Scheduling of Plausible Task Sequences

ν : the corresponding first or second counterpart task of ν.

4.3 Scheduling of Plausible Task Sequences

Discussion in the previous section introduces the notion of _n+1¹ (_2n

n

)plausible task sequence

for a given job sequence. This section is dedicated to the development of a polynomial time algorithm for determining the makespan of a plausible task sequence, if it is feasible.

Given a plausible task sequence, it could be non-trivial to determine its feasibility and a schedule with the minimum makespan, if feasible.

Denote a plausible task sequence by π = (o₁, o₂, . . . , o_2n), where o_h stands for the task assigned to the h-th position in π. If no confusion would arise, hereafter this study simply mentions sequences to indicate plausible task sequences. Notice that in any schedule considered hereafter, the constraint of exact delays is satisfied. In other words, the interval between each pair of coupled tasks a_j and b_j in any schedule is exactly l_j, j ∈ Nn. Denote the starting time and the completion time of job J_j in a schedule σ by s_j(σ) and C_j(σ), respectively. It is obvious that C_j(σ) = s_j(σ) + a_j + l_j + b_j. A schedule σ is feasible if and only if at any time, at most one task is processed in σ, i.e. no overlap between tasks occurs. Sequence π is called feasible if and only if there exists a feasible schedule whose permutation of tasks agrees with π, i.e. a schedule in which the processing of any task oh

for h∈ N2n−1 completes earlier than or exactly at the starting time of task o_h+1.

Consider first how to determine the feasibility of a given sequence π. For any feasible sequence π, the constraint of exact delays implies that the interval induced by the exact delay l_j of any job J_j must accommodate all the tasks arranged between a_jand b_j. Namely, the following condition is necessary for the feasibility of a sequence π.

Condition (C): For any job J_j with o_ℓ = a_j and o_g = b_j, the inequality ∑g−1

h=ℓ+1p_oh ≤ lj

must hold, where po_h is the processing length of task oh.

Note that condition (C) is not sufficient to make sequence π feasible. Consider the

following instance: (a₁, l₁, b₁) = (1, 2, 1), (a₂, l₂, b₂) = (2, 5, 1), (a₃, l₃, b₃) = (2, 4, 2), (a4, l4, b4) = (2, 3, 1). Condition (C) holds for sequence π = (a1, a2, b1, a3, a4, b2, b3, b4).

However, π is infeasible since an overlap between b3 and b4 (Figure 4.4(a)) or between a4

and b₂ (Figure 4.4(b)) is inevitable in any attempt to create a feasible schedule of π.

0 1 2 3 4 5 6 7 8 9 10 11 12

a₁ b₃

a₄ b₄

b₂ a₂ b₁ a₃

(a) (b)

0 1 2 3 4 5 6 7 8 9 10 11 12 13

a₁ b₃

a₄ b₄

b₂ a₂ b₁ a₃

Figure 4.4: An inevitable task overlap happens (a) between b₃ and b₄ or (b) between a₄ and b₂.

Condition (C) only partially verifies the feasibility of π because in a schedule whether an idle time or overlap exists between o_h and o_h+1 cannot be detected before assigning each task a starting time. Therefore, we turn to develop a procedure for constructing a schedule for sequence π and prove that the feasibility of π can be determined by the constructed schedule. If sequence π is indeed feasible, it can be further proved that the constructed schedule attains the minimum makespan among those of feasible schedules.

Consider the subsequences of a particular permutation pattern

(a_i1, a_i1+1, . . . , a_i2, b_i1, b_i1+1, . . . , b_i2),

1 ≤ i1 ≤ i2 ≤ n, where all its first (respectively, second) tasks are consecutively se-quenced without any second (respectively, first) task inserted. A given sequence π is derived by merging several subsequences of this pattern. Consider the sequence π = (a₁, a₂, b₁, a₃, a₄, a₅, b₂, b₃, a₆, b₄, a₇, a₈, b₅, b₆, b₇, b₈) as an example. As shown in Figure 4.5, sequence π can be regarded as the outcome of four interleaved subsequences (a₁, a₂, b₁, b₂), (a₃, a₄, a₅, b₃, b₄, b₅), (a₆, b₆) and (a₇, a₈, b₇, b₈). Such particular subsequences are regarded as the maximal fundamental clusters of a given sequence π and these subsequences are scheduled individually. Scheduling these fundamental clusters is the first attempt to ex-amine the feasibility of sequence π. Later it will be elucidated that the infeasibility of

any fundamental cluster leads to the infeasibility of π. If all these fundamental clusters are feasible, then we proceed to schedule π by interleaving those obtained subschedules.

π= (a1, a2, b1, a3, a4, a5, b2, b3, a6, b4, a7, a8, b5, b6, b7, b8)

ß

(a₁,a₂,b₁, a3, a4, a5,b₂, b3, a6, b4, a7, a8, b5, b6, b7, b8) (a1, a2, b1,a₃,a₄,a₅, b2,b₃, a6,b₄, a7, a8,b₅, b6, b7, b8) (a1, a2, b1, a3, a4, a5, b2, b3,a₆, b4, a7, a8, b5,b₆, b7, b8) (a1, a2, b1, a3, a4, a5, b2, b3, a6, b4,a₇,a₈, b5, b6,b₇,b₈)

( )

( )

( )

( )

Figure 4.5: Four subsequences of a particular pattern in sequence π.

Now some notations for collating fundamental clusters from sequence π are defined.

Given a sequence π, a segment is defined as a maximal, by inclusion, subsequence of tasks {aj} without inserted tasks {bi}. Assume that the task sequence (a1, a₂, . . . , a_n) is parti-tioned into k disjoint segments for 1≤ k ≤ n. To facilitate discussion, we denote by X the sequence of subscripts (1, 2, . . . , n) and by H a k-subsequence partition of X corresponding to the k-segment partition of (a₁, a₂, . . . , a_n). Partitioning X into k disjoint subsequences, we have H ={X1, X₂, . . . , X_k} and X = X1⊕X2⊕· · ·⊕Xk, where X_rdenotes the r-th sub-sequence in X and⊕ is a sequence concatenation operator. For r ∈ Nk, the last element of subsequence Xr is denoted by nr, and we have Xr = (nr−1 + 1, . . . , nr), where n0 = 0 and n_k= n. Denote by ˜X_r the set of elements in sequence X_r. |Xr| = nr− nr−1 indicates the length of X_r. Denote by b_n′

r the immediate predecessor of a_nr+1 in π for r ∈ Nk−1 and n^′_r−1+ 1≤ n^′_r ≤ nr, where n^′₀ = 0. Notice that any single task a_nr−1+1 = a_nr, which is sur-rounded by two tasks b_n′

r−1 and b_n′

r−1+1 in π, forms a segment, i.e. |Xr| = 1. According to the assumption of k segments, sequence π consists of k fundamental clusters in which the r-th one contains jobs{Jj| j ∈ ˜X_r}, r ∈ Nk. Denote the r-th fundamental cluster in π by πr= (anr−1+1, . . . , anr, bnr−1+1, . . . , bnr), r∈ Nk. The subsequence obtained by eliminating the jobs of{Jj| j ∈ ˜X_r+1∪· · ·∪ ˜X_k} from π is denoted by χr, r ∈ Nk−1. Note that χ_k = π.

To construct a schedule of fundamental cluster π_r, we proposes a recursive procedure to augment the subschedule job by job, instead of task by task. Namely, coupled tasks a_j

and b_j are simultaneously added into the subschedule of jobs (J_nr−1+1, J_nr−1+2, . . . , J_j−1), j ∈ {nr−1 + 2, . . . , nr}, in each recursion step. In the proposed procedure, job Jj is interleaved with job Jj−1 by conjoining two first tasks aj−1 and aj (Figure 4.6(a)) or two second tasks b_j−1 and b_j (Figure 4.6(b)). The obtained schedule is denoted by σ_r, and the recursive formula for the job starting times is given as follows:

s_j(σ_r) =









0, j = n_r−1+ 1;

s_j−1(σ_r) + a_j−1

+ max{0, lj−1+ b_j−1− aj− lj}, nr−1+ 2 ≤ j ≤ nr.

(4.1)

a_j-1 a_j b_j-1 b_j a_j-1 a_j b_j-1 b_j

(a) (b)

lj-1

l_j

lj-1

lj

Figure 4.6: J_j is interleaved with J_j−1 by conjoining (a) a_j−1 and a_j or (b) b_j−1 and b_j. Eq. (4.1) implies that in σr task aj (respectively, bj) is started later than or exactly at the completion of aj−1 (respectively, bj−1). Schedule σr is a feasible schedule of πr if task a_nr completes earlier than or exactly at the start of task b_nr−1+1. The following lemma gives structural properties of fundamental clusters.

Lemma 4.1. Given a subsequence π_r = (a_nr−1+1, . . . , a_nr, b_nr−1+1, . . . , b_nr), the following three properties hold: (i) If s_nr(σ_r)+a_nr > C_nr−1+1(σ_r)−bnr−1+1, then π_r is infeasible. (ii) If s_nr(σ_r) + a_nr ≤ Cnr−1+1(σ_r)− bnr−1+1, then π_r is feasible and σ_r is a feasible schedule attaining the minimum makespan among those of all feasible schedules of πr. (iii) The feasibility and the shortest schedule, if feasible, can be determined in O(|Xr|) time.

Proof. If s_nr(σ_r)+a_nr > C_nr−1+1(σ_r)−bnr−1+1, then task a_nr completes later than the start of task bnr−1+1 in σr. The only possible way to find a feasible schedule of πr is to process a_nr earlier or process b_nr−1+1later. In schedule σ_r, task a_j starts exactly at the completion of a_j−1, or task b_j starts exactly at the completion of b_j−1, j ∈ {nr−1+ 2, . . . , n_r}. Starting J_nr earlier or J_nr−1+1 later finally results in the shifting of the whole schedule, which is

futile. Therefore, a feasible schedule of π_r does not exist and π_r is infeasible. Property (i) is proved.

Property (ii) is concerned about the feasibility of π_r and the optimality of σ_r. As for the feasibility of π_r, the inequality s_nr(σ_r) + a_nr ≤ Cnr−1+1(σ_r)− bnr−1+1 indicates that task a_nr completes earlier than or exactly at the starting time of task b_nr−1+1 in σ_r. Therefore, a feasible schedule of π_r, i.e. σ_r, exists and π_r is feasible. Next, we will show that schedule σ_r attains the minimum makespan for sequence π_r, i.e. C_nr(σ_r)≤ Cnr(σ_πr), where σ_πr denotes any feasible schedule of π_r. This inequality can be proved by induction on nr. We derive the following recursive equation for Cj(σr) by adapting Eq. (4.1).

C_j(σ_r) =









a_nr−1+1+ l_nr−1+1+ b_nr−1+1, j = n_r−1+ 1;

Cj−1(σr) + bj

+ max{0, aj+ l_j − lj−1− bj−1}, nr−1+ 2 ≤ j ≤ nr.

(4.2)

Consider the induction base nr = nr−1+ 2 and πr = (anr−1+1, anr−1+2, bnr−1+1, bnr−1+2).

If a_nr−1+2 + l_nr−1+2 > l_nr−1+1 + b_nr−1+1, then it is impossible to interleave J_nr−1+1 and J_nr−1+2 with a completion time less than a_nr−1+1 + a_nr−1+2 + l_nr−1+2 + b_nr−1+2. On the other hand, if a_nr−1+2+ l_nr−1+2 ≤ ln_r−1+1+ b_nr−1+1, then there exists no feasible schedule σ_πr where J_nr−1+2 completes earlier than a_nr−1+1 + l_nr−1+1 + b_nr−1+1 + b_nr−1+2. Since C_nr−1+2(σ_r) = a_nr−1+1+ a_nr−1+2+ l_nr−1+2+ b_nr−1+2 for a_nr−1+2+ l_nr−1+2 > l_nr−1+1+ b_nr−1+1 and C_nr−1+2(σ_r) = a_nr−1+1 + l_nr−1+1+ b_nr−1+1+ b_nr−1+2 for a_nr−1+2+ l_nr−1+2 ≤ lnr−1+1+ bnr−1+1 from Eq. (4.2), we have the induction base Cnr−1+2(σr)≤ Cnr−1+2(σπr).

Assume, as the induction hypothesis, that the inequality holds for n_r= i > n_r−1+ 2, i.e. C_i(σ_r) ≤ Ci(σ_πr). If a_i+1+ l_i+1 > l_i + b_i, then the minimum time span from the completion time of J_i to that of J_i+1 is a_i+1+ l_i+1+ b_i+1− li− bi. In case of a_i+1+ l_i+1≤ l_i+ b_i, the minimum aforementioned time span is b_i+1. By Eq. (4.2), we have C_i+1(σ_r) = C_i(σ_r) + b_i+1+ a_i+1+ l_i+1− li− bi for a_i+1+ l_i+1 > l_i+ b_i and C_i+1(σ_r) = C_i(σ_r) + b_i+1 for a_i+1+ l_i+1 ≤ li + b_i. Hence, C_i+1(σ_r) is less than or equal to the completion time of Ji+1 in any feasible schedule σπr, i.e. Ci+1(σr) ≤ Ci+1(σπr). By induction, the inequality C_nr(σ_r)≤ Cnr(σ_πr) is established. The proof of property (ii) is done.

For schedule π_r, the feasibility and the shortest schedule, if feasible, can be obtained by the values of sj(σr) for j = nr−1+ 1, . . . , nr. By virtue of Eq. (4.1), the calculation involves |Xr| − 1 iterations, each of which requires constant time. Therefore, either a feasible schedule with the minimum makespan or the infeasibility of sequence π_r can be determined in O(|Xr|) time.

The infeasibility of fundamental cluster πr implies that there exists no feasible sub-schedule whose task permutation agrees with subsequence π_r, r ∈ Nk. Since the sub-sequence π_r is a part of complete sequence π, no feasible complete schedule of π exists either. We therefore have the following property.

Property 4.1. If any fundamental cluster π_r, r ∈ Nk, is infeasible, then sequence π is infeasible.

Before presenting a step-wise procedure for determining the infeasibility or the shortest feasible schedule of sequence π, we define some notations. For each schedule σ_r (r ∈ Nk), denote by α_r the time span from the start of a_nr−1+1 to the completion of a_nr, and γ_r the idle time between a_nr and b_nr−1+1. S_r denotes a subschedule constructed by arranging the first r subschedules, σ₁, . . . , σ_r. Note that S_k is the constructed complete schedule of sequence π, which consists of k clusters. Denote by β_r¹ the idle time between b_n′

r and b_n′ r+1

in subschedule S_r, and by β_r² the time span from the start of b_n′

r+1to the completion of b_nr.

Algorithm Plausible-Task-Sequence

Step 1. Scan π to obtain the partition H = (X₁, . . . , X_k) and keep track of n_r, n^′_r and χ_r for each r = 1, . . . , k− 1.

Step 2. Schedule π_r by Eq. (4.1) for each r = 1, . . . , k. If any π_r is infeasible, then go to Step 7. Otherwise, set I = 1 and S_I = σ₁.

Step 3. If n^′_I = nI, then merge SI with σI+1 by appending an_I+1 to the end of bn_I and go to Step 6.

Step 4. If β_I¹ ≥ αI+1∧ βI² ≤ γI+1, then go to Step 4(a). If β_I¹ < α_I+1∧ βI² ≤ γI+1, then go to Step 4(b). If β_I¹ ≥ αI+1∧ βI² > γI+1, then go to Step 4(c). Otherwise, go to Step 4(d).

Step 4(a). If β_I¹+ β_I² ≤ αI+1+ γ_I+1, then merge S_I with σ_I+1 by appending a_nI+1 to the end of b_n′

I. Otherwise, merge S_I with σ_I+1 by appending b_nI+1 to the end of b_nI. Go to Step 6.

Step 4(b). Shift b_n′

I+1 (and a_n′

I+1 will be simultaneously shifted, i.e. shift J_n′ I+1) to extend β_I¹ such that β_I¹ = α_I+1 and merge S_I with σ_I+1 by appending a_nI+1 to the end of b_n^′

I. Go to Step 5.

Step 4(c). Shift J_n′

I+1 to shorten β_I² such that β_I² = γ_I+1 and merge S_I with σ_I+1 by appending b_nI+1 to the end of b_nI. Go to Step 5.

Step 4(d). If β_I¹+ β_I² ≤ αI+1+ γ_I+1, then shift J_n′

I+1 to extend β_I¹ such that β_I¹ = α_I+1 and merge S_I with σ_I+1 by appending a_nI+1 to the end of b_n′

I. Otherwise, shift J_n′ I+1

to shorten β_I² such that β_I² = γ_I+1 and merge S_I with σ_I+1 by appending b_nI+1 to the end of bn_I. Go to Step 5.

Step 5. Call Checking routine with input b_n′

I+1. Call Checking routine with input a_n′

I+1.

Step 6. Let S_I+1 be the obtained schedule. If I = k− 1, then output the schedule SI+1

and stop. Otherwise, set I = I + 1 and go to Step 3.