# 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 which are to be balanced. These tasks are partially ordered as shown for example in the precedence diagram below.

Method. The five heuristic rules which can be chosen are:

longest operation time

ranked positional weight

shortest operation time and

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).

Time unit. The time unit for the tasks is given by this dropdown 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.

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 we have typed both 'a' and 'A'. As mentioned previously, the case of the letters is irrelevant.

Example 1

In this example we have 6 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 which 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 screen above 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 which is chosen for the balance. The information is as follows.

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

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

Task names. The tasks which are loaded into the station are listed in the second column. In this example tasks b, e and a are in station 1, tasks d and c are in station 2 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 which 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 1 second of idle time at station 1, 1 second at station 2, and 2 seconds at station 3 for a total of 4 seconds of idle time per cycle.

Time allocated. The total time allocated for making each unit is printed. This time is the product of the number of stations and the cycle time at each station. In this example we have 3 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 found in the column of time left.

The 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

The 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 then a note will appear at the top indicating that the balance can be improved (by reducing the cycle time).

### 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

as illustrated in the balance in the screen below.

## Other rules

A common way to choose tasks is by using the task with the most tasks following. Notice from the diagram that a has three tasks following it and b has 2 tasks following it. Therefore the first task scheduled is task a when using this rule.

### Ranked positional weight method

The ranked positional weight computes the sum of the task and all tasks which 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.

### Shortest operation time

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

### Least number of followers

The last rule which 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 then the task with the second longest time should be placed in the station if it will fit.

The balance appears above for a cycle time of 5 seconds. After task a is completed tasks b and c are ready. Task b is longer but will not fit in the 4 seconds which 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 then there is a problem. We perform what is termed task splitting but 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 6 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 below

## Graphs

Two different graphs are available. The first is a precedence graph as shown below. 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)

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