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Multi-processor Search and Scheduling Problems with Setup Cost

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Abstract

We study two optimization problems in a multiprocessor environment in the presence of set-up costs. The first problem involves multiple parallel searchers (e.g., robots) that must locate a target which lies in one of many concurrent rays, and at an unknown position from their common origin. Every time a searcher turns direction, it incurs a turn cost. The objective is to derive a search strategy for locating the target as efficiently as possible. The second problem involves contract algorithms, namely algorithms in which the available computation time is specified prior to their execution. In particular, we seek a schedule of executions of contract algorithms for several different problems in identical parallel processors so as to efficiently simulate an interruptible algorithm, assuming that each execution incurs a given set-up cost. The performance of the search and scheduling strategies are evaluated by means of well-established measures, namely the competitive ratio and the acceleration ratio, respectively. In this paper we provide near-optimal strategies for the above problems, using an approach based on infinite linear-programming formulations. More precisely, we present a search strategy (resp. schedule) which is optimal when the number of rays (resp. problems) is a multiple of the number of searchers (resp. processors). For the general case, we show that the corresponding solutions are very close to optimal.

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Notes

  1. Note that all turn costs are incorporated in times T and T , respectively.

  2. A full turn consists of a turn in the origin (as the searcher enters a new ray, and a turn at the end of its exploration on the said ray. Thus the cost of a full turn is equal to d.

  3. Note that for the purposes of the proof we will have to establish properties of some y i ’s, with i > k (e.g. Corollary 1.) This does not imply that such values are part of the dual solution.

  4. We emphasize that despite the similarity of the lower-bound LP formulations for the two problems, there is no obvious way to derive Lemma 14 from a reduction from the p-searcher m-ray problem. Instead, one needs to first establish normalization properties which are tailored to the scheduling problem, and which are not readily derivable from the normalization properties in the context of ray searching.

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Correspondence to Spyros Angelopoulos.

Additional information

This work was supported by project ANR-11-BS02-0015 “New Techniques in Online Computation” (NeTOC).

Appendices

Appendix A: Omitted Proofs of Section 3

Here, we provide complete justifications for Lemmas 7, 8 and 9 from Section 3, whose proofs have been omitted therein.

Proof of Lemma 7

We argue by induction. Thus, let us suppose that

$$Y_{i+1}\geq t Y_{i} \geq 0, \qquad\text{for all} ~~i\leq N, $$

for some integer Nl − 2. We only have to show that the above inequalities remain true for i = N + 1.

To this end, note first that

$$Y_{i}\geq t^{i-j} Y_{j}\geq 0, \qquad\text{for all} j\leq i\leq N+1. $$

Therefore, employing the recurrence relation (3.1), we find that

$$\begin{array}{@{}rcl@{}} Y_{N+2} & =& Y_{N+1} - \frac{1}{M} Y_{N+1-(l-1)} \geq Y_{N+1} - \frac{1}{Mt^{l-1}} Y_{N+1} \\ & =& tY_{N+1}-\frac{t^{l}-t^{l-1}+\frac{1}{M}}{t^{l-1}}Y_{N+1} \geq tY_{N+1}, \end{array} $$

which concludes the proof of the lemma. □

Proof of Lemma 8

For each \(i\in \mathbb {N}\), we denote by \(Y_{i}^{\prime }\) the derivative (or any finite difference quotient) of Y i with respect to z. Then, clearly, by linearity, the sequence \(\left \{Y_{i}^{\prime }\right \}_{i=0}^{\infty }\) verifies the same recurrence relation

$$Y_{i}^{\prime} = Y_{i-1}^{\prime} - \frac{1}{M} Y_{i-l}^{\prime}, \qquad\text{for all~~} i\in \mathbb{N}. $$

Furthermore, it is readily seen that \(Y_{i}^{\prime }=-\frac {1}{M}\), for every i = l − 1, l,…,2l − 2. Therefore, in view of Lemma 7, we conclude that \(Y_{i}^{\prime }<0\), for every il − 1. □

Proof of Lemma 9

Let us denote by \(U^{+}\subset \left [0,\frac {1}{p}\right ]\) the set of values of \(z\in \left [0,\frac {1}{p}\right ]\) such that the corresponding sequence \(\left \{Y_{i}\right \}_{i=0}^{\infty }\) remains always non-negative, and by \(U^{-}\subset \left [0,\frac {1}{p}\right ]\) its complement, that is the set of values of \(z\in \left [0,\frac {1}{p}\right ]\) such that the corresponding sequence \(\left \{Y_{i}\right \}_{i=0}^{\infty }\) eventually becomes negative.

Then, in view of Lemma 7, we find that if

$$0\leq z\leq \frac{1}{pl} , $$

then the whole sequence \(\left \{Y_{i}\right \}_{i=0}^{\infty }\) remains non-negative. Hence \(\left [0,\frac {1}{pl}\right ]\subset U^{+}\). Furthermore, for any given \(i\in \mathbb {N}\), the value of Y i clearly depends continuously on z. Therefore, it is readily seen that the set U + is closed. Finally, since each Y i ≥ 0 is monotonic with respect to z according to Lemma 8, one shows easily that the set U + is convex.

Thus, on the whole, the set U + is closed, convex and contains \(\left [0,\frac {1}{pl}\right ]\). It follows that U + is a closed interval and so, there exists \(\zeta \in \left [\frac {1}{pl}, \frac {1}{p}\right ]\) such that U +=[0, ζ]. Moreover, it is readily seen that Y l <0 whenever \(z > \frac {M-1}{pM}\), whence \(\zeta \in \left [\frac {1}{pl}, \frac {M-1}{pM} \right ]\).

Finally, let us consider any zU . Then, there exists \(j\in \mathbb {N}\) such that Y i ≥ 0, for all ij, and Y j + 1 <0. It is then readily seen that

$$0>Y_{j+1}\geq Y_{j+2}\geq {\ldots} \geq Y_{j+l-1} \geq Y_{j+l}. $$

Therefore, applying Lemma 7, we conclude that Y i <0, for all i > j.

The proof of the lemma is complete. □

Appendix B: Omitted Proofs of Section 4

Before proceeding with the proof of Lemma 14 we show how to derive the key lemmas that describe the same properties of optimal normalized schedules as in [24]. Namely, we argue that one can derive essentially the same normalization properties of optimal schedules, in the presence of setup costs and for the (N, B) definition of competitiveness, as for the standard version (no setup costs, and considering the acceleration ratio as performance measure).

We begin with some preliminary notation and definitions. Let Σ be a given schedule of contracts. We will follow the convention of denoting by lower-case c and upper case C the lengths of consecutively completed contracts for a given problem p, respectively. More precisely, if (p i , c i ) denotes a contract of length c i for a problem p i , then the earliest contract in Σ for problem p i which is completed after contract (p i , c i ) finishes will be denoted by (p i , C i ).

Given a schedule Σ with associated set of contracts \(\mathcal {C}\), and two problem instances p 1, p 2, let \(\mathcal {C}_{1},\mathcal {C}_{2}\) denote two (potentially infinite) subsets of \(\mathcal {C}\), such that all contracts in \(\mathcal {C}_{1}\) have problem tag p 1, and all contracts in \(\mathcal {C}_{2}\) have problem tag \(\mathcal {C}_{2}\). Consider a new set of contracts \(\mathcal {C}^{\prime }\) which is identical to \(\mathcal {C}\), with the exception that every contract in \(\mathcal {C}_{1}\) acquires problem tag p 2 instead of p 1 and every contract in \(\mathcal {C}_{2}\) acquires problem tag p 1 instead of p 2. Consider also the schedule Σ which is otherwise identical to S (except for the problem tag swaps described above). We say that Σ is obtained from Σ by a swap for sets \(\mathcal {C}_{1}\) and \(\mathcal {C}_{2}\).

Lemma 15

Suppose at time T i a processor is to start working on contract (p i ,C i ), and that there is another problem p j such that the contracts (p j ,c j ) and (p j ,C j ) appear in X. Let T j denote the time at which a processor is to start working on contract (p j ,C j ).

Given a schedule Σ, and two problems p i and p j as described above for which c j <c i and T j >T i , then either C j <C i or we can define a new schedule such that c j <c i , and which is no worse than that of the original schedule.

Proof

The proof follows the lines of Lemma 1 in [24]. Suppose Σ is such that c j < c i and T j > T i , and suppose that C j > C i . Execute a swap of program tags for all contracts on p i that complete after (p i , c i ) and all contracts on p j that complete after (p j , c j ), so as to obtain a new schedule Σ.

Observe that all contracts before the swapping remain untouched. Likewise contracts for problems not involved in the swap are also untouched. These contracts do not have any effect on the performance of the schedule.

Now consider contracts for the swapped problems once their very first swapped contract is completed. I.e. let c t r for r > i be a contract for problem p i in the original schedule. Since the schedule is (N, B)-competitive we have that T r + C r N c r + B. After the swap this expression still relates the performance, but now corresponding to problem p j , since contracts c r and C r are now ran on problem p j , which implies that the schedule remains unaffected by the swapping for those contracts as well.

Thus the only point in time where the swapping may have an effect is at the time in which the previously longest completed contract comes from the old schedule while the new contract about to be completed comes from the new schedule. We will show that the competitiveness on those contracts is no worse than that of the original schedule. In particular, for the original schedule we have

$$T_{i}+C_{i}\leq N c_{i} + B, \ \text{ and} \ T_{j}+C_{j}\leq N c_{j} + B.$$

Hence

$$\max \left\{ \frac{T_{i}+C_{i}-B}{c_{i}}, \frac{T_{j}+C_{j}-B}{c_{j}} \right\} \leq N. $$

Now observe that for Σ at C i , C j we have

$$\max \left\{ \frac{T_{i}+C_{i}-B}{b_{j}}, \frac{T_{j}+C_{j}-B}{b_{i}} \right\} \leq N. $$

since T j > T i and C j > c i , imply \(\frac {T_{j}+C_{j}}{c_{j}} \geq \frac {T_{i}+C_{i}}{c_{j}}\); moreover since c j < c i , we have \(\frac {T_{j}+C_{j}}{c_{j}} \geq \frac {T_{j}+C_{j}}{c_{i}}\) which implies T i + C i N c j + B and T j + C j N c i + B. Therefore, we conclude that Σ is also (N, B)-competitive. □

Corollary 2

Given a schedule Σ there is another schedule of no worse competitive ratio with the property that for any two problems p i and p j , with b j <b i , and T j >T i , it is always the case that C j <C i .

Proof

Same as in [24]. □

We introduce some notation that will be needed in the statement and proof of Lemma 16. For a schedule Σ, let S T denote the set of all contracts completed by time T. Also let S T be the complement of S T , namely all contracts in Σ−S T . Fix a contract c t 0 = (p 0, C 0), which is scheduled to start at time T 0. For any problem p j let \(C_{j}= \min \{ C: (p_{j},C) \in S^{T_{0}+C_{0}} \}\). Observe that in the context of the above definitions, we have \(c_{j}=\max \{c: (p_{j},c) \in S_{T_{0}+C_{0}}\}\).

Lemma 16

Let ct 0 =(p 0 ,C 0 ) be a contract scheduled by Σ at time T 0 and ct j =(p j ,C j ) be any contract in \(S^{T_{0}}\) . Then there exists a schedule Σ of no worse competitive ratio such that if c 0 ≥c j for a problem p j ≠p 0 then T 0 +C 0 ≥T j +C j .

Proof sketch

The proof follows along the lines of the proof of Lemma 2 in [24]. More precisely, we use the concept of a swap of problem tags, similar to the proof of Lemma 15. This swapping affects the competitiveness of the schedule only for two possible interruption times, and we can show, using the same argument as in Lemma 15 that the inequalities hold in the presence of setup costs and the schedule remains (N, B)-competitive. □

A schedule Σ is said to be normalized if it satisfies the conditions of Corollary 2 and Lemma 16. Using the same strategy as in [24], we can show the following:

Lemma 17

If a schedule is (N,B)-competitive, then there exists an normalized schedule that is also (N,B)-competitive. In particular, there is an optimal normalized schedule.

Having established the properties of optimal normalized schedules, we now proceed with the proof of Lemma 14. Note that similar to the ray-searching problem, we can assume that n/qc, for some constant c, when qn since otherwise the desired bounds follows straightforwardly.

Proof of Lemma 14

We will now proceed to set-up the family of lower-bound LPs for our problem. Let Σ denote an optimal normalized schedule, and let X = x 1, x 2,… denote the sequence of the sorted contract lengths in Σ. Consider a time T 0 such that processor M 0 is about to schedule a new contract c t 0, of length denoted by C 0. Since S is a normalized schedule, c t 0 must be a contract for problem p 0 that satisfies the conditions of Corollary 2. Let the longest completed contract at time T 0 + C 0 for problem p 0 be c 0.

Observe now that because of Lemma 15 and Lemma 16, every contract of length strictly less than c 0 must complete within the open interval (0, T 0 + C 0) and hence at the end of this interval every processor is engaged in a contract of length at least c 0 and every problem p j , with 0 ≤ ln − 1, has completed a contract of length at least c 0. Let us denote by c l these lengths (and note that c l c 0). Observe that the lengths c l are elements in the sequence X, thus we can write \(c_{l}=x_{i_{l}}\), for all such l. For a given processor M j , let us denote by I j the set of indices in X of all the contracts completed on processor M j up to time T 0 + C 0, not inclusively, for 0 ≤ jm − 1. Note that \(c_{0}=x_{i_{0}}\), for some i 0 ≥ 0.

Let j denote the number of contracts that have been scheduled in processor M j by time T 0 + D 0. Also, let C j be the last completed contract on processor M j , say for a problem p l , such that the previous completed contract c j for p l is less than c 0. We can write the following condition concerning the competitive ratio of the schedule, assuming an interruption occurring at time right before C j is completed in M j :

$$ \sum\limits_{t \in I_{j}} x_{t} +d\ell_{j}\leq N x_{k_{j}}+B, $$
(B.1)

Summing up all inequalities (B.1) concerning every processor M j , with j ∈ {0,…q − 1} we obtain that

$$ \sum\limits_{j=0}^{q-1} \sum\limits_{t \in I_{j}} x_{t} + d \sum\limits_{j=0}^{q-1} \ell_{j} \leq N\sum\limits_{j=0}^{q-1} x_{i_{j}} +Bq. $$
(B.2)

From [24] we know that \({\sum }_{j=0}^{q-1} {\sum }_{t \in I_{j}} x_{t} \geq {\sum }_{j=1}^{i_{0}+n+1}x_{j}\), and that \({\sum }_{j=0}^{q-1} x_{i_{j}} \leq {\sum }_{j=i_{0}-q+2}^{i_{0}+1} x_{j}\). We also know that by time T + C 0 at least i 0 + n + 1 contracts have been completed; therefore \({\sum }_{j=0}^{q-1} \ell _{j} \geq i_{0}+n+1\). Using the above, we obtain from (2.3) that

$$ \sum\limits_{j=1}^{i_{0}+n+1}x_{j} + d(i_{0}+n+1) \leq N \sum\limits_{j=i_{0}-q+2}^{i_{0}+1} x_{j} +Bq, $$
(B.3)

for all i 0q − 1.

Last, we need to add one more constraint in order to obtain our lower-bound LP. □

Lemma 18

For any normalized schedule, it must be that \({\sum }_{j=1}^{n} x_{j}+dn \leq Bq\).

Proof

Suppose that the adversary chooses time \(t=\frac {{\sum }_{j=1}^{n} x_{j}+dn}{q}-\epsilon \) as an interruption (for arbitrarily small 𝜖). It is easy to see that there is at least one problem for which no associated contract has completed its execution by time t. Indeed, if this were not the case, then at least n different contracts (each with a setup cost equal to d) would have finished by time t, a contradiction since x 1, x 2,…x n are the smallest possible contract lengths. From the definition of acceleration ratio, this implies that B must be such that Bt, or equivalently, that \({\sum }_{j=1}^{n} x_{j}+dn \leq Bq\). □

We can combine the inequalities (B.3) (for all i 0q − 1) as well as the inequality of Lemma 18 into a linear program, with the objective of minimizing B. We substitute i 0q + 1 with k, so as to obtain a simplified expression.

$$\begin{array}{rrrlrr} \min & B \\ \text{s.t.} & \sum\limits_{j = 1}^{n} x_{j} - qB &\leqslant& - dn & &\qquad(P^{\prime})\\ & 2\sum\limits_{j = 1}^{i+n+q} x_{j} - N \sum\limits_{j=i+1}^{i+q} x_{j} - qB &\leqslant& - d(n+q + i)& \qquad\forall i = 0 {\ldots} k \\ & B, x_{1}, \ldots, x_{q+k} &\geqslant& 0, \end{array} $$

with corresponding dual:

$$\begin{array}{rrrrrc} \max & d \left( nz+\sum\limits_{i=0}^{k} y_{i}(i+n+q) \right)&&&& \qquad(D^{\prime}) \\ \text{s.t.} & z + \sum\limits_{i = 0}^{k} y_{i} &\leqslant& \frac{1}{q}& &\qquad(B) \\ & \left\{ \begin{array}{llll} z & \ j \leq n \\ 0 & \text{\ otherwise} \end{array} \right. + \sum\limits^{k}_{i = \max (0, j - n-q)} y_{i} - N \sum\limits_{i = \max (0, j - q)}^{\min (j - 1, k)}y_{i} &\geqslant& 0& \qquad\forall j = 1 {\ldots} ,n+q+k&\qquad(x_{j})\\ & z, y_{0}, \ldots, y_{k} &\geqslant& 0,& \end{array} $$

which completes the proof of the lemma.

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Angelopoulos, S., Arsénio, D., Dürr, C. et al. Multi-processor Search and Scheduling Problems with Setup Cost. Theory Comput Syst 60, 637–670 (2017). https://doi.org/10.1007/s00224-016-9691-3

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