxi 1Þ þ ð:61 :28Þðwi 2:49ui xi þ 2:49Þ; i ¼ 1; 2; 3; wi P xi þ 999ðui 1Þ; i ¼ 1; 2; 3; where d 1 P 0; d 2 P 0; d 3 P 0; wi P 0 and ui are 0–1 variables:
ð34Þ ð35Þ ð36Þ ð37Þ ð38Þ ð39Þ ð40Þ
Example 1 is then converted into the following linear mixed 0–1 program: Minimize
z þ 2x1
Subject to
ð28Þ–ð31Þ; ð34Þ–ð40Þ:
Solve this program by LINDO [12], the solution is (z, y, y1, y2, y3, x1, x2, x3, d1, d2, d3, u1, u2, u3) = (7.54, 1.92, 0, .305, 1.47, 1, 1.5, 4.5, 0, .94, .83, 1, 1, 0). Since jy2 ln x2j = 0.1 > E3, more break points required to be chosen. For the next iteration we increase three new break points between b0 = 1 and b1 = 1.75. The new set of breaks points and associated slopes values for z becomes: (b0(1), b1(1), b11(1), b12(1), b13(1), b2(1), b3(1)) = (1, 1.75, 2.07, 2.36, 2.62, 2.86, 3.69),(s1(1), s11(1), s12(1), s13(1), s14(1), s3(1), s4(1)) = (2.72, 6.78, 9.2, 12.1, 15.52, 27.21, 56.72). Similarly, a new break point is added for each ln xi (i = 1, 2, 3). The new set of break points for ln xi becomes ai1 ð1Þ ¼
2:49 1 ¼ 1:63 ln 2:49 ln 1
for i ¼ 1; 2; 3:
The associated slope values becomes ln 1:63 ln 1 ¼ :78; 1:63 1 ln 5 ln 2:49 ¼ :28: si3 ð1Þ ¼ 5 2:49
si1 ð1Þ ¼
si2 ð1Þ ¼
ln 2:49 ln 1:63 ¼ :49; 2:49 1:63
Using these new break points to update z, y1, y2, and y3 as follows (Fig. 1): 6:78 2:72 9:2 6:78 ðjy 1:75j þ y 1:75Þ þ 2 2 12:1 9:2 ðjy 2:36j þ y 2:36Þ ðjy 2:07j þ y 2:07Þ þ 2 15:53 12:1 27:21 15:53 þ ðjy 2:62j þ y 2:62Þ þ 2 2 56:72 27:21 ðjy 3:69j þ y 3:69Þ; ðjy 2:86j þ y 2:86Þ þ 2
z ffi 1 þ 2:72y þ
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:49 :78 :28 :49 ðjxi 1:63j þ xi 1:63Þ þ 2 2 ðjxi 2:49j þ xi 2:49Þ; i ¼ 1; 2; 3:
y i ffi :78ðxi 1Þ þ
Solve the following linear mixed 0–1 program by LINDO [12]: Maximize
z þ 2x1
Subject to
z ffi 1 þ 2:72y þ 4:06ðy 1:75 þ d 1 Þ þ 2:42ðy 2:07 þ d 1 þ d 2 Þ þ 2:9ðy 2:36 þ d 1 þ d 2 þ d 3 Þ þ 3:43ðy 2:62 þ d 1 þ d 2 þ d 3 þ d 4 Þ þ 11:68ðy 2:86 þ d 1 þ d 2 þ d 3 þ d 4 þ d 5 Þ þ 29:51ðy 3:69 þ d 1 þ d 2 þ d 3 þ d 4 þ d 5 þ d 6 Þ; y þ d 1 þ d 2 þ d 3 þ d 4 þ d 5 þ d 6 ¼ 3:69; d 1 6 1:75;
d 2 6 0:32;
d 3 6 0:29;
d 4 6 0:26;
d 5 6 0:24;
d 6 6 0:83;
y ¼ y 1 þ :5y 2 þ 1:2y 3 ; y i ¼ :78ðxi 1Þ þ :29ðwi 1:63ui xi þ 1:63Þ þ :21ðw3þi 2:49u3þi xi þ 2:49Þ; wi P xi þ 999ðui 1Þ;
i ¼ 1; 2; 3;
i ¼ 1; 2; 3;
w3þi P xi þ 999ðu3þi 1Þ;
i ¼ 1; 2; 3;
ð28Þ–ð31Þ; d i P 0;
wi P 0;
w3þi P 0;
ui and u3þi are 0–1 variables:
The new solution is (z, y, y1, y2, y3, x1, x2, x3, d1, d2, d3, d4, d5, d6, u1, u2, u3, u4, u5, u6) = (7.25, 1.97, 0, .39, 1.475, 1, 1.5, 4.5, 0, .104, .29, .26, .24, .83, 1, 1:2 1, 1, 1, 0, 0). Since jx1 x:5 2 x3 zj 6 0:3 and jyi ln xij 6 0.03, we terminate the process. The optimal solution is ðx 1 ; x 2 ; x 3 Þ ¼ ð1; 1:5; 4:5Þ and the objective value is 9.45. Example 5 (Taken from [10]). In practical design optimization problems, integer and discrete variables occur quite frequently. Here we take a pressure vessel design optimization problem from Sandgren [10] depicted in Fig. 2 for instance to illustrate MISP problem with continuous and discrete variables. Sandgren solved this problem by quadratic integer programming method, Fu et al. [4] solved this problem by integer penalty method. Li and Chou [7] solved this problem by multi-level single linkage technique. Tsai et al. [13] solved this problem by convexification techniques. The problem is formulated below:
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
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y1 = lnx1
z 77
1.61
S4
S3 S2
S1 1
1.75 ↑
0
2.86
3.69
2.00
(a)
4.3
1.63
y=lnz
5
x1
y3 = lnx3
y2 = lnx2
1.61
1.61
1
(c)
2.09
(b) 1
1.63
2.49
5
1.5
x2
1
2.49
3.6
5
x3
4.5
(d)
Fig. 1. Break points of Example 4.
x1
x4
x3
x2
x3
Fig. 2. Tube and pressure vessel [10].
Minimize f ðxÞ ¼ 0:6224x1 x3 x4 þ 1:7781x2 x23 þ 3:1661x21 x4 þ 19:84x21 x3 Subject to x1 0:0193x3 P 0; x2 0:00954x3 P 0; 4 px23 x4 þ px33 P 750 1728; 3
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C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
1:000 6 x1 6 1:375 ðdiscrete variable with discreteness 0:0625Þ; 0:625 6 x2 6 1:000 ðdiscrete variable with discreteness 0:0625Þ; 45 6 x3 6 55 ðcontinuous variableÞ; 80 6 x4 6 110 ðcontinuous variableÞ;
where x1 is the spherical head thickness, x2 is the shell thickness. x3 is radius and x4 is the length of shell. First denote z1, z2, z3, and z4 as z1 = x1x3x4, z2 ¼ x2 x23 ; z3 ¼ and z4 ¼ x21 x3 , The choosing the break points of z1, z2, z3, and z4 are similar to Example 1. Since x1 and x2 are discrete variables, ln x1 and ln x2 can be completely expressed by binary variables without error, as shown below: x21 x4 ;
x1 ¼ 1 þ :0625u1 þ :125u2 þ :25u3 ; x2 ¼ :625 þ :0625u4 þ :125u5 þ :25u6 ; where ui ði ¼ 1; . . . ; 6Þ 2 f0; 1g: Solving this problem by LINDO [12], we obtain the optimal solution as x1 = 1, x2 = .625, x3 = 51.81347, x4 = 84.57862, and f(x) = 7006.367. The solution for pressure vessel problem is listed on Table 2. Table 2 indicates that the proposed method can find a solution better than what Sandgren [10], Li and Chou [7], and Tsai et al. [13] have found.
4. Conclusions The MISP problem is particularly complex and important both for theoretical studies and practical applications. However, no work has been done for solving this problem. The proposed method first approximately converts a signomial term into the sum of the absolute terms. Employing the linearization techniques then linearizes these absolute terms. A mixed integer signomial program is finally transformed into a linear mixed 0–1 program solvable for finding a solution close to the global optimum. In addition, the undefined problem
Table 2 The solutions for pressure vessel problem
x1 x2 x3 x4 f(x)
Sandgren [10]
Fu et al. [4]
Li and Chou [7]
Tsai et al. [13]
The proposed method
1.125 0.625 48.97 106.72 7982.5
1.125 0.625 48.38 111.745 8048.62
1 0.625 51.252 90.9913 7127.3
1 0.625 51 91 7079.037
1 0.625 51.81347 84.57862 7006.367
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of the logarithmic piecewise technique is solved, which leads to more usefulness of the proposed method.
References [1] M.S. Bazara, H.D. Shereli, C.M. Shetty, Nonlinear Programming, John Wiley & Sons, Inc, 1993 (Chapter 11). [2] C.S. Beightler, D.T. Phllips, Applied Geometric Programming, John Wiley, Inc, 1976. [3] C.S. Beightler, D.T. Phillips, D.J. Wilde, Foundation of Optimization, Prentice Hall, 1979. [4] J.F. Fu, R.G. Fenton, W.L. Cleghon, A mixed integer discrete continuous programming method and its application to engineering design optimization, Engineering Optimization 17 (1991) 263–280. [5] P. Hansen, B. Jaumard, S.H. Lu, A framework for algorithms in globally optimal design, ASME Journal of Mechanisms, Transmissions, and Automation in Design 111 (1989) 353–360. [6] R. Horst, H. Tuy, Global Optimization: Deterministic Approaches, Springer, Heidelberg, 1990. [7] H.-L. Li, C.T. Chou, A global approach for nonlinear mixed discrete programming in design optimization, Engineering Optimization 22 (1994) 109–122. [8] H.-L. Li, C.-T. Chang, An approximately global optimization method for assortment problems, European Journal of Operational Research 105 (1998) 604–612. [9] K. Rinnooy, G. Timmer, Towards global optimization methods (I and II), Mathematical Programming 39 (1987) 27–28. [10] E. Sandgren, Nonlinear integer and discrete programming in mechanical design optimization, Journal of Mechanical Design 112 (1990) 223–229. [11] H. Sherali, C. Tuncbilek, A global optimization algorithm for polynomial programming problems using a reformulation—linearization technique, Journal of Global Optimization 2 (1992) 101–112. [12] L. Schrage, Optimization Modeling with LINDO, Duxbury Press, 1997. [13] J.-F. Tsai, H.-L. Li, N.-Z. Hu, Global optimization for signomial discrete programming problems in engineering design, Engineering Optimization 34 (2002) 613–622. [14] C.-T. Chang, On the posinomial fractional programming problems, European Journal of Operational Research 143 (2002) 42–52. [15] P. Michelon, N. Maculan, Lagrangean decomposition for integer nonlinear programming with linear constraints, Mathematical Programming 52 (1991) 303–313. [16] C.-T. Chang, An efficient linearization approach for mixed integer problems, European Journal of Operational Research 123 (2000) 652–659.
www.elsevier.com/locate/amc
On the mixed integer signomial programming problems Ching-Ter Chang Department of Information Management, Changhua University of Education, Paisa Village, Changhua 50058, Taiwan, ROC
Abstract This paper proposes an approximate method to solve the mixed integer signomial programming problem, for which the objective function and the constraints may contain product terms with exponents and decision variables, which could be continuous or integral. A linear programming relaxation is derived for the problem based on piecewise linearization techniques, which first convert a signomial term into the sum of absolute terms; these absolute terms are then linearized by linearization strategies. In addition, a novel approach is included for solving integer and undefined problems in the logarithmic piecewise technique, which leads to more usefulness of the proposed method. The proposed method could reach a solution as close as possible to the global optimum. 2005 Elsevier Inc. All rights reserved. Keywords: Piecewise linear function; Signomial programming; Linearization technique
E-mail address: [email protected] 0096-3003/$ - see front matter 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.01.034
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1. Introduction This paper addresses the mixed integer signomial programming (MISP) problem, which seeks a global optimum to a mixed integer signomial objective function subject to a set of mixed integer signomial constraint functions, each decision variable could be continuous or integral. Such optimization problems occur quite frequently in engineering design, inventory control, scheduling, etc. [1,3,4,10]. Some approaches in the literature for solving the related problem called the polynomial programming (PP) problem are discussed as follows. Horst and Tuy [6], Hanson et al. [5] derived outer approximation techniques and interval analysis based on sufficient conditions for convergence and then proposed a variety of methods for solving PP problems. These methods, however, are only convergent in the absence of blocking sub-problems. In addition, these methods exploit the pure 0–1 structure of the problem, and are therefore not directly applicable for MISP problems. Sherali and Tuncbilek [11] derived a generic algorithm called Reformulation Linearization Technique (RLT) to solve PP problems, their algorithm generates non-linear implied constraints by taking the products of bounding terms in the constraints set to a suitable order. Defining new variables, one for each polynomial term appearing in the problem, subsequently linearizes the resulting problem. Incorporating appropriate bound factor products in their RLT scheme, and employing a suitable partitioning technique have developed a convergent branch and bound algorithm. Although the RLT algorithm is very promising in solving a PP problem, however, there are some difficulties for applying that algorithm to solve a MISP Problem, as listed below: (i) RLT algorithm cannot be directly applied to solve problems containing integer variables. (ii) There are various approaches in designing a suitable RLT process for solving a PP problem. There is no adequate understanding of how best to choose and implement an approach. (iii) RLT algorithm required generating several types of implied constraints at the expense of a huge increase in problem size, which might cause a heavy computation burden. Li and Chou [7] proposed a method for solving the PP problem, based on the multi-level single linkage technique of Rinnooy and Timmer [9], to obtain a global optimum at a pre-specified confidence level. The difficulty of their method is that in order to reach a global optimum at a sufficiently high confidence level, their method the solving of a huge amount of non-linear optimization problems based on various starting points. This always causes a heavy computational burden. Michelon and Maculan [15] proposed a lagrangean decomposition technique for solving SP problems. Li and Chang [8] derived a 0–1 linearization approach for solving PP problems. However, it can only solve a small size of PP problems. Suppose all variables in the problem are continuous, then a PP problem can be reformulated as a geometric programming problem [2,3]. For some special cases it is simpler computationally to reformulate a PP problem as a geometric program. However, current
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geometric programming methods only exploit the continuous structure of the problem, and are incapable of treating MISP problems. In addition, only if the degree of difficulty is less than two could a geometric programming method be applied to solve a PP problem. Tsai et al. [13] proposed some convexification techniques for solving discrete signomial engineering design problems. The mathematical formulation of MISP problem is given below. 1.1. MISP problem Minimize
ZðxÞ ¼
n X p¼1
Subject to
m X q¼1
hq
Y
cp
Y
xai
i2J
xbi 6 lk ;
k ¼ 1; 2; . . . ; K;
i2J
where 0 < xi 6 xi 6 xi (xi and xi are, respectively, the upper and lower bound of xai ); cp, hq, and lk are constants and unrestricted in sign; xi could be continuous or integral variables indexed by some set J; a and b could be an integer or noninteger exponent. This paper proposes a novel method for solving the MISP problem. What follows are the advantages of the proposed approach: (i) The proposed method can treat a MISP problem where a and b are allowed to be non-integer, and xi could be continuous or integral variables. In addition, there is no restriction on the degree of difficulty. (ii) The proposed method can handle a MISP problem without solving a huge amount of non-linear programs. (iii) The proposed method could reach a solution as close as possible to the global optimum, in which the error can be pre-specified by a decision maker. This study first converted the MISP problem into a piecewise linear function (PLF) problem. The non-linear integer formulation strategy is then derived for solving the problem containing the integer variable. Following that, some numerical examples of generalized SP and engineering design problems are solved to demonstrate the usefulness of the proposed method.
2. Formulation of MISP problem 2.1. Formulation of the non-linear real term in MISP problem By refereeing to Chang [14], any signomial term, xai , in MISP can be represented by the following PLF.
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
f ðxÞ ¼ a0 þ s1 ðx a1 Þ þ
n1 X si si1 ðjx ai j þ x ai Þ; 2 i¼2
1439
ð1Þ
where ai, i = 1, 2, . . . , n are break points of f(x) and si, i = 1, 2, . . . , n are slopes of line segments between ai and ai+1. For more concise expression of (1), let us consider a simple SP problem below. Example 1 Minimize
1:3 z ¼ x1 x1:7 2 x3
Subject to X 2 F
ðF is a feasible setÞ:
We choose break points at a1 = 1, a2 = 2, a3 = 3, and a4 = 4. By referring to (1), Example 1 can be approximated as follows: Minimize z Subject to ln z ¼ ln x1 þ 1:7 ln x2 þ 1:3 ln x3 ; ln x1 ¼ 0:69314ðx1 1Þ 0:14383ðjx1 2j þ x1 2Þ 0:0589ðjx1 3j þ x1 3Þ; ln x2 ¼ 0:69314ðx2 1Þ 0:14383ðjx2 2j þ x2 2Þ 0:0589ðjx2 3j þ x2 3Þ; ln x3 ¼ 0:69314ðx3 1Þ 0:14383ðjx3 2j þ x3 2Þ 0:0589ðjx3 3j þ x3 3Þ; z ¼ eln z ; X 2 F ðF is a feasible setÞ: If x1 = 2, x2 = 3, and x4 = 4 in this instance, the tolerance error approximated below ln x1 ¼ 0:69314ð2 1Þ ¼ 0:69314; ln x2 ¼ 0:69314ð3 1Þ 0:14383ðj3 2j þ 3 2Þ ¼ 1:09862; ln x3 ¼ 0:69314ð4 1Þ 0:14383ðj4 2j þ 4 2Þ 0:0589ðj4 3j þ 4 3Þ ¼ 1:3863; ln z ¼ ln x1 þ 1:7 ln x2 þ 1:3 ln x3 ¼ 4:362984; z ¼ e4:362984 ¼ 78:49100247; error ¼ 0:00104073: 1:3 Now, we really transform a SP term x1 x1:7 2 x3 into an approximate PLF form successful. The error of this approximation is 0.00104073.
By referring to [14], all absolute terms in (1) can be linearized by the following programs P1–P3.
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Program P1 (for linearizing an absolute term with non-positive coefficient in the objective function/constraints). Function Subject to
z ¼ 2g þ 2uai Mðu 1Þ < x ai 6 Mu; x þ ðu 1ÞM 6 g 6 x þ ð1 uÞM; g 6 Mu; g P 0;
ð2Þ ð3Þ ð4Þ ð5Þ
where u is a 0–1 variable; g is a continuous variable. Program P2 (for linearizing an absolute term with positive coefficient in the objective function/constraints). Function
z ¼ 2g 2uai
Subject to
ð2Þ–ð5Þ;
where all variables are as defined in P1. Program P3 (for linearizing an absolute term with positive coefficient in the objective function). ! j m X X Minimize z¼2 x aj þ dk j¼1
Subject to
xþ
m X
k¼1
d k P am ;
k¼1
0 6 d 1 6 a1 ; 0 6 d k 6 ak ak1
for k ¼ 2; 3; . . . ; m;
x 2 F ðF is a feasible setÞ: Notably, the accuracy of the PLF heavily depends upon the number of break points for each variable. However, when the number of break points is increased, the number of 0–1 variables in the PLF is also increased. This will cause a heavy computation burden in the solution process of the problem. In order to solve this problem, a way of choosing the optimal break points has been proposed by Chang [14]. 2.2. Formulation of the non-linear integer term in MISP problem a a a For an be expressed as P integer variable xi ; 0 < xi 6 xi 6 xi ; xi can P xai ¼ xi þ Kk¼0 2k uk , where K is a minimal integer satisfying Kk¼0 2k P xi xi , and uk is a 0–1 variable, then ln xai can be expressed as
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
ln xai ¼ ln xi þ
1441
K K X Ki X X ½lnð2k þ xi Þ ln xi uk þ ½lnð2k þ 2i Þ þ ln xi k¼0
i¼0
k¼0
lnð2k þ xi Þ lnð2i þ xi Þ uik : For instance, consider an integer x, 2 6 x 6 9, denoting x as x = 2 + u0 + 2u1 + 4u2, ln x can then be expressed as ln x ¼ ln 2 þ ðln 3 ln 2Þu0 þ ðln 4 ln 2Þu1 þ ðln 6 ln 2Þu2 þ ðln 5 þ ln 2 ln 3 ln 4Þu01 þ ðln 7 þ ln 2 ln 3 ln 6Þu02 þ ðln 8 þ ln 2 ln 4 ln 6Þu12 ; where u0, u1, u2 2 {0, 1}, and u01, u02, and u12 are additional continuous variables satisfied the following inequalities: ðu0 þ u1 2Þ þ 1 6 u01 6 ð2 u0 u1 Þ þ 1; u0 6 u01 6 u0 ; u1 6 u01 6 u1 ; ðu0 þ u2 2Þ þ 1 6 u02 6 ð2 u0 u2 Þ þ 1; u0 6 u02 6 u0 ; u2 6 u02 6 u2 ; ðu1 þ u2 2Þ þ 1 6 u12 6 ð2 u1 u2 Þ þ 1; u1 6 u12 6 u1 ; u2 6 u12 6 u2 : If x = 5 in this instance, this leads to u0 = u1 = 1, u2 = 0, u01 = 1, and then ln 5 ¼ ln 2 þ ðln 3 ln 2Þ þ ðln 4 ln 2Þ þ ðln 5 þ ln 2 ln 3 ln 4Þ ¼ 1:609437912: Based on the above strategy, the MISP with non-integer exponent variable can be easily solved. 2.3. The undefined problem of the logarithmic piecewise technique Assume that x1, x2, and x3 are a positive value in Example 1; otherwise (value zero or negative), the logarithmic LPF is undefined. Chang [14] proposed a bounded method to solve the problem of a logarithm of zero. However, the problem of the logarithm being negative can still not be solved by the current published methods. It can be observed that the problem can be a solved if the problem with a negative variable can be converted into the equivalent problem with a positive variable. For more simplicity, but without loss of generality, the following example is given an idea on how to solve the logarithm being negative problem in the LPF. Example 2 Minimize Subject to
x1 x2 x3 X 2 F ðF is a feasible setÞ; 4 6 x1 6 4;
3 6 x2 6 3;
3 6 x3 6 3:
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Example 2 cannot be solved using logarithmic PLF because the problem contains three negative undefined variables as x1 2 [4, 4], x2 2 [3, 3], and x3 2 [3, 3], In order to overcome the undefined problem, the following equivalent program with positive variables is then introduced. Program P4 Minimize
ð Þz
Subject to
y 1 ¼ jx1 j;
y 2 ¼ jx2 j;
ð6Þ
y 3 ¼ jx3 j;
ln u ¼ ln y 1 þ ln y 2 þ ln y 3 ;
ð7Þ
ln y 1 ¼ 0:69314ðy 1 1Þ 0:14383ðjy 1 2j þ y 1 2Þ 0:0589ðjy 1 3j þ y 1 3Þ;
ð8Þ
ln y 2 ¼ 0:69314ðy 2 1Þ 0:14383ðjy 2 2j þ y 2 2Þ;
ð9Þ
ln y 3 ¼ 0:69314ðy 3 1Þ 0:14383ðjy 3 2j þ y 3 2Þ;
ð10Þ
z ¼ eln u ;
ð11Þ
X 2 F ðF is a feasible setÞ;
ð12Þ
0 6 y 1 6 4;
0 6 y 2 6 3;
4 6 x1 6 4;
0 6 y 3 6 3;
3 6 x2 6 3;
3 6 x3 6 3;
ð13Þ ð14Þ
where (•) is a sign determinator. Assume negative variable, xi, as ÔoddÕ numbers and positive variable, xi, as ÔevenÕ numbers. The rules of multiplying follows the same kind of pattern: 2 negatives gives you a positive product, 2 positives gives you a positive product, and one of each gives you a negative product and so on. Thus, the behavior of the sign determinator (•) can be expressed as follows: þ; number of negative variables; xi ; are even numbers; ð Þ ¼ ; number of negative variables; xi ; are odd numbers: If x1 = 2, x2 = 3, and x3 = 2 in this instance, the problem can be approximated below. y 1 ¼ 2;
y 2 ¼ 3;
y 3 ¼ 2;
ln y 1 ¼ 0:69314ð2 1Þ ¼ 0:69314; ln y 2 ¼ 0:69314ð3 1Þ 0:14383ðj3 2j þ 3 2Þ ¼ 1:09862; ln y 3 ¼ 0:69314ð2 1Þ ¼ 0:69314; ln u ¼ 0:69314 þ 1:09862 þ 0:69314 ¼ 2:4849; z ¼ e2:4849 ¼ 11:99992;
error ¼ 0:00008:
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(•) is a negative in sign (2 positives (i.e., x2, x3) and 1 negative (i.e., x1) give you a negative product). Clearly, the objective value is approximated 11.99992. From P4 we can see that there have two major elements in solving the negative undefined problem: (1) yi = jxij; (2) a sign determinator (•). Use the following auxiliary constraints to reach yi = jxij. y i ¼ ð2si 1Þxi ;
ð15Þ
ðsi 1Þxi P 0; si xi P 0;
ð16Þ ð17Þ
where si is a 0–1 variable. Proposition 1. (15)–(17) and yi = jxij are equivalent in the sense that they have the same optimal solutions. Proof. If xi > 0 then si = 1 (from (16)). This forces yi = xi (from (15)). If xi < 0 then si = 0 (from (17)). This forces yi = xi (from (15)). It can be observed that (15)–(17) are equivalent to yi = jxij, This completes the proof of this proposition. h In order to determine the behavior of sign determinator (•), the logical relationships of P5 are listed in Table 1. Program P5 Minimize
ð Þz
Subject to
ð15Þ–ð17Þ;
for i ¼ 1; 2; 3;
ð18Þ
ð7Þ–ð14Þ; where s1, s2, and s3 are 0–1 variables.
Table 1 Logical relationships of sign determinator (•) in P5 s1
s2
s3
(•) (sign determinator)
0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
Negative (i.e., x1 < 0, x2 < 0, and x3 < 0) Positive (i.e., x1 < 0, x2 < 0, and x3 > 0) Positive (i.e., x1 < 0, x2 > 0, and x3 < 0) Negative (i.e., x1 < 0, x2 > 0, and x3 > 0) Positive (i.e., x1 > 0, x2 < 0, and x3 < 0) Negative (i.e., x1 > 0, x2 < 0, and x3 > 0) Negative (i.e., x1 > 0, x2 > 0, and x3 < 0) Positive (i.e., x1 > 0, x2 > 0, and x3 > 0)
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From Table 1 we can see that all cases of negative in sign of (•) can be expressed as: (1 s1)(1 s2)(1 s3) or (1 s1)s2s3 or s1(1 s2)s3 or s1s2(1 s3). Substituting (•) in the objective function of P5 by these logic relationships to obtain the following equivalent objective function. 2½ð1 s1 Þð1 s2 Þð1 s3 Þ ð1 s1 Þs2 s3 s1 ð1 s2 Þs3 s1 s2 ð1 s3 Þ z þ z;
ð19Þ
By referring to the approach of Chang [16], all cubic mixed integer terms s1s2s3 in (19) can be easily linearized as follows: s1 s2 s3 ¼ w;
ð20Þ
where s1, s2, s3, and w satisfy the following inequalities: ðs1 þ s2 þ s3 3Þ þ 1 6 w 6 ð3 s1 s2 s3 Þ þ 1;
ð21Þ
s1 6 w 6 s1 ;
ð22Þ
s2 6 w 6 s2 ;
ð23Þ
s3 6 w 6 s3 ;
ð24Þ
where w is a continuous variable. The above inequalities can be checked as follows: (i) if
Q3
(ii) if
Q3
i¼1 si
¼ 1 then this forces w to be one (from (21)).
i¼1 si
¼ 0 then this forces w to be zero (from (22)–(24)).
The same idea can be applied to quadratic mixed integer terms si sj in (19). Without loss of generality, (19) can be checked by the following cases. (i) If x1 = 2, x2 = 3, and x3 = 1 in P5 then s1 = 0, s2 = 1, and s3 = 0 (from (18)). This forces (19) to be z (i.e., 2[0]z + z). (ii) If x1 = 2, x2 = 3, and x3 = 2 in P4 then s1 = 0, s2 = 1, and s3 = 1 (from (18)). This forces (19) to be z (i.e., 2[1]z + z). Consider the following problem with negative variables. Example 3 Minimize
z ¼ x1 x2 x3 þ x1 x2
Subject to
2x1 þ x2 þ x3 P 0;
ð25Þ
x1 þ 2x2 þ x3 6 5;
ð26Þ
2 6 x1 6 2;
3 6 x2 6 3;
2 6 x3 6 3:
ð27Þ
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
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This problem contains variables with negative range, which cannot be solved by logarithmic PLF approach. In contrast, this problem can be converted into the following equivalent problem using the above strategies. Program P6 Minimize z ¼ 2½ð1 s1 Þð1 s2 Þð1 s3 Þ ð1 s1 Þs2 s3 s1 ð1 s2 Þs3 s1 s2 ð1 s3 Þ u1 þ u1 þ 2½ð1 s1 Þs2 s1 ð1 s2 Þ u2 þ u2 Subject to ð15Þ–ð17Þ for i ¼ 1;2; 3 Use to replace y 1 ¼ jx1 j; y 2 ¼ jx2 j; y 3 ¼ jx3 j 3 ln u1 ¼ ln y 1 þ ln y 2 þ ln y 3 ; 7 7 ln u2 ¼ ln y 1 þ ln y 2 ; 7 7 7 ln y 1 ¼ :69314ðy 1 1Þ; 7 7 7 Use to approximate ln y 2 ¼ :69314ðy 2 1Þ 7 7 :14383ðjy 2 2j þ y 2 2Þ; 7 7 z ¼ x1 x2 x3 þ x1 x2 ; 7 7 ln y 3 ¼ :69314ðy 3 1Þ 7 7 :14383ðjy 3 2j þ y 3 2Þ; 7 5 u1 þ u2 ¼ elnu1 þ elnu2 ; 0 6 y 1 6 2; 0 6 y 2 6 3; 0 6 y 3 6 3 Corresponding toð27Þ; ð25Þ–ð27Þ:
All absolute terms and mixed integer terms in P6 can be easily linearized by P1 and P2. Solve this problem by LINDO [12] to obtain the approximate global optimal solutions as (x1, x2, x3, y1, y2, y3, ln y1, ln y2, ln y3) = (2, 3, 3, 2, 3, 3, .69324, 1.09861, 1.09861), objective value = 23.99987, error = 0.00013.
3. Numerical examples Example 4. Solve the following problem with a signomial objective function and linear constraints. The tolerable errors are specified as E1 6 0.3 and E3 6 0.03. Minimize Subject to
1:2 x1 x0:5 2 x3 þ 2x1 2x1 þ x2 þ x3 P 8;
ð28Þ
x2 þ 2x3 P 10:5; x1 þ 2x3 6 10;
ð29Þ ð30Þ
1 6 x1 6 5;
1 6 x2 6 5;
1 6 x3 6 5:
ð31Þ
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It is noticeable that because the objective function contains a signomial term which has non-integer exponents, this problem can not be solved by RLT method (Sherali and Tuncbilek [11]). Since all coefficients in the objective function are positive and all constraints are linear, this example can be converted 1:2 into PLF problem. Denote z ¼ x1 x0:5 2 x3 and approximate z by (1) where three best initial break points are chosen as follows, by referring to Chang [14] 77:13 1 e2:86 1 ¼ 2:86; b1 ¼ ln ¼ 1:75; ln 77:13 ln 1 2:86 ln 1 77:13 e2:86 b3 ¼ ln ¼ 3:69; ln 77:13 2:86
b2 ¼ ln
where 77.13 and 1 is the upper and lower bound of z, respectively. The associated slope values are computed as s1 ¼
e1:75 e0 ¼ 2:72; 1:75 0
s3 ¼
e3:69 e2:86 ¼ 27:21; 3:69 2:86
s2 ¼
e2:86 e1:75 ¼ 10:55; 2:86 1:75
s4 ¼
e4:35 e3:69 ¼ 56:72: 4:35 3:69
z is then expressed approximately as 10:55 2:72 ðjy 1:75j þ y 1:75Þ 2 27:21 10:55 56:72 27:21 þ ðjy 2:86j þ y 2:86Þ þ 2 2 ðjy 3:69j þ y 3:69Þ;
z ffi 1 þ 2:72y þ
ð32Þ
where y = y1 + 1.2y2 + 1.2y3 = ln x1 + 1.2 ln x2 + 1.2 ln x3. Since ln xi is a concave function of xi, only one break point is chosen for linearizing ln xi in order to save the number of new added 0–1 variables. This break point is chosen following, by referring to Chang [14] ai1 ð0Þ ¼
51 ¼ 2:49; ln 5 ln 1
for i ¼ 1; 2; 3:
The associated slope values becomes si1 ¼
ln 2:49 ln 1 ¼ :61; 2:49 1
si2 ¼
ln 5 ln 2:49 ¼ :28: 5 2:49
Denote yi as yi = ln xi, which is approximately expressed as y i ffi :61ðxi 1Þ þ
:28 :61 ðjxi 2:49j þ xi 2:49Þ; 2
i ¼ 1; 2; 3:
ð33Þ
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
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Expression (32) and (33) can then be linearized by P1 through P3 as follows: z ffi 1 þ 2:72y þ ð10:55 2:72Þðy 1:75 þ d 1 Þ þ ð27:21 10:55Þ ðy 2:86 þ d 1 þ d 2 Þ þ ð56:72 27:21Þ ðy 3:69 þ d 1 þ d 2 þ d 3 Þ; y þ d 1 þ d 2 þ d 3 P 3:69; d 1 6 1:75; d 2 6 1:11; d 3 6 :83; y ¼ y 1 þ :5y 2 þ 1:2y 3 ; y i ¼ :61ðxi 1Þ þ ð:61 :28Þðwi 2:49ui xi þ 2:49Þ; i ¼ 1; 2; 3; wi P xi þ 999ðui 1Þ; i ¼ 1; 2; 3; where d 1 P 0; d 2 P 0; d 3 P 0; wi P 0 and ui are 0–1 variables:
ð34Þ ð35Þ ð36Þ ð37Þ ð38Þ ð39Þ ð40Þ
Example 1 is then converted into the following linear mixed 0–1 program: Minimize
z þ 2x1
Subject to
ð28Þ–ð31Þ; ð34Þ–ð40Þ:
Solve this program by LINDO [12], the solution is (z, y, y1, y2, y3, x1, x2, x3, d1, d2, d3, u1, u2, u3) = (7.54, 1.92, 0, .305, 1.47, 1, 1.5, 4.5, 0, .94, .83, 1, 1, 0). Since jy2 ln x2j = 0.1 > E3, more break points required to be chosen. For the next iteration we increase three new break points between b0 = 1 and b1 = 1.75. The new set of breaks points and associated slopes values for z becomes: (b0(1), b1(1), b11(1), b12(1), b13(1), b2(1), b3(1)) = (1, 1.75, 2.07, 2.36, 2.62, 2.86, 3.69),(s1(1), s11(1), s12(1), s13(1), s14(1), s3(1), s4(1)) = (2.72, 6.78, 9.2, 12.1, 15.52, 27.21, 56.72). Similarly, a new break point is added for each ln xi (i = 1, 2, 3). The new set of break points for ln xi becomes ai1 ð1Þ ¼
2:49 1 ¼ 1:63 ln 2:49 ln 1
for i ¼ 1; 2; 3:
The associated slope values becomes ln 1:63 ln 1 ¼ :78; 1:63 1 ln 5 ln 2:49 ¼ :28: si3 ð1Þ ¼ 5 2:49
si1 ð1Þ ¼
si2 ð1Þ ¼
ln 2:49 ln 1:63 ¼ :49; 2:49 1:63
Using these new break points to update z, y1, y2, and y3 as follows (Fig. 1): 6:78 2:72 9:2 6:78 ðjy 1:75j þ y 1:75Þ þ 2 2 12:1 9:2 ðjy 2:36j þ y 2:36Þ ðjy 2:07j þ y 2:07Þ þ 2 15:53 12:1 27:21 15:53 þ ðjy 2:62j þ y 2:62Þ þ 2 2 56:72 27:21 ðjy 3:69j þ y 3:69Þ; ðjy 2:86j þ y 2:86Þ þ 2
z ffi 1 þ 2:72y þ
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:49 :78 :28 :49 ðjxi 1:63j þ xi 1:63Þ þ 2 2 ðjxi 2:49j þ xi 2:49Þ; i ¼ 1; 2; 3:
y i ffi :78ðxi 1Þ þ
Solve the following linear mixed 0–1 program by LINDO [12]: Maximize
z þ 2x1
Subject to
z ffi 1 þ 2:72y þ 4:06ðy 1:75 þ d 1 Þ þ 2:42ðy 2:07 þ d 1 þ d 2 Þ þ 2:9ðy 2:36 þ d 1 þ d 2 þ d 3 Þ þ 3:43ðy 2:62 þ d 1 þ d 2 þ d 3 þ d 4 Þ þ 11:68ðy 2:86 þ d 1 þ d 2 þ d 3 þ d 4 þ d 5 Þ þ 29:51ðy 3:69 þ d 1 þ d 2 þ d 3 þ d 4 þ d 5 þ d 6 Þ; y þ d 1 þ d 2 þ d 3 þ d 4 þ d 5 þ d 6 ¼ 3:69; d 1 6 1:75;
d 2 6 0:32;
d 3 6 0:29;
d 4 6 0:26;
d 5 6 0:24;
d 6 6 0:83;
y ¼ y 1 þ :5y 2 þ 1:2y 3 ; y i ¼ :78ðxi 1Þ þ :29ðwi 1:63ui xi þ 1:63Þ þ :21ðw3þi 2:49u3þi xi þ 2:49Þ; wi P xi þ 999ðui 1Þ;
i ¼ 1; 2; 3;
i ¼ 1; 2; 3;
w3þi P xi þ 999ðu3þi 1Þ;
i ¼ 1; 2; 3;
ð28Þ–ð31Þ; d i P 0;
wi P 0;
w3þi P 0;
ui and u3þi are 0–1 variables:
The new solution is (z, y, y1, y2, y3, x1, x2, x3, d1, d2, d3, d4, d5, d6, u1, u2, u3, u4, u5, u6) = (7.25, 1.97, 0, .39, 1.475, 1, 1.5, 4.5, 0, .104, .29, .26, .24, .83, 1, 1:2 1, 1, 1, 0, 0). Since jx1 x:5 2 x3 zj 6 0:3 and jyi ln xij 6 0.03, we terminate the process. The optimal solution is ðx 1 ; x 2 ; x 3 Þ ¼ ð1; 1:5; 4:5Þ and the objective value is 9.45. Example 5 (Taken from [10]). In practical design optimization problems, integer and discrete variables occur quite frequently. Here we take a pressure vessel design optimization problem from Sandgren [10] depicted in Fig. 2 for instance to illustrate MISP problem with continuous and discrete variables. Sandgren solved this problem by quadratic integer programming method, Fu et al. [4] solved this problem by integer penalty method. Li and Chou [7] solved this problem by multi-level single linkage technique. Tsai et al. [13] solved this problem by convexification techniques. The problem is formulated below:
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
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y1 = lnx1
z 77
1.61
S4
S3 S2
S1 1
1.75 ↑
0
2.86
3.69
2.00
(a)
4.3
1.63
y=lnz
5
x1
y3 = lnx3
y2 = lnx2
1.61
1.61
1
(c)
2.09
(b) 1
1.63
2.49
5
1.5
x2
1
2.49
3.6
5
x3
4.5
(d)
Fig. 1. Break points of Example 4.
x1
x4
x3
x2
x3
Fig. 2. Tube and pressure vessel [10].
Minimize f ðxÞ ¼ 0:6224x1 x3 x4 þ 1:7781x2 x23 þ 3:1661x21 x4 þ 19:84x21 x3 Subject to x1 0:0193x3 P 0; x2 0:00954x3 P 0; 4 px23 x4 þ px33 P 750 1728; 3
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C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
1:000 6 x1 6 1:375 ðdiscrete variable with discreteness 0:0625Þ; 0:625 6 x2 6 1:000 ðdiscrete variable with discreteness 0:0625Þ; 45 6 x3 6 55 ðcontinuous variableÞ; 80 6 x4 6 110 ðcontinuous variableÞ;
where x1 is the spherical head thickness, x2 is the shell thickness. x3 is radius and x4 is the length of shell. First denote z1, z2, z3, and z4 as z1 = x1x3x4, z2 ¼ x2 x23 ; z3 ¼ and z4 ¼ x21 x3 , The choosing the break points of z1, z2, z3, and z4 are similar to Example 1. Since x1 and x2 are discrete variables, ln x1 and ln x2 can be completely expressed by binary variables without error, as shown below: x21 x4 ;
x1 ¼ 1 þ :0625u1 þ :125u2 þ :25u3 ; x2 ¼ :625 þ :0625u4 þ :125u5 þ :25u6 ; where ui ði ¼ 1; . . . ; 6Þ 2 f0; 1g: Solving this problem by LINDO [12], we obtain the optimal solution as x1 = 1, x2 = .625, x3 = 51.81347, x4 = 84.57862, and f(x) = 7006.367. The solution for pressure vessel problem is listed on Table 2. Table 2 indicates that the proposed method can find a solution better than what Sandgren [10], Li and Chou [7], and Tsai et al. [13] have found.
4. Conclusions The MISP problem is particularly complex and important both for theoretical studies and practical applications. However, no work has been done for solving this problem. The proposed method first approximately converts a signomial term into the sum of the absolute terms. Employing the linearization techniques then linearizes these absolute terms. A mixed integer signomial program is finally transformed into a linear mixed 0–1 program solvable for finding a solution close to the global optimum. In addition, the undefined problem
Table 2 The solutions for pressure vessel problem
x1 x2 x3 x4 f(x)
Sandgren [10]
Fu et al. [4]
Li and Chou [7]
Tsai et al. [13]
The proposed method
1.125 0.625 48.97 106.72 7982.5
1.125 0.625 48.38 111.745 8048.62
1 0.625 51.252 90.9913 7127.3
1 0.625 51 91 7079.037
1 0.625 51.81347 84.57862 7006.367
C.-T. Chang / Appl. Math. Comput. 170 (2005) 1436–1451
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of the logarithmic piecewise technique is solved, which leads to more usefulness of the proposed method.
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