Minimization Of Makespan In Three Stage Flowshop

Published Date: 02 Nov 2017

Deepak Gupta

Prof. & Head, Dept. of Mathematics,

Maharishi Markandeshwar University, Mullana, Ambala, India

[email protected]

Harminder Singh1, Hardeep Singh2, Pooja Sharma3

Research Scholar, Dept. of Mathematics

Maharishi Markandeshwar University, Mullana, Ambala, India

[email protected], [email protected],

[email protected]

Abstract: This paper is an attempt to study n-job,3-machine flow shop scheduling problem in which processing times are associated with their respective probabilities and the transportation time is also considered .The idle/waiting time operator is being used for the equivalent job for a job block criteria.The objective of the study is to get an optimal sequence of jobs in order to minimize the total elapsed time. Algorithm is made clear with the help of numerical illustration.

Keywords:

Flow shop, Equivalent Job block, Transportation Time, Elapsed Time, idle/waiting time operator Oi,w

Introduction:

The basic study in the field of scheduling made by Johnson(1954) ,who developed an algorithm to minimize the total elapsed time in two,three stage flow shop , Smith(1967) considered minimization of mean flow time and maximum tardiness.Yashida &hitomi (1979) further considered the problem with set up time.The work was developed by Sen and Gupta (1983),Chandasekharan (1992).Bagga & Bhambani (1997) and Gupta Deepak(2011) by considering various parameters.Maggu & Dass(1977) established an equivalent theorem.Singh T.P & Gupta Deepak(2004) made an attempt to study the optimal two stage production schedule in which processing time and set up time both were associated with probabilities including job block criteria.

Further we have made an attempt to extend the study done by Gupta Deepak(2012) for n-job,3-machine flow shop scheduling to minimize the total elapsed time with the application of idle/waiting time operator for an equivalent job for a job block.The processing times are associated with their respective probabilities and the concept of transportation time is also included . The operator theorem is very useful in economical and computational point of view and gives optimal schedule in order to optimize total elapsed time . Also the concept of equivalent job block has many applications in different fields where priority of one job over the other becomes significant.

Practical situation:

Various problems occur in real life when priority of one job over other becomes significant.It usually occurs in so many fields such as hospitals,factories etc.An extra price is paid to get the production in desired sequence.And to do so in production, we can use the job block criteria.The different applications are given by researchers for the job block criteria.Maggu and dass gave an algorithm for the job block criteria,which is very useful.Also an Idle/Waiting time operator is used for finding an equivalent job for a given job block.Operator theorem becomes significant for the application of idle/waiting time operator.

Notations:

S = sequence of jobs 1,2,3,4,….n.

A,B,C: Three different machines.

Ai=Processing times of job ith on ist machine A.

Bi=Processing times of job ith on second machine B.

Ci=Processing times of job ith on third machine C.

ti=Transportation time of ith job from machine A to machine B.

gi=Transportation time of ith job from machine B to machine C.

pi=Probabilities associated with processing times Ai.0 ≤ pi ≤ 1 , 0 ≤ ∑pi ≤ 1.

qi=Probabilities associated with processing times Bi.0 ≤qi ≤1 , 0 ≤∑qi ≤ 1.

ri=Probabilities associated with processing times Ci.0≤ri ≤1 , 0≤∑ri ≤ 1.

Problem Formulation:

Let n jobs are to be processed on three machinesA,B and C.Let Ai ,Bi ,Ci (i=1,2,3,…n) be the processing times of each job on machines A,B and C respectively.Let ti and gi (i=1,2,…n) be the transportation times of job i from machine A to B and from B to C.The mathematical model of the problem is as follows:

job

Machine A

Transportation time from A to B

Machine B

Transportation time from B to C

Machine C

i

Ai

pi

ti

Bi

qi

gi

Ci

ri

1

A1

p1

t1

B1

q1

g1

C1

ri

2

A2

p2

t2

B2

q2

g2

C2

r2

3

A3

p3

t3

B3

q3

g3

C3

r3

4

A4

p4

t4

B4

q4

g4

C4

r4

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

n

An

pn

tn

Bn

qn

gn

Cn

rn

Our intension is to find an optimal sequence so as to find the minimum elapsed time.

Algorithm:

Step1: Define expected processing times ai , bi and ci on machines A ,B and C respectively as follows:

ai = Ai × pi

bi =Bi × qi

ci = Ci × ri

Step2: Compute processing times by creating two fictitious machines G and H with their processing times Gi and Hi respectively as follows:

Gi = ×€ai + bi + ti + gi ×€

And Hi = ×€ci + bi + ti + gi ×€

if , either min( ai + ti ) ≥ max( bi + ti )

OR min(ci + gi ) ≥ max ( bi + gi )

OR both,

Step3: Determine equivalent job for the job block using idle/waiting time operator Oi,w as per definition,

(Gk , Hk) Oi,w (Gm ,Hm) =( Gβ , Hβ )

Gβ =Gk + max(Gm - Hk , 0)

Hβ =Hm + max(Hm - Gm , 0)

Step 4:Define new reduced problem with two fictitious machines G and H having processing times Gi and Hi respectively,and the job block is replaced by a single job β with processing times Gβ and Hβ.

Step 5:Apply johnsons algorithm to find an optimal sequence of jobs ,which gives the minimum elapsed time. This optimal schedule will also be the optimal schedule for the original problem.

Definition: Idle/waiting Time Operator (Oiw )

Let R+ be the set of non negative numbers.Let G=R+ × R+. Then Oiw is defined as amapping from G×G→G given by

Oiw[(x1,y1),(x2,y2)]=(x1,y1)Oiw(x2,y2) ={x1 + max(x2-y1,0) , y2+ max (y1-x2 , 0)}

Where x1,x2,y1,y2 Є R.

THEOREM:

Let n jobs 1,2,3,…….n are processed through two machines A&B in order AB with processing times ai and bi (i=1,2,3,…n) on machines A &B respectively. If (ap,bp)Oi,w(aq,bq)=(aβ,bβ) then

aβ = ap+max(aq-bp , 0) and bβ = bq+ max(bq-aq , 0)

where β is an equivalent job for the job block (p,q) and p,qЄ{1,2,3,…n}.

PROOF: starting by the equivalent job block criteria theorem for β=(p,q) given by maggu and das,we have:

aβ = ap +aq – min(bq,aq) ………..(1)

bβ = bp + bq –min(bp,aq) ……….(2)

Now we prove the above theorem by following way:

Case 1 : When aq > bp

aq> bp >0

max{aq>bp ,0} = aq>bp …………(3)

and bp>aq<0 ,

max{bp>aq , 0} = 0 ………….(4)

….. aβ = ap + aq – min(bp,aq)

=ap+ aq - bp as aq > bp

=ap + max(aq - bp , 0) using (3)

(2)…..bβ = bp + bq – min(bp,aq)

=bp + bq - bp as aq>bp

=bq + (bp- bp)

=bq + 0

=bq + max (bp - aq , 0) using (4)

Case 2: when aq<bp

aq-bp<0

max(aq-bp, 0) = 0

and bp-aq >0

max (bp-aq, 0) = bp-aq

aβ = ap+aq-min (bp,aq)

= ap+aq-aq as aq>bp

= ap+0

= ap+max (aq-bq,0) using (7) (9)

bβ = bp+bq- min(bp,aq)

= bp+bq-aq as aq <bp

=bp+(bp-aq)

=bp+max (bp-aq,0) using (8) (10)

case 3: when aq=bp , aq-bp =0

max (aq-bp,0) =0 (11)

also

bp-aq = 0

max( bp-aq,0) = 0

aβ =ap+aq-min(bp,aq) (12)

= bp+aq-ap as bq =ap

= ap+0

= ap+max (aq-bp,0) (13)

bβ =bp+bq-min(bp,aq) (14)

=bp+bq-bp

=bq+(bp-bp)

= bq+0

= bq+max(bp-aq,0) using (14) (15)

By (5),(6),(9),(10),(13) and (15) we conclude:

aβ =ap+max(aq- bp,0)

bβ =bp+max (bp- aq,0) for all possible three cases.

Numerical illustration:

Consider 5 jobs and three machines A,B and C.Let Ai ,Bi and Ci be the processing times of the jobs corresponding to machines A,B and C respectively.Also let ti and gi be the transportation times of jobs from machine A to B and from machine B to C respectively.

job

Machine A

Machine B

Machine C

i

ai

pi

ti

bi

qi

gi

ci

ri

1

30

0.3

4

30

0.2

3

20

0.2

2

40

0.2

3

25

0.2

2

30

0.1

3

30

0.2

6

30

0.1

2

35

0.2

4

20

0.2

8

20

0.1

3

20

0.2

5

70

0.1

5

15

0.4

4

10

0.3

Obtain the optimal sequence of jobs so as to minimize the total elapsed time by taking (2,5) as a job block.

Solution:

STEP 1: First of all we find the expected processing times by multiplying the processing times with their respective probabilities.

jobs

Machine A

ti

Machine B

gi

Machine C

i

ai

bi

ci

1

9

4

6

3

4

2

8

3

5

2

3

3

6

6

3

2

7

4

4

8

2

3

4

5

7

5

6

4

3

STEP 2: Now we check the conditions discussed in the algorithm,

And here min(ai+ti)= max(bi+ti) satisfies.So we create two fictitious machines G & H with their processing times Gi &Hi as follows:

Gi =|ai + bi +ti +gi| and Hi = |ci +bi+ti+gi |, as follows

Jobs

Gi

Hi

1

22

17

2

18

13

3

17

18

4

17

17

5

22

18

STEP 3: Now we find an equivalent job β for a given job block (2,5) by using idle/waiting time operator,

(Gk ,Hk) Oi,ω (Gm ,Gm) =(Gβ , Hβ )

Gβ = Gk + max (Gm – Hk , 0 )

& Hβ = Hm + max( Hm – Gm , 0)

So Gβ =18+ max (22 – 13 , 0)

= 18+ max(9 , 0)

= 18 + 9

= 27

Hβ = 18 + max ( 18 – 22 , 0) = 18

STEP 4: Represent new reduced problem in table form using step 1 and step 2,

Jobs

Gi

Hi

1

22

17

β

27

18

3

17

18

4

17

17

STEP 5: Now by using Johnson’s technique , optimal sequence is

4 , 3 , β ,1

i.e, 4, 3, 2, 5, 1

The above optimal sequence will also be the optimal sequence for the original sequence.

In-Out flow table for sequence S, is as follows:

Jobs

Machine A

Machine B

Machine C

i

In - Out

In - Out

In - Out

4

0 - 4

12 - 14

17 - 21

3

4 - 10

16 - 19

21- 28

2

10 - 18

21 - 26

28 - 31

5

18 - 25

30 - 36

40 - 43

1

25 - 34

38 - 44

47 - 51

Minimum total elapsed time for the given problem is 51 units.

REMARK:

The study may further be extended by introducing different parameters such as Set up time , job weights etc.