Six jobs are performed on Machines $M_1$ and $M_2$ respectively. Time in hours taken by each job on each machine is given below: | Jobs $\rightarrow$ | A | B | C | D | E | F | | --- | --- | --- | --- | --- | --- | --- | | Machines $\downarrow$ | | | | | | | | $M_1$ | 3 | 12 | 5 | 2 | 9 | 11 | | $M_2$ | 8 | 10 | 9 | 6 | 3 | 1 | Determine the optimal sequence of jobs and find total elapsed time. Also find the idle time for machines $M_1$ and $M_2$. Solution: Given jobs can be arranged in optimal sequence as, | D | A | C | B | E | F | | --- | --- | --- | --- | --- | --- | | Jobs | Machine $M_1$ | Machine $M_2$ | | | | --- | --- | --- | --- | --- | | | In | Out | In | Out | | D | 0 | 2 | $\square$ | 8 | | A | 2 | 5 | 8 | 16 | | C | 5 | 10 | 16 | 25 | | B | 10 | 22 | 25 | 35 | | E | 22 | 31 | 35 | 38 | | F | 31 | 42 | $\square$ | 43 | Total Elapsed time = $\square$ hrs. Idle time for Machine $M_1$ = $43 - 42 = 1$ hour. Idle time for Machine $M_2$ = $\square$ hrs.

> > [!TLDR] > Apply Johnson's two-machine rule to order the jobs as $D,A,C,B,E,F$, build the in–out timing table, read the last out-time on $M_2$ as the elapsed time $43$ hrs, and get idle times $M_1=1$ hr, $M_2=6$ hrs. > > **Processing times (hours).** > > | Job | A | B | C | D | E | F | > | --- | --- | --- | --- | --- | --- | --- | > | $M_1$ | 3 | 12 | 5 | 2 | 9 | 11 | > | $M_2$ | 8 | 10 | 9 | 6 | 3 | 1 | > > **Johnson's rule.** Repeatedly pick the smallest time in the whole table: if it belongs to $M_1$ schedule that job as early as possible, if to $M_2$ schedule it as late as possible. > - Smallest $=1$ ($F$ on $M_2$) $\Rightarrow$ $F$ last. > - Next $=2$ ($D$ on $M_1$) $\Rightarrow$ $D$ first. > - Next $=3$ ($A$ on $M_1$, and $E$ on $M_2$): $A$ on $M_1\Rightarrow$ place $A$ early (2nd); $E$ on $M_2\Rightarrow$ place $E$ late (2nd from end). > - Next $=5$ ($C$ on $M_1$) $\Rightarrow$ $C$ next early (3rd). > - $B$ takes the remaining middle slot. > > Optimal sequence: $\;D\to A\to C\to B\to E\to F$. > > **In–out schedule.** On $M_1$ each job starts when the previous one leaves $M_1$; on $M_2$ a job starts at $\max(\text{its }M_1\text{ out-time},\ \text{previous }M_2\text{ out-time})$. > > | Job | $M_1$ In | $M_1$ Out | $M_2$ In | $M_2$ Out | > | --- | --- | --- | --- | --- | > | D | 0 | 2 | 2 | 8 | > | A | 2 | 5 | 8 | 16 | > | C | 5 | 10 | 16 | 25 | > | B | 10 | 22 | 25 | 35 | …

Six jobs are performed on Machines M₁ and M₂ respectively. Time in hours taken by each job on each machine is

Question 20 of 32

Six jobs are performed on Machines $M_1$ and $M_2$ respectively. Time in hours taken by each job on each machine is given below:

Jobs $\rightarrow$	A	B	C	D	E	F
Machines $\downarrow$
$M_1$	3	12	5	2	9	11
$M_2$	8	10	9	6	3	1
Determine the optimal sequence of jobs and find total elapsed time. Also find the idle time for machines $M_1$ and $M_2$ .
Solution:
Given jobs can be arranged in optimal sequence as,
D	A	C	B	E	F
---	---	---	---	---	---
Jobs	Machine $M_1$	Machine $M_2$
---	---	---	---	---
	In	Out	In	Out
D	0	2	$\square$	8
A	2	5	8	16
C	5	10	16	25
B	10	22	25	35
E	22	31	35	38
F	31	42	$\square$	43
Total Elapsed time = $\square$ hrs.
Idle time for Machine $M_1$ = $43 - 42 = 1$ hour.
Idle time for Machine $M_2$ = $\square$ hrs.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 4mImportance★★★★★

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Apply Johnson's two-machine rule to order the jobs as $D,A,C,B,E,F$ , build the in–out timing table, read the last out-time on $M_2$ as the elapsed time $43$ hrs, and get idle times $M_1=1$ hr, $M_2=6$ hrs.

Processing times (hours).

Job	A	B	C	D	E	F
$M_1$	3	12	5	2	9	11
$M_2$	8	10	9	6	3	1

Johnson's rule. Repeatedly pick the smallest time in the whole table: if it belongs to $M_1$ schedule that job as early as possible, if to $M_2$ schedule it as late as possible.

Smallest $=1$ ( $F$ on $M_2$ ) $\Rightarrow$ $F$ last.
Next $=2$ ( $D$ on $M_1$ ) $\Rightarrow$ $D$ first.
Next $=3$ ( $A$ on $M_1$ , and $E$ on $M_2$ ): $A$ on $M_1\Rightarrow$ place $A$ early (2nd); $E$ on $M_2\Rightarrow$ place $E$ late (2nd from end).
Next $=5$ ( $C$ on $M_1$ ) $\Rightarrow$ $C$ next early (3rd).
$B$ takes the remaining middle slot.

Optimal sequence: $\;D\to A\to C\to B\to E\to F$ .

In–out schedule. On $M_1$ each job starts when the previous one leaves $M_1$ ; on $M_2$ a job starts at $\max(\text{its }M_1\text{ out-time},\ \text{previous }M_2\text{ out-time})$ .

Job	$M_1$ In	$M_1$ Out	$M_2$ In	$M_2$ Out
D	0	2	2	8
A	2	5	8	16
C	5	10	16	25
B	10	22	25	35

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