DOI 10.1007/s10951-009-0151-7
Minimizing the total weighted completion time in the relocation
problem
Alexander V. Kononov· Bertrand M.T. Lin
Published online: 17 December 2009
© Springer Science+Business Media, LLC 2009
Abstract This paper studies the minimization of total weighted completion time in the relocation problem on a single machine. The relocation problem, formulated from an area redevelopment project, can be treated as a resource-constrained scheduling problem. In this paper, we show four special cases to be NP-hard in the strong sense. Problem equivalence between the unit-weighted case and the UET (unit-execution-time) case is established. For two further re-stricted special cases, we present a polynomial time approx-imation algorithm and show its performance ratio to be 2. Keywords Relocation problem· Resource-constrained scheduling· NP-hardness · Approximation algorithm
1 Introduction
Resource constraints are one of the most commonly con-sidered factors in project management and scheduling. In this paper, we study a variant of the relocation problem that involves scheduling with generalized resource constraints. Formally, there is a set of jobsJ = {J1, J2, . . . , Jn} avail-able for processing on a single machine. A pool of V0units
A.V. Kononov
Sobolev Institute of Mathematics, Novosibirsk, Russia e-mail:[email protected]
A.V. Kononov
Novosibirsk State University, Novosibirsk, Russia B.M.T. Lin (
)Institute of Information Management, Department of Information and Finance Management, National Chiao Tung University, Hsinchu 300, Taiwan
e-mail:[email protected]
of a common single-type resource is given for processing the jobs. A job Ji ∈ J requires and consumes αi units of the resource from the resource pool. That is, when the job
Ji is to be processed, there must be at least αi units of the resource in the pool. Upon its completion, the job Ji will immediately return βi units back to the resource pool. The job Ji has a processing length (time) pi and a weight wi. A schedule is said to be feasible if all jobs can be success-fully processed. As no idle time is assumed, throughout this paper, schedule and sequence are used interchangeably if no confusion would arise. Let Ci be the completion time of the job Ji in a particular schedule. The studied problem is to determine a schedule that is feasible with respect to V0and the total weighted completion timeni=1wiCi is minimum. Let us denote the problem by RPWT.
The relocation problem was first proposed and formu-lated from a redevelopment project in Boston (Kaplan1986; Kaplan and Berman1988; PHRG1986). There were sev-eral buildings to be demolished and rebuilt. Before each building was redeveloped, its tenants had to be temporar-ily housed until new capacities were available for reloca-tion. Tenants were not subjected to reside at the old site. Given a fixed budget for temporary housing during the re-development, the municipal government needed to deter-mine a reconstruction sequence of the buildings such that all tenants could be successfully evacuated and housed dur-ing the course of the project. The significance of the re-location problem could be attributed to its potential appli-cations in database management (Amir and Kaplan1988) and financial planning (Xie 1997). Moreover, the reloca-tion problem provides a generalizareloca-tion of convenreloca-tional re-source constraints (Blazewicz et al. 1983; Hammer 1986; Brucker et al.1999) by allowing that the amount of resource
βi returned by a job can be less than, equal to or greater than αi, the amount the job has required.
In a basic model of the relocation problem, the temporal parameter piis not included, i.e., given a fixed amount of the resource, the problem is to plan a feasible redevelopment se-quence of the buildings. The minimization counterpart of the feasibility problem is to determine the minimum initial bud-get required for the existence of a feasible sequence. Kaplan and Amir (1988) showed that this minimization problem is equivalent to the well-known makespan minimization in a two-machine flowshop (Johnson1954). Kaplan (1986) and Amir and Kaplan (1988) addressed the deployment of mul-tiple working crews. If there are available crews and suf-ficient resource, the development of several buildings can overlap. Kononov and Lin (2006) showed that minimizing the makespan is strongly NP-hard even when there are only two working crews, all buildings have the same process-ing time and the new capacity of each buildprocess-ing is no less than its original capacity. They also proposed approxima-tion algorithms and analyzed the associated performance ra-tios. Sevastyanov et al. (2009) investigated the relocation problem of makespan minimization subject to release dates. They analyzed the complexities of several cases and also developed a pseudo-polynomial time algorithm, based on a multi-parametric dynamic programming technique, for the case where the number of different release dates is constant. To the best of our knowledge, the problem of minimizing the total weighted completion time studied in this paper is new. In Sect.2, we study the computational complexities of several special cases. The equivalence between special cases will be given to extend the complexity results. Section3is devoted the development and analysis of an approximation algorithm for two further restricted cases. We analyze the performance ratio of the proposed algorithm. An instance is given to establish the tightness of the ratio. Concluding remarks will be presented in Sect.4.
2 Complexity results and problem equivalence
In this section, we give the complexity results of the RPWT problem. Auxiliary notations for the following discussion are required. The contribution of the job Jiis defined by δi= βi− αi. The resource level at time t is denoted by Vt. If at time t some job Jicompletes and another job Jjstarts being processed, Vt gives the resource level of the moment after Ji deposits the resource it produces and before Jj requires the resource for processing. Let σ be a particular sequence or schedule of jobs. The job in position i, 1≤ i ≤ n, is denoted by σ (i). The resource level at time t subject to schedule σ is specified by Vt(σ ). If no confusion arises, Vtwill be used for simplicity. The weighted sum of job completion times subject to a feasible schedule σ is represented by Z(σ ).
Although the WSPT (Weighted Shortest Processing Time First) rule (Smith 1956) optimally solves the 1||wiCi
problem, it cannot deal with the RPWT problem. Consider the following numerical example of 4 jobs and V0= 0.
Jobs J1 J2 J3 J4
pi 1 1 1 1
wi 1 1 1 5
αi 0 0 0 10
βi 2 3 5 10
By the WSPT rule, the job J4should be processed first. In RPWT, the lack of resource, however, defers the process-ing of J4. The example indicates the difficulty in composprocess-ing an optimal schedule as well as suggests the design of NP-hardness proofs.
The NP-hardness results start with the case where each job makes non-positive contributions and has a unit execu-tion time (UET), i.e., δi ≤ 0 and pi= 1 for all jobs Ji∈ J . The reduction is based upon the following Ordered Numer-ical 3-Dimensional Matching problem, which can be easily transformed from the well-known Numerical 3-Dimensional Matching problem (Garey and Johnson1979).
Ordered Numerical 3-Dimensional Matching (ON3M problem)
Instance. An integer bound B ∈ Z+, the sets of in-dicesA1= {1, . . . , m}, A2= {m + 1, . . . , 2m}, A3= {2m + 1, . . . , 3m}, a positive size xiof each element i, 1≤ i ≤ 3m, with 3mi=1xi = mB and x1≥ x2≥ · · · ≥ xm ≥ xm+1≥ · · · ≥ x2m≥ x2m+1≥ · · · ≥ x3m.
Question. CanA1∪ A2∪ A3be partitioned into m dis-joint sets A1, A2, . . . , Am such that each Aj,1≤ j ≤ m, contains exactly one element from each ofA1,A2,A3and
i∈Ajxi= B?
Theorem 1 The RPWT problem is strongly NP-hard, even if δi≤ 0 and pi= 1 for all jobs Ji∈ J .
Proof Given 3m, B, andA1,A2,A3as specified for ON3M, we let θk=
i∈Akxi for 1≤ k ≤ 3. Let η = 3m2B. An in-stance I of 6m− 3 jobs is constructed as follows:
• Basic jobs Ji,1≤ i ≤ 3m,
αi= mB + xi, βi= 0, wi= η + xi.
• Connecting jobs J3(m+l−1)+k, 1≤ l ≤ m − 1, 1 ≤ k ≤ 3,
α3(m+l−1)+k = β3(m+l−1)+k = (3mB + B)(m − l),
w3(m+l−1)+k= 0.
The initial resource level is V0= (3mB +B)m. Through-out the proof, we use Z∗ to denote 3mη(3m− 1) + 3mB(m− 1) +i∈A1xi+ 2 i∈A2xi+ 3 i∈A3xi. Note that Z∗<3mη(3m− 1) + 3mB(m − 1) + 3(i∈A1xi + i∈A2xi + i∈A3xi)= 3mη(3m − 1) + 3m 2B.It can be
showed that a feasible schedule of I with wiCi ≤ Z∗ exists if and only if the ON3M problem has a partition as specified. Please refer to AppendixAfor the details of the
proof.
Next we investigate the case where all jobs have non-negative contributions and a unit execution time.
Theorem 2 The RPWT problem is strongly NP-hard, even if δi≥ 0 and pi= 1 for all jobs Ji∈ J .
Proof Similarly, from an instance of the ON3M problem, we construct an instance I of 6m− 3 jobs as follows:
• Basic jobs Ji,1≤ i ≤ 3m,
αi= 0, βi= mB + xi, wi= B − xi.
• Connecting jobs J3(m+i−1)+k,1≤ i ≤ m − 1, 1 ≤ k ≤ 3,
α3(m+i−1)+k= β3(m+i−1)+k= (3mB + B)i, w3(m+i−1)+k
= η.
Because all jobs have non-negative contributions, we set the initial resource level as V0= 0. Using a similar line of reasoning as in the proof of Theorem 1, we can show that there exists a feasible schedule of I with a weighted completion time of no more than 3η(3m− 7)(m − 1) + 12m2B− 3i∈A 1xi− 2 i∈A2xi− i∈A3xi if and only
if the ON3M problem has the required partition. Following the above two theorems, we subsequently want to study the complexity status of the following two problems with arbitrary processing times and a unit weight: (1) δi ≥ 0, wi = 1; and (2) δi ≤ 0, wi = 1. In the follow-ing, we show the strong connection between the UET case (pi= 1) and the unit-weighted case (wi= 1). We adopt the definition on p. 118 from Korte and Vygen textbook (Korte and Vygen2006) for our case.
Let us consider two minimization problems P and Q with objective functions P and Q, correspondingly. We say that the problemP totally reduces to the problem Q if there are functions f and g, each computable in linear time, such that f transforms an instance I ofP to an instance ¯I ofQ, and g transforms a solution ¯σ of ¯I to a solution σ of
I and P(I, σ )= Q( ¯I ,¯σ). If P totally reduces to Q and Q totally reduces to P, then both problems are called totally equivalent.
Theorem 3 The UET case (pi= 1) and the unit-weighted case (wi= 1) are totally equivalent.
Proof Let I be an instance of the UET case containing n jobs J = {J1, J2, . . . , Jn} with αi, βi and wi given for each job Ji and an initial resource level V0. We construct an instance ¯I of the unit-weighted case having n jobs ¯J =
{ ¯J1, . . . , ¯Jn} with ¯pi= wi,¯αi= βi,and ¯βi= αi. Set the ini-tial resource level ¯V0= V0+
n
i=1(βi− αi).
Let σ = (σ (1), σ (2), . . . , σ (n)) be a feasible permuta-tion of jobs in the UET case. We show that ¯σ = (σ (n),
σ (n− 1), . . . , σ (1)) is feasible for ¯I. Indeed, let ¯Vk de-note the amount of resource after the completion of the jobs
Jσ (n), . . . , Jσ (k+1) in ¯σ . We have ¯Vk − ¯ασ (k) = ¯V0 + n i=k+1( ¯βσ (i)− ¯ασ (i))− ¯ασ (k)= V0+ k i=1(βσ (i)−ασ (i))− βσ (k)= V0+ k−1
i=1(βσ (i)− ασ (i))− ασ (k)≥ 0. The last in-equality follows from the feasibility of schedule σ . Thus, we get ¯Vk≥ ¯ασ (k), and the feasibility of ¯σ is guaranteed.
Let Ci(σ ) and Ci(¯σ ) denote the completion times of the job Ji in schedules σ and ¯σ , respectively. Then, n i=1Ci(¯σ ) = n k=1k¯p¯σ(n−k+1) = n k=1kwσ (k) = n
i=1wiCi(σ ). It follows that if we can get an optimal schedule of the UET case, then we can easily construct an optimal one of the unit-weighted case, and vice versa. Note that an instance of the UET case with δi≤ 0 is to-tally equivalent to an instance of the unit-weighted case with
δi ≥ 0, and that an instance of the UET case with δi≥ 0 is totally equivalent to an instance of the unit-weighted case with δi ≤ 0. Therefore, two results follow from Theorem1 and Theorem2.
Corollary 1 The RPWT problem is strongly NP-hard, even if δi≥ 0 and wi= 1 for all jobs Ji∈ J .
Corollary 2 The RPWT problem is strongly NP-hard, even if δi≤ 0 and wi= 1 for all jobs Ji∈ J .
3 2-Approximation algorithm
In this section, we present a 2-approximation algorithm for the UET case with δi≥ 0 for all jobs Ji. The algorithm dis-patches jobs in a greedy way. In the relocation problem, in-tuition suggests that a job is preferred if it is more impor-tant (wi is larger) or produces more resource (δi is larger). Taking into account both attributes, we therefore create a se-quence π1of all jobs in non-increasing order of weights wi and a sequence π2of all jobs in non-increasing order of con-tributions δi. In the course of execution of the algorithm, a job Jiis called available at time t if its resource requirement αi ≤ Vt. The algorithm starts by locating the first available job of sequence π1. The job, if it exists, is assigned to the first position of our schedule. If no job is available, then infeasibility arises. The same logic is applied to sequence
π2for the second position of our schedule. The dispatching process is continued, by exploiting π1and π2alternatively, until either all jobs are dispatched or infeasibility is encoun-tered.
Algorithm W
Input. An initial resource level V0and a job setJ with pi= 1, αi≤ βi for all Ji∈ J ;
Output. A feasible schedule σ or “No feasible schedule”. 1. For i= 0, 1, . . . , n2 do
2. If no job is available then stop and report “no feasible schedule”.
3. Let Jk be the first available job in π1.
4. Set σ (2i+1) = k and delete Jkfrom π1and π2. 5. Set V2i+1= V2i+ δk.
6. If all jobs are scheduled then stop and output σ . 7. If no job is available then stop and report
“no feasible schedule”.
8. Let Jk be the first available job in π2.
9. Set σ (2i+2) = k and delete Jkfrom π1and π2. 10. Set V2i+2= V2i+1+ δk.
11. End For
12. Stop and output σ
The running time of AlgorithmWis analyzed as follows. The execution consists of O(n) iterations. Step 3 and Step 8 require O(n) time to locate the first available element in the sequence π1and in the sequence π2. Therefore, the overall running time is O(n2). The following lemma is concerned with the feasibility of produced schedules.
Lemma 1 If AlgorithmWcannot generate a feasible ule of the given instance, then there exists no feasible sched-ule of the instance.
Proof Assume Algorithm Wterminates without a feasible schedule after successfully scheduling k jobs, 1≤ k < n. Let σ be some feasible schedule and the index i, 1≤ i < k, be the first position where σ (i)= σ (i). We locate the job
Jσ (i) in the schedule σ and insert it into the position i of the schedule σ . The new schedule remains feasible because the job Jσ (i) has non-negative contributions. Repeating the process, we can come up with a feasible schedule whose first
kjobs are the same as those in σ , which is, however, infeasi-ble. A contradiction arises. Therefore, no feasible schedule can exist if AlgorithmWstops without a feasible schedule
constructed.
To analyze the performance ratio, we let σ be the sched-ule obtained by Algorithm W and let σ∗ be an optimal schedule.
Lemma 2 Vt(σ∗)≤ V2t(σ )for all t= 1, . . . , n2. Proof Note that for a specific t , Vt(σ∗)= V0+
t i=1δσ∗(i) and V2t(σ )= V0+ 2t i=1δσ (i). To show t i=1δσ∗(i) ≤ 2t
i=1δσ (i) for any t = 1, . . . , n2, we prove by induction on t that for any t= 1, . . . , n2 there is a one-to-one map-ping ft : {δσ∗(1), . . . , δσ∗(t )} → {δσ (1), . . . , δσ (2t)} such that
δσ∗(i)≤ ft(δσ∗(i))for all 1≤ i ≤ t.
For t= 1, define f1(δσ∗(1))= max{δσ (1), δσ (2)} ≥ δσ∗(1). Assume there exists a mapping ft as specified for some t , 1≤ t < n2. It follows that Vt(σ∗)≤ V2t(σ ) for this t. Consider the case for t+ 1. Define ft+1(δσ∗(i))= ft(δσ∗(i)) for all 1 ≤ i ≤ t. If δσ∗(t+1) ≤ δσ (2t+2), then define
ft+1(δσ∗(t+1))= δσ (2t+2), and the proof is complete. On the other hand, if δσ∗(t+1)> δσ (2t+2), then by the logic of Algorithm W, in the schedule σ the job Jσ∗(t+1) should have been scheduled earlier than the job Jσ (2t+2), in other words, δσ∗(t+1)= δσ (j ) for some j, 1≤ j ≤ 2t + 1. Let job Jσ∗(k) be the job satisfying ft+1(δσ∗(k))= δσ (j ). If δσ∗(k)≤ δσ (2t+2), then redefine ft+1(δσ∗(k))= δσ (2t+2), and define ft+1(δσ∗(t+1))= δσ (j ) (δσ (j )= δσ∗(t+1)). The map-ping ft+1satisfies the required criterion, and thus the proof is complete. If, however, δσ∗(k) > δσ (2t+2), then repeat the above process to redefine the mapping ft+1. Because |{δσ∗(1), δσ∗(2), . . . , δσ∗(t+1)}| < |{δσ (1), δσ (2), . . . , δσ (2t+2)}| and we never use any element of σ∗twice, the process will terminate with a mapping ft+1as specified. Theorem 4 AlgorithmWfor the RPWT problem with δi≥ 0 and pi= 1 has a performance ratio of 2.
Proof Given the job weights wσ∗(1), wσ∗(2), . . . , wσ∗(n) along the positions in an optimal schedule σ∗, we have the weighted completion time Z(σ∗)=nk=1kwσ∗(k). The following discussion will construct a sequence σ out of σ such that Z(σ )≤ Z(σ )and Z(σ )≤ 2Z(σ∗). The second inequality will be established by confirming that the coef-ficient of any job weight wσ∗(k),1≤ k ≤ n2, is no greater than the coefficient of the job weight wσ (2k−1). In the proof, we focus on the coefficients of job weights and ignore the feasibility issue of the constructed intermediate sequences.
Initially, let σ = σ . For each k from n2 down to 1, we consider two cases:
Case 1. The job Jσ∗(k) with the weight wσ∗(k) occupies a position from{1, 2, . . . , 2k − 1} in σ .
In this case, the coefficient of weight wσ∗(k) is less than 2k in σ .
Case 2. The job Jσ∗(k) with the weight wσ∗(k) occupies a position from{2k, 2k + 1, . . . , n} in σ .
From Lemma2and the fact that all jobs have non-negative contributions, the inequality
Vk−1(σ∗)≤ V2k−2(σ )≤ V2k−1(σ )
will hold. Therefore, in the execution of AlgorithmW, the job Jσ∗(k)is in the candidate list for position 2k− 1. Due to the selection logic of AlgorithmW, we know that wσ∗(k)≤ wσ (2k−1). Swapping the positions of job weights wσ∗(k)and wσ (2k−1)in σ will not decrease the total weighted comple-tion time of σ . Moreover, the coefficient of wσ∗(k) is now 2k− 1.
The above iterative process will result in a sequence σ such that Z(σ )≤ n k=1 2kwσ∗(k)= 2Z(σ∗).
Moreover, Z(σ )≥ Z(σ ) is maintained in the iterative
process. Therefore, we have Z(σ )/Z(σ∗)≤ 2 and the proof
is complete.
To examine the tightness of the performance ratio, we consider an instance with V0= 0 and J = {J1, J2, . . . , Jn}, where n is even. The jobs are defined as
αi= 0, βi= 0, wi= 1, for i = 1, 2, . . . , n 2; αi= 0, βi= 1, wi= 0, for i = n 2+ 1, n 2 + 2, . . . , n. Applied to this instance, AlgorithmWproduces the sched-ule σ= 1,n 2+ 1, 2, n 2+ 2, . . . , n 2 − 1, n
with Z(σ )= 1 + 3 + · · · + (n − 1) = n24−n. The optimal schedule of the instance is
σ∗= 1, 2, . . . ,n 2, n 2 + 1, . . . , n with Z(σ∗) = 1 + 2 + · · · + n2 = n2+2n8 . Therefore, limn→∞Z(σZ(σ )∗)= 2.
By the problem equivalence given in Theorem3, Algo-rithmWcan be applied to generate an approximate solution for the problem with δi≤ 0 and wi= 1.
4 Conclusion
In this paper, we have considered the minimization of the weighted sum of completion times in the relocation prob-lem, which is a generalized resource-constrained scheduling problem. Four restricted cases were shown to be strongly NP-hard. In the proof, we have also introduced the equiv-alence between the UET case and the unit-weighted case. A polynomial-time approximation algorithm with a perfor-mance ratio of 2 was presented for the special case with
δi ≥ 0 and pi = 1. We gave an instance to establish the tightness of the ratio. By the equivalence conveyed by The-orem3, the approximation result can be applied to the spe-cial case with δi ≤ 0 and wi = 1. An intriguing phenom-enon is that the proof techniques used to establish the per-formance ratio cannot be modified to the two other cases:
δi≤ 0 and pi = 1, and δi≥ 0 and wi = 1. It could be in-teresting to develop approximation algorithms for these two
special cases. Moreover, investigating the approximability or inapproximability of the general version of the RPWT problem could be another research topic.
Acknowledgements This research was partially supported by a Russia–Taiwan joint grant under contract numbers NSC95-2416-H-009-013 (Taiwan) and RFBR08-06-92000-HHCa (Russia).
The authors are grateful for the reviewers’ comments which im-proved the presentation of an earlier version of this paper.
Appendix A: Detailed proof of Theorem1
Assume that the connecting jobs start in increasing order of their indices. If it is not the case, we let Ji and Jj be the two connecting jobs with i < j and Jj preceding Jiin some optimal schedule σ . We swap their positions. Since their net contributions and weights are equal to zero, the swap will not change the amount of resource over time and the value of the objective function. So we just need to be sure that
Ji and Jj are still available in new positions. Since all jobs have non-positive contributions, the amount of resource will not increase over time. Further, the inequality αi≥ αj holds and Jiis available in σ . It follows that both jobs are available after the swap.
IF Let the sets A1, A2, . . . , Ambe a partition as specified in the ON3M problem. Let πl be a permutation of integers of Alin increasing order. Define the sequence
σ0= π1,3m+ 1, 3m + 2, 3m + 3, π2, . . . , πl,3(m+ l − 1) + 1, 3(m + l − 1) + 2, 3(m + l − 1) + 3, πl+1, . . . , πm .
It is easy to verify that the schedule defined by the integers in σ0 as job indices is feasible. Now we calculate the to-tal weighted completion time, Z(σ0),of the schedule. Here-after, subscripts enclosed by brackets are positional indices for a particular schedule. Because the jobs indexed by the elements of set Alare processed from time 6(l− 1) to time 6l− 3 and the weights of all connecting jobs are 0, we have
Z(σ0)= m l=1 3 k=1 6(l− 1) + k(η+ x[6(l−1)+k]) = m l=1 (18l− 12)η + m l=1 3 k=1 6(l− 1)x[6(l−1)+k] + m l=1 3 k=1 kx[6(l−1)+k].
Note that the indices 6(l − 1) + 1, 6(l − 1) + 2 and 6(l− 1) + 3 correspond to the integers of the set Al.Thus,
we have 3k=1x[6(l−1)+k]= B for all l = 1, . . . , m. It fol-lows that Z(σ0)= 9m(m + 1)η − 12mη + 6 m l=1 (l− 1)B + m l=1 3 k=1 kx[6(l−1)+k] = 3mη(3m − 1) + 3mB(m − 1) + m i=1 3 k=1 kx[6(l−1)+k].
When σ0was constructed, each element ofAk, k= 1, 2, or 3, corresponds to some position 6(l− 1) + k, l = 1, 2, . . . , m in σ0. It follows that for a specific k,ml=1kx[6(l−1)+k]= ki∈Akxi, and we obtain Z(σ )= Z∗.
ONLY IF Let σ be a feasible schedule of the instance
I such that Z(σ )≤ Z∗. Firstly, we show that connecting
J3(m+l−1)+k does not start after time 6l+ k − 4, 1 ≤ l ≤
m− 1, 1 ≤ k ≤ 3. Let J3(m+l−1)+k be some connecting job that starts at time τ > 6l+ k − 4. In the schedule σ, ex-actly 3(l− 1) + k − 1 connecting jobs will complete before
J3(m+l−1)+k. It follows that at least 3l+ 1 basic jobs com-plete before τ . Then,
Vτ ≤ (3mB + B)m − (3l + 1)mB
= (3mB + B)(m − l) − B(m − l) < (3mB+ B)(m − l).
The last strict inequality follows from l≤ m − 1. Thus,
α3(m+l−1)+k= (3mB + B)(m − l) > Vτ, and we get a con-tradiction to the feasibility of schedule σ . Consequently, for all l and k, 1≤ l ≤ m − 1, 1 ≤ k ≤ 3, the completion time of job J3(m+l−1)+k must be less than or equal to 6l+ k − 4. Let J3(m+l−1)+kbe some connecting job which starts at time
τ <6l+ k − 4 in the schedule σ . We move J3(m+l−1)+k to time 6l+k −4 and shift all jobs between τ +1 and 6l +k −4 one unit earlier. Because α3(m+l−1)+k = β3(m+l−1)+k and
w3(m+l−1)+k = 0, the move will not increase the objective function value.
Thus, we proved that the connecting jobs must oc-cupy time intervals[6l − 3, 6l), l = 1, . . . , m − 1 and, cor-respondingly, the basic jobs must occupy time intervals
[6(l − 1), 6l − 3), l = 1, . . . , m. It is clear that each interval [6l − 6, 6l − 3) contains exactly three jobs J[6l−6], J[6l−5],
and J[6l−4]. Let Bl= 3
k=1x[6(l−1)+k]. Next we show that s l=1 Bl= s l=1 3 k=1 x[6(l−1)+k]≤ sB (1)
for all s= 1, . . . , m. Let s be some index such that (1) does not hold. Then,
V6s −3= V0− s l=1 3 k=1 α[6(l−1)+k] = (3mB + B)m − 3s mB+ s l=1 3 k=1 x[6(l−1)+k] < (3mB+ B)(m − s ).
It is a contradiction to the feasibility of the schedule σ because the connecting job J3(m+s −1)+1 requires (3mB+
B)(m− s )units of the resource. Finally, we have Z(σ )= m l=1 3 k=1 6(l− 1) + k(η+ x[6(l−1)+k]) = 9m(m + 1)η − 12mη + m l=1 3 k=1 6(l− 1)x[6(l−1)+k] + m l=1 3 k=1 kx[6(l−1)+k].
The second termml=13k=16(l− 1)x[6(l−1)+k]can be fur-ther elaborated as follows:
m l=1 3 k=1 6(l− 1)x[6(l−1)+k] = 6 m l=1 (l− 1)Bl = 6 m l=1 mBl− m l=1 (m− l + 1)Bl = 6m2B−mB1+ (m − 1)B2+ · · · + Bm = 6 m2B− m l=1 Bl+ m−1 l=1 Bl+ · · · + 1 l=1 Bl ≥ 6m2B− 6(mB + (m − 1)B + · · · + B) (by (1)) = 3mB(m − 1).
Incorporating the inequality into the equation of Z(σ ), we get Z(σ )≥ 3mη(3m − 1) + 3mB(m − 1) + m l=1 3 k=1 kx[6(l−1)+k]. (2)
The last term in (2) is at least i∈A1xi + 2
i∈A2xi +
3i∈A
3xi. Indeed, the equality
m l=1 3 k=1 kx[6(l−1)+k]= i∈A1 xi+ 2 i∈A2 xi+ 3 i∈A3 xi
holds if for all l, the job J[6(l−1)+1] corresponds to some el-ement from the set A1, the job J[6(l−1)+2] corresponds to some element from the setA2, and the job J[6(l−1)+3] corre-sponds to some element from the setA3. It follows that
Z≥ 3mη(3m − 1) + 3mB(m − 1) + i∈A1 xi + 2 i∈A2 xi+ 3 i∈A3 xi= Z∗.
Equations (1) and (2) together imply that Z= Z∗if and only ifsl=1Bl= sB for all s, 1 ≤ s ≤ m. It follows that Bl = B, 1 ≤ l ≤ m, and the corresponding instance of the Ordered Numerical 3-Dimensional Matching problem has the required partition. The proof is complete.
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