Assembly Line Balancing

This model is used to balance workloads on an assembly line. Five heuristic rules can be used for performing the balance. The cycle time can be given explicitly or the production rate can be given and the program will compute the cycle time. This model will not split tasks. Task splitting is discussed in more detail in a later section.

The Model

The general framework for assembly line balancing is dictated by the number of tasks that are to be balanced. These tasks are partially ordered, as shown, for example in the precedence diagram that follows.

Method. The five heuristic rules that can be chosen are

Longest operation time
Most following tasks
Ranked positional weight
Shortest operation time
Least number of following tasks

NOTE: Ties are broken in an arbitrary fashion if two tasks have the same priority based on the rule given.

The remaining parameters are

Cycle time. The cycle time can be given in one of two ways. One way is when the cycle time is given directly as shown above. While this is the easiest method, it is more common to determine the cycle time from the demand rate. The cycle time is converted into the same units as the times for the tasks. (See example 2).

Task time unit. The time unit for the tasks is given by this drop-down box. You must choose either seconds, hours or minutes. Notice that the column heading for the task times will change as you select different time units.

Task names. The task names are essential for assembly line balancing since they determine the precedences . Case does not matter.

Task times. The task times are given.

Precedences . Enter the precedences , one per cell. If there are two precedences they must be entered in two cells. Do not enter 'a, b'. In fact, a comma will not be accepted. Notice that in the precedence list in the screen above we have typed both a and A. As mentioned previously, the case of the letters is irrelevant.

Example 1

In this example we have six tasks named a through f. The precedence diagram for this problem appears above. The time to perform each task is above the task. Also, note that the tasks that are ready at the beginning of the balance are tasks a and b. Finally, in this first example we use a cycle time of 10.

Solution

The following screen contains the solution to our first example. The solution screen will always have the same appearance and contain the same information regardless of the rule that is chosen for the balance. The information is as follows.

Station numbers. The station numbers appear in the far left column. They are displayed only for the first task that is loaded into each station. In this example, three stations are required.

Task names. The tasks that are loaded into the station are listed in the second column. In this example, tasks b, e and a are in station one; tasks d and c are in station two; and task f is in station 3.

Task times. The length of time for each task appears in this column.

Time left. The length of time that remains at the station is listed in this column. The last number at each station is, of course, the idle time at that station. For example, there is one second of idle time at station 1, one second at station 2, and two seconds at station 3 for a total of four seconds of idle time per cycle.

Ready tasks. The tasks that are ready appear here. A ready task is any task that has had its precedences met. We emphasize this because some books do not list a task as ready if its time exceeds the time remaining at the station. Also, if the number of characters in the ready task list is very long, you might want to widen that column.

Cycle time. The cycle time that was used appears below the balance. This cycle time was either given directly or computed. In this example, the cycle time was given directly as 10 seconds.

Time allocated. The total time allocated for making each unit is displayed. This time is the product of the number of stations and the cycle time at each station. In this example we have three stations each with a cycle time of 10 seconds, for a total work time of 30 station-seconds.

The time needed to make one unit. This is simply the sum of the task times. In the example we have 1 + 5 + 2 + 7 + 3 + 8 = 26 seconds.

Idle time. This is the time needed subtracted from the time allocated. This example has 30 - 26 = 4, which matches the 4 seconds.

Efficiency. Efficiency is defined as the time needed divided by the time allocated. In this example the efficiency is computed as 26/30, which is .8667.

Balance delay. The balance delay is the percentage of wasted time or 100% - the efficiency. In this example, it is 4 (the idle time)/30 or .1333, which is also determined by 1-.8667.

Minimum theoretical number of stations. This is the total time to make one unit divided by the cycle time and rounded up to the nearest integer. In this example, we have 26 seconds required to make one unit divided by a 10-second cycle time for an answer of 2.6, which we round up to 3 stations.

The precedence graph can be displayed, as well as a bar graph indicating how much time was used at each station. These are shown at the end of this section. In addition, if there is idle time at every station, a note will appear at the top indicating that the balance can be improved by reducing the cycle time. For example, since there are idle times of 1, 1 ans 2 seconds at the three stations we could reduce the cycle time by 1 second.

Example 2 - Computing the cycle time

Suppose that for the same data we require a production of 2250 units in 7.5 hours.

We assume full minutes and hours and compute the cycle time as

(7.5 hrs /2,250units )*60 min/hr*60 sec/min = 27,000/2,250 = 12 seconds.

The solution screen follows.

Other Rules

We mention the other rules that may be used although we do not display the results.

Please note that this is one of the modules where if you change the method (using the dropdown box) from the solution screen the problem will immediately be resolved.

Most following tasks

A common way to choose tasks is by using the task with the most following tasks. Notice from the diagram that a has three tasks following it and b has two tasks following it. Therefore, the first task scheduled is task a when using this rule. The results display indicates the number of following tasks in the column named 'ready tasks'. This is displayed in the results screen for example 4.

Ranked positional weight method

The ranked positional weight computes the sum of the task and all tasks that follow. For example, for task a the ranked positional weight is 1 + 2 + 7 + 8 = 18 while for task b the weight is 5 + 3 + 8 = 16. The task with the largest weight is scheduled first (if it will fit in the remaining time). Notice that e has a higher ranked positional weight than c. The results display indicates the ranked positional weight in the column named 'ready tasks'.

Shortest operation time

Another rule that is used sometimes is to give priority to the task that takes the least amount of time.

Least number of followers

The last rule that is available is the least number of followers.

Example 3 - What to do if longest operation time will not fit

Some books and some software do not apply the longest operation time rule properly. If the task with the longest time will not fit into the station, the task with the second longest time should be placed in the station if it will fit.

In the following screen we present data for eight tasks. Notice that tasks b, c, d and e follow task a.

The balance appears above for a cycle time of 5 seconds. After task a is completed, tasks b, c, d and e are ready. Task b is longest but will not fit in the 4 seconds that remain at station 1. Therefore, task c is inserted into the balance. We caution you that if the answer in your book differs from the program to check if the book has neglected to put in the task with the longest operation time that will fit.

Example 4 - Splitting tasks

If the cycle time is less than the amount of time to perform a specific task, there is a problem. We perform what is termed task splitting but which in reality is actually duplication. For example, suppose that the cycle time is 2 minutes and a task takes 5 minutes. Then we have the task performed three times (by three people at three machines independent of one another). The effect is that three units will be done every 5 minutes, which is equivalent to one unit every 1.33 minutes, which fits into the 2 minute cycle.

Now, the actual way that the three people work may vary. While other programs will split tasks, the assumptions vary from program to program. Rather than making assumptions, we leave it to you to split the tasks by dividing the task time appropriately.

Suppose that in example 1 we wanted to use a cycle time of 5 seconds. Then it is necessary to replicate both tasks d and f since they will not fit in the cycle time. The approach to use is to solve the problem by dividing the task times by two, since this replication is needed. We present the results in the following screen.

Graphs

Two different graphs are available. The first is a precedence graph as shown in the example that follows. Please note that there may be several different ways to draw a precedence graph.

The second graph is of time used at each station. In a perfect world these would all be the same (a perfect balance).