With a cycle time of 80 seconds, how many windows are produced daily? If the operation runs for one 8-hour shift each day, the available productive time each day is 28,800 seconds (8 hours * 3600 seconds/hour). Therefore maximum daily output can be as follows: (available time)/(cycle time/unit) = 28800 seconds / 80 seconds/unit = 360 units.
Since this assembly line can generate more than the required 320 units daily, capacity is adequate.
An alternative method for determining whether capacity is adequate calculates the maximum allowable cycle time given a desired capacity (320 units/day).
Maximum allowable cycle time to produce desired output
= (time available)/ (desired number of units)
= 28,800 seconds /320 units = 90 seconds /unit.
This calculation shows that a layout whose cycle time is 90 seconds or less will yield the desired capacity. A layout whose cycle time is greater than 90 seconds will not yield adequate capacity.
Is the Sequence of Tasks Feasible?
For now, we will assume that the proposed sequence of tasks is feasible. We will return to this question soon.
Is the line efficient?
The proposed layout has six stations, each manned by one employee. All six workers are paid for 8 hours daily. How much of our employees’ time is spent productively, and how much idly? It depends on the pace of the line that management sets. The pace can be set anywhere between the 80-second cycle time and 90-second maximum allowable cycle time. In Table-II, we have calculated the efficiency of labor for cycle times of 90 and 80 seconds.
As you can see idleness is higher for the 90-second cycle, 29.6 % of the employees’ total time, than for 80-second cycle, 20.8 %. Idle time amounts to 10 labor hours daily, for the 80-second cycle. If the hourly wage is $10, each day $100 is paid for unnecessary idleness. These excessive costs would eventually have to be passed on to the customer.
Table-II: Calculation of labor-utilization efficiency for proposed 80- and 90-second lines
As you can see idleness is higher for the 90-second cycle, 29.6 % of the employees’ total time, than for 80-second cycle, 20.8 %. Idle time amounts to 10 labor hours daily, for the 80-second cycle. If the hourly wage is $10, each day $100 is paid for unnecessary idleness.
Balancing the Line
How can the cost of idleness be reduced? Perhaps the eight tasks (A to H in Table-I) can be reassigned so that more available employee time is used. Notice that if every station used up an equal amount of task time, no time would be idle time. The problem of equalizing stations in this way is called the line balancing problem, and solving it takes six steps:
Define tasks
- Identify precedence requirements
- Calculate minimum number of work stations required to produce desired output
- Apply an assignment heuristic to assign tasks to each station
- Evaluate effectiveness and efficiency
- Seek further improvement
For the example of the aluminum storm window facility, we have already taken the first step, defining tasks, shown in Table-I.
The second step reminds us that tasks must be done in a specific sequence. Certainly the window units can not be packed until they are completely assembled. These sequence requirements are listed in Table-I under the heading Task’s required Predecessor.
Once the desired output is specified, we can calculate the theoretical minimum number of stations required, the third step in our solution. We do so 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 all the task times in Table-II: 380 seconds. And we have calculated the time allowable, as the maximum allowable cycle time : 90 seconds. Since just 90 seconds are allowed to produce one unit, 4.22 stations must operate simultaneously, each contributing 90 seconds, so that the required 380 seconds are made available, as required: (90) * (4.22) = 380, or 4.22 = 380 / 90.
Since only whole stations are possible, at least five 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-I uses six stations.
The fourth step assigns tasks to each station. The designer must assign eight tasks to five or more stations. Several assignment combinations are possible. For larger problems with thousands of tasks and hundreds of stations, we often use heuristics. We will apply a longest-operation-time heuristic to find a balance for the 90-second cycle time.
The steps in the longest-operation-time (LOT)rule are:
LOT 1. Assign the task that takes most time to the first station. Maintain precedence sequence.
LOT 2. After assigning a task, determine how much time the station has to contribute (time-task times).
LOT 3. If the station can contribute more time, assign it a task requiring as much time as possible. Maintain precedence relationship.
Otherwise, return to LOT 1. Continue until all tasks have been assigned to stations.
To apply the rule, we first array the tasks in descending order of time. Tasks with their time in parentheses are: B (80), A (70), G (50), H (50), C (40), E (40), F (30), and D (20).
Following LOT1 we try to assign B to station 1, since B has the longest time. However, B is ineligible because it must follow A (precedence requirement). Leave B and choose a, the task taking the next highest time. A is not proceeded by any task, so assign it to station 1. After A is assigned, station ! has 20 seconds left to (LOT2). Following LOT3, we see that D is the first task on the list that takes 20 seconds or less, so we assign it to this station. Therefore, station 1 consists of tasks A and D for a total 90 seconds operation time.
Now we begin again with LOT1 and station 2. B takes the longest time, meets precedence requirements and is therefore assigned to station 2. In LOT2, we find the station has 10 seconds ( 90 – 80 ) left to contribute. Since all other tasks require more than 10 seconds, none is eligible to be assigned to station 2.
To station 3 we may assign C or E. Other choices would violate precedence requirements. We arbitrarily select C, with an operation time of 40 seconds. Remaining time at station 3 is therefore 50 seconds ( 90 – 40 ). Now E and G become eligible at this station. Since G takes longest time, it is assigned. The station 3 consists of tasks C and G with a total performance time of 90 seconds ( 40 + 50 ).
This entire process, carried to completion, is summarized in Table-III and shown in Figure-II: a five station assembly line comprising 8 tasks.
Table-III: Assigning tasks to stations using the longest-operation-time heuristic achieving 90-second cycle time
A D B C G E F H
Figure-II: Revised storm window assembly
This layout is effective if it yields the desired capacity. It is efficient if it minimizes idle time. In the fifth step, we want to check both measures of performance. Table-IV shows calculations of efficiency and effectiveness. The layout is more efficient than the one where the efficiency is calculated in Table-II.
Table-IV: Assembly line design for 90-second cycle.
At this stage, we may be able to improve the layout by trial and error, step 6 of our station. In addition many other heuristics may be used instead of LOT approach. Several computerized heuristics are available, and since different heuristics can lead to different layouts, managers may want to try more than one approach.
There are occasions when effectiveness and efficiency can be increased by deviating from the procedure we have presented. Task sharing, for example, occurs when there are three stations manned by workers, all of whom are sometimes idle. We can reduce idleness by eliminating one worker, and letting the other two take turns doing the task at the third station. Other improvements are possible if more than one worker can be assigned to a single station. Finally, if the desired output exceeds capacity, bottlenecks may be reexamined.