Two-machine flow shop scheduling with deteriorating jobs: minimizing ...

Journal of the Operational Research Society (2014), 1–11

© 2014 Operational Research Society Ltd. All rights reserved. 0160-5682/14 www.palgrave-journals.com/jors/

Two-machine flow shop scheduling with deteriorating jobs: minimizing the weighted sum of makespan and total completion time Mingbao Cheng1*, Pandu R Tadikamalla2, Jennifer Shang2 and Bixi Zhang1 1

Guangdong University of Technology, Guangzhou, P.R, China; and 2University of Pittsburgh, Pittsburgh, PA, USA This paper considers a two-machine flow shop scheduling problem with deteriorating jobs in which the processing times of jobs are dependent on their starting times in the sequence. The objective is to minimize the weighted sum of makespan and total completion time. To analyse the problem, we propose a mixed integer programming model, and discuss several polynomially solvable special cases. We also present a branch-and-bound algorithm with several dominance rules, an upper bound and a lower bound. Finally, we present results of computational experiments conducted to evaluate the performance of the proposed model and the exact algorithm. Journal of the Operational Research Society advance online publication, 28 May 2014; doi:10.1057/jors.2014.58 Keywords: scheduling; flow shop; bi-criteria; deteriorating jobs; branch-and-bound; heuristics

1. Introduction In the classical machine scheduling problems, it is assumed that the processing times of jobs are constant. However, this assumption is inadequate in modelling many modern industrial processes, in which a job executed under the same condition has a variable processing time. Production scheduling problems with deteriorating jobs have gained attention in recent years and have been studied by academicians and practitioners. Such scheduling issue was introduced by Gupta and Gupta (1988), who address a steel rolling mills scheduling problem, in which the heating time of the ingot depends on its current temperature which in turn depends on its waiting time. Other examples can be found in maintenance scheduling, where in a delay in processing a job may result in additional time for completing the job. Mosheiov (1994) studied simple linear deterioration cases in a single-machine setting. He considered several classical performance metrics and showed that these type of problems can be solved in polynomial-time. Oron (2014) studied a single-machine setting with general linear deterioration and convex resource functions to minimize the makespan. Kononov and Gawiejnowicz (2001) considered NPhard cases in scheduling problems with deteriorating jobs on dedicated machines to minimize the maximum completion time. Since then, many related topics have been examined from various perspectives. More scheduling problems of deteriorating jobs are surveyed by Cheng et al (2004), Gawiejnowicz (2008) *Correspondence: Mingbao Cheng, School of Management, Guangdong University of Technology, Guangzhou 510520, P.R. China. E-mail: [email protected]

and Huang et al (2010). Flow shop scheduling problems with deteriorating times were also studied by Lee et al (2010), Gawiejnowicz et al (2011), Kuo and Yang (2011) and Liu et al (2013). For flow shop scheduling problems with deteriorating jobs, Mosheiov (2002) introduced a polynomial-time algorithm for the two-machine flowshop scheduling problem and proved its NP-hardness for three- or more-machine settings. Wang et al (2006), Wu and Lee (2006) and Shiau et al (2007) considered a two-machine flow shop scheduling problem with simple linear deterioration to minimize the mean flow time. They all used the branch-and-bound algorithm with several dominance conditions to search for the optimal solution. Yang and Wang (2011) studied the total weighted completion time in a two-machine flow shop setting with simple linear deterioration. Other related papers include Cheng et al (2007) and Gawiejnowicz and Kononov (2010). However, the literature discussed above only considered deteriorating jobs in flow shops with single objective (such as makespan, total completion time, total weighted completion time, etc). In this research, we consider both the makespan and the total completion time as objectives. Among the objective functions of machine scheduling problems, makespan is a popularly adopted metric for measuring machine utilization, while the total completion time, defined as the durations of all jobs spend in the system, is an important measure for understanding the system throughputs. Both measures are crucial to the management of resources and service quality. At times, it may be more useful and practical for a decision maker to jointly consider two or more opposing or conflicting criteria. One way of dealing with a multi-criteria

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Journal of the Operational Research Society

problem is to linearly combine two or more objectives into a single measure, that is, w1 f1 + w2 f2, where f1 and f2 denote two criteria, and ω1 and ω2 are nonnegative real numbers. Such scheduling problem can be denoted as α|β|Fw( f1, f2), where Fw( f1, f2) = w1 f1 + w2 f2 represents the weighted sum of the two criteria f1 and f2. Nagar et al (1995a) proposed a branch-andbound algorithm to solve a two-machine flow shop scheduling problem with a weighted sum of flow time and makespan as the objective function. For the same bi-criteria scheduling problem, Nagar et al (1996) developed a hybrid algorithm by integrating a branch-and-bound procedure and a genetic algorithm. Based on their work, Yeh (2001) improved the branch-and-bound algorithm by incorporating additional properties and implementation skills. By formulating an integer programming model and a heuristic algorithm, Chou and Lee (1999) studied a twomachine flow shop scheduling problem with release dates to minimize a weighed sum of total flow time and makespan. Lin and Wu (2006) studied a production scheduling problem in a two-machine flow shop to minimize a weighted sum of the makespan and the total completion time. Chiang et al (2011) proposed an integrated multi-objective evolutionary algorithm to tackle a flow shop scheduling problem. Their goal is to minimize the makespan and total flow time. Chung and Tong (2012) considered a bi-criteria scheduling problem in m-machine flowshop environment with machine-based learning effects to minimize the weighted sum of the total completion time and the makespan. They proposed a branch-and-bound algorithm to derive the optimal/near-optimal solutions. Naderia et al (2012) studied multi-objective no-wait flow shop scheduling problems to minimize both makespan and total tardiness by using two effective multi-objective mixed integer linear programming models. Dang (2013) proposed two algorithms for solving multi-objective planning problems with multi alternate resources and alternative routes. Bozorgirad and Logendran (2013) considered a group scheduling problem in a hybrid flow shop to simultaneously minimize the total weighted completion time and the total weighted tardiness. Lu and Logendran (2013) studied a bi-criteria group flow shop scheduling problem with sequence-dependent setup time. They seek to minimize the weighted sum of total weighted completion time and total weighted tardiness. The survey literature of multi-objective flowshop scheduling problems can be found in Nagar et al (1995b), Allahverd (2003), Dugardin et al (2010), Sun et al (2011) and Xu et al (2013). Recently, researchers began to focus on multi-objective scheduling problems with changeable job processing times. Gawiejnowicz et al (2006) considered a single-machine bi-criteria scheduling problem with deteriorating jobs to minimize a convex combination of the total completion time and the makespan criteria. Eren and Güner (2008) analysed a flow shop scheduling problem with the learning effect to minimize a weighted sum of total completion time and makespan. KarimiNasaba and Fatemi Ghomib (2012) studied a single-stage production scheduling problem subject to controllable process times to minimize the total costs and the total variations in

production volumes simultaneously. Liu et al (2014) considered a single-machine scheduling problem with deteriorating jobs to minimize the weighted sum of earliness, tardiness and duewindow location penalties. Cheng et al (2014) studied a hierarchical optimization of two-machine flow shop scheduling problem with time-dependent deteriorating jobs. In this research, we aim to minimize the weighted sum of the makespan and the total completion time in a two-machine flow shop scheduling problem with deteriorating jobs. We propose a mixed integer programming (MIP) model and a branch-andbound algorithm combined with an upper bound (or a lower bound) to tackle the problem. It is well known that the flow shop scheduling to minimize total completion time is NP-hard even if there are no deteriorating jobs (Garey et al, 1976). Wang et al (2006) proved that minimizing total completion time with simple linear deterioration on two-machine flow shop is NPhard. Therefore, the problem we considered is also NP-hard. The reminder of the paper is organized as follows. Section 2 formalizes the basic problem, and a MIP model for the problem is presented in Section 3. In Section 4, we study the polynomially solvable special cases. Several dominance conditions are developed in Section 5. In Section 6, we propose a branchand-bound algorithm based on the upper bound and the lower bound, and gives an example to illustrate how the heuristic works. Numerical studies are conducted in Section 7 to examine the effectiveness and efficiency of the proposed algorithms. Finally, we make concluding remarks in Section 8.

2. Problem formulation The fundamental notations used in this paper are given as below. Jj: N: Mi: bi,j: αi: βi: []: λ:

job j, where j = 1, 2, …, n set of jobs which contains n jobs, that is, N = {J1, J2, …, Jn} ith machine, where i = 1, 2 the deterioration rate of job Ji on Mj (j = 1, 2) the deterioration rates at M1 for job Ji ∈ N the deterioration rates at M2 for job Ji ∈ N The symbol which denotes the job order in a sequence the weight factor associated with makespan and the total completion time with 0 ⩽ λ ⩽ 1

There is a set of jobs (ie N = {J1, J2, …, Jn}) to be processed on a non-preemptive two-machine flow shop. Each job has two operations and requires its first operation on machine M1 and its second operation on machine M2. The actual processing time of each job is pi,j = bi,jt, where t (>0) denotes the operation starting time of job Ji. Suppose all jobs are available at t0 = 1. The objective is to find a sequence so as to minimize a weighted sum of makespan and the total complete time. We denote such a problem as F2|pi,j = bi,jt|λCmax + (1 − λ)∑Cj.

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Mingbao Cheng et al—Two-machine flow shop scheduling with deteriorating jobs

3. Mixed integer programming model In this section, we develop a MIP model for the problem F2|pi,j = bi,jt|λCmax + (1 − λ)∑Cj. Let C[i],j(σ) denote the completion time of job J[i] in a sequence σ = (J[1], J[2], …, J[n]) on machine Mj(j = 1, 2); thus the completion time of the ith job on machine M2 is calculated by C[i],2(σ) = max1 ⩽ k ⩽ i{∏kj = 1(1 + α[j])∏ij = k(1 + β[j])}. Suppose that δij is a binary variable that equals to 1 if job Ji is processed in jth position in a sequence and 0, otherwise. In addition, let C[j] = C[j],2 be the completion time of a job processed in jth position in a sequence on machine M2. Thus, the MIP model can be described as follows. Its objective function is to minimize the weighted sum of the makespan and total flow time. n X min Z ¼ λC½n + ð1 - λÞ C½ j j¼1

s:t:

n X

δi;j ¼ 1;

1⩽ j⩽n

(1)

obtained by sequencing all the jobs in non-decreasing order of αj. Proof We consider three cases as follows: (1) If αmin ⩾ β, the completion time of the jth job is Cj = (1 + β)∏ji = 1(1 + αi). Thus, we have

λCmax + ð1 - λÞ

1⩽i⩽n

Cj ¼ ð1 + βÞ λ

n Y

1 + αj

j¼1

j¼1 i¼1

n X

Cj ¼ 1 + α½1

"

n-1 X ð1 + βÞj + ð1 + βÞn ð1 - λÞ

j¼1

(2)

C½ j;1 ¼ C½ j - 1;1 1 +

! δi;j α½i ;

1⩽ j⩽n

(3)

i¼1

C½ j;2 ⩾C½ j;1 1 +

n X

! δi;j β½i ;

1⩽ j⩽n

(4)

i¼1

C½j;2 ⩾C½j - 1;2 1 +

n X

! δi;j β½i ;

1⩽ j⩽n

(5)

i¼1

C½0;1 ¼ C½0;2 ¼ 1 δi;j 2 f0; 1g;

1⩽i; j⩽n

(6) (7)

Constraints (1) specify that only one job can be scheduled at one position. Constraints (2) indicate that only one position can be occupied by one job. Constraints (3) ensure that M1 will not be idle. Constraints (4) and (5) assure the starting time of the jth job at M2 is greater than or equal to the jth job’s completion time at M1 and the completion time of ( j − 1)th job on M2. Constraints (6) and (7) confirm that the boundary conditions for the decision variables are respected. The model contains n2 + 2n variables and 7n + 1 constraints.

j¼1

Obviously, if (1 + α[1]) = min1⩽ i ⩽ n{(1 + αi)}, then Equation (8) can be minimized. Therefore, we can obtain an optimal schedule by sequencing all the jobs in nondecreasing order of αj. (3) If αmin ⩽ β ⩽ αmax, the completion time of jth job can be described as the following subcases: (i) If there dose not exist any job such that its completion time at M1 is later than its predecessor’s completion time at M2, then we have C[j] = (1 + α[1])(1 + β) j for all jobs, and it is similar to case (2) above. (ii) If there exists a job J[r], with its completion time at M1 is later than its predecessor’s completion time at M2, then we have C½ j ¼ 1 + α½1 ð1 + βÞj ; j ¼ 1; 2; ¼ ; r - 1 and C½ j ¼ ð1 + βÞ

j Y 1 + α½i ;

j ¼ r; r + 1; ¼ ; n

i¼1

Therefore, we have n X j¼1

( C j ¼ ð1 + β Þ

n Y

1 + α½ j

j¼1

4. Polynomially solvable special cases

Theorem 1 For problem F2|pi,j = bi,jt|λCmax + (1 − λ)∑Cj, if βj ≡ β for all jobs, then an optimal schedule can be

#

(8)

λCmax + ð1 - λÞ In this section, we consider several special cases of F2|pi,j = bi,jt| λCmax + (1 − λ)∑Cj problem and derive properties of the optimal solution.

#

Since the term ∏ni = 1(1 + αj) is a constant, using pairwise job interchange, an optimal schedule can be obtained by sequencing all the jobs in non-decreasing order of αj. (2) If β ⩾ αmax, the completion time of the jth job is C[j] = (1 + α[1])(1 + β)j. Thus, we have

j¼1 n X

j n Y X ð1 + α i Þ + ð1 - λÞ

λCmax + ð1 - λÞ

δi;j ¼ 1;

"

j¼1

i¼1 n X

n X

+ ð1 - λÞ 1 + α½1

" r Y

ð1 + β Þi - 1

i¼1

#) j n-1 Y X + 1 + α½i j¼r + 1 i¼2

4


Using pairwise job interchange, it is obvious that we can obtain an optimal schedule by sequencing the jobs in nondecreasing order of αj. □ Theorem 2 For problem F2|pi,j = bi,jt|λCmax + (1 − λ)∑Cj, if αj ≡ α we have: (1) if α ⩽ βmin, an optimal schedule can be obtained by sequencing all the jobs in non-decreasing order of βj; (2) if α ⩾ βmax, an optimal schedule can be obtained by sequencing all the jobs in non-increasing order of βj. Proof (1) If α ⩽ βmin, then the completion time of jth job is C[ j] = (1 + α)∏ji = 1(1 + β[i]). Therefore, we have " n n Y X λCmax + ð1 - λÞ Cj ¼ ð1 + αÞ λ 1 + β½i j¼1

i¼1

+ ð1 - λ Þ

j n Y X

1 + β½ j

#

j¼1 i¼1

Obviously, the term is constant and the term Σnj = 1∏ji = 1(1 + β[j]) can be minimized by sequencing all the jobs in non-decreasing order of βj. (2) If α ⩾ βmax, then the completion time of jth is C[j] = (1 + β[j])(1 + α)j. Therefore, we have n X λCmax + ð1 - λÞ Cj ¼ 1 + β½n ð1 + αÞn j¼1

+ ð1 - λ Þ

ð1 + αÞj 1 + β½ j

j¼1

It is easy to see that an optimal schedule can be obtained by sequencing the jobs in non-increasing order of βj. □ Theorem 3 For problem F2|pi,j = bi,jt|λCmax + (1 − λ)∑Cj, if αj = βj for all jobs, then an optimal schedule can be obtained by sequencing all the jobs in non-decreasing order of αj or βj. Proof The proof can be easily obtained by pairwise job interchange and is omitted here. □ Theorem 4 For problem F2|pi,j = bi,jt|λCmax + (1 − λ)∑Cj, if the first job to be processed is fixed and αmax ⩽ βmin, then an optimal schedule can be obtained by sequencing the remaining (n − 1) jobs in non-decreasing order of βj. Proof By using the condition of αmax ⩽ βmin, we have j Y C½ j ¼ 1 + α½1 1 + β½1 1 + β½i i¼2

Therefore, λCmax + ð1 - λÞ

n X

Cj ¼ 1 + α½1

1 + β½1

j¼1

" λ

n Y

1 + β½ j

Theorem 5 For problem F2|pi,j = bi,jt|λCmax + (1 − λ)∑Cj, if the last processed job is fixed and αmin ⩾ βmax, then an optimal schedule can be obtained by sequencing the remaining (n − 1) jobs in non-decreasing order of (αj)/((1 + αj)(1 + βj)). Proof Under the condition of αmin ⩾ βmax, we have j-1 Y 1 + α½i C½ j ¼ 1 + β½ j 1 + α½ j i¼1

∏ni = 1(1 + β[i])

n-1 X

Since the first processed job is given and fixed, that is, (1 + α[1])(1 + β[1]) is constant, and the term ∏nj = 2(1 + β[j]) is also constant. Obviously, Σnj = 2∏ji = 2(1 + β[j]) can be minimized by sequencing the remaining (n − 1) jobs in nondecreasing order of βj. □

Therefore, λCmax + ð1 - λÞ

n X

"

# j n-1 Y X Cj ¼ ð1 - λÞ 1 + α½i 1 + β½ j

j¼1

j¼1 i¼1

+

n Y

1 + α½i

1 + β½n

i¼1

The term ∏ni = 1(1 + α[i])(1 + β[n]) is constant since the last processed job is given and fixed, and term Σnj =−11∏ji = 1 (1 + α[i])(1 + β[ j]) can be minimized by sequencing the remaining (n − 1) jobs not containing job J[n] in nondecreasing order of (αj)/((1 + αj)(1 + βj)). □

5. Dominance properties Dominance properties can be used to eliminate nodes in the branching tree of branch-and-bound algorithm in solving NP-hard scheduling problems. In the following, several dominance properties are proposed to search near optimal or optimal solution fast. Suppose there are two schedules σ1 = (π, Ji, Jj) and σ2 = (π, Jj, Ji), where π is a partial sequence of jobs. Let A and B denote the completion times for the last job in π on machine M1 and M2, respectively. Then, we can obtain the following dominance properties. Property 1 For any two consecutively scheduled jobs Ji and Jj, if their deterioration rate satisfies the following conditions: αi ⩽ βi and αi ⩽ αj implies βi ⩽ βj (agreeable condition), then σ1 dominates σ2. Proof The completion times of job Ji and Jj in σ is

j¼2

+ ð1 - λ Þ

j n Y X j¼2 i¼2

1 + β½ j

#

C½i ðσÞ ¼ maxfAð1 + αi Þ; Bgð1 + βi Þ ¼ maxfAð1 + αi Þð1 + βi Þ; Bð1 + βi Þg

ð9Þ

Mingbao Cheng et al—Two-machine flow shop scheduling with deteriorating jobs

C½ j ðσÞ ¼ maxfAð1 + αi Þ 1 + αj ;

Combining Equations (15) and (16), we have

Að1 + αi Þð1 + βi Þ; Bð1 + βi Þg 1 + βj max Að1 + αi Þ 1 + αj 1 + βj ; Að1 + αi Þð1 + βi Þ 1 + βj ; Bð1 + βi Þ 1 + βj ð10Þ Similarly, the completion times of job Jj and Ji in σ′ is C½i ðσ 0 Þ ¼ max Að1 + αj Þ; B ð1 + βj Þ ¼ max Að1 + aj Þð1 + βj Þ; Bð1 + βj Þ

ð11Þ

Combining Equations (9), (11) and the agreeable conditions of the Property, we have C½i ðσÞ - C½i ðσ′Þ⩽0

(13)

Combining Equations (10), (12) and the agreeable conditions of Property, we have C½ j ðσÞ - C½ j ðσ′Þ⩽0

X

i C½ j ðσ′Þ ¼ ð1 - λÞB βi - βj ⩽0

Property 3 For any two consecutively scheduled jobs Ji and Jj, if their deterioration rate satisfies the following conditions: B ⩽ A(αi + 1), αi < αj < βj and αi + βi + αiβi ⩽ αj + βj + αjβj, then σ1 dominates σ2. Property 4 For any two consecutively scheduled jobs Ji and Jj, if their deterioration rate satisfies the following conditions: A(1 + αi) ⩽ B, A(1 + αj) ⩾ B and βj ⩾ βi ⩾ αj, then σ1 dominates σ2. The proofs of Properties 3 and 4 are omitted since they are similar to that of Property 2.

6. Branch-and-bound algorithm In this section, a greedy heuristic algorithm (GHA) for the upper-bound of the problem is proposed and a lower bound is also considered. Then, a branch-and-bound algorithm is used to solve the problem.

6.1. The algorithm for upper-bound The GHA is a very simple method to solve some complex problems. In this subsection, we propose such a heuristic algorithm to identify the upper bound of the studied problem. The procedure of GHA is given as follows. Initialize αi, βi for i = 1, 2, …, n and λ (randomly generated), and t0 = 1.

h i X λC½ j ðσ Þ + ð1 - λÞ C½ j ðσ Þ h i - λC½ j ðσ′Þ + ð1 - λÞ½ j ðσ′Þ ¼ ð1 - λÞ C½i ðσ Þ - C½i ðσ′Þ + C½ j ðσ Þ - C½ j ðσ′Þ ⩽0 and

+ ð1 - λ Þ

(14)

Similarly, combining inequalities (13) and (14), we have

σ2

h i λC½ j ðσ Þ + ð1 - λÞ½ j ðσ Þ - λC½ j ðσ′Þ

Therefore, σ1 dominates σ2 and the proof is completed. □

C½ j ðσ 0 Þ ¼ max A 1 + αj ð1 + αi Þ; A 1 + α j 1 + β j ; B 1 + β j ð1 + β i Þ ¼ max A 1 + αj ð1 + αi Þ 1 + βj ; A 1 + αj 1 + βj ð1 + βi Þ; B 1 + βj ð1 + βi Þ ð12Þ

Therefore, σ1 dominates completed. □

5

the

Step 1: Step 2: proof

is

Property 2 For any two consecutively scheduled jobs Ji and Jj, if their deterioration rate satisfies the following conditions: A(1 + αi)(1 + αj) ⩽ B(1 + βi) and βi ⩽ βj, then σ1 dominates σ2. Proof The completion times of job Ji and Jj in σ is C½i ðσ Þ ¼ Bð1 + βi Þ; C½ j ðσ Þ ¼ Bð1 + βi Þ 1 + βj

Step 3:

(15)

Similarly, the completion times of job Jj and Ji in σ′ is C½i ðσ′Þ ¼ B 1 + βj ; C½ j ðσ′Þ ¼ B 1 + βj ð1 + βi Þ (16)

Step 4: Step 5:

Set k = 1 and N = {J1, J2, …, Jn}. Choose a job Jp ∈ N with the smallest (1 + αp) (1 + βp) and schedule it in the kth position. Let A[k] = (1 + αp)t0, C[k] = (1 + αp)(1 + βp)t0, and set N = N − {Jp}. If there exists job Ji ∈ N with A[k](1 + αi) ⩽ C[k], assign the job Ji with the smallest deterioration rate βi to the (k + 1)th position. Let A[k + 1] = A[k](1 + αi) and C[k + 1] = C[k](1 + βi). Otherwise, assign the job Ji ∈ N with smallest (1 + αi)(1 + βi) to the (k + 1)th position. Let A[k + 1] = A[k](1 + αi) and C[k + 1] = A[k + 1](1 + βi). Set N = N − {Ji} and k = k + 1. If k < n, go to Step 3. Let σ0 be an original schedule obtained from the above procedure, and reset k = 1.

6


Step 6: Step 7:

Step 8: Step 9: Step 10:

Set j = k + 1. Create a new schedule σ by pairwise job interchanging in the kth and jth positions from σ0. If the value of the objective function of σ is smaller than that of σ0, then replace σ0 with σ. If j < n, set j = j + 1 and go to Step 7. Otherwise, go to Step 9. If k < n − 1, set k = k + 1 and go to Step 6. Otherwise, stop. Calculate the value of schedule σ, that is, UB = λCmax(σ) + (1 − λ)Σnj = 1Cj(σ).

In the following, we use a five-job, two-machine flow shop example to illustrate the decision process of the proposed GHA heuristic. The objective function is assumed to be 0.5Cmax + 0.5∑Cj. Table 1 shows the pertinent data of the studied problem. An upper bound of the case is given by using GHA as follows. Stage 1

Stage 3 A½2 ð1 + α1 Þ ¼ 1:773 ´ ð1 + 0:37Þ ¼ 2:429C½2 A½2 ð1 + α4 Þ ¼ 1:773 ´ ð1 + 0:60Þ ¼ 2:837>C½2 By steps 3 and 4 and assigning J1 to the third position, we have A[3] = 2.429 and C[3] = 3.638. Stage 4 A½3 ð1 + α3 Þ ¼ 2:429 ´ ð1 + 0:65Þ ¼ 4:008>C½3

A½3 ð1 + α4 Þ ¼ 2:429 ´ ð1 + 0:60Þ ¼ 3:887>C½3 Since (1 + α3)(1 + β3) < (1 + α4)(1 + β4), by steps 3 and 4 and assigning J3 to the fourth position, we have A[4] = 4.008 and C[4] = 5.612. Stage 5

ð1 + α1 Þð1 + β1 Þ ¼ ð1 + 0:37Þ ´ ð1 + 0:41Þ ¼ 1:932 ð1 + α2 Þð1 + β2 Þ ¼ ð1 + 0:24Þ ´ ð1 + 0:53Þ ¼ 1:897 ð1 + α3 Þð1 + β3 Þ ¼ ð1 + 0:65Þ ´ ð1 + 0:40Þ ¼ 2:310

A½4 ð1 + α4 Þ ¼ 4:008 ´ ð1 + 0:42Þ ¼ 6:413>C½4 and C½5

ð1 + α4 Þð1 + β4 Þ ¼ ð1 + 0:60Þ ´ ð1 + 0:64Þ ¼ 2:624

¼ 10:518

ð1 + α5 Þð1 + β5 Þ ¼ ð1 + 0:43Þ ´ ð1 + 0:36Þ ¼ 1:945 By step 2, we assign J2 to the first position; we then have A[1] = 1.240 and C[1] = 1.897. Stage 2 A½1 ð1 + α1 Þ ¼ 1:24 ´ ð1 + 0:37Þ ¼ 1:699C½1 A½1 ð1 + α4 Þ ¼ 1:24 ´ ð1 + 0:60Þ ¼ 1:984>C½1

M1

A½1 ð1 + α5 Þ ¼ 1:24 ´ ð1 + 0:43Þ ¼ 1:773