Page 271 - DMGT206_PRODUCTION_AND_OPERATIONS_MANAGEMENT

Production and Operations Management

Notes
Example: The contact breaker facility, we have already taken the first step, defining
tasks, shown in Table above The second step requires identifying a specific sequence. These
sequence requirements are also listed in Table 13.1 in column 4.
Once the desired output is specified, we can calculate the theoretical minimum number of
stations required. This is done by contrasting the time required to produce one unit with the
time we can allow, given the daily output requirements. We have already calculated the time
required, as the sum of the task times in Table above and we have calculated the time allowable,
as the maximum allowable cycle time.
Since just 0.098 hours are allowed to produce one unit, 5.56 stations must operate simultaneously,
each contributing 0.098 hours, so that the required 0.356 hours are made available.
Theoretical minimum Number of stations = time required/(unit time allowed/unit)
To produce 1 unit = 0.356 hours / (0.098 hours/ unit) = 3.63 stations

Since only whole stations are possible, at least four stations are needed. The actual layout may
use more than the minimum number of stations, depending on the precedence requirements.
The initial layout in Table above uses nine stations.
The fourth step assigns tasks to each station. The designer must assign ten tasks to six or more
stations. Several assignment combinations are possible. In the example given earlier, TAMS
designed a system that provided a rectangular platen system manned by only five operators. All
assembly was completed on the platen with the sub-assemblies being transferred to a central
position on the platen for riveting.
For larger problems with thousands of tasks and hundreds of stations, we often use heuristics.
We will apply a Longest Operation Time (LOT) heuristic to find a balance for the 0.098 hours/
unit cycle time. The LOT steps are:

Heuristic Step 1: Longest operation time (LOT) gives the top priority of assignment to the task
requiring the longest operation time. Assign first the task that takes the most time to the first
station. However, the precedence requirements have to be maintained. In our example, task ‘K’
requires the longest operation time of 5 minutes (the bottleneck operation); therefore, this task
has the highest priority of assignment at the first workstation. Table 13.1 shows that task ‘K’ has
precedence requirement of other tasks, i.e., there is a need for other tasks to be competed for the
execution of task ‘K’. Therefore, task ‘K’ cannot be assigned to the first workstation. We have to
assign task ‘A’ as the first task.
Heuristic Step 2: In the first rule, task ‘A’ is the eligible task for the first workstation and is
assigned to it. As the task time of ‘A’ is 0.010 hours, and the bottleneck task is 0.098 hours,
additional tasks can be assigned to the station. Therefore tasks ‘B’, ‘C’, and ‘D’ which require a
total time of 0.080 hours can also be assigned to this station. The time available on station 1 after
completing these tasks is 0.008 hours. As there is no other task that has this timing, no more tasks
can be assigned to this station.
Heuristic Step 3: For workstation 3, we see that task ‘H’ requires the longest task time of 0.050
hours. From Table below, notice that tasks ‘I’ and ‘J’ require 0.008 and 0.040 hours respectively.
In keeping with the precedence requirement, tasks H, I and J can be assigned to workstation 3 as
the total of the time required to complete these tasks is 0.098 hours.
Heuristic Step 4: Workstation 4 is the bottleneck station. The task ‘K’ cannot be split into parts,
this task has to be assigned to a workstation and the cycle time cannot be less than the duration
of this task. No other task can be accommodated at this workstation.

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