# The times required to complete each of eight jobs in a two-machine flow shop are shown in the...

## Question:

The times required to complete each of eight jobs in a two-machine flow shop are shown in the table that follows. Each job must follow the same sequence, beginning with machine A and moving to machine B.

Time(hours)
Job Machine A Machine B
A 16 5
B 3 13
C 9 6
D 8 7
E 2 14
F 12 4
G 18 14
H 20 11

a) Determine a sequence that will minimize makespan time

b) Find machine's idle time

c) For the sequence determined in a) how much would machine B's idle time be reduced by splitting the last two jobs in half?

Task is the term which is used to refer the subpart of the big process. The one big process includes various small task which altogether constitute the whole process, that is, all the individual task needs to be completed in order to complete the whole process.

(a)

 Job Machine A Machine B Sequence A 16 5 7 B 3 13 2 C 9 6 6 D 8 7 5 E 2 14 1 F 12 4 8 G 18 14 3 H 20 11 4

In this sequence, the job will be carried out. The sequence is build up on the basis of low processing time of every job. For an example, E has the lowest processing time; therefore, it is on the first number.

(b)

The machine's B idle time is calculated below:

According to the data, Machine A will operate or work and then Machine B will work. Therefore, at task E machine B will have an idle time of 2 hours because first Machine A will finish the E task and then B will start.

Again, according to the data Machine A will finish the A task in total 76 hours and Machine B can start after that only. Therefore, B will face idle time of 9 hours at task A.

{eq}\begin{align*} {\rm{Machine}}\,{\rm{A}}\,{\rm{(task}}\,{\rm{'a'}}\,{\rm{finished}}\,{\rm{time)}} &= {\rm{e + b + g + h + d + c + a}}\\ &= 2 + 3 + 18 + 20 + 8 + 9 + 16\\ &= 76\\ {\rm{Machine}}\,{\rm{B}}\,(task\,'a'\,start\,time) &= idle\,time + e + b + g + h + d + c\\ &= 2 + 14 + 13 + 14 + 11 + 7 + 6)\\ &= 67\\ {\rm{Idle}}\,{\rm{time}} &= 76 - 67\\ &= 9 \end{align*} {/eq}

At task f, Machine will again face the idle time of 7 hours:

{eq}\begin{align*} {\rm{Idle}}\,{\rm{time}} &= {\rm{Machine}}\,{\rm{A}}\,'{\rm{f}}\,{\rm{' task}} - \,{\rm{Machine}}\,{\rm{B}}\,'{\rm{a'}}\,{\rm{task}}\\ &= 12 - 5\\ &= 7 \end{align*} {/eq}

{eq}\begin{align*} {\rm{Total}}\,{\rm{Ilde}}\,{\rm{Time}} &= 2 + 9 + 7\\ &= 18\,{\rm{hours}} \end{align*} {/eq}

c) For the sequence determined in a) how much would machine B's idle time be reduced by splitting the last two jobs in half?

The activity time will be reduced by half, that is,

{eq}\begin{align*} {\rm{Task}}\,{\rm{A}} &= 16/2\\ &= 8\\ Task\,F &= 12/2\\ &= 6 \end{align*} {/eq}

If, the duration will be reduced by half then the idle time will also be reduced by half.

{eq}\begin{align*} {\rm{Idle}}\,{\rm{time}} &= 2 + (9/2) + (7/2)\\ &= 2 + 4.5 + 3.5\\ &= 11 \end{align*} {/eq}