Johnson Cogs wants to set up a line to serve 60 customers per hour. The processi
ID: 445077 • Letter: J
Question
Johnson Cogs wants to set up a line to serve 60 customers per hour. The processing time for each task and their preceding tasks are given in the following table. Assume operating time is 8hrs in a day.
a. What is the minimum cycle time and maximum output rate for this assembly line assuming one task per workstation?
b. What is the maximum cycle time and minimum output rate for this line when all tasks are performed in one workstation?
c. Johnson Cogs desired output rate is 60 customers per hour. Draw the precedence diagram, find cycle time and use line balancing rules to assign tasks to workstations.
Work
Station
Time
Remaing
Eligible
Task
Assign
Task
Revised Time
Remaining
Station Idle
Time
d. Determine the percentage idle time and efficiency for the system you designed in part (d).
Tasks Proceeding Time(sec) Proceeding Task(s) A 40 - B 30 A C 20 A D 40 B E 25 B F 25 C G 15 C H 20 D,E I 15 F,G J 30 H,IExplanation / Answer
a. minimum cycle time and maximum output rate for this assembly line assuming one task per workstation
Formula for cycle time = Available Time / Output = 8hours per day/60 per hour that as per the given information to serve 60 customers per hours, therefore cycle time is 1 minute (time to serve one customer)
In case each task per workstation then minimum cycle time is the total sum of all critical tasks as per the network
A--B--D---H--J (40+30+40+20+30) is the longest/critical path having lenght 160seconds
Therefore maximum output rate = (3600 * 8)/160= 180 per day
b. What is the maximum cycle time and minimum output rate for this line when all tasks are performed in one workstation?
Maximum cycle time = Sum of all tasks times = 40+30+20+40+25+25+15+20+15+30= 260 seconds
Therefore minimum output rate = (3600 * 8) / 260 = 110 per day
c. Assignments of tasks to workstations given in the table as follows:
d. Determine the percentage idle time and efficiency for the system you designed in part (c )
From the above table total actual workstations time = 60 * 5 = 300seconds whereas the total time required is 260 seconds , therefore idle time is 40 seconds (13.33%) and efficiency = 260/300 = 86.67%
The corrected table is as follows, in the earlier table task I was wrongly assigned (again ) to workstation 5
Work Time Eligible Assign Revised Time Station Idle Station Remaing Task Task Remaining Time Column1 Column2 Column3 Column4 Column5 Column6 1 60 A A 20 1 20 B,C C 0 0 2 60 B,F,G G,F 20 2 20 B,I I 5 5 3 60 B B 30 3 30 D,E E 5 5 4 60 D D 20 4 20 H H 0 0 5 60 I I 45 5 45 J J 15 15Related Questions
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