Q&A Hero

Question

Station Task Time (seconds) Time left (seconds) Ready tasks
A,B,C,D
1 B 66. A,C,D
C 25. 41. A,D
A 20. 21. D,E
D 10. 11. E,F
2 E 55. 41. F,G
F 30. 11. G
3 G 25. 71. H
H 40. 31.
Summary Statistics
Cycle time 96 seconds
Time allocated (cycle time * #) 288 seconds/cycle
Time needed (sum of task times) 235 seconds/unit
Idle time (allocated-needed) 53 seconds/cycle
Efficiency (needed/allocated) 81.59722%
Balance Delay (1-efficiency) 18.40278%
Min (theoretical) # of stations 3

Key Term: Assembly-line balancing

49) A company is trying to balance production between 3 workstations on an assembly line. Currently there are 5 tasks that need to be performed. These tasks, ABCDE, have required times of 2 minutes, 4 minutes, 1 minutes, 3 minutes, and 10 minutes, respectively. The assembly line needs to produce 40 units per day to meet demand and can work for up to 8 hours each day.

(a) What is the required cycle time?

(b) What is the theoretical minimum # of workstations?

(c) Assign the tasks according to the shortest task time heuristic.

Answer:

(a) Required cycle time = Production time available/Units required = 8 hours * 60 minutes / 40 units = 12 minutes

(b) Minimum # of workstations = Sum of task times/Cycle time

= (2 + 4 + 1 + 3 + 10)/12 = 1.667 stations = 2 stations

(c) Workstation A would be assigned Task C for a total of 1 minute, Task A for a total of 3 minutes, Task D for a total of 6 minutes, Task B for a total of 10 minutes. Task E cannot be assigned because 10 + 10 > 12 minutes. Therefore Task E would be assigned to station B for a total of 10 minutes. Station C would have no assignment.

The assignment method of Part (c) creates the optimum assignment (2 workstations = theoretical minimum), so the third workstation is not needed.

Key Term: Assembly-line balancing

50) An assembly line is assigned as follows. Station 1- task A, B, and C. Station 2- task D. Station 3- task E and F. The task times are 7, 3, 2, 9, 4, and 5 minutes respective to A, B, C, D, E, and F. (Assume that there is no precedence relationships between the tasks, i.e. they can be performed in any order)

(a) Calculate the efficiency.

(b) What would the assignment of tasks to stations be using the shortest processing time heuristic (assuming the current maximum station time remains the cycle time)?

Answer:

(a) Efficiency = Sum of task time / (# Stations ∗ Longest Station duration)

= (7 + 3 + 2 + 9 + 4 + 5)/(3 ∗ 12)= 83.3%

(b) Station 1 would be assigned task C for a total of 2 minutes, task B for a total of 5 minutes, Task E for a total of 9 minutes. Task F cannot be assigned because the total would be 14 minutes which is greater than the 12 maximum. Therefore Station 2 would be assigned task F for a total of 5 minutes and task A for a total of 12 minutes. Finally station 3 would be assigned task D for a total of 9 minutes.

Key Term: Assembly-line balancing

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