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<ul><li><p>Optim LettDOI 10.1007/s11590-012-0535-z</p><p>ORIGINAL PAPER</p><p>Single machine group scheduling with time and positiondependent processing times</p><p>Xue Huang Ming-Zheng Wang</p><p>Received: 27 April 2011 / Accepted: 31 July 2012 Springer-Verlag 2012</p><p>Abstract In the paper two resource constrained single-machine group schedulingproblems with time and position dependent processing times are considered. By timeand position dependent processing times and group technology assumption, we meanthat the processing time of a job is defined by the function of its starting time andposition in the group, and the group setup times of a group is a positive strictly decreas-ing continuous function of the amount of consumed resource. We present polynomialsolutions for the makespan minimization problem under the constraint that the totalresource consumption does not exceed a given limit, and the total resource consump-tion minimization problem under the constraint that the makespan does not exceed agiven limit, respectively.</p><p>Keywords Scheduling Learning effects Resource allocation Group technology</p><p>1 Introduction</p><p>In classical scheduling theory, it is assumed that the job processing times are fixed andconstant. In practice, however, we often encounter settings in which job processingtimes may be subject to change due to the phenomenon of deterioration and/or learn-ing and/or resource allocation [26,912,1620,27]. Extensive surveys of differentscheduling models and problems involving deteriorating jobs and/or learning can be</p><p>X. Huang (B)School of Science, Shenyang Aerospace University, Shenyang 110136, Chinae-mail: huangxuebj@126.com</p><p>M.-Z. WangSchool of Management Science and Engineering, Dalian University of Technology,Dalian 116024, China</p><p>123</p></li><li><p>X. Huang, M.-Z. Wang</p><p>found in Alidaee and Womer [1], Cheng et al. [8], Gawiejnowicz [13] and Biskup [7].An extensive survey of research related to scheduling with resource allocation wasprovided by Shabtay and Steiner [29].</p><p>On the other hand, the production efficiency can be increased by grouping variousparts and products with similar designs and/or production processes. This phenomenonis known as group technology in the literature. Group technology that groups similarproducts into families helps increase the efficiency of operations and decrease therequirement of facilities [21,25,28,30,35]. To the best of our knowledge, only a fewresults concerning scheduling problems with deteriorating jobs and/or learning and/orresource allocation and group technology simultaneously are known. Kuo and Yang[22] considered the single-machine group scheduling problem with a time-dependentlearning effect. They showed that the single-machine group scheduling problem witha time-dependent learning effect remains polynomially solvable for the objectives ofminimizing the makespan and minimizing the total completion time. Wu et al. [37]and Wu and Lee [36] considered a situation where the setup time grows and jobs dete-riorate as they wait for processing, i.e., group setup times and job processing times areboth described by a linear deterioration function. For single machine group schedul-ing, they proved that the makespan minimization problem can be solved in polynomialtime. Wang et al. [33] considered the single machine makespan minimization prob-lem and total completion time minimization problem with deteriorating jobs and grouptechnology. In addition, the group technology assumes that the setup times are startingtime-dependent, and the actual processing time of a job is a proportional linear functionof its starting time. They showed that the problems can be solved in polynomial time.Wang et al. [31] considered the single machine makespan minimization problem withdeteriorating jobs and group technology. In addition, the group technology assumesthat the setup times are starting time-dependent, and the actual processing time ofa job is a general linear function of its starting time. They showed that the problemcan be solved in polynomial time. Lee and Wu [23] considered single-machine groupscheduling problems with position-based learning effect in which the learning effectnot only depends on the job position, but also depends on the group position. Theyshowed that the makespan and the total completion time problems remain polynomi-ally solvable under the proposed model. Yan et al. [38] considered the single-machinescheduling problems with the effects of learning and deterioration under the grouptechnology consumption. They proved that the single-machine makespan minimiza-tion problem and total resource minimization problem under the group technologyconsumption remain polynomially solvable. Huang et al. [15] considered two single-machine group scheduling problems with both learning effects and deteriorating jobssimultaneously. For both resource constrained problems, they proved that the prob-lems can be solved in polynomial time. Wang et al. [32] considered a single machinescheduling problem with deteriorating jobs and group technology assumption. Theyshowed that the makespan minimization problem with ready times can be solved inpolynomial time under an agreeable condition.</p><p>In this paper we study resource allocation scheduling problems with time andposition dependent processing times under group consumption, and the setup timeof each group is dependent on the resource it consumes. Motivated by observationsin March [24] and Mukhopadhyay et al. [26], slack leads to unintentional learning.</p><p>123</p></li><li><p>Single machine group scheduling</p><p>When the setup times of groups are processed, a worker (i.e., a machine) has slacktime to explore different ways to improve performance (i.e., the time dependent learn-ing effect). The remaining part of this paper is organized as follows. In Sect. 2 wedescribe and formulate the problem. In Sects. 3 and 4 we consider two resource-dependent scheduling problems and present polynomial algorithms to solve them.The last section includes conclusions.</p><p>2 Problem formulation</p><p>There are n jobs grouped into m groups, and these n jobs are to be processed on asingle machine. All the jobs are available at time 0. A setup time is incurred beforethe process of jobs in each group. No setup time is required between jobs in the samegroup and jobs in the same group must be processed consecutively. The machine canhandle one job at a time and job preemption is not allowed. Let ni be the number ofjobs belonging to group Gi , thus, n1 + n2 + + nm = n. Let Ji j denote the j th jobin group Gi , i = 1, 2, . . . , m; j = 1, 2, . . . , ni . Let pi jk(t) be the (actual) processingtime of job Ji j if it is started at time t and scheduled in position k in the group Gi , si bethe (actual) setup time of group Gi . As in Zhang et al. [39], we consider the followingmodel</p><p>pi jk(t) = pi j (1 t)k1i , (1)</p><p>where pi j denotes the basic (normal) processing time of job Ji j , 0 < 1 and0 < i 1 is a constant learning index of group Gi . We assume that the setup timeof group Gi is:</p><p>si = f (ui ), u ui u, i = 1, 2, . . . , m, (2)</p><p>where ui is the amount of a doubly-constrained non-renewable resource allocated togroup Gi , with u ui u, u and u are known technological constraints (withoutloss of generality, we assume u = 0) and f : R+ R+ (R+ is the set of non-negative real numbers) is a strictly decreasing continuous function with f (u) 0.The values of these parameters can be estimated empirically. In practical setting,the value of i should not be very small so the processing times of jobs decreaserather slowly along the sequence. We also reasonably assume that each job has apositive processing time, so we have the requirement (m f (u) + mi=1</p><p>nij=1 pi j </p><p>mini=1,2...,m, j=1,2,...,ni pi j ) < 1, which implies that is fairly small number [34,39].Let u = (u1, u2, . . . , un) be the resource allocation vector satisfying the resourceconstraint (i.e., 0 u j u, j = 1, 2, . . . , n,nj=1 u j U, where U is the totalamount of the resource available).</p><p>For a given schedule , let Ci j () be the completion time of job Ji j in groupGi under schedule . Cmax = max{Ci j |i = 1, 2, . . . , m; j = 1, 2, . . . , ni }represents makespan of a given schedule. We consider the problem of minimizingthe makespan with a resource consumption constraint and the total resource consump-tion with a makespan constraint. In the three-field notation scheme || introduced</p><p>123</p></li><li><p>X. Huang, M.-Z. Wang</p><p>by Graham et al. [14], the two problems are denoted as 1|si = f (ui ), pi jk(t) =pi j (1 t)k1i , GT,</p><p>nj=1 u j U |Cmax and 1|si = f (ui ), pi jk(t) = pi j (1 </p><p>t)k1i , GT, Cmax C |</p><p>u j , respectively.</p><p>3 Minimizing the makespan</p><p>We first give a lemma, which will be applied further on.</p><p>Lemma 1 For a given schedule = [Ji1, Ji2, . . . , Ji,ni ] of 1|pi jk(t) = pi j (1 t)k1i |Cmax, if the first job starts at time t0 0, then the completion time Ci j of jobJi j is equal to</p><p>Ci j = (t0 1</p><p>)</p><p>j</p><p>l=1</p><p>(1 pill1i</p><p>)+ 1</p><p>, j = 1, 2, . . . , ni . (3)</p><p>Proof :(by induction).</p><p>Ci1 = t0 + pi1(1 t0)11i =(</p><p>t0 1</p><p>)(1 pi111i</p><p>)+ 1</p><p>,</p><p>Ci2 = C1 + pi2(1 Ci1)21i =(</p><p>t0 1</p><p>)(1pi111i</p><p>) (1pi221i</p><p>)+ 1</p><p>.</p><p>Suppose Lemma 1 holds for job Ji j , i.e.,</p><p>Ci j =(</p><p>t0 1</p><p>) j</p><p>l=1</p><p>(1 pill1i</p><p>)+ 1</p><p>.</p><p>Consider job J j+1.</p><p>Ci, j+1 = Ci j + pi, j+1(1 Ci j ) ji =(</p><p>t0 1</p><p>) j+1</p><p>l=1</p><p>(1 pill1i</p><p>)+ 1</p><p>.</p><p>Hence, Lemma 1 holds for Ji, j+1. This completes the proof of Lemma 1. </p><p>Theorem 1 For a given schedule = [G1, G2, . . . , Gm] = [J11,J12, . . . , J1,n1 , J21, J22, . . . , J2,n2 , . . . , Jm1, Jm2, . . . , Jm,nm ]of 1|si = f (ui ), pi jk(t)= pi j (1 t)k1i , GT |Cmax, the makespan is equal to</p><p>Cmax =m</p><p>i=1</p><p>si</p><p>m</p><p>l=i</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)</p><p> 1</p><p>m</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1 pi j j1i</p><p>)+ 1</p><p>. (4)</p><p>123</p></li><li><p>Single machine group scheduling</p><p>Proof (by induction). From Lemma 1, the completion times of groups G1 and G2 are</p><p>C1,n1 =(</p><p>s1 1</p><p>) n1</p><p>j=1</p><p>(1 p1 j j11</p><p>)+ 1</p><p>,</p><p>C2,n2 =(</p><p>C1,n1 + s2 1</p><p>) n2</p><p>j=1</p><p>(1 p2 j j12</p><p>)+ 1</p><p>=(</p><p>s1 1</p><p>) n1</p><p>j=1</p><p>(1 p1 ja j11</p><p>) n2</p><p>j=1</p><p>(1 p2 j j12</p><p>)</p><p>+s2n2</p><p>j=1</p><p>(1 p2 j j12</p><p>)+ 1</p><p>=2</p><p>i=1</p><p>si</p><p>2</p><p>l=i</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)</p><p> 1</p><p>2</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1 pi j j1i</p><p>)+ 1</p><p>.</p><p>Suppose Theorem 1 holds for group Gk , i.e.,</p><p>Ck,nk =k</p><p>i=1</p><p>si</p><p>k</p><p>l=i</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)</p><p> 1</p><p>k</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1 pi j j1i</p><p>)+ 1</p><p>.</p><p>Con sider group Gk+1.</p><p>Ck+1,nk+1</p><p>=(</p><p>Ck,nk + sk+1 1</p><p>) nk+1</p><p>j=1</p><p>(1 pk+1, j j1k+1</p><p>)+ 1</p><p>=</p><p>k</p><p>i=1</p><p>si</p><p>k</p><p>l=i</p><p>nl</p><p>j=1</p><p>(1 pl ja j1l</p><p>)</p><p> 1</p><p>k</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1 pi j j1i</p><p>)+ 1</p><p>+sk+1 1</p><p>]</p><p>nk+1</p><p>j=1</p><p>(1 pk+1, j j1k+1</p><p>)+ 1</p><p>=k+1</p><p>i=1</p><p>si</p><p>k+1</p><p>l=i</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)</p><p> 1</p><p>k+1</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1 pi j j1i</p><p>)+ 1</p><p>.</p><p>Hence, Theorem 1 holds for group Gk+1. This completes the proof of Theorem 1.unionsq</p><p>Theorem 2 For the problem 1|si = f (ui ), pi jk(t) = pi j (1 t)k1i , GT |Cmax,the optimal job sequence within a group is arranged jobs in non-decreasing order ofpi j (i.e., the smallest processing time (SPT) first rule).</p><p>123</p></li><li><p>X. Huang, M.-Z. Wang</p><p>Proof : The result can be obtained by a standard pair-wise interchange technique (see[39]). unionsq</p><p>Theorem 3 For any given schedule of the groups of the problem 1|si = f (ui ), pi jk(t)= pi j (1 t)k1i , GT |Cmax, the groups scheduled in the later position should begiven the priority in resource allocation decisions.</p><p>Proof : Without loss of generality, we assume = [G1, G2, . . . , Gm] is an anyschedule of all the groups. Based on (4), first allocate the resource to the setup time siwith bigger</p><p>ml=i</p><p>nlj=1(1 pl j j1l ), the makespan can be minimized by the same</p><p>amount of resource. Obviouslyni</p><p>j=1(1 pi j j1i ) 1. Thus</p><p>m</p><p>l=1</p><p>nl</p><p>j=1</p><p>(1pl j j1l</p><p>)</p><p>m</p><p>l=2</p><p>nl</p><p>j=1</p><p>(1pi j j1l</p><p>) </p><p>m</p><p>l=m</p><p>nl</p><p>j=1</p><p>(1pl j j1l</p><p>).</p><p>Then, the groups scheduled in the later position should be given the priority to resourceallocation to minimize the objective Cmax. unionsq</p><p>Theorem 4 For the problem 1|si = f (ui ), pi jk(t) = pi j (1 t)k1i , GT |Cmax, ifresource amount of the groups in each position is fixed, and in each group the jobsare scheduled in non-decreasing order of their normal processing time, then the opti-mal schedule can be obtained by scheduling the groups in non-increasing order ofGi =</p><p>nij=1(1 pi j j1i ).</p><p>Proof : Without loss of generality, we assume that the resource amount of the groupscheduled at the i th position is denoted as u[i](i = 1, 2, . . . , m), Gv and Gv+1 aretwo adjacent groups in the schedule = [G1, G2, . . . , Gv, Gv+1, . . . , Gm]. If wechange the sequence of Gv and Gv+1, then we can obtain a new schedule =[G1, G2, . . . , Gv+1, Gv, . . . , Gm]. In , Gv is scheduled in the (v + 1)th position,then the resource allocation amount of Gv is u[v+1]; Gv+1 is scheduled in the vthposition, then the resource allocation amount of Gv+1 is u[v]. Then the objective valueof and are</p><p>Cmax() = f (u[1])m</p><p>l=1</p><p>nl</p><p>j=1</p><p>(1pl j j1l</p><p>)+ f (u[2])</p><p>m</p><p>l=2</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)+ </p><p>+ f (u[v])m</p><p>l=v</p><p>nl</p><p>j=1</p><p>(1pl j j1l</p><p>)+ f (u[v+1])</p><p>m</p><p>l=v+1</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)</p><p>+ + f (u[m])m</p><p>l=m</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>) 1</p><p>m</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1 pi ji j1</p><p>)</p><p>+ 1</p><p>.</p><p>123</p></li><li><p>Single machine group scheduling</p><p>and</p><p>Cmax( ) = f (u[1])m</p><p>l=1</p><p>nl</p><p>j=1</p><p>(1pl j j1l</p><p>)+ f (u[2])</p><p>m</p><p>l=2</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)+ </p><p>+ f (u[v])m</p><p>l=v</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)</p><p>+ f (u[v+1])nv</p><p>j=1</p><p>(1 pv j j1v</p><p>) m</p><p>l=v+2</p><p>nl</p><p>j=1</p><p>(1 pl j j1l</p><p>)+ </p><p>+ f (u[m])m</p><p>l=m</p><p>nl</p><p>j=1</p><p>(1 pl jl j1</p><p>) 1</p><p>m</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1 pi j j1i</p><p>)</p><p>+ 1</p><p>.</p><p>Then, we have</p><p>Cmax() Cmax( )</p><p>= f (u[v+1])m</p><p>l=v+2</p><p>nl</p><p>j=1</p><p>(1 pl jl j1</p><p>)</p><p>nv+1</p><p>j=1</p><p>(1 pv+1, jv+1 j1</p><p>)</p><p>nv</p><p>j=1</p><p>(1 pv jv j1</p><p>)</p><p> 0</p><p>if and only ifnv+1</p><p>j=1 (1 pv+1, jv+1 j1) nv</p><p>j=1(1 pv jv j1). This completesthe proof. unionsqBased on Theorems 2, 3 and 4, we propose an algorithm to solve the problem 1|si =f (ui ), pi jk(t) = pi j (1 t)k1i , GT,</p><p>nj=1 u j U |Cmax.</p><p>Algorithm 1.Step 1. Jobs in each group scheduled in non-decreasing order of pi j , i.e.,</p><p>pi1 pi2 pi,ni , i = 1, 2, . . . , m.</p><p>Step 2. Groups scheduled in non-increasing order of Gi =ni</p><p>j=1(1pi ji j1),i.e.,</p><p>G1 G2 Gm .</p><p>Step 3. For the obtained schedule = [G1, G2, . . . , Gm], set ui = 0(i =1, 2, . . . , m) and l = m.</p><p>Step 4. Set ul = min{u, U }, U = U ul and l = l 1.Step 5. If U = 0 or l = 0, stop; else go to Step 4.Obviously the total time for Algorithm 1 is O(n log n).</p><p>123</p></li><li><p>X. Huang, M.-Z. Wang</p><p>4 Minimizing the resource consumption</p><p>From the results of Sect. 3, for the problem 1|si = f (ui ), pi jk(t) = pi j (1 t)k1i , GT, Cmax C |</p><p>u j , we also only need to consider the schedule in which</p><p>the jobs in each group scheduled in non-decreasing order of pi j (i.e., the SPT rule), thegroups scheduled in non-increasing order of Gi =</p><p>nij=1(1 pi ji j1) and the</p><p>groups scheduled in the later position should be given the priority to resource alloca-tion.</p><p>From (4), the minimum makespan of the schedule is</p><p>Cmax =m</p><p>i=1</p><p> f (u)m</p><p>l=i</p><p>nl</p><p>j=1</p><p>(1 pl ja j1l</p><p>)</p><p> 1</p><p>m</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1 pi ji j1</p><p>)+ 1</p><p>.</p><p>So, a sequence is feasible only if</p><p>Cmax =m</p><p>i=1</p><p> f (u)m</p><p>l=i</p><p>nl</p><p>j=1</p><p>(1pl jl j1</p><p>)</p><p> 1</p><p>m</p><p>i=1</p><p>ni</p><p>j=1</p><p>(1pi ji j1</p><p>)+ 1</p><p> C.</p><p>Since the premise is minimizing the total resource consumption with a makespanconstraint, resource consumption of groups should be given as few as possible and thecompletion time of the last group is C , i.e., Cmax = C .</p><p>Without loss of generality, let = [G1, G2, . . . , Gm] = [J11, J12, . . . , J1,n1 , J21,J22, . . . , J2,n2 , . . . , Jm...</p></li></ul>