Semi-on-line scheduling with ordinal data on two uniform machines

Operations Research Letters 28 (2001) 221–231www.elsevier.com/locate/dsw

Semi-on-line scheduling with ordinal data ontwo uniform machines

Zhiyi Tan, Yong He ∗

Department of Mathematics, Zhejiang University, Hangzhou 310027, People’s Republic of China

Received 26 June 2000; received in revised form 1 February 2001

Abstract

We investigate the problem of semi-on-line scheduling of jobs on two uniform machines where the order of jobs bytheir processing times is known as a priori. We propose an algorithm for any machine speed ratio s¿ 1. We also presenta comprehensive lower bound, which is a piecewise function of s. The algorithm is optimal for the majority of s∈ [1;∞).The total length of the intervals of s where the competitive ratio does not match the lower bound is less than 0.7784 andthe biggest gap between them never exceeds 0:0521. c© 2001 Elsevier Science B.V. All rights reserved.

MSC: 90B35; 90C27

Keywords: Online; Scheduling; Uniform machines; Analysis of algorithm; Competitive ratio; Makespan

1. Introduction

In this paper, we consider the following scheduling problem: we are given n independent jobs J = {p1; : : : ;pn}, which must be assigned to two uniform machines M = {M1; M2} with the objective to minimize themakespan (i.e., the latest job completion time). We identify the jobs with their processing times. MachineM1 has speed s1 = 1 and machine M2 has speed s2 = s¿ 1. If pi is assigned to machine Mj, then pi=sj timeunits are required to process this job. Both machines and jobs are available at time zero, and no preemptionis allowed. We further assume that jobs arrive one by one and we know nothing about the values of theprocessing times but the order of the jobs by their processing times. Hence without loss of generality, wesuppose p1¿p2¿ · · ·¿pn. We are asked to decide the assignment of all jobs at time zero by utilizing onlyordinal (rank) data rather than the actual magnitudes.

Competitive analysis is a type of worst-case analysis where the performance of an online or semi-on-linealgorithm is compared to that of the optimal oCine algorithm [12]. For an online or semi-on-line algorithm A,let CA denote the makespan of a solution produced by the algorithm A and COPT denote the minimal makespanin an oCine solution. Then the competitive ratio of the algorithm A is deFned as the smallest number c such

∗ Corresponding author.E-mail address: [email protected] (Y. He).

0167-6377/01/$ - see front matter c© 2001 Elsevier Science B.V. All rights reserved.PII: S 0167 -6377(01)00071 -2

222 Z. Tan, Y. He / Operations Research Letters 28 (2001) 221–231

that CA6 cCOPT for all instances. An algorithm with a competitive ratio c is called a c-competitive algorithm.An online or semi-on-line deterministic (randomized) algorithm A is called the best possible (or optimal)algorithm if there is no deterministic (randomized) online or semi-on-line algorithm for the discussed problemwith a competitive ratio smaller than that of A.Our discussed problem belongs to a kind of semi-on-line scheduling, a variant of online where we do have

some partial knowledge on the job set which makes the problem easier to solve than standard online schedulingproblems. In our problem, we know the order of jobs by their processing times. Although, there are manyresults on semi-on-line scheduling problems on identical machines [2,5–7,11], to the authors’ knowledge, littleis known about uniform machine scheduling.On the other hand, problems with ordinal data exist in many Felds of combinatorial optimization such as

matroid, bin-packing and scheduling problem etc. [1,8–10]. Algorithms which utilize only ordinal data ratherthan actual magnitudes are also called Ordinal algorithms. For ordinal scheduling, Liu et al. [10] gave athorough study on Pm ‖ Cmax. Because ordinal scheduling is, to some extent, stricter than the non-clairvoyantscheduling in that we must assign all jobs at time zero, classical online algorithms depending on the machineloads, such as list scheduling (LS), are of no use. For m=2; 3 [10], presented respective optimal ordinalalgorithms. In the same paper, an algorithm with a competitive ratio 101=70 was given while the lower boundis 23=16 for m=4 and an algorithm with a competitive ratio 1 + (m− 1)=(m+ �m=2�)6 5=3 was developedwhile the lower bound is 3=2 for general m¿ 4.Due to the above motivation, this paper considers semi-on-line scheduling with ordinal date on two uniform

machines. We present an algorithm Ordinal and analyze its competitive ratio. We give a lower bound of anyordinal algorithm, which is, denoted by clow, a piecewise function dependent on machine speed ratio s.Comparing the competitive ratio and the lower bound, we claim that the algorithm is optimal for the majorityof the values of s∈ [1;∞). The total length of the intervals of s where the competitive ratio does not matchthe lower bound is less than 0.7784, and the biggest gap between them never exceeds 0:0521.A related problem is the classical clairvoyant online uniform machine scheduling problem Q2 ‖ Cmax, which

is deeply studied. Cho and Sahni [3] have shown that the competitive ratio of LS algorithm is (1 +√5)=2.

Epstein et al. [4] have proved that the parametric competitive ratio of LS is min{(2s + 1)=(s + 1); 1 + 1=s},and LS is thus the best possible on-line algorithm for any s. They further proved that randomization does nothelp for s¿ 2, presented a simple memoryless randomized algorithm with a competitive ratio of (4− s)(1 +s)=46 1:5625, and devised barely random algorithms with a competitive ratio at most 1.53 for any s¡ 2.The paper is organized as follows: Section 2 presents the algorithm Ordinal and proves its exact competitive

ratio. Section 3 gives the lower bound dependent on s and shows the optimality of Ordinal for all but a smallpart of s∈ [1;∞).

2. The algorithm Ordinal

In this section, we present the algorithm Ordinal and study its competitive ratio. The algorithm consists ofan inFnite series of procedures. For any s, it chooses exactly one procedure to assign jobs. First, we give thedeFnition of procedures.

Procedure (0): Assign all jobs to M2.Procedure (1): Assign jobs in the subset {p3i+2; p3i+3 | i¿ 0} to M1; assign jobs in the subset

{p3i+1 | i¿ 0} to M2.Procedure (l); l¿ 2: Assign jobs in the subset {pli+2 | i¿ 0} to M1; assign jobs in the subset {p1} ∪

{pli+3; pli+4; : : : ; pli+l+1 | i¿ 0} to M2.

Algorithm. Ordinal

1. If s¿ 1 +√3, assign jobs by Procedure (0).

Z. Tan, Y. He / Operations Research Letters 28 (2001) 221–231 223

2. If s∈ [s(l− 1); s(l)); l¿ 1, assign jobs by Procedure (l), where

s(l)=

1; l=0;1 +

√13

4; l=1;

l2 − 1 +√

(l2 − 1)2 + 2l3(l+ 1)l(l+ 1)

; l¿ 2:

In the above deFnition, s(l) is the positive root of the equation ((l + 1)s + l)=(l + 1)s=(ls + 2)=2l forany l¿ 2. Because it is a monotone increasing sequence of l and liml→∞ s(l)= 1 +

√3, the algorithm is

well-deFned for all s¿ 1.

Theorem 2.1. The parametric competitive ratio of the algorithm Ordinal is

cOrdinal(s)=

2s+ 23

; s∈ [s(0); s(1));

ls+ l− 1ls

; s∈ [s(l− 1); s′(l)); l=2; 3; : : : ;

ls+ 22l

; s∈ [s′(l); s(l)); l=2; 3; : : : ;

s+ 1s; s¿ 1 +

√3;

where s′(l)= (l− 1 +√

(l− 1)2 + 2l(l− 1))=l (l¿ 2) is the positive root of (ls+ l− 1)=ls=(ls+ 2)=2l.

Proof. We Frst show that COrdinal=COPT6 cOrdinal(s). Let b be the total processing time of all jobs. Let �1and �2 be the completion times of M1 and M2 after processing all jobs by our algorithm, respectively, thenCOrdinal =max{�1; �2}. Obviously, COPT¿ b=(s+ 1) and COPT¿p1=s. To estimate the ratio between COrdinal

and COPT, we distinguish four cases with respect to the speed ratio s.Case 1: s¿ 1 +

√3. In this case, Ordinal chooses Procedure (0). It is clear that �1 = 0 and �2 = b=s=

(s+ 1)=s× b=(s+ 1)6 (s+ 1)=sCOPT. This is the desired ratio.Case 2: s∈ [s(0); s(1)). In this case, Ordinal chooses Procedure (1). We prove the subcase of n=3k;

the other subcases of n=3k + 1 and 3k + 2 can be proved similarly. By the rule of the procedure andp1¿ · · ·¿pn, we have

�1 =k−1∑i=0

(p3i+2 + p3i+3)623

k−1∑i=0

(p3i+1 + p3i+2 + p3i+3)

=2b3

=2(s+ 1)

3b

s+ 16

2(s+ 1)3

COPT;

�2 =1s

k−1∑i=0

p3i+16p1

s+

13s

n∑i=2

pi=2p1

3s+b3s6

3s+ 13s

COPT:

Note that for s∈ [s(0); s(1)), we have (3s+ 1)=(3s)6 2(s+ 1)=3, we thus get the desired ratio.Case 3: s∈ [s(1); s(2)). At this time, we need to analyze Procedure (2). We prove the subcase of n=2k,

the subcase of n=2k + 1 can be proved similarly. In fact, we have

�1 =k−1∑i=0

p2i+26b26s+ 12COPT;


and

�2 =1s

(p1 +

k−2∑i=0

p2i+3

)6p1

s+

12s

n∑i=2

pi=p1

2s+b2s6

2s+ 12s

COPT:

By comparing (s+ 1)=2 and (2s+ 1)=(2s), we have

COrdinal

COPT 6

2s+ 12s

s∈ [s(1); s′(2));

s+ 12

s∈ [s′(2); s(2));

where s′(2) is deFned in the theorem.Case 4: s∈ [s(l−1); s(l)); l¿ 3. Consider Procedure (l) as follows. We only prove the subcase of n=1+kl

as above. Obviously we have

�1 =k−1∑i=0

pil+26p2 +1l

n∑i=3

pi:

If p1 and p2 are processed by the same machine in the optimal schedule, we have COPT¿ (p1 + p2)=s:Substituting it to the above inequality, we get

�16l− 22l

(p1 + p2) +bl6ls+ 22l

COPT:

Otherwise, since p1 and p2 are not processed by the same machine in the optimum, we have COPT¿p2 andhence

�16l− 2lp2 +

bl6l+ s− 1

lCOPT:

Next, we consider the ratio between �2 and COPT. By straight computation, we have

�2 =1s

p1 +

k−1∑i=1

l+1∑j=3

pil+j

6 p1

s+l− 1ls

n∑i=2

pi6p1

ls+

(l− 1)bls

6(1l+

(l− 1)(s+ 1)ls

)COPT =

ls+ l− 1ls

COPT:

Note that for l=3 and s¡ 2, we have (3s+2)=(3s)¿ (s+2)=3 and s′(3)= 2, which imply COrdinal=COPT6(3s+2)=(3s) when s∈ [s(2); s′(3)). For l¿ 3 and s¿ 2, we have (ls+2)=(2l)¿ (l+ s− 1)=l, which impliesthat COrdinal=COPT6max{(ls+ l− 1)=(ls); (ls+ 2)=(2l)} when s∈ [2; 1 +

√3). The desired ratio can thus be

obtained.Now, we have Fnally proved that COrdinal=COPT6 cOrdinal(s). The instances described in Table 1 can show

that the ratio is exact for any s¿ 1.

Table 1

s Instance COrdinal=COpt

s∈ [s(0); s(1)) p1 =p2 =p3 = 1, p4 =p5 =p6 = 1=s (2s + 2)=3s∈ [s(l− 1); s′(l)), l¿ 2 p1 = 1, p2 = · · ·=pl =1=(ls) (ls + l− 1)=(ls)s∈ [s′(l); s(l)), l¿ 2 p1 =p2 = 1, p3 = · · ·=pl =2=(ls) (ls + 2)=(2l)s¿ 1 +

√3 p1 = 1, p2 = 1=s (s + 1)=s


Table 2

s clow(s) s clow(s)[1;

5 +√265

20

)≈ [1; 1:064)

2s + 23

[1 +

√13

2;5 +

√345

10

)≈ [2:303; 2:357)

s + 34

[1 +

√7

3;1 +

√5

2

)≈ [1:215; 1:618)

2s + 12s

[5 +

√345

10;1 +

√17

2

)≈ [2:357; 2:562)

5s + 45s

[1 +

√5

2;3 +

√57

6

)≈ [1:618; 1:758)

s + 12

[1 +

√17

2;3 +

√109

6

)≈ [2:562; 2:602)

s + 45

[3 +

√57

6; 2

)≈ [1:758; 2)

3s + 23s

[3 +

√109

6;3 + 2

√6

3

)≈ [2:602; 2:633)

6s + 56s

[2;

1 +√10

2

)≈ [2; 2:081)

s + 23

[3 + 2

√6

3; 1 +

√3

)≈ [2:633; 2:732)

s2

[1 +

√10

2;1 +

√13

2

)≈ [2:081; 2:303)

4s + 34s

[1 +√3;∞) ≈ [2:732;∞)

s + 1s

3. Parametric lower bound

This section presents a parametric lower bound for any deterministic ordinal algorithm applied to ourproblem. We generalize the results in the following two theorems.

Theorem 3.1. For any s∈ [1; (5 +√265)=20) ∪ [(1 +

√7)=3;∞); the competitive ratio of any deterministic

ordinal algorithm is at least clow(s) where clow(s) is de8ned in Table 2.

Theorem 3.2. For any s∈ [5 +√265=20; (1 +

√7)=3); the competitive ratio of any deterministic ordinal

algorithm is at least clow(s); where

clow(s)=

5s+ 25s

; s∈[5 +

√265

20; s2(2)

);

(3k − 4)s+ (2k − 2)4k − 5

; s∈ [s2(k − 1); s1(k)); k =3; 4; : : : ;

(2k + 1)s+ k(2k + 1)s

; s∈ [s1(k); s2(k)); k =3; 4; : : : ;

and for each k¿ 3;

s1(k)=(2k + 1)(2k − 3) +

√(2k + 1)2(2k − 3)2 + 4k(2k + 1)(3k − 4)(4k − 5)

2(2k + 1)(3k − 4);

is the positive root of the equation

(2k + 1)s+ k(2k + 1)s

=(3k − 4)s+ 2k − 2

4k − 5


and

s2(k)=(2k − 1)(2k + 1) +

√(2k + 1)2(2k − 1)2 + 4k(2k + 1)(3k − 1)(4k − 1)

2(2k + 1)(3k − 1);

is the positive root of the equation

(2k + 1)s+ k(2k + 1)s

=(3k − 1)s+ 2k

4k − 1:

From the deFnition of s1(k) and s2(k), we can see that s1(k − 1)¡s2(k − 1)¡s1(k)¡s2(k); k¿ 3 andlimk→∞ s2(k)= (1+

√7)=3, so the lower bound is well-deFned for any s¿ 1 through Theorems 3:1 and 3:2.

The proofs will be completed by the adversary method. To prove Theorem 3.1, we consider an adversarywho presents the online algorithm with one of these 19 job sequences.

(I1) {1} (I8){1; 1;

1s;1s

}(I14)

{1;

15s;15s;15s;15s;15s

}

(I2) {1; 1} (I9){1; 1; 1;

2− ss

}(I15)

{1; 1;

s− 14;s− 14;s− 14;s− 14

}

(I3){1;

1s

}(I10)

{1;

14s;14s;14s;14s

}(I16)

{1; 1; 1;

1s;1s;1s

}

(I4){1;

12s;12s

}(I11)

{1; 1;

s− 13;s− 13;s− 13

}(I17) {1; 1; 1; 1; 1; 2s− 3}

(I5){1;

1s− 1

;1

s− 1

}(I12)

{1; 1; 1;

s− 22s

;s− 22s

}(I18)

{1;

16s;16s;16s;16s;16s;16s

}

(I6) {1; 1; s− 1} (I13) {1; 1; 1; 1; 2s− 2} (I19){1; 1;

s− 15;s− 15;s− 15;s− 15;s− 15

}

(I7){1;

13s;13s;13s

}

A lower bound obtained in this restricted situation is a lower bound on the competitive ratio in general.For easy reading and understanding, we show Theorem 3.1 by distinguishing seven cases with respect to

the value of s. We prove the cases of s∈ [1; (5 +√265)=20) and s∈ [(1 +

√7)=3; (1 +

√5)=2) in detail. The

remaining cases of s∈ [(1 +√5)=2;∞) can be veriFed essentially by the same arguments, hence, we only

list the schedules of algorithm A, and the corresponding adversarial sequences for all possible situations inTables 3–7 case by case afterwards.

Case 1: For any s∈ [1; (5 +√265)=20) and any ordinal algorithm A, there exists some instance such that

CA=COPT¿ clow(s)= (2s+ 2)=3.

Proof. If an algorithm A assigns p1 to M1 and p2; p3; p4 to M2, consider the instance (I8). It follows thatCA=COPT¿ (s+ 2)=(2s)¿ (2s+ 2)=3. If A assigns both p1 and at least one of p2; p3; p4 to M1, consider theinstance (I7). It implies that CA=COPT¿ (3s+1)=3¿ (2s+2)=3. So, we know that the competitive ratio willbe greater than (2s+ 2)=3 as long as A assigns p1 to M1.Next, we analyze the case that A assigns p1 to M2. Firstly, we claim that both p2 and p3 can not be assigned

to M2. Otherwise consider the instance (I4). It implies that CA=COPT¿ (2s+ 1)=(2s)¿ (2s+ 2)=3. Secondly,p4 must be assigned to M2. Otherwise, we consider the instance (I9), which follows that CA=COPT¿ (s +2)=2¿ (2s + 2)=3. Thirdly, if A assigns at least one of p5 and p6 to M2, consider the instance (I14). It is


Table 3

Schedule by A Adversary CA

COPTSchedule by A Adversary CA

COPTinstance instance

M1 M2 M1 M2

{p1} ∅ (I1) s {p2} {p1; p3; p4} (I7)3s + 23s

∅ {p1; p2} (I3)s + 1s

{p2; p4} {p1; p3} (I8)s + 12

{p2; p3} {p1} (I6) s

Table 4




M1 M2 M1 M2

{p1} ∅ (I1) s {p2; p4} {p1; p3} (I8)s + 12

∅ {p1; p2} (I3)s + 1s

{p2; p5} {p1; p3; p4} (I11)s + 23

{p2; p3} {p1} (I5) 2 {p2} {p1; p3; p4; p5} (I10)4s + 34s

Table 5


COPT Schedule by A Adversary CA


M1 M2 M1 M2

{p1} ∅ (I1) s {p2; p5} {p1; p3; p4} (I11)s + 23

∅ {p1; p2} (I3)s + 1s

{p2; p6} {p1; p3; p4; p5} (I15)s + 34

{p2; p3} {p1} (I5) 2 {p2} {p1; p3; p4; p5; p6} (I14)5s + 45s

{p2; p4} {p1; p3} (I8)s + 12

clear that CA=COPT¿ (5s + 2)=(5s)¿ (2s + 2)=3. Lastly, we are left to consider the case where A assignsp1; p4 to M2 and p2; p3; p5; p6 to M1. At this time, we choose (I16). We have CA=COPT¿ (2s+ 2)=3, too.

In summary, we have proved for any s in the given interval, any algorithm A is such that CA=COPT¿(2s+ 2)=3 on some instance.

Case 2: For any s∈ [(1+√7)=3; (1+

√5)=2) and any ordinal algorithm A, there exists some instance such

that CA=COPT¿ clow(s)= (2s+ 1)=(2s):

Proof. Suppose that the algorithm A assigns p1 to M1. If A further assigns any job of {p2; : : : ; p5} to M1, theinstance (I10) implies that CA=COPT¿ (4s + 1)=4¿ (2s + 1)=(2s). It follows that every job of {p2; : : : ; p5}


Table 6




M1 M2 M1 M2

{p1} ∅ (I1) s {p2; p5} {p1; p3; p4} (I11)s + 23

∅ {p1; p2} (I3)s + 1s

{p2; p6} {p1; p3; p4; p5} (I15)s + 34

{p2; p3} {p1} (I5) 2 {p2; p7} {p1; p3; p4; p5; p6} (I19)s + 45

{p2; p4} {p1; p3} (I8)s + 12

{p2} {p1; p3; p4; p5; p6; p7} (I18)6s + 56s

Table 7




M1 M2 M1 M2

{p1} ∅ (I1) s {p2} {p1} (I2)s2

∅ {p1; p2} (I3)s + 1s

must be assigned to M2. For s6 3=2, by considering the instance (I13), we obtain CA=COPT¿ (2s+ 1)=(2s).For s¿ 3=2, we need to analyze the assignment of job p6. If p6 is assigned to M1, the instance (I14) impliesthat CA=COPT¿ (5s+1)=5¿ (2s+1)=(2s); otherwise the instance (I17) also shows CA=COPT¿ (2s+1)=(2s).Therefore, we conclude that p1 should be assigned to M2 in the case that the lower bound is less than(2s+ 1)=(2s).By the same arguments as in Case 1, we can further claim that algorithm A should assign p2 and p3 to

M1, and p4 to M2 in order to achieve a competitive ratio at most (2s+ 1)=(2s). We can Fnish the proof byconsidering the assignment of p5: If p5 is assigned to M1, (I12) implies CA=COPT¿ (3s+2)=4¿ (2s+1)=(2s).If p5 is assigned to M2, (I10) implies CA=COPT¿ (2s+ 1)=(2s).

Case 3: For any s∈ [(1 +√5)=2; 2) and any ordinal algorithm A, there exists some instance such that

CA=COPT¿ clow(s)=min{(s+ 1)=2; (3s+ 2)=(3s)}.Case 4: For any s∈ [2; (1 +

√13)=2) and any ordinal algorithm A, there exists some instance such that

CA=COPT¿ clow(s)=min{(s+ 2)=3; (4s+ 3)=(4s)}.Case 5: For any s∈ [(1 +

√13)=2; (1 +

√17)=2) and any ordinal algorithm A, there exists some instance

such that CA=COPT¿ clow(s)=min{(s+ 3)=4; (5s+ 4)=(5s)}.Case 6: For any s∈ [(1 +

√17)=2; (2 + 2

√6)=3) and any ordinal algorithm A, there exists some instance

such that CA=COPT¿ clow(s)=min{(s+ 4)=5; (6s+ 5)=(6s)}.Case 7: For any s∈ [(2 + 2

√6)=3;∞) and any ordinal algorithm A, there exists some instance such that

CA=COPT¿ clow(s)=min{s=2; (s+ 1)=s}.To prove Theorem 3.2, we rewrite it as follows:

Theorem 3.2′. For any l¿ 3 and s∈ [(5+√265)=20; (1+

√7)=3); the competitive ratio of any deterministic

ordinal algorithm is at least clow(s; l); where clow(s; l) is shown in Table 8.


Table 8

s clow(s; l) s clow(s; l)[5 +

√265

20; s2(2)

)5s + 25s

[s1(k); s2(k)); k =3; : : : ; l(2k + 1)s + k(2k + 1)s

[s2(k − 1); s1(k)); k =3; : : : ; l(3k − 4)s + (2k − 2)

4k − 5[s2(l);

1 +√7

3)

3s + 24

Because {((3k − 4)s+2k − 2)=(4k − 5)} is a monotone increasing sequence of k and limk→∞ ((3k − 4)s+2k − 2)=(4k − 5)= (3s+ 2)=4, one can easily verify the equivalence of these two versions as l→ ∞.

Proof of Theorem 3.2′. We prove the result by induction. For l=3, by using a similar method in the proofof Cases 1 and 2 of Theorem 3:1, it is not diMcult to prove

CA

COPT ¿

5s+ 25s

; s∈[5 +

√265

20;5 +

√145

15

);

3s+ 24

; s∈[5 +

√145

15;1 +

√7

3

):

But it is a little bit diNerent from the desired bound described in the theorem. To reach the goal, we needmore careful analysis. By the same arguments as in the proof of Cases 1 and 2 of Theorem 3:1, we only needto consider the following assignment: A assigns p2; p3; p5 to M1, and p1; p4; p6 to M2. Because for otherassignments, the ratio is no less than clow(s) already.Instead of using the instances which we did in the proof of Case 2 of Theorem 3:1 to achieve the lower

bound of (3s+2)=4, we now distinguish three cases according to the assignment of the next two jobs p7; p8.Case 1: Algorithm A assigns p7 to M2. By considering the instance (I18), we have CA=COPT¿ (2s+1)=(2s).Case 2: Algorithm A assigns p7; p8 to M1. By considering the instance {p1 =p2 =p3 = 1; p4 = · · ·=p8 = (2−

s)=(4s− 1)}, we have CA=COPT¿ (5s+ 4)=7.Case 3: Algorithm A assigns p7 to M1 and p8 to M2. The instance {p1 = 1; p2 = · · ·=p8 = 1=(7s)} follows

that CA=COPT¿ (7s+ 3)=(7s).By straight arithmetic calculation, thus, we have that

CA

COPT ¿ clow(s; 3)=

5s+ 25s

; s∈[5 +

√265

20;3 +

√65

10

)=

[5 +

√265

20; s2(2)

);

5s+ 47

; s∈[3 +

√65

10;3 +

√69

10

)= [s2(2); s1(3));

7s+ 37s

; s∈[3 +

√69

10;7 +

√301

21

)= [s1(3); s2(3));

3s+ 24

; s∈[7 +

√301

21;1 +

√7

3

)=

[s2(3);

1 +√7

3

):

Hence the theorem holds for l=3.By induction, we assume that the theorem is true for all l¡k, and further, we can assume that M1 processes

{p2; p3; p5; p7; : : : ; p2k−1} and M2 processes {p1; p4; p6; : : : ; p2k} in any algorithm A with the desired ratio. Forl= k, three cases are considered separately with respect to the assignment of the next two jobs p2k+1; p2k+2.


Fig. 1. Solid curve denotes the lower bound and dashed curve denotes the competitive ratio of our algorithm.

Case 1: Algorithm A assigns p2k+1 to M2. By considering the instance {p1 = 1; p2 = · · ·=p2k+1 =1=(2ks)},we have CA=COPT¿ (2s+ 1)=(2s).

Case 2: Algorithm A assigns p2k+1; p2k+2 to M1. By considering {p1 =p2 =p3 = 1; p4 = · · ·=p2k+2 =(2− s)=((2k − 2)s− 1)}, we have that CA=COPT¿ ((3k − 4)s+ (2k − 2))=(4k − 5).

Case 3: Algorithm A assigns p2k+1 to M1 and p2k+2 to M2. By considering {p1 = 1; p2 = · · ·=p2k+2 =1=((2k + 1)s)},we have that CA=COPT¿ ((2k + 1)s+ k)=((2k + 1)s).Thus, we have proved that

CA=COPT¿min{(3k − 4)s+ (2k − 2)

4k − 5;(2k + 1)s+ k(2k + 1)s

}is valid for l= k, and the proof is completed.

Fig. 1 illustrates the lower bound and the competitive ratio of our algorithm Ordinal. We conclude thatOrdinal is optimal in the following intervals of s[

1;5 +

√265

20

]∪[1 +

√7

3; 2

]∪[1 +

√10

2;3 +

√33

4

]∪ [1 +

√3;∞):

The total length of the intervals where the competitive ratio does not match the lower bound is less than0:7784 and the biggest gap between the competitive ratio and the lower bound is approximately 0.0520412which occurs at s=(35 +

√8617)=112 ≈ 1:14132.

4. Acknowledgements

A preliminary version of this paper appeared in the proceedings of the 6th Annual International Computingand Combinatorics Conference, Lecture Notes of Computer Science 1858. This research was supported by NSFof China (19701028) and 973 National Fundamental Research Project of China. The authors would like toacknowledge the constructive comments by the referee and Associate Editor which improved the presentationof the paper.

References

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Semi-on-line scheduling with ordinal data on two uniform machines