11.optimal three stage flow shop scheduling in which processing time, set up time, each associated with probabilities including transportation time

Journal of Natural Sciences Research www.iiste.org
ISSN 2224-3186 (Paper) ISSN 2225-0921 (Online)
Vol.1, No.3, 2011

Optimal Three Stage Flow Shop Scheduling in which
Processing Time, Set Up Time, Each Associated With
Probabilities Including Transportation Time and Jobs in a
String of Disjoint Job-Blocks
Sameer Sharma*
Department of Mathematics, Maharishi Markandeshwar University, Mullana, Ambala, India
* E-mail of the corresponding author: samsharma31@yahoo.com

Abstract
The paper deals with n jobs, 3 machines flow shop production scheduling in which processing times and set
up times are associated with their respective probabilities involving transportation time and jobs are
processed in two disjoint job blocks in a string. A heuristic method with an objective to minimize the total
time elapsed time/idle time of the jobs/machines is discussed. A computer program followed by numerical
illustration is given to clarify the algorithm.
Keywords: Disjoint job-block, Set up time, Processing time, Transportation time, Equivalent job

1. Introduction
In today’s world of global competition, scheduling problem has become vital in order to meet customer
requirements as promptly as possible while maximizing the profits. A scheduling problem is to find
sequences of jobs on given machines with the objective of minimizing some function of the job completion
times. In flow shop scheduling all jobs passes through all machines in the same order without any pre-
emission. Johnson (1954) gave a heuristic technique for production schedule in which n jobs are processed
on two or three machines in an ordered manner to minimize the total idle time of machines. The work was
developed by Ignall & Scharge (1965), Camphell (1970) , Maggu & Das (1977) , Yoshida & Hitomi (1979)
, Singh (1985) , Anup (2002) ,Chandramouli (2005) , Khodadadi (2008), Pandian & Rajenderan (2010) by
considering various parameters. Heydari (2003) dealt with a flow shop scheduling problem where the jobs
are processed in two disjoint job blocks in a string consists of one block in which order of jobs is fixed &
other block in which order of job is arbitrary.
Gupta, Sharma & Gulati (2011) studied n×3 machine flow shop schedule in which processing time, set up
time, each associated with probabilities along with jobs in a string of disjoint job-blocks. Most machine
scheduling models assume that jobs are delivered instantaneously from one location to another without
considering significant transportation time. However, there are many situations where the transportation
times are quite significant and can not be simply neglected. As example, when the machines on which jobs
are to be processed are planted at different stations and these jobs require form of loading-time of jobs,
moving time and then unloading-time of jobs. In this paper, we have studied machines scheduling problems
with explicit transportation considerations. We have extended the study made by Gupta, Sharma & Gulati
(2011) by introducing the concept of transportation time. The problem discussed here is wider and
practically more applicable and has significant results in process industries

2. Practical Situation
Many applied and experimental situations exist in our day-to-day working in factories and industrial
production concerns. In many manufacturing / production companies different jobs are processed on
various machines. These jobs are required to process in a machine shop A, B, C, ---- in a specified order.
When the machines on which jobs are to be processed are planted at different places, the transportation

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time (which includes loading time, moving time and unloading time etc.) has a significant role in
production concern. Setup includes work to prepare the machine, process or bench for product parts or the
cycle. This includes obtaining tools, positioning work-in-process material, return tooling, cleaning up,
setting the required jigs and fixtures, adjusting tools and inspecting material and hence significant. The idea
of job block has practical significance to create a balance between a cost of providing priority in service to
the customer and cost of giving service with non priority, .i.e. how much is to be charged from the priority
customer(s) as compared to non priority customer(s).

3. Notations
S : Sequence of jobs 1, 2, 3… n
Sk : Sequence obtained by applying Johnson’s procedure, k = 1, 2, 3, -------
Mj : Machine j, j= 1, 2, 3
M : Minimum makespan
aij : Processing time of ith job on machine Mj
pij : Probability associated to the processing time aij
sij : Set up time of ith job on machine Mj
qij : Probability associated to the set up time sij
Aij : Expected processing time of ith job on machine Mj
Sij : Expected set up time of ith job on machine Mj
Iij(Sk): Idle time of machine Mj for job i in the sequence Sk
Ti , j →k : Transportation time of ith job from jth machine to kth machine

4. Assumptions
1. n jobs be processed through three machines M1, M2 & M3 in the order M1M2M3 i.e. no passing is
allowed.
2. A sequence of k jobs i1, i2 …… ik as a block or group-job in the order ( i1, i2 ……ik ) shows priority
of job i1 over i2, etc.
3. Jobs may be held in inventory before going to a machine.
4. The storage space is available and the cost of holding inventory for each job is either same or
negligible.
5. Time intervals for processing are independent of the order in which operations are performed.
6. A job is an entity i.e. even though the job represents a lot of individual part; no job may be
processed by more than one machine at a time.
7. Each operation once started must performed till completion.

4. Problem Formulation
Let some job i (i = 1,2,……..,n) is to be processed on three machines Mj ( j = 1,2,3) in the way such that no
passing is allowed. Let aij be the processing time of ith job on jth machine with probabilities pij and sij be the
setup time of ith job on jth machine with probabilities qij. Let Aij be the expected processing time and Si,j be
the expected setup time of ith job on jth machine. Let Ti,j→k be the transportation time of ith job from jth
machine to kth machine..Let α =(ik, im) be an equivalent job for job block in which job ik is given priority
over job im Take two job blocks α and β such that block α consists of m jobs out of n jobs in which the
order of jobs is fixed and β consists of r jobs out of n in which order of jobs is arbitrary such that m + r = n.
let α ∩ β = Φ i.e. the two job blocks α & β form a disjoint set in the sense that the two blocks have no job
in common. A string S of job blocks α and β is defined as S = (α, β). Our objective is to find an optimal

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Vol.1, No.3, 2011
schedule for all the jobs minimizing the total elapsed time.
The mathematical model of the problem in matrix form can be stated as:

Jobs Machine A Ti ,1→2 Machine B Ti ,2→3 Machine C
i ai1 pi1 si1 qi1 ai 2 pi 2 si2 qi2 ai 3 pi 3 si3 qi3
1 a11 p11 s11 q11 T1,1→2 a12 p12 s12 q12 T1,2→3 a13 p13 s13 q13
2 a21 p21 s21 q21 T2,1→2 a22 p22 s22 q22 T2,2→3 a23 p23 s23 q23
3 a31 p31 s31 q31 T3,1→2 a32 p32 s32 q32 T3,2→3 a33 p33 s33 q33
4 a41 p41 s41 q41 T4,1→2 a42 p42 s42 q42 T4,2→3 a43 p43 s43 q43
- - - - - - - - - - - - - - -
n an1 pn1 sn1 qn1 Tn,1→2 an 2 pn 2 sn2 qn2 Tn,2→3 an 3 pn3 sn3 qn3
(Tableau 1)

5. Algorithm:
Step 1: Calculate the expected processing times and expected set up times as follows
Aij = aij × pij and Sij = sij × qij ∀i, j =1,2,3
Also we consider the following structure relation holds good if
Step 2: Check the condition
Either Min {Ai1 + Ti,1→2 – Si2} ≥ Max{Ai2 + Ti,1→2 – Si1}
or Min{Ai3 + Ti,2→3 – Si2} ≥ Max{Ai2 + Ti,2→3 – Si3} or both for all i
If the conditions are satisfied then go to step 3, else the data is not in the standard form.
Step 3: Introduce the two fictitious machines G and H with processing times Gi and Hi as
Gi = Ai1 + Ai2 + max (Si1 , Si2) + Ti,1→2 and Hi = Ai2 + Ai3 - Si3 + Ti,2→3
Step 4: Take equivalent job α = (ik , im) for the given job block (ik , im ) and define its processing time on the
lines of Maggu & Das (1977) defined as follows:
Gα = Gk + Gm – min(Gm ,H k)
Hα = Hk+ Hm – min(Gm ,H k )
Step 5: Obtain the order of jobs in the job block β in an optimal manner using Johnson’s (1954) technique
by treating job block β as sub flow shop scheduling problem of the main problem. Let β be the new job
'

block. Define its processing time Gβ΄ & Hβ΄ on the lines of Maggu & Das (1977) as defined in step 4.
Now, the given problem reduce into new problem replacing m jobs by job block α with processing times Gα
& Hα on machine G & H respectively as defined in step 4 and r jobs of job block β by β΄ with processing
times Gβ΄ & Hβ΄ on machine G & H respectively as defined in step 5.

The new problem can be represented as –
Jobs (i) Machine G ( Gi ) Machine H (Hi)
α Gα Hα
β' Gβ’ Hβ’
(Tableau – 2)

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Vol.1, No.3, 2011
Step 6: Consider S1 of all the processing time Gi when Gi ≤ H i and let S2 denote the set of processing times
which are not covered in set S1.
Step 7: Let S'1 denote a suboptimal sequence of jobs corresponding to non decreasing times in set S1 & let
S'2 denote a suboptimal sequence of jobs corresponding to non-decreasing times in set S2.
Step 8: The augmented ordered sequence (S'1, S'2) gives optimal sequence for processing the jobs for the
original problem.

6. Programme
#include<iostream.h>
#include<stdio.h>
#include<conio.h>
#include<process.h>

int n;
float a1[16],b1[16],c1[16],g[16],h[16],sa1[16],sb1[16],sc1[16];
float macha[16],machb[16],machc[16];
int e;int group[16];//variables to store two job blocks
float minval;int gg=0;float gcal;float hcal;float gbeta=0.0,hbeta=0.0;float galfa=0.0,halfa=0.0;
char s1[5];char s2[5];
void ghcal(float k,float m)
{
float minv;
if(g[m]>h[k])
minv=h[k];
else
minv=g[m];gcal=g[k]+g[m]-minv;hcal=h[k]+h[m]-minv;
//return(c);}
void main()
{
clrscr();
int a[16],b[16],c[16],sa[16],sb[16],sc[16],T12[16],T23[16];
float p[16],q[16],r[16],u[16],v[16],w[16];float maxv;
cout<<"How many Jobs (<=15) : ";cin>>n;
if(n<1 || n>15)
{ cout<<endl<<"Wrong input, No. of jobs should be less than 15..n Exitting"; getch();
exit(0);}
for(int i=1;i<=n;i++)
{cout<<"nEnter the processing time and its probability, Setup time and its probability of
"<<i<<" job for machine A and Transportation time from A to B : ";
cin>>a[i]>>p[i]>>sa[i]>>u[i]>>T12[i];
cout<<"nEnter the processing time and its probability, Setup time and its probability of
"<<i<<" job for machine B and Transportation time from B to C : ";
cin>>b[i]>>q[i]>>sb[i]>>v[i]>>T23[i];
cout<<"nEnter the processing time and its probability, Setup time and its probability of
"<<i<<"job for machine C: ";cin>>c[i]>>r[i]>>sc[i]>>w[i];

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Vol.1, No.3, 2011
//Calculate the expected processing times of the jobs for the machines:
a1[i] = a[i]*p[i]; b1[i] = b[i]*q[i]; c1[i] = c[i]*r[i];
//Calculate the expected setup times of the jobs for the machines:
sa1[i] = sa[i]*u[i];sb1[i] = sb[i]*v[i];sc1[i] = sc[i]*w[i];}
cout<<endl<<"Expected processing time of machine A, B and C: n";
for(i=1;i<=n;i++)
{cout<<a1[i]<<"t"<<sa1[i]<<"t"<<T12[i]<<"t"<<b1[i]<<"t"<<sb1[i]<<"t"<<T23[i]<<"t"<<c1
[i]<<"t"<<sc1[i];cout<<endl;}
//Finding smallest in a1
float mina1;mina1=a1[1]+T12[1]-sb1[1];
for(i=2;i<n;i++)
{if((a1[i]+T12[i]-sb1[i])<mina1)
mina1=a1[i]+T12[i]-sb1[i];}
//For finding largest in b1
float maxb1;maxb1=b1[1]+T12[1]-sa1[1];
for(i=2;i<n;i++)
{ if(b1[i]+T12[i]-sa1[i]>maxb1)
maxb1=b1[i]+T12[i]-sa1[i];}
float maxb2;maxb2=b1[1]+T23[1]-sc1[i];
for(i=2;i<n;i++)
{if((b1[i]+T23[i]-sc1[i])>maxb2)
maxb2=b1[i]+T23[i]-sc1[i];}
//Finding smallest in c1
float minc1;minc1=c1[1]+T23[1]-sb1[i];
for(i=2;i<n;i++)
{if((c1[i]+T23[i]-sb1[i])<minc1)
minc1=c1[i]+T23[i]-sb1[i];}
if(mina1<=maxb1||minc1<=maxb2)
{g[i]=a1[i]+b1[i]+maxv;h[i]=b1[i]+c1[i]-sc1[i];}
else {cout<<"n data is not in Standard Form...nExitting";getch();exit(0);}
//Function for two ficticious machine G and H
for(i=1;i<=n;i++)
{if(sa1[i]>sb1[i])
{maxv= sa1[i];}
else {maxv=sb1[i];}
g[i]=a1[i]+b1[i]+maxv+T12[i];h[i]=b1[i]+c1[i]-sc1[i]+T23[i]; }
cout<<endl<<"Expected processing time for two fictious machines G and H: n";
for(i=1;i<=n;i++)
{ cout<<endl;cout<<g[i]<<"t"<<h[i];cout<<endl;}
cout<<"nEnter the number of fixed jobs in job block alpha <="<<n<<":";cin>>e;
cout<<"nEnter the fixed job blocks ("<<e<<" numbers from 1 to "<<n<<") alpha : ";
for(int y=1;y<=e;y++)
{cin>>group[y];}

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Vol.1, No.3, 2011
cout<<"nEnter the jobs having disjoint job block ( numbers from 1 to "<<n<<" other than the
fixed job block) beta:";
for(int j=e+1;j<=n;j++)
{cin>>group[j];}
float btj[16],btg[16],bth[16];
cout<<"Expected processing time for two fictious machines G and H for Beta: n";
for(i=1,j=e+1;j<=n;i++,j++)
{btj[i]=group[j];btg[i]=g[group[j]];bth[i]=h[group[j]];
cout<<endl<<btj[i]<<"t"<<btg[i]<<"t"<<bth[i];}
float mingh[16];char ch[16];
for(i=1;i<=n-e;i++)
{
if(btg[i]<bth[i])
{
mingh[i]=btg[i];ch[i]='g';
}
else
{
mingh[i]=bth[i];ch[i]='h';
}}
for(i=1;i<=n-e;i++)
{for(j=1;j<=n-e;j++)
if(mingh[i]<mingh[j])
{float temp=mingh[i]; int temp1=btj[i]; char d=ch[i];
mingh[i]=mingh[j]; btj[i]=btj[j]; ch[i]=ch[j];
mingh[j]=temp; btj[j]=temp1; ch[j]=d;}}
// calculate beta scheduling
float sbeta[16];int t=1,s=0;
for(i=1;i<=n-e;i++)
{if(ch[i]=='h')
{ sbeta[(n-s-e)]=btj[i];s++;}
else if(ch[i]=='g')
{sbeta[t]=btj[i];t++;}
}
cout<<endl<<endl<<"Beta Scheduling:"<<"t";
for(i=1;i<=n-e;i++)
{cout<<sbeta[i]<<" ";}
//calculate G_Alfa and H_Alfa
ghcal(group[1],group[2]);
galfa=gcal;halfa=hcal;i=3;
while(i<=e)
{
if(i>e)

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Vol.1, No.3, 2011
break;
else
{
if(g[group[i]]<halfa)
minval=g[group[i]];
else
minval=halfa;galfa=galfa+g[group[i]]-minval;halfa=halfa+h[group[i]]-minval;
}
i++;
}
cout<<endl<<endl<<"G_Alfa="<<galfa;cout<<endl<<"H_Alfa="<<halfa;
//calculate G_Beta and H_Beta
ghcal(sbeta[1],sbeta[2]);gbeta=gcal;hbeta=hcal;i=3;
while(i<=(n-e))
{
if(i>(n-e))
break;
else
{
if(g[sbeta[i]]<hbeta)
minval=g[sbeta[i]];
else
minval=hbeta;gbeta=gbeta+g[sbeta[i]]-minval;hbeta=hbeta+h[sbeta[i]]-minval;}
i++;}
cout<<endl<<endl<<"G_Beta="<<gbeta;cout<<endl<<"H_Beta="<<hbeta;
//calculate optimal sequence
if(galfa<=halfa)
{
s1[1]='a';s2[1]='0';
}
else
{
s2[1]='a';s1[1]='0';
}
if(gbeta<=hbeta)
{
s1[2]='b';s2[2]='0';
}
else
{
s2[2]='b';s1[2]='0';
}
//cout<<endl<<endl<<"Optimal Sequence:"<<"t";

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Vol.1, No.3, 2011
int arr[16];
if(s1[1]=='a')
{
//cout<<"n a";
for(i=1;i<=e;i++)
{
//cout<<group[i]<<"t";
arr[i]=group[i];
}
gg=gg+e;
}
if(s1[2]=='b')
{
//cout<<"n b";
for(i=1;i<=n-e;i++)
{
//cout<<endl<<sbeta[i]<<"t";
arr[i+gg]=sbeta[i];
}
gg=gg+(n-e)+1;
}
if(s2[1]=='a')
{
//cout<<"n a";
for(i=1;i<=e;i++)
{
//cout<<endl<<group[i]<<"t";
arr[i+gg]=group[i];
}
gg=gg+e;
}
if(s2[2]=='b')
{
//cout<<"n b";
for(i=1;i<=(n-e);i++)
{
//cout<<sbeta[i]<<"t";
arr[i+gg]=sbeta[i];
}}
//calculating total computation sequence;
float time=0.0,macha1[16] ;float maxv1[16],maxv2[16];
macha[1]=time+a1[arr[1]];
for(i=2;i<=n;i++)

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Vol.1, No.3, 2011
{macha1[i]=macha[i-1]+sa1[arr[i-1]];macha[i]=macha1[i]+a1[arr[i]];}
machb[1]=macha[1]+b1[arr[1]]+T12[arr[1]];
for(i=2;i<=n;i++)
{if((machb[i-1]+sb1[arr[i-1]])>(macha[i]+T12[arr[i]]))
{maxv1[i]=machb[i-1]+sb1[arr[i-1]];}
else
{maxv1[i]=macha[i]+T12[arr[i]];}machb[i]=maxv1[i]+b1[arr[i]];}
machc[1]=machb[1]+c1[arr[1]]+T23[arr[1]];
for(i=2;i<=n;i++)
{if((machc[i-1]+sc1[arr[i-1]])>(machb[i]+T23[arr[i]]))
maxv2[i]=machc[i-1]+sc1[arr[i-1]];
else
maxv2[i]=machb[i]+T23[arr[i]];machc[i]=maxv2[i]+c1[arr[i]];}
//displaying solution
cout<<"nnnnnttt #####THE SOLUTION##### ";
cout<<"nnt***************************************************************";
cout<<"nnnt Optimal Sequence is : ";
for(i=1;i<=n;i++)
{cout<<" "<<arr[i];}
cout<<endl<<endl<<"In-Out Table is:"<<endl<<endl;
cout<<"Jobs"<<"t"<<"Machine M1"<<"t"<<"t"<<"Machine M2" <<"t"<<"t"<<"Machine M3"<<endl;
cout<<arr[1]<<"t"<<time<<"--"<<macha[1]<<" t"<<"t"<<macha[1]+T12[arr[1]]<<"--"<<machb[1]<<"
t"<<"t"<<machb[1]+T23[arr[1]]<<"--"<<machc[1]<<endl;
for(i=2;i<=n;i++)
{cout<<arr[i]<<"t"<<macha1[i]<<"--"<<macha[i]<<" "<<"t"<<maxv1[i]<<"--"<<machb[i]<<"
"<<"t"<<maxv2[i]<<"--"<<machc[i]<<endl;}
cout<<"nnnTotal Computation Time (T) = "<<machc[n];
float sum1=0.0,sum2=0.0,sum3=0.0;
for(i=1;i<=n;i++)
{sum1=sum1+a1[i];sum2=sum2+b1[i];sum3=sum3+c1[i];}
float tt=machc[n];
cout<<endl<<endl<<endl<<"Total Expected Idle Time on Machine A: "<<(tt-sum1);
cout<<endl<<"Total Expected Idle Time on Machine B: "<<(tt-sum2);
cout<<endl<<"Total Expected Idle Time on Machine C: "<<(tt-sum3);
cout<<"nnt***************************************************************";
getch();
}

7. Numerical Illustration
Consider 5 jobs, 3 machine flow shop problem with processing time ,setup time associated with their
respective probabilities and transportation time as given in table
Jobs Machine M1 Machine M2 Machine M3
Ti ,1→2 Ti ,2→3
i ai1 pi1 si1 qi1 ai2 pi2 si2 qi2 ai3 pi3 si3 qi3

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Vol.1, No.3, 2011
1 30 0.2 3 0.3 2 10 0.3 3 0.2 2 20 0.2 4 0.2
2 32 0.2 2 0.1 1 22 0.2 2 0.2 1 19 0.3 3 0.2
3 43 0.1 2 0.3 2 22 0.2 1 0.3 2 15 0.2 3 0.3
4 25 0.2 4 0.1 3 25 0.1 3 0.1 3 24 0.1 4 0.2
5 23 0.3 2 0.2 4 10 0.2 2 0.2 1 26 0.2 5 0.1
(Tableau 3 )
Our objective is to obtain an optimal schedule for above said problem to minimize the total production time
/ total elapsed time in which jobs 2,5 are to be processed as a group job in a fixed order and remaining jobs
as a disjoint string in any random order.
Solution: As per Step 1: the expected processing times and expected setup times for machines M1, M2 and
M3 are as shown in table 4.
As per step 2: The expected processing time for two fictitious machine G & H is as shown in table 5.
As per step 3: Here β = (1, 3, 4)
Now, using Johnson (1954) technique by treating job block β as sub flow shop scheduling problem of the
main problem. Let β΄ be the new job block. Here we get β΄ = (3, 1, 4)
As per step 4: Here α = (2, 5)
Therefore, Gα = 12.2 + 13.3 – 10.5 = 15 and Hα = 10.5 + 7.7 – 10.5 = 7.7

Also β' = (3,1,4) = ((3, 1), 4) = (α
'
, 4) , where α ' = (3, 1)
Therefore, G ' = 11.3 + 11.9 – 8.5 =14.7 and H ' = 8.5 + 8.2 – 8.5 = 8.2
α α
G ' = 14.7 + 10.9 – 8.2 = 17.4 and H ' = 8.2 + 7.1 – 8.2 = 7.1
β β
Now problem reduces to jobs α and β΄ as shown in table 6
As per step 5: S1 = φ, S2 = [15, 17.4]
As per step 6: S’1 = φ, S’2 = (α, β΄)
As per step 7: The Optimal sequence is S = 2 – 5 – 3 – 1 – 4 .The In-Out flow table for the optimal
sequence S is as shown in table 7.
Total expected idle time on machine A = 0.2+0.4+0.6+0.9+10.9 = 13hrs
Total expected idle time on machine B = 7.4+5.7+0.7+2.2+3.9+5.4= 25.3 hrs
Total expected idle time on machine C = 12.8+2+0.9+2.2+3.4 = 21.3 hrs

8. Conclusions
This paper provides a new heuristic method with an objective to minimize the total time elapsed time / idle
time of the jobs / machines for n jobs, 3 machines flow shop production scheduling in which processing
times and set up times are associated with their respective probabilities involving transportation time and
jobs are processed in two disjoint job blocks in a string. The study may further be extended by considering
various parameters such as weights in jobs, arbitrary time, break down interval etc.

References
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14(4), 277-281.
Rajenderan, C.(1993), “Heuristic algorithm for scheduling in flow shop to minimize the total flow time”,
International Journal of Production Economics, 29(1), 65-73.
Singh, T.P.(1985), “On n x 2 flow shop problem involving job block, transportation times & break- down
machine times”, PAMS XXI, 1-2.
Singh, T P, Kumar, R. & Gupta, D. (2005), “Optimal three stage production schedule, the processing and
set up times associated with probabilities including job block criteria”, Proceedings of National Conference
on FACM, 463-470.
Pandian, P. &.Rajendran, P.(2010), “Solving constrained flow-shop scheduling problems with three
machines”, Int. J. Contemp. Math. Sciences, 5(19), 921-929.
Yoshida & Hitomi (1979), “Optimal two stage production scheduling with set up times separated”, AIIE
Transactions, Vol . II, 261-263

28

Vol.1, No.3, 2011

Tables
Table 4: The expected processing times and expected setup times for machines M1, M2 and M3 are

Jobs Ai1 Si1 Ti ,1→2 Ai2 Si2 Ti ,2→3 Ai3 Si3
1 6.0 0.9 2 3.0 0.6 2 4.0 0.8
2 6.4 0.2 1 4.4 0.4 1 5.7 0.6
3 4.3 0.6 2 4.4 0.3 2 3.0 0.9
4 5.0 0.4 3 2.5 0.3 3 2.4 0.8
5 6.9 0.4 4 2.0 0.4 1 5.2 0.5

Table 5: The expected processing time for two fictitious machine G & H is
Jobs Gi Hi
1 11.9 8.2
2 12.2 10.5
3 11.3 8.5
4 10.9 7.1
5 13.3 7.7

Table 6: The new reduced problem is
Jobs(i) Machine G(Gi) Machine H(Hi)
α '
15 7.7
β' 17.4 7.1

Table 7: The In-Out flow table for the optimal sequence S is
Jobs Machine M1 Ti ,1→2 Machine M2 Ti ,2→3 Machine M3

i In – Out In – Out In - Out

2 0 – 6.4 1 7.4 – 11.8 1 12.8 – 18.5

5 6.6 – 13.5 4 17.5 – 19.5 1 20.5 – 25.7

3 13.9 – 18.2 2 20.2 – 24.6 2 26.6 – 29.6

1 18.8 – 24.8 2 26.8 – 29.8 2 31.8 – 35.8

4 25.7 – 30.7 3 33.7 – 36.2 3 39.2 – 41.6

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