balance the assembly line shown below for (a) a shift output of 30 pieces and (b) a shift output of 24…

Explanation:

Step1: Calculate cycle time for 30 - piece output

Assume an hour - long shift (60 minutes). Cycle time (C_1=\frac{60}{30}=2) minutes per piece.

Step2: Analyze tasks for workstation 2

Available tasks and their times:

• For task E (4 minutes), task G (4 minutes), task C (2 minutes), task H (2 minutes). We need to assign tasks such that the total time does not exceed the cycle time. We first consider the longest - processing - time rule.
• If we take task E and task H, total time (t = 4 + 2=6) minutes. But we need to fit within the cycle time. Since the cycle time is 2 minutes, we note that we need to balance tasks more carefully.
• Let's list all possible combinations of tasks that can be assigned to workstation 2 while respecting the precedence relationships. The tasks that can follow the tasks in workstation 1 (A, B, D) are E, F, G. The longest - processing - time among them is E (4 minutes), but it exceeds the cycle time alone. We can combine E with a shorter task. If we take E and C, the total time is (4 + 2=6) minutes which is not valid. If we consider tasks that can be combined within the cycle time, we find that we can assign task F (5 minutes) alone (as it is the only task that can be assigned without violating precedence and coming close to the cycle time). So the total time for workstation 2 is 5 minutes and idle time (=2\times3 - 5 = 1) minute (assuming 3 pieces are processed in this workstation within the cycle).

Step3: Analyze tasks for workstation 3

The remaining tasks are G, I, H, J, K. The longest - processing - time task available is G (4 minutes). We can combine G with H (2 minutes) to get a total time of (4+2 = 6) minutes. But if we consider the cycle time, we can assign G and J (3 minutes) for a total time of (4 + 3=7) minutes which is not valid. So we assign G alone for a total time of 4 minutes and idle time (=2\times2 - 4 = 0) minutes.

Step4: Analyze tasks for workstation 4

The remaining tasks are I, H, J, K. We can assign I (3 minutes), H (2 minutes), K (2 minutes) for a total time of (3+2 + 2=7) minutes which is not valid considering the cycle time. We can assign I and H for a total time of (3+2 = 5) minutes and idle time (=2\times3 - 5 = 1) minute.

Step5: Calculate line efficiency for 30 - piece output

Total task time (T=5 + 4+4 + 3+2+5+4+3+2=32) minutes. Number of workstations (n = 4). Cycle time (C_1 = 2) minutes. Efficiency (E_1=\frac{\sum t_i}{n\times C_1\times30}\times100=\frac{32}{4\times2\times30}\times100\approx13.3%) (This is wrong. Let's start over for a correct approach).

Let's start from the beginning for a more standard approach:

Step1: Calculate cycle time for 30 - piece output

Cycle time (C_1=\frac{60}{30}=2) minutes per piece. Total task time (\sum t=5 + 3+4+4+2+5+4+3+2+3+2=37) minutes.

Step2: Assign tasks to workstations

Workstation 1: We can assign tasks A (5 minutes), D (3 minutes), B (4 minutes) for a total time of (5 + 3+4 = 12) minutes. Since the cycle time is 2 minutes per piece and we assume the workstation processes pieces continuously, the number of pieces processed in this workstation in 60 minutes is (\frac{60}{12}\times1 = 5) pieces. Workstation 2: We can assign tasks E (4 minutes), F (5 minutes) for a total time of (4 + 5=9) minutes. Idle time (=2\times5 - 9 = 1) minute. Workstation 3: We can assign tasks G (4 minutes), H (2 minutes) for a total time of (4+2 = 6) minutes. Idle time (=2\times3 - 6 = 0) minutes. Workstation 4: We can assign tasks I (3 minutes), J (3 minutes), K (2 minutes) for a total time of (3+3 + 2=8) minutes. Idle time (=2\times4 - 8 = 0) minutes.

Step3: Calculate line efficiency for 30 - piece output

Efficiency (E_1=\frac{\sum t}{n\times C_1\times30}\times100=\frac{37}{4\times2\times30}\times100\approx15.4%)

Step4: Calculate cycle time for 24 - piece output

Cycle time (C_2=\frac{60}{24}=2.5) minutes per piece.

Step5: Re - assign tasks to workstations for 24 - piece output

We repeat the task - assignment process with the new cycle time. Total task time (\sum t = 37) minutes. Workstation 1: We can still assign A, B, D with a total time of 12 minutes. Number of pieces processed in this workstation in 60 minutes is (\frac{60}{12}\times1 = 5) pieces. Workstation 2: We can assign E and C for a total time of (4 + 2=6) minutes. Idle time (=2.5\times3 - 6 = 1.5) minutes. Workstation 3: We can assign G and J for a total time of (4+3 = 7) minutes. Idle time (=2.5\times3 - 7 = 0.5) minutes. Workstation 4: We can assign I, H, K for a total time of (3+2 + 2=7) minutes. Idle time (=2.5\times3 - 7 = 0.5) minutes.

Step6: Calculate line efficiency for 24 - piece output

Efficiency (E_2=\frac{\sum t}{n\times C_2\times24}\times100=\frac{37}{4\times2.5\times24}\times100=\frac{37}{240}\times100\approx15.4%)

Workstation	Tasks	Total Time	Idle Time
1	A, B, D	12	0
2	E, F	5	1
3	G	4	0
4	I, H, K	7	1

For 30 - piece output:

Workstation	Tasks	Total Time	Idle Time
1	A, B, D	12	0
2	F	5	1
3	G	4	0
4	I, H, K	7	1
Efficiency: (\approx15.4%)

For 24 - piece output:

Workstation	Tasks	Total Time	Idle Time
1	A, B, D	12	0
2	E, C	6	1.5
3	G, J	7	0.5
4	I, H, K	7	0.5
Efficiency: (\approx15.4%)

Answer:

For 30 - piece output:

Workstation	Tasks	Total Time	Idle Time
1	A, B, D	12	0
2	F	5	1
3	G	4	0
4	I, H, K	7	1
Efficiency: (15.4%)

For 24 - piece output:

Workstation	Tasks	Total Time	Idle Time
1	A, B, D	12	0
2	E, C	6	1.5
3	G, J	7	0.5
4	I, H, K	7	0.5
Efficiency: (15.4%)