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In this section, we consider stability radius %s(p) of an optimal schedule for problem G//PCi with criterionPCi. If Φ =PCi, conditions (4.17) and (1.6) for the general shop problem G//Φ are converted to the following conditions (1.30) and (1.31), respectively.

n

Obviously, the value Ci for a digraph Gs(p) is equal to the largest weight of a path from the set Hsi, and hence, to solve problem G//PCi, it is sufficient to find a digraph Gs(p) such that equality (1.30) holds.

Due to equality (1.31) to find the stability radius %s(p), it is sufficient to construct a vector x ∈ Rq+ that satisfies the following three conditions.

(a’) There exists a digraph Gk(p) ∈ Λ(G), k 6= s, such that

(b’) For any given real  > 0, which may be as small as desired, there exists a vector

(c’) The distance d(p, x) achieves its minimal value among the distances between the vector p and the other vectors in the space Rq+ which satisfy both above conditions (a’) and (b’).

Similarly as in the previous section (see conditions (a), (b) and (c)), after having constructed such a vector x ∈ Rq+ one can define the stability radius of digraph Gs(p):

%s(p) = d(p, x). Thus, due to (1.32) and (1.33) the calculation of the stability radius may again be reduced to an extremal problem on the set of weighted digraphs Λ(G).

However, in this case we are forced to consider sets of representatives of the family of sets Hki, 1 ≤ i ≤ n, which may be defined as follows.

Similarly to the proof of Theorem 1.4, one can find vector x satisfying conditions (a’) and (b’) in the form x(r) = (x1(r), x2(r), . . ., xq(r)) with the components xi(r) from the set {pi, pi + r, pi − r} on the basis of a direct comparison of the set Ωus ∈ Ωs,k of representatives of the family of sets (Hsi)1≤i≤n, and the set Ωuk of representatives of the family of sets (Hki)1≤i≤n, where k = 1, 2, . . . , λ, k 6= s, and

s,k = {Ωvs : There does not exist a set Ωuk such that ni(Ωvs) ≤ ni(Ωuk) for each i = 1, 2, . . . , q}.

As a result, the following lower bound of the stability radius has been obtained:

%s(p) ≥ r = min

It is easy to see that it may happen that the above vector x(r) ∈ Rq does not belong to set R+q since substraction value r from some component of vector p may result a negative number. To obtain the exact value of %s(p), one can use the vector x(r) =

In contrast to vector x(r), vector x(r) necessary has no negative components.

Let the set of operations Q be ordered in the following way:

i1, i2, . . . , im, im+1, . . . , iq, (1.35) where niα(Ωuk) ≤ niα(Ωvs) for each α = 1, 2, . . . , m and niα(Ωuk) > niα(Ωvs) for each α = m + 1, m + 2, . . . , q. Moreover, for the sequence (1.35) the inequalities

pim+1 ≤ pim+2 ≤ . . . ≤ piq

have to be satisfied. Using sequence of operations 1.35 it is easy to derive the following formula for calculating %s(p). If only a subset of the processing times can be changed but the other ones cannot be changed, a formulas similar to (1.36) and (1.37) can be derived (see the remark after Theorem 1.4).

Zero Stability Radius

We consider the case of %s(p) = 0. Similarly to the notion of a critical path and the critical weight, which is important for problem G//Cmax (see Section 1.3), we introduce the notion of a critical set of paths Ωuk and the critical sum of weights of digraph Gk(p) ∈ Λ(G) for

for the weighted digraph Gk(p) is reached on this set:

X

The value Lpk defined in (1.38) is called critical sum of weights for digraph Gk(p).

Obviously, a critical set Ωuk may include a path ν ∈ Hki, i = 1, 2, . . . , n, if and only if

Lemma 1.4 There exists a real  > 0 such that the set Ωk\Ωk(p) contains no critical set of digraph Gk(p) ∈ Λ(G) for any vector p0 ∈ O(p) ∩ Rq+ of the processing times, i.e.

k(p0) ⊆ Ωk(p).

Proof. After having calculated the value

k = 1

2qnminnLpkX

ν∈Ωuk

lp(ν) : Ωuk ∈ Ωk\Ωk(p)o, (1.39)

one can verify that for any real , which satisfies the inequalities 0 <  < k, the difference in the right side of equality (1.39) remains positive when vector p is replaced by any vector p0 ∈ O(p) ∩ R+q. Indeed, for any u ∈ {1, 2, . . . , ωk}, the cardinality of set Ωuk may be at most equal to qn. Thus, the difference LpkPν∈Ωu

k lp(ν) may not be ’overcome’ by a vector p0 if d(p, p0) < k.

3 Next, we prove the following necessary and sufficient conditions for equality %s(p) = 0.

Theorem 1.6 Let Gs be an optimal digraph for problem G//PCi with positive processing times pi > 0 of all operations i ∈ Q. The equality %s(p) = 0 holds if and only if the following three conditions hold:

1) there exists another optimal schedule k ∈ SΦ(p), Φ =PCi, k 6= s,

2) there exists a set Ωvs ∈ Ωs(p) such that for any set Ωuk ∈ Ωk(p), there exists an operation i ∈ Q for which the condition

ni(Ωvs) ≥ ni(Ωuk), Ωuk ∈ Ωk(p), (1.40) holds (or the condition

ni(Ωvs) ≤ ni(Ωuk), Ωuk ∈ Ωk(p), (1.41) holds) and

3) inequality (1.40) (or inequality (1.41), respectively) is satisfied as a strict one for the set Ωuk.

Proof. We prove necessity by contradiction. Assume that %s(p) = 0 but the conditions of the theorem are not satisfied. We consider three cases (j), (jj) and (jjj) of violating these conditions.

(j) Assume that there does not exist another optimal schedule, i.e., we have SΦ(p) = {s} with Φ =PCi. Then we consider a real  such that

0 <  < 1 2qnmin

t6=s(Lpt − Lps)

holds. Similarly to the proof of Lemma 1.4, we can show that digraph Gsremains optimal for any vector p0 = (p01, p02, . . . , p0q) ∈ Rq+of the processing times provided that d(p, p0) ≤ .

Therefore, we have %s(p) ≥  > 0 which contradicts the assumption %s(p) = 0.

(jj) Assume that |SΦ(p)| > 1 and for any optimal schedule k ∈ SΦ(p) with k 6= s, and for any set Ωvs ∈ Ωs(p), there exists a set Ωuk ∈ Ωk(p) such that ni(Ωvs) = ni(Ωuk) for any operation i ∈ Q.

In this case, we can take any  that satisfies the inequalities 0 <  < minns, k, 1

2qn min

t6∈φΣ(p)(Lpt − Lps)o. (1.42) From Lemma 1.4, due to inequality  < s, we get that equality

Lps0 = max

Sufficiency. We show that, if the conditions of Theorem 1.6 are satisfied, then %s(p) <  for any given  > 0.

We construct a vector p = (p1, p2, . . . , pq) ∈ Rq+with components pi ∈ {pi, pi+ , pi

}, where  = min{k, , mini∈Qpi}, using the following rule: For each Ωuk ∈ Ωk(p), mentioned in Theorem 1.6, we set pi = pi + , if inequalities (1.40) hold, or we set pi = pi− , if inequalities (1.41) hold. Note that  > 0 since pi > 0, i ∈ Q.

After changing at most |Ωk(p)| components of vector p according to this rule, we obtain a vector p of processing times for which inequality

X

Thus, we conclude that s /∈ SΦ(p) with d(p, p) =  which implies %s(p) <  ≤ .

3 Theorem 1.6 directly implies the following assertion.

Corollary 1.5 If s ∈ S is a unique optimal schedule for problem G//PCi, then %s(p) > 0.

It is easy to prove the following upper bound for the stability radius of an optimal schedule for problem G//PCi.

Theorem 1.7 If s ∈ S is an optimal schedule for problem G//PCi with λ > 1 and pi > 0 for at least one operation i ∈ Q, then

%s(p) ≤ max

i∈Q pi.

Proof. We consider vector p0 ∈ Rq+ with zero components: p0i = 0 for each i ∈ Q. For this vector of processing times, each feasible digraph Gt ∈ Λ(G) is optimal and each set of representatives Ωut is critical. We can take a schedule k ∈ SΦ(p0) which has only one arc (j, i) ∈ Ek different from the arcs in Es, i.e. (i, j) ∈ Es and Es\{(i, j)} = Ek\{(j, i)}.

It is easy to see that there exist sets Ωvs ∈ Ωs(p) and Ωuk ∈ Ωk(p) such that ni(Ωvs) > ni(Ωuk).

Setting pi =  > 0 and pl = 0 for each l ∈ Q\{i}, we obtain s /∈ SΦ(p) and d(p, p) <

max{pi : i ∈ Q}.

3 Remark 1.1 As it follows from Theorem 1.7, problem J //PCi with λ > 1 cannot have an optimal schedule with an infinitely large stability radius in contrast to problem J //Cmax and problem J //Lmax (see Section 1.3).

Note that all of the results in this section and in Section ?? are valid for any general shop scheduling problem. However, we use the partition of the set of operations Q into n chains Q(i), i = 1, 2, . . . , n, (which is necessary for the job shop and flow shop but is not necessary for the general shop) for a better presentation of the results. Table 1.1 collects special cases of the shop scheduling problem which are characterized by the machine service and the technological routes of the jobs. In scheduling theory often the classical job shop is considered for which each job has to be processed exactly once on each machine (see problem in the third row in Table 1.1). For our consideration this restriction is not important. We will consider the job shop problem J m//Φ with recirculation (see [Pin95a]), which may occur when a job may visit a machine more than once.

To illustrate the above notations and some results, we consider in Section 6 an example of a job shop scheduling problem with two jobs and two machines.

Example 1.5 The job shop problem is specified by the mixed graph G = (Q, A, E) given in Figure 1.5. The first job consists of operations 1 and 2, and the second job consists of operations 3 and 4. So we have the precedence constraints 1 → 2 and 3 → 4. The

assignment of the operations to the machines is as follows: Q1 = {1, 4}, Q2 = {2, 3}. The vector p = (10, 30, 20, 40) defines the processing times of the operations Q = {1, 2, 3, 4}.

For this problem we get Λ(G) = {G1, G2, G3} with the following signatures of semi-active schedules: E1 = {(1, 4), (3, 2)}, E2 = {(1, 4), (2, 3)} and E3 = {(4, 1), (3, 2)}.

The corresponding sets of dominant paths are as follows: H11 = {(1, 2), (3, 2)}, H12 = {(1, 4), (3, 4)}, H21 = {(1, 2)}, H22 = {(1, 2, 3, 4)}, H31 = {(3, 4, 1, 2)} and H32 = {(3, 4)}.

The optimal digraph G1 = (Q, A ∪ E1, ∅) is represented in Figure 1.5 and it defines the unique optimal semiactive schedule (10, 50, 20, 60) for both criteria Cmax andPCi. There-fore, due to Corollary 1.4 and Corollary 1.5 we have %bs(p) > 0 and %s(p) > 0.

















-l

l l

l l

l

1 2

3 4

p1 = 10 p2 = 30

p3 = 20 p4 = 40

Figure 1: Mixed graph G = (Q, A, E) for example 1.5

We can calculate the exact value of %b1(p) on the basis of Theorem 1.5. First, we compare the digraphs G1 and G2. We have four sets of representatives for di-graph G1, namely: Ω11 = {(1, 2), (1, 4)}, Ω21 = {(1, 2), (3, 4)}, Ω31 = {(3, 2), (1, 4)} and Ω41 = {(3, 2), (3, 4)}. We can calculate vectors n(Ω11) = (2, 1, 0, 1), n(Ω21) = (1, 1, 1, 1), n(Ω31) = (1, 1, 1, 1) and n(Ω41) = (0, 1, 2, 1). Digraph G2 has only one set of representatives:

12 = {(1, 2), (1, 2, 3, 4)} with n(Ω12) = (2, 2, 1, 1). Obviously, we have ni(Ωu1) ≤ ni(Ω12) for u ∈ {1, 2, 3} and for all i ∈ {1, 2, 3, 4}. Thus, we have to calculate only

r1

2,Ω41 = (40 + 100) − (50 + 60)

| 2 − 0 | + | 2 − 1 | + | 1 − 2 | + | 1 − 1 | = 30

4 = 7.5.

Next, we compare the digraphs G1 and G3. Digraph G3 has only one set of representatives:

13 = {(3, 4, 1, 2), (3, 4)} with n(Ω13) = (1, 1, 2, 2). Obviously, we have ni(Ωu1) ≤ n(Ω13) for u ∈ {2, 3, 4} and for all i ∈ {1, 2, 3, 4}. Thus, we have to calculate only

r1

3,Ω11 = (100 + 60) − (40 + 50)

| 1 − 2 | + | 1 − 1 | + | 2 − 0 | + | 2 − 1 | = 70

4 = 17.5.

Since 7.5 < 10 = min{p1, p2, p3, p4}, the bound (1.34) is reached and we obtain %Σ1(p) = min{7.5; 17.5} = 7.5 and G2 is a competitive digraph for digraph G1, which is optimal for p = (10, 30, 20, 40).

















-1 2

3 4

c1(1) = 10 c2(1) = 50

c3(1) = 20 c4(1) = 60

1

@

@

@

@

@ R

Figure 2: The optimal digraph G1 = (Q, A ∪ E1, ∅)

Next we consider the stability radius of the optimal digraph G1 with respect to crite-rion Cmax. We have the following sets of dominant paths: H = {(1, 2), (3, 4)}, H1 = {(1, 2), (1, 4), (3, 2), (3, 4)}, H2 = {(1, 2, 3, 4)} and H3 = {(3, 4, 1, 2)}. It is clear that any path µ ∈ H1\H is dominated by the path (1, 2, 3, 4) ∈ H2 and by the path (3, 4, 1, 2) ∈ H3. Thus, due to Theorem 1.1 we have %b1(p) = ∞.

In scheduling theory open shop problem is also considered when technological routes for processing jobs are not fixed before scheduling, i.e., for each job Ji ∈ J only set of operations QJi is given but order of these operations is not fixed. While solving open shop problem O//Φ it is necessary to find optimal orders of operations QJi for each job Ji ∈ J along with optimal orders of operations on each machine Mk ∈ M . Similarly to flow shop problem, each job Ji ∈ J has to be processed by each machine exactly once.

The same property of the technological routes are assumed for so-called classical job shop. Classification of shop scheduling problems is given in Table 1.1.

Table 1.1: Different shop scheduling problems

Characterization Shop scheduling Technological routes

of machine service problem of the jobs

Open shop Different jobs may have O//Φ different routes, which are Each job Ji ∈ J has not fixed before scheduling to be processed by Flow shop Jobs have the same route, each machine Mk ∈ M F //Φ which is fixed before

exactly once scheduling

Classical Different jobs may have job shop different routes, which A job may be processed Job shop are fixed before

by a machine more J //Φ scheduling

than once