Question

Sue Helms Appliances wants to establish an assembly line to manufacture its new product, the Micro Popcorn Popper.





Sue Helms Appliances wants to establish an assembly line to manufacture its new product, the Micro Popcorn Popper. The goal is to produce five poppers per hour. The tasks, task times, and   immediate predecessors for producing one Micro Popcorn Popper are as follows:






















TaskTime (min)Immediate   Predecessors








A10----








B12A








C8A,B








D6B,C








E6C








F6D,E




















a) What is the theoretical minimum for the smallest number of workstations that Helms can achieve in this assembly line?












b) Graph the assembly line and assign workers to workstations. Can you assign them with the theoretical minimum?












c) What is the efficiency of your assignment?








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Answer #1

a)

Cycle time = 12 minutes

Sum of task times = 10 + 12 + 8 + 6 + 6 + 6 = 48

b)

To assign the tasks to the work stations, we have to consider both the precedence of the tasks and the time required for completion of each task.

Task A is the first task and can be assigned to work station 1. We cannot assign more tasks to work station 1 because it will exceed the cycle time.

Task B is the next task and will assigned to work station 2 and we cannot assign any more tasks to work station 2.

Task C is assigned to workstation 3 and we cannot assign any more tasks to station 3.

Task D & E is assigned to workstation 4.

We cannot assign task F to any station as no more stations are available.

Task Time Workstation
A 10 1
B 12 2
C 8 3
D 6 4
E 6 4
F 6 _

With the theoretical number of workstations, it is not possible to assign all the tasks to the workstations. Task F remains unassigned.

c)

The actual number of workstations required will be 5. With the theoretical minimum number of workstations, we cannot assign task F. Hence we require an additional workstation and in total there are 5 workstations.

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Answer #3

(C.)

Efficiency=Tasks times total/# of workstations*largest assigned cycle time

(10+12+8+6+6+6)/(5*12)=0.8

0.8*100=80%

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