A group of six jobs is to be processed through a two-machine flow shop. The first operation involves cleaning and the second involves painting. Processing times in hours are as follows: (a) Use Johnson’s rule to determine a sequence that will minimize the total completion time for this group of jobs. (b) Draw a Gant Chart and determine the throughput time and idle times at the work centres.
(a) We see that the job consuming lowest time is with D ( In workstation 1), so we place D at the leftmost position.
Further, the next lowest value is with B ( In workstation 2)so we place it to the rightmost position.
Next lowest value is with A, we can put it at No. 2 from the left.
Next lowest value is with E ( Workstation 1) which can be placed at position 3 from left.
Next lowest value is at C, which can be placed at position 4 from the left.
Remaining value F can be placed at 5th position from left.
we get the sequence as D>A>C>E>F>B
WS I – D(2) > A(7)>C(15) >E(21)>f(33)>B(37)
WS II – D(9) > A(14)>C(24)>E(32)>F(48)>B(51)
Idle hours at Workstation I -0
Idle hours at Workstation II -4
Gantt chart ( idle hours are marked in red)