Tropical Optimization Problems in Project Scheduling

arXiv:1502.06222v1 [math.OC] 22 Feb 2015

Tropical Optimization Problems in Project Scheduling N. Krivulin∗

Abstract We consider a project that consists of activities operating in parallel under various temporal constraints, including start-start, startfinish and finish-start precedence relations, early start, late start and late finish time boundaries, and due dates. Scheduling problems are formulated to find optimal schedules for the project under different optimality criteria to minimize, including the project makespan, the maximum deviation from the due dates, the maximum flow-time, and the maximum deviation of finish times. We represent the problems as optimization problems in terms of tropical mathematics, and then solve these problems by applying direct solution methods of tropical optimization. As a result, new direct solutions of the problems are obtained in a compact vector form, which is ready for further analysis and practical implementation. Key-Words: tropical mathematics, idempotent semifield, optimization problem, project scheduling, precedence relationship, due date, makespan, flow-time. MSC (2010): 65K10, 15A80, 65K05, 90C48, 90B35

1

Introduction

Tropical optimization problems, which are formulated and solved in the framework of tropical mathematics, find increasing application in various fields of operations research, including project scheduling. As an applied mathematical discipline concentrated on the theory and applications of idempotent semirings, tropical (idempotent) mathematics dates back to a few seminal papers by Pandit [1], Cuninghame-Green [2], Giffler [3], Hoffman [4], Vorob’ev [5], and Romanovski˘ı [6], including the papers [2] and [3] concerned with optimization problems drawn from machine scheduling. ∗ Faculty of Mathematics and Mechanics, Saint Petersburg State University, 28 Universitetsky Ave., St. Petersburg, 198504, Russia, [email protected].

1

In succeeding years, tropical optimization problems were investigated in many publications, such as the monographs by Cuninghame-Green [7], Zimmermann [8] and Butkoviˇc [9], where scheduling problems frequently serve to motivate and illustrate the study. The optimization problems are usually formulated in the tropical mathematics setting to minimize a linear or nonlinear function defined on vectors over an idempotent semifield, and calculated using conjugate vector transposition. For some problems, complete direct solutions are obtained in a closed form under fairly general assumptions. Other problems are solved algorithmically by means of iterative computational procedures. In this paper, we examine several problems drawn from project scheduling [10, 11, 12] to provide new solutions to the problems on the basis of recent results in tropical optimization. We consider scheduling problems, which are to find an optimal schedule for a project that involves a set of activities operating in parallel under various temporal constraints, including start-start, start-finish, finish-start, early start, late start, late finish, and due dates constraints. As optimality criteria to minimize, we take the project makespan, the maximum deviation from due dates, the maximum flow-time, and the maximum deviation of finish times. These problems are known to have algorithmic solutions in the form of iterative computational procedures, and can also be solved as linear programming problems by an appropriate linear programming algorithm. We represent these scheduling problems as tropical optimization problems, which are then solved by applying direct solution methods developed in [13, 14, 15, 16, 17]. As a result, we offer new direct solutions to the scheduling problems under consideration, which, in contrast to the conventional algorithmic solutions, provide results in a compact explicit vector form, which is immediately ready for further analysis and practical implementation. The rest of the paper is organized as follows. In Section 2, we present scheduling problems that motivate and illustrate the study, and formulate these problems by using the standard notation. Section 3 includes a brief overview of preliminary definitions and results of tropical mathematics to be used in the subsequent sections. In Section 4, we describe some tropical optimization problems together with their solutions. In Section 5, we first rewrite the scheduling problems as tropical optimization problems, and then solve them by applying the results from Section 4.

2

Project scheduling model and example problems

We start with the description of a project scheduling model and the formulation of example problems of optimal scheduling in a general form, which strives to follow the usage of the common terminology in the area (see, eg, [10, 11, 12]).

2

Consider a project that consists of n activities operating in parallel under start-start, start-finish and finish-start precedence relations, due dates, and time boundaries in the form of early start, late start and late finish constraints. To describe the temporal constraints and scheduling objectives, we use, for each activity i = 1, . . . , n, the notation xi and yi to represent, respectively, the start time and the finish time of the activity.

2.1

Temporal constraints

We now examine constraints on the start and finish times of each activity i = 1, . . . , n. First, we represent precedence relations, which link activity i with other activities. Let aij be the lower limit on the time lag between the start of activity j and the finish of i. If there is no limit defined, we assume aij = −∞. The start-finish constraints take the form of the inequalities aij + xj ≤ yi for all j = 1, . . . , n. The activity is assumed to finish immediately after all start-finish constraints are satisfied, and thus at least one of the inequalities holds as an equality. Then, these inequalities are equivalent to one equality max(ai1 + x1 , . . . , xin + xn ) = yi . Furthermore, we denote by bij the lower bound on the time interval between the start of activity j and the start of i, and put bij = −∞ if the bound is not indicated. The start-start constraints are given by the inequalities bij + xj ≤ xi for all j , which can be readily rewritten as one inequality max(bi1 + x1 , . . . , bin + xn ) ≤ xi . Let the minimum allowed time between the finish of activity j and the start of i be denoted by cij , with cij = −∞ if undefined. The finish-start constraints are written as the inequalities cij + yj ≤ xi for all j , or as one inequality max(ci1 + y1 , . . . , cin + yn ) ≤ xi . Finally, we introduce due dates and time boundaries constraints. The due date indicates the time when the activity is expected to finish, and therefore, is not actually a strict constraint. For activity i, we denote the due date by di . Let gi and hi be the earliest and latest possible times to start, and fi be the latest possible time to finish. The early start, late start, and late finish constraints provide strict boundaries for the start and finish times, given by gi ≤ xi ≤ hi ,

3

yi ≤ fi .

2.2

Optimality criteria

To describe different scheduling objectives, we use several criteria that commonly arise in the development of optimal schedules in real-world problems. The criteria are written below in the form ready for further translation into terms of tropical mathematics. We begin with the makespan, which is the interval between the earliest start time and the latest finish time of activities in a project. The makespan describes the total duration of the project and finds wide application as a natural objective function to be minimized in scheduling problems. With the above introduced notation, the makespan is given by max yi − min xi = max yi + max (−xi ).

1≤i≤n

1≤i≤n

1≤i≤n

1≤i≤n

Another important criterion takes the form of the maximum absolute deviation of finish times of activities from given due dates for a project. The minimum value of this criterion corresponds to the minimal violation of due dates, which can be accomplished in the project. The maximum deviation from due dates is written as max |yi − di | = max max(yi − di , di − yi ).

1≤i≤n

1≤i≤n

The flow time of an activity is defined as the difference between its start and finish times, and also referred to as the system, processing, throughput and turn-around time. In many real-world problems, the objective is formulated to minimize the maximum flow time taken over all activities. The maximum flow time is readily described by the expression max (yi − xi ).

1≤i≤n

Finally, we consider the maximum deviation of completion times of all activities. The minimization of this criterion is equivalent to finding a schedule, where all activities have to finish simultaneously as much as possible, and arises in scheduling problems in just-in-time manufacturing. The maximum deviation of completion times is given by max yi − min yi = max yi + max (−yi ).

1≤i≤n

2.3

1≤i≤n

1≤i≤n

1≤i≤n

Examples of problems

We conclude with typical examples of scheduling problems, which are to serve to both motivate and illustrate the results in the rest of the paper. To formulate the problems, we use the notation and formulae introduced above to represent the unknown variables and given parameters, as well as to write the constraints and objectives for scheduling. 4

2.3.1

Minimization of maximum flow time

First, we consider a problem of minimization of maximum flow time under start-finish, start-start, finish-start and early start constraints. Given aij , bij , cij and gi for all i, j = 1, . . . , n, the problem is to find the start and finish times xi and yi for each activity i = 1, . . . , n, that minimize subject to

max (yi − xi ),

1≤i≤n

max (aij + xj ) = yi ,

1≤j≤n

max (cij + yj ) ≤ xi ,

1≤j≤n

2.3.2

max (bij + xj ) ≤ xi ,

1≤j≤n

gi ≤ xi ,

(1)

i = 1, . . . , n.

Minimization of maximum deviation from due dates

We now formulate a problem of minimization of maximum deviation from due dates under start-finish, start-start, finish-start and due dates constraints. Given parameters aij , bij , cij and di , the problem is to find the unknowns xi and yi , that minimize subject to

max max(yi − di , di − yi ),

1≤i≤n

max (aij + xj ) = yi ,

1≤j≤n

max (bij + xj ) ≤ xi ,

1≤j≤n

max (cij + yj ) ≤ xi ,

i = 1, . . . , n.

1≤j≤n

2.3.3

(2)

Minimization of makespan

Suppose that we need to minimize the makespan under start-finish, early start, late start and late finish constraints. Given parameters aij , gi , hi and fi , the problem is to find xi and yi to minimize

max yi + max (−xi ),

1≤i≤n

subject to

max (aij + xj ) = yi ,

1≤j≤n

yi ≤ fi , 2.3.4

1≤i≤n

gi ≤ xi ≤ hi ,

(3)

i = 1, . . . , n.

Minimization of maximum deviation of finish times

Consider the problem of minimizing the maximum deviation of finish times under start-finish, start-start, finish-start, and late finish constraints: given aij , bij , cij and fi , find xi and yi , that minimize subject to

max yi + max (−yi ),

1≤i≤n

1≤i≤n

max (aij + xj ) = yi ,

1≤j≤n

max (cij + yj ) ≤ xi ,

1≤j≤n

5

max (bij + xj ) ≤ xi ,

1≤j≤n

yi ≤ fi ,

i = 1, . . . , n.

(4)

Note that such problems can normally be solved using iterative computational algorithms (see [10, 11, 12] for overviews of available solutions). In addition, these problems can be reformulated as linear programming problems to solve them by applying one of the computational methods of linear programming. Below, we provide new solutions to the problems, which are based on optimization methods in tropical mathematics, and offer results in a compact explicit vector form rather than in the form of a numerical algorithm.

3

Preliminary algebraic definitions and results

In this section, we give a brief overview of preliminary definitions, notation and results of tropical algebra to provide a formal basis for describing and solving tropical optimization problems as well as for applying them in project scheduling in the next sections. Both introductory and advanced material on tropical mathematics is provided in many publications, including [18, 19, 20, 21, 22, 23, 24, 9] to name only a few. The overview below is mainly based on the presentation of results in [13, 14, 15, 17], which provides a useful framework to obtain direct solutions to the problems under study in a compact vector form.

3.1

Idempotent semifield

We consider a system (X, ⊕, ⊗, 0, 1), where X is a set closed under addition ⊕ and multiplication ⊗ with zero 0 and identity 1 such that (X, ⊕, 0) is a commutative idempotent monoid, (X \ {0}, ⊗, 1) is an abelian group, multiplication is distributive over addition, and 0 is absorbing for multiplication. This system is usually called the idempotent semifield. Addition is idempotent, which means that x ⊕ x = x for each x ∈ X. The idempotent addition induces on X a partial order such that x ≤ y if and only if x ⊕ y = y . It follows directly from the definition that x ⊕ y ≥ x and x ⊕ y ≥ y for all x, y ∈ X. Moreover, the inequality x ⊕ y ≤ z appears to be equivalent to two inequalities x ≤ z and y ≤ z , and both addition and multiplication are monotone in each argument. Finally, we assume that the partial order is extendable to a total order, and thus consider X to be linearly ordered. Multiplication is invertible to let each nonzero x ∈ X have the inverse x−1 such that x ⊗ x−1 = 1. The power notation with integer exponents is used to represent repeated multiplication defined as x0 = 1, xp = xp−1 ⊗ x and x−p = (x−1 )p for all nonzero x and positive integer p. The integer power is assumed to extend to rational exponents to make X algebraically closed (algebraically complete, radicable). In the algebraic expressions below, we omit the multiplication sign to save writing, and read the exponents only in the above sense. 6

A typical example of the idempotent semifield under consideration is the real semifield Rmax,+ = (R ∪ {−∞}, max, +, −∞, 0), where the addition ⊕ is defined as maximum and the multiplication ⊗ as ordinary addition, with the zero 0 given by −∞ and the identity 1 by 0. Each number x ∈ R has the inverse x−1 , which is equal to the opposite number −x in the conventional notation. For all x, y ∈ R, the power xy is defined and coincides, in ordinary arithmetic, with the product xy . The partial order induced by the idempotent addition corresponds to the standard linear order given on R. As another example, consider the semifield Rmin,× = (R+ ∪{+∞}, min, ×, +∞, 1), where R+ is the set of positive real numbers, ⊕ = min, ⊗ = ×, 0 = +∞ and 1 = 1. In this semifield, the notation of inverses and exponents has the standard interpretation. The partial order defined by addition extends to a linear order that is opposite to the standard order on R.

3.2

Matrices and vectors

We now examine matrices and vectors over the idempotent semifield introduced above. The set of matrices that have m rows and n columns with entries from X is denoted Xm×n . A matrix, in which all entries are 0, is the zero matrix denoted by 0. A matrix is called row-regular (column-regular), if it has no row (column) consisting entirely of 0. Provided that a matrix is both row-regular and column-regular, it is called regular. For any matrices A, B ∈ Xm×n and C ∈ Xn×l , and a scalar x ∈ X, the matrix addition, matrix multiplication and scalar multiplication follow the standard rules with the scalar operations ⊕ and ⊗ in the place of the ordinary addition and multiplication, given by {A⊕B}ij = {A}ij ⊕{B}ij ,

n M {A}ik {C}kj , {AC}ij =

{xA}ij = x{A}ij .

k=1

The partial order associated with the idempotent addition and its properties extend to that on the set of matrices, where the relations are expanded entry-wise. Consider any matrix A = (aij ) ∈ Xm×n . The transpose of A is the matrix AT ∈ Xn×m . The multiplicative conjugate transpose of A is the matrix A− = (a− ij ), − −1 − where aij = aji if aji 6= 0, and aij = 0 otherwise. Consider square matrices of order n, which form the set Xn×n . A matrix which has the diagonal entries equal to 1 and all off-diagonal entries equal to 0 is the identity matrix. The power notation with nonnegative integer exponent serves to represent iterated products as A0 = I and Ap = Ap−1 A for any square matrix A and positive integer p. For any matrix A = (aij ) ∈ Xn×n , the trace is given by tr A = a11 ⊕ · · · ⊕ ann . 7

Every matrix that has only one row (column) is considered a row (column) vector. All vectors below are column vectors unless otherwise specified. The set of column vectors of order n is denoted Xn . A vector with all elements equal to 0 is the zero vector. If a vector has no zero elements, it is called regular. Let A ∈ Xn×n be a row regular matrix and x ∈ Xn be a regular vector. Then, the vector Ax is clearly regular. If the matrix A is column regular, then the row vector xT A is also regular. The multiplicative conjugate transpose of a nonzero column vector x = − −1 (xi ) ∈ Xn is a row vector x− = (x− if xi 6= 0, i ) with the entries xi = xi and xi = 0 otherwise. It is not difficult to verify that the conjugate transposition has the following useful properties. First, the equality x− x = 1 is valid for any nonzero vector x. Furthermore, suppose that x and y are regular vectors of the same order. Then, it is easy to see that the element-wise inequality x ≤ y is equivalent to x− ≥ y − . In addition, the matrix inequality xy − ≥ (x− y)−1 I holds, which becomes xx− ≥ I when y = x. Finally, consider a square matrix A ∈ Xn×n . A scalar λ ∈ X is an eigenvalue of A, if there exists a nonzero vector x ∈ Xn to satisfy the equality Ax = λx. The maximum eigenvalue with respect to the order defined on X is called the spectral radius and given by λ=

n M

tr1/k (Ak ).

k=1

3.3

Solution to linear inequalities

We conclude the overview of the preliminary definitions and results with the solution of linear inequalities to be used below in the analysis of tropical optimization problems. Suppose that, given a matrix A ∈ Xm×n and a regular vector d ∈ Xm , we need to find regular vectors x ∈ Xn that satisfy the inequality Ax ≤ d.

(5)

A direct complete solution to the problem under fairly general assumptions is obtained in [13, 17] in the following form. Lemma 1. For any column-regular matrix A and regular vector d, all regular solutions to inequality (5) are given by x ≤ (d− A)− .

8

Furthermore, we consider the problem: given a matrix A ∈ Xn×n , find regular vectors x ∈ Xn to satisfy the inequality Ax ≤ x.

(6)

To represent a solution to the problem for any matrix A ∈ Xn×n , we define a function that takes A to produce the scalar Tr(A) = tr A ⊕ · · · ⊕ tr An , and use the asterate operator (the Kleene star), which maps A to the matrix A∗ = I ⊕ A ⊕ · · · ⊕ An−1 . The next result obtained in [25, 13, 14] offers a direct complete solution to inequality (6). Theorem 2. For any matrix A, the following statements hold: 1. If Tr(A) ≤ 1, then all regular solutions to (6) are given by x = A∗ u, where u is any regular vector. 2. If Tr(A) > 1, then there is no regular solution.

4

Tropical optimization problems

Tropical optimization problems present an area in tropical mathematics, which is of both theoretical interest and practical importance (see, eg, [26] for an overview). The problems are formulated in the framework of tropical mathematics to minimize or maximize nonlinear functions defined on vectors over idempotent semifields and calculated using multiplicative conjugate transposition of vectors. The problems may have constraints given by linear inequalities and equalities. There are problems that can be solved directly in a rather general setting. For other problems, only algorithmic solution are known in the form of an iterative computational scheme to produce a particular solution or signify that no solution exist. The purpose of this section is twofold: first, to offer representative examples to demonstrate a variety of optimization problems under study, and second, to provide an efficient basis for the solution of scheduling problems in the next section. We consider examples of both unconstrained and constrained optimization problems with different objective functions defined in the common setting in terms of a general idempotent semifield. For all problems, direct solutions are given in a compact vector form ready for further analysis and straightforward computations. For some problems, the solutions obtained are complete solutions.

9

We start with the following problem: given matrices A, B ∈ Xn×n and a vector g ∈ Xn , find regular vectors x ∈ Xn that x− Ax,

minimize

subject to Bx ⊕ g ≤ x.

(7)

A direct complete solution to the problem is given in [14, 15] as follows. Theorem 3. Let A be a matrix with spectral radius λ > 0 and B a matrix with Tr(B) ≤ 1. Then, the minimum value in problem (7) is equal to θ =λ⊕

n−1 M

M

tr1/k (AB i1 · · · AB ik ),

k=1 1≤i1 +···+ik ≤n−k

and all regular solutions are given by x = (θ −1 A ⊕ B)∗ u,

u ≥ g.

Furthermore, suppose that, given a matrix A ∈ Xn×n and vectors f , g, h ∈ Xn , we need to find regular vectors x ∈ Xn that solve the problem x− Ax,

minimize

subject to g ≤ x ≤ h.

(8)

The following direct exact solution is proposed in [15]. Theorem 4. Let A be a matrix with spectral radius λ > 0, and h be a regular vector such that h− g ≤ 1. Then the minimum value in problem (8) is equal to n−1 M (h− Ak g)1/k , θ =λ⊕ k=1

and all regular solutions are given by x = (θ −1 A)∗ u,

g ≤ u ≤ (h− (θ −1 A)∗ )− .

Given a matrix A ∈ Xm×n and a vector d ∈ Xm , consider the problem to find regular vectors x ∈ Xn that minimize d− Ax ⊕ (Ax)− d.

(9)

A direct solution obtained to the problem in [27, 13, 28] is as follows. Theorem 5. Let A be a row-regular matrix and d be a regular vector. Then, the minimum value in problem (9) is equal to ∆ = ((A(d− A)− )− d)1/2 , and the maximum solution is given by x = ∆(d− A)− . 10

Finally, we present a solution to the problem: given matrices A, B ∈ Xm×n and vectors p, q ∈ Xm , find regular vectors x ∈ Xn such that minimize q − Bx(Ax)− p.

(10)

The next statement offers a direct solution to the problem [29]. Theorem 6. Let A be row-regular and B be column-regular matrices, p be nonzero and q be regular vectors. Then the minimum value in problem (10) is equal to ∆ = (A(q − B)− )− p, and attained at any vector x = α(q− B)− ,

5

α > 0.

Application to project scheduling

We are now in a position to derive solutions to the scheduling problems formulated in the beginning of the paper. In this section, we first represent each problem in the framework of the idempotent semifield Rmax,+ in both scalar and vector forms, and then solve this problem by reducing to an optimization problem in the previous section.

5.1

Minimization of maximum flow time

Consider problem (1) and describe it in the terms of the semifield Rmax,+ . By replacing the usual operations by those of Rmax,+ , we obtain the problem to find the unknowns xi and yi for all i = 1, . . . , n that minimize

n M

subject to

n M

x−1 i yi ,

i=1 n M

aij xj = yi ,

bij xj ≤ xi ,

j=1

j=1

gi ≤ xi ,

n M

cij yj ≤ xi ,

j=1

i = 1, . . . , n.

To represent the problem in a compact vector form, we introduce the following matrix-vector notation: A = (aij ),

B = (bij ),

C = (cij ),

g = (gi ),

x = (xi ).

With this notation, the problem is to find vectors x and y that minimize

x− y,

subject to Ax = y,

Bx ≤ x,

Cy ≤ x,

g ≤ x. A direct complete solution of the problem is given as follows. 11

(11)

Theorem 7. Let A be a matrix with spectral radius λ > 0, B and C be matrices such that Tr(B ⊕ CA) ≤ 1. Then the minimum value in problem (11) is equal to θ =λ⊕

n−1 M

M

tr1/k (A(B ⊕ CA)i1 · · · A(B ⊕ CA)ik ),

k=1 1≤i1 +···+ik ≤n−k

and all regular solutions are given by x = (θ −1 A ⊕ B ⊕ CA)∗ u,

y = A(θ −1 A ⊕ B ⊕ CA)∗ u,

u ≥ g.

Proof. By substitution of the first equality constraint y = Ax, we eliminate the vector y . Then, we combine all inequality constraints into one, to write the problem minimize x− Ax, subject to (B ⊕ CA)x ⊕ g ≤ x. Clearly, this problem has the form of that at (7), where B is replaced by B ⊕ CA. Thus, a direct application of Theorem 3 yields the desired result.

5.2

Minimization of maximum deviation from due dates

Consider problem (2), which, in terms of the semifield Rmax,+ , takes the form minimize

n M

subject to

n M

−1 (d−1 i yi ⊕ yi di ),

i=1

aij xj = yi ,

j=1

n M

bij xj ≤ xi ,

n M

cij yj ≤ xi ,

i = 1, . . . , n.

j=1

j=1

In addition to the previously introduced matrix-vector notation, we define the vector d = (di ), and represent the problem in the form minimize

d− y ⊕ y − d,

subject to Ax = y,

Bx ≤ x,

Cy ≤ x.

(12)

The next result offers a solution to the problem. Theorem 8. Let A be a row-regular matrix, B and C matrices such that Tr(B ⊕ CA) ≤ 1, and d be a regular vector. Then, the minimum value in problem (12) is equal to ∆ = ((A(B ⊕ CA)∗ (d− A(B ⊕ CA)∗ )− )− d)1/2 , and the maximum solution is given by x = ∆(B⊕CA)(d− A(B⊕CA)∗ )− ,

y = ∆A(B⊕CA)(d− A(B⊕CA)∗ )− . 12

Proof. After substitution y = Ax, we combine both inequality constraints into one inequality (B ⊕CA)x ≤ x. Then, we apply Theorem 2 to solve the last inequality as x = (B ⊕ CA)u, where u is any regular vector. Finally, by substitution of the solution, we reduce problem (12) to the unconstrained problem minimize d− A(B ⊕ CA)∗ u ⊕ (A(B ⊕ CA)∗ u)− d. This problem has the form of (9) with A replaced by A(B ⊕ CA)∗ . Thus, we apply Theorem 5 to obtain a solution in terms of the vector u. Turning back to the vectors x and y , we complete the proof.

5.3

Minimization of makespan

In the framework of the semifield Rmax,+ , problem (3) is rewritten as minimize

n M

subject to

n M

i=1

yi +

n M

x−1 i ,

i=1

aij xj = yi ,

gi ≤ xi ≤ hi ,

yi ≤ fi ,

i = 1, . . . , n.

j=1

By adding the vector notation f = (fi ), we represent the problem as minimize

1T yx− 1,

subject to Ax = y,

g ≤ x ≤ h,

(13)

y ≤ f. Theorem 9. Let A be a nonzero matrix, h and f be regular vectors such that (h− ⊕ f − A)g ≤ 1. Then, the minimum makespan in problem (13) is equal to θ = 1T A(I ⊕ gh− )1, and all regular solutions are given by x = (I ⊕ θ −1 11T A)u,

y = A(I ⊕ θ −1 11T A)u,

where g ≤ u ≤ ((h− ⊕ f − A)(I ⊕ θ −1 11T A))− . Proof. As before, we first substitute y = Ax. An application of Lemma 1 to solve the inequality Ax ≤ f yields x ≤ (f − A)− . Then, we take two upper boundaries x ≤ h and x ≤ (f − A)− , and apply conjugate transposition to rewrite them as x− ≥ h− and x− ≥ f − A. By coupling both inequalities into one, and again taking the conjugate transposition, we obtain one upper 13

bound x ≤ (h− ⊕ f − A)− . Finally, we represent the objective function x− 11T Ax in its equivalent form x− 11T Ax to write the problem as minimize

x− 11T Ax,

(14)

subject to g ≤ x ≤ (h− ⊕ f − A)− .

The problem obtained is of the form of (8), where A is replaced by 11T A and h by (h− ⊕ f − A)− . To apply Theorem 4, we first calculate (11T A)k = (1T A1)k−1 11T A,

tr(11T A)k = (1T A1)k ,

k = 1, . . . , n;

from which is directly follows that the spectral radius of the matrix 11T A is equal to λ = 1T A1 > 0. Furthermore, we consider the minimum, which is given by θ = 1T A1 ⊕ (1T A1)

n−1 M ((1T A1)−1 h− 11T Ag)1/k . k=1

Suppose that 1T A1 ≤ h− 11T Ag . Since the inequality (1T A1)−1 h− 11T Ag ≥ 1 holds, we have ((1T A1)−1 h− 11T Ag)1/k ≤ (1T A1)−1 h− 11T Ag , and hence, θ = h− 11T Ag . On the other hand, if 1T A1 > h− 11T Ag , then we immediately find that θ = 1T A1. By combining both results, we finally obtain θ = 1T A1 ⊕ h− 11T Ag = 1T A1 ⊕ 1T Agh− 1 = 1T A(I ⊕ gh− )1. To describe the solution set according to Theorem 4, we examine the matrix (θ

−1

n−1 n−1 M M −1 T k −1 (θ −1 1T A1)k−1 11T A. (θ 11 A) = I ⊕ θ 11 A) = T

∗

k=1

k=0

Considering that θ ≥ 1T A1, we obtain (θ −1 11T A)∗ = I ⊕ θ −1 11T A. Substitution into the solution provided by Theorem 4 yields x = (I ⊕ θ −1 11T A)u,

g ≤ u ≤ ((h− ⊕ f − A)(I ⊕ θ −1 11T A))− .

Finally, we represent the vector y = Ax, which completes the proof.

14

5.4

Minimization of maximum deviation of finish times

After rewriting problem (4) in terms of Rmax,+ , the problem takes the form minimize

n M

subject to

n M

i=1

j=1 n M

yi

n M

yj−1 ,

j=1

aij xj = yi ,

n M

bij xj ≤ xi ,

j=1

cij yj ≤ xi ,

yi ≤ fi ,

i = 1, . . . , n.

j=1

Switching to matrix-vector notation, the problem becomes minimize

1T yy − 1,

subject to Ax = y,

Bx ≤ x,

Cy ≤ x,

(15)

y ≤ f. The following result offers a solution to the problem. Theorem 10. Let A be row-regular and B column-regular matrices, p be nonzero and q regular vectors. Then, the minimum value in problem (15) is equal to ∆ = (A(B ⊕ CA)∗ (1T A(B ⊕ CA)∗ )− )− 1, and attained if x = α(B⊕CA)∗ (1T A(B⊕CA)∗ )− ,

y = αA(B⊕CA)∗ (1T A(B⊕CA)∗ )− ,

where α ≤ (f − A(B ⊕ CA)∗ (1T A(B ⊕ CA)∗ )− )− . Proof. After substitution y = Ax, we combine the first two inequalities into one inequality (B ⊕ CA)x ≤ x, which is then solved by using Theorem 2 to obtain x = (B ⊕ CA)∗ u, where u is a regular vector. Furthermore, we take the last inequality A(B ⊕ CA)∗ u ≤ f , and apply Lemma 1 to find u ≤ (f − A(B ⊕ CA)∗ )− . The problem takes the form minimize

1T A(B ⊕ CA)∗ u(A(B ⊕ CA)∗ u)− 1,

subject to u ≤ (f − A(B ⊕ CA)∗ )− . First, we remove the constraints and solve the obtained unconstrained problem. By applying Theorem 6, where both matrices A and B are replaced by A(B ⊕ CA)∗ , and both vectors p and q by 1, we find the minimum ∆ = (A(B ⊕ CA)∗ (1T A(B ⊕ CA)∗ )− )− 1, 15

which is attained at the vector u = α(1T A(B ⊕ CA)∗ )− ,

α > 0.

To find the values of α, which satisfy the constraint u ≤ (f − A(B ⊕ CA)∗ )− , we solve the inequality α(1T A(B ⊕ CA)∗ )− ≤ (f − A(B ⊕ CA)∗ )− . By applying Lemma 1 with α as the unknown, we have α ≤ (f − A(B ⊕ CA)∗ (1T A(B ⊕ CA)∗ )− )− . It remains to turn back to vectors x and y to complete the proof.

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