The Holiday Charity Group uses volunteers to assemble care packages for needy families for the coming holiday season. The group would like to organize the work as efficiently as possible. A list of tasks, task times, and precedence requirements are given in the following chart.

 

Task Precedence Relationships Task Time (Minutes)

A 6

B A 4

C B 6

D B 5

E C,D 4

F E 5

 

a.     Draw the precedence diagram. What is/are the bottleneck task(s) of the assembly line if the operation manager assigns one volunteer per task? What is the hourly capacity of the line (in units per hour)?

b.     What are the minimum and maximum cycle times? What hourly range of theoretical output is possible for this assembly line?

c.     If the group wants to finish a care package every 10 minutes, what is the theoretical minimum number of volunteers needed? Balance the assembly line (i.e., assign tasks to workstations). How many volunteers should be called in?  Calculate the efficiency of the assembly line. How many packages can be assembled in a five-hour shift?

d.     Suppose that volunteers are plentiful. Balance the assembly line to maximize output. What is the efficiency of the assembly line? How many care packages can be assembled in a five-hour shift?

e.     Suppose that volunteers are plentiful and the charity can add the second assembly line with an identical configuration. How many care packages can be assembled in a five-hour shift?

 

Bottleneck tasks are the tasks with the longest task time.

Tasks A and C have the longest task time of 6 minutes.

Hence, tasks A and C are the bottlenecks.

 

Hourly Capacity of the line = 60 (minutes/hour) / Task time of the bottleneck task (minutes/unit)

Hourly Capacity of the line = 60 / 6

Hourly Capacity of the line = 10 units/hour

 

b) Minimum cycle time = Longest task time among all the tasks

Minimum cycle time = 6 minutes

 

Maximum cycle time is when all the tasks are assigned to a single volunteer (i.e. each volunteer performs all the tasks), thereby making the total task time = sum of task times of all the tasks

Maximum cycle time = Total time for all the tasks

Maximum cycle time = 6 + 4 + 6 + 5 + 4 + 5

Maximum cycle time = 30 minutes

 

Lower the cycle time, higher will be the output.

 

Minimum Theoretical Output is for maximum cycle time

Minimum Theoretical Output = 60 (minutes/hour) / 30 (minutes/unit)

Minimum Theoretical Output = 2 units/hour

 

Maximum Theoretical Output is for minimum cycle time

Maximum Theoretical Output = 60 (minutes/hour) / 6 (minutes/unit)

Maximum Theoretical Output = 10 units/hour

 

Range of Theoretical Output is 2 units/hour to 10 units/hour

 

c) Cycle Time = Time required to finish a care package

Cycle Time = 10 minutes

 

Theoretical minimum number of volunteers needed = Total task time / Cycle time

Theoretical minimum number of volunteers needed = 30 / 10

Theoretical minimum number of volunteers needed = 3

 

We assign the task to the workstations such that no task gets assigned before its predecessor task and the total task time of no workstation exceeds the cycle time of 10 minutes. We break the tie using the longest task time first rule.

 

We first assign A to workstation 1.

Balance time remaining at workstation 1 after assigning A = Cycle time - Task time of A

Balance time remaining at workstation 1 after assigning A = 10 - 6

Balance time remaining at workstation 1 after assigning A = 4 minutes

 

We then assign B to workstation 1.

Balance time remaining at workstation 1 after assigning B = Balance time remaining at workstation 1 after assigning A - Task time of B

Balance time remaining at workstation 1 after assigning B = 4 - 4

Balance time remaining at workstation 1 after assigning B = 0

 

The next eligible task is task C (as task time of C is longer than that of D) but we cannot assign it to workstation 1 as task time of C is longer than the balance time remaining at workstation 1 after assigning task B.

Hence, we assign C to workstation 2

Balance time remaining at workstation 2 after assigning C = Cycle time - Task time of C

Balance time remaining at workstation 2 after assigning C = 10 - 6

Balance time remaining at workstation 2 after assigning C = 4 minutes

 

The next eligible task is D but we cannot assign it to workstation 2 as task time of D is longer than the balance time remaining at workstation 2 after assigning task C.

Hence, we assign D to workstation 3

Balance time remaining at workstation 3 after assigning D = Cycle time - Task time of D

Balance time remaining at workstation 3 after assigning D = 10 - 5

Balance time remaining at workstation 3 after assigning D = 5 minutes

 

We assign E to workstation 3.

Balance time remaining at workstation 3 after assigning E = Balance time remaining at workstation 1 after assigning D - Task time of E

Balance time remaining at workstation 3 after assigning E = 5 - 4

Balance time remaining at workstation 3 after assigning E = 1 minute

 

We cannot assign task F to workstation 3.

Hence, we assign task F to workstation 4.

Balance time remaining at workstation 4 after assigning F = Cycle time - Task time of F

Balance time remaining at workstation 4 after assigning F = 10 - 5

Balance time remaining at workstation 4 after assigning F = 5 minutes

 

Assignment of tasks:

Workstation 1: A and B

Workstation 2: C

Workstation 3: D and E

Workstation 4: F

 

Number of volunteers to be called in = 4

 

Total time at the workstations = number of workstations * cycle time

Total time at the workstations = 4 * 10

Total time at the workstations = 40 minutes

 

Idle time = Idle time for stations (1 + 2 + 3 + 4)

Idle time = 0 + 4 + 1 + 5

Idle time = 10 minutes

 

Balance Delay % = Idle time / Total time at the workstations * 100

Balance Delay % = 10 / 40 * 100

Balance Delay % = 25%

 

Efficiency % = 100 - Balance Delay %

Efficiency % = 100 - 25

Efficiency % = 75%

 

Cycle time = 10 minutes

 

Output in a 5-hour shift = 5 (hours/shift) * 60 (minutes/hour) / Cycle time

Output in a 5-hour shift = 5 * 60 / 10

Output in a 5-hour shift = 30 units

 

d) To maximize the output, we set the cycle time to the minimum i.e. 6 minutes.

This also means that each task is performed by a different volunteer and thereby performed on a different workstation.

 

Assignment of tasks:

Workstation 1: A

Workstation 2: B

Workstation 3: C

Workstation 4: D

Workstation 5: E

Workstation 6: F

 

Total time at the workstations = number of workstations * cycle time

Total time at the workstations = 6 * 6

Total time at the workstations = 36 minutes

 

Idle time = Idle time for stations (1 + 2 + 3 + 4)

Idle time = (6 - 6) + (6 - 4) + (6 - 6) + (6 - 5) + (6 - 4) + (6 - 5)

Idle time = 6 minutes

 

Balance Delay % = Idle time / Total time at the workstations * 100

Balance Delay % = 6 / 36 * 100

Balance Delay % = 16.67%

 

Efficiency % = 100 - Balance Delay %

Efficiency % = 100 - 16.67

Efficiency % = 83.33%

 

Output in a 5-hour shift = 5 (hours/shift) * 60 (minutes/hour) / Cycle time

Output in a 5-hour shift = 5 * 60 / 6

Output in a 5-hour shift = 50 units

 

e) With two parallel workstation lines,

Output in a 5-hour shift = Output per line in a 5-hour shift * 2

Output in a 5-hour shift = 50 * 2

Output in a 5-hour shift = 100 care packages

 

(Note: here, units = Care packages)

Please see the attached file for the complete solution