Single machine scheduling with deadlines and increasing rates of processing times

Acta Informatica 36, 673–692 (2000)

c© Springer-Verlag 2000

Single machine scheduling with deadlinesand increasing rates of processing times

T.C.E. Cheng, Q. Ding

The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong,(e-mail: [email protected])

Received: 17 November 1997 / 24 August 1999

Abstract. The makespan, flow time and maximum lateness problems ofscheduling a set of tasks with deadlines and increasing rates of processingtimes on a single machine are considered in this paper. We first show that,when the increasing rates of processing time are identical, the makespanproblem is equivalent to the corresponding flow time problem. Both prob-lems are solvable inO(n5) time by a dynamic programming algorithm. Asan application of the dynamic programming algorithm, we demonstrate thatthe corresponding maximum lateness problem can be solved inO(n6 log n)time. We then show that the general makespan problem is strongly NP-complete. Thus, both the corresponding flow time problem and maximumlateness problem are also strongly NP-complete.

1 Introduction

Machine scheduling problems with time dependent processing times havereceived increasing attention in recent years. Formally, these time-dependentproblems can be stated as follows. A task system consists ofn independenttasks and is denoted byTS = (Ti, di, ai, wi). Each taskTi isassociated with a deadlinedi and characterized by a normal processing timeai ≥ 0 and a decreasing / increasing processing ratewi > 0, dependingon the task starting timesi. The actual processing time of taskTi is pi =ai ± wisi ≥ 0. Similar to the classical scheduling problems,ai and di

are assumed to be integers. For all tasks, the release time is0. Since theprocessing rates are not integers in many practical cases,wi is allowed tobe a rational number.

674 T.C.E. Cheng, Q. Ding

For a given scheduleS, let si(S) denote the starting time ofTi. Thus,the actual processing time ofTi is given bypi(S) = ai ± wisi(S) ≥ 0and the completion time ofTi is given byci(S) = (1 ± wi)si(S) + ai. Anonpreemptive schedule isfeasibleif eachTi is completely processed in theinterval [0, di]. A task system isfeasibleif there is a feasible schedule forit. Let G ≥ 0 denote a given threshold. The makespan problem is to decidewhether there is a feasible schedule forTS on a single machine withCmax =maxTi∈TS

ci(S) ≤ G. Adopting the three-field notation proposed by Graham

et al [5] to describe classical scheduling problems, we denote the makespanproblem as1/pi = ai ±wisi, di/Cmax. Similarly, the flow time problem isto decide whether there is a feasible schedule forTS on a single machinewith

∑Ti∈TS ci(S) ≤ G and is denoted by1/pi = ai±wisi, di/ΣCi. The

lateness problem is to decide whether there is a feasible schedule forTS ona single machine withLmax = max

Ti∈TSci(S) − di ≤ G and is denoted by

1/pi = ai ± wisi, di/Lmax.Several papers have considered the makespan and flow time problems of

single machine scheduling with deadlines and time-dependent processingtimes. For the modelpi = ai − wisi ≥ 0, anO(n log n) time algorithmfor the makespan problem has been given for a set of tasks with identicaldeadlines by Ho et al [8]. Cheng and Ding (see [2] and [3]) have shownthat the case withwi = w and arbitrary deadlines is strongly NP-complete,while the case withwi = w and two distinct deadlines is NP-complete.

For the modelpi = ai + wisi ≥ 0 without deadlines, Gupta and Gupta[7] have shown that tasks arranged in nondecreasing order of the ratiosai

wiis

optimal for the makespan problem, while Browne and Yechiali [1] obtaineda similar result for the stochastic version. Mosheiov [9] considered the flowtime problem withai = a. He showed that there exists an optimal schedulethat is V-shaped with respect to the increasing processing rates; that is, thetasks appearing before the task with the smallestwi are sequenced in non-increasing order ofwi, and the ones after it are sequenced in nondecreasingorder ofwi.

In this paper, we focus on the modelpi = ai + wisi ≥ 0 with dead-lines. Our contribution is first to show that the problem withwi = w andarbitrary deadlines, denoted as1/pi = ai + wsi, di/Cmax, is equivalent tothe corresponding flow time problem, denoted as1/pi = ai +wsi, di/ΣCi,and both problems are solvable inO(n5) time by a dynamic programmingalgorithm. As an application of this dynamic programming algorithm, wedemonstrate that the corresponding maximum lateness problem, denoted as1/pi = ai+wsi, di/Lmax, can be solved inO(n6 log n) time. We then showthat the general makespan problem, denoted as1/pi = ai +wisi, di/Cmax,is strongly NP-complete. Thus, both the corresponding flow time problem,

Single machine scheduling with deadlines 675

denoted as1/pi = ai + wisi, di/ΣCi, and the maximum lateness problem,denoted as1/pi = ai + wisi, di/Lmax, are strongly NP-complete.

This class of scheduling problems has many real-world applications, in-cluding fire fighting, steel production, scheduling of resources to controlepidemics, and maintenance scheduling, where any delay may result in anadditional effort (time, cost, etc.) to accomplish a task (see, for example,[1], [4], [9], [10], [11]). When the additional effort depends only on circum-stances, e.g. the machine, it is reasonable to use a model with an identicalprocessing rate to approximate such a situation.

2 Preliminaries

We will make a distinction between a sequence and a schedule defined inthis paper. A sequence is an order for a group of tasks (i.e. a task order)with arbitrary task starting times. A schedule is a sequence withsi+1 = Ci.For a given task orderT1, T2, · · · , Tn, we will use(T1, T2, · · · , Tn) and[T1, T2, · · · , Tn] to denote the sequence and the schedule, respectively, ofthe task order.

Given a scheduleS for an instance of1/pi = ai +wsi, di/Cmax, letCi,ai andpi denote the completion time, normal processing time and actualprocessing time of the task in thei-th position ofS, respectively. Lettingα = 1 + w, we obtain

Ci = ai + αai−1 + α2ai−2 + · · ·+ αi−1a1 =i∑

k=1

αi−kak. (1)

Let si denote the starting time of thei-th task inS. Sincepi = ai +wsi,Ci = αsi + ai, we obtain

Ci =i∑

j=1

pj

=i∑

j=1

aj + wi∑

j=1

sj

=i∑

j=1

aj + w

i∑j=1

1α

(Cj − aj)

=1

1 + w

i∑j=1

aj +w

1 + w

i∑j=1

Cj . (2)

In the sense that an algorithm for one problem can be used to solve the otherproblem, the following theorem is readily established.


Theorem 1. The problem1/pi = ai + wsi, di/Cmax is equivalent to theproblem1/pi = ai + wsi, di/ΣCi.

Let D(i) denote the set of tasks among the firsti tasks inS whosedeadlines are no less than the completion time ofTi, i.e.D(i) = Tj |Ci ≤dj for 1 ≤ j ≤ i, for 1 ≤ i ≤ n. If aj ≤ al, for eachj < l ≤ i, Tj ∈ D(i),Tl ∈ D(i) and1 ≤ i ≤ n, thenS is called acanonical schedule. Based onthe argument of interchanging two tasks and the principle of optimality fordynamic programming, the following theorem is easy to verify.

Theorem 2. For an instance of1/pi = ai + wsi, di/Cmax, each optimalschedule[T1, T2, · · · , Tn] is canonical and the corresponding partial sched-ule [T1, T2, · · · , Ti] is also optimal for the tasks inT1, T2, · · · , Ti.

3 Dynamic program

Given an instanceI of 1/pi = ai + wsi, di/Cmax, without loss of thegenerality, assume thatai 6= aj , for eachi 6= j. Now we introduce adynamic program forI as follows.

Construct an initial scheduleΦ = [T[1], T[2], · · · , T[n]] by the ShortestProcessing Time (SPT) rule, where the subscript[i] denotes the index of thetask in thei-th position inΦ. Let C[i] anda[i] denote the completion timeand normal processing time of thei-th taskT[i] in Φ, respectively. From (1),

we obtain thatC[i] =∑i

j=1(1 + w)i−ja[j].We need to generate fromΦ an iterative sequenceΦ′ = (T(1), T(2), · · · ,

T(n)), where the subscript(i) denotes the index of the task in thei-th positionin Φ′. Let s(i) anda(i) denote the starting time and normal processing timeof thei-th taskT(i) in Φ′, respectively. By induction, we can determine thei-th choice pointt(i), which is a means to help determine thei-th taskT(i)in Φ′, from n to 1.

First, sets(n+1) = C(n). We determinet(n) = maxs(n+1), C(n) =s(n+1). From among all tasksTj with dj ≥ t(n), choose the taskT(n)with the largest normal processing time. This leaves a problem instancewith n − 1 tasks. Sinces(n) = t(n)−a(n)

1+w , t(n−1) can be determined asmaxs(n), C[n−1]. From among the remaining tasksTj with dj ≥ t(n−1),choose a taskT(n−1) with the largest normal processing time. The proce-dure is again applied to the problem instance with then − 2 remainingtasks. Thus, we can determinet(i) andT(i) from n − 2 to 1. If there ex-ists no task that can be chosen at a choice point or if the condition that∑n

j=1(1+w)n−ja(j) = C[n] is satisfied after the entire sequenceΦ′ is gen-erated, then stop. ReplaceΦ = [T(1), T(2), · · · , T(n)] by Φ′ if it exists. Usethe new scheduleΦ = [T(1), T(2), · · · , T(n)] to generate the next iterativesequenceΦ′. This dynamic procedure is calledRule DP.


Given a scheduleΦ = [T[1], T[2], · · · , T[n]], generate its iterative se-quenceΦ = (T(1), T(2), · · · , T(n)) by Rule DP. AfterT(i) is determined, wedefine two task setsG(i) andH(i) as follows:

G(i) = T(i), · · · , T(n) − T[i], · · · , T[n],

H(i) = T[i], · · · , T[n] − T(i), · · · , T(n).From the symmetry of the definitions, we get the following remark im-

mediately.

Remark 1.The number of tasks inG(i) is equal to that inH(i).

Let g(i) denote the number of tasks inG(i) or H(i). If G(i) and H(i)can be partitioned intog(i) disjoint pairs of tasksTbjc andTdje such thatabjc < adje, whereabjc andadje are the normal processing times ofTbjcandTdje, respectively, forTbjc ∈ G(i), Tdje ∈ H(i) and1 ≤ j ≤ g(i),thenM(i) = Tb1c, Td1e, Tb2c, Td2e, · · · , Tbg(i)c, Tdg(i)e is called amatchingfor G(i) andH(i), for 1 ≤ i ≤ n.

If there exists an optimal schedule for an instanceI of 1/pi = ai +wsi, di/Cmax, then from the principle of optimality for dynamic program-

ming, there exists a feasible schedule for each sub-instance ofI. Let C [i]max

andC(i)max denote the makespans of the sub-instancesI[i] andI(i) for the

tasks inT[1], T[2], · · · , T[i] and T(1), T(2), · · · , T(i), respectively, for1 ≤ i ≤ n. We obtain the following lemma.

Lemma 1. Given a feasible instanceI of 1/pi = ai + wsi, di/Cmax and

a canonical scheduleΦ = [T[1], T[2], · · · , T[n]], if C[i] ≤ C[i]max, then the

iterative sequenceΦ′ = (T(1), T(2), · · · , T(n)) generated by Rule DP satis-

fiest(i) ≤ C(i)max and there exists a matchingM(i) for G(i) andH(i), for

1 ≤ i ≤ n.

Proof. Note that eitherΦ orΦ′ is not necessary to be feasible. Now we showthis lemma by induction as follows.

First, consider the casei = n. SinceI is feasible,C [n]max andC

(n)max exist

andC[n]max = C

(n)max = Cmax, whereCmax is the makespan ofI. The choice

point for this case ist(n) = C[n]. Since, from the assumption of this lemma,

C[n] ≤ C[n]max, there exists a taskT(n) with d(n) ≥ t(n) andt(n) = C[n] ≤

C[n]max = C

(n)max. If T(n) = T[n], then we obtain thatG(n) = H(n) = ∅

andM(n) = ∅ is a matching forG(n) andH(n). Otherwise, assume thatT(n) = T[l]. Note thatl < n andT(n) = T[l] ∈ D(n) in Φ. SinceΦ iscanonical, we havea(n) = a[l] < a[n]. Therefore, we obtainG(n) = T(n)andH(n) = T[n]. Moreover,M(n) = T(n), T[n] is a matching for


G(n) andH(n). Suppose that the result holds fori, k+1 ≤ i ≤ n. Considerthe casei = k in the following.

First, we showt(k) ≤ C(k)max. Since the tasks inT(k+1), T(k+2), · · · ,

T(n) have been determined, the sub-instanceI(k) = I − T(k+1), T(k+2),

· · · , T(n) is identified. Suppose thatt(k) > C(k)max, then there exists a fea-

sible scheduleΨ(k) for I(k) with makespan less thant(k) = maxs(k+1),C[k].

For the caset(k) = s(k+1), scheduleT(k+1) at the end ofΨ(k), thenthe new schedule with makespan less thant(k+1) is a feasible schedule

for I(k + 1), contradicting the conditiont(k+1) ≤ C(k+1)max . For the case

t(k) = C[k], the deadline of each task inG(k + 1) is larger thanC[k] = t(k).Note thatI[k] = I(k) − H(k + 1) + G(k + 1) and there is a matchingM(k + 1) for G(k + 1) andH(k + 1). Based on scheduleΨ(k), swapeach task inH(k + 1) with its counterpart inG(k + 1). Since each task inG(k+1) has a smaller normal processing time than that of its counterpart inH(k+1), the new schedule with makespan less thanC[k] = t(k) is a feasible

schedule forI[k], contradicting the conditionC[k] ≤ C[k]max. Therefore, we

obtaint(k) ≤ C(k)max. Besides, since there exists a feasible schedule forI(k),

there exists at least a taskTk /∈ T(k+1), T(k+2), · · · , T(n) with dk ≥ t(k).As toG(k), H(k) andM(k), there are five cases to consider.

Case 1.T(k) = T[k]. It is easy to see thatG(k) = G(k + 1) andH(k) =H(k + 1). We obtain thatM(k) = M(k + 1) is a matching forG(k) andH(k).

Case 2.T(k) ∈ H(k + 1) andT[k] /∈ G(k + 1). This is the case that wechoose a task remaining inT[k+1], · · · , T[n] as thek-th task inΦ′, whileT[k] is not inT(k+1), · · · , T(n) yet. It is easy to see thatG(k) = G(k +1)andH(k) = H(k + 1) − T(k) + T[k]. Now we match the counterpartof T(k) in M(k + 1), denoted asT(r) = T[l] ∈ G(k + 1), with T[k]. Fromthe structure ofΦ andΦ′, we havel < k < r and t(r) > t(k) ≥ C[k].SinceT(r) is chosen by Rule DP att(r) in Φ′, we haveT[l] ∈ D(k) in Φ.SinceΦ is canonical, we obtain thata[k] > a[l] = a(r). DefineM(k) =M(k + 1) − T[l], T(k) + T[l], T[k]. M(k) is a matching forG(k) andH(k).

Case 3.T(k) ∈ H(k + 1) andT[k] ∈ G(k + 1). This is the case withT(k) ∈T[k+1], · · · , T[n] − T(k+1), · · · , T(n) andT[k] ∈ T(k+1), · · · , T(n). Itis easy to see thatG(k) = G(k+1)−T[k]andH(k) = H(k+1)−T(k).Let T(r) = T[l] ∈ G(k + 1) denote the counterpart ofT(k) in M(k + 1).From the same discussion as in Case 2, we obtaina[k] > a[l] = a(r).Let T[u] ∈ H(k + 1) denote the counterpart ofT[k] in M(k + 1). Fromthe definition ofM(k + 1), we obtaina[u] > a[k] > a[l] = a(r). Define


M(k) = M(k + 1) − T[l], T(k) − T[k], T[u] + T[l], T[u]. We obtainthatM(k) is a matching forG(k) andH(k).

Case 4.T(k) = T[l], l < k andT[k] /∈ G(k + 1). This is the case choos-ing a task inT[1], · · · , T[k] as thek-th task inΦ, while T[k] is not inT(k+1), · · · , T(n) yet. It is easy to see thatG(k) = G(k + 1) + T(k)andH(k) = H(k + 1) + T[k]. Similar to the discussion in Case 2, wehavea(k) ∈ D(k) in Φ anda(k) < a[k]. Define. Then,M(k) is a matchingfor G(k) andH(k).

Case 5.T(k) = T[l], l < k andT[k] ∈ G(k + 1). This is the case withT(k) ∈ T[1], · · · , T[k] andT[k] ∈ T(k+1), · · · , T(n). It is easy to seethat G(k) = G(k + 1) − T[k] + T(k) andH(k) = H(k + 1). Thesame as Case 4, we havea(k) ∈ D(k) in Φ anda(k) < a[k]. Match thecounterpart ofT[k] in M(k + 1), denoted asT[u] ∈ H(k + 1), with T(k).DefineM(k) = M(k + 1) − T[k], T[u] + T(k), T[u]. We obtain thatM(k) is a matching forG(k) andH(k).

The scheduleΦ′ = [T(1), T(2), · · · , T(n)] corresponding to the iterative se-

quenceΦ′ obtained from Rule DP is called theiterative scheduleof Φ. LetC(i) denote the completion time ofT(i) in Φ, for 1 ≤ i ≤ n. We give thefollowing lemma.

Lemma 2. Given a feasible instanceI of 1/pi = ai + wsi, di/Cmax and

a canonical scheduleΦ = [T[1], T[2], · · · , T[n]], if C[i] ≤ C[i]max, then the

iterative scheduleΦ′ = [T(1), T(2), · · · , T(n)] obtained from applying Rule

DP is canonical andt(i) ≤ C(i) ≤ C(i)max, where1 ≤ i ≤ n. If C(n) = C[n],

thenΦ′

is an optimal schedule.

Proof. SinceΦ is a canonical and feasible schedule forI with C[i] ≤ C[i]max,

from Lemma 1, there exists an iterative sequenceΦ = (T(1), T(2), · · · , T(n))

with t(i) ≤ C(i)max. Now we constructΦ

′from Φ.

It is easy to see thatt(1) ≤ C(1)max = a(1) = C(1) and there exists no idle

time in [0, t(1)] in Φ′. Sincet(i) = maxs(i+1), C[i] ≥ s(i+1), there existsno idle time in the interval[t(1), t(n)] in Φ′. Hence, all tasks are shifted from

left to right whenΦ′

is constructed fromΦ′. Thus, we obtaint(i) ≤ C(i).If t(n) = C(n), then there is no task really shifted. We getC(i) = t(i),

for 1 ≤ i ≤ n. This means thatΦ′

is a feasible schedule with makespanC(n) = t(n) = C[n] ≤ C

[n]max. Therefore,Φ

′is an optimal schedule.

Now we showC(i) ≤ C(i)max by induction as follows. As the basic case,

consider the casei = 1. SinceI(1) = T(1) andC(1)max = a(1), we obtain

C(1) = a(1) = C(1)max. Suppose thatC(i) ≤ C

(i)max holds, for1 ≤ i ≤ k − 1.


For the casei = k, suppose that[T1, T2, · · · , Tk] is an optimal schedule forI(k). There are two cases to consider.

Case 1.T(k) = Tk: from Theorem 2,[T1, T2, · · · , Tk−1] is also an opti-

mal schedule forI(k − 1). Note thatC(k−1) andC(k−1)max are the starting

times ofT(k) = Tk in schedules[T[1], T[2], · · · , T[k]] and[T1, T2, · · · , Tk],

respectively. SinceC(k−1) ≤ C(k−1)max , we obtainC(k) ≤ C

(k)max.

Case 2.T(k) 6= Tk: Sincet(k) ≤ C(k)max, we obtain thatTk is a candidate

at t(k) in Φ′. Thus, we haveak < a(k) andd(k) ≤ C(k)max. Extendd(k) to

C(k)max. I(k) becomes a new instanceII(k). Let C(k)

max denote the makespanof II(k). Since each feasible schedule forI(k) is also feasible forII(k),we obtain thatC(k)

max ≤ C(k)max. From Theorem 2, each optimal schedule is

canonical, it is easy to verify that there exists an optimal schedule forII(k),in which T(k) is the last task. Note that the sequence, which is generatedby Rule DP with thek-th choice point att(k) for II(k), is the same as

that for I(k). From Case 1, we obtainC(k) ≤ C(k)′max. Therefore, we have

C(k) ≤ C(k)max.

For a feasible instanceI of 1/pi = ai + wsi, di/Cmax, the initial sched-ule Φ is initiated by the SPT rule. It is obvious that the SPT schedule iscanonical andC[i] ≤ C

[i]max. From Lemma 2, the iterative scheduleΦ

′is

also canonical andC(i) ≤ C(i)max. By induction, we obtain that each sched-

ule generated by Rule DP is canonical andC[i] ≤ C[i]max, for 1 ≤ i ≤ n.

For an iterative sequenceΦ′, when the choice pointt(i) and the task setT(i+1), T(i+2), · · · , T(n) are determined, similar to the definition ofD(i)in Φ, let

D′(i) = Tj |dj ≥ t(i) and Tj ∈ I − T(i+1), T(i+2), . . . , T(n)

denote the set of candidates at the choice pointt(i). Since the instance is

feasible, from Lemma 2, we havet(i) ≤ C(i)max andD(i) 6= ∅, for 1 ≤ i ≤ n.

Therefore the following lemma is established.

Lemma 3. For any instanceI of1/pi = ai +wsi, di/Cmax, if I is feasible,then the iterative scheduleΦ = [T(1), T(2), · · · , T(n)] generated by Rule DP

is always canonical withC(i) ≤ C(i)max, for 1 ≤ i ≤ n. On the other hand, if

in an iterative sequence there exists a setD(i) = ∅ in Rule DP, then thereexists no feasible schedule forI.


Now we determine the number of iterations performed by Rule DP inthe worst case. For a scheduleΦ = [T1, T2, · · · , Tn], we define a set

E = 〈Ti, Tj〉 |1 ≤ i < j ≤ nand a function

e (〈Ti, Tj〉 , Φ) =

0 ai < aj

1 ai > aj

.

Let T1, T2, · · · , Tn be the task order witha1 < a2 < · · · < anin Φ. Define

⟨Tu, Tv

⟩=

〈Ti, Tj〉 i < j

〈Tj , Ti〉 i > j,

whereu = i, v = j, 1 ≤ u < v ≤ n. The above set and function canbe rewritten as

E =⟨

Tu, Tv⟩ |1 ≤ u < v ≤ n

and

e(⟨

Tu, Tv⟩, Φ)

=

0 u < v

1 u > v.

It is easy to see that there are12n(n− 1) elements inE and

0 ≤∑

〈Ti,Tj〉∈E

e (〈Ti, Tj〉 , Φ)

=∑

〈Tu,Tv〉∈E

e(⟨

Tu, Tv⟩, Φ) ≤ 1

2n(n− 1).

DefineE(i) =⟨

Tu, Ti⟩ | 1 ≤ u < i

, for 1 < i ≤ n. It is obvious

that there arei− 1 elements inE(i), E =n⋃

i=2E(i) andE(i) ∩ E(j) = ∅,

for eachi 6= j.

Lemma 4. For any instanceI of1/pi = ai+wsi, di/Cmax, letΦ andΦ′be

a pair of schedules generated by Rule DP. LetT1, T2, · · · , Tn denotethe task order witha1 < a2 < · · · < an. Then, we have

∆(i) =∑

〈Tu,Ti〉∈E(i)

e(⟨

Tu, Ti⟩, Φ

′)

−∑

〈Tu,Ti〉∈E(i)

e(⟨

Tu, Ti⟩, Φ) ≥ 0,


for 2 ≤ i ≤ n.∑n

i=2 ∆(i) = 0 if and only ifΦ = Φ′is an optimal schedule.

Proof. First we show∆(i) ≥ 0 by the following induction. For the casei = 2, there is only one task pair

⟨T1, T2

⟩in E(2). We only need to

show that, ifT1 is scheduled afterT2 in Φ, then it is still afterT2 in Φ′.

Assume thatT1 is scheduled afterT2 in Φ andT1 = T[r]. Sincea1is the smallest normal processing time andΦ is canonical,T1 is the only

one task inD(r) of Φ. The choice pointt(r) in Φ′

is larger thanC[r], thecompletion time ofT1 in Φ. Limiting by deadline, no task beforeT1 in Φ

can be scheduled at or after ther-th position inΦ′. Since there are onlyn−r

tasks afterT1 in Φ, T1 must be scheduled at or after ther-th position in

Φ′. Thus, we obtain thatT1 is still afterT2 in Φ

′and∆(2) ≥ 0.

Now we consider the casei ≥ 3. From the structure of the task order, wehaveax < ai, for each

⟨Tx, Ti

⟩ ∈ E(i). AssumingTi = T[r] = T(l)andk = min(r, l), whereT[r] is the task at ther-th position inΦ andT(l) is

the task at thel-th position inΦ′. There are two different cases as follows.

Case 1.k = l ≤ r: We haveTi = T(k). For each⟨Tx, Ti

⟩ ∈ E(i) with

e(⟨

Tx, Ti⟩, Φ

′)−e(⟨

Tx, Ti⟩, Φ)

= −1, we havee(⟨

Tx, Ti⟩, Φ)

= 1

ande(⟨

Tx, T(i)⟩, Φ

′) = 0. This means thatTx is scheduled after ther-th

position inΦ and before thek = l-th position inΦ′. We getTx ∈ X =

Tx|ax < ai and Tx ∈ H(k) ⊂ H(k).

For such a taskTx ∈ X, there exists a counterpart taskTy ∈ M(k)such thatTy ∈ G(k) in Φ

′and ay < ax < a(k). It is easy to see that⟨

Ty, Ti⟩ ∈ E(i) ande

(⟨Ty, Ti

⟩, Φ

′) − e(⟨

Ty, Ti⟩, Φ)

= 1. Thus,

we obtain∆(i) ≥ 0.

Case 2.k = r < l: We haveTi = T[k]. Similar to case 1, for each

Tx, Ti ∈ E(i) with e(⟨

Tx, Ti⟩, Φ

′) − e(⟨

Tx, Ti⟩, Φ)

= −1, we

haveTx scheduled after thek = r-th position inΦ and before thel-thposition inΦ

′. Sinceax < ai andΦ is canonical, the completion time of

Tx in Φ is larger than the deadline ofTi as well as the completion time of

Ti in Φ′. We get thatTx is actually scheduled later than thel-th position

in Φ andTx ∈ X =Tx|ax < ai and Tx ∈ H(l)

⊂ H(l).For such a taskTx ∈ X, there exists a counterpart taskTy ∈ M(l)

such thatTy ∈ G(l) and ay < ax < ai. SinceΦ is canonical, fromTy ∈ G(l) andTi = T[k] = T(l), it is easy to see thatTy, Ti ∈ D(l)in Φ; furthermore,Ty is scheduled beforeTi = T[k]. Thus, we obtain that

e(⟨

Ty, Ti⟩, Φ

′)− e(⟨

Ty, Ti⟩, Φ)

= 1 and∆(i) ≥ 0.


By induction, we obtain∆(i) ≥ 0, for 2 ≤ i ≤ n.If Φ is an optimal schedule, then, from Rule DP,Φ

′has the same task

order asΦ. Therefore, we obtain∑n

i=2 ∆(i) = 0.If∑n

i=2 ∆(i) = 0, then, from∆(i) ≥ 0, the result just proved in thislemma, we have∆(i) = 0, for 2 ≤ i ≤ n. There is only one pair oftasks in the casei = 2. It is easy to see that, if∆(2) = 0, then thepair of tasks inE(2) have the same order inΦ

′andΦ. Assume that the

result holds for the casesi ≤ k. For the casei = k + 1, if there ex-ists a pair of tasks withe

(⟨Tx, Tk+1

⟩, Φ

′) − e(⟨

Tx, Tk+1⟩, Φ)

=

−1, then Tx is after Tk+1 in Φ and before it inΦ′. From the above

proof in this lemma, there exists a taskTy such thatay < ak+1 and

e(⟨

Ty, Tk+1⟩, Φ

′)−e(⟨

Ty, Tk+1⟩, Φ)

= 1. We have thatTy is before

Tk+1 in Φ and after it inΦ′. Comparing the order ofTx, Ty andTk+1 in

each schedule, we get that the task pair〈Ty, Tx〉 ∈k⋃

i=2E(i) with different

orders betweenΦ′

andΦ, contradicting the induction assumption. We get

e(⟨

Tx, Tk+1⟩, Φ

′)− e(⟨

Tx, Tk+1⟩, Φ) ≥ 0, for each pair of tasks in

E(k + 1). Sincee(⟨

Tx, Tk+1⟩, Φ

′)− e(⟨

Tx, Tk+1⟩, Φ)

= 0. This is

the case that the task pair have the same task order inΦ′andΦ. By induction,

we get that each pair of tasks in task system have the same task order. FromRule DP,Φ

′andΦ are a pair of feasible schedules withC(n) = C[n]. From

Lemma 2, we obtain thatΦ = Φ′

is an optimal schedule.

4 Implementation

To give a formal description of Rule DP, we present in the following animplementation scheme for it in the form of an algorithm.

Algorithm A: for 1/pi = ai + wsi, di/Cmax

1. InitiateΦ = [T[1], T[2], · · · , T[n]] asa[1] < a[2] < · · · < a[n]; Φ′ := ∅;2. C[0] := 0; C := 0;3. FORi := 1 TO n DO C[i] = (1 + w)C[i−1] + a[i];4. WHILE C 6= C[n] DO

BEGIN5. s := C[n];6. FORi := n DOWN TO 1 DO

BEGIN7. t := max(C[i], s);


8. Find taskT(i) ∈ Φ with d(i) ≥ t and maximala(i);9. IF there exists no suchT(i)

THEN BEGIN10. “OUTUT there exists no feasible schedule”; STOP;

ENDELSE BEGIN

11. s := t−a(i)(1+w) ; Φ := Φ \ T(i);Φ′ := T(i) + Φ;

END;END;

12. C := C[n];Φ := Φ′;13. FORi := 1 TO n DO C[i] = (1 + w)C[i−1] + a[i];

END;14. OUTPUT “Φ is an optimal schedule”;15. Stop.

It is easy to determine the complexity of Algorithm A. Let|x| denote theinput length ofI in a reasonable encoding. The longest input length of thenumber in Algorithm A is in a form similar toC[n]. It is bounded by

(n− 1) log2

(u + v

u

)+ log2

(n∑

i=1

ai

)

+ log2

(max1≤i≤n

di

)≤ O(|x| log |x|),

wherew = vu , u andv are integers as small as possible. Thus, each number

can be input inO(n) time. There areO(n) operations in Steps 8 and 13,which takeO(n2) time. Meanwhile, there areO(n) iterations in Step 6and, from Lemma 4, there areO(n2) iterations in Step 4. Therefore, thecomplexity of Algorithm A isO(n5) (see [6]).

If there exists no taskT(i) in Step 9, then, from Lemma 3, there existsno feasible schedule forI. If the caseC = C[n] occurs in Step 4, then, from

Rule DP,Φ′ = Φ is a feasible schedule. From Lemma 2 or 4,Φ is an optimal

schedule. Therefore, the following theorem is established.

Theorem 3. The problem1/pi = ai+wsi, di/Cmax can be solved inO(n5)time.

5 The maximum lateness problem

For any instanceI of 1/pi = ai + wsi, di/Lmax, modify each deadline ofIby a number∆ and construct a group of new deadlinesdi +∆. We obtain anew task system.TS′ = (Ti, di +∆, ai, w). Note that there exists


a feasible schedule for the makespan problem ofTS′ if and only if Lmax isthe minimum value of the maximum lateness forTS. Since the completiontime of a task forTS may be a rational number, the maximum lateness forTS may be a rational number. Hence,∆ is allowed to be a rational number.It is easy to check that Algorithm A is still applicable toTS′, though thedue dates forTS′ are allowed to be rational numbers. Thus, we obtain thefollowing lemma.

Lemma 5. For any instanceI of 1/pi = ai + wsi, di/Lmax, Lmax is thesmallest rational number.

Use a method of equinoctial point to choose the bound of∆ and applyAlgorithm A to identify whether there exists a feasible schedule forTS′.We present an algorithm for1/pi = ai + wsi, di/Lmax as follows.

Sincew is the unique rational original parameter of the task system, alldenominators of the rational numbers are in the form of1

ui , for1 ≤ i ≤ n−1,wherew = v

u , u andv are integers as small as possible. Thus,un−1Lmax isan integer. From the definition of maximum lateness and (1), we have

−mindi − ai|1 ≤ i ≤ n ≤ Lmax ≤ C[n]

≤ (1 + w)n−1n∑

i=1

ai =(u + v)n−1

un−1

n∑i=1

ai.

DefineB1 = un−1(−mindi − ai|1 ≤ i ≤ n) andB2 = (u + v)n−1∑ni=1 ai. Then,B1 andB2 are a pair of initial values for the lower bound

and upper bound ofun−1Lmax, respectively. Now we give an algorithm forthe problem1/pi = ai + wsi, di/Lmax as follows.

Algorithm B: for 1/pi = ai + wsi, di/Lmax

1. B1 = un−1(−mindi − ai|1 ≤ i ≤ n); B2 = (u + v)n−1∑ni=1 ai;

2. WHILE B2 −B1 > 1 DOBEGIN

3. L′max :=

⌈B1+B2

2

⌉; Lmax = L′

maxun−1 ;

4. FORi := 1 TO n DO d′i = di + Lmax;

5. Use AlgorithmA to find an optimal schedule forTS′;6. IF there exists an optimal schedule7. THENB1 := L′

max8. ELSEB2 := L′

max;END;

9. Lmax = B1un−1 ;

10. STOP.


The value ofLmax can be determined inlog2(B2 −B1) time in Step 2.It is easy to see that

log2(B2 −B1) ≤ (n− 1) log2(u + v) + log2

n∑i=1

ai + (n− 1) log2 u

+ log2 max(di) ≤ O(|x| log |x|).Thus, the value ofLmax can be determined in at mostO(n log n) time (see[6]). In each iteration of Step 2, the most complex part is Step 5, whichtakesO(n5) time to determine whether there exists a feasible schedule forTS′. Therefore, the complexity of Algorithm B isO(n6 log n). We haveestablished the following theorem.

Theorem 4. The problem1/pi = ai + wisi, di/Lmax can be solved inO(n6 log n) time.

6 NP-completeness of the problem1/pi = ai + wisi, di/Cmax

The strongly NP-complete 3-Partition problem can be reduced to1/pi =ai + wisi, di/Cmax. The 3-Partition problem is defined as follows.

3-Partition. Given a listH = h1, h2, · · · , h3m of 3m integers such that∑3mi=1 hi = mB and B

4 < hi < B2 for each1 ≤ i ≤ 3m, canH be

partitioned intoH1, H2, · · · , Hm such that∑

hi∈Hjhi = B for each1 ≤

j ≤ m?Given an instanceI of 3-partition with a listH = h1, h2, · · · , h3m

andB, let v denote an integer larger than28m3B3. Construct an instanceII of 1/pi = ai + wisi, di/Cmax as follows:

Set of tasks:TS = R ∪ Q, whereR = T1, T2, · · · , T3m andQ =T 0

1 , T 02 , · · · , T 0

m

, consisting of4m tasks, with the normal processing

times:

ai = vhi 1 ≤ i ≤ 3m,

a0i = v 1 ≤ i ≤ m,

the increasing rates of processing times:

wi = hiv 1 ≤ i ≤ 3m,

w0i = 0 1 ≤ i ≤ m ,

the deadlines:

di = D = vm(B + 1) +n∑

i=1

i−1∑j=1

hihj +12m(m− 1)B + 1 1 ≤ i ≤ 3m,

d0i = iv + (i− 1)(vB + 4mB2) 1 ≤ i ≤ m,


and the threshold:

G = D.

It is easy to see that the above construction can be performed in polyno-mial time andII is an available instance of1/pi = ai +wisi, di/Cmax. Thefirst 3m tasks are thepartition tasks, while the lastm tasks are theenforcertasks. Now we show thatI has a solution if and only ifII has a solution.

Given a scheduleS = [T[1], T[2], · · · , T[4m]] for II, let C[i], a[i], p[i] andw[i] denote the completion time, normal processing time, actual processingtime and processing rate ofT[i] in S, respectively, for1 ≤ i ≤ 4m. Sincep[1] = a[1] andp[i] = a[i] + w[i]C[i−1], for 2 ≤ i ≤ 4m, we obtain

C[4m] =4m∑i=1

(a[i] + w[i]C[i−1])

=4m∑i=1

a[i] +4m−1∑i=1

w[i+1]C[i]

=4m∑i=1

a[i] +4m−1∑i=1

w[i+1](i∑

j=1

a[j] +i−1∑j=1

w[j+1]C[j])

=4m∑i=1

ai +4m−1∑i=1

i∑j=1

w[i+1]a[j] +4m−1∑i=1

i−1∑j=1

w[i+1]w[j+1]C[j]. (3)

Without loss of generality, suppose that the enforcer tasks are scheduledin increasing order of their deadlines inS. Sincea0

1 = d01 = v, S can be

expressed in the form[T 01 , R1, T

02 , R2, · · · , T 0

m−1, Rm−1, T0m, Rm], where

Ri is the set of partition tasks scheduled betweenT 0i andT 0

i+1, 1 ≤ i ≤ m−1, andRm is the set of partition tasks scheduled afterT 0

m. Sincew0i = 0, for

1 ≤ i ≤ m, we define two task setsΘ[i] = T[j]|T[j] ∈ R and 1 ≤ j < iandΩ[i] = T[j]|T[j] ∈ S and 1 ≤ j < i, for eachT[i] ∈ R in S. From(3), we obtain

C[4m] =∑

T[i]∈TS

a[i] +∑

T[i]∈R

∑T[j]∈Θ[i]

w[i]a[j] +∑

T[i]∈R

∑T[j]∈Ω[i]

w[i]a[j]

+4m−1∑i=1

i−1∑j=1

w[i+1]w[j+1]C[j]. (4)

Let Ti denote thei-th partition task inS, where the subscriptidenotes the index of the task in thei-th position of the partition tasks inS.Let Ci, ai, pi andwi denote the completion time, normal processing


time, actual processing time and processing rate ofTi in S, respectively,for 1 ≤ i ≤ 3m. We obtain

∑T[i]∈R

∑T[j]∈Θ[i]

w[i]a[j] =3m∑i=1

i−1∑j=1

wiaj =3m∑i=1

i−1∑j=1

hiv

vhj

=3m∑i=1

i−1∑j=1

hihj . (5)

Use∆i to denote∑

Tj∈Rihj , for 1 ≤ i ≤ m. DefineΨ[i] = T[j]|T[j] ∈

R and i < j ≤ 4m, for each enforcer taskT[i] ∈ Q in S. Sincea0i = v,

for 1 ≤ i ≤ m, andw[i] = h[i]v , for eachT[i] ∈ R in S, we obtain

∑T[i]∈R

∑T[j]∈Ω[i]

w[i]a[j] =∑

T[i]∈Q

∑T[j]∈Ψ[i]

a[i]w[j] =∑

T[i]∈Q

∑T[j]∈Ψ[i]

v

(h[j]

v

)

=m∑

i=1

m∑j=i

∆i =m∑

i=1

i∆i. (6)

An upper bound for∑4m−1

i=1∑i−1

j=1 w[i+1]w[j+1]C[j] can easily deter-

mined. Sincew[i] ≤ Bv , C[i] ≤ C[4m], for 1 ≤ i ≤ 4m, and

∑T[i]∈TS a[i] =

vm(B + 1), from (3), we have

C[4m] =4m∑i=1

(a[i] + w[i]C[i−1]) ≤ vm(B + 1) +4mB

vC[4m]. (7)

Sincev > 28m3B3, from (7), we have

0 < C[4m] ≤ vm(B + 1)v

v − 4mB≤ 4vmB. (8)

Sincew[i] ≤ Bv , for 1 ≤ i ≤ 4m, andv > 28m3B3, from (7) and (8), we

obtain

0 <

4m−1∑i=1

i−1∑j=1

w[i+1]w[j+1]C[j] ≤ 4m · 4m · Bv· B

v· 4vmB

=26m3B3

v< 1. (9)

Since∑

T[i]∈TS a[i] = vm(B + 1), from (4), (5), (6) and (9), we obtain

C[4m] < vm(B + 1) +3m∑i=1

i−1∑j=1

hihj +m∑

i=1

i∆i + 1, (10)


and

C[4m] > vm(B + 1) +3m∑i=1

i−1∑j=1

hihj +m∑

i=1

i∆i. (11)

Lemma 6. If I has a solution, thenII also has a solution.

Proof. Let H1, H2, · · · , Hm be a solution toI. Let T 0i start atd0

i − a0i , for

1 ≤ i ≤ m. The time interval[0, m(vB + v + 4mB2)] is partitioned intom intervalsI1, I2, · · · , Im with the length of each interval exactly equal tovB +4mB2. For eachhj ∈ Hi, scheduleTj in Ii. The tasks assigned to thesame interval can be scheduled in an arbitrary order. LetPi denote the totalprocessing time of all the tasks assigned toIi. We have

Pi =∑

Tj∈Ii

(aj + wjsj) ≤∑

hj∈Hi

[vhj +

hj

vm(vB + v + 4mB2)

]

≤ vB + 4mB2.

Thus, them enforcer tasks can all meet their deadlines.Since∆i =

∑Tj∈Ii

hj = B, for 1 ≤ i ≤ m, we obtain from (10),

C[4m] ≤ vm(B + 1) +3m∑i=1

i−1∑j=1

hihj +m∑

i=1

iB + 1 = D = G

Thus, the resulting schedule is a solution toII.

Lemma 7. If II has a solution, thenI also has a solution.

Proof. Let S = [T 01 , R1, T

02 , R2, · · · , T 0

m−1, Rm−1, T0m, Rm] be a feasi-

ble schedule forTS with C[4m] ≤ G. The listH can be partitioned intoH1, H2, · · · , Hm asHi ← hj |Tj ∈ Rj. Use∆i to denote

∑Tj∈Ri

hj ,

for1 ≤ i ≤ m. LetC0i denote the completion time ofT 0

i inS, for1 ≤ i ≤ m.Sincea0

i = v,d0i = iv+(i−1)(vB+4mB2), andai = vhi, for1 ≤ i ≤ 3m,

we obtain


C0i − d0

i >

i∑

j=1

a0i +

i−1∑j=1

∑Tk∈Rj

ak

− d0

i

=

iv +

i−1∑j=1

v∆j

− [iv + (i− 1)(vB + 4mB2)]

= vi−1∑j=1

∆j − (i− 1)vB − 4(i− 1)mB2

≥ v

i−1∑

j=1

∆j − (i− 1)B − 12

. (12)

SinceS is feasible, from (12), we have∑i

j=1 ∆j − iB ≤ 0, for 1 ≤ i ≤m− 1. Thus, from

∑mi=1 ∆j = mB, it is easy to see that

m∑j=i

∆j ≥ (m− i + 1)B, for 1 ≤ i ≤ m. (13)

From (11), we obtain

C[4m] −G >

vm(B + 1) +

3m∑i=1

i−1∑j=1

hihj +m∑

i=1

i∆i

−vm(B + 1) +

3m∑i=1

i−1∑j=1

hihj +12m(m− 1)B + 1

=m∑

i=1

i∆i − 12m(m− 1)B − 1

=m∑

i=1

m∑

j=i

∆j − (m− k + 1)B

− 1. (14)

SinceC[4m] ≤ G, from (14), we have∑m

i=1

(∑mj=i ∆j − (m− k + 1)B

)≤ 0. From (13), it is easy to see that

∑mj=i ∆j = (m − i + 1)B. Thus,

we obtain∆i = B, for 1 ≤ i ≤ m. Furthermore, the 3-Partition listH1, H2, · · · , Hm is a solution toI.

Lemma 6 and Lemma 7 immediately lead to the following theorem.

Theorem 5. The problem1/pi = ai + wisi, di/Cmax is strongly NP-complete.


SinceG = D = di, the equivalent recognition version ofII can be statedas: Does there exist a feasible schedule forII. TakingII as an instance of1/pi = ai + wisi, di/ΣCi and extendingG to a large enough value,IIbecomes an instanceIII of 1/pi = ai + wisi, di/ΣCi such thatI has asolution if and only ifIII has a solution. Considering the deadlines as duedates, we see thatII becomes an instanceIV of 1/pi = ai +wisi, di/Lmaxsuch thatI has a solution if and only ifIV has a solution. Thus we obtainthe following theorem.

Theorem 6. The problem1/pi = ai + wisi, di/ΣCi and the problem1/pi = ai + wisi, di/Lmax are both strongly NP-complete.

7 Conclusions

The problems of scheduling a set of tasks with deadlines and increasingrates of processing times on a single machine to minimize the makespan,flow time and maximum lateness have many industrial applications. Theseproblems have remained untackled in the literature. In this paper, we showthat the makespan problem1/pi = ai + wsi, di/Cmax is solvable inO(n5)time by a dynamic programming algorithm. We also show that the corre-sponding flow time problem1/pi = ai + wsi, di/ΣCi is equivalent to themakespan problem. As an application of this dynamic programming algo-rithm, we further show that the problem1/pi = ai +wsi, di/Lmax is solvedin O(n6 log n) time.

On the other hand, we show that the general makespan problem1/pi =ai+wisi, di/Cmax is strongly NP-complete by a reduction from 3-Partition.So, the corresponding flow time problem1/pi = ai + wisi, di/ΣCi andmaximum lateness problem1/pi = ai + wisi, di/Lmax are both stronglyNP-complete.

Thus, we have resolved the computational complexities of six open prob-lems for this class of scheduling model. However, the complexity of thecorresponding problems with identical normal processing times is yet to bestudied.

Acknowledgements.The research was supported in part by The Hong Kong PolytechnicUniversity under grant number 350/239. We are grateful to an anonymous referee for hisconstructive comments on an earlier version of the paper.


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Single machine scheduling with deadlines and increasing rates of processing times