Some tasks and the order in which they must be performed according to their asse

Question

Some tasks and the order in which they must be performed according to their assembly requirements are shown in the following table. These are to be combined into workstations to create an assembly line. The assembly line operates 6.25 hours per day. The output requirement is 750 units per day.

  
a. What is the required workstation cycle time to meet the desired output rate?

Workstation cycle time             seconds per unit

b. Balance the line using the longest task time based on the 750-unit forecast, stating which tasks would be done in each workstation. (Leave no cells blank - be certain to enter "0" wherever required.)

c. What is the efficiency of your line balance, assuming it is running at the cycle time determined in part a? (Round your answer to 1 decimal place.)

Efficiency             %

d. After production was started, Marketing realized that they understated demand and must increase output to 850 units. What action would you take? (Round your answer for cycle time down to the nearest whole number. Round your answer for overtime up to the nearest whole number.)

(Click to select)IncreaseReduce cycle time to  seconds or work  minutes of overtime.

TASK PRECEDING
TASKS TIME
(SECONDS) TASK PRECEDING
TASKS TIME
(SECONDS) A — 19 G C 13 B A 15 H D 10 C A 17 I E 12 D B 10 J F, G 7 E B 12 K H, I 12 F C 14 L J, K 7

Explanation / Answer

a) Available production time = 6.25 hrs = 6.25 * 60 * 60 = 22500 secs

Customer demand = 750 units/day

Cycle time = 22500/750 = 30 secs/unit

b) Assigning tasks to workstation using longest operating time rule:

Longest operating time order and predecessor rule: A-C-B-F-G-E-I-D-J-H-K-L

The total station time is nothing but the cycle time. So, total station time of each station is 30 secs.

               Workstation 1:

                        First task =A

                        Time left= 30-19 = 11 secs

                       

So, workstation 1: A

Workstation 2:

                        First task =C

                        Time left= 30-17 = 13 secs

                       

So, workstation 2: C

Workstation 3:

                        First task =B

                        Time left= 30-15 = 15 secs

                        Second task=F

                        Time left=15-14 = 1 sec

So, workstation 3: B->F

Workstation 4:

                        First task =G

                        Time left= 30-13 = 17 secs

                        Second task=E

                        Time left=17-12 = 5 secs

So, workstation 4: G->E

Workstation 5:

                        First task =I

                        Time left= 30-12 = 18 secs

                        Second task=D

                        Time left=18-10 = 8 secs

                        Third task =J

                        Time left= 8-7 = 1 sec

So, workstation 5: I->D->J

Workstation 6:

                        First task =H

                        Time left= 30-10 = 20 secs

                        Second task=K

                        Time left=20-12 = 8 secs

                        Third task= L

                        Time left= 8-7 = 1 sec

So, workstation 6: H->K->L

Work Station

Tasks

Workstation Time (secs)

Idle Time (secs)

1

A

19

11

2

C

17

13

3

B->F

29

1

4

G->E

25

5

5

I->D->J

29

1

6

H->K->L

29

1

c) Efficiency with 6 workstations = (sum of all tasks) / (no of workstations *   Cycle time) = (148)/ (6*30) = 0.822 or 82.2 %

d) If New Demand = 850 units / day

Available production time = 6.25 hrs = 6.25 * 60 * 60 = 22500 secs

Cycle time = 22500/850 = 26.5 secs/unit = 27 secs/unit

Reduce cycle time to 27 secs/unit

Work Station

Tasks

Workstation Time (secs)

Idle Time (secs)

1

A

19

11

2

C

17

13

3

B->F

29

1

4

G->E

25

5

5

I->D->J

29

1

6

H->K->L

29

1

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